[
    {
        "id": "Chemistry_1462",
        "problem": "${ }^{131}$I is a radioactive isotope of iodine ( $\\mathrm{e}^{-}$emitter) used in nuclear medicine for analytical procedures to determine thyroid endocrine disorders by scintigraphy. The decay rate constant, $k$, of ${ }^{131} \\mathrm{I}$ is $9.93 \\times 10^{-7} \\mathrm{~s}^{-1}$.\n\nCalculate the half-life of ${ }^{131}$ I expressed in days.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n${ }^{131}$I is a radioactive isotope of iodine ( $\\mathrm{e}^{-}$emitter) used in nuclear medicine for analytical procedures to determine thyroid endocrine disorders by scintigraphy. The decay rate constant, $k$, of ${ }^{131} \\mathrm{I}$ is $9.93 \\times 10^{-7} \\mathrm{~s}^{-1}$.\n\nCalculate the half-life of ${ }^{131}$ I expressed in days.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of d, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "d"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_429",
        "problem": "可用于电动汽车的铝-空气燃料电池, 通常以 $\\mathrm{NaCl}$ 溶液或 $\\mathrm{NaOH}$ 溶液为电解液, 铝合金为负极，空气电极为正极。下列说法正确的是\nA: 以 $\\mathrm{NaCl}$ 溶液或 $\\mathrm{NaOH}$ 溶液为电解液时, 负极反应都为 $\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{e}^{-}=4 \\mathrm{OH}^{-}$\nB: 以 $\\mathrm{NaOH}$ 溶液为电解液时, 负极反应为 $\\mathrm{Al}^{-3} \\mathrm{e}^{-}=\\mathrm{Al}^{3-}$\nC: 以 $\\mathrm{NaOH}$ 溶液为电解液时, 电池在工作过程中电解液的 $\\mathrm{pH}$ 减小\nD: 电池工作时, 电子通过外电路从负极流向正极\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n可用于电动汽车的铝-空气燃料电池, 通常以 $\\mathrm{NaCl}$ 溶液或 $\\mathrm{NaOH}$ 溶液为电解液, 铝合金为负极，空气电极为正极。下列说法正确的是\n\nA: 以 $\\mathrm{NaCl}$ 溶液或 $\\mathrm{NaOH}$ 溶液为电解液时, 负极反应都为 $\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{e}^{-}=4 \\mathrm{OH}^{-}$\nB: 以 $\\mathrm{NaOH}$ 溶液为电解液时, 负极反应为 $\\mathrm{Al}^{-3} \\mathrm{e}^{-}=\\mathrm{Al}^{3-}$\nC: 以 $\\mathrm{NaOH}$ 溶液为电解液时, 电池在工作过程中电解液的 $\\mathrm{pH}$ 减小\nD: 电池工作时, 电子通过外电路从负极流向正极\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_245",
        "problem": "Which of the following molecules contains $29.67 \\%$ sulfur by mass?\nA: $\\mathrm{SF}_{4}$\nB: $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$\nC: $\\mathrm{SOCl}_{2}$\nD: $\\mathrm{SF}_{6}$\nE: $\\mathrm{S}_{2} \\mathrm{~F}_{10}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following molecules contains $29.67 \\%$ sulfur by mass?\n\nA: $\\mathrm{SF}_{4}$\nB: $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$\nC: $\\mathrm{SOCl}_{2}$\nD: $\\mathrm{SF}_{6}$\nE: $\\mathrm{S}_{2} \\mathrm{~F}_{10}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_736",
        "problem": "利用平衡移动原理, 分析一定温度下 $\\mathrm{Mg}^{2+}$ 在不同 $\\mathrm{pH}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 体系中的可能产物。已知:\n\ni. 下图中曲线表示 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 体系中各含碳粒子的物质的量分数与 $\\mathrm{pH}$ 的关系。\n\n[图1]\n\nii. 下图中曲线 $I$ 的离子浓度关系符合 $c\\left(\\mathrm{Mg}^{2+}\\right) c^{2}\\left(\\mathrm{OH}^{-}\\right)=K_{\\mathrm{sp}}\\left[\\mathrm{Mg}(\\mathrm{OH})_{2}\\right]$; 曲线II的离子浓度关系符合 $c\\left(\\mathrm{Mg}^{2+}\\right) c\\left(\\mathrm{CO}_{3}^{2-}\\right)=K_{\\mathrm{sp}}\\left[\\mathrm{MgCO}_{3}\\right]$ [注: 起始 $c\\left(\\mathrm{Na}_{2} \\mathrm{CO}_{3}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 不同 $\\mathrm{pH}$ 下 $c\\left(\\mathrm{CO}_{3}^{2-}\\right)$ 由上图得到]。\n\n[图2]\n\n下列说法正确的是\nA: 由第 1 个图, $\\mathrm{pH}=9$ 时, $c\\left(\\mathrm{HCO}_{3}^{-}\\right)<c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nB: 由第 2 个图, 初始状态 $\\mathrm{pH}=9 、 \\lg \\left[c\\left(\\mathrm{Mg}^{2+}\\right)\\right]=-2$, 平衡后溶液中存在: $c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+c\\left(\\mathrm{HCO}_{3}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: 由第 2 个图, 初始状态 $\\mathrm{pH}=11 、 \\lg \\left[c\\left(\\mathrm{Mg}^{2+}\\right)\\right]=-6$, 有沉淀生成\nD: 由两个图分析, 初始状态 $\\mathrm{pH}=8 、 \\lg \\left[c\\left(\\mathrm{Mg}^{2+}\\right)\\right]=-1$, 发生反应: $$ \\mathrm{Mg}^{2+}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{MgCO}_{3} \\downarrow+\\mathrm{CO}_{2} \\uparrow+\\mathrm{H}_{2} \\mathrm{O} $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n利用平衡移动原理, 分析一定温度下 $\\mathrm{Mg}^{2+}$ 在不同 $\\mathrm{pH}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 体系中的可能产物。已知:\n\ni. 下图中曲线表示 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 体系中各含碳粒子的物质的量分数与 $\\mathrm{pH}$ 的关系。\n\n[图1]\n\nii. 下图中曲线 $I$ 的离子浓度关系符合 $c\\left(\\mathrm{Mg}^{2+}\\right) c^{2}\\left(\\mathrm{OH}^{-}\\right)=K_{\\mathrm{sp}}\\left[\\mathrm{Mg}(\\mathrm{OH})_{2}\\right]$; 曲线II的离子浓度关系符合 $c\\left(\\mathrm{Mg}^{2+}\\right) c\\left(\\mathrm{CO}_{3}^{2-}\\right)=K_{\\mathrm{sp}}\\left[\\mathrm{MgCO}_{3}\\right]$ [注: 起始 $c\\left(\\mathrm{Na}_{2} \\mathrm{CO}_{3}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 不同 $\\mathrm{pH}$ 下 $c\\left(\\mathrm{CO}_{3}^{2-}\\right)$ 由上图得到]。\n\n[图2]\n\n下列说法正确的是\n\nA: 由第 1 个图, $\\mathrm{pH}=9$ 时, $c\\left(\\mathrm{HCO}_{3}^{-}\\right)<c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nB: 由第 2 个图, 初始状态 $\\mathrm{pH}=9 、 \\lg \\left[c\\left(\\mathrm{Mg}^{2+}\\right)\\right]=-2$, 平衡后溶液中存在: $c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+c\\left(\\mathrm{HCO}_{3}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: 由第 2 个图, 初始状态 $\\mathrm{pH}=11 、 \\lg \\left[c\\left(\\mathrm{Mg}^{2+}\\right)\\right]=-6$, 有沉淀生成\nD: 由两个图分析, 初始状态 $\\mathrm{pH}=8 、 \\lg \\left[c\\left(\\mathrm{Mg}^{2+}\\right)\\right]=-1$, 发生反应: $$ \\mathrm{Mg}^{2+}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{MgCO}_{3} \\downarrow+\\mathrm{CO}_{2} \\uparrow+\\mathrm{H}_{2} \\mathrm{O} $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-064.jpg?height=557&width=722&top_left_y=178&top_left_x=330",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-064.jpg?height=577&width=645&top_left_y=1085&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_924",
        "problem": "已知过氧化氢在强碱性溶液中主要以 $\\mathrm{HO}_{2}$-存在。我国研究的 $\\mathrm{Al}-\\mathrm{H}_{2} \\mathrm{O}_{2}$ 燃料电池可用于深海资源的勘查、军事侦察等国防科技领域，装置示意图如下。下列说法错误的是\n\n[图1]\nA: 电池工作时, 溶液中 $\\mathrm{OH}$-通过阴离子交换膜向 $\\mathrm{Al}$ 极迁移\nB: $\\mathrm{Ni}$ 极的电极反应式是 $\\mathrm{HO}_{2}^{-}+2 \\mathrm{e}^{-}+\\mathrm{H}_{2} \\mathrm{O}=3 \\mathrm{OH}^{-}$\nC: 电池工作结束后, 电解质溶液的 $\\mathrm{pH}$ 降低\nD: $\\mathrm{Al}$ 电极质量减轻 $13.5 \\mathrm{~g}$, 电路中通过 $9.03 \\times 10^{23}$ 个电子\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知过氧化氢在强碱性溶液中主要以 $\\mathrm{HO}_{2}$-存在。我国研究的 $\\mathrm{Al}-\\mathrm{H}_{2} \\mathrm{O}_{2}$ 燃料电池可用于深海资源的勘查、军事侦察等国防科技领域，装置示意图如下。下列说法错误的是\n\n[图1]\n\nA: 电池工作时, 溶液中 $\\mathrm{OH}$-通过阴离子交换膜向 $\\mathrm{Al}$ 极迁移\nB: $\\mathrm{Ni}$ 极的电极反应式是 $\\mathrm{HO}_{2}^{-}+2 \\mathrm{e}^{-}+\\mathrm{H}_{2} \\mathrm{O}=3 \\mathrm{OH}^{-}$\nC: 电池工作结束后, 电解质溶液的 $\\mathrm{pH}$ 降低\nD: $\\mathrm{Al}$ 电极质量减轻 $13.5 \\mathrm{~g}$, 电路中通过 $9.03 \\times 10^{23}$ 个电子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-48.jpg?height=362&width=351&top_left_y=1001&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_411",
        "problem": "利用 $\\mathrm{Sb}$ 的参杂来实现对 $\\mathrm{Bi}$ 的有效调制, 优化了 $\\mathrm{CO}_{2}$ 和 $\\mathrm{N}_{2}$ 偶联的限速步骤中间产物的对称性, 以促进 $\\mathrm{C}-\\mathrm{N}$ 偶联。其反应历程中能量变化如图所示。\n\n[图1]\n下列有关说法错误的是\nA: 尿素合成的主要决速步骤是 $\\mathrm{C}-\\mathrm{N}$ 偶联\nB: $\\mathrm{Sb}$ 元素的参杂降低了热力学反应的焓变\nC: $\\mathrm{Sb}$ 元素掺杂不会减少副产物 $\\mathrm{CO}$ 的产生\nD: 过程 ${ }^{*} \\mathrm{COOH} \\rightarrow{ }^{*} \\mathrm{CO}+\\mathrm{H}_{2} \\mathrm{O}$ 中存在 $\\mathrm{O}-\\mathrm{H}$ 的形成\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用 $\\mathrm{Sb}$ 的参杂来实现对 $\\mathrm{Bi}$ 的有效调制, 优化了 $\\mathrm{CO}_{2}$ 和 $\\mathrm{N}_{2}$ 偶联的限速步骤中间产物的对称性, 以促进 $\\mathrm{C}-\\mathrm{N}$ 偶联。其反应历程中能量变化如图所示。\n\n[图1]\n下列有关说法错误的是\n\nA: 尿素合成的主要决速步骤是 $\\mathrm{C}-\\mathrm{N}$ 偶联\nB: $\\mathrm{Sb}$ 元素的参杂降低了热力学反应的焓变\nC: $\\mathrm{Sb}$ 元素掺杂不会减少副产物 $\\mathrm{CO}$ 的产生\nD: 过程 ${ }^{*} \\mathrm{COOH} \\rightarrow{ }^{*} \\mathrm{CO}+\\mathrm{H}_{2} \\mathrm{O}$ 中存在 $\\mathrm{O}-\\mathrm{H}$ 的形成\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-024.jpg?height=597&width=1050&top_left_y=2103&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1163",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nAssuming that gaseous $\\mathrm{SkCl}_{2}$ has one lone pair on the stuck-at-homium, predict the bond angle around the central metal.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nAssuming that gaseous $\\mathrm{SkCl}_{2}$ has one lone pair on the stuck-at-homium, predict the bond angle around the central metal.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $^{\\circ}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$^{\\circ}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_605",
        "problem": "用微生物燃料电池作电源进行模拟消除酸性工业废水中的重铬酸根离子 $\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)$ 的示意图如下, 反应一段时间后, 在装置(2)中得到 $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Cr}(\\mathrm{OH})_{3}$ 两种沉淀。下列说法错误的是\n\n[图1]\n\n(1)\n\n质子交换膜\nA: 装置(2)中惰性电极为阴极\nB: 装置(1)中 $\\mathrm{a}$ 极区 $\\mathrm{pH}$ 逐渐减小\nC: 装置(1)中 $\\mathrm{b}$ 极上消耗的 $\\mathrm{n}\\left(\\mathrm{O}_{2}\\right)$ 与装置(2)中惰性电极上生成的 $\\mathrm{n}$ (生成物)相等\nD: 装图(2)中, 生成 $\\mathrm{Cr}(\\mathrm{OH})_{3}$ 的反应式: $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+8 \\mathrm{H}^{+}=2 \\mathrm{Cr}(\\mathrm{OH})_{3} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n用微生物燃料电池作电源进行模拟消除酸性工业废水中的重铬酸根离子 $\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)$ 的示意图如下, 反应一段时间后, 在装置(2)中得到 $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Cr}(\\mathrm{OH})_{3}$ 两种沉淀。下列说法错误的是\n\n[图1]\n\n(1)\n\n质子交换膜\n\nA: 装置(2)中惰性电极为阴极\nB: 装置(1)中 $\\mathrm{a}$ 极区 $\\mathrm{pH}$ 逐渐减小\nC: 装置(1)中 $\\mathrm{b}$ 极上消耗的 $\\mathrm{n}\\left(\\mathrm{O}_{2}\\right)$ 与装置(2)中惰性电极上生成的 $\\mathrm{n}$ (生成物)相等\nD: 装图(2)中, 生成 $\\mathrm{Cr}(\\mathrm{OH})_{3}$ 的反应式: $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+8 \\mathrm{H}^{+}=2 \\mathrm{Cr}(\\mathrm{OH})_{3} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-36.jpg?height=391&width=1285&top_left_y=1518&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_71",
        "problem": "What is observed if a piece of lead is placed in an aqueous solution of $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ ?\n\n| Half-Reaction | $E^{o}, \\mathrm{~V}$ |\n| :---: | :---: |\n| $\\mathrm{~Pb}^{2+}(a q)+2 e^{-} \\rightarrow \\mathrm{Pb}(s)$ | -0.13 |\n| $\\mathrm{Cu}^{2+}(a q)+2 e^{-} \\rightarrow \\mathrm{Cu}(s)$ | +0.34 |\n\nI. The mass of the metal piece increases.\n\nII. The blue color of the solution becomes less intense.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is observed if a piece of lead is placed in an aqueous solution of $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ ?\n\n| Half-Reaction | $E^{o}, \\mathrm{~V}$ |\n| :---: | :---: |\n| $\\mathrm{~Pb}^{2+}(a q)+2 e^{-} \\rightarrow \\mathrm{Pb}(s)$ | -0.13 |\n| $\\mathrm{Cu}^{2+}(a q)+2 e^{-} \\rightarrow \\mathrm{Cu}(s)$ | +0.34 |\n\nI. The mass of the metal piece increases.\n\nII. The blue color of the solution becomes less intense.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_127",
        "problem": "A student places $0.750 \\mathrm{~g}$ of solid sodium hydroxide $(\\mathrm{NaOH})$ into $20.00 \\mathrm{~mL}$ of water at $25.0^{\\circ} \\mathrm{C}$ inside a coffee cup calorimeter. The final temperature of the calorimeter contents is $34.6^{\\circ} \\mathrm{C}$. The density of water is $1.0 \\mathrm{~g} \\mathrm{~mL}^{-1}$. Assume the specific heat capacity of the solution approximates that of water at $4.184 \\mathrm{~J} \\mathrm{~g}^{-1} \\mathrm{C}^{-1}$ and the calorimeter has $100 \\%$ efficiency. What is the $\\Delta \\mathrm{H}_{\\mathrm{rxn}}$ for the dissolution reaction below?\n\n$$\n\\mathrm{NaOH}(\\mathrm{s}) \\rightarrow \\mathrm{NaOH}(\\mathrm{aq})\n$$\nA: $-42.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-44.4 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-803 \\mathrm{~J} \\mathrm{~mol}^{-1}$\nD: $-833 \\mathrm{~J} \\mathrm{~mol}^{-1}$\nE: $-1070 \\mathrm{~J} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student places $0.750 \\mathrm{~g}$ of solid sodium hydroxide $(\\mathrm{NaOH})$ into $20.00 \\mathrm{~mL}$ of water at $25.0^{\\circ} \\mathrm{C}$ inside a coffee cup calorimeter. The final temperature of the calorimeter contents is $34.6^{\\circ} \\mathrm{C}$. The density of water is $1.0 \\mathrm{~g} \\mathrm{~mL}^{-1}$. Assume the specific heat capacity of the solution approximates that of water at $4.184 \\mathrm{~J} \\mathrm{~g}^{-1} \\mathrm{C}^{-1}$ and the calorimeter has $100 \\%$ efficiency. What is the $\\Delta \\mathrm{H}_{\\mathrm{rxn}}$ for the dissolution reaction below?\n\n$$\n\\mathrm{NaOH}(\\mathrm{s}) \\rightarrow \\mathrm{NaOH}(\\mathrm{aq})\n$$\n\nA: $-42.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-44.4 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-803 \\mathrm{~J} \\mathrm{~mol}^{-1}$\nD: $-833 \\mathrm{~J} \\mathrm{~mol}^{-1}$\nE: $-1070 \\mathrm{~J} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_562",
        "problem": "下列溶液中有关物质的量浓度关系正确的是\nA: $\\mathrm{pH}=2$ 的 $\\mathrm{HA}$ 溶液与 $\\mathrm{pH}=12$ 的 $\\mathrm{MOH}$ 溶液任意比混合: $\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}$ $\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$\nB: pH 相等的 $\\mathrm{CH}_{3} \\mathrm{COONa} 、 \\mathrm{NaOH}$ 和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 三种溶液 $\\mathrm{c}\\left(\\mathrm{Na}_{2} \\mathrm{CO}_{3}\\right)<\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COONa}\\right)$ $<_{\\mathrm{c}}(\\mathrm{NaOH})$\nC: 物质的量浓度相等 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 和 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$ $+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 NaHA 溶液, 其 $\\mathrm{pH}=4: \\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>_{\\mathrm{c}}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>\\mathrm{c}$ $\\left(\\mathrm{A}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列溶液中有关物质的量浓度关系正确的是\n\nA: $\\mathrm{pH}=2$ 的 $\\mathrm{HA}$ 溶液与 $\\mathrm{pH}=12$ 的 $\\mathrm{MOH}$ 溶液任意比混合: $\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}$ $\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$\nB: pH 相等的 $\\mathrm{CH}_{3} \\mathrm{COONa} 、 \\mathrm{NaOH}$ 和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 三种溶液 $\\mathrm{c}\\left(\\mathrm{Na}_{2} \\mathrm{CO}_{3}\\right)<\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COONa}\\right)$ $<_{\\mathrm{c}}(\\mathrm{NaOH})$\nC: 物质的量浓度相等 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 和 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$ $+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 NaHA 溶液, 其 $\\mathrm{pH}=4: \\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>_{\\mathrm{c}}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>\\mathrm{c}$ $\\left(\\mathrm{A}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-012.jpg?height=297&width=813&top_left_y=577&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_535",
        "problem": "向某密闭容器中加入 $0.3 \\mathrm{~mol} \\mathrm{~A} 、 0.1 \\mathrm{~mol} \\mathrm{C}$ 和一定量的 $\\mathrm{B}$ 三种气体。一定条件下发生反应, 各物质的浓度随时间变化如甲图所示 $\\left[\\mathrm{t}_{0} \\sim \\mathrm{t}_{1}\\right.$ 阶段的 $\\mathrm{c}(\\mathrm{B})$ 变化未画出 $]$ 。乙图为 $\\mathrm{t}_{2}$ 时刻后改变条件平衡体系中正、逆反应速率随时间变化的情况，且四个阶段都各改变一种反应条件且互不相同， $t_{3}$ 时刻为使用催化剂。下列说法中正确的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\nA: 若 $\\mathrm{t}_{1}=15 \\mathrm{~s}$, 用 $\\mathrm{A}$ 的浓度变化表示 $\\mathrm{t}_{0} \\sim \\mathrm{t}_{1}$ 阶段的平均反应速率为 $0.004 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{s}$ $-1$\nB: $\\mathrm{t}_{4} \\sim \\mathrm{t}_{5}$ 阶段改变的条件一定为减小压强\nC: 该容器的容积为 $2 \\mathrm{~L}, \\mathrm{~B}$ 的起始物质的量为 $0.02 \\mathrm{~mol}$\nD: $\\mathrm{t}_{5} \\sim \\mathrm{t}_{6}$ 阶段, 容器内 $\\mathrm{A}$ 的物质的量减少了 $0.06 \\mathrm{~mol}$, 而此过程中容器与外界的热交换总量为 a $\\mathrm{kJ}$, 该反应的热化学方程式 $3 \\mathrm{~A}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{B}(\\mathrm{g})+2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}=-50 \\mathrm{a}$ $\\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n向某密闭容器中加入 $0.3 \\mathrm{~mol} \\mathrm{~A} 、 0.1 \\mathrm{~mol} \\mathrm{C}$ 和一定量的 $\\mathrm{B}$ 三种气体。一定条件下发生反应, 各物质的浓度随时间变化如甲图所示 $\\left[\\mathrm{t}_{0} \\sim \\mathrm{t}_{1}\\right.$ 阶段的 $\\mathrm{c}(\\mathrm{B})$ 变化未画出 $]$ 。乙图为 $\\mathrm{t}_{2}$ 时刻后改变条件平衡体系中正、逆反应速率随时间变化的情况，且四个阶段都各改变一种反应条件且互不相同， $t_{3}$ 时刻为使用催化剂。下列说法中正确的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\nA: 若 $\\mathrm{t}_{1}=15 \\mathrm{~s}$, 用 $\\mathrm{A}$ 的浓度变化表示 $\\mathrm{t}_{0} \\sim \\mathrm{t}_{1}$ 阶段的平均反应速率为 $0.004 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{s}$ $-1$\nB: $\\mathrm{t}_{4} \\sim \\mathrm{t}_{5}$ 阶段改变的条件一定为减小压强\nC: 该容器的容积为 $2 \\mathrm{~L}, \\mathrm{~B}$ 的起始物质的量为 $0.02 \\mathrm{~mol}$\nD: $\\mathrm{t}_{5} \\sim \\mathrm{t}_{6}$ 阶段, 容器内 $\\mathrm{A}$ 的物质的量减少了 $0.06 \\mathrm{~mol}$, 而此过程中容器与外界的热交换总量为 a $\\mathrm{kJ}$, 该反应的热化学方程式 $3 \\mathrm{~A}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{B}(\\mathrm{g})+2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}=-50 \\mathrm{a}$ $\\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-009.jpg?height=388&width=622&top_left_y=1302&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-009.jpg?height=340&width=491&top_left_y=1372&top_left_x=997"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1214",
        "problem": "The elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nWhat mass should theoretically be obtained?\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 \\text {; }  \\mathrm{S}: 32\n$$\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nWhat mass should theoretically be obtained?\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 \\text {; }  \\mathrm{S}: 32\n$$\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_163",
        "problem": "Which of the following accurately represents a trend in atomic radius?\nA: $\\mathrm{F}>\\mathrm{Cl}>\\mathrm{Br}$\nB: $\\mathrm{F}>\\mathrm{O}>\\mathrm{N}$\nC: $\\mathrm{Cl}^{-}>\\mathrm{Na}^{+}>\\mathrm{Mg}^{2+}$\nD: $\\mathrm{Ca}^{2+}>\\mathrm{K}^{+}>\\mathrm{Ca}$\nE: $\\mathrm{O}^{2-}>\\mathrm{S}^{2-}>\\mathrm{Cl}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following accurately represents a trend in atomic radius?\n\nA: $\\mathrm{F}>\\mathrm{Cl}>\\mathrm{Br}$\nB: $\\mathrm{F}>\\mathrm{O}>\\mathrm{N}$\nC: $\\mathrm{Cl}^{-}>\\mathrm{Na}^{+}>\\mathrm{Mg}^{2+}$\nD: $\\mathrm{Ca}^{2+}>\\mathrm{K}^{+}>\\mathrm{Ca}$\nE: $\\mathrm{O}^{2-}>\\mathrm{S}^{2-}>\\mathrm{Cl}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1442",
        "problem": "Oxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nEstimate from Figure 1 (to 2 significant figures) how many moles of $\\mathrm{O}_{2}$ are deposited in muscle tissue when one mole of $\\mathrm{Hb}$ travels from the lungs to the muscles and back again for the three different types of $\\mathrm{Hb}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nOxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nEstimate from Figure 1 (to 2 significant figures) how many moles of $\\mathrm{O}_{2}$ are deposited in muscle tissue when one mole of $\\mathrm{Hb}$ travels from the lungs to the muscles and back again for the three different types of $\\mathrm{Hb}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [Hb-T 1, Hb- 2, Hb- 3].\nTheir units are, in order, [$\\mathrm{~mol}$, $\\mathrm{~mol}$, $\\mathrm{~mol}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-418.jpg?height=823&width=1400&top_left_y=568&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "$\\mathrm{~mol}$",
            "$\\mathrm{~mol}$",
            "$\\mathrm{~mol}$"
        ],
        "answer_sequence": [
            "Hb-T 1",
            "Hb- 2",
            "Hb- 3"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1531",
        "problem": "Which of the following elements has the largest third ionization energy?\nA: $\\mathrm{B}$\nB: $\\mathrm{C}$\nC: $\\mathrm{N}$\nD: $\\mathrm{Mg}$\nE: $\\mathrm{Al}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following elements has the largest third ionization energy?\n\nA: $\\mathrm{B}$\nB: $\\mathrm{C}$\nC: $\\mathrm{N}$\nD: $\\mathrm{Mg}$\nE: $\\mathrm{Al}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1536",
        "problem": "IONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nCalculate the $p K_{a}$ of the corresponding acid-base pair.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nCalculate the $p K_{a}$ of the corresponding acid-base pair.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_722",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 某混合溶液中 $c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)=0.01 \\mathrm{~mol} / \\mathrm{L}$, 由水电离出的 $c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$的对数 $\\lg c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$与 $\\lg \\frac{c\\left(\\mathrm{HA}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)} 、 \\lg \\frac{c\\left(\\mathrm{~A}^{2-}\\right)}{c\\left(\\mathrm{HA}^{-}\\right)}$的关系如图所示。下列说法正确的是\n\n[图1]\nA: $K_{a l}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=10^{-8}$\nB: $\\mathrm{Z}$ 点时溶液的 $\\mathrm{pH}=7$\nC: $\\mathrm{M}$ 点和 $\\mathrm{N}$ 点溶液的组成完全相同\nD: 从 $\\mathrm{X}$ 点到 $\\mathrm{Y}$ 点发生的反应可能为: $\\mathrm{HA}^{-}+\\mathrm{OH}^{-}=\\mathrm{A}^{2-}+\\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 某混合溶液中 $c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)=0.01 \\mathrm{~mol} / \\mathrm{L}$, 由水电离出的 $c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$的对数 $\\lg c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$与 $\\lg \\frac{c\\left(\\mathrm{HA}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)} 、 \\lg \\frac{c\\left(\\mathrm{~A}^{2-}\\right)}{c\\left(\\mathrm{HA}^{-}\\right)}$的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: $K_{a l}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=10^{-8}$\nB: $\\mathrm{Z}$ 点时溶液的 $\\mathrm{pH}=7$\nC: $\\mathrm{M}$ 点和 $\\mathrm{N}$ 点溶液的组成完全相同\nD: 从 $\\mathrm{X}$ 点到 $\\mathrm{Y}$ 点发生的反应可能为: $\\mathrm{HA}^{-}+\\mathrm{OH}^{-}=\\mathrm{A}^{2-}+\\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-016.jpg?height=549&width=874&top_left_y=802&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-017.jpg?height=57&width=1370&top_left_y=328&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_94",
        "problem": "Which statements about the collision theory of reactions are correct?\n\nI. Molecules must have the correct spatial orientations for collisions to lead to reactions.\n\nII. Only collisions with an energy greater than a certain threshold lead to reactions.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich statements about the collision theory of reactions are correct?\n\nI. Molecules must have the correct spatial orientations for collisions to lead to reactions.\n\nII. Only collisions with an energy greater than a certain threshold lead to reactions.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_395",
        "problem": "Which solution has the greatest percent ionization?\nA: $0.010 \\mathrm{M}$ formic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-4}\\right)$\nB: $0.10 \\mathrm{M}$ formic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-4}\\right)$\nC: $0.010 \\mathrm{M}$ acetic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-5}\\right)$\nD: $0.10 \\mathrm{M}$ acetic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-5}\\right)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich solution has the greatest percent ionization?\n\nA: $0.010 \\mathrm{M}$ formic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-4}\\right)$\nB: $0.10 \\mathrm{M}$ formic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-4}\\right)$\nC: $0.010 \\mathrm{M}$ acetic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-5}\\right)$\nD: $0.10 \\mathrm{M}$ acetic acid $\\left(K_{\\mathrm{a}}=1.8 \\times 10^{-5}\\right)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_212",
        "problem": "Which of the following lists species in order of increasing ionic radius?\nA: $\\mathrm{Cs}^{+}, \\mathrm{Rb}^{+}, \\mathrm{Na}^{+}$\nB: $\\mathrm{S}^{2-}, \\mathrm{Cl}^{-}, \\mathrm{K}^{+}$\nC: $\\mathrm{O}^{2-}, \\mathrm{Na}^{+}, \\mathrm{Ba}^{2+}$\nD: $\\mathrm{I}^{-}, \\mathrm{Cl}^{-}, \\mathrm{Br}^{-}$\nE: $\\mathrm{Sr}^{2+}, \\mathrm{Rb}^{+}, \\mathrm{Br}^{-}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following lists species in order of increasing ionic radius?\n\nA: $\\mathrm{Cs}^{+}, \\mathrm{Rb}^{+}, \\mathrm{Na}^{+}$\nB: $\\mathrm{S}^{2-}, \\mathrm{Cl}^{-}, \\mathrm{K}^{+}$\nC: $\\mathrm{O}^{2-}, \\mathrm{Na}^{+}, \\mathrm{Ba}^{2+}$\nD: $\\mathrm{I}^{-}, \\mathrm{Cl}^{-}, \\mathrm{Br}^{-}$\nE: $\\mathrm{Sr}^{2+}, \\mathrm{Rb}^{+}, \\mathrm{Br}^{-}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_957",
        "problem": "一种以镍电极废料(含 $\\mathrm{Ni}$ 以及少量 $\\mathrm{Al}_{2} \\mathrm{O}_{3} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3}$ 和不溶性杂质)为原料制备 $\\mathrm{NiOOH}$的过程可表示为:\n\n[图1]\n\n“酸浸”后溶液中的金属离子除 $\\mathrm{Ni}^{2+}$ 外还有少量的 $\\mathrm{Al}^{3+}$ 和 $\\mathrm{Fe}^{2+}$ 等, 下列说法错误的是\nA: 氧化性: $\\mathrm{Fe}^{3+}>\\mathrm{Ni}^{2+}$\nB: 除杂过程仅为过滤操作\nC: 氧化过程中每生成 $1 \\mathrm{~mol} \\mathrm{NiOOH}$ 消耗 $2 \\mathrm{~mol} \\mathrm{OH}^{-}$\nD: 工业上也可电解碱性 $\\mathrm{Ni}(\\mathrm{OH})_{2}$ 悬浊液制备 $\\mathrm{NiOOH}$, 加入一定量的 $\\mathrm{KCl}$ 有助于提高生产效率\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一种以镍电极废料(含 $\\mathrm{Ni}$ 以及少量 $\\mathrm{Al}_{2} \\mathrm{O}_{3} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3}$ 和不溶性杂质)为原料制备 $\\mathrm{NiOOH}$的过程可表示为:\n\n[图1]\n\n“酸浸”后溶液中的金属离子除 $\\mathrm{Ni}^{2+}$ 外还有少量的 $\\mathrm{Al}^{3+}$ 和 $\\mathrm{Fe}^{2+}$ 等, 下列说法错误的是\n\nA: 氧化性: $\\mathrm{Fe}^{3+}>\\mathrm{Ni}^{2+}$\nB: 除杂过程仅为过滤操作\nC: 氧化过程中每生成 $1 \\mathrm{~mol} \\mathrm{NiOOH}$ 消耗 $2 \\mathrm{~mol} \\mathrm{OH}^{-}$\nD: 工业上也可电解碱性 $\\mathrm{Ni}(\\mathrm{OH})_{2}$ 悬浊液制备 $\\mathrm{NiOOH}$, 加入一定量的 $\\mathrm{KCl}$ 有助于提高生产效率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-083.jpg?height=154&width=779&top_left_y=1782&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1006",
        "problem": "How does the $\\mathrm{pH}$ of a solution change as $\\mathrm{HCl}$ is added to a solution of $\\mathrm{NaOH}$ ?\nA: The $\\mathrm{pH}$ decreases and may go below 7 .\nB: The $\\mathrm{pH}$ will not change.\nC: The $\\mathrm{pH}$ decreases until it reaches a value of 7 and then stops.\nD: The $\\mathrm{pH}$ increases until it reaches a value of 7 and then stops.\nE: The pH increases and may go above 7 .\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow does the $\\mathrm{pH}$ of a solution change as $\\mathrm{HCl}$ is added to a solution of $\\mathrm{NaOH}$ ?\n\nA: The $\\mathrm{pH}$ decreases and may go below 7 .\nB: The $\\mathrm{pH}$ will not change.\nC: The $\\mathrm{pH}$ decreases until it reaches a value of 7 and then stops.\nD: The $\\mathrm{pH}$ increases until it reaches a value of 7 and then stops.\nE: The pH increases and may go above 7 .\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_269",
        "problem": "A substance conducts electricity well when liquid but not when solid. Which of the following could this substance be? Select all that apply.\nA: copper\nB: sodium nitrate\nC: argon\nD: carbon tetrachloride\nE: boron nitride\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA substance conducts electricity well when liquid but not when solid. Which of the following could this substance be? Select all that apply.\n\nA: copper\nB: sodium nitrate\nC: argon\nD: carbon tetrachloride\nE: boron nitride\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_531",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{NaOH}$ 溶液分别滴定 $\\mathrm{HX} 、 \\mathrm{CuSO}_{4} 、 \\mathrm{FeSO}_{4}$ 三种溶液, $\\mathrm{pM}[\\mathrm{p}$ 表示负对数, $\\mathrm{M}$ 表示 $\\frac{\\mathrm{c}(\\mathrm{HX})}{\\mathrm{c}(\\mathrm{X})} 、 \\mathrm{c}\\left(\\mathrm{Cu}^{2+}\\right) 、 \\mathrm{c}\\left(\\mathrm{Fe}^{2+}\\right)$ 等 $]$ 随 $\\mathrm{pH}$ 变化关系如图所示, 已知 $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Cu}(\\mathrm{OH})_{2}\\right]$ $<\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Fe}(\\mathrm{OH})_{2}\\right]$, 下列说法正确的是\n\n[图1]\nA: (1)代表滴定 $\\mathrm{FeSO}_{4}$ 溶液的变化关系\nB: $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 不易溶于 $\\mathrm{HX}$ 溶液中，而 $\\mathrm{Fe}(\\mathrm{OH})_{2}$ 易溶于 $\\mathrm{HX}$ 溶液中\nC: 调整溶液的 $\\mathrm{pH}=7$ ，无法除去工业废水中的 $\\mathrm{Cu}^{2+}$\nD: 滴定 HX 溶液至 $a$ 点时, 溶液中一定满足 $c\\left(N^{+}\\right)>c\\left(X^{-}\\right)>c\\left(O^{-}\\right)>c\\left(H^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{NaOH}$ 溶液分别滴定 $\\mathrm{HX} 、 \\mathrm{CuSO}_{4} 、 \\mathrm{FeSO}_{4}$ 三种溶液, $\\mathrm{pM}[\\mathrm{p}$ 表示负对数, $\\mathrm{M}$ 表示 $\\frac{\\mathrm{c}(\\mathrm{HX})}{\\mathrm{c}(\\mathrm{X})} 、 \\mathrm{c}\\left(\\mathrm{Cu}^{2+}\\right) 、 \\mathrm{c}\\left(\\mathrm{Fe}^{2+}\\right)$ 等 $]$ 随 $\\mathrm{pH}$ 变化关系如图所示, 已知 $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Cu}(\\mathrm{OH})_{2}\\right]$ $<\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Fe}(\\mathrm{OH})_{2}\\right]$, 下列说法正确的是\n\n[图1]\n\nA: (1)代表滴定 $\\mathrm{FeSO}_{4}$ 溶液的变化关系\nB: $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 不易溶于 $\\mathrm{HX}$ 溶液中，而 $\\mathrm{Fe}(\\mathrm{OH})_{2}$ 易溶于 $\\mathrm{HX}$ 溶液中\nC: 调整溶液的 $\\mathrm{pH}=7$ ，无法除去工业废水中的 $\\mathrm{Cu}^{2+}$\nD: 滴定 HX 溶液至 $a$ 点时, 溶液中一定满足 $c\\left(N^{+}\\right)>c\\left(X^{-}\\right)>c\\left(O^{-}\\right)>c\\left(H^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-055.jpg?height=428&width=802&top_left_y=1428&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1274",
        "problem": "The chemical oxygen demand (COD) refers to the amount of oxidizable substance, such as organic compounds, in a sample solution, and it is used as an indication of water quality in seas, lakes, and marshes. For example, the COD of service water is kept below $1 \\mathrm{mg} \\mathrm{dm}^{-3}$. The COD $\\left(\\mathrm{mg} \\mathrm{dm}^{-3}\\right)$ is represented by mass of $\\mathrm{O}_{2}(\\mathrm{mg})$ which accepts the same amount of electrons which would be accepted by the strong oxidizing agent when $1 \\mathrm{dm}^{3}$ of a sample solution is treated with it. An example of the operation is presented below.\n\nA sample solution with a volume of $1.00 \\mathrm{dm}^{3}$ was acidified with a sufficient amount of sulphuric acid, and chloride ions were removed by the addition of silver nitrate solution. A volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of potassium permanganate solution $\\left(c=5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) was added to the sample solution, and the mixture was heated for $30 \\mathrm{~min}$. Further, a volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of disodium oxalate $\\left(\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ or $\\left.\\mathrm{NaOOC}-\\mathrm{COONa}\\right)$ standard solution ( $c=1$. $25 \\cdot 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) was added, and the mixture was stirred well. Oxalate ions that remained unreacted were titrated with potassium permanganate solution $(c=$ $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ). A volume of $3.00 \\cdot 10^{-2} \\mathrm{dm}^{3}$ of the solution was used for the titration.From the following choices, select the most appropriate reason for the removal of chloride ions:\nA: Some of the chloride ions react with potassium permanganate, resulting in an error in COD.\nB: Some of the chloride ions react with disodium oxalate, resulting in an error in COD.\nC: Some of the chloride ions react with organic compounds in the sample solution, resulting in an error in COD.\nD: A colour is developed during titration, resulting in an error in COD.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nThe chemical oxygen demand (COD) refers to the amount of oxidizable substance, such as organic compounds, in a sample solution, and it is used as an indication of water quality in seas, lakes, and marshes. For example, the COD of service water is kept below $1 \\mathrm{mg} \\mathrm{dm}^{-3}$. The COD $\\left(\\mathrm{mg} \\mathrm{dm}^{-3}\\right)$ is represented by mass of $\\mathrm{O}_{2}(\\mathrm{mg})$ which accepts the same amount of electrons which would be accepted by the strong oxidizing agent when $1 \\mathrm{dm}^{3}$ of a sample solution is treated with it. An example of the operation is presented below.\n\nA sample solution with a volume of $1.00 \\mathrm{dm}^{3}$ was acidified with a sufficient amount of sulphuric acid, and chloride ions were removed by the addition of silver nitrate solution. A volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of potassium permanganate solution $\\left(c=5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) was added to the sample solution, and the mixture was heated for $30 \\mathrm{~min}$. Further, a volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of disodium oxalate $\\left(\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ or $\\left.\\mathrm{NaOOC}-\\mathrm{COONa}\\right)$ standard solution ( $c=1$. $25 \\cdot 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) was added, and the mixture was stirred well. Oxalate ions that remained unreacted were titrated with potassium permanganate solution $(c=$ $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ). A volume of $3.00 \\cdot 10^{-2} \\mathrm{dm}^{3}$ of the solution was used for the titration.\n\nproblem:\nFrom the following choices, select the most appropriate reason for the removal of chloride ions:\n\nA: Some of the chloride ions react with potassium permanganate, resulting in an error in COD.\nB: Some of the chloride ions react with disodium oxalate, resulting in an error in COD.\nC: Some of the chloride ions react with organic compounds in the sample solution, resulting in an error in COD.\nD: A colour is developed during titration, resulting in an error in COD.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1047",
        "problem": "Former Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nLactic acid has an acid dissociation constant, $K_{\\mathrm{a}}=1.38 \\times 10^{-4}$.\n\nCalculate the $\\mathrm{p} K_{\\mathrm{a}}$ of lactic acid.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nFormer Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nLactic acid has an acid dissociation constant, $K_{\\mathrm{a}}=1.38 \\times 10^{-4}$.\n\nCalculate the $\\mathrm{p} K_{\\mathrm{a}}$ of lactic acid.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-13.jpg?height=616&width=699&top_left_y=323&top_left_x=1178",
            "https://i.postimg.cc/Vs42tGMJ/5.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1430",
        "problem": "Metallic gold frequently is found in aluminosilicate rocks and it is finely dispersed among other minerals. It may be extracted by treating the crushed rock with aerated sodium cyanide solution. During this process metallic gold is slowly converted to $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$, which is soluble in water (reaction 1 ).\n\nAfter equilibrium has been reached, the aqueous phase is pumped off and the metallic gold is recovered from it by reacting the gold complex with zinc, which is converted to $\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}$ (reaction 2).\n\n$\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$is a very stable complex under certain conditions. What concentration of sodium cyanide is required to keep $99 \\mathrm{~mol} \\%$ of the gold in solution in the form of the cyanide complex? $\\quad\\left\\{\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}: K_{f}=4 \\times 10^{28}\\right\\}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nMetallic gold frequently is found in aluminosilicate rocks and it is finely dispersed among other minerals. It may be extracted by treating the crushed rock with aerated sodium cyanide solution. During this process metallic gold is slowly converted to $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$, which is soluble in water (reaction 1 ).\n\nAfter equilibrium has been reached, the aqueous phase is pumped off and the metallic gold is recovered from it by reacting the gold complex with zinc, which is converted to $\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}$ (reaction 2).\n\n$\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$is a very stable complex under certain conditions. What concentration of sodium cyanide is required to keep $99 \\mathrm{~mol} \\%$ of the gold in solution in the form of the cyanide complex? $\\quad\\left\\{\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}: K_{f}=4 \\times 10^{28}\\right\\}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_866",
        "problem": "从中草药中提取的 calebin A（结构简式如下）可用于治疗阿尔茨海默症。下列关于 calebin A 的说法错误的是[图1]\nA: 可与 $\\mathrm{FeCl}_{3}$ 溶液发生显色反应\nB: 其酸性水解的产物均可与 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液反应\nC: 苯环上氢原子发生氯代时, 一氯代物有 3 种\nD: $1 \\mathrm{~mol}$ 该分子最多与 $8 \\mathrm{molH}_{2}$ 发生加成反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n从中草药中提取的 calebin A（结构简式如下）可用于治疗阿尔茨海默症。下列关于 calebin A 的说法错误的是[图1]\n\nA: 可与 $\\mathrm{FeCl}_{3}$ 溶液发生显色反应\nB: 其酸性水解的产物均可与 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液反应\nC: 苯环上氢原子发生氯代时, 一氯代物有 3 种\nD: $1 \\mathrm{~mol}$ 该分子最多与 $8 \\mathrm{molH}_{2}$ 发生加成反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/L6mrsCTR/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_52",
        "problem": "A $1.0-\\mathrm{L}$ container with $1.00 \\mathrm{~atm} \\mathrm{Ar}$ gas is connected by a stopcock to a 2.0-L container containing $0.5 \\mathrm{~atm} \\mathrm{Kr}$ gas. Which statement accurately describes the system after the stopcock is opened and the system is allowed to reach equilibrium? The temperature is $25^{\\circ} \\mathrm{C}$ throughout the experiment.\nA: The total pressure is $1.00 \\mathrm{~atm}$.\nB: The number of moles of Ar in the 2.0-L container is equal to the number of moles of $\\mathrm{Kr}$ in the $1.0-\\mathrm{L}$ container.\nC: The average speed of the Ar atoms is equal to the average speed of the $\\mathrm{Kr}$ atoms.\nD: The partial pressure of Ar is equal to the partial pressure of Kr.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $1.0-\\mathrm{L}$ container with $1.00 \\mathrm{~atm} \\mathrm{Ar}$ gas is connected by a stopcock to a 2.0-L container containing $0.5 \\mathrm{~atm} \\mathrm{Kr}$ gas. Which statement accurately describes the system after the stopcock is opened and the system is allowed to reach equilibrium? The temperature is $25^{\\circ} \\mathrm{C}$ throughout the experiment.\n\nA: The total pressure is $1.00 \\mathrm{~atm}$.\nB: The number of moles of Ar in the 2.0-L container is equal to the number of moles of $\\mathrm{Kr}$ in the $1.0-\\mathrm{L}$ container.\nC: The average speed of the Ar atoms is equal to the average speed of the $\\mathrm{Kr}$ atoms.\nD: The partial pressure of Ar is equal to the partial pressure of Kr.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1122",
        "problem": "Give half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{NO}_{3}{ }^{-}$to $\\mathrm{NO}_{2}$\nA: oxidation\nB: reduction\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGive half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{NO}_{3}{ }^{-}$to $\\mathrm{NO}_{2}$\n\nA: oxidation\nB: reduction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_319",
        "problem": "How many different structural isomers are there for the compound chlorobutane $\\left(\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{Cl}\\right)$ ?\nA: two\nB: three\nC: four\nD: five\nE: more than five\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many different structural isomers are there for the compound chlorobutane $\\left(\\mathrm{C}_{4} \\mathrm{H}_{9} \\mathrm{Cl}\\right)$ ?\n\nA: two\nB: three\nC: four\nD: five\nE: more than five\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_842",
        "problem": "乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right.$, 简写为 $\\left.\\mathrm{X}\\right)$ 为二元弱碱。 $25^{\\circ} \\mathrm{C}$ 时, 向 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 其盐酸盐溶液 $\\mathrm{XH}_{2} \\mathrm{Cl}_{2}$ 中加入固体 $\\mathrm{NaOH}\\left(\\right.$ 溶液体积变化忽略不计), 溶液 $\\mathrm{pH}$ 值, 体系中 $\\mathrm{XH}_{2}^{2+} 、$\n\n$\\mathrm{XH}^{+} 、 \\mathrm{X}$ 三种粒子的浓度的对数值 $(\\operatorname{lgc})$ 与 $\\mathrm{t}=\\frac{\\mathrm{n}(\\mathrm{NaOH})}{\\mathrm{n}\\left(\\mathrm{XH}_{2} \\mathrm{Cl}_{2}\\right)}$ 的关系如图所示。下列说法正确的是\n\n[图1]\nA: 乙二胺第二步电离常数的数量级为 $10^{-7}$\nB: $\\mathrm{t}_{1}$ 时, $3 \\mathrm{c}\\left(\\mathrm{XH}_{2}^{2+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nC: b 点时, $\\mathrm{c}\\left(\\mathrm{HX}^{+}\\right)>10 \\mathrm{c}\\left(\\mathrm{XH}_{2}^{2+}\\right)$\nD: $\\mathrm{c}$ 点时, $\\mathrm{t}_{2}=2$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right.$, 简写为 $\\left.\\mathrm{X}\\right)$ 为二元弱碱。 $25^{\\circ} \\mathrm{C}$ 时, 向 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 其盐酸盐溶液 $\\mathrm{XH}_{2} \\mathrm{Cl}_{2}$ 中加入固体 $\\mathrm{NaOH}\\left(\\right.$ 溶液体积变化忽略不计), 溶液 $\\mathrm{pH}$ 值, 体系中 $\\mathrm{XH}_{2}^{2+} 、$\n\n$\\mathrm{XH}^{+} 、 \\mathrm{X}$ 三种粒子的浓度的对数值 $(\\operatorname{lgc})$ 与 $\\mathrm{t}=\\frac{\\mathrm{n}(\\mathrm{NaOH})}{\\mathrm{n}\\left(\\mathrm{XH}_{2} \\mathrm{Cl}_{2}\\right)}$ 的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: 乙二胺第二步电离常数的数量级为 $10^{-7}$\nB: $\\mathrm{t}_{1}$ 时, $3 \\mathrm{c}\\left(\\mathrm{XH}_{2}^{2+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nC: b 点时, $\\mathrm{c}\\left(\\mathrm{HX}^{+}\\right)>10 \\mathrm{c}\\left(\\mathrm{XH}_{2}^{2+}\\right)$\nD: $\\mathrm{c}$ 点时, $\\mathrm{t}_{2}=2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-077.jpg?height=506&width=1059&top_left_y=2254&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_66",
        "problem": "The rate of the gas-phase reaction of nitrogen dioxide with carbon monoxide at $177^{\\circ} \\mathrm{C}$ was measured under three different sets of concentrations as shown. What is the rate law for this reaction under these conditions?\n\n| $\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\rightarrow \\mathrm{NO}(g)+\\mathrm{CO}_{2}(g)$ |  |  |  |\n| :---: | :---: | :---: | :---: |\n|  | Initial $\\left[\\mathrm{NO}_{2}\\right]$, <br> $\\mathrm{mol} \\mathrm{L}^{-1}$ | Initial [CO], <br> mol L $^{-1}$ | Initial rate, <br> $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~min}^{-1}$ |\n| Ex 1 | $4.0 \\times 10^{-4}$ | $1.7 \\times 10^{-2}$ | $1.70 \\times 10^{-7}$ |\n| Ex 2 | $4.0 \\times 10^{-4}$ | $3.4 \\times 10^{-2}$ | $1.70 \\times 10^{-7}$ |\n| Ex 3 | $1.2 \\times 10^{-3}$ | $3.4 \\times 10^{-2}$ | $1.53 \\times 10^{-6}$ |\nA: Rate $=k\\left[\\mathrm{NO}_{2}\\right]$\nB: Rate $=k\\left[\\mathrm{NO}_{2}\\right]^{2}$\nC: Rate $=k[\\mathrm{CO}]$\nD: Rate $=k\\left[\\mathrm{NO}_{2}\\right][\\mathrm{CO}]$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe rate of the gas-phase reaction of nitrogen dioxide with carbon monoxide at $177^{\\circ} \\mathrm{C}$ was measured under three different sets of concentrations as shown. What is the rate law for this reaction under these conditions?\n\n| $\\mathrm{NO}_{2}(g)+\\mathrm{CO}(g) \\rightarrow \\mathrm{NO}(g)+\\mathrm{CO}_{2}(g)$ |  |  |  |\n| :---: | :---: | :---: | :---: |\n|  | Initial $\\left[\\mathrm{NO}_{2}\\right]$, <br> $\\mathrm{mol} \\mathrm{L}^{-1}$ | Initial [CO], <br> mol L $^{-1}$ | Initial rate, <br> $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~min}^{-1}$ |\n| Ex 1 | $4.0 \\times 10^{-4}$ | $1.7 \\times 10^{-2}$ | $1.70 \\times 10^{-7}$ |\n| Ex 2 | $4.0 \\times 10^{-4}$ | $3.4 \\times 10^{-2}$ | $1.70 \\times 10^{-7}$ |\n| Ex 3 | $1.2 \\times 10^{-3}$ | $3.4 \\times 10^{-2}$ | $1.53 \\times 10^{-6}$ |\n\nA: Rate $=k\\left[\\mathrm{NO}_{2}\\right]$\nB: Rate $=k\\left[\\mathrm{NO}_{2}\\right]^{2}$\nC: Rate $=k[\\mathrm{CO}]$\nD: Rate $=k\\left[\\mathrm{NO}_{2}\\right][\\mathrm{CO}]$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_371",
        "problem": "Electrolysis of $1.00 \\mathrm{~g}$ of a copper(II) salt required passage of $0.100 \\mathrm{~A}$ of current for $123 \\mathrm{~min}$ for complete deposition of all the copper metal. What is the copper salt?\nA: $\\mathrm{CuCl}_{2}, M=134.5$\nB: $\\mathrm{CuBr}_{2}, M=223.4$\nC: $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}, M=187.6$\nD: $\\mathrm{Cu}\\left(\\mathrm{ClO}_{4}\\right)_{2}, M=262.5$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nElectrolysis of $1.00 \\mathrm{~g}$ of a copper(II) salt required passage of $0.100 \\mathrm{~A}$ of current for $123 \\mathrm{~min}$ for complete deposition of all the copper metal. What is the copper salt?\n\nA: $\\mathrm{CuCl}_{2}, M=134.5$\nB: $\\mathrm{CuBr}_{2}, M=223.4$\nC: $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}, M=187.6$\nD: $\\mathrm{Cu}\\left(\\mathrm{ClO}_{4}\\right)_{2}, M=262.5$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1057",
        "problem": "2022 was the $90^{\\text {th }}$ anniversary of Linus Pauling proposing the concept of electronegativity. His book The nature of the chemical bond is considered the most influential chemistry book of $20^{\\text {th }}$ century, and he was awarded the 1954 Nobel prize in chemistry for his work.\n\nElectronegativity, $\\chi$, is a measure of the ability of an atom to attract a pair of electrons in a covalent bond.\n\n[figure1]\n\nPauling used thermodynamic data to calculate the difference in electronegativity between two atoms A and B. All electronegativity values are positive with no units, and atom A has a higher electronegativity than atom $B$.\n\n$$\n\\chi_{A}-\\chi_{B}=0.102 \\sqrt{\\mathrm{B}_{d}(A B)-\\frac{\\mathrm{B}_{d}(A A)+\\mathrm{B}_{d}(B B)}{2}}\n$$\n\n$\\mathrm{B}_{d}$ represents the bond dissociation energies, in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, of the $\\mathrm{A}-\\mathrm{A}, \\mathrm{B}-\\mathrm{B}$ and $\\mathrm{A}-\\mathrm{B}$ bonds.\n\nAs this method gives a difference in electronegativity it is necesssary to choose a reference point in order to gain values for individual elements. Pauling originally chose hydrogen as the reference, with a value of 2.20 .\n\nOther scientists suggested other methods to determine the electronegativity of an element, which can be adjusted to be consistent with those determined by Pauling. Robert Mulliken calculated the value from the first ionisation energy $\\left(\\mathrm{E}_{i}\\right)$ and first electron affinity $\\left(\\mathrm{E}_{e a}\\right)$ of an atom.\n\n$$\n\\chi_{A}=0.00197\\left(\\mathrm{E}_{i}+\\mathrm{E}_{e a}\\right)+0.19\n$$\n\nIn this equation $\\mathrm{E}_{i}$ and $\\mathrm{E}_{e a}$ are in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, although $\\mathrm{E}_{i}$ is often measured in electronvolts (eV). $1 \\mathrm{eV}=96.49 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nCalculate the adjusted Mulliken electronegativity of nitrogen given that $\\mathrm{E}_{i}(\\mathrm{~N})=14.5 \\mathrm{eV}$ and $\\mathrm{E}_{e a}(\\mathrm{~N})=6.80 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n2022 was the $90^{\\text {th }}$ anniversary of Linus Pauling proposing the concept of electronegativity. His book The nature of the chemical bond is considered the most influential chemistry book of $20^{\\text {th }}$ century, and he was awarded the 1954 Nobel prize in chemistry for his work.\n\nElectronegativity, $\\chi$, is a measure of the ability of an atom to attract a pair of electrons in a covalent bond.\n\n[figure1]\n\nPauling used thermodynamic data to calculate the difference in electronegativity between two atoms A and B. All electronegativity values are positive with no units, and atom A has a higher electronegativity than atom $B$.\n\n$$\n\\chi_{A}-\\chi_{B}=0.102 \\sqrt{\\mathrm{B}_{d}(A B)-\\frac{\\mathrm{B}_{d}(A A)+\\mathrm{B}_{d}(B B)}{2}}\n$$\n\n$\\mathrm{B}_{d}$ represents the bond dissociation energies, in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, of the $\\mathrm{A}-\\mathrm{A}, \\mathrm{B}-\\mathrm{B}$ and $\\mathrm{A}-\\mathrm{B}$ bonds.\n\nAs this method gives a difference in electronegativity it is necesssary to choose a reference point in order to gain values for individual elements. Pauling originally chose hydrogen as the reference, with a value of 2.20 .\n\nOther scientists suggested other methods to determine the electronegativity of an element, which can be adjusted to be consistent with those determined by Pauling. Robert Mulliken calculated the value from the first ionisation energy $\\left(\\mathrm{E}_{i}\\right)$ and first electron affinity $\\left(\\mathrm{E}_{e a}\\right)$ of an atom.\n\n$$\n\\chi_{A}=0.00197\\left(\\mathrm{E}_{i}+\\mathrm{E}_{e a}\\right)+0.19\n$$\n\nIn this equation $\\mathrm{E}_{i}$ and $\\mathrm{E}_{e a}$ are in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, although $\\mathrm{E}_{i}$ is often measured in electronvolts (eV). $1 \\mathrm{eV}=96.49 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nCalculate the adjusted Mulliken electronegativity of nitrogen given that $\\mathrm{E}_{i}(\\mathrm{~N})=14.5 \\mathrm{eV}$ and $\\mathrm{E}_{e a}(\\mathrm{~N})=6.80 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-03.jpg?height=611&width=505&top_left_y=314&top_left_x=1301"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_433",
        "problem": "某课题组利用盐度梯度和氧化还原工艺成功设计了如图所示的电池装置, 将酸性废水中存在的高毒性 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 转化为 $\\mathrm{Cr}^{3+}$ 后再经后续处理转化为 $\\mathrm{Cr}(\\mathrm{OH})_{3}$ 离开溶液。设装置内温度恒定为 $298 \\mathrm{~K}$ 。\n\n[图1]\n\n已知: 对于电池正极反应 $\\mathrm{aOx}+\\mathrm{ne}^{-}=\\mathrm{bRed}$ 有 Nernst 方程\n\n$\\varphi_{+}(\\mathrm{Ox} / \\mathrm{Red})=\\varphi_{0}(\\mathrm{Ox} / \\mathrm{Red})+\\frac{2.303 \\mathrm{RT}}{\\mathrm{nF}} \\lg \\frac{[\\mathrm{Ox}]^{\\mathrm{a}}}{[\\text { Red }]^{\\mathrm{b}}}$\n\n注: (1) $298 \\mathrm{~K}$ 下, $\\frac{2.303 \\mathrm{RT}}{\\mathrm{F}}=0.0592, \\mathrm{n}$ 为转移电子数; 电势越高反应的趋势越大;\n\n(2) $\\mathrm{Ox}$ 为氧化态物质, Red 为还原态物质， $\\varphi_{0}(\\mathrm{Ox} / \\mathrm{Red})$ 为定值;\n\n(3)对于除 $\\mathrm{Ox} 、 \\mathrm{Red}$ 外还有其他如 $\\mathrm{H}^{+} 、 \\mathrm{OH}^{-}$等参加反应的物质，它们的浓度的幂次方也要代入方程。\n\n(参考数据: $\\frac{0.0592}{3} \\approx 0.0197$ )\n\n下列说法错误的是\nA: 该电池利用溶液盐度梯度产生电势, “C2in”表示低浓度盐溶液\nB: 负极产生标准状况下 $6.72 \\mathrm{LO}_{2}$, 理论上转化 $0.4 \\mathrm{molCr}_{2} \\mathrm{O}_{7}^{2-}$\nC: 若要升高正极电势, 可向正极区中加入 $\\mathrm{Na}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 固体或升高正极区 $\\mathrm{pH}$\nD: 若正极区 $\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{Cr}^{3+}\\right)=1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 恒定, 当正极区 $\\mathrm{pH}$ 由 1.0 降低为 0.5 时,\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某课题组利用盐度梯度和氧化还原工艺成功设计了如图所示的电池装置, 将酸性废水中存在的高毒性 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 转化为 $\\mathrm{Cr}^{3+}$ 后再经后续处理转化为 $\\mathrm{Cr}(\\mathrm{OH})_{3}$ 离开溶液。设装置内温度恒定为 $298 \\mathrm{~K}$ 。\n\n[图1]\n\n已知: 对于电池正极反应 $\\mathrm{aOx}+\\mathrm{ne}^{-}=\\mathrm{bRed}$ 有 Nernst 方程\n\n$\\varphi_{+}(\\mathrm{Ox} / \\mathrm{Red})=\\varphi_{0}(\\mathrm{Ox} / \\mathrm{Red})+\\frac{2.303 \\mathrm{RT}}{\\mathrm{nF}} \\lg \\frac{[\\mathrm{Ox}]^{\\mathrm{a}}}{[\\text { Red }]^{\\mathrm{b}}}$\n\n注: (1) $298 \\mathrm{~K}$ 下, $\\frac{2.303 \\mathrm{RT}}{\\mathrm{F}}=0.0592, \\mathrm{n}$ 为转移电子数; 电势越高反应的趋势越大;\n\n(2) $\\mathrm{Ox}$ 为氧化态物质, Red 为还原态物质， $\\varphi_{0}(\\mathrm{Ox} / \\mathrm{Red})$ 为定值;\n\n(3)对于除 $\\mathrm{Ox} 、 \\mathrm{Red}$ 外还有其他如 $\\mathrm{H}^{+} 、 \\mathrm{OH}^{-}$等参加反应的物质，它们的浓度的幂次方也要代入方程。\n\n(参考数据: $\\frac{0.0592}{3} \\approx 0.0197$ )\n\n下列说法错误的是\n\nA: 该电池利用溶液盐度梯度产生电势, “C2in”表示低浓度盐溶液\nB: 负极产生标准状况下 $6.72 \\mathrm{LO}_{2}$, 理论上转化 $0.4 \\mathrm{molCr}_{2} \\mathrm{O}_{7}^{2-}$\nC: 若要升高正极电势, 可向正极区中加入 $\\mathrm{Na}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 固体或升高正极区 $\\mathrm{pH}$\nD: 若正极区 $\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{Cr}^{3+}\\right)=1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 恒定, 当正极区 $\\mathrm{pH}$ 由 1.0 降低为 0.5 时,\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-03.jpg?height=776&width=1239&top_left_y=820&top_left_x=320"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1141",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhat is the oxidation state of the indium in $\\mathrm{In}_{2} \\mathrm{O}_{3}$ ?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhat is the oxidation state of the indium in $\\mathrm{In}_{2} \\mathrm{O}_{3}$ ?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_627",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 向 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{X}$ 溶液中滴入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, 溶液中由水电离出的 $c_{\\text {水 }}\\left(\\mathrm{OH}^{-}\\right)$的负对数 $\\left[-1 \\mathrm{~g} c_{\\text {水 }}\\left(\\mathrm{OH}^{-}\\right)\\right]$与所加 $\\mathrm{NaOH}$ 溶液体积的关系如图所示。\n\n[图1]\n\n下列说法中正确的是\nA: 水的电离程度: $\\mathrm{M}>\\mathrm{N}=\\mathrm{Q}>\\mathrm{P}$\nB: 图中 $\\mathrm{M} 、 \\mathrm{P} 、 \\mathrm{Q}$ 三点对应溶液中 $\\frac{\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)}$ 相等\nC: $\\mathrm{N}$ 点溶液中 $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{X}^{2-}\\right)>c\\left(\\mathrm{HX}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)$\nD: P 点溶液中 $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{HX}^{-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 向 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{X}$ 溶液中滴入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, 溶液中由水电离出的 $c_{\\text {水 }}\\left(\\mathrm{OH}^{-}\\right)$的负对数 $\\left[-1 \\mathrm{~g} c_{\\text {水 }}\\left(\\mathrm{OH}^{-}\\right)\\right]$与所加 $\\mathrm{NaOH}$ 溶液体积的关系如图所示。\n\n[图1]\n\n下列说法中正确的是\n\nA: 水的电离程度: $\\mathrm{M}>\\mathrm{N}=\\mathrm{Q}>\\mathrm{P}$\nB: 图中 $\\mathrm{M} 、 \\mathrm{P} 、 \\mathrm{Q}$ 三点对应溶液中 $\\frac{\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)}$ 相等\nC: $\\mathrm{N}$ 点溶液中 $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{X}^{2-}\\right)>c\\left(\\mathrm{HX}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)$\nD: P 点溶液中 $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{HX}^{-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-50.jpg?height=466&width=764&top_left_y=681&top_left_x=332"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_353",
        "problem": "In which of the following ways may a catalyst increase the rate of a reaction?\n\nI. It may alter the rate law.\n\nII. It may decrease the overall activation energy.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn which of the following ways may a catalyst increase the rate of a reaction?\n\nI. It may alter the rate law.\n\nII. It may decrease the overall activation energy.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_85",
        "problem": "A solution containing a divalent metal ion forms a precipitate when treated with hydrogen sulfide which does not dissolve in $1 \\mathrm{M} \\mathrm{HCl}$. Which ion is present?\nA: $\\mathrm{Ca}^{2+}$\nB: $\\mathrm{Mn}^{2+}$\nC: $\\mathrm{Cd}^{2+}$\nD: $\\mathrm{Ba}^{2+}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA solution containing a divalent metal ion forms a precipitate when treated with hydrogen sulfide which does not dissolve in $1 \\mathrm{M} \\mathrm{HCl}$. Which ion is present?\n\nA: $\\mathrm{Ca}^{2+}$\nB: $\\mathrm{Mn}^{2+}$\nC: $\\mathrm{Cd}^{2+}$\nD: $\\mathrm{Ba}^{2+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1049",
        "problem": "The volume of the $\\mathrm{c}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$ is $4.78 \\times 10^{-23} \\mathrm{~cm}^{3}$.Boron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the density of $\\mathrm{c}-\\mathrm{BN}$ in $\\mathrm{g} \\mathrm{cm}^{-3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe volume of the $\\mathrm{c}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$ is $4.78 \\times 10^{-23} \\mathrm{~cm}^{3}$.\n\nproblem:\nBoron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the density of $\\mathrm{c}-\\mathrm{BN}$ in $\\mathrm{g} \\mathrm{cm}^{-3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~g} \\mathrm{~cm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=448&width=511&top_left_y=541&top_left_x=270",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=596&width=436&top_left_y=458&top_left_x=844",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=471&width=443&top_left_y=524&top_left_x=1389",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=603&width=305&top_left_y=1206&top_left_x=1315"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~g} \\mathrm{~cm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_848",
        "problem": "常温下, 向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中滴加同浓度的 $\\mathrm{NaOH}$ 溶液, 溶液中一\n$\\mathrm{pY}\\left[\\mathrm{pY}=-\\operatorname{lgY}, \\mathrm{Y}\\right.$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)}$ 或 $\\left.\\frac{\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}\\right]$与溶液 $\\mathrm{pH}$ 的变化关系如图所示。下列叙\n\n述错误的是\n\n[图1]\nA: 直线 $\\mathrm{a}$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)}$ 与 $\\mathrm{pH}$ 的变化关系\nB: Q 点溶液中, $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $\\mathrm{N}$ 点溶液中, $3 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 常温下, $\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 的水解常数 $\\mathrm{K}_{\\mathrm{h} 1}=10^{-11.18}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中滴加同浓度的 $\\mathrm{NaOH}$ 溶液, 溶液中一\n$\\mathrm{pY}\\left[\\mathrm{pY}=-\\operatorname{lgY}, \\mathrm{Y}\\right.$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)}$ 或 $\\left.\\frac{\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}\\right]$与溶液 $\\mathrm{pH}$ 的变化关系如图所示。下列叙\n\n述错误的是\n\n[图1]\n\nA: 直线 $\\mathrm{a}$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)}$ 与 $\\mathrm{pH}$ 的变化关系\nB: Q 点溶液中, $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $\\mathrm{N}$ 点溶液中, $3 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 常温下, $\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 的水解常数 $\\mathrm{K}_{\\mathrm{h} 1}=10^{-11.18}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-104.jpg?height=377&width=551&top_left_y=423&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_996",
        "problem": "These three compounds have been isolated: $\\mathrm{NaCl}$, $\\mathrm{Na}_{2} \\mathrm{O}$, and $\\mathrm{AlCl}_{3}$. What is the formula of aluminum oxide?\nA: $\\mathrm{Al}_{2} \\mathrm{O}$\nB: $\\mathrm{Al}_{2} \\mathrm{O}_{3}$\nC: $\\mathrm{Al}_{3} \\mathrm{O}$\nD: $\\mathrm{AlO}$\nE: $\\quad \\mathrm{AlO}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThese three compounds have been isolated: $\\mathrm{NaCl}$, $\\mathrm{Na}_{2} \\mathrm{O}$, and $\\mathrm{AlCl}_{3}$. What is the formula of aluminum oxide?\n\nA: $\\mathrm{Al}_{2} \\mathrm{O}$\nB: $\\mathrm{Al}_{2} \\mathrm{O}_{3}$\nC: $\\mathrm{Al}_{3} \\mathrm{O}$\nD: $\\mathrm{AlO}$\nE: $\\quad \\mathrm{AlO}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1174",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhen Mendeleev created his first table indium had only just been discovered and its chemical properties had not been fully studied. The error in the mass arose from the fact that the correct formula for indium oxide was not known.\n\nIn one experiment to determine the atomic mass of indium, $0.5135 \\mathrm{~g}$ of indium metal was converted to $0.6243 \\mathrm{~g}$ of the oxide. Using these data, and the modern relative atomic mass of oxygen, calculate the apparent atomic mass of indium assuming:\n\nthe formula for indium oxide is $\\ln \\mathrm{O}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhen Mendeleev created his first table indium had only just been discovered and its chemical properties had not been fully studied. The error in the mass arose from the fact that the correct formula for indium oxide was not known.\n\nIn one experiment to determine the atomic mass of indium, $0.5135 \\mathrm{~g}$ of indium metal was converted to $0.6243 \\mathrm{~g}$ of the oxide. Using these data, and the modern relative atomic mass of oxygen, calculate the apparent atomic mass of indium assuming:\n\nthe formula for indium oxide is $\\ln \\mathrm{O}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_464",
        "problem": "在体积为 $2 \\mathrm{~L}$ 的密闭容器中进行下列反应: $\\mathrm{C}(\\mathrm{g})+\\mathrm{CO}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{CO}(\\mathrm{g}) ; \\Delta H=\\mathrm{Q} \\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$ 。下图为 $\\mathrm{CO}_{2} 、 \\mathrm{CO}$ 的物质的量随时间 $\\mathrm{t}$ 的变化关系图。下列说法不正确的是 ( )\n\n[图1]\nA: $\\mathrm{CO}$ 在 2-3min 和 4-5 min 时平均速率不相等\nB: 当固焦炭的质量不发生变化时, 说明反应已达平衡状态\nC: $5 \\mathrm{~min}$ 时再充入一定量的 $\\mathrm{CO}, \\mathrm{n}(\\mathrm{CO}) 、 \\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)$ 的变化可分别由 $\\mathrm{c} 、 \\mathrm{~b}$ 曲线表示\nD: $3 \\mathrm{~min}$ 时温度由 $\\mathrm{T}_{1}$ 升高到 $\\mathrm{T}_{2}$, 重新平衡时 $K\\left(\\mathrm{~T}_{2}\\right)$ 小于 $K\\left(\\mathrm{~T}_{1}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在体积为 $2 \\mathrm{~L}$ 的密闭容器中进行下列反应: $\\mathrm{C}(\\mathrm{g})+\\mathrm{CO}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{CO}(\\mathrm{g}) ; \\Delta H=\\mathrm{Q} \\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$ 。下图为 $\\mathrm{CO}_{2} 、 \\mathrm{CO}$ 的物质的量随时间 $\\mathrm{t}$ 的变化关系图。下列说法不正确的是 ( )\n\n[图1]\n\nA: $\\mathrm{CO}$ 在 2-3min 和 4-5 min 时平均速率不相等\nB: 当固焦炭的质量不发生变化时, 说明反应已达平衡状态\nC: $5 \\mathrm{~min}$ 时再充入一定量的 $\\mathrm{CO}, \\mathrm{n}(\\mathrm{CO}) 、 \\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)$ 的变化可分别由 $\\mathrm{c} 、 \\mathrm{~b}$ 曲线表示\nD: $3 \\mathrm{~min}$ 时温度由 $\\mathrm{T}_{1}$ 升高到 $\\mathrm{T}_{2}$, 重新平衡时 $K\\left(\\mathrm{~T}_{2}\\right)$ 小于 $K\\left(\\mathrm{~T}_{1}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-007.jpg?height=417&width=620&top_left_y=360&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_896",
        "problem": "相同温度和压强下, 研究氯气分别在不同浓度的盐酸和氯化钠溶液中溶解情况, 实验测得如图所示。\n[图1]\n\n图 $1 \\mathrm{NaCl}$ 溶液浓度与溶解 $\\mathrm{Cl}_{2}$ 及含氯微粒的浓度变化 图 2 盐酸浓度与溶解 $\\mathrm{Cl}_{2}$ 的浓度变化\n\n已知：氯气溶解过程发生如下反应:\n\n(1) $\\mathrm{Cl}_{2}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{HClO}(\\mathrm{aq}) \\quad K_{1}=4.2 \\times 10^{-4}$\n\n(2) $\\mathrm{Cl}_{2}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{Cl}_{3}^{-}(\\mathrm{aq}) \\quad K_{2}=0.19$\n\n(3) $\\mathrm{HClO}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{ClO}^{-}(\\mathrm{aq}) \\quad K_{\\mathrm{a}}=3.2 \\times 10^{-8}$\n\n下列说法正确的是\nA: $2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸中: $c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{Cl}_{3}^{-}\\right)>c(\\mathrm{HClO})>c\\left(\\mathrm{ClO}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)$\nB: 在 $2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaCl}$ 溶液中: $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{Cl}^{-}\\right)+c\\left(\\mathrm{Cl}_{3}^{-}\\right)-c(\\mathrm{HClO})-c\\left(\\mathrm{ClO}^{-}\\right)$\nC: 由图 2 推知在稀硫酸中, 随硫酸浓度增大 $\\mathrm{Cl}_{2}$ 的溶解度会增大\nD: $\\mathrm{NaCl}$ 曲线中, 随着 $\\mathrm{Cl}_{2}$ 溶解度的减小, 溶液中 $\\frac{n\\left(\\mathrm{ClO}^{-}\\right)}{n(\\mathrm{HClO})}$ 减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n相同温度和压强下, 研究氯气分别在不同浓度的盐酸和氯化钠溶液中溶解情况, 实验测得如图所示。\n[图1]\n\n图 $1 \\mathrm{NaCl}$ 溶液浓度与溶解 $\\mathrm{Cl}_{2}$ 及含氯微粒的浓度变化 图 2 盐酸浓度与溶解 $\\mathrm{Cl}_{2}$ 的浓度变化\n\n已知：氯气溶解过程发生如下反应:\n\n(1) $\\mathrm{Cl}_{2}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq})+\\mathrm{HClO}(\\mathrm{aq}) \\quad K_{1}=4.2 \\times 10^{-4}$\n\n(2) $\\mathrm{Cl}_{2}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{Cl}_{3}^{-}(\\mathrm{aq}) \\quad K_{2}=0.19$\n\n(3) $\\mathrm{HClO}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{H}^{+}(\\mathrm{aq})+\\mathrm{ClO}^{-}(\\mathrm{aq}) \\quad K_{\\mathrm{a}}=3.2 \\times 10^{-8}$\n\n下列说法正确的是\n\nA: $2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸中: $c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{Cl}_{3}^{-}\\right)>c(\\mathrm{HClO})>c\\left(\\mathrm{ClO}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)$\nB: 在 $2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaCl}$ 溶液中: $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{Cl}^{-}\\right)+c\\left(\\mathrm{Cl}_{3}^{-}\\right)-c(\\mathrm{HClO})-c\\left(\\mathrm{ClO}^{-}\\right)$\nC: 由图 2 推知在稀硫酸中, 随硫酸浓度增大 $\\mathrm{Cl}_{2}$ 的溶解度会增大\nD: $\\mathrm{NaCl}$ 曲线中, 随着 $\\mathrm{Cl}_{2}$ 溶解度的减小, 溶液中 $\\frac{n\\left(\\mathrm{ClO}^{-}\\right)}{n(\\mathrm{HClO})}$ 减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-025.jpg?height=394&width=1052&top_left_y=1553&top_left_x=384"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_247",
        "problem": "Which of the following has the largest atomic radius?\nA: $\\mathrm{Rb}$\nB: Xe\nC: $\\mathrm{K}$\nD: $\\mathrm{Kr}$\nE: Sb\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following has the largest atomic radius?\n\nA: $\\mathrm{Rb}$\nB: Xe\nC: $\\mathrm{K}$\nD: $\\mathrm{Kr}$\nE: Sb\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_118",
        "problem": "A student combines $75 \\mathrm{~mL}$ of $0.500 \\mathrm{~mol} \\mathrm{~L}^{-1}$ hydrochloric acid with $55 \\mathrm{~mL}$ of $0.125 \\mathrm{M} \\mathrm{KOH}$. What is the $\\mathrm{pH}$ of the resulting solution?\nA: 0.30\nB: 0.39\nC: 0.63\nD: 1.51\nE: 7.00\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student combines $75 \\mathrm{~mL}$ of $0.500 \\mathrm{~mol} \\mathrm{~L}^{-1}$ hydrochloric acid with $55 \\mathrm{~mL}$ of $0.125 \\mathrm{M} \\mathrm{KOH}$. What is the $\\mathrm{pH}$ of the resulting solution?\n\nA: 0.30\nB: 0.39\nC: 0.63\nD: 1.51\nE: 7.00\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_379",
        "problem": "The lattice energy (energy required to separate the ions in an ionic solid) of $\\mathrm{MgO}$ is much larger than that of $\\mathrm{LiF}$. What contributes the most to this difference?\nA: $\\mathrm{Mg}^{2+}$ is a smaller ion than $\\mathrm{Li}^{+}$, and $\\mathrm{O}^{2-}$ is a smaller ion than $\\mathrm{F}^{-}$.\nB: $\\mathrm{F}$ is more electronegative than $\\mathrm{O}$, and $\\mathrm{Li}$ is more electropositive than Mg.\nC: MgO contains doubly charged ions, while LiF contains singly charged ions.\nD: $\\mathrm{MgO}$ contains more electrons than LiF.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe lattice energy (energy required to separate the ions in an ionic solid) of $\\mathrm{MgO}$ is much larger than that of $\\mathrm{LiF}$. What contributes the most to this difference?\n\nA: $\\mathrm{Mg}^{2+}$ is a smaller ion than $\\mathrm{Li}^{+}$, and $\\mathrm{O}^{2-}$ is a smaller ion than $\\mathrm{F}^{-}$.\nB: $\\mathrm{F}$ is more electronegative than $\\mathrm{O}$, and $\\mathrm{Li}$ is more electropositive than Mg.\nC: MgO contains doubly charged ions, while LiF contains singly charged ions.\nD: $\\mathrm{MgO}$ contains more electrons than LiF.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_98",
        "problem": "In a reaction between two substances $A$ and $B$, the rate is found to be proportional to the concentration of $A$ at all concentrations studied. At low concentrations of B, the rate of the reaction is found to be proportional to [B], but to level off at high concentrations of $\\mathrm{B}$. What is a reasonable explanation for the observed dependence of rate on $[\\mathrm{B}]$ ?\nA: $\\mathrm{B}$ binds to $\\mathrm{A}$ prior to the rate-determining step.\nB: The reaction has an order of -1 in $\\mathrm{B}$.\nC: The reaction involves a 1:2 stoichiometry of A:B.\nD: The reaction is catalyzed by one of its products.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a reaction between two substances $A$ and $B$, the rate is found to be proportional to the concentration of $A$ at all concentrations studied. At low concentrations of B, the rate of the reaction is found to be proportional to [B], but to level off at high concentrations of $\\mathrm{B}$. What is a reasonable explanation for the observed dependence of rate on $[\\mathrm{B}]$ ?\n\nA: $\\mathrm{B}$ binds to $\\mathrm{A}$ prior to the rate-determining step.\nB: The reaction has an order of -1 in $\\mathrm{B}$.\nC: The reaction involves a 1:2 stoichiometry of A:B.\nD: The reaction is catalyzed by one of its products.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1019",
        "problem": "What is the mass percentage of copper in $\\mathrm{CuCl}_{2}$ ?\nA: $12.1 \\%$\nB: $64.2 \\%$\nC: $91.2 \\%$\nD: $25.2 \\%$\nE: $47.3 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the mass percentage of copper in $\\mathrm{CuCl}_{2}$ ?\n\nA: $12.1 \\%$\nB: $64.2 \\%$\nC: $91.2 \\%$\nD: $25.2 \\%$\nE: $47.3 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_217",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the amount of pure KHP (in $\\mathrm{mol}$ or $\\mathrm{mmol}$ ) required to react completely with $4.359 \\mathrm{~g}$ of the sodium hydroxide solution.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the amount of pure KHP (in $\\mathrm{mol}$ or $\\mathrm{mmol}$ ) required to react completely with $4.359 \\mathrm{~g}$ of the sodium hydroxide solution.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mmol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mmol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1198",
        "problem": "For sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.The Arrhenius activation energy for this catalytic hydrolysis of sodium borohydride is $E_{\\mathrm{a}}=42.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Calculate the temperature required to achieve the same rate of hydrogen evolution by using a half of the amount of ruthenium catalyst used at $25.0^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\nHere is some context information for this question, which might assist you in solving it:\nFor sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.\n\nproblem:\nThe Arrhenius activation energy for this catalytic hydrolysis of sodium borohydride is $E_{\\mathrm{a}}=42.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Calculate the temperature required to achieve the same rate of hydrogen evolution by using a half of the amount of ruthenium catalyst used at $25.0^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir units are, in order, [K, $^{\\circ} \\mathrm{C}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            "K",
            "$^{\\circ} \\mathrm{C}$"
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1020",
        "problem": "The difference between deuterium, ${ }_{1}^{2} \\mathrm{H}$, and the more common form hydrogen is that deuterium\nA: does not occur naturally.\nB: is radioactive.\nC: has one more atom per molecule.\nD: has one more proton in the nucleus.\nE: has one more neutron in the nucleus.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe difference between deuterium, ${ }_{1}^{2} \\mathrm{H}$, and the more common form hydrogen is that deuterium\n\nA: does not occur naturally.\nB: is radioactive.\nC: has one more atom per molecule.\nD: has one more proton in the nucleus.\nE: has one more neutron in the nucleus.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_754",
        "problem": "以四甲基氯化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NCl}\\right]$ 水溶液为原料, 利用光伏并网发电装置制备四甲基氢氧化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}\\right]$, 下列叙述正确的是\n\n[图1]\nA: 光伏并网发电装置中 $P$ 型半导体为负极\nB: 保持电流恒定, 升高温度, 制备 $\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}$ 的反应速率不变\nC: $\\mathrm{c}$ 为阳离子交换膜, $\\mathrm{e}$ 为阴离子交换膜\nD: b 极收集 $2.24 \\mathrm{~L}$ (标况)气体时, 溶液中共有 $0.4 \\mathrm{~mol}$ 离子透过交换膜\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n以四甲基氯化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NCl}\\right]$ 水溶液为原料, 利用光伏并网发电装置制备四甲基氢氧化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}\\right]$, 下列叙述正确的是\n\n[图1]\n\nA: 光伏并网发电装置中 $P$ 型半导体为负极\nB: 保持电流恒定, 升高温度, 制备 $\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}$ 的反应速率不变\nC: $\\mathrm{c}$ 为阳离子交换膜, $\\mathrm{e}$ 为阴离子交换膜\nD: b 极收集 $2.24 \\mathrm{~L}$ (标况)气体时, 溶液中共有 $0.4 \\mathrm{~mol}$ 离子透过交换膜\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-017.jpg?height=419&width=1259&top_left_y=156&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_385",
        "problem": "$1.000 \\mathrm{~g}$ of a weak base is titrated with $1.000 \\mathrm{M}$ aqueous $\\mathrm{HCl}$ to give the data shown. What is the identity of the base?\n\n[figure1]\nA: Ammonia, $\\mathrm{NH}_{3}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{NH}_{4}{ }^{+}=9.3\\right)$\nB: Aniline, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}=4.6\\right)$\nC: Hydroxylamine, $\\mathrm{NH}_{2} \\mathrm{OH}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{NH}_{3} \\mathrm{OH}^{+}=6.0\\right)$\nD: Trimethylamine, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}=9.8\\right)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n$1.000 \\mathrm{~g}$ of a weak base is titrated with $1.000 \\mathrm{M}$ aqueous $\\mathrm{HCl}$ to give the data shown. What is the identity of the base?\n\n[figure1]\n\nA: Ammonia, $\\mathrm{NH}_{3}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{NH}_{4}{ }^{+}=9.3\\right)$\nB: Aniline, $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{2}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{NH}_{3}{ }^{+}=4.6\\right)$\nC: Hydroxylamine, $\\mathrm{NH}_{2} \\mathrm{OH}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\mathrm{NH}_{3} \\mathrm{OH}^{+}=6.0\\right)$\nD: Trimethylamine, $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{~N}\\left(\\mathrm{p} K_{\\mathrm{a}}\\right.$ of $\\left.\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{NH}^{+}=9.8\\right)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_08_3c4530151c35262e5915g-6.jpg?height=632&width=699&top_left_y=340&top_left_x=209"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1562",
        "problem": "Lactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n$3 .00 \\times 10^{-3}$ mol of lactic acid $(\\mathrm{HL})$ is added to $1.00 \\mathrm{dm}^{3}$ of $0.024 \\mathrm{M}$ solution of $\\mathrm{NaHCO}_{3}$ (no change in volume, $\\mathrm{HL}$ completely neutralized).\n\ni) Calculate the value of $\\mathrm{pH}$ in the solution of $\\mathrm{NaHCO}_{3}$ before $\\mathrm{HL}$ is added.\n\nii) Calculate the value of $\\mathrm{pH}$ in the solution after the addition of $\\mathrm{HL}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nLactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n$3 .00 \\times 10^{-3}$ mol of lactic acid $(\\mathrm{HL})$ is added to $1.00 \\mathrm{dm}^{3}$ of $0.024 \\mathrm{M}$ solution of $\\mathrm{NaHCO}_{3}$ (no change in volume, $\\mathrm{HL}$ completely neutralized).\n\ni) Calculate the value of $\\mathrm{pH}$ in the solution of $\\mathrm{NaHCO}_{3}$ before $\\mathrm{HL}$ is added.\n\nii) Calculate the value of $\\mathrm{pH}$ in the solution after the addition of $\\mathrm{HL}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [ $\\mathrm{pH}$ in the solution of $\\mathrm{NaHCO}_{3}$ ,  the value of $\\mathrm{pH}$ in the solution after the addition of $\\mathrm{HL}$.].\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            " $\\mathrm{pH}$ in the solution of $\\mathrm{NaHCO}_{3}$ ",
            " the value of $\\mathrm{pH}$ in the solution after the addition of $\\mathrm{HL}$."
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_20",
        "problem": "An enzyme catalyzes the hydrolysis of an ester with a certain activity, but this activity is lost in a $3 \\mathrm{M}$ urea solution. What is the most likely explanation for the loss of activity?\nA: Urea binds to the active site of the enzyme competitively with the substrate.\nB: Urea causes the cleavage of the peptide bonds in the enzyme.\nC: Urea causes the enzyme to denature and lose its specific three-dimensional shape.\nD: Urea reacts with disulfide bonds in the enzyme.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn enzyme catalyzes the hydrolysis of an ester with a certain activity, but this activity is lost in a $3 \\mathrm{M}$ urea solution. What is the most likely explanation for the loss of activity?\n\nA: Urea binds to the active site of the enzyme competitively with the substrate.\nB: Urea causes the cleavage of the peptide bonds in the enzyme.\nC: Urea causes the enzyme to denature and lose its specific three-dimensional shape.\nD: Urea reacts with disulfide bonds in the enzyme.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1518",
        "problem": "Regarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe absorbance is linearly related to the wavelength of the incident light.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nRegarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe absorbance is linearly related to the wavelength of the incident light.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1271",
        "problem": "Contemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.Mark the factors that can make the efficiency of a heat engine higher:\nA: Increasing the friction in the mechanical parts of the engine\nB:  Increasing the burning temperature of the fuel in the engine\nC: Narrowing the working temperature interval of the engine\nD:  Increasing the working pressure of the gas\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nContemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nproblem:\nMark the factors that can make the efficiency of a heat engine higher:\n\nA: Increasing the friction in the mechanical parts of the engine\nB:  Increasing the burning temperature of the fuel in the engine\nC: Narrowing the working temperature interval of the engine\nD:  Increasing the working pressure of the gas\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_62",
        "problem": "A sample of a washing powder that contains a mixture of $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ and $\\mathrm{NaHCO}_{3}$ is titrated with aqueous $\\mathrm{HCl}$ and the following result is obtained:\n\n[figure1]\n\nWhat is the mole ratio of $\\mathrm{CO}_{3}{ }^{2-}$ to $\\mathrm{HCO}_{3}^{-}$in the washing powder?\nA: $2 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 1 \\mathrm{~mol} \\mathrm{HCO}_{3}{ }^{-}$\nB: $1 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 1 \\mathrm{~mol} \\mathrm{HCO}_{3}^{-}$\nC: $1 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 2 \\mathrm{~mol} \\mathrm{HCO}_{3}^{-}$\nD: $1 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 3 \\mathrm{~mol} \\mathrm{HCO}_{3}{ }^{-}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA sample of a washing powder that contains a mixture of $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ and $\\mathrm{NaHCO}_{3}$ is titrated with aqueous $\\mathrm{HCl}$ and the following result is obtained:\n\n[figure1]\n\nWhat is the mole ratio of $\\mathrm{CO}_{3}{ }^{2-}$ to $\\mathrm{HCO}_{3}^{-}$in the washing powder?\n\nA: $2 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 1 \\mathrm{~mol} \\mathrm{HCO}_{3}{ }^{-}$\nB: $1 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 1 \\mathrm{~mol} \\mathrm{HCO}_{3}^{-}$\nC: $1 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 2 \\mathrm{~mol} \\mathrm{HCO}_{3}^{-}$\nD: $1 \\mathrm{~mol} \\mathrm{CO}_{3}{ }^{2-}: 3 \\mathrm{~mol} \\mathrm{HCO}_{3}{ }^{-}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-07.jpg?height=485&width=631&top_left_y=625&top_left_x=259"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_807",
        "problem": "一种微生物燃料电池如图所示, 下列关于该电池说法正确的是\n\n[图1]\nA: a 电极为正极, 发生还原反应\nB: $\\mathrm{H}^{+}$由左室通过质子交换膜进入右室\nC: 当 $\\mathrm{b}$ 电极上产生 $1 \\mathrm{molN}_{2}$ 时, 溶液中将有 $10 \\mathrm{~mole}$ 通过\nD: $b$ 电极反应式为: $2 \\mathrm{NO}_{3}^{-}+10 \\mathrm{e}^{-}+12 \\mathrm{H}^{+}=\\mathrm{N}_{2} \\uparrow+6 \\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种微生物燃料电池如图所示, 下列关于该电池说法正确的是\n\n[图1]\n\nA: a 电极为正极, 发生还原反应\nB: $\\mathrm{H}^{+}$由左室通过质子交换膜进入右室\nC: 当 $\\mathrm{b}$ 电极上产生 $1 \\mathrm{molN}_{2}$ 时, 溶液中将有 $10 \\mathrm{~mole}$ 通过\nD: $b$ 电极反应式为: $2 \\mathrm{NO}_{3}^{-}+10 \\mathrm{e}^{-}+12 \\mathrm{H}^{+}=\\mathrm{N}_{2} \\uparrow+6 \\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-37.jpg?height=446&width=803&top_left_y=1242&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_668",
        "problem": "某二元酸 $\\mathrm{H}_{2} \\mathrm{X}$ 在水中的电离方程式是 $\\mathrm{H}_{2} \\mathrm{X} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{HX}^{-}, \\mathrm{HX}^{-} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{X}^{2-}$, 常温\n\n下, 向 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{X}$ 溶液中滴入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, $\\mathrm{pOH}_{\\text {水 }}$ 与所加 $\\mathrm{NaOH}$溶液体积的关系如图所示。下列说法错误的是\n\n[图1]\n\n[图2]\nA: 常温下, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{X}$ 溶液的 $\\mathrm{pH}$ 为 1.4\nB: 溶液中水的电离程度: $d>c>b$\nC: e 点溶液呈碱性\nD: $\\mathrm{d}$ 点溶液中存在: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某二元酸 $\\mathrm{H}_{2} \\mathrm{X}$ 在水中的电离方程式是 $\\mathrm{H}_{2} \\mathrm{X} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{HX}^{-}, \\mathrm{HX}^{-} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{X}^{2-}$, 常温\n\n下, 向 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{X}$ 溶液中滴入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, $\\mathrm{pOH}_{\\text {水 }}$ 与所加 $\\mathrm{NaOH}$溶液体积的关系如图所示。下列说法错误的是\n\n[图1]\n\n[图2]\n\nA: 常温下, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{X}$ 溶液的 $\\mathrm{pH}$ 为 1.4\nB: 溶液中水的电离程度: $d>c>b$\nC: e 点溶液呈碱性\nD: $\\mathrm{d}$ 点溶液中存在: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-061.jpg?height=63&width=1273&top_left_y=1505&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-061.jpg?height=399&width=688&top_left_y=1608&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_479",
        "problem": "肼 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 是一种二元弱碱。常温下, 向 $1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{~N}_{2} \\mathrm{H}_{4}$ 水溶液中滴加盐酸, 所得混合溶液中 $-\\lg \\frac{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)}{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)}$或 $-\\lg \\frac{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)}{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{6}^{2+}\\right)}$ 与 $\\mathrm{pH}$ 的变化关系如图所示。下列说法错误的是\n\n[图1]\nA: 常温下, $\\mathrm{N}_{2} \\mathrm{H}_{4}$ 的电离平衡常数 $K_{\\mathrm{b} 2}=10^{-15}$\nB: 常温下, $\\mathrm{N}_{2} \\mathrm{H}_{5} \\mathrm{Cl}$ 溶液的 $\\mathrm{pH}<7$\nC: $\\mathrm{M}$ 点溶液中 $c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)=c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)$\nD: 当溶液 $\\mathrm{pH}=8$ 时, $c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)-c\\left(\\mathrm{Cl}^{-}\\right)<9.9 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n肼 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 是一种二元弱碱。常温下, 向 $1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{~N}_{2} \\mathrm{H}_{4}$ 水溶液中滴加盐酸, 所得混合溶液中 $-\\lg \\frac{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)}{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)}$或 $-\\lg \\frac{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)}{c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{6}^{2+}\\right)}$ 与 $\\mathrm{pH}$ 的变化关系如图所示。下列说法错误的是\n\n[图1]\n\nA: 常温下, $\\mathrm{N}_{2} \\mathrm{H}_{4}$ 的电离平衡常数 $K_{\\mathrm{b} 2}=10^{-15}$\nB: 常温下, $\\mathrm{N}_{2} \\mathrm{H}_{5} \\mathrm{Cl}$ 溶液的 $\\mathrm{pH}<7$\nC: $\\mathrm{M}$ 点溶液中 $c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)=c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)$\nD: 当溶液 $\\mathrm{pH}=8$ 时, $c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)-c\\left(\\mathrm{Cl}^{-}\\right)<9.9 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-015.jpg?height=594&width=776&top_left_y=1482&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1390",
        "problem": "One of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, $\\gamma=C_{p} / C_{V}$, which is exactly $5 / 3(1.67 \\pm 0.01)$ for a monoatomic gas. The ratio was derived from the measurement of speed of sound $v_{\\mathrm{s}}$ by using the following equation, where $f$ and $\\lambda$ are the frequency and wavelength of the sound, and $R, T$, and $M$ denote the molar gas constant, absolute temperature, and molar mass, respectively.\n\n$$\nv_{\\mathrm{s}}=f \\lambda=\\sqrt{\\frac{\\gamma R T}{M}}\n$$\n\nFor an unknown gas sample, the wavelength of the sound was measured to be $\\lambda=0.116 \\mathrm{~m}$ at a frequency of $f=3520 \\mathrm{~Hz}\\left(\\mathrm{~Hz}=\\mathrm{s}^{-1}\\right)$ and temperature of $15.0{ }^{\\circ} \\mathrm{C}$ and\nunder atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$. The density $\\rho$ of the gas for these conditions was measured to be $0.850 \\pm 0.005 \\mathrm{~kg} \\mathrm{~m}^{-3}$.Calculate the molar mass $M\\left[\\mathrm{~kg} \\mathrm{~mol}^{-1}\\right]$ of this gas.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOne of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, $\\gamma=C_{p} / C_{V}$, which is exactly $5 / 3(1.67 \\pm 0.01)$ for a monoatomic gas. The ratio was derived from the measurement of speed of sound $v_{\\mathrm{s}}$ by using the following equation, where $f$ and $\\lambda$ are the frequency and wavelength of the sound, and $R, T$, and $M$ denote the molar gas constant, absolute temperature, and molar mass, respectively.\n\n$$\nv_{\\mathrm{s}}=f \\lambda=\\sqrt{\\frac{\\gamma R T}{M}}\n$$\n\nFor an unknown gas sample, the wavelength of the sound was measured to be $\\lambda=0.116 \\mathrm{~m}$ at a frequency of $f=3520 \\mathrm{~Hz}\\left(\\mathrm{~Hz}=\\mathrm{s}^{-1}\\right)$ and temperature of $15.0{ }^{\\circ} \\mathrm{C}$ and\nunder atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$. The density $\\rho$ of the gas for these conditions was measured to be $0.850 \\pm 0.005 \\mathrm{~kg} \\mathrm{~m}^{-3}$.\n\nproblem:\nCalculate the molar mass $M\\left[\\mathrm{~kg} \\mathrm{~mol}^{-1}\\right]$ of this gas.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kg}\\mathrm{~mol}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kg}\\mathrm{~mol}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1295",
        "problem": "Beach sand mineral, monazite, is a rich source of thorium, available in large quantities in the state of Kerala in India. A typical monazite sample contains about $9 \\%$ $\\mathrm{ThO}_{2}$ and $0.35 \\% \\mathrm{U}_{3} \\mathrm{O}_{8} \\cdot{ }^{208} \\mathrm{~Pb}$ a ${ }^{206} \\mathrm{~Pb}$ are the stable end-products in the radioactive decay series of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$, respectively. All the lead $(\\mathrm{Pb})$ found in monazite is of radiogenic origin.\n\nThe isotopic atom ratio ${ }^{208} \\mathrm{~Pb} /{ }^{232} \\mathrm{Th}$, measured mass spectrometrically, in a monazite sample was found to be 0.104 . The half-lives of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U} 1.41 \\times 10^{10}$ years and $4.47 \\times 10^{9}$ years, respectively. Assume that ${ }^{208} \\mathrm{~Pb},{ }^{206} \\mathrm{~Pb},{ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$ remained entirely in the monazite sample since the formation of monazite mineral.\n\nA freshly prepared radiochemically pure sample of ${ }^{101}$ Mo contains 5000 atoms of ${ }^{101}$ Mo initially. How many atoms of\ni) ${ }^{101} \\mathrm{Mo}$\n\nii) ${ }^{101} \\mathrm{Tc}$\n\niii) ${ }^{101} \\mathrm{Ru}$\n\nwill be present in the sample after $14.6 \\mathrm{~min}$ ?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nBeach sand mineral, monazite, is a rich source of thorium, available in large quantities in the state of Kerala in India. A typical monazite sample contains about $9 \\%$ $\\mathrm{ThO}_{2}$ and $0.35 \\% \\mathrm{U}_{3} \\mathrm{O}_{8} \\cdot{ }^{208} \\mathrm{~Pb}$ a ${ }^{206} \\mathrm{~Pb}$ are the stable end-products in the radioactive decay series of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$, respectively. All the lead $(\\mathrm{Pb})$ found in monazite is of radiogenic origin.\n\nThe isotopic atom ratio ${ }^{208} \\mathrm{~Pb} /{ }^{232} \\mathrm{Th}$, measured mass spectrometrically, in a monazite sample was found to be 0.104 . The half-lives of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U} 1.41 \\times 10^{10}$ years and $4.47 \\times 10^{9}$ years, respectively. Assume that ${ }^{208} \\mathrm{~Pb},{ }^{206} \\mathrm{~Pb},{ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$ remained entirely in the monazite sample since the formation of monazite mineral.\n\nA freshly prepared radiochemically pure sample of ${ }^{101}$ Mo contains 5000 atoms of ${ }^{101}$ Mo initially. How many atoms of\ni) ${ }^{101} \\mathrm{Mo}$\n\nii) ${ }^{101} \\mathrm{Tc}$\n\niii) ${ }^{101} \\mathrm{Ru}$\n\nwill be present in the sample after $14.6 \\mathrm{~min}$ ?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [atoms of ${ }^{101} \\mathrm{Mo}$, atoms of ${ }^{101} \\mathrm{Tc}$, atoms of ${ }^{101} \\mathrm{Ru}$].\nTheir answer types are, in order, [numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null,
            null
        ],
        "answer_sequence": [
            "atoms of ${ }^{101} \\mathrm{Mo}$",
            "atoms of ${ }^{101} \\mathrm{Tc}$",
            "atoms of ${ }^{101} \\mathrm{Ru}$"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_794",
        "problem": "某兴趣小组对化合物 $\\mathrm{M}$ 开展探究实验。\n\n[图1]\n\n已知: (1) $\\mathrm{M}$ 是强酸盐, 由三种元素组成; 两种元素组成的固体 $\\mathrm{A}$ 是纯净物, 气体 $\\mathrm{B}$ 是混合物。2酚酞溶液作指示剂, 用 $0.5000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $\\mathrm{B}_{1}$, 消耗 $20.00 \\mathrm{~mL} \\mathrm{NaOH}$溶液。下列推断正确的是\nA: 砖红色固体和固体 $\\mathrm{A}$ 的成分相同\nB: 气体 B 溶于水的反应原子利用率为 $100 \\%$\nC: $\\mathrm{M}$ 是 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$, 物质的量为 $0.2 \\mathrm{~mol}$\nD: 上述转化中包括 4 个氧化还原反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某兴趣小组对化合物 $\\mathrm{M}$ 开展探究实验。\n\n[图1]\n\n已知: (1) $\\mathrm{M}$ 是强酸盐, 由三种元素组成; 两种元素组成的固体 $\\mathrm{A}$ 是纯净物, 气体 $\\mathrm{B}$ 是混合物。2酚酞溶液作指示剂, 用 $0.5000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $\\mathrm{B}_{1}$, 消耗 $20.00 \\mathrm{~mL} \\mathrm{NaOH}$溶液。下列推断正确的是\n\nA: 砖红色固体和固体 $\\mathrm{A}$ 的成分相同\nB: 气体 B 溶于水的反应原子利用率为 $100 \\%$\nC: $\\mathrm{M}$ 是 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$, 物质的量为 $0.2 \\mathrm{~mol}$\nD: 上述转化中包括 4 个氧化还原反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-46.jpg?height=346&width=1472&top_left_y=1820&top_left_x=332"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_422",
        "problem": "金刚烷是一种重要的化工原料, 工业上可通过下列途径制备:\n\n[图1]\n\n下列说法正确的是\nA: 环戊二烯分子中所有原子可能共面\nB: 金刚烷的二氯代物有 6 种\nC: 二聚环戊二烯与 $\\mathrm{HBr}$ 加成反应最多得 7 种产物\nD: 上述四种烃均能使溴的四氯化碳溶液裉色\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n金刚烷是一种重要的化工原料, 工业上可通过下列途径制备:\n\n[图1]\n\n下列说法正确的是\n\nA: 环戊二烯分子中所有原子可能共面\nB: 金刚烷的二氯代物有 6 种\nC: 二聚环戊二烯与 $\\mathrm{HBr}$ 加成反应最多得 7 种产物\nD: 上述四种烃均能使溴的四氯化碳溶液裉色\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-65.jpg?height=203&width=1108&top_left_y=1709&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-66.jpg?height=564&width=1302&top_left_y=180&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1204",
        "problem": "The transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nThe next steps in the overall reaction involve the amine $\\left(\\mathrm{RNH}_{2}\\right)$ and ${ }^{t} \\mathrm{BuONa}$. To determine the order with respect to $\\mathrm{RNH}_{2}$ and ${ }^{t} \\mathrm{BuONa}$, the dependence of the initial rates of the reaction on the concentrations of these two reagents was carried with the other reagents present in large excess. Some results are shown in the tables below.\n\n| $\\left[\\mathrm{NaO}{ }^{t} \\mathrm{Bu}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.2 | 0.6 | 0.9 | 0.12 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.16 \\times 10^{-5}$ | $4.12 \\times 10^{-5}$ | $4.24 \\times 10^{-5}$ | $4.20 \\times 10^{-5}$ |\n\n\n| $\\left[\\mathrm{RNH}_{2}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.3 | 0.6 | 0.9 | 1.2 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $4.26 \\times 10^{-5}$ | $4.21 \\times 10^{-5}$ | $4.23 \\times 10^{-5}$ |Determine the order with $\\mathrm{[NaO^tBu]}$, assuming each is an integer. (Use the grids if you like)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nThe next steps in the overall reaction involve the amine $\\left(\\mathrm{RNH}_{2}\\right)$ and ${ }^{t} \\mathrm{BuONa}$. To determine the order with respect to $\\mathrm{RNH}_{2}$ and ${ }^{t} \\mathrm{BuONa}$, the dependence of the initial rates of the reaction on the concentrations of these two reagents was carried with the other reagents present in large excess. Some results are shown in the tables below.\n\n| $\\left[\\mathrm{NaO}{ }^{t} \\mathrm{Bu}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.2 | 0.6 | 0.9 | 0.12 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.16 \\times 10^{-5}$ | $4.12 \\times 10^{-5}$ | $4.24 \\times 10^{-5}$ | $4.20 \\times 10^{-5}$ |\n\n\n| $\\left[\\mathrm{RNH}_{2}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.3 | 0.6 | 0.9 | 1.2 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $4.26 \\times 10^{-5}$ | $4.21 \\times 10^{-5}$ | $4.23 \\times 10^{-5}$ |\n\nproblem:\nDetermine the order with $\\mathrm{[NaO^tBu]}$, assuming each is an integer. (Use the grids if you like)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_126",
        "problem": "Which of the following metals should be labelled with the WHMIS symbol for flammable substances ?\nA: $\\mathrm{Al}$\nB: $\\mathrm{Ni}$\nC: $\\mathrm{Mg}$\nD: $\\mathrm{Pb}$\nE: $\\mathrm{Hg}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following metals should be labelled with the WHMIS symbol for flammable substances ?\n\nA: $\\mathrm{Al}$\nB: $\\mathrm{Ni}$\nC: $\\mathrm{Mg}$\nD: $\\mathrm{Pb}$\nE: $\\mathrm{Hg}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1315",
        "problem": "The muon $(\\mu)$ is a subatomic particle of the lepton family which has same charge and magnetic behavior as the electron, but has a different mass and is unstable, i.e., it disintegrates into other particles within microseconds after its creation. Here you will attempt to determine the mass of the muon using two rather different approaches.\n\nMany experiments have studied the spectroscopy of atoms that have captured a muon in place of an electron. These exotic atoms are formed in a variety of excited states. The transition from the third excited state to the first excited state of an atom consisting of a ${ }^{1} \\mathrm{H}$ nucleus and a muon attached to it was observed at a wavelength of $2.615 \\mathrm{~nm}$. Determine the mass of the muon.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe muon $(\\mu)$ is a subatomic particle of the lepton family which has same charge and magnetic behavior as the electron, but has a different mass and is unstable, i.e., it disintegrates into other particles within microseconds after its creation. Here you will attempt to determine the mass of the muon using two rather different approaches.\n\nMany experiments have studied the spectroscopy of atoms that have captured a muon in place of an electron. These exotic atoms are formed in a variety of excited states. The transition from the third excited state to the first excited state of an atom consisting of a ${ }^{1} \\mathrm{H}$ nucleus and a muon attached to it was observed at a wavelength of $2.615 \\mathrm{~nm}$. Determine the mass of the muon.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kg}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kg}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1039",
        "problem": "The difference in electronegativity between $\\mathrm{Cl}$ and $\\mathrm{H}$ is 0.962.2022 was the $90^{\\text {th }}$ anniversary of Linus Pauling proposing the concept of electronegativity. His book The nature of the chemical bond is considered the most influential chemistry book of $20^{\\text {th }}$ century, and he was awarded the 1954 Nobel prize in chemistry for his work.\n\nElectronegativity, $\\chi$, is a measure of the ability of an atom to attract a pair of electrons in a covalent bond.\n\n[figure1]\n\nPauling used thermodynamic data to calculate the difference in electronegativity between two atoms A and B. All electronegativity values are positive with no units, and atom A has a higher electronegativity than atom $B$.\n\n$$\n\\chi_{A}-\\chi_{B}=0.102 \\sqrt{\\mathrm{B}_{d}(A B)-\\frac{\\mathrm{B}_{d}(A A)+\\mathrm{B}_{d}(B B)}{2}}\n$$\n\n$\\mathrm{B}_{d}$ represents the bond dissociation energies, in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, of the $\\mathrm{A}-\\mathrm{A}, \\mathrm{B}-\\mathrm{B}$ and $\\mathrm{A}-\\mathrm{B}$ bonds.\n\nAs this method gives a difference in electronegativity it is necesssary to choose a reference point in order to gain values for individual elements. Pauling originally chose hydrogen as the reference, with a value of 2.20 .\n\nUsing this reference and your knowledge of trends in electronegativity in the periodic table, calculate the electronegativity value for chlorine.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe difference in electronegativity between $\\mathrm{Cl}$ and $\\mathrm{H}$ is 0.962.\n\nproblem:\n2022 was the $90^{\\text {th }}$ anniversary of Linus Pauling proposing the concept of electronegativity. His book The nature of the chemical bond is considered the most influential chemistry book of $20^{\\text {th }}$ century, and he was awarded the 1954 Nobel prize in chemistry for his work.\n\nElectronegativity, $\\chi$, is a measure of the ability of an atom to attract a pair of electrons in a covalent bond.\n\n[figure1]\n\nPauling used thermodynamic data to calculate the difference in electronegativity between two atoms A and B. All electronegativity values are positive with no units, and atom A has a higher electronegativity than atom $B$.\n\n$$\n\\chi_{A}-\\chi_{B}=0.102 \\sqrt{\\mathrm{B}_{d}(A B)-\\frac{\\mathrm{B}_{d}(A A)+\\mathrm{B}_{d}(B B)}{2}}\n$$\n\n$\\mathrm{B}_{d}$ represents the bond dissociation energies, in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, of the $\\mathrm{A}-\\mathrm{A}, \\mathrm{B}-\\mathrm{B}$ and $\\mathrm{A}-\\mathrm{B}$ bonds.\n\nAs this method gives a difference in electronegativity it is necesssary to choose a reference point in order to gain values for individual elements. Pauling originally chose hydrogen as the reference, with a value of 2.20 .\n\nUsing this reference and your knowledge of trends in electronegativity in the periodic table, calculate the electronegativity value for chlorine.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-03.jpg?height=611&width=505&top_left_y=314&top_left_x=1301"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_274",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount of $\\mathrm{CO}_{3}{ }^{2-}$ ions present in $20.00 \\mathrm{~mL}$ of the \"dissolved mineral solution”.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount of $\\mathrm{CO}_{3}{ }^{2-}$ ions present in $20.00 \\mathrm{~mL}$ of the \"dissolved mineral solution”.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_822",
        "problem": "鉴别己烷、己烯、乙酸、乙醇时可选用的试剂组合是\nA: 溴水、金属 $\\mathrm{Na}$\nB: $\\mathrm{FeCl}_{3}$ 溶液、金属 $\\mathrm{Na}$\nC: 石荵试液、溴水\nD: 金属 $\\mathrm{Na}$ 、石荵试液\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n鉴别己烷、己烯、乙酸、乙醇时可选用的试剂组合是\n\nA: 溴水、金属 $\\mathrm{Na}$\nB: $\\mathrm{FeCl}_{3}$ 溶液、金属 $\\mathrm{Na}$\nC: 石荵试液、溴水\nD: 金属 $\\mathrm{Na}$ 、石荵试液\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_198",
        "problem": "What is the final temperature of a mixture of $50.0 \\mathrm{~g}$ of $\\mathrm{Cu}$ (specific heat capacity $=0.3845 \\mathrm{~J} \\mathrm{~g}^{-1 \\mathrm{o}} \\mathrm{C}^{-1}$ ) initially at $135.0^{\\circ} \\mathrm{C}$ and $150.0 \\mathrm{~mL}$ of water (specific heat capacity $=4.184 \\mathrm{~J} \\mathrm{~g}^{-1}{ }^{\\circ} \\mathrm{C}^{-1}$ ) initially at $21.0^{\\circ} \\mathrm{C}$ ? Assume that there is no loss of heat and the container has a negligible heat capacity.\nA: $17.4^{\\circ} \\mathrm{C}$\nB: $22.5^{\\circ} \\mathrm{C}$\nC: $24.4^{\\circ} \\mathrm{C}$\nD: $35.2^{\\circ} \\mathrm{C}$\nE: $78.0^{\\circ} \\mathrm{C}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the final temperature of a mixture of $50.0 \\mathrm{~g}$ of $\\mathrm{Cu}$ (specific heat capacity $=0.3845 \\mathrm{~J} \\mathrm{~g}^{-1 \\mathrm{o}} \\mathrm{C}^{-1}$ ) initially at $135.0^{\\circ} \\mathrm{C}$ and $150.0 \\mathrm{~mL}$ of water (specific heat capacity $=4.184 \\mathrm{~J} \\mathrm{~g}^{-1}{ }^{\\circ} \\mathrm{C}^{-1}$ ) initially at $21.0^{\\circ} \\mathrm{C}$ ? Assume that there is no loss of heat and the container has a negligible heat capacity.\n\nA: $17.4^{\\circ} \\mathrm{C}$\nB: $22.5^{\\circ} \\mathrm{C}$\nC: $24.4^{\\circ} \\mathrm{C}$\nD: $35.2^{\\circ} \\mathrm{C}$\nE: $78.0^{\\circ} \\mathrm{C}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_685",
        "problem": "$\\mathrm{T}^{\\circ} \\mathrm{C}, 2 \\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4(\\mathrm{~g})} \\Delta \\mathrm{H}<0$, 该反应正、逆反应速率与浓度的关系为 $\\mathrm{v}$ 正 $=\\mathrm{k}$ 正 $\\mathrm{c}^{2}\\left(\\mathrm{NO}_{2}\\right), \\mathrm{v}$ 逆 $=\\mathrm{k}$ 逆 $\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)\\left(\\mathrm{k}_{\\text {正、 }} 、 \\mathrm{k}\\right.$ 逆为速率常数)。结合图像, 下列说法错误的是\n\n[图1]\nA: 图中表示 $\\operatorname{lgv}_{\\text {逆 }} \\sim \\operatorname{lgc}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)$ 的线是 $\\mathrm{n}$\nB: 当 $2_{\\text {正 }} \\mathrm{v}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{V}$ 逆 $\\left(\\mathrm{NO}_{2}\\right)$ 时, 说明反应达到平衡状态\nC: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向 $2 \\mathrm{~L}$ 的容器中充入 $5 \\mathrm{molN}_{2} \\mathrm{O}_{4}$ 气体和 $1 \\mathrm{molNO}_{2}$ 气体, 此时 $\\mathrm{v}_{\\text {正 }}<\\mathrm{v}$ 逆\nD: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向刚性容器中充入一定量 $\\mathrm{NO}_{2}$ 气体, 平衡后测得 $\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)$ 为 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$,则平衡时, $\\mathrm{v}$ 正的数值为 $10^{\\mathrm{a}}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{T}^{\\circ} \\mathrm{C}, 2 \\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4(\\mathrm{~g})} \\Delta \\mathrm{H}<0$, 该反应正、逆反应速率与浓度的关系为 $\\mathrm{v}$ 正 $=\\mathrm{k}$ 正 $\\mathrm{c}^{2}\\left(\\mathrm{NO}_{2}\\right), \\mathrm{v}$ 逆 $=\\mathrm{k}$ 逆 $\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)\\left(\\mathrm{k}_{\\text {正、 }} 、 \\mathrm{k}\\right.$ 逆为速率常数)。结合图像, 下列说法错误的是\n\n[图1]\n\nA: 图中表示 $\\operatorname{lgv}_{\\text {逆 }} \\sim \\operatorname{lgc}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)$ 的线是 $\\mathrm{n}$\nB: 当 $2_{\\text {正 }} \\mathrm{v}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{V}$ 逆 $\\left(\\mathrm{NO}_{2}\\right)$ 时, 说明反应达到平衡状态\nC: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向 $2 \\mathrm{~L}$ 的容器中充入 $5 \\mathrm{molN}_{2} \\mathrm{O}_{4}$ 气体和 $1 \\mathrm{molNO}_{2}$ 气体, 此时 $\\mathrm{v}_{\\text {正 }}<\\mathrm{v}$ 逆\nD: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向刚性容器中充入一定量 $\\mathrm{NO}_{2}$ 气体, 平衡后测得 $\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{4}\\right)$ 为 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$,则平衡时, $\\mathrm{v}$ 正的数值为 $10^{\\mathrm{a}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-002.jpg?height=425&width=537&top_left_y=781&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_890",
        "problem": "在温度、容积相同的 3 个密闭容器中, 按不同方式投入反应, 保持恒温、恒容, 测得反应达到平衡时的有关数据如下 $\\left(\\right.$ 已知 $\\left.\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\quad \\Delta H=-92.4 \\mathrm{~kJ} / \\mathrm{mol}\\right)$\n\n| 容器 | 甲 | 乙 | 丙 |\n| :--- | :--- | :--- | :--- |\n| 反应物投入量 | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ | $2 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $4 \\mathrm{~mol} \\mathrm{NH}_{3}$ |\n| 的浓度 $(\\mathrm{mol} / \\mathrm{L})$ | $\\mathrm{c}_{1}$ | $\\mathrm{c}_{2}$ | $\\mathrm{c}_{3}$ |\n| 反应的能量变化 | 放出 $\\mathrm{a} \\mathrm{kJ}$ | 吸收 $\\mathrm{b} \\mathrm{kJ}$ | 吸收 $\\mathrm{c} \\mathrm{kJ}$ |\n| 体系压强 | $\\mathrm{p}_{1}$ | $\\mathrm{p}_{2}$ | $\\mathrm{p}_{3}$ |\n| 反应物转化率 | $\\alpha_{1}$ | $\\alpha_{2}$ | $\\alpha_{3}$ |\n\n下列说法正确的是\nA: $2 c_{1}>c_{3}$\nB: $\\mathrm{a}+\\mathrm{b}>92.4$\nC: $2 \\mathrm{p}_{1}<\\mathrm{p}_{3}$\nD: $\\alpha_{1}+\\alpha_{3}<1$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在温度、容积相同的 3 个密闭容器中, 按不同方式投入反应, 保持恒温、恒容, 测得反应达到平衡时的有关数据如下 $\\left(\\right.$ 已知 $\\left.\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\quad \\Delta H=-92.4 \\mathrm{~kJ} / \\mathrm{mol}\\right)$\n\n| 容器 | 甲 | 乙 | 丙 |\n| :--- | :--- | :--- | :--- |\n| 反应物投入量 | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ | $2 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $4 \\mathrm{~mol} \\mathrm{NH}_{3}$ |\n| 的浓度 $(\\mathrm{mol} / \\mathrm{L})$ | $\\mathrm{c}_{1}$ | $\\mathrm{c}_{2}$ | $\\mathrm{c}_{3}$ |\n| 反应的能量变化 | 放出 $\\mathrm{a} \\mathrm{kJ}$ | 吸收 $\\mathrm{b} \\mathrm{kJ}$ | 吸收 $\\mathrm{c} \\mathrm{kJ}$ |\n| 体系压强 | $\\mathrm{p}_{1}$ | $\\mathrm{p}_{2}$ | $\\mathrm{p}_{3}$ |\n| 反应物转化率 | $\\alpha_{1}$ | $\\alpha_{2}$ | $\\alpha_{3}$ |\n\n下列说法正确的是\n\nA: $2 c_{1}>c_{3}$\nB: $\\mathrm{a}+\\mathrm{b}>92.4$\nC: $2 \\mathrm{p}_{1}<\\mathrm{p}_{3}$\nD: $\\alpha_{1}+\\alpha_{3}<1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1042",
        "problem": "Former Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nMany varities of Swiss cheese, such as Emmental, are famous for the holes or 'eyes' that appear in the cheese. To produce the holes another species of bacteria, Propionibacterium freudenreichii is important. This bacterium carries out the reaction of lactic acid to propanoic acid, ethanoic acid, carbon dioxide and water. The production of carbon dioxide causes the bubbles to appear.\n\n[figure3]\n\nAssume during fermentation at $21^{\\circ} \\mathrm{C}$, a spherical bubble of diameter $1.5 \\mathrm{~cm}$ appears in the cheese.\n\nCalculate the volume of this bubble in $\\mathrm{m}^{3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nFormer Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nMany varities of Swiss cheese, such as Emmental, are famous for the holes or 'eyes' that appear in the cheese. To produce the holes another species of bacteria, Propionibacterium freudenreichii is important. This bacterium carries out the reaction of lactic acid to propanoic acid, ethanoic acid, carbon dioxide and water. The production of carbon dioxide causes the bubbles to appear.\n\n[figure3]\n\nAssume during fermentation at $21^{\\circ} \\mathrm{C}$, a spherical bubble of diameter $1.5 \\mathrm{~cm}$ appears in the cheese.\n\nCalculate the volume of this bubble in $\\mathrm{m}^{3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~m}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-13.jpg?height=616&width=699&top_left_y=323&top_left_x=1178",
            "https://i.postimg.cc/Vs42tGMJ/5.png",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-14.jpg?height=528&width=785&top_left_y=747&top_left_x=1047"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~m}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1397",
        "problem": "Dissociating gas cycle\n\nDinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:\n\n$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$\n\n1.00 mol of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ was put into an empty vessel with a fixed volume of $24.44 \\mathrm{dm}^{3}$. The equilibrium gas pressure at $298 \\mathrm{~K}$ was found to be 1.190 bar. When heated to $348 \\mathrm{~K}$, the gas pressure increased to its equilibrium value of 1.886 bar.Calculate $\\Delta H^{0}$  of the reaction, assuming that they do not change significantly with temperature.\n\nNote: If you cannot calculate $\\Delta H^{0}$, use $\\Delta H^{0}=30.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ for further calculations.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nDissociating gas cycle\n\nDinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:\n\n$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$\n\n1.00 mol of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ was put into an empty vessel with a fixed volume of $24.44 \\mathrm{dm}^{3}$. The equilibrium gas pressure at $298 \\mathrm{~K}$ was found to be 1.190 bar. When heated to $348 \\mathrm{~K}$, the gas pressure increased to its equilibrium value of 1.886 bar.\n\nproblem:\nCalculate $\\Delta H^{0}$  of the reaction, assuming that they do not change significantly with temperature.\n\nNote: If you cannot calculate $\\Delta H^{0}$, use $\\Delta H^{0}=30.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ for further calculations.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1} $, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1} $"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1338",
        "problem": "Clathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nThe enthalpy of this process equals $17.47 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.Deduce at which minimum temperature methane hydrate can coexist with liquid water. Choose the correct answer.\nA: 272.9 K\nB: 273.15  K\nC:  273.4 K\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nClathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nThe enthalpy of this process equals $17.47 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.\n\nproblem:\nDeduce at which minimum temperature methane hydrate can coexist with liquid water. Choose the correct answer.\n\nA: 272.9 K\nB: 273.15  K\nC:  273.4 K\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-222.jpg?height=425&width=434&top_left_y=1255&top_left_x=1459"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1486",
        "problem": "Anhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion.\n\nIn an experiment, gaseous $\\mathrm{NH}_{3}$ is burned with $\\mathrm{O}_{2}$ in a container of fixed volume according to the equation given below.\n\n$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe initial and final states are at $298 \\mathrm{~K}$. After combustion with $14.40 \\mathrm{~g}$ of $\\mathrm{O}_{2}$, some of $\\mathrm{NH}_{3}$ remains unreacted.\n\nTo determine the amount of $\\mathrm{NH}_{3}$ gas dissolved in water, produced during the combustion process, a $10.00 \\mathrm{~cm}^{3}$ sample of the aqueous solution was withdrawn from the reaction vessel and added to $15.0 \\mathrm{~cm}^{3}$ of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ solution $\\left(c=0.0100 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ). The resulting solution was titrated with a standard $\\mathrm{NaOH}$ solution $\\left(c=0.0200 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) and the equivalence point was reached at $10.64 \\mathrm{~cm}^{3}$.\n\n$$\n\\left(K_{b}\\left(\\mathrm{NH}_{3}\\right)=1.8 \\cdot 10^{-5} ; \\quad K_{a}\\left(\\mathrm{HSO}_{4}^{-}\\right)=1.1 \\cdot 10^{-2}\\right)\n$$Tick the correct statement for the $\\mathrm{pH}$ of the solution at the equivalence point.\nA: $\\mathrm{pH}>7$\nB: $\\mathrm{pH}=7$\nC: $\\mathrm{pH}<7$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nAnhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion.\n\nIn an experiment, gaseous $\\mathrm{NH}_{3}$ is burned with $\\mathrm{O}_{2}$ in a container of fixed volume according to the equation given below.\n\n$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe initial and final states are at $298 \\mathrm{~K}$. After combustion with $14.40 \\mathrm{~g}$ of $\\mathrm{O}_{2}$, some of $\\mathrm{NH}_{3}$ remains unreacted.\n\nTo determine the amount of $\\mathrm{NH}_{3}$ gas dissolved in water, produced during the combustion process, a $10.00 \\mathrm{~cm}^{3}$ sample of the aqueous solution was withdrawn from the reaction vessel and added to $15.0 \\mathrm{~cm}^{3}$ of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ solution $\\left(c=0.0100 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ). The resulting solution was titrated with a standard $\\mathrm{NaOH}$ solution $\\left(c=0.0200 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) and the equivalence point was reached at $10.64 \\mathrm{~cm}^{3}$.\n\n$$\n\\left(K_{b}\\left(\\mathrm{NH}_{3}\\right)=1.8 \\cdot 10^{-5} ; \\quad K_{a}\\left(\\mathrm{HSO}_{4}^{-}\\right)=1.1 \\cdot 10^{-2}\\right)\n$$\n\nproblem:\nTick the correct statement for the $\\mathrm{pH}$ of the solution at the equivalence point.\n\nA: $\\mathrm{pH}>7$\nB: $\\mathrm{pH}=7$\nC: $\\mathrm{pH}<7$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_448",
        "problem": "可充电水系 $\\mathrm{Zn}-\\mathrm{CO}_{2}$ 电池用锌和催化剂材料作两极, 电池工作示意图如图所示,其中双极膜是由阳膜和阴膜制成的复合膜, 在直流电场的作用下, 双极膜复合层间的 $\\mathrm{H}_{2} \\mathrm{O}$ 电离出的 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$可以分别通过膜移向两极。下列说法错误的是\n\n[图1]\nA: 放电时, 右侧电极的电极反应为 $\\mathrm{CO}_{2}+2 \\mathrm{H}^{+}-2 \\mathrm{e}^{-}=\\mathrm{HCOOH}$\nB: 放电时, 金属锌为负极, 发生氧化反应\nC: 充电时, 双极隔膜产生的 $\\mathrm{H}^{+}$向右侧正极室移动\nD: 充电时, 电池总反应为 $2 \\mathrm{Zn}(\\mathrm{OH})_{4}^{2-}=2 \\mathrm{Zn}+\\mathrm{O}_{2} \\uparrow+4 \\mathrm{OH}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n可充电水系 $\\mathrm{Zn}-\\mathrm{CO}_{2}$ 电池用锌和催化剂材料作两极, 电池工作示意图如图所示,其中双极膜是由阳膜和阴膜制成的复合膜, 在直流电场的作用下, 双极膜复合层间的 $\\mathrm{H}_{2} \\mathrm{O}$ 电离出的 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$可以分别通过膜移向两极。下列说法错误的是\n\n[图1]\n\nA: 放电时, 右侧电极的电极反应为 $\\mathrm{CO}_{2}+2 \\mathrm{H}^{+}-2 \\mathrm{e}^{-}=\\mathrm{HCOOH}$\nB: 放电时, 金属锌为负极, 发生氧化反应\nC: 充电时, 双极隔膜产生的 $\\mathrm{H}^{+}$向右侧正极室移动\nD: 充电时, 电池总反应为 $2 \\mathrm{Zn}(\\mathrm{OH})_{4}^{2-}=2 \\mathrm{Zn}+\\mathrm{O}_{2} \\uparrow+4 \\mathrm{OH}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-73.jpg?height=434&width=782&top_left_y=431&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_183",
        "problem": "Canada is a world-leader in the production of radionuclides such as technetium-99m ( $\\left.{ }^{99 \\mathrm{~m}} \\mathrm{Tc}\\right)$, which is used to diagnose bone diseases. ${ }^{99 \\mathrm{~m}} \\mathrm{Tc}$ has a relative atomic mass of $98.91 \\mathrm{amu}$ and exhibits first-order radioactive decay with a half-life of 6.00 hours. If $8.62 \\times 10^{-12} \\mathrm{~mol}$ of sodium pertechnetate $\\left(\\mathrm{Na}^{99 \\mathrm{~m}} \\mathrm{TcO}_{4}\\right)$ is injected into a $75 \\mathrm{~kg}$ adult patient, what mass of ${ }^{99 \\mathrm{~m}} \\mathrm{Tc}$ would remain after 24 hours?\nA: $1.46 \\times 10^{-9} \\mathrm{~g}$\nB: $2.13 \\times 10^{-10} \\mathrm{~g}$\nC: $3.66 \\times 10^{-11} \\mathrm{~g}$\nD: $5.33 \\times 10^{-11} \\mathrm{~g}$\nE: $9.15 \\times 10^{-11} \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCanada is a world-leader in the production of radionuclides such as technetium-99m ( $\\left.{ }^{99 \\mathrm{~m}} \\mathrm{Tc}\\right)$, which is used to diagnose bone diseases. ${ }^{99 \\mathrm{~m}} \\mathrm{Tc}$ has a relative atomic mass of $98.91 \\mathrm{amu}$ and exhibits first-order radioactive decay with a half-life of 6.00 hours. If $8.62 \\times 10^{-12} \\mathrm{~mol}$ of sodium pertechnetate $\\left(\\mathrm{Na}^{99 \\mathrm{~m}} \\mathrm{TcO}_{4}\\right)$ is injected into a $75 \\mathrm{~kg}$ adult patient, what mass of ${ }^{99 \\mathrm{~m}} \\mathrm{Tc}$ would remain after 24 hours?\n\nA: $1.46 \\times 10^{-9} \\mathrm{~g}$\nB: $2.13 \\times 10^{-10} \\mathrm{~g}$\nC: $3.66 \\times 10^{-11} \\mathrm{~g}$\nD: $5.33 \\times 10^{-11} \\mathrm{~g}$\nE: $9.15 \\times 10^{-11} \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_583",
        "problem": "一种双阴极微生物燃料电池装置如图所示。该装置可以同时进行硝化和反硝化脱氮,下列叙述正确的是\n\n[图1]\n\n进水: $\\mathrm{NO}_{3}^{-}$\n\n进水： $\\mathrm{NH}_{4}^{+}$葡萄糖\n\n进水: $\\mathrm{NH}_{4}^{+}$\nA: 电池工作时, $\\mathrm{H}^{+}$的迁移方向: 左 $\\rightarrow$ 右\nB: 电池工作时, “缺氧阴极”电极附近的溶液 $\\mathrm{pH}$ 减小\nC: “好氧阴极”存在反应: $\\mathrm{NH}_{4}^{+}-6 \\mathrm{e}^{-}+8 \\mathrm{OH}^{-}=\\mathrm{NO}_{2}^{-}+6 \\mathrm{H}_{2} \\mathrm{O}$\nD: “厌氧阳极”区质量减少 $28.8 \\mathrm{~g}$ 时, 该电极输出电子 $2.4 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一种双阴极微生物燃料电池装置如图所示。该装置可以同时进行硝化和反硝化脱氮,下列叙述正确的是\n\n[图1]\n\n进水: $\\mathrm{NO}_{3}^{-}$\n\n进水： $\\mathrm{NH}_{4}^{+}$葡萄糖\n\n进水: $\\mathrm{NH}_{4}^{+}$\n\nA: 电池工作时, $\\mathrm{H}^{+}$的迁移方向: 左 $\\rightarrow$ 右\nB: 电池工作时, “缺氧阴极”电极附近的溶液 $\\mathrm{pH}$ 减小\nC: “好氧阴极”存在反应: $\\mathrm{NH}_{4}^{+}-6 \\mathrm{e}^{-}+8 \\mathrm{OH}^{-}=\\mathrm{NO}_{2}^{-}+6 \\mathrm{H}_{2} \\mathrm{O}$\nD: “厌氧阳极”区质量减少 $28.8 \\mathrm{~g}$ 时, 该电极输出电子 $2.4 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-076.jpg?height=728&width=1459&top_left_y=161&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_981",
        "problem": "What is the sum of the coefficients when the following equation is balanced using the smallest whole number coefficients?\n\n$$\n-\\mathrm{P}_{4}+\\ldots \\mathrm{Cl}_{2} \\rightarrow-\\mathrm{PCl}_{3}\n$$\nA: 12\nB: 11\nC: 6\nD: 5\nE: 3\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the sum of the coefficients when the following equation is balanced using the smallest whole number coefficients?\n\n$$\n-\\mathrm{P}_{4}+\\ldots \\mathrm{Cl}_{2} \\rightarrow-\\mathrm{PCl}_{3}\n$$\n\nA: 12\nB: 11\nC: 6\nD: 5\nE: 3\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1479",
        "problem": "The chemical oxygen demand (COD) refers to the amount of oxidizable substance, such as organic compounds, in a sample solution, and it is used as an indication of water quality in seas, lakes, and marshes. For example, the COD of service water is kept below $1 \\mathrm{mg} \\mathrm{dm}^{-3}$. The COD $\\left(\\mathrm{mg} \\mathrm{dm}^{-3}\\right)$ is represented by mass of $\\mathrm{O}_{2}(\\mathrm{mg})$ which accepts the same amount of electrons which would be accepted by the strong oxidizing agent when $1 \\mathrm{dm}^{3}$ of a sample solution is treated with it. An example of the operation is presented below.\n\nA sample solution with a volume of $1.00 \\mathrm{dm}^{3}$ was acidified with a sufficient amount of sulphuric acid, and chloride ions were removed by the addition of silver nitrate solution. A volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of potassium permanganate solution $\\left(c=5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) was added to the sample solution, and the mixture was heated for $30 \\mathrm{~min}$. Further, a volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of disodium oxalate $\\left(\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ or $\\left.\\mathrm{NaOOC}-\\mathrm{COONa}\\right)$ standard solution ( $c=1$. $25 \\cdot 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) was added, and the mixture was stirred well. Oxalate ions that remained unreacted were titrated with potassium permanganate solution $(c=$ $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ). A volume of $3.00 \\cdot 10^{-2} \\mathrm{dm}^{3}$ of the solution was used for the titration.Calculate the COD $\\left(\\mathrm{mg} \\mathrm{~dm}^{-3}\\right)$ of the sample solution described in the analytical operation above.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe chemical oxygen demand (COD) refers to the amount of oxidizable substance, such as organic compounds, in a sample solution, and it is used as an indication of water quality in seas, lakes, and marshes. For example, the COD of service water is kept below $1 \\mathrm{mg} \\mathrm{dm}^{-3}$. The COD $\\left(\\mathrm{mg} \\mathrm{dm}^{-3}\\right)$ is represented by mass of $\\mathrm{O}_{2}(\\mathrm{mg})$ which accepts the same amount of electrons which would be accepted by the strong oxidizing agent when $1 \\mathrm{dm}^{3}$ of a sample solution is treated with it. An example of the operation is presented below.\n\nA sample solution with a volume of $1.00 \\mathrm{dm}^{3}$ was acidified with a sufficient amount of sulphuric acid, and chloride ions were removed by the addition of silver nitrate solution. A volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of potassium permanganate solution $\\left(c=5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) was added to the sample solution, and the mixture was heated for $30 \\mathrm{~min}$. Further, a volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of disodium oxalate $\\left(\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ or $\\left.\\mathrm{NaOOC}-\\mathrm{COONa}\\right)$ standard solution ( $c=1$. $25 \\cdot 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) was added, and the mixture was stirred well. Oxalate ions that remained unreacted were titrated with potassium permanganate solution $(c=$ $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ). A volume of $3.00 \\cdot 10^{-2} \\mathrm{dm}^{3}$ of the solution was used for the titration.\n\nproblem:\nCalculate the COD $\\left(\\mathrm{mg} \\mathrm{~dm}^{-3}\\right)$ of the sample solution described in the analytical operation above.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{mg} \\mathrm{~dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{mg} \\mathrm{~dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_169",
        "problem": "After a chemist stirs the contents of a beaker, the rate of a reaction between reactants A and B increases by a factor of four. Use the diagrams below to help you to determine the main reason why the rate of reaction increases.\n[figure1]\n\nBefore stirring\n\n[figure2]\n\nAfter stirring\nA: Reaction surface area increases\nB: Reactant concentrations increase\nC: Activation energy increases\nD: Total kinetic energy increases\nE: Average kinetic energy increases\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAfter a chemist stirs the contents of a beaker, the rate of a reaction between reactants A and B increases by a factor of four. Use the diagrams below to help you to determine the main reason why the rate of reaction increases.\n[figure1]\n\nBefore stirring\n\n[figure2]\n\nAfter stirring\n\nA: Reaction surface area increases\nB: Reactant concentrations increase\nC: Activation energy increases\nD: Total kinetic energy increases\nE: Average kinetic energy increases\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_de998ba2e5d42a77dc86g-1.jpg?height=126&width=348&top_left_y=680&top_left_x=1586",
            "https://cdn.mathpix.com/cropped/2024_03_06_de998ba2e5d42a77dc86g-1.jpg?height=137&width=172&top_left_y=675&top_left_x=2012"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_765",
        "problem": "常温下, 用 $\\mathrm{NaOH}$ 溶液滴加到某酒石酸 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 溶液中, 混合溶液的 $\\mathrm{pH}$ 与离子浓度变化的关系如图所示。下列叙述错误的是\n\n[图1]\nA: $\\mathrm{L}_{2}$ 表示 $\\mathrm{pH}$ 与 $-\\lg \\frac{c\\left(\\mathrm{X}^{2-}\\right)}{c\\left(\\mathrm{HX}^{-}\\right)}$的变化关系\nB: $\\mathrm{NaHX}$ 溶液中 $c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)$\nC: 线 $\\mathrm{L}_{2}$ 上每点对应的溶液中水的电离程度都大于线 $\\mathrm{L}_{1}$\nD: 当 $c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{HX}^{-}\\right)-c\\left(\\mathrm{X}^{2-}\\right)=0$ 时, $\\mathrm{pH}<7$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 用 $\\mathrm{NaOH}$ 溶液滴加到某酒石酸 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 溶液中, 混合溶液的 $\\mathrm{pH}$ 与离子浓度变化的关系如图所示。下列叙述错误的是\n\n[图1]\n\nA: $\\mathrm{L}_{2}$ 表示 $\\mathrm{pH}$ 与 $-\\lg \\frac{c\\left(\\mathrm{X}^{2-}\\right)}{c\\left(\\mathrm{HX}^{-}\\right)}$的变化关系\nB: $\\mathrm{NaHX}$ 溶液中 $c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)$\nC: 线 $\\mathrm{L}_{2}$ 上每点对应的溶液中水的电离程度都大于线 $\\mathrm{L}_{1}$\nD: 当 $c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{HX}^{-}\\right)-c\\left(\\mathrm{X}^{2-}\\right)=0$ 时, $\\mathrm{pH}<7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-11.jpg?height=465&width=548&top_left_y=430&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1343",
        "problem": "Ramsay and Clève discovered helium in cleveite (a mineral consisting of uranium oxide and oxides of lead, thorium, and rare earths; an impure variety of uraninite) independently and virtually simultaneously in 1895 . The gas extracted from the rock showed a unique spectroscopic line at around $588 \\mathrm{~nm}$ (indicated by $\\mathrm{D}_{3}$ in Figure 1), which was first observed in the spectrum of solar prominence during a total eclipse in 1868, near the well-known $D_{1}$ and $D_{2}$ lines of sodium.\n\n[figure1]\n\nFigure 1. Spectral lines around $588 \\mathrm{~nm}$Calculate the energy $E[\\mathrm{~J}]$ of a photon with the wavelength of the $\\mathrm{D}_{3}$ line of helium shown in Figure 1.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nRamsay and Clève discovered helium in cleveite (a mineral consisting of uranium oxide and oxides of lead, thorium, and rare earths; an impure variety of uraninite) independently and virtually simultaneously in 1895 . The gas extracted from the rock showed a unique spectroscopic line at around $588 \\mathrm{~nm}$ (indicated by $\\mathrm{D}_{3}$ in Figure 1), which was first observed in the spectrum of solar prominence during a total eclipse in 1868, near the well-known $D_{1}$ and $D_{2}$ lines of sodium.\n\n[figure1]\n\nFigure 1. Spectral lines around $588 \\mathrm{~nm}$\n\nproblem:\nCalculate the energy $E[\\mathrm{~J}]$ of a photon with the wavelength of the $\\mathrm{D}_{3}$ line of helium shown in Figure 1.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~J}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-058.jpg?height=403&width=700&top_left_y=1175&top_left_x=678"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~J}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_642",
        "problem": "$25^{\\circ} \\mathrm{C}$, 改变 $0.01 \\mathrm{~mol} / \\mathrm{L} \\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液的 $\\mathrm{pH}$, 溶液中 $\\mathrm{CH}_{3} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{COO}^{-} 、 \\mathrm{H}^{+} 、$ $\\mathrm{OH}-$ 浓度的对数值 $1 \\mathrm{gc}$ 与溶液 $\\mathrm{pH}$ 的变化关系如图所示。下列叙述不正确的是\n\n[图1]\nA: d 线表示 $\\mathrm{OH}^{-}$\nB: $\\mathrm{pH}=6$ 时, $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: 点 $\\mathrm{M}$ 的纵坐标与点 $\\mathrm{N}$ 的横坐标数值相等, 符号相反\nD: $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 的 $\\mathrm{pH}$ 约等于线 $\\mathrm{c}$ 与线 $\\mathrm{d}$ 交点处的横坐标值\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$, 改变 $0.01 \\mathrm{~mol} / \\mathrm{L} \\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液的 $\\mathrm{pH}$, 溶液中 $\\mathrm{CH}_{3} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{COO}^{-} 、 \\mathrm{H}^{+} 、$ $\\mathrm{OH}-$ 浓度的对数值 $1 \\mathrm{gc}$ 与溶液 $\\mathrm{pH}$ 的变化关系如图所示。下列叙述不正确的是\n\n[图1]\n\nA: d 线表示 $\\mathrm{OH}^{-}$\nB: $\\mathrm{pH}=6$ 时, $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: 点 $\\mathrm{M}$ 的纵坐标与点 $\\mathrm{N}$ 的横坐标数值相等, 符号相反\nD: $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 的 $\\mathrm{pH}$ 约等于线 $\\mathrm{c}$ 与线 $\\mathrm{d}$ 交点处的横坐标值\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-41.jpg?height=425&width=734&top_left_y=1324&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1011",
        "problem": "An element, $X$, from group 1 of the periodic table, combines to form a stable compound with an element, $Y$, from group 16. The formula of that compound is most likely to be\nA: $\\mathrm{X}_{3} \\mathrm{Y}$\nB: $\\mathrm{XY}_{3}$\nC: $X Y$\nD: $\\mathrm{X}_{2} \\mathrm{Y}$\nE: $X Y_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn element, $X$, from group 1 of the periodic table, combines to form a stable compound with an element, $Y$, from group 16. The formula of that compound is most likely to be\n\nA: $\\mathrm{X}_{3} \\mathrm{Y}$\nB: $\\mathrm{XY}_{3}$\nC: $X Y$\nD: $\\mathrm{X}_{2} \\mathrm{Y}$\nE: $X Y_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_988",
        "problem": "If 17.0 grams of sodium chloride are dissolved in water to make $0.5 \\mathrm{~L}$ of solution, then what is the final concentration of the solution? Give your answer with the correct number of significant figures.\nA: $0.6 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.58 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.581 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $0.3 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $\\quad 0.291 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf 17.0 grams of sodium chloride are dissolved in water to make $0.5 \\mathrm{~L}$ of solution, then what is the final concentration of the solution? Give your answer with the correct number of significant figures.\n\nA: $0.6 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.58 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.581 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $0.3 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $\\quad 0.291 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_475",
        "problem": "某 $\\mathrm{NaAlO}_{2} 、 \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 的混合溶液中逐滴加入 $1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸, 测得溶液中的 $\\mathrm{CO}_{3}{ }^{2-} 、$ $\\mathrm{HCO}_{3}-\\mathrm{AlO}_{2} 、 \\mathrm{Al}^{3+}$ 的物质的量与加入盐酸溶液的体积变化关系如图所示。(已知 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$的电离平衡常数 $\\mathrm{K}_{1}=4.3 \\times 10^{-7}, \\mathrm{~K}_{2}=5.6 \\times 10^{-11} ; \\mathrm{Al}(\\mathrm{OH})_{3}$ 的酸式电离平衡常数 $\\left.\\mathrm{K}=6.3 \\times 10^{-13}\\right)$\n则下列说法正确的是\n\n[图1]\nA: $\\mathrm{V}_{1}: \\mathrm{V}_{2}=1: 4$\nB: $\\mathrm{M}$ 点时生成的 $\\mathrm{CO}_{2}$ 为 $0.05 \\mathrm{~mol}$\nC: 原混合溶液中的 $\\mathrm{CO}_{3}{ }^{2-}$ 与 $\\mathrm{AlO}_{2}{ }^{-}$的物质的量之比为 $1: 3$\nD: $\\mathrm{a}$ 曲线表示的离子方程式为: $\\mathrm{AlO}_{2}{ }^{-}+4 \\mathrm{H}^{+}=\\mathrm{Al}^{3+}+2 \\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某 $\\mathrm{NaAlO}_{2} 、 \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 的混合溶液中逐滴加入 $1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸, 测得溶液中的 $\\mathrm{CO}_{3}{ }^{2-} 、$ $\\mathrm{HCO}_{3}-\\mathrm{AlO}_{2} 、 \\mathrm{Al}^{3+}$ 的物质的量与加入盐酸溶液的体积变化关系如图所示。(已知 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$的电离平衡常数 $\\mathrm{K}_{1}=4.3 \\times 10^{-7}, \\mathrm{~K}_{2}=5.6 \\times 10^{-11} ; \\mathrm{Al}(\\mathrm{OH})_{3}$ 的酸式电离平衡常数 $\\left.\\mathrm{K}=6.3 \\times 10^{-13}\\right)$\n则下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{V}_{1}: \\mathrm{V}_{2}=1: 4$\nB: $\\mathrm{M}$ 点时生成的 $\\mathrm{CO}_{2}$ 为 $0.05 \\mathrm{~mol}$\nC: 原混合溶液中的 $\\mathrm{CO}_{3}{ }^{2-}$ 与 $\\mathrm{AlO}_{2}{ }^{-}$的物质的量之比为 $1: 3$\nD: $\\mathrm{a}$ 曲线表示的离子方程式为: $\\mathrm{AlO}_{2}{ }^{-}+4 \\mathrm{H}^{+}=\\mathrm{Al}^{3+}+2 \\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-33.jpg?height=591&width=854&top_left_y=230&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_109",
        "problem": "The metric known as reaction mass efficiency (RME) provides a way to assess how much reactant material ends up in a desired product at the end of a chemical reaction. One way of expressing RME is as follows:\n\nreaction mass efficiency $=\\frac{\\text { mass of desired product }}{(\\text { total input mass }- \\text { mass of recycled material })}$\n\n0.115 moles of cholesterol $\\left(\\mathrm{C}_{27} \\mathrm{H}_{46} \\mathrm{O}\\right)$ was reacted with 0.365 moles of molecular bromine to form 0.102 moles of dibromocholesterol in an addition reaction (shown below). It was possible to recover and recycle 0.151 moles of molecular bromine from the reaction mixture.\n\n[figure1]\n\nWhat is the percentage reaction mass efficiency for this process?\nA: $43.3 \\%$\nB: $54.3 \\%$\nC: $70.9 \\%$\nD: $84.6 \\%$\nE: $89.1 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe metric known as reaction mass efficiency (RME) provides a way to assess how much reactant material ends up in a desired product at the end of a chemical reaction. One way of expressing RME is as follows:\n\nreaction mass efficiency $=\\frac{\\text { mass of desired product }}{(\\text { total input mass }- \\text { mass of recycled material })}$\n\n0.115 moles of cholesterol $\\left(\\mathrm{C}_{27} \\mathrm{H}_{46} \\mathrm{O}\\right)$ was reacted with 0.365 moles of molecular bromine to form 0.102 moles of dibromocholesterol in an addition reaction (shown below). It was possible to recover and recycle 0.151 moles of molecular bromine from the reaction mixture.\n\n[figure1]\n\nWhat is the percentage reaction mass efficiency for this process?\n\nA: $43.3 \\%$\nB: $54.3 \\%$\nC: $70.9 \\%$\nD: $84.6 \\%$\nE: $89.1 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_de998ba2e5d42a77dc86g-3.jpg?height=397&width=909&top_left_y=1338&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_738",
        "problem": "用下图所示装置除去含 $\\mathrm{CN}^{-} 、 \\mathrm{Cl}^{-}$废水中的 $\\mathrm{CN}^{-}$时, 控制溶液 $\\mathrm{PH}$ 为 $9 \\sim 10$, 阳极产生的 $\\mathrm{ClO}^{-}$将 $\\mathrm{CN}^{-}$氧化为两种无污染的气体, 下列说法不正确的是\n\n[图1]\nA: 用石墨作阳极, 铁作阴极\nB: 阳极的电极反应式为: $\\mathrm{Cl}^{-}+2 \\mathrm{OH}^{-}-2 \\mathrm{e}^{-}=\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 阴极的电极反应式为: $2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow+2 \\mathrm{OH}^{-}$\nD: 除去 $\\mathrm{CN}^{-}$的反应: $2 \\mathrm{CN}^{-}+5 \\mathrm{ClO}^{-}+2 \\mathrm{H}^{+}=\\mathrm{N}_{2} \\uparrow+2 \\mathrm{CO}_{2} \\uparrow+5 \\mathrm{Cl}^{-}+\\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用下图所示装置除去含 $\\mathrm{CN}^{-} 、 \\mathrm{Cl}^{-}$废水中的 $\\mathrm{CN}^{-}$时, 控制溶液 $\\mathrm{PH}$ 为 $9 \\sim 10$, 阳极产生的 $\\mathrm{ClO}^{-}$将 $\\mathrm{CN}^{-}$氧化为两种无污染的气体, 下列说法不正确的是\n\n[图1]\n\nA: 用石墨作阳极, 铁作阴极\nB: 阳极的电极反应式为: $\\mathrm{Cl}^{-}+2 \\mathrm{OH}^{-}-2 \\mathrm{e}^{-}=\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 阴极的电极反应式为: $2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow+2 \\mathrm{OH}^{-}$\nD: 除去 $\\mathrm{CN}^{-}$的反应: $2 \\mathrm{CN}^{-}+5 \\mathrm{ClO}^{-}+2 \\mathrm{H}^{+}=\\mathrm{N}_{2} \\uparrow+2 \\mathrm{CO}_{2} \\uparrow+5 \\mathrm{Cl}^{-}+\\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-58.jpg?height=488&width=539&top_left_y=2189&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_408",
        "problem": "常温下, 将体积均为 $V_{0}$ 的 HA 溶液和 $\\mathrm{BOH}$ 溶液分别加水稀释至 $\\mathrm{V}$, 溶液的 $\\mathrm{pH}$ 随 $\\lg \\frac{V}{V_{0}}$的变化如图所示。已知 $\\mathrm{pOH}=-\\lg c\\left(\\mathrm{OH}^{-}\\right)$, 下列说法正确的是\n\n[图1]\nA: 电离常数: $K_{\\mathrm{a}}(\\mathrm{HA})>K_{\\mathrm{b}}(\\mathrm{BOH})$\nB: 水的电离程度: $b>d$\nC: 升温, $\\mathrm{a}$ 点溶液中 $\\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HA})}$ 增大\nD: 等浓度 $\\mathrm{HA}$ 和 $\\mathrm{BOH}$ 溶液等体积混合, 溶液的 $\\mathrm{pH}<7$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 将体积均为 $V_{0}$ 的 HA 溶液和 $\\mathrm{BOH}$ 溶液分别加水稀释至 $\\mathrm{V}$, 溶液的 $\\mathrm{pH}$ 随 $\\lg \\frac{V}{V_{0}}$的变化如图所示。已知 $\\mathrm{pOH}=-\\lg c\\left(\\mathrm{OH}^{-}\\right)$, 下列说法正确的是\n\n[图1]\n\nA: 电离常数: $K_{\\mathrm{a}}(\\mathrm{HA})>K_{\\mathrm{b}}(\\mathrm{BOH})$\nB: 水的电离程度: $b>d$\nC: 升温, $\\mathrm{a}$ 点溶液中 $\\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HA})}$ 增大\nD: 等浓度 $\\mathrm{HA}$ 和 $\\mathrm{BOH}$ 溶液等体积混合, 溶液的 $\\mathrm{pH}<7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-038.jpg?height=357&width=556&top_left_y=781&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_18",
        "problem": "Triiodide reacts with acetone in acidic aqueous solution:\n\n$$\n\\begin{aligned}\n\\mathrm{I}_{3}^{-}(a q)+\\mathrm{CH}_{3} \\mathrm{COCH}_{3}(a q) \\rightarrow & \\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{I}(a q)+2 \\mathrm{I}^{-}(a q) \\\\\n& +\\mathrm{H}^{+}(a q)\n\\end{aligned}\n$$\n\nThe reaction is carried out twice, both times with $\\left[\\mathrm{I}_{3}{ }^{-}\\right]_{0}=1.0 \\times 10^{-3} \\mathrm{M}$ and $\\left[\\mathrm{H}^{+}\\right]_{0}=1.00 \\mathrm{M}$, but with two different concentrations of acetone, and the concentration of triiodide is found to vary with time as shown:\n\n[figure1]\n\nWhat are the orders of triiodide and acetone in the reaction?\nA: First-order in $\\mathrm{I}_{3}^{-}$, first-order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$\nB: Zeroth-order in $\\mathrm{I}_{3}^{-}$, first-order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$\nC: First-order in $\\mathrm{I}_{3}{ }^{-}$; the order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ cannot be determined from the available data\nD: First-order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$; the order in $\\mathrm{I}_{3}^{-}$cannot be determined from the available data\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTriiodide reacts with acetone in acidic aqueous solution:\n\n$$\n\\begin{aligned}\n\\mathrm{I}_{3}^{-}(a q)+\\mathrm{CH}_{3} \\mathrm{COCH}_{3}(a q) \\rightarrow & \\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{I}(a q)+2 \\mathrm{I}^{-}(a q) \\\\\n& +\\mathrm{H}^{+}(a q)\n\\end{aligned}\n$$\n\nThe reaction is carried out twice, both times with $\\left[\\mathrm{I}_{3}{ }^{-}\\right]_{0}=1.0 \\times 10^{-3} \\mathrm{M}$ and $\\left[\\mathrm{H}^{+}\\right]_{0}=1.00 \\mathrm{M}$, but with two different concentrations of acetone, and the concentration of triiodide is found to vary with time as shown:\n\n[figure1]\n\nWhat are the orders of triiodide and acetone in the reaction?\n\nA: First-order in $\\mathrm{I}_{3}^{-}$, first-order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$\nB: Zeroth-order in $\\mathrm{I}_{3}^{-}$, first-order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$\nC: First-order in $\\mathrm{I}_{3}{ }^{-}$; the order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$ cannot be determined from the available data\nD: First-order in $\\mathrm{CH}_{3} \\mathrm{COCH}_{3}$; the order in $\\mathrm{I}_{3}^{-}$cannot be determined from the available data\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-06.jpg?height=456&width=674&top_left_y=550&top_left_x=281"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_657",
        "problem": "四氢铝锂 $\\left(\\mathrm{LiAlH}_{4}\\right)$ 常作为有机合成的重要还原剂。工业上以辉锂矿(主要成分\n\n$\\mathrm{LiAlSi}_{2} \\mathrm{O}_{6}$, 含少量 $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ ) 为原料合成 $\\mathrm{LiAlH}_{4}$ 的流程如下:\n\n[图1]\n\n已知：金属氢氧化物沉淀的 $\\mathrm{pH}$ 如表所示:\n\n| 物质 | $\\mathrm{Fe}(\\mathrm{OH})_{3}$ | $\\mathrm{Al}(\\mathrm{OH})_{3}$ |\n| :--- | :--- | :--- |\n| 开始沉淀的 <br> $\\mathrm{pH}$ | 2.3 | 4 |\n| 完全沉淀的 <br> $\\mathrm{pH}$ | 3.7 | 6.5 |\n\n碳酸锂溶解度数据如下:\n\n| 温度 $/{ }^{\\circ} \\mathrm{C}$ | 10 | 30 | 60 | 90 |\n| :--- | :--- | :--- | :--- | :--- |\n| 浓度 $/\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)$ | 0.21 | 0.17 | 0.14 | 0.10 |\n\n下列说法错误的是\nA: 为了获取产品较为纯净, $\\mathrm{a}$ 的最小值为 6.5\nB: 操作 1 中加入足量碳酸钠后, 蒸发浓缩后趁热过滤, 能减少产品的溶解损失\nC: 流程中由 $\\mathrm{LiCl}$ 制备 $\\mathrm{Li}$ 单质时电解得到的 $\\mathrm{HCl}$ 可循环利用于上一步 $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 合成 $\\mathrm{LiCl}$\nD: 最后一步合成时原子利用率能达到 $100 \\%$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n四氢铝锂 $\\left(\\mathrm{LiAlH}_{4}\\right)$ 常作为有机合成的重要还原剂。工业上以辉锂矿(主要成分\n\n$\\mathrm{LiAlSi}_{2} \\mathrm{O}_{6}$, 含少量 $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ ) 为原料合成 $\\mathrm{LiAlH}_{4}$ 的流程如下:\n\n[图1]\n\n已知：金属氢氧化物沉淀的 $\\mathrm{pH}$ 如表所示:\n\n| 物质 | $\\mathrm{Fe}(\\mathrm{OH})_{3}$ | $\\mathrm{Al}(\\mathrm{OH})_{3}$ |\n| :--- | :--- | :--- |\n| 开始沉淀的 <br> $\\mathrm{pH}$ | 2.3 | 4 |\n| 完全沉淀的 <br> $\\mathrm{pH}$ | 3.7 | 6.5 |\n\n碳酸锂溶解度数据如下:\n\n| 温度 $/{ }^{\\circ} \\mathrm{C}$ | 10 | 30 | 60 | 90 |\n| :--- | :--- | :--- | :--- | :--- |\n| 浓度 $/\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)$ | 0.21 | 0.17 | 0.14 | 0.10 |\n\n下列说法错误的是\n\nA: 为了获取产品较为纯净, $\\mathrm{a}$ 的最小值为 6.5\nB: 操作 1 中加入足量碳酸钠后, 蒸发浓缩后趁热过滤, 能减少产品的溶解损失\nC: 流程中由 $\\mathrm{LiCl}$ 制备 $\\mathrm{Li}$ 单质时电解得到的 $\\mathrm{HCl}$ 可循环利用于上一步 $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 合成 $\\mathrm{LiCl}$\nD: 最后一步合成时原子利用率能达到 $100 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-66.jpg?height=437&width=1448&top_left_y=2209&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1495",
        "problem": "Bohemian garnet (pyrope) is a famous Czech blood coloured semi-precious stone. The chemical composition of natural garnets is expressed by the general stoichiometric formula of $\\mathrm{A}_{3} \\mathrm{~B}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$, where $A^{\\text {II }}$ is a divalent cation and $B^{\\text {III }}$ is a trivalent cation. Garnets have a cubic unit cell that contains 8 formula units. The structure comprises 3 types of polyhedra: the $A^{l l}$ cation occupies a dodecahedral position (it is surrounded with eight $O$ atoms), the $\\mathrm{B}^{\\text {III }}$ cation occupies an octahedral position (it is surrounded with six $\\mathrm{O}$ atoms) and $\\mathrm{Si}^{\\mathrm{lV}}$ is surrounded with four $\\mathrm{O}$ atoms arranged into a tetrahedron.\n\n\nAndradite is another garnet mineral; its chemical composition is $\\mathrm{Ca}_{3} \\mathrm{Fe}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$. A double cation substitution $-\\mathrm{Ti}^{\\mathrm{IV}}$ for $\\mathrm{Fe}^{\\mathrm{III}}$ in the octahedral position and $\\mathrm{Fe}^{\\mathrm{III}}$ for $\\mathrm{Si}^{\\mathrm{IV}}$ in the tetrahedral position - gives rise to black schorlomite. Its chemical composition can be expressed as $\\mathrm{Ca}_{3}[\\mathrm{Fe}, \\mathrm{Ti}]_{2}^{\\text {oct }}\\left([\\mathrm{Si}, \\mathrm{Fe}]^{\\text {tet }} \\mathrm{O}_{4}\\right)_{3}$.Calculate the percentage of $\\mathrm{Si}^{\\mathrm{V}}$ ions in a sample of schorlomite that must be substituted with $\\mathrm{Fe}^{\\mathrm{III}}$, if we know that $5 \\%$ of $\\mathrm{Fe}^{\\mathrm{III}}$ ions in octahedral position are substituted with $\\mathrm{Ti}^{\\mathrm{IV}}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nBohemian garnet (pyrope) is a famous Czech blood coloured semi-precious stone. The chemical composition of natural garnets is expressed by the general stoichiometric formula of $\\mathrm{A}_{3} \\mathrm{~B}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$, where $A^{\\text {II }}$ is a divalent cation and $B^{\\text {III }}$ is a trivalent cation. Garnets have a cubic unit cell that contains 8 formula units. The structure comprises 3 types of polyhedra: the $A^{l l}$ cation occupies a dodecahedral position (it is surrounded with eight $O$ atoms), the $\\mathrm{B}^{\\text {III }}$ cation occupies an octahedral position (it is surrounded with six $\\mathrm{O}$ atoms) and $\\mathrm{Si}^{\\mathrm{lV}}$ is surrounded with four $\\mathrm{O}$ atoms arranged into a tetrahedron.\n\n\nAndradite is another garnet mineral; its chemical composition is $\\mathrm{Ca}_{3} \\mathrm{Fe}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$. A double cation substitution $-\\mathrm{Ti}^{\\mathrm{IV}}$ for $\\mathrm{Fe}^{\\mathrm{III}}$ in the octahedral position and $\\mathrm{Fe}^{\\mathrm{III}}$ for $\\mathrm{Si}^{\\mathrm{IV}}$ in the tetrahedral position - gives rise to black schorlomite. Its chemical composition can be expressed as $\\mathrm{Ca}_{3}[\\mathrm{Fe}, \\mathrm{Ti}]_{2}^{\\text {oct }}\\left([\\mathrm{Si}, \\mathrm{Fe}]^{\\text {tet }} \\mathrm{O}_{4}\\right)_{3}$.\n\nproblem:\nCalculate the percentage of $\\mathrm{Si}^{\\mathrm{V}}$ ions in a sample of schorlomite that must be substituted with $\\mathrm{Fe}^{\\mathrm{III}}$, if we know that $5 \\%$ of $\\mathrm{Fe}^{\\mathrm{III}}$ ions in octahedral position are substituted with $\\mathrm{Ti}^{\\mathrm{IV}}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_141",
        "problem": "In their solid form, which of the following is made up of discrete particles (either atoms, ions or molecules) that only have London Dispersion forces of attraction between them?\nA: $\\mathrm{Ag}$\nB: $\\mathrm{CO}_{2}$\nC: C, graphite\nD: $\\mathrm{KCl}$\nE: $\\mathrm{NH}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn their solid form, which of the following is made up of discrete particles (either atoms, ions or molecules) that only have London Dispersion forces of attraction between them?\n\nA: $\\mathrm{Ag}$\nB: $\\mathrm{CO}_{2}$\nC: C, graphite\nD: $\\mathrm{KCl}$\nE: $\\mathrm{NH}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_922",
        "problem": "某无色稀溶液 X 中, 可能含有如表所列离子中的某几种。\n\n取该溶液适量, 向其中加入某试剂 $\\mathrm{Y}$, 产生沉淀的物质的量 $(\\mathrm{n})$ 与加入试剂体积 $(\\mathrm{Y})$ 的关系如图所示。下列说法正确的是\n\n| 阴离子 | $\\mathrm{CO}_{3}^{2-} 、 \\mathrm{SiO}_{3}^{2-} 、\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-} 、$ |\n| :--- | :--- |\n|  | $\\mathrm{Cl}^{-}$ |\n| 阳离子 | $\\mathrm{Al}^{3+} 、 \\mathrm{Fe}^{3+} 、 \\mathrm{Mg}^{2+} 、 \\mathrm{NH}_{4}^{+} 、 \\mathrm{Na}^{+}$ |\n\n[图1]\nA: 若 $Y$ 是盐酸, 则 $\\mathrm{X}$ 中一定含有 $\\mathrm{CO}_{3}^{2-} 、 \\mathrm{SiO}_{3}^{2-} 、\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$ 和 $\\mathrm{NH}_{4}{ }^{+}$\nB: 若 $\\mathrm{Y}$ 是 $\\mathrm{NaOH}$ 溶液, 则 $\\mathrm{X}$ 中一定含有 $\\mathrm{Al}^{3+} 、 \\mathrm{Fe}^{3+} 、 \\mathrm{NH}_{4}^{+} 、 \\mathrm{Cl}^{-}$\nC: 若 $\\mathrm{Y}$ 是 $\\mathrm{NaOH}$ 溶液, 则 ab 段发生反应的离子方程式为: $\\mathrm{NH}_{4}^{+}+\\mathrm{OH}^{-}=\\mathrm{NH}_{3} \\uparrow+\\mathrm{H}_{2} \\mathrm{O}$\nD: 若 $\\mathrm{Y}$ 是 $\\mathrm{NaOH}$ 溶液, 则 $\\mathrm{X}$ 中的 $\\mathrm{Al}^{3+} 、 \\mathrm{Mg}^{2+} 、 \\mathrm{NH}_{4}^{+}$物质的量之比为 2: 1: 4\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某无色稀溶液 X 中, 可能含有如表所列离子中的某几种。\n\n取该溶液适量, 向其中加入某试剂 $\\mathrm{Y}$, 产生沉淀的物质的量 $(\\mathrm{n})$ 与加入试剂体积 $(\\mathrm{Y})$ 的关系如图所示。下列说法正确的是\n\n| 阴离子 | $\\mathrm{CO}_{3}^{2-} 、 \\mathrm{SiO}_{3}^{2-} 、\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-} 、$ |\n| :--- | :--- |\n|  | $\\mathrm{Cl}^{-}$ |\n| 阳离子 | $\\mathrm{Al}^{3+} 、 \\mathrm{Fe}^{3+} 、 \\mathrm{Mg}^{2+} 、 \\mathrm{NH}_{4}^{+} 、 \\mathrm{Na}^{+}$ |\n\n[图1]\n\nA: 若 $Y$ 是盐酸, 则 $\\mathrm{X}$ 中一定含有 $\\mathrm{CO}_{3}^{2-} 、 \\mathrm{SiO}_{3}^{2-} 、\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$ 和 $\\mathrm{NH}_{4}{ }^{+}$\nB: 若 $\\mathrm{Y}$ 是 $\\mathrm{NaOH}$ 溶液, 则 $\\mathrm{X}$ 中一定含有 $\\mathrm{Al}^{3+} 、 \\mathrm{Fe}^{3+} 、 \\mathrm{NH}_{4}^{+} 、 \\mathrm{Cl}^{-}$\nC: 若 $\\mathrm{Y}$ 是 $\\mathrm{NaOH}$ 溶液, 则 ab 段发生反应的离子方程式为: $\\mathrm{NH}_{4}^{+}+\\mathrm{OH}^{-}=\\mathrm{NH}_{3} \\uparrow+\\mathrm{H}_{2} \\mathrm{O}$\nD: 若 $\\mathrm{Y}$ 是 $\\mathrm{NaOH}$ 溶液, 则 $\\mathrm{X}$ 中的 $\\mathrm{Al}^{3+} 、 \\mathrm{Mg}^{2+} 、 \\mathrm{NH}_{4}^{+}$物质的量之比为 2: 1: 4\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-86.jpg?height=277&width=440&top_left_y=2323&top_left_x=320",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-87.jpg?height=54&width=1380&top_left_y=681&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_19",
        "problem": "The vapor pressure of iodomethane, $\\mathrm{CH}_{3} \\mathrm{I}(M=141.9)$, is 110. $\\mathrm{mm} \\mathrm{Hg}$ at $266 \\mathrm{~K}$. A $0.824 \\mathrm{~g}$ sample of iodomethane is placed in a closed, evacuated $370 . \\mathrm{mL}$ container at $266 \\mathrm{~K}$. At equilibrium, what will be the pressure in the container?\nA: $96.4 \\mathrm{~mm} \\mathrm{Hg}$\nB: $110 . \\mathrm{mm} \\mathrm{Hg}$\nC: $260 . \\mathrm{mm} \\mathrm{Hg}$\nD: 292. $\\mathrm{mm} \\mathrm{Hg}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe vapor pressure of iodomethane, $\\mathrm{CH}_{3} \\mathrm{I}(M=141.9)$, is 110. $\\mathrm{mm} \\mathrm{Hg}$ at $266 \\mathrm{~K}$. A $0.824 \\mathrm{~g}$ sample of iodomethane is placed in a closed, evacuated $370 . \\mathrm{mL}$ container at $266 \\mathrm{~K}$. At equilibrium, what will be the pressure in the container?\n\nA: $96.4 \\mathrm{~mm} \\mathrm{Hg}$\nB: $110 . \\mathrm{mm} \\mathrm{Hg}$\nC: $260 . \\mathrm{mm} \\mathrm{Hg}$\nD: 292. $\\mathrm{mm} \\mathrm{Hg}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_569",
        "problem": "在相同温度下, 在三个大小完全一样的容器中发生反应(保持容器体积不变): $3 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{N}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\triangle H=-92 \\mathrm{~kJ} / \\mathrm{mol}$ 。开始时投料情况是: $\\mathrm{A}$ 中 $3 \\mathrm{molH}_{2}(\\mathrm{~g})$ 和 $1 \\mathrm{molN}_{2}(\\mathrm{~g}) ; \\mathrm{B}$ 中 $2 \\mathrm{molNH}_{3}(\\mathrm{~g}) ; \\mathrm{C}$ 中 $4 \\mathrm{molNH}_{3}(\\mathrm{~g})$, 都达到平衡后, 下列说法正确的是\nA: 三个容器中的转化率的关系是: $\\alpha(\\mathrm{A})+\\alpha(\\mathrm{B})>\\alpha(\\mathrm{A})+\\alpha(\\mathrm{C})$\nB: A 容器放出的热量为 $92 k J$, B 容器中吸收热量为 $92 k J$\nC: 达到平衡后三容器中氨气的体积分数相等\nD: A 容器中氨气的平衡浓度是 C 容器中氨气的平衡浓度的一半\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在相同温度下, 在三个大小完全一样的容器中发生反应(保持容器体积不变): $3 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{N}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\triangle H=-92 \\mathrm{~kJ} / \\mathrm{mol}$ 。开始时投料情况是: $\\mathrm{A}$ 中 $3 \\mathrm{molH}_{2}(\\mathrm{~g})$ 和 $1 \\mathrm{molN}_{2}(\\mathrm{~g}) ; \\mathrm{B}$ 中 $2 \\mathrm{molNH}_{3}(\\mathrm{~g}) ; \\mathrm{C}$ 中 $4 \\mathrm{molNH}_{3}(\\mathrm{~g})$, 都达到平衡后, 下列说法正确的是\n\nA: 三个容器中的转化率的关系是: $\\alpha(\\mathrm{A})+\\alpha(\\mathrm{B})>\\alpha(\\mathrm{A})+\\alpha(\\mathrm{C})$\nB: A 容器放出的热量为 $92 k J$, B 容器中吸收热量为 $92 k J$\nC: 达到平衡后三容器中氨气的体积分数相等\nD: A 容器中氨气的平衡浓度是 C 容器中氨气的平衡浓度的一半\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1024",
        "problem": "For E5 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 31170. For E10 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 30680.To help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll the carbon dioxide released by the combustion of ethanol is offset by the carbon dioxide captured when the plants are grown. As such, the contribution from the ethanol is not counted when the $\\mathrm{CO}_{2}$ footprints of the $\\mathrm{E} 5$ and $\\mathrm{E} 10$ fuels are compared.\n\nTaking the $\\mathrm{CO}_{2}$ produced from burning 1 litre of $\\mathrm{E} 5$ fuel as $100 \\%$, calculate the percentage of $\\mathrm{CO}_{2}$ produced from burning 1 litre of $\\mathrm{E} 10$ fuel. Consider only $\\mathrm{CO}_{2}$ formed from the combustion of octane isomers.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nFor E5 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 31170. For E10 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 30680.\n\nproblem:\nTo help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll the carbon dioxide released by the combustion of ethanol is offset by the carbon dioxide captured when the plants are grown. As such, the contribution from the ethanol is not counted when the $\\mathrm{CO}_{2}$ footprints of the $\\mathrm{E} 5$ and $\\mathrm{E} 10$ fuels are compared.\n\nTaking the $\\mathrm{CO}_{2}$ produced from burning 1 litre of $\\mathrm{E} 5$ fuel as $100 \\%$, calculate the percentage of $\\mathrm{CO}_{2}$ produced from burning 1 litre of $\\mathrm{E} 10$ fuel. Consider only $\\mathrm{CO}_{2}$ formed from the combustion of octane isomers.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %(percentage), but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-02.jpg?height=459&width=923&top_left_y=356&top_left_x=949"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%(percentage)"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1169",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidations states of elements from Groups 7 and 17 (except fluorine) are +7 , what are the valence electrons of manganese?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidations states of elements from Groups 7 and 17 (except fluorine) are +7 , what are the valence electrons of manganese?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_186",
        "problem": "Which of the following would increase the $K_{\\text {eq }}$ of the reaction below?\n\n$$\n\\mathrm{HCO}_{3}^{-}{ }_{(\\mathrm{aq})}+\\mathrm{H}_{3} \\mathrm{O}_{(\\mathrm{aq})}^{+} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{CO}_{3(\\mathrm{aq})}+\\mathrm{H}_{2} \\mathrm{O}_{(\\mathrm{l})}\n$$\nA: Increasing the $\\mathrm{pH}$\nB: Decreasing the $\\mathrm{pH}$\nC: Adding $\\mathrm{H}_{2} \\mathrm{CO}_{3}$\nD: None of the options\nE: Adding water provided\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following would increase the $K_{\\text {eq }}$ of the reaction below?\n\n$$\n\\mathrm{HCO}_{3}^{-}{ }_{(\\mathrm{aq})}+\\mathrm{H}_{3} \\mathrm{O}_{(\\mathrm{aq})}^{+} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{CO}_{3(\\mathrm{aq})}+\\mathrm{H}_{2} \\mathrm{O}_{(\\mathrm{l})}\n$$\n\nA: Increasing the $\\mathrm{pH}$\nB: Decreasing the $\\mathrm{pH}$\nC: Adding $\\mathrm{H}_{2} \\mathrm{CO}_{3}$\nD: None of the options\nE: Adding water provided\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_200",
        "problem": "A component of diesel fuel is the hydrocarbon $\\mathrm{C}_{12} \\mathrm{H}_{24}$, with density $0.790 \\mathrm{~g} \\mathrm{~mL}^{-1}$. What volume of $\\mathrm{CO}_{2}$ (measured at $25^{\\circ} \\mathrm{C}$ and $100 \\mathrm{kPa}$ ) is produced from the complete combustion of $2.00 \\mathrm{~L}$ of $\\mathrm{C}_{12} \\mathrm{H}_{24}$ in excess oxygen?\nA: $2.79 \\mathrm{~L}$\nB: $3.53 \\mathrm{~L}$\nC: $1400 \\mathrm{~L}$\nD: $2790 \\mathrm{~L}$\nE: $3530 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA component of diesel fuel is the hydrocarbon $\\mathrm{C}_{12} \\mathrm{H}_{24}$, with density $0.790 \\mathrm{~g} \\mathrm{~mL}^{-1}$. What volume of $\\mathrm{CO}_{2}$ (measured at $25^{\\circ} \\mathrm{C}$ and $100 \\mathrm{kPa}$ ) is produced from the complete combustion of $2.00 \\mathrm{~L}$ of $\\mathrm{C}_{12} \\mathrm{H}_{24}$ in excess oxygen?\n\nA: $2.79 \\mathrm{~L}$\nB: $3.53 \\mathrm{~L}$\nC: $1400 \\mathrm{~L}$\nD: $2790 \\mathrm{~L}$\nE: $3530 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_749",
        "problem": "下列实验能得出相关结论的是\n\n|  | 实验操作 | 实验结论 |\n| :--- | :--- | :--- |\n| A | 向 $\\mathrm{NaHA}$ 溶液中滴加紫色石荵溶液, 溶液变为蓝色 | $\\mathrm{K}_{\\mathrm{w}}>\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right) \\cdot \\mathrm{K}_{\\mathrm{a} 2}($ |\n| $\\left.\\mathrm{H}_{2} \\mathrm{~A}\\right)$ |  |  |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列实验能得出相关结论的是\n\n|  | 实验操作 | 实验结论 |\n| :--- | :--- | :--- |\n| A | 向 $\\mathrm{NaHA}$ 溶液中滴加紫色石荵溶液, 溶液变为蓝色 | $\\mathrm{K}_{\\mathrm{w}}>\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right) \\cdot \\mathrm{K}_{\\mathrm{a} 2}($ |\n| $\\left.\\mathrm{H}_{2} \\mathrm{~A}\\right)$ |  |  |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_964",
        "problem": "已知: $\\mathrm{A}(\\mathrm{g})+2 \\mathrm{~B}(\\mathrm{~g}) \\rightleftharpoons 3 \\mathrm{C}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}<0$, 向恒温恒容的密闭容器中充入 $1 \\mathrm{molA}$ 和\n\n$2 m o l B, t_{1}$ 时达到平衡状态 $I$, 在 $t_{2}$ 时改变某一条件, $t_{3}$ 时重新达到平衡状态II, 正反应速率随时间的变化如图所示. 下列说法正确的是\n\n[图1]\nA: $t_{2}$ 时刻改变的条件可能是升高温度\nB: $t_{2} \\sim t_{3}$ 时, 平衡向逆反应方向移动\nC: 平衡时 $\\mathrm{A}$ 的体积分数 $\\varphi: \\varphi(\\mathrm{II})=\\varphi(\\mathrm{I})$\nD: 平衡常数 $\\mathrm{K}: \\mathrm{K}(\\mathrm{I})>\\mathrm{K}(\\mathrm{II})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知: $\\mathrm{A}(\\mathrm{g})+2 \\mathrm{~B}(\\mathrm{~g}) \\rightleftharpoons 3 \\mathrm{C}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}<0$, 向恒温恒容的密闭容器中充入 $1 \\mathrm{molA}$ 和\n\n$2 m o l B, t_{1}$ 时达到平衡状态 $I$, 在 $t_{2}$ 时改变某一条件, $t_{3}$ 时重新达到平衡状态II, 正反应速率随时间的变化如图所示. 下列说法正确的是\n\n[图1]\n\nA: $t_{2}$ 时刻改变的条件可能是升高温度\nB: $t_{2} \\sim t_{3}$ 时, 平衡向逆反应方向移动\nC: 平衡时 $\\mathrm{A}$ 的体积分数 $\\varphi: \\varphi(\\mathrm{II})=\\varphi(\\mathrm{I})$\nD: 平衡常数 $\\mathrm{K}: \\mathrm{K}(\\mathrm{I})>\\mathrm{K}(\\mathrm{II})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-027.jpg?height=492&width=597&top_left_y=148&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_298",
        "problem": "Which of the following correctly characterizes the bonds and geometry of $\\mathrm{C}_{2} \\mathrm{H}_{4}$ ?\nA: four $\\sigma$ bonds, one $\\pi$ bond and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $109^{\\circ}$\nB: five $\\sigma$ bonds, no $\\pi$ bonds and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $90^{\\circ}$\nC: five $\\sigma$ bonds, one $\\pi$ bond and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $120^{\\circ}$\nD: three $\\sigma$ bonds, two $\\pi$ bonds and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $109^{\\circ}$\nE: four $\\sigma$ bonds, two $\\pi$ bonds and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $120^{\\circ}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following correctly characterizes the bonds and geometry of $\\mathrm{C}_{2} \\mathrm{H}_{4}$ ?\n\nA: four $\\sigma$ bonds, one $\\pi$ bond and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $109^{\\circ}$\nB: five $\\sigma$ bonds, no $\\pi$ bonds and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $90^{\\circ}$\nC: five $\\sigma$ bonds, one $\\pi$ bond and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $120^{\\circ}$\nD: three $\\sigma$ bonds, two $\\pi$ bonds and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $109^{\\circ}$\nE: four $\\sigma$ bonds, two $\\pi$ bonds and an $\\mathrm{H}-\\mathrm{C}-\\mathrm{C}$ bond angle very close to $120^{\\circ}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_508",
        "problem": "对固体电解质体系的研究是电化学研究的重要领域之一, 用离子交换膜 $\\mathrm{H}^{+} / \\mathrm{NH}_{4}^{+}$型 Nafion 膜作电解质, 在一定条件下实现了常温常压下电化学合成氨, 原理如图所示。下列说法不正确的是\n\n[图1]\nA: 电极 $\\mathrm{M}$ 接电源的负极\nB: 离子交换膜中 $\\mathrm{H}^{+} 、 \\mathrm{NH}_{4}^{+}$浓度均保持不变\nC: $\\mathrm{H}^{+} / \\mathrm{NH}_{4}^{+}$型离子交换膜具有较高的传导质子的能力\nD: 阴极的电极反应式: $\\mathrm{N}_{2}+6 \\mathrm{e}^{-}+6 \\mathrm{H}^{+}=2 \\mathrm{NH}_{3}, 2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n对固体电解质体系的研究是电化学研究的重要领域之一, 用离子交换膜 $\\mathrm{H}^{+} / \\mathrm{NH}_{4}^{+}$型 Nafion 膜作电解质, 在一定条件下实现了常温常压下电化学合成氨, 原理如图所示。下列说法不正确的是\n\n[图1]\n\nA: 电极 $\\mathrm{M}$ 接电源的负极\nB: 离子交换膜中 $\\mathrm{H}^{+} 、 \\mathrm{NH}_{4}^{+}$浓度均保持不变\nC: $\\mathrm{H}^{+} / \\mathrm{NH}_{4}^{+}$型离子交换膜具有较高的传导质子的能力\nD: 阴极的电极反应式: $\\mathrm{N}_{2}+6 \\mathrm{e}^{-}+6 \\mathrm{H}^{+}=2 \\mathrm{NH}_{3}, 2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-38.jpg?height=577&width=782&top_left_y=594&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1550",
        "problem": "${ }^{131}$I is a radioactive isotope of iodine ( $\\mathrm{e}^{-}$emitter) used in nuclear medicine for analytical procedures to determine thyroid endocrine disorders by scintigraphy. The decay rate constant, $k$, of ${ }^{131} \\mathrm{I}$ is $9.93 \\times 10^{-7} \\mathrm{~s}^{-1}$.\n\nCalculate the time necessary (expressed in days) for a sample of ${ }^{131}$ I to reduce its activity to $30 \\%$ of the original value.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n${ }^{131}$I is a radioactive isotope of iodine ( $\\mathrm{e}^{-}$emitter) used in nuclear medicine for analytical procedures to determine thyroid endocrine disorders by scintigraphy. The decay rate constant, $k$, of ${ }^{131} \\mathrm{I}$ is $9.93 \\times 10^{-7} \\mathrm{~s}^{-1}$.\n\nCalculate the time necessary (expressed in days) for a sample of ${ }^{131}$ I to reduce its activity to $30 \\%$ of the original value.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of d, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "d"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_867",
        "problem": "常温下, 向 $10.0 \\mathrm{~mL}$ 浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{AlCl}_{3}$ 和 $\\mathrm{FeCl}_{3}$ 混合溶液中加入 $\\mathrm{NaOH}$ 固体,溶液中金属元素有不同的存在形式, 它们的物质的量浓度与 $\\mathrm{NaOH}$ 物质的量关系如图所示，测得 $\\mathrm{a} 、 \\mathrm{~b}$ 两点溶液的 $\\mathrm{pH}$ 分别为 3.0, 4.3.\n\n[图1]\n\n已知: (1) $K_{\\mathrm{sp}}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]>K_{\\mathrm{sp}}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]$;\n\n(2) $\\mathrm{Al}^{3+}(\\mathrm{aq})+4 \\mathrm{OH}^{-}(\\mathrm{aq}) \\rightleftharpoons\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}(\\mathrm{aq}), 298 \\mathrm{~K}$,\n\n$K_{\\text {稳 }}=\\frac{c\\left\\{\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}\\right\\}}{c\\left(\\mathrm{Al}^{3+}\\right) \\times c^{4}\\left(\\mathrm{OH}^{-}\\right)}=1.1 \\times 10^{33}$ 。\n\n下列叙述正确的是\nA: 曲线II代表 $\\mathrm{Fe}^{3+}$\nB: 常温下, $K_{\\text {sp }}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]=1.0 \\times 10^{-34.1}$\nC: $c$ 点铁铝元素主要存在形式为 $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}$\nD: $\\mathrm{Al}(\\mathrm{OH})_{3}(\\mathrm{~s})+\\mathrm{OH}^{-}(\\mathrm{aq}) \\rightleftharpoons\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}(\\mathrm{aq})$ 的平衡常数 $\\mathrm{K}$ 为 $10^{-1}$ 数量级\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $10.0 \\mathrm{~mL}$ 浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{AlCl}_{3}$ 和 $\\mathrm{FeCl}_{3}$ 混合溶液中加入 $\\mathrm{NaOH}$ 固体,溶液中金属元素有不同的存在形式, 它们的物质的量浓度与 $\\mathrm{NaOH}$ 物质的量关系如图所示，测得 $\\mathrm{a} 、 \\mathrm{~b}$ 两点溶液的 $\\mathrm{pH}$ 分别为 3.0, 4.3.\n\n[图1]\n\n已知: (1) $K_{\\mathrm{sp}}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]>K_{\\mathrm{sp}}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]$;\n\n(2) $\\mathrm{Al}^{3+}(\\mathrm{aq})+4 \\mathrm{OH}^{-}(\\mathrm{aq}) \\rightleftharpoons\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}(\\mathrm{aq}), 298 \\mathrm{~K}$,\n\n$K_{\\text {稳 }}=\\frac{c\\left\\{\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}\\right\\}}{c\\left(\\mathrm{Al}^{3+}\\right) \\times c^{4}\\left(\\mathrm{OH}^{-}\\right)}=1.1 \\times 10^{33}$ 。\n\n下列叙述正确的是\n\nA: 曲线II代表 $\\mathrm{Fe}^{3+}$\nB: 常温下, $K_{\\text {sp }}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]=1.0 \\times 10^{-34.1}$\nC: $c$ 点铁铝元素主要存在形式为 $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}$\nD: $\\mathrm{Al}(\\mathrm{OH})_{3}(\\mathrm{~s})+\\mathrm{OH}^{-}(\\mathrm{aq}) \\rightleftharpoons\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}(\\mathrm{aq})$ 的平衡常数 $\\mathrm{K}$ 为 $10^{-1}$ 数量级\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-052.jpg?height=571&width=717&top_left_y=842&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_351",
        "problem": "A sample of neon gas is held at $25.0^{\\circ} \\mathrm{C}$ and $1.0 \\mathrm{~atm}$ in a cylinder with a movable piston. Under these conditions the gas occupies $5.0 \\mathrm{~L}$. What volume does the gas occupy at $12.5^{\\circ} \\mathrm{C}$ and $1.0 \\mathrm{~atm}$ ?\nA: $2.5 \\mathrm{~L}$\nB: $4.8 \\mathrm{~L}$\nC: $5.2 \\mathrm{~L}$\nD: $10 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA sample of neon gas is held at $25.0^{\\circ} \\mathrm{C}$ and $1.0 \\mathrm{~atm}$ in a cylinder with a movable piston. Under these conditions the gas occupies $5.0 \\mathrm{~L}$. What volume does the gas occupy at $12.5^{\\circ} \\mathrm{C}$ and $1.0 \\mathrm{~atm}$ ?\n\nA: $2.5 \\mathrm{~L}$\nB: $4.8 \\mathrm{~L}$\nC: $5.2 \\mathrm{~L}$\nD: $10 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_675",
        "problem": "某温度下, $\\mathrm{AgCl}$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 的沉淀溶解平衡曲线如下图 (图中 $\\mathrm{A}^{m-}$ 表示 $\\mathrm{Cl}^{-}$或 $\\left.\\mathrm{CrO}_{4}^{2-}\\right)$ 所示, 下列说法正确的是\n\n[图1]\nA: $\\mathrm{Y}$ 是 $\\mathrm{AgCl}$ 的沉淀溶解平衡曲线\nB: $\\mathrm{a}$ 点时, $\\mathrm{AgCl}$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 均生成沉淀\nC: $\\mathrm{b}$ 点时, 有 $c\\left(\\mathrm{Cl}^{-}\\right)=c\\left(\\mathrm{CrO}_{4}^{2-}\\right)$, 则 $K_{\\mathrm{sp}}(\\mathrm{AgCl})=K_{\\mathrm{sp}}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nD: 当用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{AgNO}_{3}$ 溶液滴定 $\\mathrm{Cl}^{-}$浓度约为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液时, 可用 $\\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ 作指示剂 $\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right.$ 为红色沉淀）\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某温度下, $\\mathrm{AgCl}$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 的沉淀溶解平衡曲线如下图 (图中 $\\mathrm{A}^{m-}$ 表示 $\\mathrm{Cl}^{-}$或 $\\left.\\mathrm{CrO}_{4}^{2-}\\right)$ 所示, 下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{Y}$ 是 $\\mathrm{AgCl}$ 的沉淀溶解平衡曲线\nB: $\\mathrm{a}$ 点时, $\\mathrm{AgCl}$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 均生成沉淀\nC: $\\mathrm{b}$ 点时, 有 $c\\left(\\mathrm{Cl}^{-}\\right)=c\\left(\\mathrm{CrO}_{4}^{2-}\\right)$, 则 $K_{\\mathrm{sp}}(\\mathrm{AgCl})=K_{\\mathrm{sp}}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nD: 当用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{AgNO}_{3}$ 溶液滴定 $\\mathrm{Cl}^{-}$浓度约为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液时, 可用 $\\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ 作指示剂 $\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right.$ 为红色沉淀）\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-023.jpg?height=457&width=642&top_left_y=171&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1087",
        "problem": "The enthalpy of combustion of an isomer of octane is -5144 $\\mathrm{kJ} \\mathrm{mol}^{-1}$.To help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nFor E5 fuels, calculate the energy, in kJ, released when 1 litre of the fuel is burnt.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe enthalpy of combustion of an isomer of octane is -5144 $\\mathrm{kJ} \\mathrm{mol}^{-1}$.\n\nproblem:\nTo help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nFor E5 fuels, calculate the energy, in kJ, released when 1 litre of the fuel is burnt.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kJ, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-02.jpg?height=459&width=923&top_left_y=356&top_left_x=949"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kJ"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_930",
        "problem": "天津大学研究团队以 $\\mathrm{KOH}$ 溶液为电解质, $\\mathrm{CoP}$ 和 $\\mathrm{Ni}_{2} \\mathrm{P}$ 纳米片为催化电极材料,电催化合成偶氮化合物的装置如图所示( $\\mathrm{R}$ 代表烃基)下列说法正确的是\n\n[图1]\nA: 硝基苯分子中所有原子可能共面\nB: $\\mathrm{Ni}_{2} \\mathrm{P}$ 电极反应式为 $\\mathrm{RCH}_{2} \\mathrm{NH}_{2}+4 \\mathrm{e}^{-}+4 \\mathrm{OH}=\\mathrm{RCN}+4 \\mathrm{H}_{2} \\mathrm{O}$\nC: 合成 $1 \\mathrm{~mol}$ 偶氮化合物转移电子 $8 \\mathrm{~mol}$\nD: 离子交换膜是阳离子交换膜\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n天津大学研究团队以 $\\mathrm{KOH}$ 溶液为电解质, $\\mathrm{CoP}$ 和 $\\mathrm{Ni}_{2} \\mathrm{P}$ 纳米片为催化电极材料,电催化合成偶氮化合物的装置如图所示( $\\mathrm{R}$ 代表烃基)下列说法正确的是\n\n[图1]\n\nA: 硝基苯分子中所有原子可能共面\nB: $\\mathrm{Ni}_{2} \\mathrm{P}$ 电极反应式为 $\\mathrm{RCH}_{2} \\mathrm{NH}_{2}+4 \\mathrm{e}^{-}+4 \\mathrm{OH}=\\mathrm{RCN}+4 \\mathrm{H}_{2} \\mathrm{O}$\nC: 合成 $1 \\mathrm{~mol}$ 偶氮化合物转移电子 $8 \\mathrm{~mol}$\nD: 离子交换膜是阳离子交换膜\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-63.jpg?height=634&width=802&top_left_y=1870&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1105",
        "problem": "The mass of product (in grains) that would have been produced assuming a yield of $100 \\%$ is 74.58.[figure1]\n\nUnderstanding the proportions in which the elements combine was a crucial step in developing the atomic theory of matter. The picture above shows an experiment performed in the 1660s in which antimony was heated using the sun's rays to form an oxide.\n\nIn the experiment, 'ye Artist' reported that ' 12 grains of antimony increased to 15 grains of calx', (a grain is an old measure of mass). Given the crudeness of the experiment, this value is remarkably close to the theoretical yield of 14.4 'grains'.\n\nIn another experiment published in 1673, Robert Boyle measured the increase in mass when zinc metal is heated in air. He describes the experiment thus:\n\nWe took a Drachm of filings of Zink and kept it in a Cupelling-fire about three Hours. Then it look'd as if the filings had been calcin'd. This being weigh'd in the same scales gain'd full six grains.\n\nAssuming that Boyle's measurements are accurate, only a fraction, $\\alpha$, of the zinc must have been converted to the oxide. ( $\\alpha$ is a fraction between 0 and 1; 0 meaning none of the zinc reacted and 1 meaning all reacted).\n\nCalculate the value of $\\alpha$ and hence the masses of zinc oxide and unreacted zinc metal at the end of the experiment.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nThe mass of product (in grains) that would have been produced assuming a yield of $100 \\%$ is 74.58.\n\nproblem:\n[figure1]\n\nUnderstanding the proportions in which the elements combine was a crucial step in developing the atomic theory of matter. The picture above shows an experiment performed in the 1660s in which antimony was heated using the sun's rays to form an oxide.\n\nIn the experiment, 'ye Artist' reported that ' 12 grains of antimony increased to 15 grains of calx', (a grain is an old measure of mass). Given the crudeness of the experiment, this value is remarkably close to the theoretical yield of 14.4 'grains'.\n\nIn another experiment published in 1673, Robert Boyle measured the increase in mass when zinc metal is heated in air. He describes the experiment thus:\n\nWe took a Drachm of filings of Zink and kept it in a Cupelling-fire about three Hours. Then it look'd as if the filings had been calcin'd. This being weigh'd in the same scales gain'd full six grains.\n\nAssuming that Boyle's measurements are accurate, only a fraction, $\\alpha$, of the zinc must have been converted to the oxide. ( $\\alpha$ is a fraction between 0 and 1; 0 meaning none of the zinc reacted and 1 meaning all reacted).\n\nCalculate the value of $\\alpha$ and hence the masses of zinc oxide and unreacted zinc metal at the end of the experiment.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the value of $\\alpha$, the mass of unreacted zinc, the mass of zinc oxide].\nTheir units are, in order, [None, grains, grains], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-07.jpg?height=791&width=1039&top_left_y=333&top_left_x=514"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            "grains",
            "grains"
        ],
        "answer_sequence": [
            "the value of $\\alpha$",
            "the mass of unreacted zinc",
            "the mass of zinc oxide"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1542",
        "problem": "One of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\n${ }^{228} \\mathrm{Th}$ is a member of the thorium series. What volume in $\\mathrm{cm}^{3}$ of helium at $0{ }^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ collected when 1.00 gram of ${ }^{228} \\mathrm{Th}\\left(t_{1 / 2}=1.91\\right.$ years) is stored in a container for 20.0 years. The half-lives of all intermediate nuclides are short compared to the halflife of ${ }^{228} \\mathrm{Th}$.\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOne of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\n${ }^{228} \\mathrm{Th}$ is a member of the thorium series. What volume in $\\mathrm{cm}^{3}$ of helium at $0{ }^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ collected when 1.00 gram of ${ }^{228} \\mathrm{Th}\\left(t_{1 / 2}=1.91\\right.$ years) is stored in a container for 20.0 years. The half-lives of all intermediate nuclides are short compared to the halflife of ${ }^{228} \\mathrm{Th}$.\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~cm}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~cm}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_139",
        "problem": "Predict the conditions maximizing the rate of hydrogen gas production in the steam methane reforming reaction:\n\n$$\n\\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta \\mathrm{H}=+206 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\nA: Low temperature, high pressure and a catalyst\nB: Low temperature, low pressure and a catalyst\nC: Low temperature, high pressure and no catalyst\nD: High temperature, high pressure and a catalyst\nE: High temperature, low pressure and a catalyst\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPredict the conditions maximizing the rate of hydrogen gas production in the steam methane reforming reaction:\n\n$$\n\\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta \\mathrm{H}=+206 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\n\nA: Low temperature, high pressure and a catalyst\nB: Low temperature, low pressure and a catalyst\nC: Low temperature, high pressure and no catalyst\nD: High temperature, high pressure and a catalyst\nE: High temperature, low pressure and a catalyst\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1365",
        "problem": "The concentration of carbon dioxide in the atmosphere has increased substantially during this century and is predicted to continue to increase. The $\\left[\\mathrm{CO}_{2}\\right]$ is expected to be about $440 \\mathrm{ppm}\\left(440 \\times 10^{-6} \\mathrm{~atm}\\right)$ in the year 2020 .\n\nCalculate the concentration (in $\\mathrm{mol} \\mathrm{dm}^{-3}$ ) of $\\mathrm{CO}_{2}$ dissolved in distilled water equilibrated with the atmosphere in the year 2020.\n\nData:\n\nHenry's Law constant for $\\mathrm{CO}_{2}$ at $298 \\mathrm{~K}: 0.0343 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~atm}^{-1}$\n\nThermodynamic values, in $\\mathrm{kJ} / \\mathrm{mol}$ at $298 \\mathrm{~K}$ are:\n\n|  | $\\Delta_{f} G^{0}$ | $\\Delta_{f} H^{0}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CO}_{2}(\\mathrm{aq})$ | -386.2 | -412.9 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -237.2 | -285.8 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -587.1 | -691.2 |\n| $\\mathrm{H}^{+}(\\mathrm{aq})$ | 0.00 | 0.00 |",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe concentration of carbon dioxide in the atmosphere has increased substantially during this century and is predicted to continue to increase. The $\\left[\\mathrm{CO}_{2}\\right]$ is expected to be about $440 \\mathrm{ppm}\\left(440 \\times 10^{-6} \\mathrm{~atm}\\right)$ in the year 2020 .\n\nCalculate the concentration (in $\\mathrm{mol} \\mathrm{dm}^{-3}$ ) of $\\mathrm{CO}_{2}$ dissolved in distilled water equilibrated with the atmosphere in the year 2020.\n\nData:\n\nHenry's Law constant for $\\mathrm{CO}_{2}$ at $298 \\mathrm{~K}: 0.0343 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~atm}^{-1}$\n\nThermodynamic values, in $\\mathrm{kJ} / \\mathrm{mol}$ at $298 \\mathrm{~K}$ are:\n\n|  | $\\Delta_{f} G^{0}$ | $\\Delta_{f} H^{0}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CO}_{2}(\\mathrm{aq})$ | -386.2 | -412.9 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -237.2 | -285.8 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -587.1 | -691.2 |\n| $\\mathrm{H}^{+}(\\mathrm{aq})$ | 0.00 | 0.00 |\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of M, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "M"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_960",
        "problem": "常温下, 在 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中逐滴加入 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 盐酸 $40 \\mathrm{~mL}$, 溶液的 $\\mathrm{pH}$逐渐降低, 此时溶液中含碳元素的微粒物质的量浓度的百分含量（纵轴）也发生变化 $\\left(\\mathrm{CO}_{2}\\right.$ 因逸出未画出 $)$, 如图所示。下列说法不正确的是\n\n[图1]\nA: $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 在 A 点: $c\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: 常温下 $\\mathrm{CO}_{2}$ 饱和溶液的 $\\mathrm{pH}$ 约为 5.6\nD: 当加入 $20 \\mathrm{~mL}$ 盐酸时, 混合溶液的 $\\mathrm{pH}$ 约为 8\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 在 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中逐滴加入 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 盐酸 $40 \\mathrm{~mL}$, 溶液的 $\\mathrm{pH}$逐渐降低, 此时溶液中含碳元素的微粒物质的量浓度的百分含量（纵轴）也发生变化 $\\left(\\mathrm{CO}_{2}\\right.$ 因逸出未画出 $)$, 如图所示。下列说法不正确的是\n\n[图1]\n\nA: $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 在 A 点: $c\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: 常温下 $\\mathrm{CO}_{2}$ 饱和溶液的 $\\mathrm{pH}$ 约为 5.6\nD: 当加入 $20 \\mathrm{~mL}$ 盐酸时, 混合溶液的 $\\mathrm{pH}$ 约为 8\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-003.jpg?height=346&width=531&top_left_y=495&top_left_x=363"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1408",
        "problem": "Clathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nThe enthalpy of this process equals $17.47 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.At what external pressure does decomposition of methane hydrate into methane and ice take place at $-5{ }^{\\circ} \\mathrm{C}$ ?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nClathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nThe enthalpy of this process equals $17.47 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.\n\nproblem:\nAt what external pressure does decomposition of methane hydrate into methane and ice take place at $-5{ }^{\\circ} \\mathrm{C}$ ?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of MPa, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-222.jpg?height=425&width=434&top_left_y=1255&top_left_x=1459"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "MPa"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_321",
        "problem": "What is the pressure (in $\\mathrm{mmHg}$ ) of the gas inside the apparatus below if $\\mathrm{P}_{\\mathrm{atm}}=750 \\mathrm{mmHg}, \\Delta \\mathrm{h}_{1}=20 \\mathrm{~mm}$ and $\\Delta \\mathrm{h}_{2}=50 \\mathrm{~mm}$ ?\n[figure1]\nA: $20 \\mathrm{mmHg}$\nB: $50 \\mathrm{mmHg}$\nC: $700 \\mathrm{mmHg}$\nD: $730 \\mathrm{mmHg}$\nE: $\\quad 770 \\mathrm{mmHg}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the pressure (in $\\mathrm{mmHg}$ ) of the gas inside the apparatus below if $\\mathrm{P}_{\\mathrm{atm}}=750 \\mathrm{mmHg}, \\Delta \\mathrm{h}_{1}=20 \\mathrm{~mm}$ and $\\Delta \\mathrm{h}_{2}=50 \\mathrm{~mm}$ ?\n[figure1]\n\nA: $20 \\mathrm{mmHg}$\nB: $50 \\mathrm{mmHg}$\nC: $700 \\mathrm{mmHg}$\nD: $730 \\mathrm{mmHg}$\nE: $\\quad 770 \\mathrm{mmHg}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_92a3e606264c5ee1789ag-8.jpg?height=634&width=632&top_left_y=320&top_left_x=451"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_499",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 将 $1.0 \\mathrm{~L} \\mathrm{cmol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液与 $0.1 \\mathrm{molNaOH}$ 固体混合, 使之充分反应。然后向该混合物中加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 或 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 固体(忽略体积变化和温度变化), 加入的量与溶液 $\\mathrm{pH}$ 的变化如下图所示。下列叙述不正确的是\n\n[图1]\nA: 水的电离程度: $c>b>a$\nB: 该温度下, 醋酸的电离平衡常数 $\\mathrm{K}_{\\mathrm{a}}=\\frac{2 \\times 10^{-8}}{\\mathrm{c}-0.1}$\nC: $\\mathrm{a}$ 点对应的混合溶液中, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 当混合溶液呈中性时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 将 $1.0 \\mathrm{~L} \\mathrm{cmol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液与 $0.1 \\mathrm{molNaOH}$ 固体混合, 使之充分反应。然后向该混合物中加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 或 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 固体(忽略体积变化和温度变化), 加入的量与溶液 $\\mathrm{pH}$ 的变化如下图所示。下列叙述不正确的是\n\n[图1]\n\nA: 水的电离程度: $c>b>a$\nB: 该温度下, 醋酸的电离平衡常数 $\\mathrm{K}_{\\mathrm{a}}=\\frac{2 \\times 10^{-8}}{\\mathrm{c}-0.1}$\nC: $\\mathrm{a}$ 点对应的混合溶液中, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 当混合溶液呈中性时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-085.jpg?height=540&width=611&top_left_y=1346&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_797",
        "problem": "一定条件下, 有机物 $\\mathrm{M}$ 和 $\\mathrm{N}$ 反应可生成 $\\mathrm{P}$, 反应方程式如下:\n\n[图1]\n\n下列有关说法错误的是\nA: $1 \\mathrm{~mol}$ 有机物 $\\mathrm{N}$ 能与 $6 \\mathrm{~mol} \\mathrm{H}_{2}$ 发生加成反应\nB: 有机物 $\\mathrm{N}$ 中所有碳原子不可能共平面\nC: 有机物 $\\mathrm{M} 、 \\mathrm{~N} 、 \\mathrm{P}$ 均能发生加成反应、取代反应\nD: 有机物 $\\mathrm{M}$ 的同分异构体可能属于芳香烃\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一定条件下, 有机物 $\\mathrm{M}$ 和 $\\mathrm{N}$ 反应可生成 $\\mathrm{P}$, 反应方程式如下:\n\n[图1]\n\n下列有关说法错误的是\n\nA: $1 \\mathrm{~mol}$ 有机物 $\\mathrm{N}$ 能与 $6 \\mathrm{~mol} \\mathrm{H}_{2}$ 发生加成反应\nB: 有机物 $\\mathrm{N}$ 中所有碳原子不可能共平面\nC: 有机物 $\\mathrm{M} 、 \\mathrm{~N} 、 \\mathrm{P}$ 均能发生加成反应、取代反应\nD: 有机物 $\\mathrm{M}$ 的同分异构体可能属于芳香烃\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-27.jpg?height=234&width=1306&top_left_y=337&top_left_x=335"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_804",
        "problem": "有关化合物[图1]的叙述正确的是\nA: $\\mathrm{b}$ 的一氯代物有 4 种\nB: $b 、 d 、 p$ 的分子式均为 $\\mathrm{C}_{8} \\mathrm{H}_{10}$, 均属于碳水化合物\nC: b、d、p 分子中所有碳原子均不可能在同一平面上\nD: $1 \\mathrm{mold}$ 与 $2 \\mathrm{~mol} \\mathrm{Br}_{2}$ 发生加成反应的产物最多有 4 种(不考虑立体异构)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有关化合物[图1]的叙述正确的是\n\nA: $\\mathrm{b}$ 的一氯代物有 4 种\nB: $b 、 d 、 p$ 的分子式均为 $\\mathrm{C}_{8} \\mathrm{H}_{10}$, 均属于碳水化合物\nC: b、d、p 分子中所有碳原子均不可能在同一平面上\nD: $1 \\mathrm{mold}$ 与 $2 \\mathrm{~mol} \\mathrm{Br}_{2}$ 发生加成反应的产物最多有 4 种(不考虑立体异构)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/Znmf92Fm/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1154",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the misplacement of thallium\n\nMendeleev also misplaced the highly toxic element thallium ( $\\mathrm{Tl}$ ) with the elements from Group 1 rather than in Group 13. There are good reasons for this error. Thallium can form salts in the +3 oxidation state like other members of its group, but $\\mathrm{Tl}^{3+}$ ions are oxidising and the most stable oxidation state is +1 . For example, adding iodide ions to solutions of $\\mathrm{Tl}^{3+}$ actually gives a precipitate of thallium $(\\mathrm{I})$ iodide.\n\nAn iodide of thallium with the formula $\\mathrm{THI}_{3}$ is known and has exactly the same structure as $\\mathrm{CsI}_{3}$.\n\nWhat is the bond angle in the anion present in the compounds?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the misplacement of thallium\n\nMendeleev also misplaced the highly toxic element thallium ( $\\mathrm{Tl}$ ) with the elements from Group 1 rather than in Group 13. There are good reasons for this error. Thallium can form salts in the +3 oxidation state like other members of its group, but $\\mathrm{Tl}^{3+}$ ions are oxidising and the most stable oxidation state is +1 . For example, adding iodide ions to solutions of $\\mathrm{Tl}^{3+}$ actually gives a precipitate of thallium $(\\mathrm{I})$ iodide.\n\nAn iodide of thallium with the formula $\\mathrm{THI}_{3}$ is known and has exactly the same structure as $\\mathrm{CsI}_{3}$.\n\nWhat is the bond angle in the anion present in the compounds?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $^{\\circ}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$^{\\circ}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_45",
        "problem": "Cyclopropane isomerizes to propene in an irreversible, first-order reaction. At $700 \\mathrm{~K}$, a sample of $22.0 \\mathrm{~mm} \\mathrm{Hg}$ of cyclopropane is introduced into a reaction vessel. After $1.0 \\mathrm{~min}$, the partial pressure of the product, propene, was found to be $17.5 \\mathrm{~mm} \\mathrm{Hg}$. What is the rate constant for the isomerization at this temperature?\nA: $3.8 \\times 10^{-3} \\mathrm{~s}^{-1}$\nB: $2.6 \\times 10^{-2} \\mathrm{~s}^{-1}$\nC: $0.23 \\mathrm{~s}^{-1}$\nD: $1.6 \\mathrm{~s}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCyclopropane isomerizes to propene in an irreversible, first-order reaction. At $700 \\mathrm{~K}$, a sample of $22.0 \\mathrm{~mm} \\mathrm{Hg}$ of cyclopropane is introduced into a reaction vessel. After $1.0 \\mathrm{~min}$, the partial pressure of the product, propene, was found to be $17.5 \\mathrm{~mm} \\mathrm{Hg}$. What is the rate constant for the isomerization at this temperature?\n\nA: $3.8 \\times 10^{-3} \\mathrm{~s}^{-1}$\nB: $2.6 \\times 10^{-2} \\mathrm{~s}^{-1}$\nC: $0.23 \\mathrm{~s}^{-1}$\nD: $1.6 \\mathrm{~s}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_209",
        "problem": "Which of the following compounds contains both ionic and covalent bonds?\nA: $\\mathrm{CO}_{2}$\nB: $\\mathrm{NaCl}$\nC: $\\mathrm{Na}_{2} \\mathrm{O}$\nD: $\\mathrm{H}_{2} \\mathrm{CO}$\nE: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following compounds contains both ionic and covalent bonds?\n\nA: $\\mathrm{CO}_{2}$\nB: $\\mathrm{NaCl}$\nC: $\\mathrm{Na}_{2} \\mathrm{O}$\nD: $\\mathrm{H}_{2} \\mathrm{CO}$\nE: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1556",
        "problem": "One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.Calculate the smallest $\\mathrm{Cu}-\\mathrm{Cu} $ distance in the structure?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOne of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\n\nproblem:\nCalculate the smallest $\\mathrm{Cu}-\\mathrm{Cu} $ distance in the structure?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of pm, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-139.jpg?height=410&width=1236&top_left_y=660&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "pm"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1108",
        "problem": "$[S P]=2.36 \\times 10^{-13} \\mathrm{~mol} \\mathrm{dm}^{-3}$\n\n$[S P N P]=2.29 \\times 10^{-13} \\mathrm{~mol} \\mathrm{dm}^{-3}$A typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nDuring the test a person swabs their nose/throat and places it in the extraction solution. The resulting solution is called the test solution. A test sample of a few drops $\\left(0.10 \\mathrm{~cm}^{3}\\right)$ of the test solution are placed on the sample pad.\n\nIf the person taking the test has COVID-19, there will typically be $7.1 \\times 10^{6}$ virus particles per $\\mathrm{cm}^{3}$ in the test solution. Each virus particle has approximately 20 spike proteins on its surface.\n\nOnce the sample passes the conjugate pad, it becomes saturated with red-coloured antibody-coated Au nanoparticles (NP), to a concentration $1.6 \\times 10^{12} \\mathrm{NP}$ per $\\mathrm{cm}^{3}$.\n\nAn equilibrium is set up where the NP bind to any spike proteins, SP. Assume that any binding that takes place between spike proteins and nanoparticles is 1:1.\n\n$$\n\\begin{gathered}\n\\mathrm{SP}(\\mathrm{aq})+\\mathrm{NP}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{SPNP}(\\mathrm{aq}) \\\\\nK=\\frac{[\\mathrm{SPNP}]}{[\\mathrm{NP}][\\mathrm{SP}]}=1.2 \\times 10^{10} \\mathrm{~mol}^{-1} \\mathrm{dm}^{3}\n\\end{gathered}\n$$\n\nThe resulting mixture flows over the test strip which contains surface-immobilized antibodies (AB). When considering the amount of a substance on a surface, we use surface density, $\\sigma_{\\mathrm{A}}$, instead of concentration.\n\nThe test strip is $3.0 \\mathrm{~mm}$ long, $1.0 \\mathrm{~mm}$ wide and $0.10 \\mathrm{~mm}$ deep, with a surface density of antibodies, $\\sigma_{A B}=1.2 \\times 10^{9} \\mathrm{~mm}^{-2}$.\n\n[figure2]\n\nCalculate how many surface-immobilized antibodies, $\\mathrm{AB}$, there are on the test strip surface.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n$[S P]=2.36 \\times 10^{-13} \\mathrm{~mol} \\mathrm{dm}^{-3}$\n\n$[S P N P]=2.29 \\times 10^{-13} \\mathrm{~mol} \\mathrm{dm}^{-3}$\n\nproblem:\nA typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nDuring the test a person swabs their nose/throat and places it in the extraction solution. The resulting solution is called the test solution. A test sample of a few drops $\\left(0.10 \\mathrm{~cm}^{3}\\right)$ of the test solution are placed on the sample pad.\n\nIf the person taking the test has COVID-19, there will typically be $7.1 \\times 10^{6}$ virus particles per $\\mathrm{cm}^{3}$ in the test solution. Each virus particle has approximately 20 spike proteins on its surface.\n\nOnce the sample passes the conjugate pad, it becomes saturated with red-coloured antibody-coated Au nanoparticles (NP), to a concentration $1.6 \\times 10^{12} \\mathrm{NP}$ per $\\mathrm{cm}^{3}$.\n\nAn equilibrium is set up where the NP bind to any spike proteins, SP. Assume that any binding that takes place between spike proteins and nanoparticles is 1:1.\n\n$$\n\\begin{gathered}\n\\mathrm{SP}(\\mathrm{aq})+\\mathrm{NP}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{SPNP}(\\mathrm{aq}) \\\\\nK=\\frac{[\\mathrm{SPNP}]}{[\\mathrm{NP}][\\mathrm{SP}]}=1.2 \\times 10^{10} \\mathrm{~mol}^{-1} \\mathrm{dm}^{3}\n\\end{gathered}\n$$\n\nThe resulting mixture flows over the test strip which contains surface-immobilized antibodies (AB). When considering the amount of a substance on a surface, we use surface density, $\\sigma_{\\mathrm{A}}$, instead of concentration.\n\nThe test strip is $3.0 \\mathrm{~mm}$ long, $1.0 \\mathrm{~mm}$ wide and $0.10 \\mathrm{~mm}$ deep, with a surface density of antibodies, $\\sigma_{A B}=1.2 \\times 10^{9} \\mathrm{~mm}^{-2}$.\n\n[figure2]\n\nCalculate how many surface-immobilized antibodies, $\\mathrm{AB}$, there are on the test strip surface.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-08.jpg?height=146&width=940&top_left_y=495&top_left_x=929",
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-09.jpg?height=255&width=734&top_left_y=535&top_left_x=1112"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1007",
        "problem": "Radioactive ${ }^{131}$ I is used to treat thyroid cancer. An incomplete chemical equation for the radioactive decay of ${ }^{131} \\mathrm{I}$ is given below.\n\n$$\n{ }^{131} \\mathrm{I} \\rightarrow+{ }_{-1}^{0} \\mathrm{e}\n$$\n\nWhat is the missing product in the equation above?\nA: ${ }^{130}$ I\nB: ${ }^{129}$ |\nC: ${ }^{131} \\mathrm{Xe}$\nD: ${ }^{131} \\mathrm{Te}$\nE: $\\left.^{131}\\right|^{+}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRadioactive ${ }^{131}$ I is used to treat thyroid cancer. An incomplete chemical equation for the radioactive decay of ${ }^{131} \\mathrm{I}$ is given below.\n\n$$\n{ }^{131} \\mathrm{I} \\rightarrow+{ }_{-1}^{0} \\mathrm{e}\n$$\n\nWhat is the missing product in the equation above?\n\nA: ${ }^{130}$ I\nB: ${ }^{129}$ |\nC: ${ }^{131} \\mathrm{Xe}$\nD: ${ }^{131} \\mathrm{Te}$\nE: $\\left.^{131}\\right|^{+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_728",
        "problem": "下列叙述正确的是\nA: 甘氨酸 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COOH}\\right)$ 与强酸或强碱均能反应\nB: 聚丙烯纤维比聚乙烯醇有更好的吸水性\nC: 灼烧纯羊毛和合成纤维产生相同的气味\nD: 酸性强弱: 苯磺酸 $\\left.\\mathrm{SO}_{3} \\mathrm{H}\\right)$ 苯甲酸 $\\mathrm{COOH}_{) \\text {苯酚 }}$[图1]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列叙述正确的是\n\nA: 甘氨酸 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COOH}\\right)$ 与强酸或强碱均能反应\nB: 聚丙烯纤维比聚乙烯醇有更好的吸水性\nC: 灼烧纯羊毛和合成纤维产生相同的气味\nD: 酸性强弱: 苯磺酸 $\\left.\\mathrm{SO}_{3} \\mathrm{H}\\right)$ 苯甲酸 $\\mathrm{COOH}_{) \\text {苯酚 }}$[图1]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-36.jpg?height=149&width=268&top_left_y=1896&top_left_x=400"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_272",
        "problem": "Which of the following pairs of compounds will form a precipitate when $0.1 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solutions of each are mixed?\nA: $\\mathrm{AgNO}_{3}$ and $\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}$\nB: $\\mathrm{K}_{2} \\mathrm{SO}_{4}$ and $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$\nC: $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and $\\mathrm{KBr}$\nD: $\\mathrm{NaOH}$ and $\\mathrm{CuCl}_{2}$\nE: $\\mathrm{CuCl}_{2}$ and $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following pairs of compounds will form a precipitate when $0.1 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solutions of each are mixed?\n\nA: $\\mathrm{AgNO}_{3}$ and $\\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}$\nB: $\\mathrm{K}_{2} \\mathrm{SO}_{4}$ and $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$\nC: $\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}$ and $\\mathrm{KBr}$\nD: $\\mathrm{NaOH}$ and $\\mathrm{CuCl}_{2}$\nE: $\\mathrm{CuCl}_{2}$ and $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_285",
        "problem": "A substance does not conduct electricity well when solid but does when liquid. Which of the following could this substance be? Select all that apply.\nA: neon\nB: mercury\nC: nitrogen monoxide\nD: aluminium chloride\nE: calcium fluoride\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nA substance does not conduct electricity well when solid but does when liquid. Which of the following could this substance be? Select all that apply.\n\nA: neon\nB: mercury\nC: nitrogen monoxide\nD: aluminium chloride\nE: calcium fluoride\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_317",
        "problem": "What is the molecular geometry of phosphorus pentachloride, $\\mathrm{PCl}_{5}$ ?\nA: square pyramidal\nB: trigonal bipyramidal\nC: pentagonal\nD: trigonal pyramidal\nE: octahedral\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the molecular geometry of phosphorus pentachloride, $\\mathrm{PCl}_{5}$ ?\n\nA: square pyramidal\nB: trigonal bipyramidal\nC: pentagonal\nD: trigonal pyramidal\nE: octahedral\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1145",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## Mendeleev's short-form table\n\nIn the second version of his table, Mendeleev updated the mass of indium and correctly positioned both it and thallium. He also arranged the Groups of similar elements vertically but there are too many elements within each Group; Mendeleev mixed elements from the transition metals with those from the main block of the periodic table. There is sense to this since the elements which Mendeleev places in a given Group form similar compounds - notably their oxides and their hydrides.\n\n[figure1]\n\nMendeleev's second periodic table from 1871\n\nSolid nitrogen $(\\mathrm{V})$ oxide has a different structure from the gaseous form and consists of linear cations, and trigonal-planar anions. Suggest the formula of the cation in solid nitrogen(V) oxide.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## Mendeleev's short-form table\n\nIn the second version of his table, Mendeleev updated the mass of indium and correctly positioned both it and thallium. He also arranged the Groups of similar elements vertically but there are too many elements within each Group; Mendeleev mixed elements from the transition metals with those from the main block of the periodic table. There is sense to this since the elements which Mendeleev places in a given Group form similar compounds - notably their oxides and their hydrides.\n\n[figure1]\n\nMendeleev's second periodic table from 1871\n\nSolid nitrogen $(\\mathrm{V})$ oxide has a different structure from the gaseous form and consists of linear cations, and trigonal-planar anions. Suggest the formula of the cation in solid nitrogen(V) oxide.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_bf14c3e936c3f7031d02g-05.jpg?height=742&width=1567&top_left_y=183&top_left_x=290"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_910",
        "problem": "阿司匹林(乙酰水杨酸)具有解热镇痛的作用, 是家中常备药, 其合成原理如图所示\n\n[图1]\n下列说法错误的是\nA: 阿司匹林的分子式为 $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}$ ，该药物属于 OTC 药物可以长期服用\nB: 水杨酸可以发生酯化反应、加成反应\nC: 可以用碳酸氢钠溶液区别水杨酸和阿司匹林\nD: 阿司匹林与布洛芬(退烧)药物不能同时服用\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n阿司匹林(乙酰水杨酸)具有解热镇痛的作用, 是家中常备药, 其合成原理如图所示\n\n[图1]\n下列说法错误的是\n\nA: 阿司匹林的分子式为 $\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{O}$ ，该药物属于 OTC 药物可以长期服用\nB: 水杨酸可以发生酯化反应、加成反应\nC: 可以用碳酸氢钠溶液区别水杨酸和阿司匹林\nD: 阿司匹林与布洛芬(退烧)药物不能同时服用\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/x82MFqvL/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_368",
        "problem": "Addition of small amounts of which solids to $4 \\mathrm{M} \\mathrm{HCl}$ will result in gas evolution?\nI. $\\mathrm{Zn}$\nII. $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAddition of small amounts of which solids to $4 \\mathrm{M} \\mathrm{HCl}$ will result in gas evolution?\nI. $\\mathrm{Zn}$\nII. $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1167",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nOne of the most important commercial uses of $\\mathrm{SkCl}_{4}$ is the preparation of organo-stuck-at-homium compounds which are heavily used in the plastics industry. These may be prepared using so-called Grignard reagents, $\\mathrm{RMgCl}$, where $\\mathrm{R}$ is some alkyl group. Different products may be formed depending on the proportions of the Grignard reagent used. A (partial) generalised equation where 1 mol of $\\mathrm{SkCl}_{4}$ reacts with $x$ moles of $\\mathrm{RMgCl}$ may be written:\n\n$$\n\\mathrm{SkCl}_{4}+x \\mathrm{RMgCl} \\longrightarrow \\longrightarrow+x \\mathrm{MgCl}_{2}\n$$\n\nBy balancing the above equation, give the generalised formula (using $x$ ) for the organostuck-at-homium compounds that can be formed.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nOne of the most important commercial uses of $\\mathrm{SkCl}_{4}$ is the preparation of organo-stuck-at-homium compounds which are heavily used in the plastics industry. These may be prepared using so-called Grignard reagents, $\\mathrm{RMgCl}$, where $\\mathrm{R}$ is some alkyl group. Different products may be formed depending on the proportions of the Grignard reagent used. A (partial) generalised equation where 1 mol of $\\mathrm{SkCl}_{4}$ reacts with $x$ moles of $\\mathrm{RMgCl}$ may be written:\n\n$$\n\\mathrm{SkCl}_{4}+x \\mathrm{RMgCl} \\longrightarrow \\longrightarrow+x \\mathrm{MgCl}_{2}\n$$\n\nBy balancing the above equation, give the generalised formula (using $x$ ) for the organostuck-at-homium compounds that can be formed.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_882",
        "problem": "设计如图装置回收金属钴。保持细菌所在环境 $\\mathrm{pH}$ 稳定, 借助其降解乙酸盐生成 $\\mathrm{CO}_{2}$, 将废旧锂离子电池的正极材料 $\\mathrm{LiCoO}_{2}(\\mathrm{~s})$ 转化为 $\\mathrm{Co}^{2+}$, 工作时保持厌氧环境, 并定时将乙室溶液转移至甲室。已知电极材料均为石墨材质, 右侧装置为原电池。下列说法不正确的是\n\n[图1]\nA: 装置工作时, 甲室溶液 $\\mathrm{pH}$ 最终升高\nB: 装置工作一段时间后，乙室应补充盐酸\nC: 乙室电极反应式为 $\\mathrm{LiCoO}_{2}+4 \\mathrm{H}^{+}+\\mathrm{e}^{-}=\\mathrm{Li}^{+}+\\mathrm{Co}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 若甲室 $\\mathrm{Co}^{2+}$ 减少 $200 \\mathrm{mg}$, 乙室 $\\mathrm{Co}^{2+}$ 增加 $200 \\mathrm{mg}$, 则此时已进行过溶液转移\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n设计如图装置回收金属钴。保持细菌所在环境 $\\mathrm{pH}$ 稳定, 借助其降解乙酸盐生成 $\\mathrm{CO}_{2}$, 将废旧锂离子电池的正极材料 $\\mathrm{LiCoO}_{2}(\\mathrm{~s})$ 转化为 $\\mathrm{Co}^{2+}$, 工作时保持厌氧环境, 并定时将乙室溶液转移至甲室。已知电极材料均为石墨材质, 右侧装置为原电池。下列说法不正确的是\n\n[图1]\n\nA: 装置工作时, 甲室溶液 $\\mathrm{pH}$ 最终升高\nB: 装置工作一段时间后，乙室应补充盐酸\nC: 乙室电极反应式为 $\\mathrm{LiCoO}_{2}+4 \\mathrm{H}^{+}+\\mathrm{e}^{-}=\\mathrm{Li}^{+}+\\mathrm{Co}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 若甲室 $\\mathrm{Co}^{2+}$ 减少 $200 \\mathrm{mg}$, 乙室 $\\mathrm{Co}^{2+}$ 增加 $200 \\mathrm{mg}$, 则此时已进行过溶液转移\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-05.jpg?height=425&width=691&top_left_y=1892&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_330",
        "problem": "How many structural isomers are there for $\\mathrm{C}_{4} \\mathrm{H}_{8}$ ?\nA: one\nB: two\nC: three\nD: four\nE: more than four\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many structural isomers are there for $\\mathrm{C}_{4} \\mathrm{H}_{8}$ ?\n\nA: one\nB: two\nC: three\nD: four\nE: more than four\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1345",
        "problem": "The $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\nModel B attempts to analyse the probability that the dust particles carry 0,1 or $2 \\mathrm{H}$ atoms. The three states are linked by the following reaction scheme. The assumption is made that no more than 2 atoms may be adsorbed simultaneously.\n\n[figure1]\n\n$x_{0}, x_{1}$ and $x_{2}$ are the fractions of dust particles existing in state 0,1 or 2 , respectively. These fractions may be treated in the same way as concentrations in the following kinetic analysis. For a system in state $m$ with fraction $x_{m}$, the rates of the three possible processes are\n\nAdsorption $(m \\rightarrow m+1):$ rate $=k_{\\mathrm{a}}[\\mathrm{H}] x_{m}$\n\nDesorption $(m \\rightarrow m-1):$ rate $=k_{\\mathrm{d}} m x_{m}$\n\nReaction $(m \\rightarrow m-2)$ : rate $=1 / 2 k_{\\mathrm{r}} m(m-1) x_{m}$Evaluate the rate of production of $\\mathrm{H}_{2}$ per dust particle in this model.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\nModel B attempts to analyse the probability that the dust particles carry 0,1 or $2 \\mathrm{H}$ atoms. The three states are linked by the following reaction scheme. The assumption is made that no more than 2 atoms may be adsorbed simultaneously.\n\n[figure1]\n\n$x_{0}, x_{1}$ and $x_{2}$ are the fractions of dust particles existing in state 0,1 or 2 , respectively. These fractions may be treated in the same way as concentrations in the following kinetic analysis. For a system in state $m$ with fraction $x_{m}$, the rates of the three possible processes are\n\nAdsorption $(m \\rightarrow m+1):$ rate $=k_{\\mathrm{a}}[\\mathrm{H}] x_{m}$\n\nDesorption $(m \\rightarrow m-1):$ rate $=k_{\\mathrm{d}} m x_{m}$\n\nReaction $(m \\rightarrow m-2)$ : rate $=1 / 2 k_{\\mathrm{r}} m(m-1) x_{m}$\n\nproblem:\nEvaluate the rate of production of $\\mathrm{H}_{2}$ per dust particle in this model.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-015.jpg?height=263&width=357&top_left_y=1205&top_left_x=861"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_80",
        "problem": "A reaction has $K_{\\text {eq }}=0.020$ at $300 \\mathrm{~K}$, and its $K_{\\text {eq }}$ value increases with increasing temperature. What can be inferred about the values of $\\Delta H^{\\circ}{ }_{\\text {rxn }}$ and $\\Delta S^{\\circ}{ }_{\\text {rxn }}$, assuming that they are independent of temperature?\nI. $\\Delta H^{\\circ}{ }_{\\text {rxn }}<0$\n\nII. $\\quad \\Delta S_{\\text {rxn }}^{\\circ}<0$\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA reaction has $K_{\\text {eq }}=0.020$ at $300 \\mathrm{~K}$, and its $K_{\\text {eq }}$ value increases with increasing temperature. What can be inferred about the values of $\\Delta H^{\\circ}{ }_{\\text {rxn }}$ and $\\Delta S^{\\circ}{ }_{\\text {rxn }}$, assuming that they are independent of temperature?\nI. $\\Delta H^{\\circ}{ }_{\\text {rxn }}<0$\n\nII. $\\quad \\Delta S_{\\text {rxn }}^{\\circ}<0$\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_229",
        "problem": "A. $16.51 \\mathrm{~L}$\nB. $\\quad 19.73 \\mathrm{~L}$\nC. $25.52 \\mathrm{~L}$\nD. $\\quad 33.01 \\mathrm{~L}$\nE. $\\quad 39.45 \\mathrm{~L}$\nA: $\\mathrm{NH}_{3}$\nB: $\\mathrm{NF}_{3}$\nC: $\\mathrm{NCl}_{3}$\nD: $\\mathrm{NBr}_{3}$\nE: $\\mathrm{NI}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA. $16.51 \\mathrm{~L}$\nB. $\\quad 19.73 \\mathrm{~L}$\nC. $25.52 \\mathrm{~L}$\nD. $\\quad 33.01 \\mathrm{~L}$\nE. $\\quad 39.45 \\mathrm{~L}$\n\nA: $\\mathrm{NH}_{3}$\nB: $\\mathrm{NF}_{3}$\nC: $\\mathrm{NCl}_{3}$\nD: $\\mathrm{NBr}_{3}$\nE: $\\mathrm{NI}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_470",
        "problem": "盐酸羟胺 $\\left(\\mathrm{NH}_{3} \\mathrm{OHCl}\\right)$ 是一种常见的还原剂和显像剂, 其化学性质与 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 类似。工业上主要采用电化学法制备, 装置如图 1 所示, 含 $\\mathrm{Fe}$ 的催化电极反应机理如图 2 所示,不考虑溶液体积的变化。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 电池工作时, $\\mathrm{Pt}$ 电极为正极\nB: 图 2 中, $\\mathrm{M}$ 为 $\\mathrm{H}^{+}, \\mathrm{N}$ 为 $\\mathrm{NH}_{3} \\mathrm{OH}^{+}$\nC: 电池工作时, 每消耗 2.24LNO(标准状况)左室溶液质量增加 $3.3 \\mathrm{~g}$\nD: 电池工作一段时间后, 正极区溶液的 $\\mathrm{pH}$ 减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n盐酸羟胺 $\\left(\\mathrm{NH}_{3} \\mathrm{OHCl}\\right)$ 是一种常见的还原剂和显像剂, 其化学性质与 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 类似。工业上主要采用电化学法制备, 装置如图 1 所示, 含 $\\mathrm{Fe}$ 的催化电极反应机理如图 2 所示,不考虑溶液体积的变化。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 电池工作时, $\\mathrm{Pt}$ 电极为正极\nB: 图 2 中, $\\mathrm{M}$ 为 $\\mathrm{H}^{+}, \\mathrm{N}$ 为 $\\mathrm{NH}_{3} \\mathrm{OH}^{+}$\nC: 电池工作时, 每消耗 2.24LNO(标准状况)左室溶液质量增加 $3.3 \\mathrm{~g}$\nD: 电池工作一段时间后, 正极区溶液的 $\\mathrm{pH}$ 减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-06.jpg?height=389&width=545&top_left_y=2124&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-06.jpg?height=459&width=537&top_left_y=2055&top_left_x=1019"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_251",
        "problem": "First ionisation energy is defined as the energy required to remove one mole of electrons from one mole of gaseous ions. Which of the following lists elements in order of increasing first ionisation energy?\nA: C, F, N, Li\nB: C, N, Li, F\nC: Li, C, N, F\nD: Li, N, F, C\nE: F, N, C, Li\nF: F, Li, N, C\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFirst ionisation energy is defined as the energy required to remove one mole of electrons from one mole of gaseous ions. Which of the following lists elements in order of increasing first ionisation energy?\n\nA: C, F, N, Li\nB: C, N, Li, F\nC: Li, C, N, F\nD: Li, N, F, C\nE: F, N, C, Li\nF: F, Li, N, C\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_927",
        "problem": "向 $\\mathrm{AgCl}$ 饱和溶液(有足量 $\\mathrm{AgCl}$ 固体)中滴加氨水, 发生反应 $\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$和 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}, \\lg \\left[\\mathrm{c}(\\mathrm{M}) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 与 $\\lg \\left[\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 的关系如图所示(其中 $\\mathrm{M}$ 代表 $\\mathrm{Ag}^{+} 、 \\mathrm{Cl} 、\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$或 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$)。下列说法错误的是\n\n[图1]\nA: 曲线 IV 可视为 $\\mathrm{AgCl}$ 溶解度随 $\\mathrm{NH}_{3}$ 浓度变化曲线\nB: $\\mathrm{AgCl}$ 的溶度积常数的数量级为 $10^{-9}$\nC: 反应 $\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$的平衡常数 $\\mathrm{K}$ 的值为 $10^{3.24}$\nD: 向 $\\mathrm{AgCl}$ 饱和溶液 (有足量 $\\mathrm{AgCl}$ 固体) 中滴加氨水后, 溶液中存在 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n向 $\\mathrm{AgCl}$ 饱和溶液(有足量 $\\mathrm{AgCl}$ 固体)中滴加氨水, 发生反应 $\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$和 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}, \\lg \\left[\\mathrm{c}(\\mathrm{M}) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 与 $\\lg \\left[\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 的关系如图所示(其中 $\\mathrm{M}$ 代表 $\\mathrm{Ag}^{+} 、 \\mathrm{Cl} 、\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$或 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$)。下列说法错误的是\n\n[图1]\n\nA: 曲线 IV 可视为 $\\mathrm{AgCl}$ 溶解度随 $\\mathrm{NH}_{3}$ 浓度变化曲线\nB: $\\mathrm{AgCl}$ 的溶度积常数的数量级为 $10^{-9}$\nC: 反应 $\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$的平衡常数 $\\mathrm{K}$ 的值为 $10^{3.24}$\nD: 向 $\\mathrm{AgCl}$ 饱和溶液 (有足量 $\\mathrm{AgCl}$ 固体) 中滴加氨水后, 溶液中存在 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-075.jpg?height=577&width=739&top_left_y=1436&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_883",
        "problem": "氨是农业上“氮的固定”的必需原料，随着世界人口的增长, 氨的需求量在不断增大。科研人员新发现以磷盐作质子 $\\left(\\mathrm{H}^{+}\\right)$传导体, 以四氢呋喃(无色易挥发的液体)为电解剂,利用电化学法将氮气还原成氨的原理如图所示。下列说法错误的是\n\n[图1]\nA: $\\mathrm{M}$ 电极为阳极, 电极反应式为 $\\mathrm{H}_{2}-2 \\mathrm{e}^{-}=2 \\mathrm{H}^{+}$\nB: (I) $\\rightarrow$ (II)的变化中, 磷原子的成键数目发生变化\nC: 图示中最后一步反应为 $3 \\mathrm{Li}+\\mathrm{N}_{2}+3 \\mathrm{H}^{+}=3 \\mathrm{Li}^{+}+2 \\mathrm{NH}_{3}$\nD: 该方法制备氨气所需的温度低于传统工业合成氨的温度\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n氨是农业上“氮的固定”的必需原料，随着世界人口的增长, 氨的需求量在不断增大。科研人员新发现以磷盐作质子 $\\left(\\mathrm{H}^{+}\\right)$传导体, 以四氢呋喃(无色易挥发的液体)为电解剂,利用电化学法将氮气还原成氨的原理如图所示。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{M}$ 电极为阳极, 电极反应式为 $\\mathrm{H}_{2}-2 \\mathrm{e}^{-}=2 \\mathrm{H}^{+}$\nB: (I) $\\rightarrow$ (II)的变化中, 磷原子的成键数目发生变化\nC: 图示中最后一步反应为 $3 \\mathrm{Li}+\\mathrm{N}_{2}+3 \\mathrm{H}^{+}=3 \\mathrm{Li}^{+}+2 \\mathrm{NH}_{3}$\nD: 该方法制备氨气所需的温度低于传统工业合成氨的温度\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-47.jpg?height=720&width=1108&top_left_y=148&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1070",
        "problem": "2022 was the $90^{\\text {th }}$ anniversary of Linus Pauling proposing the concept of electronegativity. His book The nature of the chemical bond is considered the most influential chemistry book of $20^{\\text {th }}$ century, and he was awarded the 1954 Nobel prize in chemistry for his work.\n\nElectronegativity, $\\chi$, is a measure of the ability of an atom to attract a pair of electrons in a covalent bond.\n\n[figure1]\n\nPauling used thermodynamic data to calculate the difference in electronegativity between two atoms A and B. All electronegativity values are positive with no units, and atom A has a higher electronegativity than atom $B$.\n\n$$\n\\chi_{A}-\\chi_{B}=0.102 \\sqrt{\\mathrm{B}_{d}(A B)-\\frac{\\mathrm{B}_{d}(A A)+\\mathrm{B}_{d}(B B)}{2}}\n$$\n\n$\\mathrm{B}_{d}$ represents the bond dissociation energies, in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, of the $\\mathrm{A}-\\mathrm{A}, \\mathrm{B}-\\mathrm{B}$ and $\\mathrm{A}-\\mathrm{B}$ bonds.\n\nCalculate the difference in electronegativity between $\\mathrm{Cl}$ and $\\mathrm{H}$, given that $\\mathrm{B}_{d}\\left(\\mathrm{H}_{2}\\right)=432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}, \\mathrm{~B}_{d}\\left(\\mathrm{Cl}_{2}\\right)=244 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and $\\mathrm{B}_{d}(\\mathrm{HCl})=427 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n2022 was the $90^{\\text {th }}$ anniversary of Linus Pauling proposing the concept of electronegativity. His book The nature of the chemical bond is considered the most influential chemistry book of $20^{\\text {th }}$ century, and he was awarded the 1954 Nobel prize in chemistry for his work.\n\nElectronegativity, $\\chi$, is a measure of the ability of an atom to attract a pair of electrons in a covalent bond.\n\n[figure1]\n\nPauling used thermodynamic data to calculate the difference in electronegativity between two atoms A and B. All electronegativity values are positive with no units, and atom A has a higher electronegativity than atom $B$.\n\n$$\n\\chi_{A}-\\chi_{B}=0.102 \\sqrt{\\mathrm{B}_{d}(A B)-\\frac{\\mathrm{B}_{d}(A A)+\\mathrm{B}_{d}(B B)}{2}}\n$$\n\n$\\mathrm{B}_{d}$ represents the bond dissociation energies, in $\\mathrm{kJ} \\mathrm{mol}^{-1}$, of the $\\mathrm{A}-\\mathrm{A}, \\mathrm{B}-\\mathrm{B}$ and $\\mathrm{A}-\\mathrm{B}$ bonds.\n\nCalculate the difference in electronegativity between $\\mathrm{Cl}$ and $\\mathrm{H}$, given that $\\mathrm{B}_{d}\\left(\\mathrm{H}_{2}\\right)=432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}, \\mathrm{~B}_{d}\\left(\\mathrm{Cl}_{2}\\right)=244 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and $\\mathrm{B}_{d}(\\mathrm{HCl})=427 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-03.jpg?height=611&width=505&top_left_y=314&top_left_x=1301"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_437",
        "problem": "一定条件下, 通过 $\\mathrm{H}_{2}$ 的还原作用, $\\mathrm{CO}_{2}$ 可转化为甲醚 $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$, 反应如下:\n\n反应 $\\mathrm{I}: \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\n反应II: $\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})$\n\n反应III: $2 \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\n将 $\\mathrm{H}_{2}$ 与 $\\mathrm{CO}_{2}$ 按照 3:1 的体积比通入 $\\mathrm{p}=4 \\mathrm{Mpa}$ 的恒压容器中, 反应过程中 $\\mathrm{CO}_{2}$ 的平衡转化率 $(\\mathrm{X})$ 与含碳物质的平衡收率 $(\\mathrm{Y})$ 随温度变化如图所示。 $520 \\mathrm{~K}$ 达到平衡时,\n\n$n\\left(\\mathrm{H}_{2}\\right): n\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)=7.7: 1$ 。下列说法正确的是\n\n已知: I. 选择性 $(\\alpha)=\\frac{\\text { 且标产物中碳原子的物质的量 }}{\\text { 消耗 } \\mathrm{CO}_{2} \\text { 的物质的量 }} \\times 100 \\%$\n\nII. (收率) $(\\mathrm{Y})=\\frac{\\text { 且标产物中碳原子的物质的量 }}{\\mathrm{CO}_{2} \\text { 总的物质量 }} \\times 100 \\%$\n\n[图1]\nA: $520 \\mathrm{~K}$ 时，甲醚的选择性约为 0.58\nB: 反应I的压强平衡常数 $K_{\\mathrm{p}}=8.5 \\times 10^{3}$\nC: 甲醚的选择性在 $520 \\mathrm{~K}$ 到 $535 \\mathrm{~K}$ 范围内随温度的升高逐渐减小\nD: 使用高效的催化剂可以提高转化率\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一定条件下, 通过 $\\mathrm{H}_{2}$ 的还原作用, $\\mathrm{CO}_{2}$ 可转化为甲醚 $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$, 反应如下:\n\n反应 $\\mathrm{I}: \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\n反应II: $\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})$\n\n反应III: $2 \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\n将 $\\mathrm{H}_{2}$ 与 $\\mathrm{CO}_{2}$ 按照 3:1 的体积比通入 $\\mathrm{p}=4 \\mathrm{Mpa}$ 的恒压容器中, 反应过程中 $\\mathrm{CO}_{2}$ 的平衡转化率 $(\\mathrm{X})$ 与含碳物质的平衡收率 $(\\mathrm{Y})$ 随温度变化如图所示。 $520 \\mathrm{~K}$ 达到平衡时,\n\n$n\\left(\\mathrm{H}_{2}\\right): n\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)=7.7: 1$ 。下列说法正确的是\n\n已知: I. 选择性 $(\\alpha)=\\frac{\\text { 且标产物中碳原子的物质的量 }}{\\text { 消耗 } \\mathrm{CO}_{2} \\text { 的物质的量 }} \\times 100 \\%$\n\nII. (收率) $(\\mathrm{Y})=\\frac{\\text { 且标产物中碳原子的物质的量 }}{\\mathrm{CO}_{2} \\text { 总的物质量 }} \\times 100 \\%$\n\n[图1]\n\nA: $520 \\mathrm{~K}$ 时，甲醚的选择性约为 0.58\nB: 反应I的压强平衡常数 $K_{\\mathrm{p}}=8.5 \\times 10^{3}$\nC: 甲醚的选择性在 $520 \\mathrm{~K}$ 到 $535 \\mathrm{~K}$ 范围内随温度的升高逐渐减小\nD: 使用高效的催化剂可以提高转化率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-033.jpg?height=663&width=851&top_left_y=1119&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1051",
        "problem": "[figure1]\n\nUnderstanding the proportions in which the elements combine was a crucial step in developing the atomic theory of matter. The picture above shows an experiment performed in the 1660s in which antimony was heated using the sun's rays to form an oxide.\n\nIn the experiment, 'ye Artist' reported that ' 12 grains of antimony increased to 15 grains of calx', (a grain is an old measure of mass). Given the crudeness of the experiment, this value is remarkably close to the theoretical yield of 14.4 'grains'.\n\nIn another experiment published in 1673, Robert Boyle measured the increase in mass when zinc metal is heated in air. He describes the experiment thus:\n\nWe took a Drachm of filings of Zink and kept it in a Cupelling-fire about three Hours. Then it look'd as if the filings had been calcin'd. This being weigh'd in the same scales gain'd full six grains.\n\nGiven that there are 60 grains in a Drachm, calculate the mass of product (in grains) that would have been produced assuming a yield of $100 \\%$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n[figure1]\n\nUnderstanding the proportions in which the elements combine was a crucial step in developing the atomic theory of matter. The picture above shows an experiment performed in the 1660s in which antimony was heated using the sun's rays to form an oxide.\n\nIn the experiment, 'ye Artist' reported that ' 12 grains of antimony increased to 15 grains of calx', (a grain is an old measure of mass). Given the crudeness of the experiment, this value is remarkably close to the theoretical yield of 14.4 'grains'.\n\nIn another experiment published in 1673, Robert Boyle measured the increase in mass when zinc metal is heated in air. He describes the experiment thus:\n\nWe took a Drachm of filings of Zink and kept it in a Cupelling-fire about three Hours. Then it look'd as if the filings had been calcin'd. This being weigh'd in the same scales gain'd full six grains.\n\nGiven that there are 60 grains in a Drachm, calculate the mass of product (in grains) that would have been produced assuming a yield of $100 \\%$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of grains, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-07.jpg?height=791&width=1039&top_left_y=333&top_left_x=514"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "grains"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_549",
        "problem": "下列溶液中微粒的关系正确的是\nA: $\\mathrm{pH}=2$ 的 $\\mathrm{HA}$ 溶液与 $\\mathrm{pH}=12$ 的 $\\mathrm{MOH}$ 溶液任意比混合: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+$ $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$\nB: 氨水和盐酸混合后溶液呈酸性, 溶液中可能存在: $\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nC: 在一定温度下, 相同 $\\mathrm{pH}$ 的硫酸和硫酸铁溶液中水电离出来的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$分别是 $1.0 \\times 10$ ${ }^{-\\mathrm{a}} \\mathrm{mol} / \\mathrm{L}$ 和是 $1.0 \\times 10^{-\\mathrm{b}} \\mathrm{mol} / \\mathrm{L}$, 在此温度时, 水的离子积为 $1.0 \\times 10^{-(\\mathrm{a}+\\mathrm{b})}$\nD: 常温下, $0.1 \\mathrm{~mol} / \\mathrm{L}$ 酸 $\\mathrm{HA} \\mathrm{pH}=3,0.1 \\mathrm{~mol} / \\mathrm{L}$ 碱 $\\mathrm{BOH} \\mathrm{pH}=13$, 则盐 $\\mathrm{BA}$ 溶液的 $\\mathrm{pH}$ $<7$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列溶液中微粒的关系正确的是\n\nA: $\\mathrm{pH}=2$ 的 $\\mathrm{HA}$ 溶液与 $\\mathrm{pH}=12$ 的 $\\mathrm{MOH}$ 溶液任意比混合: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+$ $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$\nB: 氨水和盐酸混合后溶液呈酸性, 溶液中可能存在: $\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nC: 在一定温度下, 相同 $\\mathrm{pH}$ 的硫酸和硫酸铁溶液中水电离出来的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$分别是 $1.0 \\times 10$ ${ }^{-\\mathrm{a}} \\mathrm{mol} / \\mathrm{L}$ 和是 $1.0 \\times 10^{-\\mathrm{b}} \\mathrm{mol} / \\mathrm{L}$, 在此温度时, 水的离子积为 $1.0 \\times 10^{-(\\mathrm{a}+\\mathrm{b})}$\nD: 常温下, $0.1 \\mathrm{~mol} / \\mathrm{L}$ 酸 $\\mathrm{HA} \\mathrm{pH}=3,0.1 \\mathrm{~mol} / \\mathrm{L}$ 碱 $\\mathrm{BOH} \\mathrm{pH}=13$, 则盐 $\\mathrm{BA}$ 溶液的 $\\mathrm{pH}$ $<7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_528",
        "problem": "在温度、容积相同的 3 个密闭容器中, 按不同方式投入反应物, 保持恒温、恒容,测得反应达到平衡时的有关数据如下 $\\left(\\right.$ 已知 $\\left.\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-92.4 \\mathrm{~kJ} / \\mathrm{mol}\\right)$\n\n| 容器 | 甲 | 乙 | 丙 |\n| :--- | :--- | :--- | :--- |\n| 反应物投入量 | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ | $2 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $4 \\mathrm{~mol} \\mathrm{NH}_{3}$ |\n| $\\mathrm{NH}_{3}$ 浓度 $(\\mathrm{mol} / \\mathrm{L})$ | $\\mathrm{c}_{1}$ | $\\mathrm{c}_{2}$ | $\\mathrm{c}_{3}$ |\n| 反应的能量变化 | 放出 $\\mathrm{a} \\mathrm{kJ}$ | 吸收 $\\mathrm{b} \\mathrm{kJ}$ | 吸收 $\\mathrm{c} \\mathrm{kJ}$ |\n| 体系压强 | $\\mathrm{p}_{1}$ | $\\mathrm{p}_{2}$ | $\\mathrm{p}_{3}$ |\n| 反应物转化率 | $\\mathrm{a}_{1}$ | $\\mathrm{a}_{2}$ | $\\mathrm{a}_{3}$ |\n\n下列说法正确的是\nA: $2 \\mathrm{c}_{1}>\\mathrm{c}_{3}$\nB: $a+b=92.4$\nC: $2 \\mathrm{p}_{1}<\\mathrm{p}_{3}$\nD: $\\mathrm{a}_{2}+\\mathrm{a}_{3}>1$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在温度、容积相同的 3 个密闭容器中, 按不同方式投入反应物, 保持恒温、恒容,测得反应达到平衡时的有关数据如下 $\\left(\\right.$ 已知 $\\left.\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-92.4 \\mathrm{~kJ} / \\mathrm{mol}\\right)$\n\n| 容器 | 甲 | 乙 | 丙 |\n| :--- | :--- | :--- | :--- |\n| 反应物投入量 | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ | $2 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $4 \\mathrm{~mol} \\mathrm{NH}_{3}$ |\n| $\\mathrm{NH}_{3}$ 浓度 $(\\mathrm{mol} / \\mathrm{L})$ | $\\mathrm{c}_{1}$ | $\\mathrm{c}_{2}$ | $\\mathrm{c}_{3}$ |\n| 反应的能量变化 | 放出 $\\mathrm{a} \\mathrm{kJ}$ | 吸收 $\\mathrm{b} \\mathrm{kJ}$ | 吸收 $\\mathrm{c} \\mathrm{kJ}$ |\n| 体系压强 | $\\mathrm{p}_{1}$ | $\\mathrm{p}_{2}$ | $\\mathrm{p}_{3}$ |\n| 反应物转化率 | $\\mathrm{a}_{1}$ | $\\mathrm{a}_{2}$ | $\\mathrm{a}_{3}$ |\n\n下列说法正确的是\n\nA: $2 \\mathrm{c}_{1}>\\mathrm{c}_{3}$\nB: $a+b=92.4$\nC: $2 \\mathrm{p}_{1}<\\mathrm{p}_{3}$\nD: $\\mathrm{a}_{2}+\\mathrm{a}_{3}>1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_636",
        "problem": "有一化合物 X,其水溶液为浅绿色, 有如图所示的转化关系(部分反应物、生成物已略)。其中 $\\mathrm{B} 、 \\mathrm{D} 、 \\mathrm{E} 、 \\mathrm{~F}$ 均为无色气体, $\\mathrm{M} 、 \\mathrm{~L}$ 为常见的金属单质, $\\mathrm{C}$ 为难溶于水的红褐色固体。在混合液中加入 $\\mathrm{BaCl}_{2}$ 溶液可生成不溶于稀盐酸的白色沉淀, $\\mathrm{H}$ 和 $\\mathrm{M}$ 反应可放出大量的热。(电解装置中用石墨做电极)下列说法正确的是( )\n\n[图1]\nA: 生成 $\\mathrm{G}$ 的反应中, 生成 $1 \\mathrm{molG}$ 转移 $1 \\mathrm{~mol}$ 电子\nB: 此转化关系中只有一个置换反应\nC: “混合液”电解一段时间后 $\\mathrm{pH}$ 不变\nD: 检验 $\\mathrm{X}$ 中是否存在 $\\mathrm{Cl}^{-}$, 所用试剂为 $\\mathrm{HNO}_{3}$ 酸化的 $\\mathrm{AgNO}_{3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n有一化合物 X,其水溶液为浅绿色, 有如图所示的转化关系(部分反应物、生成物已略)。其中 $\\mathrm{B} 、 \\mathrm{D} 、 \\mathrm{E} 、 \\mathrm{~F}$ 均为无色气体, $\\mathrm{M} 、 \\mathrm{~L}$ 为常见的金属单质, $\\mathrm{C}$ 为难溶于水的红褐色固体。在混合液中加入 $\\mathrm{BaCl}_{2}$ 溶液可生成不溶于稀盐酸的白色沉淀, $\\mathrm{H}$ 和 $\\mathrm{M}$ 反应可放出大量的热。(电解装置中用石墨做电极)下列说法正确的是( )\n\n[图1]\n\nA: 生成 $\\mathrm{G}$ 的反应中, 生成 $1 \\mathrm{molG}$ 转移 $1 \\mathrm{~mol}$ 电子\nB: 此转化关系中只有一个置换反应\nC: “混合液”电解一段时间后 $\\mathrm{pH}$ 不变\nD: 检验 $\\mathrm{X}$ 中是否存在 $\\mathrm{Cl}^{-}$, 所用试剂为 $\\mathrm{HNO}_{3}$ 酸化的 $\\mathrm{AgNO}_{3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-18.jpg?height=431&width=920&top_left_y=156&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1463",
        "problem": "One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.Which of the atoms (A or B) is copper?\nA: A\nB: B\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nOne of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\n\nproblem:\nWhich of the atoms (A or B) is copper?\n\nA: A\nB: B\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-139.jpg?height=410&width=1236&top_left_y=660&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_966",
        "problem": "碳酸是二元弱酸, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液 $\\mathrm{pH}$ 约为 8.3 , 向 $5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$的 $\\mathrm{NaHCO}_{3}$ 溶液中逐滴滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{BaCl}_{2}$ 溶液。[已知: 常温下碳酸的电离平衡常数 $\\mathrm{K}_{\\mathrm{a} 1}=4.3 \\times 10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}=5.6 \\times 10^{-11}, \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{BaCO}_{3}\\right)=2.6 \\times 10^{-9}$, 溶液混合后体积变化忽略不计]下列有关说法不正确的是\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中, $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right) 、 \\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$比值为: $5.6 \\times 10^{-2.7}$\nB: $\\mathrm{NaHCO}_{3}$ 溶液中存在自耦电离: $\\mathrm{HCO}_{3}^{-}+\\mathrm{HCO}_{3}^{-} \\rightleftharpoons \\mathrm{CO}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{CO}_{3}$, 其平衡常数约为 $1.3 \\times 10^{-4}$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 溶液中离子浓度大小为: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: 上述实验过程中可能发生反应: $\\mathrm{Ba}^{2+}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{BaCO}_{3} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2} \\uparrow$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n碳酸是二元弱酸, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液 $\\mathrm{pH}$ 约为 8.3 , 向 $5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$的 $\\mathrm{NaHCO}_{3}$ 溶液中逐滴滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{BaCl}_{2}$ 溶液。[已知: 常温下碳酸的电离平衡常数 $\\mathrm{K}_{\\mathrm{a} 1}=4.3 \\times 10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}=5.6 \\times 10^{-11}, \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{BaCO}_{3}\\right)=2.6 \\times 10^{-9}$, 溶液混合后体积变化忽略不计]下列有关说法不正确的是\n\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中, $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right) 、 \\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$比值为: $5.6 \\times 10^{-2.7}$\nB: $\\mathrm{NaHCO}_{3}$ 溶液中存在自耦电离: $\\mathrm{HCO}_{3}^{-}+\\mathrm{HCO}_{3}^{-} \\rightleftharpoons \\mathrm{CO}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{CO}_{3}$, 其平衡常数约为 $1.3 \\times 10^{-4}$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 溶液中离子浓度大小为: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: 上述实验过程中可能发生反应: $\\mathrm{Ba}^{2+}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{BaCO}_{3} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2} \\uparrow$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_42",
        "problem": "For which reaction is $\\Delta S^{0}<0$ ?\nA: $\\mathrm{NH}_{4} \\mathrm{Br}(\\mathrm{s}) \\rightarrow \\mathrm{NH}_{3}(g)+\\mathrm{HBr}(g)$\nB: $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l) \\rightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(g)$\nC: $2 \\mathrm{NO}_{2}(g) \\rightarrow 2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g)$\nD: $2 \\mathrm{BrCl}(g) \\rightarrow \\mathrm{Cl}_{2}(g)+\\mathrm{Br}_{2}(l)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor which reaction is $\\Delta S^{0}<0$ ?\n\nA: $\\mathrm{NH}_{4} \\mathrm{Br}(\\mathrm{s}) \\rightarrow \\mathrm{NH}_{3}(g)+\\mathrm{HBr}(g)$\nB: $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(l) \\rightarrow \\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}(g)$\nC: $2 \\mathrm{NO}_{2}(g) \\rightarrow 2 \\mathrm{NO}(g)+\\mathrm{O}_{2}(g)$\nD: $2 \\mathrm{BrCl}(g) \\rightarrow \\mathrm{Cl}_{2}(g)+\\mathrm{Br}_{2}(l)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_96",
        "problem": "How many unpaired electrons are there in a ground-state NF molecule in the gas phase?\nA: 0\nB: 1\nC: 2\nD: 4\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many unpaired electrons are there in a ground-state NF molecule in the gas phase?\n\nA: 0\nB: 1\nC: 2\nD: 4\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_955",
        "problem": "常温下, 一元碱 $\\mathrm{BOH}$ 的 $\\mathrm{K}_{\\mathrm{b}}(\\mathrm{BOH})=1.0 \\times 10^{-5}$ 。在某体系中, $\\mathrm{B}^{+}$与 $\\mathrm{OH}^{-}$离子不能穿过隔膜, 未电离的 $\\mathrm{BOH}$ 可自由穿过该膜(如图所示)。设溶液中 $\\mathrm{c}_{\\text {总 }}(\\mathrm{BOH})=\\mathrm{c}(\\mathrm{BOH})+\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)$,当达到平衡时，下列叙述正确的是\n\n[图1]\nA: 溶液 II 中 $\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 溶液 $\\mathrm{I}$ 中 $\\mathrm{BOH}$ 的电离度 $\\left[\\frac{\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)}{\\mathrm{c}_{\\text {总(BOH) }}}\\right]$ 为 $\\frac{1}{1001}$\nC: 溶液 I 和 II 中的 $\\mathrm{c}(\\mathrm{BOH})$ 相等\nD: 溶液 I 和 II 中的 $\\mathrm{c}_{\\text {总 }}(\\mathrm{BOH})$ 之比为 $10^{-4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 一元碱 $\\mathrm{BOH}$ 的 $\\mathrm{K}_{\\mathrm{b}}(\\mathrm{BOH})=1.0 \\times 10^{-5}$ 。在某体系中, $\\mathrm{B}^{+}$与 $\\mathrm{OH}^{-}$离子不能穿过隔膜, 未电离的 $\\mathrm{BOH}$ 可自由穿过该膜(如图所示)。设溶液中 $\\mathrm{c}_{\\text {总 }}(\\mathrm{BOH})=\\mathrm{c}(\\mathrm{BOH})+\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)$,当达到平衡时，下列叙述正确的是\n\n[图1]\n\nA: 溶液 II 中 $\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 溶液 $\\mathrm{I}$ 中 $\\mathrm{BOH}$ 的电离度 $\\left[\\frac{\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)}{\\mathrm{c}_{\\text {总(BOH) }}}\\right]$ 为 $\\frac{1}{1001}$\nC: 溶液 I 和 II 中的 $\\mathrm{c}(\\mathrm{BOH})$ 相等\nD: 溶液 I 和 II 中的 $\\mathrm{c}_{\\text {总 }}(\\mathrm{BOH})$ 之比为 $10^{-4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-061.jpg?height=305&width=725&top_left_y=1355&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_443",
        "problem": "常温下, 向 $100 \\mathrm{~mL} 0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KHB}$ 溶液中分别加入浓度均为 $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KOH}$溶液和盐酸, 混合溶液的 $\\mathrm{pH}$ 随所加溶液体积的变化如图所示。下列说法错误的是\n\n[图1]\nA: 常温下, $\\mathrm{H}_{2} \\mathrm{~B}$ 的 $\\mathrm{K}_{\\mathrm{a} 1}=1.0 \\times 10^{-4}, \\mathrm{~K}_{\\mathrm{a} 2}=1.0 \\times 10^{-6}$\nB: KHB 溶液中, $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HB}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{B}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $N$ 点: $c\\left(H^{+}\\right)+c\\left(H B^{-}\\right)+2 c\\left(H_{2} B\\right)=c\\left(O H^{-}\\right)$; P 点： $c\\left(K^{+}\\right)=c\\left(B^{2 \\cdot}\\right)+c\\left(H^{-}\\right)+c\\left(H_{2} B\\right)$\nD: 水的电离程度 $\\mathrm{N}>\\mathrm{M}>\\mathrm{P}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $100 \\mathrm{~mL} 0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KHB}$ 溶液中分别加入浓度均为 $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KOH}$溶液和盐酸, 混合溶液的 $\\mathrm{pH}$ 随所加溶液体积的变化如图所示。下列说法错误的是\n\n[图1]\n\nA: 常温下, $\\mathrm{H}_{2} \\mathrm{~B}$ 的 $\\mathrm{K}_{\\mathrm{a} 1}=1.0 \\times 10^{-4}, \\mathrm{~K}_{\\mathrm{a} 2}=1.0 \\times 10^{-6}$\nB: KHB 溶液中, $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HB}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{B}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $N$ 点: $c\\left(H^{+}\\right)+c\\left(H B^{-}\\right)+2 c\\left(H_{2} B\\right)=c\\left(O H^{-}\\right)$; P 点： $c\\left(K^{+}\\right)=c\\left(B^{2 \\cdot}\\right)+c\\left(H^{-}\\right)+c\\left(H_{2} B\\right)$\nD: 水的电离程度 $\\mathrm{N}>\\mathrm{M}>\\mathrm{P}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-089.jpg?height=554&width=628&top_left_y=1636&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_461",
        "problem": "人们发现生产人造羊毛和聚氯乙烯(PVC)可以用煤作为原料, 合成路线如下。下列说法正确的是\n\n[图1]\nA: PVC 塑料制品可以直接用于盛放食物\nB: A 生成 C、D 的反应类型分别为加成反应、取代反应\nC: 实验室制备 $\\mathrm{A}$ 的过程中, 常用 $\\mathrm{CuSO}_{4}$ 溶液净化气体\nD: 与 D 互为同分异构体且可发生碱性水解的物质有 4 种(不包括环状化合物和 D 本身)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n人们发现生产人造羊毛和聚氯乙烯(PVC)可以用煤作为原料, 合成路线如下。下列说法正确的是\n\n[图1]\n\nA: PVC 塑料制品可以直接用于盛放食物\nB: A 生成 C、D 的反应类型分别为加成反应、取代反应\nC: 实验室制备 $\\mathrm{A}$ 的过程中, 常用 $\\mathrm{CuSO}_{4}$ 溶液净化气体\nD: 与 D 互为同分异构体且可发生碱性水解的物质有 4 种(不包括环状化合物和 D 本身)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-51.jpg?height=343&width=1225&top_left_y=494&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-51.jpg?height=129&width=583&top_left_y=2334&top_left_x=1002"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1350",
        "problem": "The chemical oxygen demand (COD) refers to the amount of oxidizable substance, such as organic compounds, in a sample solution, and it is used as an indication of water quality in seas, lakes, and marshes. For example, the COD of service water is kept below $1 \\mathrm{mg} \\mathrm{dm}^{-3}$. The COD $\\left(\\mathrm{mg} \\mathrm{dm}^{-3}\\right)$ is represented by mass of $\\mathrm{O}_{2}(\\mathrm{mg})$ which accepts the same amount of electrons which would be accepted by the strong oxidizing agent when $1 \\mathrm{dm}^{3}$ of a sample solution is treated with it. An example of the operation is presented below.\n\nA sample solution with a volume of $1.00 \\mathrm{dm}^{3}$ was acidified with a sufficient amount of sulphuric acid, and chloride ions were removed by the addition of silver nitrate solution. A volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of potassium permanganate solution $\\left(c=5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) was added to the sample solution, and the mixture was heated for $30 \\mathrm{~min}$. Further, a volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of disodium oxalate $\\left(\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ or $\\left.\\mathrm{NaOOC}-\\mathrm{COONa}\\right)$ standard solution ( $c=1$. $25 \\cdot 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) was added, and the mixture was stirred well. Oxalate ions that remained unreacted were titrated with potassium permanganate solution $(c=$ $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ). A volume of $3.00 \\cdot 10^{-2} \\mathrm{dm}^{3}$ of the solution was used for the titration.Calculate the mass of $\\mathrm{O}_{2}$ (in $\\mathrm{mg}$ ) that will oxidize the same number of moles of oxidizable substance as $1.00 \\cdot 10^{-3} \\mathrm{dm}^{3}$ of potassium permanganate solution with a concentration of $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{~dm}^{-3}$ does.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe chemical oxygen demand (COD) refers to the amount of oxidizable substance, such as organic compounds, in a sample solution, and it is used as an indication of water quality in seas, lakes, and marshes. For example, the COD of service water is kept below $1 \\mathrm{mg} \\mathrm{dm}^{-3}$. The COD $\\left(\\mathrm{mg} \\mathrm{dm}^{-3}\\right)$ is represented by mass of $\\mathrm{O}_{2}(\\mathrm{mg})$ which accepts the same amount of electrons which would be accepted by the strong oxidizing agent when $1 \\mathrm{dm}^{3}$ of a sample solution is treated with it. An example of the operation is presented below.\n\nA sample solution with a volume of $1.00 \\mathrm{dm}^{3}$ was acidified with a sufficient amount of sulphuric acid, and chloride ions were removed by the addition of silver nitrate solution. A volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of potassium permanganate solution $\\left(c=5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) was added to the sample solution, and the mixture was heated for $30 \\mathrm{~min}$. Further, a volume of $1.00 \\cdot 10^{-1} \\mathrm{dm}^{3}$ of disodium oxalate $\\left(\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ or $\\left.\\mathrm{NaOOC}-\\mathrm{COONa}\\right)$ standard solution ( $c=1$. $25 \\cdot 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) was added, and the mixture was stirred well. Oxalate ions that remained unreacted were titrated with potassium permanganate solution $(c=$ $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{dm}^{-3}$ ). A volume of $3.00 \\cdot 10^{-2} \\mathrm{dm}^{3}$ of the solution was used for the titration.\n\nproblem:\nCalculate the mass of $\\mathrm{O}_{2}$ (in $\\mathrm{mg}$ ) that will oxidize the same number of moles of oxidizable substance as $1.00 \\cdot 10^{-3} \\mathrm{dm}^{3}$ of potassium permanganate solution with a concentration of $5.00 \\cdot 10^{-3} \\mathrm{~mol} \\mathrm{~dm}^{-3}$ does.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{mg}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{mg}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1059",
        "problem": "A typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nIf $1.0 \\mathrm{~cm}^{3}$ of $0.10 \\mathrm{~mol} \\mathrm{dm}^{-3} \\mathrm{HCl}$ were then added to the extraction solution, what $\\mathrm{pH}$ would the resulting solution be?\nA: Very Acidic\nB: Neutral\nC: pH 7.4\nD: Very Alkaline\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nIf $1.0 \\mathrm{~cm}^{3}$ of $0.10 \\mathrm{~mol} \\mathrm{dm}^{-3} \\mathrm{HCl}$ were then added to the extraction solution, what $\\mathrm{pH}$ would the resulting solution be?\n\nA: Very Acidic\nB: Neutral\nC: pH 7.4\nD: Very Alkaline\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-08.jpg?height=146&width=940&top_left_y=495&top_left_x=929"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_643",
        "problem": "水溶液锌电池 (图 1)的电极材料是研究热点之一, 一种在晶体 $\\mathrm{MnO}$ 空位中嵌入 $\\mathrm{Zn}^{2+}$的电极材料充放电过程如图 2 所示(除中心空位外, 其他空位未画出)。下列叙述中正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n○ O $\\mathrm{Mn}$ 或 $\\mathrm{Zn} \\square$ 空位\n\n图2\nA: 该材料在锌电池中作为负极材料\nB: (1)为 $\\mathrm{MnO}$ 活化过程, 其中 $\\mathrm{Mn}$ 的价态一定发生变化\nC: (2)代表电池放电过程\nD: (3) $1 \\mathrm{~mol}$ 晶胞转移的电子数为 $0.2 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n水溶液锌电池 (图 1)的电极材料是研究热点之一, 一种在晶体 $\\mathrm{MnO}$ 空位中嵌入 $\\mathrm{Zn}^{2+}$的电极材料充放电过程如图 2 所示(除中心空位外, 其他空位未画出)。下列叙述中正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n○ O $\\mathrm{Mn}$ 或 $\\mathrm{Zn} \\square$ 空位\n\n图2\n\nA: 该材料在锌电池中作为负极材料\nB: (1)为 $\\mathrm{MnO}$ 活化过程, 其中 $\\mathrm{Mn}$ 的价态一定发生变化\nC: (2)代表电池放电过程\nD: (3) $1 \\mathrm{~mol}$ 晶胞转移的电子数为 $0.2 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-15.jpg?height=388&width=440&top_left_y=434&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-15.jpg?height=320&width=971&top_left_y=494&top_left_x=814",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-15.jpg?height=97&width=1382&top_left_y=1705&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_147",
        "problem": "What is the $\\mathrm{pH}$ of a solution created by mixing $1000.0 \\mathrm{~mL}$ of $0.120 \\mathrm{M}$ $\\mathrm{HNO}_{3}(\\mathrm{aq})$ with $250.0 \\mathrm{~mL}$ of $0.750 \\mathrm{M} \\mathrm{HBr}(\\mathrm{aq})$ ?\nA: 1.268\nB: 0.000\nC: 0.060\nD: 0.512\nE: 0.609\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the $\\mathrm{pH}$ of a solution created by mixing $1000.0 \\mathrm{~mL}$ of $0.120 \\mathrm{M}$ $\\mathrm{HNO}_{3}(\\mathrm{aq})$ with $250.0 \\mathrm{~mL}$ of $0.750 \\mathrm{M} \\mathrm{HBr}(\\mathrm{aq})$ ?\n\nA: 1.268\nB: 0.000\nC: 0.060\nD: 0.512\nE: 0.609\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1413",
        "problem": "The label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\nCould it be possible that the solution contained acetic acid?\n\nAcetic acid: $\\quad \\mathrm{p} K_{\\mathrm{a}}=4.76$\n\n$\\square$ True $\\square$ False\n\nIf True, calculate the $\\mathrm{pH}$ (or at least try to estimate it) and show your work.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\nCould it be possible that the solution contained acetic acid?\n\nAcetic acid: $\\quad \\mathrm{p} K_{\\mathrm{a}}=4.76$\n\n$\\square$ True $\\square$ False\n\nIf True, calculate the $\\mathrm{pH}$ (or at least try to estimate it) and show your work.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_840",
        "problem": "室温下, 某二元碱 $\\mathrm{X}(\\mathrm{OH})_{2}$ 水溶液中相关组分的物质的量分数随溶液 $\\mathrm{pH}$ 变化的曲线如图所示, 下列说法错误的是\n\n[图1]\nA: 由图可知 $\\mathrm{X}(\\mathrm{OH})_{2}$ 一级、二级电离平衡常数\nB: $\\mathrm{X}(\\mathrm{OH}) \\mathrm{NO}_{3}$ 水溶液显碱性\nC: 等物质的量的 $\\mathrm{X}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 和 $\\mathrm{X}(\\mathrm{OH}) \\mathrm{NO}_{3}$ 混合溶液中 $c\\left(\\mathrm{X}^{2+}\\right)<c\\left[\\mathrm{X}(\\mathrm{OH})^{+}\\right]$\nD: 在 $\\mathrm{pH}=6$ 的水溶液中, $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{X}^{2+}\\right)+c\\left(\\mathrm{H}^{+}\\right)+c\\left[\\mathrm{X}(\\mathrm{OH})^{+}\\right]$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 某二元碱 $\\mathrm{X}(\\mathrm{OH})_{2}$ 水溶液中相关组分的物质的量分数随溶液 $\\mathrm{pH}$ 变化的曲线如图所示, 下列说法错误的是\n\n[图1]\n\nA: 由图可知 $\\mathrm{X}(\\mathrm{OH})_{2}$ 一级、二级电离平衡常数\nB: $\\mathrm{X}(\\mathrm{OH}) \\mathrm{NO}_{3}$ 水溶液显碱性\nC: 等物质的量的 $\\mathrm{X}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 和 $\\mathrm{X}(\\mathrm{OH}) \\mathrm{NO}_{3}$ 混合溶液中 $c\\left(\\mathrm{X}^{2+}\\right)<c\\left[\\mathrm{X}(\\mathrm{OH})^{+}\\right]$\nD: 在 $\\mathrm{pH}=6$ 的水溶液中, $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{X}^{2+}\\right)+c\\left(\\mathrm{H}^{+}\\right)+c\\left[\\mathrm{X}(\\mathrm{OH})^{+}\\right]$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-018.jpg?height=420&width=571&top_left_y=1620&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1257",
        "problem": "When the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\nThe concentration of chromate at the endpoint is $0.020 \\mathrm{M}$. Calculate the concentration of chloride ions in the solution when the red precipitate forms.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nWhen the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\nThe concentration of chromate at the endpoint is $0.020 \\mathrm{M}$. Calculate the concentration of chloride ions in the solution when the red precipitate forms.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_51",
        "problem": "Thallium(I) bromide $(M=284.3)$ adopts the cubic unit cell shown below, with an edge length $a=397 \\mathrm{pm}$. What is the density of $\\mathrm{TlBr}$ ?\n\n[figure1]\nA: $5.42 \\mathrm{~g} \\mathrm{~cm}^{-3}$\nB: $7.55 \\mathrm{~g} \\mathrm{~cm}^{-3}$\nC: $18.0 \\mathrm{~g} \\mathrm{~cm}^{-3}$\nD: $22.4 \\mathrm{~g} \\mathrm{~cm}^{-3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThallium(I) bromide $(M=284.3)$ adopts the cubic unit cell shown below, with an edge length $a=397 \\mathrm{pm}$. What is the density of $\\mathrm{TlBr}$ ?\n\n[figure1]\n\nA: $5.42 \\mathrm{~g} \\mathrm{~cm}^{-3}$\nB: $7.55 \\mathrm{~g} \\mathrm{~cm}^{-3}$\nC: $18.0 \\mathrm{~g} \\mathrm{~cm}^{-3}$\nD: $22.4 \\mathrm{~g} \\mathrm{~cm}^{-3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-05.jpg?height=404&width=629&top_left_y=337&top_left_x=301"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_619",
        "problem": "$\\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 是一种钠离子电池正极材料, 充放电过程中正极材料立方晶胞(示意图)的组成变化如图所示, 晶胞内未标出因放电产生的 0 价 $\\mathrm{Cu}$ 原子。下列说法正确的是\n\n[图1]\n\n$\\mathrm{Cu}_{2-x} \\mathrm{Se}$\n\n[图2]\n\nO Se ${ }^{2-}$\n\n[图3]\n\n$\\mathrm{Na}_{y} \\mathrm{Cu}_{2-x} \\mathrm{Se}$\n\n[图4]\n\n$\\mathrm{Na}^{+} 、 \\mathrm{Cu}^{+} 、 \\mathrm{Cu}^{2+}$ 可能占据的位置\n\n[图5]\n\n$\\mathrm{NaCuSe}$\n\n[图6]\n\n$\\mathrm{Na}_{2} \\mathrm{Se}$\n\n$\\mathrm{Na}^{+}$\nA: 每个 $\\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 晶胞中 $\\mathrm{ar}^{2+}$ 个数为 $\\mathrm{x}$\nB: 每个 $\\mathrm{Na}_{2} \\mathrm{Se}$ 晶胞完全转化为 $\\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 晶胞, 转移电子数为 8\nC: 每个 $\\mathrm{NaCuSe}$ 晶胞中 0 价 $\\mathrm{Cu}$ 原子个数为 $1-\\mathrm{x}$\nD: 当 $\\mathrm{Na}_{\\mathrm{y}} \\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 转化为 $\\mathrm{NaCuSe}$ 时, 每转移 (1-y)mol 电子, 产生 (1-x)molCu 原子\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 是一种钠离子电池正极材料, 充放电过程中正极材料立方晶胞(示意图)的组成变化如图所示, 晶胞内未标出因放电产生的 0 价 $\\mathrm{Cu}$ 原子。下列说法正确的是\n\n[图1]\n\n$\\mathrm{Cu}_{2-x} \\mathrm{Se}$\n\n[图2]\n\nO Se ${ }^{2-}$\n\n[图3]\n\n$\\mathrm{Na}_{y} \\mathrm{Cu}_{2-x} \\mathrm{Se}$\n\n[图4]\n\n$\\mathrm{Na}^{+} 、 \\mathrm{Cu}^{+} 、 \\mathrm{Cu}^{2+}$ 可能占据的位置\n\n[图5]\n\n$\\mathrm{NaCuSe}$\n\n[图6]\n\n$\\mathrm{Na}_{2} \\mathrm{Se}$\n\n$\\mathrm{Na}^{+}$\n\nA: 每个 $\\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 晶胞中 $\\mathrm{ar}^{2+}$ 个数为 $\\mathrm{x}$\nB: 每个 $\\mathrm{Na}_{2} \\mathrm{Se}$ 晶胞完全转化为 $\\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 晶胞, 转移电子数为 8\nC: 每个 $\\mathrm{NaCuSe}$ 晶胞中 0 价 $\\mathrm{Cu}$ 原子个数为 $1-\\mathrm{x}$\nD: 当 $\\mathrm{Na}_{\\mathrm{y}} \\mathrm{Cu}_{2-\\mathrm{x}} \\mathrm{Se}$ 转化为 $\\mathrm{NaCuSe}$ 时, 每转移 (1-y)mol 电子, 产生 (1-x)molCu 原子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-43.jpg?height=286&width=303&top_left_y=2013&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-43.jpg?height=263&width=117&top_left_y=2027&top_left_x=615",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-43.jpg?height=271&width=303&top_left_y=2029&top_left_x=725",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-43.jpg?height=254&width=100&top_left_y=2032&top_left_x=1024",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-43.jpg?height=254&width=386&top_left_y=2029&top_left_x=1115",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-43.jpg?height=260&width=294&top_left_y=2029&top_left_x=1498"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1370",
        "problem": "In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Later, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were measured by using a glass vessel with a known volume under atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$.\n\nTable 1. Mass of Chemical Nitrogen in the Vessel\n\n| From nitric oxide | $2.3001 \\mathrm{~g}$ |\n| :--- | :--- |\n| From nitrous oxide | $2.2990 \\mathrm{~g}$ |\n| From ammonium nitrite purified at a red heat | $2.2987 \\mathrm{~g}$ |\n| From urea | $2.2985 \\mathrm{~g}$ |\n| From ammonium nitrite purified in the cold | $2.2987 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 2 9 9 0} \\mathbf{~ g}$ |\n\nTable 2. Mass of Atmospheric Nitrogen in the Vessel\n\n| $\\mathrm{O}_{2}$ was removed by hot copper (1892) | $2.3103 \\mathrm{~g}$ |\n| :--- | :--- |\n| $\\mathrm{O}_{2}$ was removed by hot iron (1893) | $2.3100 \\mathrm{~g}$ |\n| $\\mathrm{O}_{2}$ was removed by ferrous hydrate (1894) | $2.3102 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 3 1 0 2} \\mathbf{~ g}$ |Calculate the volume $V\\left[\\mathrm{~m}^{3}\\right]$ of the vessel used by Rayleigh from the mean mass of chemical nitrogen, which must have been pure nitrogen. Assume that the measurements were carried out at a temperature of $15.0^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Later, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were measured by using a glass vessel with a known volume under atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$.\n\nTable 1. Mass of Chemical Nitrogen in the Vessel\n\n| From nitric oxide | $2.3001 \\mathrm{~g}$ |\n| :--- | :--- |\n| From nitrous oxide | $2.2990 \\mathrm{~g}$ |\n| From ammonium nitrite purified at a red heat | $2.2987 \\mathrm{~g}$ |\n| From urea | $2.2985 \\mathrm{~g}$ |\n| From ammonium nitrite purified in the cold | $2.2987 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 2 9 9 0} \\mathbf{~ g}$ |\n\nTable 2. Mass of Atmospheric Nitrogen in the Vessel\n\n| $\\mathrm{O}_{2}$ was removed by hot copper (1892) | $2.3103 \\mathrm{~g}$ |\n| :--- | :--- |\n| $\\mathrm{O}_{2}$ was removed by hot iron (1893) | $2.3100 \\mathrm{~g}$ |\n| $\\mathrm{O}_{2}$ was removed by ferrous hydrate (1894) | $2.3102 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 3 1 0 2} \\mathbf{~ g}$ |\n\nproblem:\nCalculate the volume $V\\left[\\mathrm{~m}^{3}\\right]$ of the vessel used by Rayleigh from the mean mass of chemical nitrogen, which must have been pure nitrogen. Assume that the measurements were carried out at a temperature of $15.0^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{m}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{m}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_469",
        "problem": "人工肾脏可采用间接电化学方法除去代谢产物中的尿素 $\\left(\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}\\right)$, 原理如图所示，下列有关说法正确的是\n\n[图1]\nA: 除去 $1 \\mathrm{~mol}$ 尿素, 质子交换膜中转移 $6 \\mathrm{~mol} \\mathrm{H}^{+}$\nB: 电解结束后, 阴极室溶液的 $\\mathrm{pH}$ 与电解前相比将升高\nC: 该方法与血液透析清除尿素的原理基本相同\nD: 若两极共收集到气体 $13.44 \\mathrm{~L}$ (标准状况), 则除去的尿素为 $7.2 \\mathrm{~g}$ (忽略气体的溶解)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n人工肾脏可采用间接电化学方法除去代谢产物中的尿素 $\\left(\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}\\right)$, 原理如图所示，下列有关说法正确的是\n\n[图1]\n\nA: 除去 $1 \\mathrm{~mol}$ 尿素, 质子交换膜中转移 $6 \\mathrm{~mol} \\mathrm{H}^{+}$\nB: 电解结束后, 阴极室溶液的 $\\mathrm{pH}$ 与电解前相比将升高\nC: 该方法与血液透析清除尿素的原理基本相同\nD: 若两极共收集到气体 $13.44 \\mathrm{~L}$ (标准状况), 则除去的尿素为 $7.2 \\mathrm{~g}$ (忽略气体的溶解)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-83.jpg?height=637&width=737&top_left_y=1809&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_771",
        "problem": "向 $100 \\mathrm{~mL} 0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{~A}$ 溶液中逐滴加入等浓度 $\\mathrm{B}$ 溶液, 反应为\n\n[图1]\nA: $\\Delta H=-100.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\nB: $\\mathrm{T}_{1}<\\mathrm{T}_{2}$\nC: $\\mathrm{T}_{2}$ 时, 该反应平衡常数约为 $667 \\mathrm{~L} \\cdot \\mathrm{mol}^{-1}$\nD: $\\mathrm{T}_{1}$ 时, $100 \\mathrm{~mL} 0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{AB}$ 溶液达到平衡时, 吸收热量 $\\mathrm{Q}>4.3 \\mathrm{~J}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n向 $100 \\mathrm{~mL} 0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{~A}$ 溶液中逐滴加入等浓度 $\\mathrm{B}$ 溶液, 反应为\n\n[图1]\n\nA: $\\Delta H=-100.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\nB: $\\mathrm{T}_{1}<\\mathrm{T}_{2}$\nC: $\\mathrm{T}_{2}$ 时, 该反应平衡常数约为 $667 \\mathrm{~L} \\cdot \\mathrm{mol}^{-1}$\nD: $\\mathrm{T}_{1}$ 时, $100 \\mathrm{~mL} 0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{AB}$ 溶液达到平衡时, 吸收热量 $\\mathrm{Q}>4.3 \\mathrm{~J}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-015.jpg?height=716&width=1308&top_left_y=1618&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_159",
        "problem": "Which of the following molecules has a molecular dipole?\nA: $\\mathrm{XeF}_{4}$\nB: $\\mathrm{SeF}_{4}$\nC: $\\mathrm{CF}_{4}$\nD: $\\mathrm{SiF}_{4}$\nE: $\\mathrm{KrF}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following molecules has a molecular dipole?\n\nA: $\\mathrm{XeF}_{4}$\nB: $\\mathrm{SeF}_{4}$\nC: $\\mathrm{CF}_{4}$\nD: $\\mathrm{SiF}_{4}$\nE: $\\mathrm{KrF}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_820",
        "problem": "下列关于 $\\mathrm{ClCH}_{2} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{COOH}$ 性质的描述正确且解释合理的是\n\n| 选项 | 性质 | 解释 |\n| :--- | :--- | :--- |\n| A | 酸性: <br> $\\mathrm{ClCH}_{2} \\mathrm{COOH}>$ | 氯的电负性大于氢, 导致 $\\mathrm{ClCH}_{2} \\mathrm{COOH}$ 中氢 <br> 氧键极性增强 |\n| B | $\\mathrm{ClCH}_{2} \\mathrm{COOH}$ <br> 水溶液中存在 <br> $\\mathrm{Cl}^{-}$ | $\\mathrm{ClCH}_{2} \\mathrm{COOH}$ 是离子化合物 |\n| $\\mathrm{C}$ | $\\mathrm{CH}_{3} \\mathrm{COOH}$ 沸 点比$\\mathrm{HCOOCH}_{3}$ 高很多| $\\mathrm{CH}_{3} \\mathrm{COOH}$ 是极性分子 |\n| D | 溶于水 | $\\mathrm{CH}_{3} \\mathrm{COOH}$ 与水分子形成氢键 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列关于 $\\mathrm{ClCH}_{2} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{COOH}$ 性质的描述正确且解释合理的是\n\n| 选项 | 性质 | 解释 |\n| :--- | :--- | :--- |\n| A | 酸性: <br> $\\mathrm{ClCH}_{2} \\mathrm{COOH}>$ | 氯的电负性大于氢, 导致 $\\mathrm{ClCH}_{2} \\mathrm{COOH}$ 中氢 <br> 氧键极性增强 |\n| B | $\\mathrm{ClCH}_{2} \\mathrm{COOH}$ <br> 水溶液中存在 <br> $\\mathrm{Cl}^{-}$ | $\\mathrm{ClCH}_{2} \\mathrm{COOH}$ 是离子化合物 |\n| $\\mathrm{C}$ | $\\mathrm{CH}_{3} \\mathrm{COOH}$ 沸 点比$\\mathrm{HCOOCH}_{3}$ 高很多| $\\mathrm{CH}_{3} \\mathrm{COOH}$ 是极性分子 |\n| D | 溶于水 | $\\mathrm{CH}_{3} \\mathrm{COOH}$ 与水分子形成氢键 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_347",
        "problem": "Quartz, $\\mathrm{SiO}_{2}$, is the most common mineral found on the surface of the earth. What is the best explanation for the fact that quartz is hard and has a high melting point?\nA: Quartz crystals are extended structures in which each atom forms strong covalent bonds with all of its neighboring atoms.\nB: Quartz crystals consist of positive and negative ions that are attracted to one another.\nC: Quartz crystals are formed under extremes of temperature and pressure.\nD: Silicon and oxygen atoms are especially hard because of their electronic structure.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nQuartz, $\\mathrm{SiO}_{2}$, is the most common mineral found on the surface of the earth. What is the best explanation for the fact that quartz is hard and has a high melting point?\n\nA: Quartz crystals are extended structures in which each atom forms strong covalent bonds with all of its neighboring atoms.\nB: Quartz crystals consist of positive and negative ions that are attracted to one another.\nC: Quartz crystals are formed under extremes of temperature and pressure.\nD: Silicon and oxygen atoms are especially hard because of their electronic structure.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_815",
        "problem": "常温下, 水溶液中 $\\mathrm{H}_{2} \\mathrm{~A} 、 \\mathrm{HA}^{-} 、 \\mathrm{~A}^{2-} 、 \\mathrm{HB} 、 \\mathrm{~B}$-的分布系数 $\\delta$ [如\n\n$\\left.\\delta\\left(\\mathrm{A}^{2-}\\right)=\\frac{c\\left(\\mathrm{~A}^{2-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)}\\right]$ 随 $\\mathrm{POH}$ 变化曲线如图 1 所示, 溶液中\n\n$\\mathrm{pBa}\\left[\\mathrm{pBa}=-\\lg c\\left(\\mathrm{Ba}^{2+}\\right)\\right]$ 与 $\\mathrm{pA}\\left[\\mathrm{pA}=-\\operatorname{lgc}\\left(\\mathrm{A}^{2-}\\right)\\right]$ 关系如图 2 所示。用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{BaB}_{2}$ 溶液, 若混合后溶液体积变化忽略不计, 下列说法错误的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 常温下, $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $K_{\\mathrm{a} 1}=10^{-3.2}$\nB: 当滴入 $5.00 \\mathrm{mLH}_{2} \\mathrm{~A}$ 溶液时, $c\\left(\\mathrm{~A}^{2-}\\right) \\cdot c(\\mathrm{HB})>c\\left(\\mathrm{HA}^{-}\\right) \\cdot c\\left(\\mathrm{~B}^{-}\\right)$\nC: 当滴入 $20.00 \\mathrm{mLH}_{2} \\mathrm{~A}$ 溶液时, 此时溶液的 $\\mathrm{pH}$ 约为 5.1\nD: 当溶液的 $\\mathrm{pOH}=7$ 时, $2 c\\left(\\mathrm{~A}^{2-}\\right)+c\\left(\\mathrm{HA}^{-}\\right)=c(\\mathrm{HB})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 水溶液中 $\\mathrm{H}_{2} \\mathrm{~A} 、 \\mathrm{HA}^{-} 、 \\mathrm{~A}^{2-} 、 \\mathrm{HB} 、 \\mathrm{~B}$-的分布系数 $\\delta$ [如\n\n$\\left.\\delta\\left(\\mathrm{A}^{2-}\\right)=\\frac{c\\left(\\mathrm{~A}^{2-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)}\\right]$ 随 $\\mathrm{POH}$ 变化曲线如图 1 所示, 溶液中\n\n$\\mathrm{pBa}\\left[\\mathrm{pBa}=-\\lg c\\left(\\mathrm{Ba}^{2+}\\right)\\right]$ 与 $\\mathrm{pA}\\left[\\mathrm{pA}=-\\operatorname{lgc}\\left(\\mathrm{A}^{2-}\\right)\\right]$ 关系如图 2 所示。用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{BaB}_{2}$ 溶液, 若混合后溶液体积变化忽略不计, 下列说法错误的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 常温下, $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $K_{\\mathrm{a} 1}=10^{-3.2}$\nB: 当滴入 $5.00 \\mathrm{mLH}_{2} \\mathrm{~A}$ 溶液时, $c\\left(\\mathrm{~A}^{2-}\\right) \\cdot c(\\mathrm{HB})>c\\left(\\mathrm{HA}^{-}\\right) \\cdot c\\left(\\mathrm{~B}^{-}\\right)$\nC: 当滴入 $20.00 \\mathrm{mLH}_{2} \\mathrm{~A}$ 溶液时, 此时溶液的 $\\mathrm{pH}$ 约为 5.1\nD: 当溶液的 $\\mathrm{pOH}=7$ 时, $2 c\\left(\\mathrm{~A}^{2-}\\right)+c\\left(\\mathrm{HA}^{-}\\right)=c(\\mathrm{HB})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-002.jpg?height=454&width=365&top_left_y=1806&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-002.jpg?height=297&width=331&top_left_y=1962&top_left_x=791"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_426",
        "problem": "现有三种有机化合物的结构简式[图1]下列说法错误的是\nA: b、d、p 互为同分异构体\nB: $\\mathrm{d}$ 的二氯代物有 4 种\nC: $\\mathrm{p}$ 的二氯代物有 6 种\nD: $\\mathrm{b}$ 的同分异构体只有 $\\mathrm{d}$ 和 $\\mathrm{p}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n现有三种有机化合物的结构简式[图1]下列说法错误的是\n\nA: b、d、p 互为同分异构体\nB: $\\mathrm{d}$ 的二氯代物有 4 种\nC: $\\mathrm{p}$ 的二氯代物有 6 种\nD: $\\mathrm{b}$ 的同分异构体只有 $\\mathrm{d}$ 和 $\\mathrm{p}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/P5rKg0Mp/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_492",
        "problem": "二氧化锰为黑色无定形粉末, 难溶于水, 常用于反应的氧化剂、除锈剂、催化剂。\n\n现以软锰矿 (主要成分是 $\\mathrm{MnO}_{2}$, 含少量 $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ 和 $\\mathrm{SiO}_{2}$ ) 和闪锌矿(主要成分是 $\\mathrm{ZnS}$, 含少试卷第 56 页, 共 88 页\n量 $\\mathrm{FeS}$ 等)为原料制备 $\\mathrm{MnO}_{2}$ 和 $\\mathrm{Zn}$ ，设计工艺流程如下:\n\n[图1]\n\n已知: (1) $\\mathrm{Mn}^{2+}$ 在酸性条件下比较稳定, $\\mathrm{pH}$ 高于 5.5 时易被 $\\mathrm{O}_{2}$ 氧化。\n\n(2)室温下，\n\n$\\mathrm{Ksp}\\left[\\mathrm{Mn}(\\mathrm{OH})_{2}\\right]=10^{-13}, \\mathrm{Ksp}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]=10^{-38}, \\mathrm{Ksp}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]=10^{-33}, \\mathrm{Ksp}\\left[\\mathrm{Zn}(\\mathrm{OH})_{2}\\right]=10^{-17}$ (当离子浓度 $\\leq 10^{-5}$ 可认为沉淀完全)\n\n下列说法错误的是\nA: $\\mathrm{ZnS}$ 溶解离子方程式: $\\mathrm{ZnS}+\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}=\\mathrm{Zn}^{2+}+\\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{S}$\nB: 滤渣 $\\mathrm{C}$ 主要成分: $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Al}(\\mathrm{OH})_{3}$\nC: 若滤液 $\\mathrm{B}$ 中 $\\mathrm{Mn}^{2+} 、 \\mathrm{Zn}^{2+}$ 的浓度约为 $0.1 \\mathrm{~mol} / \\mathrm{L}$, 过程(3)调节 $\\mathrm{pH}$ 的合理范围 $3 \\leq \\mathrm{pH}$ $<8$\nD: “电解”时, 阳极产物制得 $\\mathrm{MnO}_{2}$, 阴极制得 $\\mathrm{Zn}$, 余下电解质溶液经处理回“滤液 B”中可循环利用\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n二氧化锰为黑色无定形粉末, 难溶于水, 常用于反应的氧化剂、除锈剂、催化剂。\n\n现以软锰矿 (主要成分是 $\\mathrm{MnO}_{2}$, 含少量 $\\mathrm{Al}_{2} \\mathrm{O}_{3}$ 和 $\\mathrm{SiO}_{2}$ ) 和闪锌矿(主要成分是 $\\mathrm{ZnS}$, 含少试卷第 56 页, 共 88 页\n量 $\\mathrm{FeS}$ 等)为原料制备 $\\mathrm{MnO}_{2}$ 和 $\\mathrm{Zn}$ ，设计工艺流程如下:\n\n[图1]\n\n已知: (1) $\\mathrm{Mn}^{2+}$ 在酸性条件下比较稳定, $\\mathrm{pH}$ 高于 5.5 时易被 $\\mathrm{O}_{2}$ 氧化。\n\n(2)室温下，\n\n$\\mathrm{Ksp}\\left[\\mathrm{Mn}(\\mathrm{OH})_{2}\\right]=10^{-13}, \\mathrm{Ksp}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]=10^{-38}, \\mathrm{Ksp}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]=10^{-33}, \\mathrm{Ksp}\\left[\\mathrm{Zn}(\\mathrm{OH})_{2}\\right]=10^{-17}$ (当离子浓度 $\\leq 10^{-5}$ 可认为沉淀完全)\n\n下列说法错误的是\n\nA: $\\mathrm{ZnS}$ 溶解离子方程式: $\\mathrm{ZnS}+\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}=\\mathrm{Zn}^{2+}+\\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{S}$\nB: 滤渣 $\\mathrm{C}$ 主要成分: $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Al}(\\mathrm{OH})_{3}$\nC: 若滤液 $\\mathrm{B}$ 中 $\\mathrm{Mn}^{2+} 、 \\mathrm{Zn}^{2+}$ 的浓度约为 $0.1 \\mathrm{~mol} / \\mathrm{L}$, 过程(3)调节 $\\mathrm{pH}$ 的合理范围 $3 \\leq \\mathrm{pH}$ $<8$\nD: “电解”时, 阳极产物制得 $\\mathrm{MnO}_{2}$, 阴极制得 $\\mathrm{Zn}$, 余下电解质溶液经处理回“滤液 B”中可循环利用\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-57.jpg?height=715&width=1470&top_left_y=251&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_90",
        "problem": "How many stereoisomers of octahedral $\\mathrm{CoCl}_{2}\\left(\\mathrm{NH}_{3}\\right)_{2}(\\mathrm{CN})_{2}{ }^{-}$ are possible?\nA: 3\nB: 4\nC: 5\nD: 6\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many stereoisomers of octahedral $\\mathrm{CoCl}_{2}\\left(\\mathrm{NH}_{3}\\right)_{2}(\\mathrm{CN})_{2}{ }^{-}$ are possible?\n\nA: 3\nB: 4\nC: 5\nD: 6\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_110",
        "problem": "Dapsone is the active ingredient in Aczone ${ }^{\\mathrm{TM}}$ gel, a treatment for adult acne. Each gram of Aczone gel contains $50 \\mathrm{mg}$ of dapsone (molecular weight $248.3 \\mathrm{~g}$ $\\mathrm{mol}^{-1}$ ). The dapsone content of a $10.0 \\mathrm{~g}$ sample of Aczone was analyzed and found to be composed of $290.3 \\mathrm{mg}$ carbon, $64.5 \\mathrm{mg}$ sulphur, $56.4 \\mathrm{mg}$ nitrogen and $24.4 \\mathrm{mg}$ hydrogen with the remaining mass being oxygen. What is the molecular formula of dapsone?\nA: $\\mathrm{C}_{12} \\mathrm{H}_{10} \\mathrm{NOS}_{2}$\nB: $\\mathrm{C}_{10} \\mathrm{H}_{8} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{~S}$\nC: $\\mathrm{C}_{11} \\mathrm{H}_{8} \\mathrm{~N}_{2} \\mathrm{O}_{3} \\mathrm{~S}$\nD: $\\mathrm{C}_{13} \\mathrm{H}_{14} \\mathrm{NO}_{2} \\mathrm{~S}$\nE: $\\mathrm{C}_{12} \\mathrm{H}_{12} \\mathrm{~N}_{2} \\mathrm{O}_{2} \\mathrm{~S}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nDapsone is the active ingredient in Aczone ${ }^{\\mathrm{TM}}$ gel, a treatment for adult acne. Each gram of Aczone gel contains $50 \\mathrm{mg}$ of dapsone (molecular weight $248.3 \\mathrm{~g}$ $\\mathrm{mol}^{-1}$ ). The dapsone content of a $10.0 \\mathrm{~g}$ sample of Aczone was analyzed and found to be composed of $290.3 \\mathrm{mg}$ carbon, $64.5 \\mathrm{mg}$ sulphur, $56.4 \\mathrm{mg}$ nitrogen and $24.4 \\mathrm{mg}$ hydrogen with the remaining mass being oxygen. What is the molecular formula of dapsone?\n\nA: $\\mathrm{C}_{12} \\mathrm{H}_{10} \\mathrm{NOS}_{2}$\nB: $\\mathrm{C}_{10} \\mathrm{H}_{8} \\mathrm{~N}_{4} \\mathrm{O}_{2} \\mathrm{~S}$\nC: $\\mathrm{C}_{11} \\mathrm{H}_{8} \\mathrm{~N}_{2} \\mathrm{O}_{3} \\mathrm{~S}$\nD: $\\mathrm{C}_{13} \\mathrm{H}_{14} \\mathrm{NO}_{2} \\mathrm{~S}$\nE: $\\mathrm{C}_{12} \\mathrm{H}_{12} \\mathrm{~N}_{2} \\mathrm{O}_{2} \\mathrm{~S}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1279",
        "problem": "Identify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{..82}^{214} \\mathrm{~Pb} \\rightarrow{ }_{83}^{214} \\mathrm{Bi}+\\mathrm{X}$\nA: alpha\nB: beta\nC: gamma\nD: neutron\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIdentify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{..82}^{214} \\mathrm{~Pb} \\rightarrow{ }_{83}^{214} \\mathrm{Bi}+\\mathrm{X}$\n\nA: alpha\nB: beta\nC: gamma\nD: neutron\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1310",
        "problem": "Clathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nThe enthalpy of this process equals $17.47 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.The minimum temperature methane hydrate can coexist with liquid water is 272.9K. What is the minimum possible depth of pure liquid water at which methane hydrates can be stable?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nClathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nThe enthalpy of this process equals $17.47 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.\n\nproblem:\nThe minimum temperature methane hydrate can coexist with liquid water is 272.9K. What is the minimum possible depth of pure liquid water at which methane hydrates can be stable?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of m, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-222.jpg?height=425&width=434&top_left_y=1255&top_left_x=1459"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "m"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1004",
        "problem": "The volume of a gas, initially at $1 \\mathrm{~atm}$ and $20^{\\circ} \\mathrm{C}$, is increased from $40.0 \\mathrm{~mL}$ to $80.0 \\mathrm{~mL}$. If the pressure remains constant, what is the final temperature of the gas?\nA: $293 \\mathrm{~K}+\\frac{80.0}{40.0}$\nB: $20^{\\circ} \\mathrm{C} \\times \\frac{80.0}{40.0}$\nC: $293 \\mathrm{~K} \\times \\frac{80.0}{40.0}$\nD: $293 \\mathrm{~K} \\times \\frac{40.0}{80.0}$\nE: $\\quad 20^{\\circ} \\mathrm{C} \\times \\frac{40.0}{80.0}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe volume of a gas, initially at $1 \\mathrm{~atm}$ and $20^{\\circ} \\mathrm{C}$, is increased from $40.0 \\mathrm{~mL}$ to $80.0 \\mathrm{~mL}$. If the pressure remains constant, what is the final temperature of the gas?\n\nA: $293 \\mathrm{~K}+\\frac{80.0}{40.0}$\nB: $20^{\\circ} \\mathrm{C} \\times \\frac{80.0}{40.0}$\nC: $293 \\mathrm{~K} \\times \\frac{80.0}{40.0}$\nD: $293 \\mathrm{~K} \\times \\frac{40.0}{80.0}$\nE: $\\quad 20^{\\circ} \\mathrm{C} \\times \\frac{40.0}{80.0}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1058",
        "problem": "Boron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the volume of the $\\mathrm{c}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBoron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the volume of the $\\mathrm{c}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~cm}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=448&width=511&top_left_y=541&top_left_x=270",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=596&width=436&top_left_y=458&top_left_x=844",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=471&width=443&top_left_y=524&top_left_x=1389",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=603&width=305&top_left_y=1206&top_left_x=1315"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~cm}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1377",
        "problem": "Phosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nZinc is an essential micronutrient for plant growth. Plant can absorb zinc in water soluble form only. In a given soil water with $\\mathrm{pH}=7.0$, zinc phosphate was found to be the only source of zinc and phosphate. Calculate the concentration of $\\left[\\mathrm{Zn}^{2+}\\right]$ and $\\left[\\mathrm{PO}_{4}^{3-}\\right]$ ions in the solution. Ksp for zinc phosphate is $9.1 \\times 10^{-33}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nPhosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nZinc is an essential micronutrient for plant growth. Plant can absorb zinc in water soluble form only. In a given soil water with $\\mathrm{pH}=7.0$, zinc phosphate was found to be the only source of zinc and phosphate. Calculate the concentration of $\\left[\\mathrm{Zn}^{2+}\\right]$ and $\\left[\\mathrm{PO}_{4}^{3-}\\right]$ ions in the solution. Ksp for zinc phosphate is $9.1 \\times 10^{-33}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1358",
        "problem": "Regarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe transmittance is inversely proportional to the concentration of the absorbing compound.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nRegarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe transmittance is inversely proportional to the concentration of the absorbing compound.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_699",
        "problem": "常温下, 将 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KOH}$ 溶液滴加到一定浓度的二元酸 $\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)$ 溶液混合溶液的 $\\mathrm{pH}$与离子浓度变化的关系如图所示。下列叙述错误的是\n\n[图1]\nA: 线 $M$ 表示 $\\mathrm{pH}$ 与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)}{\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)}$的变化关系\nB: KHR 和 $\\mathrm{K}_{2} \\mathrm{R}$ 的 1: 1 混合溶液显酸性\nC: 混合溶液中一定有: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\nD: 当混合溶液呈中性时, $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)+\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 将 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KOH}$ 溶液滴加到一定浓度的二元酸 $\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)$ 溶液混合溶液的 $\\mathrm{pH}$与离子浓度变化的关系如图所示。下列叙述错误的是\n\n[图1]\n\nA: 线 $M$ 表示 $\\mathrm{pH}$ 与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)}{\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)}$的变化关系\nB: KHR 和 $\\mathrm{K}_{2} \\mathrm{R}$ 的 1: 1 混合溶液显酸性\nC: 混合溶液中一定有: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\nD: 当混合溶液呈中性时, $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)+\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-008.jpg?height=563&width=628&top_left_y=364&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_821",
        "problem": "常温下, 将 $\\mathrm{pH}$ 均为 3 、体积均为 $\\mathrm{V}_{0}$ 的 $H X$ 和 $H Y$ 溶液, 分别加水稀释至体积为 $\\mathrm{V}, \\mathrm{pH}$\n随 $\\lg \\frac{\\mathrm{V}}{\\mathrm{V}_{0}}$ 的变化如图所示。下列说法正确的是\n\n[图1]\nA: 等浓度的盐溶液的 $\\mathrm{pH}: \\mathrm{NaX}>\\mathrm{NaY}$\nB: 水的电离程度: $b>a>c$\nC: 若初始时将两种溶液混合: $c(H X)>c(H Y)>c\\left(H^{+}\\right)>c\\left(\\mathrm{Y}^{-}\\right)>c\\left(X^{-}\\right)$\nD: $\\frac{\\mathrm{c}_{\\mathrm{a}}\\left(\\mathrm{X}^{-}\\right)}{\\mathrm{c}_{\\mathrm{a}}(\\mathrm{HX})}>\\frac{\\mathrm{c}_{\\mathrm{b}}\\left(\\mathrm{Y}^{-}\\right)}{\\mathrm{c}_{\\mathrm{b}}(\\mathrm{HY})}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 将 $\\mathrm{pH}$ 均为 3 、体积均为 $\\mathrm{V}_{0}$ 的 $H X$ 和 $H Y$ 溶液, 分别加水稀释至体积为 $\\mathrm{V}, \\mathrm{pH}$\n随 $\\lg \\frac{\\mathrm{V}}{\\mathrm{V}_{0}}$ 的变化如图所示。下列说法正确的是\n\n[图1]\n\nA: 等浓度的盐溶液的 $\\mathrm{pH}: \\mathrm{NaX}>\\mathrm{NaY}$\nB: 水的电离程度: $b>a>c$\nC: 若初始时将两种溶液混合: $c(H X)>c(H Y)>c\\left(H^{+}\\right)>c\\left(\\mathrm{Y}^{-}\\right)>c\\left(X^{-}\\right)$\nD: $\\frac{\\mathrm{c}_{\\mathrm{a}}\\left(\\mathrm{X}^{-}\\right)}{\\mathrm{c}_{\\mathrm{a}}(\\mathrm{HX})}>\\frac{\\mathrm{c}_{\\mathrm{b}}\\left(\\mathrm{Y}^{-}\\right)}{\\mathrm{c}_{\\mathrm{b}}(\\mathrm{HY})}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-094.jpg?height=517&width=608&top_left_y=270&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1353",
        "problem": "Bohemian garnet (pyrope) is a famous Czech blood coloured semi-precious stone. The chemical composition of natural garnets is expressed by the general stoichiometric formula of $\\mathrm{A}_{3} \\mathrm{~B}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$, where $A^{\\text {II }}$ is a divalent cation and $B^{\\text {III }}$ is a trivalent cation. Garnets have a cubic unit cell that contains 8 formula units. The structure comprises 3 types of polyhedra: the $A^{l l}$ cation occupies a dodecahedral position (it is surrounded with eight $O$ atoms), the $\\mathrm{B}^{\\text {III }}$ cation occupies an octahedral position (it is surrounded with six $\\mathrm{O}$ atoms) and $\\mathrm{Si}^{\\mathrm{lV}}$ is surrounded with four $\\mathrm{O}$ atoms arranged into a tetrahedron.\n\nThe most common garnet mineral is almandine with the formula of $\\mathrm{Fe}_{3} \\mathrm{Al}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$. Its unit cell parameter is $a=11.50 \\AA$.Calculate the theoretical density of almandine.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nBohemian garnet (pyrope) is a famous Czech blood coloured semi-precious stone. The chemical composition of natural garnets is expressed by the general stoichiometric formula of $\\mathrm{A}_{3} \\mathrm{~B}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$, where $A^{\\text {II }}$ is a divalent cation and $B^{\\text {III }}$ is a trivalent cation. Garnets have a cubic unit cell that contains 8 formula units. The structure comprises 3 types of polyhedra: the $A^{l l}$ cation occupies a dodecahedral position (it is surrounded with eight $O$ atoms), the $\\mathrm{B}^{\\text {III }}$ cation occupies an octahedral position (it is surrounded with six $\\mathrm{O}$ atoms) and $\\mathrm{Si}^{\\mathrm{lV}}$ is surrounded with four $\\mathrm{O}$ atoms arranged into a tetrahedron.\n\nThe most common garnet mineral is almandine with the formula of $\\mathrm{Fe}_{3} \\mathrm{Al}_{2}\\left(\\mathrm{SiO}_{4}\\right)_{3}$. Its unit cell parameter is $a=11.50 \\AA$.\n\nproblem:\nCalculate the theoretical density of almandine.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_246",
        "problem": "$\\quad 0.39 \\mathrm{~mol}$ of magnesium chloride is dissolved in water to produce a solution with a volume of $1.5 \\mathrm{~L}$. What is the concentration, in $\\mathrm{mol} \\mathrm{L}^{-1}$, of chloride ions in this solution?\nA. $\\quad 0.098 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nA: $\\quad 0.098 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $0.20 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.26 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $0.39 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $0.52 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n$\\quad 0.39 \\mathrm{~mol}$ of magnesium chloride is dissolved in water to produce a solution with a volume of $1.5 \\mathrm{~L}$. What is the concentration, in $\\mathrm{mol} \\mathrm{L}^{-1}$, of chloride ions in this solution?\nA. $\\quad 0.098 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nA: $\\quad 0.098 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $0.20 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.26 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $0.39 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $0.52 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_813",
        "problem": "石家庄二中实验学校某化学兴趣小组同学查阅资料得知, 常温下对于任一电池反应 $a A+b B=c C+d D$, 其电动势 $E=E^{\\theta}-\\frac{(0.0592)}{n} \\cdot \\lg \\frac{c(C)^{c} \\cdot c(D)^{d}}{c(A)^{a} \\cdot c(B)^{b}}, n$ 为电池反应中转移的电子数。该小组同学设计装置(如图 1), 以 $\\mathrm{Zn}-\\mathrm{Cu}$ 原电池探究离子浓度的改变对电极电势的影响。小组同学测得初始时 $\\mathrm{Zn}_{(s)}+\\mathrm{Cu}^{2+}{ }_{(1 \\mathrm{~mol} / \\mathrm{L}}=\\mathrm{Zn}^{2+}{ }_{(1 \\mathrm{~mol} / \\mathrm{L})}+\\mathrm{Cu}_{(\\mathrm{s})} \\quad \\mathrm{E}^{\\mathrm{a}}=1.1 \\mathrm{~V}$ (该反应 $\\mathrm{n}=2$ ),随放电进行, 观察电池电动势的变化趋势并绘制了电池电动势变化示意图(如图 2)。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 电压表读数为零后, 则说明该原电池中 $\\mathrm{Cu}^{2+}$ 已经消耗完全\nB: 小组同学推测图 2 中直线与 $\\mathrm{X}$ 轴的交点坐标大约为 $(37,0)$\nC: 小组同学向 $\\mathrm{ZnSO}_{4}$ 和 $\\mathrm{CuSO}_{4}$ 溶液中同时快速加入少量相同体积和浓度的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶 液, 发现电池电动势突然减小, 则可知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS})$\nD: 小组同学推测若将初始时左侧 $1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{ZnSO}_{4}-\\mathrm{Zn}$ 半电池, 换为 $2 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{CuSO}_{4}-\\mathrm{Cu}$ 半电池, 右侧半电池保持不变, 则仍能观察到相同的电压表偏转情况,\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n石家庄二中实验学校某化学兴趣小组同学查阅资料得知, 常温下对于任一电池反应 $a A+b B=c C+d D$, 其电动势 $E=E^{\\theta}-\\frac{(0.0592)}{n} \\cdot \\lg \\frac{c(C)^{c} \\cdot c(D)^{d}}{c(A)^{a} \\cdot c(B)^{b}}, n$ 为电池反应中转移的电子数。该小组同学设计装置(如图 1), 以 $\\mathrm{Zn}-\\mathrm{Cu}$ 原电池探究离子浓度的改变对电极电势的影响。小组同学测得初始时 $\\mathrm{Zn}_{(s)}+\\mathrm{Cu}^{2+}{ }_{(1 \\mathrm{~mol} / \\mathrm{L}}=\\mathrm{Zn}^{2+}{ }_{(1 \\mathrm{~mol} / \\mathrm{L})}+\\mathrm{Cu}_{(\\mathrm{s})} \\quad \\mathrm{E}^{\\mathrm{a}}=1.1 \\mathrm{~V}$ (该反应 $\\mathrm{n}=2$ ),随放电进行, 观察电池电动势的变化趋势并绘制了电池电动势变化示意图(如图 2)。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 电压表读数为零后, 则说明该原电池中 $\\mathrm{Cu}^{2+}$ 已经消耗完全\nB: 小组同学推测图 2 中直线与 $\\mathrm{X}$ 轴的交点坐标大约为 $(37,0)$\nC: 小组同学向 $\\mathrm{ZnSO}_{4}$ 和 $\\mathrm{CuSO}_{4}$ 溶液中同时快速加入少量相同体积和浓度的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶 液, 发现电池电动势突然减小, 则可知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS})$\nD: 小组同学推测若将初始时左侧 $1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{ZnSO}_{4}-\\mathrm{Zn}$ 半电池, 换为 $2 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{CuSO}_{4}-\\mathrm{Cu}$ 半电池, 右侧半电池保持不变, 则仍能观察到相同的电压表偏转情况,\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-46.jpg?height=551&width=628&top_left_y=176&top_left_x=514",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-46.jpg?height=679&width=625&top_left_y=163&top_left_x=1161"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_911",
        "problem": "将 $2 \\mathrm{molA}$ 和 $1 \\mathrm{molB}$ 充入某密闭容器中发生反应: $2 \\mathrm{~A}(\\mathrm{~g})+\\mathrm{B}(\\mathrm{g}) \\rightleftharpoons \\mathrm{xC}(\\mathrm{g})$, 达到化学平衡后, $\\mathrm{C}$ 的体积分数为 $\\mathrm{a}$ (假设该反应的条件分别和下列各选项的条件相同), 下列判断正确的是\nA: 若在恒温恒压下, 当 $\\mathrm{x}=1$ 时, 按 $1.5 \\mathrm{molA} 、 1 \\mathrm{molC}$ 作为起始物质, 达到平衡后, $\\mathrm{C}$ 的体积分数仍为 $\\mathrm{a}$\nB: 若在恒温恒容下, 当 $\\mathrm{x}=2$ 时, 将 $2 \\mathrm{molC}$ 作起始物质, 达到平衡后, $\\mathrm{C}$ 的体积分数仍为 $\\mathrm{a}$\nC: 若在恒温恒压下, 当 $\\mathrm{x}=3$ 时, $1 \\mathrm{~mol} \\mathrm{A、} 1 \\mathrm{molB} 、 6 \\mathrm{molC}$ 作起始物质, 达到平衡后, $\\mathrm{C}$ 的体积分数仍为 $\\mathrm{a}$\nD: 若在恒温恒容下, 按 $0.6 \\mathrm{molA} 、 0.3 \\mathrm{molB} 、 1.4 \\mathrm{molC}$ 作起始物质, 达到平衡后, $\\mathrm{C}$的体积分数仍为 $\\mathrm{a}$ ，则 $\\mathrm{x}$ 一定为 2 或 3\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n将 $2 \\mathrm{molA}$ 和 $1 \\mathrm{molB}$ 充入某密闭容器中发生反应: $2 \\mathrm{~A}(\\mathrm{~g})+\\mathrm{B}(\\mathrm{g}) \\rightleftharpoons \\mathrm{xC}(\\mathrm{g})$, 达到化学平衡后, $\\mathrm{C}$ 的体积分数为 $\\mathrm{a}$ (假设该反应的条件分别和下列各选项的条件相同), 下列判断正确的是\n\nA: 若在恒温恒压下, 当 $\\mathrm{x}=1$ 时, 按 $1.5 \\mathrm{molA} 、 1 \\mathrm{molC}$ 作为起始物质, 达到平衡后, $\\mathrm{C}$ 的体积分数仍为 $\\mathrm{a}$\nB: 若在恒温恒容下, 当 $\\mathrm{x}=2$ 时, 将 $2 \\mathrm{molC}$ 作起始物质, 达到平衡后, $\\mathrm{C}$ 的体积分数仍为 $\\mathrm{a}$\nC: 若在恒温恒压下, 当 $\\mathrm{x}=3$ 时, $1 \\mathrm{~mol} \\mathrm{A、} 1 \\mathrm{molB} 、 6 \\mathrm{molC}$ 作起始物质, 达到平衡后, $\\mathrm{C}$ 的体积分数仍为 $\\mathrm{a}$\nD: 若在恒温恒容下, 按 $0.6 \\mathrm{molA} 、 0.3 \\mathrm{molB} 、 1.4 \\mathrm{molC}$ 作起始物质, 达到平衡后, $\\mathrm{C}$的体积分数仍为 $\\mathrm{a}$ ，则 $\\mathrm{x}$ 一定为 2 或 3\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1530",
        "problem": "Bridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate the equilibrium constant at $25^{\\circ} \\mathrm{C}$ for the overall cell reaction.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate the equilibrium constant at $25^{\\circ} \\mathrm{C}$ for the overall cell reaction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-345.jpg?height=951&width=1445&top_left_y=595&top_left_x=314"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_698",
        "problem": "普鲁士蓝及其衍生物可用作钠离子电池的正极材料。放电时 $\\mathrm{Na}^{+}$嵌入正极或充电时\n\n[图1]\n三者的通式为 $\\mathrm{Na}_{x} \\mathrm{Fe}_{y}(\\mathrm{CN})_{z}$, 均为立方晶胞, 如图 1 所示。嵌入的 $\\mathrm{Na}^{+}$填充在小立方体\n的体心, 图中 $\\mathrm{Na}^{+}$均未标出, 从普鲁士蓝晶胞中切出 $\\frac{1}{8}$ 的结构如图 2 所示。下列说法错误的是\n\n[图2]\n\n图1( $\\mathrm{Na}^{+}$未标出)\n\n[图3]\n\n图2( $\\mathrm{Na}^{+}$未标出)\nA: 柏林绿中, $\\mathrm{Fe}^{3+}$ 的 d 轨道参与杂化\nB: 普鲁士蓝晶胞中 $\\mathrm{Na}^{+}$填充率小于 $50 \\%$\nC: 普鲁士白的化学式为 $\\mathrm{NaFe}(\\mathrm{CN})_{3}$\nD: 放电过程中, 1 个普鲁士蓝晶胞完全转化为普鲁士白晶胞时, 转移 8 个电子\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n普鲁士蓝及其衍生物可用作钠离子电池的正极材料。放电时 $\\mathrm{Na}^{+}$嵌入正极或充电时\n\n[图1]\n三者的通式为 $\\mathrm{Na}_{x} \\mathrm{Fe}_{y}(\\mathrm{CN})_{z}$, 均为立方晶胞, 如图 1 所示。嵌入的 $\\mathrm{Na}^{+}$填充在小立方体\n的体心, 图中 $\\mathrm{Na}^{+}$均未标出, 从普鲁士蓝晶胞中切出 $\\frac{1}{8}$ 的结构如图 2 所示。下列说法错误的是\n\n[图2]\n\n图1( $\\mathrm{Na}^{+}$未标出)\n\n[图3]\n\n图2( $\\mathrm{Na}^{+}$未标出)\n\nA: 柏林绿中, $\\mathrm{Fe}^{3+}$ 的 d 轨道参与杂化\nB: 普鲁士蓝晶胞中 $\\mathrm{Na}^{+}$填充率小于 $50 \\%$\nC: 普鲁士白的化学式为 $\\mathrm{NaFe}(\\mathrm{CN})_{3}$\nD: 放电过程中, 1 个普鲁士蓝晶胞完全转化为普鲁士白晶胞时, 转移 8 个电子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-054.jpg?height=83&width=1370&top_left_y=2494&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-055.jpg?height=365&width=556&top_left_y=360&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-055.jpg?height=428&width=483&top_left_y=346&top_left_x=932"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_725",
        "problem": "常温下, $K_{\\mathrm{a}}(\\mathrm{HCOOH})=1.77 \\times 10^{-4}, K_{\\mathrm{a}}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=1.75 \\times 10^{-5}, K_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ $=1.76 \\times 10^{-5}$, 下列说法正确的是\nA: 用相同浓度的 $\\mathrm{NaOH}$ 溶液分别滴定等体积 $\\mathrm{pH}$ 均为 3 的 $\\mathrm{HCOOH}$ 和 $\\mathrm{CH}_{3} \\mathrm{COOH}$溶液至终点, 消耗 $\\mathrm{NaOH}$ 溶液的体积相等\nB: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOONa}$ 和 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中阳离子的物质的量浓度之和前者小于后者\nC: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCOOH}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 等体积混合后的溶液中: $c\\left(\\mathrm{HCOO}^{-}\\right)+$ $c\\left(\\mathrm{OH}^{-}\\right)=c(\\mathrm{HCOOH})+c\\left(\\mathrm{H}^{+}\\right)$\nD: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸等体积混合后的溶液中 $(\\mathrm{pH}<7)$ : $c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, $K_{\\mathrm{a}}(\\mathrm{HCOOH})=1.77 \\times 10^{-4}, K_{\\mathrm{a}}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=1.75 \\times 10^{-5}, K_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ $=1.76 \\times 10^{-5}$, 下列说法正确的是\n\nA: 用相同浓度的 $\\mathrm{NaOH}$ 溶液分别滴定等体积 $\\mathrm{pH}$ 均为 3 的 $\\mathrm{HCOOH}$ 和 $\\mathrm{CH}_{3} \\mathrm{COOH}$溶液至终点, 消耗 $\\mathrm{NaOH}$ 溶液的体积相等\nB: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOONa}$ 和 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中阳离子的物质的量浓度之和前者小于后者\nC: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCOOH}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 等体积混合后的溶液中: $c\\left(\\mathrm{HCOO}^{-}\\right)+$ $c\\left(\\mathrm{OH}^{-}\\right)=c(\\mathrm{HCOOH})+c\\left(\\mathrm{H}^{+}\\right)$\nD: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸等体积混合后的溶液中 $(\\mathrm{pH}<7)$ : $c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_387",
        "problem": "What is $E^{\\circ}$ for the following reaction?\n\n$$\n2 \\mathrm{Al}^{3+}(a q)+3 \\mathrm{Zn}(s) \\rightarrow 2 \\mathrm{Al}(s)+3 \\mathrm{Zn}^{2+}(a q)\n$$\n\n| Half-Reaction | $E^{0}, \\mathrm{~V}$ |\n| :---: | :---: |\n| $\\mathrm{Al}^{3+}(a q)+3 e^{-} \\rightarrow \\mathrm{Al}(s)$ | -1.66 |\n| $\\mathrm{Zn}^{2+}(a q)+2 e^{-} \\rightarrow \\mathrm{Zn}(s)$ | -0.76 |\nA: $-1.04 \\mathrm{~V}$\nB: $-0.90 \\mathrm{~V}$\nC: $0.90 \\mathrm{~V}$\nD: $1.04 \\mathrm{~V}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is $E^{\\circ}$ for the following reaction?\n\n$$\n2 \\mathrm{Al}^{3+}(a q)+3 \\mathrm{Zn}(s) \\rightarrow 2 \\mathrm{Al}(s)+3 \\mathrm{Zn}^{2+}(a q)\n$$\n\n| Half-Reaction | $E^{0}, \\mathrm{~V}$ |\n| :---: | :---: |\n| $\\mathrm{Al}^{3+}(a q)+3 e^{-} \\rightarrow \\mathrm{Al}(s)$ | -1.66 |\n| $\\mathrm{Zn}^{2+}(a q)+2 e^{-} \\rightarrow \\mathrm{Zn}(s)$ | -0.76 |\n\nA: $-1.04 \\mathrm{~V}$\nB: $-0.90 \\mathrm{~V}$\nC: $0.90 \\mathrm{~V}$\nD: $1.04 \\mathrm{~V}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_288",
        "problem": "The density of water at $4^{\\circ} \\mathrm{C}$ is $1.00 \\mathrm{~g} \\mathrm{~mL}^{-1}$. The density of ice at $0^{\\circ} \\mathrm{C}$ is $0.917 \\mathrm{~g} \\mathrm{~mL}^{-1}$.\n\nAn ice cube of mass $7.92 \\mathrm{~g}$, initially at $0^{\\circ} \\mathrm{C}$, melted to form liquid water with a final temperature of $4^{\\circ} \\mathrm{C}$. What happens to the volume of the water?\nA: The volume of water increases by $0.657 \\mathrm{~mL}$.\nB: The volume of water decreases by $0.657 \\mathrm{~mL}$.\nC: The volume of water increases by $0.717 \\mathrm{~mL}$.\nD: The volume of water decreases by $0.717 \\mathrm{~mL}$.\nE: The volume of water does not change.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe density of water at $4^{\\circ} \\mathrm{C}$ is $1.00 \\mathrm{~g} \\mathrm{~mL}^{-1}$. The density of ice at $0^{\\circ} \\mathrm{C}$ is $0.917 \\mathrm{~g} \\mathrm{~mL}^{-1}$.\n\nAn ice cube of mass $7.92 \\mathrm{~g}$, initially at $0^{\\circ} \\mathrm{C}$, melted to form liquid water with a final temperature of $4^{\\circ} \\mathrm{C}$. What happens to the volume of the water?\n\nA: The volume of water increases by $0.657 \\mathrm{~mL}$.\nB: The volume of water decreases by $0.657 \\mathrm{~mL}$.\nC: The volume of water increases by $0.717 \\mathrm{~mL}$.\nD: The volume of water decreases by $0.717 \\mathrm{~mL}$.\nE: The volume of water does not change.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_737",
        "problem": "电芬顿工艺被认为是一种很有应用前景的高级氧化技术, 可用于降解去除废水中的持久性有机污染物, 其工作原理如图 a 所示, 工作 $10 \\mathrm{~min}$ 时, $\\mathrm{Fe}^{2+} 、 \\mathrm{H}_{2} \\mathrm{O}_{2}$ 电极产生量 $\\left(\\mathrm{mmol} \\cdot \\mathrm{L}^{-1}\\right)$ 与电流强度关系如图 $\\mathrm{b}$ 所示:\n\n[图1]\n\n图a\n\n[图2]\n\n图b\n\n下列说法错误的是\nA: 电流流动方向: $\\mathrm{Pt}$ 电极 $\\rightarrow$ 电解质 $\\rightarrow \\mathrm{HMC}-3$ 电极\nB: $\\mathrm{Fe}^{3+}$ 是该电芬顿工艺的催化剂\nC: 根据图 b 可判断合适的电流强度范围为 $55-60 \\mathrm{~mA}$\nD: 若处理 $9.4 \\mathrm{~g}$ 苯酚, 理论上消耗标准状况下 $15.68 \\mathrm{LO}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n电芬顿工艺被认为是一种很有应用前景的高级氧化技术, 可用于降解去除废水中的持久性有机污染物, 其工作原理如图 a 所示, 工作 $10 \\mathrm{~min}$ 时, $\\mathrm{Fe}^{2+} 、 \\mathrm{H}_{2} \\mathrm{O}_{2}$ 电极产生量 $\\left(\\mathrm{mmol} \\cdot \\mathrm{L}^{-1}\\right)$ 与电流强度关系如图 $\\mathrm{b}$ 所示:\n\n[图1]\n\n图a\n\n[图2]\n\n图b\n\n下列说法错误的是\n\nA: 电流流动方向: $\\mathrm{Pt}$ 电极 $\\rightarrow$ 电解质 $\\rightarrow \\mathrm{HMC}-3$ 电极\nB: $\\mathrm{Fe}^{3+}$ 是该电芬顿工艺的催化剂\nC: 根据图 b 可判断合适的电流强度范围为 $55-60 \\mathrm{~mA}$\nD: 若处理 $9.4 \\mathrm{~g}$ 苯酚, 理论上消耗标准状况下 $15.68 \\mathrm{LO}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-22.jpg?height=611&width=737&top_left_y=1308&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-22.jpg?height=631&width=688&top_left_y=1301&top_left_x=1067"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_12",
        "problem": "Which element gives a red flame test?\nA: Aluminum\nB: Boron\nC: Sodium\nD: Strontium\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich element gives a red flame test?\n\nA: Aluminum\nB: Boron\nC: Sodium\nD: Strontium\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_307",
        "problem": "The reaction below was studied at $40^{\\circ} \\mathrm{C}$ using the method of initial rates. Data are given in the table below.\n\n$$\n\\mathrm{S}_{2} \\mathrm{O}_{8}^{2-}(\\mathrm{aq})+2 \\mathrm{I}^{-}(\\mathrm{aq}) \\rightarrow 2 \\mathrm{SO}_{4}^{2-}(\\mathrm{aq})+\\mathrm{I}_{2}(\\mathrm{~s})\n$$\n\n| run | $\\left[\\mathrm{S}_{2} \\mathrm{O}_{8}^{2-}\\right]$ <br> $\\left(\\right.$ in $\\left.\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | $\\left[\\mathrm{I}^{-}\\right]$ <br> $($in mol L-1 $)$ | Initial Rate <br> $($ in mol L {fa08cdf0c-7ff6-4533-96c1-b7ace9b679ff} ) |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.010 | 0.10 | $3.5 \\times 10^{-4}$ |\n| 2 | 0.020 | 0.20 | $1.4 \\times 10^{-3}$ |\n| 3 | 0.020 | 0.40 | $2.8 \\times 10^{-3}$ |\n\nWhat are the correct value and units of the rate constant, $k$ ?\nA: $\\quad 0.35 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$\nB: $3.5 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$\nC: $\\quad 0.35 \\mathrm{~mol}^{-1} \\mathrm{~L} \\mathrm{~s}^{-1}$\nD: $0.35 \\mathrm{~mol}^{-2} \\mathrm{~L}^{2} \\mathrm{~s}^{-1}$\nE: $1.8 \\times 10^{2} \\mathrm{~mol}^{-1} \\mathrm{~L} \\mathrm{~s}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction below was studied at $40^{\\circ} \\mathrm{C}$ using the method of initial rates. Data are given in the table below.\n\n$$\n\\mathrm{S}_{2} \\mathrm{O}_{8}^{2-}(\\mathrm{aq})+2 \\mathrm{I}^{-}(\\mathrm{aq}) \\rightarrow 2 \\mathrm{SO}_{4}^{2-}(\\mathrm{aq})+\\mathrm{I}_{2}(\\mathrm{~s})\n$$\n\n| run | $\\left[\\mathrm{S}_{2} \\mathrm{O}_{8}^{2-}\\right]$ <br> $\\left(\\right.$ in $\\left.\\mathrm{mol} \\mathrm{L}^{-1}\\right)$ | $\\left[\\mathrm{I}^{-}\\right]$ <br> $($in mol L-1 $)$ | Initial Rate <br> $($ in mol L {fa08cdf0c-7ff6-4533-96c1-b7ace9b679ff} ) |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.010 | 0.10 | $3.5 \\times 10^{-4}$ |\n| 2 | 0.020 | 0.20 | $1.4 \\times 10^{-3}$ |\n| 3 | 0.020 | 0.40 | $2.8 \\times 10^{-3}$ |\n\nWhat are the correct value and units of the rate constant, $k$ ?\n\nA: $\\quad 0.35 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$\nB: $3.5 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{~s}^{-1}$\nC: $\\quad 0.35 \\mathrm{~mol}^{-1} \\mathrm{~L} \\mathrm{~s}^{-1}$\nD: $0.35 \\mathrm{~mol}^{-2} \\mathrm{~L}^{2} \\mathrm{~s}^{-1}$\nE: $1.8 \\times 10^{2} \\mathrm{~mol}^{-1} \\mathrm{~L} \\mathrm{~s}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1566",
        "problem": "In crystals of ionic compounds, cations are generally arranged in the interstices of the closest packed lattice of anions. The structure of an ionic crystal such as sodium chloride becomes stable when the cations are in contact with the nearest anions.The ionic radii of $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions in the crystal of sodium chloride are $0.102 \\mathrm{~nm}$ and $0.181 \\mathrm{~nm}$, respectively.\n\nCalculate the density $\\left[\\mathrm{kg} \\mathrm{m}^{-3}\\right]$ of the sodium chloride crystal.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn crystals of ionic compounds, cations are generally arranged in the interstices of the closest packed lattice of anions. The structure of an ionic crystal such as sodium chloride becomes stable when the cations are in contact with the nearest anions.\n\nproblem:\nThe ionic radii of $\\mathrm{Na}^{+}$and $\\mathrm{Cl}^{-}$ions in the crystal of sodium chloride are $0.102 \\mathrm{~nm}$ and $0.181 \\mathrm{~nm}$, respectively.\n\nCalculate the density $\\left[\\mathrm{kg} \\mathrm{m}^{-3}\\right]$ of the sodium chloride crystal.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kg} \\mathrm{~m}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kg} \\mathrm{~m}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_600",
        "problem": "常温下, 向 $100 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{X}$ 溶液中加入过量固体单质 $\\mathrm{X}$, 用 $\\mathrm{NaOH}$ 或 $\\mathrm{HCl}$调节溶液 $\\mathrm{pH}$ (不考虑温度和体积变化, 无气体逸出), 平衡后溶液中 $\\mathrm{X}$ 微粒 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right.$ 、 $\\left.\\mathrm{HX}^{-} 、 \\mathrm{X}^{2-} 、 \\mathrm{X}_{2}^{2-} 、 \\mathrm{X}_{3}^{2-}\\right)$ 的分布系数 $\\delta\\left[\\right.$ 如 $\\left.\\delta\\left(\\mathrm{X}_{2}^{2-}\\right)=\\frac{\\mathrm{c}\\left(\\mathrm{X}_{2}^{2-}\\right)}{\\mathrm{c}_{\\text {总 }}(\\text { 含 } \\mathrm{X} \\text { 微粒 })}\\right]$ 与 $\\mathrm{pH}$ 的关系如图, 下列说法错误的是\n\n已知: $\\mathrm{X}(\\mathrm{s})+\\mathrm{X}^{2-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{X}_{2}^{2-}(\\mathrm{aq}) K_{1}, \\mathrm{X}(\\mathrm{s})+\\mathrm{X}_{2}^{2-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{X}_{3}^{2-}(\\mathrm{aq}) K_{2},\\left(K_{1}=K_{2}>1\\right)$ 。\n\n[图1]\nA: 曲线 $\\mathrm{c}$ 表示 $\\mathrm{X}_{3}^{2-}$\nB: $\\mathrm{p} K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)=6.9, \\mathrm{p} K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)=12.2$\nC: $K_{1}=K_{2}=10$\nD: $\\mathrm{pH}=11$ 时, $\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)+\\mathrm{n}\\left(\\mathrm{HX}^{-}\\right)+\\mathrm{n}\\left(\\mathrm{X}^{2-}\\right)+2 \\mathrm{n}\\left(\\mathrm{X}_{2}^{2-}\\right)+3 \\mathrm{n}\\left(\\mathrm{X}_{3}^{2-}\\right)=1.0 \\times 10^{-3} \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 向 $100 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{X}$ 溶液中加入过量固体单质 $\\mathrm{X}$, 用 $\\mathrm{NaOH}$ 或 $\\mathrm{HCl}$调节溶液 $\\mathrm{pH}$ (不考虑温度和体积变化, 无气体逸出), 平衡后溶液中 $\\mathrm{X}$ 微粒 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right.$ 、 $\\left.\\mathrm{HX}^{-} 、 \\mathrm{X}^{2-} 、 \\mathrm{X}_{2}^{2-} 、 \\mathrm{X}_{3}^{2-}\\right)$ 的分布系数 $\\delta\\left[\\right.$ 如 $\\left.\\delta\\left(\\mathrm{X}_{2}^{2-}\\right)=\\frac{\\mathrm{c}\\left(\\mathrm{X}_{2}^{2-}\\right)}{\\mathrm{c}_{\\text {总 }}(\\text { 含 } \\mathrm{X} \\text { 微粒 })}\\right]$ 与 $\\mathrm{pH}$ 的关系如图, 下列说法错误的是\n\n已知: $\\mathrm{X}(\\mathrm{s})+\\mathrm{X}^{2-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{X}_{2}^{2-}(\\mathrm{aq}) K_{1}, \\mathrm{X}(\\mathrm{s})+\\mathrm{X}_{2}^{2-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{X}_{3}^{2-}(\\mathrm{aq}) K_{2},\\left(K_{1}=K_{2}>1\\right)$ 。\n\n[图1]\n\nA: 曲线 $\\mathrm{c}$ 表示 $\\mathrm{X}_{3}^{2-}$\nB: $\\mathrm{p} K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)=6.9, \\mathrm{p} K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)=12.2$\nC: $K_{1}=K_{2}=10$\nD: $\\mathrm{pH}=11$ 时, $\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)+\\mathrm{n}\\left(\\mathrm{HX}^{-}\\right)+\\mathrm{n}\\left(\\mathrm{X}^{2-}\\right)+2 \\mathrm{n}\\left(\\mathrm{X}_{2}^{2-}\\right)+3 \\mathrm{n}\\left(\\mathrm{X}_{3}^{2-}\\right)=1.0 \\times 10^{-3} \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-042.jpg?height=491&width=737&top_left_y=1699&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-043.jpg?height=137&width=1190&top_left_y=600&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-043.jpg?height=180&width=1380&top_left_y=932&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_28",
        "problem": "Equal volumes of aqueous $1.00 \\mathrm{~m}$ glucose $\\left(\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\\right)$ and $1.00 \\mathrm{~m}$ sodium chloride solutions are placed on opposite sides of a U-tube, separated by a semipermeable membrane (through which only water can diffuse). What will the setup look like at equilibrium?\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEqual volumes of aqueous $1.00 \\mathrm{~m}$ glucose $\\left(\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\\right)$ and $1.00 \\mathrm{~m}$ sodium chloride solutions are placed on opposite sides of a U-tube, separated by a semipermeable membrane (through which only water can diffuse). What will the setup look like at equilibrium?\n\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=257&width=238&top_left_y=782&top_left_x=1255",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=266&width=244&top_left_y=1062&top_left_x=1252",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=268&width=260&top_left_y=777&top_left_x=1667",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=274&width=260&top_left_y=1061&top_left_x=1667"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_396",
        "problem": "The critical point of carbon dioxide is $304 \\mathrm{~K}$ and $73 \\mathrm{~atm}$. Under which conditions is carbon dioxide a liquid?\nI. $303 \\mathrm{~K}$ and $73 \\mathrm{~atm}$\nII. $305 \\mathrm{~K}$ and $74 \\mathrm{~atm}$\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe critical point of carbon dioxide is $304 \\mathrm{~K}$ and $73 \\mathrm{~atm}$. Under which conditions is carbon dioxide a liquid?\nI. $303 \\mathrm{~K}$ and $73 \\mathrm{~atm}$\nII. $305 \\mathrm{~K}$ and $74 \\mathrm{~atm}$\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1444",
        "problem": "Potentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nCalculate the ratio of the stability constants $\\beta\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3}\\right] / \\beta\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nPotentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nCalculate the ratio of the stability constants $\\beta\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3}\\right] / \\beta\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_680",
        "problem": "常温下, 将 $\\mathrm{CaCO}_{3}(\\mathrm{~s})$ 或 $\\mathrm{ZnCO}_{3}(\\mathrm{~s})$ 悬浊液置于分压固定的 $\\mathrm{CO}_{2}$ 气相中, 体系中\n\n$\\mathrm{pM} \\sim \\mathrm{pH}$ 关系如图所示, 已知 $\\mathrm{pM}=-\\operatorname{lgc}(\\mathrm{M}), \\mathrm{M}$ 为溶液中 $\\mathrm{H}_{2} \\mathrm{CO}_{3} 、 \\mathrm{HCO}_{3}^{-} 、 \\mathrm{CO}_{3}^{2-} 、 \\mathrm{Ca}^{2+} 、$\n\n$\\mathrm{Zn}^{2+}$ 的浓度, $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)>\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{ZnCO}_{3}\\right)$ 。下列说法正确的是\n\n[图1]\nA: $\\mathrm{L}_{1}$ 表示 $\\mathrm{Pc}\\left(\\mathrm{CO}_{3}^{2-}\\right)-\\mathrm{pH}$ 曲线\nB: 向 $\\mathrm{a}$ 点对应的 $\\mathrm{ZnCO}_{3}$ 悬浊液中加入 $\\mathrm{ZnCl}_{2}$ 固体, 可以达到 $b$ 点\nC: $\\frac{\\mathrm{c}^{2}\\left(\\mathrm{HCO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)}=1 \\times 10^{3.9}$\nD: $\\mathrm{CaCO}_{3}(\\mathrm{~s})+\\mathrm{Zn}^{2+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{ZnCO}_{3}(\\mathrm{~s})+\\mathrm{Ca}^{2+}(\\mathrm{aq}) K=1.0 \\times 10^{2.2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 将 $\\mathrm{CaCO}_{3}(\\mathrm{~s})$ 或 $\\mathrm{ZnCO}_{3}(\\mathrm{~s})$ 悬浊液置于分压固定的 $\\mathrm{CO}_{2}$ 气相中, 体系中\n\n$\\mathrm{pM} \\sim \\mathrm{pH}$ 关系如图所示, 已知 $\\mathrm{pM}=-\\operatorname{lgc}(\\mathrm{M}), \\mathrm{M}$ 为溶液中 $\\mathrm{H}_{2} \\mathrm{CO}_{3} 、 \\mathrm{HCO}_{3}^{-} 、 \\mathrm{CO}_{3}^{2-} 、 \\mathrm{Ca}^{2+} 、$\n\n$\\mathrm{Zn}^{2+}$ 的浓度, $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)>\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{ZnCO}_{3}\\right)$ 。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{L}_{1}$ 表示 $\\mathrm{Pc}\\left(\\mathrm{CO}_{3}^{2-}\\right)-\\mathrm{pH}$ 曲线\nB: 向 $\\mathrm{a}$ 点对应的 $\\mathrm{ZnCO}_{3}$ 悬浊液中加入 $\\mathrm{ZnCl}_{2}$ 固体, 可以达到 $b$ 点\nC: $\\frac{\\mathrm{c}^{2}\\left(\\mathrm{HCO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)}=1 \\times 10^{3.9}$\nD: $\\mathrm{CaCO}_{3}(\\mathrm{~s})+\\mathrm{Zn}^{2+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{ZnCO}_{3}(\\mathrm{~s})+\\mathrm{Ca}^{2+}(\\mathrm{aq}) K=1.0 \\times 10^{2.2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-020.jpg?height=631&width=662&top_left_y=1164&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-021.jpg?height=72&width=1373&top_left_y=1740&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1376",
        "problem": "The absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nPure aqueous solutions of $\\mathrm{HX}$ were used. Only the anionic species $\\mathrm{X}^{-}$absorb.\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nPure aqueous solutions of $\\mathrm{HX}$ were used. Only the anionic species $\\mathrm{X}^{-}$absorb.\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-463.jpg?height=454&width=777&top_left_y=321&top_left_x=545"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_532",
        "problem": "室温下, 改变二元弱酸 $\\mathrm{H}_{2} \\mathrm{SeO}_{3}$ 溶液的 $\\mathrm{pH}$, 溶液中 $\\mathrm{H}_{2} \\mathrm{SeO}_{3} 、 \\mathrm{HSeO}_{3}^{-} 、 \\mathrm{SeO}_{3}^{2-}$ 的物质的量分数 $\\delta(\\mathrm{x})$ 随 $\\mathrm{pH}$ 变化如图所示 $\\left[\\delta(\\mathrm{x})=\\frac{\\mathrm{c}(\\mathrm{x})}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SeO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HSeO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{SeO}_{3}^{2-}\\right)}\\right]$ 。下列说法错误的是( )\n\n[图1]\nA: $\\mathrm{H}_{2} \\mathrm{SeO}_{3}$, 的 $K_{a l}=10^{-2.62}$\nB: $\\mathrm{C}$ 点的 $\\mathrm{pH}$ 为 5.47\nC: $\\mathrm{NaHSeO}_{3}$ 溶液中离子浓度的大小顺序为: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HSeO}_{3}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{SeO}_{3}^{2-}\\right)>$ $c\\left(\\mathrm{OH}^{-}\\right)$\nD: 向 $100 \\mathrm{~mL} \\mathrm{pH}=2.62$, 含 $c\\left(\\mathrm{H}_{2} \\mathrm{SeO}_{3}\\right)=0.05 \\mathrm{~mol} / \\mathrm{L}, c\\left(\\mathrm{HSeO}_{3}^{-}\\right)=0.05 \\mathrm{~mol} / \\mathrm{L}$ 的溶液中, 加入 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaOH} \\mathrm{100mL,} c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{SeO}_{3}\\right)<c\\left(\\mathrm{SeO}_{3}^{2-}\\right)+0.05 \\mathrm{~mol} / \\mathrm{L}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 改变二元弱酸 $\\mathrm{H}_{2} \\mathrm{SeO}_{3}$ 溶液的 $\\mathrm{pH}$, 溶液中 $\\mathrm{H}_{2} \\mathrm{SeO}_{3} 、 \\mathrm{HSeO}_{3}^{-} 、 \\mathrm{SeO}_{3}^{2-}$ 的物质的量分数 $\\delta(\\mathrm{x})$ 随 $\\mathrm{pH}$ 变化如图所示 $\\left[\\delta(\\mathrm{x})=\\frac{\\mathrm{c}(\\mathrm{x})}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SeO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HSeO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{SeO}_{3}^{2-}\\right)}\\right]$ 。下列说法错误的是( )\n\n[图1]\n\nA: $\\mathrm{H}_{2} \\mathrm{SeO}_{3}$, 的 $K_{a l}=10^{-2.62}$\nB: $\\mathrm{C}$ 点的 $\\mathrm{pH}$ 为 5.47\nC: $\\mathrm{NaHSeO}_{3}$ 溶液中离子浓度的大小顺序为: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HSeO}_{3}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{SeO}_{3}^{2-}\\right)>$ $c\\left(\\mathrm{OH}^{-}\\right)$\nD: 向 $100 \\mathrm{~mL} \\mathrm{pH}=2.62$, 含 $c\\left(\\mathrm{H}_{2} \\mathrm{SeO}_{3}\\right)=0.05 \\mathrm{~mol} / \\mathrm{L}, c\\left(\\mathrm{HSeO}_{3}^{-}\\right)=0.05 \\mathrm{~mol} / \\mathrm{L}$ 的溶液中, 加入 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaOH} \\mathrm{100mL,} c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{SeO}_{3}\\right)<c\\left(\\mathrm{SeO}_{3}^{2-}\\right)+0.05 \\mathrm{~mol} / \\mathrm{L}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-16.jpg?height=335&width=465&top_left_y=152&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_852",
        "problem": "羟基自由基 $(\\cdot \\mathrm{OH})$ 是自然界中氧化性仅次于氟的氧化剂。我国科学家设计了一种能将苯酚氧化为 $\\mathrm{CO}_{2}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 的原电池-电解池组合装置(如下图所示), 该装置能实现发电、环保二位一体。下列说法错误的是\n\n[图1]\nA: 系统工作时，电流由 $\\mathrm{b}$ 极经 III、II、I 室流向 $\\mathrm{a}$ 极\nB: $\\mathrm{d}$ 极区苯酚被氧化的化学方程式为 $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OH}+28 \\cdot \\mathrm{OH}=6 \\mathrm{CO}_{2} \\uparrow+17 \\mathrm{H}_{2} \\mathrm{O}$\nC: 系统工作时，每转移 $28 \\mathrm{~mol} \\mathrm{e}$-消耗 $1 \\mathrm{~mol}$ 苯酚\nD: $\\mathrm{a}$ 电极的电极反应式: $7 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}=2 \\mathrm{Cr}(\\mathrm{OH})_{3}+8 \\mathrm{OH}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n羟基自由基 $(\\cdot \\mathrm{OH})$ 是自然界中氧化性仅次于氟的氧化剂。我国科学家设计了一种能将苯酚氧化为 $\\mathrm{CO}_{2}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 的原电池-电解池组合装置(如下图所示), 该装置能实现发电、环保二位一体。下列说法错误的是\n\n[图1]\n\nA: 系统工作时，电流由 $\\mathrm{b}$ 极经 III、II、I 室流向 $\\mathrm{a}$ 极\nB: $\\mathrm{d}$ 极区苯酚被氧化的化学方程式为 $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OH}+28 \\cdot \\mathrm{OH}=6 \\mathrm{CO}_{2} \\uparrow+17 \\mathrm{H}_{2} \\mathrm{O}$\nC: 系统工作时，每转移 $28 \\mathrm{~mol} \\mathrm{e}$-消耗 $1 \\mathrm{~mol}$ 苯酚\nD: $\\mathrm{a}$ 电极的电极反应式: $7 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}=2 \\mathrm{Cr}(\\mathrm{OH})_{3}+8 \\mathrm{OH}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-062.jpg?height=505&width=1325&top_left_y=1609&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_195",
        "problem": "Vitamin-B3, or Niacin, is an essential nutrient for humans. Its molecular structure is shown on the right. A $0.0050 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solution of Niacin in water has a $\\mathrm{pH}$ of 3.56 . What is the percentage ionization of niacin in water?\nA: $1.4 \\%$\nB: $2.8 \\%$\nC: $3.6 \\%$\nD: $5.5 \\%$\nE: $7.8 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nVitamin-B3, or Niacin, is an essential nutrient for humans. Its molecular structure is shown on the right. A $0.0050 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solution of Niacin in water has a $\\mathrm{pH}$ of 3.56 . What is the percentage ionization of niacin in water?\n\nA: $1.4 \\%$\nB: $2.8 \\%$\nC: $3.6 \\%$\nD: $5.5 \\%$\nE: $7.8 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1155",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidations states of elements from Groups 7 and 17 (except fluorine) are +7 , what are the valence electrons of bromine?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidations states of elements from Groups 7 and 17 (except fluorine) are +7 , what are the valence electrons of bromine?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1520",
        "problem": "For sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.For how many minutes will this system supply hydrogen gas at this rate?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nFor sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.\n\nproblem:\nFor how many minutes will this system supply hydrogen gas at this rate?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of min, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "min"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_143",
        "problem": "Hydrofluoric acid reacts with potassium hydroxide according to the net ionic equation:\n\n$$\n\\mathrm{HF}(\\mathrm{aq})+\\mathrm{OH}^{-}(\\mathrm{aq}) \\rightarrow \\mathrm{F}^{-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})\n$$\n\nIf $5.0 \\mathrm{~mL}$ of $1.0 \\mathrm{M} \\mathrm{KOH}$ are added to $25.0 \\mathrm{~mL}$ of $2.0 \\mathrm{M} \\mathrm{HF}$, what is the $\\mathrm{pH}$ of the resulting solution? The Ka of HF is $7.2 \\times 10^{-4}$.\nA: 1.42\nB: 1.48\nC: 1.66\nD: 2.19\nE: 2.84\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHydrofluoric acid reacts with potassium hydroxide according to the net ionic equation:\n\n$$\n\\mathrm{HF}(\\mathrm{aq})+\\mathrm{OH}^{-}(\\mathrm{aq}) \\rightarrow \\mathrm{F}^{-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})\n$$\n\nIf $5.0 \\mathrm{~mL}$ of $1.0 \\mathrm{M} \\mathrm{KOH}$ are added to $25.0 \\mathrm{~mL}$ of $2.0 \\mathrm{M} \\mathrm{HF}$, what is the $\\mathrm{pH}$ of the resulting solution? The Ka of HF is $7.2 \\times 10^{-4}$.\n\nA: 1.42\nB: 1.48\nC: 1.66\nD: 2.19\nE: 2.84\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_747",
        "problem": "一种可充放电的铝离子电池工作原理如图所示, 电解质为 $\\mathrm{Al}_{\\mathrm{x}} \\mathrm{O}_{\\mathrm{y}}$ 离子液体, $\\mathrm{CuS}$ 在电池反应后转化为 $\\mathrm{Cu}_{2} \\mathrm{~S}$ 和 $\\mathrm{Al}_{2} \\mathrm{~S}_{3}$ 。下列说法正确的是\n\n[图1]\nA: 若 $\\mathrm{CuS}$ 从电极表面脱落, 则电池单位质量释放电量减少\nB: 该电池放电时, 正极反应为 $6 \\mathrm{CuS}+8 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}+6 \\mathrm{e}-=3 \\mathrm{Cu}_{2} \\mathrm{~S}+\\mathrm{Al}_{2} \\mathrm{~S}_{3}+14 \\mathrm{AlCl}_{4}^{-}$\nC: 为提高电池效率, 可以向 $\\mathrm{CuS} @ \\mathrm{C}$ 电极附近加入适量 $\\mathrm{AlCl}_{3}$ 水溶液\nD: 充电时电池负极的反应为 $\\mathrm{Al}+7 \\mathrm{AlCl}_{4}^{-}-3 \\mathrm{e}^{-}=4 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种可充放电的铝离子电池工作原理如图所示, 电解质为 $\\mathrm{Al}_{\\mathrm{x}} \\mathrm{O}_{\\mathrm{y}}$ 离子液体, $\\mathrm{CuS}$ 在电池反应后转化为 $\\mathrm{Cu}_{2} \\mathrm{~S}$ 和 $\\mathrm{Al}_{2} \\mathrm{~S}_{3}$ 。下列说法正确的是\n\n[图1]\n\nA: 若 $\\mathrm{CuS}$ 从电极表面脱落, 则电池单位质量释放电量减少\nB: 该电池放电时, 正极反应为 $6 \\mathrm{CuS}+8 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}+6 \\mathrm{e}-=3 \\mathrm{Cu}_{2} \\mathrm{~S}+\\mathrm{Al}_{2} \\mathrm{~S}_{3}+14 \\mathrm{AlCl}_{4}^{-}$\nC: 为提高电池效率, 可以向 $\\mathrm{CuS} @ \\mathrm{C}$ 电极附近加入适量 $\\mathrm{AlCl}_{3}$ 水溶液\nD: 充电时电池负极的反应为 $\\mathrm{Al}+7 \\mathrm{AlCl}_{4}^{-}-3 \\mathrm{e}^{-}=4 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-69.jpg?height=354&width=765&top_left_y=1953&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_967",
        "problem": "室温下, 向体积均为 $20 \\mathrm{~mL}$ 且浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HX}$ 溶液、 $\\mathrm{HY}$ 溶液中分别滴加同浓度的 $\\mathrm{NaOH}$ 溶液, 反应后溶液中水电离出的 $c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$的负对数 $\\left[-\\lg c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)\\right]$与滴加氢氧化钠溶液体积的关系如图所示。下列说法正确的是\n\n[图1]\nA: 室温时, $K_{\\mathrm{a}}(\\mathrm{HX})=10^{-6}$\nB: $a 、 b 、 c$ 三点对应溶液均显中性\nC: b、d 两点对应溶液的阴离子总浓度不同\nD: $\\mathrm{c}$ 点对应溶液中 $c\\left(\\mathrm{X}^{-}\\right)+c(\\mathrm{HX})=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 向体积均为 $20 \\mathrm{~mL}$ 且浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HX}$ 溶液、 $\\mathrm{HY}$ 溶液中分别滴加同浓度的 $\\mathrm{NaOH}$ 溶液, 反应后溶液中水电离出的 $c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$的负对数 $\\left[-\\lg c_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)\\right]$与滴加氢氧化钠溶液体积的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: 室温时, $K_{\\mathrm{a}}(\\mathrm{HX})=10^{-6}$\nB: $a 、 b 、 c$ 三点对应溶液均显中性\nC: b、d 两点对应溶液的阴离子总浓度不同\nD: $\\mathrm{c}$ 点对应溶液中 $c\\left(\\mathrm{X}^{-}\\right)+c(\\mathrm{HX})=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-028.jpg?height=580&width=817&top_left_y=818&top_left_x=331"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_641",
        "problem": "由环戊二烯 $(\\mathrm{a})$ 制备金刚烷(d)的合成路线如图所示. 下列说法正确的是\n\n[图1]\nA: 反应(1)为加成反应\nB: 含苯环且苯环上只有两个链状取代基的 $\\mathrm{b}$ 的同分异构体有 12 种(不考虑顺反异构)\nC: $\\mathrm{b}$ 和 $\\mathrm{d}$ 中手性碳原子的个数均为 4\nD: $\\mathrm{c}$ 与 $\\mathrm{d}$ 的结构相似，两者互为同系物\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n由环戊二烯 $(\\mathrm{a})$ 制备金刚烷(d)的合成路线如图所示. 下列说法正确的是\n\n[图1]\n\nA: 反应(1)为加成反应\nB: 含苯环且苯环上只有两个链状取代基的 $\\mathrm{b}$ 的同分异构体有 12 种(不考虑顺反异构)\nC: $\\mathrm{b}$ 和 $\\mathrm{d}$ 中手性碳原子的个数均为 4\nD: $\\mathrm{c}$ 与 $\\mathrm{d}$ 的结构相似，两者互为同系物\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-17.jpg?height=214&width=1142&top_left_y=1646&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_787",
        "problem": "一种新型 $\\mathrm{AC} / \\mathrm{LiMn}_{2} \\mathrm{O}_{4}$ 体系, 在快速启动、电动车等领域具有广阔应用前景。其采用尖晶石结构的 $\\mathrm{LiMn}_{2} \\mathrm{O}_{4}$ 作正极(可由 $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{MnO}_{2}$ 按物质的量比 $1: 2$ 反应合成), 高比表面积活性炭 $\\mathrm{AC}$ (石墨颗粒组成)作负极, $\\mathrm{Li}_{2} \\mathrm{SO}_{4}$ 作电解液, 充电、放电的过程如图所示:\n[图1]\n\n下列说法正确的是\nA: 合成 $\\mathrm{LiMn}_{2} \\mathrm{O}_{4}$ 的过程中可能有 $\\mathrm{O}_{2}$ 产生\nB: 放电时正极的电极反应式为: $\\mathrm{LiMn}_{2} \\mathrm{O}_{4}+\\mathrm{xe}^{-}=\\mathrm{Li}_{(1-\\mathrm{x})} \\mathrm{Mn}_{2} \\mathrm{O}_{4}+\\mathrm{xLi}^{+}$\nC: 充电时 $\\mathrm{AC}$ 极应与电源负极相连\nD: 可以用 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 代替 $\\mathrm{Li}_{2} \\mathrm{SO}_{4}$ 作电解液\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种新型 $\\mathrm{AC} / \\mathrm{LiMn}_{2} \\mathrm{O}_{4}$ 体系, 在快速启动、电动车等领域具有广阔应用前景。其采用尖晶石结构的 $\\mathrm{LiMn}_{2} \\mathrm{O}_{4}$ 作正极(可由 $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{MnO}_{2}$ 按物质的量比 $1: 2$ 反应合成), 高比表面积活性炭 $\\mathrm{AC}$ (石墨颗粒组成)作负极, $\\mathrm{Li}_{2} \\mathrm{SO}_{4}$ 作电解液, 充电、放电的过程如图所示:\n[图1]\n\n下列说法正确的是\n\nA: 合成 $\\mathrm{LiMn}_{2} \\mathrm{O}_{4}$ 的过程中可能有 $\\mathrm{O}_{2}$ 产生\nB: 放电时正极的电极反应式为: $\\mathrm{LiMn}_{2} \\mathrm{O}_{4}+\\mathrm{xe}^{-}=\\mathrm{Li}_{(1-\\mathrm{x})} \\mathrm{Mn}_{2} \\mathrm{O}_{4}+\\mathrm{xLi}^{+}$\nC: 充电时 $\\mathrm{AC}$ 极应与电源负极相连\nD: 可以用 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 代替 $\\mathrm{Li}_{2} \\mathrm{SO}_{4}$ 作电解液\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-16.jpg?height=454&width=1372&top_left_y=1932&top_left_x=366"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_335",
        "problem": "An aqueous solution is $5.0 \\%$ ethanoic acid $\\left(\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}\\right)$ by mass and its density is $0.96 \\mathrm{~g} \\mathrm{~mL}^{-1}$. What is the molar concentration of ethanoic acid in this solution?\n$\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}, 60.05 \\mathrm{~g} \\mathrm{~mol}^{-1}$\nA: $0.80 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $4.8 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $12 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $\\quad 0.087 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $16 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn aqueous solution is $5.0 \\%$ ethanoic acid $\\left(\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}\\right)$ by mass and its density is $0.96 \\mathrm{~g} \\mathrm{~mL}^{-1}$. What is the molar concentration of ethanoic acid in this solution?\n$\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}, 60.05 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nA: $0.80 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $4.8 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $12 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $\\quad 0.087 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $16 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_625",
        "problem": "常温下, 将一定浓度的 $\\mathrm{AgNO}_{3}$ 溶液分别滴加到浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KCl} 、 \\mathrm{~K}_{2} \\mathrm{CrO}_{4}$\n\n[图1]\n\n[图2]\nA: (1)、(2)分别代表 $\\mathrm{AgCl} 、 \\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$\nB: $\\mathrm{K}_{\\text {sp }}\\left[\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right]=1 \\times 10^{-9}$\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right]<\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgCl})$, 说明 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 的溶解度更小\nD: $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}(\\mathrm{~s})+2 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons 2 \\mathrm{AgCl}(\\mathrm{s})+\\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})$ 的 $\\mathrm{K} \\approx 10^{7.4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 将一定浓度的 $\\mathrm{AgNO}_{3}$ 溶液分别滴加到浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KCl} 、 \\mathrm{~K}_{2} \\mathrm{CrO}_{4}$\n\n[图1]\n\n[图2]\n\nA: (1)、(2)分别代表 $\\mathrm{AgCl} 、 \\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$\nB: $\\mathrm{K}_{\\text {sp }}\\left[\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right]=1 \\times 10^{-9}$\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right]<\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgCl})$, 说明 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 的溶解度更小\nD: $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}(\\mathrm{~s})+2 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons 2 \\mathrm{AgCl}(\\mathrm{s})+\\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})$ 的 $\\mathrm{K} \\approx 10^{7.4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-065.jpg?height=69&width=1382&top_left_y=2233&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-066.jpg?height=699&width=894&top_left_y=153&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_630",
        "problem": "常温下, 在含有 $\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{~s})$ 的溶液中, 一定 $\\mathrm{pH}$ 范围内, 存在以下平衡:\n\n$\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq}) ; \\quad \\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{~A}^{-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) ;$\n\n$\\mathrm{H}_{2} \\mathrm{~A}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{HA}^{2-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) ; \\mathrm{HA}^{2-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{A}^{3-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq})$ 。平衡常数依次为 $K_{0}$ 、\n\n$K_{1} 、 K_{2} 、 K_{3}$ 。已知 $\\lg c\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right) 、 \\lg c\\left(\\mathrm{HA}^{2-}\\right) 、 \\lg c\\left(\\mathrm{~A}^{3-}\\right)$ 随 $\\mathrm{pH}$ 变化关系如图所示。下列说法错误的是\n\n[图1]\nA: $\\mathrm{pH}$ 增大时, $\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq})$ 的浓度保持不变\nB: 直线 $\\mathrm{L}$ 表示 $\\lg c\\left(\\mathrm{~A}^{3-}\\right)$ 的变化情况\nC: $-\\lg K_{2} K_{3}=2 \\mathrm{a}$\nD: $\\mathrm{pH}=\\mathrm{a}$ 时, $c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+2 c\\left(\\mathrm{HA}^{2-}\\right)+4 c\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 在含有 $\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{~s})$ 的溶液中, 一定 $\\mathrm{pH}$ 范围内, 存在以下平衡:\n\n$\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq}) ; \\quad \\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{~A}^{-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) ;$\n\n$\\mathrm{H}_{2} \\mathrm{~A}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{HA}^{2-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) ; \\mathrm{HA}^{2-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{A}^{3-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq})$ 。平衡常数依次为 $K_{0}$ 、\n\n$K_{1} 、 K_{2} 、 K_{3}$ 。已知 $\\lg c\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right) 、 \\lg c\\left(\\mathrm{HA}^{2-}\\right) 、 \\lg c\\left(\\mathrm{~A}^{3-}\\right)$ 随 $\\mathrm{pH}$ 变化关系如图所示。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{pH}$ 增大时, $\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq})$ 的浓度保持不变\nB: 直线 $\\mathrm{L}$ 表示 $\\lg c\\left(\\mathrm{~A}^{3-}\\right)$ 的变化情况\nC: $-\\lg K_{2} K_{3}=2 \\mathrm{a}$\nD: $\\mathrm{pH}=\\mathrm{a}$ 时, $c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+2 c\\left(\\mathrm{HA}^{2-}\\right)+4 c\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-031.jpg?height=480&width=460&top_left_y=1099&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_473",
        "problem": "已知: $\\mathrm{CH}_{4}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CS}_{2}(\\mathrm{~g})+4 \\mathrm{H}_{2}(\\mathrm{~g})$ 。向恒容密闭容器中充入 $0.1 \\mathrm{molCH}_{4}$ 和 $0.2 \\mathrm{molH}_{2} \\mathrm{~S}$, 下图所示:\n\n[图1]\n\n下列说法正确的是\nA: 该反应的 $\\triangle \\mathrm{H}<\\mathrm{O}$\nB: $\\mathrm{X}$ 点 $\\mathrm{CH}_{4}$ 的转化率为 $20 \\%$\nC: X 点与 $Y$ 点容器内压强比为 $55: 51$\nD: 维持 $\\mathrm{Z}$ 点温度, 向容器中再充入 $\\mathrm{CH}_{4} 、 \\mathrm{H}_{2} \\mathrm{~S} 、 \\mathrm{CS}_{2} 、 \\mathrm{H}_{2}$ 各 $0.1 \\mathrm{~mol}$ 时 $\\mathrm{v}($ 正 $)<\\mathrm{v}($ 逆 $)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知: $\\mathrm{CH}_{4}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CS}_{2}(\\mathrm{~g})+4 \\mathrm{H}_{2}(\\mathrm{~g})$ 。向恒容密闭容器中充入 $0.1 \\mathrm{molCH}_{4}$ 和 $0.2 \\mathrm{molH}_{2} \\mathrm{~S}$, 下图所示:\n\n[图1]\n\n下列说法正确的是\n\nA: 该反应的 $\\triangle \\mathrm{H}<\\mathrm{O}$\nB: $\\mathrm{X}$ 点 $\\mathrm{CH}_{4}$ 的转化率为 $20 \\%$\nC: X 点与 $Y$ 点容器内压强比为 $55: 51$\nD: 维持 $\\mathrm{Z}$ 点温度, 向容器中再充入 $\\mathrm{CH}_{4} 、 \\mathrm{H}_{2} \\mathrm{~S} 、 \\mathrm{CS}_{2} 、 \\mathrm{H}_{2}$ 各 $0.1 \\mathrm{~mol}$ 时 $\\mathrm{v}($ 正 $)<\\mathrm{v}($ 逆 $)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-098.jpg?height=551&width=759&top_left_y=758&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1012",
        "problem": "A calcium chloride solution was prepared by dissolving $11.00 \\mathrm{~g} \\mathrm{CaCl}_{2}$ in water to make $500 \\mathrm{~mL}$ of solution. What is the correct way to report the concentration of this solution?\nThe number of significant figures in the volume is ambiguous (i.e. we don't know if the zeros are significant), so we must assume the worst: that the volume is known only to 1 significant figure. Therefore, we report the concentration to 1 significant figure only. If the volume had been recorded as $0.500 \\mathrm{~L}$, then we could have given the concentration as $0.198 \\mathrm{~mol} \\mathrm{~L}^{-1}$.\n$\\mathrm{CaCl}_{2}, 110.98 \\mathrm{~g} \\mathrm{~mol}^{-1}$\nA: $0.2 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.1982 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.198 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $0.2000 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $0.20 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA calcium chloride solution was prepared by dissolving $11.00 \\mathrm{~g} \\mathrm{CaCl}_{2}$ in water to make $500 \\mathrm{~mL}$ of solution. What is the correct way to report the concentration of this solution?\nThe number of significant figures in the volume is ambiguous (i.e. we don't know if the zeros are significant), so we must assume the worst: that the volume is known only to 1 significant figure. Therefore, we report the concentration to 1 significant figure only. If the volume had been recorded as $0.500 \\mathrm{~L}$, then we could have given the concentration as $0.198 \\mathrm{~mol} \\mathrm{~L}^{-1}$.\n$\\mathrm{CaCl}_{2}, 110.98 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nA: $0.2 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.1982 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.198 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $0.2000 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $0.20 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1414",
        "problem": "One of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nOne member of thorium series, after isolation, is found to contain $1.50 \\times 10^{10}$ atoms of the nuclide and decays at the rate of 3440 disintegrations per minute. What is the half-life in years?\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOne of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nOne member of thorium series, after isolation, is found to contain $1.50 \\times 10^{10}$ atoms of the nuclide and decays at the rate of 3440 disintegrations per minute. What is the half-life in years?\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_91",
        "problem": "Two atomic orbitals have the same shape and orientation but have a different number of nodes. Which statement about the quantum numbers of these orbitals is correct?\nA: They have the same value of $l$ but different values of $n$.\nB: They have the same value of $l$ but different values of $m_{l}$.\nC: They have the same value of $m_{l}$ but different values of $l$.\nD: They have the same value of $m_{s}$ but different values of $m_{l}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTwo atomic orbitals have the same shape and orientation but have a different number of nodes. Which statement about the quantum numbers of these orbitals is correct?\n\nA: They have the same value of $l$ but different values of $n$.\nB: They have the same value of $l$ but different values of $m_{l}$.\nC: They have the same value of $m_{l}$ but different values of $l$.\nD: They have the same value of $m_{s}$ but different values of $m_{l}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1337",
        "problem": "Using the information provided on this graph, give numerical answers with appropriate units to the following questions:\n\n[figure1]\n\nWhat is the ionisation energy of the $\\mathrm{H}$ atom?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nUsing the information provided on this graph, give numerical answers with appropriate units to the following questions:\n\n[figure1]\n\nWhat is the ionisation energy of the $\\mathrm{H}$ atom?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-471.jpg?height=911&width=1279&top_left_y=590&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_189",
        "problem": "Cows are one of the most significant sources of methane $\\left(\\mathrm{CH}_{4}\\right)$ emitted into the atmosphere. The average cow produces 259 grams of methane per day. The lower explosive limit (LEL) is the lowest concentration at which there is a risk of an explosion. For methane, the LEL at SATP is 5.0\\% $\\mathrm{v} / \\mathrm{v}$. Assuming a constant rate of methane emission from cows, how long would it be safe to keep an average cow in a $10.0 \\mathrm{~m}^{3}$ enclosed space at $25^{\\circ} \\mathrm{C}$ and $100 \\mathrm{kPa}$ before there is a risk of explosion? The molar volume of a gas at SATP is $24.8 \\mathrm{~L}$.\nA: 30 hours\nB: 60 hours\nC: 90 hours\nD: 600 hours\nE: 900 hours\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCows are one of the most significant sources of methane $\\left(\\mathrm{CH}_{4}\\right)$ emitted into the atmosphere. The average cow produces 259 grams of methane per day. The lower explosive limit (LEL) is the lowest concentration at which there is a risk of an explosion. For methane, the LEL at SATP is 5.0\\% $\\mathrm{v} / \\mathrm{v}$. Assuming a constant rate of methane emission from cows, how long would it be safe to keep an average cow in a $10.0 \\mathrm{~m}^{3}$ enclosed space at $25^{\\circ} \\mathrm{C}$ and $100 \\mathrm{kPa}$ before there is a risk of explosion? The molar volume of a gas at SATP is $24.8 \\mathrm{~L}$.\n\nA: 30 hours\nB: 60 hours\nC: 90 hours\nD: 600 hours\nE: 900 hours\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_755",
        "problem": "乙二胺 $\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]$ 是一种二元弱碱, 在水中的电离方式与氨相似。 $25^{\\circ} \\mathrm{C}$ 时,\n\n用 $\\mathrm{HCl}$ 调节 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 乙二胺溶液的 $\\mathrm{pH}$, 溶液中含氮物种的分布系数 $(\\delta)$ 与 $\\mathrm{pH}$ 的关系\n\n如图所示(如乙二胺的分布系数 $\\delta\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]=\\frac{\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]}{0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}}$ 。下列说法错误\n\n的是\n\n[图1]\nA: 乙二胺第一步电离的方程式为 $\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{3}^{+}+\\mathrm{OH}^{-}$\nB: 曲线 $\\mathrm{L}_{3}$ 表示 $\\delta\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]$ 与 $\\mathrm{pH}$ 的关系\nC: $\\mathrm{Q}$ 点对应溶液的 $\\mathrm{pH}=8.39$\nD: $\\mathrm{M} 、 \\mathrm{~N}$ 点溶液中 $\\mathrm{c}\\left(\\mathrm{Cl}-\\mathrm{l}^{-}\\right.$相等\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n乙二胺 $\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]$ 是一种二元弱碱, 在水中的电离方式与氨相似。 $25^{\\circ} \\mathrm{C}$ 时,\n\n用 $\\mathrm{HCl}$ 调节 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 乙二胺溶液的 $\\mathrm{pH}$, 溶液中含氮物种的分布系数 $(\\delta)$ 与 $\\mathrm{pH}$ 的关系\n\n如图所示(如乙二胺的分布系数 $\\delta\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]=\\frac{\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]}{0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}}$ 。下列说法错误\n\n的是\n\n[图1]\n\nA: 乙二胺第一步电离的方程式为 $\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{3}^{+}+\\mathrm{OH}^{-}$\nB: 曲线 $\\mathrm{L}_{3}$ 表示 $\\delta\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{2} \\mathrm{NH}_{2}\\right]$ 与 $\\mathrm{pH}$ 的关系\nC: $\\mathrm{Q}$ 点对应溶液的 $\\mathrm{pH}=8.39$\nD: $\\mathrm{M} 、 \\mathrm{~N}$ 点溶液中 $\\mathrm{c}\\left(\\mathrm{Cl}-\\mathrm{l}^{-}\\right.$相等\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-069.jpg?height=631&width=1085&top_left_y=1032&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-069.jpg?height=63&width=1285&top_left_y=2587&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-069.jpg?height=54&width=1128&top_left_y=2686&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_892",
        "problem": "联氨 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 溶于水可以发生与氨类似的电离。常温下, 向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 联氨盐\n\n酸盐溶液 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{6} \\mathrm{Cl}_{2}\\right)$ 中加入 $\\mathrm{NaOH}$ 固体(假设溶液体积变化忽略不计)。体系中各含氮粒\n\n子的分布系数 $(\\delta)$ 随 $\\mathrm{pH}$ 的变化如图所示。下列说法正确的是\n\n[图1]\nA: 曲线 $\\mathrm{a}$ 可表示 $\\delta\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 随 $\\mathrm{pH}$ 的变化关系\nB: $\\mathrm{pH}=7$ 时, $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)+2 c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)$\nC: $\\mathrm{N}_{2} \\mathrm{H}_{5} \\mathrm{Cl}$ 溶液中: $c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)<c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{6}^{2+}\\right)$\nD: 常温下, 反应 $\\mathrm{N}_{2} \\mathrm{H}_{6}^{2+}+\\mathrm{N}_{2} \\mathrm{H}_{4} \\rightleftharpoons 2 \\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}$的平衡常数为 $10^{-3.1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n联氨 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 溶于水可以发生与氨类似的电离。常温下, 向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 联氨盐\n\n酸盐溶液 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{6} \\mathrm{Cl}_{2}\\right)$ 中加入 $\\mathrm{NaOH}$ 固体(假设溶液体积变化忽略不计)。体系中各含氮粒\n\n子的分布系数 $(\\delta)$ 随 $\\mathrm{pH}$ 的变化如图所示。下列说法正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{a}$ 可表示 $\\delta\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 随 $\\mathrm{pH}$ 的变化关系\nB: $\\mathrm{pH}=7$ 时, $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)+2 c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)$\nC: $\\mathrm{N}_{2} \\mathrm{H}_{5} \\mathrm{Cl}$ 溶液中: $c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{4}\\right)<c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{6}^{2+}\\right)$\nD: 常温下, 反应 $\\mathrm{N}_{2} \\mathrm{H}_{6}^{2+}+\\mathrm{N}_{2} \\mathrm{H}_{4} \\rightleftharpoons 2 \\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}$的平衡常数为 $10^{-3.1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-081.jpg?height=500&width=902&top_left_y=635&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_424",
        "problem": "均苯三甲酸是一种重要的有机三元弱酸, 可表示为 $\\mathrm{H}_{3} \\mathrm{~A}$ 。向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中加入 $\\mathrm{HCl}(\\mathrm{g})$ 或 $\\mathrm{NaOH}(\\mathrm{s})$ 时(忽略溶液体积的变化), 各微粒的分布系数 $\\delta(\\mathrm{X})$ 随溶液 $\\mathrm{pH}$变化的曲线如图所示:\n\n已知: (1) $\\left[\\delta(\\mathrm{X})=\\frac{\\mathrm{c}(\\mathrm{X})}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{3-}\\right)}, \\mathrm{X}\\right.$ 为 $\\mathrm{H}_{3} \\mathrm{~A} 、 \\mathrm{H}_{2} \\mathrm{~A}^{-} 、 \\mathrm{HA}^{2-}$ 或\n\n$\\left.\\mathrm{A}^{3-}\\right]$\n\n(2) $\\left(\\mathrm{H}_{3} \\mathrm{~A} \\quad K_{\\mathrm{a} 1}=1.0 \\times 10^{-3.1} 、 K_{\\mathrm{a} 2}=1.0 \\times 10^{-4.7} 、 K_{\\mathrm{a} 3}=1.0 \\times 10^{-6.3}\\right)$\n\n[图1]\n\n下列叙述错误的是\nA: 若用 $\\mathrm{NaOH}(\\mathrm{aq})$ 滴定 $\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq})$ 至恰好生成 $\\mathrm{NaH}_{2} \\mathrm{~A}$, 可选甲基橙做指示剂\nB: $\\mathrm{N}$ 点的 $\\mathrm{pH}_{\\mathrm{N}}=\\frac{\\mathrm{pH}_{\\mathrm{R}}+\\mathrm{pH}_{\\mathrm{S}}}{2}=5.5$\nC: 常温下, $\\mathrm{Na}_{2} \\mathrm{HA}$ 的水解常数 $K_{\\mathrm{h} 2}$ 的数量级为 $10^{-10}$\nD: $\\mathrm{R}$ 点满足: $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)>0.1 \\mathrm{~mol} / \\mathrm{L}+c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{HA}^{2-}\\right)+2 c\\left(\\mathrm{~A}^{3-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n均苯三甲酸是一种重要的有机三元弱酸, 可表示为 $\\mathrm{H}_{3} \\mathrm{~A}$ 。向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中加入 $\\mathrm{HCl}(\\mathrm{g})$ 或 $\\mathrm{NaOH}(\\mathrm{s})$ 时(忽略溶液体积的变化), 各微粒的分布系数 $\\delta(\\mathrm{X})$ 随溶液 $\\mathrm{pH}$变化的曲线如图所示:\n\n已知: (1) $\\left[\\delta(\\mathrm{X})=\\frac{\\mathrm{c}(\\mathrm{X})}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{3-}\\right)}, \\mathrm{X}\\right.$ 为 $\\mathrm{H}_{3} \\mathrm{~A} 、 \\mathrm{H}_{2} \\mathrm{~A}^{-} 、 \\mathrm{HA}^{2-}$ 或\n\n$\\left.\\mathrm{A}^{3-}\\right]$\n\n(2) $\\left(\\mathrm{H}_{3} \\mathrm{~A} \\quad K_{\\mathrm{a} 1}=1.0 \\times 10^{-3.1} 、 K_{\\mathrm{a} 2}=1.0 \\times 10^{-4.7} 、 K_{\\mathrm{a} 3}=1.0 \\times 10^{-6.3}\\right)$\n\n[图1]\n\n下列叙述错误的是\n\nA: 若用 $\\mathrm{NaOH}(\\mathrm{aq})$ 滴定 $\\mathrm{H}_{3} \\mathrm{~A}(\\mathrm{aq})$ 至恰好生成 $\\mathrm{NaH}_{2} \\mathrm{~A}$, 可选甲基橙做指示剂\nB: $\\mathrm{N}$ 点的 $\\mathrm{pH}_{\\mathrm{N}}=\\frac{\\mathrm{pH}_{\\mathrm{R}}+\\mathrm{pH}_{\\mathrm{S}}}{2}=5.5$\nC: 常温下, $\\mathrm{Na}_{2} \\mathrm{HA}$ 的水解常数 $K_{\\mathrm{h} 2}$ 的数量级为 $10^{-10}$\nD: $\\mathrm{R}$ 点满足: $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)>0.1 \\mathrm{~mol} / \\mathrm{L}+c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{HA}^{2-}\\right)+2 c\\left(\\mathrm{~A}^{3-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-042.jpg?height=517&width=645&top_left_y=1141&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_810",
        "problem": "科学家 McMorris 测定和计算了关于一氯化碘 (ICl) 反应的 $\\mathrm{K}$ 值与温度关系\n$2 \\mathrm{NO}(\\mathrm{g})+2 \\mathrm{ICl}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{NOCl}(\\mathrm{g})+\\mathrm{I}_{2}(\\mathrm{~g}) \\quad \\mathrm{Kp}_{1} \\quad$ 反应 1\n\n$2 \\mathrm{NOCl}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{NO}(\\mathrm{g})+\\mathrm{Cl}_{2}(\\mathrm{~g}) \\quad \\mathrm{Kp}_{2} \\quad$ 反应 2\n\n得到两个线性关系的图像，如下图所示，则下列说法不正确的是\n\n[图1]\nA: 反应 2 焓变大于 0\nB: $2 \\mathrm{ICl}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Cl}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~g})$ 的化学平衡常数 $\\mathrm{K}=\\mathrm{Kp}_{1} \\cdot \\mathrm{Kp}_{2}$\nC: $409 \\sim 453 \\mathrm{~K}$ 的温度范围内 $\\mathrm{Kp}_{1}>\\mathrm{Kp}_{2}$\nD: $2 \\mathrm{ICl}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Cl}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~g})$ 为放热反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科学家 McMorris 测定和计算了关于一氯化碘 (ICl) 反应的 $\\mathrm{K}$ 值与温度关系\n$2 \\mathrm{NO}(\\mathrm{g})+2 \\mathrm{ICl}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{NOCl}(\\mathrm{g})+\\mathrm{I}_{2}(\\mathrm{~g}) \\quad \\mathrm{Kp}_{1} \\quad$ 反应 1\n\n$2 \\mathrm{NOCl}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{NO}(\\mathrm{g})+\\mathrm{Cl}_{2}(\\mathrm{~g}) \\quad \\mathrm{Kp}_{2} \\quad$ 反应 2\n\n得到两个线性关系的图像，如下图所示，则下列说法不正确的是\n\n[图1]\n\nA: 反应 2 焓变大于 0\nB: $2 \\mathrm{ICl}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Cl}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~g})$ 的化学平衡常数 $\\mathrm{K}=\\mathrm{Kp}_{1} \\cdot \\mathrm{Kp}_{2}$\nC: $409 \\sim 453 \\mathrm{~K}$ 的温度范围内 $\\mathrm{Kp}_{1}>\\mathrm{Kp}_{2}$\nD: $2 \\mathrm{ICl}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Cl}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~g})$ 为放热反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-051.jpg?height=817&width=1222&top_left_y=485&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1033",
        "problem": "Former Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nMany varities of Swiss cheese, such as Emmental, are famous for the holes or 'eyes' that appear in the cheese. To produce the holes another species of bacteria, Propionibacterium freudenreichii is important. This bacterium carries out the reaction of lactic acid to propanoic acid, ethanoic acid, carbon dioxide and water. The production of carbon dioxide causes the bubbles to appear.\n\n[figure3]\n\nNote: the state symbol for cheese is (ch).\n\nCarbon dioxide dissolved in cheese can exist in two forms: dissolved gaseous carbon dioxide, $\\mathrm{CO}_{2(\\mathrm{ch})}$, or dissolved hydrogen carbonate, $\\mathrm{HCO}_{3}{ }^{-}$(ch).\n\n$$\n\\mathrm{CO}_{2(\\mathrm{ch})}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{HCO}_{3}{ }^{-}{ }_{(\\mathrm{ch})}+\\mathrm{H}^{+}{ }_{(\\mathrm{ch})}, \\quad K=\\frac{\\left[\\mathrm{H}_{(\\mathrm{ch})}^{+}\\right)\\left[\\mathrm{HCO}_{3(\\mathrm{ch})}^{-}\\right]}{\\left[\\mathrm{CO}_{2(\\mathrm{ch})}\\right]}=4.47 \\times 10^{-7} \\mathrm{~mol} \\mathrm{dm}^{-3},\n$$\n\nAt the end of the fermentation, $\\left[\\mathrm{CO}_{2(\\mathrm{ch})}\\right]+\\left[\\mathrm{HCO}_{3}{ }^{-}{ }_{(\\mathrm{ch})}\\right]=3.70 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~dm}^{-3}$, and $\\mathrm{pH}=5.20$\n\nCalculate the equilibrium concentration of carbon dioxide dissolved in cheese, $\\left[\\mathrm{CO}_{2(\\mathrm{ch})}\\right]$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nFormer Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nMany varities of Swiss cheese, such as Emmental, are famous for the holes or 'eyes' that appear in the cheese. To produce the holes another species of bacteria, Propionibacterium freudenreichii is important. This bacterium carries out the reaction of lactic acid to propanoic acid, ethanoic acid, carbon dioxide and water. The production of carbon dioxide causes the bubbles to appear.\n\n[figure3]\n\nNote: the state symbol for cheese is (ch).\n\nCarbon dioxide dissolved in cheese can exist in two forms: dissolved gaseous carbon dioxide, $\\mathrm{CO}_{2(\\mathrm{ch})}$, or dissolved hydrogen carbonate, $\\mathrm{HCO}_{3}{ }^{-}$(ch).\n\n$$\n\\mathrm{CO}_{2(\\mathrm{ch})}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{HCO}_{3}{ }^{-}{ }_{(\\mathrm{ch})}+\\mathrm{H}^{+}{ }_{(\\mathrm{ch})}, \\quad K=\\frac{\\left[\\mathrm{H}_{(\\mathrm{ch})}^{+}\\right)\\left[\\mathrm{HCO}_{3(\\mathrm{ch})}^{-}\\right]}{\\left[\\mathrm{CO}_{2(\\mathrm{ch})}\\right]}=4.47 \\times 10^{-7} \\mathrm{~mol} \\mathrm{dm}^{-3},\n$$\n\nAt the end of the fermentation, $\\left[\\mathrm{CO}_{2(\\mathrm{ch})}\\right]+\\left[\\mathrm{HCO}_{3}{ }^{-}{ }_{(\\mathrm{ch})}\\right]=3.70 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~dm}^{-3}$, and $\\mathrm{pH}=5.20$\n\nCalculate the equilibrium concentration of carbon dioxide dissolved in cheese, $\\left[\\mathrm{CO}_{2(\\mathrm{ch})}\\right]$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{~dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-13.jpg?height=616&width=699&top_left_y=323&top_left_x=1178",
            "https://i.postimg.cc/Vs42tGMJ/5.png",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-14.jpg?height=528&width=785&top_left_y=747&top_left_x=1047"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{~dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1288",
        "problem": "Problem I:  Calculate the concentration of calcium ions in water in equilibrium with calcium carbonate in an atmosphere with a partial pressure of carbon dioxide of 1.00 bar.\n\nProblem II: A $0.0150 \\mathrm{~m}$ solution of calcium hydroxide is saturated with carbon dioxide gas at a partial pressure of 1.00 bar. Calculate the concentration of calcium ions in the solution by considering the equilibrium equation given above in connection with problem I.In Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nThe calcium hydroxide solution referred to in problem II is diluted to twice the volume with water before saturation with carbon dioxide gas at a partial pressure of 1.00 bar. Calculate the concentration of calcium ions in the resulting solution saturated with $\\mathrm{CO}_{2}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nProblem I:  Calculate the concentration of calcium ions in water in equilibrium with calcium carbonate in an atmosphere with a partial pressure of carbon dioxide of 1.00 bar.\n\nProblem II: A $0.0150 \\mathrm{~m}$ solution of calcium hydroxide is saturated with carbon dioxide gas at a partial pressure of 1.00 bar. Calculate the concentration of calcium ions in the solution by considering the equilibrium equation given above in connection with problem I.\n\nproblem:\nIn Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nThe calcium hydroxide solution referred to in problem II is diluted to twice the volume with water before saturation with carbon dioxide gas at a partial pressure of 1.00 bar. Calculate the concentration of calcium ions in the resulting solution saturated with $\\mathrm{CO}_{2}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-362.jpg?height=166&width=1354&top_left_y=1005&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_497",
        "problem": "已知 $\\mathrm{H}_{2} \\mathrm{~A}$ 为二元弱酸。室温时, 配制一组 $c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)=0.10 \\mathrm{~mol}^{-} \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{~A}$ 和 $\\mathrm{NaOH}$ 混合溶液,溶液中部分微粒的物质的量浓度随 $\\mathrm{pH}$ 的变化曲线如下图所示。下列指定溶液中微粒的物质的量浓度关系不正确的是\n\n[图1]\nA: $\\mathrm{pH}=7$ 的溶液中: $c\\left(\\mathrm{Na}^{+}\\right)>2 c\\left(\\mathrm{~A}^{2-}\\right)$\nB: $\\mathrm{E}$ 点溶液中: $c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{HA}^{-}\\right)<0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $c\\left(\\mathrm{Na}^{+}\\right)=0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液中: $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)$\nD: $\\mathrm{pH}=2$ 的溶液中 $c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)>c\\left(\\mathrm{HA}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知 $\\mathrm{H}_{2} \\mathrm{~A}$ 为二元弱酸。室温时, 配制一组 $c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)=0.10 \\mathrm{~mol}^{-} \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{~A}$ 和 $\\mathrm{NaOH}$ 混合溶液,溶液中部分微粒的物质的量浓度随 $\\mathrm{pH}$ 的变化曲线如下图所示。下列指定溶液中微粒的物质的量浓度关系不正确的是\n\n[图1]\n\nA: $\\mathrm{pH}=7$ 的溶液中: $c\\left(\\mathrm{Na}^{+}\\right)>2 c\\left(\\mathrm{~A}^{2-}\\right)$\nB: $\\mathrm{E}$ 点溶液中: $c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{HA}^{-}\\right)<0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $c\\left(\\mathrm{Na}^{+}\\right)=0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液中: $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)$\nD: $\\mathrm{pH}=2$ 的溶液中 $c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)>c\\left(\\mathrm{HA}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-82.jpg?height=634&width=925&top_left_y=166&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_233",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the percentage by mass of acetic acid present in the vinegar sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the percentage by mass of acetic acid present in the vinegar sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_541",
        "problem": "$500 \\mathrm{mLKNO}_{3}$ 和 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 的混合溶液中 $\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)=0.6 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 用石墨作电极电解此溶液, 当通电一段时间后, 两极均收集到 $2.24 \\mathrm{~L}$ 气体(标准状况下), 假定电解后溶液体积仍为 $500 \\mathrm{~mL}$ ，下列说法正确的是\nA: 原混合溶液中 $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)$为 $0.4 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: 上述电解过程中共转移 $0.2 \\mathrm{~mol}$ 电子\nC: 电解得到的 $\\mathrm{Cu}$ 的物质的量为 $0.05 \\mathrm{molD}$. 电解后溶液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$为 $0.4 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$500 \\mathrm{mLKNO}_{3}$ 和 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 的混合溶液中 $\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)=0.6 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 用石墨作电极电解此溶液, 当通电一段时间后, 两极均收集到 $2.24 \\mathrm{~L}$ 气体(标准状况下), 假定电解后溶液体积仍为 $500 \\mathrm{~mL}$ ，下列说法正确的是\n\nA: 原混合溶液中 $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)$为 $0.4 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: 上述电解过程中共转移 $0.2 \\mathrm{~mol}$ 电子\nC: 电解得到的 $\\mathrm{Cu}$ 的物质的量为 $0.05 \\mathrm{molD}$. 电解后溶液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$为 $0.4 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_696",
        "problem": "节能减排是保护环境的有力手段, 利用如图所示的装置处理工业尾气中的硫化氢具有明显优势。下列有关说法错误的是\n\n[图1]\nA: 该装置可将化学能转化为电能\nB: 在 $\\mathrm{Fe}^{3+}$ 存在的条件下可获得硫单质\nC: 该装置工作时, $\\mathrm{H}^{+}$从乙电极移动到甲电极\nD: 甲电极上的电极反应式为 $\\mathrm{H}_{2} \\mathrm{~S}+2 \\mathrm{H}_{2} \\mathrm{O}-6 \\mathrm{e}=\\mathrm{SO}_{2}+6 \\mathrm{H}^{+}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n节能减排是保护环境的有力手段, 利用如图所示的装置处理工业尾气中的硫化氢具有明显优势。下列有关说法错误的是\n\n[图1]\n\nA: 该装置可将化学能转化为电能\nB: 在 $\\mathrm{Fe}^{3+}$ 存在的条件下可获得硫单质\nC: 该装置工作时, $\\mathrm{H}^{+}$从乙电极移动到甲电极\nD: 甲电极上的电极反应式为 $\\mathrm{H}_{2} \\mathrm{~S}+2 \\mathrm{H}_{2} \\mathrm{O}-6 \\mathrm{e}=\\mathrm{SO}_{2}+6 \\mathrm{H}^{+}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-78.jpg?height=406&width=645&top_left_y=1930&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1560",
        "problem": "Write all reactions as balanced equations.\n\n$ \\mathrm{SO}_{2}+\\mathrm{CaCO}_{3}+\\frac{1}{2} \\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2}$\n\n$\\mathrm{SO}_{2}+\\mathrm{Ca}(\\mathrm{OH})_{2}+1 / 2 \\mathrm{O}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$Sulphur dioxide is removed from waste gases of coal power stations by washing with aqueous suspensions of calcium carbonate or calcium hydroxide. The residue formed is recovered.\n\nAssuming that the sulphur dioxide is not being removed and equally spread in an atmospheric liquid water pool of $5000 \\mathrm{~m}^{3}$ and fully returned on earth as rain, what is the expected $\\mathrm{pH}$ of the condensed water?\n\n\nNote:\n\nProtolysis of sulphur dioxide in aqueous solutions can be described by the first step dissociation of sulphurous acid. The dissociation constant $K_{a, 1}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)=10^{-2.25}$.\n\nAssume ideal gases and a constant temperature of $0^{\\circ} \\mathrm{C}$ at standard pressure.\n\n$M\\left(\\mathrm{CaCO}_{3}\\right)=100 \\mathrm{~g} \\mathrm{~mol}^{-1} ; M\\left(\\mathrm{CaSO}_{4}\\right)=172 \\mathrm{~g} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nWrite all reactions as balanced equations.\n\n$ \\mathrm{SO}_{2}+\\mathrm{CaCO}_{3}+\\frac{1}{2} \\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2}$\n\n$\\mathrm{SO}_{2}+\\mathrm{Ca}(\\mathrm{OH})_{2}+1 / 2 \\mathrm{O}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$\n\nproblem:\nSulphur dioxide is removed from waste gases of coal power stations by washing with aqueous suspensions of calcium carbonate or calcium hydroxide. The residue formed is recovered.\n\nAssuming that the sulphur dioxide is not being removed and equally spread in an atmospheric liquid water pool of $5000 \\mathrm{~m}^{3}$ and fully returned on earth as rain, what is the expected $\\mathrm{pH}$ of the condensed water?\n\n\nNote:\n\nProtolysis of sulphur dioxide in aqueous solutions can be described by the first step dissociation of sulphurous acid. The dissociation constant $K_{a, 1}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)=10^{-2.25}$.\n\nAssume ideal gases and a constant temperature of $0^{\\circ} \\mathrm{C}$ at standard pressure.\n\n$M\\left(\\mathrm{CaCO}_{3}\\right)=100 \\mathrm{~g} \\mathrm{~mol}^{-1} ; M\\left(\\mathrm{CaSO}_{4}\\right)=172 \\mathrm{~g} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1266",
        "problem": "Nitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample.\n\n$\\quad 0.2515 \\mathrm{~g}$ of a grain sample was treated with sulphuric acid, sodium hydroxide was then added and the ammonia distilled into $50.00 \\mathrm{~cm}^{3}$ of $0.1010 \\mathrm{M}$ hydrochloric acid. The excess acid was back-titrated with $19.30 \\mathrm{~cm}^{3}$ of $0.1050 \\mathrm{M}$ sodium hydroxide.\n\nCalculate the concentration of nitrogen in the sample, in percent by mass.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nNitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample.\n\n$\\quad 0.2515 \\mathrm{~g}$ of a grain sample was treated with sulphuric acid, sodium hydroxide was then added and the ammonia distilled into $50.00 \\mathrm{~cm}^{3}$ of $0.1010 \\mathrm{M}$ hydrochloric acid. The excess acid was back-titrated with $19.30 \\mathrm{~cm}^{3}$ of $0.1050 \\mathrm{M}$ sodium hydroxide.\n\nCalculate the concentration of nitrogen in the sample, in percent by mass.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1222",
        "problem": "The energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.Calculate the threshold energy $E_{\\mathrm{E}}(\\mathrm{eV})$ of the following dissociative ionization reaction to the first decimal place:\n\n$$\n\\mathrm{H}_{2} \\rightarrow \\mathrm{H}^{*}(n=2)+\\mathrm{H}^{+}+\\mathrm{e}^{-} .\n$$\n\nIf you were unable to determine the values for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, then use $15.0 \\mathrm{eV}$ and 5.0 $\\mathrm{eV}$ for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, respectively.\n\nWhen $\\mathrm{H}_{2}$ absorbs monochromatic light of $21.2 \\mathrm{eV}$, the following dissociation process occurs at the same time.\n\n\n\n$$\n\\mathrm{H}_{2} \\stackrel{21.2 \\mathrm{eV}}{\\longrightarrow} \\mathrm{H}(n=1)+\\mathrm{H}(n=1)\n$$\n\nTwo hydrogen atoms move in opposite directions with the same speed.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.\n\nproblem:\nCalculate the threshold energy $E_{\\mathrm{E}}(\\mathrm{eV})$ of the following dissociative ionization reaction to the first decimal place:\n\n$$\n\\mathrm{H}_{2} \\rightarrow \\mathrm{H}^{*}(n=2)+\\mathrm{H}^{+}+\\mathrm{e}^{-} .\n$$\n\nIf you were unable to determine the values for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, then use $15.0 \\mathrm{eV}$ and 5.0 $\\mathrm{eV}$ for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, respectively.\n\nWhen $\\mathrm{H}_{2}$ absorbs monochromatic light of $21.2 \\mathrm{eV}$, the following dissociation process occurs at the same time.\n\n\n\n$$\n\\mathrm{H}_{2} \\stackrel{21.2 \\mathrm{eV}}{\\longrightarrow} \\mathrm{H}(n=1)+\\mathrm{H}(n=1)\n$$\n\nTwo hydrogen atoms move in opposite directions with the same speed.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of eV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-075.jpg?height=645&width=962&top_left_y=1956&top_left_x=547"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "eV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1482",
        "problem": "Oxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nUsing the relation between $K$ and the standard Gibbs energy $\\Delta G^{0}$ for a reaction, calculate the difference between the $\\Delta G^{0}$ values for the heme reactions (1) and (2).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nUsing the relation between $K$ and the standard Gibbs energy $\\Delta G^{0}$ for a reaction, calculate the difference between the $\\Delta G^{0}$ values for the heme reactions (1) and (2).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-418.jpg?height=823&width=1400&top_left_y=568&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1048",
        "problem": "For E5 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 31170. For E10 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 30680.To help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nTaking the energy released from burning 1 litre of E5 as $100 \\%$, calculate the percentage of the energy released from burning 1 litre of E10 fuel.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nFor E5 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 31170. For E10 fuels, the energy, in kJ, released when 1 litre of the fuel is burnt is 30680.\n\nproblem:\nTo help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nTaking the energy released from burning 1 litre of E5 as $100 \\%$, calculate the percentage of the energy released from burning 1 litre of E10 fuel.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of % (percentage), but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-02.jpg?height=459&width=923&top_left_y=356&top_left_x=949"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "% (percentage)"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_444",
        "problem": "我们将有离子参与的反应叫做离子反应, 下列离子方程式中正确的是\nA: 足量的碳酸氢钠与氢氧化钙溶液反应: $\\mathrm{HCO}_{3}^{-}+\\mathrm{Ca}^{2+}+\\mathrm{OH}^{-}=\\mathrm{CaCO}_{3} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}$\nB: 氢氧化铁溶于氢碘酸: $\\mathrm{Fe}(\\mathrm{OH})_{3}+3 \\mathrm{H}^{+}=\\mathrm{Fe}^{3+}+3 \\mathrm{H}_{2} \\mathrm{O}$\nC: 将使 84 消毒液有刺鼻性气味的气体通入水中: $\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{H}^{+}+\\mathrm{Cl}^{-}+\\mathrm{HClO}$\nD: 将高铁酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{FeO}_{4}\\right)$ 加入水中: $4 \\mathrm{FeO}_{4}^{2-}+10 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{Fe}(\\mathrm{OH})_{3}$ (胶体) $+3 \\mathrm{O}_{2} \\uparrow+8 \\mathrm{OH}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n我们将有离子参与的反应叫做离子反应, 下列离子方程式中正确的是\n\nA: 足量的碳酸氢钠与氢氧化钙溶液反应: $\\mathrm{HCO}_{3}^{-}+\\mathrm{Ca}^{2+}+\\mathrm{OH}^{-}=\\mathrm{CaCO}_{3} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}$\nB: 氢氧化铁溶于氢碘酸: $\\mathrm{Fe}(\\mathrm{OH})_{3}+3 \\mathrm{H}^{+}=\\mathrm{Fe}^{3+}+3 \\mathrm{H}_{2} \\mathrm{O}$\nC: 将使 84 消毒液有刺鼻性气味的气体通入水中: $\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{H}^{+}+\\mathrm{Cl}^{-}+\\mathrm{HClO}$\nD: 将高铁酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{FeO}_{4}\\right)$ 加入水中: $4 \\mathrm{FeO}_{4}^{2-}+10 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{Fe}(\\mathrm{OH})_{3}$ (胶体) $+3 \\mathrm{O}_{2} \\uparrow+8 \\mathrm{OH}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_211",
        "problem": "Select all molecules that have the same molecular shape.\nA: $\\mathrm{CO}_{2}$\nB: $\\mathrm{BF}_{3}$\nC: $\\mathrm{Cl}_{2} \\mathrm{O}$\nD: $\\mathrm{H}_{2} \\mathrm{~S}$\nE: $\\mathrm{CH}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nSelect all molecules that have the same molecular shape.\n\nA: $\\mathrm{CO}_{2}$\nB: $\\mathrm{BF}_{3}$\nC: $\\mathrm{Cl}_{2} \\mathrm{O}$\nD: $\\mathrm{H}_{2} \\mathrm{~S}$\nE: $\\mathrm{CH}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_458",
        "problem": "化合物 X 是合成一种治疗直肠癌和小细胞肺癌药物的中间体，其结构简式如图所\n示，下列说法正确的是[图1]\nA: 化合物 $\\mathrm{X}$ 分子中含有两种官能团\nB: 化合物 $\\mathrm{X}$ 分子中所有碳原子可能共平面\nC: 化合物 $\\mathrm{X}$ 分子中含有 1 个手性碳原子\nD: $1 \\mathrm{~mol}$ 化合物 $\\mathrm{X}$ 最多能与 $5 \\mathrm{molNaOH}$ 反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化合物 X 是合成一种治疗直肠癌和小细胞肺癌药物的中间体，其结构简式如图所\n示，下列说法正确的是[图1]\n\nA: 化合物 $\\mathrm{X}$ 分子中含有两种官能团\nB: 化合物 $\\mathrm{X}$ 分子中所有碳原子可能共平面\nC: 化合物 $\\mathrm{X}$ 分子中含有 1 个手性碳原子\nD: $1 \\mathrm{~mol}$ 化合物 $\\mathrm{X}$ 最多能与 $5 \\mathrm{molNaOH}$ 反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/zBCHxX6s/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_415",
        "problem": "我国科学家发现, 将纳米级 $\\mathrm{FeF}_{3}$ 嵌入电极材料, 能大大提高可充电铝离子电池的容量。其中有机离子导体主要含 $\\mathrm{Al}_{x} \\mathrm{Cl}_{y}^{n-}$, 隔膜仅允许含铝元素的微粒通过。工作原理如图所示:\n\n[图1]\n\n入电极\n\n下列说法正确的是\nA: 若 $\\mathrm{FeF}_{3}$ 从电极表面脱落, 则电池单位质量释放电量减少\nB: 为了提高电导效率, 左极室采用酸性 $\\mathrm{Al}_{x} \\mathrm{Cl}_{y}^{n-}$ 水溶液\nC: 放电时, $\\mathrm{AlCl}_{4}^{-}$离子可经过隔膜进入右极室中\nD: 充电时, 电池的阳极反应为 $\\mathrm{Al}+7 \\mathrm{AlCl}_{4}^{-}-3 \\mathrm{e}^{-}=4 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n我国科学家发现, 将纳米级 $\\mathrm{FeF}_{3}$ 嵌入电极材料, 能大大提高可充电铝离子电池的容量。其中有机离子导体主要含 $\\mathrm{Al}_{x} \\mathrm{Cl}_{y}^{n-}$, 隔膜仅允许含铝元素的微粒通过。工作原理如图所示:\n\n[图1]\n\n入电极\n\n下列说法正确的是\n\nA: 若 $\\mathrm{FeF}_{3}$ 从电极表面脱落, 则电池单位质量释放电量减少\nB: 为了提高电导效率, 左极室采用酸性 $\\mathrm{Al}_{x} \\mathrm{Cl}_{y}^{n-}$ 水溶液\nC: 放电时, $\\mathrm{AlCl}_{4}^{-}$离子可经过隔膜进入右极室中\nD: 充电时, 电池的阳极反应为 $\\mathrm{Al}+7 \\mathrm{AlCl}_{4}^{-}-3 \\mathrm{e}^{-}=4 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-25.jpg?height=508&width=942&top_left_y=714&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_847",
        "problem": "常温下, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCl}$ 溶液分别滴定 $\\mathrm{V}_{\\mathrm{x}} \\mathrm{mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaX}$ 溶液和 $\\mathrm{V}_{\\mathrm{Y}} \\mathrm{mL}$ $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaY}$ 溶液, 二者的 $\\mathrm{pH}$ 随着加入盐酸体积的变化曲线如图所示, 已知: 酸性 $H X>H Y$ 。下列判断正确的是\n\n[图1]\nA: $\\mathrm{V}_{\\mathrm{x}}>\\mathrm{V}_{\\mathrm{Y}}$\nB: $\\mathrm{M}$ 点: $c\\left(\\mathrm{X}^{-}\\right)=c\\left(\\mathrm{Y}^{-}\\right)$\nC: b 表示盐酸滴定 $\\mathrm{NaX}$ 溶液的曲线\nD: 常温下, $K_{a}(H X)=9.0 \\times 10^{-12}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCl}$ 溶液分别滴定 $\\mathrm{V}_{\\mathrm{x}} \\mathrm{mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaX}$ 溶液和 $\\mathrm{V}_{\\mathrm{Y}} \\mathrm{mL}$ $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaY}$ 溶液, 二者的 $\\mathrm{pH}$ 随着加入盐酸体积的变化曲线如图所示, 已知: 酸性 $H X>H Y$ 。下列判断正确的是\n\n[图1]\n\nA: $\\mathrm{V}_{\\mathrm{x}}>\\mathrm{V}_{\\mathrm{Y}}$\nB: $\\mathrm{M}$ 点: $c\\left(\\mathrm{X}^{-}\\right)=c\\left(\\mathrm{Y}^{-}\\right)$\nC: b 表示盐酸滴定 $\\mathrm{NaX}$ 溶液的曲线\nD: 常温下, $K_{a}(H X)=9.0 \\times 10^{-12}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-105.jpg?height=494&width=577&top_left_y=798&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1398",
        "problem": "Phosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nWrite down the conjugate base of dihydrogen phosphate ion and determine its $p K_{b}$ value.\n\nSmall quantities of phosphoric acid are extensively used to impart the sour or tart taste to many soft drinks such as colas and root beers. A cola having a density of 1.00 $\\mathrm{g} \\mathrm{cm}^{-3}$ contains $0.05 \\%$ by weight of phosphoric acid.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nPhosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nWrite down the conjugate base of dihydrogen phosphate ion and determine its $p K_{b}$ value.\n\nSmall quantities of phosphoric acid are extensively used to impart the sour or tart taste to many soft drinks such as colas and root beers. A cola having a density of 1.00 $\\mathrm{g} \\mathrm{cm}^{-3}$ contains $0.05 \\%$ by weight of phosphoric acid.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_578",
        "problem": "电位滴定法是根据滴定过程中指示电极电位的变化来确定滴定终点的一种滴定分析方法。在一定条件下, 用 $0.1000 \\mathrm{~mol} / \\mathrm{L}$ 的高氯酸溶液滴定 $10.0 \\mathrm{~mL}$ 某弱酸酸式盐 $\\mathrm{KHA}$ 溶液, 测得电压变化与滴入 $\\mathrm{HClO}_{4}$ 溶液的体积关系如图所示。做空白对照实验, 消耗 $\\mathrm{HClO}_{4}$溶液的体积为 $0.25 \\mathrm{~mL}$ 。已知 $\\mathrm{N}$ 为滴定终点。下列说法正确的是\n[图1]\nA: 初始时该酸式盐 KHA 的浓度为 $0.1725 \\mathrm{~mol} / \\mathrm{L}$\nB: 水的电离程度: $N>M$\nC: 已知 $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $\\mathrm{K}_{\\mathrm{a} 1} \\gg \\mathrm{K}_{\\mathrm{a} 2}$, 若 $\\mathrm{M}$ 点处为中性, 则有 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}$ 约为 0\nD: 若 $\\mathrm{M}$ 点时溶液呈中性, 则存在: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{ClO}_{4}^{-}\\right)=2\\left[\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)\\right]$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n电位滴定法是根据滴定过程中指示电极电位的变化来确定滴定终点的一种滴定分析方法。在一定条件下, 用 $0.1000 \\mathrm{~mol} / \\mathrm{L}$ 的高氯酸溶液滴定 $10.0 \\mathrm{~mL}$ 某弱酸酸式盐 $\\mathrm{KHA}$ 溶液, 测得电压变化与滴入 $\\mathrm{HClO}_{4}$ 溶液的体积关系如图所示。做空白对照实验, 消耗 $\\mathrm{HClO}_{4}$溶液的体积为 $0.25 \\mathrm{~mL}$ 。已知 $\\mathrm{N}$ 为滴定终点。下列说法正确的是\n[图1]\n\nA: 初始时该酸式盐 KHA 的浓度为 $0.1725 \\mathrm{~mol} / \\mathrm{L}$\nB: 水的电离程度: $N>M$\nC: 已知 $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $\\mathrm{K}_{\\mathrm{a} 1} \\gg \\mathrm{K}_{\\mathrm{a} 2}$, 若 $\\mathrm{M}$ 点处为中性, 则有 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}$ 约为 0\nD: 若 $\\mathrm{M}$ 点时溶液呈中性, 则存在: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{ClO}_{4}^{-}\\right)=2\\left[\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)\\right]$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-048.jpg?height=378&width=834&top_left_y=530&top_left_x=354"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_466",
        "problem": "以 $0.10 \\mathrm{~mol} / \\mathrm{L}$ 的氢氧化钠溶液滴定同浓度某一元酸 $\\mathrm{HA}$ 的滴定曲线如图所示 (滴定分数 $\\left.=\\frac{\\text { 滴定用量 }}{\\text { 总滴定用量 }}\\right)$ 。下列表述错误的是\n\n[图1]\nA: $\\mathrm{z}$ 点后存在某点, 溶液中的水的电离程度和 $\\mathrm{y}$ 点的相同\nB: $\\mathrm{a}$ 约为 3.5\nC: $\\mathrm{z}$ 点处, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$\nD: $\\mathrm{x}$ 点处的溶液中离子满足: $\\mathrm{c}(\\mathrm{HA})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n以 $0.10 \\mathrm{~mol} / \\mathrm{L}$ 的氢氧化钠溶液滴定同浓度某一元酸 $\\mathrm{HA}$ 的滴定曲线如图所示 (滴定分数 $\\left.=\\frac{\\text { 滴定用量 }}{\\text { 总滴定用量 }}\\right)$ 。下列表述错误的是\n\n[图1]\n\nA: $\\mathrm{z}$ 点后存在某点, 溶液中的水的电离程度和 $\\mathrm{y}$ 点的相同\nB: $\\mathrm{a}$ 约为 3.5\nC: $\\mathrm{z}$ 点处, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$\nD: $\\mathrm{x}$ 点处的溶液中离子满足: $\\mathrm{c}(\\mathrm{HA})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-39.jpg?height=663&width=931&top_left_y=1744&top_left_x=331"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1348",
        "problem": "The rechargeable lithium ion battery has been developed in Japan.\n\nThe standard electromotive force of the battery is $3.70 \\mathrm{~V}$. Assume that the halfreaction at the cathode is\n\n$$\n\\mathrm{CoO}_{2}+\\mathrm{Li}^{+}+\\mathrm{e}^{-} \\rightarrow \\mathrm{LiCoO}_{2}\n$$\n\nand the half-reaction at the anode is\n\n$$\n\\mathrm{LiC}_{6} \\rightarrow 6 \\mathrm{C}+\\mathrm{Li}^{+}+\\mathrm{e}^{-} .\n$$\n\nThe battery cell is constructed using $\\mathrm{LiCoO}_{2}$ and graphite (C) as the electrode materials.Calculate the maximum energy generated per mass of the lithium ion battery cell $\\left[\\mathrm{kJ} \\mathrm{~kg}^{-1}\\right]$. Assume that the correct ratio for complete reaction between the cathode and anode materials is used and the sum of the mass of electrodes is $50.0 \\%$ of the total mass of the battery cell. In comparison, the energy density of lead-acid batteries used for vehicles is about $200 \\mathrm{~kJ} \\mathrm{~kg}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe rechargeable lithium ion battery has been developed in Japan.\n\nThe standard electromotive force of the battery is $3.70 \\mathrm{~V}$. Assume that the halfreaction at the cathode is\n\n$$\n\\mathrm{CoO}_{2}+\\mathrm{Li}^{+}+\\mathrm{e}^{-} \\rightarrow \\mathrm{LiCoO}_{2}\n$$\n\nand the half-reaction at the anode is\n\n$$\n\\mathrm{LiC}_{6} \\rightarrow 6 \\mathrm{C}+\\mathrm{Li}^{+}+\\mathrm{e}^{-} .\n$$\n\nThe battery cell is constructed using $\\mathrm{LiCoO}_{2}$ and graphite (C) as the electrode materials.\n\nproblem:\nCalculate the maximum energy generated per mass of the lithium ion battery cell $\\left[\\mathrm{kJ} \\mathrm{~kg}^{-1}\\right]$. Assume that the correct ratio for complete reaction between the cathode and anode materials is used and the sum of the mass of electrodes is $50.0 \\%$ of the total mass of the battery cell. In comparison, the energy density of lead-acid batteries used for vehicles is about $200 \\mathrm{~kJ} \\mathrm{~kg}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{kJ} \\mathrm{~kg}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{kJ} \\mathrm{~kg}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_331",
        "problem": "What is the most accurate and precise way to measure one litre of water?\nA: Use a 1-L graduated cylinder.\nB: Use a 1-L volumetric flask.\nC: Use a 100-mL volumetric flask ten times.\nD: Use a $100-\\mathrm{mL}$ pipette ten times.\nE: Weigh $1 \\mathrm{~kg}$ of water using a balance that weighs to $\\pm 1 \\mathrm{~g}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the most accurate and precise way to measure one litre of water?\n\nA: Use a 1-L graduated cylinder.\nB: Use a 1-L volumetric flask.\nC: Use a 100-mL volumetric flask ten times.\nD: Use a $100-\\mathrm{mL}$ pipette ten times.\nE: Weigh $1 \\mathrm{~kg}$ of water using a balance that weighs to $\\pm 1 \\mathrm{~g}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_718",
        "problem": "有机物 $\\mathrm{A}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 。 $\\mathrm{A}$ 的核磁共振氢谱有 4 个峰且面积之比为 $1: 2: 2: 3$, A 分子中只含一个苯环且苯环上只有一个取代基, 其核磁共振氢谱与红外光谱如下图。关于 $\\mathrm{A}$ 的下列说法中, 正确的是\n[图1]\nA: 有机物 A 分子中最多有 16 个原子共平面\nB: 有机物 A 能使紫色石荵试液变红\nC: 有机物 $\\mathrm{A}$ 的结构简式为 $-\\mathrm{COOCH}_{3}$\nD: 与有机物 A 属于同类化合物的同分异构体只有 2 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有机物 $\\mathrm{A}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 。 $\\mathrm{A}$ 的核磁共振氢谱有 4 个峰且面积之比为 $1: 2: 2: 3$, A 分子中只含一个苯环且苯环上只有一个取代基, 其核磁共振氢谱与红外光谱如下图。关于 $\\mathrm{A}$ 的下列说法中, 正确的是\n[图1]\n\nA: 有机物 A 分子中最多有 16 个原子共平面\nB: 有机物 A 能使紫色石荵试液变红\nC: 有机物 $\\mathrm{A}$ 的结构简式为 $-\\mathrm{COOCH}_{3}$\nD: 与有机物 A 属于同类化合物的同分异构体只有 2 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-19.jpg?height=360&width=1382&top_left_y=1739&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_164",
        "problem": "A closed $2.0 \\mathrm{~L}$ container initially holds $4.0 \\mathrm{~mol}$ of oxygen and $2.0 \\mathrm{~mol}$ of nitrogen at temperature $T$. If the pressure remains constant when $2.0 \\mathrm{~mol}$ of oxygen are removed, which expression correctly describes the final temperature of the system in terms of the initial temperature, T? Assume Ideal Gas behaviour.\nA: $3 T / 2$\nB: $2 \\mathrm{~T}$\nC: $2 \\mathrm{~T} / 3$\nD: $3 \\mathrm{~T}$\nE: $\\mathrm{T}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA closed $2.0 \\mathrm{~L}$ container initially holds $4.0 \\mathrm{~mol}$ of oxygen and $2.0 \\mathrm{~mol}$ of nitrogen at temperature $T$. If the pressure remains constant when $2.0 \\mathrm{~mol}$ of oxygen are removed, which expression correctly describes the final temperature of the system in terms of the initial temperature, T? Assume Ideal Gas behaviour.\n\nA: $3 T / 2$\nB: $2 \\mathrm{~T}$\nC: $2 \\mathrm{~T} / 3$\nD: $3 \\mathrm{~T}$\nE: $\\mathrm{T}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_459",
        "problem": "肉桂酸是一种香料, 具有很好的保香作用。利用苯甲醛与乙酸䣫发生 Perkin 反应可制备肉桂酸，方法如下:\n\n[图1]\n\n下列说法错误的是\nA: 苯甲醛和肉桂酸均能使酸性高锰酸钾溶液褪色\nB: 肉桂酸与安息香酸(C-COOH )互为同系物\nC: 可以利用银镜反应检验粗产品中是否含有苯甲醛\nD: 该反应属于取代反应, 碱性环境有利于反应的进行\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n肉桂酸是一种香料, 具有很好的保香作用。利用苯甲醛与乙酸䣫发生 Perkin 反应可制备肉桂酸，方法如下:\n\n[图1]\n\n下列说法错误的是\n\nA: 苯甲醛和肉桂酸均能使酸性高锰酸钾溶液褪色\nB: 肉桂酸与安息香酸(C-COOH )互为同系物\nC: 可以利用银镜反应检验粗产品中是否含有苯甲醛\nD: 该反应属于取代反应, 碱性环境有利于反应的进行\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-58.jpg?height=131&width=1202&top_left_y=928&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-58.jpg?height=160&width=897&top_left_y=2461&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-59.jpg?height=152&width=1064&top_left_y=164&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-59.jpg?height=120&width=1182&top_left_y=348&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_674",
        "problem": "在体积为 $2 \\mathrm{~L}$ 的恒容密闭容器中充入一定量的 $\\mathrm{H}_{2} \\mathrm{~S}$ 气体, 发生反应\n\n$2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{S}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2}(\\mathrm{~g})$, 平衡时三种组分的物质的量与温度的关系如图所示, 下列说法正确的是\n\n[图1]\nA: 反应 $2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{S}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2}(\\mathrm{~g})$ 的 $\\Delta \\mathrm{H}<0$\nB: $\\mathrm{X}$ 点和 $\\mathrm{Y}$ 点气体的压强之比为 $15: 16$\nC: $\\mathrm{T}_{1}$ 时, 向 $\\mathrm{X}$ 点容器中再充入 $2 \\mathrm{molH}_{2} \\mathrm{~S}$ 和 $2 \\mathrm{molH}_{2}$, 上述反应平衡正向移动\nD: $T_{2}$ 时, 若起始时向容器中充入 $8 \\mathrm{molH}_{2} \\mathrm{~S}$ 气体, 则平衡时 $\\mathrm{H}_{2} \\mathrm{~S}$ 的转化率小于 $66.7 \\%$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在体积为 $2 \\mathrm{~L}$ 的恒容密闭容器中充入一定量的 $\\mathrm{H}_{2} \\mathrm{~S}$ 气体, 发生反应\n\n$2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{S}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2}(\\mathrm{~g})$, 平衡时三种组分的物质的量与温度的关系如图所示, 下列说法正确的是\n\n[图1]\n\nA: 反应 $2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{S}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2}(\\mathrm{~g})$ 的 $\\Delta \\mathrm{H}<0$\nB: $\\mathrm{X}$ 点和 $\\mathrm{Y}$ 点气体的压强之比为 $15: 16$\nC: $\\mathrm{T}_{1}$ 时, 向 $\\mathrm{X}$ 点容器中再充入 $2 \\mathrm{molH}_{2} \\mathrm{~S}$ 和 $2 \\mathrm{molH}_{2}$, 上述反应平衡正向移动\nD: $T_{2}$ 时, 若起始时向容器中充入 $8 \\mathrm{molH}_{2} \\mathrm{~S}$ 气体, 则平衡时 $\\mathrm{H}_{2} \\mathrm{~S}$ 的转化率小于 $66.7 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-025.jpg?height=471&width=691&top_left_y=570&top_left_x=366"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1226",
        "problem": "[figure1]\n\nFigure 2.\n\nEnergy diagram of atomic orbitals of helium when an electron resides in the 1 s orbital.\n\nFigure 2 shows an energy diagram of the atomic orbitals of helium. The arrows indicate the \"allowed\" transitions according to the spectroscopic principle.Which equation explains the occurrence of helium in cleveite among $[A]$ to [D] below? Mark one.\nA: $\\quad{ }^{238} \\mathrm{U} \\rightarrow{ }^{234} \\mathrm{Th}+\\alpha$\nB: $\\quad \\mathrm{UHe}_{2} \\rightarrow \\mathrm{U}+2 \\mathrm{He}$\nC: $\\quad{ }^{240} U \\rightarrow{ }^{240} \\mathrm{~Np}+\\beta^{-}$\nD: $\\quad{ }^{235} \\mathrm{U}+\\mathrm{n} \\rightarrow{ }^{95} \\mathrm{Y}+{ }^{139} \\mathrm{I}+2 \\mathrm{n}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n[figure1]\n\nFigure 2.\n\nEnergy diagram of atomic orbitals of helium when an electron resides in the 1 s orbital.\n\nFigure 2 shows an energy diagram of the atomic orbitals of helium. The arrows indicate the \"allowed\" transitions according to the spectroscopic principle.\n\nproblem:\nWhich equation explains the occurrence of helium in cleveite among $[A]$ to [D] below? Mark one.\n\nA: $\\quad{ }^{238} \\mathrm{U} \\rightarrow{ }^{234} \\mathrm{Th}+\\alpha$\nB: $\\quad \\mathrm{UHe}_{2} \\rightarrow \\mathrm{U}+2 \\mathrm{He}$\nC: $\\quad{ }^{240} U \\rightarrow{ }^{240} \\mathrm{~Np}+\\beta^{-}$\nD: $\\quad{ }^{235} \\mathrm{U}+\\mathrm{n} \\rightarrow{ }^{95} \\mathrm{Y}+{ }^{139} \\mathrm{I}+2 \\mathrm{n}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-058.jpg?height=771&width=876&top_left_y=1816&top_left_x=196"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1037",
        "problem": "This question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nCalculate the bond enthalpy of the $\\mathrm{O}=\\mathrm{O}$ bond in $\\mathrm{kJ} \\mathrm{mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nCalculate the bond enthalpy of the $\\mathrm{O}=\\mathrm{O}$ bond in $\\mathrm{kJ} \\mathrm{mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-02.jpg?height=434&width=619&top_left_y=317&top_left_x=1181"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_270",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount of $\\mathrm{HCl}$ consumed in reaction $\\mathbf{A}$. (in mol or mmol)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount of $\\mathrm{HCl}$ consumed in reaction $\\mathbf{A}$. (in mol or mmol)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_451",
        "problem": "我国科学家在太阳能光电催化一化学耦合分解硫化氢研究中获得新进展, 相关装置如图所示。下列说法不正确的是\n\n[图1]\nA: 该装置的总反应为 $\\mathrm{H}_{2} \\mathrm{~S} \\xlongequal[\\text { 催化剂 }]{\\text { 光照 }} H_{2}+\\mathrm{S}$\nB: 能量转化方式主要为“光能 $\\rightarrow$ 电能 $\\rightarrow$ 化学能”\nC: $\\mathrm{a}$ 极上发生的电极反应为 $\\mathrm{Fe}^{3+}-\\mathrm{e}^{-}=\\mathrm{Fe}^{2+}$\nD: $\\mathrm{a}$ 极区需不断补充含 $\\mathrm{Fe}^{3+}$ 和 $\\mathrm{Fe}^{2+}$ 的溶液\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n我国科学家在太阳能光电催化一化学耦合分解硫化氢研究中获得新进展, 相关装置如图所示。下列说法不正确的是\n\n[图1]\n\nA: 该装置的总反应为 $\\mathrm{H}_{2} \\mathrm{~S} \\xlongequal[\\text { 催化剂 }]{\\text { 光照 }} H_{2}+\\mathrm{S}$\nB: 能量转化方式主要为“光能 $\\rightarrow$ 电能 $\\rightarrow$ 化学能”\nC: $\\mathrm{a}$ 极上发生的电极反应为 $\\mathrm{Fe}^{3+}-\\mathrm{e}^{-}=\\mathrm{Fe}^{2+}$\nD: $\\mathrm{a}$ 极区需不断补充含 $\\mathrm{Fe}^{3+}$ 和 $\\mathrm{Fe}^{2+}$ 的溶液\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-74.jpg?height=415&width=737&top_left_y=158&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_266",
        "problem": "Which of the following are molecules with tetrahedral electron pair geometry around the central atom? Select all that apply.\nA: $\\mathrm{CS}_{2}$\nB: $\\mathrm{BF}_{3}$\nC: $\\mathrm{NCl}_{3}$\nD: $\\mathrm{H}_{2} \\mathrm{~S}$\nE: $\\mathrm{CH}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhich of the following are molecules with tetrahedral electron pair geometry around the central atom? Select all that apply.\n\nA: $\\mathrm{CS}_{2}$\nB: $\\mathrm{BF}_{3}$\nC: $\\mathrm{NCl}_{3}$\nD: $\\mathrm{H}_{2} \\mathrm{~S}$\nE: $\\mathrm{CH}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_312",
        "problem": "For the reaction below, $\\Delta H^{\\circ}=-518.02 \\mathrm{~kJ}$ per mole of $\\mathrm{H}_{2} \\mathrm{~S}$. What is $\\Delta H_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{H}_{2} \\mathrm{~S}(g)$ ?\n\n$$\n\\mathrm{H}_{2} \\mathrm{~S}(g)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\rightarrow \\mathrm{SO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n[figure1]\nA: $-20.63 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $41.26 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $20.63 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-497.39 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nE: $-41.26 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the reaction below, $\\Delta H^{\\circ}=-518.02 \\mathrm{~kJ}$ per mole of $\\mathrm{H}_{2} \\mathrm{~S}$. What is $\\Delta H_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{H}_{2} \\mathrm{~S}(g)$ ?\n\n$$\n\\mathrm{H}_{2} \\mathrm{~S}(g)+\\frac{3}{2} \\mathrm{O}_{2}(g) \\rightarrow \\mathrm{SO}_{2}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n[figure1]\n\nA: $-20.63 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $41.26 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $20.63 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-497.39 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nE: $-41.26 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_92a3e606264c5ee1789ag-7.jpg?height=263&width=423&top_left_y=920&top_left_x=561"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_789",
        "problem": "$\\mathrm{H}_{2} \\mathrm{~A}$ 是一种二元弱酸, 常温下向 $\\mathrm{H}_{2} \\mathrm{~A}$ 溶液中滴加 $\\mathrm{NaOH}$ 溶液, 混合溶液中 $\\operatorname{lgX}[\\mathrm{X}$ 表\n示 $\\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}$ 或 $\\left.\\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}\\right]$随 $\\mathrm{pH}$ 的变化关系如图所示, 下列说法不正确的是\n\n[图1]\nA: 直线II中 $X$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}$\nB: 由图可知 $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $K_{l}=10^{-1.22}$\nC: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$ 时, $\\mathrm{pH}=7$\nD: $0.1 \\mathrm{~mol} / \\mathrm{LNaHA}$ 溶液中, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{H}_{2} \\mathrm{~A}$ 是一种二元弱酸, 常温下向 $\\mathrm{H}_{2} \\mathrm{~A}$ 溶液中滴加 $\\mathrm{NaOH}$ 溶液, 混合溶液中 $\\operatorname{lgX}[\\mathrm{X}$ 表\n示 $\\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}$ 或 $\\left.\\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}\\right]$随 $\\mathrm{pH}$ 的变化关系如图所示, 下列说法不正确的是\n\n[图1]\n\nA: 直线II中 $X$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}$\nB: 由图可知 $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $K_{l}=10^{-1.22}$\nC: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$ 时, $\\mathrm{pH}=7$\nD: $0.1 \\mathrm{~mol} / \\mathrm{LNaHA}$ 溶液中, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-04.jpg?height=465&width=502&top_left_y=333&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_764",
        "problem": "我国学者在 Science 报道了一种氯离子介导的电化学合成方法, 能将乙烯高效清洁、选择性地转化为环氧乙烷, 电化学反应的具体过程如图所示。在电解结束后, 将阴、阳极电解液输出混合, 便可反应生成环氧乙烷。下列说法错误的是\n\n[图1]\n物质转化原理:\n\n[图2]\nA: Pt 电极与电源正极相连\nB: 电解时右侧电解池溶液 $\\mathrm{pH}$ 减小\nC: 该过程的总反应为: [图3]\nD: 当电路中通过 $2 \\mathrm{~mol} \\mathrm{e}$-时, 消耗的 $\\mathrm{CH}_{2}=\\mathrm{CH}_{2}$ 为 $22.4 \\mathrm{~L}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n我国学者在 Science 报道了一种氯离子介导的电化学合成方法, 能将乙烯高效清洁、选择性地转化为环氧乙烷, 电化学反应的具体过程如图所示。在电解结束后, 将阴、阳极电解液输出混合, 便可反应生成环氧乙烷。下列说法错误的是\n\n[图1]\n物质转化原理:\n\n[图2]\n\nA: Pt 电极与电源正极相连\nB: 电解时右侧电解池溶液 $\\mathrm{pH}$ 减小\nC: 该过程的总反应为: [图3]\nD: 当电路中通过 $2 \\mathrm{~mol} \\mathrm{e}$-时, 消耗的 $\\mathrm{CH}_{2}=\\mathrm{CH}_{2}$ 为 $22.4 \\mathrm{~L}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-47.jpg?height=411&width=857&top_left_y=2250&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-48.jpg?height=203&width=1103&top_left_y=161&top_left_x=585",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-48.jpg?height=169&width=720&top_left_y=555&top_left_x=791"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1323",
        "problem": "In quantum mechanics, the movement of $\\pi$ electrons along a neutral chain of conjugated carbon atoms may be modeled using the 'particle in a box' method. The energy of the $\\pi$ electrons is given by the following equation:\n\n$$\nE_{\\mathrm{n}}=\\frac{n^{2} h^{2}}{8 m L^{2}}\n$$\n\nwhere $n$ is the quantum number $(n=1,2,3, \\ldots), h$ is Planck's constant, $m$ is the mass of electron, and $L$ is the length of the box which may be approximated by $L=(k+2) \\times 1.40 \\AA$ ( $k$ being the number of conjugated double bonds along the carbon chain in the molecule). A photon with the appropriate wavelength $\\lambda$ may promote $a$ m electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). An approximate semi-empirical formula based on this model which relates the wavelength $\\lambda$, to the number of double bonds $k$ and constant $B$ is as follows:\n\n$$\n\\lambda(\\mathrm{nm})=B \\times \\frac{(k+2)^{2}}{(2 k+1)}\n$$Derive equation 1 (an expression for the wavelength $\\lambda(\\mathrm{nm})$ corresponding to the transfer of an electron from the HOMO to the LUMO) in terms of $k$ and the fundamental constants, and hence calculate theoretical value of the constant $B_{\\text {calc. }}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn quantum mechanics, the movement of $\\pi$ electrons along a neutral chain of conjugated carbon atoms may be modeled using the 'particle in a box' method. The energy of the $\\pi$ electrons is given by the following equation:\n\n$$\nE_{\\mathrm{n}}=\\frac{n^{2} h^{2}}{8 m L^{2}}\n$$\n\nwhere $n$ is the quantum number $(n=1,2,3, \\ldots), h$ is Planck's constant, $m$ is the mass of electron, and $L$ is the length of the box which may be approximated by $L=(k+2) \\times 1.40 \\AA$ ( $k$ being the number of conjugated double bonds along the carbon chain in the molecule). A photon with the appropriate wavelength $\\lambda$ may promote $a$ m electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). An approximate semi-empirical formula based on this model which relates the wavelength $\\lambda$, to the number of double bonds $k$ and constant $B$ is as follows:\n\n$$\n\\lambda(\\mathrm{nm})=B \\times \\frac{(k+2)^{2}}{(2 k+1)}\n$$\n\nproblem:\nDerive equation 1 (an expression for the wavelength $\\lambda(\\mathrm{nm})$ corresponding to the transfer of an electron from the HOMO to the LUMO) in terms of $k$ and the fundamental constants, and hence calculate theoretical value of the constant $B_{\\text {calc. }}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of nm, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "nm"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_805",
        "problem": "向某 $\\mathrm{Na}_{2} \\mathrm{~A} 、 \\mathrm{NaHA}$ 混合液中加入 $\\mathrm{CaCl}_{2}$ 固体(忽略溶液体积、温度的变化), 测得溶液中离子浓度变化如图所示。已知: $\\mathrm{H}_{2} \\mathrm{~A}$ 为二元弱酸, $K_{\\mathrm{sp}}(\\mathrm{CaA})=2 \\times 10^{-9}, \\mathrm{Ca}(\\mathrm{HA})_{2}$易溶于水且溶液呈碱性。下列说法正确的是\n\n[图1]\nA: X 点溶液可能呈酸性\nB: $\\mathrm{X} 、 \\mathrm{Y} 、 \\mathrm{Z}$ 三点对应溶液 $\\mathrm{pH}$ 大小顺序为: $\\mathrm{X}<\\mathrm{Y}<\\mathrm{Z}$\nC: $\\mathrm{Y}$ 点溶液中 $c\\left(\\mathrm{HA}^{-}\\right)=2 \\times 10^{-2} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 向 $2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中加入等体积 $2 \\times 10^{-4} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CaCl}_{2}$ 溶液有 $\\mathrm{CaA}$ 沉淀生成\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n向某 $\\mathrm{Na}_{2} \\mathrm{~A} 、 \\mathrm{NaHA}$ 混合液中加入 $\\mathrm{CaCl}_{2}$ 固体(忽略溶液体积、温度的变化), 测得溶液中离子浓度变化如图所示。已知: $\\mathrm{H}_{2} \\mathrm{~A}$ 为二元弱酸, $K_{\\mathrm{sp}}(\\mathrm{CaA})=2 \\times 10^{-9}, \\mathrm{Ca}(\\mathrm{HA})_{2}$易溶于水且溶液呈碱性。下列说法正确的是\n\n[图1]\n\nA: X 点溶液可能呈酸性\nB: $\\mathrm{X} 、 \\mathrm{Y} 、 \\mathrm{Z}$ 三点对应溶液 $\\mathrm{pH}$ 大小顺序为: $\\mathrm{X}<\\mathrm{Y}<\\mathrm{Z}$\nC: $\\mathrm{Y}$ 点溶液中 $c\\left(\\mathrm{HA}^{-}\\right)=2 \\times 10^{-2} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 向 $2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中加入等体积 $2 \\times 10^{-4} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CaCl}_{2}$ 溶液有 $\\mathrm{CaA}$ 沉淀生成\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-005.jpg?height=368&width=645&top_left_y=1135&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_161",
        "problem": "Levobunolol (structure below) is used topically to treat glaucoma, an eye disorder which causes damage to the optic nerve.\n\n[figure1]\n\n## levobunolol\n\nWhich of the following functional groups are contained within the structure of levobunolol?\nA: amine, ketone, ether, phenol\nB: amine, ketone, ester, alcohol\nC: amine, aldehyde, ether, alcohol\nD: amide, ketone, ether, alcohol\nE: amine, ketone, ether, alcohol\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nLevobunolol (structure below) is used topically to treat glaucoma, an eye disorder which causes damage to the optic nerve.\n\n[figure1]\n\n## levobunolol\n\nWhich of the following functional groups are contained within the structure of levobunolol?\n\nA: amine, ketone, ether, phenol\nB: amine, ketone, ester, alcohol\nC: amine, aldehyde, ether, alcohol\nD: amide, ketone, ether, alcohol\nE: amine, ketone, ether, alcohol\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_de998ba2e5d42a77dc86g-2.jpg?height=281&width=569&top_left_y=866&top_left_x=389"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_483",
        "problem": "常温下, 浓度相同的一元弱酸 $\\mathrm{HA}$ 和一元弱碱 $\\mathrm{BOH}$ 相互滴定体系中, 溶液中 $\\mathrm{pH}$ 与 $\\lg \\frac{c\\left(A^{-}\\right)}{c(H A)}$ 或 $\\lg \\frac{c\\left(B^{+}\\right)}{c(B O H)}$ 的关系如图所示。已知 $H A$ 和 $B O H$ 的电离常数为 $K_{a}$ 和 $K_{b}$, 且 $b$和 $\\mathrm{b}^{\\prime}$ 点的纵坐标之和为 14 . 下列说法错误的是\n\n[图1]\nA: $\\mathrm{c}$ 和 $\\mathrm{c}^{\\prime}$ 点的纵坐标之和小于 $\\mathrm{a}$ 和 $\\mathrm{a}^{\\prime}$ 纵坐标之和 $\\quad$.. abcd 表示 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)}{\\mathrm{c}(\\mathrm{HA})}$ 与 $\\mathrm{pH}$ 的关系\nB: $\\mathrm{K}_{\\mathrm{a}}=\\mathrm{K}_{\\mathrm{b}}>1 \\times 10^{-5}$\nC: 水的电离程度: $a^{\\prime}<b^{\\prime}<c<d$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 浓度相同的一元弱酸 $\\mathrm{HA}$ 和一元弱碱 $\\mathrm{BOH}$ 相互滴定体系中, 溶液中 $\\mathrm{pH}$ 与 $\\lg \\frac{c\\left(A^{-}\\right)}{c(H A)}$ 或 $\\lg \\frac{c\\left(B^{+}\\right)}{c(B O H)}$ 的关系如图所示。已知 $H A$ 和 $B O H$ 的电离常数为 $K_{a}$ 和 $K_{b}$, 且 $b$和 $\\mathrm{b}^{\\prime}$ 点的纵坐标之和为 14 . 下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{c}$ 和 $\\mathrm{c}^{\\prime}$ 点的纵坐标之和小于 $\\mathrm{a}$ 和 $\\mathrm{a}^{\\prime}$ 纵坐标之和 $\\quad$.. abcd 表示 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)}{\\mathrm{c}(\\mathrm{HA})}$ 与 $\\mathrm{pH}$ 的关系\nB: $\\mathrm{K}_{\\mathrm{a}}=\\mathrm{K}_{\\mathrm{b}}>1 \\times 10^{-5}$\nC: 水的电离程度: $a^{\\prime}<b^{\\prime}<c<d$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-097.jpg?height=703&width=828&top_left_y=154&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_925",
        "problem": "已知 $\\mathrm{H}_{2} \\mathrm{X}$ 的电离常数 $K_{\\mathrm{a} 1}=2 \\times 10^{-8} 、 K_{\\mathrm{a} 2}=3 \\times 10^{-17}$ 。常温下, 难溶物 $\\mathrm{BaX}$ 在不同浓度盐酸(足量)中恰好不再溶解时, 测得混合液中 $\\lg c\\left(\\mathrm{Ba}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的关系如图所示:下列说法正确的是\n\n[图1]\nA: 直线上任一点均满足: $c\\left(\\mathrm{H}^{+}\\right)+2 c\\left(\\mathrm{Ba}^{2+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{Cl}^{-}\\right)+2 c\\left(\\mathrm{X}^{2-}\\right)$\nB: $\\mathrm{M}$ 点: $c\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nC: $\\mathrm{N}$ 点: $\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)$约为 $2 \\times 10^{-7} \\mathrm{~mol} / \\mathrm{L}$\nD: $K \\operatorname{sp}(\\mathrm{BaX})$ 约为 $1.2 \\times 10^{-21}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知 $\\mathrm{H}_{2} \\mathrm{X}$ 的电离常数 $K_{\\mathrm{a} 1}=2 \\times 10^{-8} 、 K_{\\mathrm{a} 2}=3 \\times 10^{-17}$ 。常温下, 难溶物 $\\mathrm{BaX}$ 在不同浓度盐酸(足量)中恰好不再溶解时, 测得混合液中 $\\lg c\\left(\\mathrm{Ba}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的关系如图所示:下列说法正确的是\n\n[图1]\n\nA: 直线上任一点均满足: $c\\left(\\mathrm{H}^{+}\\right)+2 c\\left(\\mathrm{Ba}^{2+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{Cl}^{-}\\right)+2 c\\left(\\mathrm{X}^{2-}\\right)$\nB: $\\mathrm{M}$ 点: $c\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nC: $\\mathrm{N}$ 点: $\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)$约为 $2 \\times 10^{-7} \\mathrm{~mol} / \\mathrm{L}$\nD: $K \\operatorname{sp}(\\mathrm{BaX})$ 约为 $1.2 \\times 10^{-21}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-059.jpg?height=514&width=554&top_left_y=171&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1098",
        "problem": "The standard molar enthalpy change for the reaction in $\\mathrm{CaO}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}$ is $-82 \\mathbf{k J ~ m o l}^{-1}$.\n\nThe energy needed to warm $210 \\mathrm{ml}$ of coffee by $40^{\\circ} \\mathrm{C}$ is $35.1 \\mathrm{~kJ}$.Nescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nThe standard enthalpies of formation of calcium hydroxide, calcium oxide and water are $-1003,-635$ and $-286 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ respectively. The heat capacity for the coffee is the same as that of water, $4.18 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~g}^{-1}$.\n\nCalculate the minimum mass of calcium oxide needed in the can to function as specified.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe standard molar enthalpy change for the reaction in $\\mathrm{CaO}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}$ is $-82 \\mathbf{k J ~ m o l}^{-1}$.\n\nThe energy needed to warm $210 \\mathrm{ml}$ of coffee by $40^{\\circ} \\mathrm{C}$ is $35.1 \\mathrm{~kJ}$.\n\nproblem:\nNescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nThe standard enthalpies of formation of calcium hydroxide, calcium oxide and water are $-1003,-635$ and $-286 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ respectively. The heat capacity for the coffee is the same as that of water, $4.18 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~g}^{-1}$.\n\nCalculate the minimum mass of calcium oxide needed in the can to function as specified.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-03.jpg?height=417&width=699&top_left_y=385&top_left_x=1021"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1477",
        "problem": "Palindromic sequences are an interesting class of DNA. In a palindromic double-stranded DNA (dsDNA) species, the sequence of one strand read in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction matches the $5^{\\prime} \\rightarrow 3^{\\prime}$ reading on the complementary strand. Hence, a palindromic dsDNA consists of two identical strands that are complementary to each other. An example is the so-called\n\nDrew-Dickerson dodecanucleotide (1):\n\n[figure1]\n\nThe melting temperature of dsDNA, $T_{\\mathrm{m}}$ is defined as the temperature at which $50 \\%$ of the original amount of DNA double strands are dissociated into separate strands.Consider the Drew-Dickerson dodecanucleotide (1). Assume that a G-C nucleobase pair contributes to the DNA duplex stability more than an A-T pair does.\n\nWhat is the probability that its $T_{m}$ increases when a single randomly selected base pair is replaced by a G-C pair?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\nHere is some context information for this question, which might assist you in solving it:\nPalindromic sequences are an interesting class of DNA. In a palindromic double-stranded DNA (dsDNA) species, the sequence of one strand read in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction matches the $5^{\\prime} \\rightarrow 3^{\\prime}$ reading on the complementary strand. Hence, a palindromic dsDNA consists of two identical strands that are complementary to each other. An example is the so-called\n\nDrew-Dickerson dodecanucleotide (1):\n\n[figure1]\n\nThe melting temperature of dsDNA, $T_{\\mathrm{m}}$ is defined as the temperature at which $50 \\%$ of the original amount of DNA double strands are dissociated into separate strands.\n\nproblem:\nConsider the Drew-Dickerson dodecanucleotide (1). Assume that a G-C nucleobase pair contributes to the DNA duplex stability more than an A-T pair does.\n\nWhat is the probability that its $T_{m}$ increases when a single randomly selected base pair is replaced by a G-C pair?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir units are, in order, [None, %], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-260.jpg?height=140&width=440&top_left_y=1729&top_left_x=814"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            null,
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1276",
        "problem": "Natural antimony consists of the following 2 stable isotopes: ${ }^{121} \\mathrm{Sb},{ }^{123} \\mathrm{Sb}$. Natural chlorine consists of the following 2 stable isotopes: ${ }^{35} \\mathrm{Cl},{ }^{37} \\mathrm{Cl}$. Natural hydrogen consists of the following 2 stable isotopes: ${ }^{1} \\mathrm{H},{ }^{2} \\mathrm{H}$. How many peaks are expected in a low resolution mass spectrum for the ionic fragment $\\mathrm{SbHCl}^{+}$?\nA: 4\nB: 5\nC: 6\nD: 7\nE: 8\nF: 9\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNatural antimony consists of the following 2 stable isotopes: ${ }^{121} \\mathrm{Sb},{ }^{123} \\mathrm{Sb}$. Natural chlorine consists of the following 2 stable isotopes: ${ }^{35} \\mathrm{Cl},{ }^{37} \\mathrm{Cl}$. Natural hydrogen consists of the following 2 stable isotopes: ${ }^{1} \\mathrm{H},{ }^{2} \\mathrm{H}$. How many peaks are expected in a low resolution mass spectrum for the ionic fragment $\\mathrm{SbHCl}^{+}$?\n\nA: 4\nB: 5\nC: 6\nD: 7\nE: 8\nF: 9\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1472",
        "problem": "The unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nAssuming that the values of $\\Delta H^{\\circ}$ and $\\Delta S^{\\circ}$ for the protein unfolding reaction are constant with temperature then:\n\n$$\n\\ln K=-\\frac{\\Delta H^{\\circ}}{R T}+C\n$$\n\nwhere $C$ is a constant.$\\Delta H^{o}=330 \\mathrm{~kJ} \\mathrm{~mol}^{-1} ; \\quad \\Delta S^{O}=980 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$\n\nCalculate the equilibrium constant for the unfolding reaction at $25^{\\circ} \\mathrm{C}$.\n\n[If you have been unable to calculate a value for $K$, you should use the following incorrect value for the subsequent parts of the problem: $K=3.6 \\cdot 10^{-6}$ ]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nAssuming that the values of $\\Delta H^{\\circ}$ and $\\Delta S^{\\circ}$ for the protein unfolding reaction are constant with temperature then:\n\n$$\n\\ln K=-\\frac{\\Delta H^{\\circ}}{R T}+C\n$$\n\nwhere $C$ is a constant.\n\nproblem:\n$\\Delta H^{o}=330 \\mathrm{~kJ} \\mathrm{~mol}^{-1} ; \\quad \\Delta S^{O}=980 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$\n\nCalculate the equilibrium constant for the unfolding reaction at $25^{\\circ} \\mathrm{C}$.\n\n[If you have been unable to calculate a value for $K$, you should use the following incorrect value for the subsequent parts of the problem: $K=3.6 \\cdot 10^{-6}$ ]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1313",
        "problem": "The transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nTo determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at $298 \\mathrm{~K}$ are shown in the tables below. (Use the grids if you like.)\n\n| $[\\mathrm{ArCl}]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.1 | 0.2 | 0.4 | 0.6 |\n| :--- | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $1.88 \\times 10^{-5}$ | $4.13 \\times 10^{-5}$ | $9.42 \\times 10^{-5}$ | $1.50 \\times 10^{-4}$ |\n\n[figure1]\n\n| $\\left[\\mathrm{NiLL}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | $6 \\times 10^{-3}$ | $9 \\times 10^{-3}$ | $1.2 \\times 10^{-2}$ | $1.5 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $6.01 \\times 10^{-5}$ | $7.80 \\times 10^{-5}$ | $1.10 \\times 10^{-4}$ |\n\n\n| $\\left[\\mathrm{L}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.06 | 0.09 | 0.12 | 0.15 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $5.8 \\times 10^{-5}$ | $4.3 \\times 10^{-5}$ | $3.4 \\times 10^{-5}$ | $2.8 \\times 10^{-5}$ |Determine the order with respect to $\\mathrm{[ArCl]}$ assuming it is integer.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nTo determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at $298 \\mathrm{~K}$ are shown in the tables below. (Use the grids if you like.)\n\n| $[\\mathrm{ArCl}]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.1 | 0.2 | 0.4 | 0.6 |\n| :--- | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $1.88 \\times 10^{-5}$ | $4.13 \\times 10^{-5}$ | $9.42 \\times 10^{-5}$ | $1.50 \\times 10^{-4}$ |\n\n[figure1]\n\n| $\\left[\\mathrm{NiLL}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | $6 \\times 10^{-3}$ | $9 \\times 10^{-3}$ | $1.2 \\times 10^{-2}$ | $1.5 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $6.01 \\times 10^{-5}$ | $7.80 \\times 10^{-5}$ | $1.10 \\times 10^{-4}$ |\n\n\n| $\\left[\\mathrm{L}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.06 | 0.09 | 0.12 | 0.15 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $5.8 \\times 10^{-5}$ | $4.3 \\times 10^{-5}$ | $3.4 \\times 10^{-5}$ | $2.8 \\times 10^{-5}$ |\n\nproblem:\nDetermine the order with respect to $\\mathrm{[ArCl]}$ assuming it is integer.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-032.jpg?height=834&width=902&top_left_y=1553&top_left_x=583"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_430",
        "problem": "向 $1.00 \\mathrm{~L}$ 浓度均为 $0.0100 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{SO}_{3} 、 \\mathrm{NaOH}$ 混合溶液中通入 $\\mathrm{HCl}$ 气体调节溶液 $\\mathrm{pH}$ (忽略溶液体积变化)。其中比 $\\mathrm{H}_{2} \\mathrm{SO}_{3} 、 \\mathrm{HSO}_{3}-\\mathrm{SO}_{3}{ }^{2-}$ 平衡时的分布系数(各含硫物种的浓度与含硫物种总浓度的比)随 $\\mathrm{HCl}$ 气体体积(标况下)的变化关系如图所示(忽略 $\\mathrm{SO}_{2}$气体的逸出); 已知 $\\mathrm{K}_{\\mathrm{a} 1}$ 代表 $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ 在该实验条件下的一级电离常数。下列说法正确的是\n\n[图1]\nA: $\\mathrm{Z}$ 点处的 $\\mathrm{pH}=-\\operatorname{lgK}_{\\mathrm{al}}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\nB: 从 $\\mathrm{X}$ 点到 $\\mathrm{Y}$ 点发生的主要反应为 $\\mathrm{SO}_{3}{ }^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{HSO}_{3}{ }^{-}+\\mathrm{OH}^{-}$\nC: 当 $\\mathrm{V}(\\mathrm{HCl}) \\geq 672 \\mathrm{~mL}$ 时, $\\mathrm{c}\\left(\\mathrm{HSO}_{3}{ }^{-}\\right)=\\mathrm{c}\\left(\\mathrm{SO}_{3}{ }^{2-}\\right)=0 \\mathrm{~mol} / \\mathrm{L}$\nD: 若将 $\\mathrm{HCl}$ 改为 $\\mathrm{NO}_{2}, \\mathrm{Y}$ 点对应位置不变\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n向 $1.00 \\mathrm{~L}$ 浓度均为 $0.0100 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{SO}_{3} 、 \\mathrm{NaOH}$ 混合溶液中通入 $\\mathrm{HCl}$ 气体调节溶液 $\\mathrm{pH}$ (忽略溶液体积变化)。其中比 $\\mathrm{H}_{2} \\mathrm{SO}_{3} 、 \\mathrm{HSO}_{3}-\\mathrm{SO}_{3}{ }^{2-}$ 平衡时的分布系数(各含硫物种的浓度与含硫物种总浓度的比)随 $\\mathrm{HCl}$ 气体体积(标况下)的变化关系如图所示(忽略 $\\mathrm{SO}_{2}$气体的逸出); 已知 $\\mathrm{K}_{\\mathrm{a} 1}$ 代表 $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ 在该实验条件下的一级电离常数。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{Z}$ 点处的 $\\mathrm{pH}=-\\operatorname{lgK}_{\\mathrm{al}}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\nB: 从 $\\mathrm{X}$ 点到 $\\mathrm{Y}$ 点发生的主要反应为 $\\mathrm{SO}_{3}{ }^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{HSO}_{3}{ }^{-}+\\mathrm{OH}^{-}$\nC: 当 $\\mathrm{V}(\\mathrm{HCl}) \\geq 672 \\mathrm{~mL}$ 时, $\\mathrm{c}\\left(\\mathrm{HSO}_{3}{ }^{-}\\right)=\\mathrm{c}\\left(\\mathrm{SO}_{3}{ }^{2-}\\right)=0 \\mathrm{~mol} / \\mathrm{L}$\nD: 若将 $\\mathrm{HCl}$ 改为 $\\mathrm{NO}_{2}, \\mathrm{Y}$ 点对应位置不变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-31.jpg?height=825&width=1279&top_left_y=153&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1312",
        "problem": "The label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\n1.4 Could it be possible that the solution contained EDTA (ethylene diamino tetraacetic acid)? You may use reasonable approximations.\n\nEDTA: $\\mathrm{p} K_{\\mathrm{a} 1}=1.70, \\mathrm{p} K_{\\mathrm{a} 2}=2.60, \\mathrm{p} K_{\\mathrm{a} 3}=6.30, \\mathrm{p} K_{\\mathrm{a} 4}=10.60$\n\n$\\square$ True $\\square$ False\n\nIf True, calculate the concentration.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\n1.4 Could it be possible that the solution contained EDTA (ethylene diamino tetraacetic acid)? You may use reasonable approximations.\n\nEDTA: $\\mathrm{p} K_{\\mathrm{a} 1}=1.70, \\mathrm{p} K_{\\mathrm{a} 2}=2.60, \\mathrm{p} K_{\\mathrm{a} 3}=6.30, \\mathrm{p} K_{\\mathrm{a} 4}=10.60$\n\n$\\square$ True $\\square$ False\n\nIf True, calculate the concentration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_907",
        "problem": "我国科学家制备了一种新型低毒性和较好亲水性的聚合物 W, 制备过程如下。\n\n[图1]\n\n下列说法正确的是\nA: X 分子中不同化学环境的氢原子有 1 种\nB: $\\mathrm{Z}$ 在酸性条件下充分水解能生成乳酸\nC: 合成 $\\mathrm{Z}$ 的过程中，理论上应控制 $\\mathrm{X} 、 \\mathrm{Y}$ 的物质的量之比为 $\\mathrm{m}: \\mathrm{n}$\nD: 由 $\\mathrm{Z}$ 合成 $\\mathrm{W}$ 的过程中发生了加成反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n我国科学家制备了一种新型低毒性和较好亲水性的聚合物 W, 制备过程如下。\n\n[图1]\n\n下列说法正确的是\n\nA: X 分子中不同化学环境的氢原子有 1 种\nB: $\\mathrm{Z}$ 在酸性条件下充分水解能生成乳酸\nC: 合成 $\\mathrm{Z}$ 的过程中，理论上应控制 $\\mathrm{X} 、 \\mathrm{Y}$ 的物质的量之比为 $\\mathrm{m}: \\mathrm{n}$\nD: 由 $\\mathrm{Z}$ 合成 $\\mathrm{W}$ 的过程中发生了加成反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-13.jpg?height=402&width=1464&top_left_y=2072&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_673",
        "problem": "某温度下, 改变 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液的 $\\mathrm{pH}$ 时, 各种含铬元素微粒及 $\\mathrm{OH}^{-}$浓度变化如图所示(已知 $\\mathrm{H}_{2} \\mathrm{CrO}_{4}$ 是二元酸), 下列有关说法中, 正确的是\n\n[图1]\nA: 该温度下的 $\\mathrm{K}_{\\mathrm{w}}=10^{-13}$\nB: 溶液中存在平衡 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons 2 \\mathrm{CrO}_{4}^{2-}+2 \\mathrm{H}^{+}$, 该温度下此反应的 $\\mathrm{K}=10^{-13.2}$\nC: 向 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中加入一定量 $\\mathrm{NaOH}$ 固体, 溶液橙色变浅\nD: $\\mathrm{E}$ 点溶液中存在 $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<2 \\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)+3 \\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某温度下, 改变 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液的 $\\mathrm{pH}$ 时, 各种含铬元素微粒及 $\\mathrm{OH}^{-}$浓度变化如图所示(已知 $\\mathrm{H}_{2} \\mathrm{CrO}_{4}$ 是二元酸), 下列有关说法中, 正确的是\n\n[图1]\n\nA: 该温度下的 $\\mathrm{K}_{\\mathrm{w}}=10^{-13}$\nB: 溶液中存在平衡 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons 2 \\mathrm{CrO}_{4}^{2-}+2 \\mathrm{H}^{+}$, 该温度下此反应的 $\\mathrm{K}=10^{-13.2}$\nC: 向 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中加入一定量 $\\mathrm{NaOH}$ 固体, 溶液橙色变浅\nD: $\\mathrm{E}$ 点溶液中存在 $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<2 \\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)+3 \\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-064.jpg?height=760&width=1211&top_left_y=154&top_left_x=331"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_802",
        "problem": "$\\mathrm{ROH}$ 是一元弱碱, 难溶盐 $\\mathrm{RA}$ 的饱和溶液中 $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$随 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$而变化, $\\mathrm{A}^{-}$不发生水解; $298 \\mathrm{~K}$ 时, $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$有如图所示线性关系, 下列叙述错误的是\n\n[图1]\nA: $\\mathrm{pH}=6, \\mathrm{c}\\left(\\mathrm{A}^{-}\\right)<2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)=\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{RA})+\\frac{\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{RA})}{\\mathrm{K}_{\\mathrm{b}}(\\mathrm{ROH})} \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: RA 的溶度积 $K_{\\mathrm{sp}}(\\mathrm{RA})=2 \\times 10^{-10}$\nD: $\\mathrm{ROH}$ 的电离平衡常数 $K_{\\mathrm{b}}(\\mathrm{ROH})=2 \\times 10^{-6}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{ROH}$ 是一元弱碱, 难溶盐 $\\mathrm{RA}$ 的饱和溶液中 $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$随 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$而变化, $\\mathrm{A}^{-}$不发生水解; $298 \\mathrm{~K}$ 时, $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$有如图所示线性关系, 下列叙述错误的是\n\n[图1]\n\nA: $\\mathrm{pH}=6, \\mathrm{c}\\left(\\mathrm{A}^{-}\\right)<2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)=\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{RA})+\\frac{\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{RA})}{\\mathrm{K}_{\\mathrm{b}}(\\mathrm{ROH})} \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: RA 的溶度积 $K_{\\mathrm{sp}}(\\mathrm{RA})=2 \\times 10^{-10}$\nD: $\\mathrm{ROH}$ 的电离平衡常数 $K_{\\mathrm{b}}(\\mathrm{ROH})=2 \\times 10^{-6}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-082.jpg?height=605&width=900&top_left_y=1294&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_875",
        "problem": "相同金属在其不同浓度盐溶液中可形成浓差电池。如图所示装置是利用浓差电池电解 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液 ( $\\mathrm{a} 、 \\mathrm{~b}$ 电极均为石墨电极), 可以制得 $\\mathrm{O}_{2} 、 \\mathrm{H}_{2} 、 \\mathrm{H}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{NaOH}$ 。下列说法正确的是\n\n[图1]\n\n阴离子交换膜\n\n[图2]\n\n离子交换膜\nA: $\\mathrm{a}$ 电极的电极反应为 $4 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{e}^{-}=2 \\mathrm{H}_{2} \\uparrow+4 \\mathrm{OH}$\nB: c、d 离子交换膜依次为阴离子交换膜和阳离子交换膜\nC: 电池放电过程中, $\\mathrm{Cu}(2)$ 电极上的电极反应为 $\\mathrm{Cu}-2 \\mathrm{e}^{-}=\\mathrm{Cu}^{2+}$\nD: 电池从开始工作到停止放电, 电解池理论上可制得 $320 \\mathrm{gNaOH}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n相同金属在其不同浓度盐溶液中可形成浓差电池。如图所示装置是利用浓差电池电解 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液 ( $\\mathrm{a} 、 \\mathrm{~b}$ 电极均为石墨电极), 可以制得 $\\mathrm{O}_{2} 、 \\mathrm{H}_{2} 、 \\mathrm{H}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{NaOH}$ 。下列说法正确的是\n\n[图1]\n\n阴离子交换膜\n\n[图2]\n\n离子交换膜\n\nA: $\\mathrm{a}$ 电极的电极反应为 $4 \\mathrm{H}_{2} \\mathrm{O}+4 \\mathrm{e}^{-}=2 \\mathrm{H}_{2} \\uparrow+4 \\mathrm{OH}$\nB: c、d 离子交换膜依次为阴离子交换膜和阳离子交换膜\nC: 电池放电过程中, $\\mathrm{Cu}(2)$ 电极上的电极反应为 $\\mathrm{Cu}-2 \\mathrm{e}^{-}=\\mathrm{Cu}^{2+}$\nD: 电池从开始工作到停止放电, 电解池理论上可制得 $320 \\mathrm{gNaOH}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-68.jpg?height=386&width=460&top_left_y=595&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-68.jpg?height=365&width=803&top_left_y=614&top_left_x=775"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_279",
        "problem": "How many nitrogen atoms are there in $0.25 \\mathrm{~mol}$ of ammonium nitrate?\nA: $1.5 \\times 10^{23}$\nB: $2.4 \\times 10^{23}$\nC: $3.0 \\times 10^{23}$\nD: $6.0 \\times 10^{23}$\nE: $4.8 \\times 10^{24}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many nitrogen atoms are there in $0.25 \\mathrm{~mol}$ of ammonium nitrate?\n\nA: $1.5 \\times 10^{23}$\nB: $2.4 \\times 10^{23}$\nC: $3.0 \\times 10^{23}$\nD: $6.0 \\times 10^{23}$\nE: $4.8 \\times 10^{24}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1208",
        "problem": "In the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.Estimate the density of this sample of YBCO (with $\\delta=0.25$ ) in $\\mathrm{g} \\mathrm{cm}^{-3}$. If you were unable to calculate the values for $a$ and $c$ from part 4.2, then use $a=500 \\mathrm{pm}$ and $c=1500 \\mathrm{pm}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.\n\nproblem:\nEstimate the density of this sample of YBCO (with $\\delta=0.25$ ) in $\\mathrm{g} \\mathrm{cm}^{-3}$. If you were unable to calculate the values for $a$ and $c$ from part 4.2, then use $a=500 \\mathrm{pm}$ and $c=1500 \\mathrm{pm}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~g} \\mathrm{~cm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~g} \\mathrm{~cm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_763",
        "problem": "下列说法正确的是\nA: 某元素基态原子最外层的电子排布式为 $n s^{2} n p^{1}$, 该元素一定为IIIA 族元素\nB: 苯和氯气生成 $\\mathrm{C}_{6} \\mathrm{H}_{6} \\mathrm{Cl}_{6}$ 的反应是取代反应\nC: 由分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}$ 的醇在浓硫酸加热的条件下发生消去反应得到的烯烃有(不含立体异构)4 种\nD: 分子式为 $\\mathrm{C}_{6} \\mathrm{H}_{12}$ 的烃, 其核磁共振氢谱可能只出现一个峰\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列说法正确的是\n\nA: 某元素基态原子最外层的电子排布式为 $n s^{2} n p^{1}$, 该元素一定为IIIA 族元素\nB: 苯和氯气生成 $\\mathrm{C}_{6} \\mathrm{H}_{6} \\mathrm{Cl}_{6}$ 的反应是取代反应\nC: 由分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{12} \\mathrm{O}$ 的醇在浓硫酸加热的条件下发生消去反应得到的烯烃有(不含立体异构)4 种\nD: 分子式为 $\\mathrm{C}_{6} \\mathrm{H}_{12}$ 的烃, 其核磁共振氢谱可能只出现一个峰\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_460",
        "problem": "近些年,地表水和地下水域 中的高氯酸盐污染及其降解受到环境工作者的关注。某科研小组研究了一定条件 下温度、酸碱性对其降解的影响(初始质量浓度均为 $100 \\mathrm{mg} / \\mathrm{L}$ ), 测得数据如图所示, 下列说法不正确的是\n\n[图1]\n\n图1 $\\mathrm{pH}=7.5$ 时, 温度对降解的影响\n\n[图2]\n\n图2 温度为 $30^{\\circ} \\mathrm{C}$ 时, $\\mathrm{pH}$ 对降解的影响\nA: 在 $\\mathrm{pH}=7.5$ 、温度为 $35^{\\circ} \\mathrm{C}$ 时, $0 \\sim 60 \\mathrm{~h}$ 内高氯酸盐的降解平均速率为 $1 \\mathrm{mg} /(\\mathrm{L} \\cdot \\mathrm{h})$\nB: 当 $\\mathrm{pH}=7.5$ 时,高氯酸盐降解最适宜温度为 $30^{\\circ} \\mathrm{C}$\nC: 当温度一定时, 随 $\\mathrm{pH}$ 的减小, 高氯酸盐的降解速率不一定增大\nD: 当降解时间为 $100 \\mathrm{~h}$ 时, $\\mathrm{pH}=7.5$ 、温度为 $28^{\\circ} \\mathrm{C}$ 与 $\\mathrm{pH}=7.3$ 、温度为 $30^{\\circ} \\mathrm{C}$ 两种条件下高氯酸盐的降解率可能相等\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n近些年,地表水和地下水域 中的高氯酸盐污染及其降解受到环境工作者的关注。某科研小组研究了一定条件 下温度、酸碱性对其降解的影响(初始质量浓度均为 $100 \\mathrm{mg} / \\mathrm{L}$ ), 测得数据如图所示, 下列说法不正确的是\n\n[图1]\n\n图1 $\\mathrm{pH}=7.5$ 时, 温度对降解的影响\n\n[图2]\n\n图2 温度为 $30^{\\circ} \\mathrm{C}$ 时, $\\mathrm{pH}$ 对降解的影响\n\nA: 在 $\\mathrm{pH}=7.5$ 、温度为 $35^{\\circ} \\mathrm{C}$ 时, $0 \\sim 60 \\mathrm{~h}$ 内高氯酸盐的降解平均速率为 $1 \\mathrm{mg} /(\\mathrm{L} \\cdot \\mathrm{h})$\nB: 当 $\\mathrm{pH}=7.5$ 时,高氯酸盐降解最适宜温度为 $30^{\\circ} \\mathrm{C}$\nC: 当温度一定时, 随 $\\mathrm{pH}$ 的减小, 高氯酸盐的降解速率不一定增大\nD: 当降解时间为 $100 \\mathrm{~h}$ 时, $\\mathrm{pH}=7.5$ 、温度为 $28^{\\circ} \\mathrm{C}$ 与 $\\mathrm{pH}=7.3$ 、温度为 $30^{\\circ} \\mathrm{C}$ 两种条件下高氯酸盐的降解率可能相等\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-70.jpg?height=452&width=562&top_left_y=1987&top_left_x=339",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-70.jpg?height=437&width=511&top_left_y=2003&top_left_x=984"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_482",
        "problem": "利用光能源可以将 $\\mathrm{CO}_{2}$ 转化为重要的化工原料 $\\mathrm{C}_{2} \\mathrm{H}_{4}$ (电解质溶液为稀硫酸), 同时可为制备次磷酸 $\\left(\\mathrm{H}_{3} \\mathrm{PO}_{2}\\right)$ 提供电能, 其工作原理如图所示。下列说法错误的是\n\n[图1]\nA: $Y$ 极为阳极\nB: 标准状况下, 当 $\\mathrm{Z}$ 极产生 $11.2 \\mathrm{LO}_{2}$ 时, 可生成 $\\mathrm{H}_{3} \\mathrm{PO}_{2}$ 的数目为 $1 \\mathrm{~N}_{\\mathrm{A}}$\nC: $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{~d}$ 为阳离子交换膜, $\\mathrm{c}$ 为阴离子交换膜\nD: $\\mathrm{W}$ 极的电极反应式为 $2 \\mathrm{CO}_{2}+12 \\mathrm{H}^{+}+12 \\mathrm{e}^{-}=\\mathrm{C}_{2} \\mathrm{H}_{4}+4 \\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用光能源可以将 $\\mathrm{CO}_{2}$ 转化为重要的化工原料 $\\mathrm{C}_{2} \\mathrm{H}_{4}$ (电解质溶液为稀硫酸), 同时可为制备次磷酸 $\\left(\\mathrm{H}_{3} \\mathrm{PO}_{2}\\right)$ 提供电能, 其工作原理如图所示。下列说法错误的是\n\n[图1]\n\nA: $Y$ 极为阳极\nB: 标准状况下, 当 $\\mathrm{Z}$ 极产生 $11.2 \\mathrm{LO}_{2}$ 时, 可生成 $\\mathrm{H}_{3} \\mathrm{PO}_{2}$ 的数目为 $1 \\mathrm{~N}_{\\mathrm{A}}$\nC: $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{~d}$ 为阳离子交换膜, $\\mathrm{c}$ 为阴离子交换膜\nD: $\\mathrm{W}$ 极的电极反应式为 $2 \\mathrm{CO}_{2}+12 \\mathrm{H}^{+}+12 \\mathrm{e}^{-}=\\mathrm{C}_{2} \\mathrm{H}_{4}+4 \\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-42.jpg?height=437&width=1311&top_left_y=318&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_669",
        "problem": "我国学者最近研发出一种以铝为负极、石墨烯薄膜 $\\left(\\mathrm{C}_{n}\\right)$ 为正极的新型铝一石墨烯电池, $\\mathrm{AlCl}_{4}^{-}$可在石墨烯薄膜上嵌入或脱嵌, 离子液体 $\\mathrm{AlCl}_{3} /[\\mathrm{EMIM}] \\mathrm{Cl}$ 作电解质, 其中\n阴离子有 $\\mathrm{AlCl}_{4}^{-} 、 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$, 阳离子为 $\\mathrm{EMIM}^{+}\\left(\\left\\langle_{\\mathrm{N}^{\\prime}}{ }^{\\prime}{ }^{+}\\right.\\right.$, 其结构中存在大 $\\pi$ 键 $)$, 放电机理如图所示。已知: 大 $\\pi$ 键可用符号 $\\Pi_{\\mathrm{m}}^{\\mathrm{n}}$ 表示, 其中 $\\mathrm{m}$ 代表参与形成大 $\\pi$ 键的原子数, $\\mathrm{n}$ 代表参与形成大 $\\pi$ 键的电子数, 如苯分子中的大 $\\pi$ 键可表示为 $\\Pi_{6}^{6}$ 。下列说法错误的是\n\n[图1]\nA: $\\mathrm{EMIM}^{+}$中的大 $\\pi$ 键可表示为 $\\Pi_{5}^{6}$\nB: $\\mathrm{AlCl}_{4}^{-}$和 $\\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$中各原子最外层均达到 8 电子结构\nC: 充电时, 阴极反应式为 $4 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}+3 \\mathrm{e}^{-}=\\mathrm{Al}+7 \\mathrm{AlCl}_{4}^{-}$\nD: 放电时, 总反应式为 $\\mathrm{Al}+4 \\mathrm{AlCl}_{4}^{-}+\\mathrm{C}_{\\mathrm{n}}\\left[\\mathrm{AlCl}_{4}\\right]=3 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}+\\mathrm{C}_{\\mathrm{n}}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n我国学者最近研发出一种以铝为负极、石墨烯薄膜 $\\left(\\mathrm{C}_{n}\\right)$ 为正极的新型铝一石墨烯电池, $\\mathrm{AlCl}_{4}^{-}$可在石墨烯薄膜上嵌入或脱嵌, 离子液体 $\\mathrm{AlCl}_{3} /[\\mathrm{EMIM}] \\mathrm{Cl}$ 作电解质, 其中\n阴离子有 $\\mathrm{AlCl}_{4}^{-} 、 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$, 阳离子为 $\\mathrm{EMIM}^{+}\\left(\\left\\langle_{\\mathrm{N}^{\\prime}}{ }^{\\prime}{ }^{+}\\right.\\right.$, 其结构中存在大 $\\pi$ 键 $)$, 放电机理如图所示。已知: 大 $\\pi$ 键可用符号 $\\Pi_{\\mathrm{m}}^{\\mathrm{n}}$ 表示, 其中 $\\mathrm{m}$ 代表参与形成大 $\\pi$ 键的原子数, $\\mathrm{n}$ 代表参与形成大 $\\pi$ 键的电子数, 如苯分子中的大 $\\pi$ 键可表示为 $\\Pi_{6}^{6}$ 。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{EMIM}^{+}$中的大 $\\pi$ 键可表示为 $\\Pi_{5}^{6}$\nB: $\\mathrm{AlCl}_{4}^{-}$和 $\\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}$中各原子最外层均达到 8 电子结构\nC: 充电时, 阴极反应式为 $4 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}+3 \\mathrm{e}^{-}=\\mathrm{Al}+7 \\mathrm{AlCl}_{4}^{-}$\nD: 放电时, 总反应式为 $\\mathrm{Al}+4 \\mathrm{AlCl}_{4}^{-}+\\mathrm{C}_{\\mathrm{n}}\\left[\\mathrm{AlCl}_{4}\\right]=3 \\mathrm{Al}_{2} \\mathrm{Cl}_{7}^{-}+\\mathrm{C}_{\\mathrm{n}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-027.jpg?height=829&width=1017&top_left_y=636&top_left_x=331"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_162",
        "problem": "Health Canada recommends that women between the ages of 12 and 45 consume 400 micrograms of folic acid $\\left(\\mathrm{C}_{19} \\mathrm{H}_{19} \\mathrm{~N}_{7} \\mathrm{O}_{6}\\right)$ per day, which reduces the risk of neural tube defects during pregnancy. How many moles of folic acid are equivalent to 400 micrograms?\nA: $1.10 \\mathrm{mmol}$\nB: $0.177 \\mathrm{~mol}$\nC: $0.906 \\mathrm{~mol}$\nD: $0.906 \\mathrm{mmol}$\nE: $9.06 \\times 10^{-7} \\mathrm{~mol}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHealth Canada recommends that women between the ages of 12 and 45 consume 400 micrograms of folic acid $\\left(\\mathrm{C}_{19} \\mathrm{H}_{19} \\mathrm{~N}_{7} \\mathrm{O}_{6}\\right)$ per day, which reduces the risk of neural tube defects during pregnancy. How many moles of folic acid are equivalent to 400 micrograms?\n\nA: $1.10 \\mathrm{mmol}$\nB: $0.177 \\mathrm{~mol}$\nC: $0.906 \\mathrm{~mol}$\nD: $0.906 \\mathrm{mmol}$\nE: $9.06 \\times 10^{-7} \\mathrm{~mol}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_860",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaOH}$ 溶液和盐酸分别滴定体积均为 $20 \\mathrm{~mL}$, 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HA}$ 溶液与 $\\mathrm{BOH}$ 溶液。滴定过程中溶液的 $\\mathrm{pH}$ 随滴加溶液的体积变化关系如图所示。下列说法中正确的是\n\n[图1]\nA: HA 为弱酸, $\\mathrm{BOH}$ 为强碱\nB: $\\mathrm{a}$ 点时, 溶液中粒子浓度存在关系: $\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}(\\mathrm{BOH})$\nC: b 点时两种溶液中水的电离程度相同, 且 $\\mathrm{V}=20$\nD: $\\mathrm{c} 、 \\mathrm{~d}$ 两点溶液混合后微粒之间存在关系： $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}(\\mathrm{BOH})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaOH}$ 溶液和盐酸分别滴定体积均为 $20 \\mathrm{~mL}$, 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HA}$ 溶液与 $\\mathrm{BOH}$ 溶液。滴定过程中溶液的 $\\mathrm{pH}$ 随滴加溶液的体积变化关系如图所示。下列说法中正确的是\n\n[图1]\n\nA: HA 为弱酸, $\\mathrm{BOH}$ 为强碱\nB: $\\mathrm{a}$ 点时, 溶液中粒子浓度存在关系: $\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}(\\mathrm{BOH})$\nC: b 点时两种溶液中水的电离程度相同, 且 $\\mathrm{V}=20$\nD: $\\mathrm{c} 、 \\mathrm{~d}$ 两点溶液混合后微粒之间存在关系： $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}(\\mathrm{BOH})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-093.jpg?height=517&width=516&top_left_y=872&top_left_x=353"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_130",
        "problem": "7) Table 1: Successive Ionization Energies of 3rd Period Elements\n\n| Element | IE 1 <br> $\\left(\\mathbf{k j}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE2 <br> $\\left(\\mathbf{k j}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE3 <br> $\\left(\\mathbf{k j}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE4 <br> $\\left(\\mathbf{k}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE5 <br> $\\left(\\mathbf{k J}^{-1} \\mathbf{m o l}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\mathbf{V}$ | 787 | 1577 | 3231 | 4356 | 16091 |\n| $\\mathbf{W}$ | 738 | 1451 | 7733 | 10540 | 13630 |\n| $\\mathbf{X}$ | 1251 | 2297 | 3822 | 5158 | 6540 |\n| $\\mathbf{Y}$ | 496 | 4562 | 6912 | 9543 | 13353 |\n| $\\mathbf{Z}$ | 578 | 1817 | 2745 | 11575 | 14830 |\n\nBased on this information, which of the following statements is FALSE?\nA: $Y$ is sodium\nB: $\\mathrm{X}$ has the smallest radius\nC: W is an alkaline earth metal\nD: Z forms the largest cation\nE: $\\mathrm{V}$ is a semi-metal\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n7) Table 1: Successive Ionization Energies of 3rd Period Elements\n\n| Element | IE 1 <br> $\\left(\\mathbf{k j}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE2 <br> $\\left(\\mathbf{k j}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE3 <br> $\\left(\\mathbf{k j}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE4 <br> $\\left(\\mathbf{k}^{-1} \\mathbf{m o l}^{-1}\\right)$ | IE5 <br> $\\left(\\mathbf{k J}^{-1} \\mathbf{m o l}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\mathbf{V}$ | 787 | 1577 | 3231 | 4356 | 16091 |\n| $\\mathbf{W}$ | 738 | 1451 | 7733 | 10540 | 13630 |\n| $\\mathbf{X}$ | 1251 | 2297 | 3822 | 5158 | 6540 |\n| $\\mathbf{Y}$ | 496 | 4562 | 6912 | 9543 | 13353 |\n| $\\mathbf{Z}$ | 578 | 1817 | 2745 | 11575 | 14830 |\n\nBased on this information, which of the following statements is FALSE?\n\nA: $Y$ is sodium\nB: $\\mathrm{X}$ has the smallest radius\nC: W is an alkaline earth metal\nD: Z forms the largest cation\nE: $\\mathrm{V}$ is a semi-metal\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_320",
        "problem": "Consider the reaction below.\n\n$$\n2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)\n$$\n\nIn an experiment, $0.10 \\mathrm{~mol}$ of $\\mathrm{O}_{2}$ and $0.10 \\mathrm{~mol}$ of $\\mathrm{SO}_{3}$ are added to an empty 1.0-L flask and then the flask is sealed. Which of the following must be true at equilibrium?\nA: $\\left[\\mathrm{SO}_{2}\\right]=\\left[\\mathrm{O}_{2}\\right]=\\left[\\mathrm{SO}_{3}\\right]$\nB: $\\left[\\mathrm{O}_{2}\\right]<\\left[\\mathrm{SO}_{3}\\right]$\nC: $\\left[\\mathrm{O}_{2}\\right]=2\\left[\\mathrm{SO}_{2}\\right]$ The reaction must go $\\leftarrow$ to establish equilibrium.\nD: $\\left[\\mathrm{O}_{2}\\right]=\\left[\\mathrm{SO}_{2}\\right]$\nE: $\\left[\\mathrm{SO}_{3}\\right]<\\left[\\mathrm{O}_{2}\\right]$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nConsider the reaction below.\n\n$$\n2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{SO}_{3}(g)\n$$\n\nIn an experiment, $0.10 \\mathrm{~mol}$ of $\\mathrm{O}_{2}$ and $0.10 \\mathrm{~mol}$ of $\\mathrm{SO}_{3}$ are added to an empty 1.0-L flask and then the flask is sealed. Which of the following must be true at equilibrium?\n\nA: $\\left[\\mathrm{SO}_{2}\\right]=\\left[\\mathrm{O}_{2}\\right]=\\left[\\mathrm{SO}_{3}\\right]$\nB: $\\left[\\mathrm{O}_{2}\\right]<\\left[\\mathrm{SO}_{3}\\right]$\nC: $\\left[\\mathrm{O}_{2}\\right]=2\\left[\\mathrm{SO}_{2}\\right]$ The reaction must go $\\leftarrow$ to establish equilibrium.\nD: $\\left[\\mathrm{O}_{2}\\right]=\\left[\\mathrm{SO}_{2}\\right]$\nE: $\\left[\\mathrm{SO}_{3}\\right]<\\left[\\mathrm{O}_{2}\\right]$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_962",
        "problem": "傲娇, 来源日本 ACGN 的一个词语, 指人物为了掩饰害羞腼腆而做出态度强硬高傲表里不一言行的代名词, 常用来形容平常说话带刺, 态度强硬高傲, 但在一定的条件下害臊地黏淢在身边的人物。在化学中也有这样的原理, 那就是勒夏特列原理。勒夏特列对于可逆反应平衡的移动总结出的规则在后人的二次总结下总结成了一条: 平衡总是朝着能减小外界条件改变对其产生的影响的方向移动。已知相同条件下, 每个产物浓度系数次幂的连乘积与每个反应物浓度系数次幂的连乘积之比是个常数。\n\n初中的时候, 我们学到的无机基本反应类型共 4 种: 化合反应、分解反应、置换反应和复分解反应。其中, 复分解反应发生的条件是: 产物中必须存在沉淀、气体或 $\\mathrm{H}_{2} \\mathrm{O}$, 那么, 按照我们初中的知识, 有两种物质在水溶液中混合后应该不反应, 他们是我们平时吃的食盐 $\\mathrm{A}$ 和波尔多液制备中属于盐的物质 B。\n\n可是，他们混合后真的不发生反应吗?\n\n让我们来试验一下:\n\n实验 I: 分别取 $25 \\mathrm{~mL}$ 饱和 A 溶液与饱和 B 溶液, 倒入小烧杯中, 观察到溶液最终呈绿色; 经过笔式酸度计测量可知, 在此条件下测得酸度计在饱和 A 溶液溶液中的示数为 7.01 , 在饱和 B 溶液中的示数为 3.22 , 而在混合溶液中的示数为 3.07 。\n\n实验 II: 将实验 $\\mathrm{I}$ 中所得溶液均匀分成 2 份, 分别标记成 $\\mathrm{a} 、 \\mathrm{~b}$ 两组, $\\mathrm{a}$ 组放入冰水中,而 $\\mathrm{b}$ 组仍保存在小烧杯中。将小烧杯放在加热器上加热, 一段时间后观察到 $\\mathrm{a}$ 组溶液从绿色变成蓝色， $\\mathrm{b}$ 组溶液绿色加深。\n\n实验 III：向两个洁净的试管中分别加入等浓度的 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 溶液和 $\\mathrm{CuCl}_{2}$ 溶液, 使用同一热源均匀加热，观察到加入 $\\mathrm{CuCl}_{2}$ 溶液的试管内溶液绿色逐渐加深。\n\n下列说法正确的是\nA: 由实验 I 可知, A 和 B 能够发生复分解反应。\nB: 由实验 II 可知, 从蓝色向绿色的转化可逆，且此反应是吸热反应。\nC: 由实验 III 可知, 反应中影响颜色的因素中有 $\\mathrm{Cl}^{-}$和 $\\mathrm{Cu}^{2+}$ 在溶液中的浓度。\nD: 若已知铜盐水溶液中 $\\mathrm{Cu}$ 元素以 $\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right]^{2+}$ 存在, 且反应后溶液中没有新离子产生, 可推测产物为 $\\left[\\mathrm{CuCl}_{4}\\right]^{2-}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 。\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n傲娇, 来源日本 ACGN 的一个词语, 指人物为了掩饰害羞腼腆而做出态度强硬高傲表里不一言行的代名词, 常用来形容平常说话带刺, 态度强硬高傲, 但在一定的条件下害臊地黏淢在身边的人物。在化学中也有这样的原理, 那就是勒夏特列原理。勒夏特列对于可逆反应平衡的移动总结出的规则在后人的二次总结下总结成了一条: 平衡总是朝着能减小外界条件改变对其产生的影响的方向移动。已知相同条件下, 每个产物浓度系数次幂的连乘积与每个反应物浓度系数次幂的连乘积之比是个常数。\n\n初中的时候, 我们学到的无机基本反应类型共 4 种: 化合反应、分解反应、置换反应和复分解反应。其中, 复分解反应发生的条件是: 产物中必须存在沉淀、气体或 $\\mathrm{H}_{2} \\mathrm{O}$, 那么, 按照我们初中的知识, 有两种物质在水溶液中混合后应该不反应, 他们是我们平时吃的食盐 $\\mathrm{A}$ 和波尔多液制备中属于盐的物质 B。\n\n可是，他们混合后真的不发生反应吗?\n\n让我们来试验一下:\n\n实验 I: 分别取 $25 \\mathrm{~mL}$ 饱和 A 溶液与饱和 B 溶液, 倒入小烧杯中, 观察到溶液最终呈绿色; 经过笔式酸度计测量可知, 在此条件下测得酸度计在饱和 A 溶液溶液中的示数为 7.01 , 在饱和 B 溶液中的示数为 3.22 , 而在混合溶液中的示数为 3.07 。\n\n实验 II: 将实验 $\\mathrm{I}$ 中所得溶液均匀分成 2 份, 分别标记成 $\\mathrm{a} 、 \\mathrm{~b}$ 两组, $\\mathrm{a}$ 组放入冰水中,而 $\\mathrm{b}$ 组仍保存在小烧杯中。将小烧杯放在加热器上加热, 一段时间后观察到 $\\mathrm{a}$ 组溶液从绿色变成蓝色， $\\mathrm{b}$ 组溶液绿色加深。\n\n实验 III：向两个洁净的试管中分别加入等浓度的 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 溶液和 $\\mathrm{CuCl}_{2}$ 溶液, 使用同一热源均匀加热，观察到加入 $\\mathrm{CuCl}_{2}$ 溶液的试管内溶液绿色逐渐加深。\n\n下列说法正确的是\n\nA: 由实验 I 可知, A 和 B 能够发生复分解反应。\nB: 由实验 II 可知, 从蓝色向绿色的转化可逆，且此反应是吸热反应。\nC: 由实验 III 可知, 反应中影响颜色的因素中有 $\\mathrm{Cl}^{-}$和 $\\mathrm{Cu}^{2+}$ 在溶液中的浓度。\nD: 若已知铜盐水溶液中 $\\mathrm{Cu}$ 元素以 $\\left[\\mathrm{Cu}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{4}\\right]^{2+}$ 存在, 且反应后溶液中没有新离子产生, 可推测产物为 $\\left[\\mathrm{CuCl}_{4}\\right]^{2-}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 。\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_170",
        "problem": "Naphthalene is a white crystalline solid, traditionally used as the primary ingredient in mothballs. It has the chemical formula $\\mathrm{C}_{10} \\mathrm{H}_{8}$, and its structure is shown to the right. Rank the solubility of naphthalene in the following solvents from most to least soluble.\nI. water\nII. hexane\nIII. ethanol\nIV. hexanol\n\n[figure1]\nA: II $>$ IV $>$ III $>$ I\nB: IV $>$ II $>$ III $>$ I\nC: I $>$ III $>$ II $>$ IV\nD: II $>$ III $>$ IV $>$ I\nE: IV $>$ III $>$ II $>$ I\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNaphthalene is a white crystalline solid, traditionally used as the primary ingredient in mothballs. It has the chemical formula $\\mathrm{C}_{10} \\mathrm{H}_{8}$, and its structure is shown to the right. Rank the solubility of naphthalene in the following solvents from most to least soluble.\nI. water\nII. hexane\nIII. ethanol\nIV. hexanol\n\n[figure1]\n\nA: II $>$ IV $>$ III $>$ I\nB: IV $>$ II $>$ III $>$ I\nC: I $>$ III $>$ II $>$ IV\nD: II $>$ III $>$ IV $>$ I\nE: IV $>$ III $>$ II $>$ I\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_de998ba2e5d42a77dc86g-3.jpg?height=203&width=323&top_left_y=1207&top_left_x=2287"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_871",
        "problem": "Boger 吡啶合成反应是指 1, 2, 4-三唑和亲双烯体在四氢吡咯的催化作用下, 经杂\n原子 Diels-Alder 反应再脱氮生成吡啶的反应\n\n[图1]\n如下:\n\n[图2]\n\n下列说法错误的是\nA: 步骤 1,3 为加成反应\nB: 步骤 $2,4,5$ 为消去反应\nC: 步骤 1 的产物中有 1 个手性碳原子\nD: 步骤 5 的产物中最多有 5 个碳原子共平面\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\nBoger 吡啶合成反应是指 1, 2, 4-三唑和亲双烯体在四氢吡咯的催化作用下, 经杂\n原子 Diels-Alder 反应再脱氮生成吡啶的反应\n\n[图1]\n如下:\n\n[图2]\n\n下列说法错误的是\n\nA: 步骤 1,3 为加成反应\nB: 步骤 $2,4,5$ 为消去反应\nC: 步骤 1 的产物中有 1 个手性碳原子\nD: 步骤 5 的产物中最多有 5 个碳原子共平面\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-46.jpg?height=157&width=645&top_left_y=156&top_left_x=1068",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-46.jpg?height=245&width=1426&top_left_y=406&top_left_x=355",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-46.jpg?height=271&width=817&top_left_y=1235&top_left_x=631"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_646",
        "problem": "常温下, $\\mathrm{Ka}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=1.0 \\times 10^{-5}$, 向某含有 $\\mathrm{ZnSO}_{4}$ 酸性废液加入一定量 $\\mathrm{CH}_{3} \\mathrm{COONa}$后, 再通入 $\\mathrm{H}_{2} \\mathrm{~S}$ 生成 $\\mathrm{ZnS}$ 沉淀, 始终保持 $\\mathrm{H}_{2} \\mathrm{~S}$ 饱和, 即 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=0.1 \\mathrm{~mol} / \\mathrm{L}$, 体系中\n\n$\\mathrm{pX}\\left[\\mathrm{pX}=-1 \\mathrm{gX}, \\mathrm{X}\\right.$ 为 $\\frac{\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)} 、 \\frac{\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)}$或 $\\mathrm{c}\\left(\\mathrm{Zn}^{2+}\\right)$, 单位为 $\\left.\\mathrm{mol} / \\mathrm{L}\\right]$ 与 $\\mathrm{p} \\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)}$ 关系如图。下列说法错误的是\n\n[图1]\nA: (2) $\\mathrm{X}$ 为 $\\frac{\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)}$\nB: A 点溶液的 $\\mathrm{pH}$ 为 4\nC: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$ 的数量级为 $10^{-7}$\nD: $\\operatorname{Ksp}(\\mathrm{ZnS})=10^{-21.7}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, $\\mathrm{Ka}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=1.0 \\times 10^{-5}$, 向某含有 $\\mathrm{ZnSO}_{4}$ 酸性废液加入一定量 $\\mathrm{CH}_{3} \\mathrm{COONa}$后, 再通入 $\\mathrm{H}_{2} \\mathrm{~S}$ 生成 $\\mathrm{ZnS}$ 沉淀, 始终保持 $\\mathrm{H}_{2} \\mathrm{~S}$ 饱和, 即 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=0.1 \\mathrm{~mol} / \\mathrm{L}$, 体系中\n\n$\\mathrm{pX}\\left[\\mathrm{pX}=-1 \\mathrm{gX}, \\mathrm{X}\\right.$ 为 $\\frac{\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)} 、 \\frac{\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)}$或 $\\mathrm{c}\\left(\\mathrm{Zn}^{2+}\\right)$, 单位为 $\\left.\\mathrm{mol} / \\mathrm{L}\\right]$ 与 $\\mathrm{p} \\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)}$ 关系如图。下列说法错误的是\n\n[图1]\n\nA: (2) $\\mathrm{X}$ 为 $\\frac{\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)}$\nB: A 点溶液的 $\\mathrm{pH}$ 为 4\nC: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$ 的数量级为 $10^{-7}$\nD: $\\operatorname{Ksp}(\\mathrm{ZnS})=10^{-21.7}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-001.jpg?height=477&width=874&top_left_y=898&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1478",
        "problem": "For sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.The Arrhenius activation energy for this catalytic hydrolysis of sodium borohydride is $E_{\\mathrm{a}}=42.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Calculate the temperature required to achieve the same rate of hydrogen evolution by using a half of the amount of ruthenium catalyst used at $25.0^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\nHere is some context information for this question, which might assist you in solving it:\nFor sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.\n\nproblem:\nThe Arrhenius activation energy for this catalytic hydrolysis of sodium borohydride is $E_{\\mathrm{a}}=42.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Calculate the temperature required to achieve the same rate of hydrogen evolution by using a half of the amount of ruthenium catalyst used at $25.0^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir units are, in order, [K, $^{\\circ} \\mathrm{C}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            "K",
            "$^{\\circ} \\mathrm{C}$"
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_861",
        "problem": "下图为常温 $\\mathrm{Al}(\\mathrm{OH})_{3} 、 \\mathrm{Mn}(\\mathrm{OH})_{2} 、 \\mathrm{Cu}(\\mathrm{OH})_{2}$ 在水中达沉淀溶解平衡时的 $\\mathrm{pM}-\\mathrm{pH}$ 关系图 $\\left(\\mathrm{pM}=-\\lg \\left[\\mathrm{c}(\\mathrm{M}) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right), \\mathrm{M}\\right.\\right.$ 为 $\\mathrm{Al}^{3+} 、 \\mathrm{Mn}^{2+}$ 或 $\\mathrm{Cu}^{2+} ; \\mathrm{c}(\\mathrm{M}) \\leq 10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 可认为 $\\mathrm{M}$ 离子沉淀完全)。下列叙述正确的是\n\n[图1]\nA: 常温下, $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]=10^{-22.3}$\nB: 加适量 $\\mathrm{CuCl}_{2}$ 固体可使溶液由 $\\mathrm{B}$ 点变到 $\\mathrm{C}$ 点\nC: 若 $\\mathrm{A}$ 点为含 $\\mathrm{Al}^{3+} 、 \\mathrm{Cu}^{2+}$ 的混合溶液, 则 $\\frac{\\mathrm{c}\\left(\\mathrm{O}^{2+}\\right)}{\\mathrm{c}\\left(\\mathrm{A}^{3+}\\right)}=10^{-0.7}$\nD: 浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Cu}^{2+}$ 和 $\\mathrm{Mn}^{2+}$ 可通过分步沉淀进行分离\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下图为常温 $\\mathrm{Al}(\\mathrm{OH})_{3} 、 \\mathrm{Mn}(\\mathrm{OH})_{2} 、 \\mathrm{Cu}(\\mathrm{OH})_{2}$ 在水中达沉淀溶解平衡时的 $\\mathrm{pM}-\\mathrm{pH}$ 关系图 $\\left(\\mathrm{pM}=-\\lg \\left[\\mathrm{c}(\\mathrm{M}) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right), \\mathrm{M}\\right.\\right.$ 为 $\\mathrm{Al}^{3+} 、 \\mathrm{Mn}^{2+}$ 或 $\\mathrm{Cu}^{2+} ; \\mathrm{c}(\\mathrm{M}) \\leq 10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 可认为 $\\mathrm{M}$ 离子沉淀完全)。下列叙述正确的是\n\n[图1]\n\nA: 常温下, $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]=10^{-22.3}$\nB: 加适量 $\\mathrm{CuCl}_{2}$ 固体可使溶液由 $\\mathrm{B}$ 点变到 $\\mathrm{C}$ 点\nC: 若 $\\mathrm{A}$ 点为含 $\\mathrm{Al}^{3+} 、 \\mathrm{Cu}^{2+}$ 的混合溶液, 则 $\\frac{\\mathrm{c}\\left(\\mathrm{O}^{2+}\\right)}{\\mathrm{c}\\left(\\mathrm{A}^{3+}\\right)}=10^{-0.7}$\nD: 浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Cu}^{2+}$ 和 $\\mathrm{Mn}^{2+}$ 可通过分步沉淀进行分离\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-049.jpg?height=466&width=605&top_left_y=595&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_969",
        "problem": "A The energy of the electron in a hydrogen atom is restricted to certain values.\n\nB The energy of the electron in a hydrogen atom is not restricted in any way.\n\nC The electron in a hydrogen atom is restricted to one of only four possible circular orbits.\n\nD The distance between the electron and the nucleus in a hydrogen atom is restricted to certain values.\n\nE none of the above\nA: seawater\nB: table sugar\nC: brass\nD: cement\nE: smoke\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA The energy of the electron in a hydrogen atom is restricted to certain values.\n\nB The energy of the electron in a hydrogen atom is not restricted in any way.\n\nC The electron in a hydrogen atom is restricted to one of only four possible circular orbits.\n\nD The distance between the electron and the nucleus in a hydrogen atom is restricted to certain values.\n\nE none of the above\n\nA: seawater\nB: table sugar\nC: brass\nD: cement\nE: smoke\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_204",
        "problem": "One step in the manufacture of aluminium is the production of aluminium hydroxide by the following process:\n\n$2 \\mathrm{NaAlO}_{2}(\\mathrm{aq})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+\\mathrm{CO}_{2}(\\mathrm{~g}) \\rightarrow 2 \\mathrm{Al}(\\mathrm{OH})_{3}(\\mathrm{~s})+\\mathrm{Na}_{2} \\mathrm{CO}_{3}(\\mathrm{aq})$\n\nWhat mass of aluminium hydroxide can be produced from a mixture of $40 \\mathrm{~g} \\mathrm{NaAlO}_{2}$, $15 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ and $10 \\mathrm{~g} \\mathrm{CO}_{2}(\\mathrm{~g})$ ?\nA: $17.7 \\mathrm{~g}$\nB: $21.6 \\mathrm{~g}$\nC: $35.4 \\mathrm{~g}$\nD: $38.1 \\mathrm{~g}$\nE: $43.3 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOne step in the manufacture of aluminium is the production of aluminium hydroxide by the following process:\n\n$2 \\mathrm{NaAlO}_{2}(\\mathrm{aq})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+\\mathrm{CO}_{2}(\\mathrm{~g}) \\rightarrow 2 \\mathrm{Al}(\\mathrm{OH})_{3}(\\mathrm{~s})+\\mathrm{Na}_{2} \\mathrm{CO}_{3}(\\mathrm{aq})$\n\nWhat mass of aluminium hydroxide can be produced from a mixture of $40 \\mathrm{~g} \\mathrm{NaAlO}_{2}$, $15 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ and $10 \\mathrm{~g} \\mathrm{CO}_{2}(\\mathrm{~g})$ ?\n\nA: $17.7 \\mathrm{~g}$\nB: $21.6 \\mathrm{~g}$\nC: $35.4 \\mathrm{~g}$\nD: $38.1 \\mathrm{~g}$\nE: $43.3 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_74",
        "problem": "Which compound has the highest normal boiling point?\nA: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}$\nB: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$\nC: $\\mathrm{CH}_{3} \\mathrm{COOCH}_{3}$\nD: $\\mathrm{HCOOCH}_{2} \\mathrm{CH}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich compound has the highest normal boiling point?\n\nA: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}$\nB: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$\nC: $\\mathrm{CH}_{3} \\mathrm{COOCH}_{3}$\nD: $\\mathrm{HCOOCH}_{2} \\mathrm{CH}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_950",
        "problem": "乙醛酸是一种重要的化工中间体, 可用如下图所示的电化学装置合成。图中的双极膜中间层中的 $\\mathrm{H}_{2} \\mathrm{O}$ 可解离为 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$, 并在直流电场作用下分别向两极迁移。下列说法错误的是\n\n[图1]\nA: $\\mathrm{KBr}$ 在上述电化学合成过程中起电解质和反应物的作用\nB: 阳极上的电极反应式为[图2][图3]\nC: 制得 $1 \\mathrm{~mol}$ 乙醛酸, 理论上外电路中迁移了 $2 \\mathrm{~mol}$ 电子\nD: 双极膜中间层中的 $\\mathrm{H}^{+}$在外电场作用下向铅电极方向迁移\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n乙醛酸是一种重要的化工中间体, 可用如下图所示的电化学装置合成。图中的双极膜中间层中的 $\\mathrm{H}_{2} \\mathrm{O}$ 可解离为 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$, 并在直流电场作用下分别向两极迁移。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{KBr}$ 在上述电化学合成过程中起电解质和反应物的作用\nB: 阳极上的电极反应式为[图2][图3]\nC: 制得 $1 \\mathrm{~mol}$ 乙醛酸, 理论上外电路中迁移了 $2 \\mathrm{~mol}$ 电子\nD: 双极膜中间层中的 $\\mathrm{H}^{+}$在外电场作用下向铅电极方向迁移\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-41.jpg?height=634&width=1060&top_left_y=160&top_left_x=338",
            "https://i.postimg.cc/wBGRMr20/image.png",
            "https://i.postimg.cc/Xq8VGG34/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1160",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nHow many neutrons  does an iodide ion contain",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nHow many neutrons  does an iodide ion contain\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1217",
        "problem": "Propene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nA better reaction to produce large quantity of propene is the oxidative dehydrogenation $(O D H)$ using solid catalysts, such as vanadium oxides, under molecular oxygen gas. Although this type of reaction is still under intense research development, its promise toward the production of propene at an industrial scale eclipses that of the direct dehydrogenation.The overall rate of propane consumption in the reaction is\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=\\frac{1}{\\frac{p^{0}}{k_{\\text {red }} p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}+\\frac{p^{0}}{k_{o x} p_{\\mathrm{O}_{2}}}}\n$$\n\nwhere $k_{\\text {red }}$ and $k_{o x}$ are the rate constants for the reduction of metal oxide catalyst by propane and for the oxidation of the catalyst by molecular oxygen, respectively, and $p^{o}$ is the standard pressure of 1 bar. Some experiments found that the rate of\noxidation of the catalyst is 100,000 times faster than that of the propane oxidation. The experimental\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=k_{o b s} \\frac{p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}{p^{0}}\n$$\n\nwhere $k_{o b s}$ is the observed rate constant $\\left(0.062 \\mathrm{~mol} \\mathrm{~s}^{-1}\\right)$. If the reactor containing the catalyst is continuously passed through with propane and oxygen at a total pressure of 1 bar, determine the value of $k_{\\text {red }}$ when the partial pressure of propane is 0.10 bar. Assume that the partial pressure of propene is negligible.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nPropene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nA better reaction to produce large quantity of propene is the oxidative dehydrogenation $(O D H)$ using solid catalysts, such as vanadium oxides, under molecular oxygen gas. Although this type of reaction is still under intense research development, its promise toward the production of propene at an industrial scale eclipses that of the direct dehydrogenation.\n\nproblem:\nThe overall rate of propane consumption in the reaction is\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=\\frac{1}{\\frac{p^{0}}{k_{\\text {red }} p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}+\\frac{p^{0}}{k_{o x} p_{\\mathrm{O}_{2}}}}\n$$\n\nwhere $k_{\\text {red }}$ and $k_{o x}$ are the rate constants for the reduction of metal oxide catalyst by propane and for the oxidation of the catalyst by molecular oxygen, respectively, and $p^{o}$ is the standard pressure of 1 bar. Some experiments found that the rate of\noxidation of the catalyst is 100,000 times faster than that of the propane oxidation. The experimental\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=k_{o b s} \\frac{p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}{p^{0}}\n$$\n\nwhere $k_{o b s}$ is the observed rate constant $\\left(0.062 \\mathrm{~mol} \\mathrm{~s}^{-1}\\right)$. If the reactor containing the catalyst is continuously passed through with propane and oxygen at a total pressure of 1 bar, determine the value of $k_{\\text {red }}$ when the partial pressure of propane is 0.10 bar. Assume that the partial pressure of propene is negligible.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $ \\mathrm{~mol} \\mathrm{~s}^{-1}$., but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$ \\mathrm{~mol} \\mathrm{~s}^{-1}$."
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1229",
        "problem": "The production of zinc from zinc sulphide proceeds in two stages: the roasting of zinc sulphide in the air and the reduction of the zinc oxide formed by carbon monoxide. In this problem we will consider the roasting of zinc sulphide.\n\nThis operation consists in burning zinc sulphide in the air. The equation of the reaction taking place is as follows:\n\n$$\n\\mathrm{ZnS}(\\mathrm{s})+3 / 2 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{ZnO}(\\mathrm{s})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\Delta_{r} H_{1350}^{0}=-448.98 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\n\nIndustrially this reaction is carried out at $1350 \\mathrm{~K}$.\nSuppose that the zinc containing mineral contains only zinc sulphide, ZnS.\n\nStarting with a stoichiometric mixture of one mole zinc blend only and a necessary quantity of the air at $298 \\mathrm{~K}$, calculate the temperature to which the mixture will raise by the heat evolved during the roasting of the mineral at $1350 \\mathrm{~K}$ under standard pressure. Is the reaction self-sustaining? Air is considered to be a mixture of oxygen and nitrogen in a volume ratio equal to $1: 4$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe production of zinc from zinc sulphide proceeds in two stages: the roasting of zinc sulphide in the air and the reduction of the zinc oxide formed by carbon monoxide. In this problem we will consider the roasting of zinc sulphide.\n\nThis operation consists in burning zinc sulphide in the air. The equation of the reaction taking place is as follows:\n\n$$\n\\mathrm{ZnS}(\\mathrm{s})+3 / 2 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{ZnO}(\\mathrm{s})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\Delta_{r} H_{1350}^{0}=-448.98 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\n\nIndustrially this reaction is carried out at $1350 \\mathrm{~K}$.\nSuppose that the zinc containing mineral contains only zinc sulphide, ZnS.\n\nStarting with a stoichiometric mixture of one mole zinc blend only and a necessary quantity of the air at $298 \\mathrm{~K}$, calculate the temperature to which the mixture will raise by the heat evolved during the roasting of the mineral at $1350 \\mathrm{~K}$ under standard pressure. Is the reaction self-sustaining? Air is considered to be a mixture of oxygen and nitrogen in a volume ratio equal to $1: 4$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1439",
        "problem": "Which second period (row) element has the first six ionization energies (IE in electron volts, eV) listed below?\n\n| $I E_{1}$ | $I E_{2}$ | $I E_{3}$ | $I E_{4}$ | $I E_{5}$ | $I E_{6}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 11 | 24 | 48 | 64 | 392 | 490 |\nA: $\\mathrm{B}$\nB: $\\mathrm{C}$\nC: $\\mathrm{N}$\nD: $\\mathrm{O}$\nE: $\\mathrm{F}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich second period (row) element has the first six ionization energies (IE in electron volts, eV) listed below?\n\n| $I E_{1}$ | $I E_{2}$ | $I E_{3}$ | $I E_{4}$ | $I E_{5}$ | $I E_{6}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 11 | 24 | 48 | 64 | 392 | 490 |\n\nA: $\\mathrm{B}$\nB: $\\mathrm{C}$\nC: $\\mathrm{N}$\nD: $\\mathrm{O}$\nE: $\\mathrm{F}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1524",
        "problem": "Clathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nLarge methane hydrate stocks on the floor of Baikal lake, the largest freshwater lake in Russia and in the world, have been discovered in July 2009 by the crew of a deepsubmergence vehicle Mir-2. During the ascent from the depth of $1400 \\mathrm{~m}$ methane hydrate samples started to decompose at the depth of $372 \\mathrm{~m}$.Total amount of methane in hydrates on the Earth is no less than $5 \\cdot 10^{11}$ tons.\n\nBy how many degrees would the Earth atmosphere heat up, if such amount of methane is burned by reacting with atmospheric oxygen?\n\nThe enthalpy of combustion of methane is $-889 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$, the total heat capacity of the Earth's atmosphere is about $4 \\cdot 10^{21} \\mathrm{~J} \\mathrm{~K}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nClathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nLarge methane hydrate stocks on the floor of Baikal lake, the largest freshwater lake in Russia and in the world, have been discovered in July 2009 by the crew of a deepsubmergence vehicle Mir-2. During the ascent from the depth of $1400 \\mathrm{~m}$ methane hydrate samples started to decompose at the depth of $372 \\mathrm{~m}$.\n\nproblem:\nTotal amount of methane in hydrates on the Earth is no less than $5 \\cdot 10^{11}$ tons.\n\nBy how many degrees would the Earth atmosphere heat up, if such amount of methane is burned by reacting with atmospheric oxygen?\n\nThe enthalpy of combustion of methane is $-889 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$, the total heat capacity of the Earth's atmosphere is about $4 \\cdot 10^{21} \\mathrm{~J} \\mathrm{~K}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of K, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-222.jpg?height=425&width=434&top_left_y=1255&top_left_x=1459"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "K"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_284",
        "problem": "Acrylonitrile $\\left(\\mathrm{C}_{3} \\mathrm{H}_{3} \\mathrm{~N}\\right)$ can be synthesised industrially according to the following chemical equation:\n\n$2 \\mathrm{C}_{3} \\mathrm{H}_{6}(\\mathrm{~g})+2 \\mathrm{NH}_{3}(\\mathrm{~g})+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow 2 \\mathrm{C}_{3} \\mathrm{H}_{3} \\mathrm{~N}(\\mathrm{~g})+6 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\nWhen $100 \\mathrm{~kg}$ of $\\mathrm{C}_{3} \\mathrm{H}_{6}, 50 \\mathrm{~kg}$ of $\\mathrm{NH}_{3}$ and $125 \\mathrm{~kg}$ of $\\mathrm{O}_{2}$ are mixed, which of these reactants is present in excess? Select all that apply.\nA: $\\mathrm{C}_{3} \\mathrm{H}_{6}$\nB: $\\mathrm{NH}_{3}$\nC: $\\mathrm{O}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nAcrylonitrile $\\left(\\mathrm{C}_{3} \\mathrm{H}_{3} \\mathrm{~N}\\right)$ can be synthesised industrially according to the following chemical equation:\n\n$2 \\mathrm{C}_{3} \\mathrm{H}_{6}(\\mathrm{~g})+2 \\mathrm{NH}_{3}(\\mathrm{~g})+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow 2 \\mathrm{C}_{3} \\mathrm{H}_{3} \\mathrm{~N}(\\mathrm{~g})+6 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\nWhen $100 \\mathrm{~kg}$ of $\\mathrm{C}_{3} \\mathrm{H}_{6}, 50 \\mathrm{~kg}$ of $\\mathrm{NH}_{3}$ and $125 \\mathrm{~kg}$ of $\\mathrm{O}_{2}$ are mixed, which of these reactants is present in excess? Select all that apply.\n\nA: $\\mathrm{C}_{3} \\mathrm{H}_{6}$\nB: $\\mathrm{NH}_{3}$\nC: $\\mathrm{O}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1246",
        "problem": "Using the information provided on this graph, give numerical answers with appropriate units to the following questions:\n\n[figure1]\n\nWhat are the binding energies of $\\mathrm{H}_{2}$ and $\\mathrm{H}_{2}^{+}$?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nUsing the information provided on this graph, give numerical answers with appropriate units to the following questions:\n\n[figure1]\n\nWhat are the binding energies of $\\mathrm{H}_{2}$ and $\\mathrm{H}_{2}^{+}$?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [$E_{\\text {bond }}$, $E_{\\text {bond }}\\left(\\mathrm{H}_{2}^{+}\\right)$].\nTheir units are, in order, [$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-471.jpg?height=911&width=1279&top_left_y=590&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$",
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": [
            "$E_{\\text {bond }}$",
            "$E_{\\text {bond }}\\left(\\mathrm{H}_{2}^{+}\\right)$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_786",
        "problem": "下列方案设计、现象和结论都正确的是\n\n| 目的 | 方案设计 | 现象和结论 |\n| :---: | :---: | :---: |\n| 除去碳酸钠 <br> 固体中碳酸 <br> 氢钠 | 将固体放置在坩埚中加热, 一段时间后, <br> 在干燥器中冷却、称量整个坩埚与固体的 <br> 质量为 $\\mathrm{m}_{1}$, 再次加热、冷却、称量质量 <br> 为 $\\mathrm{m}_{2}$ | 若 $\\mathrm{m}_{1}=\\mathrm{m}_{2}$, 则说明杂质已 <br> 除尽 |\n| 检验苯中的 <br> 苯酚 | 取少量样品, 加入适量的浓溴水, 观察现 <br> 象 | 滴加溴水后, 若未出现白 <br> 色沉淀, 则说明苯中不存 <br> 在苯酚 |\n| 探究铁与水 <br> 蒸气反应后 <br> 固体物质 | 取少量固体粉末, 用磁铁吸引, 观察现象 | 若磁铁能吸引, 则固体物 <br> 质为 $\\mathrm{Fe}_{3} \\mathrm{O}_{4}$ |\n| 检验含氟化 <br> 钠的牙膏中 <br> 的 $\\mathrm{F}^{-}$ | 向 $\\mathrm{Fe}(\\mathrm{SCN})_{3}$ 稀溶液中加过量含氟牙膏的 <br> 浸泡液 | 血红色褪去说明含 $F^{-}$ |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列方案设计、现象和结论都正确的是\n\n| 目的 | 方案设计 | 现象和结论 |\n| :---: | :---: | :---: |\n| 除去碳酸钠 <br> 固体中碳酸 <br> 氢钠 | 将固体放置在坩埚中加热, 一段时间后, <br> 在干燥器中冷却、称量整个坩埚与固体的 <br> 质量为 $\\mathrm{m}_{1}$, 再次加热、冷却、称量质量 <br> 为 $\\mathrm{m}_{2}$ | 若 $\\mathrm{m}_{1}=\\mathrm{m}_{2}$, 则说明杂质已 <br> 除尽 |\n| 检验苯中的 <br> 苯酚 | 取少量样品, 加入适量的浓溴水, 观察现 <br> 象 | 滴加溴水后, 若未出现白 <br> 色沉淀, 则说明苯中不存 <br> 在苯酚 |\n| 探究铁与水 <br> 蒸气反应后 <br> 固体物质 | 取少量固体粉末, 用磁铁吸引, 观察现象 | 若磁铁能吸引, 则固体物 <br> 质为 $\\mathrm{Fe}_{3} \\mathrm{O}_{4}$ |\n| 检验含氟化 <br> 钠的牙膏中 <br> 的 $\\mathrm{F}^{-}$ | 向 $\\mathrm{Fe}(\\mathrm{SCN})_{3}$ 稀溶液中加过量含氟牙膏的 <br> 浸泡液 | 血红色褪去说明含 $F^{-}$ |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-39.jpg?height=48&width=1391&top_left_y=2483&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_782",
        "problem": "羟基自由基 $(\\cdot \\mathrm{OH})$ 具有极强的氧化性, 一种能将苯酚氧化为 $\\mathrm{CO}_{2}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 的组合装置如图所示。下列说法错误的是\n\n[图1]\nA: 离子交换膜 $\\mathrm{n}$ 为阴离子交换膜\nB: $\\mathrm{a}$ 极的电极反应式为 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{-}+6 \\mathrm{e}^{-}+7 \\mathrm{H}_{2} \\mathrm{O}=2 \\mathrm{Cr}(\\mathrm{OH})_{3} \\downarrow+8 \\mathrm{OH}^{-}$\nC: b、c 两极产生气体的体积比为 $3: 7$ (标准状况)\nD: 电池工作时, $\\mathrm{d}$ 极室溶液 $\\mathrm{pH}$ 逐渐增大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n羟基自由基 $(\\cdot \\mathrm{OH})$ 具有极强的氧化性, 一种能将苯酚氧化为 $\\mathrm{CO}_{2}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 的组合装置如图所示。下列说法错误的是\n\n[图1]\n\nA: 离子交换膜 $\\mathrm{n}$ 为阴离子交换膜\nB: $\\mathrm{a}$ 极的电极反应式为 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{-}+6 \\mathrm{e}^{-}+7 \\mathrm{H}_{2} \\mathrm{O}=2 \\mathrm{Cr}(\\mathrm{OH})_{3} \\downarrow+8 \\mathrm{OH}^{-}$\nC: b、c 两极产生气体的体积比为 $3: 7$ (标准状况)\nD: 电池工作时, $\\mathrm{d}$ 极室溶液 $\\mathrm{pH}$ 逐渐增大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-19.jpg?height=458&width=1149&top_left_y=522&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_514",
        "problem": "将 $0.8 \\mathrm{~mol} \\mathrm{~S}$ 放入一个装满 $\\mathrm{O}_{2}$ 的恒压密闭容器中, 在 $\\mathrm{T}^{\\circ} \\mathrm{C} 、 1 \\times 10^{5} \\mathrm{~Pa}$ 的条件下使之充分反应至 $\\mathrm{t}_{1}$ 时刻, 此时容器中无 $\\mathrm{S}$ 剩余, 剩余 $0.2 \\mathrm{~mol} \\mathrm{O}_{2}$. 再向该体系中迅速投入 $1 \\mathrm{~mol}$无水 $\\mathrm{FeSO}_{4}$ 固体, 此过程容器与外界未发生气体交换, 充分反应至 $\\mathrm{t}_{2}$ 时刻。已知此条件下可以发生如下反应:\n\n$\\mathrm{S}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g})=\\mathrm{SO}_{2}(\\mathrm{~g})$\n\n$2 \\mathrm{FeSO}_{4}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Fe}_{2} \\mathrm{O}_{3}(\\mathrm{~s})+\\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{SO}_{3}(\\mathrm{~g}) \\quad \\mathrm{K}_{\\mathrm{P} 1}=8.1 \\times 10^{7} \\mathrm{~Pa}^{2}$\n\n$2 \\mathrm{SO}_{3}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\quad \\mathrm{K}_{\\mathrm{P} 2}=7.2 \\times 10^{3} \\mathrm{~Pa}$\n\n下列说法不正确的是\nA: $\\mathrm{t}_{1}$ 时刻, 容器中气体总物质的量为 $1.0 \\mathrm{~mol}$\nB: $t_{1}$ 时刻, 容器中 $\\mathrm{SO}_{2}(\\mathrm{~g})$ 的分压为 $3 \\times 10^{4} \\mathrm{~Pa}$\nC: 起始时, 容器中 $n\\left(\\mathrm{O}_{2}\\right)=1.25 \\mathrm{~mol}$\nD: $\\mathrm{t}_{2}$ 时刻, 容器中的固体为 $\\mathrm{FeSO}_{4}$ 与 $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ 的混合物\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将 $0.8 \\mathrm{~mol} \\mathrm{~S}$ 放入一个装满 $\\mathrm{O}_{2}$ 的恒压密闭容器中, 在 $\\mathrm{T}^{\\circ} \\mathrm{C} 、 1 \\times 10^{5} \\mathrm{~Pa}$ 的条件下使之充分反应至 $\\mathrm{t}_{1}$ 时刻, 此时容器中无 $\\mathrm{S}$ 剩余, 剩余 $0.2 \\mathrm{~mol} \\mathrm{O}_{2}$. 再向该体系中迅速投入 $1 \\mathrm{~mol}$无水 $\\mathrm{FeSO}_{4}$ 固体, 此过程容器与外界未发生气体交换, 充分反应至 $\\mathrm{t}_{2}$ 时刻。已知此条件下可以发生如下反应:\n\n$\\mathrm{S}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g})=\\mathrm{SO}_{2}(\\mathrm{~g})$\n\n$2 \\mathrm{FeSO}_{4}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Fe}_{2} \\mathrm{O}_{3}(\\mathrm{~s})+\\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{SO}_{3}(\\mathrm{~g}) \\quad \\mathrm{K}_{\\mathrm{P} 1}=8.1 \\times 10^{7} \\mathrm{~Pa}^{2}$\n\n$2 \\mathrm{SO}_{3}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\quad \\mathrm{K}_{\\mathrm{P} 2}=7.2 \\times 10^{3} \\mathrm{~Pa}$\n\n下列说法不正确的是\n\nA: $\\mathrm{t}_{1}$ 时刻, 容器中气体总物质的量为 $1.0 \\mathrm{~mol}$\nB: $t_{1}$ 时刻, 容器中 $\\mathrm{SO}_{2}(\\mathrm{~g})$ 的分压为 $3 \\times 10^{4} \\mathrm{~Pa}$\nC: 起始时, 容器中 $n\\left(\\mathrm{O}_{2}\\right)=1.25 \\mathrm{~mol}$\nD: $\\mathrm{t}_{2}$ 时刻, 容器中的固体为 $\\mathrm{FeSO}_{4}$ 与 $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ 的混合物\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-010.jpg?height=219&width=670&top_left_y=667&top_left_x=342"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_939",
        "problem": "将下图所示实验装置的 $\\mathrm{K}$ 闭合, 下列判断正确的是\n\n[图1]\nA: 片刻后甲池中 $\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)$ 增大\nB: 电子沿 $\\mathrm{Zn} \\rightarrow \\mathrm{a} \\rightarrow \\mathrm{b} \\rightarrow \\mathrm{Cu}$ 路径流动\nC: $\\mathrm{Cu}$ 电极上发生还原反应\nD: 片刻后可观察到滤纸 a 点变红色\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将下图所示实验装置的 $\\mathrm{K}$ 闭合, 下列判断正确的是\n\n[图1]\n\nA: 片刻后甲池中 $\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)$ 增大\nB: 电子沿 $\\mathrm{Zn} \\rightarrow \\mathrm{a} \\rightarrow \\mathrm{b} \\rightarrow \\mathrm{Cu}$ 路径流动\nC: $\\mathrm{Cu}$ 电极上发生还原反应\nD: 片刻后可观察到滤纸 a 点变红色\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-57.jpg?height=525&width=649&top_left_y=180&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_876",
        "problem": "利用电解原理实现乙烯高效合成环氧乙烷, 电解装置如图所示:\n\n已知: $\\mathrm{CH}_{2}=\\mathrm{CH}_{2} \\xrightarrow{\\mathrm{HClO}} \\mathrm{HOCH}_{2}-\\mathrm{CH}_{2} \\mathrm{Cl} \\xrightarrow{\\mathrm{OH}^{-}}{ }^{\\mathrm{O}}$, 下列说法正确的是\n\n[图1]\n\n$\\mathrm{Cl}^{-}$离子交换膜\nA: 反应一段时间后，阴极区 $\\mathrm{pH}$ 升高\nB: $\\mathrm{Pt}$ 电极反应式为: $\\mathrm{Cl}^{-}-2 \\mathrm{e}^{-}+2 \\mathrm{OH}^{-}=\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 生成 $1 \\mathrm{~mol}$ 环氧乙烷, 有 $1 \\mathrm{molCl}^{-}$通过交换膜\nD: 电解完成后, 将阴极区和阳极区溶液混合后可得到环氧乙烷\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用电解原理实现乙烯高效合成环氧乙烷, 电解装置如图所示:\n\n已知: $\\mathrm{CH}_{2}=\\mathrm{CH}_{2} \\xrightarrow{\\mathrm{HClO}} \\mathrm{HOCH}_{2}-\\mathrm{CH}_{2} \\mathrm{Cl} \\xrightarrow{\\mathrm{OH}^{-}}{ }^{\\mathrm{O}}$, 下列说法正确的是\n\n[图1]\n\n$\\mathrm{Cl}^{-}$离子交换膜\n\nA: 反应一段时间后，阴极区 $\\mathrm{pH}$ 升高\nB: $\\mathrm{Pt}$ 电极反应式为: $\\mathrm{Cl}^{-}-2 \\mathrm{e}^{-}+2 \\mathrm{OH}^{-}=\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 生成 $1 \\mathrm{~mol}$ 环氧乙烷, 有 $1 \\mathrm{molCl}^{-}$通过交换膜\nD: 电解完成后, 将阴极区和阳极区溶液混合后可得到环氧乙烷\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-12.jpg?height=308&width=625&top_left_y=817&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_812",
        "problem": "研究表明, 用 $\\mathrm{V}_{2} \\mathrm{O}_{5}$ 作催化剂促进水分解时存在两种不同的路径, 分解过程中的部分反应历程如图所示(物质中原子之间的距离单位为 $\\AA)$ 。下列说法错误的是\n\n[图1]\nA: 水的分解反应为放热反应\nB: 反应历程中, 钒原子的杂化方式发生改变\nC: IM 2 中, 距离为“ $2.174 \\AA$ 和“ $2.390 \\AA$ ”的原子之间作用力是氢键\nD: 适当升高温度, IM2 $\\rightarrow \\mathrm{FS} 3$ 的正反应速率增大的程度小于逆反应速率增大的程度\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n研究表明, 用 $\\mathrm{V}_{2} \\mathrm{O}_{5}$ 作催化剂促进水分解时存在两种不同的路径, 分解过程中的部分反应历程如图所示(物质中原子之间的距离单位为 $\\AA)$ 。下列说法错误的是\n\n[图1]\n\nA: 水的分解反应为放热反应\nB: 反应历程中, 钒原子的杂化方式发生改变\nC: IM 2 中, 距离为“ $2.174 \\AA$ 和“ $2.390 \\AA$ ”的原子之间作用力是氢键\nD: 适当升高温度, IM2 $\\rightarrow \\mathrm{FS} 3$ 的正反应速率增大的程度小于逆反应速率增大的程度\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-052.jpg?height=554&width=780&top_left_y=177&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1298",
        "problem": "If the density of a liquid compound $\\mathrm{B}$ is $\\rho$ (in $\\mathrm{g} \\mathrm{cm}^{-3}$ ), $M$ is the molar mass of $\\mathrm{B}$ and $N_{A}$ is the Avogadro constant, then the number of molecules of $B$ in $1 \\mathrm{dm}^{3}$ of this compound is:\nA: $(1000 \\times \\rho) /\\left(M \\times N_{A}\\right)$\nB: $\\left(1000 \\times \\rho \\times N_{A}\\right) / M$\nC: $\\left(N_{A} \\times \\rho\\right) /(M \\times 1000)$\nD: $\\left(N_{A} \\times \\rho \\times M\\right) / 1000$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf the density of a liquid compound $\\mathrm{B}$ is $\\rho$ (in $\\mathrm{g} \\mathrm{cm}^{-3}$ ), $M$ is the molar mass of $\\mathrm{B}$ and $N_{A}$ is the Avogadro constant, then the number of molecules of $B$ in $1 \\mathrm{dm}^{3}$ of this compound is:\n\nA: $(1000 \\times \\rho) /\\left(M \\times N_{A}\\right)$\nB: $\\left(1000 \\times \\rho \\times N_{A}\\right) / M$\nC: $\\left(N_{A} \\times \\rho\\right) /(M \\times 1000)$\nD: $\\left(N_{A} \\times \\rho \\times M\\right) / 1000$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_357",
        "problem": "In which of the following substances is chlorine in the lowest oxidation state?\nA: $\\mathrm{Cl}_{2}$\nB: $\\mathrm{KCl}$\nC: $\\mathrm{KClO}$\nD: $\\mathrm{KClO}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn which of the following substances is chlorine in the lowest oxidation state?\n\nA: $\\mathrm{Cl}_{2}$\nB: $\\mathrm{KCl}$\nC: $\\mathrm{KClO}$\nD: $\\mathrm{KClO}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_44",
        "problem": "A reaction initially contains pure reactant $\\mathbf{R}$, which forms product $\\mathbf{P}$ reversibly. Which reaction coordinate diagram corresponds to the reaction in which $\\mathbf{P}$ achieves a concentration within $5 \\%$ of its equilibrium value the fastest?\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA reaction initially contains pure reactant $\\mathbf{R}$, which forms product $\\mathbf{P}$ reversibly. Which reaction coordinate diagram corresponds to the reaction in which $\\mathbf{P}$ achieves a concentration within $5 \\%$ of its equilibrium value the fastest?\n\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-06.jpg?height=282&width=328&top_left_y=932&top_left_x=1256",
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-06.jpg?height=284&width=328&top_left_y=1224&top_left_x=1256",
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-06.jpg?height=285&width=320&top_left_y=931&top_left_x=1683",
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-06.jpg?height=284&width=314&top_left_y=1224&top_left_x=1689"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_533",
        "problem": "利用间接成对电化学合成间氨基苯甲酸的工作原理如图所示。下列说法错误的是\n[图1]\n离子交换膜\nA: 阳极的电极反应式为: $2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}-6 \\mathrm{e}^{-}=\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+}$\nB: 阳极槽外氧化反应为[图2]\nC: 通电时阳极区 $\\mathrm{pH}$ 增大\nD: 当电路中转移 $1 \\mathrm{~mole}-$ 时, 理论上可得到 $1 \\mathrm{~mol}$ 间氨基苯甲酸\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用间接成对电化学合成间氨基苯甲酸的工作原理如图所示。下列说法错误的是\n[图1]\n离子交换膜\n\nA: 阳极的电极反应式为: $2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}-6 \\mathrm{e}^{-}=\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+14 \\mathrm{H}^{+}$\nB: 阳极槽外氧化反应为[图2]\nC: 通电时阳极区 $\\mathrm{pH}$ 增大\nD: 当电路中转移 $1 \\mathrm{~mole}-$ 时, 理论上可得到 $1 \\mathrm{~mol}$ 间氨基苯甲酸\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-70.jpg?height=458&width=1330&top_left_y=1528&top_left_x=338",
            "https://i.postimg.cc/CKXY91W7/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_324",
        "problem": "When a 10.0-g sample of a mixture of $\\mathrm{CH}_{4}$ and $\\mathrm{C}_{2} \\mathrm{H}_{6}$ is burned excess oxygen, exactly $525 \\mathrm{~kJ}$ of heat is produced. What is the percentage by mass of $\\mathrm{CH}_{4}$ in the original mixture?\n\n$$\n\\begin{aligned}\n& \\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\rightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n& \\Delta H=-890.4 \\mathrm{~kJ}\\left(\\text { per mol } \\mathrm{CH}_{4}\\right)\n\\end{aligned}\n$$\n$\\mathrm{CH}_{4}, 16.042 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n$\\mathrm{C}_{2} \\mathrm{H}_{6}, 30.068 \\mathrm{~g} \\mathrm{~mol}^{-1}$\nA: $17 \\%$\nB: $21 \\%$\nC: $34 \\%$\nD: $59 \\%$\nE: $87 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen a 10.0-g sample of a mixture of $\\mathrm{CH}_{4}$ and $\\mathrm{C}_{2} \\mathrm{H}_{6}$ is burned excess oxygen, exactly $525 \\mathrm{~kJ}$ of heat is produced. What is the percentage by mass of $\\mathrm{CH}_{4}$ in the original mixture?\n\n$$\n\\begin{aligned}\n& \\mathrm{CH}_{4}(g)+2 \\mathrm{O}_{2}(g) \\rightarrow \\mathrm{CO}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l) \\\\\n& \\Delta H=-890.4 \\mathrm{~kJ}\\left(\\text { per mol } \\mathrm{CH}_{4}\\right)\n\\end{aligned}\n$$\n$\\mathrm{CH}_{4}, 16.042 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n$\\mathrm{C}_{2} \\mathrm{H}_{6}, 30.068 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nA: $17 \\%$\nB: $21 \\%$\nC: $34 \\%$\nD: $59 \\%$\nE: $87 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1038",
        "problem": "Diiodine pentoxide, $\\mathrm{I}_{2} \\mathrm{O}_{5}$, is a white crystalline powder that has the useful property of reacting quantitatively with carbon monoxide to yield iodine and one other product.\n\n[figure1]\n\nA $150 \\mathrm{~cm}^{3}$ sample of gas (at room temperature and pressure, r.t.p.) that was known to contain carbon monoxide was repeatedly passed over excess $\\mathrm{I}_{2} \\mathrm{O}_{5}$ at $170{ }^{\\circ} \\mathrm{C}$. The $\\mathrm{I}_{2} \\mathrm{O}_{5}$ became coloured with iodine. The iodine generated required exactly $8.00 \\mathrm{~cm}^{3}$ of 0.100 $\\mathrm{mol} \\mathrm{dm}^{-3}$ sodium thiosulfate solution to react with it. This reaction is:\n\n$$\n\\mathrm{I}_{2(\\mathrm{aq})}+2 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3(\\mathrm{aq})} \\longrightarrow 2 \\mathrm{NaI}_{(\\mathrm{aq})}+\\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6(\\mathrm{aq})}\n$$\n\nCalculate the percentage by volume of carbon monoxide present in the sample of gas. [Assume 1 mol of any gas occupies $24.0 \\mathrm{dm}^{3}$ at r.t.p.]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nDiiodine pentoxide, $\\mathrm{I}_{2} \\mathrm{O}_{5}$, is a white crystalline powder that has the useful property of reacting quantitatively with carbon monoxide to yield iodine and one other product.\n\n[figure1]\n\nA $150 \\mathrm{~cm}^{3}$ sample of gas (at room temperature and pressure, r.t.p.) that was known to contain carbon monoxide was repeatedly passed over excess $\\mathrm{I}_{2} \\mathrm{O}_{5}$ at $170{ }^{\\circ} \\mathrm{C}$. The $\\mathrm{I}_{2} \\mathrm{O}_{5}$ became coloured with iodine. The iodine generated required exactly $8.00 \\mathrm{~cm}^{3}$ of 0.100 $\\mathrm{mol} \\mathrm{dm}^{-3}$ sodium thiosulfate solution to react with it. This reaction is:\n\n$$\n\\mathrm{I}_{2(\\mathrm{aq})}+2 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3(\\mathrm{aq})} \\longrightarrow 2 \\mathrm{NaI}_{(\\mathrm{aq})}+\\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6(\\mathrm{aq})}\n$$\n\nCalculate the percentage by volume of carbon monoxide present in the sample of gas. [Assume 1 mol of any gas occupies $24.0 \\mathrm{dm}^{3}$ at r.t.p.]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %(percentage), but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_4408eb29f8e936fbb54dg-04.jpg?height=851&width=694&top_left_y=368&top_left_x=1044"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%(percentage)"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_131",
        "problem": "Assuming the oxidation number of chlorine is -1 , the oxidation numbers of iodine and antimony in the compound $\\left[\\mathrm{Cl}_{2}\\right]^{+}\\left[\\mathrm{SbCl}_{6}\\right]^{-}$are respectively\nA: +2 and +6\nB: +2 and +5\nC: +1 and +7\nD: +3 and +5\nE: +3 and +7\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAssuming the oxidation number of chlorine is -1 , the oxidation numbers of iodine and antimony in the compound $\\left[\\mathrm{Cl}_{2}\\right]^{+}\\left[\\mathrm{SbCl}_{6}\\right]^{-}$are respectively\n\nA: +2 and +6\nB: +2 and +5\nC: +1 and +7\nD: +3 and +5\nE: +3 and +7\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1481",
        "problem": "Sulphuric acid is produced by catalytic oxidation of $\\mathrm{SO}_{2}$ to $\\mathrm{SO}_{3}$, absorption of $\\mathrm{SO}_{3}$ in concentrated sulphuric acid forming oleum (containing $20 \\% \\mathrm{SO}_{3}$ by mass) and appropriate dilution hereafter. The gas leaving the catalyst chamber contains nitrogen, oxygen, a trace of $\\mathrm{SO}_{2}$ and $10 \\%$ (by volume) of $\\mathrm{SO}_{3}$. Sulphur trioxide, $\\mathrm{SO}_{3}$, is converted into sulphuric acid (98\\% by mass) and/or oleum.\n\nAssuming that oleum is the only product formed, calculate the mass of water which is required for $1000 \\mathrm{~m}^{3}$ of gas leaving the catalyst chamber ( $273 \\mathrm{~K}, 101.3 \\mathrm{kPa}$ ).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nSulphuric acid is produced by catalytic oxidation of $\\mathrm{SO}_{2}$ to $\\mathrm{SO}_{3}$, absorption of $\\mathrm{SO}_{3}$ in concentrated sulphuric acid forming oleum (containing $20 \\% \\mathrm{SO}_{3}$ by mass) and appropriate dilution hereafter. The gas leaving the catalyst chamber contains nitrogen, oxygen, a trace of $\\mathrm{SO}_{2}$ and $10 \\%$ (by volume) of $\\mathrm{SO}_{3}$. Sulphur trioxide, $\\mathrm{SO}_{3}$, is converted into sulphuric acid (98\\% by mass) and/or oleum.\n\nAssuming that oleum is the only product formed, calculate the mass of water which is required for $1000 \\mathrm{~m}^{3}$ of gas leaving the catalyst chamber ( $273 \\mathrm{~K}, 101.3 \\mathrm{kPa}$ ).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_691",
        "problem": "下列有关实验操作、现象和结论的叙述均正确的是\n\n| 选 | 实验操作 | 实验现象 | 结论 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{A}$ | 向盛有 $\\mathrm{Na}_{2} \\mathrm{~S}$ 固体的试管中滴加浓硫酸| 产生有臭鸡蛋气味的 气体 | 酸性: 硫酸>硫化 <br> 氢 | \n| $\\mathrm{B}$ | 充满试管, 倒置在饱和食盐水中, 用 <br> 可见光照射 | 和食盐水进入到试管 <br> 中体 | 中某些产物呈液态 <br> 或易溶于水 |\n| $\\mathrm{C}$ | 常温下, 用胶头滴管吸取某盐溶液滴 在 $\\mathrm{pH}$ 试纸中央, 然后与标准比色卡对 <br> 比| 试纸颜色与标准比色卡中 $\\mathrm{pH}=7$ 的颜色相同  | 该盐属于强酸强碱盐 |\n| $\\mathrm{D}$ | 镀铜铁的镀层破损后, 浸泡在盐酸中 <br> 一段时间, 加入几滴 $\\mathrm{KSCN}$ 溶液 | 溶液未显血红色 | 无法说明整个过 <br> 程是否生成 $\\mathrm{Fe}(\\mathrm{III})$ |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列有关实验操作、现象和结论的叙述均正确的是\n\n| 选 | 实验操作 | 实验现象 | 结论 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{A}$ | 向盛有 $\\mathrm{Na}_{2} \\mathrm{~S}$ 固体的试管中滴加浓硫酸| 产生有臭鸡蛋气味的 气体 | 酸性: 硫酸>硫化 <br> 氢 | \n| $\\mathrm{B}$ | 充满试管, 倒置在饱和食盐水中, 用 <br> 可见光照射 | 和食盐水进入到试管 <br> 中体 | 中某些产物呈液态 <br> 或易溶于水 |\n| $\\mathrm{C}$ | 常温下, 用胶头滴管吸取某盐溶液滴 在 $\\mathrm{pH}$ 试纸中央, 然后与标准比色卡对 <br> 比| 试纸颜色与标准比色卡中 $\\mathrm{pH}=7$ 的颜色相同  | 该盐属于强酸强碱盐 |\n| $\\mathrm{D}$ | 镀铜铁的镀层破损后, 浸泡在盐酸中 <br> 一段时间, 加入几滴 $\\mathrm{KSCN}$ 溶液 | 溶液未显血红色 | 无法说明整个过 <br> 程是否生成 $\\mathrm{Fe}(\\mathrm{III})$ |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1064",
        "problem": "Increasing concerns over the use and generation of hazardous substances in chemical processes has encouraged some chemists to look for more environmentally friendly ways to make chemical products. To help evaluate a process environmentally, chemists often use the term 'percentage atom economy', where\n\n$\\%$ Atom Economy $\\quad=\\quad$ RMM of desired product $\\times 100$ RMM of all products\n\n[figure1]\n\nAn environmentally friendly chemical process would normally be expected to have a high $\\%$ atom economy, indicating that a high proportion of the starting materials end up as part of the final product, hence reducing the amount of waste. Efforts are constantly being made to increase the \\% atom economy of chemical processes. As an example, the manufacture of ethene oxide $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\\right)$ for many years was via the classical chlorohydrin route:\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\Longleftrightarrow \\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{HCl} \\\\\n\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{Ca}(\\mathrm{OH})_{2}+\\mathrm{HCl} \\Longleftrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}+\\mathrm{CaCl}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nCalculate the $\\%$ atom economy for this process.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIncreasing concerns over the use and generation of hazardous substances in chemical processes has encouraged some chemists to look for more environmentally friendly ways to make chemical products. To help evaluate a process environmentally, chemists often use the term 'percentage atom economy', where\n\n$\\%$ Atom Economy $\\quad=\\quad$ RMM of desired product $\\times 100$ RMM of all products\n\n[figure1]\n\nAn environmentally friendly chemical process would normally be expected to have a high $\\%$ atom economy, indicating that a high proportion of the starting materials end up as part of the final product, hence reducing the amount of waste. Efforts are constantly being made to increase the \\% atom economy of chemical processes. As an example, the manufacture of ethene oxide $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\\right)$ for many years was via the classical chlorohydrin route:\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\Longleftrightarrow \\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{HCl} \\\\\n\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{Ca}(\\mathrm{OH})_{2}+\\mathrm{HCl} \\Longleftrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}+\\mathrm{CaCl}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nCalculate the $\\%$ atom economy for this process.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-05.jpg?height=454&width=343&top_left_y=316&top_left_x=1482"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_732",
        "problem": "乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 与 $\\mathrm{NH}_{3}$ 相似, 水溶液呈碱性, $25^{\\circ} \\mathrm{C}$ 时, 向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$乙二胺溶液中滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸, 各组分的物质的量分数 $\\delta$ 随溶液 $\\mathrm{pH}$ 的变化曲线如图所示, 下列说法错误的是\n\n[图1]\nA: 混合溶液呈中性时, 滴加盐酸的体积小于 $10 \\mathrm{~mL}$\nB: $\\mathrm{K}_{\\mathrm{b} 2}\\left[\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right]$ 的数量级为 $10^{-8}$\nC: $\\mathrm{a}$ 点所处的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+3 \\mathrm{c}\\left(\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{2+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+(\\mathrm{Cl})$\nD: 水的电离程度: $\\mathrm{a}>\\mathrm{b}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 与 $\\mathrm{NH}_{3}$ 相似, 水溶液呈碱性, $25^{\\circ} \\mathrm{C}$ 时, 向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$乙二胺溶液中滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸, 各组分的物质的量分数 $\\delta$ 随溶液 $\\mathrm{pH}$ 的变化曲线如图所示, 下列说法错误的是\n\n[图1]\n\nA: 混合溶液呈中性时, 滴加盐酸的体积小于 $10 \\mathrm{~mL}$\nB: $\\mathrm{K}_{\\mathrm{b} 2}\\left[\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right]$ 的数量级为 $10^{-8}$\nC: $\\mathrm{a}$ 点所处的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+3 \\mathrm{c}\\left(\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{2+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+(\\mathrm{Cl})$\nD: 水的电离程度: $\\mathrm{a}>\\mathrm{b}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-67.jpg?height=517&width=805&top_left_y=2009&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_664",
        "problem": "天然水体中的 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与空气中的 $\\mathrm{CO}_{2}$ 保持平衡。某地溶洞水体中 $\\lg c(\\mathrm{X})(\\mathrm{X}$ 为\n\n$\\mathrm{H}_{2} \\mathrm{CO}_{3} 、 \\mathrm{HCO}_{3}^{-} 、 \\mathrm{CO}_{3}^{2-}$ 或 $\\left.\\mathrm{Ca}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的线性关系如图所示。下列说法不正确的是\n\n[图1]\nA: 直线(1)代表 $\\mathrm{HCO}_{3}^{-}$, 直线(3)代表 $\\mathrm{Ca}^{2+}$\nB: 图中 $\\lg c\\left(\\mathrm{CO}_{3}^{2-}\\right)=2 \\mathrm{pH}+\\lg K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\lg K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)-5$\nC: $K_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)=10^{-8.6}$\nD: 若空气中的 $\\mathrm{CO}_{2}$ 浓度增加, 则水体中的 $\\mathrm{Ca}^{2+}$ 浓度减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n天然水体中的 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与空气中的 $\\mathrm{CO}_{2}$ 保持平衡。某地溶洞水体中 $\\lg c(\\mathrm{X})(\\mathrm{X}$ 为\n\n$\\mathrm{H}_{2} \\mathrm{CO}_{3} 、 \\mathrm{HCO}_{3}^{-} 、 \\mathrm{CO}_{3}^{2-}$ 或 $\\left.\\mathrm{Ca}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的线性关系如图所示。下列说法不正确的是\n\n[图1]\n\nA: 直线(1)代表 $\\mathrm{HCO}_{3}^{-}$, 直线(3)代表 $\\mathrm{Ca}^{2+}$\nB: 图中 $\\lg c\\left(\\mathrm{CO}_{3}^{2-}\\right)=2 \\mathrm{pH}+\\lg K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\lg K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)-5$\nC: $K_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)=10^{-8.6}$\nD: 若空气中的 $\\mathrm{CO}_{2}$ 浓度增加, 则水体中的 $\\mathrm{Ca}^{2+}$ 浓度减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-110.jpg?height=572&width=697&top_left_y=1413&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1339",
        "problem": "The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nChoose the correct option for change in the entropy of the system $\\left(\\Delta S_{\\text {sys }}\\right)$ and of the surroundings $\\left(\\Delta S_{\\text {sur }}\\right)$ :\nA: $\\Delta S_{\\text {sys }}>0 \\quad \\Delta S_{\\text {sur }}=0$\nB: $\\Delta S_{\\text {sys }}<0 \\quad \\Delta S_{\\text {sur }}>0$\nC: $\\Delta S_{\\text {sys }}>0 \\quad \\Delta S_{\\text {sur }}<0$\nD: $\\Delta S_{\\text {sys }}=0 \\quad \\Delta S_{\\text {sur }}=0$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nChoose the correct option for change in the entropy of the system $\\left(\\Delta S_{\\text {sys }}\\right)$ and of the surroundings $\\left(\\Delta S_{\\text {sur }}\\right)$ :\n\nA: $\\Delta S_{\\text {sys }}>0 \\quad \\Delta S_{\\text {sur }}=0$\nB: $\\Delta S_{\\text {sys }}<0 \\quad \\Delta S_{\\text {sur }}>0$\nC: $\\Delta S_{\\text {sys }}>0 \\quad \\Delta S_{\\text {sur }}<0$\nD: $\\Delta S_{\\text {sys }}=0 \\quad \\Delta S_{\\text {sur }}=0$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_829",
        "problem": "$\\mathrm{ClO}_{2}$ 是一种高效、低毒的消毒剂。实验室用 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 、盐酸、 $\\mathrm{NaClO}_{2}$ (亚氯酸钠)为原料, 制备 $\\mathrm{ClO}_{2}$ 的流程如图:\n\n[图1]\n已知: (1)电解过程中氯元素被氧化。\n\n(2) $\\mathrm{ClO}_{2}$ 是一种黄绿色易溶于水的气体; 三氯化氮为黄色油状液体, 熔点较低, 很不稳定，受热 $90^{\\circ} \\mathrm{C}$ 以上或受震动时发生猛烈爆炸。下列说法正确的是\nA: “电解”过程中的阳极电极反应式为 $3 \\mathrm{Cl}^{-}-6 \\mathrm{e}^{-}+\\mathrm{NH}_{4}^{+}=\\mathrm{NCl}_{3}+4 \\mathrm{H}^{+}$\nB: “反应”过程中可快速搅拌反应混合液以加快反应速率\nC: “反应”过程中的还原产物存在于溶液 X 中\nD: 可用稀盐酸除去 $\\mathrm{ClO}_{2}$ 中的氨气\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{ClO}_{2}$ 是一种高效、低毒的消毒剂。实验室用 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 、盐酸、 $\\mathrm{NaClO}_{2}$ (亚氯酸钠)为原料, 制备 $\\mathrm{ClO}_{2}$ 的流程如图:\n\n[图1]\n已知: (1)电解过程中氯元素被氧化。\n\n(2) $\\mathrm{ClO}_{2}$ 是一种黄绿色易溶于水的气体; 三氯化氮为黄色油状液体, 熔点较低, 很不稳定，受热 $90^{\\circ} \\mathrm{C}$ 以上或受震动时发生猛烈爆炸。下列说法正确的是\n\nA: “电解”过程中的阳极电极反应式为 $3 \\mathrm{Cl}^{-}-6 \\mathrm{e}^{-}+\\mathrm{NH}_{4}^{+}=\\mathrm{NCl}_{3}+4 \\mathrm{H}^{+}$\nB: “反应”过程中可快速搅拌反应混合液以加快反应速率\nC: “反应”过程中的还原产物存在于溶液 X 中\nD: 可用稀盐酸除去 $\\mathrm{ClO}_{2}$ 中的氨气\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-20.jpg?height=285&width=1034&top_left_y=2476&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_777",
        "problem": "MFC一电芬顿技术不需要外加能即可发生, 通过产生羟基自由基 $(\\cdot \\mathrm{OH})$ 处理有机污染物, 可获得高效的废水净化效果。其耦合系统原理示意图如图, 下列说法不正确的是\n\n[图1]\nA: 电子移动方向为 $\\mathrm{a} \\rightarrow \\mathrm{Y}, \\mathrm{X} \\rightarrow \\mathrm{b}$\nB: $\\mathrm{Y}$ 电极上得到双氧水的反应为 $\\mathrm{O}_{2}+2 \\mathrm{e}^{-}+2 \\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}_{2}$\nC: 乙池可在酸性较弱的环境中使用\nD: 理论上当乙池中产生 $1 \\mathrm{~mol}$ 羟基自由基时, 甲池中有 $4 \\mathrm{molH}^{+}$从 $\\mathrm{M}$ 室移动到 $\\mathrm{N}$ 室\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\nMFC一电芬顿技术不需要外加能即可发生, 通过产生羟基自由基 $(\\cdot \\mathrm{OH})$ 处理有机污染物, 可获得高效的废水净化效果。其耦合系统原理示意图如图, 下列说法不正确的是\n\n[图1]\n\nA: 电子移动方向为 $\\mathrm{a} \\rightarrow \\mathrm{Y}, \\mathrm{X} \\rightarrow \\mathrm{b}$\nB: $\\mathrm{Y}$ 电极上得到双氧水的反应为 $\\mathrm{O}_{2}+2 \\mathrm{e}^{-}+2 \\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}_{2}$\nC: 乙池可在酸性较弱的环境中使用\nD: 理论上当乙池中产生 $1 \\mathrm{~mol}$ 羟基自由基时, 甲池中有 $4 \\mathrm{molH}^{+}$从 $\\mathrm{M}$ 室移动到 $\\mathrm{N}$ 室\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-40.jpg?height=574&width=1108&top_left_y=413&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_899",
        "problem": "常温下, 在含 $\\mathrm{MnCl}_{2}$ 和 $\\mathrm{HCN}$ 的混合溶液中滴加 $\\mathrm{NaOH}$ 溶液, 得到的浊液中 $\\mathrm{pX}\\left[\\mathrm{pX}=-\\operatorname{lgc}\\left(\\mathrm{Mn}^{2+}\\right)\\right.$ 或 $\\left.-1 \\mathrm{~g} \\frac{\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)}{\\mathrm{c}(\\mathrm{HCN})}\\right]$ 与由的关系如图所示。已知, $10^{0.3=}=2$ 。下列复述正确的是\n\n[图1]\nA: $\\mathrm{L}_{1}$ 代表 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)}{\\mathrm{c}(\\mathrm{HCN})}$ 与 $\\mathrm{pH}$ 的关系\nB: 常温下, $\\mathrm{K}_{\\mathrm{a}}(\\mathrm{HCN})=5.0 \\times 10^{-11}$\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Mn}(\\mathrm{OH})_{2}\\right]$ 的数量级为 $10^{-13}$\nD: $\\mathrm{Mn}(\\mathrm{OH})_{2}+2 \\mathrm{HCN}=\\mathrm{Mn}^{2+}+2 \\mathrm{CN}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}$ 平衡常数 $\\mathrm{K}=5.0 \\times 10^{4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 在含 $\\mathrm{MnCl}_{2}$ 和 $\\mathrm{HCN}$ 的混合溶液中滴加 $\\mathrm{NaOH}$ 溶液, 得到的浊液中 $\\mathrm{pX}\\left[\\mathrm{pX}=-\\operatorname{lgc}\\left(\\mathrm{Mn}^{2+}\\right)\\right.$ 或 $\\left.-1 \\mathrm{~g} \\frac{\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)}{\\mathrm{c}(\\mathrm{HCN})}\\right]$ 与由的关系如图所示。已知, $10^{0.3=}=2$ 。下列复述正确的是\n\n[图1]\n\nA: $\\mathrm{L}_{1}$ 代表 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)}{\\mathrm{c}(\\mathrm{HCN})}$ 与 $\\mathrm{pH}$ 的关系\nB: 常温下, $\\mathrm{K}_{\\mathrm{a}}(\\mathrm{HCN})=5.0 \\times 10^{-11}$\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Mn}(\\mathrm{OH})_{2}\\right]$ 的数量级为 $10^{-13}$\nD: $\\mathrm{Mn}(\\mathrm{OH})_{2}+2 \\mathrm{HCN}=\\mathrm{Mn}^{2+}+2 \\mathrm{CN}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}$ 平衡常数 $\\mathrm{K}=5.0 \\times 10^{4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-004.jpg?height=523&width=626&top_left_y=435&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_869",
        "problem": "一种具有广谱抗菌活性的有机物结构简式如图, 下列关于该有机物的说法错误的是\n\n[图1]\nA: 其分子式为 $\\mathrm{C}_{16} \\mathrm{H}_{15} \\mathrm{NO}_{7}$\nB: 其水解产物均为反式结构\nC: 该分子中有 6 种官能团\nD: $1 \\mathrm{~mol}$ 该物质与足量 $\\mathrm{NaOH}$ 溶液反应时消耗 $5 \\mathrm{molNaOH}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种具有广谱抗菌活性的有机物结构简式如图, 下列关于该有机物的说法错误的是\n\n[图1]\n\nA: 其分子式为 $\\mathrm{C}_{16} \\mathrm{H}_{15} \\mathrm{NO}_{7}$\nB: 其水解产物均为反式结构\nC: 该分子中有 6 种官能团\nD: $1 \\mathrm{~mol}$ 该物质与足量 $\\mathrm{NaOH}$ 溶液反应时消耗 $5 \\mathrm{molNaOH}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-15.jpg?height=497&width=737&top_left_y=777&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-15.jpg?height=338&width=988&top_left_y=1778&top_left_x=585"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_671",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 将 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液滴加到 $10 \\mathrm{~mL} 0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液中, $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)}$ 与 $\\mathrm{pH}$ 的关系如图所示, $\\mathrm{C}$ 点坐标是 $(6,1.7)$ 。下列说法不正确的是\n\n[图1]\nA: $\\mathrm{pH}=11$ 时, $\\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)}$ 的数量级为 $10^{6}$\nB: 溶液中, 水的电离程度: $\\mathrm{C}>\\mathrm{A}$ [图2]\nC: $\\mathrm{CH}_{3} \\mathrm{COONa}$ 的水解常数 $\\mathrm{K}_{\\mathrm{h}}=10^{-9.7}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 将 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液滴加到 $10 \\mathrm{~mL} 0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液中, $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)}$ 与 $\\mathrm{pH}$ 的关系如图所示, $\\mathrm{C}$ 点坐标是 $(6,1.7)$ 。下列说法不正确的是\n\n[图1]\n\nA: $\\mathrm{pH}=11$ 时, $\\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)}$ 的数量级为 $10^{6}$\nB: 溶液中, 水的电离程度: $\\mathrm{C}>\\mathrm{A}$ [图2]\nC: $\\mathrm{CH}_{3} \\mathrm{COONa}$ 的水解常数 $\\mathrm{K}_{\\mathrm{h}}=10^{-9.7}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-100.jpg?height=320&width=1094&top_left_y=2079&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-100.jpg?height=48&width=846&top_left_y=2666&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1570",
        "problem": "Potentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nWhat are the units of the pre-logarithm factor 0.0590 in the Nernst equation?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nPotentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nWhat are the units of the pre-logarithm factor 0.0590 in the Nernst equation?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_599",
        "problem": "一种利用含钴废料(主要成分为 $\\mathrm{Co}_{3} \\mathrm{O}_{4}$, 还含有少量 $\\mathrm{SiO}_{2} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3} 、 \\mathrm{Al}_{2} \\mathrm{O}_{3}$ 等杂质)制备 $\\mathrm{Li} \\mathrm{COO}_{2}$ 的工艺流程如下。下列说法错误的是\n\n[图1]\n\n已知：常温下，部分金属离子沉淀的 $\\mathrm{pH}$ 如下表。\n\n| 金属离子 | $\\mathrm{Fe}^{2+}$ | $\\mathrm{Fe}^{3+}$ | $\\mathrm{Al}^{3+}$ | $\\mathrm{Co}^{2+}$ |\n| :--- | :--- | :--- | :--- | :--- |\n| 开始沉淀的 $\\mathrm{pH}$ | 8.5 | 2.2 | 3.4 | 7.8 |\n| 完全沉淀 $\\left[\\mathrm{c}(\\right.$ 金属离子 $\\left.) \\leq 1 \\times 10^{-5} \\mathrm{~mol} / \\mathrm{L}\\right]$ 的 $\\mathrm{pH}$ | 10.0 | 3.0 | 4.5 | 9.4 |\nA: “酸浸”时可用 $\\mathrm{SO}_{2}$ 代替 $\\mathrm{H}_{2} \\mathrm{O}_{2}$\nB: “调 $\\mathrm{pH}$ ”的范围为 4.5 7.8, 滤液中残留的 $c\\left(\\mathrm{Al}^{3+}\\right): c\\left(\\mathrm{Fe}^{3+}\\right)=10^{4.5}$\nC: 气体 X 可在“调 $\\mathrm{pH}$ ”操作单元回收利用\nD: “高温焙烧”时的化学方程式: $4 \\mathrm{CoCO}_{3}+2 \\mathrm{Li}_{2} \\mathrm{CO}_{3}+\\mathrm{O}_{2} \\stackrel{\\text { 高温 }}{=} 4 \\mathrm{LiCoO}_{2}+6 \\mathrm{CO}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一种利用含钴废料(主要成分为 $\\mathrm{Co}_{3} \\mathrm{O}_{4}$, 还含有少量 $\\mathrm{SiO}_{2} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3} 、 \\mathrm{Al}_{2} \\mathrm{O}_{3}$ 等杂质)制备 $\\mathrm{Li} \\mathrm{COO}_{2}$ 的工艺流程如下。下列说法错误的是\n\n[图1]\n\n已知：常温下，部分金属离子沉淀的 $\\mathrm{pH}$ 如下表。\n\n| 金属离子 | $\\mathrm{Fe}^{2+}$ | $\\mathrm{Fe}^{3+}$ | $\\mathrm{Al}^{3+}$ | $\\mathrm{Co}^{2+}$ |\n| :--- | :--- | :--- | :--- | :--- |\n| 开始沉淀的 $\\mathrm{pH}$ | 8.5 | 2.2 | 3.4 | 7.8 |\n| 完全沉淀 $\\left[\\mathrm{c}(\\right.$ 金属离子 $\\left.) \\leq 1 \\times 10^{-5} \\mathrm{~mol} / \\mathrm{L}\\right]$ 的 $\\mathrm{pH}$ | 10.0 | 3.0 | 4.5 | 9.4 |\n\nA: “酸浸”时可用 $\\mathrm{SO}_{2}$ 代替 $\\mathrm{H}_{2} \\mathrm{O}_{2}$\nB: “调 $\\mathrm{pH}$ ”的范围为 4.5 7.8, 滤液中残留的 $c\\left(\\mathrm{Al}^{3+}\\right): c\\left(\\mathrm{Fe}^{3+}\\right)=10^{4.5}$\nC: 气体 X 可在“调 $\\mathrm{pH}$ ”操作单元回收利用\nD: “高温焙烧”时的化学方程式: $4 \\mathrm{CoCO}_{3}+2 \\mathrm{Li}_{2} \\mathrm{CO}_{3}+\\mathrm{O}_{2} \\stackrel{\\text { 高温 }}{=} 4 \\mathrm{LiCoO}_{2}+6 \\mathrm{CO}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-014.jpg?height=294&width=1243&top_left_y=1012&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_184",
        "problem": "A $0.48 \\mathrm{~g}$ piece of magnesium metal is placed in hydrochloric acid. Assuming the hydrochloric acid is in excess and the magnesium reacts completely, how many grams of hydrogen gas are produced?\nA: $0.010 \\mathrm{~g}$\nB: $0.040 \\mathrm{~g}$\nC: $0.080 \\mathrm{~g}$\nD: $0.48 \\mathrm{~g}$\nE: $0.96 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $0.48 \\mathrm{~g}$ piece of magnesium metal is placed in hydrochloric acid. Assuming the hydrochloric acid is in excess and the magnesium reacts completely, how many grams of hydrogen gas are produced?\n\nA: $0.010 \\mathrm{~g}$\nB: $0.040 \\mathrm{~g}$\nC: $0.080 \\mathrm{~g}$\nD: $0.48 \\mathrm{~g}$\nE: $0.96 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_767",
        "problem": "已知：(1)离子浓度为 $10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 时可认为沉淀完全;\n\n(2) $\\mathrm{Al}(\\mathrm{OH})_{3}+\\mathrm{OH}^{-}=\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}$。向 $\\mathrm{MgCl}_{2}$ 和 $\\mathrm{AlCl}_{3}$ 混合溶液中加入 $\\mathrm{NaOH}$ 溶液, $\\mathrm{Mg}$ 和 $\\mathrm{Al}$ 两种元素在水溶液中的存在形式与 $\\mathrm{pH}$ 的关系如图所示, 纵轴 $\\lg =\\operatorname{lgc}\\left(\\mathrm{M}^{\\mathrm{n}}\\right)$ 或\n$\\operatorname{lgc}\\left(\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}\\right) 。$\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n下列说法正确的是:\nA: 图 1 中 $\\mathrm{M}$ 代表 $\\mathrm{Mg}^{2+}$\nB: $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]=10^{-33.91}$\nC: $\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$ 完全沉淀时, 溶液 $\\mathrm{pH}=6.37$\nD: 该实验中, 沉淀的物质的量随 $\\mathrm{NaOH}$ 溶液体积变化图象如图 2\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知：(1)离子浓度为 $10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 时可认为沉淀完全;\n\n(2) $\\mathrm{Al}(\\mathrm{OH})_{3}+\\mathrm{OH}^{-}=\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}$。向 $\\mathrm{MgCl}_{2}$ 和 $\\mathrm{AlCl}_{3}$ 混合溶液中加入 $\\mathrm{NaOH}$ 溶液, $\\mathrm{Mg}$ 和 $\\mathrm{Al}$ 两种元素在水溶液中的存在形式与 $\\mathrm{pH}$ 的关系如图所示, 纵轴 $\\lg =\\operatorname{lgc}\\left(\\mathrm{M}^{\\mathrm{n}}\\right)$ 或\n$\\operatorname{lgc}\\left(\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]^{-}\\right) 。$\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n下列说法正确的是:\n\nA: 图 1 中 $\\mathrm{M}$ 代表 $\\mathrm{Mg}^{2+}$\nB: $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Al}(\\mathrm{OH})_{3}\\right]=10^{-33.91}$\nC: $\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$ 完全沉淀时, 溶液 $\\mathrm{pH}=6.37$\nD: 该实验中, 沉淀的物质的量随 $\\mathrm{NaOH}$ 溶液体积变化图象如图 2\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-109.jpg?height=431&width=685&top_left_y=270&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-109.jpg?height=428&width=691&top_left_y=254&top_left_x=1094"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1527",
        "problem": "The transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nThe next steps in the overall reaction involve the amine $\\left(\\mathrm{RNH}_{2}\\right)$ and ${ }^{t} \\mathrm{BuONa}$. To determine the order with respect to $\\mathrm{RNH}_{2}$ and ${ }^{t} \\mathrm{BuONa}$, the dependence of the initial rates of the reaction on the concentrations of these two reagents was carried with the other reagents present in large excess. Some results are shown in the tables below.\n\n| $\\left[\\mathrm{NaO}{ }^{t} \\mathrm{Bu}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.2 | 0.6 | 0.9 | 0.12 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.16 \\times 10^{-5}$ | $4.12 \\times 10^{-5}$ | $4.24 \\times 10^{-5}$ | $4.20 \\times 10^{-5}$ |\n\n\n| $\\left[\\mathrm{RNH}_{2}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.3 | 0.6 | 0.9 | 1.2 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $4.26 \\times 10^{-5}$ | $4.21 \\times 10^{-5}$ | $4.23 \\times 10^{-5}$ |Determine the order with $\\mathrm{[RNH_2]}$, assuming each is an integer. (Use the grids if you like)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nThe next steps in the overall reaction involve the amine $\\left(\\mathrm{RNH}_{2}\\right)$ and ${ }^{t} \\mathrm{BuONa}$. To determine the order with respect to $\\mathrm{RNH}_{2}$ and ${ }^{t} \\mathrm{BuONa}$, the dependence of the initial rates of the reaction on the concentrations of these two reagents was carried with the other reagents present in large excess. Some results are shown in the tables below.\n\n| $\\left[\\mathrm{NaO}{ }^{t} \\mathrm{Bu}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.2 | 0.6 | 0.9 | 0.12 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.16 \\times 10^{-5}$ | $4.12 \\times 10^{-5}$ | $4.24 \\times 10^{-5}$ | $4.20 \\times 10^{-5}$ |\n\n\n| $\\left[\\mathrm{RNH}_{2}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.3 | 0.6 | 0.9 | 1.2 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $4.26 \\times 10^{-5}$ | $4.21 \\times 10^{-5}$ | $4.23 \\times 10^{-5}$ |\n\nproblem:\nDetermine the order with $\\mathrm{[RNH_2]}$, assuming each is an integer. (Use the grids if you like)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_165",
        "problem": "Canadian cities began adding fluoride to their drinking water in the mid-1950s, as a means to reducing dental caries (cavities) in children.\n\nSodium fluorosilicate ( $\\mathrm{Na}_{2} \\mathrm{SiF}_{6}$ ), a by-product of the industrial processing of phosphate minerals, can be added to drinking water to produce free fluoride ions. A drinking water analysis for a Canadian city revealed the following data:\n\n$$\n\\mathrm{pH}=7.60 \\quad[\\mathrm{~F}-]=3.2 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}\n$$\n\nGiven that the pKa of hydrofluoric acid (HF) is 3.17, determine the concentration of $\\mathrm{HF}$ in the drinking water of this Canadian city. Assume a water temperature of $25^{\\circ} \\mathrm{C}$.\nA: $5.4 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $3.2 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $6.4 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $1.9 \\times 10^{-8} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $1.2 \\times 10^{-9} \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCanadian cities began adding fluoride to their drinking water in the mid-1950s, as a means to reducing dental caries (cavities) in children.\n\nSodium fluorosilicate ( $\\mathrm{Na}_{2} \\mathrm{SiF}_{6}$ ), a by-product of the industrial processing of phosphate minerals, can be added to drinking water to produce free fluoride ions. A drinking water analysis for a Canadian city revealed the following data:\n\n$$\n\\mathrm{pH}=7.60 \\quad[\\mathrm{~F}-]=3.2 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}\n$$\n\nGiven that the pKa of hydrofluoric acid (HF) is 3.17, determine the concentration of $\\mathrm{HF}$ in the drinking water of this Canadian city. Assume a water temperature of $25^{\\circ} \\mathrm{C}$.\n\nA: $5.4 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $3.2 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $6.4 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $1.9 \\times 10^{-8} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $1.2 \\times 10^{-9} \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_364",
        "problem": "In a sample consisting of $1.00 \\mathrm{~mol} \\mathrm{NaBr}$ and $0.300 \\mathrm{~mol}$ $\\mathrm{KI}$, what is the mass percent of iodine?\nA: $24.9 \\%$\nB: $32.6 \\%$\nC: $47.2 \\%$\nD: $83.1 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a sample consisting of $1.00 \\mathrm{~mol} \\mathrm{NaBr}$ and $0.300 \\mathrm{~mol}$ $\\mathrm{KI}$, what is the mass percent of iodine?\n\nA: $24.9 \\%$\nB: $32.6 \\%$\nC: $47.2 \\%$\nD: $83.1 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_740",
        "problem": "下列对有关实验现象的解释或所得结论错误的是\n\n| 选 <br> 项 | 实验操作 | 现象 | 解释或结论 |\n| :---: | :---: | :---: | :---: |\n| A | 向 $2 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中滴加 <br> $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Zn} \\mathrm{SO}_{4}$ 溶液至不再有沉淀 <br> 产生, 再滴加几滴 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CuSO}_{4}$ <br> 溶液 | 先产生白色沉淀, <br> 又出现黑色沉淀 | $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS})$ |\n| B | 将少量铜粉加入稀硫酸中, 无明显 <br> 现象, 再加入硝酸铁溶液 | 铜粉溶解 | $\\mathrm{Fe}^{3+}$ 与铜粉反应 |\n| C | 将 $25^{\\circ} \\mathrm{C} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液加 <br> 热到 $40^{\\circ} \\mathrm{C}$, 用传感器监测溶液 $\\mathrm{pH}$ <br> 变化 | 溶液的 $\\mathrm{pH}$ 逐渐减 <br> 小 | 温度升高, $\\mathrm{SO}_{3}^{2-}$ 的水 <br> 解增大程度大于水的 <br> 电离增大程度 |\n| $\\mathrm{D}$ | 取 $2 \\mathrm{~mL}$ 某卤代烃样品于试管中, 加 <br> 入 $5 \\mathrm{~mL} 20 \\% \\mathrm{KOH}$ 溶液并加热, 冷却 <br> 到室温后加入足量稀硝酸再滴加 <br> $\\mathrm{AgNO}_{3}$ 溶液 | 产生白色沉淀 | 该卤代烃中含有氯元 <br> 素 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列对有关实验现象的解释或所得结论错误的是\n\n| 选 <br> 项 | 实验操作 | 现象 | 解释或结论 |\n| :---: | :---: | :---: | :---: |\n| A | 向 $2 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中滴加 <br> $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Zn} \\mathrm{SO}_{4}$ 溶液至不再有沉淀 <br> 产生, 再滴加几滴 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CuSO}_{4}$ <br> 溶液 | 先产生白色沉淀, <br> 又出现黑色沉淀 | $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS})$ |\n| B | 将少量铜粉加入稀硫酸中, 无明显 <br> 现象, 再加入硝酸铁溶液 | 铜粉溶解 | $\\mathrm{Fe}^{3+}$ 与铜粉反应 |\n| C | 将 $25^{\\circ} \\mathrm{C} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液加 <br> 热到 $40^{\\circ} \\mathrm{C}$, 用传感器监测溶液 $\\mathrm{pH}$ <br> 变化 | 溶液的 $\\mathrm{pH}$ 逐渐减 <br> 小 | 温度升高, $\\mathrm{SO}_{3}^{2-}$ 的水 <br> 解增大程度大于水的 <br> 电离增大程度 |\n| $\\mathrm{D}$ | 取 $2 \\mathrm{~mL}$ 某卤代烃样品于试管中, 加 <br> 入 $5 \\mathrm{~mL} 20 \\% \\mathrm{KOH}$ 溶液并加热, 冷却 <br> 到室温后加入足量稀硝酸再滴加 <br> $\\mathrm{AgNO}_{3}$ 溶液 | 产生白色沉淀 | 该卤代烃中含有氯元 <br> 素 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_192",
        "problem": "The boiling points of the compounds acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$, 1-pentanol $\\left(\\mathrm{C}_{5} \\mathrm{H}_{11} \\mathrm{OH}\\right)$, dibutyl ether $\\left(\\mathrm{C}_{8} \\mathrm{H}_{18} \\mathrm{O}\\right)$ and dodecane $\\left(\\mathrm{C}_{12} \\mathrm{H}_{26}\\right)$ increase in that order. Which of the following statements provides the best explanation for this increase?\nA: The London dispersion forces increase\nB: The hydrogen bonding increases\nC: The dipole-dipole forces increase\nD: The chemical reactivity increases\nE: The number of carbon atoms increases\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe boiling points of the compounds acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$, 1-pentanol $\\left(\\mathrm{C}_{5} \\mathrm{H}_{11} \\mathrm{OH}\\right)$, dibutyl ether $\\left(\\mathrm{C}_{8} \\mathrm{H}_{18} \\mathrm{O}\\right)$ and dodecane $\\left(\\mathrm{C}_{12} \\mathrm{H}_{26}\\right)$ increase in that order. Which of the following statements provides the best explanation for this increase?\n\nA: The London dispersion forces increase\nB: The hydrogen bonding increases\nC: The dipole-dipole forces increase\nD: The chemical reactivity increases\nE: The number of carbon atoms increases\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_572",
        "problem": "2019 年诺贝尔化学奖授予三位开发锂离子电池的科学家。某高能锂离子电池的反应方程式为 $\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{CoO}_{2}+\\mathrm{Li}_{\\mathrm{x}} \\mathrm{C}_{6} \\underset{\\text { 充电 }}{\\text { 放电 }} \\mathrm{LiCoO}_{2}+\\mathrm{C}_{6}(\\mathrm{x}<1)$ 。以该锂离子电池为电源、苯乙酮为原\n的是\n\n[图1]\nA: 电池放电时, $\\mathrm{Li}^{+}$向 $\\mathrm{a}$ 极移动\nB: $\\mathrm{M}$ 为阳离子交换膜\nC: 电池充电时, $\\mathrm{b}$ 极反应式为 $\\mathrm{LiCoO}_{2}-\\mathrm{xe}^{-}=\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{CoO}_{2} \\mathrm{~V}+\\mathrm{xLi}^{+}$\nD: 生成苯甲酸盐的反应[图2]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n2019 年诺贝尔化学奖授予三位开发锂离子电池的科学家。某高能锂离子电池的反应方程式为 $\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{CoO}_{2}+\\mathrm{Li}_{\\mathrm{x}} \\mathrm{C}_{6} \\underset{\\text { 充电 }}{\\text { 放电 }} \\mathrm{LiCoO}_{2}+\\mathrm{C}_{6}(\\mathrm{x}<1)$ 。以该锂离子电池为电源、苯乙酮为原\n的是\n\n[图1]\n\nA: 电池放电时, $\\mathrm{Li}^{+}$向 $\\mathrm{a}$ 极移动\nB: $\\mathrm{M}$ 为阳离子交换膜\nC: 电池充电时, $\\mathrm{b}$ 极反应式为 $\\mathrm{LiCoO}_{2}-\\mathrm{xe}^{-}=\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{CoO}_{2} \\mathrm{~V}+\\mathrm{xLi}^{+}$\nD: 生成苯甲酸盐的反应[图2]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-39.jpg?height=426&width=853&top_left_y=318&top_left_x=333",
            "https://i.postimg.cc/1zx8wFwZ/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_819",
        "problem": "银- Ferrozine 法检测甲醛(HCHO)的原理为: (1)在原电池装置中, 氧化银能将甲醛充分氧化为 $\\mathrm{CO}_{2}$; (2) $\\mathrm{Fe}^{3+}$ 与产生的 $\\mathrm{Ag}$ 定量反应生成 $\\mathrm{Fe}^{2+}$; (3) $\\mathrm{Fe}^{2+}$ 与 ferrozine 形成有色配合物: (4)测定溶液的吸光度(吸光度与溶液中有色物质的浓度成正比)。下列说法正确的是\nA: (1)中, 正极的电极反应式为 $\\mathrm{Ag}_{2} \\mathrm{O}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=2 \\mathrm{Ag}+\\mathrm{H}_{2} \\mathrm{O}$\nB: (1)中, 溶液中的 $\\mathrm{H}^{+}$由正极移向负极。\nC: 理论上消耗的甲醛与生成的 $\\mathrm{Fe}^{2+}$ 的物质的量之比为 1: 4\nD: (4)中, 甲醛浓度越大, 吸光度越小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n银- Ferrozine 法检测甲醛(HCHO)的原理为: (1)在原电池装置中, 氧化银能将甲醛充分氧化为 $\\mathrm{CO}_{2}$; (2) $\\mathrm{Fe}^{3+}$ 与产生的 $\\mathrm{Ag}$ 定量反应生成 $\\mathrm{Fe}^{2+}$; (3) $\\mathrm{Fe}^{2+}$ 与 ferrozine 形成有色配合物: (4)测定溶液的吸光度(吸光度与溶液中有色物质的浓度成正比)。下列说法正确的是\n\nA: (1)中, 正极的电极反应式为 $\\mathrm{Ag}_{2} \\mathrm{O}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=2 \\mathrm{Ag}+\\mathrm{H}_{2} \\mathrm{O}$\nB: (1)中, 溶液中的 $\\mathrm{H}^{+}$由正极移向负极。\nC: 理论上消耗的甲醛与生成的 $\\mathrm{Fe}^{2+}$ 的物质的量之比为 1: 4\nD: (4)中, 甲醛浓度越大, 吸光度越小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_361",
        "problem": "An orbital has the radial wavefunction shown below. What orbital is it?\n\n[figure1]\nA: $1 \\mathrm{~s}$\nB: $2 \\mathrm{~s}$\nC: $3 s$\nD: $4 \\mathrm{~s}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn orbital has the radial wavefunction shown below. What orbital is it?\n\n[figure1]\n\nA: $1 \\mathrm{~s}$\nB: $2 \\mathrm{~s}$\nC: $3 s$\nD: $4 \\mathrm{~s}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_08_3c4530151c35262e5915g-7.jpg?height=439&width=743&top_left_y=1014&top_left_x=214"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_326",
        "problem": "The following figure shows the contents and pressures of three vessels of gas which are joined by a connecting tube.\n\n[figure1]\n\nAfter the valves on the vessels are opened, the final pressure is measured and found to be $0.675 \\mathrm{~atm}$. What is the total volume of the connecting tube? All vessels are at a constant temperature of $25^{\\circ} \\mathrm{C}$.\nA: $\\quad 0.53 \\mathrm{~L}$\nB: $\\quad 0.056 \\mathrm{~L}$\nC: $\\quad 0.094 \\mathrm{~L}$\nD: $\\quad 0.040 \\mathrm{~L}$\nE: $\\quad 0.023 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following figure shows the contents and pressures of three vessels of gas which are joined by a connecting tube.\n\n[figure1]\n\nAfter the valves on the vessels are opened, the final pressure is measured and found to be $0.675 \\mathrm{~atm}$. What is the total volume of the connecting tube? All vessels are at a constant temperature of $25^{\\circ} \\mathrm{C}$.\n\nA: $\\quad 0.53 \\mathrm{~L}$\nB: $\\quad 0.056 \\mathrm{~L}$\nC: $\\quad 0.094 \\mathrm{~L}$\nD: $\\quad 0.040 \\mathrm{~L}$\nE: $\\quad 0.023 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_529737e841b31def0936g-4.jpg?height=296&width=723&top_left_y=329&top_left_x=1189"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1415",
        "problem": "\"Kinetic coupling\"\n\nThe standard Gibbs energy of the gas-phase reaction\n\n$\\mathrm{Br}+\\mathrm{H}_{2} \\underset{k_{-5}}{\\stackrel{k_{5}}{\\rightleftarrows}} \\mathrm{HBr}+\\mathrm{H}$\n\nis positive, $\\Delta G^{\\circ}(5)=66 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ at $T=600 \\mathrm{~K}$.What is the ratio of the rates of forward and reverse reactions, $r_{5} / r_{-5}$, at this temperature, standard pressures of $\\mathrm{H}_{2}$ and $\\mathrm{HBr}$ and equal pressures of $\\mathrm{H}$ and $\\mathrm{Br}$ ? (If you could not answer this question, for further calculations use reference value $r_{5} / r_{-5}=3.14 \\times 10^{-7}$.)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n\"Kinetic coupling\"\n\nThe standard Gibbs energy of the gas-phase reaction\n\n$\\mathrm{Br}+\\mathrm{H}_{2} \\underset{k_{-5}}{\\stackrel{k_{5}}{\\rightleftarrows}} \\mathrm{HBr}+\\mathrm{H}$\n\nis positive, $\\Delta G^{\\circ}(5)=66 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ at $T=600 \\mathrm{~K}$.\n\nproblem:\nWhat is the ratio of the rates of forward and reverse reactions, $r_{5} / r_{-5}$, at this temperature, standard pressures of $\\mathrm{H}_{2}$ and $\\mathrm{HBr}$ and equal pressures of $\\mathrm{H}$ and $\\mathrm{Br}$ ? (If you could not answer this question, for further calculations use reference value $r_{5} / r_{-5}=3.14 \\times 10^{-7}$.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_337",
        "problem": "A concentrated solution of ethanoic acid $\\left(\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}\\right)$ has a concentration of $17.4 \\mathrm{~mol} \\mathrm{~L}^{-1}$. What volume of this solution is needed to prepare $0.25 \\mathrm{~L}$ of $0.30 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}(\\mathrm{aq})$ ?\nA: $\\quad 4.7 \\mathrm{~mL}$\nB: $4.3 \\mathrm{~mL}$\nC: $\\quad 3.0 \\mathrm{~mL}$\nD: $2.5 \\mathrm{~mL}$\nE: $\\quad 2.2 \\mathrm{~mL}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA concentrated solution of ethanoic acid $\\left(\\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}\\right)$ has a concentration of $17.4 \\mathrm{~mol} \\mathrm{~L}^{-1}$. What volume of this solution is needed to prepare $0.25 \\mathrm{~L}$ of $0.30 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HC}_{2} \\mathrm{H}_{3} \\mathrm{O}_{2}(\\mathrm{aq})$ ?\n\nA: $\\quad 4.7 \\mathrm{~mL}$\nB: $4.3 \\mathrm{~mL}$\nC: $\\quad 3.0 \\mathrm{~mL}$\nD: $2.5 \\mathrm{~mL}$\nE: $\\quad 2.2 \\mathrm{~mL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_383",
        "problem": "Given the standard enthalpy changes for the reactions:\n\n$\\mathrm{P}_{4}(\\mathrm{~s})+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{P}_{4} \\mathrm{O}_{6}(\\mathrm{~s}) \\quad \\Delta H^{\\mathrm{o}}=-1640 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\n$\\mathrm{P}_{4}(\\mathrm{~s})+5 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(\\mathrm{~s}) \\quad \\Delta H^{\\mathrm{o}}=-2940 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nCalculate the standard enthalpy change $\\Delta H^{\\circ}$ for the following reaction:\n\n$$\n\\mathrm{P}_{4} \\mathrm{O}_{6}(s)+2 \\mathrm{O}_{2}(g) \\rightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s)\n$$\nA: $-4.58 \\times 10^{3} \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-1.30 \\times 10^{3} \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $1.79 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $4.82 \\times 10^{6} \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGiven the standard enthalpy changes for the reactions:\n\n$\\mathrm{P}_{4}(\\mathrm{~s})+3 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{P}_{4} \\mathrm{O}_{6}(\\mathrm{~s}) \\quad \\Delta H^{\\mathrm{o}}=-1640 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\n$\\mathrm{P}_{4}(\\mathrm{~s})+5 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(\\mathrm{~s}) \\quad \\Delta H^{\\mathrm{o}}=-2940 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nCalculate the standard enthalpy change $\\Delta H^{\\circ}$ for the following reaction:\n\n$$\n\\mathrm{P}_{4} \\mathrm{O}_{6}(s)+2 \\mathrm{O}_{2}(g) \\rightarrow \\mathrm{P}_{4} \\mathrm{O}_{10}(s)\n$$\n\nA: $-4.58 \\times 10^{3} \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-1.30 \\times 10^{3} \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $1.79 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $4.82 \\times 10^{6} \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_577",
        "problem": "下列有关电解质溶液说法正确的是\n\n[图1]\nA: 常温下, $\\mathrm{pH}$ 相同的三种溶液 $\\mathrm{NaF} 、 \\mathrm{Na}_{2} \\mathrm{CO}_{3} 、 \\mathrm{Na}_{2} \\mathrm{~S}$, 物质的量浓度最大的是 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$\nB: 等体积、等物质的量浓度的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 溶液混合: $\\frac{c\\left(\\mathrm{HCO}_{3}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)}<\\frac{c\\left(\\mathrm{CO}_{3}^{2-}\\right)}{c\\left(\\mathrm{HCO}_{3}^{-}\\right)}$\nC: 同温下, 等浓度的溶液中离子总浓度大小: $\\mathrm{NaF}>\\mathrm{NaCN}$\nD: 将过量 $\\mathrm{H}_{2} \\mathrm{~S}$ 通入 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液, 离子方程式为: $\\mathrm{CO}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}=\\mathrm{HS}^{-}+\\mathrm{HCO}_{3}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列有关电解质溶液说法正确的是\n\n[图1]\n\nA: 常温下, $\\mathrm{pH}$ 相同的三种溶液 $\\mathrm{NaF} 、 \\mathrm{Na}_{2} \\mathrm{CO}_{3} 、 \\mathrm{Na}_{2} \\mathrm{~S}$, 物质的量浓度最大的是 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$\nB: 等体积、等物质的量浓度的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 溶液混合: $\\frac{c\\left(\\mathrm{HCO}_{3}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)}<\\frac{c\\left(\\mathrm{CO}_{3}^{2-}\\right)}{c\\left(\\mathrm{HCO}_{3}^{-}\\right)}$\nC: 同温下, 等浓度的溶液中离子总浓度大小: $\\mathrm{NaF}>\\mathrm{NaCN}$\nD: 将过量 $\\mathrm{H}_{2} \\mathrm{~S}$ 通入 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液, 离子方程式为: $\\mathrm{CO}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}=\\mathrm{HS}^{-}+\\mathrm{HCO}_{3}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-030.jpg?height=556&width=588&top_left_y=2072&top_left_x=311"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_639",
        "problem": "已知某可逆反应: $2 \\mathrm{M}(\\mathrm{g}) \\rightleftharpoons \\mathrm{N}(\\mathrm{g}) \\Delta \\mathrm{H}<0$ 。现将 $\\mathrm{M}$ 和 $\\mathrm{N}$ 的混和气体通入容积为 $1 \\mathrm{~L}$的恒温密闭容器中, 反应体系中各物质浓度随时间变化关系如图所示。下列说法中, 正确的是\n\n[图1]\nA: $a 、 b 、 c 、 d$ 四个点中处于平衡状态的点是 $a 、 b$\nB: 反应进行至 $25 \\mathrm{~min}$ 时, 曲线发生变化的原因是加入了 $0.4 \\mathrm{~mol} \\mathrm{~N}$\nC: 若调节温度使 $35 \\mathrm{~min}$ 时体系内 $\\mathrm{N}$ 的体积分数与 $15 \\mathrm{~min}$ 时相等, 应升高温度\nD: 若在 $40 \\mathrm{~min}$ 时出现如图所示变化, 则可能是因为加入催化剂引起的\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知某可逆反应: $2 \\mathrm{M}(\\mathrm{g}) \\rightleftharpoons \\mathrm{N}(\\mathrm{g}) \\Delta \\mathrm{H}<0$ 。现将 $\\mathrm{M}$ 和 $\\mathrm{N}$ 的混和气体通入容积为 $1 \\mathrm{~L}$的恒温密闭容器中, 反应体系中各物质浓度随时间变化关系如图所示。下列说法中, 正确的是\n\n[图1]\n\nA: $a 、 b 、 c 、 d$ 四个点中处于平衡状态的点是 $a 、 b$\nB: 反应进行至 $25 \\mathrm{~min}$ 时, 曲线发生变化的原因是加入了 $0.4 \\mathrm{~mol} \\mathrm{~N}$\nC: 若调节温度使 $35 \\mathrm{~min}$ 时体系内 $\\mathrm{N}$ 的体积分数与 $15 \\mathrm{~min}$ 时相等, 应升高温度\nD: 若在 $40 \\mathrm{~min}$ 时出现如图所示变化, 则可能是因为加入催化剂引起的\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-014.jpg?height=560&width=808&top_left_y=174&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1328",
        "problem": "IONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\n\nWrite down the equation for the disproportionation of copper(I) ions and calculate the corresponding equilibrium constant.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\n\nWrite down the equation for the disproportionation of copper(I) ions and calculate the corresponding equilibrium constant.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_795",
        "problem": "对下列各溶液中, 微粒的物质的量浓度关系表述正确的是\nA: $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nC: 将 $0.2 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaA}$ 溶液和 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 盐酸等体积混合所得碱性溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}(\\mathrm{H}$ $\\left.{ }^{+}\\right)=\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nD: 在 $25^{\\circ} \\mathrm{C} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n对下列各溶液中, 微粒的物质的量浓度关系表述正确的是\n\nA: $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nC: 将 $0.2 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaA}$ 溶液和 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 盐酸等体积混合所得碱性溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}(\\mathrm{H}$ $\\left.{ }^{+}\\right)=\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nD: 在 $25^{\\circ} \\mathrm{C} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_314",
        "problem": "Examine the diagrams below carefully. Which of the burets shown below is/are ready for use?\n[figure1]\nA: (1) only\nB: (2) only\nC: (3) only\nD: (4) only\nE: (1), (3) and (4)\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nExamine the diagrams below carefully. Which of the burets shown below is/are ready for use?\n[figure1]\n\nA: (1) only\nB: (2) only\nC: (3) only\nD: (4) only\nE: (1), (3) and (4)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_92a3e606264c5ee1789ag-4.jpg?height=650&width=630&top_left_y=1494&top_left_x=236"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_116",
        "problem": "A quantity of solid material weighing $6.445 \\mathrm{~g}$ was obtained from a hazardous waste facility. A $1.545 \\mathrm{~g}$ sample of this material was analyzed for barium content by dissolving in water and then adding sodium sulfate. The insoluble barium sulfate precipitate was dried, and a total of $73.8 \\mathrm{mg}$ of $\\mathrm{BaSO}_{4}$ was collected. What percentage by mass of the sample is barium?\nA: $0.281 \\%$\nB: $0.674 \\%$\nC: $2.81 \\%$\nD: $4.02 \\%$\nE: $6.74 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA quantity of solid material weighing $6.445 \\mathrm{~g}$ was obtained from a hazardous waste facility. A $1.545 \\mathrm{~g}$ sample of this material was analyzed for barium content by dissolving in water and then adding sodium sulfate. The insoluble barium sulfate precipitate was dried, and a total of $73.8 \\mathrm{mg}$ of $\\mathrm{BaSO}_{4}$ was collected. What percentage by mass of the sample is barium?\n\nA: $0.281 \\%$\nB: $0.674 \\%$\nC: $2.81 \\%$\nD: $4.02 \\%$\nE: $6.74 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_987",
        "problem": "A colourless, odourless gas is thought to be oxygen. Which of the following experimental results would support this conclusion?\nA: Burning the gas in air produces only water.\nB: The gas extinguishes a flame.\nC: The gas turns a $\\mathrm{Ca}(\\mathrm{OH})_{2}$ solution milky.\nD: A glowing piece of wood bursts into flames in the gas.\nE: The gas tarnishes silver.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA colourless, odourless gas is thought to be oxygen. Which of the following experimental results would support this conclusion?\n\nA: Burning the gas in air produces only water.\nB: The gas extinguishes a flame.\nC: The gas turns a $\\mathrm{Ca}(\\mathrm{OH})_{2}$ solution milky.\nD: A glowing piece of wood bursts into flames in the gas.\nE: The gas tarnishes silver.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1100",
        "problem": "Polonium is a radioactive group VI element, discovered in 1898 by Marie Curie. It occurs naturally in trace amounts in some uranium ores but is now made by neutron irradiation of ${ }^{209} \\mathrm{Bi}$. This produces short-lived ${ }^{210} \\mathrm{Bi}$ which decays to polonium by the emission of a betaparticle (an electron):\n\n$$\n{ }_{83}^{209} \\mathrm{Bi}+{ }_{0}^{1} n \\rightarrow{ }_{83}^{210} \\mathrm{Bi}+\\gamma \\quad{ }_{83}^{210} \\mathrm{Bi} \\rightarrow{ }_{84}^{210} \\mathrm{Po}+{ }_{-1}^{0} \\beta\n$$\n\nPolonium-210 has a half life of 138 days and decays by emitting an alpha particle (a helium nucleus).\n\n[figure1]\n\nDue to its very short half life and the impedance of the alpha-particles it emits, metallic polonium and its compounds are self heating; $1 \\mathrm{~g}$ of metal produces $141 \\mathrm{~W}$. This led to its use in Radioisotope Heater Units (RHUs) to keep satellites warm and functioning in space, and in Radioisotope Thermal Generators (RTGs) to produce electrical power. More recently, plutonium-238 has been used instead of polonium. ${ }^{238} \\mathrm{Pu}$ has a much longer half-life but produces less power $\\left(0.56 \\mathrm{~W} \\mathrm{~g}^{-1}\\right)$.\n\nAfter 5 years, the power output of ${ }^{238} \\mathrm{Pu}$ is approximately $96 \\%$ of its initial value. Estimate the half life of plutonium-238.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nPolonium is a radioactive group VI element, discovered in 1898 by Marie Curie. It occurs naturally in trace amounts in some uranium ores but is now made by neutron irradiation of ${ }^{209} \\mathrm{Bi}$. This produces short-lived ${ }^{210} \\mathrm{Bi}$ which decays to polonium by the emission of a betaparticle (an electron):\n\n$$\n{ }_{83}^{209} \\mathrm{Bi}+{ }_{0}^{1} n \\rightarrow{ }_{83}^{210} \\mathrm{Bi}+\\gamma \\quad{ }_{83}^{210} \\mathrm{Bi} \\rightarrow{ }_{84}^{210} \\mathrm{Po}+{ }_{-1}^{0} \\beta\n$$\n\nPolonium-210 has a half life of 138 days and decays by emitting an alpha particle (a helium nucleus).\n\n[figure1]\n\nDue to its very short half life and the impedance of the alpha-particles it emits, metallic polonium and its compounds are self heating; $1 \\mathrm{~g}$ of metal produces $141 \\mathrm{~W}$. This led to its use in Radioisotope Heater Units (RHUs) to keep satellites warm and functioning in space, and in Radioisotope Thermal Generators (RTGs) to produce electrical power. More recently, plutonium-238 has been used instead of polonium. ${ }^{238} \\mathrm{Pu}$ has a much longer half-life but produces less power $\\left(0.56 \\mathrm{~W} \\mathrm{~g}^{-1}\\right)$.\n\nAfter 5 years, the power output of ${ }^{238} \\mathrm{Pu}$ is approximately $96 \\%$ of its initial value. Estimate the half life of plutonium-238.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of years, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-10.jpg?height=500&width=648&top_left_y=367&top_left_x=1132"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "years"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1003",
        "problem": "Which of the following molecules forms hydrogen bonds amongst themselves?\nA: dimethyl ether $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$\nB: methane $\\left(\\mathrm{CH}_{4}\\right)$\nC: hydrogen sulfide $\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nD: ethanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$\nE: formaldehyde $\\left(\\mathrm{H}_{2} \\mathrm{CO}\\right)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following molecules forms hydrogen bonds amongst themselves?\n\nA: dimethyl ether $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$\nB: methane $\\left(\\mathrm{CH}_{4}\\right)$\nC: hydrogen sulfide $\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nD: ethanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$\nE: formaldehyde $\\left(\\mathrm{H}_{2} \\mathrm{CO}\\right)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_329",
        "problem": "What is the ground state electron configuration of Ar?\nA: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$\nB: $1 s^{2} 2 s^{2} 2 p^{6}$\nC: $1 s^{2} 2 s^{2} 3 s^{2} 3 p^{6}$\nD: $1 s^{2} 2 s^{2} 2 p^{3} 3 s^{2} 3 p^{3}$\nE: $1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the ground state electron configuration of Ar?\n\nA: $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$\nB: $1 s^{2} 2 s^{2} 2 p^{6}$\nC: $1 s^{2} 2 s^{2} 3 s^{2} 3 p^{6}$\nD: $1 s^{2} 2 s^{2} 2 p^{3} 3 s^{2} 3 p^{3}$\nE: $1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_295",
        "problem": "In the statements below, $\\mathrm{X}$ refers to one of $\\mathrm{Ca}, \\mathrm{Fe}, \\mathrm{Pb}$, Cu or Pt. What is the identity of $X$ ?\n\n- $\\mathrm{X}(\\mathrm{s})$ reacts spontaneously in $1 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(\\mathrm{aq})$ to give $\\mathrm{XCl}_{2}(\\mathrm{aq})$ and $\\mathrm{H}_{2}(g)$.\n- The reaction $3 \\mathrm{X}^{2+}(\\mathrm{aq})+2 \\mathrm{Al}(s) \\rightarrow 3 \\mathrm{X}(\\mathrm{s})+2 \\mathrm{Al}^{3+}(\\mathrm{aq})$ is spontaneous under standard conditions.\n- $\\mathrm{X}(\\mathrm{s})$ is a better reducing agent than $\\mathrm{Co}(\\mathrm{s})$ under standard conditions.\n\n| Half-reaction | $\\mathrm{E}^{\\mathbf{o}}$ |\n| :--- | :---: |\n| $\\mathrm{Ca}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Ca}(s)$ | $-2.84 \\mathrm{~V}$ |\n| $\\mathrm{Al}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Al}(s)$ | $-1.66 \\mathrm{~V}$ |\n| $\\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe}(s)$ | $-0.44 \\mathrm{~V}$ |\n| $\\mathrm{Co}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Co}(s)$ | $-0.28 \\mathrm{~V}$ |\n| $\\mathrm{~Pb}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Pb}(s)$ | $-0.13 \\mathrm{~V}$ |\n| $2 \\mathrm{H}^{+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{H}_{2}(g)$ | $0.00 \\mathrm{~V}$ |\n| $\\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Cu}(s)$ | $0.34 \\mathrm{~V}$ |\n| $\\mathrm{Pt}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Pt}(\\mathrm{s})$ | $1.18 \\mathrm{~V}$ |\n|  |  |\nA: $\\mathrm{Ca}$\nB: Fe\nC: $\\mathrm{Pb}$\nD: Cu\nE: Pt\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the statements below, $\\mathrm{X}$ refers to one of $\\mathrm{Ca}, \\mathrm{Fe}, \\mathrm{Pb}$, Cu or Pt. What is the identity of $X$ ?\n\n- $\\mathrm{X}(\\mathrm{s})$ reacts spontaneously in $1 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(\\mathrm{aq})$ to give $\\mathrm{XCl}_{2}(\\mathrm{aq})$ and $\\mathrm{H}_{2}(g)$.\n- The reaction $3 \\mathrm{X}^{2+}(\\mathrm{aq})+2 \\mathrm{Al}(s) \\rightarrow 3 \\mathrm{X}(\\mathrm{s})+2 \\mathrm{Al}^{3+}(\\mathrm{aq})$ is spontaneous under standard conditions.\n- $\\mathrm{X}(\\mathrm{s})$ is a better reducing agent than $\\mathrm{Co}(\\mathrm{s})$ under standard conditions.\n\n| Half-reaction | $\\mathrm{E}^{\\mathbf{o}}$ |\n| :--- | :---: |\n| $\\mathrm{Ca}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Ca}(s)$ | $-2.84 \\mathrm{~V}$ |\n| $\\mathrm{Al}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightarrow \\mathrm{Al}(s)$ | $-1.66 \\mathrm{~V}$ |\n| $\\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe}(s)$ | $-0.44 \\mathrm{~V}$ |\n| $\\mathrm{Co}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Co}(s)$ | $-0.28 \\mathrm{~V}$ |\n| $\\mathrm{~Pb}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Pb}(s)$ | $-0.13 \\mathrm{~V}$ |\n| $2 \\mathrm{H}^{+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{H}_{2}(g)$ | $0.00 \\mathrm{~V}$ |\n| $\\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Cu}(s)$ | $0.34 \\mathrm{~V}$ |\n| $\\mathrm{Pt}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Pt}(\\mathrm{s})$ | $1.18 \\mathrm{~V}$ |\n|  |  |\n\nA: $\\mathrm{Ca}$\nB: Fe\nC: $\\mathrm{Pb}$\nD: Cu\nE: Pt\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_301",
        "problem": "Which of the following statements concerning the structure below is true?\n[figure1]\nA: There are eight $\\sigma$ bonds in this structure.\nB: The nitrogen atom is $s p$-hybridized.\nC: The $\\mathrm{H}-\\mathrm{C}-\\mathrm{H}$ bond angle is $90^{\\circ}$.\nD: The structure above is the most important structure for the $\\mathrm{CH}_{3} \\mathrm{NCO}$ molecule.\nE: None of the statements above are true.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements concerning the structure below is true?\n[figure1]\n\nA: There are eight $\\sigma$ bonds in this structure.\nB: The nitrogen atom is $s p$-hybridized.\nC: The $\\mathrm{H}-\\mathrm{C}-\\mathrm{H}$ bond angle is $90^{\\circ}$.\nD: The structure above is the most important structure for the $\\mathrm{CH}_{3} \\mathrm{NCO}$ molecule.\nE: None of the statements above are true.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://i.postimg.cc/JzcRXCSK/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1319",
        "problem": "The amount of many reducing agents can be determined by permanganatometric titration in alkaline medium allowing permanganate ion reduction to manganate.A volume of $10.00 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{Mn}}\\right)$ of $\\mathrm{KMnO}_{4}$ solution with a concentration of 0.0400 mol $\\mathrm{dm}^{-3}\\left(c_{\\mathrm{Mn}}\\right)$ was placed in each of flasks $\\mathbf{A}, \\mathbf{B}$, and $\\mathbf{C}$ and different reactions were conducted in each flask.\n\nA sample solution containing crotonic acid (CA) $\\mathrm{CH}_{3}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{COOH}$, an alkali and barium nitrate (both in an excess) were added to flask $\\mathbf{A}$, and the reaction mixture was incubated for $45 \\mathrm{~min}$. It is known that crotonic acid loses 10 electrons under the experiment conditions.\n\nA volume of $8.00 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{CN}}\\right)$ of potassium cyanide solution $\\left(C_{\\mathrm{CN}}=0.0100 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$ was then added to the incubated mixture. This resulted in completion of the following reaction:\n\n$$\n2 \\mathrm{Ba}^{2+}+2 \\mathrm{MnO}_{4}^{-}+\\mathrm{CN}^{-}+2 \\mathrm{OH}^{-} \\rightarrow 2 \\mathrm{BaMnO}_{4}+\\mathrm{CNO}^{-}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\n$\\mathrm{BaMnO}_{4}$ precipitate was then filtered off, and the excess of cyanide in the filtrate was titrated with $\\mathrm{AgNO}_{3}$ solution $\\left(c_{\\mathrm{Ag}}=0.0050 \\mathrm{~mol} \\mathrm{dm}{ }^{-3}\\right.$ ) till detectable precipitation was observed. Note that both $\\mathrm{CN}^{-}$and $\\mathrm{CNO}^{-}$are analogs of halide ions, but $\\mathrm{CNO}^{-}$affords soluble silver salt.\n\nCalculate the mass of crotonic acid (in $\\mathrm{mg})$ if $5.40 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{Ag}}\\right)$ of the silver salt solution was consumed for the titration to the endpoint.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe amount of many reducing agents can be determined by permanganatometric titration in alkaline medium allowing permanganate ion reduction to manganate.\n\nproblem:\nA volume of $10.00 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{Mn}}\\right)$ of $\\mathrm{KMnO}_{4}$ solution with a concentration of 0.0400 mol $\\mathrm{dm}^{-3}\\left(c_{\\mathrm{Mn}}\\right)$ was placed in each of flasks $\\mathbf{A}, \\mathbf{B}$, and $\\mathbf{C}$ and different reactions were conducted in each flask.\n\nA sample solution containing crotonic acid (CA) $\\mathrm{CH}_{3}-\\mathrm{CH}=\\mathrm{CH}-\\mathrm{COOH}$, an alkali and barium nitrate (both in an excess) were added to flask $\\mathbf{A}$, and the reaction mixture was incubated for $45 \\mathrm{~min}$. It is known that crotonic acid loses 10 electrons under the experiment conditions.\n\nA volume of $8.00 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{CN}}\\right)$ of potassium cyanide solution $\\left(C_{\\mathrm{CN}}=0.0100 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$ was then added to the incubated mixture. This resulted in completion of the following reaction:\n\n$$\n2 \\mathrm{Ba}^{2+}+2 \\mathrm{MnO}_{4}^{-}+\\mathrm{CN}^{-}+2 \\mathrm{OH}^{-} \\rightarrow 2 \\mathrm{BaMnO}_{4}+\\mathrm{CNO}^{-}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\n$\\mathrm{BaMnO}_{4}$ precipitate was then filtered off, and the excess of cyanide in the filtrate was titrated with $\\mathrm{AgNO}_{3}$ solution $\\left(c_{\\mathrm{Ag}}=0.0050 \\mathrm{~mol} \\mathrm{dm}{ }^{-3}\\right.$ ) till detectable precipitation was observed. Note that both $\\mathrm{CN}^{-}$and $\\mathrm{CNO}^{-}$are analogs of halide ions, but $\\mathrm{CNO}^{-}$affords soluble silver salt.\n\nCalculate the mass of crotonic acid (in $\\mathrm{mg})$ if $5.40 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{Ag}}\\right)$ of the silver salt solution was consumed for the titration to the endpoint.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_480",
        "problem": "亚砷酸 $\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)$ 可用于白血病的治疗。室温下, 配制一组 $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}{ }^{-}\\right)+$ $\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{AsO}_{3}{ }^{3-}\\right)=0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{3} \\mathrm{AsO}_{3}$ 和 $\\mathrm{NaOH}$ 混合溶液, 溶液中部分微粒的物质的量浓度随 $\\mathrm{pH}$ 的变化关系曲线如图所示。下列指定溶液中微粒的物质的量浓度关系正确的是 ( )\n\n[图1]\nA: $\\mathrm{pH}=11$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{AsO}_{3}{ }^{3-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}{ }^{-}\\right)$\nB: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)$\nC: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=0.200 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{AsO}_{3}{ }^{3-}\\right)$\nD: $\\mathrm{pH}=12.8$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>4 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n亚砷酸 $\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)$ 可用于白血病的治疗。室温下, 配制一组 $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}{ }^{-}\\right)+$ $\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{AsO}_{3}{ }^{3-}\\right)=0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{3} \\mathrm{AsO}_{3}$ 和 $\\mathrm{NaOH}$ 混合溶液, 溶液中部分微粒的物质的量浓度随 $\\mathrm{pH}$ 的变化关系曲线如图所示。下列指定溶液中微粒的物质的量浓度关系正确的是 ( )\n\n[图1]\n\nA: $\\mathrm{pH}=11$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{AsO}_{3}{ }^{3-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}{ }^{-}\\right)$\nB: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)$\nC: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=0.200 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{AsO}_{3}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{AsO}_{3}{ }^{3-}\\right)$\nD: $\\mathrm{pH}=12.8$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>4 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{AsO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HAsO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-085.jpg?height=631&width=940&top_left_y=161&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_13",
        "problem": "Solid $\\mathrm{KOH}\\left(M=56.11, \\Delta H^{\\circ}\\right.$ soln $\\left.=-57.6 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\\right)$ is slowly added to distilled water contained in a wellinsulated vessel, initially at $25.0{ }^{\\circ} \\mathrm{C}$. What is the $\\mathrm{pH}$ when the temperature of the solution reaches $26.0^{\\circ} \\mathrm{C}$ ? You may assume that the density of the solution remains at $1.00 \\mathrm{~g} \\mathrm{~mL}^{-1}$ and its heat capacity remains at $4.184 \\mathrm{~J} \\mathrm{~g}^{-1}$ $\\mathrm{K}^{-1}$ throughout the experiment.\nA: 9.1\nB: 9.9\nC: 12.1\nD: 12.9\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSolid $\\mathrm{KOH}\\left(M=56.11, \\Delta H^{\\circ}\\right.$ soln $\\left.=-57.6 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\\right)$ is slowly added to distilled water contained in a wellinsulated vessel, initially at $25.0{ }^{\\circ} \\mathrm{C}$. What is the $\\mathrm{pH}$ when the temperature of the solution reaches $26.0^{\\circ} \\mathrm{C}$ ? You may assume that the density of the solution remains at $1.00 \\mathrm{~g} \\mathrm{~mL}^{-1}$ and its heat capacity remains at $4.184 \\mathrm{~J} \\mathrm{~g}^{-1}$ $\\mathrm{K}^{-1}$ throughout the experiment.\n\nA: 9.1\nB: 9.9\nC: 12.1\nD: 12.9\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_825",
        "problem": "与 $\\mathrm{CaC}_{2}$ 一样, $\\mathrm{Mg}_{3} \\mathrm{C}_{2}$ 也是离子晶体, 与水能发生复分解反应。关于 $\\mathrm{Mg}_{3} \\mathrm{C}_{2}$ 的叙述正确的是\nA: 与水反应有乙炔生成\nB: 与水反应有 $\\mathrm{Mg}(\\mathrm{OH})_{2}$ 生成\nC: 有 $\\mathrm{C}-\\mathrm{C}$ 键\nD: 有 $\\mathrm{C}_{2}{ }^{2-\\text { 离子 }}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n与 $\\mathrm{CaC}_{2}$ 一样, $\\mathrm{Mg}_{3} \\mathrm{C}_{2}$ 也是离子晶体, 与水能发生复分解反应。关于 $\\mathrm{Mg}_{3} \\mathrm{C}_{2}$ 的叙述正确的是\n\nA: 与水反应有乙炔生成\nB: 与水反应有 $\\mathrm{Mg}(\\mathrm{OH})_{2}$ 生成\nC: 有 $\\mathrm{C}-\\mathrm{C}$ 键\nD: 有 $\\mathrm{C}_{2}{ }^{2-\\text { 离子 }}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-72.jpg?height=66&width=1145&top_left_y=618&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_744",
        "problem": "常温下, 用 $\\mathrm{NaOH}$ 溶液滴定二元弱酸亚磷酸溶液, 溶液中- $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)}$ 和 $-\\operatorname{lgc}\\left(\\mathrm{H}_{2} \\mathrm{PO},\\right)$ 或 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)}$和 $-\\operatorname{lgc}\\left(\\mathrm{HO}_{3}^{-}\\right)$的关系如图所示, 下列说法错误的是 $-\\lg c\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)$或 $-\\lg c\\left(\\mathrm{HPO}_{3}^{2-}\\right)$\n\n[图1]\nA: $\\mathrm{L}_{2}$ 表示 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)}$和- $\\operatorname{lgc}\\left(\\mathrm{HPO}_{3}^{-}\\right)$的关系\nB: $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$ 的数量级为 $10^{-2}$\nC: 等浓度等体积的 $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ 溶液与 $\\mathrm{NaOH}$ 溶液充分混合, $\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{-}\\right)<\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$\nD: $\\mathrm{pH}=6.54$ 时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<3 \\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 用 $\\mathrm{NaOH}$ 溶液滴定二元弱酸亚磷酸溶液, 溶液中- $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)}$ 和 $-\\operatorname{lgc}\\left(\\mathrm{H}_{2} \\mathrm{PO},\\right)$ 或 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)}$和 $-\\operatorname{lgc}\\left(\\mathrm{HO}_{3}^{-}\\right)$的关系如图所示, 下列说法错误的是 $-\\lg c\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)$或 $-\\lg c\\left(\\mathrm{HPO}_{3}^{2-}\\right)$\n\n[图1]\n\nA: $\\mathrm{L}_{2}$ 表示 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)}$和- $\\operatorname{lgc}\\left(\\mathrm{HPO}_{3}^{-}\\right)$的关系\nB: $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$ 的数量级为 $10^{-2}$\nC: 等浓度等体积的 $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ 溶液与 $\\mathrm{NaOH}$ 溶液充分混合, $\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{-}\\right)<\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$\nD: $\\mathrm{pH}=6.54$ 时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<3 \\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-074.jpg?height=668&width=965&top_left_y=1645&top_left_x=477"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1471",
        "problem": "The unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nThe first order rate constant for the $\\mathrm{Cl} 2$ protein folding reaction can be determined by following the fluorescence intensity when a sample of unfolded protein is allowed to refold (typically the $\\mathrm{pH}$ of the solution is changed). The concentration of protein when a $1.0 \\mu \\mathrm{M}$ sample of unfolded $\\mathrm{Cl} 2$ was allowed to refold was measured at a temperature of $25{ }^{\\circ} \\mathrm{C}$ :\n\n| time / ms | 0 | 10 | 20 | 30 | 40 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| concentration $/ \\mu \\mathrm{M}$ | 1 | 0.64 | 0.36 | 0.23 | 0.14 |Determine the value of the rate constant for the protein unfolding reaction, $k_{u}$, at $25^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nThe first order rate constant for the $\\mathrm{Cl} 2$ protein folding reaction can be determined by following the fluorescence intensity when a sample of unfolded protein is allowed to refold (typically the $\\mathrm{pH}$ of the solution is changed). The concentration of protein when a $1.0 \\mu \\mathrm{M}$ sample of unfolded $\\mathrm{Cl} 2$ was allowed to refold was measured at a temperature of $25{ }^{\\circ} \\mathrm{C}$ :\n\n| time / ms | 0 | 10 | 20 | 30 | 40 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| concentration $/ \\mu \\mathrm{M}$ | 1 | 0.64 | 0.36 | 0.23 | 0.14 |\n\nproblem:\nDetermine the value of the rate constant for the protein unfolding reaction, $k_{u}$, at $25^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{s}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{s}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_405",
        "problem": "某温度下, 向一定体积 $0.1 \\mathrm{~mol} / \\mathrm{LHCl}$ 溶液中逐滴加入等浓度的氨水溶液, 溶液中 $\\mathrm{pOH}\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$与 $\\mathrm{pH}$ 的变化关系如图所示, 则下列说法错误的是\n\n[图1]\nA: $\\mathrm{Q}$ 点消耗氨水溶液的体积等于 $\\mathrm{HCl}$ 溶液的体积\nB: $\\mathrm{M}$ 点所示溶液导电能力弱于 $\\mathrm{Q}$ 点\nC: $\\mathrm{M}$ 点和 $\\mathrm{N}$ 点所示溶液中水的电离程度相同\nD: $\\mathrm{N}$ 点所示溶液中 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某温度下, 向一定体积 $0.1 \\mathrm{~mol} / \\mathrm{LHCl}$ 溶液中逐滴加入等浓度的氨水溶液, 溶液中 $\\mathrm{pOH}\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$与 $\\mathrm{pH}$ 的变化关系如图所示, 则下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{Q}$ 点消耗氨水溶液的体积等于 $\\mathrm{HCl}$ 溶液的体积\nB: $\\mathrm{M}$ 点所示溶液导电能力弱于 $\\mathrm{Q}$ 点\nC: $\\mathrm{M}$ 点和 $\\mathrm{N}$ 点所示溶液中水的电离程度相同\nD: $\\mathrm{N}$ 点所示溶液中 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-104.jpg?height=472&width=554&top_left_y=969&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_292",
        "problem": "The reaction below was studied using the method of initial rates.\n\n$\\mathrm{BrO}_{3}^{-}(a q)+5 \\mathrm{Br}^{-}(a q)+6 \\mathrm{H}^{+}(a q) \\rightarrow 3 \\mathrm{Br}_{2}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n\nThe rate law for the reaction was determined to be Rate $=k\\left[\\mathrm{BrO}_{3}^{-}\\right]\\left[\\mathrm{Br}^{-}\\right]\\left[\\mathrm{H}^{+}\\right]^{2}$, where Rate refers to the rate of consumption of $\\mathrm{BrO}_{3}^{-}$. Which of the following statements is false?\nA: If concentrations are measured in $\\mathrm{mol} \\mathrm{L}^{-1}$ and time is measured in seconds (s), then the units of $k$ are $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}$.\nB: The rate of consumption of $\\mathrm{Br}^{-}$is five times greater than the rate of consumption of $\\mathrm{BrO}_{3}^{-}$.\nC: The conversion of reactants into products must involve two or more simpler reactions.\nD: If the concentrations of all reactants are doubled, the rate of consumption of $\\mathrm{BrO}_{3}^{-}$will increase by a factor of sixteen.\nE: When the reaction reaches a state of dynamic equilibrium, $\\left[\\mathrm{BrO}_{3}^{-}\\right]$stops changing.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction below was studied using the method of initial rates.\n\n$\\mathrm{BrO}_{3}^{-}(a q)+5 \\mathrm{Br}^{-}(a q)+6 \\mathrm{H}^{+}(a q) \\rightarrow 3 \\mathrm{Br}_{2}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)$\n\nThe rate law for the reaction was determined to be Rate $=k\\left[\\mathrm{BrO}_{3}^{-}\\right]\\left[\\mathrm{Br}^{-}\\right]\\left[\\mathrm{H}^{+}\\right]^{2}$, where Rate refers to the rate of consumption of $\\mathrm{BrO}_{3}^{-}$. Which of the following statements is false?\n\nA: If concentrations are measured in $\\mathrm{mol} \\mathrm{L}^{-1}$ and time is measured in seconds (s), then the units of $k$ are $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{~s}^{-1}$.\nB: The rate of consumption of $\\mathrm{Br}^{-}$is five times greater than the rate of consumption of $\\mathrm{BrO}_{3}^{-}$.\nC: The conversion of reactants into products must involve two or more simpler reactions.\nD: If the concentrations of all reactants are doubled, the rate of consumption of $\\mathrm{BrO}_{3}^{-}$will increase by a factor of sixteen.\nE: When the reaction reaches a state of dynamic equilibrium, $\\left[\\mathrm{BrO}_{3}^{-}\\right]$stops changing.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_772",
        "problem": "$\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液中的物种分布系数与 $\\mathrm{pH}$ 的关系如图所示。对于 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液, 下列叙述正确的是\n\n[图1]\n\n[比如 A-的分布系数: $\\delta\\left(\\mathrm{A}^{-}\\right)=\\frac{c\\left(\\mathrm{~A}^{-}\\right)}{c(\\mathrm{HA})+c\\left(\\mathrm{~A}^{-}\\right)}$]\nA: 水解常数: $K_{\\mathrm{h}}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>1.0 \\times 10^{-9}$\nB: $\\delta\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)=\\frac{c\\left(\\mathrm{H}^{+}\\right)}{c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)+c\\left(\\mathrm{H}^{+}\\right)}$\nC: $\\mathrm{pH}=7.0$ 时, 混合溶液中有 $c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\nD: 采用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定，指示剂酚酞比甲基橙更准确\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液中的物种分布系数与 $\\mathrm{pH}$ 的关系如图所示。对于 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液, 下列叙述正确的是\n\n[图1]\n\n[比如 A-的分布系数: $\\delta\\left(\\mathrm{A}^{-}\\right)=\\frac{c\\left(\\mathrm{~A}^{-}\\right)}{c(\\mathrm{HA})+c\\left(\\mathrm{~A}^{-}\\right)}$]\n\nA: 水解常数: $K_{\\mathrm{h}}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>1.0 \\times 10^{-9}$\nB: $\\delta\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)=\\frac{c\\left(\\mathrm{H}^{+}\\right)}{c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)+c\\left(\\mathrm{H}^{+}\\right)}$\nC: $\\mathrm{pH}=7.0$ 时, 混合溶液中有 $c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\nD: 采用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定，指示剂酚酞比甲基橙更准确\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-079.jpg?height=549&width=665&top_left_y=148&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_378",
        "problem": "Barium chloride reacts with sodium sulfate according to the following equation:\n\n$$\n\\mathrm{BaCl}_{2}(a q)+\\mathrm{Na}_{2} \\mathrm{SO}_{4}(a q) \\rightarrow \\mathrm{BaSO}_{4}(s)+2 \\mathrm{NaCl}(a q)\n$$\n\nA student mixes a solution containing $10.0 \\mathrm{~g} \\mathrm{BaCl}_{2}$ ( $M=208.2)$ with a solution containing $10.0 \\mathrm{~g} \\mathrm{Na}_{2} \\mathrm{SO}_{4}$ $(M=142.1)$ and obtains $12.0 \\mathrm{~g} \\mathrm{BaSO}_{4}(M=233.2)$. What is the percent yield of this reaction?\nA: $60.0 \\%$\nB: $73.1 \\%$\nC: $93.3 \\%$\nD: The isolated barium sulfate is most likely wet, since the yield would otherwise be greater than $100 \\%$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nBarium chloride reacts with sodium sulfate according to the following equation:\n\n$$\n\\mathrm{BaCl}_{2}(a q)+\\mathrm{Na}_{2} \\mathrm{SO}_{4}(a q) \\rightarrow \\mathrm{BaSO}_{4}(s)+2 \\mathrm{NaCl}(a q)\n$$\n\nA student mixes a solution containing $10.0 \\mathrm{~g} \\mathrm{BaCl}_{2}$ ( $M=208.2)$ with a solution containing $10.0 \\mathrm{~g} \\mathrm{Na}_{2} \\mathrm{SO}_{4}$ $(M=142.1)$ and obtains $12.0 \\mathrm{~g} \\mathrm{BaSO}_{4}(M=233.2)$. What is the percent yield of this reaction?\n\nA: $60.0 \\%$\nB: $73.1 \\%$\nC: $93.3 \\%$\nD: The isolated barium sulfate is most likely wet, since the yield would otherwise be greater than $100 \\%$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1023",
        "problem": "Give half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{HCl}$ to $\\mathrm{H}_{2}$\nA: oxidation\nB: reduction\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGive half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{HCl}$ to $\\mathrm{H}_{2}$\n\nA: oxidation\nB: reduction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_520",
        "problem": "金属锂燃料电池是一种新型电池, 比锂离子电池具有更高的能量密度。它无电时也无需充电, 只需更换其中的某些材料即可, 其工作示意图如图, 下列说法正确的是\n\n[图1]\nA: 放电时, 通入空气的一极为负极\nB: 放电时, 电池反应为 $4 \\mathrm{Li}+\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{LiOH}$\nC: 有机电解液可以是乙醇等无水有机物\nD: 在更换锂电极的同时, 要更换水性电解液\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n金属锂燃料电池是一种新型电池, 比锂离子电池具有更高的能量密度。它无电时也无需充电, 只需更换其中的某些材料即可, 其工作示意图如图, 下列说法正确的是\n\n[图1]\n\nA: 放电时, 通入空气的一极为负极\nB: 放电时, 电池反应为 $4 \\mathrm{Li}+\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{LiOH}$\nC: 有机电解液可以是乙醇等无水有机物\nD: 在更换锂电极的同时, 要更换水性电解液\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-60.jpg?height=568&width=925&top_left_y=1418&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_712",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{HCl}$ 气体调节 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水的 $\\mathrm{pH}$, 溶液中微粒浓度的对数值 $(\\lg c)$ 、反应物的物质的量之比 $\\left[\\mathrm{t}=\\frac{\\mathrm{n}(\\mathrm{HCl})}{\\mathrm{n}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}\\right]$ 与 $\\mathrm{pH}$ 的关系如图所示。若忽略通入气体后溶\n\n[图1]\nA: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的电离平衡常数为 $10^{-4.75}$\nB: $\\mathrm{P}_{1}$ 所示溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)=0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $\\mathrm{P}_{2}$ 所示溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$\nD: $\\mathrm{P}_{3}$ 所示溶液中: $c\\left(\\mathrm{NH}_{4}^{+}\\right)+c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=c\\left(\\mathrm{Cl}^{-}\\right)+c\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{HCl}$ 气体调节 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水的 $\\mathrm{pH}$, 溶液中微粒浓度的对数值 $(\\lg c)$ 、反应物的物质的量之比 $\\left[\\mathrm{t}=\\frac{\\mathrm{n}(\\mathrm{HCl})}{\\mathrm{n}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}\\right]$ 与 $\\mathrm{pH}$ 的关系如图所示。若忽略通入气体后溶\n\n[图1]\n\nA: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的电离平衡常数为 $10^{-4.75}$\nB: $\\mathrm{P}_{1}$ 所示溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)=0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $\\mathrm{P}_{2}$ 所示溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$\nD: $\\mathrm{P}_{3}$ 所示溶液中: $c\\left(\\mathrm{NH}_{4}^{+}\\right)+c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=c\\left(\\mathrm{Cl}^{-}\\right)+c\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-066.jpg?height=545&width=1262&top_left_y=230&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_635",
        "problem": "由实验操作和现象, 可得出相应正确结论的是\n\n|  | 实验操作 | 现象 | 结论 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{A}$ | 向 $\\mathrm{NaBr}$ 溶液中滴加过量氯水, 再加入 <br> 淀粉 $\\mathrm{KI}$ 溶液 | 溶液先变橙色, 后变 <br> 蓝色 | 氧化性: <br> $\\mathrm{Cl}_{2}>\\mathrm{Br}_{2}>\\mathrm{I}_{2}$ |\n| $\\mathrm{B}$ | 向饱和 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶液中滴加鸡蛋清溶 <br> 液 | 出现白色沉淀 | 饱和 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶 <br> 液可使蛋白质变性 |\n| $\\mathrm{C}$ | 石蜡油加强热, 将产生的气体通入 $\\mathrm{Br}_{2}$ <br> 的 $\\mathrm{CCl}_{4}$ 溶液 | 溶液红棕色变无色 | 气体中含有不饱和烃 |\n| $\\mathrm{D}$ | 将封装 $\\mathrm{NO}_{2}$ 和 $\\mathrm{N}_{2} \\mathrm{O}_{4}$ 混合气体的球形玻璃容器, 浸入热水中 | 气体颜色变深 | 温度可以破坏化学平衡 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n由实验操作和现象, 可得出相应正确结论的是\n\n|  | 实验操作 | 现象 | 结论 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{A}$ | 向 $\\mathrm{NaBr}$ 溶液中滴加过量氯水, 再加入 <br> 淀粉 $\\mathrm{KI}$ 溶液 | 溶液先变橙色, 后变 <br> 蓝色 | 氧化性: <br> $\\mathrm{Cl}_{2}>\\mathrm{Br}_{2}>\\mathrm{I}_{2}$ |\n| $\\mathrm{B}$ | 向饱和 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶液中滴加鸡蛋清溶 <br> 液 | 出现白色沉淀 | 饱和 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶 <br> 液可使蛋白质变性 |\n| $\\mathrm{C}$ | 石蜡油加强热, 将产生的气体通入 $\\mathrm{Br}_{2}$ <br> 的 $\\mathrm{CCl}_{4}$ 溶液 | 溶液红棕色变无色 | 气体中含有不饱和烃 |\n| $\\mathrm{D}$ | 将封装 $\\mathrm{NO}_{2}$ 和 $\\mathrm{N}_{2} \\mathrm{O}_{4}$ 混合气体的球形玻璃容器, 浸入热水中 | 气体颜色变深 | 温度可以破坏化学平衡 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_598",
        "problem": "室温下, 用含少量 $\\mathrm{Co}^{2+}$ 和 $\\mathrm{Ni}^{2+}$ 的 $\\mathrm{FeSO}_{4}$ 溶液制备 $\\mathrm{FeC}_{2} \\mathrm{O}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ 的过程如下。\n\n[图1]\n\n已知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CoS})=1.8 \\times 10^{-22}, \\mathrm{~K}_{\\mathrm{sp}}(\\mathrm{NiS})=3.0 \\times 10^{-21} ; \\mathrm{K}_{\\mathrm{al}}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=1.1 \\times 10^{-7}$,\n\n$\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=1.3 \\times 10^{-13}$\n\n下列说法正确的是\nA: $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中： $c\\left(\\mathrm{HS}^{-}\\right)>c\\left(\\mathrm{~S}^{2-}\\right)$\nB: “除钴锦”后得到的上层清液中 $c\\left(\\mathrm{Ni}^{2+}\\right)$ 为 $1.0 \\times 10^{-6} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 则 $c\\left(\\mathrm{Co}^{2+}\\right)$ 为 $6 \\times 10^{-2} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHC}_{2} \\mathrm{O}_{4}$ 溶液中: $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nD: “沉铁”后的滤液中: $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 用含少量 $\\mathrm{Co}^{2+}$ 和 $\\mathrm{Ni}^{2+}$ 的 $\\mathrm{FeSO}_{4}$ 溶液制备 $\\mathrm{FeC}_{2} \\mathrm{O}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ 的过程如下。\n\n[图1]\n\n已知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CoS})=1.8 \\times 10^{-22}, \\mathrm{~K}_{\\mathrm{sp}}(\\mathrm{NiS})=3.0 \\times 10^{-21} ; \\mathrm{K}_{\\mathrm{al}}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=1.1 \\times 10^{-7}$,\n\n$\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=1.3 \\times 10^{-13}$\n\n下列说法正确的是\n\nA: $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中： $c\\left(\\mathrm{HS}^{-}\\right)>c\\left(\\mathrm{~S}^{2-}\\right)$\nB: “除钴锦”后得到的上层清液中 $c\\left(\\mathrm{Ni}^{2+}\\right)$ 为 $1.0 \\times 10^{-6} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 则 $c\\left(\\mathrm{Co}^{2+}\\right)$ 为 $6 \\times 10^{-2} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHC}_{2} \\mathrm{O}_{4}$ 溶液中: $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nD: “沉铁”后的滤液中: $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-040.jpg?height=211&width=1094&top_left_y=1194&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-041.jpg?height=222&width=1376&top_left_y=500&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1379",
        "problem": "Anhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion.\n\nIn an experiment, gaseous $\\mathrm{NH}_{3}$ is burned with $\\mathrm{O}_{2}$ in a container of fixed volume according to the equation given below.\n\n$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe initial and final states are at $298 \\mathrm{~K}$. After combustion with $14.40 \\mathrm{~g}$ of $\\mathrm{O}_{2}$, some of $\\mathrm{NH}_{3}$ remains unreacted.Calculate the heat released during the process.\n$$\n\\Delta_{f} H^{0}\\left(\\mathrm{NH}_{3}(g)\\right)=-46.11 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\text { and } \\Delta_{f} H^{0}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right)=-285.83 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nAnhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion.\n\nIn an experiment, gaseous $\\mathrm{NH}_{3}$ is burned with $\\mathrm{O}_{2}$ in a container of fixed volume according to the equation given below.\n\n$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe initial and final states are at $298 \\mathrm{~K}$. After combustion with $14.40 \\mathrm{~g}$ of $\\mathrm{O}_{2}$, some of $\\mathrm{NH}_{3}$ remains unreacted.\n\nproblem:\nCalculate the heat released during the process.\n$$\n\\Delta_{f} H^{0}\\left(\\mathrm{NH}_{3}(g)\\right)=-46.11 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\text { and } \\Delta_{f} H^{0}\\left(\\mathrm{H}_{2} \\mathrm{O}(l)\\right)=-285.83 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kJ, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kJ"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_498",
        "problem": "常温下, 下列说法正确的是\nA: $0.1 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{LHCN}$ 和 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaCN}$ 的等体积混合溶液 中: $\\mathrm{c}(\\mathrm{HCN})>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液滴加 $\\mathrm{HCl}$ 溶液至 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right): 3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} / \\mathrm{LNH}_{4} \\mathrm{HSO}_{4}$ 溶液滴加 $\\mathrm{NaOH}$ 溶液至 $\\mathrm{pH}=7: \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 下列说法正确的是\n\nA: $0.1 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{LHCN}$ 和 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaCN}$ 的等体积混合溶液 中: $\\mathrm{c}(\\mathrm{HCN})>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液滴加 $\\mathrm{HCl}$ 溶液至 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right): 3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} / \\mathrm{LNH}_{4} \\mathrm{HSO}_{4}$ 溶液滴加 $\\mathrm{NaOH}$ 溶液至 $\\mathrm{pH}=7: \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_468",
        "problem": "常温下, 已知 $\\mathrm{Ksp}(\\mathrm{ZnS})=1.6 \\times 10^{-24}, \\mathrm{Ksp}(\\mathrm{CuS})=1.3 \\times 10^{-36}$, 如图所示 $\\mathrm{CuS}$ 和 $\\mathrm{ZnS}$ 饱和溶液中阳离子 $\\left(\\mathrm{R}^{2+}\\right)$ 浓度与阴离子 $\\left(\\mathrm{S}^{2-}\\right)$ 浓度的负对数关系, 下列说法正确的是\n\n[图1]\nA: 曲线 $\\mathrm{A}$ 表示的是 $\\mathrm{ZnS}$, 曲线 B 表示的是 $\\mathrm{CuS}$\nB: 向含 $\\mathrm{Zn}^{2+}$ 和 $\\mathrm{Cu}^{2+}$ 的溶液中逐滴加入 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, $\\mathrm{Cu}^{2+}$ 先沉淀\nC: $\\mathrm{p}$ 点表示 $\\mathrm{CuS}$ 或 $\\mathrm{ZnS}$ 的过饱和溶液, 该温度下将析出沉淀\nD: 向曲线 $\\mathrm{A}$ 表示的溶液中加入 $1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, 不可能实现 $\\mathrm{n}$ 点到 $\\mathrm{m}$ 点的转换\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 已知 $\\mathrm{Ksp}(\\mathrm{ZnS})=1.6 \\times 10^{-24}, \\mathrm{Ksp}(\\mathrm{CuS})=1.3 \\times 10^{-36}$, 如图所示 $\\mathrm{CuS}$ 和 $\\mathrm{ZnS}$ 饱和溶液中阳离子 $\\left(\\mathrm{R}^{2+}\\right)$ 浓度与阴离子 $\\left(\\mathrm{S}^{2-}\\right)$ 浓度的负对数关系, 下列说法正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{A}$ 表示的是 $\\mathrm{ZnS}$, 曲线 B 表示的是 $\\mathrm{CuS}$\nB: 向含 $\\mathrm{Zn}^{2+}$ 和 $\\mathrm{Cu}^{2+}$ 的溶液中逐滴加入 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, $\\mathrm{Cu}^{2+}$ 先沉淀\nC: $\\mathrm{p}$ 点表示 $\\mathrm{CuS}$ 或 $\\mathrm{ZnS}$ 的过饱和溶液, 该温度下将析出沉淀\nD: 向曲线 $\\mathrm{A}$ 表示的溶液中加入 $1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, 不可能实现 $\\mathrm{n}$ 点到 $\\mathrm{m}$ 点的转换\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-050.jpg?height=471&width=556&top_left_y=570&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1052",
        "problem": "When ammonium dichromate( $\\mathrm{VI})$ is added gradually to molten ammonium thiocyanate, Reinecke's salt is formed. It has the formula $\\mathrm{NH}_{4}\\left[\\mathrm{Cr}(\\mathrm{SCN})_{4}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]$ and the following composition by mass:\n\n$$\n\\begin{array}{ll}\n\\mathrm{Cr} & 15.5 \\% \\\\\n\\mathrm{~S} & 38.15 \\% \\\\\n\\mathrm{~N} & 29.2 \\%\n\\end{array}\n$$\n\nCalculate the oxidation number of chromium in the complex.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nWhen ammonium dichromate( $\\mathrm{VI})$ is added gradually to molten ammonium thiocyanate, Reinecke's salt is formed. It has the formula $\\mathrm{NH}_{4}\\left[\\mathrm{Cr}(\\mathrm{SCN})_{4}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]$ and the following composition by mass:\n\n$$\n\\begin{array}{ll}\n\\mathrm{Cr} & 15.5 \\% \\\\\n\\mathrm{~S} & 38.15 \\% \\\\\n\\mathrm{~N} & 29.2 \\%\n\\end{array}\n$$\n\nCalculate the oxidation number of chromium in the complex.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_951",
        "problem": "常温下, 向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 溶液中不断通入 $\\mathrm{HCl}$ 气体, 溶液中三种含氮微粒的物质的量分数 $\\delta$ 与 $\\mathrm{pOH}$ 的关系图如图。已知：(1)乙胺为无色液体，有类似氨的性质, 常温下 $\\mathrm{K}_{\\mathrm{b} 1}=10^{-4.07}, \\mathrm{~K}_{\\mathrm{b} 2}=10^{-7.15}$; (2)假设溶液体积保持不变。下列说法中不正确的是 $\\left\\{\\right.$ 已知: $\\left.\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right\\}$\n\n[图1]\nA: 曲线II代表的微粒符号为 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}^{+}$\nB: $\\mathrm{b}$ 点的对应 $\\mathrm{pOH}=5.61$\nC: $\\mathrm{pOH}=\\mathrm{a}$ 时, $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 在 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3} \\mathrm{Cl}$ 溶液中: $\\mathrm{c}\\left(\\left[\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{+}\\right)>$ $\\mathrm{c}\\left(\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{2+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 溶液中不断通入 $\\mathrm{HCl}$ 气体, 溶液中三种含氮微粒的物质的量分数 $\\delta$ 与 $\\mathrm{pOH}$ 的关系图如图。已知：(1)乙胺为无色液体，有类似氨的性质, 常温下 $\\mathrm{K}_{\\mathrm{b} 1}=10^{-4.07}, \\mathrm{~K}_{\\mathrm{b} 2}=10^{-7.15}$; (2)假设溶液体积保持不变。下列说法中不正确的是 $\\left\\{\\right.$ 已知: $\\left.\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right\\}$\n\n[图1]\n\nA: 曲线II代表的微粒符号为 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}^{+}$\nB: $\\mathrm{b}$ 点的对应 $\\mathrm{pOH}=5.61$\nC: $\\mathrm{pOH}=\\mathrm{a}$ 时, $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 在 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3} \\mathrm{Cl}$ 溶液中: $\\mathrm{c}\\left(\\left[\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{+}\\right)>$ $\\mathrm{c}\\left(\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{2+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-060.jpg?height=546&width=751&top_left_y=178&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_827",
        "problem": "已知 $\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)$。向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的氨水中滴加未知浓度的稀硫酸, 测得混合溶液的温度、 $\\mathrm{pOH}$ 随加入稀硫酸体积的变化如下图所示, 下列说法不正确的是\n\n[图1]\nA: 稀硫酸中溶质的物质的量浓度为 $0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: 当溶液中 $\\mathrm{pH}=\\mathrm{pOH}$ 时, 水的电离程度最大\nC: $b$ 点时溶液中存在 $c\\left(N_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $a 、 b 、 c$ 三点对应 $\\mathrm{NH}_{4}^{+}$的水解平衡常数: $K_{h}(b)>K_{h}(a)>K_{h}(c)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知 $\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)$。向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的氨水中滴加未知浓度的稀硫酸, 测得混合溶液的温度、 $\\mathrm{pOH}$ 随加入稀硫酸体积的变化如下图所示, 下列说法不正确的是\n\n[图1]\n\nA: 稀硫酸中溶质的物质的量浓度为 $0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: 当溶液中 $\\mathrm{pH}=\\mathrm{pOH}$ 时, 水的电离程度最大\nC: $b$ 点时溶液中存在 $c\\left(N_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $a 、 b 、 c$ 三点对应 $\\mathrm{NH}_{4}^{+}$的水解平衡常数: $K_{h}(b)>K_{h}(a)>K_{h}(c)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-067.jpg?height=500&width=694&top_left_y=150&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_571",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 下列有关溶液中微粒的物质的量浓度关系正确的是\nA: 向醋酸钠溶液中加入适量醋酸, 得到的酸性混合溶液: $\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$ $>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液等体积混合: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=$ $3 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+3 \\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nC: $0.1 \\mathrm{~mol} / \\mathrm{LNH}_{4} \\mathrm{Cl}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 氨水等体积混合 $(\\mathrm{pH}>7): \\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>$ $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{LHCl}$ 溶液等体积混合 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ 为二元弱 酸): $2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 下列有关溶液中微粒的物质的量浓度关系正确的是\n\nA: 向醋酸钠溶液中加入适量醋酸, 得到的酸性混合溶液: $\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$ $>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液等体积混合: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=$ $3 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+3 \\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nC: $0.1 \\mathrm{~mol} / \\mathrm{LNH}_{4} \\mathrm{Cl}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 氨水等体积混合 $(\\mathrm{pH}>7): \\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>$ $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{LHCl}$ 溶液等体积混合 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right.$ 为二元弱 酸): $2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_111",
        "problem": "The ionization energies for period 3 element $X$ are listed in the table below.\n\n| Ionization Energies for element $X\\left(\\mathrm{~kJ} \\mathrm{~mol}^{-1}\\right)$ |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| First | Second | Third | Fourth | Fifth |\n| 580 | 1,815 | 2,740 | 11,600 | 14,800 |\n\nBased on the data, which statement about element $\\mathrm{X}$ is FALSE?\nA: Its most common oxidation state is +3\nB: It is displaced from aqueous solution by copper metal\nC: It is the most abundant metal in the Earth's crust\nD: Its oxide is insoluble in water\nE: It is a lustrous metal\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe ionization energies for period 3 element $X$ are listed in the table below.\n\n| Ionization Energies for element $X\\left(\\mathrm{~kJ} \\mathrm{~mol}^{-1}\\right)$ |  |  |  |  |\n| :---: | :---: | :---: | :---: | :---: |\n| First | Second | Third | Fourth | Fifth |\n| 580 | 1,815 | 2,740 | 11,600 | 14,800 |\n\nBased on the data, which statement about element $\\mathrm{X}$ is FALSE?\n\nA: Its most common oxidation state is +3\nB: It is displaced from aqueous solution by copper metal\nC: It is the most abundant metal in the Earth's crust\nD: Its oxide is insoluble in water\nE: It is a lustrous metal\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1157",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the misplacement of thallium\n\nMendeleev also misplaced the highly toxic element thallium ( $\\mathrm{Tl}$ ) with the elements from Group 1 rather than in Group 13. There are good reasons for this error. Thallium can form salts in the +3 oxidation state like other members of its group, but $\\mathrm{Tl}^{3+}$ ions are oxidising and the most stable oxidation state is +1 . For example, adding iodide ions to solutions of $\\mathrm{Tl}^{3+}$ actually gives a precipitate of thallium $(\\mathrm{I})$ iodide.\n\nGive the equation for the reaction of aqueous thallium(III) nitrate with potassium iodide.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the misplacement of thallium\n\nMendeleev also misplaced the highly toxic element thallium ( $\\mathrm{Tl}$ ) with the elements from Group 1 rather than in Group 13. There are good reasons for this error. Thallium can form salts in the +3 oxidation state like other members of its group, but $\\mathrm{Tl}^{3+}$ ions are oxidising and the most stable oxidation state is +1 . For example, adding iodide ions to solutions of $\\mathrm{Tl}^{3+}$ actually gives a precipitate of thallium $(\\mathrm{I})$ iodide.\n\nGive the equation for the reaction of aqueous thallium(III) nitrate with potassium iodide.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_157",
        "problem": "The equilibrium $\\mathrm{CO}(\\mathrm{g})+\\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{NO}(\\mathrm{g})$ is established in four different, but identical containers. Each container started with a different composition as follows:\n\n| Container | $\\mathrm{CO}(\\mathrm{mol})$ | $\\mathrm{NO}_{2}(\\mathrm{~mol})$ | $\\mathrm{CO}_{2}(\\mathrm{~mol})$ | $\\mathrm{NO}(\\mathrm{mol})$ |\n| :--- | :--- | :--- | :--- | :--- |\n| 1 | 1 | 1 | 0 | 0 |\n| 2 | 1 | 0 | 1 | 1 |\n| 3 | 1 | 1 | 1 | 0 |\n| 4 | 0 | 1 | 1 | 1 |\n| 5 | 1 | 1 | 1 | 1 |\n\nAfter equilibrium is established, which container would have the largest concentration of $\\mathrm{CO}(\\mathrm{g})$\nA: 1\nB: 2\nC: 3\nD: 4\nE: 5\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe equilibrium $\\mathrm{CO}(\\mathrm{g})+\\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{NO}(\\mathrm{g})$ is established in four different, but identical containers. Each container started with a different composition as follows:\n\n| Container | $\\mathrm{CO}(\\mathrm{mol})$ | $\\mathrm{NO}_{2}(\\mathrm{~mol})$ | $\\mathrm{CO}_{2}(\\mathrm{~mol})$ | $\\mathrm{NO}(\\mathrm{mol})$ |\n| :--- | :--- | :--- | :--- | :--- |\n| 1 | 1 | 1 | 0 | 0 |\n| 2 | 1 | 0 | 1 | 1 |\n| 3 | 1 | 1 | 1 | 0 |\n| 4 | 0 | 1 | 1 | 1 |\n| 5 | 1 | 1 | 1 | 1 |\n\nAfter equilibrium is established, which container would have the largest concentration of $\\mathrm{CO}(\\mathrm{g})$\n\nA: 1\nB: 2\nC: 3\nD: 4\nE: 5\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_417",
        "problem": "工业上制备 $\\mathrm{Ti}$, 采用碳氯化法将 $\\mathrm{TiO}_{2}$ 转化成 $\\mathrm{TiCl}_{4}$ 。在 $1000^{\\circ} \\mathrm{C}$ 时发生如下:\n\n(1) $\\mathrm{TiO}_{2}(\\mathrm{~s})+2 \\mathrm{Cl}_{2}(\\mathrm{~g})+2 \\mathrm{C}(\\mathrm{s})=\\mathrm{TiCl}_{4}(\\mathrm{~g})+2 \\mathrm{CO}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}_{1}=-51.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n$\\mathrm{K}_{1}=1.6 \\times 10^{14}$\n\n(2) $2 \\mathrm{CO}(\\mathrm{g})=\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{s}) \\quad \\Delta \\mathrm{H}_{2}=-172.5 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} \\quad \\mathrm{~K}_{2}=1.0 \\times 10^{-4}$\n\n(3) $2 \\mathrm{C}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g})=2 \\mathrm{CO}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}_{4}=-223.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} \\quad \\mathrm{~K}_{3}=2.5 \\times 10^{18}$ 。\n\n在 $1.0 \\times 10^{5} \\mathrm{~Pa}$, 将 $\\mathrm{TiO}_{2} 、 \\mathrm{C}_{、} \\mathrm{Cl}_{2}$ 以物质的量比 1: 2.2: 2 进行碳氯化, 平衡时体系中\n\n$\\mathrm{CO}_{2} 、 \\mathrm{CO} 、 \\mathrm{TiCl}_{4}$ 和 $\\mathrm{C}$ 的组成比(物质的量分数)随温度变化如下图所示。下列说法不正确的是\n\n[图1]\nA: $1000^{\\circ} \\mathrm{C}$ 时, 反应 $\\mathrm{TiO}_{2}(\\mathrm{~s})+2 \\mathrm{Cl}_{2}(\\mathrm{~g})=\\mathrm{TiCl}_{4}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})$ 的平衡常数 $\\mathrm{K}=6.4 \\times 10^{-5}$\nB: 曲线 III 表示平衡时 $\\mathrm{CO}_{2}$ 的物质的量分数随温度的变化\nC: 高于 $600^{\\circ} \\mathrm{C}$, 升高温度, 主要对反应(2)的平衡产生影响\nD: 为保证 $\\mathrm{TiCl}_{4}$ 的平衡产率, 选择反应温度应高于 $1000^{\\circ} \\mathrm{C}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n工业上制备 $\\mathrm{Ti}$, 采用碳氯化法将 $\\mathrm{TiO}_{2}$ 转化成 $\\mathrm{TiCl}_{4}$ 。在 $1000^{\\circ} \\mathrm{C}$ 时发生如下:\n\n(1) $\\mathrm{TiO}_{2}(\\mathrm{~s})+2 \\mathrm{Cl}_{2}(\\mathrm{~g})+2 \\mathrm{C}(\\mathrm{s})=\\mathrm{TiCl}_{4}(\\mathrm{~g})+2 \\mathrm{CO}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}_{1}=-51.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n$\\mathrm{K}_{1}=1.6 \\times 10^{14}$\n\n(2) $2 \\mathrm{CO}(\\mathrm{g})=\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{s}) \\quad \\Delta \\mathrm{H}_{2}=-172.5 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} \\quad \\mathrm{~K}_{2}=1.0 \\times 10^{-4}$\n\n(3) $2 \\mathrm{C}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g})=2 \\mathrm{CO}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}_{4}=-223.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} \\quad \\mathrm{~K}_{3}=2.5 \\times 10^{18}$ 。\n\n在 $1.0 \\times 10^{5} \\mathrm{~Pa}$, 将 $\\mathrm{TiO}_{2} 、 \\mathrm{C}_{、} \\mathrm{Cl}_{2}$ 以物质的量比 1: 2.2: 2 进行碳氯化, 平衡时体系中\n\n$\\mathrm{CO}_{2} 、 \\mathrm{CO} 、 \\mathrm{TiCl}_{4}$ 和 $\\mathrm{C}$ 的组成比(物质的量分数)随温度变化如下图所示。下列说法不正确的是\n\n[图1]\n\nA: $1000^{\\circ} \\mathrm{C}$ 时, 反应 $\\mathrm{TiO}_{2}(\\mathrm{~s})+2 \\mathrm{Cl}_{2}(\\mathrm{~g})=\\mathrm{TiCl}_{4}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})$ 的平衡常数 $\\mathrm{K}=6.4 \\times 10^{-5}$\nB: 曲线 III 表示平衡时 $\\mathrm{CO}_{2}$ 的物质的量分数随温度的变化\nC: 高于 $600^{\\circ} \\mathrm{C}$, 升高温度, 主要对反应(2)的平衡产生影响\nD: 为保证 $\\mathrm{TiCl}_{4}$ 的平衡产率, 选择反应温度应高于 $1000^{\\circ} \\mathrm{C}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-014.jpg?height=436&width=560&top_left_y=770&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_149",
        "problem": "A student performs the following reaction to make solid sulfur:\n\n$$\n\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}(\\mathrm{aq})+2 \\mathrm{HCl}(\\mathrm{aq}) \\rightarrow 2 \\mathrm{NaCl}(\\mathrm{aq})+\\mathrm{S}(\\mathrm{s})+\\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})\n$$\n\nThe student records the following data at the start of the reaction:\n\n|  | Concentration (mol L-1) | Volume (mL) |\n| :--- | :---: | :---: |\n| $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}(\\mathrm{aq})$ | 0.45 | 130 |\n| $\\mathrm{HCl}(\\mathrm{aq})$ | 0.15 | 400 |\n\nIf the student recovers $0.89 \\mathrm{~g}$ of solid sulfur from the experiment, what is the $\\%$ yield of the reaction?\nA: $46 \\%$\nB: $48 \\%$\nC: $75 \\%$\nD: $89 \\%$\nE: $93 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student performs the following reaction to make solid sulfur:\n\n$$\n\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}(\\mathrm{aq})+2 \\mathrm{HCl}(\\mathrm{aq}) \\rightarrow 2 \\mathrm{NaCl}(\\mathrm{aq})+\\mathrm{S}(\\mathrm{s})+\\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})\n$$\n\nThe student records the following data at the start of the reaction:\n\n|  | Concentration (mol L-1) | Volume (mL) |\n| :--- | :---: | :---: |\n| $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}(\\mathrm{aq})$ | 0.45 | 130 |\n| $\\mathrm{HCl}(\\mathrm{aq})$ | 0.15 | 400 |\n\nIf the student recovers $0.89 \\mathrm{~g}$ of solid sulfur from the experiment, what is the $\\%$ yield of the reaction?\n\nA: $46 \\%$\nB: $48 \\%$\nC: $75 \\%$\nD: $89 \\%$\nE: $93 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_181",
        "problem": "Given the following electrochemical cell data\n\nCell $1 \\mathrm{Cd}(\\mathrm{s})\\left|\\mathrm{Cd}^{2+}(1.0 \\mathrm{M}) \\| \\mathrm{Cu}^{2+}(1.0 \\mathrm{M})\\right| \\mathrm{Cu}(\\mathrm{s}) \\quad \\mathrm{E}^{0}=+0.74 \\mathrm{~V}$\n\nCell $2 \\mathrm{Zn}(\\mathrm{s})\\left|\\mathrm{Zn}^{2+}(1.0 \\mathrm{M}) \\| \\mathrm{Cu}^{2+}(1.0 \\mathrm{M})\\right| \\mathrm{Cu}(\\mathrm{s}) \\quad \\mathrm{E}^{0}=+1.10 \\mathrm{~V}$\n\nCell $3 \\mathrm{Zn}(\\mathrm{s})\\left|\\mathrm{Zn}^{2+}(1.0 \\mathrm{M}) \\| \\mathrm{Cd}^{2+}(1.0 \\mathrm{M})\\right| \\mathrm{Cd}(\\mathrm{s})$\n\nDetermine the standard cell potential for Cell 3.\nA: $-0.36 \\mathrm{~V}$\nB: $0.36 \\mathrm{~V}$\nC: $-1.84 \\mathrm{~V}$\nD: $-0.18 \\mathrm{~V}$\nE: $0.18 \\mathrm{~V}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGiven the following electrochemical cell data\n\nCell $1 \\mathrm{Cd}(\\mathrm{s})\\left|\\mathrm{Cd}^{2+}(1.0 \\mathrm{M}) \\| \\mathrm{Cu}^{2+}(1.0 \\mathrm{M})\\right| \\mathrm{Cu}(\\mathrm{s}) \\quad \\mathrm{E}^{0}=+0.74 \\mathrm{~V}$\n\nCell $2 \\mathrm{Zn}(\\mathrm{s})\\left|\\mathrm{Zn}^{2+}(1.0 \\mathrm{M}) \\| \\mathrm{Cu}^{2+}(1.0 \\mathrm{M})\\right| \\mathrm{Cu}(\\mathrm{s}) \\quad \\mathrm{E}^{0}=+1.10 \\mathrm{~V}$\n\nCell $3 \\mathrm{Zn}(\\mathrm{s})\\left|\\mathrm{Zn}^{2+}(1.0 \\mathrm{M}) \\| \\mathrm{Cd}^{2+}(1.0 \\mathrm{M})\\right| \\mathrm{Cd}(\\mathrm{s})$\n\nDetermine the standard cell potential for Cell 3.\n\nA: $-0.36 \\mathrm{~V}$\nB: $0.36 \\mathrm{~V}$\nC: $-1.84 \\mathrm{~V}$\nD: $-0.18 \\mathrm{~V}$\nE: $0.18 \\mathrm{~V}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1304",
        "problem": "A fuel/oxidant system consisting of $\\mathrm{N}, \\mathrm{N}$-dimethylhydrazine $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NNH}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}_{4}$ (both liquids) is commonly used in space vehicle propulsion. Components are mixed stoichiometrically so that $\\mathrm{N}_{2}, \\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$ are the only products (all gases under the same reaction conditions). How many moles of gases are produced from $1 \\mathrm{~mol}$ of $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NNH}_{2}$ ?\nA: 8\nB: 9\nC: 10\nD: 11\nE: 12\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA fuel/oxidant system consisting of $\\mathrm{N}, \\mathrm{N}$-dimethylhydrazine $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NNH}_{2}$ and $\\mathrm{N}_{2} \\mathrm{O}_{4}$ (both liquids) is commonly used in space vehicle propulsion. Components are mixed stoichiometrically so that $\\mathrm{N}_{2}, \\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$ are the only products (all gases under the same reaction conditions). How many moles of gases are produced from $1 \\mathrm{~mol}$ of $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NNH}_{2}$ ?\n\nA: 8\nB: 9\nC: 10\nD: 11\nE: 12\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1551",
        "problem": "Thiomolybdate ions are derived from molybdate ions, $\\mathrm{MoO}_{4}^{2-}$, by replacing oxygen atoms with sulfur atoms. In nature, thiomolybdate ions are found in such places as the deep waters of the Black Sea, where biological sulfate reduction generates $\\mathrm{H}_{2} \\mathrm{~S}$. The molybdate to thiomolybdate transformation leads to rapid loss of dissolved Mo from seawater to underlying sediments, depleting the ocean in Mo, a trace element essential for life.\n\nThe following equilibria control the relative concentrations of molybdate and thiomolybdate ions in dilute aqueous solution:\n\n$$\n\\begin{array}{ll}\n\\mathrm{MoS}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{1}=1.3 \\cdot 10^{-5} \\\\\n\\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{2}=1.0 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{3}=1.6 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{4}=6.5 \\cdot 10^{-6}\n\\end{array}\n$$If at equilibrium the concentrations of $\\mathrm{MoO}_{4}^{2-}$ and $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq})$ are equal to $1 \\cdot 10^{-7}$ and $1 \\cdot 10^{-6}$, respectively, what would be the equilibrium concentration of $\\mathrm{MoS}_{4}^{2-}$ ?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThiomolybdate ions are derived from molybdate ions, $\\mathrm{MoO}_{4}^{2-}$, by replacing oxygen atoms with sulfur atoms. In nature, thiomolybdate ions are found in such places as the deep waters of the Black Sea, where biological sulfate reduction generates $\\mathrm{H}_{2} \\mathrm{~S}$. The molybdate to thiomolybdate transformation leads to rapid loss of dissolved Mo from seawater to underlying sediments, depleting the ocean in Mo, a trace element essential for life.\n\nThe following equilibria control the relative concentrations of molybdate and thiomolybdate ions in dilute aqueous solution:\n\n$$\n\\begin{array}{ll}\n\\mathrm{MoS}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{1}=1.3 \\cdot 10^{-5} \\\\\n\\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{2}=1.0 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{3}=1.6 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{4}=6.5 \\cdot 10^{-6}\n\\end{array}\n$$\n\nproblem:\nIf at equilibrium the concentrations of $\\mathrm{MoO}_{4}^{2-}$ and $\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq})$ are equal to $1 \\cdot 10^{-7}$ and $1 \\cdot 10^{-6}$, respectively, what would be the equilibrium concentration of $\\mathrm{MoS}_{4}^{2-}$ ?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_654",
        "problem": "$\\mathrm{Cu}^{2+}$ 与 $\\mathrm{NH}_{3}$ 可结合生成多种络合物, 在水溶液中存在如下平衡: $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+} \\rightleftharpoons$\n\n$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{3}\\right]^{2+} \\stackrel{\\mathrm{K}_{2}}{\\rightleftharpoons}\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+} \\stackrel{\\mathrm{K}_{3}}{\\rightleftharpoons}\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)\\right]^{2+} \\stackrel{\\mathrm{K}_{4}}{\\rightleftharpoons} \\mathrm{Cu}^{2+}$ 。向某浓度的硫酸铜溶液\n中滴加浓氨水, 实验测得含 $\\mathrm{Cu}$ 微粒的物质的量分布分数 $(\\delta)$ 与溶液中游离氨的 $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)$关系如图所示。下列说法正确的是\n\n[图1]\nA: $\\frac{\\mathrm{K}_{2}}{\\mathrm{~K}_{1}}<\\frac{\\mathrm{K}_{4}}{\\mathrm{~K}_{2}}$\nB: 曲线 $\\mathrm{c}$ 表示 $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+}$\nC: $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)=-4$ 时, $\\mathrm{c}\\left(\\mathrm{Cu}{ }^{2+}\\right)>\\mathrm{c}\\left(\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)\\right]^{2+}\\right)>\\mathrm{c}\\left(\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+}\\right)$\nD: $M$ 点时, $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)=\\frac{\\lg \\mathrm{K}_{1}+\\lg \\mathrm{K}_{2}+\\lg \\mathrm{K}_{3}}{3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{Cu}^{2+}$ 与 $\\mathrm{NH}_{3}$ 可结合生成多种络合物, 在水溶液中存在如下平衡: $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+} \\rightleftharpoons$\n\n$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{3}\\right]^{2+} \\stackrel{\\mathrm{K}_{2}}{\\rightleftharpoons}\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+} \\stackrel{\\mathrm{K}_{3}}{\\rightleftharpoons}\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)\\right]^{2+} \\stackrel{\\mathrm{K}_{4}}{\\rightleftharpoons} \\mathrm{Cu}^{2+}$ 。向某浓度的硫酸铜溶液\n中滴加浓氨水, 实验测得含 $\\mathrm{Cu}$ 微粒的物质的量分布分数 $(\\delta)$ 与溶液中游离氨的 $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)$关系如图所示。下列说法正确的是\n\n[图1]\n\nA: $\\frac{\\mathrm{K}_{2}}{\\mathrm{~K}_{1}}<\\frac{\\mathrm{K}_{4}}{\\mathrm{~K}_{2}}$\nB: 曲线 $\\mathrm{c}$ 表示 $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+}$\nC: $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)=-4$ 时, $\\mathrm{c}\\left(\\mathrm{Cu}{ }^{2+}\\right)>\\mathrm{c}\\left(\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)\\right]^{2+}\\right)>\\mathrm{c}\\left(\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+}\\right)$\nD: $M$ 点时, $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)=\\frac{\\lg \\mathrm{K}_{1}+\\lg \\mathrm{K}_{2}+\\lg \\mathrm{K}_{3}}{3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-071.jpg?height=543&width=865&top_left_y=314&top_left_x=321"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_144",
        "problem": "In the emergency oxygen system on commercial passenger aircraft, sodium chlorate undergoes thermal decomposition to produce oxygen gas:\n\n$$\n2 \\mathrm{NaClO}_{3(\\mathrm{~s})} \\rightarrow 2 \\mathrm{NaCl}_{(\\mathrm{s})}+3 \\mathrm{O}_{2(\\mathrm{~g})}\n$$\n\nOn average a human under stress consumes $38.0 \\mathrm{~L}$ of $\\mathrm{O}_{2}$ through respiration every 15 minutes. Determine the minimum mass of sodium chlorate required to deliver this volume of $\\mathrm{O}_{2}$ (assume $P=100 \\mathrm{kPa}$ and $T=$ 273.15 K)\nA: $65.0 \\mathrm{~g}$\nB: $119 \\mathrm{~g}$\nC: $178 \\mathrm{~g}$\nD: $267 \\mathrm{~g}$\nE: $356 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the emergency oxygen system on commercial passenger aircraft, sodium chlorate undergoes thermal decomposition to produce oxygen gas:\n\n$$\n2 \\mathrm{NaClO}_{3(\\mathrm{~s})} \\rightarrow 2 \\mathrm{NaCl}_{(\\mathrm{s})}+3 \\mathrm{O}_{2(\\mathrm{~g})}\n$$\n\nOn average a human under stress consumes $38.0 \\mathrm{~L}$ of $\\mathrm{O}_{2}$ through respiration every 15 minutes. Determine the minimum mass of sodium chlorate required to deliver this volume of $\\mathrm{O}_{2}$ (assume $P=100 \\mathrm{kPa}$ and $T=$ 273.15 K)\n\nA: $65.0 \\mathrm{~g}$\nB: $119 \\mathrm{~g}$\nC: $178 \\mathrm{~g}$\nD: $267 \\mathrm{~g}$\nE: $356 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1340",
        "problem": "Problem I: \n$\\Delta H^{0}=172.45 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\n$\\Delta S^{0}=176 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$\n\n$\\Delta G^{0}=\\Delta \\mathrm{H}^{0}-T \\Delta S^{0}=120 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\n$\\Delta G^{0}>0$ implies $K<1$The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nThe reaction above (in Problem I) is carried out between $\\mathrm{CO}_{2}$ and excess hot graphite in a reactor maintained at about $800{ }^{\\circ} \\mathrm{C}$ and a total pres sure of 5.0 bar. The equilibrium constant $K_{p}$ under these conditions is 10.0 . Calculate the partial pressure of $\\mathrm{CO}$ at equilibrium.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nProblem I: \n$\\Delta H^{0}=172.45 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\n$\\Delta S^{0}=176 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$\n\n$\\Delta G^{0}=\\Delta \\mathrm{H}^{0}-T \\Delta S^{0}=120 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\n$\\Delta G^{0}>0$ implies $K<1$\n\nproblem:\nThe second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nThe reaction above (in Problem I) is carried out between $\\mathrm{CO}_{2}$ and excess hot graphite in a reactor maintained at about $800{ }^{\\circ} \\mathrm{C}$ and a total pres sure of 5.0 bar. The equilibrium constant $K_{p}$ under these conditions is 10.0 . Calculate the partial pressure of $\\mathrm{CO}$ at equilibrium.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of bar, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "bar"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_199",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the total mass lost when $10.00 \\mathrm{~g}$ of this mineral is thermally decomposed.A sample of a different mineral is analysed by the same methods. This mineral also contains only $\\mathrm{Pb}^{2+}, \\mathrm{CO}_{3}{ }^{2-}, \\mathrm{OH}^{-}$and $\\mathrm{O}^{2-}$ ions.\n\nWhen a $5.000 \\mathrm{~g}$ sample of this mineral is treated with $25.00 \\mathrm{~mL}$ of $2.000 \\mathrm{~mol} \\mathrm{~L}^{-1}$ nitric acid $\\left(\\mathrm{HNO}_{3}\\right), 0.5214 \\mathrm{~g}$ of carbon dioxide is released, and $0.01051 \\mathrm{~mol}$ of the acid remains.\n\nWhen subjected to thermal decomposition, $5.000 \\mathrm{~g}$ of this mineral loses $0.5926 \\mathrm{~g}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the total mass lost when $10.00 \\mathrm{~g}$ of this mineral is thermally decomposed.A sample of a different mineral is analysed by the same methods. This mineral also contains only $\\mathrm{Pb}^{2+}, \\mathrm{CO}_{3}{ }^{2-}, \\mathrm{OH}^{-}$and $\\mathrm{O}^{2-}$ ions.\n\nWhen a $5.000 \\mathrm{~g}$ sample of this mineral is treated with $25.00 \\mathrm{~mL}$ of $2.000 \\mathrm{~mol} \\mathrm{~L}^{-1}$ nitric acid $\\left(\\mathrm{HNO}_{3}\\right), 0.5214 \\mathrm{~g}$ of carbon dioxide is released, and $0.01051 \\mathrm{~mol}$ of the acid remains.\n\nWhen subjected to thermal decomposition, $5.000 \\mathrm{~g}$ of this mineral loses $0.5926 \\mathrm{~g}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_302",
        "problem": "What is the hybridization of the carbon atoms in benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$ ?\nA: $s p^{2}$ and $s p^{3}$\nB: $s p^{3}$ only\nC: $s p, s p^{2}$ and $s p^{3}$\nD: $s p^{2}$ only\nE: $s p$ only\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the hybridization of the carbon atoms in benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$ ?\n\nA: $s p^{2}$ and $s p^{3}$\nB: $s p^{3}$ only\nC: $s p, s p^{2}$ and $s p^{3}$\nD: $s p^{2}$ only\nE: $s p$ only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1333",
        "problem": "Type II electrodes that are made of a metal covered with a sparingly soluble salt of the metal are dipped into a soluble salt solution containing an anion of the sparingly soluble salt. The silver/silver chloride $(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$ and the calomel electrode $(\\mathrm{Hg}$, $\\left.\\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Cl}\\right)$ are examples of such electrodes. The standard emf of a cell built of those electrodes (-) Ag, $\\mathrm{AgCl}^{-} / \\mathrm{Cl}^{-} \\| \\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Hg}(+)$ is $E^{0}=0.0455 \\mathrm{~V}$ at $T=298 \\mathrm{~K}$. The temperature coefficient for this cell is $d E^{0} / d T=3.38 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}$.\n\nKnowing the standard potential of $\\mathrm{Ag} / \\mathrm{Ag}^{+}$electrode is $E^{0}=0.799 \\mathrm{~V}$ and the solubility product of $\\mathrm{AgCl} K_{s p}=1.73 \\times 10^{-10}$, calculate the standard electrode potential value of the silver/silver chloride electrode. Derive an expression showing the dependence between $E^{O}\\left(\\mathrm{Ag} / \\mathrm{Ag}^{+}\\right)$and $E^{O}(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nType II electrodes that are made of a metal covered with a sparingly soluble salt of the metal are dipped into a soluble salt solution containing an anion of the sparingly soluble salt. The silver/silver chloride $(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$ and the calomel electrode $(\\mathrm{Hg}$, $\\left.\\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Cl}\\right)$ are examples of such electrodes. The standard emf of a cell built of those electrodes (-) Ag, $\\mathrm{AgCl}^{-} / \\mathrm{Cl}^{-} \\| \\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Hg}(+)$ is $E^{0}=0.0455 \\mathrm{~V}$ at $T=298 \\mathrm{~K}$. The temperature coefficient for this cell is $d E^{0} / d T=3.38 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}$.\n\nKnowing the standard potential of $\\mathrm{Ag} / \\mathrm{Ag}^{+}$electrode is $E^{0}=0.799 \\mathrm{~V}$ and the solubility product of $\\mathrm{AgCl} K_{s p}=1.73 \\times 10^{-10}$, calculate the standard electrode potential value of the silver/silver chloride electrode. Derive an expression showing the dependence between $E^{O}\\left(\\mathrm{Ag} / \\mathrm{Ag}^{+}\\right)$and $E^{O}(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of V, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "V"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_706",
        "problem": "图示为一种天然产物, 具有一定的除草功效。下列有关该化合物的说法错误的是\n\n[图1]\nA: 分子中含有三种含氧官能团\nB: $1 \\mathrm{~mol}$ 该化合物最多能与 $6 \\mathrm{molNaOH}$ 反应\nC: 既可以发生取代反应, 又能够发生加成反应\nD: 能与 $\\mathrm{FeCl}_{3}$ 发生显色反应，分子中所有碳原子可能共面\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n图示为一种天然产物, 具有一定的除草功效。下列有关该化合物的说法错误的是\n\n[图1]\n\nA: 分子中含有三种含氧官能团\nB: $1 \\mathrm{~mol}$ 该化合物最多能与 $6 \\mathrm{molNaOH}$ 反应\nC: 既可以发生取代反应, 又能够发生加成反应\nD: 能与 $\\mathrm{FeCl}_{3}$ 发生显色反应，分子中所有碳原子可能共面\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-82.jpg?height=600&width=1220&top_left_y=911&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_694",
        "problem": "下列叙述中不正确的是\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{HS}$ 溶液中有: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nB: $25^{\\circ} \\mathrm{C}$ 时, 将 $\\mathrm{a} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$ 的氨水与 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸等体积混合后, $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}(\\mathrm{Cl}$ $-)$, 则 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的电离常数为 $\\frac{10^{-9}}{\\mathrm{a}-0.01}$\nC: 等浓度的 $\\mathrm{HCN}$ 和 $\\mathrm{NaCN}$ 混合溶液中有: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)+\\mathrm{c}(\\mathrm{HCN})$\nD: 等 PH 的 HA 和 HB 溶液, 分别与一定浓度的氢氧化钠溶液完全中和, HA 消耗的氢氧化钠溶液体积多, 则可证明酸性 $\\mathrm{HA}>\\mathrm{HB}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列叙述中不正确的是\n\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{HS}$ 溶液中有: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nB: $25^{\\circ} \\mathrm{C}$ 时, 将 $\\mathrm{a} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$ 的氨水与 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸等体积混合后, $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}(\\mathrm{Cl}$ $-)$, 则 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的电离常数为 $\\frac{10^{-9}}{\\mathrm{a}-0.01}$\nC: 等浓度的 $\\mathrm{HCN}$ 和 $\\mathrm{NaCN}$ 混合溶液中有: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CN}^{-}\\right)+\\mathrm{c}(\\mathrm{HCN})$\nD: 等 PH 的 HA 和 HB 溶液, 分别与一定浓度的氢氧化钠溶液完全中和, HA 消耗的氢氧化钠溶液体积多, 则可证明酸性 $\\mathrm{HA}>\\mathrm{HB}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1109",
        "problem": "Boron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the number of $\\mathrm{B}$ atoms and $\\mathrm{N}$ atoms in the unit cell of $\\mathrm{w}-\\mathrm{BN}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nBoron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the number of $\\mathrm{B}$ atoms and $\\mathrm{N}$ atoms in the unit cell of $\\mathrm{w}-\\mathrm{BN}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the number of $\\mathrm{B}$ atoms, the number of $\\mathrm{N}$ atoms].\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=448&width=511&top_left_y=541&top_left_x=270",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=596&width=436&top_left_y=458&top_left_x=844",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=471&width=443&top_left_y=524&top_left_x=1389",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=603&width=305&top_left_y=1206&top_left_x=1315"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "the number of $\\mathrm{B}$ atoms",
            "the number of $\\mathrm{N}$ atoms"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_638",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 溶液中, 不同型体的分布优势如下图所示 $(\\mathrm{X}$ 指含磷微粒的百分含量、 $\\mathrm{pH}$ 的调控使用 $\\mathrm{HCl} 、 \\mathrm{NaOH})$ 。\n\n已知: (1) $\\mathrm{H}_{2} \\mathrm{CO}_{3}\\left(\\mathrm{~K}_{\\mathrm{a}_{1}}=1.0 \\times 10^{-6.38} 、 \\mathrm{~K}_{\\mathrm{a}_{2}}=1.0 \\times 10^{-10.25}\\right)$\n\n(2)当某一型体分布优势接近 $100 \\%$ 时，另外两种相邻型体含量近似相等;\n\n[图1]\n\n则下列说法错误的是\nA: $\\mathrm{pH}_{\\mathrm{B}} \\approx\\left(\\mathrm{pH}_{\\mathrm{A}}+\\mathrm{pH}_{\\mathrm{C}}\\right) / 2 \\approx 4.7$\nB: 当使用 $\\mathrm{NaOH}$ 溶液滴定 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 溶液至 $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ 时, 可选用酚䣭作指示剂\nC: D 点满足: $c\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+0.1$\nD: 不可用 $\\mathrm{NaHCO}_{3}$ 与 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 反应生成 $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 溶液中, 不同型体的分布优势如下图所示 $(\\mathrm{X}$ 指含磷微粒的百分含量、 $\\mathrm{pH}$ 的调控使用 $\\mathrm{HCl} 、 \\mathrm{NaOH})$ 。\n\n已知: (1) $\\mathrm{H}_{2} \\mathrm{CO}_{3}\\left(\\mathrm{~K}_{\\mathrm{a}_{1}}=1.0 \\times 10^{-6.38} 、 \\mathrm{~K}_{\\mathrm{a}_{2}}=1.0 \\times 10^{-10.25}\\right)$\n\n(2)当某一型体分布优势接近 $100 \\%$ 时，另外两种相邻型体含量近似相等;\n\n[图1]\n\n则下列说法错误的是\n\nA: $\\mathrm{pH}_{\\mathrm{B}} \\approx\\left(\\mathrm{pH}_{\\mathrm{A}}+\\mathrm{pH}_{\\mathrm{C}}\\right) / 2 \\approx 4.7$\nB: 当使用 $\\mathrm{NaOH}$ 溶液滴定 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 溶液至 $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ 时, 可选用酚䣭作指示剂\nC: D 点满足: $c\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+0.1$\nD: 不可用 $\\mathrm{NaHCO}_{3}$ 与 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 反应生成 $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-070.jpg?height=517&width=1145&top_left_y=164&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_249",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the percentage by mass of ammonium ions in the $1.988 \\mathrm{~g}$ ammonium salt sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the percentage by mass of ammonium ions in the $1.988 \\mathrm{~g}$ ammonium salt sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_406",
        "problem": "$\\mathrm{LiFePO}_{4}$ 的晶胞结构示意图如(a)所示。其中 $\\mathrm{O}$ 围绕 $\\mathrm{Fe}$ 和 $\\mathrm{P}$ 分别形成正八面体和四面体。电池充放电时, $\\mathrm{LiFePO}_{4}$ 脱出或嵌入 $\\mathrm{Li}^{+}$形成结构示意图如(b)所示。下列说法正确的是\n\n[图1]\n\n(a) $\\mathrm{LiFePO}_{4}$\n\n[图2]\n\n(b) $\\mathrm{Li}_{1-x} \\mathrm{FePO}_{4}$\n\n[图3]\n\n(c) $\\mathrm{FePO}_{4}$\nA: 每个 $\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}$ 晶胞中 $\\mathrm{Li}^{+}$个数为 $1-\\mathrm{x}$\nB: 图(b)中 $x=0.1875$\nC: 图(b)中 $n\\left(\\mathrm{Fe}^{2+}\\right): n\\left(\\mathrm{Fe}^{3+}\\right)=3: 13$\nD: 当 $\\mathrm{FePO}_{4}$ 转化为 $\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}$ 时, 每转移电子 $(1-\\mathrm{x}) \\mathrm{mol}$, 嵌入 $4(1-\\mathrm{x}) \\mathrm{mol} \\mathrm{Li}^{+}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{LiFePO}_{4}$ 的晶胞结构示意图如(a)所示。其中 $\\mathrm{O}$ 围绕 $\\mathrm{Fe}$ 和 $\\mathrm{P}$ 分别形成正八面体和四面体。电池充放电时, $\\mathrm{LiFePO}_{4}$ 脱出或嵌入 $\\mathrm{Li}^{+}$形成结构示意图如(b)所示。下列说法正确的是\n\n[图1]\n\n(a) $\\mathrm{LiFePO}_{4}$\n\n[图2]\n\n(b) $\\mathrm{Li}_{1-x} \\mathrm{FePO}_{4}$\n\n[图3]\n\n(c) $\\mathrm{FePO}_{4}$\n\nA: 每个 $\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}$ 晶胞中 $\\mathrm{Li}^{+}$个数为 $1-\\mathrm{x}$\nB: 图(b)中 $x=0.1875$\nC: 图(b)中 $n\\left(\\mathrm{Fe}^{2+}\\right): n\\left(\\mathrm{Fe}^{3+}\\right)=3: 13$\nD: 当 $\\mathrm{FePO}_{4}$ 转化为 $\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}$ 时, 每转移电子 $(1-\\mathrm{x}) \\mathrm{mol}$, 嵌入 $4(1-\\mathrm{x}) \\mathrm{mol} \\mathrm{Li}^{+}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-013.jpg?height=342&width=477&top_left_y=460&top_left_x=333",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-013.jpg?height=334&width=278&top_left_y=467&top_left_x=783",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-013.jpg?height=317&width=436&top_left_y=475&top_left_x=1067"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_334",
        "problem": "A $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solution of which of the following salts has the highest $\\mathrm{pH}$ at $298 \\mathrm{~K}$ ?\n| lonization constants <br> $($ at $298 \\mathrm{~K})$ |  |\n| :--- | :--- |\n| $\\mathrm{HIO}_{3}$, | $K_{a}=1.7 \\times 10^{-2}$ |\n| $\\mathrm{HF}$, | $K_{a}=6.3 \\times 10^{-4}$ |\n| $\\mathrm{HCN}$, | $K_{a}=6.2 \\times 10^{-10}$ |\n| $\\mathrm{NH}_{3}$, | $K_{b}=1.8 \\times 10^{-5}$ |\n| $\\mathrm{H}_{2} \\mathrm{O}$, | $K_{w}=1.0 \\times 10^{-14}$ |\nA: $\\mathrm{NaF}$\nB: $\\mathrm{NaIO}_{3}$\nC: $\\mathrm{NaCN}$\nD: $\\mathrm{NH}_{4} \\mathrm{~F}$\nE: $\\mathrm{NH}_{4} \\mathrm{IO}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solution of which of the following salts has the highest $\\mathrm{pH}$ at $298 \\mathrm{~K}$ ?\n| lonization constants <br> $($ at $298 \\mathrm{~K})$ |  |\n| :--- | :--- |\n| $\\mathrm{HIO}_{3}$, | $K_{a}=1.7 \\times 10^{-2}$ |\n| $\\mathrm{HF}$, | $K_{a}=6.3 \\times 10^{-4}$ |\n| $\\mathrm{HCN}$, | $K_{a}=6.2 \\times 10^{-10}$ |\n| $\\mathrm{NH}_{3}$, | $K_{b}=1.8 \\times 10^{-5}$ |\n| $\\mathrm{H}_{2} \\mathrm{O}$, | $K_{w}=1.0 \\times 10^{-14}$ |\n\nA: $\\mathrm{NaF}$\nB: $\\mathrm{NaIO}_{3}$\nC: $\\mathrm{NaCN}$\nD: $\\mathrm{NH}_{4} \\mathrm{~F}$\nE: $\\mathrm{NH}_{4} \\mathrm{IO}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_846",
        "problem": "恒容条件下, $1 \\mathrm{~mol} \\mathrm{SiHCl}_{3}$ 发生如下反应: $2 \\mathrm{SiHCl}_{3}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{SiH}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})+\\mathrm{SiCl}_{4}(\\mathrm{~g})$ 。已知 $\\mathrm{v}_{\\text {正 }}=\\mathrm{v}_{\\text {消耗 }}\\left(\\mathrm{SiHCl}_{3}\\right)=\\mathrm{k}_{\\text {正 }} \\mathrm{x}^{2}\\left(\\mathrm{SiHCl}_{3}\\right), \\mathrm{v}_{\\text {逆 }}=2 \\mathrm{v}$ 消耗 $\\left(\\mathrm{SiH}_{2} \\mathrm{Cl}_{2}\\right)=\\mathrm{k}$ 逆 $\\mathrm{x}\\left(\\mathrm{SiH}_{2} \\mathrm{Cl}_{2}\\right) \\mathrm{x}\\left(\\mathrm{SiCl}_{4}\\right), \\mathrm{k}_{\\text {正 }}$ $\\mathrm{k}$ 逆分别为正、逆向反应速率常数 (仅与温度有关), $\\mathrm{x}$ 为物质的量分数。如图是不同温度下 $\\mathrm{x}\\left(\\mathrm{SiHCl}_{3}\\right)$ 随时间的变化。下列说法正确的是\n\n[图1]\nA: 该反应为放热反应, $\\mathrm{v}_{\\text {正, }} \\mathrm{a}^{<} \\mathrm{v}$ 逆, $\\mathrm{b}$\nB: $\\mathrm{T}_{1} \\mathrm{~K}$ 时平衡体系中可通过移走 $\\mathrm{SiCl}_{4}$ 提高 $\\mathrm{SiHCl}_{3}$ 的转化率\nC: 当反应进行到 $\\mathrm{a}$ 处时, $\\mathrm{v}$ 正 $/ \\mathrm{v}$ 逆 $=16 / 9$\nD: $\\mathrm{T}_{2} \\mathrm{~K}$ 时平衡体系中再充入 $1 \\mathrm{~mol} \\mathrm{SiHCl}_{3}$, 平衡正向移动, $\\mathrm{x}\\left(\\mathrm{SiH}_{2} \\mathrm{Cl}_{2}\\right)$ 增大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n恒容条件下, $1 \\mathrm{~mol} \\mathrm{SiHCl}_{3}$ 发生如下反应: $2 \\mathrm{SiHCl}_{3}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{SiH}_{2} \\mathrm{Cl}_{2}(\\mathrm{~g})+\\mathrm{SiCl}_{4}(\\mathrm{~g})$ 。已知 $\\mathrm{v}_{\\text {正 }}=\\mathrm{v}_{\\text {消耗 }}\\left(\\mathrm{SiHCl}_{3}\\right)=\\mathrm{k}_{\\text {正 }} \\mathrm{x}^{2}\\left(\\mathrm{SiHCl}_{3}\\right), \\mathrm{v}_{\\text {逆 }}=2 \\mathrm{v}$ 消耗 $\\left(\\mathrm{SiH}_{2} \\mathrm{Cl}_{2}\\right)=\\mathrm{k}$ 逆 $\\mathrm{x}\\left(\\mathrm{SiH}_{2} \\mathrm{Cl}_{2}\\right) \\mathrm{x}\\left(\\mathrm{SiCl}_{4}\\right), \\mathrm{k}_{\\text {正 }}$ $\\mathrm{k}$ 逆分别为正、逆向反应速率常数 (仅与温度有关), $\\mathrm{x}$ 为物质的量分数。如图是不同温度下 $\\mathrm{x}\\left(\\mathrm{SiHCl}_{3}\\right)$ 随时间的变化。下列说法正确的是\n\n[图1]\n\nA: 该反应为放热反应, $\\mathrm{v}_{\\text {正, }} \\mathrm{a}^{<} \\mathrm{v}$ 逆, $\\mathrm{b}$\nB: $\\mathrm{T}_{1} \\mathrm{~K}$ 时平衡体系中可通过移走 $\\mathrm{SiCl}_{4}$ 提高 $\\mathrm{SiHCl}_{3}$ 的转化率\nC: 当反应进行到 $\\mathrm{a}$ 处时, $\\mathrm{v}$ 正 $/ \\mathrm{v}$ 逆 $=16 / 9$\nD: $\\mathrm{T}_{2} \\mathrm{~K}$ 时平衡体系中再充入 $1 \\mathrm{~mol} \\mathrm{SiHCl}_{3}$, 平衡正向移动, $\\mathrm{x}\\left(\\mathrm{SiH}_{2} \\mathrm{Cl}_{2}\\right)$ 增大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-094.jpg?height=514&width=648&top_left_y=1576&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-095.jpg?height=49&width=1374&top_left_y=661&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-095.jpg?height=103&width=1302&top_left_y=754&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1153",
        "problem": "This question is about benzene and its isomers\n\nWorking out the structure of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, was one of the great achievements of $19^{\\text {th }}$ century chemists. The carcinogenic hydrocarbon was first discovered by Michael Faraday at the Royal Institution in 1825 during the repeated fractional distillation of some oil residues.\n\nFaraday analysed the pure benzene by passing its vapour over heated copper(II) oxide to produce carbon dioxide, water and copper. He collected and measured the volume of liquid water and gaseous $\\mathrm{CO}_{2}$ produced from a given mass of benzene.\n\nAssuming $1.00 \\mathrm{~mol}$ of a gas occupies $24.0 \\mathrm{dm}^{3}$ at room temperature and pressure, calculate the volume of carbon dioxide that could be produced from $0.644 \\mathrm{mmol}$ of benzene. Give your answer in $\\mathrm{cm}^{3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about benzene and its isomers\n\nWorking out the structure of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, was one of the great achievements of $19^{\\text {th }}$ century chemists. The carcinogenic hydrocarbon was first discovered by Michael Faraday at the Royal Institution in 1825 during the repeated fractional distillation of some oil residues.\n\nFaraday analysed the pure benzene by passing its vapour over heated copper(II) oxide to produce carbon dioxide, water and copper. He collected and measured the volume of liquid water and gaseous $\\mathrm{CO}_{2}$ produced from a given mass of benzene.\n\nAssuming $1.00 \\mathrm{~mol}$ of a gas occupies $24.0 \\mathrm{dm}^{3}$ at room temperature and pressure, calculate the volume of carbon dioxide that could be produced from $0.644 \\mathrm{mmol}$ of benzene. Give your answer in $\\mathrm{cm}^{3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of cm^3, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "cm^3"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_115",
        "problem": "In 1962, at the University of British Columbia, Neil Bartlett shattered conventional chemistry wisdom and synthesized xenon tetrafluoride, the first binary compound of a noble gas. Which of the following statements is/are true about a molecule of xenon tetrafluoride?\n\nI) The molecule has no lone pairs on the central atom\n\nII) The molecule has a tetrahedral geometry\n\nIII) The molecule has no net molecular dipole\nA: I only\nB: II only\nC: III only\nD: II and III\nE: I, II and III\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn 1962, at the University of British Columbia, Neil Bartlett shattered conventional chemistry wisdom and synthesized xenon tetrafluoride, the first binary compound of a noble gas. Which of the following statements is/are true about a molecule of xenon tetrafluoride?\n\nI) The molecule has no lone pairs on the central atom\n\nII) The molecule has a tetrahedral geometry\n\nIII) The molecule has no net molecular dipole\n\nA: I only\nB: II only\nC: III only\nD: II and III\nE: I, II and III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_419",
        "problem": "应用电解法对煤进行脱硫处理具有脱硫效率高、经济效益好等优点。电解脱硫的基本原理如图所示, 下列说法错误的是\n\n[图1]\nA: $\\mathrm{FeS}_{2}$ 化学名称为二硫化亚铁\nB: 阳极反应为 $\\mathrm{Mn}^{2+}-\\mathrm{e}^{-}=\\mathrm{Mn}^{3+}$\nC: 随着电解, 混合液 $\\mathrm{pH}$ 逐渐增大\nD: 混合液中每增加 $1 \\mathrm{molSO}_{4}^{2-}$, 理论上通过电路的电子数为 $7 N_{A}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n应用电解法对煤进行脱硫处理具有脱硫效率高、经济效益好等优点。电解脱硫的基本原理如图所示, 下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{FeS}_{2}$ 化学名称为二硫化亚铁\nB: 阳极反应为 $\\mathrm{Mn}^{2+}-\\mathrm{e}^{-}=\\mathrm{Mn}^{3+}$\nC: 随着电解, 混合液 $\\mathrm{pH}$ 逐渐增大\nD: 混合液中每增加 $1 \\mathrm{molSO}_{4}^{2-}$, 理论上通过电路的电子数为 $7 N_{A}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-80.jpg?height=586&width=645&top_left_y=498&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_254",
        "problem": "Element A has 3 valence electrons and element B has 6 valence electrons. Elements A and $\\mathrm{B}$ are in the same period of the Periodic Table.\n\nWhat is the likely formula of the compound that elements A and B form together?\nA: $\\mathrm{AB}$\nB: $\\mathrm{A}_{2} \\mathrm{~B}_{3}$\nC: $\\mathrm{AB}_{2}$\nD: $\\mathrm{A}_{2} \\mathrm{~B}$\nE: $\\mathrm{A}_{3} \\mathrm{~B}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nElement A has 3 valence electrons and element B has 6 valence electrons. Elements A and $\\mathrm{B}$ are in the same period of the Periodic Table.\n\nWhat is the likely formula of the compound that elements A and B form together?\n\nA: $\\mathrm{AB}$\nB: $\\mathrm{A}_{2} \\mathrm{~B}_{3}$\nC: $\\mathrm{AB}_{2}$\nD: $\\mathrm{A}_{2} \\mathrm{~B}$\nE: $\\mathrm{A}_{3} \\mathrm{~B}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_850",
        "problem": "$\\mathrm{ROH}$ 是一元弱碱, 难溶盐 $\\mathrm{RA} 、 \\mathrm{RB}$ 的两饱和溶液中 $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$或 $\\mathrm{c}\\left(\\mathrm{B}^{-}\\right)$随 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$而变化, $\\mathrm{A}^{-}$和 $\\mathrm{B}$-不发生水解。实验发现, $298 \\mathrm{~K}$ 时 $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right) 、 \\mathrm{c}^{2}\\left(\\mathrm{~B}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$的关系如图所示, 甲表示 $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$关系。下列叙述错误的是\n\n[图1]\nA: RA 饱和溶液 $\\mathrm{pH}=6$ 时, $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)<2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $R B$ 的溶度积 $K_{s p}(R B)$ 的数值为 $5 \\times 10^{-11}$\nC: $\\mathrm{ROH}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{b}}(\\mathrm{ROH})$ 的数值为 $2 \\times 10^{-6}$\nD: $\\mathrm{RB}$ 饱和溶液中 $\\mathrm{pH}=7$ 时, $\\mathrm{c}\\left(\\mathrm{R}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{B}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{ROH}$ 是一元弱碱, 难溶盐 $\\mathrm{RA} 、 \\mathrm{RB}$ 的两饱和溶液中 $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)$或 $\\mathrm{c}\\left(\\mathrm{B}^{-}\\right)$随 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$而变化, $\\mathrm{A}^{-}$和 $\\mathrm{B}$-不发生水解。实验发现, $298 \\mathrm{~K}$ 时 $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right) 、 \\mathrm{c}^{2}\\left(\\mathrm{~B}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$的关系如图所示, 甲表示 $\\mathrm{c}^{2}\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$关系。下列叙述错误的是\n\n[图1]\n\nA: RA 饱和溶液 $\\mathrm{pH}=6$ 时, $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)<2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $R B$ 的溶度积 $K_{s p}(R B)$ 的数值为 $5 \\times 10^{-11}$\nC: $\\mathrm{ROH}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{b}}(\\mathrm{ROH})$ 的数值为 $2 \\times 10^{-6}$\nD: $\\mathrm{RB}$ 饱和溶液中 $\\mathrm{pH}=7$ 时, $\\mathrm{c}\\left(\\mathrm{R}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{B}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-070.jpg?height=511&width=577&top_left_y=413&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_7",
        "problem": "$0.422 \\mathrm{~g}$ of an element $\\mathrm{Z}$ reacts with oxygen to form 0.797 $\\mathrm{g}$ of the oxide $\\mathrm{Z}_{2} \\mathrm{O}_{3}$. What is the element $\\mathrm{Z}$ ?\nA: $\\mathrm{Al}$\nB: $\\mathrm{Sc}$\nC: $\\mathrm{Cr}$\nD: $\\mathrm{Ga}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n$0.422 \\mathrm{~g}$ of an element $\\mathrm{Z}$ reacts with oxygen to form 0.797 $\\mathrm{g}$ of the oxide $\\mathrm{Z}_{2} \\mathrm{O}_{3}$. What is the element $\\mathrm{Z}$ ?\n\nA: $\\mathrm{Al}$\nB: $\\mathrm{Sc}$\nC: $\\mathrm{Cr}$\nD: $\\mathrm{Ga}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_452",
        "problem": "$\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 在密闭容器中进行的反应: $\\mathrm{COCl}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+\\mathrm{Cl}_{2}(\\mathrm{~g}) \\triangle \\mathrm{H}=+108 \\mathrm{KJ} / \\mathrm{mol}$, 在 $4 \\mathrm{~min} 、 10 \\mathrm{~min} 、 14 \\mathrm{~min}$ 时均只改变影响平衡的一个条件, 各物质的浓度变化如图所示,下列说法正确的是\n\n[图1]\nA: $4 \\mathrm{~min}$ 时降低温度, $10 \\mathrm{~min}$ 时充入 $\\mathrm{Cl}_{2}, 14 \\mathrm{~min}$ 时增大容器的体积\nB: 平衡常数 $\\mathrm{K}_{2 \\mathrm{~min}}<\\mathrm{K}_{9 \\text { min }}<\\mathrm{K}_{13 \\text { min }}$\nC: $4 \\mathrm{~min} \\sim 8 \\mathrm{~min} \\mathrm{COCl}_{2}$ 平均反应速率为 $0.01 \\mathrm{~mol}(\\mathrm{~L} \\cdot \\mathrm{min})$\nD: 若 $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 起始向 $2 \\mathrm{~L}$ 的恒容密闭容器中充入 $\\mathrm{COCl}_{2} 、 \\mathrm{Cl}_{2} 、 \\mathrm{CO}$ 均为 $0.20 \\mathrm{~mol}$,则达到平衡前, $\\mathrm{v}$ 正 $>\\mathrm{v}$ 逆\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 在密闭容器中进行的反应: $\\mathrm{COCl}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+\\mathrm{Cl}_{2}(\\mathrm{~g}) \\triangle \\mathrm{H}=+108 \\mathrm{KJ} / \\mathrm{mol}$, 在 $4 \\mathrm{~min} 、 10 \\mathrm{~min} 、 14 \\mathrm{~min}$ 时均只改变影响平衡的一个条件, 各物质的浓度变化如图所示,下列说法正确的是\n\n[图1]\n\nA: $4 \\mathrm{~min}$ 时降低温度, $10 \\mathrm{~min}$ 时充入 $\\mathrm{Cl}_{2}, 14 \\mathrm{~min}$ 时增大容器的体积\nB: 平衡常数 $\\mathrm{K}_{2 \\mathrm{~min}}<\\mathrm{K}_{9 \\text { min }}<\\mathrm{K}_{13 \\text { min }}$\nC: $4 \\mathrm{~min} \\sim 8 \\mathrm{~min} \\mathrm{COCl}_{2}$ 平均反应速率为 $0.01 \\mathrm{~mol}(\\mathrm{~L} \\cdot \\mathrm{min})$\nD: 若 $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 起始向 $2 \\mathrm{~L}$ 的恒容密闭容器中充入 $\\mathrm{COCl}_{2} 、 \\mathrm{Cl}_{2} 、 \\mathrm{CO}$ 均为 $0.20 \\mathrm{~mol}$,则达到平衡前, $\\mathrm{v}$ 正 $>\\mathrm{v}$ 逆\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-22.jpg?height=536&width=834&top_left_y=1077&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_748",
        "problem": "向 21 恒温恒容容器中加入 $2 \\mathrm{molMgSO}_{4}$ 并充入 $2 \\mathrm{molCO}$, 发生反应 $\\mathrm{MgSO}_{4}(\\mathrm{~s})+\\mathrm{CO}$ $(\\mathrm{g}) \\rightleftharpoons \\mathrm{MgO}(\\mathrm{s})+\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\Delta \\mathrm{H}>0$ 。测得反应过程中残留固体的质量随时间变化如图所示，下列说法不正确的是\n\n[图1]\nA: $0 \\sim 2 \\mathrm{~min}$ 内平均反应速率 $\\mathrm{v}\\left(\\mathrm{SO}_{2}\\right)=0.3 \\mathrm{~mol} \\cdot \\mathrm{l}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: $2 \\sim 4 \\min$ 内容器内气体的密度没有变化\nC: 该温度下, 反应的平衡常数为 1.8\nD: 保持其他条件不变, 起始时向容器中充入 $1.00 \\mathrm{~mol} \\mathrm{MgSO}_{4}$ 和 $1.00 \\mathrm{molCO}$, 反应达到平衡时, $\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)>0.60 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n向 21 恒温恒容容器中加入 $2 \\mathrm{molMgSO}_{4}$ 并充入 $2 \\mathrm{molCO}$, 发生反应 $\\mathrm{MgSO}_{4}(\\mathrm{~s})+\\mathrm{CO}$ $(\\mathrm{g}) \\rightleftharpoons \\mathrm{MgO}(\\mathrm{s})+\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\Delta \\mathrm{H}>0$ 。测得反应过程中残留固体的质量随时间变化如图所示，下列说法不正确的是\n\n[图1]\n\nA: $0 \\sim 2 \\mathrm{~min}$ 内平均反应速率 $\\mathrm{v}\\left(\\mathrm{SO}_{2}\\right)=0.3 \\mathrm{~mol} \\cdot \\mathrm{l}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: $2 \\sim 4 \\min$ 内容器内气体的密度没有变化\nC: 该温度下, 反应的平衡常数为 1.8\nD: 保持其他条件不变, 起始时向容器中充入 $1.00 \\mathrm{~mol} \\mathrm{MgSO}_{4}$ 和 $1.00 \\mathrm{molCO}$, 反应达到平衡时, $\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)>0.60 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-81.jpg?height=320&width=488&top_left_y=191&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1403",
        "problem": "Iron $(\\mathrm{Fe})$ is the fourth most abundant element in the Earth's crust and has been used for more than 5,000 years.\n\nPure iron is easily oxidized, which limits its utilization. Element $\\mathbf{X}$ is one of the alloying elements that is added to improve the oxidation resistance property of iron.Both $\\mathrm{Fe}$ and $\\mathrm{X}$ crystallize in the body centered cubic structure. Approximating the Fe atoms as hard-spheres, the volume taken up by the Fe atoms inside the unit cell is $1.59 \\times 10^{-23} \\mathrm{~cm}^{3}$. The volume of the unit cell of $\\mathbf{X}$ is $0.0252 \\mathrm{~nm}^{3}$. A complete substitutional solid solution usually occurs when\n\n$$\n\\Delta R=\\left(\\frac{\\left|R_{x}-R_{F e}\\right|}{R_{F e}}\\right) \\times 100\n$$\n\nis less than or equal to 15 , where $R_{X}$ and $R_{F e}$ are the atomic radii of $\\mathbf{X}$ and $\\mathrm{Fe}$, respectively. Can $\\mathbf{X}$ and Fe form a complete substitutional solid solution?\n\nShow your calculation. The volume of sphere is $4 / 3 \\pi r^{3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\nHere is some context information for this question, which might assist you in solving it:\nIron $(\\mathrm{Fe})$ is the fourth most abundant element in the Earth's crust and has been used for more than 5,000 years.\n\nPure iron is easily oxidized, which limits its utilization. Element $\\mathbf{X}$ is one of the alloying elements that is added to improve the oxidation resistance property of iron.\n\nproblem:\nBoth $\\mathrm{Fe}$ and $\\mathrm{X}$ crystallize in the body centered cubic structure. Approximating the Fe atoms as hard-spheres, the volume taken up by the Fe atoms inside the unit cell is $1.59 \\times 10^{-23} \\mathrm{~cm}^{3}$. The volume of the unit cell of $\\mathbf{X}$ is $0.0252 \\mathrm{~nm}^{3}$. A complete substitutional solid solution usually occurs when\n\n$$\n\\Delta R=\\left(\\frac{\\left|R_{x}-R_{F e}\\right|}{R_{F e}}\\right) \\times 100\n$$\n\nis less than or equal to 15 , where $R_{X}$ and $R_{F e}$ are the atomic radii of $\\mathbf{X}$ and $\\mathrm{Fe}$, respectively. Can $\\mathbf{X}$ and Fe form a complete substitutional solid solution?\n\nShow your calculation. The volume of sphere is $4 / 3 \\pi r^{3}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-221.jpg?height=297&width=334&top_left_y=1045&top_left_x=313"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1082",
        "problem": "The electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\n\nCalculate the ionization energy of a helium ion, $\\mathrm{He}^{+}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\n\nCalculate the ionization energy of a helium ion, $\\mathrm{He}^{+}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~J}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_801f28df4b84ba4c94fag-09.jpg?height=568&width=571&top_left_y=327&top_left_x=1234"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~J}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1206",
        "problem": "Anhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion.\n\nIn an experiment, gaseous $\\mathrm{NH}_{3}$ is burned with $\\mathrm{O}_{2}$ in a container of fixed volume according to the equation given below.\n\n$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe initial and final states are at $298 \\mathrm{~K}$. After combustion with $14.40 \\mathrm{~g}$ of $\\mathrm{O}_{2}$, some of $\\mathrm{NH}_{3}$ remains unreacted.\n\nTo determine the amount of $\\mathrm{NH}_{3}$ gas dissolved in water, produced during the combustion process, a $10.00 \\mathrm{~cm}^{3}$ sample of the aqueous solution was withdrawn from the reaction vessel and added to $15.0 \\mathrm{~cm}^{3}$ of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ solution $\\left(c=0.0100 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ). The resulting solution was titrated with a standard $\\mathrm{NaOH}$ solution $\\left(c=0.0200 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) and the equivalence point was reached at $10.64 \\mathrm{~cm}^{3}$.\n\n$$\n\\left(K_{b}\\left(\\mathrm{NH}_{3}\\right)=1.8 \\cdot 10^{-5} ; \\quad K_{a}\\left(\\mathrm{HSO}_{4}^{-}\\right)=1.1 \\cdot 10^{-2}\\right)\n$$Calculate $\\mathrm{pH}$ of the solution in the container after combustion.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nAnhydrous ammonia is an ultra-clean, energy-dense alternative liquid fuel. It produces no greenhouse gases on combustion.\n\nIn an experiment, gaseous $\\mathrm{NH}_{3}$ is burned with $\\mathrm{O}_{2}$ in a container of fixed volume according to the equation given below.\n\n$$\n4 \\mathrm{NH}_{3}(g)+3 \\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{~N}_{2}(g)+6 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nThe initial and final states are at $298 \\mathrm{~K}$. After combustion with $14.40 \\mathrm{~g}$ of $\\mathrm{O}_{2}$, some of $\\mathrm{NH}_{3}$ remains unreacted.\n\nTo determine the amount of $\\mathrm{NH}_{3}$ gas dissolved in water, produced during the combustion process, a $10.00 \\mathrm{~cm}^{3}$ sample of the aqueous solution was withdrawn from the reaction vessel and added to $15.0 \\mathrm{~cm}^{3}$ of $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ solution $\\left(c=0.0100 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ). The resulting solution was titrated with a standard $\\mathrm{NaOH}$ solution $\\left(c=0.0200 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) and the equivalence point was reached at $10.64 \\mathrm{~cm}^{3}$.\n\n$$\n\\left(K_{b}\\left(\\mathrm{NH}_{3}\\right)=1.8 \\cdot 10^{-5} ; \\quad K_{a}\\left(\\mathrm{HSO}_{4}^{-}\\right)=1.1 \\cdot 10^{-2}\\right)\n$$\n\nproblem:\nCalculate $\\mathrm{pH}$ of the solution in the container after combustion.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1055",
        "problem": "Increasing concerns over the use and generation of hazardous substances in chemical processes has encouraged some chemists to look for more environmentally friendly ways to make chemical products. To help evaluate a process environmentally, chemists often use the term 'percentage atom economy', where\n\n$\\%$ Atom Economy $\\quad=\\quad$ RMM of desired product $\\times 100$ RMM of all products\n\n[figure1]\n\nAn environmentally friendly chemical process would normally be expected to have a high $\\%$ atom economy, indicating that a high proportion of the starting materials end up as part of the final product, hence reducing the amount of waste. Efforts are constantly being made to increase the \\% atom economy of chemical processes. As an example, the manufacture of ethene oxide $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\\right)$ for many years was via the classical chlorohydrin route:\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\Longleftrightarrow \\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{HCl} \\\\\n\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{Ca}(\\mathrm{OH})_{2}+\\mathrm{HCl} \\Longleftrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}+\\mathrm{CaCl}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nThe modern petrochemical route involves the following reaction:\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{4}+1 / 2 \\mathrm{O}_{2} \\stackrel{\\mathbf{A g}}{ } \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\n$$\n\nCalculate the $\\%$ atom efficiency of this process.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIncreasing concerns over the use and generation of hazardous substances in chemical processes has encouraged some chemists to look for more environmentally friendly ways to make chemical products. To help evaluate a process environmentally, chemists often use the term 'percentage atom economy', where\n\n$\\%$ Atom Economy $\\quad=\\quad$ RMM of desired product $\\times 100$ RMM of all products\n\n[figure1]\n\nAn environmentally friendly chemical process would normally be expected to have a high $\\%$ atom economy, indicating that a high proportion of the starting materials end up as part of the final product, hence reducing the amount of waste. Efforts are constantly being made to increase the \\% atom economy of chemical processes. As an example, the manufacture of ethene oxide $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\\right)$ for many years was via the classical chlorohydrin route:\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\Longleftrightarrow \\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{HCl} \\\\\n\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{Ca}(\\mathrm{OH})_{2}+\\mathrm{HCl} \\Longleftrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}+\\mathrm{CaCl}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nThe modern petrochemical route involves the following reaction:\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{4}+1 / 2 \\mathrm{O}_{2} \\stackrel{\\mathbf{A g}}{ } \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\n$$\n\nCalculate the $\\%$ atom efficiency of this process.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of % percentage, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-05.jpg?height=454&width=343&top_left_y=316&top_left_x=1482"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "% percentage"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_704",
        "problem": "常温下, 二元弱酸 $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ 溶液中含磷物种的浓度之和为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 溶液中各含磷物种的 $\\mathrm{pc}-\\mathrm{pOH}$ 关系如图所示。图中 $\\mathrm{pc}$ 表示各含磷物种的浓度负对数 $(\\mathrm{pc}=-\\operatorname{lgc}), \\mathrm{pOH}$表示 $\\mathrm{OH}$ 的浓度的负对数 $\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$。下列有关说法正确的是\n\n[图1]\nA: 若反应 $\\mathrm{HPO}_{3}^{2-}+\\mathrm{H}_{3} \\mathrm{PO}_{3} \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}$可以发生, 其平衡常数值为 $10^{-5.3}$\nB: 在浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaH}_{2} \\mathrm{PO}_{3}$ 和 $\\mathrm{Na}_{2} \\mathrm{HPO}_{3}$ 混合溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $b$ 点时, $x=9.95$\nD: $\\mathrm{d}$ 点溶液中存在关系式 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=0.1+\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 二元弱酸 $\\mathrm{H}_{3} \\mathrm{PO}_{3}$ 溶液中含磷物种的浓度之和为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 溶液中各含磷物种的 $\\mathrm{pc}-\\mathrm{pOH}$ 关系如图所示。图中 $\\mathrm{pc}$ 表示各含磷物种的浓度负对数 $(\\mathrm{pc}=-\\operatorname{lgc}), \\mathrm{pOH}$表示 $\\mathrm{OH}$ 的浓度的负对数 $\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$。下列有关说法正确的是\n\n[图1]\n\nA: 若反应 $\\mathrm{HPO}_{3}^{2-}+\\mathrm{H}_{3} \\mathrm{PO}_{3} \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}$可以发生, 其平衡常数值为 $10^{-5.3}$\nB: 在浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaH}_{2} \\mathrm{PO}_{3}$ 和 $\\mathrm{Na}_{2} \\mathrm{HPO}_{3}$ 混合溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $b$ 点时, $x=9.95$\nD: $\\mathrm{d}$ 点溶液中存在关系式 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=0.1+\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-071.jpg?height=569&width=714&top_left_y=1326&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-072.jpg?height=131&width=1151&top_left_y=2499&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1548",
        "problem": "Identify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{..52}^{130} \\mathrm{Te}+{ }_{1}^{2} \\mathrm{H} \\rightarrow{ }_{..53}^{131} \\mathrm{I}+\\mathrm{X}$\nA: alpha\nB: beta\nC: gamma\nD: neutron\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIdentify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{..52}^{130} \\mathrm{Te}+{ }_{1}^{2} \\mathrm{H} \\rightarrow{ }_{..53}^{131} \\mathrm{I}+\\mathrm{X}$\n\nA: alpha\nB: beta\nC: gamma\nD: neutron\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_56",
        "problem": "The isotope actinium-226 undergoes both beta decay, with a half-life of $35.4 \\mathrm{~d}$, and electron capture, with a half-life of $173 \\mathrm{~d}$. What is the overall half-life for the radioactive decay of ${ }^{226} \\mathrm{Ac}$ ?\nA: $29.4 \\mathrm{~d}$\nB: $78.3 \\mathrm{~d}$\nC: $104 \\mathrm{~d}$\nD: $208 \\mathrm{~d}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe isotope actinium-226 undergoes both beta decay, with a half-life of $35.4 \\mathrm{~d}$, and electron capture, with a half-life of $173 \\mathrm{~d}$. What is the overall half-life for the radioactive decay of ${ }^{226} \\mathrm{Ac}$ ?\n\nA: $29.4 \\mathrm{~d}$\nB: $78.3 \\mathrm{~d}$\nC: $104 \\mathrm{~d}$\nD: $208 \\mathrm{~d}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_41",
        "problem": "In a hydrogen atom, which transition requires the greatest input of energy?\nA: $n=7 \\rightarrow n=3$\nB: $n=2 \\rightarrow n=1$\nC: $n=3 \\rightarrow n=7$\nD: $n=1 \\rightarrow n=2$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn a hydrogen atom, which transition requires the greatest input of energy?\n\nA: $n=7 \\rightarrow n=3$\nB: $n=2 \\rightarrow n=1$\nC: $n=3 \\rightarrow n=7$\nD: $n=1 \\rightarrow n=2$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_104",
        "problem": "Over $80 \\%$ of global methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$ production is converted into further synthetic chemicals. Consequently, methanol is an economically significant chemical compound. Methanex, a Vancouver-based company, is the world's largest producer and distributor of methanol. Methanol is produced according to the following balanced chemical reaction:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})\n$$\n\nGiven initial partial pressures of $P_{\\mathrm{CO}}=26$ bar and $P_{\\mathrm{H} 2}=65$ bar, and an equilibrium partial pressure of $P_{\\mathrm{CH} 3 \\mathrm{OH}}=16 \\mathrm{bar}$, determine $K_{\\mathrm{p}}$. Assume constant container volume and ideal gas behaviour.\nA: $1.5 \\times 10^{-4}$\nB: $6.7 \\times 10^{-4}$\nC: $1.5 \\times 10^{-3}$\nD: $3.3 \\times 10^{-2}$\nE: $4.8 \\times 10^{-2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOver $80 \\%$ of global methanol $\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)$ production is converted into further synthetic chemicals. Consequently, methanol is an economically significant chemical compound. Methanex, a Vancouver-based company, is the world's largest producer and distributor of methanol. Methanol is produced according to the following balanced chemical reaction:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})\n$$\n\nGiven initial partial pressures of $P_{\\mathrm{CO}}=26$ bar and $P_{\\mathrm{H} 2}=65$ bar, and an equilibrium partial pressure of $P_{\\mathrm{CH} 3 \\mathrm{OH}}=16 \\mathrm{bar}$, determine $K_{\\mathrm{p}}$. Assume constant container volume and ideal gas behaviour.\n\nA: $1.5 \\times 10^{-4}$\nB: $6.7 \\times 10^{-4}$\nC: $1.5 \\times 10^{-3}$\nD: $3.3 \\times 10^{-2}$\nE: $4.8 \\times 10^{-2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1138",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nHow many electrons does an iodide ion contain?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nHow many electrons does an iodide ion contain?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_16",
        "problem": "How many unpaired electrons does a ground-state gasphase $\\mathrm{Mn}^{2+}$ ion have?\nA: 1\nB: 3\nC: 5\nD: 7\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many unpaired electrons does a ground-state gasphase $\\mathrm{Mn}^{2+}$ ion have?\n\nA: 1\nB: 3\nC: 5\nD: 7\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_388",
        "problem": "For an exothermic reaction, which of the following best describes the effect of increasing the temperature on the forward and reverse reactions?\nA: Both the forward and reverse rates increase, but the forward rate increases more than the reverse rate.\nB: Both the forward and reverse rates increase, but the reverse rate increases more than the forward rate.\nC: The forward rate increases while the reverse rate decreases.\nD: The reverse rate increases while the forward rate decreases.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor an exothermic reaction, which of the following best describes the effect of increasing the temperature on the forward and reverse reactions?\n\nA: Both the forward and reverse rates increase, but the forward rate increases more than the reverse rate.\nB: Both the forward and reverse rates increase, but the reverse rate increases more than the forward rate.\nC: The forward rate increases while the reverse rate decreases.\nD: The reverse rate increases while the forward rate decreases.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_759",
        "problem": "某温度下, 改变 $0.10 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{X}_{2} \\mathrm{O}_{7}$ 溶液的 $\\mathrm{pH}$ 时, 各种含 $\\mathrm{X}$ 元素粒子及 $\\mathrm{OH}^{-}$浓度变化如图所示。下列有关说法正确的是\n\n[图1]\nA: $\\frac{\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{4}\\right)}{\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{4}\\right)}=1 \\times 10^{5.76}$\nB: $\\mathrm{E}$ 点溶液中存在: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<5 \\mathrm{c}\\left(\\mathrm{X}_{2} \\mathrm{O}_{7}^{2-}\\right)$\nC: $\\mathrm{Kw}=1 \\times 10^{-13}$\nD: 反应 $\\mathrm{X}_{2} \\mathrm{O}_{7}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons 2 \\mathrm{XO}_{4}^{2-}+2 \\mathrm{H}^{+}$的平衡常数 $\\mathrm{K}=1 \\times 10^{-13.2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某温度下, 改变 $0.10 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{X}_{2} \\mathrm{O}_{7}$ 溶液的 $\\mathrm{pH}$ 时, 各种含 $\\mathrm{X}$ 元素粒子及 $\\mathrm{OH}^{-}$浓度变化如图所示。下列有关说法正确的是\n\n[图1]\n\nA: $\\frac{\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{4}\\right)}{\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{4}\\right)}=1 \\times 10^{5.76}$\nB: $\\mathrm{E}$ 点溶液中存在: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<5 \\mathrm{c}\\left(\\mathrm{X}_{2} \\mathrm{O}_{7}^{2-}\\right)$\nC: $\\mathrm{Kw}=1 \\times 10^{-13}$\nD: 反应 $\\mathrm{X}_{2} \\mathrm{O}_{7}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons 2 \\mathrm{XO}_{4}^{2-}+2 \\mathrm{H}^{+}$的平衡常数 $\\mathrm{K}=1 \\times 10^{-13.2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-034.jpg?height=531&width=811&top_left_y=2030&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_121",
        "problem": "The limit for lead in drinking water is $0.015 \\mathrm{ppm}$ according to the Environmental Protection Agency (EPA). A $100.0 \\mathrm{~mL}$ sample of well water tested for dissolved $\\mathrm{Pb}^{2+}$ with a saturated potassium iodide solution produced $1.7 \\mathrm{mg}$ of yellow lead iodide precipitate. How many times higher than the EPA limit was the concentration of lead in the well water?\nA: 110 times\nB: 1100 times\nC: 510 times\nD: 700 times\nE: 51 times\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe limit for lead in drinking water is $0.015 \\mathrm{ppm}$ according to the Environmental Protection Agency (EPA). A $100.0 \\mathrm{~mL}$ sample of well water tested for dissolved $\\mathrm{Pb}^{2+}$ with a saturated potassium iodide solution produced $1.7 \\mathrm{mg}$ of yellow lead iodide precipitate. How many times higher than the EPA limit was the concentration of lead in the well water?\n\nA: 110 times\nB: 1100 times\nC: 510 times\nD: 700 times\nE: 51 times\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_839",
        "problem": "将 $2.00 \\mathrm{~mol} \\mathrm{X}$ 和 $1.00 \\mathrm{~mol} \\mathrm{Y}$ 充入体积不变的密闭容器中, 在一定条件下发生反应: $\\mathrm{X}(\\mathrm{g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Z}(\\mathrm{g})$, 达到平衡时 $\\mathrm{X}$ 为 $1.50 \\mathrm{~mol}$, 如果此时移走 $1.00 \\mathrm{~mol} \\mathrm{X}$ 和 $0.50 \\mathrm{~mol} \\mathrm{Y}$,保持温度和体积不变, 再次达到平衡时, $\\mathrm{X}$ 的物质的量可能为\nA: $0.75 \\mathrm{~mol}$\nB: $0.55 \\mathrm{~mol}$\nC: $0.80 \\mathrm{~mol}$\nD: $1.00 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将 $2.00 \\mathrm{~mol} \\mathrm{X}$ 和 $1.00 \\mathrm{~mol} \\mathrm{Y}$ 充入体积不变的密闭容器中, 在一定条件下发生反应: $\\mathrm{X}(\\mathrm{g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Z}(\\mathrm{g})$, 达到平衡时 $\\mathrm{X}$ 为 $1.50 \\mathrm{~mol}$, 如果此时移走 $1.00 \\mathrm{~mol} \\mathrm{X}$ 和 $0.50 \\mathrm{~mol} \\mathrm{Y}$,保持温度和体积不变, 再次达到平衡时, $\\mathrm{X}$ 的物质的量可能为\n\nA: $0.75 \\mathrm{~mol}$\nB: $0.55 \\mathrm{~mol}$\nC: $0.80 \\mathrm{~mol}$\nD: $1.00 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1367",
        "problem": "The muon $(\\mu)$ is a subatomic particle of the lepton family which has same charge and magnetic behavior as the electron, but has a different mass and is unstable, i.e., it disintegrates into other particles within microseconds after its creation. Here you will attempt to determine the mass of the muon using two rather different approaches.\n\nThe most common spontaneous disintegration reaction for the muon is:\n\n$\\mu \\rightarrow \\mathrm{e}+\\bar{v}_{\\mathrm{e}}+\\mathrm{v}_{\\mu}$,\n\nwhere $\\bar{v}_{\\mathrm{e}}$ is the electron antineutrino, and $\\mathrm{v}_{\\mu}$ the muon neutrino. In a given\n\nexperiment using a stationary muon, $\\bar{v}_{\\mathrm{e}}+\\mathrm{v}_{\\mu}$, carried away a total energy of $2.000 \\times 10^{-12} \\mathrm{~J}$, while the electron was moving with a kinetic energy of $1.4846 \\times 10^{-11} \\mathrm{~J}$. \n\nDetermine the mass of the muon.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe muon $(\\mu)$ is a subatomic particle of the lepton family which has same charge and magnetic behavior as the electron, but has a different mass and is unstable, i.e., it disintegrates into other particles within microseconds after its creation. Here you will attempt to determine the mass of the muon using two rather different approaches.\n\nThe most common spontaneous disintegration reaction for the muon is:\n\n$\\mu \\rightarrow \\mathrm{e}+\\bar{v}_{\\mathrm{e}}+\\mathrm{v}_{\\mu}$,\n\nwhere $\\bar{v}_{\\mathrm{e}}$ is the electron antineutrino, and $\\mathrm{v}_{\\mu}$ the muon neutrino. In a given\n\nexperiment using a stationary muon, $\\bar{v}_{\\mathrm{e}}+\\mathrm{v}_{\\mu}$, carried away a total energy of $2.000 \\times 10^{-12} \\mathrm{~J}$, while the electron was moving with a kinetic energy of $1.4846 \\times 10^{-11} \\mathrm{~J}$. \n\nDetermine the mass of the muon.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kg}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kg}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1101",
        "problem": "The energy of an electron in a $2 p$ orbital in an excited hydrogen atom is $-{5 . 4 5} \\times {1 0}^{-19} \\mathrm{~J}$.The electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\n\nCalculate the energy needed to promote the electron in a hydrogen atom from the $1 \\mathrm{~s}$ orbital to the $2 p$ orbital.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe energy of an electron in a $2 p$ orbital in an excited hydrogen atom is $-{5 . 4 5} \\times {1 0}^{-19} \\mathrm{~J}$.\n\nproblem:\nThe electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\n\nCalculate the energy needed to promote the electron in a hydrogen atom from the $1 \\mathrm{~s}$ orbital to the $2 p$ orbital.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~J}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_801f28df4b84ba4c94fag-09.jpg?height=568&width=571&top_left_y=327&top_left_x=1234"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~J}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_504",
        "problem": "有甲酸、乙酸和乙二酸组成的混合物 $\\mathrm{ag}$, 经完全燃烧可生成 $\\mathrm{CO}_{2} 0.2 \\mathrm{~mol}$; 完全中和等质量的该混合物消耗 $80 \\mathrm{~mL} 2 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaOH}$ 溶液, 则 $\\mathrm{a}$ 值可能为\nA: 8.00\nB: 7.92\nC: 7.90\nD: 7.86\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有甲酸、乙酸和乙二酸组成的混合物 $\\mathrm{ag}$, 经完全燃烧可生成 $\\mathrm{CO}_{2} 0.2 \\mathrm{~mol}$; 完全中和等质量的该混合物消耗 $80 \\mathrm{~mL} 2 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaOH}$ 溶液, 则 $\\mathrm{a}$ 值可能为\n\nA: 8.00\nB: 7.92\nC: 7.90\nD: 7.86\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_350",
        "problem": "A white ionic solid is dissolved in water. Addition of a solution of sodium chloride to this solution results in a white precipitate. What was the cation in the original ionic solid?\nA: $\\mathrm{Na}^{+}$\nB: $\\mathrm{Fe}^{3+}$\nC: $\\mathrm{Ag}^{+}$\nD: $\\mathrm{Sr}^{2+}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA white ionic solid is dissolved in water. Addition of a solution of sodium chloride to this solution results in a white precipitate. What was the cation in the original ionic solid?\n\nA: $\\mathrm{Na}^{+}$\nB: $\\mathrm{Fe}^{3+}$\nC: $\\mathrm{Ag}^{+}$\nD: $\\mathrm{Sr}^{2+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1211",
        "problem": "Identify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{11}^{23} \\mathrm{Na}+{ }_{0}^{1} \\mathrm{n} \\rightarrow{ }_{11}^{24} \\mathrm{Na}+\\mathrm{X}$\nA: alpha\nB: beta\nC: gamma\nD: neutron\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIdentify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{11}^{23} \\mathrm{Na}+{ }_{0}^{1} \\mathrm{n} \\rightarrow{ }_{11}^{24} \\mathrm{Na}+\\mathrm{X}$\n\nA: alpha\nB: beta\nC: gamma\nD: neutron\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1015",
        "problem": "Which of the following is not a gas at $298 \\mathrm{~K}$ ?\nA: $\\mathrm{Ar}$\nB: $\\mathrm{He}$\nC: $\\mathrm{Br}_{2}$\nD: $\\mathrm{H}_{2}$\nE: $\\mathrm{O}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is not a gas at $298 \\mathrm{~K}$ ?\n\nA: $\\mathrm{Ar}$\nB: $\\mathrm{He}$\nC: $\\mathrm{Br}_{2}$\nD: $\\mathrm{H}_{2}$\nE: $\\mathrm{O}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_120",
        "problem": "To safely and efficiently prepare $1.0 \\mathrm{~L}$ of a $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solution of $\\mathrm{HCl}$ from a $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1}$ stock solution of $\\mathrm{HCl}$, what should a chemist add to a 1 L volumetric flask first?\nA: $500 \\mathrm{~mL}$ of distilled water with a beaker\nB: $120.0 \\mathrm{~mL}$ of $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ with a graduated cylinder\nC: $83.3 \\mathrm{~mL}$ of $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ with a graduated cylinder\nD: $12.0 \\mathrm{~mL}$ of $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ with a graduated cylinder\nE: $10.0 \\mathrm{~mL}$ of distilled water with a graduated cylinder\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTo safely and efficiently prepare $1.0 \\mathrm{~L}$ of a $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solution of $\\mathrm{HCl}$ from a $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1}$ stock solution of $\\mathrm{HCl}$, what should a chemist add to a 1 L volumetric flask first?\n\nA: $500 \\mathrm{~mL}$ of distilled water with a beaker\nB: $120.0 \\mathrm{~mL}$ of $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ with a graduated cylinder\nC: $83.3 \\mathrm{~mL}$ of $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ with a graduated cylinder\nD: $12.0 \\mathrm{~mL}$ of $12.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ with a graduated cylinder\nE: $10.0 \\mathrm{~mL}$ of distilled water with a graduated cylinder\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_421",
        "problem": "下列方案设计、现象和结论都正确的是\n\n|  | 目的 | 方案设计 | 现象和结论 |\n| :---: | :---: | :---: | :---: |\n| A | 检验 $\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$ 中 <br> 铁元素的存在形式 | 向盛有少量蒸馏水的试管中滴加 <br> 2 滴 $\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$ 溶液, 然后再滴 <br> 加 2 滴 $\\mathrm{KSCN}$ 溶液 | 溶液不变血红色, 铁元 <br> 素以 $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$ 的形式 <br> 存在 |\n| B | 验证苯环对羟基是否有活化作用 | 向苯酚浊液加入碳酸钠溶液, 充分振荡后观察现象 | 溶液变澄清, 苯环对羟基有活化作用 |\n| $\\mathrm{C}$ | 检验乙炔是否具有还 <br> 原性 | 将电石与水反应生成的气体通入 <br> 酸性高锰酸钾溶液 | 溶液紫色褪去, 乙炔具 <br> 有还原性 |\n| $\\mathrm{D}$ | 检验淀粉的水解程度|向淀粉溶液中加入适量 <br> $10 \\% \\mathrm{H}_{2} \\mathrm{SO}_{4}$, 加热煮沸, 加入 <br> $10 \\% \\mathrm{NaOH}$ 溶液至溶液显碱性, 再 | 溶液不变蓝, 淀粉完全 <br> 解程度 <br> 加入少量碘水 |  |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列方案设计、现象和结论都正确的是\n\n|  | 目的 | 方案设计 | 现象和结论 |\n| :---: | :---: | :---: | :---: |\n| A | 检验 $\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$ 中 <br> 铁元素的存在形式 | 向盛有少量蒸馏水的试管中滴加 <br> 2 滴 $\\mathrm{K}_{3}\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]$ 溶液, 然后再滴 <br> 加 2 滴 $\\mathrm{KSCN}$ 溶液 | 溶液不变血红色, 铁元 <br> 素以 $\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}\\right]^{3-}$ 的形式 <br> 存在 |\n| B | 验证苯环对羟基是否有活化作用 | 向苯酚浊液加入碳酸钠溶液, 充分振荡后观察现象 | 溶液变澄清, 苯环对羟基有活化作用 |\n| $\\mathrm{C}$ | 检验乙炔是否具有还 <br> 原性 | 将电石与水反应生成的气体通入 <br> 酸性高锰酸钾溶液 | 溶液紫色褪去, 乙炔具 <br> 有还原性 |\n| $\\mathrm{D}$ | 检验淀粉的水解程度|向淀粉溶液中加入适量 <br> $10 \\% \\mathrm{H}_{2} \\mathrm{SO}_{4}$, 加热煮沸, 加入 <br> $10 \\% \\mathrm{NaOH}$ 溶液至溶液显碱性, 再 | 溶液不变蓝, 淀粉完全 <br> 解程度 <br> 加入少量碘水 |  |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_512",
        "problem": "某温度下, 分别向 $10 \\mathrm{~mL}$ 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{CuCl}_{2}$ 和 $\\mathrm{ZnCl}_{2}$ 溶液中滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, 滴加过程中溶液中 $\\mathrm{pM}\\left[\\mathrm{pM}=-\\operatorname{lgc}\\left(\\mathrm{M}^{2+}\\right), \\mathrm{M}^{2+}\\right.$ 代表 $\\mathrm{Cu}^{2+}$ 或 $\\left.\\mathrm{Zn}^{2+}\\right] \\mathrm{Na}_{2} \\mathrm{~S}$ 溶液体积 $(\\mathrm{V})$ 的关系如图所示 [已知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS}), \\lg 3 \\approx 0.5$ ] 。下列有关说法正确的是\n\n[图1]\nA: $\\mathrm{d}$ 点纵坐标小于 33.9\nB: $\\mathrm{CuS}$ 的 $\\mathrm{K}_{\\mathrm{sp}}$ 为 $1 \\times 10^{-25.4}$\nC: 溶液的 $\\mathrm{pH}: \\mathrm{a}>\\mathrm{b}>\\mathrm{e}$\nD: $\\mathrm{a}$ 点的 $\\mathrm{ZnCl}_{2}$ 溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)<2\\left[c\\left(\\mathrm{Zn}^{2+}\\right)+c\\left(\\mathrm{H}^{+}\\right)\\right]$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某温度下, 分别向 $10 \\mathrm{~mL}$ 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{CuCl}_{2}$ 和 $\\mathrm{ZnCl}_{2}$ 溶液中滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, 滴加过程中溶液中 $\\mathrm{pM}\\left[\\mathrm{pM}=-\\operatorname{lgc}\\left(\\mathrm{M}^{2+}\\right), \\mathrm{M}^{2+}\\right.$ 代表 $\\mathrm{Cu}^{2+}$ 或 $\\left.\\mathrm{Zn}^{2+}\\right] \\mathrm{Na}_{2} \\mathrm{~S}$ 溶液体积 $(\\mathrm{V})$ 的关系如图所示 [已知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS}), \\lg 3 \\approx 0.5$ ] 。下列有关说法正确的是\n\n[图1]\n\nA: $\\mathrm{d}$ 点纵坐标小于 33.9\nB: $\\mathrm{CuS}$ 的 $\\mathrm{K}_{\\mathrm{sp}}$ 为 $1 \\times 10^{-25.4}$\nC: 溶液的 $\\mathrm{pH}: \\mathrm{a}>\\mathrm{b}>\\mathrm{e}$\nD: $\\mathrm{a}$ 点的 $\\mathrm{ZnCl}_{2}$ 溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)<2\\left[c\\left(\\mathrm{Zn}^{2+}\\right)+c\\left(\\mathrm{H}^{+}\\right)\\right]$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-098.jpg?height=548&width=442&top_left_y=554&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_785",
        "problem": "常温下, 下列有关叙述正确的是\nA: 在 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nB: 向 $10 \\mathrm{~mL} \\mathrm{pH}=12$ 的 $\\mathrm{NaOH}$ 溶液中滴加等体积 $\\mathrm{pH}=2$ 的 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液: $\\mathrm{c}\\left(\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)\\right.$\nC: 浓度均为 $0.1 \\mathrm{~mol} \\cdot L^{-1}$ 的小苏打溶液与烧碱溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}$ $\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nD: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的硫酸氢铵溶液与氢氧化钠溶液等体积混合 $\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)=$ $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 下列有关叙述正确的是\n\nA: 在 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nB: 向 $10 \\mathrm{~mL} \\mathrm{pH}=12$ 的 $\\mathrm{NaOH}$ 溶液中滴加等体积 $\\mathrm{pH}=2$ 的 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液: $\\mathrm{c}\\left(\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)\\right.$\nC: 浓度均为 $0.1 \\mathrm{~mol} \\cdot L^{-1}$ 的小苏打溶液与烧碱溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}$ $\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nD: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的硫酸氢铵溶液与氢氧化钠溶液等体积混合 $\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)=$ $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_539",
        "problem": "在一恒容密闭容器中发生反应: $\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g})$ 。起始时,\n\n$\\mathrm{c}\\left(\\mathrm{N}_{2}\\right)=0.15 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}, \\mathrm{c}\\left(\\mathrm{H}_{2}\\right)=0.60 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}, \\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.40 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 。在一定条件下, 当反应达到平衡时，下列各物质的浓度关系可能正确的是\nA: $\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.64 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.85 \\mathrm{~mol} \\cdot \\mathrm{I}^{-1}$\nC: $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{N}_{2}\\right)=1.55 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{N}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.95 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n在一恒容密闭容器中发生反应: $\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g})$ 。起始时,\n\n$\\mathrm{c}\\left(\\mathrm{N}_{2}\\right)=0.15 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}, \\mathrm{c}\\left(\\mathrm{H}_{2}\\right)=0.60 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}, \\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.40 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 。在一定条件下, 当反应达到平衡时，下列各物质的浓度关系可能正确的是\n\nA: $\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.64 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.85 \\mathrm{~mol} \\cdot \\mathrm{I}^{-1}$\nC: $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{N}_{2}\\right)=1.55 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{N}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.95 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1071",
        "problem": "The number of moles of hydrogen molecules in $1 \\mathrm{dm}^{3}$ of liquid hydrogen is $35.2 \\mathrm{~mol}$.This question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nThe density of liquid hydrogen is $0.071 \\mathrm{~g} \\mathrm{~cm}^{-3}$.\n\nCalculate the energy released when the gas formed from $1 \\mathrm{dm}^{3}$ of liquid hydrogen is combusted.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe number of moles of hydrogen molecules in $1 \\mathrm{dm}^{3}$ of liquid hydrogen is $35.2 \\mathrm{~mol}$.\n\nproblem:\nThis question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nThe density of liquid hydrogen is $0.071 \\mathrm{~g} \\mathrm{~cm}^{-3}$.\n\nCalculate the energy released when the gas formed from $1 \\mathrm{dm}^{3}$ of liquid hydrogen is combusted.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-02.jpg?height=434&width=619&top_left_y=317&top_left_x=1181"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_652",
        "problem": "已知金属离子 $\\mathrm{M}^{2+}, 25^{\\circ} \\mathrm{C}$ 时在水中存在 $\\mathrm{M}^{2+}(\\mathrm{aq}) 、 \\mathrm{M}(\\mathrm{OH})^{+}(\\mathrm{aq}) 、 \\mathrm{M}(\\mathrm{OH})_{2}(\\mathrm{~s}) 、 \\mathrm{M}(\\mathrm{OH})_{3}$ ${ }^{-}(\\mathrm{aq}) 、 \\mathrm{M}(\\mathrm{OH})_{4}{ }^{2-}(\\mathrm{aq})$ 五种形态, 该体系中各形态的物质的量分数 $(\\alpha)$ 随 $\\mathrm{pH}$ 的变化关系如图, 下列叙述中错误的是\n\n[图1]\nA: $\\mathrm{P}$ 点的 $\\mathrm{pH}$ 为 12 , 则 $\\mathrm{M}(\\mathrm{OH})_{3}{ }^{-} \\rightleftharpoons \\mathrm{M}(\\mathrm{OH})_{2}+\\mathrm{OH}^{-}$的平衡常数为 $10^{-2}$\nB: $\\mathrm{M}(\\mathrm{OH})_{2}$ 完全沉淀后, 增大溶液的 $\\mathrm{pH}$, 沉淀不会立即开始溶解\nC: 溶液 $\\mathrm{pH}$ 达到 14 之前, 沉淀 $\\mathrm{M}(\\mathrm{OH})_{2}$ 已完全溶解\nD: $\\mathrm{M}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 溶液显酸性, 其水解的离子方程式为: $\\mathrm{M}^{2+}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{M}(\\mathrm{OH})^{+}+\\mathrm{H}^{+}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知金属离子 $\\mathrm{M}^{2+}, 25^{\\circ} \\mathrm{C}$ 时在水中存在 $\\mathrm{M}^{2+}(\\mathrm{aq}) 、 \\mathrm{M}(\\mathrm{OH})^{+}(\\mathrm{aq}) 、 \\mathrm{M}(\\mathrm{OH})_{2}(\\mathrm{~s}) 、 \\mathrm{M}(\\mathrm{OH})_{3}$ ${ }^{-}(\\mathrm{aq}) 、 \\mathrm{M}(\\mathrm{OH})_{4}{ }^{2-}(\\mathrm{aq})$ 五种形态, 该体系中各形态的物质的量分数 $(\\alpha)$ 随 $\\mathrm{pH}$ 的变化关系如图, 下列叙述中错误的是\n\n[图1]\n\nA: $\\mathrm{P}$ 点的 $\\mathrm{pH}$ 为 12 , 则 $\\mathrm{M}(\\mathrm{OH})_{3}{ }^{-} \\rightleftharpoons \\mathrm{M}(\\mathrm{OH})_{2}+\\mathrm{OH}^{-}$的平衡常数为 $10^{-2}$\nB: $\\mathrm{M}(\\mathrm{OH})_{2}$ 完全沉淀后, 增大溶液的 $\\mathrm{pH}$, 沉淀不会立即开始溶解\nC: 溶液 $\\mathrm{pH}$ 达到 14 之前, 沉淀 $\\mathrm{M}(\\mathrm{OH})_{2}$ 已完全溶解\nD: $\\mathrm{M}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 溶液显酸性, 其水解的离子方程式为: $\\mathrm{M}^{2+}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{M}(\\mathrm{OH})^{+}+\\mathrm{H}^{+}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-60.jpg?height=488&width=1259&top_left_y=750&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1424",
        "problem": "Distribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nA $100 \\mathrm{~cm}^{3}$ of aqueous sample of ammonium phosphomolybdate $\\left(\\left(\\mathrm{NH}_{4}\\right)_{3} \\mathrm{PMo}_{12} \\mathrm{O}_{40}\\right)$ compound is extracted with $5.0 \\mathrm{~cm}^{3}$ of an organic solvent. The organic-water partition coefficient (Kow) is defined as the ratio of the concentration of the compound in the organic phase $\\left(c_{0}\\right)$ to that in the water phase $\\left(c_{w}\\right)$. Kow of the ammonium phosphomolybdate is 5.0 .\n\nThe molar absorptivity of ammonium phosphomolybdate in the organic phase is $5000 \\mathrm{dm}^{3}$ $\\mathrm{mol}^{-1} \\mathrm{~cm}^{-1}$.If the absorbance in the organic phase is 0.200 , calculate the total mass of phosphorus (in $\\mathrm{mg}$ unit) in the original aqueous sample solution. The optical pathlength of the cuvette is $1.00 \\mathrm{~cm}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nDistribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nA $100 \\mathrm{~cm}^{3}$ of aqueous sample of ammonium phosphomolybdate $\\left(\\left(\\mathrm{NH}_{4}\\right)_{3} \\mathrm{PMo}_{12} \\mathrm{O}_{40}\\right)$ compound is extracted with $5.0 \\mathrm{~cm}^{3}$ of an organic solvent. The organic-water partition coefficient (Kow) is defined as the ratio of the concentration of the compound in the organic phase $\\left(c_{0}\\right)$ to that in the water phase $\\left(c_{w}\\right)$. Kow of the ammonium phosphomolybdate is 5.0 .\n\nThe molar absorptivity of ammonium phosphomolybdate in the organic phase is $5000 \\mathrm{dm}^{3}$ $\\mathrm{mol}^{-1} \\mathrm{~cm}^{-1}$.\n\nproblem:\nIf the absorbance in the organic phase is 0.200 , calculate the total mass of phosphorus (in $\\mathrm{mg}$ unit) in the original aqueous sample solution. The optical pathlength of the cuvette is $1.00 \\mathrm{~cm}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_986",
        "problem": "When a sample of atomic hydrogen gas is heated, it emits violet, blue, green and red light. Which of the following statements best explains this observation?\nA: The energy of the electron in a hydrogen atom is restricted to certain values.\nB: The energy of the electron in a hydrogen atom is not restricted in any way.\nC: The electron in a hydrogen atom is restricted to one of only four possible circular orbits.\nD: The distance between the electron and the nucleus in a hydrogen atom is restricted to certain values.\nE: none of the above\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen a sample of atomic hydrogen gas is heated, it emits violet, blue, green and red light. Which of the following statements best explains this observation?\n\nA: The energy of the electron in a hydrogen atom is restricted to certain values.\nB: The energy of the electron in a hydrogen atom is not restricted in any way.\nC: The electron in a hydrogen atom is restricted to one of only four possible circular orbits.\nD: The distance between the electron and the nucleus in a hydrogen atom is restricted to certain values.\nE: none of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1128",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWith hot nitric acid, or with more concentrated nitric acid, the only gases formed are oxides of nitrogen. Give the equation for the reaction of indium metal with nitric acid forming nitric oxide, NO.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWith hot nitric acid, or with more concentrated nitric acid, the only gases formed are oxides of nitrogen. Give the equation for the reaction of indium metal with nitric acid forming nitric oxide, NO.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_515",
        "problem": "从砷化镓废料(主要成分为 $\\mathrm{GaAs} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3} 、 \\mathrm{SiO}_{2}$ 和 $\\mathrm{CaCO}_{3}$ )中回收镓和砷的工艺流程如图所示, 已知: $\\mathrm{Ga}(\\mathrm{OH})_{3}$ 是两性氢氧化物, 其酸式电离和碱式电离的方程式如下:\n\n$\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}^{+}+\\mathrm{GaO}_{2}^{-} \\rightleftharpoons \\mathrm{Ga}(\\mathrm{OH})_{3} \\rightleftharpoons 3 \\mathrm{OH}^{-}+\\mathrm{Ga}^{3+}, 25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{Ga}(\\mathrm{OH})_{3}$ 的溶度积\n\n$K_{\\mathrm{sp}}\\left[\\mathrm{Ga}(\\mathrm{OH})_{3}\\right]=1.6 \\times 10^{-34}$, 电离常数 $K_{\\mathrm{a}}=1 \\times 10^{-7}$ 。下列说法不正确的是\n\n[图1]\nA: “中和”过程中为提高镓的回收率, 加硫酸调 $\\mathrm{pH}$ 的最大值约是 2\nB: “碱浸”时, $\\mathrm{H}_{2} \\mathrm{O}_{2}$ 既做氧化剂又做还原剂\nC: “电解”时, 阴极电极反应式: $\\mathrm{GaO}_{2}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}+3 \\mathrm{e}^{-}=\\mathrm{Ga}+4 \\mathrm{OH}^{-}$\nD: 电解尾液中溶质主要为 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, 可以循环利用, 提高经济效益\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n从砷化镓废料(主要成分为 $\\mathrm{GaAs} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3} 、 \\mathrm{SiO}_{2}$ 和 $\\mathrm{CaCO}_{3}$ )中回收镓和砷的工艺流程如图所示, 已知: $\\mathrm{Ga}(\\mathrm{OH})_{3}$ 是两性氢氧化物, 其酸式电离和碱式电离的方程式如下:\n\n$\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}^{+}+\\mathrm{GaO}_{2}^{-} \\rightleftharpoons \\mathrm{Ga}(\\mathrm{OH})_{3} \\rightleftharpoons 3 \\mathrm{OH}^{-}+\\mathrm{Ga}^{3+}, 25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{Ga}(\\mathrm{OH})_{3}$ 的溶度积\n\n$K_{\\mathrm{sp}}\\left[\\mathrm{Ga}(\\mathrm{OH})_{3}\\right]=1.6 \\times 10^{-34}$, 电离常数 $K_{\\mathrm{a}}=1 \\times 10^{-7}$ 。下列说法不正确的是\n\n[图1]\n\nA: “中和”过程中为提高镓的回收率, 加硫酸调 $\\mathrm{pH}$ 的最大值约是 2\nB: “碱浸”时, $\\mathrm{H}_{2} \\mathrm{O}_{2}$ 既做氧化剂又做还原剂\nC: “电解”时, 阴极电极反应式: $\\mathrm{GaO}_{2}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}+3 \\mathrm{e}^{-}=\\mathrm{Ga}+4 \\mathrm{OH}^{-}$\nD: 电解尾液中溶质主要为 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, 可以循环利用, 提高经济效益\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-10.jpg?height=283&width=1019&top_left_y=150&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_585",
        "problem": "常温下, 含铝的常规电解质溶液中, $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{18}\\right]^{3+} \\rightleftharpoons\\left[\\mathrm{Al}(\\mathrm{OH})\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{2+}+\\mathrm{H}^{+}$, $K_{\\mathrm{a}}=10^{-4.8} ;\\left[\\mathrm{Al}(\\mathrm{OH})\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{2+} \\rightleftharpoons\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{3+}+\\mathrm{OH}^{-}, \\quad K_{\\mathrm{b}}=10^{-11.0}$ 。 $\\mathrm{AlAc}(\\mathrm{OH})_{2}$ (羟基乙酸铝)溶液中, $\\left[\\mathrm{AlAc}(\\mathrm{OH})_{2}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right] \\rightleftharpoons\\left[\\mathrm{AlAc}(\\mathrm{OH})_{3}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{14}\\right]^{-}+\\mathrm{H}^{+}+\\mathrm{H}_{2} \\mathrm{O}$, $K_{\\mathrm{a}}=10^{-5.0} ;\\left[\\operatorname{AlAc}(\\mathrm{OH})_{2}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right] \\rightleftharpoons\\left[\\operatorname{AlAc}(\\mathrm{OH})\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right]^{+}+\\mathrm{OH}^{-}, \\quad K_{\\mathrm{b}}=10^{-3.0} 。$下列说法错误的是\nA: $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{18}\\right]^{3+}$ 不能直接发生碱式电离\nB: $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{18}\\right]^{3+} \\rightleftharpoons\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{3+}+\\mathrm{H}_{2} \\mathrm{O}$, 平衡常数的值为 $10^{-1.8}$\nC: $\\left[\\mathrm{AlAc}(\\mathrm{OH})_{2}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right]$ 发生碱式电离的程度大于酸式电离的程度\nD: 羟基乙酸铝溶液中加入少量硫酸铝固体, 溶液的 $\\mathrm{pH}$ 增大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 含铝的常规电解质溶液中, $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{18}\\right]^{3+} \\rightleftharpoons\\left[\\mathrm{Al}(\\mathrm{OH})\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{2+}+\\mathrm{H}^{+}$, $K_{\\mathrm{a}}=10^{-4.8} ;\\left[\\mathrm{Al}(\\mathrm{OH})\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{2+} \\rightleftharpoons\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{3+}+\\mathrm{OH}^{-}, \\quad K_{\\mathrm{b}}=10^{-11.0}$ 。 $\\mathrm{AlAc}(\\mathrm{OH})_{2}$ (羟基乙酸铝)溶液中, $\\left[\\mathrm{AlAc}(\\mathrm{OH})_{2}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right] \\rightleftharpoons\\left[\\mathrm{AlAc}(\\mathrm{OH})_{3}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{14}\\right]^{-}+\\mathrm{H}^{+}+\\mathrm{H}_{2} \\mathrm{O}$, $K_{\\mathrm{a}}=10^{-5.0} ;\\left[\\operatorname{AlAc}(\\mathrm{OH})_{2}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right] \\rightleftharpoons\\left[\\operatorname{AlAc}(\\mathrm{OH})\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right]^{+}+\\mathrm{OH}^{-}, \\quad K_{\\mathrm{b}}=10^{-3.0} 。$下列说法错误的是\n\nA: $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{18}\\right]^{3+}$ 不能直接发生碱式电离\nB: $\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{18}\\right]^{3+} \\rightleftharpoons\\left[\\mathrm{Al}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{17}\\right]^{3+}+\\mathrm{H}_{2} \\mathrm{O}$, 平衡常数的值为 $10^{-1.8}$\nC: $\\left[\\mathrm{AlAc}(\\mathrm{OH})_{2}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{16}\\right]$ 发生碱式电离的程度大于酸式电离的程度\nD: 羟基乙酸铝溶液中加入少量硫酸铝固体, 溶液的 $\\mathrm{pH}$ 增大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1185",
        "problem": "Palindromic sequences are an interesting class of DNA. In a palindromic double-stranded DNA (dsDNA) species, the sequence of one strand read in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction matches the $5^{\\prime} \\rightarrow 3^{\\prime}$ reading on the complementary strand. Hence, a palindromic dsDNA consists of two identical strands that are complementary to each other. An example is the so-called\n\nDrew-Dickerson dodecanucleotide (1):\n\n[figure1]How many different palindromic double-stranded DNA dodecanucleotides (i. e. dsDNA species with twelve base pairs) exist?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nPalindromic sequences are an interesting class of DNA. In a palindromic double-stranded DNA (dsDNA) species, the sequence of one strand read in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction matches the $5^{\\prime} \\rightarrow 3^{\\prime}$ reading on the complementary strand. Hence, a palindromic dsDNA consists of two identical strands that are complementary to each other. An example is the so-called\n\nDrew-Dickerson dodecanucleotide (1):\n\n[figure1]\n\nproblem:\nHow many different palindromic double-stranded DNA dodecanucleotides (i. e. dsDNA species with twelve base pairs) exist?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-260.jpg?height=140&width=440&top_left_y=1729&top_left_x=814"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_928",
        "problem": "$\\mathrm{T}^{\\circ} \\mathrm{C}$, 反应 $\\mathrm{CaCO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{CaO}(\\mathrm{s})+\\mathrm{CO}_{2}(\\mathrm{~g})$, 在密闭容器中达到平衡时的体积为\n\n$100 \\mathrm{~mL}$ 。已知密度: $\\mathrm{CaO}(\\mathrm{s}): 3.35 \\mathrm{~g} \\cdot \\mathrm{cm}^{-3} 、 \\mathrm{CaCO}_{3}(\\mathrm{~s}): 2.93 \\mathrm{~g} \\cdot \\mathrm{cm}^{-3}$ 。下列有关说法正确的是\nA: 恒温、恒容, 体系中再加入 $56 \\mathrm{~g} \\mathrm{CaO}(\\mathrm{s})$, 平衡不移动\nB: 恒温、恒压, 体系中再加入 $100 \\mathrm{~g} \\mathrm{CaCO}_{3}(\\mathrm{~s})$, 平衡不移动\nC: 恒温, 压缩容器体积, 平衡后, 反应速率加快\nD: 恒温、恒压, 体系中通入一定量 $\\mathrm{N}_{2}$, 平衡正向移动, 建立新的平衡状态\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{T}^{\\circ} \\mathrm{C}$, 反应 $\\mathrm{CaCO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{CaO}(\\mathrm{s})+\\mathrm{CO}_{2}(\\mathrm{~g})$, 在密闭容器中达到平衡时的体积为\n\n$100 \\mathrm{~mL}$ 。已知密度: $\\mathrm{CaO}(\\mathrm{s}): 3.35 \\mathrm{~g} \\cdot \\mathrm{cm}^{-3} 、 \\mathrm{CaCO}_{3}(\\mathrm{~s}): 2.93 \\mathrm{~g} \\cdot \\mathrm{cm}^{-3}$ 。下列有关说法正确的是\n\nA: 恒温、恒容, 体系中再加入 $56 \\mathrm{~g} \\mathrm{CaO}(\\mathrm{s})$, 平衡不移动\nB: 恒温、恒压, 体系中再加入 $100 \\mathrm{~g} \\mathrm{CaCO}_{3}(\\mathrm{~s})$, 平衡不移动\nC: 恒温, 压缩容器体积, 平衡后, 反应速率加快\nD: 恒温、恒压, 体系中通入一定量 $\\mathrm{N}_{2}$, 平衡正向移动, 建立新的平衡状态\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_425",
        "problem": "将 $\\mathrm{Cl}_{2}$ 缓慢通入 $1 \\mathrm{~L} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中, 反应过程中无 $\\mathrm{CO}_{2}$ 逸出, 用数字传感器测得溶液中 $\\mathrm{pH}$ 与 $\\mathrm{c}(\\mathrm{HClO})$ 的变化如图所示。已知 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=10^{-6.3}$,\n\n$\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=10^{-10.3}$ 。下列说法错误的是\n\n[图1]\nA: 曲线 1 表示溶液中 $\\mathrm{pH}$ 的变化\nB: 整个过程中, 水的电离程度逐渐减小\nC: $\\mathrm{a}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $\\mathrm{pH}=10.3$ 时: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>3 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{ClO}^{-}\\right)+\\mathrm{c}(\\mathrm{HClO})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n将 $\\mathrm{Cl}_{2}$ 缓慢通入 $1 \\mathrm{~L} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中, 反应过程中无 $\\mathrm{CO}_{2}$ 逸出, 用数字传感器测得溶液中 $\\mathrm{pH}$ 与 $\\mathrm{c}(\\mathrm{HClO})$ 的变化如图所示。已知 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=10^{-6.3}$,\n\n$\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=10^{-10.3}$ 。下列说法错误的是\n\n[图1]\n\nA: 曲线 1 表示溶液中 $\\mathrm{pH}$ 的变化\nB: 整个过程中, 水的电离程度逐渐减小\nC: $\\mathrm{a}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $\\mathrm{pH}=10.3$ 时: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>3 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{ClO}^{-}\\right)+\\mathrm{c}(\\mathrm{HClO})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-012.jpg?height=528&width=737&top_left_y=156&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_553",
        "problem": "常温下, 在 $20.00 \\mathrm{~mL} 0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 溶液中逐滴滴加 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$溶液，溶液 $\\mathrm{pH}$ 随滴入 $\\mathrm{HCl}$ 溶液体积的变化曲线如图所示。下列说法正确的是\n\n[图1]\nA: (1) 溶液: $c\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: (2) 溶液： $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: (1)、(2)、(3)三点所示的溶液中水的电离程度(2)>(3)> (1)\nD: 滴定过程中不可能出现： $\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 在 $20.00 \\mathrm{~mL} 0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 溶液中逐滴滴加 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$溶液，溶液 $\\mathrm{pH}$ 随滴入 $\\mathrm{HCl}$ 溶液体积的变化曲线如图所示。下列说法正确的是\n\n[图1]\n\nA: (1) 溶液: $c\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: (2) 溶液： $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: (1)、(2)、(3)三点所示的溶液中水的电离程度(2)>(3)> (1)\nD: 滴定过程中不可能出现： $\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-012.jpg?height=311&width=531&top_left_y=1055&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_134",
        "problem": "Lead (II) sulfate can decompose into lead (II) sulfite and oxygen gas when heated. If the reaction generates $2.25 \\mathrm{~g}$ of oxygen gas, what mass of lead (II) sulfate reacted? Assume 100\\% yield in this reaction.\nA: $42.6 \\mathrm{~g}$\nB: $21.3 \\mathrm{~g}$\nC: $20.2 \\mathrm{~g}$\nD: $10.7 \\mathrm{~g}$\nE: $4.50 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nLead (II) sulfate can decompose into lead (II) sulfite and oxygen gas when heated. If the reaction generates $2.25 \\mathrm{~g}$ of oxygen gas, what mass of lead (II) sulfate reacted? Assume 100\\% yield in this reaction.\n\nA: $42.6 \\mathrm{~g}$\nB: $21.3 \\mathrm{~g}$\nC: $20.2 \\mathrm{~g}$\nD: $10.7 \\mathrm{~g}$\nE: $4.50 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_686",
        "problem": "现有含 $\\mathrm{SrCO}_{3}(\\mathrm{~s})$ 的 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} 、 1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\quad \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液, 含 $\\mathrm{SrSO}_{4}(\\mathrm{~s})$ 的 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}$ ${ }^{-1} 、 1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\quad \\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液。在一定 $\\mathrm{pH}$ 范围内, 四种溶液中 $\\lg \\left[c\\left(\\mathrm{Sr}^{2+}\\right) / \\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right]$ 随 $\\mathrm{pH}$的变化关系如图所示。下列说法错误的是\n\n[图1]\nA: 可以 $\\mathrm{SrSO}_{4}(\\mathrm{~s})$ 为原料生产 $\\mathrm{SrCO}_{3}(\\mathrm{~s})$\nB: $a=-6.5$\nC: 曲线(4)表示含 $\\mathrm{SrCO}_{3}(\\mathrm{~s})$ 的 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\quad \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液的变化\nD: 对含 $\\mathrm{SrSO}_{4}(\\mathrm{~s})$ 且 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 初始浓度均为 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的混合溶液, $\\mathrm{pH} \\geq 7.7$ 时才发生沉淀转化\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现有含 $\\mathrm{SrCO}_{3}(\\mathrm{~s})$ 的 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} 、 1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\quad \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液, 含 $\\mathrm{SrSO}_{4}(\\mathrm{~s})$ 的 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}$ ${ }^{-1} 、 1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\quad \\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液。在一定 $\\mathrm{pH}$ 范围内, 四种溶液中 $\\lg \\left[c\\left(\\mathrm{Sr}^{2+}\\right) / \\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right]$ 随 $\\mathrm{pH}$的变化关系如图所示。下列说法错误的是\n\n[图1]\n\nA: 可以 $\\mathrm{SrSO}_{4}(\\mathrm{~s})$ 为原料生产 $\\mathrm{SrCO}_{3}(\\mathrm{~s})$\nB: $a=-6.5$\nC: 曲线(4)表示含 $\\mathrm{SrCO}_{3}(\\mathrm{~s})$ 的 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\quad \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液的变化\nD: 对含 $\\mathrm{SrSO}_{4}(\\mathrm{~s})$ 且 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 初始浓度均为 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的混合溶液, $\\mathrm{pH} \\geq 7.7$ 时才发生沉淀转化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-107.jpg?height=620&width=805&top_left_y=1683&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1075",
        "problem": "Balanced symbol equations for:\n\ni) the oxidation of $\\mathrm{Mn}(\\mathrm{OH})_{2}$ to $\\mathrm{Mn}(\\mathrm{OH})_{3}$ by aqueous oxygen is $4 \\mathrm{Mn}(\\mathrm{OH})_{2}+\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow 4 \\mathrm{Mn}(\\mathrm{OH})_{3}$.\n\nii) the oxidation of $\\mathrm{KI}$ by $\\mathrm{Mn}(\\mathrm{OH})_{3}$ is $\\mathrm{KI}+\\mathrm{Mn}(\\mathrm{OH})_{3} \\rightarrow 1 / 2 \\mathrm{I}_{2}+\\mathrm{Mn}(\\mathrm{OH})_{2}+\\mathrm{KOH}$.Aquatic life can only survive because of the oxygen gas dissolved in the water; without it, the water rapidly becomes toxic due to decaying organic matter.\n\nHence it is important to monitor the dissolved oxygen concentration (DOC) in rivers and lakes - if this falls below $5 \\mathrm{mg} \\mathrm{dm}^{-3}$, most species of fish cannot survive.\n\n[figure1]\n\nOne of the most accurate methods for measuring the DOC in water is the Winkler method. Under alkaline conditions $\\mathrm{Mn}^{2+}$ is rapidly oxidised to $\\mathrm{Mn}^{3+}$ by dissolved oxygen, producing a pale brown precipitate of $\\mathrm{Mn}(\\mathrm{OH})_{3}$. A sample of river water is shaken with excess alkaline $\\mathrm{Mn}^{2+}$, and the resulting pale brown precipitate is then reacted with an excess of potassium iodide, which it oxidises to iodine. The iodine is then determined by a titration with sodium thiosulfate $\\left(\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)$ solution of known concentration.\n\nThe equation for the reaction between sodium thiosulfate and iodine is:\n\n$$\n2 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}(\\mathrm{aq})+\\mathrm{I}_{2}(\\mathrm{aq}) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6}(\\mathrm{aq})+2 \\mathrm{Nal}(\\mathrm{aq})\n$$\n\n$25.0 \\mathrm{~cm}^{3}$ of a sample of river water treated in this way required $25.0 \\mathrm{~cm}^{3}$ of 0.00100 mol $\\mathrm{dm}^{-3}$ sodium thiosulfate solution.\n\nCalculate the concentration of the dissolved oxygen in $\\mathrm{mg} \\mathrm{dm}^{-3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nBalanced symbol equations for:\n\ni) the oxidation of $\\mathrm{Mn}(\\mathrm{OH})_{2}$ to $\\mathrm{Mn}(\\mathrm{OH})_{3}$ by aqueous oxygen is $4 \\mathrm{Mn}(\\mathrm{OH})_{2}+\\mathrm{O}_{2}+2 \\mathrm{H}_{2} \\mathrm{O} \\rightarrow 4 \\mathrm{Mn}(\\mathrm{OH})_{3}$.\n\nii) the oxidation of $\\mathrm{KI}$ by $\\mathrm{Mn}(\\mathrm{OH})_{3}$ is $\\mathrm{KI}+\\mathrm{Mn}(\\mathrm{OH})_{3} \\rightarrow 1 / 2 \\mathrm{I}_{2}+\\mathrm{Mn}(\\mathrm{OH})_{2}+\\mathrm{KOH}$.\n\nproblem:\nAquatic life can only survive because of the oxygen gas dissolved in the water; without it, the water rapidly becomes toxic due to decaying organic matter.\n\nHence it is important to monitor the dissolved oxygen concentration (DOC) in rivers and lakes - if this falls below $5 \\mathrm{mg} \\mathrm{dm}^{-3}$, most species of fish cannot survive.\n\n[figure1]\n\nOne of the most accurate methods for measuring the DOC in water is the Winkler method. Under alkaline conditions $\\mathrm{Mn}^{2+}$ is rapidly oxidised to $\\mathrm{Mn}^{3+}$ by dissolved oxygen, producing a pale brown precipitate of $\\mathrm{Mn}(\\mathrm{OH})_{3}$. A sample of river water is shaken with excess alkaline $\\mathrm{Mn}^{2+}$, and the resulting pale brown precipitate is then reacted with an excess of potassium iodide, which it oxidises to iodine. The iodine is then determined by a titration with sodium thiosulfate $\\left(\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}\\right)$ solution of known concentration.\n\nThe equation for the reaction between sodium thiosulfate and iodine is:\n\n$$\n2 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}(\\mathrm{aq})+\\mathrm{I}_{2}(\\mathrm{aq}) \\longrightarrow \\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6}(\\mathrm{aq})+2 \\mathrm{Nal}(\\mathrm{aq})\n$$\n\n$25.0 \\mathrm{~cm}^{3}$ of a sample of river water treated in this way required $25.0 \\mathrm{~cm}^{3}$ of 0.00100 mol $\\mathrm{dm}^{-3}$ sodium thiosulfate solution.\n\nCalculate the concentration of the dissolved oxygen in $\\mathrm{mg} \\mathrm{dm}^{-3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{mg} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_801f28df4b84ba4c94fag-05.jpg?height=660&width=922&top_left_y=321&top_left_x=927"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{mg} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1371",
        "problem": "Thiomolybdate ions are derived from molybdate ions, $\\mathrm{MoO}_{4}^{2-}$, by replacing oxygen atoms with sulfur atoms. In nature, thiomolybdate ions are found in such places as the deep waters of the Black Sea, where biological sulfate reduction generates $\\mathrm{H}_{2} \\mathrm{~S}$. The molybdate to thiomolybdate transformation leads to rapid loss of dissolved Mo from seawater to underlying sediments, depleting the ocean in Mo, a trace element essential for life.\n\nThe following equilibria control the relative concentrations of molybdate and thiomolybdate ions in dilute aqueous solution:\n\n$$\n\\begin{array}{ll}\n\\mathrm{MoS}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{1}=1.3 \\cdot 10^{-5} \\\\\n\\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{2}=1.0 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{3}=1.6 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{4}=6.5 \\cdot 10^{-6}\n\\end{array}\n$$\n\nSolutions containing $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}, \\mathrm{MoOS}_{3}^{2-}$ and $\\mathrm{MoS}_{4}^{2-}$ display absorption peaks in the visible wavelength range at 395 and $468 \\mathrm{~nm}$. The other ions, as well as $\\mathrm{H}_{2} \\mathrm{~S}$, absorb negligibly in the visible wavelength range. The molar absorptivities $(\\varepsilon)$ at these two wavelengths are given in the following table:\n\n|  | $\\varepsilon$ at $468 \\mathrm{~nm}$ <br> $\\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~cm}^{-1}$ | $\\varepsilon$ at $395 \\mathrm{~nm}$ <br> $\\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~cm}^{-1}$ |\n| :---: | :---: | :---: |\n| $\\mathrm{MoS}_{4}^{2-}$ | 11870 | 120 |\n| $\\mathrm{MoOS}_{3}^{2-}$ | 0 | 9030 |\n| $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}$ | 0 | 3230 |A solution not at equilibrium contains a mixture of $\\mathrm{MoS}_{4}^{2-}, \\mathrm{MoOS}_{3}^{2-}$ and $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}$ and no other Mo-containing species. The total concentration of all species containing Mo is $6.0 \\cdot 10^{-6} \\mathrm{~mol} \\mathrm{~dm}^{-3}$. In a $10.0 \\mathrm{~cm}$ absorption cell, the absorbance of the solution at $468 \\mathrm{~nm}$ is 0.365 and that at $395 \\mathrm{~nm}$ is 0.213 . Calculate the concentrations of all three Mo-containing anions in this mixture.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nThiomolybdate ions are derived from molybdate ions, $\\mathrm{MoO}_{4}^{2-}$, by replacing oxygen atoms with sulfur atoms. In nature, thiomolybdate ions are found in such places as the deep waters of the Black Sea, where biological sulfate reduction generates $\\mathrm{H}_{2} \\mathrm{~S}$. The molybdate to thiomolybdate transformation leads to rapid loss of dissolved Mo from seawater to underlying sediments, depleting the ocean in Mo, a trace element essential for life.\n\nThe following equilibria control the relative concentrations of molybdate and thiomolybdate ions in dilute aqueous solution:\n\n$$\n\\begin{array}{ll}\n\\mathrm{MoS}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{1}=1.3 \\cdot 10^{-5} \\\\\n\\mathrm{MoOS}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{2}=1.0 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{3}=1.6 \\cdot 10^{-5} \\\\\n\\mathrm{MoO}_{3} \\mathrm{~S}^{2-}+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{MoO}_{4}^{2-}+\\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{aq}) & K_{4}=6.5 \\cdot 10^{-6}\n\\end{array}\n$$\n\nSolutions containing $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}, \\mathrm{MoOS}_{3}^{2-}$ and $\\mathrm{MoS}_{4}^{2-}$ display absorption peaks in the visible wavelength range at 395 and $468 \\mathrm{~nm}$. The other ions, as well as $\\mathrm{H}_{2} \\mathrm{~S}$, absorb negligibly in the visible wavelength range. The molar absorptivities $(\\varepsilon)$ at these two wavelengths are given in the following table:\n\n|  | $\\varepsilon$ at $468 \\mathrm{~nm}$ <br> $\\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~cm}^{-1}$ | $\\varepsilon$ at $395 \\mathrm{~nm}$ <br> $\\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~cm}^{-1}$ |\n| :---: | :---: | :---: |\n| $\\mathrm{MoS}_{4}^{2-}$ | 11870 | 120 |\n| $\\mathrm{MoOS}_{3}^{2-}$ | 0 | 9030 |\n| $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}$ | 0 | 3230 |\n\nproblem:\nA solution not at equilibrium contains a mixture of $\\mathrm{MoS}_{4}^{2-}, \\mathrm{MoOS}_{3}^{2-}$ and $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}$ and no other Mo-containing species. The total concentration of all species containing Mo is $6.0 \\cdot 10^{-6} \\mathrm{~mol} \\mathrm{~dm}^{-3}$. In a $10.0 \\mathrm{~cm}$ absorption cell, the absorbance of the solution at $468 \\mathrm{~nm}$ is 0.365 and that at $395 \\mathrm{~nm}$ is 0.213 . Calculate the concentrations of all three Mo-containing anions in this mixture.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the concentrations of $\\mathrm{MoS}_{4}^{2-}$, the concentrations of $\\mathrm{MoOS}_{3}^{2-}$ , the concentrations of $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}$].\nTheir units are, in order, [ $\\mathrm{~mol} \\mathrm{~dm}^{-3}$,  $\\mathrm{~mol} \\mathrm{~dm}^{-3}$,  $\\mathrm{~mol} \\mathrm{~dm}^{-3}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            " $\\mathrm{~mol} \\mathrm{~dm}^{-3}$",
            " $\\mathrm{~mol} \\mathrm{~dm}^{-3}$",
            " $\\mathrm{~mol} \\mathrm{~dm}^{-3}$"
        ],
        "answer_sequence": [
            "the concentrations of $\\mathrm{MoS}_{4}^{2-}$",
            "the concentrations of $\\mathrm{MoOS}_{3}^{2-}$ ",
            "the concentrations of $\\mathrm{MoO}_{2} \\mathrm{~S}_{2}^{2-}$"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_403",
        "problem": "常温下, 保持某含少量 $\\mathrm{CaCO}_{3}$ 浊液的水体中 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与空气中 $\\mathrm{CO}_{2}$ 的平衡, 调节水体 $\\mathrm{pH}$, 水体中 $-\\operatorname{g}[\\mathrm{c}(\\mathrm{X})]$ 与 $\\mathrm{pH}$ 的关系如下图所示(其中 $\\mathrm{X}$ 为 $\\mathrm{H}_{2} \\mathrm{CO}_{3} 、 \\mathrm{HCO}_{3} 、 \\mathrm{CO}_{3}^{2-}$ 或 $\\left.\\mathrm{Ca}^{2+}\\right)$. 下列说法错误的是\n\n[图1]\nA: 曲线II表示 $-\\lg \\left[\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)\\right]$ 与 $\\mathrm{pH}$ 的关系\nB: 该温度下, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离常数 $\\mathrm{K}_{\\mathrm{al}}$ 的数量级为 $10^{-6}$\nC: $\\mathrm{a}$ 点的水体中: $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: 向水体中加入适量 $\\mathrm{Ca}(\\mathrm{OH})_{2}$ 固体, 可使溶液由 $\\mathrm{b}$ 点变到 $\\mathrm{c}$ 点\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 保持某含少量 $\\mathrm{CaCO}_{3}$ 浊液的水体中 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与空气中 $\\mathrm{CO}_{2}$ 的平衡, 调节水体 $\\mathrm{pH}$, 水体中 $-\\operatorname{g}[\\mathrm{c}(\\mathrm{X})]$ 与 $\\mathrm{pH}$ 的关系如下图所示(其中 $\\mathrm{X}$ 为 $\\mathrm{H}_{2} \\mathrm{CO}_{3} 、 \\mathrm{HCO}_{3} 、 \\mathrm{CO}_{3}^{2-}$ 或 $\\left.\\mathrm{Ca}^{2+}\\right)$. 下列说法错误的是\n\n[图1]\n\nA: 曲线II表示 $-\\lg \\left[\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)\\right]$ 与 $\\mathrm{pH}$ 的关系\nB: 该温度下, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离常数 $\\mathrm{K}_{\\mathrm{al}}$ 的数量级为 $10^{-6}$\nC: $\\mathrm{a}$ 点的水体中: $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: 向水体中加入适量 $\\mathrm{Ca}(\\mathrm{OH})_{2}$ 固体, 可使溶液由 $\\mathrm{b}$ 点变到 $\\mathrm{c}$ 点\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-056.jpg?height=443&width=754&top_left_y=178&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_798",
        "problem": "下列化学过程的表述或数据说明, 明显不符合事实的是（ )\nA: 向含有 $1 \\mathrm{~mol} \\mathrm{KAl}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 的溶液中加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液至沉淀质量最大时, 沉淀的总的物质的量却并非是最大值\nB: 将 $0.12 \\mathrm{~mol} \\mathrm{Cl}_{2}$ 通入到 $100 \\mathrm{~mL} 1 \\mathrm{~mol} / \\mathrm{L}^{2}$ 的 $\\mathrm{FeI}_{2}$ 溶液中, 离子方程式是: $12 \\mathrm{Cl}_{2}+10 \\mathrm{Fe}^{2+}+14 \\mathrm{I}^{-}=10 \\mathrm{Fe}^{3+}+7 \\mathrm{I}_{2}+24 \\mathrm{Cl}^{-}$\nC: 较高的温度下, 可以测得 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 氨水的电离度为 $55 \\%$ 左右\nD: 十六烷的裂化产物中不饱和烃的分子数可能会超过 $50 \\%$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列化学过程的表述或数据说明, 明显不符合事实的是（ )\n\nA: 向含有 $1 \\mathrm{~mol} \\mathrm{KAl}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 的溶液中加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液至沉淀质量最大时, 沉淀的总的物质的量却并非是最大值\nB: 将 $0.12 \\mathrm{~mol} \\mathrm{Cl}_{2}$ 通入到 $100 \\mathrm{~mL} 1 \\mathrm{~mol} / \\mathrm{L}^{2}$ 的 $\\mathrm{FeI}_{2}$ 溶液中, 离子方程式是: $12 \\mathrm{Cl}_{2}+10 \\mathrm{Fe}^{2+}+14 \\mathrm{I}^{-}=10 \\mathrm{Fe}^{3+}+7 \\mathrm{I}_{2}+24 \\mathrm{Cl}^{-}$\nC: 较高的温度下, 可以测得 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 氨水的电离度为 $55 \\%$ 左右\nD: 十六烷的裂化产物中不饱和烃的分子数可能会超过 $50 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_23",
        "problem": "The $K_{\\text {sp }}$ of $\\mathrm{BaCrO}_{4}$ is $1.2 \\times 10^{-10}$. What is the minimum volume of $1.5 \\times 10^{-5} \\mathrm{M} \\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ that would need to be added to $1.0 \\mathrm{~L}$ of $1.5 \\times 10^{-5} \\mathrm{M} \\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution to induce precipitation of $\\mathrm{BaCrO}_{4}$ ?\nA: $0.53 \\mathrm{~L}$\nB: $0.73 \\mathrm{~L}$\nC: $1.0 \\mathrm{~L}$\nD: No amount of this $\\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ solution will cause precipitation of $\\mathrm{BaCrO}_{4}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe $K_{\\text {sp }}$ of $\\mathrm{BaCrO}_{4}$ is $1.2 \\times 10^{-10}$. What is the minimum volume of $1.5 \\times 10^{-5} \\mathrm{M} \\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ that would need to be added to $1.0 \\mathrm{~L}$ of $1.5 \\times 10^{-5} \\mathrm{M} \\mathrm{Ba}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution to induce precipitation of $\\mathrm{BaCrO}_{4}$ ?\n\nA: $0.53 \\mathrm{~L}$\nB: $0.73 \\mathrm{~L}$\nC: $1.0 \\mathrm{~L}$\nD: No amount of this $\\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ solution will cause precipitation of $\\mathrm{BaCrO}_{4}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_373",
        "problem": "What is the formal charge on the central nitrogen in the Lewis structure of $\\mathrm{N}_{2} \\mathrm{O}$ ?\nA: 0\nB: +1\nC: -1\nD: 0 in some resonance structures, -1 in other resonance structures\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the formal charge on the central nitrogen in the Lewis structure of $\\mathrm{N}_{2} \\mathrm{O}$ ?\n\nA: 0\nB: +1\nC: -1\nD: 0 in some resonance structures, -1 in other resonance structures\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1127",
        "problem": "This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nWhat is the combining molar ratio of ferrate ions: Mn(II) ions? (Give your answer in whole numbers.)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nWhat is the combining molar ratio of ferrate ions: Mn(II) ions? (Give your answer in whole numbers.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_81",
        "problem": "A unit cell of a crystalline compound with the formula $\\mathrm{ABC}_{3}$ is shown below. What is the arrangement of nearest neighbors around the atoms of type $\\mathrm{C}$ ?\n[figure1]\nA: Linear arrangement of $\\mathrm{B}$ atoms\nB: Square planar arrangement of $\\mathrm{A}$ atoms\nC: Cubic arrangement of A atoms\nD: Cubic arrangement of $\\mathrm{C}$ atoms\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA unit cell of a crystalline compound with the formula $\\mathrm{ABC}_{3}$ is shown below. What is the arrangement of nearest neighbors around the atoms of type $\\mathrm{C}$ ?\n[figure1]\n\nA: Linear arrangement of $\\mathrm{B}$ atoms\nB: Square planar arrangement of $\\mathrm{A}$ atoms\nC: Cubic arrangement of A atoms\nD: Cubic arrangement of $\\mathrm{C}$ atoms\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-05.jpg?height=432&width=694&top_left_y=304&top_left_x=279"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1464",
        "problem": "Clathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nLarge methane hydrate stocks on the floor of Baikal lake, the largest freshwater lake in Russia and in the world, have been discovered in July 2009 by the crew of a deepsubmergence vehicle Mir-2. During the ascent from the depth of $1400 \\mathrm{~m}$ methane hydrate samples started to decompose at the depth of $372 \\mathrm{~m}$.Determine the temperature in Baikal lake at the depth of $372 \\mathrm{~m}$. The enthalpy of fusion of ice is equal to $6.01 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nClathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nLarge methane hydrate stocks on the floor of Baikal lake, the largest freshwater lake in Russia and in the world, have been discovered in July 2009 by the crew of a deepsubmergence vehicle Mir-2. During the ascent from the depth of $1400 \\mathrm{~m}$ methane hydrate samples started to decompose at the depth of $372 \\mathrm{~m}$.\n\nproblem:\nDetermine the temperature in Baikal lake at the depth of $372 \\mathrm{~m}$. The enthalpy of fusion of ice is equal to $6.01 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of K, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-222.jpg?height=425&width=434&top_left_y=1255&top_left_x=1459"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "K"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_923",
        "problem": "室温下, $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=4.3 \\times 10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=5.6 \\times 10^{-11}$, 通过下列实验探究碳酸钠溶液的性质。下列有关说法不正确的是。\n\n| 实验 | 实验操作和现象 |\n| :--- | :--- |\n| 1 | 将碳酸钠溶于蒸馏水, 配制碳酸钠溶液, 测得溶液的 $\\mathrm{pH}>7$ |\n| 2 | 向碳酸钠溶液中加入等浓度等体积的盐酸, 测得溶液的 <br> $\\mathrm{pH}>7$ |\n| 3 | 向碳酸钠溶液中加入等浓度等体积的溶液, 产生白色沉淀 |\n| 4 | 向碳酸钠溶液中滴入稀盐酸至溶液 $\\mathrm{pH}=4$ |\nA: 实验 1 的溶液中存在: $4.3 \\times 10^{-21}>\\mathrm{c}^{2}\\left(\\mathrm{H}^{+}\\right) \\times\\left[4.3 \\times 10^{-7}+2 \\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)\\right]$\nB: 实验 2 的溶液中存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+4 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: 实验 3 的清液中存在: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right) \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}^{2}\\left(\\mathrm{CO}_{3}^{2-}\\right)=\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)$\nD: 实验 4 过程中溶液不存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=4.3 \\times 10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=5.6 \\times 10^{-11}$, 通过下列实验探究碳酸钠溶液的性质。下列有关说法不正确的是。\n\n| 实验 | 实验操作和现象 |\n| :--- | :--- |\n| 1 | 将碳酸钠溶于蒸馏水, 配制碳酸钠溶液, 测得溶液的 $\\mathrm{pH}>7$ |\n| 2 | 向碳酸钠溶液中加入等浓度等体积的盐酸, 测得溶液的 <br> $\\mathrm{pH}>7$ |\n| 3 | 向碳酸钠溶液中加入等浓度等体积的溶液, 产生白色沉淀 |\n| 4 | 向碳酸钠溶液中滴入稀盐酸至溶液 $\\mathrm{pH}=4$ |\n\nA: 实验 1 的溶液中存在: $4.3 \\times 10^{-21}>\\mathrm{c}^{2}\\left(\\mathrm{H}^{+}\\right) \\times\\left[4.3 \\times 10^{-7}+2 \\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)\\right]$\nB: 实验 2 的溶液中存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+4 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: 实验 3 的清液中存在: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right) \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}^{2}\\left(\\mathrm{CO}_{3}^{2-}\\right)=\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)$\nD: 实验 4 过程中溶液不存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1013",
        "problem": "Which of these chloride salts is least likely to exist?\nA: $\\mathrm{NaCl}$\nB: $\\mathrm{CuCl}$\nC: $\\mathrm{CaCl}_{2}$\nD: $\\mathrm{FeCl}_{3}$\nE: $\\mathrm{MgCl}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of these chloride salts is least likely to exist?\n\nA: $\\mathrm{NaCl}$\nB: $\\mathrm{CuCl}$\nC: $\\mathrm{CaCl}_{2}$\nD: $\\mathrm{FeCl}_{3}$\nE: $\\mathrm{MgCl}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_188",
        "problem": "Given the following equilibria,\n$\\mathrm{NH}_{3}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{NH}_{4}{ }^{+}(\\mathrm{aq})+\\mathrm{OH}^{-}(\\mathrm{aq})$ $K_{b}=1.8 \\times 10^{-5}$\n$\\mathrm{CH}_{3} \\mathrm{COOH}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{COO}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq}) \\quad K_{a}=1.8 \\times 10^{-5}$\n$\\mathrm{HCN}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{CN}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq})$\n$K_{a}=6.2 \\times 10^{-10}$\n\nWhich salt is the most alkaline from among the options below?\nA: $\\mathrm{NaCN}$\nB: $\\mathrm{NaCH}_{3} \\mathrm{COO}$\nC: $\\mathrm{NH}_{4} \\mathrm{Cl}$\nD: $\\mathrm{NH}_{4} \\mathrm{CH}_{3} \\mathrm{COO}$\nE: $\\mathrm{NH}_{4} \\mathrm{CN}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGiven the following equilibria,\n$\\mathrm{NH}_{3}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{NH}_{4}{ }^{+}(\\mathrm{aq})+\\mathrm{OH}^{-}(\\mathrm{aq})$ $K_{b}=1.8 \\times 10^{-5}$\n$\\mathrm{CH}_{3} \\mathrm{COOH}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{COO}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq}) \\quad K_{a}=1.8 \\times 10^{-5}$\n$\\mathrm{HCN}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{CN}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq})$\n$K_{a}=6.2 \\times 10^{-10}$\n\nWhich salt is the most alkaline from among the options below?\n\nA: $\\mathrm{NaCN}$\nB: $\\mathrm{NaCH}_{3} \\mathrm{COO}$\nC: $\\mathrm{NH}_{4} \\mathrm{Cl}$\nD: $\\mathrm{NH}_{4} \\mathrm{CH}_{3} \\mathrm{COO}$\nE: $\\mathrm{NH}_{4} \\mathrm{CN}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_983",
        "problem": "How many neutrons are there in the nucleus of ${ }^{131}$ I?\nA: 44\nB: 53\nC: 78\nD: 131\nE: 184\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many neutrons are there in the nucleus of ${ }^{131}$ I?\n\nA: 44\nB: 53\nC: 78\nD: 131\nE: 184\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1001",
        "problem": "Which of the following has the highest concentration in air at STP?\nA: $\\mathrm{He}$\nB: $\\mathrm{H}_{2} \\mathrm{O}$\nC: $\\mathrm{CO}_{2}$\nD: $\\mathrm{N}_{2}$\nE: $\\mathrm{O}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following has the highest concentration in air at STP?\n\nA: $\\mathrm{He}$\nB: $\\mathrm{H}_{2} \\mathrm{O}$\nC: $\\mathrm{CO}_{2}$\nD: $\\mathrm{N}_{2}$\nE: $\\mathrm{O}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_593",
        "problem": "一定温度下, 三种碳酸盐 $\\mathrm{MCO}_{3}\\left(\\mathrm{M}: \\mathrm{Mg}^{2+} 、 \\mathrm{Ca}^{2+} 、 \\mathrm{Mn}^{2+}\\right)$ 的沉淀溶解平衡曲线如图所示。已知 $\\mathrm{pM}=-\\operatorname{lgc}(\\mathrm{M}), \\mathrm{p}\\left(\\mathrm{CO}_{3}^{2-}\\right)=-\\operatorname{lgc}\\left(\\mathrm{CO}_{3}^{2-}\\right)$ 。下列说法正确的是 $(\\quad)$\n\n[图1]\nA: $\\mathrm{MgCO}_{3} 、 \\mathrm{CaCO}_{3} 、 \\mathrm{MnCO}_{3}$ 的 $\\mathrm{K}_{\\mathrm{sp}}$ 依次增大\nB: $\\mathrm{a}$ 点可表示 $\\mathrm{MnCO}_{3}$ 的饱和溶液, 且 $\\mathrm{c}\\left(\\mathrm{Mn}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: b 点可表示 $\\mathrm{CaCO}_{3}$ 的饱和溶液, 且 $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: $\\mathrm{c}$ 点可表示 $\\mathrm{MgCO}_{3}$ 的不饱和溶液, 且 $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一定温度下, 三种碳酸盐 $\\mathrm{MCO}_{3}\\left(\\mathrm{M}: \\mathrm{Mg}^{2+} 、 \\mathrm{Ca}^{2+} 、 \\mathrm{Mn}^{2+}\\right)$ 的沉淀溶解平衡曲线如图所示。已知 $\\mathrm{pM}=-\\operatorname{lgc}(\\mathrm{M}), \\mathrm{p}\\left(\\mathrm{CO}_{3}^{2-}\\right)=-\\operatorname{lgc}\\left(\\mathrm{CO}_{3}^{2-}\\right)$ 。下列说法正确的是 $(\\quad)$\n\n[图1]\n\nA: $\\mathrm{MgCO}_{3} 、 \\mathrm{CaCO}_{3} 、 \\mathrm{MnCO}_{3}$ 的 $\\mathrm{K}_{\\mathrm{sp}}$ 依次增大\nB: $\\mathrm{a}$ 点可表示 $\\mathrm{MnCO}_{3}$ 的饱和溶液, 且 $\\mathrm{c}\\left(\\mathrm{Mn}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: b 点可表示 $\\mathrm{CaCO}_{3}$ 的饱和溶液, 且 $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: $\\mathrm{c}$ 点可表示 $\\mathrm{MgCO}_{3}$ 的不饱和溶液, 且 $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-43.jpg?height=554&width=597&top_left_y=774&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_997",
        "problem": "What is the oxidation state of $\\mathrm{N}$ in $\\mathrm{HNO}_{2}$ ?\nA: +5\nB: +3\nC: +1\nD: -1\nE: -3\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the oxidation state of $\\mathrm{N}$ in $\\mathrm{HNO}_{2}$ ?\n\nA: +5\nB: +3\nC: +1\nD: -1\nE: -3\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_108",
        "problem": "Calcium fluoride, $\\mathrm{CaF}_{2}$, occurs in nature principally as the mineral fluorite and is the primary source of industrial hydrogen fluoride. Determine the concentration of fluoride ions in a saturated solution of calcium fluoride with $\\left[\\mathrm{Ca}^{2+}\\right]=0.0250 \\mathrm{~mol} \\mathrm{~L}^{-1} . K_{s p}$ for calcium fluoride is $3.45 \\times 10^{-11}$.\nA: $1.38 \\times 10^{-9} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $2.76 \\times 10^{-9} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $5.87 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $3.71 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $1.85 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCalcium fluoride, $\\mathrm{CaF}_{2}$, occurs in nature principally as the mineral fluorite and is the primary source of industrial hydrogen fluoride. Determine the concentration of fluoride ions in a saturated solution of calcium fluoride with $\\left[\\mathrm{Ca}^{2+}\\right]=0.0250 \\mathrm{~mol} \\mathrm{~L}^{-1} . K_{s p}$ for calcium fluoride is $3.45 \\times 10^{-11}$.\n\nA: $1.38 \\times 10^{-9} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $2.76 \\times 10^{-9} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $5.87 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $3.71 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $1.85 \\times 10^{-5} \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_816",
        "problem": "室温下, 往 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水中滴入 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的醋酸, 溶液的 $\\mathrm{PH}$ 与所加醋酸体积的关系如图所示。下列说法错误的是[已知:\n\n$\\left.\\mathrm{K}_{\\mathrm{a}}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right) \\approx 1.8 \\times 10^{-5}, \\lg 2 \\approx 0.3, \\lg 3 \\approx 0.5\\right]$\n\n[图1]\nA: $\\mathrm{x}=20.00$\nB: $\\mathrm{a}$ 点的值约为 10.85\nC: b 点溶液中, $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)<0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: a, b, c 三点溶液中, 水的电离度由大到小的顺序为 $b>c>a$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 往 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水中滴入 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的醋酸, 溶液的 $\\mathrm{PH}$ 与所加醋酸体积的关系如图所示。下列说法错误的是[已知:\n\n$\\left.\\mathrm{K}_{\\mathrm{a}}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right) \\approx 1.8 \\times 10^{-5}, \\lg 2 \\approx 0.3, \\lg 3 \\approx 0.5\\right]$\n\n[图1]\n\nA: $\\mathrm{x}=20.00$\nB: $\\mathrm{a}$ 点的值约为 10.85\nC: b 点溶液中, $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)<0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: a, b, c 三点溶液中, 水的电离度由大到小的顺序为 $b>c>a$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-072.jpg?height=488&width=811&top_left_y=2072&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_662",
        "problem": "下列有关同分异构体数目(不考虑立体异构)的叙述正确的是\nA: 与 [图1]\nB: [图2]\nC: [图3]\nD: 分子式为 $\\mathrm{C}_{7} \\mathrm{H}_{16}$ 的烃, 分子中有 4 个甲基的同分异构体有 4 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列有关同分异构体数目(不考虑立体异构)的叙述正确的是\n\nA: 与 [图1]\nB: [图2]\nC: [图3]\nD: 分子式为 $\\mathrm{C}_{7} \\mathrm{H}_{16}$ 的烃, 分子中有 4 个甲基的同分异构体有 4 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-53.jpg?height=125&width=666&top_left_y=2099&top_left_x=495",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-53.jpg?height=163&width=859&top_left_y=2243&top_left_x=467",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-53.jpg?height=214&width=928&top_left_y=2406&top_left_x=473"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1239",
        "problem": "One of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, $\\gamma=C_{p} / C_{V}$, which is exactly $5 / 3(1.67 \\pm 0.01)$ for a monoatomic gas. The ratio was derived from the measurement of speed of sound $v_{\\mathrm{s}}$ by using the following equation, where $f$ and $\\lambda$ are the frequency and wavelength of the sound, and $R, T$, and $M$ denote the molar gas constant, absolute temperature, and molar mass, respectively.\n\n$$\nv_{\\mathrm{s}}=f \\lambda=\\sqrt{\\frac{\\gamma R T}{M}}\n$$\n\nFor an unknown gas sample, the wavelength of the sound was measured to be $\\lambda=0.116 \\mathrm{~m}$ at a frequency of $f=3520 \\mathrm{~Hz}\\left(\\mathrm{~Hz}=\\mathrm{s}^{-1}\\right)$ and temperature of $15.0{ }^{\\circ} \\mathrm{C}$ and\nunder atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$. The density $\\rho$ of the gas for these conditions was measured to be $0.850 \\pm 0.005 \\mathrm{~kg} \\mathrm{~m}^{-3}$.Calculate the heat capacity ratio $\\gamma$ for this gas sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOne of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, $\\gamma=C_{p} / C_{V}$, which is exactly $5 / 3(1.67 \\pm 0.01)$ for a monoatomic gas. The ratio was derived from the measurement of speed of sound $v_{\\mathrm{s}}$ by using the following equation, where $f$ and $\\lambda$ are the frequency and wavelength of the sound, and $R, T$, and $M$ denote the molar gas constant, absolute temperature, and molar mass, respectively.\n\n$$\nv_{\\mathrm{s}}=f \\lambda=\\sqrt{\\frac{\\gamma R T}{M}}\n$$\n\nFor an unknown gas sample, the wavelength of the sound was measured to be $\\lambda=0.116 \\mathrm{~m}$ at a frequency of $f=3520 \\mathrm{~Hz}\\left(\\mathrm{~Hz}=\\mathrm{s}^{-1}\\right)$ and temperature of $15.0{ }^{\\circ} \\mathrm{C}$ and\nunder atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$. The density $\\rho$ of the gas for these conditions was measured to be $0.850 \\pm 0.005 \\mathrm{~kg} \\mathrm{~m}^{-3}$.\n\nproblem:\nCalculate the heat capacity ratio $\\gamma$ for this gas sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1187",
        "problem": "The energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.Considering an energy cycle, determine the bond energy $E_{\\mathrm{D}}(\\mathrm{eV})$ of $\\mathrm{H}_{2}^{+}$to the first decimal place. If you were unable to determine the values for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, then use $15.0 \\mathrm{eV}$ and $5.0 \\mathrm{eV}$ for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, respectively.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.\n\nproblem:\nConsidering an energy cycle, determine the bond energy $E_{\\mathrm{D}}(\\mathrm{eV})$ of $\\mathrm{H}_{2}^{+}$to the first decimal place. If you were unable to determine the values for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, then use $15.0 \\mathrm{eV}$ and $5.0 \\mathrm{eV}$ for $E_{\\mathrm{B}}$ and $E_{\\mathrm{C}}$, respectively.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of eV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-075.jpg?height=645&width=962&top_left_y=1956&top_left_x=547"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "eV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_610",
        "problem": "利用下列实验药品, 能达到实验目的的是\n\n| 选项 | 实验目的 | 实验药品 |\n| :--- | :--- | :--- |\n| $\\mathrm{A}$ | 与铜离子结合能力: $\\mathrm{NH}_{3}>\\mathrm{OH}^{-}$ | $\\mathrm{CuSO}_{4}$ 溶液, $\\mathrm{NaOH}$ 溶液, 氨水 |\n| B | 蔗糖是否发生水解 | 蔗糖, 稀硫酸, 新制的 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ |\n| $\\mathrm{C}$ | 检验氯乙烷消去的有机产物 | 悬浊液 |\n| D | 用标准浓度的盐酸测定未知浓度的 | 氯乙烷, $\\mathrm{NaOH}$ 溶液, 乙醇, 溴水 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用下列实验药品, 能达到实验目的的是\n\n| 选项 | 实验目的 | 实验药品 |\n| :--- | :--- | :--- |\n| $\\mathrm{A}$ | 与铜离子结合能力: $\\mathrm{NH}_{3}>\\mathrm{OH}^{-}$ | $\\mathrm{CuSO}_{4}$ 溶液, $\\mathrm{NaOH}$ 溶液, 氨水 |\n| B | 蔗糖是否发生水解 | 蔗糖, 稀硫酸, 新制的 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ |\n| $\\mathrm{C}$ | 检验氯乙烷消去的有机产物 | 悬浊液 |\n| D | 用标准浓度的盐酸测定未知浓度的 | 氯乙烷, $\\mathrm{NaOH}$ 溶液, 乙醇, 溴水 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_310",
        "problem": "For the reaction below, $\\Delta H^{\\circ}=-879.6 \\mathrm{~kJ}$.\n\n$$\n3 \\mathrm{~N}_{2} \\mathrm{O}(g)+2 \\mathrm{NH}_{3}(g) \\rightarrow 4 \\mathrm{~N}_{2}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nGiven that $\\Delta H_{f}^{0}=-45.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ for $\\mathrm{NH}_{3}(g)$ and $\\Delta H_{\\mathrm{f}}^{\\circ}=-241.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ for $\\mathrm{H}_{2} \\mathrm{O}(g)$, what is $\\Delta H_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{N}_{2} \\mathrm{O}(g)$ ?\nA: $684 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-504 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-684 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $82.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nE: The answer cannot be determined with the information provided.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the reaction below, $\\Delta H^{\\circ}=-879.6 \\mathrm{~kJ}$.\n\n$$\n3 \\mathrm{~N}_{2} \\mathrm{O}(g)+2 \\mathrm{NH}_{3}(g) \\rightarrow 4 \\mathrm{~N}_{2}(g)+3 \\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nGiven that $\\Delta H_{f}^{0}=-45.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ for $\\mathrm{NH}_{3}(g)$ and $\\Delta H_{\\mathrm{f}}^{\\circ}=-241.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ for $\\mathrm{H}_{2} \\mathrm{O}(g)$, what is $\\Delta H_{\\mathrm{f}}^{\\circ}$ for $\\mathrm{N}_{2} \\mathrm{O}(g)$ ?\n\nA: $684 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-504 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-684 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $82.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nE: The answer cannot be determined with the information provided.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1183",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nUse the periodic table to write down the full electronic configuration of oganesson $(\\mathrm{Og})$, the last element of the periodic table.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nUse the periodic table to write down the full electronic configuration of oganesson $(\\mathrm{Og})$, the last element of the periodic table.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_35",
        "problem": "A $0.100 \\mathrm{M}$ solution of a weak monoprotic acid $\\mathrm{HX}$ is $3.5 \\%$ ionized. What is the $\\mathrm{pH}$ of a $0.500 \\mathrm{M}$ solution of HX?\nA: 0.30\nB: 1.76\nC: 2.10\nD: 2.46\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $0.100 \\mathrm{M}$ solution of a weak monoprotic acid $\\mathrm{HX}$ is $3.5 \\%$ ionized. What is the $\\mathrm{pH}$ of a $0.500 \\mathrm{M}$ solution of HX?\n\nA: 0.30\nB: 1.76\nC: 2.10\nD: 2.46\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_936",
        "problem": "目前, 应用电渗析法进行海水淡化是解决淡水资源员乏的有效途径之一，该方法是利用电解原理通过多组电渗析膜实现，工作原理如图所示(以 $\\mathrm{NaCl}$ 溶液代替海水), 同时获得产品 $\\mathrm{Cl}_{2} 、 \\mathrm{H}_{2} 、 \\mathrm{NaOH}$ 。下列说法错误的是\n\n[图1]\nA: $\\mathrm{a}$ 是直流电源的正极\nB: 电极 $\\mathrm{M}$ 处产生的气体 $\\mathrm{X}$ 为氯气\nC: 膜 pq、st 分别为阳离子交换膜、阴离子交换膜\nD: 出口 I、II分别接收 $\\mathrm{NaCl}$ 浓溶液、淡水\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n目前, 应用电渗析法进行海水淡化是解决淡水资源员乏的有效途径之一，该方法是利用电解原理通过多组电渗析膜实现，工作原理如图所示(以 $\\mathrm{NaCl}$ 溶液代替海水), 同时获得产品 $\\mathrm{Cl}_{2} 、 \\mathrm{H}_{2} 、 \\mathrm{NaOH}$ 。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{a}$ 是直流电源的正极\nB: 电极 $\\mathrm{M}$ 处产生的气体 $\\mathrm{X}$ 为氯气\nC: 膜 pq、st 分别为阳离子交换膜、阴离子交换膜\nD: 出口 I、II分别接收 $\\mathrm{NaCl}$ 浓溶液、淡水\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-23.jpg?height=503&width=742&top_left_y=1873&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1434",
        "problem": "A feathery, greenish solid precipitate can be observed if chlorine gas is bubbled into water close to its freezing point. Similar precipitates form with other gases such as methane and noble gases. These materials are interesting because vast quantities of the so-called methane-hydrates are supposed to exist in nature (comparable in quantity with other natural gas deposits).\n\nThese precipitates all have related structures. The molecules of water just above its freezing point form a hydrogen-bonded structure. The gas molecules stabilize this framework by filling in the rather large cavities in the water structure forming clathrates. The crystals of chlorine and methane clathrates have the same structure. Their main characteristics are dodecahedra formed from 20 water molecules. The unit cell of the crystal can be thought as a body-centered cubic arrangement built from these dodecahedra which are almost spherical objects. The dodecahedra are connected via additional water molecules located on the faces of the unit cell. Two water molecules can be found on each face of the unit cell. The unit cell has an edge dimension of $1.182 \\mathrm{~nm}$. There are two types of cavities in this structure. One is the internal space in the dodecahedra (A). These are somewhat smaller than the other type of voids (B), of which there are 6 for each unit cell.\n\nMethane hydrate is formed with the structure in c) at temperatures between $0-10{ }^{\\circ} \\mathrm{C}$. What is the density of the clathrate?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nA feathery, greenish solid precipitate can be observed if chlorine gas is bubbled into water close to its freezing point. Similar precipitates form with other gases such as methane and noble gases. These materials are interesting because vast quantities of the so-called methane-hydrates are supposed to exist in nature (comparable in quantity with other natural gas deposits).\n\nThese precipitates all have related structures. The molecules of water just above its freezing point form a hydrogen-bonded structure. The gas molecules stabilize this framework by filling in the rather large cavities in the water structure forming clathrates. The crystals of chlorine and methane clathrates have the same structure. Their main characteristics are dodecahedra formed from 20 water molecules. The unit cell of the crystal can be thought as a body-centered cubic arrangement built from these dodecahedra which are almost spherical objects. The dodecahedra are connected via additional water molecules located on the faces of the unit cell. Two water molecules can be found on each face of the unit cell. The unit cell has an edge dimension of $1.182 \\mathrm{~nm}$. There are two types of cavities in this structure. One is the internal space in the dodecahedra (A). These are somewhat smaller than the other type of voids (B), of which there are 6 for each unit cell.\n\nMethane hydrate is formed with the structure in c) at temperatures between $0-10{ }^{\\circ} \\mathrm{C}$. What is the density of the clathrate?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~g} / \\mathrm{cm}^{3}$., but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~g} / \\mathrm{cm}^{3}$."
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_536",
        "problem": "常温下, 向 $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2} 、 \\mathrm{~Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 和 $\\mathrm{HR}$ 的混合液中加入 $\\mathrm{NaOH}$ 固体(忽略溶液体积变化), $\\mathrm{pM}$ 与 $\\mathrm{pH}$ 的部分关系如图所示。已知: $\\mathrm{pM}=-\\operatorname{lgc}(\\mathrm{M}), \\mathrm{c}(\\mathrm{M})$ 代表 $\\mathrm{c}\\left(\\mathrm{Co}^{2+}\\right) 、 \\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right)$ 或 $\\frac{\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)}{\\mathrm{c}(\\mathrm{HR})}, \\mathrm{Ksp}\\left[\\mathrm{Co}(\\mathrm{OH})_{2}\\right]>\\mathrm{Ksp}\\left[\\mathrm{Pb}(\\mathrm{OH})_{2}\\right]$ 。下列叙述错误的是\n\n[图1]\nA: $\\mathrm{Y}$ 代表 $-\\operatorname{lgc}\\left(\\mathrm{Co}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的关系\nB: $\\mathrm{Ka}$ 为弱酸 $\\mathrm{HR}$ 的电离常数, 则常温下 $\\mathrm{pK}_{\\mathrm{a}}=5$\nC: 常温下, $\\mathrm{Co}(\\mathrm{OH})_{2}$ 和 $\\mathrm{Pb}(\\mathrm{OH})_{2}$ 共存时: $\\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right): c\\left(\\mathrm{Co}^{2+}\\right)=10^{-5}: 1$\nD: $\\mathrm{Co}(\\mathrm{OH})_{2}+2 \\mathrm{HR} \\rightleftharpoons \\mathrm{Co}^{2+}+2 \\mathrm{R}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}$ 的平衡常数 $\\mathrm{K}=0.1$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2} 、 \\mathrm{~Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 和 $\\mathrm{HR}$ 的混合液中加入 $\\mathrm{NaOH}$ 固体(忽略溶液体积变化), $\\mathrm{pM}$ 与 $\\mathrm{pH}$ 的部分关系如图所示。已知: $\\mathrm{pM}=-\\operatorname{lgc}(\\mathrm{M}), \\mathrm{c}(\\mathrm{M})$ 代表 $\\mathrm{c}\\left(\\mathrm{Co}^{2+}\\right) 、 \\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right)$ 或 $\\frac{\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)}{\\mathrm{c}(\\mathrm{HR})}, \\mathrm{Ksp}\\left[\\mathrm{Co}(\\mathrm{OH})_{2}\\right]>\\mathrm{Ksp}\\left[\\mathrm{Pb}(\\mathrm{OH})_{2}\\right]$ 。下列叙述错误的是\n\n[图1]\n\nA: $\\mathrm{Y}$ 代表 $-\\operatorname{lgc}\\left(\\mathrm{Co}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的关系\nB: $\\mathrm{Ka}$ 为弱酸 $\\mathrm{HR}$ 的电离常数, 则常温下 $\\mathrm{pK}_{\\mathrm{a}}=5$\nC: 常温下, $\\mathrm{Co}(\\mathrm{OH})_{2}$ 和 $\\mathrm{Pb}(\\mathrm{OH})_{2}$ 共存时: $\\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right): c\\left(\\mathrm{Co}^{2+}\\right)=10^{-5}: 1$\nD: $\\mathrm{Co}(\\mathrm{OH})_{2}+2 \\mathrm{HR} \\rightleftharpoons \\mathrm{Co}^{2+}+2 \\mathrm{R}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}$ 的平衡常数 $\\mathrm{K}=0.1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-021.jpg?height=572&width=711&top_left_y=1847&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_935",
        "problem": "已知甲、乙为单质, 丙为化合物。能实现下述转化关系的是\n\n甲 + 乙 $\\xrightarrow{\\text { 点燃 }}$ 丙 $\\xrightarrow{\\text { 溶于水 }}$ 丙溶液 $\\xrightarrow[\\text { 忤性电极 }]{\\longrightarrow}$ 甲 + 乙\nA: 若丙溶液中滴加 $\\mathrm{NaOH}$ 溶液有蓝色沉淀生成, 则乙可能为 $\\mathrm{Cu}$\nB: 若丙溶液遇 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 放出气体 $\\mathrm{CO}_{2}$, 则甲可能是 $\\mathrm{H}_{2}$\nC: 若丙溶液中滴加 $\\mathrm{KSCN}$ 溶液显红色, 则乙可能为 $\\mathrm{Fe}$\nD: 若丙溶于水后得到强碱溶液, 则甲可能是 $\\mathrm{Na}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知甲、乙为单质, 丙为化合物。能实现下述转化关系的是\n\n甲 + 乙 $\\xrightarrow{\\text { 点燃 }}$ 丙 $\\xrightarrow{\\text { 溶于水 }}$ 丙溶液 $\\xrightarrow[\\text { 忤性电极 }]{\\longrightarrow}$ 甲 + 乙\n\nA: 若丙溶液中滴加 $\\mathrm{NaOH}$ 溶液有蓝色沉淀生成, 则乙可能为 $\\mathrm{Cu}$\nB: 若丙溶液遇 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 放出气体 $\\mathrm{CO}_{2}$, 则甲可能是 $\\mathrm{H}_{2}$\nC: 若丙溶液中滴加 $\\mathrm{KSCN}$ 溶液显红色, 则乙可能为 $\\mathrm{Fe}$\nD: 若丙溶于水后得到强碱溶液, 则甲可能是 $\\mathrm{Na}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_716",
        "problem": "某研究机构使用 $\\mathrm{Li}-\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$ 电池电解制备 $\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{2}\\right)_{2}$, 其工作原理如图所示。已知电池反应为 $2 \\mathrm{Li}+\\mathrm{SO}_{2} \\mathrm{Cl}_{2}=2 \\mathrm{LiCl}+\\mathrm{SO}_{2} \\uparrow$, 下列说法错误的是\n[图1]\nA: $\\mathrm{Li}$ 电极的电极反应式为 $\\mathrm{Li}^{-}-\\mathrm{e}^{-}=\\mathrm{Li}^{+}$\nB: $f$ 接口连接 $h$\nC: 膜 $\\mathrm{a} 、 \\mathrm{c}$ 是阴离子交换膜, 膜 $\\mathrm{b}$ 是阳离子交换膜\nD: 不锈钢电极附近溶液的 $\\mathrm{pH}$ 增大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某研究机构使用 $\\mathrm{Li}-\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$ 电池电解制备 $\\mathrm{Ni}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{2}\\right)_{2}$, 其工作原理如图所示。已知电池反应为 $2 \\mathrm{Li}+\\mathrm{SO}_{2} \\mathrm{Cl}_{2}=2 \\mathrm{LiCl}+\\mathrm{SO}_{2} \\uparrow$, 下列说法错误的是\n[图1]\n\nA: $\\mathrm{Li}$ 电极的电极反应式为 $\\mathrm{Li}^{-}-\\mathrm{e}^{-}=\\mathrm{Li}^{+}$\nB: $f$ 接口连接 $h$\nC: 膜 $\\mathrm{a} 、 \\mathrm{c}$ 是阴离子交换膜, 膜 $\\mathrm{b}$ 是阳离子交换膜\nD: 不锈钢电极附近溶液的 $\\mathrm{pH}$ 增大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-82.jpg?height=468&width=1302&top_left_y=163&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1261",
        "problem": "Acid-base equilibria in water\n\nA solution $(\\mathbf{X}$ ) contains two weak monoprotic acids (those having one acidic proton); HA with the acid dissociation constant of $K_{H A}=1.74 \\times 10^{-7}$, and $H B$ with the acid dissociation constant of $K_{H B}=1.34 \\times 10^{-7}$. The solution $\\mathbf{X}$ has a $\\mathrm{pH}$ of 3.75.A buffer solution is added to solution $\\mathbf{Y}$ to maintain a $\\mathrm{pH}$ of 10.0. Assume no change in volume of the resulting solution $\\mathbf{Z}$. Calculate the solubility (in $\\mathrm{mol} \\cdot \\mathrm{dm}^{-3}$ ) of the substance $\\mathrm{M}(\\mathrm{OH})_{2}$ in $\\mathbf{Z}$, taking into consideration that the anions $\\mathrm{A}^{-}$and $\\mathrm{B}^{-}$can form complexes with $\\mathrm{M}^{2+}$ :\n\n$\\mathrm{M}(\\mathrm{OH})_{2} \\leftrightharpoons \\mathrm{M}^{2+}+2 \\mathrm{OH}^{-} \\quad K_{s p}=3.10 \\times 10^{-12}$\n\n$\\mathrm{M}^{2+}+\\mathrm{A}^{-} \\leftrightharpoons[\\mathrm{MA}]^{+} \\quad K_{1}=2.1 \\times 10^{3}$\n\n$[\\mathrm{MA}]^{+}+\\mathrm{A}^{-} \\leftrightharpoons\\left[\\mathrm{MA}_{2}\\right] \\quad K_{2}=5.0 \\times 10^{2}$\n\n$\\mathrm{M}^{2+}+\\mathrm{B}^{-} \\leftrightharpoons[\\mathrm{MB}]^{+} \\quad \\mathrm{K}_{1}^{\\prime}=6.2 \\times 10^{3}$\n\n$[\\mathrm{MB}]^{+}+\\mathrm{B}^{-} \\leftrightharpoons\\left[\\mathrm{MB}_{2}\\right] \\quad \\mathrm{K}_{2}^{\\prime}=3.3 \\times 10^{2}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nAcid-base equilibria in water\n\nA solution $(\\mathbf{X}$ ) contains two weak monoprotic acids (those having one acidic proton); HA with the acid dissociation constant of $K_{H A}=1.74 \\times 10^{-7}$, and $H B$ with the acid dissociation constant of $K_{H B}=1.34 \\times 10^{-7}$. The solution $\\mathbf{X}$ has a $\\mathrm{pH}$ of 3.75.\n\nproblem:\nA buffer solution is added to solution $\\mathbf{Y}$ to maintain a $\\mathrm{pH}$ of 10.0. Assume no change in volume of the resulting solution $\\mathbf{Z}$. Calculate the solubility (in $\\mathrm{mol} \\cdot \\mathrm{dm}^{-3}$ ) of the substance $\\mathrm{M}(\\mathrm{OH})_{2}$ in $\\mathbf{Z}$, taking into consideration that the anions $\\mathrm{A}^{-}$and $\\mathrm{B}^{-}$can form complexes with $\\mathrm{M}^{2+}$ :\n\n$\\mathrm{M}(\\mathrm{OH})_{2} \\leftrightharpoons \\mathrm{M}^{2+}+2 \\mathrm{OH}^{-} \\quad K_{s p}=3.10 \\times 10^{-12}$\n\n$\\mathrm{M}^{2+}+\\mathrm{A}^{-} \\leftrightharpoons[\\mathrm{MA}]^{+} \\quad K_{1}=2.1 \\times 10^{3}$\n\n$[\\mathrm{MA}]^{+}+\\mathrm{A}^{-} \\leftrightharpoons\\left[\\mathrm{MA}_{2}\\right] \\quad K_{2}=5.0 \\times 10^{2}$\n\n$\\mathrm{M}^{2+}+\\mathrm{B}^{-} \\leftrightharpoons[\\mathrm{MB}]^{+} \\quad \\mathrm{K}_{1}^{\\prime}=6.2 \\times 10^{3}$\n\n$[\\mathrm{MB}]^{+}+\\mathrm{B}^{-} \\leftrightharpoons\\left[\\mathrm{MB}_{2}\\right] \\quad \\mathrm{K}_{2}^{\\prime}=3.3 \\times 10^{2}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_793",
        "problem": "银铜合金广泛应用于航空工业。从银铜合金的切割废料中回收银和制备 $\\mathrm{CuAlO}_{2}$ 的流程如下。\n\n[图1]\n\n已知: $\\mathrm{Al}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 开始分解的温度分别为 $450^{\\circ} \\mathrm{C}$ 和 $80^{\\circ} \\mathrm{C}$ 。下列说法错误的是\nA: 电解精炼时, 粗银做阳极, 纯银做阴极\nB: 为提高原料利用率, 流程中应加过量的稀 $\\mathrm{NaOH}$\nC: 滤渣 $\\mathrm{B}$ 煅烧时发生的反应为 $4 \\mathrm{CuO}+4 \\mathrm{Al}(\\mathrm{OH})_{3} \\stackrel{\\text { 高温 }}{=} 4 \\mathrm{CuAlO}_{2}+\\mathrm{O}_{2} \\uparrow+6 \\mathrm{H}_{2} \\mathrm{O}$\nD: 若用 $1.0 \\mathrm{~kg}$ 银铜合金(铜的质量分数为 $64 \\%$ )最多可生成 $1.0 \\mathrm{~mol} \\mathrm{CuAlO}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n银铜合金广泛应用于航空工业。从银铜合金的切割废料中回收银和制备 $\\mathrm{CuAlO}_{2}$ 的流程如下。\n\n[图1]\n\n已知: $\\mathrm{Al}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 开始分解的温度分别为 $450^{\\circ} \\mathrm{C}$ 和 $80^{\\circ} \\mathrm{C}$ 。下列说法错误的是\n\nA: 电解精炼时, 粗银做阳极, 纯银做阴极\nB: 为提高原料利用率, 流程中应加过量的稀 $\\mathrm{NaOH}$\nC: 滤渣 $\\mathrm{B}$ 煅烧时发生的反应为 $4 \\mathrm{CuO}+4 \\mathrm{Al}(\\mathrm{OH})_{3} \\stackrel{\\text { 高温 }}{=} 4 \\mathrm{CuAlO}_{2}+\\mathrm{O}_{2} \\uparrow+6 \\mathrm{H}_{2} \\mathrm{O}$\nD: 若用 $1.0 \\mathrm{~kg}$ 银铜合金(铜的质量分数为 $64 \\%$ )最多可生成 $1.0 \\mathrm{~mol} \\mathrm{CuAlO}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-88.jpg?height=295&width=1442&top_left_y=492&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1456",
        "problem": "The unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nThe first order rate constant for the $\\mathrm{Cl} 2$ protein folding reaction can be determined by following the fluorescence intensity when a sample of unfolded protein is allowed to refold (typically the $\\mathrm{pH}$ of the solution is changed). The concentration of protein when a $1.0 \\mu \\mathrm{M}$ sample of unfolded $\\mathrm{Cl} 2$ was allowed to refold was measured at a temperature of $25{ }^{\\circ} \\mathrm{C}$ :\n\n| time / ms | 0 | 10 | 20 | 30 | 40 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| concentration $/ \\mu \\mathrm{M}$ | 1 | 0.64 | 0.36 | 0.23 | 0.14 |Determine the value of the rate constant for the protein folding reaction, $k_{f}$, at $25^{\\circ} \\mathrm{C}$.\n\n[If you have been unable to calculate the value for $k_{\\mathrm{f}}$, you should use the following incorrect value for the subsequent parts of the question: $k_{\\mathrm{f}}=60 \\mathrm{~s}^{-1}$.]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nThe first order rate constant for the $\\mathrm{Cl} 2$ protein folding reaction can be determined by following the fluorescence intensity when a sample of unfolded protein is allowed to refold (typically the $\\mathrm{pH}$ of the solution is changed). The concentration of protein when a $1.0 \\mu \\mathrm{M}$ sample of unfolded $\\mathrm{Cl} 2$ was allowed to refold was measured at a temperature of $25{ }^{\\circ} \\mathrm{C}$ :\n\n| time / ms | 0 | 10 | 20 | 30 | 40 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| concentration $/ \\mu \\mathrm{M}$ | 1 | 0.64 | 0.36 | 0.23 | 0.14 |\n\nproblem:\nDetermine the value of the rate constant for the protein folding reaction, $k_{f}$, at $25^{\\circ} \\mathrm{C}$.\n\n[If you have been unable to calculate the value for $k_{\\mathrm{f}}$, you should use the following incorrect value for the subsequent parts of the question: $k_{\\mathrm{f}}=60 \\mathrm{~s}^{-1}$.]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{s}^{-1}$., but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{s}^{-1}$."
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1517",
        "problem": "The energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.Determine the bond energy $E_{\\mathrm{C}}(\\mathrm{eV})$ of $\\mathrm{H}_{2}$ to the first decimal place.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.\n\nproblem:\nDetermine the bond energy $E_{\\mathrm{C}}(\\mathrm{eV})$ of $\\mathrm{H}_{2}$ to the first decimal place.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of eV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "eV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_70",
        "problem": "The hypochlorite concentration in bleach can be determined by treating it with excess iodide ion in acidic solution, which causes the formation of triiodide ion according to the following balanced reaction:\n\n$$\n\\begin{gathered}\n\\mathrm{OCl}^{-}(a q)+3 \\mathrm{I}^{-}(a q)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\rightarrow \\\\\n\\mathrm{Cl}^{-}(a q)+\\mathrm{I}_{3}^{-}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nThe triiodide can then be titrated with sodium thiosulfate:\n\n$$\n\\mathrm{I}_{3}^{-}(a q)+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}(a q) \\rightarrow 3 \\mathrm{I}^{-}(a q)+\\mathrm{S}_{4} \\mathrm{O}_{6}{ }^{2-}(a q)\n$$\n\nA 75.0-mL sample of liquid bleach is treated with excess $\\mathrm{KI}$ and acid, and then titrated with $0.0235 \\mathrm{M}$ sodium thiosulfate solution. The endpoint is observed on the addition of $21.23 \\mathrm{~mL}$ of this solution. What is the hypochlorite concentration in the bleach?\nA: $4.99 \\times 10^{-4} \\mathrm{M}$\nB: $1.66 \\times 10^{-3} \\mathrm{M}$\nC: $3.33 \\times 10^{-3} \\mathrm{M}$\nD: $6.65 \\times 10^{-3} \\mathrm{M}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe hypochlorite concentration in bleach can be determined by treating it with excess iodide ion in acidic solution, which causes the formation of triiodide ion according to the following balanced reaction:\n\n$$\n\\begin{gathered}\n\\mathrm{OCl}^{-}(a q)+3 \\mathrm{I}^{-}(a q)+2 \\mathrm{H}_{3} \\mathrm{O}^{+}(a q) \\rightarrow \\\\\n\\mathrm{Cl}^{-}(a q)+\\mathrm{I}_{3}^{-}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nThe triiodide can then be titrated with sodium thiosulfate:\n\n$$\n\\mathrm{I}_{3}^{-}(a q)+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}(a q) \\rightarrow 3 \\mathrm{I}^{-}(a q)+\\mathrm{S}_{4} \\mathrm{O}_{6}{ }^{2-}(a q)\n$$\n\nA 75.0-mL sample of liquid bleach is treated with excess $\\mathrm{KI}$ and acid, and then titrated with $0.0235 \\mathrm{M}$ sodium thiosulfate solution. The endpoint is observed on the addition of $21.23 \\mathrm{~mL}$ of this solution. What is the hypochlorite concentration in the bleach?\n\nA: $4.99 \\times 10^{-4} \\mathrm{M}$\nB: $1.66 \\times 10^{-3} \\mathrm{M}$\nC: $3.33 \\times 10^{-3} \\mathrm{M}$\nD: $6.65 \\times 10^{-3} \\mathrm{M}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_374",
        "problem": "Calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}(M=128.1)$, dissolves to the extent of $0.67 \\mathrm{mg} \\mathrm{L}^{-1}$. What is its $K_{\\mathrm{sp}}$ ?\nA: $6.7 \\times 10^{-4}$\nB: $4.5 \\times 10^{-7}$\nC: $2.7 \\times 10^{-11}$\nD: $5.7 \\times 10^{-16}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCalcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}(M=128.1)$, dissolves to the extent of $0.67 \\mathrm{mg} \\mathrm{L}^{-1}$. What is its $K_{\\mathrm{sp}}$ ?\n\nA: $6.7 \\times 10^{-4}$\nB: $4.5 \\times 10^{-7}$\nC: $2.7 \\times 10^{-11}$\nD: $5.7 \\times 10^{-16}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1395",
        "problem": "The radioactive decay series of ${ }^{226} \\mathrm{Ra}$ is as follows:\n\n$$\n\\begin{aligned}\n& { }^{226} \\mathrm{Ra} \\xrightarrow{t}{ }^{222} \\mathrm{Rn} \\xrightarrow{3.825 \\mathrm{~d}}{ }^{218} \\mathrm{Po} \\xrightarrow{3.10 \\mathrm{~m}}{ }^{214} \\mathrm{~Pb} \\xrightarrow{26.8 \\mathrm{~m}}{ }^{214} \\mathrm{Bi} \\xrightarrow{19.9 \\mathrm{~m}} \\\\\n& { }^{214} \\mathrm{Po} \\xrightarrow{164.3 \\mu \\mathrm{s}} 210 \\mathrm{~Pb} \\xrightarrow{22.3 \\mathrm{y}} 210 \\mathrm{Bi} \\xrightarrow{5.013 \\mathrm{~d}} 210 \\mathrm{Po} \\xrightarrow{138.4 \\mathrm{~d}}{ }^{206 \\mathrm{~Pb}}\n\\end{aligned}\n$$\n\nThe times indicated are half-lives, the units are $y=$ years, $d=$ days, $m=$ minutes. The first decay, marked $t$ above, has a much longer half-life than the others.The total rate of $\\alpha$-decay from the sample was then determined by scintillation to be 27.7 GBq (where $1 \\mathrm{~Bq}=1$ count $\\mathrm{s}^{-1}$ ). The sample was then sealed for 163 days. Calculate the number of $\\alpha$ particles produced.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe radioactive decay series of ${ }^{226} \\mathrm{Ra}$ is as follows:\n\n$$\n\\begin{aligned}\n& { }^{226} \\mathrm{Ra} \\xrightarrow{t}{ }^{222} \\mathrm{Rn} \\xrightarrow{3.825 \\mathrm{~d}}{ }^{218} \\mathrm{Po} \\xrightarrow{3.10 \\mathrm{~m}}{ }^{214} \\mathrm{~Pb} \\xrightarrow{26.8 \\mathrm{~m}}{ }^{214} \\mathrm{Bi} \\xrightarrow{19.9 \\mathrm{~m}} \\\\\n& { }^{214} \\mathrm{Po} \\xrightarrow{164.3 \\mu \\mathrm{s}} 210 \\mathrm{~Pb} \\xrightarrow{22.3 \\mathrm{y}} 210 \\mathrm{Bi} \\xrightarrow{5.013 \\mathrm{~d}} 210 \\mathrm{Po} \\xrightarrow{138.4 \\mathrm{~d}}{ }^{206 \\mathrm{~Pb}}\n\\end{aligned}\n$$\n\nThe times indicated are half-lives, the units are $y=$ years, $d=$ days, $m=$ minutes. The first decay, marked $t$ above, has a much longer half-life than the others.\n\nproblem:\nThe total rate of $\\alpha$-decay from the sample was then determined by scintillation to be 27.7 GBq (where $1 \\mathrm{~Bq}=1$ count $\\mathrm{s}^{-1}$ ). The sample was then sealed for 163 days. Calculate the number of $\\alpha$ particles produced.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1021",
        "problem": "The reaction $2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\rightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)$ is an example of\nA: a precipitation reaction\nB: an acid-base reaction\nC: a decomposition reaction\nD: an oxidation-reduction reaction\nE: an isomerization reaction\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction $2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\rightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)$ is an example of\n\nA: a precipitation reaction\nB: an acid-base reaction\nC: a decomposition reaction\nD: an oxidation-reduction reaction\nE: an isomerization reaction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_58",
        "problem": "A copper ore consists of a mixture of $\\mathrm{Cu}_{2} \\mathrm{~S}$ and $\\mathrm{CuS}$. An $89.0 \\mathrm{~g}$ sample of this ore is smelted to produce $67.5 \\mathrm{~g}$ of elemental copper. What is the mass of $\\mathrm{CuS}$ in the ore sample?\nA: $13.3 \\mathrm{~g}$\nB: $26.6 \\mathrm{~g}$\nC: $37.5 \\mathrm{~g}$\nD: $64.1 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA copper ore consists of a mixture of $\\mathrm{Cu}_{2} \\mathrm{~S}$ and $\\mathrm{CuS}$. An $89.0 \\mathrm{~g}$ sample of this ore is smelted to produce $67.5 \\mathrm{~g}$ of elemental copper. What is the mass of $\\mathrm{CuS}$ in the ore sample?\n\nA: $13.3 \\mathrm{~g}$\nB: $26.6 \\mathrm{~g}$\nC: $37.5 \\mathrm{~g}$\nD: $64.1 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1293",
        "problem": "Which structure describes best the crystal system of iron in which the coordination number is 8 ?\nA: simple cubic\nB: body-centered cubic\nC: cubic closest packed\nD: hexagonal closest packed\nE: none of the above\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich structure describes best the crystal system of iron in which the coordination number is 8 ?\n\nA: simple cubic\nB: body-centered cubic\nC: cubic closest packed\nD: hexagonal closest packed\nE: none of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_333",
        "problem": "In acidic solution, zinc metal reacts spontaneously with $\\mathrm{ReO}_{4}{ }^{-}$. The unbalanced chemical equation for the reaction is given below.\n\n$\\mathrm{Zn}(s)+\\mathrm{ReO}_{4}{ }^{-}(a q)+\\mathrm{H}^{+}(a q) \\rightarrow \\mathrm{Re}(s)+\\mathrm{Zn}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n\nFor the reaction above, what element or ion is the reducing agent?\nA: $\\operatorname{Re}(s)$\nB: $\\mathrm{Zn}(s)$\nC: $\\mathrm{ReO}_{4}^{-}(a q)$\nD: $\\mathrm{Zn}^{2+}(a q)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn acidic solution, zinc metal reacts spontaneously with $\\mathrm{ReO}_{4}{ }^{-}$. The unbalanced chemical equation for the reaction is given below.\n\n$\\mathrm{Zn}(s)+\\mathrm{ReO}_{4}{ }^{-}(a q)+\\mathrm{H}^{+}(a q) \\rightarrow \\mathrm{Re}(s)+\\mathrm{Zn}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n\nFor the reaction above, what element or ion is the reducing agent?\n\nA: $\\operatorname{Re}(s)$\nB: $\\mathrm{Zn}(s)$\nC: $\\mathrm{ReO}_{4}^{-}(a q)$\nD: $\\mathrm{Zn}^{2+}(a q)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_182",
        "problem": "Vinegar used as a cooking ingredient, or in pickling, is a solution of $5 \\%$ acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$ by mass in solution with water. The $\\mathrm{pKa}=4.76$ for acetic acid at $25^{\\circ} \\mathrm{C}$. Evaluate the $\\mathbf{p H}$ of vinegar at $25^{\\circ} \\mathrm{C}$.\nA: 4.80\nB: 4.15\nC: 2.92\nD: 2.42\nE: 2.24\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nVinegar used as a cooking ingredient, or in pickling, is a solution of $5 \\%$ acetic acid $\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$ by mass in solution with water. The $\\mathrm{pKa}=4.76$ for acetic acid at $25^{\\circ} \\mathrm{C}$. Evaluate the $\\mathbf{p H}$ of vinegar at $25^{\\circ} \\mathrm{C}$.\n\nA: 4.80\nB: 4.15\nC: 2.92\nD: 2.42\nE: 2.24\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1014",
        "problem": "Which drawing shows a pipet correctly filled for delivery?\n\n[figure1]\nA: 1\nB: 2\nC: 3\nD: 4\nE: none of the above\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich drawing shows a pipet correctly filled for delivery?\n\n[figure1]\n\nA: 1\nB: 2\nC: 3\nD: 4\nE: none of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_08_f4bf9378e99f2103753dg-5.jpg?height=487&width=575&top_left_y=1250&top_left_x=1298"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1193",
        "problem": "Identify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{..9}^{19} \\mathrm{~F}+{ }_{0}^{1} \\mathrm{n} \\rightarrow{ }_{. .9}^{20} \\mathrm{~F}+\\mathrm{X}$\nA: alpha\nB: beta\nC: gamma\nD: neutron\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIdentify particle $\\mathrm{X}$ in each of the following nuclear reactions:\n\n${ }_{..9}^{19} \\mathrm{~F}+{ }_{0}^{1} \\mathrm{n} \\rightarrow{ }_{. .9}^{20} \\mathrm{~F}+\\mathrm{X}$\n\nA: alpha\nB: beta\nC: gamma\nD: neutron\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_901",
        "problem": "已知阿仑尼乌斯公式是反应速率常数随温度变化关系的经验公式, 可写作\n\n$R \\ln k=-\\frac{E_{a}}{T}+C\\left(k\\right.$ 为反应速率常数, $E_{a}$ 为反应活化能, $R$ 和 $C$ 为大于 0 的常数), 为探究 $\\mathrm{m} 、 \\mathrm{n}$ 两种催化剂对某反应的催化效能, 进行了实验探究, 依据实验数据获得曲线如图所示。下列说法错误的是\n\n[图1]\nA: 在 $\\mathrm{m}$ 催化剂作用下, 该反应的活化能 $\\mathrm{E}_{\\mathrm{a}}=9.6 \\times 10^{4} \\mathrm{~J} \\cdot \\mathrm{mol}^{-1}$\nB: 对该反应催化效能较高的催化剂是 $\\mathrm{m}$\nC: 不改变其他条件, 只升高温度, 反应的活化能不变\nD: 无法根据该图像判断升高温度时平衡移动的方向\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知阿仑尼乌斯公式是反应速率常数随温度变化关系的经验公式, 可写作\n\n$R \\ln k=-\\frac{E_{a}}{T}+C\\left(k\\right.$ 为反应速率常数, $E_{a}$ 为反应活化能, $R$ 和 $C$ 为大于 0 的常数), 为探究 $\\mathrm{m} 、 \\mathrm{n}$ 两种催化剂对某反应的催化效能, 进行了实验探究, 依据实验数据获得曲线如图所示。下列说法错误的是\n\n[图1]\n\nA: 在 $\\mathrm{m}$ 催化剂作用下, 该反应的活化能 $\\mathrm{E}_{\\mathrm{a}}=9.6 \\times 10^{4} \\mathrm{~J} \\cdot \\mathrm{mol}^{-1}$\nB: 对该反应催化效能较高的催化剂是 $\\mathrm{m}$\nC: 不改变其他条件, 只升高温度, 反应的活化能不变\nD: 无法根据该图像判断升高温度时平衡移动的方向\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-094.jpg?height=537&width=551&top_left_y=888&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_166",
        "problem": "What is the correctly balanced form of the chemical reaction depicted in the figure below?\n\n[figure1]\nA: $4 \\mathrm{~A}+6 \\mathrm{~B} \\rightarrow \\mathrm{A}_{4} \\mathrm{~B}_{6}$\nB: $\\mathrm{A}_{4}+\\mathrm{B}_{6} \\rightarrow \\mathrm{A}_{4} \\mathrm{~B}_{6}$\nC: $\\mathrm{A}_{4}+3 \\mathrm{~B}_{2} \\rightarrow 2 \\mathrm{~A}_{2}+2 \\mathrm{~B}_{3}$\nD: $\\mathrm{A}_{4}+3 \\mathrm{~B}_{2} \\rightarrow 2 \\mathrm{~A}_{2} \\mathrm{~B}_{3}$\nE: $4 \\mathrm{~A}+3 \\mathrm{~B}_{2} \\rightarrow 2 \\mathrm{~A}_{2} \\mathrm{~B}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the correctly balanced form of the chemical reaction depicted in the figure below?\n\n[figure1]\n\nA: $4 \\mathrm{~A}+6 \\mathrm{~B} \\rightarrow \\mathrm{A}_{4} \\mathrm{~B}_{6}$\nB: $\\mathrm{A}_{4}+\\mathrm{B}_{6} \\rightarrow \\mathrm{A}_{4} \\mathrm{~B}_{6}$\nC: $\\mathrm{A}_{4}+3 \\mathrm{~B}_{2} \\rightarrow 2 \\mathrm{~A}_{2}+2 \\mathrm{~B}_{3}$\nD: $\\mathrm{A}_{4}+3 \\mathrm{~B}_{2} \\rightarrow 2 \\mathrm{~A}_{2} \\mathrm{~B}_{3}$\nE: $4 \\mathrm{~A}+3 \\mathrm{~B}_{2} \\rightarrow 2 \\mathrm{~A}_{2} \\mathrm{~B}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_08_895edf4f0e41ca808d97g-1.jpg?height=182&width=803&top_left_y=1105&top_left_x=328"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1505",
        "problem": "In ionic inorganic compounds such as sodium chloride, the heat of lattice formation from gaseous ions is very high, and the contribution of the change in entropy is small. Therefore, the lattice formation enthalpy is estimated from enthalpy data by using a BornHaber cycle.Calculate the enthalpy of the lattice formation of $\\mathrm{NaCl}\\left[\\mathrm{kJ} \\mathrm{mol}^{-1}\\right]$ by using the following enthalpy data of the respective steps in the above Born-Haber cycle.\n\n| Formation of <br> $\\mathrm{NaCl}(s)$ | Sublimation <br> of $\\mathrm{Na}(s)$ | Ionization of <br> $\\mathrm{Na}(g)$ | Dissociation <br> of $\\mathrm{Cl}_{2}(g)$ | Electron gain <br> by Cl $(g)$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $-411 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $109 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $496 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $242 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $-349 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ |",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn ionic inorganic compounds such as sodium chloride, the heat of lattice formation from gaseous ions is very high, and the contribution of the change in entropy is small. Therefore, the lattice formation enthalpy is estimated from enthalpy data by using a BornHaber cycle.\n\nproblem:\nCalculate the enthalpy of the lattice formation of $\\mathrm{NaCl}\\left[\\mathrm{kJ} \\mathrm{mol}^{-1}\\right]$ by using the following enthalpy data of the respective steps in the above Born-Haber cycle.\n\n| Formation of <br> $\\mathrm{NaCl}(s)$ | Sublimation <br> of $\\mathrm{Na}(s)$ | Ionization of <br> $\\mathrm{Na}(g)$ | Dissociation <br> of $\\mathrm{Cl}_{2}(g)$ | Electron gain <br> by Cl $(g)$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $-411 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $109 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $496 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $242 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ | $-349 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ |\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_707",
        "problem": "数字化滴定实验应用普遍。常温下, 电控恒速滴定并数据采集等浓度 $\\mathrm{HCl}$ 溶液和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液的反应曲线如图所示, 已知碳酸的 $\\mathrm{Ka}_{1}=4 \\times 10^{-7}, \\mathrm{Ka}_{2}=5 \\times 10^{-11}$, 下列选项正确的是\n\n[图1]\nA: $\\mathrm{HCl}$ 溶液的浓度约为 $0.05 \\mathrm{~mol} / \\mathrm{L}$\nB: b 点溶液存在: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $\\mathrm{cd}$ 段发生的主要反应为 $\\mathrm{HCO}_{3}^{-}+\\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2}$\nD: e 点溶液中溶质的主要成分是 $\\mathrm{NaCl}$ 和 $\\mathrm{HCl}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n数字化滴定实验应用普遍。常温下, 电控恒速滴定并数据采集等浓度 $\\mathrm{HCl}$ 溶液和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液的反应曲线如图所示, 已知碳酸的 $\\mathrm{Ka}_{1}=4 \\times 10^{-7}, \\mathrm{Ka}_{2}=5 \\times 10^{-11}$, 下列选项正确的是\n\n[图1]\n\nA: $\\mathrm{HCl}$ 溶液的浓度约为 $0.05 \\mathrm{~mol} / \\mathrm{L}$\nB: b 点溶液存在: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $\\mathrm{cd}$ 段发生的主要反应为 $\\mathrm{HCO}_{3}^{-}+\\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2}$\nD: e 点溶液中溶质的主要成分是 $\\mathrm{NaCl}$ 和 $\\mathrm{HCl}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-077.jpg?height=608&width=808&top_left_y=1552&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_617",
        "problem": "化合物 M 具有广谱抗菌活性, 合成 M 的反应可表示如下:\n[图1]下列说法正确的是\nA: X 分子中有 2 个手性碳原子\nB: $\\mathrm{Y}$ 分子中所有原子可能在同一平面内\nC: 可用 $\\mathrm{FeCl}_{3}$ 溶液或 $\\mathrm{NaHCO}_{3}$ 溶液鉴别 $\\mathrm{X}$ 和 $\\mathrm{Y}$\nD: 在 $\\mathrm{NaOH}$ 溶液中, $1 \\mathrm{~mol} \\mathrm{M}$ 最多可与 $5 \\mathrm{~mol} \\mathrm{NaOH}$ 发生反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化合物 M 具有广谱抗菌活性, 合成 M 的反应可表示如下:\n[图1]下列说法正确的是\n\nA: X 分子中有 2 个手性碳原子\nB: $\\mathrm{Y}$ 分子中所有原子可能在同一平面内\nC: 可用 $\\mathrm{FeCl}_{3}$ 溶液或 $\\mathrm{NaHCO}_{3}$ 溶液鉴别 $\\mathrm{X}$ 和 $\\mathrm{Y}$\nD: 在 $\\mathrm{NaOH}$ 溶液中, $1 \\mathrm{~mol} \\mathrm{M}$ 最多可与 $5 \\mathrm{~mol} \\mathrm{NaOH}$ 发生反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/s2h9YzcW/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_868",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定同浓度的 $\\mathrm{H}_{2} \\mathrm{~A}$ 溶液, $\\mathrm{H}_{2} \\mathrm{~A}$ 被滴定分数 $\\left[\\frac{\\mathrm{n}(\\mathrm{NaOH})}{\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}\\right] 、 \\mathrm{pH}$ 及微粒分布分数 $\\delta\\left[\\delta(\\mathrm{X})=\\frac{\\mathrm{n}(\\mathrm{X})}{\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{n}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{n}\\left(\\mathrm{A}^{2-}\\right)}, \\mathrm{X}\\right.$ 表示 $\\mathrm{H}_{2} \\mathrm{~A}$ 、\n\n$\\mathrm{HA}^{-}$或 $\\left.\\mathrm{A}^{2-}\\right]$ 的关系如图所示:\n\n[图1]\nA: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{~A}$ 第一步电离平衡常数 $\\mathrm{K}_{\\mathrm{a} 1} \\approx 10^{-4}$\nB: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nC: $a 、 b 、 c 、 d$ 四点溶液中水的电离程度: $c>d>b>a$\nD: b 点溶液中: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HA}^{-}\\right)>c\\left(\\mathrm{~A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定同浓度的 $\\mathrm{H}_{2} \\mathrm{~A}$ 溶液, $\\mathrm{H}_{2} \\mathrm{~A}$ 被滴定分数 $\\left[\\frac{\\mathrm{n}(\\mathrm{NaOH})}{\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}\\right] 、 \\mathrm{pH}$ 及微粒分布分数 $\\delta\\left[\\delta(\\mathrm{X})=\\frac{\\mathrm{n}(\\mathrm{X})}{\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{n}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{n}\\left(\\mathrm{A}^{2-}\\right)}, \\mathrm{X}\\right.$ 表示 $\\mathrm{H}_{2} \\mathrm{~A}$ 、\n\n$\\mathrm{HA}^{-}$或 $\\left.\\mathrm{A}^{2-}\\right]$ 的关系如图所示:\n\n[图1]\n\nA: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{~A}$ 第一步电离平衡常数 $\\mathrm{K}_{\\mathrm{a} 1} \\approx 10^{-4}$\nB: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nC: $a 、 b 、 c 、 d$ 四点溶液中水的电离程度: $c>d>b>a$\nD: b 点溶液中: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HA}^{-}\\right)>c\\left(\\mathrm{~A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-045.jpg?height=540&width=813&top_left_y=2214&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-046.jpg?height=616&width=928&top_left_y=1311&top_left_x=561"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_271",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount of $\\mathrm{HCO}_{3}{ }^{-}$ions present in $20.00 \\mathrm{~mL}$ of the \"dissolved mineral solution”.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount of $\\mathrm{HCO}_{3}{ }^{-}$ions present in $20.00 \\mathrm{~mL}$ of the \"dissolved mineral solution”.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_993",
        "problem": "Which atom has the most neutrons?\nA: $\\quad{ }_{9}^{18} \\mathrm{~F}$\nB: ${ }_{8}^{18} \\mathrm{O}$\nC: $\\quad{ }_{6}^{14} \\mathrm{C}$\nD: $\\quad{ }_{7}^{15} \\mathrm{~N}$\nE: $\\quad{ }_{5}^{11} B$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich atom has the most neutrons?\n\nA: $\\quad{ }_{9}^{18} \\mathrm{~F}$\nB: ${ }_{8}^{18} \\mathrm{O}$\nC: $\\quad{ }_{6}^{14} \\mathrm{C}$\nD: $\\quad{ }_{7}^{15} \\mathrm{~N}$\nE: $\\quad{ }_{5}^{11} B$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_580",
        "problem": "磷酸是生产磷肥和饲料营养剂的原料, 同时还是常用的食品添加剂。常温下 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$在水中各级电离的过程如下图所示。已知: $\\mathrm{pK}=-1 \\mathrm{gK}$\n\n[图1]下列相关说法正确的是\nA: 反应 $\\mathrm{H}_{3} \\mathrm{PO}_{4}+2 \\mathrm{PO}_{4}^{3-} \\rightleftharpoons 3 \\mathrm{HPO}_{4}^{2-} \\quad \\mathrm{pK}=16$\nB: $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ 溶液中: $\\left[\\mathrm{PO}_{4}^{3-}\\right]>2\\left[\\mathrm{H}_{3} \\mathrm{PO}_{4}\\right]+\\left[\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right]$\nC: 用 $\\mathrm{NaOH}$ 溶液滴定 $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 溶液时宜选用甲基橙做指示剂\nD: $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 和 $\\mathrm{Na}_{3} \\mathrm{PO}_{4}$ 组成的混合溶液 $\\mathrm{pH}=6$ 时, 溶液中: $\\left[\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right]>\\left[\\mathrm{HPO}_{4}^{2-}\\right]>$ $\\left[\\mathrm{H}_{3} \\mathrm{PO}_{4}\\right]$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n磷酸是生产磷肥和饲料营养剂的原料, 同时还是常用的食品添加剂。常温下 $\\mathrm{H}_{3} \\mathrm{PO}_{4}$在水中各级电离的过程如下图所示。已知: $\\mathrm{pK}=-1 \\mathrm{gK}$\n\n[图1]下列相关说法正确的是\n\nA: 反应 $\\mathrm{H}_{3} \\mathrm{PO}_{4}+2 \\mathrm{PO}_{4}^{3-} \\rightleftharpoons 3 \\mathrm{HPO}_{4}^{2-} \\quad \\mathrm{pK}=16$\nB: $\\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ 溶液中: $\\left[\\mathrm{PO}_{4}^{3-}\\right]>2\\left[\\mathrm{H}_{3} \\mathrm{PO}_{4}\\right]+\\left[\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right]$\nC: 用 $\\mathrm{NaOH}$ 溶液滴定 $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 溶液时宜选用甲基橙做指示剂\nD: $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 和 $\\mathrm{Na}_{3} \\mathrm{PO}_{4}$ 组成的混合溶液 $\\mathrm{pH}=6$ 时, 溶液中: $\\left[\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right]>\\left[\\mathrm{HPO}_{4}^{2-}\\right]>$ $\\left[\\mathrm{H}_{3} \\mathrm{PO}_{4}\\right]$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-093.jpg?height=180&width=1448&top_left_y=338&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_259",
        "problem": "Calculate the volume of $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ that reacts with $1 \\mathrm{~mL}$ of $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ under these conditions.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the volume of $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ that reacts with $1 \\mathrm{~mL}$ of $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ under these conditions.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_929",
        "problem": "若定义 $\\mathrm{pAg}=-\\operatorname{lgc}\\left(\\mathrm{Ag}^{+}\\right), \\mathrm{pCl}=-\\operatorname{lgc}\\left(\\mathrm{Cl}^{-}\\right)$, 根据不同温度下氯化银饱和溶液的 $\\mathrm{PAg}$ 和 $\\mathrm{pCl}$ 可绘制图象如右图所示, 且已知氯化银的溶解度随温度的升高而增大, 根据该图象,下列表述正确的是\n\n[图1]\nA: $T_{3}>T_{2}>T_{1}$\nB: 将 $\\mathrm{A}$ 点的溶液降温, 可能得到 $\\mathrm{C}$ 点的饱和溶液\nC: 向 B 点所表示的溶液中加入氯化钠溶液, 溶液可能改变至 D 点\nD: A 点表示的是 $T_{1}$ 温度下的不饱和溶液\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n若定义 $\\mathrm{pAg}=-\\operatorname{lgc}\\left(\\mathrm{Ag}^{+}\\right), \\mathrm{pCl}=-\\operatorname{lgc}\\left(\\mathrm{Cl}^{-}\\right)$, 根据不同温度下氯化银饱和溶液的 $\\mathrm{PAg}$ 和 $\\mathrm{pCl}$ 可绘制图象如右图所示, 且已知氯化银的溶解度随温度的升高而增大, 根据该图象,下列表述正确的是\n\n[图1]\n\nA: $T_{3}>T_{2}>T_{1}$\nB: 将 $\\mathrm{A}$ 点的溶液降温, 可能得到 $\\mathrm{C}$ 点的饱和溶液\nC: 向 B 点所表示的溶液中加入氯化钠溶液, 溶液可能改变至 D 点\nD: A 点表示的是 $T_{1}$ 温度下的不饱和溶液\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-72.jpg?height=417&width=583&top_left_y=1051&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1035",
        "problem": "Some self-test breathalyser kits use the redox reaction between ethanol and acidified potassium dichromate to estimate blood alcohol.\n\nThe alcohol gets into the blood by absorption through the stomach wall; most is broken down in the liver to carbon dioxide and water - the rest leaves the body through sweat, in the breath and by excretion in urine.\n\n[figure1]\n\nAlcohol concentration in the blood can be estimated by analysing the alcohol in the breath because an equilibrium is set up between the alcohol in the blood and the alcohol in the air in the lungs:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{\\text {(Blood) }} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{\\text {(Breath) }}\n$$\n\nAt body temperature, the concentration of alcohol in the blood is about 2300 times that in the breath.\n\nThe half equation for the oxidation of ethanol to ethanoic acid is:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+4 \\mathrm{H}^{+}+4 \\mathrm{e}^{-}\n$$\n\nThe equation for the reduction of the dichromate ion in acid solution is:\n\n$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+14 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}$\n\nThe overall equation for this reaction is:\n\n$3 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}+2 \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2--}+16 \\mathrm{H}^{+} \\longrightarrow 3 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+4 \\mathrm{Cr}^{3+}+11 \\mathrm{H}_{2} \\mathrm{O}$\n\nAssuming that the acid used is sulfuric acid and the dichromate salt is potassium dichromate, the balanced equation for the reaction is:\n\n$3 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}+2 \\mathrm{~K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}+8 \\mathrm{H}_{2} \\mathrm{SO}_{4} ? 3 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+2 \\mathrm{Cr}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}+2 \\mathrm{~K}_{2} \\mathrm{SO}_{4}+$ $11 \\mathrm{H}_{2} \\mathrm{O}$\n\nThe breathalyser kit consists of a plastic bag that is inflated with $1000 \\mathrm{~cm}^{3}$ of breath and a glass tube containing the dichromate crystals. When the bag is connected to the tube and the breath is expelled through the tube the crystals change colour as they are reduced. The proportion of the crystals that change colour indicates the amount of alcohol present.\n\nThe current legal maximum blood alcohol concentration when driving in Britain is $80 \\mathrm{mg}$ per $100 \\mathrm{~cm}^{3}$ of blood.\n\nThen the corresponding breath alcohol concentration in $\\mu \\mathrm{g}$ per $1000 \\mathrm{~cm}^{3}$ of breath is 348.\n\nAssuming that the tube must be able to test breath at least three times over the legal limit, what mass of potassium dichromate should the tube contain?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nSome self-test breathalyser kits use the redox reaction between ethanol and acidified potassium dichromate to estimate blood alcohol.\n\nThe alcohol gets into the blood by absorption through the stomach wall; most is broken down in the liver to carbon dioxide and water - the rest leaves the body through sweat, in the breath and by excretion in urine.\n\n[figure1]\n\nAlcohol concentration in the blood can be estimated by analysing the alcohol in the breath because an equilibrium is set up between the alcohol in the blood and the alcohol in the air in the lungs:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{\\text {(Blood) }} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{\\text {(Breath) }}\n$$\n\nAt body temperature, the concentration of alcohol in the blood is about 2300 times that in the breath.\n\nThe half equation for the oxidation of ethanol to ethanoic acid is:\n\n$$\n\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{H}_{2} \\mathrm{O} \\longrightarrow \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+4 \\mathrm{H}^{+}+4 \\mathrm{e}^{-}\n$$\n\nThe equation for the reduction of the dichromate ion in acid solution is:\n\n$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}+6 \\mathrm{e}^{-}+14 \\mathrm{H}^{+} \\longrightarrow 2 \\mathrm{Cr}^{3+}+7 \\mathrm{H}_{2} \\mathrm{O}$\n\nThe overall equation for this reaction is:\n\n$3 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}+2 \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2--}+16 \\mathrm{H}^{+} \\longrightarrow 3 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+4 \\mathrm{Cr}^{3+}+11 \\mathrm{H}_{2} \\mathrm{O}$\n\nAssuming that the acid used is sulfuric acid and the dichromate salt is potassium dichromate, the balanced equation for the reaction is:\n\n$3 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}+2 \\mathrm{~K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}+8 \\mathrm{H}_{2} \\mathrm{SO}_{4} ? 3 \\mathrm{CH}_{3} \\mathrm{CO}_{2} \\mathrm{H}+2 \\mathrm{Cr}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}+2 \\mathrm{~K}_{2} \\mathrm{SO}_{4}+$ $11 \\mathrm{H}_{2} \\mathrm{O}$\n\nThe breathalyser kit consists of a plastic bag that is inflated with $1000 \\mathrm{~cm}^{3}$ of breath and a glass tube containing the dichromate crystals. When the bag is connected to the tube and the breath is expelled through the tube the crystals change colour as they are reduced. The proportion of the crystals that change colour indicates the amount of alcohol present.\n\nThe current legal maximum blood alcohol concentration when driving in Britain is $80 \\mathrm{mg}$ per $100 \\mathrm{~cm}^{3}$ of blood.\n\nThen the corresponding breath alcohol concentration in $\\mu \\mathrm{g}$ per $1000 \\mathrm{~cm}^{3}$ of breath is 348.\n\nAssuming that the tube must be able to test breath at least three times over the legal limit, what mass of potassium dichromate should the tube contain?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_4408eb29f8e936fbb54dg-08.jpg?height=588&width=711&top_left_y=323&top_left_x=1112"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_236",
        "problem": "Which of the following formulas is an empirical formula?\nA: $\\mathrm{C}_{2} \\mathrm{H}_{6}$\nB: $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}$\nC: $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$\nD: $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}_{2}$\nE: $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following formulas is an empirical formula?\n\nA: $\\mathrm{C}_{2} \\mathrm{H}_{6}$\nB: $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}$\nC: $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$\nD: $\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}_{2}$\nE: $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_346",
        "problem": "Sulfuryl chloride is in equilibrium with sulfur dioxide and chlorine gas:\n\n$$\n\\mathrm{SO}_{2} \\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{SO}_{2}(g)+\\mathrm{Cl}_{2}(g)\n$$\n\nA system with a volume of $1.00 \\mathrm{~L}$ is in equilibrium at a certain temperature with $p\\left(\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right)=1.00$ bar and $p\\left(\\mathrm{SO}_{2}\\right)$ $=p\\left(\\mathrm{Cl}_{2}\\right)=0.10$ bar. By how much will the number of moles of $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$ at equilibrium change if the volume is reduced to $0.50 \\mathrm{~L}$ ?\nA: Increase $1-10 \\%$\nB: Increase $11-50 \\%$\nC: Decrease 1-10\\%\nD: Decrease 11-50\\%\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSulfuryl chloride is in equilibrium with sulfur dioxide and chlorine gas:\n\n$$\n\\mathrm{SO}_{2} \\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{SO}_{2}(g)+\\mathrm{Cl}_{2}(g)\n$$\n\nA system with a volume of $1.00 \\mathrm{~L}$ is in equilibrium at a certain temperature with $p\\left(\\mathrm{SO}_{2} \\mathrm{Cl}_{2}\\right)=1.00$ bar and $p\\left(\\mathrm{SO}_{2}\\right)$ $=p\\left(\\mathrm{Cl}_{2}\\right)=0.10$ bar. By how much will the number of moles of $\\mathrm{SO}_{2} \\mathrm{Cl}_{2}$ at equilibrium change if the volume is reduced to $0.50 \\mathrm{~L}$ ?\n\nA: Increase $1-10 \\%$\nB: Increase $11-50 \\%$\nC: Decrease 1-10\\%\nD: Decrease 11-50\\%\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_509",
        "problem": "已知在溶液中可发生反应 $\\mathrm{A}+2 \\mathrm{~B}+3 \\mathrm{H}^{+} \\rightleftharpoons \\mathrm{D}+2 \\mathrm{E}+\\mathrm{H}_{2} \\mathrm{O}$ 。在一定温度下, 某实验小组用 $\\mathrm{A} 、 \\mathrm{~B}$ 的混合酸性溶液进行了三组实验, 三组实验中 $\\mathrm{B}$ 和 $\\mathrm{H}^{+}$的浓度相等, $\\mathrm{A}$ 的起始浓度为 $2.0 \\mathrm{~mol} / \\mathrm{L} 、 2.5 \\mathrm{~mol} / \\mathrm{L} 、 3.0 \\mathrm{~mol} / \\mathrm{L}$ 。测得 $\\mathrm{c}(\\mathrm{B})$ 随时间 $\\mathrm{t}$ 的变化曲线如下图,下列说法不正确的是\n\n[图1]\nA: 曲线III起始时 $c(A)=3.0 \\mathrm{~mol} / \\mathrm{L}$\nB: 该条件下三组实验的反应速率均随反应进程先增后减\nC: 起始时 $\\mathrm{c}(\\mathrm{A})=2.5 \\mathrm{~mol} / \\mathrm{L}$ 时, $1 \\sim 2 \\mathrm{~min}$ 内, $\\mathrm{A}$ 的平均反应速率为 $2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nD: 若曲线 $I$ 平衡时的 $\\mathrm{pH}=3$, 则该温度下该反应的平衡常数为 $\\frac{0.65 \\times 1.3^{2}}{1.35 \\times 0.2^{2} \\times 10^{-9}}(\\mathrm{~mol} / \\mathrm{L})^{-3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知在溶液中可发生反应 $\\mathrm{A}+2 \\mathrm{~B}+3 \\mathrm{H}^{+} \\rightleftharpoons \\mathrm{D}+2 \\mathrm{E}+\\mathrm{H}_{2} \\mathrm{O}$ 。在一定温度下, 某实验小组用 $\\mathrm{A} 、 \\mathrm{~B}$ 的混合酸性溶液进行了三组实验, 三组实验中 $\\mathrm{B}$ 和 $\\mathrm{H}^{+}$的浓度相等, $\\mathrm{A}$ 的起始浓度为 $2.0 \\mathrm{~mol} / \\mathrm{L} 、 2.5 \\mathrm{~mol} / \\mathrm{L} 、 3.0 \\mathrm{~mol} / \\mathrm{L}$ 。测得 $\\mathrm{c}(\\mathrm{B})$ 随时间 $\\mathrm{t}$ 的变化曲线如下图,下列说法不正确的是\n\n[图1]\n\nA: 曲线III起始时 $c(A)=3.0 \\mathrm{~mol} / \\mathrm{L}$\nB: 该条件下三组实验的反应速率均随反应进程先增后减\nC: 起始时 $\\mathrm{c}(\\mathrm{A})=2.5 \\mathrm{~mol} / \\mathrm{L}$ 时, $1 \\sim 2 \\mathrm{~min}$ 内, $\\mathrm{A}$ 的平均反应速率为 $2 \\times 10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nD: 若曲线 $I$ 平衡时的 $\\mathrm{pH}=3$, 则该温度下该反应的平衡常数为 $\\frac{0.65 \\times 1.3^{2}}{1.35 \\times 0.2^{2} \\times 10^{-9}}(\\mathrm{~mol} / \\mathrm{L})^{-3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-068.jpg?height=491&width=556&top_left_y=882&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1535",
        "problem": "Hydrogen is more energy-dense than carbon, by mass. Thus, historically there has been a move toward fuel with higher hydrogen content: coal $\\rightarrow$ oil $\\rightarrow$ natural gas $\\rightarrow$ hydrogen. Cost-effective production and safe storage of hydrogen are two major hurdles to the successful inauguration of a hydrogen economy.\n\nConsider hydrogen in a cylinder of $80 \\mathrm{MPa}$ at $25^{\\circ} \\mathrm{C}$. Using the ideal gas law, estimate the density of hydrogen in the cylinder in $\\mathrm{kg} \\mathrm{m}^{-3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nHydrogen is more energy-dense than carbon, by mass. Thus, historically there has been a move toward fuel with higher hydrogen content: coal $\\rightarrow$ oil $\\rightarrow$ natural gas $\\rightarrow$ hydrogen. Cost-effective production and safe storage of hydrogen are two major hurdles to the successful inauguration of a hydrogen economy.\n\nConsider hydrogen in a cylinder of $80 \\mathrm{MPa}$ at $25^{\\circ} \\mathrm{C}$. Using the ideal gas law, estimate the density of hydrogen in the cylinder in $\\mathrm{kg} \\mathrm{m}^{-3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $ \\mathrm{~kg} \\mathrm{~m}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$ \\mathrm{~kg} \\mathrm{~m}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_454",
        "problem": "下列说法正确的是\nA: 分子式为 $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 的有机物在酸性条件下可水解为酸和醇, 若不考虑立体异构,这些醇和酸重新组合可形成的酯共有 12 种\nB: 四联苯\nC: 与 $\\mathrm{CH}_{2}=\\stackrel{\\mathrm{C}_{\\mathrm{C}}}{\\mathrm{CH}_{3}}-\\mathrm{COOH}$ 具有相同官能团的同分异构体的结构简式为 $\\mathrm{CH}_{2}=\\mathrm{CHCH}_{2} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCOOH}$\nD: 兴奋剂乙基雌烯醇( [图1]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列说法正确的是\n\nA: 分子式为 $\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 的有机物在酸性条件下可水解为酸和醇, 若不考虑立体异构,这些醇和酸重新组合可形成的酯共有 12 种\nB: 四联苯\nC: 与 $\\mathrm{CH}_{2}=\\stackrel{\\mathrm{C}_{\\mathrm{C}}}{\\mathrm{CH}_{3}}-\\mathrm{COOH}$ 具有相同官能团的同分异构体的结构简式为 $\\mathrm{CH}_{2}=\\mathrm{CHCH}_{2} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CHCOOH}$\nD: 兴奋剂乙基雌烯醇( [图1]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-56.jpg?height=248&width=942&top_left_y=2309&top_left_x=774"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_457",
        "problem": "$700^{\\circ} \\mathrm{C}$ 时, 向容积为 $2 \\mathrm{~L}$ 的密闭容器中充入一定量的 $\\mathrm{CO}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$, 发生反应:\n\n$\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$ 。反应过程中测定的部分数据见下表 (表中 $\\left.\\mathrm{t}_{2}>\\mathrm{t}_{1}\\right):$\n\n| 反应时间 $/$ min | $\\mathrm{n}(\\mathrm{CO}) /$ mol | $\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{O} /\\right.$ mol $)$ |\n| :--- | :--- | :--- |\n| 0 | 1.20 | 0.60 |\n| $\\mathrm{t}_{1}$ | 0.80 |  |\n| $\\mathrm{t}_{2}$ |  | 0.20 |\nA: 反应在 $\\mathrm{t}_{1} \\mathrm{~min}$ 内的平均速率为 $\\mathrm{v}\\left(\\mathrm{H}_{2}\\right)=\\frac{0.40}{t_{1}} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 保持其他条件不变, 起始时向容器中充入 $0.60 \\mathrm{~mol} \\mathrm{CO}$ 和 $1.20 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}$, 到达平衡时 $\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)=0.40 \\mathrm{~mol}$\nC: 保持其他条件不变, 向平衡体系中再通入 $0.20 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}), \\Delta \\mathrm{H}$ 增大\nD: 温度升高至 $800^{\\circ} \\mathrm{C}$, 上述反应平衡常数为 0.64 , 则正反应为吸热反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$700^{\\circ} \\mathrm{C}$ 时, 向容积为 $2 \\mathrm{~L}$ 的密闭容器中充入一定量的 $\\mathrm{CO}$ 和 $\\mathrm{H}_{2} \\mathrm{O}$, 发生反应:\n\n$\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$ 。反应过程中测定的部分数据见下表 (表中 $\\left.\\mathrm{t}_{2}>\\mathrm{t}_{1}\\right):$\n\n| 反应时间 $/$ min | $\\mathrm{n}(\\mathrm{CO}) /$ mol | $\\mathrm{n}\\left(\\mathrm{H}_{2} \\mathrm{O} /\\right.$ mol $)$ |\n| :--- | :--- | :--- |\n| 0 | 1.20 | 0.60 |\n| $\\mathrm{t}_{1}$ | 0.80 |  |\n| $\\mathrm{t}_{2}$ |  | 0.20 |\n\nA: 反应在 $\\mathrm{t}_{1} \\mathrm{~min}$ 内的平均速率为 $\\mathrm{v}\\left(\\mathrm{H}_{2}\\right)=\\frac{0.40}{t_{1}} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 保持其他条件不变, 起始时向容器中充入 $0.60 \\mathrm{~mol} \\mathrm{CO}$ 和 $1.20 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}$, 到达平衡时 $\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)=0.40 \\mathrm{~mol}$\nC: 保持其他条件不变, 向平衡体系中再通入 $0.20 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}), \\Delta \\mathrm{H}$ 增大\nD: 温度升高至 $800^{\\circ} \\mathrm{C}$, 上述反应平衡常数为 0.64 , 则正反应为吸热反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_234",
        "problem": "Which of the following lists elements in order of increasing first ionisation energy?\nA: $\\mathrm{Na}, \\mathrm{F}, \\mathrm{O}, \\mathrm{N}$\nB: $\\mathrm{Na}, \\mathrm{N}, \\mathrm{O}, \\mathrm{F}$ (half marks)\nC: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{N}, \\mathrm{F}$\nD: $\\mathrm{N}, \\mathrm{O}, \\mathrm{F}, \\mathrm{Na}$\nE: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{F}, \\mathrm{N}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhich of the following lists elements in order of increasing first ionisation energy?\n\nA: $\\mathrm{Na}, \\mathrm{F}, \\mathrm{O}, \\mathrm{N}$\nB: $\\mathrm{Na}, \\mathrm{N}, \\mathrm{O}, \\mathrm{F}$ (half marks)\nC: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{N}, \\mathrm{F}$\nD: $\\mathrm{N}, \\mathrm{O}, \\mathrm{F}, \\mathrm{Na}$\nE: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{F}, \\mathrm{N}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_963",
        "problem": "氨基酸在水溶液中可通过得到或失去 $\\mathrm{H}^{+}$发生如下反应:\n\n[图1]\n\n[图2]\n\n下列说法错误的是\nA: 曲线(1)为 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}$ 的浓度与 $\\mathrm{pH}$ 的关系图\nB: $\\mathrm{pH}=7$ 时, $c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)$\nC: $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}+\\mathrm{H}^{+}$平衡常数的数量级为 $10^{-3}$\nD: $\\mathrm{C}$ 点溶液中满足: $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n氨基酸在水溶液中可通过得到或失去 $\\mathrm{H}^{+}$发生如下反应:\n\n[图1]\n\n[图2]\n\n下列说法错误的是\n\nA: 曲线(1)为 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}$ 的浓度与 $\\mathrm{pH}$ 的关系图\nB: $\\mathrm{pH}=7$ 时, $c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}^{-}\\right)>c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)$\nC: $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}+\\mathrm{H}^{+}$平衡常数的数量级为 $10^{-3}$\nD: $\\mathrm{C}$ 点溶液中满足: $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)=c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-095.jpg?height=180&width=1448&top_left_y=2446&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-096.jpg?height=703&width=1331&top_left_y=274&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_933",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 向一定浓度的 $\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中逐滴加入 $\\mathrm{HCl}$, 以 $\\mathrm{X}$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}$或 $\\frac{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)}$ 。已知 $\\mathrm{pX}=-\\lg \\mathrm{X}$, 混合溶液 $\\mathrm{pX}$ 与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\nA: 曲线II表示 $\\mathrm{p} \\frac{\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}$与 $\\mathrm{pH}$ 的关系\nB: $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 第一步电离常数的数量级为 $10^{-2}$\nC: $\\mathrm{pH}=2.71$ 时, $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: $\\mathrm{pH}=1.23$ 时, $3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 向一定浓度的 $\\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中逐滴加入 $\\mathrm{HCl}$, 以 $\\mathrm{X}$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}$或 $\\frac{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)}$ 。已知 $\\mathrm{pX}=-\\lg \\mathrm{X}$, 混合溶液 $\\mathrm{pX}$ 与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: 曲线II表示 $\\mathrm{p} \\frac{\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)}$与 $\\mathrm{pH}$ 的关系\nB: $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 第一步电离常数的数量级为 $10^{-2}$\nC: $\\mathrm{pH}=2.71$ 时, $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: $\\mathrm{pH}=1.23$ 时, $3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-062.jpg?height=385&width=697&top_left_y=1581&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_688",
        "problem": "7-ACCA 是合成头孢克罗的关键中间体, 其结构如图所示。下列说法正确的是\n[图1]\nA: 该分子中存在 3 个手性碳\nB: 分子中 $\\mathrm{N}$ 原子有 2 种杂化方式\nC: 具有两性, 可发生水解反应\nD: 不存在分子中含有苯环的同分异构体\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n7-ACCA 是合成头孢克罗的关键中间体, 其结构如图所示。下列说法正确的是\n[图1]\n\nA: 该分子中存在 3 个手性碳\nB: 分子中 $\\mathrm{N}$ 原子有 2 种杂化方式\nC: 具有两性, 可发生水解反应\nD: 不存在分子中含有苯环的同分异构体\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/wjwhw64G/image.png",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-21.jpg?height=234&width=597&top_left_y=1990&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_358",
        "problem": "What is $\\Delta G^{\\mathrm{o}}$ of $\\mathrm{HOCl}(g)$ at $298 \\mathrm{~K}$ ?.\n\n$$\n\\begin{aligned}\n\\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{Cl}_{2} \\mathrm{O}(g) \\rightleftharpoons & 2 \\mathrm{HOCl}(g) \\\\\n& K_{\\text {eq }}(298 \\mathrm{~K})=0.089\n\\end{aligned}\n$$\nA: $-6.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-3.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $3.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: It cannot be determined from the information given.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is $\\Delta G^{\\mathrm{o}}$ of $\\mathrm{HOCl}(g)$ at $298 \\mathrm{~K}$ ?.\n\n$$\n\\begin{aligned}\n\\mathrm{H}_{2} \\mathrm{O}(g)+\\mathrm{Cl}_{2} \\mathrm{O}(g) \\rightleftharpoons & 2 \\mathrm{HOCl}(g) \\\\\n& K_{\\text {eq }}(298 \\mathrm{~K})=0.089\n\\end{aligned}\n$$\n\nA: $-6.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-3.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $3.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: It cannot be determined from the information given.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_220",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the amount (in mol or mmol) of potassium hydroxide added in the original $50.00 \\mathrm{~mL}$ sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the amount (in mol or mmol) of potassium hydroxide added in the original $50.00 \\mathrm{~mL}$ sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_100",
        "problem": "If the rate law for an equation $A+B+C \\rightarrow A B C$ is Rate $=k[A]^{0}[B][C]^{2}$ and the reactant concentration of all of the reactants doubles, by what factor does the rate of reaction increase?\nA: 3\nB: 4\nC: 6\nD: 8\nE: 10\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf the rate law for an equation $A+B+C \\rightarrow A B C$ is Rate $=k[A]^{0}[B][C]^{2}$ and the reactant concentration of all of the reactants doubles, by what factor does the rate of reaction increase?\n\nA: 3\nB: 4\nC: 6\nD: 8\nE: 10\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_210",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the mass of water (in g) that must have been added to the KHP to make the 118.48 g solution.\n\nA solution of sodium hydroxide is also prepared. $4.471 \\mathrm{~g}$ of this sodium hydroxide solution reacts completely with $5.979 \\mathrm{~g}$ of the KHP solution above. Sodium hydroxide reacts in a 1:1 mole ratio with KHP.\n\nIn a similar reaction, $4.359 \\mathrm{~g}$ of the sodium hydroxide solution reacts completely with a $5.925 \\mathrm{~g}$ sample of vinegar (containing acetic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$ ). Sodium hydroxide reacts in a 1:1 mole ratio with acetic acid.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the mass of water (in g) that must have been added to the KHP to make the 118.48 g solution.\n\nA solution of sodium hydroxide is also prepared. $4.471 \\mathrm{~g}$ of this sodium hydroxide solution reacts completely with $5.979 \\mathrm{~g}$ of the KHP solution above. Sodium hydroxide reacts in a 1:1 mole ratio with KHP.\n\nIn a similar reaction, $4.359 \\mathrm{~g}$ of the sodium hydroxide solution reacts completely with a $5.925 \\mathrm{~g}$ sample of vinegar (containing acetic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$ ). Sodium hydroxide reacts in a 1:1 mole ratio with acetic acid.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1528",
        "problem": "The equilibrium constant of the reaction:\n\n$\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}(\\mathrm{~s})+2 \\mathrm{Cl}(\\mathrm{aq})^{-} \\rightleftharpoons 2 \\mathrm{AgCl}(\\mathrm{s})+\\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})$\n\nis given by the equation:\nA: $K=K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right) / K_{s p}(\\mathrm{AgCl})^{2}$\nB: $K=K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right) \\times K_{s p}(\\mathrm{AgCl})^{2}$\nC: $K=K_{s p}(\\mathrm{AgCl}) / K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nD: $K=K_{s p}(\\mathrm{AgCl})^{2} / K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nE: $K=K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right) / K_{\\text {sp }}(\\mathrm{AgCl})$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe equilibrium constant of the reaction:\n\n$\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}(\\mathrm{~s})+2 \\mathrm{Cl}(\\mathrm{aq})^{-} \\rightleftharpoons 2 \\mathrm{AgCl}(\\mathrm{s})+\\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})$\n\nis given by the equation:\n\nA: $K=K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right) / K_{s p}(\\mathrm{AgCl})^{2}$\nB: $K=K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right) \\times K_{s p}(\\mathrm{AgCl})^{2}$\nC: $K=K_{s p}(\\mathrm{AgCl}) / K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nD: $K=K_{s p}(\\mathrm{AgCl})^{2} / K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nE: $K=K_{s p}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right) / K_{\\text {sp }}(\\mathrm{AgCl})$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_601",
        "problem": "下列装置可分离废水中的 $\\mathrm{Co}^{2+}$ 和 $\\mathrm{Ni}^{2+}$ 。已知 $\\mathrm{Ni}^{2+}$ 和 $\\mathrm{Co}^{2+}$ 性质相似, $\\mathrm{Co}^{2+}$ 和乙酰丙酮不反应。下列说法正确的是\n[图1]\nA: $\\mathrm{M}$ 电极接太阳能电池的 $\\mathrm{P}$ 电极\nB: 通电一段时间后, I、IV 室内溶液 $\\mathrm{pH}$ 均减小\nC: 膜 $\\mathrm{a} 、$ 膜 b 分别为阳离子交换膜和阴离子交换膜\nD: 每生成 $1 \\mathrm{molSO}_{4}^{2-}$, 理论上双极膜至少解离 $7 \\mathrm{molH}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列装置可分离废水中的 $\\mathrm{Co}^{2+}$ 和 $\\mathrm{Ni}^{2+}$ 。已知 $\\mathrm{Ni}^{2+}$ 和 $\\mathrm{Co}^{2+}$ 性质相似, $\\mathrm{Co}^{2+}$ 和乙酰丙酮不反应。下列说法正确的是\n[图1]\n\nA: $\\mathrm{M}$ 电极接太阳能电池的 $\\mathrm{P}$ 电极\nB: 通电一段时间后, I、IV 室内溶液 $\\mathrm{pH}$ 均减小\nC: 膜 $\\mathrm{a} 、$ 膜 b 分别为阳离子交换膜和阴离子交换膜\nD: 每生成 $1 \\mathrm{molSO}_{4}^{2-}$, 理论上双极膜至少解离 $7 \\mathrm{molH}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-047.jpg?height=1130&width=1108&top_left_y=670&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_315",
        "problem": "12 In the galvanic cell shown below, what is the reaction that occurs at the cathode?\n\n[figure1]\nA: $\\mathrm{H}_{2}(g) \\rightarrow 2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-}$\nB: $2 \\mathrm{H}^{+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{H}_{2}(g)$\nC: $\\mathrm{Cu}(\\mathrm{s}) \\rightarrow \\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-}$\nD: $\\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Cu}(s)$\nE: $\\mathrm{Pt}(s)+\\mathrm{H}_{2}(g)+4 \\mathrm{Cl}^{-}(a q)$ $$ \\rightarrow \\mathrm{PtCl}_{4}^{2-}(\\mathrm{aq})+2 \\mathrm{H}^{+}(\\mathrm{aq})+4 \\mathrm{e}^{-} $$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n12 In the galvanic cell shown below, what is the reaction that occurs at the cathode?\n\n[figure1]\n\nA: $\\mathrm{H}_{2}(g) \\rightarrow 2 \\mathrm{H}^{+}(a q)+2 \\mathrm{e}^{-}$\nB: $2 \\mathrm{H}^{+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{H}_{2}(g)$\nC: $\\mathrm{Cu}(\\mathrm{s}) \\rightarrow \\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-}$\nD: $\\mathrm{Cu}^{2+}(a q)+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Cu}(s)$\nE: $\\mathrm{Pt}(s)+\\mathrm{H}_{2}(g)+4 \\mathrm{Cl}^{-}(a q)$ $$ \\rightarrow \\mathrm{PtCl}_{4}^{2-}(\\mathrm{aq})+2 \\mathrm{H}^{+}(\\mathrm{aq})+4 \\mathrm{e}^{-} $$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_92a3e606264c5ee1789ag-3.jpg?height=537&width=804&top_left_y=295&top_left_x=1146"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_579",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 改变 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 甲酸 $(\\mathrm{HCOOH})$ 与丙酸 $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)$ 溶液的 $\\mathrm{pH}$, 溶液中 $\\mathrm{RCOOH}$ 分子的物质的量分数 $\\delta(\\mathrm{RCOOH})$ 随之改变 $[$ 已知\n\n$\\left.\\delta(\\mathrm{RCOOH})=\\frac{c(\\mathrm{RCOOH})}{c(\\mathrm{RCOOH})+c\\left(\\mathrm{RCOO}^{-}\\right)}\\right], \\delta(\\mathrm{RCOOH})$ 与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\nA: 若弱酸 $\\mathrm{pH}$ 增大是通过向弱酸中加入 $\\mathrm{NaOH}$ 固体实现的, 则图中 $\\mathrm{M} 、 \\mathrm{~N}$ 两点对应溶液中的 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$相等\nB: 对于甲酸和丙酸, 当 $\\lg \\frac{c(\\mathrm{RCOOH})}{c\\left(\\mathrm{RCOO}^{-}\\right)}>0$ 时, 溶液都为酸性\nC: 等浓度的 $\\mathrm{HCOONa}$ 和 $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COONa}$ 两种溶液的 $\\mathrm{pH}$ ：前者>后者\nD: 将 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOOH}$ 溶液与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOONa}$ 溶液等体积混合, 所得溶液中: $c\\left(\\mathrm{Na}^{+}\\right)>c(\\mathrm{HCOOH})>c\\left(\\mathrm{HCOO}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 改变 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 甲酸 $(\\mathrm{HCOOH})$ 与丙酸 $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)$ 溶液的 $\\mathrm{pH}$, 溶液中 $\\mathrm{RCOOH}$ 分子的物质的量分数 $\\delta(\\mathrm{RCOOH})$ 随之改变 $[$ 已知\n\n$\\left.\\delta(\\mathrm{RCOOH})=\\frac{c(\\mathrm{RCOOH})}{c(\\mathrm{RCOOH})+c\\left(\\mathrm{RCOO}^{-}\\right)}\\right], \\delta(\\mathrm{RCOOH})$ 与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: 若弱酸 $\\mathrm{pH}$ 增大是通过向弱酸中加入 $\\mathrm{NaOH}$ 固体实现的, 则图中 $\\mathrm{M} 、 \\mathrm{~N}$ 两点对应溶液中的 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$相等\nB: 对于甲酸和丙酸, 当 $\\lg \\frac{c(\\mathrm{RCOOH})}{c\\left(\\mathrm{RCOO}^{-}\\right)}>0$ 时, 溶液都为酸性\nC: 等浓度的 $\\mathrm{HCOONa}$ 和 $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{COONa}$ 两种溶液的 $\\mathrm{pH}$ ：前者>后者\nD: 将 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOOH}$ 溶液与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOONa}$ 溶液等体积混合, 所得溶液中: $c\\left(\\mathrm{Na}^{+}\\right)>c(\\mathrm{HCOOH})>c\\left(\\mathrm{HCOO}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-022.jpg?height=369&width=711&top_left_y=1483&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_538",
        "problem": "C- $\\mathrm{C}$ 单键可以绕键轴旋转, 其结构简式如下可表示为的烃, 下列说法中正确的是\n\n[图1]\nA: 该烃是苯的同系物\nB: 分子中最多有 6 个碳原子处于同一直 线上\nC: 该烃的一氯代物最多有四种\nD: 分子中至少有 10 个碳原子处于同一平面上\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\nC- $\\mathrm{C}$ 单键可以绕键轴旋转, 其结构简式如下可表示为的烃, 下列说法中正确的是\n\n[图1]\n\nA: 该烃是苯的同系物\nB: 分子中最多有 6 个碳原子处于同一直 线上\nC: 该烃的一氯代物最多有四种\nD: 分子中至少有 10 个碳原子处于同一平面上\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-79.jpg?height=117&width=465&top_left_y=798&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_27",
        "problem": "What is the bond order of carbon monoxide, $\\mathrm{CO}$ ?\nA: 1.5\nB: 2.0\nC: 2.5\nD: 3.0\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the bond order of carbon monoxide, $\\mathrm{CO}$ ?\n\nA: 1.5\nB: 2.0\nC: 2.5\nD: 3.0\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_206",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the molar mass of the mineral with formula $\\mathrm{Pb}_{10} \\mathrm{O}_{3}\\left(\\mathrm{CO}_{3}\\right)_{6}(\\mathrm{OH})_{2}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the molar mass of the mineral with formula $\\mathrm{Pb}_{10} \\mathrm{O}_{3}\\left(\\mathrm{CO}_{3}\\right)_{6}(\\mathrm{OH})_{2}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g/mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g/mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_872",
        "problem": "化合物 $\\mathrm{A}$ 经李比希法和质谱法分析得知其相对分子质量为 136 , 分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 。\n\n$A$ 的核磁共振氢谱有 4 个峰且面积之比为 1: 2: $2: 3$, A 分子中只含一个苯环且苯环\n\n上只有一个取代基, 其核磁共振氢谱与红外光谱如图。关于 $\\mathrm{A}$ 的下列说法中, 正确的\n\n是\n[图1]\nA: 分子属于酯类化合物, 在一定条件下能发生水解反应\nB: A 在一定条件下可与 $4 \\mathrm{~mol} \\mathrm{H}_{2}$ 发生加成反应\nC: 符合题中 A 分子结构特征的有机物有两种\nD: 与 $\\mathrm{A}$ 属于同类化合物的同分异构体有 5 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化合物 $\\mathrm{A}$ 经李比希法和质谱法分析得知其相对分子质量为 136 , 分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 。\n\n$A$ 的核磁共振氢谱有 4 个峰且面积之比为 1: 2: $2: 3$, A 分子中只含一个苯环且苯环\n\n上只有一个取代基, 其核磁共振氢谱与红外光谱如图。关于 $\\mathrm{A}$ 的下列说法中, 正确的\n\n是\n[图1]\n\nA: 分子属于酯类化合物, 在一定条件下能发生水解反应\nB: A 在一定条件下可与 $4 \\mathrm{~mol} \\mathrm{H}_{2}$ 发生加成反应\nC: 符合题中 A 分子结构特征的有机物有两种\nD: 与 $\\mathrm{A}$ 属于同类化合物的同分异构体有 5 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-59.jpg?height=376&width=1264&top_left_y=886&top_left_x=332"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_560",
        "problem": "已知 $T^{\\circ} \\mathrm{C}$ 时, $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液的 $\\mathrm{pH}=a, 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HA}$ 溶液中\n$\\frac{c\\left(\\mathrm{H}^{+}\\right)}{c\\left(\\mathrm{OH}^{-}\\right)}=10^{b}$, 则 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HA}$ 溶液中水电离出的 $c\\left(\\mathrm{OH}^{-}\\right)$为\nA: $10^{-(13-a+b)} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\nB: $10^{-(11+a-b)} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\nC: $10^{-\\frac{a+b+2}{2}} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $10^{-\\frac{b+a}{2}} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知 $T^{\\circ} \\mathrm{C}$ 时, $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液的 $\\mathrm{pH}=a, 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HA}$ 溶液中\n$\\frac{c\\left(\\mathrm{H}^{+}\\right)}{c\\left(\\mathrm{OH}^{-}\\right)}=10^{b}$, 则 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HA}$ 溶液中水电离出的 $c\\left(\\mathrm{OH}^{-}\\right)$为\n\nA: $10^{-(13-a+b)} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\nB: $10^{-(11+a-b)} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\nC: $10^{-\\frac{a+b+2}{2}} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $10^{-\\frac{b+a}{2}} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_889",
        "problem": "现有 $200 \\mathrm{~mL}$ 含 $\\mathrm{KNO}_{3}$ 和 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 的混合溶液, 其中 $\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)=3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 用石墨作电极电解此溶液, 当通电一段时间后, 两极均收集到 $2.24 \\mathrm{~L}$ 气体（标准状况）。假定电解后溶液体积仍为 $200 \\mathrm{~mL}$ ，下列说法不正确的是\nA: 电解过程中共转移 $0.4 \\mathrm{~mol}$ 电子\nB: 原混合液中 $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)$为 $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: 电解得到的 $\\mathrm{Cu}$ 的质量为 $6.4 \\mathrm{~g}$\nD: 电解后溶液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$为 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n现有 $200 \\mathrm{~mL}$ 含 $\\mathrm{KNO}_{3}$ 和 $\\mathrm{Cu}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 的混合溶液, 其中 $\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)=3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 用石墨作电极电解此溶液, 当通电一段时间后, 两极均收集到 $2.24 \\mathrm{~L}$ 气体（标准状况）。假定电解后溶液体积仍为 $200 \\mathrm{~mL}$ ，下列说法不正确的是\n\nA: 电解过程中共转移 $0.4 \\mathrm{~mol}$ 电子\nB: 原混合液中 $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)$为 $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: 电解得到的 $\\mathrm{Cu}$ 的质量为 $6.4 \\mathrm{~g}$\nD: 电解后溶液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$为 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_297",
        "problem": "Which of the following is a valid set of quantum numbers for an electron in a $p$ orbital?\nA: $n=1, l=1, m_{l}=0, m_{s}=1 / 2$\nB: $n=3, I=1, m_{l}=2, m_{s}=1 / 2$\nC: $n=2, l=1, m_{l}=-1, m_{s}=1 / 2$\nD: $n=2, l=0, m_{l}=0, m_{s}=1 / 2$\nE: $n=2, I=2, m_{l}=0, m_{s}=1 / 2$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is a valid set of quantum numbers for an electron in a $p$ orbital?\n\nA: $n=1, l=1, m_{l}=0, m_{s}=1 / 2$\nB: $n=3, I=1, m_{l}=2, m_{s}=1 / 2$\nC: $n=2, l=1, m_{l}=-1, m_{s}=1 / 2$\nD: $n=2, l=0, m_{l}=0, m_{s}=1 / 2$\nE: $n=2, I=2, m_{l}=0, m_{s}=1 / 2$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_296",
        "problem": "In which of the following series are the atomic orbitals given in order of increasing energy?\nA: $3 \\mathrm{~d}, 4 \\mathrm{~s}, 4 \\mathrm{p}, 4 \\mathrm{~d}, 4 \\mathrm{f}, 5 \\mathrm{~s}$\nB: $2 \\mathrm{~s}, 3 \\mathrm{~s}, 2 \\mathrm{p}, 3 \\mathrm{p}, 3 \\mathrm{~d}, 4 \\mathrm{~s}$\nC: $4 \\mathrm{~s}, 3 \\mathrm{~d}, 4 \\mathrm{p}, 4 \\mathrm{~d}, 4 \\mathrm{f}, 5 \\mathrm{~s}$\nD: 4s, 3d, 4p, 5s, 4d, 5p\nE: $1 \\mathrm{~s}, 2 \\mathrm{~s}, 3 \\mathrm{~s}, 4 \\mathrm{~s}, 2 \\mathrm{p}, 3 \\mathrm{p}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn which of the following series are the atomic orbitals given in order of increasing energy?\n\nA: $3 \\mathrm{~d}, 4 \\mathrm{~s}, 4 \\mathrm{p}, 4 \\mathrm{~d}, 4 \\mathrm{f}, 5 \\mathrm{~s}$\nB: $2 \\mathrm{~s}, 3 \\mathrm{~s}, 2 \\mathrm{p}, 3 \\mathrm{p}, 3 \\mathrm{~d}, 4 \\mathrm{~s}$\nC: $4 \\mathrm{~s}, 3 \\mathrm{~d}, 4 \\mathrm{p}, 4 \\mathrm{~d}, 4 \\mathrm{f}, 5 \\mathrm{~s}$\nD: 4s, 3d, 4p, 5s, 4d, 5p\nE: $1 \\mathrm{~s}, 2 \\mathrm{~s}, 3 \\mathrm{~s}, 4 \\mathrm{~s}, 2 \\mathrm{p}, 3 \\mathrm{p}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_30",
        "problem": "How many isomers are there with the formula $\\mathrm{C}_{6} \\mathrm{H}_{14}$ ?\nA: 3\nB: 4\nC: 5\nD: 6\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many isomers are there with the formula $\\mathrm{C}_{6} \\mathrm{H}_{14}$ ?\n\nA: 3\nB: 4\nC: 5\nD: 6\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_841",
        "problem": "化合物 $Z_{3}\\left[M\\left(X_{2} Y_{4}\\right)_{3}\\right] \\cdot 3 W_{2} Y$ 由位于前四周期且原子序数依次增大的 $W 、 X 、 Y 、 Z 、$ $M$ 五种元素组成。 $M$ 在地壳中的含量居第四位且其原子序数等于 $W 、 X 、 Z$ 的原子序数之和。该化合物的热重曲线如图所示, $150^{\\circ} \\mathrm{C} \\sim 600^{\\circ} \\mathrm{C}$ 阶段, 只有 $\\mathrm{XY}$ 和 $\\mathrm{XY}$ 两种无刺激性气体逸出, $600^{\\circ} \\mathrm{C}$ 所得固体为 $\\mathrm{Z}_{2} \\mathrm{XY}_{3}$, 和氧化物 $\\mathrm{Q}$ 的混合物。下列说法错误的是\n\n[图1]\nA: X 的含氧酸不能使酸性高锰酸钾溶液裉色\nB: $150^{\\circ} \\mathrm{C} \\sim 600^{\\circ} \\mathrm{C}$ 阶段热分解生成的 $\\mathrm{XY}$ 和 $\\mathrm{XY}{ }_{2}$ 的物质的量之比为 $5: 4$\nC: 氧化物 $\\mathrm{Q}$ 为 $\\mathrm{MY}$\nD: 由元素 $\\mathrm{Y}$ 和 Z 组成的化合物可能含有共价键\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n化合物 $Z_{3}\\left[M\\left(X_{2} Y_{4}\\right)_{3}\\right] \\cdot 3 W_{2} Y$ 由位于前四周期且原子序数依次增大的 $W 、 X 、 Y 、 Z 、$ $M$ 五种元素组成。 $M$ 在地壳中的含量居第四位且其原子序数等于 $W 、 X 、 Z$ 的原子序数之和。该化合物的热重曲线如图所示, $150^{\\circ} \\mathrm{C} \\sim 600^{\\circ} \\mathrm{C}$ 阶段, 只有 $\\mathrm{XY}$ 和 $\\mathrm{XY}$ 两种无刺激性气体逸出, $600^{\\circ} \\mathrm{C}$ 所得固体为 $\\mathrm{Z}_{2} \\mathrm{XY}_{3}$, 和氧化物 $\\mathrm{Q}$ 的混合物。下列说法错误的是\n\n[图1]\n\nA: X 的含氧酸不能使酸性高锰酸钾溶液裉色\nB: $150^{\\circ} \\mathrm{C} \\sim 600^{\\circ} \\mathrm{C}$ 阶段热分解生成的 $\\mathrm{XY}$ 和 $\\mathrm{XY}{ }_{2}$ 的物质的量之比为 $5: 4$\nC: 氧化物 $\\mathrm{Q}$ 为 $\\mathrm{MY}$\nD: 由元素 $\\mathrm{Y}$ 和 Z 组成的化合物可能含有共价键\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-04.jpg?height=511&width=782&top_left_y=658&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_873",
        "problem": "科研工作者用金催化电极实现了常温、常压条件下合成氨, 其工作原理示意图如图所示。下列说法错误的是\n\n[图1]\nA: $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{LiCF}_{3} \\mathrm{SO}_{3}$ 的主要作用均为增强溶液导电性\nB: 交换膜适合选择质子交换膜\nC: 工作时，两极区溶液的浓度均保持不变\nD: 理论上反应消耗的 $\\mathrm{N}_{2}$ 与惰性电极生成的气体的物质的量之比为 $2: 3$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n科研工作者用金催化电极实现了常温、常压条件下合成氨, 其工作原理示意图如图所示。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{LiCF}_{3} \\mathrm{SO}_{3}$ 的主要作用均为增强溶液导电性\nB: 交换膜适合选择质子交换膜\nC: 工作时，两极区溶液的浓度均保持不变\nD: 理论上反应消耗的 $\\mathrm{N}_{2}$ 与惰性电极生成的气体的物质的量之比为 $2: 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-003.jpg?height=388&width=848&top_left_y=637&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1437",
        "problem": "In Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nThe total concentration of carbon dioxide in water which is saturated with carbon dioxide at a carbon dioxide partial pressure of 1.00 bar is $0.0752 \\mathrm{~mol} \\mathrm{dm}^{-3}$. Calculate the volume of carbon dioxide gas which can be dissolved in one litre of water under these conditions.\n\nThe gas constant $\\quad R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIn Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nThe total concentration of carbon dioxide in water which is saturated with carbon dioxide at a carbon dioxide partial pressure of 1.00 bar is $0.0752 \\mathrm{~mol} \\mathrm{dm}^{-3}$. Calculate the volume of carbon dioxide gas which can be dissolved in one litre of water under these conditions.\n\nThe gas constant $\\quad R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{dm}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-362.jpg?height=166&width=1354&top_left_y=1005&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{dm}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_567",
        "problem": "5-羟色胺对人体睡眠具有调节作用, 其结构简式如图所示。\n\n[图1]\n\n下列有关 5 -羟色胺说法错误的是\nA: 该有机物分子式为 $\\mathrm{C}_{11} \\mathrm{H}_{10} \\mathrm{~N}_{2} \\mathrm{O}_{3}$\nB: 该有机物既能和 $\\mathrm{NaOH}$ 反应, 也能和 $\\mathrm{HCl}$ 反应\nC: 分子中含有 1 个手性碳原子\nD: $1 \\mathrm{~mol}$ 该物质最多可与 $4 \\mathrm{molH}_{2}$ 或 $1 \\mathrm{molBr}_{2}$ 发生反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n5-羟色胺对人体睡眠具有调节作用, 其结构简式如图所示。\n\n[图1]\n\n下列有关 5 -羟色胺说法错误的是\n\nA: 该有机物分子式为 $\\mathrm{C}_{11} \\mathrm{H}_{10} \\mathrm{~N}_{2} \\mathrm{O}_{3}$\nB: 该有机物既能和 $\\mathrm{NaOH}$ 反应, 也能和 $\\mathrm{HCl}$ 反应\nC: 分子中含有 1 个手性碳原子\nD: $1 \\mathrm{~mol}$ 该物质最多可与 $4 \\mathrm{molH}_{2}$ 或 $1 \\mathrm{molBr}_{2}$ 发生反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-80.jpg?height=231&width=311&top_left_y=2092&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_681",
        "problem": "在 $25 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaOH}$ 溶液中逐滴加入 $0.2 \\mathrm{~mol} / \\mathrm{L}$ 醋酸溶液, 曲线如图所示,有关粒子的浓度关系正确的是\n\n[图1]\nA: 在 A、B 间任一点, 溶液中一定都有 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 在 B 点, $a>12.5$, 且有 $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)=c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)$\nC: 在 $\\mathrm{C}$ 点: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: 在 D 点: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在 $25 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaOH}$ 溶液中逐滴加入 $0.2 \\mathrm{~mol} / \\mathrm{L}$ 醋酸溶液, 曲线如图所示,有关粒子的浓度关系正确的是\n\n[图1]\n\nA: 在 A、B 间任一点, 溶液中一定都有 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 在 B 点, $a>12.5$, 且有 $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)=c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)$\nC: 在 $\\mathrm{C}$ 点: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: 在 D 点: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-73.jpg?height=357&width=505&top_left_y=1740&top_left_x=364"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1480",
        "problem": "In Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nThe solubility of calcium carbonate in water at $0^{\\circ} \\mathrm{C}$ is $0.0012 \\mathrm{~g}$ per $100 \\mathrm{~cm} 3$ of water. Calculate the concentration of calcium ions in a saturated solution of calcium carbonate in water.\n\nThe hard groundwater in Denmark is formed via contact of water with limestone in the subsoil which reacts with carbon dioxide dissolved in the groundwater according to the equilibrium equation:\n\n$\\mathrm{CaCO}_{3}(\\mathrm{~s})+\\mathrm{CO}_{2}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{Ca}^{2+}(\\mathrm{aq})+2 \\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$\n\nThe equilibrium constant, $K$, for this reaction is $10^{-4.25}$ at $0^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIn Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nThe solubility of calcium carbonate in water at $0^{\\circ} \\mathrm{C}$ is $0.0012 \\mathrm{~g}$ per $100 \\mathrm{~cm} 3$ of water. Calculate the concentration of calcium ions in a saturated solution of calcium carbonate in water.\n\nThe hard groundwater in Denmark is formed via contact of water with limestone in the subsoil which reacts with carbon dioxide dissolved in the groundwater according to the equilibrium equation:\n\n$\\mathrm{CaCO}_{3}(\\mathrm{~s})+\\mathrm{CO}_{2}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{Ca}^{2+}(\\mathrm{aq})+2 \\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$\n\nThe equilibrium constant, $K$, for this reaction is $10^{-4.25}$ at $0^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-362.jpg?height=166&width=1354&top_left_y=1005&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1582",
        "problem": "In the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.\n\nTwo identical samples of YBCO with an unknown value of $\\delta$ were prepared. The first sample was dissolved in $5 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$, evolving $\\mathrm{O}_{2}$. After boiling to expel gases, cooling, and addition of $10 \\mathrm{~cm}^{3}$ of $\\mathrm{KI}$ solution ( $c=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) under Ar, titration with thiosulfate to the starch endpoint required $1.542 \\cdot 10^{-4}$ mol thiosulfate. The second sample of YBCO was added under Ar directly to $7 \\mathrm{~cm}^{3}$ of a solution in which $c(\\mathrm{KI})=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}$ and $\\mathrm{c}(\\mathrm{HCl})=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$. Titration of this solution required $1.696 \\cdot 10^{-4} \\mathrm{~mol}$ thiosulfate to reach the endpoint.Calculate the value of $\\delta$ for these samples of YBCO.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.\n\nTwo identical samples of YBCO with an unknown value of $\\delta$ were prepared. The first sample was dissolved in $5 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$, evolving $\\mathrm{O}_{2}$. After boiling to expel gases, cooling, and addition of $10 \\mathrm{~cm}^{3}$ of $\\mathrm{KI}$ solution ( $c=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) under Ar, titration with thiosulfate to the starch endpoint required $1.542 \\cdot 10^{-4}$ mol thiosulfate. The second sample of YBCO was added under Ar directly to $7 \\mathrm{~cm}^{3}$ of a solution in which $c(\\mathrm{KI})=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}$ and $\\mathrm{c}(\\mathrm{HCl})=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$. Titration of this solution required $1.696 \\cdot 10^{-4} \\mathrm{~mol}$ thiosulfate to reach the endpoint.\n\nproblem:\nCalculate the value of $\\delta$ for these samples of YBCO.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_684",
        "problem": "使用镍基电催化剂、二茂铁(简写为 $\\mathrm{Cp}_{2} \\mathrm{Fe}$ ) 作牺牲电子供体, 同时使用 $\\mathrm{NH}_{4} \\mathrm{PF}_{6}$ 作电解质和质子供体, 乙腈 $\\left(\\mathrm{CH}_{3} \\mathrm{CN}\\right)$ 作溶剂, 可实现连续非水流通池中有效的催化 $\\mathrm{CO}_{2}$ 还原,装置如下图所示:\n\n[图1]\n\n已知:\n[图2]\n\n二茂铁\n\n下列说法错误的是\nA: 选用乙腈作溶剂, 与 $\\mathrm{CO}_{2} 、 \\mathrm{NH}_{4} \\mathrm{PF}_{6}$ 等在其中的溶解性有关\nB: 使用的交换膜是质子交换膜\nC: 该装置工作时, 阴极上还可能有 $\\mathrm{H}_{2}$ 产生\nD: 该电池反应为 $2 \\mathrm{Cp}_{2} \\mathrm{Fe}+\\mathrm{CO}_{2}+2 \\mathrm{NH}_{4} \\mathrm{PF}_{6} \\stackrel{\\text { 电催化剂 }}{=} 2 \\mathrm{Cp}_{2} \\mathrm{FePF}_{6}+\\mathrm{CO}+2 \\mathrm{NH}_{3}+\\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n使用镍基电催化剂、二茂铁(简写为 $\\mathrm{Cp}_{2} \\mathrm{Fe}$ ) 作牺牲电子供体, 同时使用 $\\mathrm{NH}_{4} \\mathrm{PF}_{6}$ 作电解质和质子供体, 乙腈 $\\left(\\mathrm{CH}_{3} \\mathrm{CN}\\right)$ 作溶剂, 可实现连续非水流通池中有效的催化 $\\mathrm{CO}_{2}$ 还原,装置如下图所示:\n\n[图1]\n\n已知:\n[图2]\n\n二茂铁\n\n下列说法错误的是\n\nA: 选用乙腈作溶剂, 与 $\\mathrm{CO}_{2} 、 \\mathrm{NH}_{4} \\mathrm{PF}_{6}$ 等在其中的溶解性有关\nB: 使用的交换膜是质子交换膜\nC: 该装置工作时, 阴极上还可能有 $\\mathrm{H}_{2}$ 产生\nD: 该电池反应为 $2 \\mathrm{Cp}_{2} \\mathrm{Fe}+\\mathrm{CO}_{2}+2 \\mathrm{NH}_{4} \\mathrm{PF}_{6} \\stackrel{\\text { 电催化剂 }}{=} 2 \\mathrm{Cp}_{2} \\mathrm{FePF}_{6}+\\mathrm{CO}+2 \\mathrm{NH}_{3}+\\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-070.jpg?height=414&width=371&top_left_y=1432&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-070.jpg?height=208&width=646&top_left_y=1853&top_left_x=430"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1468",
        "problem": "Beach sand mineral, monazite, is a rich source of thorium, available in large quantities in the state of Kerala in India. A typical monazite sample contains about $9 \\%$ $\\mathrm{ThO}_{2}$ and $0.35 \\% \\mathrm{U}_{3} \\mathrm{O}_{8} \\cdot{ }^{208} \\mathrm{~Pb}$ a ${ }^{206} \\mathrm{~Pb}$ are the stable end-products in the radioactive decay series of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$, respectively. All the lead $(\\mathrm{Pb})$ found in monazite is of radiogenic origin.\n\nThe isotopic atom ratio ${ }^{208} \\mathrm{~Pb} /{ }^{232} \\mathrm{Th}$, measured mass spectrometrically, in a monazite sample was found to be 0.104 . The half-lives of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U} 1.41 \\times 10^{10}$ years and $4.47 \\times 10^{9}$ years, respectively. Assume that ${ }^{208} \\mathrm{~Pb},{ }^{206} \\mathrm{~Pb},{ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$ remained entirely in the monazite sample since the formation of monazite mineral.\n\nEstimate the isotopic atom ratio ${ }^{206} \\mathrm{~Pb} /{ }^{238} \\mathrm{U}$ in the monazite sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBeach sand mineral, monazite, is a rich source of thorium, available in large quantities in the state of Kerala in India. A typical monazite sample contains about $9 \\%$ $\\mathrm{ThO}_{2}$ and $0.35 \\% \\mathrm{U}_{3} \\mathrm{O}_{8} \\cdot{ }^{208} \\mathrm{~Pb}$ a ${ }^{206} \\mathrm{~Pb}$ are the stable end-products in the radioactive decay series of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$, respectively. All the lead $(\\mathrm{Pb})$ found in monazite is of radiogenic origin.\n\nThe isotopic atom ratio ${ }^{208} \\mathrm{~Pb} /{ }^{232} \\mathrm{Th}$, measured mass spectrometrically, in a monazite sample was found to be 0.104 . The half-lives of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U} 1.41 \\times 10^{10}$ years and $4.47 \\times 10^{9}$ years, respectively. Assume that ${ }^{208} \\mathrm{~Pb},{ }^{206} \\mathrm{~Pb},{ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$ remained entirely in the monazite sample since the formation of monazite mineral.\n\nEstimate the isotopic atom ratio ${ }^{206} \\mathrm{~Pb} /{ }^{238} \\mathrm{U}$ in the monazite sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_597",
        "problem": "某浓度的二元弱酸 $\\mathrm{H}_{2} \\mathrm{~B}$ 溶液在不同 $\\mathrm{pH}$ 下, 测得 $\\mathrm{p} c(\\mathrm{M})$ 变化如图所示, (已知: $\\mathrm{p} c(\\mathrm{M})=-\\lg c(\\mathrm{M}), \\mathrm{M}$ 代指 $\\mathrm{H}_{2} \\mathrm{~B}$ 或 $\\mathrm{HB}^{-}$或 $\\left.\\mathrm{B}^{2-}\\right)$, 下列说法正确的是\n\n[图1]\nA: 曲线 II 表示 $\\mathrm{p} c\\left(\\mathrm{HB}^{-}\\right)$与 $\\mathrm{pH}$ 关系\nB: $\\mathrm{pH}=5$ 时, $\\mathrm{c}\\left(\\mathrm{HB}^{-}\\right)>c\\left(\\mathrm{~B}^{2-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)$\nC: 由图象数据可以计算出 $\\frac{c^{2}\\left(\\mathrm{HB}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right) \\cdot c\\left(\\mathrm{~B}^{2-}\\right)}$ 的值\nD: 在 $\\mathrm{pH}$ 增大的过程中, $c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right) 、 c\\left(\\mathrm{HB}^{-}\\right) 、 c\\left(\\mathrm{~B}^{2-}\\right)$ 三者浓度和先减小后增大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某浓度的二元弱酸 $\\mathrm{H}_{2} \\mathrm{~B}$ 溶液在不同 $\\mathrm{pH}$ 下, 测得 $\\mathrm{p} c(\\mathrm{M})$ 变化如图所示, (已知: $\\mathrm{p} c(\\mathrm{M})=-\\lg c(\\mathrm{M}), \\mathrm{M}$ 代指 $\\mathrm{H}_{2} \\mathrm{~B}$ 或 $\\mathrm{HB}^{-}$或 $\\left.\\mathrm{B}^{2-}\\right)$, 下列说法正确的是\n\n[图1]\n\nA: 曲线 II 表示 $\\mathrm{p} c\\left(\\mathrm{HB}^{-}\\right)$与 $\\mathrm{pH}$ 关系\nB: $\\mathrm{pH}=5$ 时, $\\mathrm{c}\\left(\\mathrm{HB}^{-}\\right)>c\\left(\\mathrm{~B}^{2-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)$\nC: 由图象数据可以计算出 $\\frac{c^{2}\\left(\\mathrm{HB}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right) \\cdot c\\left(\\mathrm{~B}^{2-}\\right)}$ 的值\nD: 在 $\\mathrm{pH}$ 增大的过程中, $c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right) 、 c\\left(\\mathrm{HB}^{-}\\right) 、 c\\left(\\mathrm{~B}^{2-}\\right)$ 三者浓度和先减小后增大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-068.jpg?height=391&width=851&top_left_y=176&top_left_x=357",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-068.jpg?height=60&width=1370&top_left_y=1329&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_276",
        "problem": "An element, $Z$, has 5 valence electrons. Which of the following species is likely to be the most stable?\nA: $\\mathrm{Z}_{3} \\mathrm{O}$\nB: $\\mathrm{ZH}_{3}$\nC: $\\mathrm{Z}^{3+}$\nD: $\\mathrm{Z}^{5+}$\nE: $\\mathrm{Mg}_{2} \\mathrm{Z}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn element, $Z$, has 5 valence electrons. Which of the following species is likely to be the most stable?\n\nA: $\\mathrm{Z}_{3} \\mathrm{O}$\nB: $\\mathrm{ZH}_{3}$\nC: $\\mathrm{Z}^{3+}$\nD: $\\mathrm{Z}^{5+}$\nE: $\\mathrm{Mg}_{2} \\mathrm{Z}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_644",
        "problem": "常温下, 下列关于电解质溶液的说法正确的是\nA: $\\mathrm{pH}=12$ 的氨水和氢氧化钠溶液按体积比 1:9 混合后, $\\mathrm{pH}$ 值不变\nB: 等体积、等浓度的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 溶液混合后, $\\frac{\\mathrm{c}^{2}\\left(\\mathrm{HCO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)}<1$\nC: 用氢氧化钠标准溶液滴定稀磷酸, 使用甲基橙作为指示剂, 则达到滴定终点所得溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HPO}_{4}^{2-}\\right)+3 \\mathrm{c}\\left(\\mathrm{PO}_{4}^{3-}\\right)$\nD: 等浓度的氨水和硫酸按体积比 $3: 1$ 混合后溶液显碱性, 则混合溶液 $$ 2 \\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right) $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 下列关于电解质溶液的说法正确的是\n\nA: $\\mathrm{pH}=12$ 的氨水和氢氧化钠溶液按体积比 1:9 混合后, $\\mathrm{pH}$ 值不变\nB: 等体积、等浓度的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 溶液混合后, $\\frac{\\mathrm{c}^{2}\\left(\\mathrm{HCO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)}<1$\nC: 用氢氧化钠标准溶液滴定稀磷酸, 使用甲基橙作为指示剂, 则达到滴定终点所得溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HPO}_{4}^{2-}\\right)+3 \\mathrm{c}\\left(\\mathrm{PO}_{4}^{3-}\\right)$\nD: 等浓度的氨水和硫酸按体积比 $3: 1$ 混合后溶液显碱性, 则混合溶液 $$ 2 \\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right) $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1072",
        "problem": "Nescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nGiven the standard enthalpies of formation of calcium hydroxide, calcium oxide and water are $-1003,-635$ and $-286 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ respectively, calculate the standard molar enthalpy change for the reaction in $\\mathrm{CaO}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nNescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nGiven the standard enthalpies of formation of calcium hydroxide, calcium oxide and water are $-1003,-635$ and $-286 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ respectively, calculate the standard molar enthalpy change for the reaction in $\\mathrm{CaO}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{Ca}(\\mathrm{OH})_{2}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathbf{k J ~ m o l}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-03.jpg?height=417&width=699&top_left_y=385&top_left_x=1021"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathbf{k J ~ m o l}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_540",
        "problem": "某羧酸的衍生物 $\\mathrm{X}$, 能发生如图所示的反应 $(\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C} 、 \\mathrm{D}$ 均为有机物 $)$, 下列说法不正确的是\n\n[图1]\nA: 若 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{2}, \\mathrm{BC}$ 均不能发生银镜反应, 则 $\\mathrm{X}$ 可能的结构有 2 种\nB: 若 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{2}, \\mathrm{BC}$ 均可以发生银镜反应, 则 $\\mathrm{X}$ 可能的结构有 4 种\nC: 若 $\\mathrm{B}$ 依次与新制 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 、稀硫酸反应后得到 $\\mathrm{C}$, 且 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{O}_{2}$, 则 $\\mathrm{X}$ 可能的结构有 2 种\nD: 若 $\\mathrm{B}$ 依次与新制 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 、稀硫酸反应后得到 $\\mathrm{C}$ 的同分异构体 $\\mathrm{E}$, 且 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{O}_{2}$, 则 $\\mathrm{X}$ 可能的结构有 3 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某羧酸的衍生物 $\\mathrm{X}$, 能发生如图所示的反应 $(\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C} 、 \\mathrm{D}$ 均为有机物 $)$, 下列说法不正确的是\n\n[图1]\n\nA: 若 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{2}, \\mathrm{BC}$ 均不能发生银镜反应, 则 $\\mathrm{X}$ 可能的结构有 2 种\nB: 若 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{2}, \\mathrm{BC}$ 均可以发生银镜反应, 则 $\\mathrm{X}$ 可能的结构有 4 种\nC: 若 $\\mathrm{B}$ 依次与新制 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 、稀硫酸反应后得到 $\\mathrm{C}$, 且 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{O}_{2}$, 则 $\\mathrm{X}$ 可能的结构有 2 种\nD: 若 $\\mathrm{B}$ 依次与新制 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 、稀硫酸反应后得到 $\\mathrm{C}$ 的同分异构体 $\\mathrm{E}$, 且 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{O}_{2}$, 则 $\\mathrm{X}$ 可能的结构有 3 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-74.jpg?height=206&width=600&top_left_y=1893&top_left_x=405"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_160",
        "problem": "What is the mole percent of a solution of ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)$ which consists of $71.0 \\mathrm{~g}$ of ethanol for every $12.8 \\mathrm{~g}$ of water present?\nA: $2.17 \\%$\nB: $12.3 \\%$\nC: $31.6 \\%$\nD: $68.4 \\%$\nE: $84.7 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the mole percent of a solution of ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)$ which consists of $71.0 \\mathrm{~g}$ of ethanol for every $12.8 \\mathrm{~g}$ of water present?\n\nA: $2.17 \\%$\nB: $12.3 \\%$\nC: $31.6 \\%$\nD: $68.4 \\%$\nE: $84.7 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_414",
        "problem": "双极膜在电渗析中应用广泛, 它是由阳离子交换膜和阴离子交换膜复合而成。双极膜内层为水层, 工作时水层中的 $\\mathrm{H}_{2} \\mathrm{O}$ 解离成 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$, 并分别通过离子交换膜向两侧发生迁移。下图为 $\\mathrm{NaBr}$ 溶液的电渗析装置示意图。\n\n$\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液 循环液 $\\mathrm{NaBr}$ 溶液 循环液 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液\n\n[图1]\n\n下列说法错误的是\nA: 出口 2 的物质为 $\\mathrm{NaOH}$ 溶液\nB: 出口 5 的物质为硫酸钠溶液\nC: $\\mathrm{Br}^{-}$可从盐室最终进入阳极液中\nD: 阳极电极反应式为 $2 \\mathrm{H}_{2} \\mathrm{O}-4 \\mathrm{e}^{-}=\\mathrm{O}_{2} \\uparrow+4 \\mathrm{H}^{+}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n双极膜在电渗析中应用广泛, 它是由阳离子交换膜和阴离子交换膜复合而成。双极膜内层为水层, 工作时水层中的 $\\mathrm{H}_{2} \\mathrm{O}$ 解离成 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$, 并分别通过离子交换膜向两侧发生迁移。下图为 $\\mathrm{NaBr}$ 溶液的电渗析装置示意图。\n\n$\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液 循环液 $\\mathrm{NaBr}$ 溶液 循环液 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液\n\n[图1]\n\n下列说法错误的是\n\nA: 出口 2 的物质为 $\\mathrm{NaOH}$ 溶液\nB: 出口 5 的物质为硫酸钠溶液\nC: $\\mathrm{Br}^{-}$可从盐室最终进入阳极液中\nD: 阳极电极反应式为 $2 \\mathrm{H}_{2} \\mathrm{O}-4 \\mathrm{e}^{-}=\\mathrm{O}_{2} \\uparrow+4 \\mathrm{H}^{+}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-24.jpg?height=546&width=1196&top_left_y=1206&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_36",
        "problem": "The reaction below is exothermic.\n\n$$\n\\mathrm{PCl}_{3}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{PCl}_{5}(g)\n$$\n\nWhich change will increase the number of moles of $\\mathrm{PCl}_{5}(g)$ present at equilibrium?\nA: The volume of the reaction vessel is tripled.\nB: The reaction vessel is cooled.\nC: Some of the $\\mathrm{Cl}_{2}(g)$ is removed.\nD: Krypton gas is added to the reaction vessel.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction below is exothermic.\n\n$$\n\\mathrm{PCl}_{3}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{PCl}_{5}(g)\n$$\n\nWhich change will increase the number of moles of $\\mathrm{PCl}_{5}(g)$ present at equilibrium?\n\nA: The volume of the reaction vessel is tripled.\nB: The reaction vessel is cooled.\nC: Some of the $\\mathrm{Cl}_{2}(g)$ is removed.\nD: Krypton gas is added to the reaction vessel.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_788",
        "problem": "常温下, 向 $20 \\mathrm{~mL} 0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液中滴加 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, 有关微粒的物质的量变化如图, 根据图示判断, 下列说法错误的是（ ）\n\n[图1]\nA: 在 P 点时, $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: 当 $\\mathrm{V}(\\mathrm{NaOH})=20 \\mathrm{~mL}$ 时, $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\nC: 当 $\\mathrm{V}(\\mathrm{NaOH})=30 \\mathrm{~mL}$ 时, $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=3\\left[\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)\\right]$\nD: 当 $\\mathrm{V}(\\mathrm{NaOH})=40 \\mathrm{~mL}$ 时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 向 $20 \\mathrm{~mL} 0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液中滴加 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, 有关微粒的物质的量变化如图, 根据图示判断, 下列说法错误的是（ ）\n\n[图1]\n\nA: 在 P 点时, $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: 当 $\\mathrm{V}(\\mathrm{NaOH})=20 \\mathrm{~mL}$ 时, $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\nC: 当 $\\mathrm{V}(\\mathrm{NaOH})=30 \\mathrm{~mL}$ 时, $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=3\\left[\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)\\right]$\nD: 当 $\\mathrm{V}(\\mathrm{NaOH})=40 \\mathrm{~mL}$ 时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-092.jpg?height=434&width=699&top_left_y=1131&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_672",
        "problem": "已知, $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中存在以下平衡:\n\n(1) $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons 2 \\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})+2 \\mathrm{H}^{+}(\\mathrm{aq}) \\mathrm{K}_{1}=3.27 \\times 10^{-15}$\n\n(2) $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons 2 \\mathrm{HCrO}_{4}^{-}(\\mathrm{aq}) \\mathrm{K}_{2}=3.0 \\times 10^{-2}$\n\n(3) $\\mathrm{HCrO}_{4}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) \\mathrm{K}_{3}$\n\n$25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} / \\mathrm{LK}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中, $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)}$ 随 $\\mathrm{pH}$ 的变化关系如图所示 $(\\lg 3.27=0.52$, $\\left.10^{-0.78}=0.17\\right)$ 。下列说法错误的是\n\n[图1]\nA: $0.1 \\mathrm{~mol} / \\mathrm{LK}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中, 加入少量的 $\\mathrm{HI}$ 气体, $\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)$ 增大 B．溶液颜色不再变化，可以判断该体系达到平衡\nB: $\\mathrm{a}$ 点溶液中离子浓度关系: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: 反应(3)的化学平衡常数 $\\mathrm{K}_{3}=\\sqrt{\\frac{\\mathrm{K}_{1}}{\\mathrm{~K}_{2}}}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知, $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中存在以下平衡:\n\n(1) $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons 2 \\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})+2 \\mathrm{H}^{+}(\\mathrm{aq}) \\mathrm{K}_{1}=3.27 \\times 10^{-15}$\n\n(2) $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons 2 \\mathrm{HCrO}_{4}^{-}(\\mathrm{aq}) \\mathrm{K}_{2}=3.0 \\times 10^{-2}$\n\n(3) $\\mathrm{HCrO}_{4}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) \\mathrm{K}_{3}$\n\n$25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} / \\mathrm{LK}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中, $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)}$ 随 $\\mathrm{pH}$ 的变化关系如图所示 $(\\lg 3.27=0.52$, $\\left.10^{-0.78}=0.17\\right)$ 。下列说法错误的是\n\n[图1]\n\nA: $0.1 \\mathrm{~mol} / \\mathrm{LK}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中, 加入少量的 $\\mathrm{HI}$ 气体, $\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)$ 增大 B．溶液颜色不再变化，可以判断该体系达到平衡\nB: $\\mathrm{a}$ 点溶液中离子浓度关系: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: 反应(3)的化学平衡常数 $\\mathrm{K}_{3}=\\sqrt{\\frac{\\mathrm{K}_{1}}{\\mathrm{~K}_{2}}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-008.jpg?height=412&width=560&top_left_y=2304&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_177",
        "problem": "Potassium iodate, $\\mathrm{KIO}_{3}$, has applications as both source of dietary iodine in table salt and as protection against the accumulation of radioactive iodine in the human thyroid gland. The following reaction produces potassium iodate:\n\n$$\n2 \\mathrm{KClO}_{3}(\\mathrm{~s})+\\mathrm{I}_{2}(\\mathrm{~s}) \\rightarrow 2 \\mathrm{KIO}_{3}(\\mathrm{~s})+\\mathrm{Cl}_{2}(\\mathrm{~g})\n$$\n\nWhat is the theoretical yield of $\\mathrm{KIO}_{3}$ if the limiting reactant is $51.0 \\mathrm{~g} \\mathrm{I}_{2}$ ?\nA: $43.0 \\mathrm{~g}$\nB: $46.2 \\mathrm{~g}$\nC: $86.0 \\mathrm{~g}$\nD: $172 \\mathrm{~g}$\nE: $185 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPotassium iodate, $\\mathrm{KIO}_{3}$, has applications as both source of dietary iodine in table salt and as protection against the accumulation of radioactive iodine in the human thyroid gland. The following reaction produces potassium iodate:\n\n$$\n2 \\mathrm{KClO}_{3}(\\mathrm{~s})+\\mathrm{I}_{2}(\\mathrm{~s}) \\rightarrow 2 \\mathrm{KIO}_{3}(\\mathrm{~s})+\\mathrm{Cl}_{2}(\\mathrm{~g})\n$$\n\nWhat is the theoretical yield of $\\mathrm{KIO}_{3}$ if the limiting reactant is $51.0 \\mathrm{~g} \\mathrm{I}_{2}$ ?\n\nA: $43.0 \\mathrm{~g}$\nB: $46.2 \\mathrm{~g}$\nC: $86.0 \\mathrm{~g}$\nD: $172 \\mathrm{~g}$\nE: $185 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1147",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\n Give the full electron configurations of zinc in s, p, d notation.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\n Give the full electron configurations of zinc in s, p, d notation.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_607",
        "problem": "二元有机酸 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}=1.67 \\times 10^{-8} 、 \\mathrm{~K}_{\\mathrm{a} 2}=3.34 \\times 10^{-17} 。 \\mathrm{BaX}$ 难溶于水, 常温下, 将 $\\mathrm{BaX}$ 溶解在一定浓度的 $\\mathrm{HY}$ 溶液中, 直至不再溶解, 测得混合液中 $\\mathrm{c}^{2}\\left(\\mathrm{H}^{+}\\right)$与 $\\mathrm{c}^{2}\\left(\\mathrm{Ba}^{2+}\\right)$的关系如图所示。下列说法错误的是\n\n[图1]\n\n已知: (1) $\\mathrm{HY}$ 是一元强酸, $\\mathrm{BaY}_{2}$ 易溶于水\n\n(2) $0.19^{2}=0.0361,0.38^{2}=0.1444$\nA: NaHX 溶液显碱性\nB: 溶度积 $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{BaX}) \\approx 6.18 \\times 10^{-21} \\mathrm{~mol}^{2} \\cdot \\mathrm{L}^{-2}$\nC: b 点: $2 \\mathrm{c}\\left(\\mathrm{Ba}^{2+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Y}^{-}\\right)$\nD: 若 $0.1 \\mathrm{molBaX}$ 溶于 $25 \\mathrm{mLxmol} \\cdot \\mathrm{L}^{-1} \\mathrm{HY}$ 溶液中得到 $\\mathrm{c}$ 点溶液, 则 $\\mathrm{x} \\approx 8.38$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n二元有机酸 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}=1.67 \\times 10^{-8} 、 \\mathrm{~K}_{\\mathrm{a} 2}=3.34 \\times 10^{-17} 。 \\mathrm{BaX}$ 难溶于水, 常温下, 将 $\\mathrm{BaX}$ 溶解在一定浓度的 $\\mathrm{HY}$ 溶液中, 直至不再溶解, 测得混合液中 $\\mathrm{c}^{2}\\left(\\mathrm{H}^{+}\\right)$与 $\\mathrm{c}^{2}\\left(\\mathrm{Ba}^{2+}\\right)$的关系如图所示。下列说法错误的是\n\n[图1]\n\n已知: (1) $\\mathrm{HY}$ 是一元强酸, $\\mathrm{BaY}_{2}$ 易溶于水\n\n(2) $0.19^{2}=0.0361,0.38^{2}=0.1444$\n\nA: NaHX 溶液显碱性\nB: 溶度积 $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{BaX}) \\approx 6.18 \\times 10^{-21} \\mathrm{~mol}^{2} \\cdot \\mathrm{L}^{-2}$\nC: b 点: $2 \\mathrm{c}\\left(\\mathrm{Ba}^{2+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{Y}^{-}\\right)$\nD: 若 $0.1 \\mathrm{molBaX}$ 溶于 $25 \\mathrm{mLxmol} \\cdot \\mathrm{L}^{-1} \\mathrm{HY}$ 溶液中得到 $\\mathrm{c}$ 点溶液, 则 $\\mathrm{x} \\approx 8.38$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-111.jpg?height=360&width=531&top_left_y=2033&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1002",
        "problem": "A compound of carbon and hydrogen is found to be $85.6 \\%$ carbon, by mass, and $14.38 \\%$ hydrogen. What is the simplest formula of the compound?\nA: $\\mathrm{CH}$\nB: $\\mathrm{CH}_{2}$\nC: $\\mathrm{CH}_{3}$\nD: $\\mathrm{CH}_{4}$\nE: $\\mathrm{C}_{3} \\mathrm{H}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA compound of carbon and hydrogen is found to be $85.6 \\%$ carbon, by mass, and $14.38 \\%$ hydrogen. What is the simplest formula of the compound?\n\nA: $\\mathrm{CH}$\nB: $\\mathrm{CH}_{2}$\nC: $\\mathrm{CH}_{3}$\nD: $\\mathrm{CH}_{4}$\nE: $\\mathrm{C}_{3} \\mathrm{H}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_223",
        "problem": "What is the percentage by mass of fluorine in $\\mathrm{NF}_{3}$ ?\nA: $31.13 \\%$\nB: $57.56 \\%$\nC: $73.06 \\%$\nD: $80.27 \\%$\nE: $87.15 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the percentage by mass of fluorine in $\\mathrm{NF}_{3}$ ?\n\nA: $31.13 \\%$\nB: $57.56 \\%$\nC: $73.06 \\%$\nD: $80.27 \\%$\nE: $87.15 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_824",
        "problem": "含磷有机物应用广泛。电解法可实现由白磷直接制备 $\\mathrm{Li}\\left[\\mathrm{P}(\\mathrm{CN})_{2}\\right]$, 过程如图所示 $(\\mathrm{Me}$为甲基)。下列说法正确的是\n\n[图1]\nA: 生成 $1 \\mathrm{molLi}\\left[\\mathrm{P}(\\mathrm{CN})_{2}\\right]^{-}$, 理论上外电路需要转移 $2 \\mathrm{~mol}$ 电子\nB: 阴极上的电极反应为: $\\mathrm{P}_{4}+8 \\mathrm{CN}^{-}-4 \\mathrm{e}^{-}=4\\left[\\mathrm{P}(\\mathrm{CN})_{2}\\right]^{-}$\nC: 在电解过程中 $\\mathrm{CN}^{-}$向石墨电极移动\nD: 电解产生的 $\\mathrm{H}_{2}$ 中的氢元素来自于 $\\mathrm{LiOH}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n含磷有机物应用广泛。电解法可实现由白磷直接制备 $\\mathrm{Li}\\left[\\mathrm{P}(\\mathrm{CN})_{2}\\right]$, 过程如图所示 $(\\mathrm{Me}$为甲基)。下列说法正确的是\n\n[图1]\n\nA: 生成 $1 \\mathrm{molLi}\\left[\\mathrm{P}(\\mathrm{CN})_{2}\\right]^{-}$, 理论上外电路需要转移 $2 \\mathrm{~mol}$ 电子\nB: 阴极上的电极反应为: $\\mathrm{P}_{4}+8 \\mathrm{CN}^{-}-4 \\mathrm{e}^{-}=4\\left[\\mathrm{P}(\\mathrm{CN})_{2}\\right]^{-}$\nC: 在电解过程中 $\\mathrm{CN}^{-}$向石墨电极移动\nD: 电解产生的 $\\mathrm{H}_{2}$ 中的氢元素来自于 $\\mathrm{LiOH}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-08.jpg?height=448&width=717&top_left_y=1261&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1291",
        "problem": "The unit cell is the smallest repeating unit in a crystal structure. The unit cell of a gold crystal is found by X-ray diffraction to have the face-centred cubic unit structure (i.e. where the centre of an atom is located at each corner of a cube and in the middle of each face). The side of the unit cell is found to be $0.408 \\mathrm{~nm}$.The density of $\\mathrm{Au}$ is $1.93 \\cdot 10^{4} \\mathrm{~kg} \\mathrm{~m}^{-3}$. Calculate the volume and mass of the cubic unit cell.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe unit cell is the smallest repeating unit in a crystal structure. The unit cell of a gold crystal is found by X-ray diffraction to have the face-centred cubic unit structure (i.e. where the centre of an atom is located at each corner of a cube and in the middle of each face). The side of the unit cell is found to be $0.408 \\mathrm{~nm}$.\n\nproblem:\nThe density of $\\mathrm{Au}$ is $1.93 \\cdot 10^{4} \\mathrm{~kg} \\mathrm{~m}^{-3}$. Calculate the volume and mass of the cubic unit cell.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1223",
        "problem": "There is $0.23 \\mathrm{~cm}^{3}$ of dissolved $\\mathrm{CO}_{2}$ per one litre sea water (at $24{ }^{\\circ} \\mathrm{C}$ and $101 \\mathrm{kPa}$ ).\n\ni) If diatoms can completely remove carbon dioxide from the water they process, what volume of water would they process to produce the carbohydrates required by a blue whale during the first five years of life?\n\nii) What fraction of the total volume of the oceans will be needed to supply the carbon dioxide for the first five years of growth of 1000 blue whales? The volume of the oceans is $1.37 \\times 10^{18} \\mathrm{~m}^{3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThere is $0.23 \\mathrm{~cm}^{3}$ of dissolved $\\mathrm{CO}_{2}$ per one litre sea water (at $24{ }^{\\circ} \\mathrm{C}$ and $101 \\mathrm{kPa}$ ).\n\ni) If diatoms can completely remove carbon dioxide from the water they process, what volume of water would they process to produce the carbohydrates required by a blue whale during the first five years of life?\n\nii) What fraction of the total volume of the oceans will be needed to supply the carbon dioxide for the first five years of growth of 1000 blue whales? The volume of the oceans is $1.37 \\times 10^{18} \\mathrm{~m}^{3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [The amount of water, The total ocean volume].\nTheir units are, in order, [$\\mathrm{dm}^{3}$, None], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "$\\mathrm{dm}^{3}$",
            null
        ],
        "answer_sequence": [
            "The amount of water",
            "The total ocean volume"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_653",
        "problem": "向 $\\mathrm{AgCl}$ 饱和溶液(有足量 $\\mathrm{AgCl}$ 固体)中滴加氨水, 发生反应\n\n$\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$和 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$,\n\n$\\lg \\left[\\mathrm{c}(\\mathrm{M}) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 与 $\\lg \\left[\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 的关系如下图所示(其中 $\\mathrm{M}$ 代表 $\\mathrm{Ag}^{+} 、 \\mathrm{Cl}^{-}$、 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$或 $\\left.\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right)$。\n\n[图1]\n\n下列说法错误的是\nA: 曲线IV可视为 $\\mathrm{AgCl}$ 溶解度随 $\\mathrm{NH}_{3}$ 浓度变化曲线\nB: $\\mathrm{AgCl}$ 的溶度积常数 $\\mathrm{K}_{\\mathrm{sp}}=\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=10^{9.75}$\nC: 反应 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$的平衡常数 $\\mathrm{K}$ 的值为 $10^{3.81}$\nD: $\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 时, 溶液中 $\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right)>\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n向 $\\mathrm{AgCl}$ 饱和溶液(有足量 $\\mathrm{AgCl}$ 固体)中滴加氨水, 发生反应\n\n$\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$和 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$,\n\n$\\lg \\left[\\mathrm{c}(\\mathrm{M}) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 与 $\\lg \\left[\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right) /\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)\\right]$ 的关系如下图所示(其中 $\\mathrm{M}$ 代表 $\\mathrm{Ag}^{+} 、 \\mathrm{Cl}^{-}$、 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}$或 $\\left.\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right)$。\n\n[图1]\n\n下列说法错误的是\n\nA: 曲线IV可视为 $\\mathrm{AgCl}$ 溶解度随 $\\mathrm{NH}_{3}$ 浓度变化曲线\nB: $\\mathrm{AgCl}$ 的溶度积常数 $\\mathrm{K}_{\\mathrm{sp}}=\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=10^{9.75}$\nC: 反应 $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$的平衡常数 $\\mathrm{K}$ 的值为 $10^{3.81}$\nD: $\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)=0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 时, 溶液中 $\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right)>\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-044.jpg?height=560&width=711&top_left_y=1068&top_left_x=364"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_517",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时在 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中逐滴加入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$ 溶液 $20 \\mathrm{~mL}$, 溶液中部分含碳微粒的物质的量随溶液 $\\mathrm{pH}$ 的变化如图所示。下列说法不正确的是\n\n[图1]\nA: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2\\left[\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)\\right]$\nB: $\\mathrm{a} 、 \\mathrm{~b}$ 曲线分别代表的是 $\\mathrm{HCO}_{3}{ }^{-} 、 \\mathrm{H}_{2} \\mathrm{CO}_{3}$ 量的变化情况\nC: A 点时: $c\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: B 点时加入盐酸的体积为 $10 \\mathrm{~mL}$ ，两者恰好完全反应生成 $\\mathrm{NaHCO}_{3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时在 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中逐滴加入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$ 溶液 $20 \\mathrm{~mL}$, 溶液中部分含碳微粒的物质的量随溶液 $\\mathrm{pH}$ 的变化如图所示。下列说法不正确的是\n\n[图1]\n\nA: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2\\left[\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)\\right]$\nB: $\\mathrm{a} 、 \\mathrm{~b}$ 曲线分别代表的是 $\\mathrm{HCO}_{3}{ }^{-} 、 \\mathrm{H}_{2} \\mathrm{CO}_{3}$ 量的变化情况\nC: A 点时: $c\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: B 点时加入盐酸的体积为 $10 \\mathrm{~mL}$ ，两者恰好完全反应生成 $\\mathrm{NaHCO}_{3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-014.jpg?height=406&width=688&top_left_y=2144&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_661",
        "problem": "用镁次氯酸钠燃料电池作电源模拟消除工业酸性废水中的 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 的过程 (将 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 还原为 $\\mathrm{Cr}^{3+}$ )，装置如图所示。下列说法正确的是 ( )\n\n[图1]\nA: 装置中电子的流动路线是 $\\mathrm{C}$ 电极 $\\rightarrow$ 惰性电极 $\\rightarrow$ 金属铁电极 $\\rightarrow \\mathrm{D}$ 电极\nB: 燃料电池负极反应式为 $\\mathrm{Mg}+\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{Cl}^{-}+\\mathrm{Mg}(\\mathrm{OH})_{2}$\nC: 将 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 处理后的废水比原工业废水的 $\\mathrm{PH}$ 增大\nD: 装置工作过程中消耗 $14.4 \\mathrm{~g} \\mathrm{Mg}$ ，理论上可消除 $0.1 \\mathrm{molCr}_{2} \\mathrm{O}_{7}^{2-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n用镁次氯酸钠燃料电池作电源模拟消除工业酸性废水中的 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 的过程 (将 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 还原为 $\\mathrm{Cr}^{3+}$ )，装置如图所示。下列说法正确的是 ( )\n\n[图1]\n\nA: 装置中电子的流动路线是 $\\mathrm{C}$ 电极 $\\rightarrow$ 惰性电极 $\\rightarrow$ 金属铁电极 $\\rightarrow \\mathrm{D}$ 电极\nB: 燃料电池负极反应式为 $\\mathrm{Mg}+\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{Cl}^{-}+\\mathrm{Mg}(\\mathrm{OH})_{2}$\nC: 将 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 处理后的废水比原工业废水的 $\\mathrm{PH}$ 增大\nD: 装置工作过程中消耗 $14.4 \\mathrm{~g} \\mathrm{Mg}$ ，理论上可消除 $0.1 \\mathrm{molCr}_{2} \\mathrm{O}_{7}^{2-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-86.jpg?height=445&width=1043&top_left_y=937&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_465",
        "problem": "利用光伏电池提供电能处理废水中的污染物 (有机酸阴离子用 $\\mathrm{R}^{-}$表示), 并回收有机酸 HR，装置如图所示。下列说法错误的是\n\n[图1]\nA: 在光伏电池中 $\\mathrm{a}$ 极为正极\nB: 石墨(2)极附近溶液的 $\\mathrm{pH}$ 降低\nC: HR 溶液: $\\mathrm{c}_{1}<\\mathrm{c}_{2}$\nD: 若两极共收集 $3 \\mathrm{~mol}$ 气体, 则理论上转移 $4 \\mathrm{~mol}$ 电子\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用光伏电池提供电能处理废水中的污染物 (有机酸阴离子用 $\\mathrm{R}^{-}$表示), 并回收有机酸 HR，装置如图所示。下列说法错误的是\n\n[图1]\n\nA: 在光伏电池中 $\\mathrm{a}$ 极为正极\nB: 石墨(2)极附近溶液的 $\\mathrm{pH}$ 降低\nC: HR 溶液: $\\mathrm{c}_{1}<\\mathrm{c}_{2}$\nD: 若两极共收集 $3 \\mathrm{~mol}$ 气体, 则理论上转移 $4 \\mathrm{~mol}$ 电子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-84.jpg?height=577&width=1091&top_left_y=1782&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_155",
        "problem": "How many carbon $(\\mathrm{C})$ and hydrogen $(\\mathrm{H})$ atoms are present in a molecule of Tigan?\nA: $21 \\mathrm{C}$ and $28 \\mathrm{H}$ atoms\nB: $21 \\mathrm{C}$ and $27 \\mathrm{H}$ atoms\nC: $16 \\mathrm{C}$ and $28 \\mathrm{H}$ atoms\nD: $20 \\mathrm{C}$ and $28 \\mathrm{H}$ atoms\nE: $21 \\mathrm{C}$ and $29 \\mathrm{H}$ atoms\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many carbon $(\\mathrm{C})$ and hydrogen $(\\mathrm{H})$ atoms are present in a molecule of Tigan?\n\nA: $21 \\mathrm{C}$ and $28 \\mathrm{H}$ atoms\nB: $21 \\mathrm{C}$ and $27 \\mathrm{H}$ atoms\nC: $16 \\mathrm{C}$ and $28 \\mathrm{H}$ atoms\nD: $20 \\mathrm{C}$ and $28 \\mathrm{H}$ atoms\nE: $21 \\mathrm{C}$ and $29 \\mathrm{H}$ atoms\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_766",
        "problem": "如图装置中, 容器甲内充入 $0.1 \\mathrm{~mol} \\mathrm{NO}$ 气体, 干燥管内装有一定量 $\\mathrm{Na}_{2} \\mathrm{O}_{2}$, 从 $\\mathrm{A}$ 处缓慢通入 $\\mathrm{CO}_{2}$ 气体. 恒温下, 容器甲中活塞缓慢由 $\\mathrm{D}$ 向左移动, 当移至 $\\mathrm{C}$ 处时容器体积缩小至最小, 为原体积的 $\\frac{9}{10}$, 干燥管中物质的质量增加 $2.24 \\mathrm{~g}$ 随着 $\\mathrm{CO}_{2}$ 的继续通入,活塞又逐渐向右移动. 已知: $2 \\mathrm{Na}_{2} \\mathrm{O}_{2}+2 \\mathrm{CO}_{2}=2 \\mathrm{Na}_{2} \\mathrm{CO}_{3}+\\mathrm{O}_{2} \\quad 2 \\mathrm{NO}+\\mathrm{O}_{2}=2 \\mathrm{NO}_{2} \\quad 2 \\mathrm{NO}_{2} \\rightleftharpoons$ $\\mathrm{N}_{2} \\mathrm{O}_{4}$ (不考虑活塞的摩擦) 下列说法中正确的是（）\n[图1]\nA: 活塞从 $\\mathrm{D}$ 处移动到 $\\mathrm{C}$ 处的过程中, 通入 $\\mathrm{CO}_{2}$ 体积为 $2.24 \\mathrm{~L}$ （标准状况）\nB: $\\mathrm{NO}_{2}$ 转化为 $\\mathrm{N}_{2} \\mathrm{O}_{4}$ 的转换率为 $20 \\%$\nC: 活塞移至 $\\mathrm{C}$ 处后, 继续通入 $0.01 \\mathrm{~mol} \\mathrm{CO}_{2}$, 此时活塞恰好回到 D 处\nD: 若改变干燥管中 $\\mathrm{Na}_{2} \\mathrm{O}_{2}$ 的量, 要通过调节甲容器的温度及通入 $\\mathrm{CO}_{2}$ 的量, 使活塞发生从 $\\mathrm{D}$ 到 $\\mathrm{C}$, 又从 $\\mathrm{C}$ 到 $\\mathrm{D}$ 的移动, 则 $\\mathrm{Na}_{2} \\mathrm{O}_{2}$ 的质量应大于 $1.56 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图装置中, 容器甲内充入 $0.1 \\mathrm{~mol} \\mathrm{NO}$ 气体, 干燥管内装有一定量 $\\mathrm{Na}_{2} \\mathrm{O}_{2}$, 从 $\\mathrm{A}$ 处缓慢通入 $\\mathrm{CO}_{2}$ 气体. 恒温下, 容器甲中活塞缓慢由 $\\mathrm{D}$ 向左移动, 当移至 $\\mathrm{C}$ 处时容器体积缩小至最小, 为原体积的 $\\frac{9}{10}$, 干燥管中物质的质量增加 $2.24 \\mathrm{~g}$ 随着 $\\mathrm{CO}_{2}$ 的继续通入,活塞又逐渐向右移动. 已知: $2 \\mathrm{Na}_{2} \\mathrm{O}_{2}+2 \\mathrm{CO}_{2}=2 \\mathrm{Na}_{2} \\mathrm{CO}_{3}+\\mathrm{O}_{2} \\quad 2 \\mathrm{NO}+\\mathrm{O}_{2}=2 \\mathrm{NO}_{2} \\quad 2 \\mathrm{NO}_{2} \\rightleftharpoons$ $\\mathrm{N}_{2} \\mathrm{O}_{4}$ (不考虑活塞的摩擦) 下列说法中正确的是（）\n[图1]\n\nA: 活塞从 $\\mathrm{D}$ 处移动到 $\\mathrm{C}$ 处的过程中, 通入 $\\mathrm{CO}_{2}$ 体积为 $2.24 \\mathrm{~L}$ （标准状况）\nB: $\\mathrm{NO}_{2}$ 转化为 $\\mathrm{N}_{2} \\mathrm{O}_{4}$ 的转换率为 $20 \\%$\nC: 活塞移至 $\\mathrm{C}$ 处后, 继续通入 $0.01 \\mathrm{~mol} \\mathrm{CO}_{2}$, 此时活塞恰好回到 D 处\nD: 若改变干燥管中 $\\mathrm{Na}_{2} \\mathrm{O}_{2}$ 的量, 要通过调节甲容器的温度及通入 $\\mathrm{CO}_{2}$ 的量, 使活塞发生从 $\\mathrm{D}$ 到 $\\mathrm{C}$, 又从 $\\mathrm{C}$ 到 $\\mathrm{D}$ 的移动, 则 $\\mathrm{Na}_{2} \\mathrm{O}_{2}$ 的质量应大于 $1.56 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-76.jpg?height=140&width=431&top_left_y=2072&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_153",
        "problem": "Ronald James Gillespie, who developed Valence Shell Electron Pair Repulsion (VSEPR) theory, isolated the superacid fluorosulfuric acid $\\left(\\mathrm{HSO}_{3} \\mathrm{~F}\\right)$ when he combined fluorinated compounds with concentrated sulfuric acid, creating brightly coloured solutions. Which of the following best describes the molecular shape of fluorosulfuric acid, as predicted by VSEPR theory?\nA: See saw\nB: Trigonal bipyramidal\nC: Tetrahedral\nD: Square planar\nE: Trigonal pyramidal\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRonald James Gillespie, who developed Valence Shell Electron Pair Repulsion (VSEPR) theory, isolated the superacid fluorosulfuric acid $\\left(\\mathrm{HSO}_{3} \\mathrm{~F}\\right)$ when he combined fluorinated compounds with concentrated sulfuric acid, creating brightly coloured solutions. Which of the following best describes the molecular shape of fluorosulfuric acid, as predicted by VSEPR theory?\n\nA: See saw\nB: Trigonal bipyramidal\nC: Tetrahedral\nD: Square planar\nE: Trigonal pyramidal\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_175",
        "problem": "Tigan (structure below) is an antiemetic drug used to prevent nausea and vomiting. It is often prescribed for patients with gastroenteritis, medication-induced nausea, and other illnesses.\n\n[figure1]\n\nWhich of the following functional groups are present within Tigan?\nA: amine, ether, amide\nB: ether, aldehyde, amine\nC: amine, ketone, ether, alcohol\nD: ketone, amine, ether\nE: alcohol, amine, ether, amide\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTigan (structure below) is an antiemetic drug used to prevent nausea and vomiting. It is often prescribed for patients with gastroenteritis, medication-induced nausea, and other illnesses.\n\n[figure1]\n\nWhich of the following functional groups are present within Tigan?\n\nA: amine, ether, amide\nB: ether, aldehyde, amine\nC: amine, ketone, ether, alcohol\nD: ketone, amine, ether\nE: alcohol, amine, ether, amide\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_343d5676bd7739089c6eg-2.jpg?height=211&width=543&top_left_y=403&top_left_x=1784"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_836",
        "problem": "如图所示, 隔板I固定不动, 活塞 II可自由移动, $\\mathrm{M} 、 \\mathrm{~N}$ 两个容器中均发生反应: $\\mathrm{A}(\\mathrm{g})+3 \\mathrm{~B}(\\mathrm{~g})=2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}=-210 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$ 。 向 $\\mathrm{M} 、 \\mathrm{~N}$ 中, 分别通入 $\\mathrm{xmolA}$ 和 ymolB 的混合气体, 初始时 $\\mathrm{M} 、 \\mathrm{~N}$ 容积相同。下列说法不正确的是\n\n[图1]\nA: 若平衡时 $\\mathrm{A}$ 气体在两容器中的体积分数相等, 则 $\\mathrm{x}: \\mathrm{y}=1: 1$\nB: 若 $x: y=1: 3$, 当 $M$ 中放出热量 $168 \\mathrm{~kJ}$ 时, $A$ 的转化率为 $80 \\%$\nC: 若 $x: y=1: 1$, 平衡时测得 $N$ 中含 $A 、 C$ 分别为 $1.2 \\mathrm{~mol} 、 0.2 \\mathrm{~mol}$ 。保持温度不变,再向 $\\mathrm{N}$ 中通入 $0.6 \\mathrm{molA}$, 则平衡向逆反应方向移动。\nD: 若 $x=1, y=1.8$, 在 $\\mathrm{N}$ 中达到平衡时容积为 $24 \\mathrm{~L}, \\mathrm{C}$ 为 $0.4 \\mathrm{~mol}$, 则反应起始时 $\\mathrm{N}$的容积为 $28 \\mathrm{~L}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n如图所示, 隔板I固定不动, 活塞 II可自由移动, $\\mathrm{M} 、 \\mathrm{~N}$ 两个容器中均发生反应: $\\mathrm{A}(\\mathrm{g})+3 \\mathrm{~B}(\\mathrm{~g})=2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}=-210 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$ 。 向 $\\mathrm{M} 、 \\mathrm{~N}$ 中, 分别通入 $\\mathrm{xmolA}$ 和 ymolB 的混合气体, 初始时 $\\mathrm{M} 、 \\mathrm{~N}$ 容积相同。下列说法不正确的是\n\n[图1]\n\nA: 若平衡时 $\\mathrm{A}$ 气体在两容器中的体积分数相等, 则 $\\mathrm{x}: \\mathrm{y}=1: 1$\nB: 若 $x: y=1: 3$, 当 $M$ 中放出热量 $168 \\mathrm{~kJ}$ 时, $A$ 的转化率为 $80 \\%$\nC: 若 $x: y=1: 1$, 平衡时测得 $N$ 中含 $A 、 C$ 分别为 $1.2 \\mathrm{~mol} 、 0.2 \\mathrm{~mol}$ 。保持温度不变,再向 $\\mathrm{N}$ 中通入 $0.6 \\mathrm{molA}$, 则平衡向逆反应方向移动。\nD: 若 $x=1, y=1.8$, 在 $\\mathrm{N}$ 中达到平衡时容积为 $24 \\mathrm{~L}, \\mathrm{C}$ 为 $0.4 \\mathrm{~mol}$, 则反应起始时 $\\mathrm{N}$的容积为 $28 \\mathrm{~L}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-044.jpg?height=205&width=489&top_left_y=477&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-045.jpg?height=206&width=1382&top_left_y=157&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_354",
        "problem": "A sample of $54.0 \\mathrm{~g}$ of methanol is heated from $25.0^{\\circ} \\mathrm{C}$ to $35.0^{\\circ} \\mathrm{C}$. How much heat is required? The specific heat capacity of methanol is $2.48 \\mathrm{~J} \\mathrm{~g}^{-1} \\mathrm{~K}^{-1}$.\nA: $0.00459 \\mathrm{~J}$\nB: $0.0747 \\mathrm{~J}$\nC: $1340 \\mathrm{~J}$\nD: $4690 \\mathrm{~J}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA sample of $54.0 \\mathrm{~g}$ of methanol is heated from $25.0^{\\circ} \\mathrm{C}$ to $35.0^{\\circ} \\mathrm{C}$. How much heat is required? The specific heat capacity of methanol is $2.48 \\mathrm{~J} \\mathrm{~g}^{-1} \\mathrm{~K}^{-1}$.\n\nA: $0.00459 \\mathrm{~J}$\nB: $0.0747 \\mathrm{~J}$\nC: $1340 \\mathrm{~J}$\nD: $4690 \\mathrm{~J}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1433",
        "problem": "The energy of stable states of the hydrogen atom is given by: $E_{n}=-2.18 \\times 10^{-18} / \\mathrm{n}^{2}[\\mathrm{~J}]$ where $\\mathrm{n}$ denotes the principal quantum number.\n\nCalculate the energy differences between $n=2$ (first excited state) and $n=1$ (ground state) and between $\\mathrm{n}=7$ and $\\mathrm{n}=1$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThe energy of stable states of the hydrogen atom is given by: $E_{n}=-2.18 \\times 10^{-18} / \\mathrm{n}^{2}[\\mathrm{~J}]$ where $\\mathrm{n}$ denotes the principal quantum number.\n\nCalculate the energy differences between $n=2$ (first excited state) and $n=1$ (ground state) and between $\\mathrm{n}=7$ and $\\mathrm{n}=1$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [$\\Delta E_{2 \\rightarrow 1}$, $\\Delta E_{7 \\rightarrow 1}$].\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "$\\Delta E_{2 \\rightarrow 1}$",
            "$\\Delta E_{7 \\rightarrow 1}$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_88",
        "problem": "Which metal resists oxidation in air because it forms a protective oxide coating?\nA: $\\mathrm{Al}$\nB: $\\mathrm{Ca}$\nC: $\\mathrm{Fe}$\nD: $\\mathrm{Sr}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich metal resists oxidation in air because it forms a protective oxide coating?\n\nA: $\\mathrm{Al}$\nB: $\\mathrm{Ca}$\nC: $\\mathrm{Fe}$\nD: $\\mathrm{Sr}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_392",
        "problem": "How many $\\sigma$ and $\\pi$ bonds are in 1,3-butadiene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}$ ?\nA: $7 \\sigma$ and $2 \\pi$ bonds\nB: $2 \\sigma$ and $7 \\pi$ bonds\nC: $9 \\sigma$ and $2 \\pi$ bonds\nD: $2 \\sigma$ and $9 \\pi$ bonds\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many $\\sigma$ and $\\pi$ bonds are in 1,3-butadiene, $\\mathrm{H}_{2} \\mathrm{C}=\\mathrm{CH}-\\mathrm{CH}=\\mathrm{CH}_{2}$ ?\n\nA: $7 \\sigma$ and $2 \\pi$ bonds\nB: $2 \\sigma$ and $7 \\pi$ bonds\nC: $9 \\sigma$ and $2 \\pi$ bonds\nD: $2 \\sigma$ and $9 \\pi$ bonds\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_137",
        "problem": "Molecular hydrogen is an essential feedstock for the industrial production of ammonia. Due to the impracticality of transporting molecular hydrogen, it is produced at the site of ammonia production through steam methane reforming, as follows:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{CH}_{4}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g})\n$$\n\nGiven a starting steam $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$ to methane $\\left(\\mathrm{CH}_{4}\\right)$ mole ratio of 2.5:1.0, an initial pressure of $28 \\mathrm{~atm}$, no starting carbon monoxide or molecular hydrogen, determine the $K_{\\mathrm{p}}$ if $62.5 \\%$ of the initial methane is converted to products. Assume ideal gas behaviour.\nA: $1.7 \\times 10^{0}$\nB: $1.4 \\times 10^{1}$\nC: $2.1 \\times 10^{2}$\nD: $3.8 \\times 10^{2}$\nE: $5.6 \\times 10^{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMolecular hydrogen is an essential feedstock for the industrial production of ammonia. Due to the impracticality of transporting molecular hydrogen, it is produced at the site of ammonia production through steam methane reforming, as follows:\n\n$$\n\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{CH}_{4}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g})\n$$\n\nGiven a starting steam $\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)$ to methane $\\left(\\mathrm{CH}_{4}\\right)$ mole ratio of 2.5:1.0, an initial pressure of $28 \\mathrm{~atm}$, no starting carbon monoxide or molecular hydrogen, determine the $K_{\\mathrm{p}}$ if $62.5 \\%$ of the initial methane is converted to products. Assume ideal gas behaviour.\n\nA: $1.7 \\times 10^{0}$\nB: $1.4 \\times 10^{1}$\nC: $2.1 \\times 10^{2}$\nD: $3.8 \\times 10^{2}$\nE: $5.6 \\times 10^{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_954",
        "problem": "化学性质类似 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 的盐酸羟胺 $\\left(\\mathrm{NH}_{3} \\mathrm{OHCl}\\right)$ 是一种常见的还原剂和显像剂。工业\n\n上主要采用图 1 所示的方法制备, 其电池装置中含 $\\mathrm{Fe}$ 的催化电极反应机理如图 2 所示。\n\n图 3 是用图 1 的电池电解处理含有 $\\mathrm{Cl}^{-} 、 \\mathrm{NO}_{3}^{-}$的酸性废水的装置。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n电解法处理酸性废水\n\n图3\nA: 图 1 电池工作时, $\\mathrm{Pt}$ 电极是正极\nB: 图 2 中, A 为 $\\mathrm{H}^{+}$和 $\\mathrm{e}^{-}, \\mathrm{B}$ 为 $\\mathrm{NH}_{3} \\mathrm{OH}^{+}$\nC: 电极 $\\mathrm{b}$ 接电源负极, 处理 $1 \\mathrm{~mol} \\mathrm{NO}_{3}^{-}$, 酸性废水中 $\\mathrm{Cl}^{-}$减少 $2.5 \\mathrm{~mol}$\nD: 电池工作时, 每消耗 $3.36 \\mathrm{~L} \\mathrm{H}_{2}$ (标准状况下), 左室溶液质量增加 $3.3 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化学性质类似 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 的盐酸羟胺 $\\left(\\mathrm{NH}_{3} \\mathrm{OHCl}\\right)$ 是一种常见的还原剂和显像剂。工业\n\n上主要采用图 1 所示的方法制备, 其电池装置中含 $\\mathrm{Fe}$ 的催化电极反应机理如图 2 所示。\n\n图 3 是用图 1 的电池电解处理含有 $\\mathrm{Cl}^{-} 、 \\mathrm{NO}_{3}^{-}$的酸性废水的装置。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n[图3]\n\n电解法处理酸性废水\n\n图3\n\nA: 图 1 电池工作时, $\\mathrm{Pt}$ 电极是正极\nB: 图 2 中, A 为 $\\mathrm{H}^{+}$和 $\\mathrm{e}^{-}, \\mathrm{B}$ 为 $\\mathrm{NH}_{3} \\mathrm{OH}^{+}$\nC: 电极 $\\mathrm{b}$ 接电源负极, 处理 $1 \\mathrm{~mol} \\mathrm{NO}_{3}^{-}$, 酸性废水中 $\\mathrm{Cl}^{-}$减少 $2.5 \\mathrm{~mol}$\nD: 电池工作时, 每消耗 $3.36 \\mathrm{~L} \\mathrm{H}_{2}$ (标准状况下), 左室溶液质量增加 $3.3 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-13.jpg?height=345&width=479&top_left_y=684&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-13.jpg?height=426&width=492&top_left_y=621&top_left_x=859",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-13.jpg?height=229&width=380&top_left_y=728&top_left_x=1409"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_677",
        "problem": "某温度下, 两种难溶盐 $\\operatorname{Ag}_{\\mathrm{x}} \\mathrm{X} 、 \\operatorname{Ag}_{\\mathrm{y}} \\mathrm{Y}$ 的饱和溶液中 $-\\lg \\mathrm{c}\\left(\\mathrm{X}^{\\mathrm{x}-}\\right)$ 或 $-\\lg \\mathrm{c}\\left(\\mathrm{Y}^{\\mathrm{y}-}\\right)$ 与 $-\\lg \\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$的关系如图所示。下列说法错误的是\n\n[图1]\nA: $x: y=3: 1$\nB: 若混合溶液中各离子浓度如 $\\mathrm{J}$ 点所示, 加入 $\\mathrm{AgNO}_{3}(\\mathrm{~s})$, 则平衡时 $\\frac{\\mathrm{c}\\left(\\mathrm{X}^{\\mathrm{x}}\\right)}{\\mathrm{c}\\left(\\mathrm{Y}^{\\mathrm{y}}\\right)}$ 变小\nC: 向 $\\mathrm{Ag}_{\\mathrm{x}} \\mathrm{X}$ 固体中加入 $\\mathrm{Na}_{\\mathrm{y}} \\mathrm{Y}$ 溶液, 可发生 $\\mathrm{Ag}_{\\mathrm{x}} \\mathrm{X} \\rightarrow \\mathrm{Ag}_{\\mathrm{y}} \\mathrm{Y}$ 的转化\nD: 若混合溶液中各离子起始浓度如 $\\mathrm{T}$ 点所示, 待平衡时 $\\mathrm{c}\\left(\\mathrm{X}^{\\mathrm{x}}\\right)+\\mathrm{c}\\left(\\mathrm{Y}^{\\mathrm{y}}\\right)<2 \\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某温度下, 两种难溶盐 $\\operatorname{Ag}_{\\mathrm{x}} \\mathrm{X} 、 \\operatorname{Ag}_{\\mathrm{y}} \\mathrm{Y}$ 的饱和溶液中 $-\\lg \\mathrm{c}\\left(\\mathrm{X}^{\\mathrm{x}-}\\right)$ 或 $-\\lg \\mathrm{c}\\left(\\mathrm{Y}^{\\mathrm{y}-}\\right)$ 与 $-\\lg \\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$的关系如图所示。下列说法错误的是\n\n[图1]\n\nA: $x: y=3: 1$\nB: 若混合溶液中各离子浓度如 $\\mathrm{J}$ 点所示, 加入 $\\mathrm{AgNO}_{3}(\\mathrm{~s})$, 则平衡时 $\\frac{\\mathrm{c}\\left(\\mathrm{X}^{\\mathrm{x}}\\right)}{\\mathrm{c}\\left(\\mathrm{Y}^{\\mathrm{y}}\\right)}$ 变小\nC: 向 $\\mathrm{Ag}_{\\mathrm{x}} \\mathrm{X}$ 固体中加入 $\\mathrm{Na}_{\\mathrm{y}} \\mathrm{Y}$ 溶液, 可发生 $\\mathrm{Ag}_{\\mathrm{x}} \\mathrm{X} \\rightarrow \\mathrm{Ag}_{\\mathrm{y}} \\mathrm{Y}$ 的转化\nD: 若混合溶液中各离子起始浓度如 $\\mathrm{T}$ 点所示, 待平衡时 $\\mathrm{c}\\left(\\mathrm{X}^{\\mathrm{x}}\\right)+\\mathrm{c}\\left(\\mathrm{Y}^{\\mathrm{y}}\\right)<2 \\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-056.jpg?height=708&width=691&top_left_y=1805&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1410",
        "problem": "Nitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample.\n\nThe Kjeldahl method can also be used to determine the molecular weight of amino acids. In a given experiment, the molecular weight of a naturally occurring amino acid was determined by digesting $0.2345 \\mathrm{~g}$ of the pure acid and distilling ammonia released into $50.00 \\mathrm{~cm}^{3}$ of $0.1010 \\mathrm{M}$ hydrochloric acid. A titration volume of 17.50 $\\mathrm{cm}^{3}$ was obtained for the back titration with $0.1050 \\mathrm{M}$ sodium hydroxide.\n\nCalculate the molecular weight of the amino acid based on one and two nitrogen groups in the molecule, respectively.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nNitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample.\n\nThe Kjeldahl method can also be used to determine the molecular weight of amino acids. In a given experiment, the molecular weight of a naturally occurring amino acid was determined by digesting $0.2345 \\mathrm{~g}$ of the pure acid and distilling ammonia released into $50.00 \\mathrm{~cm}^{3}$ of $0.1010 \\mathrm{M}$ hydrochloric acid. A titration volume of 17.50 $\\mathrm{cm}^{3}$ was obtained for the back titration with $0.1050 \\mathrm{M}$ sodium hydroxide.\n\nCalculate the molecular weight of the amino acid based on one and two nitrogen groups in the molecule, respectively.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the molecular weight of the amino acid based on one nitrogen groups in the molecule, the molecular weight of the amino acid based on two nitrogen groups in the molecule].\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "the molecular weight of the amino acid based on one nitrogen groups in the molecule",
            "the molecular weight of the amino acid based on two nitrogen groups in the molecule"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_299",
        "problem": "Ethanoic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$, is a weak acid in water. Which substance, when added to an aqueous solution of ethanoic acid, causes both the $\\mathrm{pH}$ and the percentage ionization of $\\mathrm{CH}_{3} \\mathrm{COOH}$ to decrease?\nA: $\\mathrm{NaCH}_{3} \\mathrm{COO}$\nB: $\\mathrm{NaCl}$\nC: $\\mathrm{CH}_{3} \\mathrm{COOH}$\nD: $\\quad \\mathrm{NaNO}_{3}$\nE: $\\mathrm{AgCl}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nEthanoic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$, is a weak acid in water. Which substance, when added to an aqueous solution of ethanoic acid, causes both the $\\mathrm{pH}$ and the percentage ionization of $\\mathrm{CH}_{3} \\mathrm{COOH}$ to decrease?\n\nA: $\\mathrm{NaCH}_{3} \\mathrm{COO}$\nB: $\\mathrm{NaCl}$\nC: $\\mathrm{CH}_{3} \\mathrm{COOH}$\nD: $\\quad \\mathrm{NaNO}_{3}$\nE: $\\mathrm{AgCl}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_575",
        "problem": "已知: $\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)$。室温下, 将稀盐酸滴加到某一元碱 $(\\mathrm{BOH})$ 溶液中, 测得混合溶液的 $\\mathrm{pOH}$ 与离子浓度的变化关系如图所示。下列叙述错误的是\n\n[图1]\nA: BOH 属于弱碱\nB: $\\mathrm{BOH}$ 的电离常数 $\\mathrm{K}=1 \\times 10^{-4.8}$\nC: $\\mathrm{P}$ 点所示的溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)$\nD: $\\mathrm{N}$ 点所示的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}(\\mathrm{BOH})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知: $\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)$。室温下, 将稀盐酸滴加到某一元碱 $(\\mathrm{BOH})$ 溶液中, 测得混合溶液的 $\\mathrm{pOH}$ 与离子浓度的变化关系如图所示。下列叙述错误的是\n\n[图1]\n\nA: BOH 属于弱碱\nB: $\\mathrm{BOH}$ 的电离常数 $\\mathrm{K}=1 \\times 10^{-4.8}$\nC: $\\mathrm{P}$ 点所示的溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)$\nD: $\\mathrm{N}$ 点所示的溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}(\\mathrm{BOH})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-52.jpg?height=774&width=834&top_left_y=144&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_278",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amounts (in mol or mmol) of carbon dioxide and water released in the thermal decomposition.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amounts (in mol or mmol) of carbon dioxide and water released in the thermal decomposition.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_859",
        "problem": "下列溶液中有关微粒的物质的量浓度关系正确的是（ )\nA: 物质的量浓度相等的 (1) $\\mathrm{NH}_{4} \\mathrm{HSO}_{4}$ 溶液、(2) $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液、(3) $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中, $\\mathrm{pH}$ 的大小关系: (2) $>$ (1) $>$ (3)\nB: 常温下, 将 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液和稀盐酸混合至溶液 $\\mathrm{pH}=7$ 时 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right.$ $\\left.{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: 将 $1 \\mathrm{molNaClO}$ 和 $2 \\mathrm{molNaHCO}_{3}$ 配制成 $1 \\mathrm{~L}$ 混合溶液: $\\mathrm{c}(\\mathrm{HClO})+\\mathrm{c}\\left(\\mathrm{ClO}^{-}\\right)=2 \\mathrm{c}\\left(\\mathrm{HClO}_{3}{ }^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nD: 浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 和 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液等体积混合后, 若溶液呈碱性, 则溶液中: $\\mathrm{C}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列溶液中有关微粒的物质的量浓度关系正确的是（ )\n\nA: 物质的量浓度相等的 (1) $\\mathrm{NH}_{4} \\mathrm{HSO}_{4}$ 溶液、(2) $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液、(3) $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中, $\\mathrm{pH}$ 的大小关系: (2) $>$ (1) $>$ (3)\nB: 常温下, 将 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液和稀盐酸混合至溶液 $\\mathrm{pH}=7$ 时 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right.$ $\\left.{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: 将 $1 \\mathrm{molNaClO}$ 和 $2 \\mathrm{molNaHCO}_{3}$ 配制成 $1 \\mathrm{~L}$ 混合溶液: $\\mathrm{c}(\\mathrm{HClO})+\\mathrm{c}\\left(\\mathrm{ClO}^{-}\\right)=2 \\mathrm{c}\\left(\\mathrm{HClO}_{3}{ }^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nD: 浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 和 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液等体积混合后, 若溶液呈碱性, 则溶液中: $\\mathrm{C}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_225",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the percentage by mass of acetic acid present in the vinegar sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the percentage by mass of acetic acid present in the vinegar sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_525",
        "problem": "从废旧磷酸铁锂电极材料 $\\left(\\mathrm{LiFePO}_{4}\\right.$ 、导电石墨、铝箔) 中回收锂的工艺流程如图, 下列说法错误的是\n\n[图1]\n\n已知: $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 在水中的溶解度随温度升高而降低, 但煮沸时与水发生反应\nA: 旧电池拆解前进行充分放电是因为放电可使 $\\mathrm{Li}^{+}$在负极富集\nB: “氧化”时, $\\mathrm{H}_{2} \\mathrm{O}_{2}$ 可用 $\\mathrm{KClO}_{3}$ 代替\nC: “氧化”时发生的化学反应方程式为: $2 \\mathrm{LiFePO}_{4}+\\mathrm{H}_{2} \\mathrm{O}_{2}+2 \\mathrm{HCl}=2 \\mathrm{LiCl}+2 \\mathrm{FePO}_{4}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: “系列操作”具体包括水浴加热、趁热过滤、洗涤、干燥\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n从废旧磷酸铁锂电极材料 $\\left(\\mathrm{LiFePO}_{4}\\right.$ 、导电石墨、铝箔) 中回收锂的工艺流程如图, 下列说法错误的是\n\n[图1]\n\n已知: $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 在水中的溶解度随温度升高而降低, 但煮沸时与水发生反应\n\nA: 旧电池拆解前进行充分放电是因为放电可使 $\\mathrm{Li}^{+}$在负极富集\nB: “氧化”时, $\\mathrm{H}_{2} \\mathrm{O}_{2}$ 可用 $\\mathrm{KClO}_{3}$ 代替\nC: “氧化”时发生的化学反应方程式为: $2 \\mathrm{LiFePO}_{4}+\\mathrm{H}_{2} \\mathrm{O}_{2}+2 \\mathrm{HCl}=2 \\mathrm{LiCl}+2 \\mathrm{FePO}_{4}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: “系列操作”具体包括水浴加热、趁热过滤、洗涤、干燥\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-02.jpg?height=297&width=1451&top_left_y=1599&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_311",
        "problem": "Which of the following equilibria shifts to the left when the external pressure is increased and shifts to the right when the temperature is increased?\nA: $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g) \\quad \\Delta H>0$\nB: $2 \\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2}(g) \\quad \\Delta H<0$\nC: $\\mathrm{PCl}_{3}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{PCl}_{5}(g) \\quad \\Delta H>0$\nD: $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g) \\quad \\Delta H<0$\nE: $2 \\mathrm{CO}_{2}(g) \\rightleftharpoons 2 \\mathrm{CO}(g)+\\mathrm{O}_{2}(g) \\quad \\Delta H>0$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following equilibria shifts to the left when the external pressure is increased and shifts to the right when the temperature is increased?\n\nA: $\\mathrm{N}_{2}(g)+\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{NO}(g) \\quad \\Delta H>0$\nB: $2 \\mathrm{H}_{2} \\mathrm{O}(g) \\rightleftharpoons \\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2}(g) \\quad \\Delta H<0$\nC: $\\mathrm{PCl}_{3}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{PCl}_{5}(g) \\quad \\Delta H>0$\nD: $\\mathrm{N}_{2}(g)+3 \\mathrm{H}_{2}(g) \\rightleftharpoons 2 \\mathrm{NH}_{3}(g) \\quad \\Delta H<0$\nE: $2 \\mathrm{CO}_{2}(g) \\rightleftharpoons 2 \\mathrm{CO}(g)+\\mathrm{O}_{2}(g) \\quad \\Delta H>0$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_612",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $0.10 \\mathrm{~mol} / \\mathrm{L}$ 的氨水滴定 $10.00 \\mathrm{~mL} 0.05 \\mathrm{~mol} / \\mathrm{L}$ 的二元酸 $\\mathrm{H}_{2} \\mathrm{~A}$ 的溶液, 滴定过程中加入氨水的体积 $(V)$ 与溶液中 $\\lg \\frac{c\\left(\\mathrm{OH}^{-}\\right)}{c\\left(\\mathrm{H}^{+}\\right)}$的关系如图所示。下列说法错误的是\n\n[图1]\nA: $\\mathrm{H}_{2} \\mathrm{~A}$ 为强电解质, 是一种强酸\nB: $\\mathrm{M} 、 \\mathrm{~N} 、 \\mathrm{P}$ 三点的溶液中, 水电离程度最大的是 $\\mathrm{N}$ 点\nC: P 点溶液中, $c\\left(\\mathrm{NH}_{4}^{+}\\right)+c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>2 c\\left(\\mathrm{~A}^{2-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, 氨水的电离平衡常数: $K=10^{-7} /(0.2 \\mathrm{x}-1)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $0.10 \\mathrm{~mol} / \\mathrm{L}$ 的氨水滴定 $10.00 \\mathrm{~mL} 0.05 \\mathrm{~mol} / \\mathrm{L}$ 的二元酸 $\\mathrm{H}_{2} \\mathrm{~A}$ 的溶液, 滴定过程中加入氨水的体积 $(V)$ 与溶液中 $\\lg \\frac{c\\left(\\mathrm{OH}^{-}\\right)}{c\\left(\\mathrm{H}^{+}\\right)}$的关系如图所示。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{H}_{2} \\mathrm{~A}$ 为强电解质, 是一种强酸\nB: $\\mathrm{M} 、 \\mathrm{~N} 、 \\mathrm{P}$ 三点的溶液中, 水电离程度最大的是 $\\mathrm{N}$ 点\nC: P 点溶液中, $c\\left(\\mathrm{NH}_{4}^{+}\\right)+c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>2 c\\left(\\mathrm{~A}^{2-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, 氨水的电离平衡常数: $K=10^{-7} /(0.2 \\mathrm{x}-1)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-21.jpg?height=428&width=660&top_left_y=1048&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1448",
        "problem": "Sulphuric acid is produced by catalytic oxidation of $\\mathrm{SO}_{2}$ to $\\mathrm{SO}_{3}$, absorption of $\\mathrm{SO}_{3}$ in concentrated sulphuric acid forming oleum (containing $20 \\% \\mathrm{SO}_{3}$ by mass) and appropriate dilution hereafter. The gas leaving the catalyst chamber contains nitrogen, oxygen, a trace of $\\mathrm{SO}_{2}$ and $10 \\%$ (by volume) of $\\mathrm{SO}_{3}$. Sulphur trioxide, $\\mathrm{SO}_{3}$, is converted into sulphuric acid (98\\% by mass) and/or oleum.\n\nAssuming that only $98 \\%$ sulphuric acid is produced, calculate the necessary mass of water and the mass of product produced thereby.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nSulphuric acid is produced by catalytic oxidation of $\\mathrm{SO}_{2}$ to $\\mathrm{SO}_{3}$, absorption of $\\mathrm{SO}_{3}$ in concentrated sulphuric acid forming oleum (containing $20 \\% \\mathrm{SO}_{3}$ by mass) and appropriate dilution hereafter. The gas leaving the catalyst chamber contains nitrogen, oxygen, a trace of $\\mathrm{SO}_{2}$ and $10 \\%$ (by volume) of $\\mathrm{SO}_{3}$. Sulphur trioxide, $\\mathrm{SO}_{3}$, is converted into sulphuric acid (98\\% by mass) and/or oleum.\n\nAssuming that only $98 \\%$ sulphuric acid is produced, calculate the necessary mass of water and the mass of product produced thereby.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the necessary mass of water , the mass of product produced thereby: $ 98 \\% \\mathrm{H}_{2} \\mathrm{SO}_{4}$].\nTheir units are, in order, [kg, kg], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "kg",
            "kg"
        ],
        "answer_sequence": [
            "the necessary mass of water ",
            "the mass of product produced thereby: $ 98 \\% \\mathrm{H}_{2} \\mathrm{SO}_{4}$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1068",
        "problem": "Give half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{K}_{2} \\mathrm{MnO}_{4}$ to $\\mathrm{KMnO}_{4}$\nA: oxidation\nB: reduction\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGive half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{K}_{2} \\mathrm{MnO}_{4}$ to $\\mathrm{KMnO}_{4}$\n\nA: oxidation\nB: reduction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_773",
        "problem": "在容积可变的密闭容器中充入 $2 \\mathrm{~mol} \\mathrm{~A}$ 和 $1 \\mathrm{~mol} \\mathrm{~B}$ 发生反应 $\\mathrm{mA}(\\mathrm{g})+\\mathrm{nB}(\\mathrm{g}) \\rightleftharpoons \\mathrm{pC}(\\mathrm{g})$ 。在一定温度下达到平衡时, 分别得到 $\\mathrm{A}$ 的物质的量浓度如下表, 以下说法正确的是\n\n| 压强 $p / \\mathrm{Pa}$ | $2 \\times 10^{5}$ | $5 \\times 10^{5}$ | $1 \\times 10^{6}$ |\n| :---: | :---: | :---: | :---: |\n| $c(\\mathrm{~A}) / \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$ | 0.08 | 0.20 | 0.44 |\nA: 维持压强 $2 \\times 10^{5} \\mathrm{~Pa}$, 若反应开始后 $5 \\mathrm{~min}$ 时达到平衡, 则 $\\mathrm{v}(\\mathrm{A})=0.016 \\mathrm{~mol} /(\\mathrm{L} \\cdot \\mathrm{min})$\nB: $\\mathrm{P}$ 从 $2 \\times 10^{5} \\mathrm{~Pa}$ 增加到 $5 \\times 10^{5} \\mathrm{~Pa}$ 时, 平衡逆向移动\nC: $P$ 为 $1 \\times 10^{6} \\mathrm{~Pa}$ 时, 平衡常数表达式 $K=\\frac{c^{p}(C)}{c^{m}(A) \\cdot c^{n}(B)}$\nD: 其它条件相同时, 在上述三个压强下分别发生该反应。A 的转化率随时间变化曲线如图所示 [图1]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在容积可变的密闭容器中充入 $2 \\mathrm{~mol} \\mathrm{~A}$ 和 $1 \\mathrm{~mol} \\mathrm{~B}$ 发生反应 $\\mathrm{mA}(\\mathrm{g})+\\mathrm{nB}(\\mathrm{g}) \\rightleftharpoons \\mathrm{pC}(\\mathrm{g})$ 。在一定温度下达到平衡时, 分别得到 $\\mathrm{A}$ 的物质的量浓度如下表, 以下说法正确的是\n\n| 压强 $p / \\mathrm{Pa}$ | $2 \\times 10^{5}$ | $5 \\times 10^{5}$ | $1 \\times 10^{6}$ |\n| :---: | :---: | :---: | :---: |\n| $c(\\mathrm{~A}) / \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$ | 0.08 | 0.20 | 0.44 |\n\nA: 维持压强 $2 \\times 10^{5} \\mathrm{~Pa}$, 若反应开始后 $5 \\mathrm{~min}$ 时达到平衡, 则 $\\mathrm{v}(\\mathrm{A})=0.016 \\mathrm{~mol} /(\\mathrm{L} \\cdot \\mathrm{min})$\nB: $\\mathrm{P}$ 从 $2 \\times 10^{5} \\mathrm{~Pa}$ 增加到 $5 \\times 10^{5} \\mathrm{~Pa}$ 时, 平衡逆向移动\nC: $P$ 为 $1 \\times 10^{6} \\mathrm{~Pa}$ 时, 平衡常数表达式 $K=\\frac{c^{p}(C)}{c^{m}(A) \\cdot c^{n}(B)}$\nD: 其它条件相同时, 在上述三个压强下分别发生该反应。A 的转化率随时间变化曲线如图所示 [图1]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-61.jpg?height=408&width=688&top_left_y=955&top_left_x=404"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_904",
        "problem": "据报道某科研团队在常温常压下实现了通过电化学的方法还原氮得到火箭燃料\n\n肼, 其中选取 POM 电解液作为电子与质子的载体, $\\mathrm{TiO}_{2}$ 作催化剂, 已知 $\\mathrm{A}$ 极上的电极反应为 $\\mathrm{POM}^{\\mathrm{Ox}}+\\mathrm{ne}=\\mathrm{POM}^{\\mathrm{Re}}$, 反应原理如图所示。下列有关说法错误的是\n\n[图1]\nA: $\\mathrm{H}^{+}$通过质子交换膜从左侧移向右侧\nB: 反应器中的离子方程式为 $4 \\mathrm{POM}^{\\mathrm{Re}}+\\mathrm{nN}_{2}+4 \\mathrm{nH}^{+}=\\mathrm{nN}_{2} \\mathrm{H}_{4}+4 \\mathrm{POM}^{\\mathrm{Ox}}$\nC: 若使用铅蓄电池作电源, 则 $\\mathrm{B}$ 电极应与电池中 $\\mathrm{Pb}$ 电极相连\nD: 装置工作时反应消耗的 $\\mathrm{N}_{2}$ 与产生的 $\\mathrm{O}_{2}$ 的物质的量之比为 $1: 1$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n据报道某科研团队在常温常压下实现了通过电化学的方法还原氮得到火箭燃料\n\n肼, 其中选取 POM 电解液作为电子与质子的载体, $\\mathrm{TiO}_{2}$ 作催化剂, 已知 $\\mathrm{A}$ 极上的电极反应为 $\\mathrm{POM}^{\\mathrm{Ox}}+\\mathrm{ne}=\\mathrm{POM}^{\\mathrm{Re}}$, 反应原理如图所示。下列有关说法错误的是\n\n[图1]\n\nA: $\\mathrm{H}^{+}$通过质子交换膜从左侧移向右侧\nB: 反应器中的离子方程式为 $4 \\mathrm{POM}^{\\mathrm{Re}}+\\mathrm{nN}_{2}+4 \\mathrm{nH}^{+}=\\mathrm{nN}_{2} \\mathrm{H}_{4}+4 \\mathrm{POM}^{\\mathrm{Ox}}$\nC: 若使用铅蓄电池作电源, 则 $\\mathrm{B}$ 电极应与电池中 $\\mathrm{Pb}$ 电极相连\nD: 装置工作时反应消耗的 $\\mathrm{N}_{2}$ 与产生的 $\\mathrm{O}_{2}$ 的物质的量之比为 $1: 1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-53.jpg?height=569&width=1448&top_left_y=1366&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_885",
        "problem": "甘氨酸 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COOH}\\right.$, 在溶液中以 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}$-形式存在, 用 $\\mathrm{HR}$ 表示)为人体必需氨基酸, 在水溶液中存在如下平衡: $\\mathrm{H}_{2} \\mathrm{R}^{+} \\stackrel{\\mathrm{K}_{1}}{\\rightleftharpoons} \\mathrm{HR} \\stackrel{\\mathrm{K}_{2}}{\\rightleftharpoons} \\mathrm{R}^{-} \\circ 0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HR}$ 溶液中 $\\mathrm{H}_{2} \\mathrm{R}^{+} 、 \\mathrm{HR}$ 和 $\\mathrm{R}$ 的浓度对数值与 $\\mathrm{pH}$ 的关系如图所示。下列说法不正确的是\n\n[图1]\nA: 曲线 1 表示 $\\mathrm{H}_{2} \\mathrm{R}^{+}$\nB: 甘氨酸溶液显弱酸性\nC: $\\mathrm{M}$ 点, $\\mathrm{pH}=\\frac{-\\lg \\mathrm{K}_{1}-\\lg \\mathrm{K}_{2}}{2}$\nD: $\\mathrm{NaR}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}(\\mathrm{HR})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n甘氨酸 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COOH}\\right.$, 在溶液中以 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}$-形式存在, 用 $\\mathrm{HR}$ 表示)为人体必需氨基酸, 在水溶液中存在如下平衡: $\\mathrm{H}_{2} \\mathrm{R}^{+} \\stackrel{\\mathrm{K}_{1}}{\\rightleftharpoons} \\mathrm{HR} \\stackrel{\\mathrm{K}_{2}}{\\rightleftharpoons} \\mathrm{R}^{-} \\circ 0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HR}$ 溶液中 $\\mathrm{H}_{2} \\mathrm{R}^{+} 、 \\mathrm{HR}$ 和 $\\mathrm{R}$ 的浓度对数值与 $\\mathrm{pH}$ 的关系如图所示。下列说法不正确的是\n\n[图1]\n\nA: 曲线 1 表示 $\\mathrm{H}_{2} \\mathrm{R}^{+}$\nB: 甘氨酸溶液显弱酸性\nC: $\\mathrm{M}$ 点, $\\mathrm{pH}=\\frac{-\\lg \\mathrm{K}_{1}-\\lg \\mathrm{K}_{2}}{2}$\nD: $\\mathrm{NaR}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}(\\mathrm{HR})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-099.jpg?height=574&width=963&top_left_y=627&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_339",
        "problem": "The reaction below reaches equilibrium in a closed reaction vessel.\n\n$\\mathrm{CaCO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{CaO}(\\mathrm{s})+\\mathrm{CO}_{2}(\\mathrm{~g}), \\quad \\Delta H=178 \\mathrm{~kJ}$\n\nWhich of the following actions cause(s) an increase in the partial pressure of $\\mathrm{CO}_{2}(g)$ ?\n\n(i) increasing the temperature\n\n(ii) adding some $\\mathrm{CaCO}_{3}(\\mathrm{~s})$\n\n(iii) increasing the volume of the reaction vessel\nA: (i) only\nB: (i) and (ii)\nC: (i), (ii) and (iii)\nD: (ii) only\nE: (i) and (iii)\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction below reaches equilibrium in a closed reaction vessel.\n\n$\\mathrm{CaCO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{CaO}(\\mathrm{s})+\\mathrm{CO}_{2}(\\mathrm{~g}), \\quad \\Delta H=178 \\mathrm{~kJ}$\n\nWhich of the following actions cause(s) an increase in the partial pressure of $\\mathrm{CO}_{2}(g)$ ?\n\n(i) increasing the temperature\n\n(ii) adding some $\\mathrm{CaCO}_{3}(\\mathrm{~s})$\n\n(iii) increasing the volume of the reaction vessel\n\nA: (i) only\nB: (i) and (ii)\nC: (i), (ii) and (iii)\nD: (ii) only\nE: (i) and (iii)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_145",
        "problem": "Given the following set of equilibria and their respective constants\n\n$$\n\\begin{aligned}\n& \\mathrm{NH}_{4}^{+} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{NH}_{3} \\quad \\mathrm{~K}_{\\mathrm{a}}=5.6 \\times 10^{-10} \\\\\n& \\mathrm{H}_{2} \\mathrm{CO}_{3} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{HCO}_{3}^{-} \\quad \\mathrm{K}_{\\mathrm{a} 1}=4.2 \\times 10^{-7} \\\\\n& \\mathrm{HCO}_{3}^{-} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{CO}_{3} 2^{-} \\quad \\mathrm{K}_{\\mathrm{a} 2}=2.4 \\times 10^{-8}\n\\end{aligned}\n$$\n\nwhat would the equilibrium constant be for the reaction below?\n\n$$\n\\mathrm{H}_{2} \\mathrm{CO}_{3}+2 \\mathrm{NH}_{3} \\rightleftharpoons\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3} \\quad \\mathrm{~K}=?\n$$\nA: $1.8 \\times 10^{-5}$\nB: $4.4 \\times 10^{-7}$\nC: $9.0 \\times 10^{-6}$\nD: $3.2 \\times 10^{4}$\nE: $3.1 \\times 10^{-5}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGiven the following set of equilibria and their respective constants\n\n$$\n\\begin{aligned}\n& \\mathrm{NH}_{4}^{+} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{NH}_{3} \\quad \\mathrm{~K}_{\\mathrm{a}}=5.6 \\times 10^{-10} \\\\\n& \\mathrm{H}_{2} \\mathrm{CO}_{3} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{HCO}_{3}^{-} \\quad \\mathrm{K}_{\\mathrm{a} 1}=4.2 \\times 10^{-7} \\\\\n& \\mathrm{HCO}_{3}^{-} \\rightleftharpoons \\mathrm{H}^{+}+\\mathrm{CO}_{3} 2^{-} \\quad \\mathrm{K}_{\\mathrm{a} 2}=2.4 \\times 10^{-8}\n\\end{aligned}\n$$\n\nwhat would the equilibrium constant be for the reaction below?\n\n$$\n\\mathrm{H}_{2} \\mathrm{CO}_{3}+2 \\mathrm{NH}_{3} \\rightleftharpoons\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3} \\quad \\mathrm{~K}=?\n$$\n\nA: $1.8 \\times 10^{-5}$\nB: $4.4 \\times 10^{-7}$\nC: $9.0 \\times 10^{-6}$\nD: $3.2 \\times 10^{4}$\nE: $3.1 \\times 10^{-5}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_830",
        "problem": ". $25^{\\circ} \\mathrm{C}$ 时, 一水合二甲胺 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right]$ 的 $\\mathrm{p} K_{\\mathrm{b}}=3.7$, 叠氮酸 $\\left(\\mathrm{HN}_{3}\\right)$ 的 $\\mathrm{p} K_{\\mathrm{a}}=4.7(\\mathrm{~K}$为电离常数)。 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2} \\mathrm{~N}_{3}$ 溶液中, $\\lg c\\left(\\mathrm{HN}_{3}\\right) 、 \\lg c\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right]$ 、 $\\lg c\\left(\\mathrm{H}^{+}\\right) 、 \\lg c\\left(\\mathrm{OH}^{-}\\right)$随 $\\mathrm{pH}$ 变化(加入盐酸或 $\\mathrm{NaOH}$ 溶液)的关系如图所示。下列说法正确的是\n\n[图1]\nA: 曲线c 代表 $\\lg c\\left(\\mathrm{HN}_{3}\\right)$ 随 $\\mathrm{pH}$ 的变化\nB: 原溶液中 $c\\left(\\mathrm{H}^{+}\\right)-c\\left(\\mathrm{HN}_{3}\\right)=c\\left(\\mathrm{OH}^{-}\\right)-c\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right]$\nC: 原溶液中 $\\frac{0.1 K_{\\mathrm{a}}}{K_{\\mathrm{a}}+c\\left(\\mathrm{H}^{+}\\right)}+c\\left(\\mathrm{H}^{+}\\right)=\\frac{0.1 K_{\\mathrm{b}}}{K_{\\mathrm{b}}+c\\left(\\mathrm{OH}^{-}\\right)}+c\\left(\\mathrm{OH}^{-}\\right)$\nD: 图中 $\\mathrm{P}$ 点对应的溶液 $\\mathrm{pH}=7.5$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n. $25^{\\circ} \\mathrm{C}$ 时, 一水合二甲胺 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right]$ 的 $\\mathrm{p} K_{\\mathrm{b}}=3.7$, 叠氮酸 $\\left(\\mathrm{HN}_{3}\\right)$ 的 $\\mathrm{p} K_{\\mathrm{a}}=4.7(\\mathrm{~K}$为电离常数)。 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH}_{2} \\mathrm{~N}_{3}$ 溶液中, $\\lg c\\left(\\mathrm{HN}_{3}\\right) 、 \\lg c\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right]$ 、 $\\lg c\\left(\\mathrm{H}^{+}\\right) 、 \\lg c\\left(\\mathrm{OH}^{-}\\right)$随 $\\mathrm{pH}$ 变化(加入盐酸或 $\\mathrm{NaOH}$ 溶液)的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: 曲线c 代表 $\\lg c\\left(\\mathrm{HN}_{3}\\right)$ 随 $\\mathrm{pH}$ 的变化\nB: 原溶液中 $c\\left(\\mathrm{H}^{+}\\right)-c\\left(\\mathrm{HN}_{3}\\right)=c\\left(\\mathrm{OH}^{-}\\right)-c\\left[\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NH} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right]$\nC: 原溶液中 $\\frac{0.1 K_{\\mathrm{a}}}{K_{\\mathrm{a}}+c\\left(\\mathrm{H}^{+}\\right)}+c\\left(\\mathrm{H}^{+}\\right)=\\frac{0.1 K_{\\mathrm{b}}}{K_{\\mathrm{b}}+c\\left(\\mathrm{OH}^{-}\\right)}+c\\left(\\mathrm{OH}^{-}\\right)$\nD: 图中 $\\mathrm{P}$ 点对应的溶液 $\\mathrm{pH}=7.5$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-042.jpg?height=394&width=599&top_left_y=174&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_39",
        "problem": "Which is the safest method for performing a flame test?\nA: Dissolve the metal salt in methanol, then squirt the solution into a lit Bunsen burner from at least 1 meter away.\nB: Dissolve the metal salt in methanol, then pour the methanol into a crystallizing dish, igniting it with the flame from a Bunsen burner.\nC: Soak a wooden splint in an aqueous solution of the metal salt, then burn the splint in the flame from a Bunsen burner.\nD: Soak a wooden splint in an aqueous solution of the metal salt, then heat the splint on top of a ceramic hotplate.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich is the safest method for performing a flame test?\n\nA: Dissolve the metal salt in methanol, then squirt the solution into a lit Bunsen burner from at least 1 meter away.\nB: Dissolve the metal salt in methanol, then pour the methanol into a crystallizing dish, igniting it with the flame from a Bunsen burner.\nC: Soak a wooden splint in an aqueous solution of the metal salt, then burn the splint in the flame from a Bunsen burner.\nD: Soak a wooden splint in an aqueous solution of the metal salt, then heat the splint on top of a ceramic hotplate.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_511",
        "problem": "通过 $\\mathrm{CH}_{4}$ 超干重整 $\\mathrm{CO}_{2}$ 技术可得到富含 $\\mathrm{CO}$ 的化工原料，其过程如下图所示。下列说法错误的是（ ）\n\n[图1]\n\n过程 II\nA: 过程I的化学方程式为 $\\mathrm{CH}_{4}+\\mathrm{CO}_{2} \\stackrel{\\text { 催化剂 }}{=} 2 \\mathrm{CO}+2 \\mathrm{H}_{2}$\nB: 过程II的化学方程式为 $\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\stackrel{\\text { 催化剂 }}{=} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{CO}$\nC: 该技术实现了含碳物质与含氢物质的分离\nD: $\\mathrm{Ni} 、 \\mathrm{Fe} 、 \\mathrm{CaCO}_{3}$ 均是上述过程中用到的催化剂\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n通过 $\\mathrm{CH}_{4}$ 超干重整 $\\mathrm{CO}_{2}$ 技术可得到富含 $\\mathrm{CO}$ 的化工原料，其过程如下图所示。下列说法错误的是（ ）\n\n[图1]\n\n过程 II\n\nA: 过程I的化学方程式为 $\\mathrm{CH}_{4}+\\mathrm{CO}_{2} \\stackrel{\\text { 催化剂 }}{=} 2 \\mathrm{CO}+2 \\mathrm{H}_{2}$\nB: 过程II的化学方程式为 $\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\stackrel{\\text { 催化剂 }}{=} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{CO}$\nC: 该技术实现了含碳物质与含氢物质的分离\nD: $\\mathrm{Ni} 、 \\mathrm{Fe} 、 \\mathrm{CaCO}_{3}$ 均是上述过程中用到的催化剂\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-45.jpg?height=357&width=1137&top_left_y=1146&top_left_x=448"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_623",
        "problem": "以下是制备环丁基甲酸的反应过程, 下列说法错误的是[图1]\nA: $\\mathrm{a} \\rightarrow \\mathrm{b}$ 的反应属于取代反应\nB: $\\mathrm{b}$ 的一氯代物有 3 种(不考虑立体异构)\nC: $a 、 b 、 c$ 均能与 $\\mathrm{NaOH}$ 溶液反应\nD: 利用 1-氯-3-溴丙烷合成 $\\mathrm{a}$ 时使用的试剂依次是氢氧化钠醇溶液 $\\rightarrow$ 酸性高锰酸钾溶液 $\\rightarrow$ 浓硫酸和乙醇\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n以下是制备环丁基甲酸的反应过程, 下列说法错误的是[图1]\n\nA: $\\mathrm{a} \\rightarrow \\mathrm{b}$ 的反应属于取代反应\nB: $\\mathrm{b}$ 的一氯代物有 3 种(不考虑立体异构)\nC: $a 、 b 、 c$ 均能与 $\\mathrm{NaOH}$ 溶液反应\nD: 利用 1-氯-3-溴丙烷合成 $\\mathrm{a}$ 时使用的试剂依次是氢氧化钠醇溶液 $\\rightarrow$ 酸性高锰酸钾溶液 $\\rightarrow$ 浓硫酸和乙醇\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/QMmDwzZC/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1268",
        "problem": "Type II electrodes that are made of a metal covered with a sparingly soluble salt of the metal are dipped into a soluble salt solution containing an anion of the sparingly soluble salt. The silver/silver chloride $(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$ and the calomel electrode $(\\mathrm{Hg}$, $\\left.\\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Cl}\\right)$ are examples of such electrodes. The standard emf of a cell built of those electrodes (-) Ag, $\\mathrm{AgCl}^{-} / \\mathrm{Cl}^{-} \\| \\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Hg}(+)$ is $E^{0}=0.0455 \\mathrm{~V}$ at $T=298 \\mathrm{~K}$. The temperature coefficient for this cell is $d E^{0} / d T=3.38 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}$.\n\nCalculate the enthalpy change for the process taking place at $298 \\mathrm{~K}$.\n\n$\\Delta S=n F \\Delta E / \\Delta T$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nType II electrodes that are made of a metal covered with a sparingly soluble salt of the metal are dipped into a soluble salt solution containing an anion of the sparingly soluble salt. The silver/silver chloride $(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$ and the calomel electrode $(\\mathrm{Hg}$, $\\left.\\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Cl}\\right)$ are examples of such electrodes. The standard emf of a cell built of those electrodes (-) Ag, $\\mathrm{AgCl}^{-} / \\mathrm{Cl}^{-} \\| \\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Hg}(+)$ is $E^{0}=0.0455 \\mathrm{~V}$ at $T=298 \\mathrm{~K}$. The temperature coefficient for this cell is $d E^{0} / d T=3.38 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}$.\n\nCalculate the enthalpy change for the process taking place at $298 \\mathrm{~K}$.\n\n$\\Delta S=n F \\Delta E / \\Delta T$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_61",
        "problem": "Which elements are most similar in atomic size?\nA: $\\mathrm{H}(Z=1)$ and $\\mathrm{Li}(Z=3)$\nB: $\\mathrm{C}(Z=6)$ and $\\mathrm{Si}(Z=14)$\nC: $\\mathrm{Mn}(Z=25)$ and $\\mathrm{Tc}(Z=43)$\nD: $\\operatorname{Zr}(Z=40)$ and $\\operatorname{Hf}(Z=72)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich elements are most similar in atomic size?\n\nA: $\\mathrm{H}(Z=1)$ and $\\mathrm{Li}(Z=3)$\nB: $\\mathrm{C}(Z=6)$ and $\\mathrm{Si}(Z=14)$\nC: $\\mathrm{Mn}(Z=25)$ and $\\mathrm{Tc}(Z=43)$\nD: $\\operatorname{Zr}(Z=40)$ and $\\operatorname{Hf}(Z=72)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_384",
        "problem": "For the reaction:\n\n$$\n\\begin{aligned}\n\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}(a q)+ & \\mathrm{SCN}^{-}(a q) \\rightarrow \\\\\n& \\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5} \\mathrm{SCN}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{aligned}\n$$\n\nThe following data were collected:\n\n| $\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right], \\mathrm{M}$ | $\\left[\\mathrm{SCN}^{-}\\right], \\mathrm{M}$ | Rate, $\\mathrm{M} \\mathrm{hr}^{-1}$ |\n| :---: | :---: | :---: |\n| 0.028 | 0.040 | $8.1 \\times 10^{-6}$ |\n| 0.028 | 0.055 | $1.1 \\times 10^{-5}$ |\n| 0.037 | 0.055 | $1.5 \\times 10^{-5}$ |\n\nWhat is the rate law for the reaction?\nA: Rate $=\\left(9.1 \\times 10^{-9} \\mathrm{M}^{-1} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]$\nB: Rate $=\\left(7.2 \\times 10^{-3} \\mathrm{M}^{-1} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]$\nC: Rate $=\\left(2.9 \\times 10^{-4} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]$\nD: Rate $=\\left(3.9 \\times 10^{-4} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the reaction:\n\n$$\n\\begin{aligned}\n\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}(a q)+ & \\mathrm{SCN}^{-}(a q) \\rightarrow \\\\\n& \\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{5} \\mathrm{SCN}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{aligned}\n$$\n\nThe following data were collected:\n\n| $\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right], \\mathrm{M}$ | $\\left[\\mathrm{SCN}^{-}\\right], \\mathrm{M}$ | Rate, $\\mathrm{M} \\mathrm{hr}^{-1}$ |\n| :---: | :---: | :---: |\n| 0.028 | 0.040 | $8.1 \\times 10^{-6}$ |\n| 0.028 | 0.055 | $1.1 \\times 10^{-5}$ |\n| 0.037 | 0.055 | $1.5 \\times 10^{-5}$ |\n\nWhat is the rate law for the reaction?\n\nA: Rate $=\\left(9.1 \\times 10^{-9} \\mathrm{M}^{-1} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]$\nB: Rate $=\\left(7.2 \\times 10^{-3} \\mathrm{M}^{-1} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]$\nC: Rate $=\\left(2.9 \\times 10^{-4} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]$\nD: Rate $=\\left(3.9 \\times 10^{-4} \\mathrm{hr}^{-1}\\right)\\left[\\mathrm{Cr}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}{ }^{3+}\\right]$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1451",
        "problem": "Methanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |Calculate $\\Delta G^{0}$ for the reaction at $298 \\mathrm{~K}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nMethanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |\n\nproblem:\nCalculate $\\Delta G^{0}$ for the reaction at $298 \\mathrm{~K}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~J} \\mathrm{~K}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~J} \\mathrm{~K}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_932",
        "problem": "常温下, 向一定浓度的 $\\mathrm{NaA}$ 溶液中加适量强酸或强碱(忽略溶液体积变化), 溶液中 $\\mathrm{c}(\\mathrm{HA}) 、 \\mathrm{c}\\left(\\mathrm{A}^{-}\\right) 、 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right) 、 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$的负对数 $(-\\lg \\mathrm{c})$ 随溶液 $\\mathrm{pH}$ 的变化关系如下图所示。下列叙述不正确的是\n\n[图1]\nA: 曲线(1)表示 $-\\operatorname{lgc}\\left(\\mathrm{H}^{+}\\right)$随溶液 $\\mathrm{pH}$ 的变化情况\nB: 曲线(1)和曲线(2)的交点对应的溶液中存在 $c\\left(N^{+}\\right)=c\\left(A^{-}\\right)$\nC: 常温下, 将 A 点对应溶液加水稀释, $\\frac{\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)}{\\mathrm{c}(\\mathrm{HA}) \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)}$不变\nD: 等物质的量浓度、等体积的 $\\mathrm{NaA}$ 溶液与 $\\mathrm{HA}$ 溶液混合后: $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}(\\mathrm{HA})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向一定浓度的 $\\mathrm{NaA}$ 溶液中加适量强酸或强碱(忽略溶液体积变化), 溶液中 $\\mathrm{c}(\\mathrm{HA}) 、 \\mathrm{c}\\left(\\mathrm{A}^{-}\\right) 、 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right) 、 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$的负对数 $(-\\lg \\mathrm{c})$ 随溶液 $\\mathrm{pH}$ 的变化关系如下图所示。下列叙述不正确的是\n\n[图1]\n\nA: 曲线(1)表示 $-\\operatorname{lgc}\\left(\\mathrm{H}^{+}\\right)$随溶液 $\\mathrm{pH}$ 的变化情况\nB: 曲线(1)和曲线(2)的交点对应的溶液中存在 $c\\left(N^{+}\\right)=c\\left(A^{-}\\right)$\nC: 常温下, 将 A 点对应溶液加水稀释, $\\frac{\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)}{\\mathrm{c}(\\mathrm{HA}) \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)}$不变\nD: 等物质的量浓度、等体积的 $\\mathrm{NaA}$ 溶液与 $\\mathrm{HA}$ 溶液混合后: $\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}(\\mathrm{HA})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-092.jpg?height=746&width=879&top_left_y=181&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1428",
        "problem": "The amount of many reducing agents can be determined by permanganatometric titration in alkaline medium allowing permanganate ion reduction to manganate.Another sample of crotonic acid and alkali (in an excess) were added to flask B, this mixture lacking barium salt. An excess of KI (instead of cyanide) was added as a reducing agent. The mixture was then acidified and the iodine evolved was titrated with a thiosulfate solution $\\left(c_{\\mathrm{S}}=0.1000 \\mathrm{~mol} \\mathrm{dm}{ }^{-3}\\right) .4 .90 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{S} 1}\\right)$ of the titrant was used to reach the endpoint.\n\nCalculate the mass of crotonic acid (in $\\mathrm{mg}$ ).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe amount of many reducing agents can be determined by permanganatometric titration in alkaline medium allowing permanganate ion reduction to manganate.\n\nproblem:\nAnother sample of crotonic acid and alkali (in an excess) were added to flask B, this mixture lacking barium salt. An excess of KI (instead of cyanide) was added as a reducing agent. The mixture was then acidified and the iodine evolved was titrated with a thiosulfate solution $\\left(c_{\\mathrm{S}}=0.1000 \\mathrm{~mol} \\mathrm{dm}{ }^{-3}\\right) .4 .90 \\mathrm{~cm}^{3}\\left(V_{\\mathrm{S} 1}\\right)$ of the titrant was used to reach the endpoint.\n\nCalculate the mass of crotonic acid (in $\\mathrm{mg}$ ).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_376",
        "problem": "The sequence of amino acids in a protein is known as its\nA: primary structure.\nB: secondary structure.\nC: tertiary structure.\nD: quaternary structure.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe sequence of amino acids in a protein is known as its\n\nA: primary structure.\nB: secondary structure.\nC: tertiary structure.\nD: quaternary structure.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_634",
        "problem": "$\\mathrm{MgCO}_{3}$ 和 $\\mathrm{CaCO}_{3}$ 的能量关系如图所示 $(\\mathrm{M}=\\mathrm{Ca} 、 \\mathrm{Mg})$ :\n\n[图1]\n\n已知：离子电荷相同时, 半径越小, 离子键越强。下列说法不正确的是\nA: $\\Delta H_{l}\\left(\\mathrm{MgCO}_{3}\\right)>\\Delta H_{l}\\left(\\mathrm{CaCO}_{3}\\right)>0$\nB: $\\Delta H_{2}\\left(\\mathrm{MgCO}_{3}\\right)=\\Delta H_{2}\\left(\\mathrm{CaCO}_{3}\\right)>0$\nC: $\\Delta H_{1}\\left(\\mathrm{CaCO}_{3}\\right)-\\Delta H_{l}\\left(\\mathrm{MgCO}_{3}\\right)=\\Delta H_{3}(\\mathrm{CaO})-\\Delta H_{3}(\\mathrm{MgO})$\nD: 对于 $\\mathrm{MgCO}_{3}$ 和 $\\mathrm{CaCO}_{3}, \\Delta H_{1}+\\Delta H_{2}>\\Delta H_{3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{MgCO}_{3}$ 和 $\\mathrm{CaCO}_{3}$ 的能量关系如图所示 $(\\mathrm{M}=\\mathrm{Ca} 、 \\mathrm{Mg})$ :\n\n[图1]\n\n已知：离子电荷相同时, 半径越小, 离子键越强。下列说法不正确的是\n\nA: $\\Delta H_{l}\\left(\\mathrm{MgCO}_{3}\\right)>\\Delta H_{l}\\left(\\mathrm{CaCO}_{3}\\right)>0$\nB: $\\Delta H_{2}\\left(\\mathrm{MgCO}_{3}\\right)=\\Delta H_{2}\\left(\\mathrm{CaCO}_{3}\\right)>0$\nC: $\\Delta H_{1}\\left(\\mathrm{CaCO}_{3}\\right)-\\Delta H_{l}\\left(\\mathrm{MgCO}_{3}\\right)=\\Delta H_{3}(\\mathrm{CaO})-\\Delta H_{3}(\\mathrm{MgO})$\nD: 对于 $\\mathrm{MgCO}_{3}$ 和 $\\mathrm{CaCO}_{3}, \\Delta H_{1}+\\Delta H_{2}>\\Delta H_{3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-53.jpg?height=232&width=939&top_left_y=1466&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_558",
        "problem": "图(I) 和图(II) 分别为二元酸 $\\mathrm{H}_{2} \\mathrm{~A}$ 和乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 溶液中各微粒的百分含量 $\\delta$ (即物质的量百分数)随溶液 $\\mathrm{pH}$ 的变化曲线 $\\left(25^{\\circ} \\mathrm{C}\\right)$ 。下列说法正确的是\n\n[图1]\n\n图 ( I )\n\n[图2]\n\n图 ( II)\nA: $\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right] \\mathrm{A}$ 溶液显碱性\nB: 乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 的 $\\mathrm{K}_{\\mathrm{b} 2}=10^{-7.15}$\nC: $\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right][\\mathrm{HA}]$ 溶液中各离子浓度大小关系为: $\\mathrm{c}\\left(\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right]^{+}>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)\\right.$\nD: 向 $\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right][\\mathrm{HA}]$ 溶液中通入一定量的 $\\mathrm{HCl}$ 气体, 则 $\\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right) \\cdot c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}{\\mathrm{c}\\left(\\left[\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{+}\\right) \\cdot c\\left(\\mathrm{HA}^{-}\\right)}$可能增大也可能减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n图(I) 和图(II) 分别为二元酸 $\\mathrm{H}_{2} \\mathrm{~A}$ 和乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 溶液中各微粒的百分含量 $\\delta$ (即物质的量百分数)随溶液 $\\mathrm{pH}$ 的变化曲线 $\\left(25^{\\circ} \\mathrm{C}\\right)$ 。下列说法正确的是\n\n[图1]\n\n图 ( I )\n\n[图2]\n\n图 ( II)\n\nA: $\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right] \\mathrm{A}$ 溶液显碱性\nB: 乙二胺 $\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)$ 的 $\\mathrm{K}_{\\mathrm{b} 2}=10^{-7.15}$\nC: $\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right][\\mathrm{HA}]$ 溶液中各离子浓度大小关系为: $\\mathrm{c}\\left(\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right]^{+}>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)\\right.$\nD: 向 $\\left[\\mathrm{H}_{3} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right][\\mathrm{HA}]$ 溶液中通入一定量的 $\\mathrm{HCl}$ 气体, 则 $\\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}\\right) \\cdot c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}{\\mathrm{c}\\left(\\left[\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{3}\\right]^{+}\\right) \\cdot c\\left(\\mathrm{HA}^{-}\\right)}$可能增大也可能减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-75.jpg?height=392&width=506&top_left_y=181&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-75.jpg?height=466&width=737&top_left_y=144&top_left_x=979",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-75.jpg?height=54&width=1319&top_left_y=1252&top_left_x=346",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-75.jpg?height=51&width=1151&top_left_y=1985&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1303",
        "problem": "Extraction of gold using sodium cyanide, a very poisonous chemical, causes environmental problems and gives rise to serious public concern about the use of this so called \"cyanide process\". Thiosulfate leaching of gold has been considered as an alternative. In this process, the main reagent is ammonium thiosulfate, $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$, which is relatively non-toxic. Although this process appears to be environmentally benign, the chemistry involved is very complex and needs to be studied thoroughly. The solution used for leaching gold contains $\\mathrm{S}_{2} \\mathrm{O}_{3}^{2-}, \\mathrm{Cu}^{2+}, \\mathrm{NH}_{3}$, and dissolved $\\mathrm{O}_{2}$. The solution must have a $\\mathrm{pH}$ greater than 8.5 to allow free ammonia to be present.\n\nAccording to the proposed mechanism, a local voltaic micro-cell is formed on the surface of gold particles during the leaching process and operates as follows:\n\nAnode:\n\n$\\mathrm{Au}(\\mathrm{s})+2 \\mathrm{NH}_{3}(\\mathrm{aq}) \\rightarrow\\left[\\mathrm{Au}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(\\mathrm{aq})+e^{-}$\n\n$\\left[\\mathrm{Au}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q)+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}(a q) \\rightarrow\\left[\\mathrm{Au}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}\\right]^{3-}(a q)+2 \\mathrm{NH}_{3}(a q)$\n\nCathode:\n\n$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}(a q)+e^{-} \\rightarrow\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q)+2 \\mathrm{NH}_{3}(a q)$\n\n$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q)+3 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}(a q) \\rightarrow\\left[\\mathrm{Cu}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{3}\\right]^{5-}(a q)+2 \\mathrm{NH}_{3}(a q)$\n\nThe formation constants, $K_{f}$, of $\\left[\\mathrm{Au}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$and $\\left[\\mathrm{Au}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}\\right]^{3-}$ complexes are $1.00 \\cdot 10^{26}$ and $1.00 \\cdot 10^{28}$, respectively. Consider a leaching solution in which the equilibrium concentrations of the species are as follows:\n\n$\\left[\\mathrm{S}_{2} \\mathrm{O}_{3}^{2 \\cdot}\\right]=0.100 ;\\left[\\mathrm{NH}_{3}\\right]=0.100$ and the total concentration of gold $(\\mathrm{I})$ species $=$ $5.50 \\cdot 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$.Calculate the percentage of gold $(I)$ ion that exists in the form of thiosulfate complex.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nExtraction of gold using sodium cyanide, a very poisonous chemical, causes environmental problems and gives rise to serious public concern about the use of this so called \"cyanide process\". Thiosulfate leaching of gold has been considered as an alternative. In this process, the main reagent is ammonium thiosulfate, $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$, which is relatively non-toxic. Although this process appears to be environmentally benign, the chemistry involved is very complex and needs to be studied thoroughly. The solution used for leaching gold contains $\\mathrm{S}_{2} \\mathrm{O}_{3}^{2-}, \\mathrm{Cu}^{2+}, \\mathrm{NH}_{3}$, and dissolved $\\mathrm{O}_{2}$. The solution must have a $\\mathrm{pH}$ greater than 8.5 to allow free ammonia to be present.\n\nAccording to the proposed mechanism, a local voltaic micro-cell is formed on the surface of gold particles during the leaching process and operates as follows:\n\nAnode:\n\n$\\mathrm{Au}(\\mathrm{s})+2 \\mathrm{NH}_{3}(\\mathrm{aq}) \\rightarrow\\left[\\mathrm{Au}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(\\mathrm{aq})+e^{-}$\n\n$\\left[\\mathrm{Au}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q)+2 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}(a q) \\rightarrow\\left[\\mathrm{Au}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}\\right]^{3-}(a q)+2 \\mathrm{NH}_{3}(a q)$\n\nCathode:\n\n$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}\\right]^{2+}(a q)+e^{-} \\rightarrow\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q)+2 \\mathrm{NH}_{3}(a q)$\n\n$\\left[\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}(a q)+3 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}(a q) \\rightarrow\\left[\\mathrm{Cu}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{3}\\right]^{5-}(a q)+2 \\mathrm{NH}_{3}(a q)$\n\nThe formation constants, $K_{f}$, of $\\left[\\mathrm{Au}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$and $\\left[\\mathrm{Au}\\left(\\mathrm{S}_{2} \\mathrm{O}_{3}\\right)_{2}\\right]^{3-}$ complexes are $1.00 \\cdot 10^{26}$ and $1.00 \\cdot 10^{28}$, respectively. Consider a leaching solution in which the equilibrium concentrations of the species are as follows:\n\n$\\left[\\mathrm{S}_{2} \\mathrm{O}_{3}^{2 \\cdot}\\right]=0.100 ;\\left[\\mathrm{NH}_{3}\\right]=0.100$ and the total concentration of gold $(\\mathrm{I})$ species $=$ $5.50 \\cdot 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$.\n\nproblem:\nCalculate the percentage of gold $(I)$ ion that exists in the form of thiosulfate complex.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_537",
        "problem": "20 世纪初, 科学家设计出质子膜 $\\mathrm{H}_{2} \\mathrm{~S}$ 燃料电池, 实现了利用 $\\mathrm{H}_{2} \\mathrm{~S}$ 废气回收能量并得到单质硫。质子膜 $\\mathrm{H}_{2} \\mathrm{~S}$ 燃料电池结构示意图如图所示。下列说法正确的是\n\n[图1]\nA: 该电池的总反应式为: $2 \\mathrm{H}_{2} \\mathrm{~S}+\\mathrm{O}_{2}=\\mathrm{S}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\nB: 电路中每通过 $2 \\mathrm{~mol}$ 电子, 在正极消耗 $22.4 \\mathrm{LH}_{2} \\mathrm{~S}$\nC: 电极 $\\mathrm{b}$ 上发生的电极反应为: $\\mathrm{O}_{2}+4 \\mathrm{H}^{+}+4 \\mathrm{e}=2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 每 $34 \\mathrm{gH}_{2} \\mathrm{~S}$ 参与反应, 有 $1 \\mathrm{molH}^{+}$经质子膜进入正极区\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n20 世纪初, 科学家设计出质子膜 $\\mathrm{H}_{2} \\mathrm{~S}$ 燃料电池, 实现了利用 $\\mathrm{H}_{2} \\mathrm{~S}$ 废气回收能量并得到单质硫。质子膜 $\\mathrm{H}_{2} \\mathrm{~S}$ 燃料电池结构示意图如图所示。下列说法正确的是\n\n[图1]\n\nA: 该电池的总反应式为: $2 \\mathrm{H}_{2} \\mathrm{~S}+\\mathrm{O}_{2}=\\mathrm{S}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\nB: 电路中每通过 $2 \\mathrm{~mol}$ 电子, 在正极消耗 $22.4 \\mathrm{LH}_{2} \\mathrm{~S}$\nC: 电极 $\\mathrm{b}$ 上发生的电极反应为: $\\mathrm{O}_{2}+4 \\mathrm{H}^{+}+4 \\mathrm{e}=2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 每 $34 \\mathrm{gH}_{2} \\mathrm{~S}$ 参与反应, 有 $1 \\mathrm{molH}^{+}$经质子膜进入正极区\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-69.jpg?height=440&width=506&top_left_y=180&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_447",
        "problem": "硬水除垢可以让循环冷却水系统稳定运行。某科研团队改进了主动式电化学硬水处理技术，原理如图所示(其中 $\\mathrm{R}$ 为有机物)。下列说法不正确的是\n\n[图1]\nA: 电极 A 为阳极, 发生还原反应\nB: 处理过程中 $\\mathrm{Cl}^{-}$可循环利用\nC: 处理后的水垢主要沉降在阴极附近\nD: 若 $\\mathrm{R}$ 为 $\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$, 当消耗 $1 \\mathrm{~mol} \\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$ 生成 $\\mathrm{N}_{2}$ 时, 则电极 $\\mathrm{B}$ 处产生的 $\\mathrm{H}_{2}$为 $3 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n硬水除垢可以让循环冷却水系统稳定运行。某科研团队改进了主动式电化学硬水处理技术，原理如图所示(其中 $\\mathrm{R}$ 为有机物)。下列说法不正确的是\n\n[图1]\n\nA: 电极 A 为阳极, 发生还原反应\nB: 处理过程中 $\\mathrm{Cl}^{-}$可循环利用\nC: 处理后的水垢主要沉降在阴极附近\nD: 若 $\\mathrm{R}$ 为 $\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$, 当消耗 $1 \\mathrm{~mol} \\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}$ 生成 $\\mathrm{N}_{2}$ 时, 则电极 $\\mathrm{B}$ 处产生的 $\\mathrm{H}_{2}$为 $3 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-31.jpg?height=731&width=917&top_left_y=1713&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1499",
        "problem": "Contemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.Calculate the mass of hydrogen required to drive the distance between Prague and Bratislava $\\left(330 \\mathrm{~km}\\right.$ ) at the average speed of $100 \\mathrm{~km} \\mathrm{~h}^{-1}$ with a car fitted with a 310 $\\mathrm{kW}$ electric engine running on average at a $15 \\%$ rate of its maximum power. Assume that the efficiency of the hydrogen cell producing electrical energy is $75 \\%$, the efficiency of the electric engine is $95 \\%$, and the Gibbs energy change for combustion of hydrogen fuel is $\\Delta_{\\mathrm{r}} G=-226 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nContemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.\n\nproblem:\nCalculate the mass of hydrogen required to drive the distance between Prague and Bratislava $\\left(330 \\mathrm{~km}\\right.$ ) at the average speed of $100 \\mathrm{~km} \\mathrm{~h}^{-1}$ with a car fitted with a 310 $\\mathrm{kW}$ electric engine running on average at a $15 \\%$ rate of its maximum power. Assume that the efficiency of the hydrogen cell producing electrical energy is $75 \\%$, the efficiency of the electric engine is $95 \\%$, and the Gibbs energy change for combustion of hydrogen fuel is $\\Delta_{\\mathrm{r}} G=-226 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_280",
        "problem": "Of the following elements, which has the highest third ionisation energy?\nA: $\\mathrm{Ar}$\nB: Si\nC: $\\mathrm{Mg}$\nD: $\\mathrm{Al}$\nE: $\\mathrm{Cl}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOf the following elements, which has the highest third ionisation energy?\n\nA: $\\mathrm{Ar}$\nB: Si\nC: $\\mathrm{Mg}$\nD: $\\mathrm{Al}$\nE: $\\mathrm{Cl}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1",
        "problem": "Which statement best describes the properties of the activation energy of a reaction?\nA: It increases in the presence of a catalyst.\nB: It decreases with increasing temperature.\nC: It must be larger than the energy change for forming any intermediates in the reaction.\nD: It must be positive for endothermic reactions and negative for exothermic reactions.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich statement best describes the properties of the activation energy of a reaction?\n\nA: It increases in the presence of a catalyst.\nB: It decreases with increasing temperature.\nC: It must be larger than the energy change for forming any intermediates in the reaction.\nD: It must be positive for endothermic reactions and negative for exothermic reactions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_226",
        "problem": "Which of the following pairs lists elements in order of increasing electronegativity?\nA: $\\mathrm{Na}, \\mathrm{F}, \\mathrm{O}, \\mathrm{N}$\nB: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{F}, \\mathrm{N}$\nC: Na, N, O, F\nD: $\\mathrm{N}, \\mathrm{O}, \\mathrm{F}, \\mathrm{Na}$\nE: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{N}, \\mathrm{F}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following pairs lists elements in order of increasing electronegativity?\n\nA: $\\mathrm{Na}, \\mathrm{F}, \\mathrm{O}, \\mathrm{N}$\nB: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{F}, \\mathrm{N}$\nC: Na, N, O, F\nD: $\\mathrm{N}, \\mathrm{O}, \\mathrm{F}, \\mathrm{Na}$\nE: $\\mathrm{Na}, \\mathrm{O}, \\mathrm{N}, \\mathrm{F}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1135",
        "problem": "This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nEach of the following elements forms an oxide exhibiting the theoretical maximum oxidation state for that element. Give the formula for each oxide.\n\n| xenon $(\\mathrm{Xe})$ | polonium $(\\mathrm{Po})$ | chlorine $(\\mathrm{Cl})$ |\n| :--- | :--- | :--- |\n| niobium $(\\mathrm{Nb})$ | osmium $(\\mathrm{Os})$ | yttrium $(\\mathrm{Y})$ |\n\nA commonly occurring unit in chemistry consists of an element associated with four oxygen atoms, and possibly a charge. An example is the tetrahedral sulfate ion, $\\mathrm{SO}_{4}{ }^{2-}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nEach of the following elements forms an oxide exhibiting the theoretical maximum oxidation state for that element. Give the formula for each oxide.\n\n| xenon $(\\mathrm{Xe})$ | polonium $(\\mathrm{Po})$ | chlorine $(\\mathrm{Cl})$ |\n| :--- | :--- | :--- |\n| niobium $(\\mathrm{Nb})$ | osmium $(\\mathrm{Os})$ | yttrium $(\\mathrm{Y})$ |\n\nA commonly occurring unit in chemistry consists of an element associated with four oxygen atoms, and possibly a charge. An example is the tetrahedral sulfate ion, $\\mathrm{SO}_{4}{ }^{2-}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [Xe, Cl, Po, Nb, Os, Y].\nTheir answer types are, in order, [expression, expression, expression, expression, expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null,
            null,
            null,
            null,
            null
        ],
        "answer_sequence": [
            "Xe",
            "Cl",
            "Po",
            "Nb",
            "Os",
            "Y"
        ],
        "type_sequence": [
            "EX",
            "EX",
            "EX",
            "EX",
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_637",
        "problem": "下列比较中正确的是\nA: 相同浓度的溶液, (1)氨水、(2) $\\mathrm{NaOH}$ 溶液、(3) $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$: (3) $>$(2) $>$(1)\nB: 相同浓度的三种溶液: (1) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$ 溶液、(2) $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液、(3) $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right):(3)>$ (1) $>$ (2)\nC: 相同 $\\mathrm{pH}$ 的溶液: (1) $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液、(2) $\\mathrm{NaHCO}_{3}$ 溶液、(3) $\\mathrm{NaClO}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right):(1)>(2)>$ (3)\nD: 同浓度、同体积的溶液: (1) $\\mathrm{NaAlO}_{2}$ 溶液、(2) $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液、(3) $\\mathrm{NaNO}_{3}$ 溶液中 $\\mathrm{pH}:(2)>(3)>(1)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列比较中正确的是\n\nA: 相同浓度的溶液, (1)氨水、(2) $\\mathrm{NaOH}$ 溶液、(3) $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$: (3) $>$(2) $>$(1)\nB: 相同浓度的三种溶液: (1) $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$ 溶液、(2) $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液、(3) $\\mathrm{NH}_{4} \\mathrm{NO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right):(3)>$ (1) $>$ (2)\nC: 相同 $\\mathrm{pH}$ 的溶液: (1) $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液、(2) $\\mathrm{NaHCO}_{3}$ 溶液、(3) $\\mathrm{NaClO}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right):(1)>(2)>$ (3)\nD: 同浓度、同体积的溶液: (1) $\\mathrm{NaAlO}_{2}$ 溶液、(2) $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液、(3) $\\mathrm{NaNO}_{3}$ 溶液中 $\\mathrm{pH}:(2)>(3)>(1)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1129",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidation states of elements in Groups 2 and 12 are +2, what are the valence electrons of mercury?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidation states of elements in Groups 2 and 12 are +2, what are the valence electrons of mercury?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_404",
        "problem": "下列有关同分异构体的叙述不正确的是\nA: 乙醇和乙醚互为同分异构体\nB: 已知丙烷的二氯代物有 4 种同分异构体，则其六氯代物的同分异构体数目为 4\nC: 甲苯在一定条件下与氢气完全加成后的产物, 与氯气发生取代反应, 生成的一氯代物有 4 种\nD: 分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}_{2}$ 的同分异构体中属于羧酸的有 4 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列有关同分异构体的叙述不正确的是\n\nA: 乙醇和乙醚互为同分异构体\nB: 已知丙烷的二氯代物有 4 种同分异构体，则其六氯代物的同分异构体数目为 4\nC: 甲苯在一定条件下与氢气完全加成后的产物, 与氯气发生取代反应, 生成的一氯代物有 4 种\nD: 分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}_{2}$ 的同分异构体中属于羧酸的有 4 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_102",
        "problem": "WHMIS 2015 is the updated labelling system which conforms to the new international standard of classifying hazardous materials. Although all of the compounds below would require some labelling for other characteristics, which compound would NOT require the oxidizer label shown to the right?\n[figure1]\nA: potassium permanganate\nB: sodium chlorate\nC: ethanol\nD: nitric acid\nE: hydrogen peroxide\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWHMIS 2015 is the updated labelling system which conforms to the new international standard of classifying hazardous materials. Although all of the compounds below would require some labelling for other characteristics, which compound would NOT require the oxidizer label shown to the right?\n[figure1]\n\nA: potassium permanganate\nB: sodium chlorate\nC: ethanol\nD: nitric acid\nE: hydrogen peroxide\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_de998ba2e5d42a77dc86g-1.jpg?height=141&width=157&top_left_y=666&top_left_x=1072"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_818",
        "problem": "还原沉淀法是处理含铬(含 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$ 和 $\\mathrm{CrO}_{4}{ }^{2-}$ ) 工业废水的常用方法, 过程如下:\n\n$\\mathrm{CrO}_{4}{ }^{2-} \\xrightarrow[\\text { 毒霜 }]{\\mathrm{H}^{+}} \\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-} \\xrightarrow[\\text { 还是 }]{\\mathrm{Fe}^{2+}} \\mathrm{Cr}^{3+} \\xrightarrow[\\text { 讯设 }]{\\mathrm{OH}^{-}} \\mathrm{Cr}(\\mathrm{OH})_{3} \\downarrow$\n\n已知转化过程中的反应为: $2 \\mathrm{CrO}_{4}{ }^{2-}(\\mathrm{aq})+2 \\mathrm{H}^{+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(1)$ 。转化后所得溶液中铬元素含量为 $28.6 \\mathrm{~g} / \\mathrm{L}, \\mathrm{CrO}_{4}{ }^{2-}$ 有 $10 / 11$ 转化为 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$ 。下列说法不正确的是\nA: 转化过程中, 增大 $\\mathrm{c}(\\mathrm{H}+)$, 平衡向正反应方向移动, $\\mathrm{CrO}_{4}{ }^{2-}$ 的转化率提高\nB: 常温下 $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Cr}(\\mathrm{OH})_{3}\\right]=1 \\times 10^{-32}$, 要使处理后废水中 $\\mathrm{c}\\left(\\mathrm{Cr}^{3+}\\right)$ 降至 $1 \\times 10^{-5} \\mathrm{~mol} / \\mathrm{L}$, 应调溶液的 $\\mathrm{pH}=5$\nC: 若用绿矾 $\\left(\\mathrm{FeSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}\\right)(\\mathrm{M}=278)$ 作还原剂, 处理 $1 \\mathrm{~L}$ 废水, 至少需要 $917.4 \\mathrm{~g}$\nD: 常温下转化反应的平衡常数 $\\mathrm{K}=10^{4}$, 则转化后所得溶液的 $\\mathrm{pH}=1$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n还原沉淀法是处理含铬(含 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$ 和 $\\mathrm{CrO}_{4}{ }^{2-}$ ) 工业废水的常用方法, 过程如下:\n\n$\\mathrm{CrO}_{4}{ }^{2-} \\xrightarrow[\\text { 毒霜 }]{\\mathrm{H}^{+}} \\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-} \\xrightarrow[\\text { 还是 }]{\\mathrm{Fe}^{2+}} \\mathrm{Cr}^{3+} \\xrightarrow[\\text { 讯设 }]{\\mathrm{OH}^{-}} \\mathrm{Cr}(\\mathrm{OH})_{3} \\downarrow$\n\n已知转化过程中的反应为: $2 \\mathrm{CrO}_{4}{ }^{2-}(\\mathrm{aq})+2 \\mathrm{H}^{+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(1)$ 。转化后所得溶液中铬元素含量为 $28.6 \\mathrm{~g} / \\mathrm{L}, \\mathrm{CrO}_{4}{ }^{2-}$ 有 $10 / 11$ 转化为 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}{ }^{2-}$ 。下列说法不正确的是\n\nA: 转化过程中, 增大 $\\mathrm{c}(\\mathrm{H}+)$, 平衡向正反应方向移动, $\\mathrm{CrO}_{4}{ }^{2-}$ 的转化率提高\nB: 常温下 $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Cr}(\\mathrm{OH})_{3}\\right]=1 \\times 10^{-32}$, 要使处理后废水中 $\\mathrm{c}\\left(\\mathrm{Cr}^{3+}\\right)$ 降至 $1 \\times 10^{-5} \\mathrm{~mol} / \\mathrm{L}$, 应调溶液的 $\\mathrm{pH}=5$\nC: 若用绿矾 $\\left(\\mathrm{FeSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}\\right)(\\mathrm{M}=278)$ 作还原剂, 处理 $1 \\mathrm{~L}$ 废水, 至少需要 $917.4 \\mathrm{~g}$\nD: 常温下转化反应的平衡常数 $\\mathrm{K}=10^{4}$, 则转化后所得溶液的 $\\mathrm{pH}=1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_801",
        "problem": "下列有关物质的浓度关系正确的是\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$ 溶液与 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水等体积混合 $(\\mathrm{pH}>7)$ : $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}(\\mathrm{Cl})>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nC: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)$相等的 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4} 、 \\mathrm{NH}_{4} \\mathrm{HSO}_{4} 、 \\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中: $c\\left(\\mathrm{NH}_{4} \\mathrm{HSO}_{4}\\right)>c\\left(\\mathrm{NH}_{4} \\mathrm{Cl}\\right)>\\mathrm{c}\\left(\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Fe}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{Fe}^{2+}\\right)=0.3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列有关物质的浓度关系正确的是\n\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$ 溶液与 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水等体积混合 $(\\mathrm{pH}>7)$ : $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}(\\mathrm{Cl})>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nC: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)$相等的 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4} 、 \\mathrm{NH}_{4} \\mathrm{HSO}_{4} 、 \\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中: $c\\left(\\mathrm{NH}_{4} \\mathrm{HSO}_{4}\\right)>c\\left(\\mathrm{NH}_{4} \\mathrm{Cl}\\right)>\\mathrm{c}\\left(\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Fe}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{Fe}^{2+}\\right)=0.3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_114",
        "problem": "The thermite reaction is the reaction of aluminum metal and iron (III) oxide:\n\n$$\n2 \\mathrm{Al}(\\mathrm{s})+\\mathrm{Fe}_{2} \\mathrm{O}_{3}(\\mathrm{~s}) \\rightarrow \\mathrm{Al}_{2} \\mathrm{O}_{3}(\\mathrm{~s})+2 \\mathrm{Fe}(\\mathrm{s}) \\quad \\Delta \\mathrm{H}=-852 \\mathrm{~kJ}\n$$\n\nA teacher does a demonstration with $1.00 \\mathrm{~mol}$ of iron (III) oxide and $2.00 \\mathrm{~mol}$ of aluminum metal both initially at $25.0^{\\circ} \\mathrm{C}$. If the combined specific heat of the products is $0.800 \\mathrm{~J} \\mathrm{~g}^{-1}{ }^{\\circ} \\mathrm{C}^{-1}$ over a wide range of temperatures, what is the final temperature of the products?\nA: $3550^{\\circ} \\mathrm{C}$\nB: $4960{ }^{\\circ} \\mathrm{C}$\nC: $5010^{\\circ} \\mathrm{C}$\nD: $6470{ }^{\\circ} \\mathrm{C}$\nE: $6500^{\\circ} \\mathrm{C}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe thermite reaction is the reaction of aluminum metal and iron (III) oxide:\n\n$$\n2 \\mathrm{Al}(\\mathrm{s})+\\mathrm{Fe}_{2} \\mathrm{O}_{3}(\\mathrm{~s}) \\rightarrow \\mathrm{Al}_{2} \\mathrm{O}_{3}(\\mathrm{~s})+2 \\mathrm{Fe}(\\mathrm{s}) \\quad \\Delta \\mathrm{H}=-852 \\mathrm{~kJ}\n$$\n\nA teacher does a demonstration with $1.00 \\mathrm{~mol}$ of iron (III) oxide and $2.00 \\mathrm{~mol}$ of aluminum metal both initially at $25.0^{\\circ} \\mathrm{C}$. If the combined specific heat of the products is $0.800 \\mathrm{~J} \\mathrm{~g}^{-1}{ }^{\\circ} \\mathrm{C}^{-1}$ over a wide range of temperatures, what is the final temperature of the products?\n\nA: $3550^{\\circ} \\mathrm{C}$\nB: $4960{ }^{\\circ} \\mathrm{C}$\nC: $5010^{\\circ} \\mathrm{C}$\nD: $6470{ }^{\\circ} \\mathrm{C}$\nE: $6500^{\\circ} \\mathrm{C}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_506",
        "problem": "下列图示与对应的叙述相符的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\nA: 图甲表示在恒容密闭容器中, 恒温条件下发生的可逆反应 $2 \\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons$ $\\mathrm{N}_{2} \\mathrm{O}_{4}(\\mathrm{~g})$, 各物质的浓度与其消耗速率之间的关系, 其中交点 $\\mathrm{A}$ 对应的状态为化学平衡状态\nB: 图乙曲线表示反应 $2 \\mathrm{~A}(\\mathrm{~g})+\\mathrm{B}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}>0$, 正、逆反应的平衡常数 $\\mathrm{K}$ 随 温度的变化\nC: 图丙表示对反应 $2 \\mathrm{X}(\\mathrm{g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Z}(\\mathrm{s})+3 \\mathrm{~W}(\\mathrm{~g}) \\Delta \\mathrm{H}<0$ 在 $\\mathrm{t} 1$ 时刻增大压强\nD: 图丁表示常温下, 稀释 HA、HB 两种酸的稀溶液时, 溶液 $\\mathrm{pH}$ 随加水量的变化,则同浓度的 $\\mathrm{NaA}$ 溶液的 $\\mathrm{pH}$ 小于 $\\mathrm{NaB}$ 溶液 $\\mathrm{pH}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列图示与对应的叙述相符的是\n\n[图1]\n\n甲\n\n[图2]\n\n乙\n\n[图3]\n\n丙\n\n[图4]\n\n丁\n\nA: 图甲表示在恒容密闭容器中, 恒温条件下发生的可逆反应 $2 \\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons$ $\\mathrm{N}_{2} \\mathrm{O}_{4}(\\mathrm{~g})$, 各物质的浓度与其消耗速率之间的关系, 其中交点 $\\mathrm{A}$ 对应的状态为化学平衡状态\nB: 图乙曲线表示反应 $2 \\mathrm{~A}(\\mathrm{~g})+\\mathrm{B}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}>0$, 正、逆反应的平衡常数 $\\mathrm{K}$ 随 温度的变化\nC: 图丙表示对反应 $2 \\mathrm{X}(\\mathrm{g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons \\mathrm{Z}(\\mathrm{s})+3 \\mathrm{~W}(\\mathrm{~g}) \\Delta \\mathrm{H}<0$ 在 $\\mathrm{t} 1$ 时刻增大压强\nD: 图丁表示常温下, 稀释 HA、HB 两种酸的稀溶液时, 溶液 $\\mathrm{pH}$ 随加水量的变化,则同浓度的 $\\mathrm{NaA}$ 溶液的 $\\mathrm{pH}$ 小于 $\\mathrm{NaB}$ 溶液 $\\mathrm{pH}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-91.jpg?height=311&width=348&top_left_y=2043&top_left_x=360",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-91.jpg?height=294&width=374&top_left_y=2052&top_left_x=704",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-91.jpg?height=291&width=348&top_left_y=2053&top_left_x=1094",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-91.jpg?height=291&width=345&top_left_y=2053&top_left_x=1435"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1475",
        "problem": "The undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe distribution ratio (D) of the acid $\\mathrm{HA}$ depends on the $\\mathrm{pH}$ of the aqueous phase.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe distribution ratio (D) of the acid $\\mathrm{HA}$ depends on the $\\mathrm{pH}$ of the aqueous phase.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-461.jpg?height=185&width=788&top_left_y=2209&top_left_x=634"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1369",
        "problem": "Acid-base equilibria in water\n\nA solution $(\\mathbf{X}$ ) contains two weak monoprotic acids (those having one acidic proton); HA with the acid dissociation constant of $K_{H A}=1.74 \\times 10^{-7}$, and $H B$ with the acid dissociation constant of $K_{H B}=1.34 \\times 10^{-7}$. The solution $\\mathbf{X}$ has a $\\mathrm{pH}$ of 3.75.Adding large amounts of distilled water to solution $\\mathbf{X}$ gives a very (infinitely) dilute solution where the total concentrations of the acids are close to zero. Calculate the percentage of dissociation of each acid in this dilute solution.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nAcid-base equilibria in water\n\nA solution $(\\mathbf{X}$ ) contains two weak monoprotic acids (those having one acidic proton); HA with the acid dissociation constant of $K_{H A}=1.74 \\times 10^{-7}$, and $H B$ with the acid dissociation constant of $K_{H B}=1.34 \\times 10^{-7}$. The solution $\\mathbf{X}$ has a $\\mathrm{pH}$ of 3.75.\n\nproblem:\nAdding large amounts of distilled water to solution $\\mathbf{X}$ gives a very (infinitely) dilute solution where the total concentrations of the acids are close to zero. Calculate the percentage of dissociation of each acid in this dilute solution.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the percentage of dissociation of HA, the percentage of dissociation of HB].\nTheir units are, in order, [%, %], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "%",
            "%"
        ],
        "answer_sequence": [
            "the percentage of dissociation of HA",
            "the percentage of dissociation of HB"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_305",
        "problem": "The enthalpy change for the reaction below is $\\Delta H=-58 \\mathrm{~kJ}$ (per mole of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ formed).\n\n$$\n2 \\mathrm{NO}_{2}(g) \\stackrel{k_{1}}{\\underset{k_{-1}}{\\rightleftarrows}} \\mathrm{N}_{2} \\mathrm{O}_{4}(g)\n$$\n\nIf $k_{1}$ and $k_{-1}$ are the rate constants for the forward and reverse reactions, respectively, and $K_{\\mathrm{c}}$ is the equilibrium constant for the reaction as written, then what is the effect of adding a catalyst on the values of $k_{1}, k_{-1}$ and $K_{\\mathrm{c}}$ ?\nA: $k_{1}$ increases, $k_{-1}$ increases, $K_{c}$ increases\nB: $k_{1}$ decreases, $k_{-1}$ decreases, $K_{\\mathrm{c}}$ decreases\nC: $k_{1}$ increases, $k_{-1}$ increases, $K_{\\mathrm{c}}$ remains the same\nD: $\\quad k_{1}$ decreases, $k_{-1}$ decreases, $K_{\\mathrm{c}}$ remains the same\nE: $k_{1}$ remains the same, $k_{-1}$ remains the same, $K_{\\mathrm{c}}$ remains the same\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe enthalpy change for the reaction below is $\\Delta H=-58 \\mathrm{~kJ}$ (per mole of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ formed).\n\n$$\n2 \\mathrm{NO}_{2}(g) \\stackrel{k_{1}}{\\underset{k_{-1}}{\\rightleftarrows}} \\mathrm{N}_{2} \\mathrm{O}_{4}(g)\n$$\n\nIf $k_{1}$ and $k_{-1}$ are the rate constants for the forward and reverse reactions, respectively, and $K_{\\mathrm{c}}$ is the equilibrium constant for the reaction as written, then what is the effect of adding a catalyst on the values of $k_{1}, k_{-1}$ and $K_{\\mathrm{c}}$ ?\n\nA: $k_{1}$ increases, $k_{-1}$ increases, $K_{c}$ increases\nB: $k_{1}$ decreases, $k_{-1}$ decreases, $K_{\\mathrm{c}}$ decreases\nC: $k_{1}$ increases, $k_{-1}$ increases, $K_{\\mathrm{c}}$ remains the same\nD: $\\quad k_{1}$ decreases, $k_{-1}$ decreases, $K_{\\mathrm{c}}$ remains the same\nE: $k_{1}$ remains the same, $k_{-1}$ remains the same, $K_{\\mathrm{c}}$ remains the same\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_665",
        "problem": "互为同分异构体的二元弱酸在催化剂作用下的相互转化可表示为 $\\mathrm{H}_{2} \\mathrm{~A} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{~B}$ $K=10$ (无催化剂不考虑转化)。常温下, 催化剂存在时, 调节 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液的 $\\mathrm{pH}$ (催化剂自身对 $\\mathrm{pH}$ 无影响), $\\mathrm{H}_{2} \\mathrm{~B} 、 \\mathrm{HB}^{-} 、 \\mathrm{~B}^{2}$-的平衡分布系数 $\\delta$ 变化如图。例如 $\\mathrm{H}_{2} \\mathrm{~B}$ 的分布系数为 $\\delta\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)=\\frac{c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{HB}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)+c\\left(\\mathrm{~B}^{2-}\\right)}$ 。下列说法正确的是\n\n[图1]\nA: 无催化剂存在时, NaHB 溶液显碱性\nB: 有催化剂存在且 $\\mathrm{pH}=3.6$ 时, $c\\left(\\mathrm{HA}^{-}\\right)>c\\left(\\mathrm{HB}^{-}\\right)>c\\left(\\mathrm{~B}^{2-}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)$\nC: $\\mathrm{p} K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right) \\approx 3.0, \\mathrm{p} K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)<3.0$\nD: $\\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中加入催化剂后, 溶液碱性增强\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n互为同分异构体的二元弱酸在催化剂作用下的相互转化可表示为 $\\mathrm{H}_{2} \\mathrm{~A} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{~B}$ $K=10$ (无催化剂不考虑转化)。常温下, 催化剂存在时, 调节 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液的 $\\mathrm{pH}$ (催化剂自身对 $\\mathrm{pH}$ 无影响), $\\mathrm{H}_{2} \\mathrm{~B} 、 \\mathrm{HB}^{-} 、 \\mathrm{~B}^{2}$-的平衡分布系数 $\\delta$ 变化如图。例如 $\\mathrm{H}_{2} \\mathrm{~B}$ 的分布系数为 $\\delta\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)=\\frac{c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{HB}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)+c\\left(\\mathrm{~B}^{2-}\\right)}$ 。下列说法正确的是\n\n[图1]\n\nA: 无催化剂存在时, NaHB 溶液显碱性\nB: 有催化剂存在且 $\\mathrm{pH}=3.6$ 时, $c\\left(\\mathrm{HA}^{-}\\right)>c\\left(\\mathrm{HB}^{-}\\right)>c\\left(\\mathrm{~B}^{2-}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)$\nC: $\\mathrm{p} K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right) \\approx 3.0, \\mathrm{p} K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)<3.0$\nD: $\\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中加入催化剂后, 溶液碱性增强\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-018.jpg?height=611&width=894&top_left_y=180&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1262",
        "problem": "IONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\n\nConsider the stability of copper(I) oxide, $\\mathrm{Cu}_{2} \\mathrm{O}$, in contact with a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of $\\mathrm{Cu}^{2+}$ ions. The solubility product of copper $(\\mathrm{I})$ oxide is $K_{s p}=\\left[\\mathrm{Cu}^{+}\\right][\\mathrm{OH}]=1 \\times 10^{-15}$\n\nComplex formation involving $\\mathrm{Cu}^{+}$and $\\mathrm{Cu}^{2+}$ ions\n\nThe dissociation constant of the complex ion $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}{3}\\right){2}\\right]^{+}$is $K_{D}=1 \\times 10^{-11}$. Calculate the standard electrode potential of the couple:",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\n\nConsider the stability of copper(I) oxide, $\\mathrm{Cu}_{2} \\mathrm{O}$, in contact with a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of $\\mathrm{Cu}^{2+}$ ions. The solubility product of copper $(\\mathrm{I})$ oxide is $K_{s p}=\\left[\\mathrm{Cu}^{+}\\right][\\mathrm{OH}]=1 \\times 10^{-15}$\n\nComplex formation involving $\\mathrm{Cu}^{+}$and $\\mathrm{Cu}^{2+}$ ions\n\nThe dissociation constant of the complex ion $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}{3}\\right){2}\\right]^{+}$is $K_{D}=1 \\times 10^{-11}$. Calculate the standard electrode potential of the couple:\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of V, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "V"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_293",
        "problem": "Which of the following is an acceptable Lewis structure for the thiocyanate ion, $\\mathrm{SCN}^{-}$?\nA: $: \\ddot{S}-\\ddot{C}-\\ddot{Ì£}:$\nB: $: \\mathrm{s} \\equiv \\mathrm{C} \\equiv \\mathrm{N}:$\nC: $: \\ddot{:} \\rightarrow-\\ddot{c}=\\ddot{+}$\nD: $\\ddot{\\mathrm{S}}=\\mathrm{C}=\\ddot{\\mathrm{N}}$\nE: $\\ddot{\\mathrm{S}}=\\ddot{\\mathrm{C}}-\\ddot{\\mathrm{N}:}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is an acceptable Lewis structure for the thiocyanate ion, $\\mathrm{SCN}^{-}$?\n\nA: $: \\ddot{S}-\\ddot{C}-\\ddot{Ì£}:$\nB: $: \\mathrm{s} \\equiv \\mathrm{C} \\equiv \\mathrm{N}:$\nC: $: \\ddot{:} \\rightarrow-\\ddot{c}=\\ddot{+}$\nD: $\\ddot{\\mathrm{S}}=\\mathrm{C}=\\ddot{\\mathrm{N}}$\nE: $\\ddot{\\mathrm{S}}=\\ddot{\\mathrm{C}}-\\ddot{\\mathrm{N}:}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1503",
        "problem": "One of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nCalculate the rate of production of energy (power) in watts ( $\\left.1 \\mathrm{~W}=\\mathrm{J} \\mathrm{s}^{-1}\\right)$ produced by 1.00 kilogram of ${ }^{232} \\mathrm{Th}\\left(\\mathrm{t}_{1 / 2}=1.40 \\times 10^{10}\\right.$ years).\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOne of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nCalculate the rate of production of energy (power) in watts ( $\\left.1 \\mathrm{~W}=\\mathrm{J} \\mathrm{s}^{-1}\\right)$ produced by 1.00 kilogram of ${ }^{232} \\mathrm{Th}\\left(\\mathrm{t}_{1 / 2}=1.40 \\times 10^{10}\\right.$ years).\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~W}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~W}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_105",
        "problem": "A compound is composed of element $X$ and hydrogen. Analysis shows the compound to be $79.89 \\% \\mathrm{X}$ by mass, with three times as many hydrogen atoms as $\\mathrm{X}$ atoms per molecule. Which element is element $\\mathrm{X}$ ?\nA: $\\mathrm{He}$\nB: C\nC: $\\mathrm{N}$\nD: P\nE: $S$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA compound is composed of element $X$ and hydrogen. Analysis shows the compound to be $79.89 \\% \\mathrm{X}$ by mass, with three times as many hydrogen atoms as $\\mathrm{X}$ atoms per molecule. Which element is element $\\mathrm{X}$ ?\n\nA: $\\mathrm{He}$\nB: C\nC: $\\mathrm{N}$\nD: P\nE: $S$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1210",
        "problem": "lodine deficiency is of special concern in Georgia because it occupies a region where iodine is scarce in soil and water. Iodine deficiency can be effectively and inexpensively prevented if salt for human consumption is fortified with small amounts of iodine. Methods for analyzing salt for iodine content are thus important. Current regulations in Georgia are that iodized salt must contain between 25 - 55 ppm iodine (1 ppm = $1 \\mathrm{mg}$ iodine/kg salt).\n\nMost salt is iodized by fortification with potassium iodate $\\left(\\mathrm{KIO}_{3}\\right)$. lodate content can be determined in salt samples using iodometric titration. In a typical procedure, $10.000 \\mathrm{~g}$ of an iodized salt sample is dissolved in $100 \\mathrm{~cm}^{3}$ aqueous $\\mathrm{HCl}\\left(c=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$ to which $1.0 \\mathrm{~g} \\mathrm{KI}$ has been added. The solution is then titrated with aqueous sodium thiosulfate solution $\\left(c=0.00235 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to a starch endpoint; this requires $7.50 \\mathrm{~cm}^{3}$ of titrant.Calculate the iodization level, in ppm, of this salt sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nlodine deficiency is of special concern in Georgia because it occupies a region where iodine is scarce in soil and water. Iodine deficiency can be effectively and inexpensively prevented if salt for human consumption is fortified with small amounts of iodine. Methods for analyzing salt for iodine content are thus important. Current regulations in Georgia are that iodized salt must contain between 25 - 55 ppm iodine (1 ppm = $1 \\mathrm{mg}$ iodine/kg salt).\n\nMost salt is iodized by fortification with potassium iodate $\\left(\\mathrm{KIO}_{3}\\right)$. lodate content can be determined in salt samples using iodometric titration. In a typical procedure, $10.000 \\mathrm{~g}$ of an iodized salt sample is dissolved in $100 \\mathrm{~cm}^{3}$ aqueous $\\mathrm{HCl}\\left(c=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$ to which $1.0 \\mathrm{~g} \\mathrm{KI}$ has been added. The solution is then titrated with aqueous sodium thiosulfate solution $\\left(c=0.00235 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to a starch endpoint; this requires $7.50 \\mathrm{~cm}^{3}$ of titrant.\n\nproblem:\nCalculate the iodization level, in ppm, of this salt sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_613",
        "problem": "已知 $\\mathrm{H}_{2} \\mathrm{R}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}=2 \\times 10^{-8}, \\mathrm{~K}_{\\mathrm{a} 2}=3 \\times 10^{-17}$ 。常温下, 难溶物 $\\mathrm{BaR}$ 在不同浓\n\n度盐酸(足量)中恰好不再溶解时, 测得混合液中 $\\operatorname{lgc}\\left(\\mathrm{Ba}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的关系如图所示, 下列说法错误的是\n\n[图1]\nA: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{BaR})$ 约为 $6 \\times 10^{-22}$\nB: $M$ 点: $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{R}^{2-}\\right)$\nC: $\\mathrm{N}$ 点: $\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)$约为 $2 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 直线上任一点均满足: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{Ba}^{2+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知 $\\mathrm{H}_{2} \\mathrm{R}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}=2 \\times 10^{-8}, \\mathrm{~K}_{\\mathrm{a} 2}=3 \\times 10^{-17}$ 。常温下, 难溶物 $\\mathrm{BaR}$ 在不同浓\n\n度盐酸(足量)中恰好不再溶解时, 测得混合液中 $\\operatorname{lgc}\\left(\\mathrm{Ba}^{2+}\\right)$ 与 $\\mathrm{pH}$ 的关系如图所示, 下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{BaR})$ 约为 $6 \\times 10^{-22}$\nB: $M$ 点: $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{R}^{2-}\\right)$\nC: $\\mathrm{N}$ 点: $\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)$约为 $2 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 直线上任一点均满足: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}\\left(\\mathrm{Ba}^{2+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-016.jpg?height=468&width=508&top_left_y=1819&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_64",
        "problem": "Hydrocyanic acid, HCN, is a weak acid with $K_{\\mathrm{a}}=4.9 \\times 10^{-10}$. Nickel(II) ion complexes strongly with cyanide ion to form $\\mathrm{Ni}(\\mathrm{CN})_{4}{ }^{2-}, K_{\\mathrm{f}}=1.0 \\times 10^{22}$. What is the $\\mathrm{pH}$ of $1.00 \\mathrm{~L}$ of a $0.100 \\mathrm{M}$ solution of $\\mathrm{HCN}$ to which $0.025 \\mathrm{~mol} \\mathrm{NiCl}_{2}$ has been added?\nA: 1.00\nB: 4.05\nC: 5.15\nD: 8.92\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHydrocyanic acid, HCN, is a weak acid with $K_{\\mathrm{a}}=4.9 \\times 10^{-10}$. Nickel(II) ion complexes strongly with cyanide ion to form $\\mathrm{Ni}(\\mathrm{CN})_{4}{ }^{2-}, K_{\\mathrm{f}}=1.0 \\times 10^{22}$. What is the $\\mathrm{pH}$ of $1.00 \\mathrm{~L}$ of a $0.100 \\mathrm{M}$ solution of $\\mathrm{HCN}$ to which $0.025 \\mathrm{~mol} \\mathrm{NiCl}_{2}$ has been added?\n\nA: 1.00\nB: 4.05\nC: 5.15\nD: 8.92\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1158",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nTo prepare $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$, an oxidising agent must be used. One preparation uses $\\mathrm{HgCl}_{2}$. Mercury forms two chlorides: $\\mathrm{HgCl}_{2}$ and $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n\nGive TWO different equations for the preparation of $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$ and $\\mathrm{HgCl}_{2}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nTo prepare $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$, an oxidising agent must be used. One preparation uses $\\mathrm{HgCl}_{2}$. Mercury forms two chlorides: $\\mathrm{HgCl}_{2}$ and $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n\nGive TWO different equations for the preparation of $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$ and $\\mathrm{HgCl}_{2}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir answer types are, in order, [expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            null,
            null
        ],
        "answer_sequence": null,
        "type_sequence": [
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_60",
        "problem": "The isotope ${ }^{69} \\mathrm{Zn}$ undergoes what mode of radioactive decay?\nA: Alpha emission\nB: Beta emission\nC: Gamma emission\nD: Positron emission\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe isotope ${ }^{69} \\mathrm{Zn}$ undergoes what mode of radioactive decay?\n\nA: Alpha emission\nB: Beta emission\nC: Gamma emission\nD: Positron emission\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_231",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the molar mass of compound $\\mathbf{A}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the molar mass of compound $\\mathbf{A}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g/mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g/mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_491",
        "problem": "将质量相等的铜片和铂片插入硫酸铜溶液中(足量), 铜片与电源正极相连, 铂片与电源负极相连, 以电流强度 $1 \\mathrm{~A}$ 通电 $10 \\mathrm{~min}$, 然后反接电源, 以电流强度 $2 \\mathrm{~A}$ 继续通电 $10 \\mathrm{~min}$ 。下列表示铜电极铂电极、电解池中产生气体的质量和电解时间的关系图正确的是\nA: [图1]\nB: [图2]\nC: [图3]\nD: [图4]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n将质量相等的铜片和铂片插入硫酸铜溶液中(足量), 铜片与电源正极相连, 铂片与电源负极相连, 以电流强度 $1 \\mathrm{~A}$ 通电 $10 \\mathrm{~min}$, 然后反接电源, 以电流强度 $2 \\mathrm{~A}$ 继续通电 $10 \\mathrm{~min}$ 。下列表示铜电极铂电极、电解池中产生气体的质量和电解时间的关系图正确的是\n\nA: [图1]\nB: [图2]\nC: [图3]\nD: [图4]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-61.jpg?height=254&width=352&top_left_y=655&top_left_x=475",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-61.jpg?height=249&width=349&top_left_y=652&top_left_x=1116",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-61.jpg?height=259&width=322&top_left_y=927&top_left_x=770",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-61.jpg?height=266&width=404&top_left_y=1189&top_left_x=769"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_180",
        "problem": "When dissolved in aqueous solution, $\\mathrm{Al}^{3+}$ forms a six coordinate complex with water that can undergo dissociation according to the following equation:\n\n$\\mathrm{Al}\\left(\\mathrm{OH}_{2}\\right)_{6}{ }^{3+}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{Al}\\left(\\mathrm{OH}_{2}\\right)_{5}(\\mathrm{OH})^{2+}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq}) K_{\\mathrm{a}}=1.0 \\times 10^{-5}$\n\nIf the initial concentration of $\\mathrm{Al}\\left(\\mathrm{OH}_{2}\\right){ }_{6}{ }^{3+}$ is $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}$, what is the approximate $\\mathrm{pH}$ of the solution?\nA: 1.00\nB: 3.00\nC: 4.00\nD: 5.00\nE: 7.00\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen dissolved in aqueous solution, $\\mathrm{Al}^{3+}$ forms a six coordinate complex with water that can undergo dissociation according to the following equation:\n\n$\\mathrm{Al}\\left(\\mathrm{OH}_{2}\\right)_{6}{ }^{3+}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{Al}\\left(\\mathrm{OH}_{2}\\right)_{5}(\\mathrm{OH})^{2+}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq}) K_{\\mathrm{a}}=1.0 \\times 10^{-5}$\n\nIf the initial concentration of $\\mathrm{Al}\\left(\\mathrm{OH}_{2}\\right){ }_{6}{ }^{3+}$ is $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}$, what is the approximate $\\mathrm{pH}$ of the solution?\n\nA: 1.00\nB: 3.00\nC: 4.00\nD: 5.00\nE: 7.00\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_640",
        "problem": "为研究反应 $X(\\mathrm{~g})+2 \\mathrm{Y}(\\mathrm{g})=2 Z(\\mathrm{~g})$, 向一固定体积的容器中加入足量 $1 \\mathrm{molX}$ 和 $2 \\mathrm{molY}$发生反应。一定条件下, 在甲、乙两种催化剂作用下, 反应相同时间, 测得 $Y$ 的转化率与温度的关系如下图所示。下列说法错误的是\n\n[图1]\nA: 反应达到平衡后, 恒温时, 再加入少量 $\\mathrm{Z}$, 达到新平衡时 $\\mathrm{Y}$ 的体积分数减小\nB: 在相同温度下，与催化剂甲相比，乙使反应活化能更低\nC: $\\mathrm{M}$ 点可能是该反应的平衡点\nD: 反应达到平衡后, 恒温时, 再加入少量 $\\mathrm{Y}, \\mathrm{Z}$ 的体积分数减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n为研究反应 $X(\\mathrm{~g})+2 \\mathrm{Y}(\\mathrm{g})=2 Z(\\mathrm{~g})$, 向一固定体积的容器中加入足量 $1 \\mathrm{molX}$ 和 $2 \\mathrm{molY}$发生反应。一定条件下, 在甲、乙两种催化剂作用下, 反应相同时间, 测得 $Y$ 的转化率与温度的关系如下图所示。下列说法错误的是\n\n[图1]\n\nA: 反应达到平衡后, 恒温时, 再加入少量 $\\mathrm{Z}$, 达到新平衡时 $\\mathrm{Y}$ 的体积分数减小\nB: 在相同温度下，与催化剂甲相比，乙使反应活化能更低\nC: $\\mathrm{M}$ 点可能是该反应的平衡点\nD: 反应达到平衡后, 恒温时, 再加入少量 $\\mathrm{Y}, \\mathrm{Z}$ 的体积分数减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-065.jpg?height=463&width=562&top_left_y=1276&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1507",
        "problem": "A less common agent for iodizing salt is potassium iodide, which cannot be easily measured by iodometric titration.\n\nOne possible method for analyzing iodide in the presence of chloride is potentiometric titration. However, this method is not very precise in the presence of large amounts of chloride. In this method, a silver wire is immersed in the solution (containing iodide and chloride) to be analyzed and silver ion is gradually added to the solution. The potential of the silver wire is measured relative to a reference electrode consisting of a silver wire in a solution of $\\mathrm{AgNO}_{3}\\left(c=1.000\\right.$ mol dm$\\left.{ }^{-3}\\right)$. The measured potentials are negative and the absolute values of these potentials are reported. The solution to be analysed has a volume of $1.000 \\mathrm{dm}^{3}$ (which you may assume does not change as silver ion is added) and $t=25.0^{\\circ} \\mathrm{C}$.\n\nThe results of this experiment are governed by three equilibria: the solubility of $\\mathrm{Agl}(\\mathrm{s})\\left[K_{\\mathrm{spl}}\\right]$ and $\\mathrm{AgCl}(\\mathrm{s})\\left[K_{\\mathrm{spCl}}\\right]$ and the formation of $\\mathrm{AgCl}_{2}{ }^{-}(\\mathrm{aq})\\left[K_{\\mathrm{f}}\\right]$. (Iodide also forms complex ions with silver but this may be neglected at the very low concentrations of iodide present in this experiment).\n\n$$\n\\begin{array}{ll}\n\\mathrm{Agl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{I}^{-}(\\mathrm{aq}) & K_{\\mathrm{spl}} \\\\\n\\mathrm{AgCl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) & K_{\\mathrm{spCl}} \\\\\n\\mathrm{Ag}^{+}(\\mathrm{aq})+2 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{AgCl}_{2}^{-}(\\mathrm{aq}) & K_{\\mathrm{f}}\n\\end{array}\n$$\n\nBelow are shown the results of two experiments measuring the observed potential as a function of added number of moles of silver ion. Experiment $\\mathbf{A}$ (solid circles) was carried out with $1.000 \\mathrm{dm}^{3}$ of iodide solution with concentration $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and containing no chloride ion. Experiment $\\mathbf{B}$ (open circles) was done using $1.000 \\mathrm{dm}^{3}$ of solution with a concentration of iodide $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and that of chloride equal to $1.00 \\times 10^{-1} \\mathrm{~mol} \\mathrm{dm}^{-3}$.\n\n[figure1]\n\n| $\\mu \\mathrm{mol} \\mathrm{Ag}^{+}$ <br> added | $\\|E\\|, \\mathrm{V}$ <br> exp. A | $\\|E\\|, \\mathrm{V}$ <br> exp. B |\n| :--- | :--- | :--- |\n| 1.00 | 0.637 | 0.637 |\n| 3.00 | 0.631 | 0.631 |\n| 5.00 | 0.622 | 0.622 |\n| 7.00 | 0.609 | 0.610 |\n| 9.00 | 0.581 | 0.584 |\n| 10.0 | 0.468 | 0.558 |\n| 11.0 | 0.355 | 0.531 |\n| 12.0 | 0.337 | 0.517 |\n| 13.0 | 0.327 | 0.517 |\n| 15.0 | 0.313 | 0.517 |Select an appropriate data from the experiments and use it to calculate the solubility product of $\\mathrm{AgCl}$, $\\left(K_{\\mathrm{spCl}}\\right)$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nA less common agent for iodizing salt is potassium iodide, which cannot be easily measured by iodometric titration.\n\nOne possible method for analyzing iodide in the presence of chloride is potentiometric titration. However, this method is not very precise in the presence of large amounts of chloride. In this method, a silver wire is immersed in the solution (containing iodide and chloride) to be analyzed and silver ion is gradually added to the solution. The potential of the silver wire is measured relative to a reference electrode consisting of a silver wire in a solution of $\\mathrm{AgNO}_{3}\\left(c=1.000\\right.$ mol dm$\\left.{ }^{-3}\\right)$. The measured potentials are negative and the absolute values of these potentials are reported. The solution to be analysed has a volume of $1.000 \\mathrm{dm}^{3}$ (which you may assume does not change as silver ion is added) and $t=25.0^{\\circ} \\mathrm{C}$.\n\nThe results of this experiment are governed by three equilibria: the solubility of $\\mathrm{Agl}(\\mathrm{s})\\left[K_{\\mathrm{spl}}\\right]$ and $\\mathrm{AgCl}(\\mathrm{s})\\left[K_{\\mathrm{spCl}}\\right]$ and the formation of $\\mathrm{AgCl}_{2}{ }^{-}(\\mathrm{aq})\\left[K_{\\mathrm{f}}\\right]$. (Iodide also forms complex ions with silver but this may be neglected at the very low concentrations of iodide present in this experiment).\n\n$$\n\\begin{array}{ll}\n\\mathrm{Agl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{I}^{-}(\\mathrm{aq}) & K_{\\mathrm{spl}} \\\\\n\\mathrm{AgCl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) & K_{\\mathrm{spCl}} \\\\\n\\mathrm{Ag}^{+}(\\mathrm{aq})+2 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{AgCl}_{2}^{-}(\\mathrm{aq}) & K_{\\mathrm{f}}\n\\end{array}\n$$\n\nBelow are shown the results of two experiments measuring the observed potential as a function of added number of moles of silver ion. Experiment $\\mathbf{A}$ (solid circles) was carried out with $1.000 \\mathrm{dm}^{3}$ of iodide solution with concentration $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and containing no chloride ion. Experiment $\\mathbf{B}$ (open circles) was done using $1.000 \\mathrm{dm}^{3}$ of solution with a concentration of iodide $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and that of chloride equal to $1.00 \\times 10^{-1} \\mathrm{~mol} \\mathrm{dm}^{-3}$.\n\n[figure1]\n\n| $\\mu \\mathrm{mol} \\mathrm{Ag}^{+}$ <br> added | $\\|E\\|, \\mathrm{V}$ <br> exp. A | $\\|E\\|, \\mathrm{V}$ <br> exp. B |\n| :--- | :--- | :--- |\n| 1.00 | 0.637 | 0.637 |\n| 3.00 | 0.631 | 0.631 |\n| 5.00 | 0.622 | 0.622 |\n| 7.00 | 0.609 | 0.610 |\n| 9.00 | 0.581 | 0.584 |\n| 10.0 | 0.468 | 0.558 |\n| 11.0 | 0.355 | 0.531 |\n| 12.0 | 0.337 | 0.517 |\n| 13.0 | 0.327 | 0.517 |\n| 15.0 | 0.313 | 0.517 |\n\nproblem:\nSelect an appropriate data from the experiments and use it to calculate the solubility product of $\\mathrm{AgCl}$, $\\left(K_{\\mathrm{spCl}}\\right)$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-145.jpg?height=968&width=1051&top_left_y=1361&top_left_x=180"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1440",
        "problem": "The label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\nGive the formulae of four acids that could have been in the solution if the $\\mathrm{pH}$ changed one unit after a tenfold dilution.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\n\nproblem:\nThe label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\nGive the formulae of four acids that could have been in the solution if the $\\mathrm{pH}$ changed one unit after a tenfold dilution.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir answer types are, in order, [expression, expression, expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            null,
            null,
            null,
            null
        ],
        "answer_sequence": null,
        "type_sequence": [
            "EX",
            "EX",
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_99",
        "problem": "A vessel contains $2.50 \\mathrm{~mol}$ of $\\mathrm{O}_{2}$ gas, $0.50 \\mathrm{~mol}$ of $\\mathrm{N}_{2}$ gas and $1.00 \\mathrm{~mol}$ of $\\mathrm{CO}_{2}$ gas. The total pressure is $200 \\mathrm{kPa}$. The partial pressure exerted by the $\\mathrm{O}_{2}$ in the mixture is:\nA: $25 \\mathrm{kPa}$\nB: $50 \\mathrm{kPa}$\nC: $100 \\mathrm{kPa}$\nD: $125 \\mathrm{kPa}$\nE: $150 \\mathrm{kPa}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA vessel contains $2.50 \\mathrm{~mol}$ of $\\mathrm{O}_{2}$ gas, $0.50 \\mathrm{~mol}$ of $\\mathrm{N}_{2}$ gas and $1.00 \\mathrm{~mol}$ of $\\mathrm{CO}_{2}$ gas. The total pressure is $200 \\mathrm{kPa}$. The partial pressure exerted by the $\\mathrm{O}_{2}$ in the mixture is:\n\nA: $25 \\mathrm{kPa}$\nB: $50 \\mathrm{kPa}$\nC: $100 \\mathrm{kPa}$\nD: $125 \\mathrm{kPa}$\nE: $150 \\mathrm{kPa}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_363",
        "problem": "Which of the following gas-phase ions has the largest number of unpaired electrons in its ground state?\nA: $\\mathrm{Cr}^{3+}$\nB: $\\mathrm{Co}^{3+}$\nC: $\\mathrm{Ni}^{2+}$\nD: $\\mathrm{Cu}^{2+}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following gas-phase ions has the largest number of unpaired electrons in its ground state?\n\nA: $\\mathrm{Cr}^{3+}$\nB: $\\mathrm{Co}^{3+}$\nC: $\\mathrm{Ni}^{2+}$\nD: $\\mathrm{Cu}^{2+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_84",
        "problem": "The kinetics of the reaction of crystal violet with excess sodium hydroxide to form colorless products is monitored using a spectrophotometer. A plot of the natural logarithm of the absorbance of green light by the system as a function of time is linear with a negative slope. Which conclusions may be drawn from the data?\n\nI. The reaction is first-order in crystal violet.\n\nII. The reaction is first-order in hydroxide ion.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe kinetics of the reaction of crystal violet with excess sodium hydroxide to form colorless products is monitored using a spectrophotometer. A plot of the natural logarithm of the absorbance of green light by the system as a function of time is linear with a negative slope. Which conclusions may be drawn from the data?\n\nI. The reaction is first-order in crystal violet.\n\nII. The reaction is first-order in hydroxide ion.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_781",
        "problem": "某实验小组用 $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 作催化剂, 以乙炔为原料制备草酸 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$ 的装置如下。\n\n[图1]\n\n下列说法错误的是\nA: 装置 A 中生成草酸的反应为 $\\mathrm{C}_{2} \\mathrm{H}_{2}+8 \\mathrm{HNO}_{3}($ 浓 $) \\xlongequal{\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}+8 \\mathrm{NO}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}$\nB: 向装置 $\\mathrm{C}$ 通入氧气目的是排除装置内的空气\nC: 装置 $\\mathrm{E}$ 的作用是吸收氮氧化物和乙炔\nD: 装置 $\\mathrm{A}$ 使用多孔球泡可增大反应速率\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某实验小组用 $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 作催化剂, 以乙炔为原料制备草酸 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$ 的装置如下。\n\n[图1]\n\n下列说法错误的是\n\nA: 装置 A 中生成草酸的反应为 $\\mathrm{C}_{2} \\mathrm{H}_{2}+8 \\mathrm{HNO}_{3}($ 浓 $) \\xlongequal{\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}+8 \\mathrm{NO}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}$\nB: 向装置 $\\mathrm{C}$ 通入氧气目的是排除装置内的空气\nC: 装置 $\\mathrm{E}$ 的作用是吸收氮氧化物和乙炔\nD: 装置 $\\mathrm{A}$ 使用多孔球泡可增大反应速率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-45.jpg?height=511&width=1442&top_left_y=161&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1400",
        "problem": "Rotational energy levels of diatomic molecules are well described by the formula $E_{\\mathrm{J}}=B \\mathrm{~J}(\\mathrm{~J}+1)$, where $\\mathrm{J}$ is the rotational quantum number of the molecule and $B$ its rotational constant. Constant $B$ is related to the reduced mass $\\mu$ and the bond length $R$ of the molecule through the equation\n\n$$\nB=\\frac{h^{2}}{8 \\pi^{2} \\mu R^{2}} .\n$$\n\nIn general, spectroscopic transitions appear at photon energies which are equal to the energy difference between appropriate states of a molecule ( $h \\nu=\\Delta E$ ). The observed rotational transitions occur between adjacent rotational levels, hence $\\Delta E=E_{J+1}-E_{J}=$ $2 B(\\mathrm{~J}+1)$. Consequently, successive rotational transitions that appear on the spectrum (such as the one shown here) follow the equation $h(\\Delta v)=2 B$.\n\nBy inspecting the spectrum provided, determine the following quantities for ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$ with appropriate units:\n\n$\\B$\n\n[figure1]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nRotational energy levels of diatomic molecules are well described by the formula $E_{\\mathrm{J}}=B \\mathrm{~J}(\\mathrm{~J}+1)$, where $\\mathrm{J}$ is the rotational quantum number of the molecule and $B$ its rotational constant. Constant $B$ is related to the reduced mass $\\mu$ and the bond length $R$ of the molecule through the equation\n\n$$\nB=\\frac{h^{2}}{8 \\pi^{2} \\mu R^{2}} .\n$$\n\nIn general, spectroscopic transitions appear at photon energies which are equal to the energy difference between appropriate states of a molecule ( $h \\nu=\\Delta E$ ). The observed rotational transitions occur between adjacent rotational levels, hence $\\Delta E=E_{J+1}-E_{J}=$ $2 B(\\mathrm{~J}+1)$. Consequently, successive rotational transitions that appear on the spectrum (such as the one shown here) follow the equation $h(\\Delta v)=2 B$.\n\nBy inspecting the spectrum provided, determine the following quantities for ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$ with appropriate units:\n\n$\\B$\n\n[figure1]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~J}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-469.jpg?height=995&width=1525&top_left_y=1664&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~J}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_93",
        "problem": "When $6 \\mathrm{M}$ sodium hydroxide is added to an unknown white solid, the solid dissolves. What is a possible identity for this solid?\nA: $\\mathrm{Mg}(\\mathrm{OH})_{2}$\nB: $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\nC: $\\mathrm{BaCO}_{3}$\nD: $\\mathrm{AgBr}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen $6 \\mathrm{M}$ sodium hydroxide is added to an unknown white solid, the solid dissolves. What is a possible identity for this solid?\n\nA: $\\mathrm{Mg}(\\mathrm{OH})_{2}$\nB: $\\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$\nC: $\\mathrm{BaCO}_{3}$\nD: $\\mathrm{AgBr}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_510",
        "problem": "一定温度下, 向容积为 $2 \\mathrm{~L}$ 的恒容密闭容器中充入 $6 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 $8 \\mathrm{~mol} \\mathrm{H}_{2}$, 发生反应:\n\n$\\mathrm{CO}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\Delta \\mathrm{H}=-49.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n测得 $\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)$ 随时间变化如曲线I所示。下列说法正确的是\n\n[图1]\nA: 该反应在 $0 \\sim 8 \\mathrm{~min}$ 内 $\\mathrm{CO}_{2}$ 的平均反应速率是 $0.375 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 若起始时向上述容器中充入 $3 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 $4 \\mathrm{~mol} \\mathrm{H}_{2}$, 则平衡时 $\\mathrm{H}_{2}$ 的体积分数大于 $20 \\%$\nC: 若起始时向上述容器中充入 $4 \\mathrm{~mol} \\mathrm{CO}_{2} 、 2 \\mathrm{~mol} \\mathrm{H}_{2} 、 2 \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{OH}$ 和 $1 \\mathrm{~mol}$ $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, 则此时反应向正反应方向进行\nD: 改变条件得到曲线II、III, 则曲线II、III改变的条件分别是升高温度、充入氦气\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一定温度下, 向容积为 $2 \\mathrm{~L}$ 的恒容密闭容器中充入 $6 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 $8 \\mathrm{~mol} \\mathrm{H}_{2}$, 发生反应:\n\n$\\mathrm{CO}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\Delta \\mathrm{H}=-49.0 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n测得 $\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)$ 随时间变化如曲线I所示。下列说法正确的是\n\n[图1]\n\nA: 该反应在 $0 \\sim 8 \\mathrm{~min}$ 内 $\\mathrm{CO}_{2}$ 的平均反应速率是 $0.375 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 若起始时向上述容器中充入 $3 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 $4 \\mathrm{~mol} \\mathrm{H}_{2}$, 则平衡时 $\\mathrm{H}_{2}$ 的体积分数大于 $20 \\%$\nC: 若起始时向上述容器中充入 $4 \\mathrm{~mol} \\mathrm{CO}_{2} 、 2 \\mathrm{~mol} \\mathrm{H}_{2} 、 2 \\mathrm{~mol} \\mathrm{CH}_{3} \\mathrm{OH}$ 和 $1 \\mathrm{~mol}$ $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, 则此时反应向正反应方向进行\nD: 改变条件得到曲线II、III, 则曲线II、III改变的条件分别是升高温度、充入氦气\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-101.jpg?height=452&width=711&top_left_y=1256&top_left_x=387",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-102.jpg?height=60&width=957&top_left_y=327&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_959",
        "problem": "纳米碗 $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 是一种奇特的碗状共轭体系, 高温条件下, $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 可以由 $\\mathrm{C}_{40} \\mathrm{H}_{20}$ 分子经过连续 5 步氢抽提和闭环脱氢反应生成, $\\mathrm{C}_{40} \\mathrm{H}_{20}(\\mathrm{~g}) \\underset{\\mathrm{H}}{\\rightarrow} \\mathrm{C}_{40} \\mathrm{H}_{18}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$ 的反应机理和能量变化如图所示. 下列说法错误的是\n[图1]\n\n$\\mathrm{C}_{40} \\mathrm{H}_{19}$\n\n$\\mathrm{C}_{40} \\mathrm{H}_{18}$\n\n[图2]\nA: 图示历程包括 3 个基元反应，其中第三个反应的反应速率最慢\nB: 该过程中既有极性键的断裂和形成又有非极性键的断裂和形成\nC: $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 纳米碗中有 6 个五元环和 10 个六元环结构\nD: 加入催化剂既能提高反应物的平衡转化率, 又能增大生成 $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 的反应速率\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n纳米碗 $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 是一种奇特的碗状共轭体系, 高温条件下, $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 可以由 $\\mathrm{C}_{40} \\mathrm{H}_{20}$ 分子经过连续 5 步氢抽提和闭环脱氢反应生成, $\\mathrm{C}_{40} \\mathrm{H}_{20}(\\mathrm{~g}) \\underset{\\mathrm{H}}{\\rightarrow} \\mathrm{C}_{40} \\mathrm{H}_{18}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$ 的反应机理和能量变化如图所示. 下列说法错误的是\n[图1]\n\n$\\mathrm{C}_{40} \\mathrm{H}_{19}$\n\n$\\mathrm{C}_{40} \\mathrm{H}_{18}$\n\n[图2]\n\nA: 图示历程包括 3 个基元反应，其中第三个反应的反应速率最慢\nB: 该过程中既有极性键的断裂和形成又有非极性键的断裂和形成\nC: $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 纳米碗中有 6 个五元环和 10 个六元环结构\nD: 加入催化剂既能提高反应物的平衡转化率, 又能增大生成 $\\mathrm{C}_{40} \\mathrm{H}_{10}$ 的反应速率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-02.jpg?height=312&width=1240&top_left_y=182&top_left_x=342",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-02.jpg?height=479&width=990&top_left_y=520&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_413",
        "problem": "绿茶中含有的一种物质 EGCG 具有抗癌作用, 能使血癌（白血病）中癌细胞自杀性死亡, 已知 EGCG 的一种衍生物 A 的结构如图所示。有关衍生物 A 说法错误的是[图1]\nA: A 能与碳酸氢钠溶液反应放出二氧化碳\nB: A 在空气中易氧化, 遇 $\\mathrm{FeCl}_{3}$ 溶液能发生显色反应\nC: $1 \\mathrm{molA}$ 与足量的浓溴水反应最多可与 $6 \\mathrm{~mol} \\mathrm{Br}_{2}$ 作用\nD: $1 \\mathrm{molA}$ 最多可与含 $10 \\mathrm{~mol}$ 氢氧化钠的溶液完全作用\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n绿茶中含有的一种物质 EGCG 具有抗癌作用, 能使血癌（白血病）中癌细胞自杀性死亡, 已知 EGCG 的一种衍生物 A 的结构如图所示。有关衍生物 A 说法错误的是[图1]\n\nA: A 能与碳酸氢钠溶液反应放出二氧化碳\nB: A 在空气中易氧化, 遇 $\\mathrm{FeCl}_{3}$ 溶液能发生显色反应\nC: $1 \\mathrm{molA}$ 与足量的浓溴水反应最多可与 $6 \\mathrm{~mol} \\mathrm{Br}_{2}$ 作用\nD: $1 \\mathrm{molA}$ 最多可与含 $10 \\mathrm{~mol}$ 氢氧化钠的溶液完全作用\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/LshrkJYs/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_809",
        "problem": "在常温下, 向 $20 \\mathrm{~mL} 0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中滴加 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸, 溶液 $\\mathrm{pH}$随滴定百分率(\\%)的变化如图所示。下列说法不正确的是\n\n[图1]\nA: 在 a 点的溶液中, $c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)-c\\left(\\mathrm{CO}_{3}^{2-}\\right)=c\\left(\\mathrm{OH}^{-}\\right)-c\\left(\\mathrm{Cl}^{-}\\right)-c\\left(\\mathrm{H}^{+}\\right)$\nB: 在 $\\mathrm{b}$ 点的溶液中, $2 n\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+n\\left(\\mathrm{HCO}_{3}^{-}\\right)<0.002 \\mathrm{~mol}$\nC: 煮沸的目的是除去 $\\mathrm{CO}_{2}$, 使突跃变大, 冷却后继续滴定到终点\nD: 若将 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸换成同浓度的 $\\mathrm{CH}_{3} \\mathrm{COOH}$, 当滴至溶液的 $\\mathrm{pH}=7$ 时, $$ \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOO}^{-}\\right) $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在常温下, 向 $20 \\mathrm{~mL} 0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中滴加 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸, 溶液 $\\mathrm{pH}$随滴定百分率(\\%)的变化如图所示。下列说法不正确的是\n\n[图1]\n\nA: 在 a 点的溶液中, $c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)-c\\left(\\mathrm{CO}_{3}^{2-}\\right)=c\\left(\\mathrm{OH}^{-}\\right)-c\\left(\\mathrm{Cl}^{-}\\right)-c\\left(\\mathrm{H}^{+}\\right)$\nB: 在 $\\mathrm{b}$ 点的溶液中, $2 n\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+n\\left(\\mathrm{HCO}_{3}^{-}\\right)<0.002 \\mathrm{~mol}$\nC: 煮沸的目的是除去 $\\mathrm{CO}_{2}$, 使突跃变大, 冷却后继续滴定到终点\nD: 若将 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的盐酸换成同浓度的 $\\mathrm{CH}_{3} \\mathrm{COOH}$, 当滴至溶液的 $\\mathrm{pH}=7$ 时, $$ \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOO}^{-}\\right) $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-090.jpg?height=431&width=462&top_left_y=1943&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_446",
        "problem": "香草醛是一种食品添加剂, 可由愈创木酚作原料合成, 合成路线如图:\n\n\n[图1]\n\n已知： $\\mathrm{R}-\\mathrm{COOH}$ 具有与乙酸相似的化学性质。下列说法错误的是\nA: $1 \\mathrm{~mol}$ 乙与足量 $\\mathrm{Na}$ 反应产生 $1 \\mathrm{molH}_{2}$\nB: 可用 $\\mathrm{NaHCO}_{3}$ 溶液鉴别化合物甲和乙\nC: 检验制得的丁(香草醛)中是否混有化合物丙, 可选用浓澳水\nD: 等物质的量的甲、乙分别与足量 $\\mathrm{NaOH}$ 反应, 消耗 $\\mathrm{NaOH}$ 的物质的量之比为 1 : 2\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n香草醛是一种食品添加剂, 可由愈创木酚作原料合成, 合成路线如图:\n\n\n[图1]\n\n已知： $\\mathrm{R}-\\mathrm{COOH}$ 具有与乙酸相似的化学性质。下列说法错误的是\n\nA: $1 \\mathrm{~mol}$ 乙与足量 $\\mathrm{Na}$ 反应产生 $1 \\mathrm{molH}_{2}$\nB: 可用 $\\mathrm{NaHCO}_{3}$ 溶液鉴别化合物甲和乙\nC: 检验制得的丁(香草醛)中是否混有化合物丙, 可选用浓澳水\nD: 等物质的量的甲、乙分别与足量 $\\mathrm{NaOH}$ 反应, 消耗 $\\mathrm{NaOH}$ 的物质的量之比为 1 : 2\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/g2hWDJ0Y/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1161",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the misplacement of thallium\n\nMendeleev also misplaced the highly toxic element thallium ( $\\mathrm{Tl}$ ) with the elements from Group 1 rather than in Group 13. There are good reasons for this error. Thallium can form salts in the +3 oxidation state like other members of its group, but $\\mathrm{Tl}^{3+}$ ions are oxidising and the most stable oxidation state is +1 . For example, adding iodide ions to solutions of $\\mathrm{Tl}^{3+}$ actually gives a precipitate of thallium $(\\mathrm{I})$ iodide.\n\nAn iodide of thallium with the formula $\\mathrm{THI}_{3}$ is known and has exactly the same structure as $\\mathrm{CsI}_{3}$.\n\nWhat is the average oxidation state of the iodine in these compounds?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the misplacement of thallium\n\nMendeleev also misplaced the highly toxic element thallium ( $\\mathrm{Tl}$ ) with the elements from Group 1 rather than in Group 13. There are good reasons for this error. Thallium can form salts in the +3 oxidation state like other members of its group, but $\\mathrm{Tl}^{3+}$ ions are oxidising and the most stable oxidation state is +1 . For example, adding iodide ions to solutions of $\\mathrm{Tl}^{3+}$ actually gives a precipitate of thallium $(\\mathrm{I})$ iodide.\n\nAn iodide of thallium with the formula $\\mathrm{THI}_{3}$ is known and has exactly the same structure as $\\mathrm{CsI}_{3}$.\n\nWhat is the average oxidation state of the iodine in these compounds?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_73",
        "problem": "When a helium-filled balloon is immersed in liquid nitrogen, which statements accurately describe the changes as the balloon deflates?\n\nI. The nitrogen does work on the balloon.\n\nII. The entropy of the nitrogen increases.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen a helium-filled balloon is immersed in liquid nitrogen, which statements accurately describe the changes as the balloon deflates?\n\nI. The nitrogen does work on the balloon.\n\nII. The entropy of the nitrogen increases.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_449",
        "problem": "羟基自由基 $(\\cdot \\mathrm{OH})$ 是自然界中氧化性仅次于氟的氧化剂, 在直流电源的作用下, 利用如图所示装置处理含苯酚废水和含甲醛废水 (已知: $\\mathrm{H}_{2} \\mathrm{O}$ 在膜 $\\mathrm{a} 、 \\mathrm{~b}$ 间解离为 $\\mathrm{H}^{+}$和 $\\left.\\mathrm{OH}^{-}\\right)$。下列说法错误的是\n\n[图1]\nA: 直流电源的电极电势：右端高于左端\nB: 阳极的电极反应式为: $\\mathrm{OH}^{-}-\\mathrm{e}^{-}=\\cdot \\mathrm{OH}$\nC: 电解质 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 的作用是可以增强溶液的导电能力\nD: 每处理 $9.4 \\mathrm{~g}$ 苯酚, 理论上有 $2.8 \\mathrm{~mol} \\mathrm{OH}^{-}$透过膜 b\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n羟基自由基 $(\\cdot \\mathrm{OH})$ 是自然界中氧化性仅次于氟的氧化剂, 在直流电源的作用下, 利用如图所示装置处理含苯酚废水和含甲醛废水 (已知: $\\mathrm{H}_{2} \\mathrm{O}$ 在膜 $\\mathrm{a} 、 \\mathrm{~b}$ 间解离为 $\\mathrm{H}^{+}$和 $\\left.\\mathrm{OH}^{-}\\right)$。下列说法错误的是\n\n[图1]\n\nA: 直流电源的电极电势：右端高于左端\nB: 阳极的电极反应式为: $\\mathrm{OH}^{-}-\\mathrm{e}^{-}=\\cdot \\mathrm{OH}$\nC: 电解质 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 的作用是可以增强溶液的导电能力\nD: 每处理 $9.4 \\mathrm{~g}$ 苯酚, 理论上有 $2.8 \\mathrm{~mol} \\mathrm{OH}^{-}$透过膜 b\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-007.jpg?height=599&width=1102&top_left_y=503&top_left_x=317"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_650",
        "problem": "一定温度下, 分别将两种亚硫酸氢盐加热分解, 达平衡时的压强分别为 $\\mathrm{p}_{1} 、 \\mathrm{p}_{2}$ 。\n\ni. $\\mathrm{NH}_{4} \\mathrm{HSO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{NH}_{3}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\mathrm{p}_{1}=\\mathrm{aPa}$\n\nii. $2 \\mathrm{NaHSO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Na}_{2} \\mathrm{SO}_{3}(\\mathrm{~s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\mathrm{p}_{2}=\\mathrm{bPa}$\n\n在该温度下, 将一定质量的 $\\mathrm{NH}_{4} \\mathrm{HSO}_{3}$ 与 $\\mathrm{NaHSO}_{3}$ 加入某密闭容器中, 平衡时, 三种固体均存在。下列说法不正确的是\nA: 平衡时, 总压为 $\\left(b+\\frac{4 a^{3}}{27 b^{2}}\\right) \\mathrm{Pa}$\nB: 保持恒温、恒压, 若向体系中再通入一定量 $\\mathrm{N}_{2}$, 容器内固体的质量将减小\nC: 保持恒温、恒容, 若再通入一定量的 $\\mathrm{NH}_{3}$, 平衡时 $\\mathrm{NH}_{3}$ 的物质的量比原平衡大\nD: 保持恒温将容器体积压缩, 再次达平衡后各气体的浓度均保持不变\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一定温度下, 分别将两种亚硫酸氢盐加热分解, 达平衡时的压强分别为 $\\mathrm{p}_{1} 、 \\mathrm{p}_{2}$ 。\n\ni. $\\mathrm{NH}_{4} \\mathrm{HSO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{NH}_{3}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\mathrm{p}_{1}=\\mathrm{aPa}$\n\nii. $2 \\mathrm{NaHSO}_{3}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Na}_{2} \\mathrm{SO}_{3}(\\mathrm{~s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\mathrm{p}_{2}=\\mathrm{bPa}$\n\n在该温度下, 将一定质量的 $\\mathrm{NH}_{4} \\mathrm{HSO}_{3}$ 与 $\\mathrm{NaHSO}_{3}$ 加入某密闭容器中, 平衡时, 三种固体均存在。下列说法不正确的是\n\nA: 平衡时, 总压为 $\\left(b+\\frac{4 a^{3}}{27 b^{2}}\\right) \\mathrm{Pa}$\nB: 保持恒温、恒压, 若向体系中再通入一定量 $\\mathrm{N}_{2}$, 容器内固体的质量将减小\nC: 保持恒温、恒容, 若再通入一定量的 $\\mathrm{NH}_{3}$, 平衡时 $\\mathrm{NH}_{3}$ 的物质的量比原平衡大\nD: 保持恒温将容器体积压缩, 再次达平衡后各气体的浓度均保持不变\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1567",
        "problem": "Determination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\nCalculate the $T$ values of iodine solution in each of the following cases:\n\n$12.20 \\mathrm{~cm}^{3}$ of Fischer reagent solution were used for titration of $1.352 \\mathrm{~g}$ of sodium tartrate dihydrate $\\mathrm{Na}_{2} \\mathrm{C}_{4} \\mathrm{H}_{4} \\mathrm{O}_{6} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nDetermination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\nCalculate the $T$ values of iodine solution in each of the following cases:\n\n$12.20 \\mathrm{~cm}^{3}$ of Fischer reagent solution were used for titration of $1.352 \\mathrm{~g}$ of sodium tartrate dihydrate $\\mathrm{Na}_{2} \\mathrm{C}_{4} \\mathrm{H}_{4} \\mathrm{O}_{6} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{mg} \\mathrm{cm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-651.jpg?height=213&width=900&top_left_y=776&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{mg} \\mathrm{cm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_952",
        "problem": "室温下进行下列实验, 根据实验操作和现象所得到的结论正确的是\n\n| 选 <br> 项 | 实验操作和现象 | 结论 |\n| :---: | :---: | :---: |\n| A | 室温下, 向苯酚钠溶液中通足量 $\\mathrm{CO}_{2}$, 溶液变浑浊 | 碳酸的酸性比苯酚 <br> 的强 |\n| B | 加热乙醇与浓硫酸的混合溶液, 将产生的气体通入少量酸性 <br> $\\mathrm{KMnO}_{4}$ 溶液, 溶液紫红色褪去 | 有乙烯生成 |\n| C | 向 $5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KI}$ 溶液中加入 $1 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{FeCl}_{3}$ 溶 <br> 液, 充分反应后, 萃取分液, 向水层中滴加 $\\mathrm{KSCN}$ 溶液, <br> 溶液呈血红色 | $\\mathrm{I}^{-}$与 $\\mathrm{Fe}^{3+\\text { 的反应有 }}$ <br> 一定限度 |\n| D | 向 $\\mathrm{NaHCO}_{3}$ 溶液中滴加紫色石荵试液, 溶液变蓝 | $K_{w}<$ <br> $K_{a l}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\times K_{a 2}\\left(\\mathrm{H}_{2}\\right.$ <br> $\\left.\\mathrm{CO}_{3}\\right)$ |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n室温下进行下列实验, 根据实验操作和现象所得到的结论正确的是\n\n| 选 <br> 项 | 实验操作和现象 | 结论 |\n| :---: | :---: | :---: |\n| A | 室温下, 向苯酚钠溶液中通足量 $\\mathrm{CO}_{2}$, 溶液变浑浊 | 碳酸的酸性比苯酚 <br> 的强 |\n| B | 加热乙醇与浓硫酸的混合溶液, 将产生的气体通入少量酸性 <br> $\\mathrm{KMnO}_{4}$ 溶液, 溶液紫红色褪去 | 有乙烯生成 |\n| C | 向 $5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KI}$ 溶液中加入 $1 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{FeCl}_{3}$ 溶 <br> 液, 充分反应后, 萃取分液, 向水层中滴加 $\\mathrm{KSCN}$ 溶液, <br> 溶液呈血红色 | $\\mathrm{I}^{-}$与 $\\mathrm{Fe}^{3+\\text { 的反应有 }}$ <br> 一定限度 |\n| D | 向 $\\mathrm{NaHCO}_{3}$ 溶液中滴加紫色石荵试液, 溶液变蓝 | $K_{w}<$ <br> $K_{a l}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\times K_{a 2}\\left(\\mathrm{H}_{2}\\right.$ <br> $\\left.\\mathrm{CO}_{3}\\right)$ |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_551",
        "problem": "常温下, 向 $25 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液中逐滴加入 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液,所得滴定曲线如图所示, 下列说法错误的是\n\n[图1]\n\n已知：不考虑溶液混合时体积和温度的变化。\nA: a 点溶液中, 所含微粒种类共有 4 种\nB: b 点溶液中, $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$\nC: $\\mathrm{c}$ 点溶液中, $c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{a} \\rightarrow \\mathrm{c}$ 的过程中, 水的电离程度先增大后减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $25 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液中逐滴加入 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液,所得滴定曲线如图所示, 下列说法错误的是\n\n[图1]\n\n已知：不考虑溶液混合时体积和温度的变化。\n\nA: a 点溶液中, 所含微粒种类共有 4 种\nB: b 点溶液中, $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$\nC: $\\mathrm{c}$ 点溶液中, $c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{a} \\rightarrow \\mathrm{c}$ 的过程中, 水的电离程度先增大后减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-028.jpg?height=474&width=759&top_left_y=2142&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_730",
        "problem": "常温下, 向 $20 \\mathrm{~mL} 0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 某一元碱 $\\mathrm{MOH}$ 溶液中逐滴加入 $0.25 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCl}$ 溶\n液, 溶液中 $\\lg \\frac{\\mathrm{c}(\\mathrm{MOH})}{\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)} 、 \\mathrm{pOH}$ 、中和率的变化如图所示。 $\\mathrm{K}_{\\mathrm{b}}$ 为 $\\mathrm{MOH}$ 的电离常数, $\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)$, 中和率 $=\\frac{\\text { 被中和的 } \\mathrm{MOH} \\text { 的物质的量 }}{\\text { 反应前 } \\mathrm{MOH} \\text { 的总物质的量 }} \\times 100 \\%$ 。下列说法正确的是\n\n[图1]\nA: 根据 $\\mathrm{a}$ 点数据计算, 常温下 $\\mathrm{K}_{\\mathrm{b}}=10^{-4.2}$\nB: 溶液中水的电离程度从 $\\mathrm{a}$ 点到 $\\mathrm{d}$ 点逐渐增大\nC: b 点时, $c\\left(\\mathrm{M}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}(\\mathrm{MOH})>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $\\mathrm{c}$ 点溶液中 $\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)+\\mathrm{c}(\\mathrm{MOH})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 向 $20 \\mathrm{~mL} 0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 某一元碱 $\\mathrm{MOH}$ 溶液中逐滴加入 $0.25 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCl}$ 溶\n液, 溶液中 $\\lg \\frac{\\mathrm{c}(\\mathrm{MOH})}{\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)} 、 \\mathrm{pOH}$ 、中和率的变化如图所示。 $\\mathrm{K}_{\\mathrm{b}}$ 为 $\\mathrm{MOH}$ 的电离常数, $\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)$, 中和率 $=\\frac{\\text { 被中和的 } \\mathrm{MOH} \\text { 的物质的量 }}{\\text { 反应前 } \\mathrm{MOH} \\text { 的总物质的量 }} \\times 100 \\%$ 。下列说法正确的是\n\n[图1]\n\nA: 根据 $\\mathrm{a}$ 点数据计算, 常温下 $\\mathrm{K}_{\\mathrm{b}}=10^{-4.2}$\nB: 溶液中水的电离程度从 $\\mathrm{a}$ 点到 $\\mathrm{d}$ 点逐渐增大\nC: b 点时, $c\\left(\\mathrm{M}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}(\\mathrm{MOH})>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $\\mathrm{c}$ 点溶液中 $\\mathrm{c}\\left(\\mathrm{M}^{+}\\right)+\\mathrm{c}(\\mathrm{MOH})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-065.jpg?height=417&width=759&top_left_y=431&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_29",
        "problem": "The natural logarithm of the molar solubility $S$ of sodium iodate monohydrate as a function of the reciprocal of the absolute temperature is plotted below.\n\n[figure1]\n\nWhat is the standard enthalpy of dissolution for $\\mathrm{NaIO}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ ?\nA: $2.1 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $17.4 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $26.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $34.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe natural logarithm of the molar solubility $S$ of sodium iodate monohydrate as a function of the reciprocal of the absolute temperature is plotted below.\n\n[figure1]\n\nWhat is the standard enthalpy of dissolution for $\\mathrm{NaIO}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ ?\n\nA: $2.1 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $17.4 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $26.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $34.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-06.jpg?height=504&width=550&top_left_y=342&top_left_x=346"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_391",
        "problem": "What is the concentration of chloride ions in a solution formed by mixing $150 . \\mathrm{mL}$ of a $1.50 \\mathrm{M} \\mathrm{NaCl}$ solution with 250 . $\\mathrm{mL}$ of a $0.750 \\mathrm{M} \\mathrm{MgCl}_{2}$ solution?\nA: $0.563 \\mathrm{M}$\nB: $1.03 \\mathrm{M}$\nC: $1.50 \\mathrm{M}$\nD: $2.25 \\mathrm{M}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the concentration of chloride ions in a solution formed by mixing $150 . \\mathrm{mL}$ of a $1.50 \\mathrm{M} \\mathrm{NaCl}$ solution with 250 . $\\mathrm{mL}$ of a $0.750 \\mathrm{M} \\mathrm{MgCl}_{2}$ solution?\n\nA: $0.563 \\mathrm{M}$\nB: $1.03 \\mathrm{M}$\nC: $1.50 \\mathrm{M}$\nD: $2.25 \\mathrm{M}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_142",
        "problem": "Which of the following compounds has the smallest bond angle?\nA: $\\mathrm{BF} 3$\nB: $\\mathrm{CF}_{4}$\nC: $\\mathrm{NF} 3$\nD: $\\mathrm{OF} 2$\nE: $\\mathrm{ClF} 3$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following compounds has the smallest bond angle?\n\nA: $\\mathrm{BF} 3$\nB: $\\mathrm{CF}_{4}$\nC: $\\mathrm{NF} 3$\nD: $\\mathrm{OF} 2$\nE: $\\mathrm{ClF} 3$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_891",
        "problem": "异丁醇催化脱水制备异丁烯主要涉及以下 2 个反应。研究一定压强下不同含水量的异丁醇在恒压反应器中的脱水反应, 得到了异丁烯的平衡产率随温度的变化结果如图。\n\n[图1]\n\n(1)\n\n下列说法不正确的是\nA: 其他条件不变时, 在催化剂的活性温度内, 升高温度有利于异丁烯的制备\nB: 高于 $190^{\\circ} \\mathrm{C}$ 时, 温度对异丁烯的平衡产率影响不大的原因是 $K_{1}>10^{4} 、 K_{2}<0.1$\nC: $190^{\\circ} \\mathrm{C}$ 时, 增大 $n\\left(\\mathrm{H}_{2} \\mathrm{O}\\right): n$ (异丁醇), 不利于反应(2)的进行\nD: 若只有异丁烯、水和二聚异丁烯生成, 则初始物质浓度 $c_{0}$ 与流出物质浓度 $c$ 之 间存在: $c_{0}($ 异丁醇 $)=c($ 异丁烯 $)+2 c($ 二聚异丁烯 $)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n异丁醇催化脱水制备异丁烯主要涉及以下 2 个反应。研究一定压强下不同含水量的异丁醇在恒压反应器中的脱水反应, 得到了异丁烯的平衡产率随温度的变化结果如图。\n\n[图1]\n\n(1)\n\n下列说法不正确的是\n\nA: 其他条件不变时, 在催化剂的活性温度内, 升高温度有利于异丁烯的制备\nB: 高于 $190^{\\circ} \\mathrm{C}$ 时, 温度对异丁烯的平衡产率影响不大的原因是 $K_{1}>10^{4} 、 K_{2}<0.1$\nC: $190^{\\circ} \\mathrm{C}$ 时, 增大 $n\\left(\\mathrm{H}_{2} \\mathrm{O}\\right): n$ (异丁醇), 不利于反应(2)的进行\nD: 若只有异丁烯、水和二聚异丁烯生成, 则初始物质浓度 $c_{0}$ 与流出物质浓度 $c$ 之 间存在: $c_{0}($ 异丁醇 $)=c($ 异丁烯 $)+2 c($ 二聚异丁烯 $)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-035.jpg?height=714&width=785&top_left_y=500&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_914",
        "problem": ".一种染料敏化太阳能电池如图, 其成本便宜且对环境无污染。敏化染料 $(S)$ 吸附在 $\\mathrm{TiO}_{2}$ 纳米空心球表面。光照时: $\\mathrm{S}($ 基态 $) \\xrightarrow{\\mathrm{hv}} \\mathrm{S}^{*}$ (激发态), $\\mathrm{S}^{*}-\\mathrm{e}^{-} \\xrightarrow{\\mathrm{TiO}_{2}} \\mathrm{~S}^{+}$, 下列说法正确的是\n\n[图1]\nA: 该装置能将太阳能转化成电能\nB: 正极的电极反应式为 $\\mathrm{I}_{3}^{-}+2 \\mathrm{e}^{-}=3 \\mathrm{I}^{-}$\nC: 敏化染料还原过程为 $\\mathrm{S}^{+}+3 \\mathrm{I}^{-}=\\mathrm{S}+\\mathrm{I}_{3}^{-}$\nD: 该电池工作一段时间后需要补充电解质\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n.一种染料敏化太阳能电池如图, 其成本便宜且对环境无污染。敏化染料 $(S)$ 吸附在 $\\mathrm{TiO}_{2}$ 纳米空心球表面。光照时: $\\mathrm{S}($ 基态 $) \\xrightarrow{\\mathrm{hv}} \\mathrm{S}^{*}$ (激发态), $\\mathrm{S}^{*}-\\mathrm{e}^{-} \\xrightarrow{\\mathrm{TiO}_{2}} \\mathrm{~S}^{+}$, 下列说法正确的是\n\n[图1]\n\nA: 该装置能将太阳能转化成电能\nB: 正极的电极反应式为 $\\mathrm{I}_{3}^{-}+2 \\mathrm{e}^{-}=3 \\mathrm{I}^{-}$\nC: 敏化染料还原过程为 $\\mathrm{S}^{+}+3 \\mathrm{I}^{-}=\\mathrm{S}+\\mathrm{I}_{3}^{-}$\nD: 该电池工作一段时间后需要补充电解质\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-30.jpg?height=419&width=737&top_left_y=156&top_left_x=320"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1541",
        "problem": "Nitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.Under what condition does this rate law reduce to the experimentally determined rate law found in Part 1.1? Tick the relevant answer.\nA:  If $k_{-1} \\ll k_{2} p_{\\mathrm{H} 2}$\nB: If $k_{-1} \\gg k_{2} p_{H 2}$\nC:  If $k_{-1}>k_{2}$\nD: If $k_{1}>k_{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nNitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.\n\nproblem:\nUnder what condition does this rate law reduce to the experimentally determined rate law found in Part 1.1? Tick the relevant answer.\n\nA:  If $k_{-1} \\ll k_{2} p_{\\mathrm{H} 2}$\nB: If $k_{-1} \\gg k_{2} p_{H 2}$\nC:  If $k_{-1}>k_{2}$\nD: If $k_{1}>k_{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_949",
        "problem": "工业上常用 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{3}$ 溶液吸收废气中的 $\\mathrm{SO}_{2}$, 室温下测得溶液中 $\\lg \\mathrm{Y}$\n\n[图1]\n\n[图2]\nA: 通入少量 $\\mathrm{SO}_{2}$ 的过程中, 直线II中的 $\\mathrm{N}$ 点向 $\\mathrm{M}$ 点移动\nB: $\\alpha_{1}=\\alpha_{2}$ 不一定等于 $45^{\\circ}$\nC: 当对应溶液的 $\\mathrm{pH}$ 处于 $1.81<\\mathrm{pH}<6.91$ 时, 溶液中的微粒浓度一定存在 $c_{\\text {平 }}\\left(\\mathrm{HSO}_{3}^{-}\\right)>c_{\\text {平 }}\\left(\\mathrm{SO}_{3}^{2-}\\right)>c_{\\text {平 }}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\nD: 直线II中 $\\mathrm{M} 、 \\mathrm{~N}$ 点一定存在 $c_{\\text {平 }}^{2}\\left(\\mathrm{HSO}_{3}^{-}\\right)>c_{\\text {平 }}\\left(\\mathrm{SO}_{3}^{2-}\\right) \\cdot c_{\\text {平 }}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n工业上常用 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{3}$ 溶液吸收废气中的 $\\mathrm{SO}_{2}$, 室温下测得溶液中 $\\lg \\mathrm{Y}$\n\n[图1]\n\n[图2]\n\nA: 通入少量 $\\mathrm{SO}_{2}$ 的过程中, 直线II中的 $\\mathrm{N}$ 点向 $\\mathrm{M}$ 点移动\nB: $\\alpha_{1}=\\alpha_{2}$ 不一定等于 $45^{\\circ}$\nC: 当对应溶液的 $\\mathrm{pH}$ 处于 $1.81<\\mathrm{pH}<6.91$ 时, 溶液中的微粒浓度一定存在 $c_{\\text {平 }}\\left(\\mathrm{HSO}_{3}^{-}\\right)>c_{\\text {平 }}\\left(\\mathrm{SO}_{3}^{2-}\\right)>c_{\\text {平 }}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\nD: 直线II中 $\\mathrm{M} 、 \\mathrm{~N}$ 点一定存在 $c_{\\text {平 }}^{2}\\left(\\mathrm{HSO}_{3}^{-}\\right)>c_{\\text {平 }}\\left(\\mathrm{SO}_{3}^{2-}\\right) \\cdot c_{\\text {平 }}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-027.jpg?height=111&width=1304&top_left_y=2126&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-027.jpg?height=434&width=460&top_left_y=2302&top_left_x=387",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-028.jpg?height=123&width=1373&top_left_y=904&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-028.jpg?height=119&width=1382&top_left_y=1314&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_676",
        "problem": "一种制备 $\\mathrm{Cu}_{2} \\mathrm{O}$ 的工艺路线如图所示, 反应II所得溶液 $\\mathrm{pH}$ 在 3 4 之间, 反应III需及时补加 $\\mathrm{NaOH}$ 以保持反应在 $\\mathrm{pH}=5$ 条件下进行。常温下, $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{al}}=1.3 \\times 10^{-2}, \\mathrm{~K}_{\\mathrm{a} 2}=6.3 \\times 10^{-8}$ 。下列说法正确的是\n\n[图1]\nA: 反应I、II、III均为氧化还原反应\nB: 低温真空蒸发主要目的是防止 $\\mathrm{NaHSO}_{3}$ 被氧化\nC: 溶液 $Y$ 可循环用于反应II所在操作单元吸收气体I\nD: 若 $\\mathrm{Cu}_{2} \\mathrm{O}$ 产量不变, 参与反应 $\\mathrm{III}$ 的 $\\mathrm{X}$ 与 $\\mathrm{CuSO}_{4}$ 物质的量之比 $\\frac{\\mathrm{n}(\\mathrm{X})}{\\mathrm{n}\\left(\\mathrm{CuSO}_{4}\\right)}$ 增大时,需补加 $\\mathrm{NaOH}$ 的量减少\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种制备 $\\mathrm{Cu}_{2} \\mathrm{O}$ 的工艺路线如图所示, 反应II所得溶液 $\\mathrm{pH}$ 在 3 4 之间, 反应III需及时补加 $\\mathrm{NaOH}$ 以保持反应在 $\\mathrm{pH}=5$ 条件下进行。常温下, $\\mathrm{H}_{2} \\mathrm{SO}_{3}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{al}}=1.3 \\times 10^{-2}, \\mathrm{~K}_{\\mathrm{a} 2}=6.3 \\times 10^{-8}$ 。下列说法正确的是\n\n[图1]\n\nA: 反应I、II、III均为氧化还原反应\nB: 低温真空蒸发主要目的是防止 $\\mathrm{NaHSO}_{3}$ 被氧化\nC: 溶液 $Y$ 可循环用于反应II所在操作单元吸收气体I\nD: 若 $\\mathrm{Cu}_{2} \\mathrm{O}$ 产量不变, 参与反应 $\\mathrm{III}$ 的 $\\mathrm{X}$ 与 $\\mathrm{CuSO}_{4}$ 物质的量之比 $\\frac{\\mathrm{n}(\\mathrm{X})}{\\mathrm{n}\\left(\\mathrm{CuSO}_{4}\\right)}$ 增大时,需补加 $\\mathrm{NaOH}$ 的量减少\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-036.jpg?height=751&width=1019&top_left_y=156&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_729",
        "problem": "丙烷与溴原子能发生以下两种反应:\n\n(1) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}(\\mathrm{~g})+\\mathrm{Br} \\cdot(\\mathrm{g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\cdot(\\mathrm{g})+\\mathrm{HBr}(\\mathrm{g})$\n\n(2) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}(\\mathrm{~g})+\\mathrm{Br} \\cdot(\\mathrm{g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{CH} \\cdot \\mathrm{CH}_{3}(\\mathrm{~g})+\\mathrm{HBr}(\\mathrm{g})$\n\n反应过程的能量变化如图所示。下列说法正确的是\n\n[图1]\nA: 反应(1)是放热反应\nB: 反应(2)使用了催化剂\nC: 产物中 $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\cdot(\\mathrm{g})$ 含量比 $\\mathrm{CH}_{3} \\mathrm{CH} \\cdot \\mathrm{CH}_{3}(\\mathrm{~g})$ 低\nD: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\cdot(\\mathrm{g})$ 转变为 $\\mathrm{CH}_{3} \\mathrm{CH} \\cdot \\mathrm{CH}_{3}(\\mathrm{~g})$ 时需要吸热\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n丙烷与溴原子能发生以下两种反应:\n\n(1) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}(\\mathrm{~g})+\\mathrm{Br} \\cdot(\\mathrm{g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\cdot(\\mathrm{g})+\\mathrm{HBr}(\\mathrm{g})$\n\n(2) $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}(\\mathrm{~g})+\\mathrm{Br} \\cdot(\\mathrm{g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{CH} \\cdot \\mathrm{CH}_{3}(\\mathrm{~g})+\\mathrm{HBr}(\\mathrm{g})$\n\n反应过程的能量变化如图所示。下列说法正确的是\n\n[图1]\n\nA: 反应(1)是放热反应\nB: 反应(2)使用了催化剂\nC: 产物中 $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\cdot(\\mathrm{g})$ 含量比 $\\mathrm{CH}_{3} \\mathrm{CH} \\cdot \\mathrm{CH}_{3}(\\mathrm{~g})$ 低\nD: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\cdot(\\mathrm{g})$ 转变为 $\\mathrm{CH}_{3} \\mathrm{CH} \\cdot \\mathrm{CH}_{3}(\\mathrm{~g})$ 时需要吸热\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-14.jpg?height=566&width=551&top_left_y=1222&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_693",
        "problem": "某柔性屏手机的柔性电池以碳纳米管做电极材料, 以吸收 $\\mathrm{ZnSO}_{4}$ 溶液的有机高聚物做固态电解质, 其电池总反应为:\n\n$\\mathrm{MnO}_{2}+\\frac{1}{2} \\mathrm{Zn}+\\left(1+\\frac{\\mathrm{x}}{6}\\right) \\mathrm{H}_{2} \\mathrm{O}+\\frac{1}{6} \\mathrm{ZnSO}_{4} \\underset{\\text { 放电 }}{\\stackrel{\\text { 充电 }}{\\rightleftarrows}} \\mathrm{MnOOH}+\\frac{1}{6} \\mathrm{ZnSO}_{4}\\left[\\mathrm{Zn}(\\mathrm{OH})_{2}\\right]_{3} \\cdot \\mathrm{xH}_{2} \\mathrm{O}$\n\n其电池结构如图 1 所示, 图 2 是有机高聚物的结构片段。\n\n[图1]\n\n图1\n\n图2\n\n下列说法中, 正确的是 ( )\nA: 碳纳米管具有导电性, 可用作电极材料\nB: 放电时, 电池的正极反应为 $\\mathrm{MnO}_{2}-\\mathrm{e}^{-}+\\mathrm{H}^{+}=\\mathrm{MnOOH}$\nC: 充电时, $\\mathrm{Zn}^{2+}$ 移向 $\\mathrm{MnO}_{2}$\nD: 合成有机高聚物的单体是[图2]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某柔性屏手机的柔性电池以碳纳米管做电极材料, 以吸收 $\\mathrm{ZnSO}_{4}$ 溶液的有机高聚物做固态电解质, 其电池总反应为:\n\n$\\mathrm{MnO}_{2}+\\frac{1}{2} \\mathrm{Zn}+\\left(1+\\frac{\\mathrm{x}}{6}\\right) \\mathrm{H}_{2} \\mathrm{O}+\\frac{1}{6} \\mathrm{ZnSO}_{4} \\underset{\\text { 放电 }}{\\stackrel{\\text { 充电 }}{\\rightleftarrows}} \\mathrm{MnOOH}+\\frac{1}{6} \\mathrm{ZnSO}_{4}\\left[\\mathrm{Zn}(\\mathrm{OH})_{2}\\right]_{3} \\cdot \\mathrm{xH}_{2} \\mathrm{O}$\n\n其电池结构如图 1 所示, 图 2 是有机高聚物的结构片段。\n\n[图1]\n\n图1\n\n图2\n\n下列说法中, 正确的是 ( )\n\nA: 碳纳米管具有导电性, 可用作电极材料\nB: 放电时, 电池的正极反应为 $\\mathrm{MnO}_{2}-\\mathrm{e}^{-}+\\mathrm{H}^{+}=\\mathrm{MnOOH}$\nC: 充电时, $\\mathrm{Zn}^{2+}$ 移向 $\\mathrm{MnO}_{2}$\nD: 合成有机高聚物的单体是[图2]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-72.jpg?height=336&width=1038&top_left_y=1317&top_left_x=335",
            "https://i.postimg.cc/YSgTDJRc/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_554",
        "problem": "利用 $\\mathrm{CH}_{4}$ 燃料电池电解制备 $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, 装置如图所示。下列说法正确的是\n[图1]\nA: 该装置能得到副产物 $\\mathrm{NaOH} 、 \\mathrm{H}_{2} 、 \\mathrm{Cl}_{2}$\nB: $\\mathrm{A}$ 膜和 $\\mathrm{C}$ 膜均为阴离子交换膜, $\\mathrm{B}$ 为阳离子交换膜\nC: $\\mathrm{a}$ 极的电极反应式: $\\mathrm{CH}_{4}+2 \\mathrm{H}_{2} \\mathrm{O}-8 \\mathrm{e}^{-}=\\mathrm{CO}_{2}+8 \\mathrm{H}^{+}$\nD: 理论上, $\\mathrm{b}$ 极上通入标况下 $2.24 \\mathrm{LO}_{2}$, 产品室增加 $0.2 \\mathrm{~mol} \\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用 $\\mathrm{CH}_{4}$ 燃料电池电解制备 $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, 装置如图所示。下列说法正确的是\n[图1]\n\nA: 该装置能得到副产物 $\\mathrm{NaOH} 、 \\mathrm{H}_{2} 、 \\mathrm{Cl}_{2}$\nB: $\\mathrm{A}$ 膜和 $\\mathrm{C}$ 膜均为阴离子交换膜, $\\mathrm{B}$ 为阳离子交换膜\nC: $\\mathrm{a}$ 极的电极反应式: $\\mathrm{CH}_{4}+2 \\mathrm{H}_{2} \\mathrm{O}-8 \\mathrm{e}^{-}=\\mathrm{CO}_{2}+8 \\mathrm{H}^{+}$\nD: 理论上, $\\mathrm{b}$ 极上通入标况下 $2.24 \\mathrm{LO}_{2}$, 产品室增加 $0.2 \\mathrm{~mol} \\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-81.jpg?height=390&width=1378&top_left_y=498&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1031",
        "problem": "$[S P]=2.36 \\times 10^{-13} \\mathrm{~mol} \\mathrm{dm}^{-3}$A typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nDuring the test a person swabs their nose/throat and places it in the extraction solution. The resulting solution is called the test solution. A test sample of a few drops $\\left(0.10 \\mathrm{~cm}^{3}\\right)$ of the test solution are placed on the sample pad.\n\nIf the person taking the test has COVID-19, there will typically be $7.1 \\times 10^{6}$ virus particles per $\\mathrm{cm}^{3}$ in the test solution. Each virus particle has approximately 20 spike proteins on its surface.\n\nOnce the sample passes the conjugate pad, it becomes saturated with red-coloured antibody-coated Au nanoparticles (NP), to a concentration $1.6 \\times 10^{12} \\mathrm{NP}$ per $\\mathrm{cm}^{3}$.\n\nAn equilibrium is set up where the NP bind to any spike proteins, SP. Assume that any binding that takes place between spike proteins and nanoparticles is 1:1.\n\n$$\n\\begin{gathered}\n\\mathrm{SP}(\\mathrm{aq})+\\mathrm{NP}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{SPNP}(\\mathrm{aq}) \\\\\nK=\\frac{[\\mathrm{SPNP}]}{[\\mathrm{NP}][\\mathrm{SP}]}=1.2 \\times 10^{10} \\mathrm{~mol}^{-1} \\mathrm{dm}^{3}\n\\end{gathered}\n$$\n\nAs there is a large excess of nanoparticles, you can assume that the concentration of nanoparticles at equilibrium is still $1.6 \\times 10^{12} \\mathrm{NP}$ per $\\mathrm{cm}^{3}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n$[S P]=2.36 \\times 10^{-13} \\mathrm{~mol} \\mathrm{dm}^{-3}$\n\nproblem:\nA typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nDuring the test a person swabs their nose/throat and places it in the extraction solution. The resulting solution is called the test solution. A test sample of a few drops $\\left(0.10 \\mathrm{~cm}^{3}\\right)$ of the test solution are placed on the sample pad.\n\nIf the person taking the test has COVID-19, there will typically be $7.1 \\times 10^{6}$ virus particles per $\\mathrm{cm}^{3}$ in the test solution. Each virus particle has approximately 20 spike proteins on its surface.\n\nOnce the sample passes the conjugate pad, it becomes saturated with red-coloured antibody-coated Au nanoparticles (NP), to a concentration $1.6 \\times 10^{12} \\mathrm{NP}$ per $\\mathrm{cm}^{3}$.\n\nAn equilibrium is set up where the NP bind to any spike proteins, SP. Assume that any binding that takes place between spike proteins and nanoparticles is 1:1.\n\n$$\n\\begin{gathered}\n\\mathrm{SP}(\\mathrm{aq})+\\mathrm{NP}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{SPNP}(\\mathrm{aq}) \\\\\nK=\\frac{[\\mathrm{SPNP}]}{[\\mathrm{NP}][\\mathrm{SP}]}=1.2 \\times 10^{10} \\mathrm{~mol}^{-1} \\mathrm{dm}^{3}\n\\end{gathered}\n$$\n\nAs there is a large excess of nanoparticles, you can assume that the concentration of nanoparticles at equilibrium is still $1.6 \\times 10^{12} \\mathrm{NP}$ per $\\mathrm{cm}^{3}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-08.jpg?height=146&width=940&top_left_y=495&top_left_x=929"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1336",
        "problem": "For sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.Calculate the volume of air at a temperature of $25^{\\circ} \\mathrm{C}$ and a pressure of $1.0 \\mathrm{~atm}$ needed to generate a constant current of $2.5 \\mathrm{~A}$ for 3.0 hours in this fuel cell. Assume that air contains $20 \\%$ by volume $\\mathrm{O}_{2}(\\mathrm{g})$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nFor sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.\n\nproblem:\nCalculate the volume of air at a temperature of $25^{\\circ} \\mathrm{C}$ and a pressure of $1.0 \\mathrm{~atm}$ needed to generate a constant current of $2.5 \\mathrm{~A}$ for 3.0 hours in this fuel cell. Assume that air contains $20 \\%$ by volume $\\mathrm{O}_{2}(\\mathrm{g})$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{dm}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{dm}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_589",
        "problem": "“一锅法”用 $[\\mathrm{Ru}]$ 催化硝基苯 $\\left(\\mathrm{PhNO}_{2}, \\mathrm{Ph}\\right.$ 一表示苯基)与醇 $\\left(\\mathrm{RCH}_{2} \\mathrm{OH}\\right)$ 反应为仲胺\n\n$\\left(\\mathrm{RCH}_{2} \\mathrm{NHPh}\\right)$, 反应过程如图所示。下列叙述错误的是\n\n[图1]\nA: 反应原料中的 $\\mathrm{RCH}_{2} \\mathrm{OH}$ 能用 $\\mathrm{RCH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{OH}$ 代替\nB: 历程中存在反应 $\\mathrm{PhNH}_{2}+\\mathrm{RCHO} \\rightarrow \\mathrm{RCH}=\\mathrm{NPh}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 有机物还原反应的氢都来自于 $\\mathrm{RCH}_{2} \\mathrm{OH}$\nD: 该反应过程结束后 RCHO 没有剩余\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n“一锅法”用 $[\\mathrm{Ru}]$ 催化硝基苯 $\\left(\\mathrm{PhNO}_{2}, \\mathrm{Ph}\\right.$ 一表示苯基)与醇 $\\left(\\mathrm{RCH}_{2} \\mathrm{OH}\\right)$ 反应为仲胺\n\n$\\left(\\mathrm{RCH}_{2} \\mathrm{NHPh}\\right)$, 反应过程如图所示。下列叙述错误的是\n\n[图1]\n\nA: 反应原料中的 $\\mathrm{RCH}_{2} \\mathrm{OH}$ 能用 $\\mathrm{RCH}\\left(\\mathrm{CH}_{3}\\right) \\mathrm{OH}$ 代替\nB: 历程中存在反应 $\\mathrm{PhNH}_{2}+\\mathrm{RCHO} \\rightarrow \\mathrm{RCH}=\\mathrm{NPh}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 有机物还原反应的氢都来自于 $\\mathrm{RCH}_{2} \\mathrm{OH}$\nD: 该反应过程结束后 RCHO 没有剩余\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-059.jpg?height=391&width=808&top_left_y=1044&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_624",
        "problem": "室温下, 通过下列实验探究 $\\mathrm{Na}_{2} \\mathrm{SO}_{3} 、 \\mathrm{NaHSO}_{3}$ 溶液的性质。下列说法错误的是\n\n| 实验 | 实验操作和现象 |\n| :--- | :--- |\n| (1) | 用 $\\mathrm{pH}$ 试纸测定 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液的 $\\mathrm{pH}$, 测得 $\\mathrm{pH}$ 约为 10 |\n| (2) | 向 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液中通入少量 $\\mathrm{SO}_{2}$, 测得溶液 $\\mathrm{pH}$ 约为 8 |\n| (3) | 用 $\\mathrm{pH}$ 试纸测定 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaHSO}_{3}$ 溶液的 $\\mathrm{pH}$, 测得 $\\mathrm{pH}$ 约为 5 |\n| (4) | 向 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液中加入等体积等浓度的 $\\mathrm{BaCl}_{2}$ 溶液, 产生白色沉淀 |\nA: $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液中存在: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)$\nB: 实验(2)得到的溶液中存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{SO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)$\nC: 实验(3)可以得出: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right) \\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)>\\mathrm{K}_{\\mathrm{w}}$\nD: 实验(4)中可推测 $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{BaSO}_{3}\\right)>2.5 \\times 10^{-3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 通过下列实验探究 $\\mathrm{Na}_{2} \\mathrm{SO}_{3} 、 \\mathrm{NaHSO}_{3}$ 溶液的性质。下列说法错误的是\n\n| 实验 | 实验操作和现象 |\n| :--- | :--- |\n| (1) | 用 $\\mathrm{pH}$ 试纸测定 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液的 $\\mathrm{pH}$, 测得 $\\mathrm{pH}$ 约为 10 |\n| (2) | 向 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液中通入少量 $\\mathrm{SO}_{2}$, 测得溶液 $\\mathrm{pH}$ 约为 8 |\n| (3) | 用 $\\mathrm{pH}$ 试纸测定 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaHSO}_{3}$ 溶液的 $\\mathrm{pH}$, 测得 $\\mathrm{pH}$ 约为 5 |\n| (4) | 向 $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液中加入等体积等浓度的 $\\mathrm{BaCl}_{2}$ 溶液, 产生白色沉淀 |\n\nA: $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液中存在: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)$\nB: 实验(2)得到的溶液中存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{SO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)$\nC: 实验(3)可以得出: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right) \\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)>\\mathrm{K}_{\\mathrm{w}}$\nD: 实验(4)中可推测 $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{BaSO}_{3}\\right)>2.5 \\times 10^{-3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_172",
        "problem": "Through the Workplace Hazardous Materials Information System (WHMIS), labs use eight labels to identify the dangers chemists encounter from the six classes of hazardous materials. Using your knowledge of the properties of chemical substances, identify the list below which contains three chemicals that should ALL be identified with the flammable materials label?\nA: Copper wire, sodium chloride and helium\nB: Hydrogen, magnesium wire, ethanol\nC: Platinum wire, iron (III) oxide, carbon dioxide\nD: Neon, potassium iodide, silver wire\nE: Liquid mercury, calcium bromide, nitrogen\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThrough the Workplace Hazardous Materials Information System (WHMIS), labs use eight labels to identify the dangers chemists encounter from the six classes of hazardous materials. Using your knowledge of the properties of chemical substances, identify the list below which contains three chemicals that should ALL be identified with the flammable materials label?\n\nA: Copper wire, sodium chloride and helium\nB: Hydrogen, magnesium wire, ethanol\nC: Platinum wire, iron (III) oxide, carbon dioxide\nD: Neon, potassium iodide, silver wire\nE: Liquid mercury, calcium bromide, nitrogen\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_919",
        "problem": "用甲醚 $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$ 燃料电池电解 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液制取 $\\mathrm{NaOH}$ 溶液和硫酸的装置如下图所示。下列说法正确的是\n\n[图1]\nA: $M$ 为硫酸、 $N$ 为 $\\mathrm{NaOH}$\nB: 膜 $\\mathrm{a} 、$ 膜 b 依次为阳离子交换膜和阴离子交换膜\nC: 燃料电池的负极反应式为 $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}-12 \\mathrm{e}^{-}+3 \\mathrm{H}_{2} \\mathrm{O}=2 \\mathrm{CO}_{2} \\uparrow+12 \\mathrm{H}^{+}$\nD: 反应的甲醚与生成气体 $\\mathrm{Y}$ 的物质的量之比为 $1: 3$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n用甲醚 $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$ 燃料电池电解 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液制取 $\\mathrm{NaOH}$ 溶液和硫酸的装置如下图所示。下列说法正确的是\n\n[图1]\n\nA: $M$ 为硫酸、 $N$ 为 $\\mathrm{NaOH}$\nB: 膜 $\\mathrm{a} 、$ 膜 b 依次为阳离子交换膜和阴离子交换膜\nC: 燃料电池的负极反应式为 $\\mathrm{CH}_{3} \\mathrm{OCH}_{3}-12 \\mathrm{e}^{-}+3 \\mathrm{H}_{2} \\mathrm{O}=2 \\mathrm{CO}_{2} \\uparrow+12 \\mathrm{H}^{+}$\nD: 反应的甲醚与生成气体 $\\mathrm{Y}$ 的物质的量之比为 $1: 3$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-85.jpg?height=508&width=1433&top_left_y=1071&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_77",
        "problem": "Carbon monoxide reacts with hydrogen in the presence of a catalyst to give methanol:\n\n$$\n\\mathrm{CO}(g)+2 \\mathrm{H}_{2}(g) \\leftrightarrows \\mathrm{CH}_{3} \\mathrm{OH}(g)\n$$\n\nA metal container at $400 \\mathrm{~K}$ is charged with 2.00 bar of an equimolar mixture of $\\mathrm{CO}$ and $\\mathrm{H}_{2}$. A catalyst for the reaction is introduced and the pressure falls to 1.29 bar once equilibrium is achieved. What is $K_{p}$ for this reaction at $400 \\mathrm{~K}$ ?\nA: 1.32\nB: 6.54\nC: 8.44\nD: 29.1\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCarbon monoxide reacts with hydrogen in the presence of a catalyst to give methanol:\n\n$$\n\\mathrm{CO}(g)+2 \\mathrm{H}_{2}(g) \\leftrightarrows \\mathrm{CH}_{3} \\mathrm{OH}(g)\n$$\n\nA metal container at $400 \\mathrm{~K}$ is charged with 2.00 bar of an equimolar mixture of $\\mathrm{CO}$ and $\\mathrm{H}_{2}$. A catalyst for the reaction is introduced and the pressure falls to 1.29 bar once equilibrium is achieved. What is $K_{p}$ for this reaction at $400 \\mathrm{~K}$ ?\n\nA: 1.32\nB: 6.54\nC: 8.44\nD: 29.1\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1431",
        "problem": "Calculation of ZPE and KIE\n\nKinetic isotope effect (KIE) is a phenomenon associated with a change in rate constant of the reaction when one of the atoms is replaced by its isotope. KIE can be used to confirm whether a particular bond to hydrogen is broken in the reaction. Harmonic oscillator model is used to estimate the difference in the rate between $\\mathrm{C}-\\mathrm{H}$ and ${ }^{1} \\mathrm{C}-\\mathrm{D}$ bond activation ( $\\mathrm{D}-$ deuterium).\n\nThe vibrational frequency $(v)$ represented by harmonic oscillator model is\n\n$$\nv=\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{\\mu}}\n$$\n\nwhere $k$ is the force constant and $\\mu$ is the reduced mass.\n\nThe vibrational energies of the molecule are given by\n\n$$\nE_{n}=\\left(n+\\frac{1}{2}\\right) h v\n$$\n\nwhere $n$ is vibrational quantum number with possible values of $0,1,2, \\ldots$ The energy of the lowest vibrational energy level ( $\\mathrm{E}_{n}$ at $n=0$ ) is called zero-point vibrational energy (ZPE).Calculate the reduced mass of $\\mathrm{C}-\\mathrm{D}\\left(\\mu_{\\mathrm{CD}}\\right)$ in atomic mass unit. Assume that the mass of deuterium is twice that of hydrogen.\n\nNote: If a student is unable to calculate the values for $\\mu_{\\text { }}$ and $\\mu_{\\mathrm{CD}}$ he/she can use $\\mu_{C H}=1.008$ and $\\mu_{C D}=2.016$ for the subsequent parts of the question. The given values are not necessarily be close to the correct values.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nCalculation of ZPE and KIE\n\nKinetic isotope effect (KIE) is a phenomenon associated with a change in rate constant of the reaction when one of the atoms is replaced by its isotope. KIE can be used to confirm whether a particular bond to hydrogen is broken in the reaction. Harmonic oscillator model is used to estimate the difference in the rate between $\\mathrm{C}-\\mathrm{H}$ and ${ }^{1} \\mathrm{C}-\\mathrm{D}$ bond activation ( $\\mathrm{D}-$ deuterium).\n\nThe vibrational frequency $(v)$ represented by harmonic oscillator model is\n\n$$\nv=\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{\\mu}}\n$$\n\nwhere $k$ is the force constant and $\\mu$ is the reduced mass.\n\nThe vibrational energies of the molecule are given by\n\n$$\nE_{n}=\\left(n+\\frac{1}{2}\\right) h v\n$$\n\nwhere $n$ is vibrational quantum number with possible values of $0,1,2, \\ldots$ The energy of the lowest vibrational energy level ( $\\mathrm{E}_{n}$ at $n=0$ ) is called zero-point vibrational energy (ZPE).\n\nproblem:\nCalculate the reduced mass of $\\mathrm{C}-\\mathrm{D}\\left(\\mu_{\\mathrm{CD}}\\right)$ in atomic mass unit. Assume that the mass of deuterium is twice that of hydrogen.\n\nNote: If a student is unable to calculate the values for $\\mu_{\\text { }}$ and $\\mu_{\\mathrm{CD}}$ he/she can use $\\mu_{C H}=1.008$ and $\\mu_{C D}=2.016$ for the subsequent parts of the question. The given values are not necessarily be close to the correct values.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of amu, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "amu"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_17",
        "problem": "Which species contains three sigma bonds and one pi bond?\nA: $\\mathrm{PF}_{3}$\nB: $\\mathrm{NH}_{4}^{+}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{2}$\nD: $\\mathrm{CO}_{3}{ }^{2-}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich species contains three sigma bonds and one pi bond?\n\nA: $\\mathrm{PF}_{3}$\nB: $\\mathrm{NH}_{4}^{+}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{2}$\nD: $\\mathrm{CO}_{3}{ }^{2-}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_931",
        "problem": "一定温度下, 在 3 个体积均为 $1.00 \\mathrm{~L}$ 的恒容密闭容器中发生反应\n$\\mathrm{C}(\\mathrm{s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{CO}(\\mathrm{g}) \\Delta \\mathrm{H}>0$, 相关数据如下表(已知: 炭粉足量)。\n\n| 容器 | $\\mathrm{T} /{ }^{\\circ} \\mathrm{C}$ | 物质的起始浓度 <br> $/\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)$ |  |  | 物质的平衡浓度 <br> $\\mathrm{c}(\\mathrm{CO})$ <br> $\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  |  | $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right.$ | $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)$ | $\\mathrm{c}(\\mathrm{CO})$ |  |\n| I | $\\mathrm{T}_{1}$ | 1.00 | 0 | 0 | 0.85 |\n| II | $\\mathrm{T}_{1}$ | 0 | 1.00 | 1.00 | $\\mathrm{x}$ |\n| III | $\\mathrm{T}_{2}$ | 2.00 | 0 | 0 | 1.60 |\n\n下列说法错误的是\nA: $\\mathrm{T}_{2}>\\mathrm{T}_{1}$\nB: 达到平衡所需时间： $\\mathrm{t}(\\mathrm{I})<\\mathrm{t}(\\mathrm{III})$\nC: $\\mathrm{x}=0.85$\nD: $\\mathrm{T}_{2}{ }^{\\circ} \\mathrm{C}$, 该反应的化学平衡常数 $\\mathrm{K}=6.4$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一定温度下, 在 3 个体积均为 $1.00 \\mathrm{~L}$ 的恒容密闭容器中发生反应\n$\\mathrm{C}(\\mathrm{s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{CO}(\\mathrm{g}) \\Delta \\mathrm{H}>0$, 相关数据如下表(已知: 炭粉足量)。\n\n| 容器 | $\\mathrm{T} /{ }^{\\circ} \\mathrm{C}$ | 物质的起始浓度 <br> $/\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)$ |  |  | 物质的平衡浓度 <br> $\\mathrm{c}(\\mathrm{CO})$ <br> $\\left(\\mathrm{mol} \\cdot \\mathrm{L}^{-1}\\right)$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n|  |  | $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right.$ | $\\mathrm{c}\\left(\\mathrm{H}_{2}\\right)$ | $\\mathrm{c}(\\mathrm{CO})$ |  |\n| I | $\\mathrm{T}_{1}$ | 1.00 | 0 | 0 | 0.85 |\n| II | $\\mathrm{T}_{1}$ | 0 | 1.00 | 1.00 | $\\mathrm{x}$ |\n| III | $\\mathrm{T}_{2}$ | 2.00 | 0 | 0 | 1.60 |\n\n下列说法错误的是\n\nA: $\\mathrm{T}_{2}>\\mathrm{T}_{1}$\nB: 达到平衡所需时间： $\\mathrm{t}(\\mathrm{I})<\\mathrm{t}(\\mathrm{III})$\nC: $\\mathrm{x}=0.85$\nD: $\\mathrm{T}_{2}{ }^{\\circ} \\mathrm{C}$, 该反应的化学平衡常数 $\\mathrm{K}=6.4$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1107",
        "problem": "Nescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nAssuming that the heat capacity for the coffee is the same as that of water, $4.18 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~g}^{-1}$, calculate the energy needed to warm $210 \\mathrm{ml}$ of coffee by $40^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nNescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nAssuming that the heat capacity for the coffee is the same as that of water, $4.18 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~g}^{-1}$, calculate the energy needed to warm $210 \\mathrm{ml}$ of coffee by $40^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-03.jpg?height=417&width=699&top_left_y=385&top_left_x=1021"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_125",
        "problem": "A solution is $0.0240 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{KI}$ and $0.0146 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{MgI}_{2}$. What volume of water should be added to $100.0 \\mathrm{~mL}$ of this solution to produce a solution with [I-] concentration of $0.0500 \\mathrm{~mol} \\mathrm{~L}^{-1}$ ?\nA: $106.4 \\mathrm{~mL}$\nB: $53.2 \\mathrm{~mL}$\nC: $26.6 \\mathrm{~mL}$\nD: $13.3 \\mathrm{~mL}$\nE: $6.4 \\mathrm{~mL}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA solution is $0.0240 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{KI}$ and $0.0146 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{MgI}_{2}$. What volume of water should be added to $100.0 \\mathrm{~mL}$ of this solution to produce a solution with [I-] concentration of $0.0500 \\mathrm{~mol} \\mathrm{~L}^{-1}$ ?\n\nA: $106.4 \\mathrm{~mL}$\nB: $53.2 \\mathrm{~mL}$\nC: $26.6 \\mathrm{~mL}$\nD: $13.3 \\mathrm{~mL}$\nE: $6.4 \\mathrm{~mL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_439",
        "problem": "在 $25^{\\circ} \\mathrm{C}$ 时, 向 $50.00 \\mathrm{~mL}$ 未知浓度的氨水中逐滴加入 $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCl}$ 溶液。滴定过程中, 溶液的 $\\mathrm{pOH}\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$与滴入 $\\mathrm{HCl}$ 溶液体积的关系如图所示, 则下列说法中正确的是\n\n[图1]\nA: 图中(2)点所示溶液的导电能力弱于(1)点\nB: (3)点处水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=1 \\times 10^{-8} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: 图中点(1)所示溶液中, $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时氨水的 $\\mathrm{K}_{\\mathrm{b}}$ 约为 $5 \\times 10^{-5.6} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在 $25^{\\circ} \\mathrm{C}$ 时, 向 $50.00 \\mathrm{~mL}$ 未知浓度的氨水中逐滴加入 $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCl}$ 溶液。滴定过程中, 溶液的 $\\mathrm{pOH}\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$与滴入 $\\mathrm{HCl}$ 溶液体积的关系如图所示, 则下列说法中正确的是\n\n[图1]\n\nA: 图中(2)点所示溶液的导电能力弱于(1)点\nB: (3)点处水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=1 \\times 10^{-8} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: 图中点(1)所示溶液中, $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时氨水的 $\\mathrm{K}_{\\mathrm{b}}$ 约为 $5 \\times 10^{-5.6} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-49.jpg?height=343&width=417&top_left_y=911&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1238",
        "problem": "When the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\nUsing the selected buffer system, calculate the mass (in g) of weak acid and of conjugated base you would need to dissolve in distilled water to prepare $500 \\mathrm{~cm}^{3}$ of a stock solution buffered at a $\\mathrm{pH}$ of 7.2 .",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nWhen the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\nUsing the selected buffer system, calculate the mass (in g) of weak acid and of conjugated base you would need to dissolve in distilled water to prepare $500 \\mathrm{~cm}^{3}$ of a stock solution buffered at a $\\mathrm{pH}$ of 7.2 .\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_622",
        "problem": "反应: $\\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~s}) \\rightleftharpoons 2 \\mathrm{HI}(\\mathrm{g})$ 的 $\\mathrm{K}_{\\mathrm{p}}$ 为 $6.4 \\times 10^{-4} \\mathrm{~atm}$ 。仔细观察发现, 在相同温度\n下以蒸汽态存在的碘的分压为 $1.6 \\times 10^{-4} \\mathrm{~atm}$ 。 反应 $\\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{HI}(\\mathrm{g})$ 的 $\\mathrm{K}_{\\mathrm{p}}$ 为\nA: 0.25\nB: 4\nC: $1.024 \\times 10^{-7}$\nD: $9.76 \\times 10^{6}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n反应: $\\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~s}) \\rightleftharpoons 2 \\mathrm{HI}(\\mathrm{g})$ 的 $\\mathrm{K}_{\\mathrm{p}}$ 为 $6.4 \\times 10^{-4} \\mathrm{~atm}$ 。仔细观察发现, 在相同温度\n下以蒸汽态存在的碘的分压为 $1.6 \\times 10^{-4} \\mathrm{~atm}$ 。 反应 $\\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{I}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{HI}(\\mathrm{g})$ 的 $\\mathrm{K}_{\\mathrm{p}}$ 为\n\nA: 0.25\nB: 4\nC: $1.024 \\times 10^{-7}$\nD: $9.76 \\times 10^{6}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1025",
        "problem": "The enthalpy of combustion of an isomer of octane is -5144 $\\mathrm{kJ} \\mathrm{mol}^{-1}$.To help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nFor E10 fuels, calculate the energy, in kJ, released when 1 litre of the fuel is burnt.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe enthalpy of combustion of an isomer of octane is -5144 $\\mathrm{kJ} \\mathrm{mol}^{-1}$.\n\nproblem:\nTo help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nAssume that 1 litre of $\\mathrm{E} 10$ fuel contains $100 \\mathrm{~mL}$ of ethanol and $900 \\mathrm{~mL}$ of octane isomers, and that 1 litre of $\\mathrm{E} 5$ fuel contains $50 \\mathrm{~mL}$ of ethanol and $950 \\mathrm{~mL}$ of octane isomers. The density of pure ethanol is $0.789 \\mathrm{~g} \\mathrm{~cm}^{-3}$ and the density of pure octane isomers is $0.703 \\mathrm{~g} \\mathrm{~cm}^{-3}$. You can assume that there is no volume change on mixing.\n\nThe enthalpy of combustion of ethanol using average bond enthalpies is $-1276 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nFor E10 fuels, calculate the energy, in kJ, released when 1 litre of the fuel is burnt.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kJ, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-02.jpg?height=459&width=923&top_left_y=356&top_left_x=949"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kJ"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_37",
        "problem": "A student wishes to prepare ethyl acetate from the reaction of ethanol and acetic acid. To be successful, this reaction requires\nA: an acidic catalyst.\nB: a basic catalyst.\nC: an oxidizing agent.\nD: a reducing agent.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student wishes to prepare ethyl acetate from the reaction of ethanol and acetic acid. To be successful, this reaction requires\n\nA: an acidic catalyst.\nB: a basic catalyst.\nC: an oxidizing agent.\nD: a reducing agent.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_219",
        "problem": "How many atoms are present in a $1.0 \\mathrm{~kg}$ sample of $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}$ ?\nA: $1.4 \\times 10^{22}$\nB: $9.6 \\times 10^{22}$\nC: $1.4 \\times 10^{25}$\nD: $9.6 \\times 10^{25}$\nE: $9.6 \\times 10^{28}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many atoms are present in a $1.0 \\mathrm{~kg}$ sample of $\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}$ ?\n\nA: $1.4 \\times 10^{22}$\nB: $9.6 \\times 10^{22}$\nC: $1.4 \\times 10^{25}$\nD: $9.6 \\times 10^{25}$\nE: $9.6 \\times 10^{28}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1218",
        "problem": "Propene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nPropene can be synthesized via a direct dehydrogenation of propane in the presence of a heterogeneous catalyst. However, such a reaction is not economically feasible due to the nature of the reaction itself. Provide a concise explanation to each of the questions below. Additional information: $H_{\\text {bond }}(\\mathrm{C}=\\mathrm{C})=1.77 \\mathrm{H}_{\\text {bond }}(\\mathrm{C}-\\mathrm{C}), \\mathrm{H}_{\\text {bond }}(\\mathrm{H}-\\mathrm{H})=$ $1.05 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})$, and $H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})=1.19 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$, where $H_{\\text {bond }}$ refers to average bond enthalpy of the indicated chemical bond.What is the enthalpy change of the direct dehydrogenation of propane? Show your calculation and express your answer in terms of $H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nPropene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nPropene can be synthesized via a direct dehydrogenation of propane in the presence of a heterogeneous catalyst. However, such a reaction is not economically feasible due to the nature of the reaction itself. Provide a concise explanation to each of the questions below. Additional information: $H_{\\text {bond }}(\\mathrm{C}=\\mathrm{C})=1.77 \\mathrm{H}_{\\text {bond }}(\\mathrm{C}-\\mathrm{C}), \\mathrm{H}_{\\text {bond }}(\\mathrm{H}-\\mathrm{H})=$ $1.05 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})$, and $H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})=1.19 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$, where $H_{\\text {bond }}$ refers to average bond enthalpy of the indicated chemical bond.\n\nproblem:\nWhat is the enthalpy change of the direct dehydrogenation of propane? Show your calculation and express your answer in terms of $H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_733",
        "problem": "$298 \\mathrm{~K}$ 时, 向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Fe}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 溶液中滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液,\n\n溶液中 $\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$ 的负对数 $\\mathrm{pX}\\left[\\mathrm{pX}=-\\operatorname{lgc}\\left(\\mathrm{SO}_{4}^{2-}\\right)\\right]$ 与 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液体积的关系如图所示。已知:常温下, $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{BaSO}_{4}\\right)=1.0 \\times 10^{-10}$ 。下列说法正确的是\n\n[图1]\nA: 在 $b 、 c 、 d$ 点中, $c$ 点对应的 $c\\left(\\mathrm{Fe}^{2+}\\right)$ 最小\nB: 在 b、c、d 点中, $d$ 点对应的水的电离程度最大\nC: 图中 $\\mathrm{m}=5, \\mathrm{c}$ 点溶液的 $\\mathrm{pH}>7$\nD: b 点溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>2 \\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$298 \\mathrm{~K}$ 时, 向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{Fe}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 溶液中滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液,\n\n溶液中 $\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$ 的负对数 $\\mathrm{pX}\\left[\\mathrm{pX}=-\\operatorname{lgc}\\left(\\mathrm{SO}_{4}^{2-}\\right)\\right]$ 与 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液体积的关系如图所示。已知:常温下, $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{BaSO}_{4}\\right)=1.0 \\times 10^{-10}$ 。下列说法正确的是\n\n[图1]\n\nA: 在 $b 、 c 、 d$ 点中, $c$ 点对应的 $c\\left(\\mathrm{Fe}^{2+}\\right)$ 最小\nB: 在 b、c、d 点中, $d$ 点对应的水的电离程度最大\nC: 图中 $\\mathrm{m}=5, \\mathrm{c}$ 点溶液的 $\\mathrm{pH}>7$\nD: b 点溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>2 \\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-090.jpg?height=569&width=736&top_left_y=1874&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_864",
        "problem": "体积均为 $2.0 \\mathrm{~L}$ 的恒容密闭容器甲、乙、丙中, 发生反应: $\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{s}) \\rightleftharpoons 2 \\mathrm{CO}(\\mathrm{g})$,起始投料量如表, 在不同温度下 $\\mathrm{CO}$ 平衡浓度随温度变化如下图。下列说法正确的是\n\n[图1]\n\n| 容器 | $n\\left(\\mathrm{CO}_{2}\\right) / \\mathrm{mol}$ | $n(\\mathrm{C}) / \\mathrm{mol}$ | $n(\\mathrm{CO}) / \\mathrm{mol}$ |\n| :--- | :--- | :--- | :--- |\n| 甲 | 0.2 | 0.6 | 0 |\n| 乙 | 0.4 | 0.8 | 0 |\n\n[图2]\nA: 曲线I、II、III对应的容器分别是乙、甲、丙\nB: $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{c}$ 三点所处状态的压强大小关系: $\\mathrm{p}_{\\mathrm{c}}>\\mathrm{p}_{\\mathrm{b}}>\\mathrm{p}_{\\mathrm{a}}$\nC: $1100 \\mathrm{~K}$ 时, 平衡时容器中混合气体的平均分子量大小关系: 甲 $>$ 乙 $>$ 丙\nD: $1000 \\mathrm{~K}$ 时, 若起始向容器乙中加入 $\\mathrm{CO} 、 \\mathrm{CO}_{2} 、 \\mathrm{C}$ 各 $1 \\mathrm{~mol}$, 则 $\\mathrm{v}_{(\\text {正 })}>\\mathrm{v}_{(\\text {逆 })}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n体积均为 $2.0 \\mathrm{~L}$ 的恒容密闭容器甲、乙、丙中, 发生反应: $\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{s}) \\rightleftharpoons 2 \\mathrm{CO}(\\mathrm{g})$,起始投料量如表, 在不同温度下 $\\mathrm{CO}$ 平衡浓度随温度变化如下图。下列说法正确的是\n\n[图1]\n\n| 容器 | $n\\left(\\mathrm{CO}_{2}\\right) / \\mathrm{mol}$ | $n(\\mathrm{C}) / \\mathrm{mol}$ | $n(\\mathrm{CO}) / \\mathrm{mol}$ |\n| :--- | :--- | :--- | :--- |\n| 甲 | 0.2 | 0.6 | 0 |\n| 乙 | 0.4 | 0.8 | 0 |\n\n[图2]\n\nA: 曲线I、II、III对应的容器分别是乙、甲、丙\nB: $\\mathrm{a} 、 \\mathrm{~b} 、 \\mathrm{c}$ 三点所处状态的压强大小关系: $\\mathrm{p}_{\\mathrm{c}}>\\mathrm{p}_{\\mathrm{b}}>\\mathrm{p}_{\\mathrm{a}}$\nC: $1100 \\mathrm{~K}$ 时, 平衡时容器中混合气体的平均分子量大小关系: 甲 $>$ 乙 $>$ 丙\nD: $1000 \\mathrm{~K}$ 时, 若起始向容器乙中加入 $\\mathrm{CO} 、 \\mathrm{CO}_{2} 、 \\mathrm{C}$ 各 $1 \\mathrm{~mol}$, 则 $\\mathrm{v}_{(\\text {正 })}>\\mathrm{v}_{(\\text {逆 })}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-010.jpg?height=483&width=591&top_left_y=1917&top_left_x=333",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-011.jpg?height=131&width=771&top_left_y=143&top_left_x=300"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1231",
        "problem": "$\\mathbf{H}$ is stable to acids, but is hydrolysed in alkali. A $0.7934 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) was heated with excess aqueous sodium hydroxide. Cobalt(III) oxide was formed and ammonia gas given off. The ammonia produced was distilled off and absorbed into $50.0 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c_{\\mathrm{HCl}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$. The residual $\\mathrm{HCl}$ required $24.8 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{KOH}$ solution $\\left(c_{\\mathrm{KOH}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to be neutralized.\n\nThe remaining suspension of cobalt(III) oxide was allowed to cool, approximately $1 \\mathrm{~g}$ of potassium iodide was added, and then the mixture was acidified with aqueous $\\mathrm{HCl}$. The liberated iodine was then titrated with aqueous solution of sodium thiosulfate $(c=0.200$ mol $\\mathrm{dm}^{-3}$ ) and required $21.0 \\mathrm{~cm}^{3}$ for complete reaction.Calculate the percentage, by mass, of ammonia in $\\mathbf{H}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n$\\mathbf{H}$ is stable to acids, but is hydrolysed in alkali. A $0.7934 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) was heated with excess aqueous sodium hydroxide. Cobalt(III) oxide was formed and ammonia gas given off. The ammonia produced was distilled off and absorbed into $50.0 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c_{\\mathrm{HCl}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$. The residual $\\mathrm{HCl}$ required $24.8 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{KOH}$ solution $\\left(c_{\\mathrm{KOH}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to be neutralized.\n\nThe remaining suspension of cobalt(III) oxide was allowed to cool, approximately $1 \\mathrm{~g}$ of potassium iodide was added, and then the mixture was acidified with aqueous $\\mathrm{HCl}$. The liberated iodine was then titrated with aqueous solution of sodium thiosulfate $(c=0.200$ mol $\\mathrm{dm}^{-3}$ ) and required $21.0 \\mathrm{~cm}^{3}$ for complete reaction.\n\nproblem:\nCalculate the percentage, by mass, of ammonia in $\\mathbf{H}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_723",
        "problem": "络合平衡遵循化学平衡移动原理。已知: (1) $\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+} \\mathrm{K}_{1}$\n\n(2) $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+} \\mathrm{K}_{2}\\left(\\mathrm{~K}_{2}>\\mathrm{K}_{1}\\right) 。$\n\n向饱和 $\\mathrm{AgCl}$ 溶液中滴加氨水, $\\operatorname{lgX}\\left[\\mathrm{X}=\\frac{\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)}\\right.$或 $\\frac{\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right)}{\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}\\right)}$与 $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)$ 关系如图所示:\n\n[图1]\n\n下列说法错误的是\nA: 直线 $\\mathrm{L}_{2}$ 代表反应(1)中离子浓度关系\nB: 络合平衡常数 $\\mathrm{K}_{2}=1.0 \\times 10^{4}$\nC: 在上述混合液中加入足量盐酸, 无明显现象\nD: $\\mathrm{Ag}^{+}+2 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$的平衡常数 $\\mathrm{K}=10^{7.4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n络合平衡遵循化学平衡移动原理。已知: (1) $\\mathrm{Ag}^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+} \\mathrm{K}_{1}$\n\n(2) $\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}+\\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{2+} \\mathrm{K}_{2}\\left(\\mathrm{~K}_{2}>\\mathrm{K}_{1}\\right) 。$\n\n向饱和 $\\mathrm{AgCl}$ 溶液中滴加氨水, $\\operatorname{lgX}\\left[\\mathrm{X}=\\frac{\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)}\\right.$或 $\\frac{\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right)}{\\mathrm{c}\\left(\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)\\right]^{+}\\right)}$与 $\\operatorname{lgc}\\left(\\mathrm{NH}_{3}\\right)$ 关系如图所示:\n\n[图1]\n\n下列说法错误的是\n\nA: 直线 $\\mathrm{L}_{2}$ 代表反应(1)中离子浓度关系\nB: 络合平衡常数 $\\mathrm{K}_{2}=1.0 \\times 10^{4}$\nC: 在上述混合液中加入足量盐酸, 无明显现象\nD: $\\mathrm{Ag}^{+}+2 \\mathrm{NH}_{3} \\rightleftharpoons\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$的平衡常数 $\\mathrm{K}=10^{7.4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-011.jpg?height=511&width=605&top_left_y=1875&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1126",
        "problem": "This question is about benzene and its isomers\n\nWorking out the structure of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, was one of the great achievements of $19^{\\text {th }}$ century chemists. The carcinogenic hydrocarbon was first discovered by Michael Faraday at the Royal Institution in 1825 during the repeated fractional distillation of some oil residues.\n\nFaraday analysed the pure benzene by passing its vapour over heated copper(II) oxide to produce carbon dioxide, water and copper. He collected and measured the volume of liquid water and gaseous $\\mathrm{CO}_{2}$ produced from a given mass of benzene.\nCalculate the relative molar mass of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, and hence the number of moles in $1.00 \\mathrm{~g}$ of benzene.\n\nThe amount used by Faraday was tiny, just 0.776 grains - an old unit of mass. This equates to $0.644 \\mathrm{mmol}$. $[1 \\mathrm{mmol}=1$ millimole $=0.001 \\mathrm{~mol}$ ]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about benzene and its isomers\n\nWorking out the structure of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, was one of the great achievements of $19^{\\text {th }}$ century chemists. The carcinogenic hydrocarbon was first discovered by Michael Faraday at the Royal Institution in 1825 during the repeated fractional distillation of some oil residues.\n\nFaraday analysed the pure benzene by passing its vapour over heated copper(II) oxide to produce carbon dioxide, water and copper. He collected and measured the volume of liquid water and gaseous $\\mathrm{CO}_{2}$ produced from a given mass of benzene.\nCalculate the relative molar mass of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, and hence the number of moles in $1.00 \\mathrm{~g}$ of benzene.\n\nThe amount used by Faraday was tiny, just 0.776 grains - an old unit of mass. This equates to $0.644 \\mathrm{mmol}$. $[1 \\mathrm{mmol}=1$ millimole $=0.001 \\mathrm{~mol}$ ]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of moles, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "moles"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1133",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nThe heaviest of the isotopes found in naturally-occurring tellurium is tellurium- 130 which has a relative mass of 129.906223 . Technically, tellurium-130 is very slightly radioactive and if there were none in the naturally occurring element, the relative atomic mass of tellurium would be 126.412449 (which would make it less than iodine).\n\nCalculate the percentage of tellurium-130 present in naturally-occurring tellurium.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nThe heaviest of the isotopes found in naturally-occurring tellurium is tellurium- 130 which has a relative mass of 129.906223 . Technically, tellurium-130 is very slightly radioactive and if there were none in the naturally occurring element, the relative atomic mass of tellurium would be 126.412449 (which would make it less than iodine).\n\nCalculate the percentage of tellurium-130 present in naturally-occurring tellurium.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_323",
        "problem": "Which of the following statements about the group 17 elements is false?\nA: The ionization energy decreases down the group.\nB: The group contains both metals and non-metals.\nC: Electronegativity decreases down the group.\nD: The melting point increases down the group.\nE: The most common ion formed by these elements is $X^{-}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following statements about the group 17 elements is false?\n\nA: The ionization energy decreases down the group.\nB: The group contains both metals and non-metals.\nC: Electronegativity decreases down the group.\nD: The melting point increases down the group.\nE: The most common ion formed by these elements is $X^{-}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_239",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\n Calculate the amount (in mol or mmol) of hydrochloric acid added.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\n Calculate the amount (in mol or mmol) of hydrochloric acid added.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1079",
        "problem": "Give half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ to $\\mathrm{Cr}^{3+}$\nA: oxidation\nB: reduction\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGive half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ to $\\mathrm{Cr}^{3+}$\n\nA: oxidation\nB: reduction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_552",
        "problem": "镁电池作为一种低成本、高安全的储能装置, 正受到国内外广大科研人员的关注。一种以固态含 $\\mathrm{Mg}^{2+}$ 的化合物为电解质的镁电池的总反应如下。下列说法错误的是 $\\mathrm{xMg}+\\mathrm{V}_{2} \\mathrm{O}_{5} \\underset{\\text { 充电 }}{\\stackrel{\\text { 放电 }}{\\rightleftharpoons}} \\mathrm{Mg}_{\\mathrm{x}} \\mathrm{V}_{2} \\mathrm{O}_{5}$\nA: 充电时，阳极质量减小\nB: 充电时, 阴极反应式: $\\mathrm{Mg}^{2+}+2 \\mathrm{e}^{-}=\\mathrm{Mg}$\nC: 放电时, 正极反应式为: $\\mathrm{V}_{2} \\mathrm{O}_{5}+\\mathrm{xMg}^{2+}+2 \\mathrm{xe}^{-}=\\mathrm{Mg}_{\\mathrm{x}} \\mathrm{V}_{2} \\mathrm{O}_{5}$\nD: 放电时, 电路中每流过 $2 \\mathrm{~mol}$ 电子, 固体电解质中有 $2 \\mathrm{molMg}^{2+}$ 迁移至正极\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n镁电池作为一种低成本、高安全的储能装置, 正受到国内外广大科研人员的关注。一种以固态含 $\\mathrm{Mg}^{2+}$ 的化合物为电解质的镁电池的总反应如下。下列说法错误的是 $\\mathrm{xMg}+\\mathrm{V}_{2} \\mathrm{O}_{5} \\underset{\\text { 充电 }}{\\stackrel{\\text { 放电 }}{\\rightleftharpoons}} \\mathrm{Mg}_{\\mathrm{x}} \\mathrm{V}_{2} \\mathrm{O}_{5}$\n\nA: 充电时，阳极质量减小\nB: 充电时, 阴极反应式: $\\mathrm{Mg}^{2+}+2 \\mathrm{e}^{-}=\\mathrm{Mg}$\nC: 放电时, 正极反应式为: $\\mathrm{V}_{2} \\mathrm{O}_{5}+\\mathrm{xMg}^{2+}+2 \\mathrm{xe}^{-}=\\mathrm{Mg}_{\\mathrm{x}} \\mathrm{V}_{2} \\mathrm{O}_{5}$\nD: 放电时, 电路中每流过 $2 \\mathrm{~mol}$ 电子, 固体电解质中有 $2 \\mathrm{molMg}^{2+}$ 迁移至正极\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_917",
        "problem": "常温下, $\\mathrm{pH}=3$ 的 $\\mathrm{NaHR}$ 溶液稀释过程中 $\\delta\\left(\\mathrm{H}_{2} \\mathrm{R}\\right) 、 \\delta\\left(\\mathrm{HR}^{-}\\right) 、 \\delta\\left(\\mathrm{R}^{2-}\\right)$ 与 $\\mathrm{pc}\\left(\\mathrm{Na}^{+}\\right)$的关系如图所示。已知 $\\mathrm{pc}\\left(\\mathrm{Na}^{+}\\right)=-\\operatorname{lgc}\\left(\\mathrm{Na}^{+}\\right), \\quad \\mathrm{HR}^{-}$的分布系数 $\\delta\\left(\\mathrm{HR}^{-}\\right)=\\frac{\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)+\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)}$, $\\lg 2=0.3$ 。下列说法正确的是\n\n[图1]\nA: 曲线 $\\mathrm{L}_{2}$ 代表 $\\delta\\left(\\mathrm{HR}^{-}\\right)$\nB: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)=2 \\times 10^{-2}$\nC: $\\mathrm{a}$ 点溶液的 $\\mathrm{pH}=3.7$\nD: $\\mathrm{b}$ 点溶液中, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>2 \\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, $\\mathrm{pH}=3$ 的 $\\mathrm{NaHR}$ 溶液稀释过程中 $\\delta\\left(\\mathrm{H}_{2} \\mathrm{R}\\right) 、 \\delta\\left(\\mathrm{HR}^{-}\\right) 、 \\delta\\left(\\mathrm{R}^{2-}\\right)$ 与 $\\mathrm{pc}\\left(\\mathrm{Na}^{+}\\right)$的关系如图所示。已知 $\\mathrm{pc}\\left(\\mathrm{Na}^{+}\\right)=-\\operatorname{lgc}\\left(\\mathrm{Na}^{+}\\right), \\quad \\mathrm{HR}^{-}$的分布系数 $\\delta\\left(\\mathrm{HR}^{-}\\right)=\\frac{\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)+\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)}$, $\\lg 2=0.3$ 。下列说法正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{L}_{2}$ 代表 $\\delta\\left(\\mathrm{HR}^{-}\\right)$\nB: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)=2 \\times 10^{-2}$\nC: $\\mathrm{a}$ 点溶液的 $\\mathrm{pH}=3.7$\nD: $\\mathrm{b}$ 点溶液中, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>2 \\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-019.jpg?height=494&width=468&top_left_y=1129&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_709",
        "problem": "某药物中间体的合成路线如下$\\cdot$下列说法正确的是 ( )\n\n[图1]\nA: 对苯二酚在空气中能稳定存在\nB: $1 \\mathrm{~mol}$ 该中间体最多可与 $11 \\mathrm{molH}_{2}$ 反应\nC: 2,5-二羟基苯乙酮能发生加成、取代、缩聚反应\nD: 该中间体分子中含有 1 个手性碳原子\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某药物中间体的合成路线如下$\\cdot$下列说法正确的是 ( )\n\n[图1]\n\nA: 对苯二酚在空气中能稳定存在\nB: $1 \\mathrm{~mol}$ 该中间体最多可与 $11 \\mathrm{molH}_{2}$ 反应\nC: 2,5-二羟基苯乙酮能发生加成、取代、缩聚反应\nD: 该中间体分子中含有 1 个手性碳原子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-76.jpg?height=414&width=1085&top_left_y=1055&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_831",
        "problem": "下列有关实验操作、现象和结论均正确的是\n\n| 选 | 实验操作 | 现象 | 结论 |\n| :--- | :--- | :--- | :--- |\n\n\n| A | 向硝酸铁溶液中加入稀硫酸 | 溶液黄色裉去 | $\\mathrm{Fe}^{3+}$ 被还原 |\n| :---: | :---: | :---: | :---: |\n| B | 向含有同浓度的 $\\mathrm{KBr}$ 和 $\\mathrm{KI}$ 混合溶液 <br> 中依次加入少量氯水和 $\\mathrm{CCl}_{4}$, 振荡, <br> 静置 | 溶液分层, 下层呈 <br> 紫红色 | 氧化性: $\\mathrm{Cl}_{2}>\\mathrm{I}_{2}$, <br> $\\mathrm{Br}_{2}>\\mathrm{I}_{2}$ |\n| $\\mathrm{C}$ | 取补铁剂(琥珀酸亚铁)水溶液 <br> $(\\mathrm{pH}=1.7)$ 少量, 加入几滴邻二氮菲溶 <br> 液 | 补铁剂溶液不变橙 <br> 红色 | 补铁剂(琥珀酸亚铁) <br> 完全变质 |\n| D | 将粗品苯甲酸进行加热溶解、趁热过 <br> 滤、冷却结晶、过滤、洗涤、干燥 | 冷却结晶时析出大 <br> 量白色晶体苯甲酸 | 苯甲酸在水中的溶解 <br> 度受温度的影响较大 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列有关实验操作、现象和结论均正确的是\n\n| 选 | 实验操作 | 现象 | 结论 |\n| :--- | :--- | :--- | :--- |\n\n\n| A | 向硝酸铁溶液中加入稀硫酸 | 溶液黄色裉去 | $\\mathrm{Fe}^{3+}$ 被还原 |\n| :---: | :---: | :---: | :---: |\n| B | 向含有同浓度的 $\\mathrm{KBr}$ 和 $\\mathrm{KI}$ 混合溶液 <br> 中依次加入少量氯水和 $\\mathrm{CCl}_{4}$, 振荡, <br> 静置 | 溶液分层, 下层呈 <br> 紫红色 | 氧化性: $\\mathrm{Cl}_{2}>\\mathrm{I}_{2}$, <br> $\\mathrm{Br}_{2}>\\mathrm{I}_{2}$ |\n| $\\mathrm{C}$ | 取补铁剂(琥珀酸亚铁)水溶液 <br> $(\\mathrm{pH}=1.7)$ 少量, 加入几滴邻二氮菲溶 <br> 液 | 补铁剂溶液不变橙 <br> 红色 | 补铁剂(琥珀酸亚铁) <br> 完全变质 |\n| D | 将粗品苯甲酸进行加热溶解、趁热过 <br> 滤、冷却结晶、过滤、洗涤、干燥 | 冷却结晶时析出大 <br> 量白色晶体苯甲酸 | 苯甲酸在水中的溶解 <br> 度受温度的影响较大 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1197",
        "problem": "Bridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate the standard electromotive force (the standard cell voltage), $E$, at $25^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate the standard electromotive force (the standard cell voltage), $E$, at $25^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of V, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-345.jpg?height=951&width=1445&top_left_y=595&top_left_x=314"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "V"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1089",
        "problem": "Polonium is a radioactive group VI element, discovered in 1898 by Marie Curie. It occurs naturally in trace amounts in some uranium ores but is now made by neutron irradiation of ${ }^{209} \\mathrm{Bi}$. This produces short-lived ${ }^{210} \\mathrm{Bi}$ which decays to polonium by the emission of a betaparticle (an electron):\n\n$$\n{ }_{83}^{209} \\mathrm{Bi}+{ }_{0}^{1} n \\rightarrow{ }_{83}^{210} \\mathrm{Bi}+\\gamma \\quad{ }_{83}^{210} \\mathrm{Bi} \\rightarrow{ }_{84}^{210} \\mathrm{Po}+{ }_{-1}^{0} \\beta\n$$\n\nPolonium-210 has a half life of 138 days and decays by emitting an alpha particle (a helium nucleus).\n\n[figure1]\n\nDue to its very short half life and the impedance of the alpha-particles it emits, metallic polonium and its compounds are self heating; $1 \\mathrm{~g}$ of metal produces $141 \\mathrm{~W}$. This led to its use in Radioisotope Heater Units (RHUs) to keep satellites warm and functioning in space, and in Radioisotope Thermal Generators (RTGs) to produce electrical power. More recently, plutonium-238 has been used instead of polonium. ${ }^{238} \\mathrm{Pu}$ has a much longer half-life but produces less power $\\left(0.56 \\mathrm{~W} \\mathrm{~g}^{-1}\\right)$.\n\nPolonium is unique amongst the elements in being the only one to have a simple cubic structure with each atom lying at the corner of a cube.\n\nGiven that the density of polonium-210 is $9.142 \\mathrm{~g} \\mathrm{~cm}^{-3}$, calculate its atomic radius.\n\n[The Avogadro number, $N_{\\mathrm{A}}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$.]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nPolonium is a radioactive group VI element, discovered in 1898 by Marie Curie. It occurs naturally in trace amounts in some uranium ores but is now made by neutron irradiation of ${ }^{209} \\mathrm{Bi}$. This produces short-lived ${ }^{210} \\mathrm{Bi}$ which decays to polonium by the emission of a betaparticle (an electron):\n\n$$\n{ }_{83}^{209} \\mathrm{Bi}+{ }_{0}^{1} n \\rightarrow{ }_{83}^{210} \\mathrm{Bi}+\\gamma \\quad{ }_{83}^{210} \\mathrm{Bi} \\rightarrow{ }_{84}^{210} \\mathrm{Po}+{ }_{-1}^{0} \\beta\n$$\n\nPolonium-210 has a half life of 138 days and decays by emitting an alpha particle (a helium nucleus).\n\n[figure1]\n\nDue to its very short half life and the impedance of the alpha-particles it emits, metallic polonium and its compounds are self heating; $1 \\mathrm{~g}$ of metal produces $141 \\mathrm{~W}$. This led to its use in Radioisotope Heater Units (RHUs) to keep satellites warm and functioning in space, and in Radioisotope Thermal Generators (RTGs) to produce electrical power. More recently, plutonium-238 has been used instead of polonium. ${ }^{238} \\mathrm{Pu}$ has a much longer half-life but produces less power $\\left(0.56 \\mathrm{~W} \\mathrm{~g}^{-1}\\right)$.\n\nPolonium is unique amongst the elements in being the only one to have a simple cubic structure with each atom lying at the corner of a cube.\n\nGiven that the density of polonium-210 is $9.142 \\mathrm{~g} \\mathrm{~cm}^{-3}$, calculate its atomic radius.\n\n[The Avogadro number, $N_{\\mathrm{A}}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$.]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of pm, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-10.jpg?height=500&width=648&top_left_y=367&top_left_x=1132",
            "https://cdn.mathpix.com/cropped/2024_03_14_606b3fce14c802af99e4g-6.jpg?height=340&width=419&top_left_y=1109&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "pm"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_336",
        "problem": "The phase diagram for carbon dioxide is shown below. The temperature and pressure at the triple point (TP) and the critical point (CP) are shown.\n\n[figure1]\n\nWhich of the following accounts for the fact that liquid $\\mathrm{CO}_{2}$ is not observed when a piece of solid $\\mathrm{CO}_{2}$ (dry ice) is placed on a lab bench at $25^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ ?\nA: The triple point temperature is less than the critical point temperature.\nB: The critical temperature is greater than $25^{\\circ} \\mathrm{C}$.\nC: The triple point temperature is less than $25^{\\circ} \\mathrm{C}$.\nD: The critical pressure is greater than $1 \\mathrm{~atm}$.\nE: The triple point pressure is greater than $1 \\mathrm{~atm}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe phase diagram for carbon dioxide is shown below. The temperature and pressure at the triple point (TP) and the critical point (CP) are shown.\n\n[figure1]\n\nWhich of the following accounts for the fact that liquid $\\mathrm{CO}_{2}$ is not observed when a piece of solid $\\mathrm{CO}_{2}$ (dry ice) is placed on a lab bench at $25^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ ?\n\nA: The triple point temperature is less than the critical point temperature.\nB: The critical temperature is greater than $25^{\\circ} \\mathrm{C}$.\nC: The triple point temperature is less than $25^{\\circ} \\mathrm{C}$.\nD: The critical pressure is greater than $1 \\mathrm{~atm}$.\nE: The triple point pressure is greater than $1 \\mathrm{~atm}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_529737e841b31def0936g-3.jpg?height=436&width=681&top_left_y=335&top_left_x=1210"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1489",
        "problem": "For sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.What should be the temperature of the hot reservoir, $T_{\\mathrm{H}}$, of a Carnot heat engine to maintain the efficiency of the fuel cell calculated in part 4.6 , if the temperature of cold reservoir $T_{\\mathrm{C}}$ is $40^{\\circ} \\mathrm{C}$ ?\n\n(If you have been unable to calculate the value for the efficiency then use the value 0.80 for the calculation.)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\nHere is some context information for this question, which might assist you in solving it:\nFor sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.\n\nproblem:\nWhat should be the temperature of the hot reservoir, $T_{\\mathrm{H}}$, of a Carnot heat engine to maintain the efficiency of the fuel cell calculated in part 4.6 , if the temperature of cold reservoir $T_{\\mathrm{C}}$ is $40^{\\circ} \\mathrm{C}$ ?\n\n(If you have been unable to calculate the value for the efficiency then use the value 0.80 for the calculation.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir units are, in order, [K, $^{\\circ}\\mathrm{C}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            "K",
            "$^{\\circ}\\mathrm{C}$"
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1221",
        "problem": "The radioactive decay series of ${ }^{226} \\mathrm{Ra}$ is as follows:\n\n$$\n\\begin{aligned}\n& { }^{226} \\mathrm{Ra} \\xrightarrow{t}{ }^{222} \\mathrm{Rn} \\xrightarrow{3.825 \\mathrm{~d}}{ }^{218} \\mathrm{Po} \\xrightarrow{3.10 \\mathrm{~m}}{ }^{214} \\mathrm{~Pb} \\xrightarrow{26.8 \\mathrm{~m}}{ }^{214} \\mathrm{Bi} \\xrightarrow{19.9 \\mathrm{~m}} \\\\\n& { }^{214} \\mathrm{Po} \\xrightarrow{164.3 \\mu \\mathrm{s}} 210 \\mathrm{~Pb} \\xrightarrow{22.3 \\mathrm{y}} 210 \\mathrm{Bi} \\xrightarrow{5.013 \\mathrm{~d}} 210 \\mathrm{Po} \\xrightarrow{138.4 \\mathrm{~d}}{ }^{206 \\mathrm{~Pb}}\n\\end{aligned}\n$$\n\nThe times indicated are half-lives, the units are $y=$ years, $d=$ days, $m=$ minutes. The first decay, marked $t$ above, has a much longer half-life than the others.Given that thee relative isotopic mass of ${ }^{226} \\mathrm{Ra}$ measured by mass spectrometry is 226.25, use the textbook value of the Avogadro constant $\\left(6.022 \\cdot 10^{23} \\mathrm{~mol}^{-1}\\right)$ to calculate the number of ${ }^{226} \\mathrm{Ra}$ atoms in the original sample, $n_{\\mathrm{Ra}}$, the decay rate constant, $\\lambda$, and the half-life, $t$, of ${ }^{226} \\mathrm{Ra}$ (in years). You need only consider the decays up to but not including the isotope identified in 1.5.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nThe radioactive decay series of ${ }^{226} \\mathrm{Ra}$ is as follows:\n\n$$\n\\begin{aligned}\n& { }^{226} \\mathrm{Ra} \\xrightarrow{t}{ }^{222} \\mathrm{Rn} \\xrightarrow{3.825 \\mathrm{~d}}{ }^{218} \\mathrm{Po} \\xrightarrow{3.10 \\mathrm{~m}}{ }^{214} \\mathrm{~Pb} \\xrightarrow{26.8 \\mathrm{~m}}{ }^{214} \\mathrm{Bi} \\xrightarrow{19.9 \\mathrm{~m}} \\\\\n& { }^{214} \\mathrm{Po} \\xrightarrow{164.3 \\mu \\mathrm{s}} 210 \\mathrm{~Pb} \\xrightarrow{22.3 \\mathrm{y}} 210 \\mathrm{Bi} \\xrightarrow{5.013 \\mathrm{~d}} 210 \\mathrm{Po} \\xrightarrow{138.4 \\mathrm{~d}}{ }^{206 \\mathrm{~Pb}}\n\\end{aligned}\n$$\n\nThe times indicated are half-lives, the units are $y=$ years, $d=$ days, $m=$ minutes. The first decay, marked $t$ above, has a much longer half-life than the others.\n\nproblem:\nGiven that thee relative isotopic mass of ${ }^{226} \\mathrm{Ra}$ measured by mass spectrometry is 226.25, use the textbook value of the Avogadro constant $\\left(6.022 \\cdot 10^{23} \\mathrm{~mol}^{-1}\\right)$ to calculate the number of ${ }^{226} \\mathrm{Ra}$ atoms in the original sample, $n_{\\mathrm{Ra}}$, the decay rate constant, $\\lambda$, and the half-life, $t$, of ${ }^{226} \\mathrm{Ra}$ (in years). You need only consider the decays up to but not including the isotope identified in 1.5.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the number of ${ }^{226} \\mathrm{Ra}$ atoms in the original sample , $n_{\\mathrm{Ra}}$, the number of the decay rate constant, $\\lambda$, the number of the half-life, $t$, of ${ }^{226} \\mathrm{Ra}$ (in years)].\nTheir units are, in order, [atoms, $\\mathrm{~s}^{-1}$, years], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "atoms",
            "$\\mathrm{~s}^{-1}$",
            "years"
        ],
        "answer_sequence": [
            "the number of ${ }^{226} \\mathrm{Ra}$ atoms in the original sample , $n_{\\mathrm{Ra}}$",
            "the number of the decay rate constant, $\\lambda$",
            "the number of the half-life, $t$, of ${ }^{226} \\mathrm{Ra}$ (in years)"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1356",
        "problem": "Using the information provided on this graph, give numerical answers with appropriate units to the following questions:\n\n[figure1]\n\nWhat is the ionisation energy of the $\\mathrm{H}_{2}$ molecule?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nUsing the information provided on this graph, give numerical answers with appropriate units to the following questions:\n\n[figure1]\n\nWhat is the ionisation energy of the $\\mathrm{H}_{2}$ molecule?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $ \\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-471.jpg?height=911&width=1279&top_left_y=590&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$ \\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_59",
        "problem": "Which statements about the Lewis structure of the fulminate ion, $\\mathrm{CNO}^{-}$, are correct?\n\nI. The nitrogen atom has a positive formal charge.\n\nII. The nitrogen atom has a lone pair of electrons.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich statements about the Lewis structure of the fulminate ion, $\\mathrm{CNO}^{-}$, are correct?\n\nI. The nitrogen atom has a positive formal charge.\n\nII. The nitrogen atom has a lone pair of electrons.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1282",
        "problem": "Methanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |Calculate $\\Delta S^{0}$ for the reaction at $298 \\mathrm{~K}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nMethanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |\n\nproblem:\nCalculate $\\Delta S^{0}$ for the reaction at $298 \\mathrm{~K}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~J} \\mathrm{~K}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~J} \\mathrm{~K}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1018",
        "problem": "A $\\alpha$-particle\n\nB $\\beta$-particle\n\nC electron\n\nD proton\n\nE neutron\nA: $0.090 \\mathrm{~L}$\nB: $\\quad 0.44 \\mathrm{~L}$\nC: $\\quad 0.090 \\mathrm{~mL}$\nD: $\\quad 0.045 \\mathrm{~L}$\nE: $\\quad 2.22 \\mathrm{~mL}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $\\alpha$-particle\n\nB $\\beta$-particle\n\nC electron\n\nD proton\n\nE neutron\n\nA: $0.090 \\mathrm{~L}$\nB: $\\quad 0.44 \\mathrm{~L}$\nC: $\\quad 0.090 \\mathrm{~mL}$\nD: $\\quad 0.045 \\mathrm{~L}$\nE: $\\quad 2.22 \\mathrm{~mL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1255",
        "problem": "Using the first and second equations write down an overall reaction (3) so that the net enthalpy change is zero.\n\n$$ \\begin{array}{ll}  6 \\mathrm{CH}_{4}+3 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}+12 \\mathrm{H}_{2} & \\Delta H=-216 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\ \\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}+3 \\mathrm{H}_{2} & \\Delta H=216 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\ 7 \\mathrm{CH}_{4}+3 \\mathrm{O}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow 7 \\mathrm{CO}+15 \\mathrm{H}_{2} & \\Delta H=0 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\end{array} $$A mixture of gases containing mainly carbon monoxide and hydrogen is produced by the reaction of alkanes with steam:\n\n$$\n\\begin{array}{ll}\n\\mathrm{CH}_{4}+1 / 2 \\mathrm{O}_{2} \\rightarrow \\mathrm{CO}+2 \\mathrm{H}_{2} & \\Delta H=36 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\\n\\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}+3 \\mathrm{H}_{2} & \\Delta H=216 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n\\end{array}\n$$\n\nThe synthesis of methanol from carbon monoxide and hydrogen is carried out either a) in two steps, where the starting mixture corresponding to equation (3) is compressed from $0.1 \\times 10^{6} \\mathrm{~Pa}$ to $3 \\times 10^{6} \\mathrm{~Pa}$, and the mixture of products thereof compressed again from $3 \\times 10^{6} \\mathrm{~Pa}$ to $6 \\times 10^{6} \\mathrm{~Pa}$\n\nor\n\nb) in one step, where the mixture of products corresponding to equation (3) is compressed from $0.1 \\times 10^{6} \\mathrm{~Pa}$ to $6 \\times 10^{6} \\mathrm{~Pa}$.\n\nCalculate the work of compression, $W_{a}$, according to the two step reaction for $100 \\mathrm{~cm}^{3}$ of starting mixture and calculate the difference in the work of compression between the reactions 1 and 2 .\n\nAssume for calculations a complete reaction at constant pressure. Temperature remains constant at $500 \\mathrm{~K}$, ideal gas behaviour is assumed.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nUsing the first and second equations write down an overall reaction (3) so that the net enthalpy change is zero.\n\n$$ \\begin{array}{ll}  6 \\mathrm{CH}_{4}+3 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}+12 \\mathrm{H}_{2} & \\Delta H=-216 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\ \\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}+3 \\mathrm{H}_{2} & \\Delta H=216 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\ 7 \\mathrm{CH}_{4}+3 \\mathrm{O}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow 7 \\mathrm{CO}+15 \\mathrm{H}_{2} & \\Delta H=0 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\end{array} $$\n\nproblem:\nA mixture of gases containing mainly carbon monoxide and hydrogen is produced by the reaction of alkanes with steam:\n\n$$\n\\begin{array}{ll}\n\\mathrm{CH}_{4}+1 / 2 \\mathrm{O}_{2} \\rightarrow \\mathrm{CO}+2 \\mathrm{H}_{2} & \\Delta H=36 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\\n\\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}+3 \\mathrm{H}_{2} & \\Delta H=216 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n\\end{array}\n$$\n\nThe synthesis of methanol from carbon monoxide and hydrogen is carried out either a) in two steps, where the starting mixture corresponding to equation (3) is compressed from $0.1 \\times 10^{6} \\mathrm{~Pa}$ to $3 \\times 10^{6} \\mathrm{~Pa}$, and the mixture of products thereof compressed again from $3 \\times 10^{6} \\mathrm{~Pa}$ to $6 \\times 10^{6} \\mathrm{~Pa}$\n\nor\n\nb) in one step, where the mixture of products corresponding to equation (3) is compressed from $0.1 \\times 10^{6} \\mathrm{~Pa}$ to $6 \\times 10^{6} \\mathrm{~Pa}$.\n\nCalculate the work of compression, $W_{a}$, according to the two step reaction for $100 \\mathrm{~cm}^{3}$ of starting mixture and calculate the difference in the work of compression between the reactions 1 and 2 .\n\nAssume for calculations a complete reaction at constant pressure. Temperature remains constant at $500 \\mathrm{~K}$, ideal gas behaviour is assumed.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of MJ, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "MJ"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_711",
        "problem": "常温下, 草酸 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$ 的 $K_{\\mathrm{a} 1}=5.6 \\times 10^{-2}, K_{\\mathrm{a} 2}=5.42 \\times 10^{-5}, \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的 $K_{\\mathrm{b}}=1.77 \\times 10^{-5}, \\quad \\mathrm{NH}_{4}^{+}$的水解平衡常数为 $\\frac{1.0 \\times 10^{-14}}{1.77 \\times 10^{-5}}=5.6 \\times 10^{-10}$ 。工业上利用 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$沉淀法和 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 沉淀法提取稀土金属。若溶液混合引起的体积变化可忽略, 室温时下列指定溶液中微粒物质的量浓度关系正确的是\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ : c( $\\left.\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nC: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 和 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 和 $0.3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)-\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)=0.1+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 草酸 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$ 的 $K_{\\mathrm{a} 1}=5.6 \\times 10^{-2}, K_{\\mathrm{a} 2}=5.42 \\times 10^{-5}, \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的 $K_{\\mathrm{b}}=1.77 \\times 10^{-5}, \\quad \\mathrm{NH}_{4}^{+}$的水解平衡常数为 $\\frac{1.0 \\times 10^{-14}}{1.77 \\times 10^{-5}}=5.6 \\times 10^{-10}$ 。工业上利用 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$沉淀法和 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 沉淀法提取稀土金属。若溶液混合引起的体积变化可忽略, 室温时下列指定溶液中微粒物质的量浓度关系正确的是\n\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ : c( $\\left.\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nC: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 和 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 和 $0.3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)-\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)=0.1+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_316",
        "problem": "Why is the boiling point of iodine chloride (I-CI) greater than that of bromine $\\left(\\mathrm{Br}_{2}\\right)$ ?\nA: $\\mathrm{ICl}$ is heavier than $\\mathrm{Br}_{2}$.\nB: $\\mathrm{ICl}$ is a covalent compound and $\\mathrm{Br}_{2}$ is not.\nC: The $\\mathrm{I}-\\mathrm{Cl}$ bond is stronger than the $\\mathrm{Br}-\\mathrm{Br}$ bond.\nD: ICl is a polar molecule and $\\mathrm{Br}_{2}$ is nonpolar.\nE: $\\mathrm{ICl}$ is an ionic compound and $\\mathrm{Br}_{2}$ is not.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhy is the boiling point of iodine chloride (I-CI) greater than that of bromine $\\left(\\mathrm{Br}_{2}\\right)$ ?\n\nA: $\\mathrm{ICl}$ is heavier than $\\mathrm{Br}_{2}$.\nB: $\\mathrm{ICl}$ is a covalent compound and $\\mathrm{Br}_{2}$ is not.\nC: The $\\mathrm{I}-\\mathrm{Cl}$ bond is stronger than the $\\mathrm{Br}-\\mathrm{Br}$ bond.\nD: ICl is a polar molecule and $\\mathrm{Br}_{2}$ is nonpolar.\nE: $\\mathrm{ICl}$ is an ionic compound and $\\mathrm{Br}_{2}$ is not.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_656",
        "problem": "䒬与三氧化硫反应同时生成 $\\alpha$-萗磺酸 2 与 $\\beta$ 萗磺酸\n\n), 稳定性 $\\beta$-萫磺酸 $>\\alpha$-萫磺酸，反应过程中的能量变化如图所示， $1 、 2 、 3 、 4$ 是中间产物, $\\mathrm{m} 、 \\mathrm{n}$ 各代表一种产物。下列说法正确的是\n\n[图1]\n\n[图2]\n\n[图3]\n\n3\n\n[图4]\n\n4\nA: 较高温度下, 反应的主要产物是 $\\beta$-葈磺酸\nB: 升高相同温度, 生成 $\\beta$-葈磺酸的反应速率变化更大\nC: 实验中测得 2 的浓度大于 4 , 是因为生成 $m$ 的反应焓变更大\nD: 选择不同催化剂, 对产物中 $\\mathrm{m} 、 \\mathrm{n}$ 的含量不会产生影响\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n䒬与三氧化硫反应同时生成 $\\alpha$-萗磺酸 2 与 $\\beta$ 萗磺酸\n\n), 稳定性 $\\beta$-萫磺酸 $>\\alpha$-萫磺酸，反应过程中的能量变化如图所示， $1 、 2 、 3 、 4$ 是中间产物, $\\mathrm{m} 、 \\mathrm{n}$ 各代表一种产物。下列说法正确的是\n\n[图1]\n\n[图2]\n\n[图3]\n\n3\n\n[图4]\n\n4\n\nA: 较高温度下, 反应的主要产物是 $\\beta$-葈磺酸\nB: 升高相同温度, 生成 $\\beta$-葈磺酸的反应速率变化更大\nC: 实验中测得 2 的浓度大于 4 , 是因为生成 $m$ 的反应焓变更大\nD: 选择不同催化剂, 对产物中 $\\mathrm{m} 、 \\mathrm{n}$ 的含量不会产生影响\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-040.jpg?height=466&width=531&top_left_y=972&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-040.jpg?height=188&width=172&top_left_y=1057&top_left_x=902",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-040.jpg?height=123&width=197&top_left_y=1252&top_left_x=907",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-040.jpg?height=317&width=212&top_left_y=1052&top_left_x=1116"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_21",
        "problem": "If water is electrolyzed for $2.0 \\mathrm{hr}$ with a current of 10.0 A, what volume of dry oxygen gas is collected at STP?\nA: $4.2 \\mathrm{~L}$\nB: $4.6 \\mathrm{~L}$\nC: $8.4 \\mathrm{~L}$\nD: $17 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf water is electrolyzed for $2.0 \\mathrm{hr}$ with a current of 10.0 A, what volume of dry oxygen gas is collected at STP?\n\nA: $4.2 \\mathrm{~L}$\nB: $4.6 \\mathrm{~L}$\nC: $8.4 \\mathrm{~L}$\nD: $17 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_176",
        "problem": "The subshell filling order used for the quantum mechanical model of the atom is an approximation of the relative subshell energies, which assumes the energies remain fixed. However, there are exceptions to the Aufbau Principle. Which of the following is the correct ground state configuration of an element found on the periodic table?\nA: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{1} 3 \\mathrm{~d}^{5}$\nB: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{2} 3 \\mathrm{~d}^{4}$\nC: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{2} 4 \\mathrm{~d}^{4}$\nD: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{2} 4 \\mathrm{p}^{4}$\nE: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{1} 4 \\mathrm{p}^{5}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe subshell filling order used for the quantum mechanical model of the atom is an approximation of the relative subshell energies, which assumes the energies remain fixed. However, there are exceptions to the Aufbau Principle. Which of the following is the correct ground state configuration of an element found on the periodic table?\n\nA: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{1} 3 \\mathrm{~d}^{5}$\nB: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{2} 3 \\mathrm{~d}^{4}$\nC: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{2} 4 \\mathrm{~d}^{4}$\nD: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{2} 4 \\mathrm{p}^{4}$\nE: $[\\mathrm{Ar}] 4 \\mathrm{~s}^{1} 4 \\mathrm{p}^{5}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_101",
        "problem": "Which element would have its highest energy valence electron correspond to the following quantum numbers? $n=4, \\quad \\ell=2$\nA: Sc\nB: $\\mathrm{Y}$\nC: $\\mathrm{K}$\nD: Zn\nE: Ga\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich element would have its highest energy valence electron correspond to the following quantum numbers? $n=4, \\quad \\ell=2$\n\nA: Sc\nB: $\\mathrm{Y}$\nC: $\\mathrm{K}$\nD: Zn\nE: Ga\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_26",
        "problem": "A hot metal coin is dropped into cold water in a wellinsulated container. Which statements are true as the system approaches equilibrium?\n\n$$\n\\begin{aligned}\n& \\text { I. }\\left|q_{\\text {metal }}\\right|=\\left|q_{\\text {water }}\\right| \\\\\n& \\text { II. }\\left|\\Delta S_{\\text {metal }}\\right|=\\left|\\Delta S_{\\text {water }}\\right|\n\\end{aligned}\n$$\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA hot metal coin is dropped into cold water in a wellinsulated container. Which statements are true as the system approaches equilibrium?\n\n$$\n\\begin{aligned}\n& \\text { I. }\\left|q_{\\text {metal }}\\right|=\\left|q_{\\text {water }}\\right| \\\\\n& \\text { II. }\\left|\\Delta S_{\\text {metal }}\\right|=\\left|\\Delta S_{\\text {water }}\\right|\n\\end{aligned}\n$$\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_286",
        "problem": "Calculate the concentration of the $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution, in $\\mathrm{mol} \\mathrm{L}^{-1}$.\nOnce the concentration of the mercury(II) nitrate solution is known, it should be straightforward to calculate the concentration of a sodium tetraphenylborate solution, using the following reaction.\n\n$\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}+2 \\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4} \\rightarrow \\mathrm{Hg}\\left(\\mathrm{B}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}\\right)_{2}+2 \\mathrm{NaNO}_{3}$\n\nUnfortunately, this reaction is slightly non-stoichiometric, meaning that under constant conditions the two reactants combine in a consistent but non-integer mole ratio, rather than the expected 1:2 mole ratio.\n\n$20.00 \\mathrm{~mL}$ of the mercury(II) nitrate solution from the previous question requires $17.97 \\mathrm{~mL}$ of a sodium tetraphenylborate solution for complete reaction.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the concentration of the $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution, in $\\mathrm{mol} \\mathrm{L}^{-1}$.\nOnce the concentration of the mercury(II) nitrate solution is known, it should be straightforward to calculate the concentration of a sodium tetraphenylborate solution, using the following reaction.\n\n$\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}+2 \\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4} \\rightarrow \\mathrm{Hg}\\left(\\mathrm{B}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}\\right)_{2}+2 \\mathrm{NaNO}_{3}$\n\nUnfortunately, this reaction is slightly non-stoichiometric, meaning that under constant conditions the two reactants combine in a consistent but non-integer mole ratio, rather than the expected 1:2 mole ratio.\n\n$20.00 \\mathrm{~mL}$ of the mercury(II) nitrate solution from the previous question requires $17.97 \\mathrm{~mL}$ of a sodium tetraphenylborate solution for complete reaction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol/l, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol/l"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_486",
        "problem": "$\\mathrm{pC}$ 类似于 $\\mathrm{pH}$, 是指极稀溶液中的溶质浓度的常用对数的负值。如某溶液中某溶质的浓度为 $1 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 则该溶液中该溶质的 $\\mathrm{pC}=-\\lg \\left(1 \\times 10^{-3}\\right)=3$ 。如图为 $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 溶液的 $\\mathrm{pC}-\\mathrm{pH}$ 图(若离子浓度小于 $10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 可认为该离子不存在)。下列说法不正确的是\n\n[图1]\nA: 某温度下, $\\mathrm{CO}_{2}$ 饱和溶液的浓度是 $0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 其中 $1 / 5$ 的 $\\mathrm{CO}_{2}$ 转变为 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, 若此时溶液的 $\\mathrm{pH}$ 约为 5 , 据此可得该温度下 $\\mathrm{CO}_{2}$ 饱和溶液中 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离度约为 $0.1 \\%$,\nB: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 一级电离平衡常数的数值 $\\mathrm{K}_{\\mathrm{a} 1}=10^{-6}$\nC: 向 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中滴加盐酸至 $\\mathrm{pH}$ 等于 11 时, 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 中 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)$比 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 中 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{pC}$ 类似于 $\\mathrm{pH}$, 是指极稀溶液中的溶质浓度的常用对数的负值。如某溶液中某溶质的浓度为 $1 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 则该溶液中该溶质的 $\\mathrm{pC}=-\\lg \\left(1 \\times 10^{-3}\\right)=3$ 。如图为 $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 溶液的 $\\mathrm{pC}-\\mathrm{pH}$ 图(若离子浓度小于 $10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 可认为该离子不存在)。下列说法不正确的是\n\n[图1]\n\nA: 某温度下, $\\mathrm{CO}_{2}$ 饱和溶液的浓度是 $0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 其中 $1 / 5$ 的 $\\mathrm{CO}_{2}$ 转变为 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$, 若此时溶液的 $\\mathrm{pH}$ 约为 5 , 据此可得该温度下 $\\mathrm{CO}_{2}$ 饱和溶液中 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离度约为 $0.1 \\%$,\nB: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 一级电离平衡常数的数值 $\\mathrm{K}_{\\mathrm{a} 1}=10^{-6}$\nC: 向 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中滴加盐酸至 $\\mathrm{pH}$ 等于 11 时, 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 中 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)$比 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 中 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-100.jpg?height=457&width=642&top_left_y=1225&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_845",
        "problem": "常温下, 向一定浓度的 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中加入 $\\mathrm{KOH}$ 固体, 所得溶液中 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} 、 \\mathrm{HC}_{2} \\mathrm{O}_{4}^{-} 、 \\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$ 三种微粒的物质的量分数 $(\\delta)$ 与溶液 $\\mathrm{pH}$ 关系如图所示 [已知 $\\mathrm{K}_{\\text {sp }}\\left(\\mathrm{CaC}_{2} \\mathrm{O}_{4}\\right)=4.0 \\times 10^{-9}$ 。 下列说法不正确的是\n\n[图1]\nA: 曲线 III 代表 $\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$\nB: B 点 $\\mathrm{pH}$ 为 2.7\nC: $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$ 饱和溶液中, $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)=\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nD: $0.1 \\mathrm{~mol} \\mathrm{CaC}_{2} \\mathrm{O}_{4}$ 固体不能全部溶于 $1 \\mathrm{~L} 0.2 \\mathrm{~mol} / \\mathrm{LHCl}$ 中\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向一定浓度的 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中加入 $\\mathrm{KOH}$ 固体, 所得溶液中 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} 、 \\mathrm{HC}_{2} \\mathrm{O}_{4}^{-} 、 \\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$ 三种微粒的物质的量分数 $(\\delta)$ 与溶液 $\\mathrm{pH}$ 关系如图所示 [已知 $\\mathrm{K}_{\\text {sp }}\\left(\\mathrm{CaC}_{2} \\mathrm{O}_{4}\\right)=4.0 \\times 10^{-9}$ 。 下列说法不正确的是\n\n[图1]\n\nA: 曲线 III 代表 $\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$\nB: B 点 $\\mathrm{pH}$ 为 2.7\nC: $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$ 饱和溶液中, $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)=\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nD: $0.1 \\mathrm{~mol} \\mathrm{CaC}_{2} \\mathrm{O}_{4}$ 固体不能全部溶于 $1 \\mathrm{~L} 0.2 \\mathrm{~mol} / \\mathrm{LHCl}$ 中\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-092.jpg?height=599&width=786&top_left_y=186&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_87",
        "problem": "Silver ion forms the complex $\\mathrm{Ag}(\\mathrm{CN})_{2}{ }^{-}$with $K_{\\mathrm{f}}=9.8 \\times$ $10^{21}$. What is the minimum amount of $\\mathrm{HCN}$ that would need to be added to $1.00 \\mathrm{~L}$ of a suspension of $0.010 \\mathrm{~mol}$ $\\mathrm{AgCl}$ in order to dissolve all the solid? The $K_{\\text {sp }}$ of $\\mathrm{AgCl}$ is $1.8 \\times 10^{-10}$ and the $K_{\\mathrm{a}}$ of $\\mathrm{HCN}$ is $6.2 \\times 10^{-10}$.\nA: $0.020 \\mathrm{~mol}$\nB: $0.079 \\mathrm{~mol}$\nC: $0.26 \\mathrm{~mol}$\nD: $1.7 \\mathrm{~mol}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSilver ion forms the complex $\\mathrm{Ag}(\\mathrm{CN})_{2}{ }^{-}$with $K_{\\mathrm{f}}=9.8 \\times$ $10^{21}$. What is the minimum amount of $\\mathrm{HCN}$ that would need to be added to $1.00 \\mathrm{~L}$ of a suspension of $0.010 \\mathrm{~mol}$ $\\mathrm{AgCl}$ in order to dissolve all the solid? The $K_{\\text {sp }}$ of $\\mathrm{AgCl}$ is $1.8 \\times 10^{-10}$ and the $K_{\\mathrm{a}}$ of $\\mathrm{HCN}$ is $6.2 \\times 10^{-10}$.\n\nA: $0.020 \\mathrm{~mol}$\nB: $0.079 \\mathrm{~mol}$\nC: $0.26 \\mathrm{~mol}$\nD: $1.7 \\mathrm{~mol}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1165",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhich species is oxidised？",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhich species is oxidised？\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1224",
        "problem": "Propene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nA better reaction to produce large quantity of propene is the oxidative dehydrogenation $(O D H)$ using solid catalysts, such as vanadium oxides, under molecular oxygen gas. Although this type of reaction is still under intense research development, its promise toward the production of propene at an industrial scale eclipses that of the direct dehydrogenation.The overall rate of propane consumption in the reaction is\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=\\frac{1}{\\frac{p^{0}}{k_{\\text {red }} p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}+\\frac{p^{0}}{k_{o x} p_{\\mathrm{O}_{2}}}}\n$$\n\nwhere $k_{\\text {red }}$ and $k_{o x}$ are the rate constants for the reduction of metal oxide catalyst by propane and for the oxidation of the catalyst by molecular oxygen, respectively, and $p^{o}$ is the standard pressure of 1 bar. Some experiments found that the rate of\noxidation of the catalyst is 100,000 times faster than that of the propane oxidation. The experimental\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=k_{o b s} \\frac{p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}{p^{0}}\n$$\n\nwhere $k_{o b s}$ is the observed rate constant $\\left(0.062 \\mathrm{~mol} \\mathrm{~s}^{-1}\\right)$. If the reactor containing the catalyst is continuously passed through with propane and oxygen at a total pressure of 1 bar, determine the value of $k_{\\text {ox}}$ when the partial pressure of propane is 0.10 bar. Assume that the partial pressure of propene is negligible.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nPropene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nA better reaction to produce large quantity of propene is the oxidative dehydrogenation $(O D H)$ using solid catalysts, such as vanadium oxides, under molecular oxygen gas. Although this type of reaction is still under intense research development, its promise toward the production of propene at an industrial scale eclipses that of the direct dehydrogenation.\n\nproblem:\nThe overall rate of propane consumption in the reaction is\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=\\frac{1}{\\frac{p^{0}}{k_{\\text {red }} p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}+\\frac{p^{0}}{k_{o x} p_{\\mathrm{O}_{2}}}}\n$$\n\nwhere $k_{\\text {red }}$ and $k_{o x}$ are the rate constants for the reduction of metal oxide catalyst by propane and for the oxidation of the catalyst by molecular oxygen, respectively, and $p^{o}$ is the standard pressure of 1 bar. Some experiments found that the rate of\noxidation of the catalyst is 100,000 times faster than that of the propane oxidation. The experimental\n\n$$\nr_{\\mathrm{C}_{3} \\mathrm{H}_{6}}=k_{o b s} \\frac{p_{\\mathrm{C}_{3} \\mathrm{H}_{6}}}{p^{0}}\n$$\n\nwhere $k_{o b s}$ is the observed rate constant $\\left(0.062 \\mathrm{~mol} \\mathrm{~s}^{-1}\\right)$. If the reactor containing the catalyst is continuously passed through with propane and oxygen at a total pressure of 1 bar, determine the value of $k_{\\text {ox}}$ when the partial pressure of propane is 0.10 bar. Assume that the partial pressure of propene is negligible.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\times 10^{2} \\mathrm{~mol} \\mathrm{~s}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\times 10^{2} \\mathrm{~mol} \\mathrm{~s}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1563",
        "problem": "Dissociating gas cycle\n\nDinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:\n\n$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$\n\n1.00 mol of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ was put into an empty vessel with a fixed volume of $24.44 \\mathrm{dm}^{3}$. The equilibrium gas pressure at $298 \\mathrm{~K}$ was found to be 1.190 bar. When heated to $348 \\mathrm{~K}$, the gas pressure increased to its equilibrium value of 1.886 bar.Calculate $\\Delta G^{0}$ of the reaction at $298 \\mathrm{~K}$, assuming the gases are ideal.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nDissociating gas cycle\n\nDinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide:\n\n$$\n\\mathrm{N}_{2} \\mathrm{O}_{4}(g) \\rightleftharpoons 2 \\mathrm{NO}_{2}(g)\n$$\n\n1.00 mol of $\\mathrm{N}_{2} \\mathrm{O}_{4}$ was put into an empty vessel with a fixed volume of $24.44 \\mathrm{dm}^{3}$. The equilibrium gas pressure at $298 \\mathrm{~K}$ was found to be 1.190 bar. When heated to $348 \\mathrm{~K}$, the gas pressure increased to its equilibrium value of 1.886 bar.\n\nproblem:\nCalculate $\\Delta G^{0}$ of the reaction at $298 \\mathrm{~K}$, assuming the gases are ideal.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1580",
        "problem": "\"Second law of thermodynamics restricts coupling\"\n\nAccording to the Second Law of thermodynamics, two simultaneously occurring chemical reactions should decrease the system's Gibbs energy $G_{\\text {syst }}$,\n\n$$\n\\frac{\\Delta G_{\\text {syst }}}{\\Delta t}<0\n$$\n\nOne of these reactions may have positive Gibbs energy and still proceed in the forward direction due to the coupling with the second reaction. This second reaction must have negative Gibbs energy and the requirements of the Second law must be fulfilled! Consider the example.\n\nThe synthesis of urea under specific conditions\n\n$$\n2 \\mathrm{NH}_{3}+\\mathrm{CO}_{2} \\rightarrow\\left(\\mathrm{NH}_{2}\\right)_{2} \\mathrm{CO}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\n$\\left(\\Delta G(7)=46.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\\right)$\n\nis supposed to be coupled with the complete oxidation of glucose (under the same conditions)\n\n$$\n1/6 \\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+\\mathrm{O}_{2} \\rightarrow \\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\n$\\Delta G(8)=-481.2 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$,\n\n$r(8)=6.0 \\times 10^{-8} \\mathrm{~mol} \\mathrm{dm}^{-3} \\mathrm{~min}^{-1}$.\n\nBoth reactions are presented schematically. No other reactions are considered.What is the maximum rate of the above reaction permitted by the Second Law if this reaction is coupled to the below reaction?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n\"Second law of thermodynamics restricts coupling\"\n\nAccording to the Second Law of thermodynamics, two simultaneously occurring chemical reactions should decrease the system's Gibbs energy $G_{\\text {syst }}$,\n\n$$\n\\frac{\\Delta G_{\\text {syst }}}{\\Delta t}<0\n$$\n\nOne of these reactions may have positive Gibbs energy and still proceed in the forward direction due to the coupling with the second reaction. This second reaction must have negative Gibbs energy and the requirements of the Second law must be fulfilled! Consider the example.\n\nThe synthesis of urea under specific conditions\n\n$$\n2 \\mathrm{NH}_{3}+\\mathrm{CO}_{2} \\rightarrow\\left(\\mathrm{NH}_{2}\\right)_{2} \\mathrm{CO}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\n$\\left(\\Delta G(7)=46.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\\right)$\n\nis supposed to be coupled with the complete oxidation of glucose (under the same conditions)\n\n$$\n1/6 \\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+\\mathrm{O}_{2} \\rightarrow \\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\n$\\Delta G(8)=-481.2 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$,\n\n$r(8)=6.0 \\times 10^{-8} \\mathrm{~mol} \\mathrm{dm}^{-3} \\mathrm{~min}^{-1}$.\n\nBoth reactions are presented schematically. No other reactions are considered.\n\nproblem:\nWhat is the maximum rate of the above reaction permitted by the Second Law if this reaction is coupled to the below reaction?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3} \\mathrm{~min}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3} \\mathrm{~min}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1294",
        "problem": "Contemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.Calculate the standard electromotive force (EMF) of a fuel cell using gaseous oxygen and hydrogen, both ideal gases at $100 \\mathrm{kPa}$ and $323.15 \\mathrm{~K}$, to produce liquid water.\n\nUse the following entropy data for $323.15 \\mathrm{~K}: \\mathrm{S}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{l}\\right)=70 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$,\n\n$S^{\\circ}\\left(\\mathrm{H}_{2}, \\mathrm{~g}\\right)=131 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$,\n\n$S^{\\circ}\\left(\\mathrm{O}_{2}, \\mathrm{~g}\\right)=205 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nContemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.\n\nproblem:\nCalculate the standard electromotive force (EMF) of a fuel cell using gaseous oxygen and hydrogen, both ideal gases at $100 \\mathrm{kPa}$ and $323.15 \\mathrm{~K}$, to produce liquid water.\n\nUse the following entropy data for $323.15 \\mathrm{~K}: \\mathrm{S}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{l}\\right)=70 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$,\n\n$S^{\\circ}\\left(\\mathrm{H}_{2}, \\mathrm{~g}\\right)=131 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$,\n\n$S^{\\circ}\\left(\\mathrm{O}_{2}, \\mathrm{~g}\\right)=205 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_428",
        "problem": "合成导电高分子材料 PPV 的反应\n\n[图1]\n\nPPV\n\n$+(2 \\mathrm{n}-1) \\mathrm{HI}$, 下列说法不正确的是\nA: 合成 PPV 的反应为缩聚反应\nB: PPV 与聚苯乙烯重复结构单元不相同\nC: PPV 与聚苯乙烯均能使溴水反应螁色\nD: $1 \\mathrm{molPPV}$ 可以与 $2 \\mathrm{n} \\mathrm{molH}_{2}$ 加成\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n合成导电高分子材料 PPV 的反应\n\n[图1]\n\nPPV\n\n$+(2 \\mathrm{n}-1) \\mathrm{HI}$, 下列说法不正确的是\n\nA: 合成 PPV 的反应为缩聚反应\nB: PPV 与聚苯乙烯重复结构单元不相同\nC: PPV 与聚苯乙烯均能使溴水反应螁色\nD: $1 \\mathrm{molPPV}$ 可以与 $2 \\mathrm{n} \\mathrm{molH}_{2}$ 加成\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-77.jpg?height=197&width=1451&top_left_y=141&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_982",
        "problem": "Which one of the following solutions will be the best electrical conductor at $25^{\\circ} \\mathrm{C}$ ?\nA: $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{Na}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$\nB: $\\quad 0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NaCl}(a q)$\nC: $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$\nD: $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{3}(a q)$\nE: $\\quad 0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{CsCl}(a q)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following solutions will be the best electrical conductor at $25^{\\circ} \\mathrm{C}$ ?\n\nA: $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{Na}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$\nB: $\\quad 0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NaCl}(a q)$\nC: $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$\nD: $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{3}(a q)$\nE: $\\quad 0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{CsCl}(a q)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_774",
        "problem": "维纶(聚乙烯醇缩甲醛纤维)可用于生产服装、绳索等。其合成路线如下:\n\n[图1]\n\n聚乙烯醇缩甲醛纤维\n\n下列说法正确的是\nA: 反应(1)是加聚反应\nB: 高分子 $\\mathrm{A}$ 的链节中含有两种官能团\nC: 通过质谱法测定高分子 B 的平均相对分子质量, 可得其聚合度\nD: 反应(3)的化学方程式为: [图2]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n维纶(聚乙烯醇缩甲醛纤维)可用于生产服装、绳索等。其合成路线如下:\n\n[图1]\n\n聚乙烯醇缩甲醛纤维\n\n下列说法正确的是\n\nA: 反应(1)是加聚反应\nB: 高分子 $\\mathrm{A}$ 的链节中含有两种官能团\nC: 通过质谱法测定高分子 B 的平均相对分子质量, 可得其聚合度\nD: 反应(3)的化学方程式为: [图2]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-57.jpg?height=172&width=1439&top_left_y=1299&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-57.jpg?height=160&width=1343&top_left_y=1913&top_left_x=402"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1078",
        "problem": "Nescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nThe rate of the heating may be controlled by altering the $\\mathrm{pH}$ of the solution. How would you expect the rate of reaction to vary in acidic, basic and neutral conditions?\nA: acidic, basic, neutral\nB: acidic, neutral, basic\nC: neutral, basic, acidic\nD: basic, neutral, acidic\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNescafé have recently launched a selfheating can of coffee. To heat up the coffee, a button is pressed which mixes the heating ingredients - a very dilute solution of sodium / potassium hydroxide and calcium oxide. The can then warms up $210 \\mathrm{ml}\\left(210 \\mathrm{~cm}^{3}\\right)$ of coffee by approximately $40^{\\circ} \\mathrm{C}$.\n\n[figure1]\n\nThe rate of the heating may be controlled by altering the $\\mathrm{pH}$ of the solution. How would you expect the rate of reaction to vary in acidic, basic and neutral conditions?\n\nA: acidic, basic, neutral\nB: acidic, neutral, basic\nC: neutral, basic, acidic\nD: basic, neutral, acidic\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-03.jpg?height=417&width=699&top_left_y=385&top_left_x=1021"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_566",
        "problem": "若用 $A G$ 表示溶液的酸度, $A G$ 的定义为 $A G=1 \\mathrm{~g} \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}(\\mathrm{OH})}$, 室温下实验室中用 0.01\n\n[图1]\n正确的是\n\n[图2]\nA: 室温下, 醋酸的电离常数约为 $10^{-5}$\nB: A 点时加入氢氧化钠溶液的体积为 $20.00 \\mathrm{~mL}$\nC: 若 $\\mathrm{B}$ 点为 $40 \\mathrm{~mL}$, 所得溶液中: $2 c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)+c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\nD: 滴定过程中一定存在: $0.01 \\mathrm{~mol} / \\mathrm{L}+c\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n若用 $A G$ 表示溶液的酸度, $A G$ 的定义为 $A G=1 \\mathrm{~g} \\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}(\\mathrm{OH})}$, 室温下实验室中用 0.01\n\n[图1]\n正确的是\n\n[图2]\n\nA: 室温下, 醋酸的电离常数约为 $10^{-5}$\nB: A 点时加入氢氧化钠溶液的体积为 $20.00 \\mathrm{~mL}$\nC: 若 $\\mathrm{B}$ 点为 $40 \\mathrm{~mL}$, 所得溶液中: $2 c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)+c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\nD: 滴定过程中一定存在: $0.01 \\mathrm{~mol} / \\mathrm{L}+c\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-12.jpg?height=60&width=1382&top_left_y=438&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-12.jpg?height=326&width=714&top_left_y=585&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_779",
        "problem": "已知: 常温下, $K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=4.4 \\times 10^{-7}, K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=5.61 \\times 10^{-11}$,\n\n$K_{\\mathrm{a}}(\\mathrm{HClO})=3.0 \\times 10^{-8}$ 。下列说法正确的是\nA: 等浓度 $\\mathrm{NaClO} 、 \\mathrm{NaHCO}_{3}$ 溶液混合后: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{ClO}^{-}\\right)>c\\left(\\mathrm{HCO}_{3}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nB: $1.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaClO}$ 溶液 $\\mathrm{pH}$ 约为 10.8 (已知: $\\lg 2=0.3$ )\nC: 向 $\\mathrm{NaClO}$ 溶液中通入少量二氧化碳, 反应的离子方程式为 $\\mathrm{CO}_{2}+\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{HClO}+\\mathrm{HCO}_{3}^{-}$\nD: $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中 $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{HCO}_{3}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)<0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知: 常温下, $K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=4.4 \\times 10^{-7}, K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=5.61 \\times 10^{-11}$,\n\n$K_{\\mathrm{a}}(\\mathrm{HClO})=3.0 \\times 10^{-8}$ 。下列说法正确的是\n\nA: 等浓度 $\\mathrm{NaClO} 、 \\mathrm{NaHCO}_{3}$ 溶液混合后: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{ClO}^{-}\\right)>c\\left(\\mathrm{HCO}_{3}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nB: $1.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaClO}$ 溶液 $\\mathrm{pH}$ 约为 10.8 (已知: $\\lg 2=0.3$ )\nC: 向 $\\mathrm{NaClO}$ 溶液中通入少量二氧化碳, 反应的离子方程式为 $\\mathrm{CO}_{2}+\\mathrm{ClO}^{-}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{HClO}+\\mathrm{HCO}_{3}^{-}$\nD: $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中 $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{HCO}_{3}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)<0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_776",
        "problem": "1, 3-丙二胺 $\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{3} \\mathrm{NH}_{2}\\right.$ ]是一种二元弱碱[可简写为 $\\left.\\mathrm{M}(\\mathrm{OH})_{2}\\right], 25^{\\circ} \\mathrm{C}$ 时, 向一定浓度的 $\\mathrm{MCl}_{2}$ 溶液中逐滴加入 $\\mathrm{NaOH}$ 溶液, 以 $\\mathrm{X}$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{M}^{2+}\\right)}{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]}$或 $\\frac{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]}{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})_{2}\\right]}$ 已知 $\\mathrm{pX}=-\\operatorname{lgX}$, 混合溶液中 $\\mathrm{pX}$ 与 $\\mathrm{pOH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\nA: 曲线II表示 $\\mathrm{p}\\left\\{\\frac{\\mathrm{c}\\left(\\mathrm{M}^{2+}\\right)}{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]}\\right\\}$与 $\\mathrm{pOH}$ 的关系\nB: $\\mathrm{M}(\\mathrm{OH})_{2}$ 第一步电离的电离常数为 $1 \\times 10^{-1.51}$\nC: $\\mathrm{pOH}=4.01$ 时, $\\mathrm{c}\\left(\\mathrm{M}^{2+}\\right)=\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})_{2}\\right]>\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]$\nD: $\\mathrm{pOH}=3.51$ 时, $3 \\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})_{2}\\right]+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n1, 3-丙二胺 $\\left[\\mathrm{H}_{2} \\mathrm{~N}\\left(\\mathrm{CH}_{2}\\right)_{3} \\mathrm{NH}_{2}\\right.$ ]是一种二元弱碱[可简写为 $\\left.\\mathrm{M}(\\mathrm{OH})_{2}\\right], 25^{\\circ} \\mathrm{C}$ 时, 向一定浓度的 $\\mathrm{MCl}_{2}$ 溶液中逐滴加入 $\\mathrm{NaOH}$ 溶液, 以 $\\mathrm{X}$ 表示 $\\frac{\\mathrm{c}\\left(\\mathrm{M}^{2+}\\right)}{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]}$或 $\\frac{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]}{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})_{2}\\right]}$ 已知 $\\mathrm{pX}=-\\operatorname{lgX}$, 混合溶液中 $\\mathrm{pX}$ 与 $\\mathrm{pOH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: 曲线II表示 $\\mathrm{p}\\left\\{\\frac{\\mathrm{c}\\left(\\mathrm{M}^{2+}\\right)}{\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]}\\right\\}$与 $\\mathrm{pOH}$ 的关系\nB: $\\mathrm{M}(\\mathrm{OH})_{2}$ 第一步电离的电离常数为 $1 \\times 10^{-1.51}$\nC: $\\mathrm{pOH}=4.01$ 时, $\\mathrm{c}\\left(\\mathrm{M}^{2+}\\right)=\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})_{2}\\right]>\\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})^{+}\\right]$\nD: $\\mathrm{pOH}=3.51$ 时, $3 \\mathrm{c}\\left[\\mathrm{M}(\\mathrm{OH})_{2}\\right]+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-037.jpg?height=368&width=434&top_left_y=810&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_628",
        "problem": "脯氨酸( ${ }^{\\mathrm{H}} \\mathrm{OH}^{\\mathrm{O}}$ )是构成蛋白质的氨基酸中唯一的亚氨基酸, 它可催化分子间的羟醛缩合反应, 机理如图所示。下列说法正确的是\n\n[图1]\nA: 有机物(2)和(7)中采取 $\\mathrm{sp}^{3}$ 杂化的原子数之比为 $1: 2$\nB: 脯氨酸具有两性, 理论上 $1 \\mathrm{~mol}$ 脯氨酸与足量 $\\mathrm{NaHCO}_{3}$ 反应生成 $44 \\mathrm{gCO}_{2}$\nC: 反应中涉及的氢离子均符合 2 电子稳定结构\nD: 若原料为 2-丁酮和甲醛, 则产物可能为 $\\mathrm{OH}^{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n脯氨酸( ${ }^{\\mathrm{H}} \\mathrm{OH}^{\\mathrm{O}}$ )是构成蛋白质的氨基酸中唯一的亚氨基酸, 它可催化分子间的羟醛缩合反应, 机理如图所示。下列说法正确的是\n\n[图1]\n\nA: 有机物(2)和(7)中采取 $\\mathrm{sp}^{3}$ 杂化的原子数之比为 $1: 2$\nB: 脯氨酸具有两性, 理论上 $1 \\mathrm{~mol}$ 脯氨酸与足量 $\\mathrm{NaHCO}_{3}$ 反应生成 $44 \\mathrm{gCO}_{2}$\nC: 反应中涉及的氢离子均符合 2 电子稳定结构\nD: 若原料为 2-丁酮和甲醛, 则产物可能为 $\\mathrm{OH}^{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-34.jpg?height=882&width=1154&top_left_y=147&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1501",
        "problem": "This result is due to the presence of products that are not expected to be found in the residue. Give two of them that under these experimental conditions can plausibly account for the data.\n\nTraditionally, in industry the analysis and the yield are expressed as percentage of oxide. The phosphorous content is expressed as if it were $\\mathrm{P}_{2} \\mathrm{O}_{5}$.\n\nIf $n_{2}$ is the amount of a soluble product obtained, $n_{1}$ the amount of a substance added as acid, $n_{0}$ the amount of apatite added, the yield is:\n\n$$\n\\begin{aligned}\n& r_{\\exp }=\\frac{n_{2}}{n_{1}+n_{0}} 100 \\\\\n& m_{2}=0.144 \\mathrm{~g} \\text { of residue is obtained on the filter. }\n\\end{aligned}\n$$The elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nCalculate $r_{\\text {exp. }}$.\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 ; \\mathrm{S}: 32.\n$$\n\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThis result is due to the presence of products that are not expected to be found in the residue. Give two of them that under these experimental conditions can plausibly account for the data.\n\nTraditionally, in industry the analysis and the yield are expressed as percentage of oxide. The phosphorous content is expressed as if it were $\\mathrm{P}_{2} \\mathrm{O}_{5}$.\n\nIf $n_{2}$ is the amount of a soluble product obtained, $n_{1}$ the amount of a substance added as acid, $n_{0}$ the amount of apatite added, the yield is:\n\n$$\n\\begin{aligned}\n& r_{\\exp }=\\frac{n_{2}}{n_{1}+n_{0}} 100 \\\\\n& m_{2}=0.144 \\mathrm{~g} \\text { of residue is obtained on the filter. }\n\\end{aligned}\n$$\n\nproblem:\nThe elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nCalculate $r_{\\text {exp. }}$.\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 ; \\mathrm{S}: 32.\n$$\n\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1234",
        "problem": "Potentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\n Calculate the concentrations of\n\na) $\\mathrm{Fe}^{2+}$ in solution $\\mathbf{I}$;\n\nb) $\\mathrm{Fe}^{3+}$ in solution II.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nPotentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\n Calculate the concentrations of\n\na) $\\mathrm{Fe}^{2+}$ in solution $\\mathbf{I}$;\n\nb) $\\mathrm{Fe}^{3+}$ in solution II.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [$\\mathrm{Fe}^{3+}$ in solution II., $\\mathrm{Fe}^{2+}$ in solution I].\nTheir units are, in order, [M, M], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "M",
            "M"
        ],
        "answer_sequence": [
            "$\\mathrm{Fe}^{3+}$ in solution II.",
            "$\\mathrm{Fe}^{2+}$ in solution I"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_140",
        "problem": "For a spontaneous reaction, which relationship below is always true?\nA: $\\Delta G^{0}{ }_{r x n}>0$\nB: $\\Delta \\mathrm{S}$ universe $>0$\nC: $\\Delta \\mathrm{S}$ universe $<0$\nD: $\\Delta \\mathrm{H}^{0}{ }_{\\mathrm{rxn}}>0$\nE: $\\Delta \\mathrm{H}^{0}{ }_{r x n}<0$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor a spontaneous reaction, which relationship below is always true?\n\nA: $\\Delta G^{0}{ }_{r x n}>0$\nB: $\\Delta \\mathrm{S}$ universe $>0$\nC: $\\Delta \\mathrm{S}$ universe $<0$\nD: $\\Delta \\mathrm{H}^{0}{ }_{\\mathrm{rxn}}>0$\nE: $\\Delta \\mathrm{H}^{0}{ }_{r x n}<0$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_281",
        "problem": "Solutions of calcium nitrate and sodium carbonate react to form a precipitate of calcium carbonate, as shown in the following equation:\n\n$\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(\\mathrm{aq})+\\mathrm{Na}_{2} \\mathrm{CO}_{3}(\\mathrm{aq}) \\rightarrow \\mathrm{CaCO}_{3}(\\mathrm{~s})+2 \\mathrm{NaNO}_{3}(\\mathrm{aq})$\n\nCalculate the concentration of carbonate ions in the resulting solution when $75.0 \\mathrm{~mL}$ of $1.065 \\mathrm{~mol} \\mathrm{~L}^{-1}$ calcium nitrate solution is mixed with $125.0 \\mathrm{~mL}$ of $1.445 \\mathrm{~mol} \\mathrm{~L}^{-1}$ sodium carbonate solution, assuming that there is no change in volume.\nA: $\\quad 0.190 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.380 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.504 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $\\quad 0.723 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $\\quad 1.445 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSolutions of calcium nitrate and sodium carbonate react to form a precipitate of calcium carbonate, as shown in the following equation:\n\n$\\mathrm{Ca}\\left(\\mathrm{NO}_{3}\\right)_{2}(\\mathrm{aq})+\\mathrm{Na}_{2} \\mathrm{CO}_{3}(\\mathrm{aq}) \\rightarrow \\mathrm{CaCO}_{3}(\\mathrm{~s})+2 \\mathrm{NaNO}_{3}(\\mathrm{aq})$\n\nCalculate the concentration of carbonate ions in the resulting solution when $75.0 \\mathrm{~mL}$ of $1.065 \\mathrm{~mol} \\mathrm{~L}^{-1}$ calcium nitrate solution is mixed with $125.0 \\mathrm{~mL}$ of $1.445 \\mathrm{~mol} \\mathrm{~L}^{-1}$ sodium carbonate solution, assuming that there is no change in volume.\n\nA: $\\quad 0.190 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.380 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.504 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $\\quad 0.723 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $\\quad 1.445 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1335",
        "problem": "At a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.\n\nThe total energy of $\\mathrm{H}_{2}$ molecule in its ground state is $-31.675 \\mathrm{eV}$, relative to the same reference as that of hydrogen atom.Calculate the dissociation energy, DE, in $\\mathrm{eV}$ of a hydrogen molecule in its ground state such that both $\\mathrm{H}$ atoms are produced in their ground states.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nAt a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.\n\nThe total energy of $\\mathrm{H}_{2}$ molecule in its ground state is $-31.675 \\mathrm{eV}$, relative to the same reference as that of hydrogen atom.\n\nproblem:\nCalculate the dissociation energy, DE, in $\\mathrm{eV}$ of a hydrogen molecule in its ground state such that both $\\mathrm{H}$ atoms are produced in their ground states.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of eV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "eV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_979",
        "problem": "Which of the following particles is the most massive?\nA: $\\alpha$-particle\nB: $\\beta$-particle\nC: electron\nD: proton\nE: neutron\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following particles is the most massive?\n\nA: $\\alpha$-particle\nB: $\\beta$-particle\nC: electron\nD: proton\nE: neutron\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_38",
        "problem": "A pure sample of benzene $\\left(\\mathrm{C}_{6} \\mathrm{H}_{6}\\right)$ melts sharply at $5.5^{\\circ} \\mathrm{C}$, while a pure sample of hexafluorobenzene $\\left(\\mathrm{C}_{6} \\mathrm{~F}_{6}\\right)$ melts sharply at $5.2^{\\circ} \\mathrm{C}$. However, an equimolar mixture of $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ melts sharply at $23.7^{\\circ} \\mathrm{C}$. Which is the best explanation for this observation?\nA: $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ are immiscible in the liquid phase.\nB: $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ form a crystal containing equal amounts of the two substances.\nC: Fluorocarbons characteristically show negative freezing point depression constants.\nD: The enthalpy of mixing of $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ is negative.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA pure sample of benzene $\\left(\\mathrm{C}_{6} \\mathrm{H}_{6}\\right)$ melts sharply at $5.5^{\\circ} \\mathrm{C}$, while a pure sample of hexafluorobenzene $\\left(\\mathrm{C}_{6} \\mathrm{~F}_{6}\\right)$ melts sharply at $5.2^{\\circ} \\mathrm{C}$. However, an equimolar mixture of $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ melts sharply at $23.7^{\\circ} \\mathrm{C}$. Which is the best explanation for this observation?\n\nA: $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ are immiscible in the liquid phase.\nB: $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ form a crystal containing equal amounts of the two substances.\nC: Fluorocarbons characteristically show negative freezing point depression constants.\nD: The enthalpy of mixing of $\\mathrm{C}_{6} \\mathrm{H}_{6}$ and $\\mathrm{C}_{6} \\mathrm{~F}_{6}$ is negative.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_152",
        "problem": "Select the arrangement below that lists the bonds in order of increasing polarity (least polar to most polar):\nA: $\\mathrm{C}-\\mathrm{F}, \\mathrm{O}-\\mathrm{F}, \\mathrm{Be}-\\mathrm{F}$\nB: $\\mathrm{O}-\\mathrm{F}, \\mathrm{C}-\\mathrm{F}, \\mathrm{Be}-\\mathrm{F}$\nC: Be-F, O-F, C-F\nD: Be-F, C-F, O-F\nE: $\\mathrm{O}-\\mathrm{F}, \\mathrm{Be}-\\mathrm{F}, \\mathrm{C}-\\mathrm{F}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSelect the arrangement below that lists the bonds in order of increasing polarity (least polar to most polar):\n\nA: $\\mathrm{C}-\\mathrm{F}, \\mathrm{O}-\\mathrm{F}, \\mathrm{Be}-\\mathrm{F}$\nB: $\\mathrm{O}-\\mathrm{F}, \\mathrm{C}-\\mathrm{F}, \\mathrm{Be}-\\mathrm{F}$\nC: Be-F, O-F, C-F\nD: Be-F, C-F, O-F\nE: $\\mathrm{O}-\\mathrm{F}, \\mathrm{Be}-\\mathrm{F}, \\mathrm{C}-\\mathrm{F}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1178",
        "problem": "This question is about thiocyanates\n\nAmmonium thiocyanate is formed when ammonia reacts with carbon disulfide, $\\mathrm{CS}_{2}$.\n\nThe thiocyanate anion, $[\\mathrm{SCN}]^{-}$, can form bonds either via the sulfur (to give thiocyanate compounds), or via the nitrogen (to give isothiocyanates).\n\nProtonation of the nitrogen in $[\\mathrm{SCN}]^{-}$forms isothiocyanic acid. Its isomer, thiocyanic acid was first detected in a large gas cloud in space in 2009 using radio astronomy.\nSagittarius B2 (Sgr B2) is one of the largest clouds of molecular gas in the Milky Way. It is located close to the supermassive black hole at centre of the galaxy.\n\nWhen ammonia and carbon disulfide react, two moles of ammonia react with one of carbon disulfide to form one mole of an intermediate compound, A. Intermediate compound $\\mathbf{A}$ decomposes on warming to form ammonium thiocyanate and hydrogen sulfide.\n\nGive the molecular formula for intermediate compound $\\mathbf{A}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about thiocyanates\n\nAmmonium thiocyanate is formed when ammonia reacts with carbon disulfide, $\\mathrm{CS}_{2}$.\n\nThe thiocyanate anion, $[\\mathrm{SCN}]^{-}$, can form bonds either via the sulfur (to give thiocyanate compounds), or via the nitrogen (to give isothiocyanates).\n\nProtonation of the nitrogen in $[\\mathrm{SCN}]^{-}$forms isothiocyanic acid. Its isomer, thiocyanic acid was first detected in a large gas cloud in space in 2009 using radio astronomy.\nSagittarius B2 (Sgr B2) is one of the largest clouds of molecular gas in the Milky Way. It is located close to the supermassive black hole at centre of the galaxy.\n\nWhen ammonia and carbon disulfide react, two moles of ammonia react with one of carbon disulfide to form one mole of an intermediate compound, A. Intermediate compound $\\mathbf{A}$ decomposes on warming to form ammonium thiocyanate and hydrogen sulfide.\n\nGive the molecular formula for intermediate compound $\\mathbf{A}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1256",
        "problem": "Metallic gold frequently is found in aluminosilicate rocks and it is finely dispersed among other minerals. It may be extracted by treating the crushed rock with aerated sodium cyanide solution. During this process metallic gold is slowly converted to $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$, which is soluble in water (reaction 1 ).\n\nAfter equilibrium has been reached, the aqueous phase is pumped off and the metallic gold is recovered from it by reacting the gold complex with zinc, which is converted to $\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}$ (reaction 2).\n\nFive hundred litres $(500 \\mathrm{~L})$ of a solution $0.0100 \\mathrm{M}$ in $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$and $0.0030 \\mathrm{M}$ in $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$was evaporated to one third of the original volume and was treated with zinc ( $40 \\mathrm{~g}$ ). Assuming that deviation from standard conditions is unimportant and that all these redox reactions go essentially to completion, calculate the concentrations of $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$and of $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$after reaction has ceased.\n\n$$\n\\begin{array}{ll}\n{\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Zn}+4 \\mathrm{CN}^{-}} & E^{\\circ}=-1.26 \\mathrm{~V} \\\\\n{\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}+\\mathrm{e}^{-} \\rightarrow \\mathrm{Au}+2 \\mathrm{CN}^{-}} & E^{\\circ}=-0.60 \\mathrm{~V} \\\\\n{\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}+\\mathrm{e}^{-} \\rightarrow \\mathrm{Ag}+2 \\mathrm{CN}^{-}} & E^{\\circ}=-0.31 \\mathrm{~V}\n\\end{array}\n$$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nMetallic gold frequently is found in aluminosilicate rocks and it is finely dispersed among other minerals. It may be extracted by treating the crushed rock with aerated sodium cyanide solution. During this process metallic gold is slowly converted to $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$, which is soluble in water (reaction 1 ).\n\nAfter equilibrium has been reached, the aqueous phase is pumped off and the metallic gold is recovered from it by reacting the gold complex with zinc, which is converted to $\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}$ (reaction 2).\n\nFive hundred litres $(500 \\mathrm{~L})$ of a solution $0.0100 \\mathrm{M}$ in $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$and $0.0030 \\mathrm{M}$ in $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$was evaporated to one third of the original volume and was treated with zinc ( $40 \\mathrm{~g}$ ). Assuming that deviation from standard conditions is unimportant and that all these redox reactions go essentially to completion, calculate the concentrations of $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$and of $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$after reaction has ceased.\n\n$$\n\\begin{array}{ll}\n{\\left[\\mathrm{Zn}(\\mathrm{CN})_{4}\\right]^{2-}+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Zn}+4 \\mathrm{CN}^{-}} & E^{\\circ}=-1.26 \\mathrm{~V} \\\\\n{\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}+\\mathrm{e}^{-} \\rightarrow \\mathrm{Au}+2 \\mathrm{CN}^{-}} & E^{\\circ}=-0.60 \\mathrm{~V} \\\\\n{\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}+\\mathrm{e}^{-} \\rightarrow \\mathrm{Ag}+2 \\mathrm{CN}^{-}} & E^{\\circ}=-0.31 \\mathrm{~V}\n\\end{array}\n$$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [ the concentrations of $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$,   the concentrations of $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$].\nTheir units are, in order, [M, M], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "M",
            "M"
        ],
        "answer_sequence": [
            " the concentrations of $\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right]^{-}$",
            "  the concentrations of $\\left[\\mathrm{Ag}(\\mathrm{CN})_{2}\\right]^{-}$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_719",
        "problem": "某有机物 M 常用于生产胶黏剂, 存在如下转化关系。\n\n[图1]\n\n已知:\n\n(1) $\\mathrm{RCHO}+\\mathrm{CH}_{3} \\mathrm{CHO} \\xrightarrow{\\mathrm{NaOH}} \\mathrm{RCH}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{CHO}$;\n\n(2) 2 个一 $\\mathrm{OH}$ 连在同一个 $\\mathrm{C}$ 原子上不稳定。\n\n下列说法错误的是\nA: $\\mathrm{N}$ 分子中含有 2 个含氧官能团\nB: $\\mathrm{H}$ 是 $\\mathrm{P}$ 的同分异构体, $1 \\mathrm{~mol} \\mathrm{H}$ 与足量 $\\mathrm{Na}$ 反应生成 $2 \\mathrm{~mol} \\mathrm{H}_{2}$, 则 $\\mathrm{H}$ 有 8 种(不考虑立体异构)\nC: $\\mathrm{Q}$ 的核磁共振氢谱有一种峰\nD: 根据 $\\mathrm{M} \\rightarrow \\mathrm{Q}$ 反应原理, $\\mathrm{M}$ 可生成 $\\mathrm{C}_{\\mathrm{n}} \\mathrm{H}_{2 \\mathrm{n}} \\mathrm{O}_{\\mathrm{n}}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某有机物 M 常用于生产胶黏剂, 存在如下转化关系。\n\n[图1]\n\n已知:\n\n(1) $\\mathrm{RCHO}+\\mathrm{CH}_{3} \\mathrm{CHO} \\xrightarrow{\\mathrm{NaOH}} \\mathrm{RCH}(\\mathrm{OH}) \\mathrm{CH}_{2} \\mathrm{CHO}$;\n\n(2) 2 个一 $\\mathrm{OH}$ 连在同一个 $\\mathrm{C}$ 原子上不稳定。\n\n下列说法错误的是\n\nA: $\\mathrm{N}$ 分子中含有 2 个含氧官能团\nB: $\\mathrm{H}$ 是 $\\mathrm{P}$ 的同分异构体, $1 \\mathrm{~mol} \\mathrm{H}$ 与足量 $\\mathrm{Na}$ 反应生成 $2 \\mathrm{~mol} \\mathrm{H}_{2}$, 则 $\\mathrm{H}$ 有 8 种(不考虑立体异构)\nC: $\\mathrm{Q}$ 的核磁共振氢谱有一种峰\nD: 根据 $\\mathrm{M} \\rightarrow \\mathrm{Q}$ 反应原理, $\\mathrm{M}$ 可生成 $\\mathrm{C}_{\\mathrm{n}} \\mathrm{H}_{2 \\mathrm{n}} \\mathrm{O}_{\\mathrm{n}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-43.jpg?height=314&width=900&top_left_y=1573&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_89",
        "problem": "At constant temperature and pressure, $5.0 \\mathrm{~L}$ of $\\mathrm{SO}_{2}$ is combined with $3.0 \\mathrm{~L}$ of $\\mathrm{O}_{2}$ according to the equation:\n\n$$\n2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{SO}_{3}(g)\n$$\n\nAfter $\\mathrm{SO}_{3}$ formation is complete, what is the volume of the mixture?\nA: $5.5 \\mathrm{~L}$\nB: $6.0 \\mathrm{~L}$\nC: $6.5 \\mathrm{~L}$\nD: $8.0 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAt constant temperature and pressure, $5.0 \\mathrm{~L}$ of $\\mathrm{SO}_{2}$ is combined with $3.0 \\mathrm{~L}$ of $\\mathrm{O}_{2}$ according to the equation:\n\n$$\n2 \\mathrm{SO}_{2}(g)+\\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{SO}_{3}(g)\n$$\n\nAfter $\\mathrm{SO}_{3}$ formation is complete, what is the volume of the mixture?\n\nA: $5.5 \\mathrm{~L}$\nB: $6.0 \\mathrm{~L}$\nC: $6.5 \\mathrm{~L}$\nD: $8.0 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_462",
        "problem": "室温下, $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 体系中各含碳微粒的物质的量分数与 $\\mathrm{pH}$ 的关系如图 1 所示. 在 $\\mathrm{c}_{\\text {起始 }}\\left(\\mathrm{Na}_{2} \\mathrm{CO}_{3}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的体系中, 研究 $\\mathrm{Mg}^{2+}$ 在不同 $\\mathrm{pH}$ 时的可能产物, $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right)$ 与\n$\\mathrm{pH}$ 的关系如图 2 所示, 曲线 $\\mathrm{I}$ 的离子浓度关系符合 $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{MgCO}_{3}\\right)$,曲线II的离子浓度关系符合 $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right) \\cdot \\mathrm{c}^{2}(\\mathrm{OH})=\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Mg}(\\mathrm{OH})_{2}\\right]$ 。\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n下列说法正确的是\nA: 由 $M$ 点可求得 $\\mathrm{Ka}_{2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=1 \\times 10^{-6.37}$\nB: $\\mathrm{pH}=11$ 的体系中: $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)<\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nC: $\\mathrm{Q}$ 点的体系中, 发生反应 $\\mathrm{Mg}^{2+}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{Mg}(\\mathrm{OH})_{2} \\downarrow+2 \\mathrm{CO}_{2} \\uparrow$\nD: $\\mathrm{P}$ 点的体系中, $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 体系中各含碳微粒的物质的量分数与 $\\mathrm{pH}$ 的关系如图 1 所示. 在 $\\mathrm{c}_{\\text {起始 }}\\left(\\mathrm{Na}_{2} \\mathrm{CO}_{3}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的体系中, 研究 $\\mathrm{Mg}^{2+}$ 在不同 $\\mathrm{pH}$ 时的可能产物, $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right)$ 与\n$\\mathrm{pH}$ 的关系如图 2 所示, 曲线 $\\mathrm{I}$ 的离子浓度关系符合 $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{MgCO}_{3}\\right)$,曲线II的离子浓度关系符合 $\\mathrm{c}\\left(\\mathrm{Mg}^{2+}\\right) \\cdot \\mathrm{c}^{2}(\\mathrm{OH})=\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Mg}(\\mathrm{OH})_{2}\\right]$ 。\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\n下列说法正确的是\n\nA: 由 $M$ 点可求得 $\\mathrm{Ka}_{2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=1 \\times 10^{-6.37}$\nB: $\\mathrm{pH}=11$ 的体系中: $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)<\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nC: $\\mathrm{Q}$ 点的体系中, 发生反应 $\\mathrm{Mg}^{2+}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{Mg}(\\mathrm{OH})_{2} \\downarrow+2 \\mathrm{CO}_{2} \\uparrow$\nD: $\\mathrm{P}$ 点的体系中, $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-019.jpg?height=454&width=580&top_left_y=387&top_left_x=381",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-019.jpg?height=429&width=483&top_left_y=411&top_left_x=1072"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1209",
        "problem": "Problem I: Calculate the equilibrium pressure (in bar) of water vapour in a closed vessel containing $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ at $25^{\\circ} \\mathrm{C}$.Bridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\n Calculate the temperature at which the equilibrium water vapour pressure is $1.00 \\mathrm{bar}$ in the system described in problem I. Assume that $\\Delta H$ and $\\Delta S$ are temperature independent.\n\nCorrosion of metals is associated with electrochemical reactions. This also applies for the formation of rust on iron surfaces, where the initial electrode reactions usually are:\n\n$$\n\\begin{aligned}\n& \\mathrm{Fe}(\\mathrm{s}) \\rightarrow \\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\\\\n& \\mathrm{O}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+4 \\mathrm{e}^{-} \\rightarrow 4 \\mathrm{OH}^{-}(\\mathrm{aq})\n\\end{aligned}\n$$\n\nAn electrochemical cell in which these electrode reactions take place is constructed. The temperature is $25^{\\circ} \\mathrm{C}$. The cell is represented by the following cell diagram:\n\n$\\mathrm{Fe}(\\mathrm{s}) \\mathrm{Fe}^{2+}(\\mathrm{aq}) \\quad \\mathrm{OH}^{-}(\\mathrm{aq}), \\mathrm{O}_{2}(\\mathrm{~g}) \\quad \\mathrm{Pt}(\\mathrm{s})$\n\nStandard electrode potentials (at $25^{\\circ} \\mathrm{C}$ ):\n\n$$\n\\begin{array}{ll}\n\\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe}(\\mathrm{s}) & E=-0.44 \\mathrm{~V} \\\\\n\\mathrm{O}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+4 \\mathrm{e}^{-} \\rightarrow 4 \\mathrm{OH}^{-}(\\mathrm{aq}) & E=0.40 \\mathrm{~V}\n\\end{array}\n$$\n\nNernst factor: $\\quad R T \\ln 10 / F=0.05916$ volt (at $25^{\\circ} \\mathrm{C}$ )\n\nFaraday constant: $F=96485 \\mathrm{C} \\mathrm{mol}^{-1}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\nHere is some context information for this question, which might assist you in solving it:\nProblem I: Calculate the equilibrium pressure (in bar) of water vapour in a closed vessel containing $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ at $25^{\\circ} \\mathrm{C}$.\n\nproblem:\nBridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\n Calculate the temperature at which the equilibrium water vapour pressure is $1.00 \\mathrm{bar}$ in the system described in problem I. Assume that $\\Delta H$ and $\\Delta S$ are temperature independent.\n\nCorrosion of metals is associated with electrochemical reactions. This also applies for the formation of rust on iron surfaces, where the initial electrode reactions usually are:\n\n$$\n\\begin{aligned}\n& \\mathrm{Fe}(\\mathrm{s}) \\rightarrow \\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\\\\n& \\mathrm{O}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+4 \\mathrm{e}^{-} \\rightarrow 4 \\mathrm{OH}^{-}(\\mathrm{aq})\n\\end{aligned}\n$$\n\nAn electrochemical cell in which these electrode reactions take place is constructed. The temperature is $25^{\\circ} \\mathrm{C}$. The cell is represented by the following cell diagram:\n\n$\\mathrm{Fe}(\\mathrm{s}) \\mathrm{Fe}^{2+}(\\mathrm{aq}) \\quad \\mathrm{OH}^{-}(\\mathrm{aq}), \\mathrm{O}_{2}(\\mathrm{~g}) \\quad \\mathrm{Pt}(\\mathrm{s})$\n\nStandard electrode potentials (at $25^{\\circ} \\mathrm{C}$ ):\n\n$$\n\\begin{array}{ll}\n\\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{Fe}(\\mathrm{s}) & E=-0.44 \\mathrm{~V} \\\\\n\\mathrm{O}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+4 \\mathrm{e}^{-} \\rightarrow 4 \\mathrm{OH}^{-}(\\mathrm{aq}) & E=0.40 \\mathrm{~V}\n\\end{array}\n$$\n\nNernst factor: $\\quad R T \\ln 10 / F=0.05916$ volt (at $25^{\\circ} \\mathrm{C}$ )\n\nFaraday constant: $F=96485 \\mathrm{C} \\mathrm{mol}^{-1}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir units are, in order, [$ \\mathrm{~K}$, $^{\\circ} \\mathrm{C}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-345.jpg?height=951&width=1445&top_left_y=595&top_left_x=314"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            "$ \\mathrm{~K}$",
            "$^{\\circ} \\mathrm{C}$"
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_237",
        "problem": "Calculate the percentage by mass of potassium in the original potassium salt.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the percentage by mass of potassium in the original potassium salt.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_915",
        "problem": "制备某杀虫剂中间体的合成路线如下:\n\n$\\mathrm{Q}\\left(\\mathrm{C}_{7} \\mathrm{H}_{4} \\mathrm{OCl}_{2}\\right) \\xrightarrow[\\mathrm{AlCl}_{3}]{\\mathrm{CH}_{2}=\\mathrm{CH}_{2}} \\mathrm{R}\\left(\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{OCl}_{2}\\right) \\xrightarrow{\\mathrm{H}_{2} \\mathrm{SO}_{4}} \\mathrm{~S}$\n\n已知: $\\mathrm{Ar}-\\mathrm{H}-\\xrightarrow[\\mathrm{H}_{2} \\mathrm{SO}_{4}]{\\mathrm{Cl}} \\mathrm{Ar}-\\mathrm{R}$ ( $\\mathrm{Ar}$ 为芳基)\n\n下列说法正确的是\nA: Q 中最多有 14 个原子共面\nB: $\\mathrm{R}$ 能发生加成、取代、消去反应\nC: $\\mathrm{S}$ 完全氢化后的产物中有 5 个手性碳原子\nD: $\\mathrm{Q} \\rightarrow \\mathrm{R}$ 与 $\\mathrm{R} \\rightarrow \\mathrm{S}$ 的反应类型相同\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n制备某杀虫剂中间体的合成路线如下:\n\n$\\mathrm{Q}\\left(\\mathrm{C}_{7} \\mathrm{H}_{4} \\mathrm{OCl}_{2}\\right) \\xrightarrow[\\mathrm{AlCl}_{3}]{\\mathrm{CH}_{2}=\\mathrm{CH}_{2}} \\mathrm{R}\\left(\\mathrm{C}_{9} \\mathrm{H}_{8} \\mathrm{OCl}_{2}\\right) \\xrightarrow{\\mathrm{H}_{2} \\mathrm{SO}_{4}} \\mathrm{~S}$\n\n已知: $\\mathrm{Ar}-\\mathrm{H}-\\xrightarrow[\\mathrm{H}_{2} \\mathrm{SO}_{4}]{\\mathrm{Cl}} \\mathrm{Ar}-\\mathrm{R}$ ( $\\mathrm{Ar}$ 为芳基)\n\n下列说法正确的是\n\nA: Q 中最多有 14 个原子共面\nB: $\\mathrm{R}$ 能发生加成、取代、消去反应\nC: $\\mathrm{S}$ 完全氢化后的产物中有 5 个手性碳原子\nD: $\\mathrm{Q} \\rightarrow \\mathrm{R}$ 与 $\\mathrm{R} \\rightarrow \\mathrm{S}$ 的反应类型相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-35.jpg?height=716&width=1382&top_left_y=1613&top_left_x=334",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-35.jpg?height=208&width=668&top_left_y=2323&top_left_x=1048",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-36.jpg?height=257&width=325&top_left_y=157&top_left_x=1408",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-36.jpg?height=260&width=1048&top_left_y=595&top_left_x=687",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-36.jpg?height=249&width=786&top_left_y=949&top_left_x=932"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1461",
        "problem": "When an atom $\\mathrm{X}$ absorbs radiation with a photon energy greater than the ionization energy of the atom, the atom is ionized to generate an ion $\\mathrm{X}^{+}$and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is,\n\nPhoton energy $(h v)=$ ionization energy (IE) of $X+$ kinetic energy of photoelectron.\n\nWhen a molecule, for example, $\\mathrm{H}_{2}$, absorbs short-wavelength light, the photoelectron is ejected and an $\\mathrm{H}_{2}{ }^{+}$ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure 2 shows a typical photoelectron spectrum when $\\mathrm{H}_{2}$ in the lowest vibrational level is irradiated by monochromatic light of $21.2 \\mathrm{eV}$. No photoelectrons are detected above $6.0 \\mathrm{eV}$. (eV is a unit of energy and $1.0 \\mathrm{eV}$ is equal to $1.6 \\cdot 10^{-19} \\mathrm{~J}$.)\n\n[figure1]\n\nFigure 1. Schematic diagram of photoelectron spectroscopy.\n\n[figure2]\n\nFigure 2. Photoelectron spectrum of $\\mathrm{H}_{2}$. The energy of the incident light is $21.2 \\mathrm{eV}$.Determine the energy difference $\\Delta E_{\\mathrm{A} 2}(\\mathrm{eV})$ between $\\mathrm{H}_{2}^{+}\\left(v_{\\text {ion }}=0\\right)$ and $\\mathrm{H}_{2}^{+}\\left(v_{\\text {ion }}=3\\right)$ to the first decimal place.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nWhen an atom $\\mathrm{X}$ absorbs radiation with a photon energy greater than the ionization energy of the atom, the atom is ionized to generate an ion $\\mathrm{X}^{+}$and the electron (called a photoelectron) is ejected at the same time. In this event, the energy is conserved as shown in Figure 1, that is,\n\nPhoton energy $(h v)=$ ionization energy (IE) of $X+$ kinetic energy of photoelectron.\n\nWhen a molecule, for example, $\\mathrm{H}_{2}$, absorbs short-wavelength light, the photoelectron is ejected and an $\\mathrm{H}_{2}{ }^{+}$ion with a variety of vibrational states is produced. A photoelectron spectrum is a plot of the number of photoelectrons as a function of the kinetic energy of the photoelectrons. Figure 2 shows a typical photoelectron spectrum when $\\mathrm{H}_{2}$ in the lowest vibrational level is irradiated by monochromatic light of $21.2 \\mathrm{eV}$. No photoelectrons are detected above $6.0 \\mathrm{eV}$. (eV is a unit of energy and $1.0 \\mathrm{eV}$ is equal to $1.6 \\cdot 10^{-19} \\mathrm{~J}$.)\n\n[figure1]\n\nFigure 1. Schematic diagram of photoelectron spectroscopy.\n\n[figure2]\n\nFigure 2. Photoelectron spectrum of $\\mathrm{H}_{2}$. The energy of the incident light is $21.2 \\mathrm{eV}$.\n\nproblem:\nDetermine the energy difference $\\Delta E_{\\mathrm{A} 2}(\\mathrm{eV})$ between $\\mathrm{H}_{2}^{+}\\left(v_{\\text {ion }}=0\\right)$ and $\\mathrm{H}_{2}^{+}\\left(v_{\\text {ion }}=3\\right)$ to the first decimal place.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of eV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-072.jpg?height=639&width=643&top_left_y=1505&top_left_x=224",
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-072.jpg?height=671&width=940&top_left_y=1549&top_left_x=935"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "eV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_15",
        "problem": "Combining which solutions would produce a color change?\nA: $\\mathrm{NaCl}(a q)$ and $\\mathrm{Br}_{2}(a q)$\nB: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)$ and $\\mathrm{HNO}_{3}(a q)$\nC: $\\mathrm{K}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}(a q)$ and $\\mathrm{FeCl}_{3}(a q)$\nD: $\\mathrm{CH}_{3} \\mathrm{COOH}(a q)$ and $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCombining which solutions would produce a color change?\n\nA: $\\mathrm{NaCl}(a q)$ and $\\mathrm{Br}_{2}(a q)$\nB: $\\mathrm{Na}_{2} \\mathrm{CO}_{3}(a q)$ and $\\mathrm{HNO}_{3}(a q)$\nC: $\\mathrm{K}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}(a q)$ and $\\mathrm{FeCl}_{3}(a q)$\nD: $\\mathrm{CH}_{3} \\mathrm{COOH}(a q)$ and $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2}(a q)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1432",
        "problem": "Write the sequence of balanced equations for the above described reactions.To determine the solubility product of copper(II) iodate, $\\mathrm{Cu}\\left(\\mathrm{IO}_{3}\\right)_{2}$, by iodometric titration in an acidic solution $\\left(25^{\\circ} \\mathrm{C}\\right) 30.00 \\mathrm{~cm}^{3}$ of a 0.100 molar sodium thiosulphate solution are needed to titrate $20.00 \\mathrm{~cm}^{3}$ of a saturated aqueous solution $\\mathrm{Cu}\\left(\\mathrm{IO}_{3}\\right)_{2}$.\n\nCalculate the initial concentration of $\\mathrm{Cu}^{2+}$ and the solubility product of copper(II) iodate. Activity coefficients can be neglected.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nWrite the sequence of balanced equations for the above described reactions.\n\nproblem:\nTo determine the solubility product of copper(II) iodate, $\\mathrm{Cu}\\left(\\mathrm{IO}_{3}\\right)_{2}$, by iodometric titration in an acidic solution $\\left(25^{\\circ} \\mathrm{C}\\right) 30.00 \\mathrm{~cm}^{3}$ of a 0.100 molar sodium thiosulphate solution are needed to titrate $20.00 \\mathrm{~cm}^{3}$ of a saturated aqueous solution $\\mathrm{Cu}\\left(\\mathrm{IO}_{3}\\right)_{2}$.\n\nCalculate the initial concentration of $\\mathrm{Cu}^{2+}$ and the solubility product of copper(II) iodate. Activity coefficients can be neglected.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_467",
        "problem": "某温度下, 分别向 $10.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{KCl}$ 和 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 溶液中滴加 $0.1 \\mathrm{~mol} / \\mathrm{LAgNO}_{3}$溶液,滴加过程中- $\\operatorname{lgc}(\\mathrm{M})\\left(\\mathrm{M}\\right.$ 为 $\\mathrm{Cl}^{-}$或 $\\left.\\mathrm{CrO}_{4}^{2-}\\right)$ 与 $\\mathrm{AgNO}_{3}$ 溶液体积 $(\\mathrm{V})$ 的变化关系如图所示(忽略溶液体积变化)。下列说法正确的是\n\n[图1]\nA: 曲线 $\\mathrm{L}_{2}$ 表示 $-\\operatorname{lgc}\\left(\\mathrm{Cl}^{-}\\right)$与 $\\mathrm{V}\\left(\\mathrm{AgNO}_{3}\\right)$ 的变化关系\nB: 该温度下, $\\mathrm{Ksp}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)=2.0 \\times 10^{-8}$\nC: $\\mathrm{M}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 相同实验条件下, 若改为 $0.05 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{KCl}$ 和 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 溶液, 则曲线 $\\mathrm{L}_{2}$ 中 $\\mathrm{N}$ 点移到 $\\mathrm{Q}$ 点\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某温度下, 分别向 $10.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{KCl}$ 和 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 溶液中滴加 $0.1 \\mathrm{~mol} / \\mathrm{LAgNO}_{3}$溶液,滴加过程中- $\\operatorname{lgc}(\\mathrm{M})\\left(\\mathrm{M}\\right.$ 为 $\\mathrm{Cl}^{-}$或 $\\left.\\mathrm{CrO}_{4}^{2-}\\right)$ 与 $\\mathrm{AgNO}_{3}$ 溶液体积 $(\\mathrm{V})$ 的变化关系如图所示(忽略溶液体积变化)。下列说法正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{L}_{2}$ 表示 $-\\operatorname{lgc}\\left(\\mathrm{Cl}^{-}\\right)$与 $\\mathrm{V}\\left(\\mathrm{AgNO}_{3}\\right)$ 的变化关系\nB: 该温度下, $\\mathrm{Ksp}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)=2.0 \\times 10^{-8}$\nC: $\\mathrm{M}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 相同实验条件下, 若改为 $0.05 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{KCl}$ 和 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 溶液, 则曲线 $\\mathrm{L}_{2}$ 中 $\\mathrm{N}$ 点移到 $\\mathrm{Q}$ 点\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-047.jpg?height=451&width=765&top_left_y=454&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1326",
        "problem": "Contemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.A polymer membrane electrolyzer facility operates at the voltage of $2.00 \\mathrm{~V}$ and is powered by a $10.0 \\mathrm{MW}$ wind turbine plant which was running at full power from 10 $\\mathrm{pm}$ to $6 \\mathrm{am}$. The electrolysis yielded $1090 \\mathrm{~kg}$ of pure hydrogen. Calculate the electrolysis yield defined as the mass of produced hydrogen divided by its theoretical produced mass.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nContemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.\n\nproblem:\nA polymer membrane electrolyzer facility operates at the voltage of $2.00 \\mathrm{~V}$ and is powered by a $10.0 \\mathrm{MW}$ wind turbine plant which was running at full power from 10 $\\mathrm{pm}$ to $6 \\mathrm{am}$. The electrolysis yielded $1090 \\mathrm{~kg}$ of pure hydrogen. Calculate the electrolysis yield defined as the mass of produced hydrogen divided by its theoretical produced mass.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1186",
        "problem": "IONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\n\nConsider the stability of copper(I) oxide, $\\mathrm{Cu}_{2} \\mathrm{O}$, in contact with a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of $\\mathrm{Cu}^{2+}$ ions. The solubility product of copper $(\\mathrm{I})$ oxide is $K_{s p}=\\left[\\mathrm{Cu}^{+}\\right][\\mathrm{OH}]=1 \\times 10^{-15}$\n\nComplex formation involving $\\mathrm{Cu}^{+}$and $\\mathrm{Cu}^{2+}$ ions\n\nCalculate the dissociation constant for the complex ion $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}{3}\\right){4}\\right]^{2+}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\n\nConsider the stability of copper(I) oxide, $\\mathrm{Cu}_{2} \\mathrm{O}$, in contact with a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of $\\mathrm{Cu}^{2+}$ ions. The solubility product of copper $(\\mathrm{I})$ oxide is $K_{s p}=\\left[\\mathrm{Cu}^{+}\\right][\\mathrm{OH}]=1 \\times 10^{-15}$\n\nComplex formation involving $\\mathrm{Cu}^{+}$and $\\mathrm{Cu}^{2+}$ ions\n\nCalculate the dissociation constant for the complex ion $\\left[\\mathrm{Cu}\\left(\\mathrm{NH}{3}\\right){4}\\right]^{2+}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of V, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "V"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_557",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{HCl}$ 调节 $\\mathrm{Na}_{3} \\mathrm{R}$ 溶液的 $\\mathrm{pH}$, 混合溶液的 $\\mathrm{pH}$ 与离子浓度的关系如图所示。下列说法正确的是\n\n[图1]\n\n$-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HR}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{R}^{3-}\\right)}$ 或 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{HR}^{2-}\\right)}$ 或 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{R}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{-}\\right)}$\nA: $\\mathrm{L}_{3}$ 表示 $\\mathrm{pH}$ 与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HR}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{R}^{3-}\\right)}$ 的关系曲线\nB: $\\mathrm{H}_{3} \\mathrm{R}$ 的电离常数 $\\left[\\mathrm{Ka}_{3}\\left(\\mathrm{H}_{3} \\mathrm{R}\\right)\\right]$ 的数量级为 $10^{-8}$\nC: $\\mathrm{NaH}_{2} \\mathrm{R}$ 溶液中水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>10^{-7} \\mathrm{~mol} / \\mathrm{L}$\nD: 混合溶液的 $\\mathrm{pH}=7$ 时, 存在 $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{R}\\right)<\\mathrm{c}\\left(\\mathrm{R}^{3-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{HCl}$ 调节 $\\mathrm{Na}_{3} \\mathrm{R}$ 溶液的 $\\mathrm{pH}$, 混合溶液的 $\\mathrm{pH}$ 与离子浓度的关系如图所示。下列说法正确的是\n\n[图1]\n\n$-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HR}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{R}^{3-}\\right)}$ 或 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{HR}^{2-}\\right)}$ 或 $-\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{R}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{-}\\right)}$\n\nA: $\\mathrm{L}_{3}$ 表示 $\\mathrm{pH}$ 与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HR}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{R}^{3-}\\right)}$ 的关系曲线\nB: $\\mathrm{H}_{3} \\mathrm{R}$ 的电离常数 $\\left[\\mathrm{Ka}_{3}\\left(\\mathrm{H}_{3} \\mathrm{R}\\right)\\right]$ 的数量级为 $10^{-8}$\nC: $\\mathrm{NaH}_{2} \\mathrm{R}$ 溶液中水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>10^{-7} \\mathrm{~mol} / \\mathrm{L}$\nD: 混合溶液的 $\\mathrm{pH}=7$ 时, 存在 $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{R}\\right)<\\mathrm{c}\\left(\\mathrm{R}^{3-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-104.jpg?height=479&width=871&top_left_y=1571&top_left_x=318"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1050",
        "problem": "Increasing concerns over the use and generation of hazardous substances in chemical processes has encouraged some chemists to look for more environmentally friendly ways to make chemical products. To help evaluate a process environmentally, chemists often use the term 'percentage atom economy', where\n\n$\\%$ Atom Economy $\\quad=\\quad$ RMM of desired product $\\times 100$ RMM of all products\n\n[figure1]\n\nAn environmentally friendly chemical process would normally be expected to have a high $\\%$ atom economy, indicating that a high proportion of the starting materials end up as part of the final product, hence reducing the amount of waste. Efforts are constantly being made to increase the \\% atom economy of chemical processes. As an example, the manufacture of ethene oxide $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\\right)$ for many years was via the classical chlorohydrin route:\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\Longleftrightarrow \\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{HCl} \\\\\n\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{Ca}(\\mathrm{OH})_{2}+\\mathrm{HCl} \\Longleftrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}+\\mathrm{CaCl}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nThe modern petrochemical route involves the following reaction:\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{4}+1 / 2 \\mathrm{O}_{2} \\stackrel{\\mathbf{A g}}{ } \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\n$$\n\nIbuprofen, a non-steroidal anti-inflammatory drug, was first synthesised by Boots using a sixstep process, with a \\% atom economy of $40 \\%$. When the patent expired in the 1980's, several companies began developing new methods for the preparation of ibuprofen. The BHC Company synthesis, which proved highly successful, is shown below:\n\n[figure2]\n\nStep 1 involves the use of ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$.\n\nCalculate the \\% atom economy of the BHC Company process.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIncreasing concerns over the use and generation of hazardous substances in chemical processes has encouraged some chemists to look for more environmentally friendly ways to make chemical products. To help evaluate a process environmentally, chemists often use the term 'percentage atom economy', where\n\n$\\%$ Atom Economy $\\quad=\\quad$ RMM of desired product $\\times 100$ RMM of all products\n\n[figure1]\n\nAn environmentally friendly chemical process would normally be expected to have a high $\\%$ atom economy, indicating that a high proportion of the starting materials end up as part of the final product, hence reducing the amount of waste. Efforts are constantly being made to increase the \\% atom economy of chemical processes. As an example, the manufacture of ethene oxide $\\left(\\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\\right)$ for many years was via the classical chlorohydrin route:\n\n$$\n\\begin{gathered}\n\\mathrm{C}_{2} \\mathrm{H}_{4}+\\mathrm{Cl}_{2}+\\mathrm{H}_{2} \\mathrm{O} \\Longleftrightarrow \\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{HCl} \\\\\n\\mathrm{ClCH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}+\\mathrm{Ca}(\\mathrm{OH})_{2}+\\mathrm{HCl} \\Longleftrightarrow \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}+\\mathrm{CaCl}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}\n\\end{gathered}\n$$\n\nThe modern petrochemical route involves the following reaction:\n\n$$\n\\mathrm{C}_{2} \\mathrm{H}_{4}+1 / 2 \\mathrm{O}_{2} \\stackrel{\\mathbf{A g}}{ } \\mathrm{C}_{2} \\mathrm{H}_{4} \\mathrm{O}\n$$\n\nIbuprofen, a non-steroidal anti-inflammatory drug, was first synthesised by Boots using a sixstep process, with a \\% atom economy of $40 \\%$. When the patent expired in the 1980's, several companies began developing new methods for the preparation of ibuprofen. The BHC Company synthesis, which proved highly successful, is shown below:\n\n[figure2]\n\nStep 1 involves the use of ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$.\n\nCalculate the \\% atom economy of the BHC Company process.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-05.jpg?height=454&width=343&top_left_y=316&top_left_x=1482",
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-05.jpg?height=482&width=1559&top_left_y=1872&top_left_x=246"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_394",
        "problem": "When the following reaction is balanced, what is the ratio of coefficients of $\\mathrm{H}^{+}(a q)$ to $\\mathrm{NO}(g)$ ?\n\n$\\mathrm{Cu}(s)+\\mathrm{H}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q) \\rightarrow \\mathrm{NO}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{Cu}^{2+}(a q)$\nA: $1: 1$\nB: $2: 1$\nC: $3: 1$\nD: $4: 1$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen the following reaction is balanced, what is the ratio of coefficients of $\\mathrm{H}^{+}(a q)$ to $\\mathrm{NO}(g)$ ?\n\n$\\mathrm{Cu}(s)+\\mathrm{H}^{+}(a q)+\\mathrm{NO}_{3}{ }^{-}(a q) \\rightarrow \\mathrm{NO}(g)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{Cu}^{2+}(a q)$\n\nA: $1: 1$\nB: $2: 1$\nC: $3: 1$\nD: $4: 1$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_345",
        "problem": "Which of the following molecules has polar bonds but is nonpolar?\nA: $\\mathrm{N}_{2} \\mathrm{H}_{4}$\nB: $\\mathrm{CCl}_{4}$\nC: $\\mathrm{HNO}_{3}$\nD: $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\nE: $\\mathrm{F}_{2} \\mathrm{O}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following molecules has polar bonds but is nonpolar?\n\nA: $\\mathrm{N}_{2} \\mathrm{H}_{4}$\nB: $\\mathrm{CCl}_{4}$\nC: $\\mathrm{HNO}_{3}$\nD: $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\nE: $\\mathrm{F}_{2} \\mathrm{O}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1540",
        "problem": "The energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.Calculate the speed $u\\left(\\mathrm{~m} \\mathrm{~s}^{-1}\\right)$ of the hydrogen atoms generated in the above reaction. $\\mathrm{H}_{2}$ is assumed to be at rest. If you were unable to determine the value for $E_{\\mathrm{C}}$, then use $5.0 \\mathrm{eV}$ for $E_{\\mathrm{C}}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe energy threshold for the generation of two electronically excited hydrogen atoms $\\mathrm{H}^{\\star}(n=2)$ from $\\mathrm{H}_{2}(v=0)$ has been derived to be $24.9 \\mathrm{eV}$ by an experiment.\n\nproblem:\nCalculate the speed $u\\left(\\mathrm{~m} \\mathrm{~s}^{-1}\\right)$ of the hydrogen atoms generated in the above reaction. $\\mathrm{H}_{2}$ is assumed to be at rest. If you were unable to determine the value for $E_{\\mathrm{C}}$, then use $5.0 \\mathrm{eV}$ for $E_{\\mathrm{C}}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~m} \\mathrm{~s}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~m} \\mathrm{~s}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_401",
        "problem": "2021 年 7 月 29 日消息, 宁德时代正式推出钠离子电池。我国科学家研究出一种钠离子可充电电池的工作示意图如下, 该电池主要依靠钠离子在正极和负极之间移动来工作，电池反应为[图1]\n下列有关说法不正确的是\n[图2]\nA: 钠离子电池相比于锂离子电池, 具有原料储量丰富, 价格低廉的优点\nB: 放电时, a 极为负极, 发生氧化反应\nC: 充电时, 阴极发生反应为[图3]\nD: 充电时, 当电路中转移 $0.3 \\mathrm{~mol}$ 电子时, $\\mathrm{P}$ 极质量减少 $6.9 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n2021 年 7 月 29 日消息, 宁德时代正式推出钠离子电池。我国科学家研究出一种钠离子可充电电池的工作示意图如下, 该电池主要依靠钠离子在正极和负极之间移动来工作，电池反应为[图1]\n下列有关说法不正确的是\n[图2]\n\nA: 钠离子电池相比于锂离子电池, 具有原料储量丰富, 价格低廉的优点\nB: 放电时, a 极为负极, 发生氧化反应\nC: 充电时, 阴极发生反应为[图3]\nD: 充电时, 当电路中转移 $0.3 \\mathrm{~mol}$ 电子时, $\\mathrm{P}$ 极质量减少 $6.9 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/d30htgSV/image.png",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-51.jpg?height=1020&width=946&top_left_y=1106&top_left_x=338",
            "https://i.postimg.cc/vBCmbD93/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_611",
        "problem": "下列比较中, 正确的是\nA: 同温度同物质的量浓度时, $\\mathrm{HF}$ 比 $\\mathrm{HCN}$ 易电离, $\\mathrm{NaF}$ 溶液的 $\\mathrm{pH}$ 比 $\\mathrm{NaCN}$ 溶液大\nB: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ 和 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合后: $c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nC: 同浓度的下列溶液中, (1) $\\mathrm{NH}_{4} \\mathrm{Al}\\left(\\mathrm{SO}_{4}\\right)_{2} 、(2) \\mathrm{NH}_{4} \\mathrm{Cl} 、(3) \\mathrm{CH}_{3} \\mathrm{COONH}_{4}$ 、 (4) $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O} ; c\\left(\\mathrm{NH}_{4}^{+}\\right)$由大到小的顺序是: (1)>(2)>(3)>(4)\nD: 物质的量浓度相等的 $\\mathrm{H}_{2} \\mathrm{~S}$ 和 NaHS 混合溶液中: $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{~S}^{2-}\\right)+c\\left(\\mathrm{HS}^{-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列比较中, 正确的是\n\nA: 同温度同物质的量浓度时, $\\mathrm{HF}$ 比 $\\mathrm{HCN}$ 易电离, $\\mathrm{NaF}$ 溶液的 $\\mathrm{pH}$ 比 $\\mathrm{NaCN}$ 溶液大\nB: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ 和 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合后: $c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nC: 同浓度的下列溶液中, (1) $\\mathrm{NH}_{4} \\mathrm{Al}\\left(\\mathrm{SO}_{4}\\right)_{2} 、(2) \\mathrm{NH}_{4} \\mathrm{Cl} 、(3) \\mathrm{CH}_{3} \\mathrm{COONH}_{4}$ 、 (4) $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O} ; c\\left(\\mathrm{NH}_{4}^{+}\\right)$由大到小的顺序是: (1)>(2)>(3)>(4)\nD: 物质的量浓度相等的 $\\mathrm{H}_{2} \\mathrm{~S}$ 和 NaHS 混合溶液中: $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{~S}^{2-}\\right)+c\\left(\\mathrm{HS}^{-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1512",
        "problem": "Under certain conditions of concentration and temperature $\\mathrm{HNO}_{3}$ reacts with $\\mathrm{Zn}$ and its reduction products are $\\mathrm{NO}_{2}$ and $\\mathrm{NO}$ in a molar ratio $1: 3$. How many moles of $\\mathrm{HNO}_{3}$ are consumed by $1 \\mathrm{~mol}$ of $\\mathrm{Zn}$ ?\nA: 2.2\nB: 2.4\nC: 2.6\nD: 2.8\nE: 3.0\nF: 3.2\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nUnder certain conditions of concentration and temperature $\\mathrm{HNO}_{3}$ reacts with $\\mathrm{Zn}$ and its reduction products are $\\mathrm{NO}_{2}$ and $\\mathrm{NO}$ in a molar ratio $1: 3$. How many moles of $\\mathrm{HNO}_{3}$ are consumed by $1 \\mathrm{~mol}$ of $\\mathrm{Zn}$ ?\n\nA: 2.2\nB: 2.4\nC: 2.6\nD: 2.8\nE: 3.0\nF: 3.2\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_690",
        "problem": "下图为 $\\mathrm{Ca}-\\mathrm{LiFePO}_{4}$ 可充电电池的工作示意图, 锂离子导体膜只允许 $\\mathrm{Li}^{+}$通过, 电池反应为: $\\mathrm{xCa}^{2+}+2 \\mathrm{LiFePO}_{4} \\underset{\\text { 充电 }}{\\text { 放电 }} \\mathrm{xCa}+2 \\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}+2 \\mathrm{xLi}^{+}$。下列说法错误的是\n\n[图1]\nA: $\\mathrm{LiPF}_{6}-\\mathrm{LiAsF}_{6}$ 为非水电解质，主要作用都是传递离子，构成闭合回路\nB: 充电时, $\\mathrm{Li}_{1-{ }_{\\mathrm{x}}} \\mathrm{FePO}_{4} / \\mathrm{LiFePO}_{4}$ 电极发生 $\\mathrm{Li}^{+}$嵌入, 放电时发生 $\\mathrm{Li}^{+}$脱嵌\nC: 放电时, 负极反应为: $\\mathrm{LiFePO}_{4}-\\mathrm{xe}^{-}=\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}+\\mathrm{xLi}^{+}$\nD: 充电时, 当转移 $0.2 \\mathrm{~mol}$ 电子时, 左室中电解质的质量减轻 $2.6 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下图为 $\\mathrm{Ca}-\\mathrm{LiFePO}_{4}$ 可充电电池的工作示意图, 锂离子导体膜只允许 $\\mathrm{Li}^{+}$通过, 电池反应为: $\\mathrm{xCa}^{2+}+2 \\mathrm{LiFePO}_{4} \\underset{\\text { 充电 }}{\\text { 放电 }} \\mathrm{xCa}+2 \\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}+2 \\mathrm{xLi}^{+}$。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{LiPF}_{6}-\\mathrm{LiAsF}_{6}$ 为非水电解质，主要作用都是传递离子，构成闭合回路\nB: 充电时, $\\mathrm{Li}_{1-{ }_{\\mathrm{x}}} \\mathrm{FePO}_{4} / \\mathrm{LiFePO}_{4}$ 电极发生 $\\mathrm{Li}^{+}$嵌入, 放电时发生 $\\mathrm{Li}^{+}$脱嵌\nC: 放电时, 负极反应为: $\\mathrm{LiFePO}_{4}-\\mathrm{xe}^{-}=\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}+\\mathrm{xLi}^{+}$\nD: 充电时, 当转移 $0.2 \\mathrm{~mol}$ 电子时, 左室中电解质的质量减轻 $2.6 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-87.jpg?height=691&width=1020&top_left_y=842&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1243",
        "problem": "If two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\\mathrm{H}_{2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\\mathrm{H}_{2}$ rapidly desorbs. This question examines two kinetic models for $\\mathrm{H}_{2}$ formation on the surface of a dust particle.\n\nIn both models, the rate constant for adsorption of $\\mathrm{H}$ atoms onto the surface of dust particles is $k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1}$. The typical number density of $\\mathrm{H}$ atoms (number of $\\mathrm{H}$ atoms per unit volume) in interstellar space is $[\\mathrm{H}]=10 \\mathrm{~cm}^{-3}$.\n\n[Note: In the following, you may treat numbers of surface-adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations. As a result, the units of the rate constants may be unfamiliar to you. Reaction rates have units of numbers of atoms or molecules per unit time.]\n\nThe $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$Reaction to form $\\mathrm{H}_{2}$ is assumed to be second order. On a dust particle the rate of removal of $\\mathrm{H}$ atoms by reaction is $k_{r} N^{2}$.\nCalculate the rate of production of $\\mathrm{H}_{2}$ per dust particle in this model.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIf two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\\mathrm{H}_{2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\\mathrm{H}_{2}$ rapidly desorbs. This question examines two kinetic models for $\\mathrm{H}_{2}$ formation on the surface of a dust particle.\n\nIn both models, the rate constant for adsorption of $\\mathrm{H}$ atoms onto the surface of dust particles is $k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1}$. The typical number density of $\\mathrm{H}$ atoms (number of $\\mathrm{H}$ atoms per unit volume) in interstellar space is $[\\mathrm{H}]=10 \\mathrm{~cm}^{-3}$.\n\n[Note: In the following, you may treat numbers of surface-adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations. As a result, the units of the rate constants may be unfamiliar to you. Reaction rates have units of numbers of atoms or molecules per unit time.]\n\nThe $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\nproblem:\nReaction to form $\\mathrm{H}_{2}$ is assumed to be second order. On a dust particle the rate of removal of $\\mathrm{H}$ atoms by reaction is $k_{r} N^{2}$.\nCalculate the rate of production of $\\mathrm{H}_{2}$ per dust particle in this model.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~s}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~s}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_168",
        "problem": "The exothermic thermite reaction between iron (III) oxide and aluminum metal occurs as follows:\n\n$$\n\\mathrm{Fe}_{2} \\mathrm{O}_{3}(\\mathrm{~s})+2 \\mathrm{Al}(\\mathrm{s}) \\rightarrow 2 \\mathrm{Fe}(\\mathrm{s})+\\mathrm{Al}_{2} \\mathrm{O}_{3}(\\mathrm{~s})\n$$\n\nIf $8.0 \\mathrm{~g}$ of iron (III) oxide is combined with $5.4 \\mathrm{~g}$ of aluminum metal, what mass of iron metal will be produced?\nA: $2.8 \\mathrm{~g}$\nB: $5.6 \\mathrm{~g}$\nC: $8.0 \\mathrm{~g}$\nD: $11 \\mathrm{~g}$\nE: $14 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe exothermic thermite reaction between iron (III) oxide and aluminum metal occurs as follows:\n\n$$\n\\mathrm{Fe}_{2} \\mathrm{O}_{3}(\\mathrm{~s})+2 \\mathrm{Al}(\\mathrm{s}) \\rightarrow 2 \\mathrm{Fe}(\\mathrm{s})+\\mathrm{Al}_{2} \\mathrm{O}_{3}(\\mathrm{~s})\n$$\n\nIf $8.0 \\mathrm{~g}$ of iron (III) oxide is combined with $5.4 \\mathrm{~g}$ of aluminum metal, what mass of iron metal will be produced?\n\nA: $2.8 \\mathrm{~g}$\nB: $5.6 \\mathrm{~g}$\nC: $8.0 \\mathrm{~g}$\nD: $11 \\mathrm{~g}$\nE: $14 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_851",
        "problem": "连二次硝酸 $\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)$ 是一种二元酸, 常温下, 用 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaOH}$ 溶液滴定\n\n$10 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}$ 溶液, 测得溶液 $\\mathrm{pH}$ 与 $\\mathrm{NaOH}$ 溶液体积的关系如图。常温下,试卷第 18 页，共 92 页\n\n[图1]\nA: a 点: $\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{2}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=9.9 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: b 点: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{~N}_{2} \\mathrm{O}_{2}^{2-}\\right)+c\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)$\nC: c 点: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)>\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{2}^{2-}\\right)$\nD: d 点: $c\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n连二次硝酸 $\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)$ 是一种二元酸, 常温下, 用 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaOH}$ 溶液滴定\n\n$10 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}$ 溶液, 测得溶液 $\\mathrm{pH}$ 与 $\\mathrm{NaOH}$ 溶液体积的关系如图。常温下,试卷第 18 页，共 92 页\n\n[图1]\n\nA: a 点: $\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{2}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=9.9 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: b 点: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{~N}_{2} \\mathrm{O}_{2}^{2-}\\right)+c\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)$\nC: c 点: $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)>\\mathrm{c}\\left(\\mathrm{N}_{2} \\mathrm{O}_{2}^{2-}\\right)$\nD: d 点: $c\\left(\\mathrm{H}_{2} \\mathrm{~N}_{2} \\mathrm{O}_{2}\\right)+\\mathrm{c}\\left(\\mathrm{HN}_{2} \\mathrm{O}_{2}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-19.jpg?height=719&width=717&top_left_y=226&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_185",
        "problem": "The smell often associated with public swimming pools comes from chloramines. The reaction of hypochlorous acid with ammonia from human urine will produce monochloramine $\\left(\\mathrm{NH}_{2} \\mathrm{Cl}\\right)$ as follows:\n\n$\\mathrm{HOCl}(\\mathrm{aq})+\\mathrm{NH}_{3}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{NH}_{2} \\mathrm{Cl}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\quad K_{\\mathrm{eq}}=1.47 \\times 10^{11}$\n\nOne part per million ( $\\mathrm{ppm}$ ) is a $\\mathrm{mg} \\mathrm{L}^{-1}$. A typical public swimming pool volume $(750,000 \\mathrm{~L})$ contains $75.0 \\mathrm{~L}$ of urine. The concentration of ammonia in $1 \\mathrm{~L}$ of urine is $0.200 \\mathrm{M}$. If the concentration of hypochlorous acid in pool water is $1.00 \\mathrm{ppm}$, determine the concentration (in ppm) of monochloramine $\\left(\\mathrm{NH}_{2} \\mathrm{Cl}\\right.$ ) in a typical public swimming pool. Assume the mass of $1 \\mathrm{~L}$ of pool water is $1 \\mathrm{~kg}$.\nA: 1.91\nB: 1.64\nC: 1.03\nD: 1.00\nE: 0.98\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe smell often associated with public swimming pools comes from chloramines. The reaction of hypochlorous acid with ammonia from human urine will produce monochloramine $\\left(\\mathrm{NH}_{2} \\mathrm{Cl}\\right)$ as follows:\n\n$\\mathrm{HOCl}(\\mathrm{aq})+\\mathrm{NH}_{3}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{NH}_{2} \\mathrm{Cl}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\quad K_{\\mathrm{eq}}=1.47 \\times 10^{11}$\n\nOne part per million ( $\\mathrm{ppm}$ ) is a $\\mathrm{mg} \\mathrm{L}^{-1}$. A typical public swimming pool volume $(750,000 \\mathrm{~L})$ contains $75.0 \\mathrm{~L}$ of urine. The concentration of ammonia in $1 \\mathrm{~L}$ of urine is $0.200 \\mathrm{M}$. If the concentration of hypochlorous acid in pool water is $1.00 \\mathrm{ppm}$, determine the concentration (in ppm) of monochloramine $\\left(\\mathrm{NH}_{2} \\mathrm{Cl}\\right.$ ) in a typical public swimming pool. Assume the mass of $1 \\mathrm{~L}$ of pool water is $1 \\mathrm{~kg}$.\n\nA: 1.91\nB: 1.64\nC: 1.03\nD: 1.00\nE: 0.98\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_900",
        "problem": "电催化合成氨有远端加氢和交替加氢两条路径。我国科学家研制催化剂 $\\mathrm{ZnS}-\\mathrm{rGO}$实现常温、常压, 高选择性合成氨, 结合实验与计算机模拟结果研究其反应路径如下图所示, 其中吸附在催化剂表面上的物种用*标注。下列有关说法正确的是\n\n[图1]\nA: 产物从催化剂上脱附会释放能量\nB: 该过程决速步骤的反应可表示为 $* \\mathrm{NN}+\\mathrm{H} \\rightarrow * \\mathrm{NNH}$\nC: 该反应在电解池的阳极上发生\nD: $\\mathrm{ZnS}-\\mathrm{rGO}$ 电催化合成氨交替加氢反应路径更容易发生\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n电催化合成氨有远端加氢和交替加氢两条路径。我国科学家研制催化剂 $\\mathrm{ZnS}-\\mathrm{rGO}$实现常温、常压, 高选择性合成氨, 结合实验与计算机模拟结果研究其反应路径如下图所示, 其中吸附在催化剂表面上的物种用*标注。下列有关说法正确的是\n\n[图1]\n\nA: 产物从催化剂上脱附会释放能量\nB: 该过程决速步骤的反应可表示为 $* \\mathrm{NN}+\\mathrm{H} \\rightarrow * \\mathrm{NNH}$\nC: 该反应在电解池的阳极上发生\nD: $\\mathrm{ZnS}-\\mathrm{rGO}$ 电催化合成氨交替加氢反应路径更容易发生\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-55.jpg?height=551&width=1256&top_left_y=593&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_710",
        "problem": "电位滴定法是在滴定过程中通过测量电位变化以确定滴定终点的方法, 当电极电位 (ERC)产生了突跃, 被测离子浓度产生突跃, 进而确定滴定终点。向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~A}$溶液 $\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right.$ 为二元弱酸) 中逐滴滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 二元酸 $\\mathrm{H}_{2} \\mathrm{~B}$ 溶液, 第一个电位突跃时消耗 $\\mathrm{amL}_{2} \\mathrm{~B}$ 溶液, 第二个电位突跃时消耗 $b m L \\mathrm{H}_{2} \\mathrm{~B}$ 溶液, 根据消耗 $\\mathrm{H}_{2} \\mathrm{~B}$ 溶液体积, 可判断 $\\mathrm{H}_{2} \\mathrm{~A}$ 和 $\\mathrm{H}_{2} \\mathrm{~B}$ 各级电离常数大小关系。下列说法错误的是\nA: 若 $\\mathrm{a}=5 \\mathrm{~mL} 、 \\mathrm{~b}=5 \\mathrm{~mL}, \\mathrm{H}_{2} \\mathrm{~B}$ 可能为强酸\nB: 若 $\\mathrm{a}=5 \\mathrm{~mL} 、 \\mathrm{~b}=15 \\mathrm{~mL}$, 则 $K_{\\mathrm{a}_{1}}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)>K_{\\mathrm{a}_{1}}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>K_{\\mathrm{a}_{2}}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)>K_{\\mathrm{a}_{2}}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\nC: 若 $\\mathrm{a}=5 \\mathrm{~mL} 、 \\mathrm{~b}=15 \\mathrm{~mL}$, 第一个电位突跃时, 溶液中 $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)+c\\left(\\mathrm{HB}^{-}\\right)+c\\left(\\mathrm{~B}^{2-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)$\nD: 若 $\\mathrm{a}=10 \\mathrm{~mL} 、 \\mathrm{~b}=10 \\mathrm{~mL}$, 则用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液滴定 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~B}$ 溶液,有两个电位突跃\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n电位滴定法是在滴定过程中通过测量电位变化以确定滴定终点的方法, 当电极电位 (ERC)产生了突跃, 被测离子浓度产生突跃, 进而确定滴定终点。向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~A}$溶液 $\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right.$ 为二元弱酸) 中逐滴滴加 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 二元酸 $\\mathrm{H}_{2} \\mathrm{~B}$ 溶液, 第一个电位突跃时消耗 $\\mathrm{amL}_{2} \\mathrm{~B}$ 溶液, 第二个电位突跃时消耗 $b m L \\mathrm{H}_{2} \\mathrm{~B}$ 溶液, 根据消耗 $\\mathrm{H}_{2} \\mathrm{~B}$ 溶液体积, 可判断 $\\mathrm{H}_{2} \\mathrm{~A}$ 和 $\\mathrm{H}_{2} \\mathrm{~B}$ 各级电离常数大小关系。下列说法错误的是\n\nA: 若 $\\mathrm{a}=5 \\mathrm{~mL} 、 \\mathrm{~b}=5 \\mathrm{~mL}, \\mathrm{H}_{2} \\mathrm{~B}$ 可能为强酸\nB: 若 $\\mathrm{a}=5 \\mathrm{~mL} 、 \\mathrm{~b}=15 \\mathrm{~mL}$, 则 $K_{\\mathrm{a}_{1}}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)>K_{\\mathrm{a}_{1}}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)>K_{\\mathrm{a}_{2}}\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)>K_{\\mathrm{a}_{2}}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\nC: 若 $\\mathrm{a}=5 \\mathrm{~mL} 、 \\mathrm{~b}=15 \\mathrm{~mL}$, 第一个电位突跃时, 溶液中 $c\\left(\\mathrm{Na}^{+}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+c\\left(\\mathrm{HA}^{-}\\right)+c\\left(\\mathrm{~A}^{2-}\\right)+c\\left(\\mathrm{HB}^{-}\\right)+c\\left(\\mathrm{~B}^{2-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~B}\\right)$\nD: 若 $\\mathrm{a}=10 \\mathrm{~mL} 、 \\mathrm{~b}=10 \\mathrm{~mL}$, 则用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液滴定 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~B}$ 溶液,有两个电位突跃\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_196",
        "problem": "A gaseous anesthetic with an unknown molecular formula is $85.63 \\%$ carbon and $14.37 \\%$ hydrogen by mass. What is the molecular formula of the unknown if $0.45 \\mathrm{~L}$ of the compound combusts with excess oxygen at $120.0^{\\circ} \\mathrm{C}$ at $72.93 \\mathrm{kPa}$ to form $2.70 \\mathrm{~L}$ of an equimolar mixture of carbon dioxide and water vapour?\nA: $\\mathrm{C}_{3} \\mathrm{H}_{6}$\nB: $\\mathrm{C}_{4} \\mathrm{H}_{8}$\nC: $\\mathrm{C}_{5} \\mathrm{H}_{10}$\nD: $\\mathrm{C}_{6} \\mathrm{H}_{12}$\nE: $\\mathrm{C}_{7} \\mathrm{H}_{14}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA gaseous anesthetic with an unknown molecular formula is $85.63 \\%$ carbon and $14.37 \\%$ hydrogen by mass. What is the molecular formula of the unknown if $0.45 \\mathrm{~L}$ of the compound combusts with excess oxygen at $120.0^{\\circ} \\mathrm{C}$ at $72.93 \\mathrm{kPa}$ to form $2.70 \\mathrm{~L}$ of an equimolar mixture of carbon dioxide and water vapour?\n\nA: $\\mathrm{C}_{3} \\mathrm{H}_{6}$\nB: $\\mathrm{C}_{4} \\mathrm{H}_{8}$\nC: $\\mathrm{C}_{5} \\mathrm{H}_{10}$\nD: $\\mathrm{C}_{6} \\mathrm{H}_{12}$\nE: $\\mathrm{C}_{7} \\mathrm{H}_{14}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_113",
        "problem": "Given the following heats of formation, what is the enthalpy of combustion for toluene shown in the following equation?\n\n$$\n\\mathrm{C}_{7} \\mathrm{H}_{8}(\\mathrm{l})+9 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow 7 \\mathrm{CO}_{2}(\\mathrm{~g})+4 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})\n$$\n\n| Substance | $\\Delta \\mathbf{H}_{\\mathbf{f}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ |\n| :---: | :---: |\n| $\\mathrm{C}_{7} \\mathrm{H}_{8}(\\mathrm{l})$ | +12.0 |\n| $\\mathrm{CO}_{2}(\\mathrm{~g})$ | -394 |\n\n\n| Substance | $\\Delta \\mathbf{H}_{\\mathbf{f}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ |\n| :---: | :---: |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -286 |\n| $\\mathrm{O}_{2}(\\mathrm{~g})$ | 0 |\nA: $680 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $692 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-692 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-3890 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nE: $-3914 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGiven the following heats of formation, what is the enthalpy of combustion for toluene shown in the following equation?\n\n$$\n\\mathrm{C}_{7} \\mathrm{H}_{8}(\\mathrm{l})+9 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow 7 \\mathrm{CO}_{2}(\\mathrm{~g})+4 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})\n$$\n\n| Substance | $\\Delta \\mathbf{H}_{\\mathbf{f}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ |\n| :---: | :---: |\n| $\\mathrm{C}_{7} \\mathrm{H}_{8}(\\mathrm{l})$ | +12.0 |\n| $\\mathrm{CO}_{2}(\\mathrm{~g})$ | -394 |\n\n\n| Substance | $\\Delta \\mathbf{H}_{\\mathbf{f}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ |\n| :---: | :---: |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -286 |\n| $\\mathrm{O}_{2}(\\mathrm{~g})$ | 0 |\n\nA: $680 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $692 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-692 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-3890 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nE: $-3914 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_971",
        "problem": "How many moles of gas are present in a 15.0-L scuba tank, if the pressure in the tank is $23.0 \\mathrm{MPa}$ and the temperature is $298 \\mathrm{~K}$ ? Assume the gas behaves ideally.\n$1 \\mathrm{MPa}=1 \\times 10^{3} \\mathrm{kPa}$\nA: $23 \\mathrm{~mol}$\nB: $72 \\mathrm{~mol}$\nC: $44 \\mathrm{~mol}$\nD: $14.1 \\mathrm{~mol}$\nE: $139 \\mathrm{~mol}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many moles of gas are present in a 15.0-L scuba tank, if the pressure in the tank is $23.0 \\mathrm{MPa}$ and the temperature is $298 \\mathrm{~K}$ ? Assume the gas behaves ideally.\n$1 \\mathrm{MPa}=1 \\times 10^{3} \\mathrm{kPa}$\n\nA: $23 \\mathrm{~mol}$\nB: $72 \\mathrm{~mol}$\nC: $44 \\mathrm{~mol}$\nD: $14.1 \\mathrm{~mol}$\nE: $139 \\mathrm{~mol}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_262",
        "problem": "Calculate the mass of phenylmagnesium bromide $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{MgBr}\\right)$ required to produce $12.73 \\mathrm{~g}$ of sodium tetraphenylborate (molar mass $342.2 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ).\nAnalysis for potassium using tetraphenylborate is accomplished using an indirect method. Firstly, the concentration of a solution of mercury(II) nitrate is determined by reaction with an ammonium thiocyanate $\\left(\\mathrm{NH}_{4} \\mathrm{SCN}\\right)$ solution of known concentration, as shown:\n\n$\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}+2 \\mathrm{NH}_{4} \\mathrm{SCN} \\rightarrow \\mathrm{Hg}(\\mathrm{SCN})_{2}+2 \\mathrm{NH}_{4} \\mathrm{NO}_{3}$\n\n$17.53 \\mathrm{~mL}$ of $0.1511 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NH}_{4} \\mathrm{SCN}$ are required for complete reaction with $25.00 \\mathrm{~mL}$ of a $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the mass of phenylmagnesium bromide $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{MgBr}\\right)$ required to produce $12.73 \\mathrm{~g}$ of sodium tetraphenylborate (molar mass $342.2 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ).\nAnalysis for potassium using tetraphenylborate is accomplished using an indirect method. Firstly, the concentration of a solution of mercury(II) nitrate is determined by reaction with an ammonium thiocyanate $\\left(\\mathrm{NH}_{4} \\mathrm{SCN}\\right)$ solution of known concentration, as shown:\n\n$\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}+2 \\mathrm{NH}_{4} \\mathrm{SCN} \\rightarrow \\mathrm{Hg}(\\mathrm{SCN})_{2}+2 \\mathrm{NH}_{4} \\mathrm{NO}_{3}$\n\n$17.53 \\mathrm{~mL}$ of $0.1511 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NH}_{4} \\mathrm{SCN}$ are required for complete reaction with $25.00 \\mathrm{~mL}$ of a $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_814",
        "problem": "维纶(聚乙烯醇缩甲醛纤维)可用于生产服装、绳索等。其合成路线如图所示(反应条件已略去):\n\n[图1]\n\n下列说法正确的是\nA: 反应(1)是加聚反应\nB: 高分子 $\\mathrm{B}$ 的链节中含有两种官能团\nC: 通过质谱法测定高分子 $\\mathrm{B}$ 的平均相对分子质量, 可得其聚合度\nD: [图2]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n维纶(聚乙烯醇缩甲醛纤维)可用于生产服装、绳索等。其合成路线如图所示(反应条件已略去):\n\n[图1]\n\n下列说法正确的是\n\nA: 反应(1)是加聚反应\nB: 高分子 $\\mathrm{B}$ 的链节中含有两种官能团\nC: 通过质谱法测定高分子 $\\mathrm{B}$ 的平均相对分子质量, 可得其聚合度\nD: [图2]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-23.jpg?height=163&width=1453&top_left_y=1089&top_left_x=336",
            "https://i.postimg.cc/50gY2pck/image.png",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-23.jpg?height=240&width=1000&top_left_y=2144&top_left_x=722",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-23.jpg?height=157&width=1359&top_left_y=2400&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_858",
        "problem": "根据下列实验操作和现象所得到的结论正确的是\n\n| 选项 | 实验操作和现象 | 结论 |\n| :--- | :--- | :--- |\n|  $\\mathrm{A}$| 向含酚酞的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中, 加入少量的 $\\mathrm{BaCl}_{2}$ 固体, 溶 <br> 液的红色变浅 | $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中存在 <br> 水解平衡 |\n| $\\mathrm{B}$ | 将铜片与锌片用导线连接后, 插入稀硫酸中, 铜片上有气 <br> 泡产生 | 铜将硫酸还原产生 <br> 氢气 |\n| $\\mathrm{C}$ | 向浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KCl}$ 和 $\\mathrm{KI}$ 混合溶液中滴加 2 滴 0.1 <br> $\\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{AgNO} \\mathrm{O}_{3}$ 溶液, 振荡, 沉淀呈黄色 <br> $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgI})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgCl})$ |  |\n| $\\mathrm{D}$ | 室温下, 用 $\\mathrm{pH}$ 试纸测得: $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液的 $\\mathrm{pH}$ 约为 9, <br> $\\mathrm{NaNO}_{2}$ 溶液的 $\\mathrm{pH}$ 约为 8 | $\\mathrm{HNO}_{2}$ 电离 $\\mathrm{H}^{+}$的能 <br> 力比 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 的强 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n根据下列实验操作和现象所得到的结论正确的是\n\n| 选项 | 实验操作和现象 | 结论 |\n| :--- | :--- | :--- |\n|  $\\mathrm{A}$| 向含酚酞的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中, 加入少量的 $\\mathrm{BaCl}_{2}$ 固体, 溶 <br> 液的红色变浅 | $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液中存在 <br> 水解平衡 |\n| $\\mathrm{B}$ | 将铜片与锌片用导线连接后, 插入稀硫酸中, 铜片上有气 <br> 泡产生 | 铜将硫酸还原产生 <br> 氢气 |\n| $\\mathrm{C}$ | 向浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KCl}$ 和 $\\mathrm{KI}$ 混合溶液中滴加 2 滴 0.1 <br> $\\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{AgNO} \\mathrm{O}_{3}$ 溶液, 振荡, 沉淀呈黄色 <br> $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgI})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgCl})$ |  |\n| $\\mathrm{D}$ | 室温下, 用 $\\mathrm{pH}$ 试纸测得: $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液的 $\\mathrm{pH}$ 约为 9, <br> $\\mathrm{NaNO}_{2}$ 溶液的 $\\mathrm{pH}$ 约为 8 | $\\mathrm{HNO}_{2}$ 电离 $\\mathrm{H}^{+}$的能 <br> 力比 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 的强 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_507",
        "problem": "已知 $\\mathrm{H}_{2} \\mathrm{R}$ 为二元弱酸, $K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)=5.4 \\times 10^{-2}, K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)=5.4 \\times 10^{-5}$ 。室温下, 下列说法不正确的是\nA: $0.1 \\mathrm{~mol} / \\mathrm{LNaHR}$ 溶液 $\\mathrm{pH}<7$\nB: 用 $\\mathrm{NaOH}$ 溶液中和一定量的 $\\mathrm{H}_{2} \\mathrm{R}$ 溶液至呈中性时, 溶液中 $c\\left(\\mathrm{HR}^{-}\\right)<c\\left(\\mathrm{R}^{2-}\\right)$\nC: $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{H}_{2} \\mathrm{R}$ 溶液: $0.2 \\mathrm{~mol} / \\mathrm{L}+c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)+c\\left(\\mathrm{HR}^{-}\\right)$\nD: 将 $0.01 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{H}_{2} \\mathrm{R}$ 溶液与 $\\mathrm{pH}=12$ 的 $\\mathrm{NaOH}$ 溶液混合恰好中和时, 消耗 $\\mathrm{NaOH}$溶液体积大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知 $\\mathrm{H}_{2} \\mathrm{R}$ 为二元弱酸, $K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)=5.4 \\times 10^{-2}, K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)=5.4 \\times 10^{-5}$ 。室温下, 下列说法不正确的是\n\nA: $0.1 \\mathrm{~mol} / \\mathrm{LNaHR}$ 溶液 $\\mathrm{pH}<7$\nB: 用 $\\mathrm{NaOH}$ 溶液中和一定量的 $\\mathrm{H}_{2} \\mathrm{R}$ 溶液至呈中性时, 溶液中 $c\\left(\\mathrm{HR}^{-}\\right)<c\\left(\\mathrm{R}^{2-}\\right)$\nC: $0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{H}_{2} \\mathrm{R}$ 溶液: $0.2 \\mathrm{~mol} / \\mathrm{L}+c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{R}\\right)+c\\left(\\mathrm{HR}^{-}\\right)$\nD: 将 $0.01 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{H}_{2} \\mathrm{R}$ 溶液与 $\\mathrm{pH}=12$ 的 $\\mathrm{NaOH}$ 溶液混合恰好中和时, 消耗 $\\mathrm{NaOH}$溶液体积大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_178",
        "problem": "What is the correct relationship between the following two organic substances?\n[figure1]\nA: non-superimposable mirror images\nB: identical\nC: Structural, non-geometric isomers\nD: hydrocarbons with different molecular formulae\nE: geometric isomers\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the correct relationship between the following two organic substances?\n[figure1]\n\nA: non-superimposable mirror images\nB: identical\nC: Structural, non-geometric isomers\nD: hydrocarbons with different molecular formulae\nE: geometric isomers\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_c5d3bbe4814b746daee8g-2.jpg?height=172&width=672&top_left_y=286&top_left_x=1796"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_958",
        "problem": "二氧化碳的再利用是实现温室气体减排的重要途径之一, 在稀 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ 中利用电催化可将 $\\mathrm{CO}_{2}$ 同时转化为多种燃料, 其原理如图所示。下列说法正确的是\n\n[图1]\nA: 离子交换膜为阳离子交换膜\nB: 一段时间后，阴极区溶液质量会减少\nC: $\\mathrm{Cu}$ 电极上的电极反应式为 $2 \\mathrm{H}_{2} \\mathrm{O}-4 \\mathrm{e}^{-}=\\mathrm{O}_{2} \\uparrow+4 \\mathrm{H}^{+}$\nD: 若阴极只生成 $0.15 \\mathrm{molCO}$ 和 $0.35 \\mathrm{~mol} \\mathrm{HCOOH}$, 则电路中转移电子的物质的量为 $0.50 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n二氧化碳的再利用是实现温室气体减排的重要途径之一, 在稀 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ 中利用电催化可将 $\\mathrm{CO}_{2}$ 同时转化为多种燃料, 其原理如图所示。下列说法正确的是\n\n[图1]\n\nA: 离子交换膜为阳离子交换膜\nB: 一段时间后，阴极区溶液质量会减少\nC: $\\mathrm{Cu}$ 电极上的电极反应式为 $2 \\mathrm{H}_{2} \\mathrm{O}-4 \\mathrm{e}^{-}=\\mathrm{O}_{2} \\uparrow+4 \\mathrm{H}^{+}$\nD: 若阴极只生成 $0.15 \\mathrm{molCO}$ 和 $0.35 \\mathrm{~mol} \\mathrm{HCOOH}$, 则电路中转移电子的物质的量为 $0.50 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-084.jpg?height=452&width=668&top_left_y=2124&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_136",
        "problem": "Calculate $\\Delta G^{\\circ}$ at $25^{\\circ} \\mathrm{C}$ for the reaction given the data below $4 \\mathrm{HCl}(\\mathrm{g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow 2 \\mathrm{Cl}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}^{\\circ}=-114.4 \\mathrm{~kJ}$ $S_{C_{2}}^{O}=223.0 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\quad S_{\\mathrm{H}_{2} \\mathrm{O}}^{O}=188.7 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$, $S_{O_{2}}^{O}=205.0 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\quad S_{H C l}^{O}=186.8 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$\nA: $+14.4 \\mathrm{~kJ}$\nB: $-111.18 \\mathrm{~kJ}$\nC: $+3105.6 \\mathrm{~kJ}$\nD: $+38 \\mathrm{~kJ}$\nE: $-76.0 \\mathrm{~kJ}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCalculate $\\Delta G^{\\circ}$ at $25^{\\circ} \\mathrm{C}$ for the reaction given the data below $4 \\mathrm{HCl}(\\mathrm{g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow 2 \\mathrm{Cl}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}^{\\circ}=-114.4 \\mathrm{~kJ}$ $S_{C_{2}}^{O}=223.0 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\quad S_{\\mathrm{H}_{2} \\mathrm{O}}^{O}=188.7 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$, $S_{O_{2}}^{O}=205.0 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} \\quad S_{H C l}^{O}=186.8 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}$\n\nA: $+14.4 \\mathrm{~kJ}$\nB: $-111.18 \\mathrm{~kJ}$\nC: $+3105.6 \\mathrm{~kJ}$\nD: $+38 \\mathrm{~kJ}$\nE: $-76.0 \\mathrm{~kJ}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_705",
        "problem": "药物 E 具有抗癌抑菌功效, 其合成路线如下:\n[图1]\n\n下列说法正确的是\nA: 反应(1)实际经历了两步反应, 分别是加成反应和消去反应\nB: 反应(1)的目的是保护醛基不被氧化, 而反应(3)是为了脱去对醛基的保护\nC: 物质 $\\mathrm{D}$ 是非电解质\nD: 分子 $\\mathrm{E}$ 中所有碳原子不可能都共平面\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n药物 E 具有抗癌抑菌功效, 其合成路线如下:\n[图1]\n\n下列说法正确的是\n\nA: 反应(1)实际经历了两步反应, 分别是加成反应和消去反应\nB: 反应(1)的目的是保护醛基不被氧化, 而反应(3)是为了脱去对醛基的保护\nC: 物质 $\\mathrm{D}$ 是非电解质\nD: 分子 $\\mathrm{E}$ 中所有碳原子不可能都共平面\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-41.jpg?height=464&width=1302&top_left_y=246&top_left_x=342"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_853",
        "problem": "联氨 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 是二元弱碱, 联氨溶液中存在:\n\n$$\n\\begin{aligned}\n& \\mathrm{N}_{2} \\mathrm{H}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{H}_{5}^{+}+\\mathrm{OH}^{-} \\quad K_{\\mathrm{b} 1} \\\\\n& \\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{H}_{6}^{2+}+\\mathrm{OH}^{-} \\quad K_{\\mathrm{b} 2}\n\\end{aligned}\n$$\n\n$25^{\\circ} \\mathrm{C}$ 时, 溶液中 $\\lg c(\\mathrm{x})$ 与 $\\mathrm{pH}$ 的关系如图所示 $(\\mathrm{x}$ 表示含氮微粒, 已知 $\\lg 2=0.3)$ 。\n\n[图1]\n\n下列说法错误的是\nA: $K_{\\mathrm{b} 1}=8.0 \\times 10^{-7}$\nB: $\\mathrm{M}$ 点溶液中 $\\mathrm{pH}=0.3$\nC: 水的电离程度: $\\mathrm{P}>\\mathrm{N}>\\mathrm{M}$\nD: $\\mathrm{N}_{2} \\mathrm{H}_{5} \\mathrm{Cl}$ 溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{6}^{2+}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n联氨 $\\left(\\mathrm{N}_{2} \\mathrm{H}_{4}\\right)$ 是二元弱碱, 联氨溶液中存在:\n\n$$\n\\begin{aligned}\n& \\mathrm{N}_{2} \\mathrm{H}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{H}_{5}^{+}+\\mathrm{OH}^{-} \\quad K_{\\mathrm{b} 1} \\\\\n& \\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{H}_{6}^{2+}+\\mathrm{OH}^{-} \\quad K_{\\mathrm{b} 2}\n\\end{aligned}\n$$\n\n$25^{\\circ} \\mathrm{C}$ 时, 溶液中 $\\lg c(\\mathrm{x})$ 与 $\\mathrm{pH}$ 的关系如图所示 $(\\mathrm{x}$ 表示含氮微粒, 已知 $\\lg 2=0.3)$ 。\n\n[图1]\n\n下列说法错误的是\n\nA: $K_{\\mathrm{b} 1}=8.0 \\times 10^{-7}$\nB: $\\mathrm{M}$ 点溶液中 $\\mathrm{pH}=0.3$\nC: 水的电离程度: $\\mathrm{P}>\\mathrm{N}>\\mathrm{M}$\nD: $\\mathrm{N}_{2} \\mathrm{H}_{5} \\mathrm{Cl}$ 溶液中: $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{5}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{~N}_{2} \\mathrm{H}_{6}^{2+}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-041.jpg?height=488&width=642&top_left_y=750&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_659",
        "problem": "缓冲溶液体系是维持生命活动的基础。配制 $\\mathrm{pH}=5$ 的磷酸盐缓冲溶液的实验步骤如下:\n\n步骤 1: 称取 $2.4 \\mathrm{~g} \\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 固体, 在小烧杯中加 $10 \\mathrm{~mL}$ 水溶解, 静置。\n\n步骤 2：将步骤 1 所得溶液转移至于容量瓶中，定容至 $100 \\mathrm{~mL}$, 震荡, 静置。\n\n步骤 3: 取步骤 2 所得溶液 $45 \\mathrm{~mL}$, 向其中滴加某浓度 $\\mathrm{NaOH}$ 溶液, 至 $\\mathrm{pH}=5$, 溶液的体积恰好为 $50 \\mathrm{~mL}$ 。\n\n步骤 4: 将步骤 3 所得溶液分成两等份, 向其中一份滴加 $0.5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液,向另一份溶液中滴加 $0.5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸, 充分震荡, 测得两溶液的 $\\mathrm{pH}=5$ 。\n\n$\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 的电离平衡常数: $\\mathrm{K}_{\\mathrm{a} 1}=7.1 \\times 10^{-3} ; \\mathrm{K}_{\\mathrm{a} 2}=6.2 \\times 10^{-8} ; \\mathrm{K}_{\\mathrm{a} 3}=4.5 \\times 10^{-13}$ 。下列说法正确的是\nA: 步骤 2 所配制的溶液中: $c\\left(\\mathrm{NaH}_{2} \\mathrm{PO}_{4}\\right)=2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: 在 $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{4}\\right)<\\mathrm{c}\\left(\\mathrm{HPO}_{4}^{2-}\\right)$\nC: 在步骤 3 所得溶液中: $\\mathrm{c}\\left(\\mathrm{HPO}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{PO}_{4}^{3-}\\right)=0.18 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 从步骤 4 的实验数据可得出: $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 与 $\\mathrm{NaOH}$ 和 $\\mathrm{HCl}$ 均不反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n缓冲溶液体系是维持生命活动的基础。配制 $\\mathrm{pH}=5$ 的磷酸盐缓冲溶液的实验步骤如下:\n\n步骤 1: 称取 $2.4 \\mathrm{~g} \\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 固体, 在小烧杯中加 $10 \\mathrm{~mL}$ 水溶解, 静置。\n\n步骤 2：将步骤 1 所得溶液转移至于容量瓶中，定容至 $100 \\mathrm{~mL}$, 震荡, 静置。\n\n步骤 3: 取步骤 2 所得溶液 $45 \\mathrm{~mL}$, 向其中滴加某浓度 $\\mathrm{NaOH}$ 溶液, 至 $\\mathrm{pH}=5$, 溶液的体积恰好为 $50 \\mathrm{~mL}$ 。\n\n步骤 4: 将步骤 3 所得溶液分成两等份, 向其中一份滴加 $0.5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液,向另一份溶液中滴加 $0.5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸, 充分震荡, 测得两溶液的 $\\mathrm{pH}=5$ 。\n\n$\\mathrm{H}_{3} \\mathrm{PO}_{4}$ 的电离平衡常数: $\\mathrm{K}_{\\mathrm{a} 1}=7.1 \\times 10^{-3} ; \\mathrm{K}_{\\mathrm{a} 2}=6.2 \\times 10^{-8} ; \\mathrm{K}_{\\mathrm{a} 3}=4.5 \\times 10^{-13}$ 。下列说法正确的是\n\nA: 步骤 2 所配制的溶液中: $c\\left(\\mathrm{NaH}_{2} \\mathrm{PO}_{4}\\right)=2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: 在 $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{4}\\right)<\\mathrm{c}\\left(\\mathrm{HPO}_{4}^{2-}\\right)$\nC: 在步骤 3 所得溶液中: $\\mathrm{c}\\left(\\mathrm{HPO}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{PO}_{4}^{3-}\\right)=0.18 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 从步骤 4 的实验数据可得出: $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ 与 $\\mathrm{NaOH}$ 和 $\\mathrm{HCl}$ 均不反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_606",
        "problem": "下列说法错误的是\nA: 分子式为 $\\mathrm{C}_{10} \\mathrm{H}_{20} \\mathrm{O}_{2}$ 的有机物 $\\mathrm{A}$, 它能在酸性条件下水解生成 $\\mathrm{B}$ 和 C, 且 $\\mathrm{B}$ 在一定条件下能转化成 C, 则有机物 $\\mathrm{A}$ 的可能结构有 4 种\nB: 对二甲苯的二溴代物有 6 种\nC: [图1] 中最多 19 个碳原子共平面、最多 6 个 原子共直线\nD: $0.1 \\mathrm{~mol}$ 某醛与足量的银氨溶液反应, 产生银的质量为 $43.2 \\mathrm{~g}$, 则该醛一定为二元醛\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列说法错误的是\n\nA: 分子式为 $\\mathrm{C}_{10} \\mathrm{H}_{20} \\mathrm{O}_{2}$ 的有机物 $\\mathrm{A}$, 它能在酸性条件下水解生成 $\\mathrm{B}$ 和 C, 且 $\\mathrm{B}$ 在一定条件下能转化成 C, 则有机物 $\\mathrm{A}$ 的可能结构有 4 种\nB: 对二甲苯的二溴代物有 6 种\nC: [图1] 中最多 19 个碳原子共平面、最多 6 个 原子共直线\nD: $0.1 \\mathrm{~mol}$ 某醛与足量的银氨溶液反应, 产生银的质量为 $43.2 \\mathrm{~g}$, 则该醛一定为二元醛\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/Mp0rwWyD/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_658",
        "problem": "铜-铈氧化物 $\\left(\\mathrm{xCuO} \\cdot \\mathrm{yCeO}_{2}\\right)$ 可除去 $\\mathrm{H}_{2}$ 中少量 $\\mathrm{CO}$, 反应机理如图。 $\\mathrm{Ce}$ 是一种活泼金属, 价电子为 $4 \\mathrm{f}^{1} 5 \\mathrm{~d}^{1} 6 \\mathrm{~s}^{2}$, 步骤(i)中 $\\mathrm{Cu} 、 \\mathrm{Ce}$ 的化合价均发生变化。\n\n[图1]\n\n将 $n(\\mathrm{CO}): n\\left(\\mathrm{O}_{2}\\right) ： n\\left(\\mathrm{H}_{2}\\right) ： n\\left(\\mathrm{~N}_{2}\\right)=1: 1: 49: 49$ 的混合气体以一定流速通过装有 $\\mathrm{xCuO} \\cdot \\mathrm{yCeO}_{2}$ 催化剂的反应器, $\\mathrm{CO}$ 转化率随温度变化的曲线如图所示, 下列说法错误的是\n\n[图2]\nA: 反应 $\\mathrm{i}$ 中, $\\mathrm{Cu}$ 元素的化合价从 +2 变成 +1\nB: 反应 $i$ 中, $\\mathrm{Ce}$ 元素的化合价从 +4 变成 +2\nC: 当温度超过 $150^{\\circ} \\mathrm{C}$ 时, $\\mathrm{CO}$ 转化率明显下降, 可能是因为催化剂的催化活性下降\nD: 温度过高, 催化剂中 $\\mathrm{Cu}^{2+}$ (或 $\\mathrm{Cu}^{+}$)可能被 $\\mathrm{H}_{2}$ (或 $\\mathrm{CO}$ )还原为 $\\mathrm{Cu}$ 造成催化活性下降\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n铜-铈氧化物 $\\left(\\mathrm{xCuO} \\cdot \\mathrm{yCeO}_{2}\\right)$ 可除去 $\\mathrm{H}_{2}$ 中少量 $\\mathrm{CO}$, 反应机理如图。 $\\mathrm{Ce}$ 是一种活泼金属, 价电子为 $4 \\mathrm{f}^{1} 5 \\mathrm{~d}^{1} 6 \\mathrm{~s}^{2}$, 步骤(i)中 $\\mathrm{Cu} 、 \\mathrm{Ce}$ 的化合价均发生变化。\n\n[图1]\n\n将 $n(\\mathrm{CO}): n\\left(\\mathrm{O}_{2}\\right) ： n\\left(\\mathrm{H}_{2}\\right) ： n\\left(\\mathrm{~N}_{2}\\right)=1: 1: 49: 49$ 的混合气体以一定流速通过装有 $\\mathrm{xCuO} \\cdot \\mathrm{yCeO}_{2}$ 催化剂的反应器, $\\mathrm{CO}$ 转化率随温度变化的曲线如图所示, 下列说法错误的是\n\n[图2]\n\nA: 反应 $\\mathrm{i}$ 中, $\\mathrm{Cu}$ 元素的化合价从 +2 变成 +1\nB: 反应 $i$ 中, $\\mathrm{Ce}$ 元素的化合价从 +4 变成 +2\nC: 当温度超过 $150^{\\circ} \\mathrm{C}$ 时, $\\mathrm{CO}$ 转化率明显下降, 可能是因为催化剂的催化活性下降\nD: 温度过高, 催化剂中 $\\mathrm{Cu}^{2+}$ (或 $\\mathrm{Cu}^{+}$)可能被 $\\mathrm{H}_{2}$ (或 $\\mathrm{CO}$ )还原为 $\\mathrm{Cu}$ 造成催化活性下降\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-026.jpg?height=528&width=877&top_left_y=1272&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-026.jpg?height=428&width=666&top_left_y=2053&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1121",
        "problem": "The 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nWhen you are bitten by an ant it does not inject you with all of its methanoic acid but keeps a little in reserve. Assuming a 'typical ant' injects $80 \\%$ of its methanoic acid, what is the total volume of pure methanoic acid contained in a 'typical ant'?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nWhen you are bitten by an ant it does not inject you with all of its methanoic acid but keeps a little in reserve. Assuming a 'typical ant' injects $80 \\%$ of its methanoic acid, what is the total volume of pure methanoic acid contained in a 'typical ant'?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~cm}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_4408eb29f8e936fbb54dg-05.jpg?height=622&width=654&top_left_y=237&top_left_x=1089"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~cm}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_762",
        "problem": "化学家们合成了如图所示的一系列的星烷, 如三星烷、四星烷、五星烷等。下列说法正确的是\n[图1]\n\n三星烷四星烷五星烷\nA: 它们之间互为同系物\nB: 六星烷的化学式为 $\\mathrm{C}_{18} \\mathrm{H}_{24}$\nC: 三星烷与丙苯互为同分异构体, 四星烷与 $\\square$ 互为同分异构体\nD: 它们的一氯代物均只有两种, 而三星烷的二氯代物有四种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化学家们合成了如图所示的一系列的星烷, 如三星烷、四星烷、五星烷等。下列说法正确的是\n[图1]\n\n三星烷四星烷五星烷\n\nA: 它们之间互为同系物\nB: 六星烷的化学式为 $\\mathrm{C}_{18} \\mathrm{H}_{24}$\nC: 三星烷与丙苯互为同分异构体, 四星烷与 $\\square$ 互为同分异构体\nD: 它们的一氯代物均只有两种, 而三星烷的二氯代物有四种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-81.jpg?height=226&width=748&top_left_y=1394&top_left_x=346",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-81.jpg?height=194&width=268&top_left_y=2247&top_left_x=560",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-82.jpg?height=83&width=720&top_left_y=163&top_left_x=999"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1142",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\n Give the full electron configurations of calcium in s, p, d notation.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\n Give the full electron configurations of calcium in s, p, d notation.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1215",
        "problem": "A feathery, greenish solid precipitate can be observed if chlorine gas is bubbled into water close to its freezing point. Similar precipitates form with other gases such as methane and noble gases. These materials are interesting because vast quantities of the so-called methane-hydrates are supposed to exist in nature (comparable in quantity with other natural gas deposits).\n\nThese precipitates all have related structures. The molecules of water just above its freezing point form a hydrogen-bonded structure. The gas molecules stabilize this framework by filling in the rather large cavities in the water structure forming clathrates. The crystals of chlorine and methane clathrates have the same structure. Their main characteristics are dodecahedra formed from 20 water molecules. The unit cell of the crystal can be thought as a body-centered cubic arrangement built from these dodecahedra which are almost spherical objects. The dodecahedra are connected via additional water molecules located on the faces of the unit cell. Two water molecules can be found on each face of the unit cell. The unit cell has an edge dimension of $1.182 \\mathrm{~nm}$. There are two types of cavities in this structure. One is the internal space in the dodecahedra (A). These are somewhat smaller than the other type of voids (B), of which there are 6 for each unit cell.\n\nHow many type A cavities can be found in a unit cell?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nA feathery, greenish solid precipitate can be observed if chlorine gas is bubbled into water close to its freezing point. Similar precipitates form with other gases such as methane and noble gases. These materials are interesting because vast quantities of the so-called methane-hydrates are supposed to exist in nature (comparable in quantity with other natural gas deposits).\n\nThese precipitates all have related structures. The molecules of water just above its freezing point form a hydrogen-bonded structure. The gas molecules stabilize this framework by filling in the rather large cavities in the water structure forming clathrates. The crystals of chlorine and methane clathrates have the same structure. Their main characteristics are dodecahedra formed from 20 water molecules. The unit cell of the crystal can be thought as a body-centered cubic arrangement built from these dodecahedra which are almost spherical objects. The dodecahedra are connected via additional water molecules located on the faces of the unit cell. Two water molecules can be found on each face of the unit cell. The unit cell has an edge dimension of $1.182 \\mathrm{~nm}$. There are two types of cavities in this structure. One is the internal space in the dodecahedra (A). These are somewhat smaller than the other type of voids (B), of which there are 6 for each unit cell.\n\nHow many type A cavities can be found in a unit cell?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_112",
        "problem": "In which of the following situations could hydrogen bonding occur between $\\mathrm{H}_{2} \\mathrm{O}$ and the solute?\nA: ammonia gas dissolved in water\nB: hydrogen gas dissolved in water\nC: carbon dioxide gas in water\nD: methane gas dissolved in water\nE: hydrogen sulfide gas dissolved in water\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn which of the following situations could hydrogen bonding occur between $\\mathrm{H}_{2} \\mathrm{O}$ and the solute?\n\nA: ammonia gas dissolved in water\nB: hydrogen gas dissolved in water\nC: carbon dioxide gas in water\nD: methane gas dissolved in water\nE: hydrogen sulfide gas dissolved in water\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_522",
        "problem": "下列离子方程式书写正确的是\nA: 用情性电极电解 $\\mathrm{MgCl}_{2}$ 溶液: $2 \\mathrm{Cl}^{-}+2 \\mathrm{H}_{2} \\mathrm{O} \\xlongequal{\\text { 电解 }} 2 \\mathrm{OH}^{-}+\\mathrm{H}_{2} \\uparrow+\\mathrm{Cl}_{2} \\uparrow$\nB: 用银氨溶液检验乙醛中的醛基: $\\mathrm{CH}_{3} \\mathrm{CHO}+2 \\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}+2 \\mathrm{OH}^{-} \\xrightarrow{\\text { 水浴加热 }}$ $\\mathrm{CH}_{3} \\mathrm{COO}^{-}+\\mathrm{NH}_{4}^{+}+3 \\mathrm{NH}_{3}+2 \\mathrm{Ag} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}$\nC: [图1]\nD: 向含 $\\mathrm{NaOH}$ 的 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 悬浊液中加入乙醛溶液并加热 $\\mathrm{CH}_{3} \\mathrm{CHO}+2 \\mathrm{Cu}(\\mathrm{OH})_{2}+\\mathrm{OH}^{-}$ $$ \\xrightarrow{\\Delta} \\mathrm{CH}_{3} \\mathrm{COO}^{-}+\\mathrm{Cu}_{2} \\mathrm{O} \\downarrow+3 \\mathrm{H}_{2} \\mathrm{O} $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列离子方程式书写正确的是\n\nA: 用情性电极电解 $\\mathrm{MgCl}_{2}$ 溶液: $2 \\mathrm{Cl}^{-}+2 \\mathrm{H}_{2} \\mathrm{O} \\xlongequal{\\text { 电解 }} 2 \\mathrm{OH}^{-}+\\mathrm{H}_{2} \\uparrow+\\mathrm{Cl}_{2} \\uparrow$\nB: 用银氨溶液检验乙醛中的醛基: $\\mathrm{CH}_{3} \\mathrm{CHO}+2 \\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}+2 \\mathrm{OH}^{-} \\xrightarrow{\\text { 水浴加热 }}$ $\\mathrm{CH}_{3} \\mathrm{COO}^{-}+\\mathrm{NH}_{4}^{+}+3 \\mathrm{NH}_{3}+2 \\mathrm{Ag} \\downarrow+\\mathrm{H}_{2} \\mathrm{O}$\nC: [图1]\nD: 向含 $\\mathrm{NaOH}$ 的 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 悬浊液中加入乙醛溶液并加热 $\\mathrm{CH}_{3} \\mathrm{CHO}+2 \\mathrm{Cu}(\\mathrm{OH})_{2}+\\mathrm{OH}^{-}$ $$ \\xrightarrow{\\Delta} \\mathrm{CH}_{3} \\mathrm{COO}^{-}+\\mathrm{Cu}_{2} \\mathrm{O} \\downarrow+3 \\mathrm{H}_{2} \\mathrm{O} $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/Cx8Z9kNX/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1202",
        "problem": "Bridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate $E$ for the cell at $25{ }^{\\circ} \\mathrm{C}$ for the following conditions :\n\n$\\left[\\mathrm{Fe}^{2+}\\right]=0.015 \\mathrm{M}, p H_{\\text {right-hand half-cell }}=9.00, p\\left(\\mathrm{O}_{2}\\right)=0.700$ bar.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate $E$ for the cell at $25{ }^{\\circ} \\mathrm{C}$ for the following conditions :\n\n$\\left[\\mathrm{Fe}^{2+}\\right]=0.015 \\mathrm{M}, p H_{\\text {right-hand half-cell }}=9.00, p\\left(\\mathrm{O}_{2}\\right)=0.700$ bar.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of V, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-345.jpg?height=951&width=1445&top_left_y=595&top_left_x=314"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "V"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_974",
        "problem": "Which group of elements contains no metals or metalloids?\nA: group 13\nB: group 14\nC: group 15\nD: group 16\nE: group 17\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich group of elements contains no metals or metalloids?\n\nA: group 13\nB: group 14\nC: group 15\nD: group 16\nE: group 17\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_370",
        "problem": "Which molecule has a dipole moment of zero?\nA: $\\mathrm{CO}$\nB: $\\mathrm{CO}_{2}$\nC: $\\mathrm{CH}_{2} \\mathrm{O}$\nD: $\\mathrm{CH}_{3} \\mathrm{OH}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich molecule has a dipole moment of zero?\n\nA: $\\mathrm{CO}$\nB: $\\mathrm{CO}_{2}$\nC: $\\mathrm{CH}_{2} \\mathrm{O}$\nD: $\\mathrm{CH}_{3} \\mathrm{OH}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_197",
        "problem": "In which of the following four compounds is intermolecular hydrog(I) 2-propanol, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHOH}$\n\n(II) triethylamine, $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{3} \\mathrm{~N}$\n\n(III) dimethyl ether, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{O}$\n\n(IV) $n$-butylamine, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$enbonding present?\nA: (I), (III) and (IV)\nB: (I) and (III)\nC: (I) only\nD: (I) and (IV)\nE: (II) and (IV)\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn which of the following four compounds is intermolecular hydrog(I) 2-propanol, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{CHOH}$\n\n(II) triethylamine, $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)_{3} \\mathrm{~N}$\n\n(III) dimethyl ether, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{O}$\n\n(IV) $n$-butylamine, $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{NH}_{2}$enbonding present?\n\nA: (I), (III) and (IV)\nB: (I) and (III)\nC: (I) only\nD: (I) and (IV)\nE: (II) and (IV)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1265",
        "problem": "Distribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of $\\mathrm{PO}_{4}{ }_{4}^{3-}$ and $\\mathrm{SiO}_{4}{ }^{4-}$ in alkaline extract\n\nBoth phosphate and silicate ions can react with molybdate in alkaline solution, producing the yellow molybdophosphate and molybdatosilicate. Further reduction with ascorbic acid produces intense color molybdenum blue compounds. Both complexes exhibit maximum absorption at $800 \\mathrm{~nm}$. Addition of tartaric acid helps preventing interference from silicate in the determination of phosphate.\n\nTwo series of phosphate standard are treated with and without tartaric acid whereas a series of silicate standard is not treated with tartaric acid. Linear equations obtained from those calibration curves are as follows:\n\n| Conditions | Linear equations |\n| :---: | :---: |\n| Phosphate with and without tartaric acid | $y=6720 \\mathrm{x}_{1}$ |\n| Silicate without tartaric acid | $\\mathrm{y}=868 \\mathrm{x}_{2}$ |\n\n$\\mathrm{y}$ is absorbance at $800 \\mathrm{~nm}$,\n\n$\\mathrm{x}_{1}$ is concentration of phosphate as $\\mathrm{mol} \\mathrm{dm}^{-3}$,\n\n$\\mathrm{x}_{2}$ is concentration of silicate as $\\mathrm{mol} \\mathrm{dm}^{-3}$\n\nAbsorbance at $800 \\mathrm{~nm}$ of the alkaline fraction of the soil extract after treated with and without tartaric acid are 0.267 and 0.510 , respectively.Calculate the silicate concentration from the soil sample in $t$ the alkaline fraction in $\\mathrm{mol} \\mathrm{~dm ^ { - 3 }}$ and calculate the corresponding silicon in $\\mathrm{mg} \\mathrm{~dm}^{-3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nDistribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of $\\mathrm{PO}_{4}{ }_{4}^{3-}$ and $\\mathrm{SiO}_{4}{ }^{4-}$ in alkaline extract\n\nBoth phosphate and silicate ions can react with molybdate in alkaline solution, producing the yellow molybdophosphate and molybdatosilicate. Further reduction with ascorbic acid produces intense color molybdenum blue compounds. Both complexes exhibit maximum absorption at $800 \\mathrm{~nm}$. Addition of tartaric acid helps preventing interference from silicate in the determination of phosphate.\n\nTwo series of phosphate standard are treated with and without tartaric acid whereas a series of silicate standard is not treated with tartaric acid. Linear equations obtained from those calibration curves are as follows:\n\n| Conditions | Linear equations |\n| :---: | :---: |\n| Phosphate with and without tartaric acid | $y=6720 \\mathrm{x}_{1}$ |\n| Silicate without tartaric acid | $\\mathrm{y}=868 \\mathrm{x}_{2}$ |\n\n$\\mathrm{y}$ is absorbance at $800 \\mathrm{~nm}$,\n\n$\\mathrm{x}_{1}$ is concentration of phosphate as $\\mathrm{mol} \\mathrm{dm}^{-3}$,\n\n$\\mathrm{x}_{2}$ is concentration of silicate as $\\mathrm{mol} \\mathrm{dm}^{-3}$\n\nAbsorbance at $800 \\mathrm{~nm}$ of the alkaline fraction of the soil extract after treated with and without tartaric acid are 0.267 and 0.510 , respectively.\n\nproblem:\nCalculate the silicate concentration from the soil sample in $t$ the alkaline fraction in $\\mathrm{mol} \\mathrm{~dm ^ { - 3 }}$ and calculate the corresponding silicon in $\\mathrm{mg} \\mathrm{~dm}^{-3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mg} \\mathrm{~dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mg} \\mathrm{~dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1332",
        "problem": "One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\n\n\nA common defect in this crystal is some copper atoms missing with the oxygen lattice unchanged. The composition of one such crystal sample was studied, and $0.2 \\%$ of all copper atoms were found to be in oxidation state +2 .What percentage of normal copper sites are empty in the crystal sample?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOne of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\n\n\nA common defect in this crystal is some copper atoms missing with the oxygen lattice unchanged. The composition of one such crystal sample was studied, and $0.2 \\%$ of all copper atoms were found to be in oxidation state +2 .\n\nproblem:\nWhat percentage of normal copper sites are empty in the crystal sample?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-139.jpg?height=410&width=1236&top_left_y=660&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_3",
        "problem": "A student has crystallized acetylsalicylic acid from a mixture of ethanol and water. Which apparatus is best suited for isolation of the crystalline material? (The point of addition of the solid/liquid mixture is indicated by the arrow.)\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student has crystallized acetylsalicylic acid from a mixture of ethanol and water. Which apparatus is best suited for isolation of the crystalline material? (The point of addition of the solid/liquid mixture is indicated by the arrow.)\n\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-04.jpg?height=469&width=185&top_left_y=1332&top_left_x=276",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-04.jpg?height=512&width=187&top_left_y=1815&top_left_x=275",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-04.jpg?height=466&width=230&top_left_y=1339&top_left_x=709",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-04.jpg?height=507&width=141&top_left_y=1817&top_left_x=705"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_300",
        "problem": "A solution is prepared by dissolving 4.50 grams of solid $\\mathrm{NaOH}$ in $1.00 \\mathrm{~L}$ of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{2}(\\mathrm{aq})$ at $298 \\mathrm{~K}$ ? What is the $\\mathrm{pH}$ of this solution? Assume that the final volume is $1.00 \\mathrm{~L}$.\nIonization constants (at $298 \\mathrm{~K}$ )\n\n$\\mathrm{HNO}_{2}, K_{a}=4.5 \\times 10^{-4}$\n\n$\\mathrm{H}_{2} \\mathrm{O}, \\quad K_{w}=1.0 \\times 10^{-14}$\nA: 7.00\nB: 1.90\nC: 2.45\nD: 12.10\nE: 13.05\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA solution is prepared by dissolving 4.50 grams of solid $\\mathrm{NaOH}$ in $1.00 \\mathrm{~L}$ of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{2}(\\mathrm{aq})$ at $298 \\mathrm{~K}$ ? What is the $\\mathrm{pH}$ of this solution? Assume that the final volume is $1.00 \\mathrm{~L}$.\nIonization constants (at $298 \\mathrm{~K}$ )\n\n$\\mathrm{HNO}_{2}, K_{a}=4.5 \\times 10^{-4}$\n\n$\\mathrm{H}_{2} \\mathrm{O}, \\quad K_{w}=1.0 \\times 10^{-14}$\n\nA: 7.00\nB: 1.90\nC: 2.45\nD: 12.10\nE: 13.05\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_34",
        "problem": "At $400.0 \\mathrm{~K}$, the rate constant of a reaction is $3.2 \\times 10^{-2} \\mathrm{~s}^{-1}$. The activation energy of the reaction is $41.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. What is the rate constant at $410.0 \\mathrm{~K}$ ?\nA: $2.4 \\times 10^{-2} \\mathrm{~s}^{-1}$\nB: $3.2 \\times 10^{-2} \\mathrm{~s}^{-1}$\nC: $4.3 \\times 10^{-2} \\mathrm{~s}^{-1}$\nD: $3.9 \\times 10^{-1} \\mathrm{~s}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAt $400.0 \\mathrm{~K}$, the rate constant of a reaction is $3.2 \\times 10^{-2} \\mathrm{~s}^{-1}$. The activation energy of the reaction is $41.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. What is the rate constant at $410.0 \\mathrm{~K}$ ?\n\nA: $2.4 \\times 10^{-2} \\mathrm{~s}^{-1}$\nB: $3.2 \\times 10^{-2} \\mathrm{~s}^{-1}$\nC: $4.3 \\times 10^{-2} \\mathrm{~s}^{-1}$\nD: $3.9 \\times 10^{-1} \\mathrm{~s}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1245",
        "problem": "The undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe distribution ratio (D) of the acid HA depends mainly on its concentration.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe distribution ratio (D) of the acid HA depends mainly on its concentration.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-461.jpg?height=185&width=788&top_left_y=2209&top_left_x=634"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1207",
        "problem": "Regarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe logarithm of transmittance is proportional to the concentration of the absorbing compound.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nRegarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe logarithm of transmittance is proportional to the concentration of the absorbing compound.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_576",
        "problem": "用石墨作电极电解 $1 \\mathrm{LAgNO}_{3}$ 溶液, 当通电一段时间后, 两极均收集到 $22.4 \\mathrm{~L}$ (标准\n状况)气体, 则原溶液中 $\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$为\nA: $2 \\mathrm{~mol} / \\mathrm{L}$\nB: $1.5 \\mathrm{~mol} / \\mathrm{L}$\nC: $1 \\mathrm{~mol} / \\mathrm{L}$\nD: $0.5 \\mathrm{~mol} / \\mathrm{L}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用石墨作电极电解 $1 \\mathrm{LAgNO}_{3}$ 溶液, 当通电一段时间后, 两极均收集到 $22.4 \\mathrm{~L}$ (标准\n状况)气体, 则原溶液中 $\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$为\n\nA: $2 \\mathrm{~mol} / \\mathrm{L}$\nB: $1.5 \\mathrm{~mol} / \\mathrm{L}$\nC: $1 \\mathrm{~mol} / \\mathrm{L}$\nD: $0.5 \\mathrm{~mol} / \\mathrm{L}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_418",
        "problem": "室温下, 在 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 三元弱酸 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中, 微粒 $\\mathrm{H}_{3} \\mathrm{~A} 、 \\mathrm{H}_{2} \\mathrm{~A}^{-} 、 \\mathrm{HA}^{2-} 、 \\mathrm{~A}^{3-}$ 的物质\n的量分数 $\\delta(\\mathrm{X})$ 随 $\\mathrm{pH}$ 的变化如图所示。下列叙述错误的是\n\n[已知: $\\delta(\\mathrm{X})=\\frac{\\mathrm{c}(\\mathrm{X})}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{3-}\\right)}$ ]\n[图1]\nA: [图2]\nB: 物质的量浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KH}_{2} \\mathrm{~A} 、 \\mathrm{~K}_{2} \\mathrm{HA}$ 混合溶液中: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{3-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: 欲用 $\\mathrm{H}_{3} \\mathrm{~A}$ 和 $\\mathrm{K}_{2} \\mathrm{HA}$ 配制 $\\mathrm{pH}=7.2$ 的缓冲溶液 $\\left(\\mathrm{KH}_{2} \\mathrm{~A}\\right.$ 和 $\\mathrm{K}_{2} \\mathrm{HA}$ 的混合溶液), 则需 $\\mathrm{n}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right): \\mathrm{n}\\left(\\mathrm{K}_{2} \\mathrm{HA}\\right)=1: 2$\nD: 将 $\\mathrm{KOH}$ 溶液还滴加入到 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中, 反应 $\\mathrm{H}_{2} \\mathrm{~A}^{-}+\\mathrm{OH}=\\mathrm{HA}^{2-+}+\\mathrm{H}_{2} \\mathrm{O}$ 发生的 $\\mathrm{pH}$ 范围是 $4.7 \\sim 9.8$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n室温下, 在 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 三元弱酸 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中, 微粒 $\\mathrm{H}_{3} \\mathrm{~A} 、 \\mathrm{H}_{2} \\mathrm{~A}^{-} 、 \\mathrm{HA}^{2-} 、 \\mathrm{~A}^{3-}$ 的物质\n的量分数 $\\delta(\\mathrm{X})$ 随 $\\mathrm{pH}$ 的变化如图所示。下列叙述错误的是\n\n[已知: $\\delta(\\mathrm{X})=\\frac{\\mathrm{c}(\\mathrm{X})}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{3-}\\right)}$ ]\n[图1]\n\nA: [图2]\nB: 物质的量浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KH}_{2} \\mathrm{~A} 、 \\mathrm{~K}_{2} \\mathrm{HA}$ 混合溶液中: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{3-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: 欲用 $\\mathrm{H}_{3} \\mathrm{~A}$ 和 $\\mathrm{K}_{2} \\mathrm{HA}$ 配制 $\\mathrm{pH}=7.2$ 的缓冲溶液 $\\left(\\mathrm{KH}_{2} \\mathrm{~A}\\right.$ 和 $\\mathrm{K}_{2} \\mathrm{HA}$ 的混合溶液), 则需 $\\mathrm{n}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right): \\mathrm{n}\\left(\\mathrm{K}_{2} \\mathrm{HA}\\right)=1: 2$\nD: 将 $\\mathrm{KOH}$ 溶液还滴加入到 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中, 反应 $\\mathrm{H}_{2} \\mathrm{~A}^{-}+\\mathrm{OH}=\\mathrm{HA}^{2-+}+\\mathrm{H}_{2} \\mathrm{O}$ 发生的 $\\mathrm{pH}$ 范围是 $4.7 \\sim 9.8$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-069.jpg?height=420&width=759&top_left_y=378&top_left_x=357",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-069.jpg?height=54&width=785&top_left_y=847&top_left_x=407",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-069.jpg?height=52&width=782&top_left_y=1499&top_left_x=354",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-069.jpg?height=49&width=1376&top_left_y=1986&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_570",
        "problem": "近日, 科学家利用有机聚合物开发出了一种铝离子电池正极材料, 其容量高于石墨。其工作原理如图所示。下列说法正确的是\n\n[图1]\n\n[图2]\nA\n\n[图3]\n\n$\\mathrm{D}$\nA: 放电时, 电流方向为有机物电极 $\\rightarrow \\mathrm{b} \\rightarrow \\mathrm{a} \\rightarrow \\mathrm{Al}$ 电极\nB: 放电时, 有机物电极可能发生反应 [图4]\nC: 充电时, $\\mathrm{Al}$ 电极电势比有机物电极的高\nD: 充电时, 当外电路通过 $3 \\mathrm{~mol}$ 电子时, 两电极质量变化量差值为 $934.5 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n近日, 科学家利用有机聚合物开发出了一种铝离子电池正极材料, 其容量高于石墨。其工作原理如图所示。下列说法正确的是\n\n[图1]\n\n[图2]\nA\n\n[图3]\n\n$\\mathrm{D}$\n\nA: 放电时, 电流方向为有机物电极 $\\rightarrow \\mathrm{b} \\rightarrow \\mathrm{a} \\rightarrow \\mathrm{Al}$ 电极\nB: 放电时, 有机物电极可能发生反应 [图4]\nC: 充电时, $\\mathrm{Al}$ 电极电势比有机物电极的高\nD: 充电时, 当外电路通过 $3 \\mathrm{~mol}$ 电子时, 两电极质量变化量差值为 $934.5 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-018.jpg?height=474&width=539&top_left_y=160&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-018.jpg?height=279&width=391&top_left_y=266&top_left_x=935",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-018.jpg?height=280&width=480&top_left_y=268&top_left_x=1322",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-018.jpg?height=377&width=896&top_left_y=808&top_left_x=403"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_756",
        "problem": "常温下, 用 $\\mathrm{HCl}$ 或 $\\mathrm{NaOH}$ 调节 $0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCOONH} \\mathrm{H}_{4}$ 的 $\\mathrm{pH}$, 忽略 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 分解及溶液体积变化, 溶液中 $\\mathrm{c}\\left(\\mathrm{HCOO}^{-}\\right) 、 \\mathrm{c}(\\mathrm{HCOOH}) 、 \\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right) 、 \\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ 四种微粒 $\\mathrm{pX}$与 $\\mathrm{pH}$ 关系如下图所示 $[\\mathrm{pX}=-\\operatorname{lgc}(\\mathrm{X})]$ 。\n\n已知: (1) $x 、 y$ 点的坐标: $x(3.75,1.3), y(9.25,1.3)$; (2) $\\mathrm{K}_{\\mathrm{a}}(\\mathrm{HCOOH})>\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ 。下列说法正确的是\n\n[图1]\nA: 曲线 1 表示 $\\mathrm{pc}\\left(\\mathrm{HCOO}^{-}\\right)$随 $\\mathrm{pH}$ 的变化\nB: $\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ 的数量级为 $10^{-10}$\nC: $z$ 点的横坐标为 6.2\nD: $\\mathrm{pH}=7$ 时, 溶液中离子浓度大小: $\\mathrm{c}\\left(\\mathrm{HCOO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n常温下, 用 $\\mathrm{HCl}$ 或 $\\mathrm{NaOH}$ 调节 $0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCOONH} \\mathrm{H}_{4}$ 的 $\\mathrm{pH}$, 忽略 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 分解及溶液体积变化, 溶液中 $\\mathrm{c}\\left(\\mathrm{HCOO}^{-}\\right) 、 \\mathrm{c}(\\mathrm{HCOOH}) 、 \\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right) 、 \\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ 四种微粒 $\\mathrm{pX}$与 $\\mathrm{pH}$ 关系如下图所示 $[\\mathrm{pX}=-\\operatorname{lgc}(\\mathrm{X})]$ 。\n\n已知: (1) $x 、 y$ 点的坐标: $x(3.75,1.3), y(9.25,1.3)$; (2) $\\mathrm{K}_{\\mathrm{a}}(\\mathrm{HCOOH})>\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ 。下列说法正确的是\n\n[图1]\n\nA: 曲线 1 表示 $\\mathrm{pc}\\left(\\mathrm{HCOO}^{-}\\right)$随 $\\mathrm{pH}$ 的变化\nB: $\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$ 的数量级为 $10^{-10}$\nC: $z$ 点的横坐标为 6.2\nD: $\\mathrm{pH}=7$ 时, 溶液中离子浓度大小: $\\mathrm{c}\\left(\\mathrm{HCOO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-060.jpg?height=611&width=645&top_left_y=822&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_616",
        "problem": "甲烷化反应即为氢气和碳氧化物反应生成甲烷, 有利于实现碳循环利用。已知涉及的反应如下:\n\n反应I： $\\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}_{1}<0$\n\n反应II: $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta \\mathrm{H}_{2}<0$\n\n在 $360^{\\circ} \\mathrm{C}$ 时, 在固定容积的容器中进行上述反应, 平衡时 $\\mathrm{CO}$ 和 $\\mathrm{H}_{2}$ 的转化率及 $\\mathrm{CH}_{4}$ 和\n\n$\\mathrm{CO}_{2}$ 的产率随 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 变化的情况如图所示。若按 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}=3$ 向恒容容器内投料, 初始压强为 $\\mathrm{p}_{0}$ [已知: (1) $\\mathrm{CH}$, 的选择性 $=\\frac{\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)}{\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)+\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)} \\times 100 \\%$; (2)平衡常数用分压表示, 分压=总压 $\\times$ 物质的量分数]。下列说法错误的是\n\n[图1]\nA: 图中表示 $\\mathrm{CO}$ 转化率变化的曲线是 $\\mathrm{b}, \\mathrm{CH}_{4}$ 产率变化的曲线是 $\\mathrm{c}$\nB: 点 $\\mathrm{C}$ 通过改变温度达到点 $\\mathrm{A}$, 则 $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 三点温度由大到小为 $\\mathrm{C}>\\mathrm{B}>\\mathrm{A}$\nC: 若两个反应达到平衡时总压为 $\\frac{3 \\mathrm{p}_{0}}{4}, \\mathrm{CO}$ 的平衡转化率为 $\\mathrm{a}$, 则 $\\mathrm{CH}_{4}$ 的选择性 $=\\frac{50}{\\mathrm{a}} \\%$\nD: 若两个反应达到平衡时总压为 $\\frac{3 \\mathrm{p}_{0}}{4}, \\mathrm{CO}$ 的平衡转化率为 $\\mathrm{a}$, 则反应 $\\mathrm{I}$ 的 $\\mathrm{K}_{\\mathrm{p}}=\\frac{8}{(1+\\mathrm{a})^{3} \\times \\mathrm{p}_{0}^{2}}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n甲烷化反应即为氢气和碳氧化物反应生成甲烷, 有利于实现碳循环利用。已知涉及的反应如下:\n\n反应I： $\\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}_{1}<0$\n\n反应II: $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta \\mathrm{H}_{2}<0$\n\n在 $360^{\\circ} \\mathrm{C}$ 时, 在固定容积的容器中进行上述反应, 平衡时 $\\mathrm{CO}$ 和 $\\mathrm{H}_{2}$ 的转化率及 $\\mathrm{CH}_{4}$ 和\n\n$\\mathrm{CO}_{2}$ 的产率随 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 变化的情况如图所示。若按 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}=3$ 向恒容容器内投料, 初始压强为 $\\mathrm{p}_{0}$ [已知: (1) $\\mathrm{CH}$, 的选择性 $=\\frac{\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)}{\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)+\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)} \\times 100 \\%$; (2)平衡常数用分压表示, 分压=总压 $\\times$ 物质的量分数]。下列说法错误的是\n\n[图1]\n\nA: 图中表示 $\\mathrm{CO}$ 转化率变化的曲线是 $\\mathrm{b}, \\mathrm{CH}_{4}$ 产率变化的曲线是 $\\mathrm{c}$\nB: 点 $\\mathrm{C}$ 通过改变温度达到点 $\\mathrm{A}$, 则 $\\mathrm{A} 、 \\mathrm{~B} 、 \\mathrm{C}$ 三点温度由大到小为 $\\mathrm{C}>\\mathrm{B}>\\mathrm{A}$\nC: 若两个反应达到平衡时总压为 $\\frac{3 \\mathrm{p}_{0}}{4}, \\mathrm{CO}$ 的平衡转化率为 $\\mathrm{a}$, 则 $\\mathrm{CH}_{4}$ 的选择性 $=\\frac{50}{\\mathrm{a}} \\%$\nD: 若两个反应达到平衡时总压为 $\\frac{3 \\mathrm{p}_{0}}{4}, \\mathrm{CO}$ 的平衡转化率为 $\\mathrm{a}$, 则反应 $\\mathrm{I}$ 的 $\\mathrm{K}_{\\mathrm{p}}=\\frac{8}{(1+\\mathrm{a})^{3} \\times \\mathrm{p}_{0}^{2}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-032.jpg?height=457&width=605&top_left_y=2224&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-033.jpg?height=228&width=1016&top_left_y=1939&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1325",
        "problem": "A mixture of gases containing mainly carbon monoxide and hydrogen is produced by the reaction of alkanes with steam:\n\n$$\n\\begin{array}{ll}\n\\mathrm{CH}_{4}+1 / 2 \\mathrm{O}_{2} \\rightarrow \\mathrm{CO}+2 \\mathrm{H}_{2} & \\Delta H=36 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\\n\\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}+3 \\mathrm{H}_{2} & \\Delta H=216 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n\\end{array}\n$$\n\nTo produce hydrogen for the synthesis of ammonia, a mixture of $40.0 \\mathrm{~mol} \\mathrm{CO}$ and $40.0 \\mathrm{~mol}$ of hydrogen, $18.0 \\mathrm{~mol}$ of carbon dioxide and $2.0 \\mathrm{~mol}$ of nitrogen are in contact with $200.0 \\mathrm{~mol}$ of steam in a reactor where the conversion equilibrium is established.\n\n$\\mathrm{CO}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}_{2}+\\mathrm{H}_{2}$\n\nCalculate the number of moles of each gas leaving the reactor.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nA mixture of gases containing mainly carbon monoxide and hydrogen is produced by the reaction of alkanes with steam:\n\n$$\n\\begin{array}{ll}\n\\mathrm{CH}_{4}+1 / 2 \\mathrm{O}_{2} \\rightarrow \\mathrm{CO}+2 \\mathrm{H}_{2} & \\Delta H=36 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\\n\\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}+3 \\mathrm{H}_{2} & \\Delta H=216 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n\\end{array}\n$$\n\nTo produce hydrogen for the synthesis of ammonia, a mixture of $40.0 \\mathrm{~mol} \\mathrm{CO}$ and $40.0 \\mathrm{~mol}$ of hydrogen, $18.0 \\mathrm{~mol}$ of carbon dioxide and $2.0 \\mathrm{~mol}$ of nitrogen are in contact with $200.0 \\mathrm{~mol}$ of steam in a reactor where the conversion equilibrium is established.\n\n$\\mathrm{CO}+\\mathrm{H}_{2} \\mathrm{O} \\rightarrow \\mathrm{CO}_{2}+\\mathrm{H}_{2}$\n\nCalculate the number of moles of each gas leaving the reactor.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [The composition of $CO$ is, The composition of $CO_2$ is, The composition of $CH_4$ is, The composition of $N_2$ is, The composition of $H_2$ is, The composition of $H_2O$ is].\nTheir units are, in order, [mol, mol, mol, mol, mol, mol], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value, numerical value, numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "mol",
            "mol",
            "mol",
            "mol",
            "mol",
            "mol"
        ],
        "answer_sequence": [
            "The composition of $CO$ is",
            "The composition of $CO_2$ is",
            "The composition of $CH_4$ is",
            "The composition of $N_2$ is",
            "The composition of $H_2$ is",
            "The composition of $H_2O$ is"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV",
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_503",
        "problem": "常温下体积为 $1 \\mathrm{ml}$ 、浓度均为 $0.10 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{XOH}$ 和 $\\mathrm{X}_{2} \\mathrm{CO}_{3}$ 溶液分别加水稀释至体积为 $\\mathrm{V}, \\mathrm{pH}$ 随 $\\operatorname{lgV}$ 的变化情况如图所示, 下列叙述中正确的是\n\n[图1]\nA: $\\mathrm{pH}=10$ 的两种溶液中的 $\\mathrm{c}\\left(\\mathrm{X}^{+}\\right): \\mathrm{XOH}$ 大于 $\\mathrm{X}_{2} \\mathrm{CO}_{3}$\nB: 已知常温下, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{XHCO}_{3}$ 溶液的 $\\mathrm{pH}=8.31$, 则 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\cdot \\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>$ $\\mathrm{K}_{\\mathrm{w}}$\nC: 已知 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{a} 1}$ 远远大于 $\\mathrm{K}_{\\mathrm{a} 2}$, 则 $\\mathrm{K}_{\\mathrm{a} 2}$ 约为 $1.0 \\times 10^{-10.2}$\nD: 当 $\\operatorname{lgV}=2$ 时, 若 $\\mathrm{X}_{2} \\mathrm{CO}_{3}$ 溶液升高温度, 溶液碱性增强, 则 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)^{--} / \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ 减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下体积为 $1 \\mathrm{ml}$ 、浓度均为 $0.10 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{XOH}$ 和 $\\mathrm{X}_{2} \\mathrm{CO}_{3}$ 溶液分别加水稀释至体积为 $\\mathrm{V}, \\mathrm{pH}$ 随 $\\operatorname{lgV}$ 的变化情况如图所示, 下列叙述中正确的是\n\n[图1]\n\nA: $\\mathrm{pH}=10$ 的两种溶液中的 $\\mathrm{c}\\left(\\mathrm{X}^{+}\\right): \\mathrm{XOH}$ 大于 $\\mathrm{X}_{2} \\mathrm{CO}_{3}$\nB: 已知常温下, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{XHCO}_{3}$ 溶液的 $\\mathrm{pH}=8.31$, 则 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right) \\cdot \\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>$ $\\mathrm{K}_{\\mathrm{w}}$\nC: 已知 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{a} 1}$ 远远大于 $\\mathrm{K}_{\\mathrm{a} 2}$, 则 $\\mathrm{K}_{\\mathrm{a} 2}$ 约为 $1.0 \\times 10^{-10.2}$\nD: 当 $\\operatorname{lgV}=2$ 时, 若 $\\mathrm{X}_{2} \\mathrm{CO}_{3}$ 溶液升高温度, 溶液碱性增强, 则 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)^{--} / \\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ 减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-57.jpg?height=634&width=516&top_left_y=1842&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1344",
        "problem": "The absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nPure aqueous solutions of $\\mathrm{HX}$ were used. Only the anionic species $\\mathrm{X}^{-}$absorb.\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nPure aqueous solutions of $\\mathrm{HX}$ were used. Only the anionic species $\\mathrm{X}^{-}$absorb.\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-463.jpg?height=454&width=777&top_left_y=321&top_left_x=545"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1090",
        "problem": "Give half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{SO}_{4}{ }^{2-}$ to $\\mathrm{S}_{2} \\mathrm{O}_{8}{ }^{2-}$\nA: oxidation\nB: reduction\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGive half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{SO}_{4}{ }^{2-}$ to $\\mathrm{S}_{2} \\mathrm{O}_{8}{ }^{2-}$\n\nA: oxidation\nB: reduction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_193",
        "problem": "The electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{8}$ corresponds to which of the following?\nA: $\\mathrm{Ni}$\nB: $\\mathrm{Ni}^{2+}$\nC: $\\mathrm{Fe}$\nD: $\\mathrm{Fe}^{2+}$\nE: $\\mathrm{Zn}^{2+}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{8}$ corresponds to which of the following?\n\nA: $\\mathrm{Ni}$\nB: $\\mathrm{Ni}^{2+}$\nC: $\\mathrm{Fe}$\nD: $\\mathrm{Fe}^{2+}$\nE: $\\mathrm{Zn}^{2+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_282",
        "problem": "Which of the following lists substances in order of increasing boiling point?\nA: $\\mathrm{CO}_{2}, \\mathrm{PCl}_{3}, \\mathrm{CaO}$\nB: $\\mathrm{PCl}_{3}, \\mathrm{CaO}, \\mathrm{CO}_{2}$\nC: $\\mathrm{CaO}, \\mathrm{CO}_{2}, \\mathrm{PCl}_{3}$\nD: $\\mathrm{CaO}, \\mathrm{PCl}_{3}, \\mathrm{CO}_{2}$\nE: $\\mathrm{CO}_{2}, \\mathrm{CaO}, \\mathrm{PCl}_{3}$\nF: $\\mathrm{PCl}_{3}, \\mathrm{CO}_{2}, \\mathrm{CaO}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following lists substances in order of increasing boiling point?\n\nA: $\\mathrm{CO}_{2}, \\mathrm{PCl}_{3}, \\mathrm{CaO}$\nB: $\\mathrm{PCl}_{3}, \\mathrm{CaO}, \\mathrm{CO}_{2}$\nC: $\\mathrm{CaO}, \\mathrm{CO}_{2}, \\mathrm{PCl}_{3}$\nD: $\\mathrm{CaO}, \\mathrm{PCl}_{3}, \\mathrm{CO}_{2}$\nE: $\\mathrm{CO}_{2}, \\mathrm{CaO}, \\mathrm{PCl}_{3}$\nF: $\\mathrm{PCl}_{3}, \\mathrm{CO}_{2}, \\mathrm{CaO}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_493",
        "problem": "已知草酸为二元中强酸, 具有较强还原性, 且 $\\mathrm{Ka}_{2}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>\\mathrm{Ka}(\\mathrm{HClO})$, 下列说法正确的是（ ）\nA: 草酸与氢氧化钠溶液混合呈中性时, 溶液中存在: $$ \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right) $$\nB: $\\mathrm{NaHC}_{2} \\mathrm{O}_{4}$ 与 $\\mathrm{NaClO}$ 溶液混合, 反应的离子方程式为: $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}+\\mathrm{ClO}^{-}=\\mathrm{HClO}+\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$\nC: $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=2 \\quad \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+2 \\quad \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 \\quad \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nD: 草酸使酸性 $\\mathrm{KMnO}_{4}$ 溶液裉色, 离子方程式为: $$ 5 \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}+2 \\mathrm{MnO}_{4}^{-}+6 \\mathrm{H}^{+}=2 \\mathrm{Mn}^{2+}+10 \\mathrm{CO}_{2} \\uparrow+8 \\mathrm{H}_{2} \\mathrm{O} $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知草酸为二元中强酸, 具有较强还原性, 且 $\\mathrm{Ka}_{2}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>\\mathrm{Ka}(\\mathrm{HClO})$, 下列说法正确的是（ ）\n\nA: 草酸与氢氧化钠溶液混合呈中性时, 溶液中存在: $$ \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right) $$\nB: $\\mathrm{NaHC}_{2} \\mathrm{O}_{4}$ 与 $\\mathrm{NaClO}$ 溶液混合, 反应的离子方程式为: $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}+\\mathrm{ClO}^{-}=\\mathrm{HClO}+\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$\nC: $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=2 \\quad \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+2 \\quad \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 \\quad \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)$\nD: 草酸使酸性 $\\mathrm{KMnO}_{4}$ 溶液裉色, 离子方程式为: $$ 5 \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}+2 \\mathrm{MnO}_{4}^{-}+6 \\mathrm{H}^{+}=2 \\mathrm{Mn}^{2+}+10 \\mathrm{CO}_{2} \\uparrow+8 \\mathrm{H}_{2} \\mathrm{O} $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_4",
        "problem": "What is the best description of the geometry of the nitrogen atoms in dimethylnitrosamine, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NNO}$ ? $\\mathrm{N}$ bonded to $\\mathrm{CH}_{3}$ groups\n\nN bonded to $\\mathrm{O}$\nA: Trigonal planar Linear\nB: Trigonal planar Bent\nC: Trigonal pyramidal Linear\nD: Trigonal pyramidal Bent\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the best description of the geometry of the nitrogen atoms in dimethylnitrosamine, $\\left(\\mathrm{CH}_{3}\\right)_{2} \\mathrm{NNO}$ ? $\\mathrm{N}$ bonded to $\\mathrm{CH}_{3}$ groups\n\nN bonded to $\\mathrm{O}$\n\nA: Trigonal planar Linear\nB: Trigonal planar Bent\nC: Trigonal pyramidal Linear\nD: Trigonal pyramidal Bent\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_222",
        "problem": "Sodium cyanide $(\\mathrm{NaCN})$ is used in the extraction of gold from its ores, according to the following equation:\n\n$4 \\mathrm{Au}(\\mathrm{s})+8 \\mathrm{NaCN}(\\mathrm{aq})+\\mathrm{O}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightarrow 4 \\mathrm{Na}\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right](\\mathrm{aq})+4 \\mathrm{NaOH}(\\mathrm{aq})$\n\nWhat volume (in L) of $0.0100 \\mathrm{~mol} \\mathrm{~L}^{-1}$ sodium cyanide solution is required for complete reaction with $11.8 \\mathrm{~g}$ of gold?\nA: $\\quad 0.749 \\mathrm{~L}$\nB: $1.50 \\mathrm{~L}$\nC: $2.99 \\mathrm{~L}$\nD: $\\quad 5.99 \\mathrm{~L}$\nE: $11.98 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSodium cyanide $(\\mathrm{NaCN})$ is used in the extraction of gold from its ores, according to the following equation:\n\n$4 \\mathrm{Au}(\\mathrm{s})+8 \\mathrm{NaCN}(\\mathrm{aq})+\\mathrm{O}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightarrow 4 \\mathrm{Na}\\left[\\mathrm{Au}(\\mathrm{CN})_{2}\\right](\\mathrm{aq})+4 \\mathrm{NaOH}(\\mathrm{aq})$\n\nWhat volume (in L) of $0.0100 \\mathrm{~mol} \\mathrm{~L}^{-1}$ sodium cyanide solution is required for complete reaction with $11.8 \\mathrm{~g}$ of gold?\n\nA: $\\quad 0.749 \\mathrm{~L}$\nB: $1.50 \\mathrm{~L}$\nC: $2.99 \\mathrm{~L}$\nD: $\\quad 5.99 \\mathrm{~L}$\nE: $11.98 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_587",
        "problem": "草酸 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$ 是一种二元弱酸。常温下, 用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液(忽略滴定过程中溶液体积变化), 得到 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 、 $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}$和 $\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$ 三种粒子的分布分数 $\\delta$ (平衡时某种含碳微粒浓度占各种含碳微粒浓度之和的分数)随溶液 $\\mathrm{pH}$ 变化的曲线如图所示。下列说法正确的是\n\n[图1]\nA: $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}$的电离程度小于其水解程度\nB: Q 点溶液中: $2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nC: $\\mathrm{pH}=4.2$ 时, 混合溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=3 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nD: 整个滴定过程中存在: $$ \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)=0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}-\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right) $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n草酸 $\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$ 是一种二元弱酸。常温下, 用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液(忽略滴定过程中溶液体积变化), 得到 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 、 $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}$和 $\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}$ 三种粒子的分布分数 $\\delta$ (平衡时某种含碳微粒浓度占各种含碳微粒浓度之和的分数)随溶液 $\\mathrm{pH}$ 变化的曲线如图所示。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}$的电离程度小于其水解程度\nB: Q 点溶液中: $2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nC: $\\mathrm{pH}=4.2$ 时, 混合溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=3 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nD: 整个滴定过程中存在: $$ \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)=0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}-\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right) $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-025.jpg?height=523&width=765&top_left_y=812&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_745",
        "problem": "体积均为 $1.0 \\mathrm{~L}$ 的甲、乙两个恒容密闭容器, 向甲中加入 $0.1 \\mathrm{molCO}_{2}$ 和 $0.3 \\mathrm{~mol}$ 碳粉,向乙中加入 $0.4 \\mathrm{molCO}$, 在不同温度下发生反应: $\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{s}) \\rightleftharpoons 2 \\mathrm{CO}(\\mathrm{g})$ 。达到平衡时 $\\mathrm{CO}$ 的物质的量浓度随温度的变化如图所示。下列说法正确的是\n\n[图1]\nA: 曲线 II 对应的是乙容器\nB: $a 、 b$ 两点对应平衡体系中的压强之比: $p_{a}: p_{b}<14: 9$\nC: b 点对应的平衡体系中, $\\mathrm{CO}$ 的体积分数小于 $\\frac{4}{7}$\nD: $900 \\mathrm{~K}$ 时, 起始时向容器乙中加入 $\\mathrm{CO} 、 \\mathrm{CO}_{2}$ 、碳粉各 $1 \\mathrm{~mol}$, 此时 $\\mathrm{v}$ 正 $>\\mathrm{v}$ 逆\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n体积均为 $1.0 \\mathrm{~L}$ 的甲、乙两个恒容密闭容器, 向甲中加入 $0.1 \\mathrm{molCO}_{2}$ 和 $0.3 \\mathrm{~mol}$ 碳粉,向乙中加入 $0.4 \\mathrm{molCO}$, 在不同温度下发生反应: $\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{C}(\\mathrm{s}) \\rightleftharpoons 2 \\mathrm{CO}(\\mathrm{g})$ 。达到平衡时 $\\mathrm{CO}$ 的物质的量浓度随温度的变化如图所示。下列说法正确的是\n\n[图1]\n\nA: 曲线 II 对应的是乙容器\nB: $a 、 b$ 两点对应平衡体系中的压强之比: $p_{a}: p_{b}<14: 9$\nC: b 点对应的平衡体系中, $\\mathrm{CO}$ 的体积分数小于 $\\frac{4}{7}$\nD: $900 \\mathrm{~K}$ 时, 起始时向容器乙中加入 $\\mathrm{CO} 、 \\mathrm{CO}_{2}$ 、碳粉各 $1 \\mathrm{~mol}$, 此时 $\\mathrm{v}$ 正 $>\\mathrm{v}$ 逆\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-13.jpg?height=980&width=1450&top_left_y=178&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-13.jpg?height=226&width=1362&top_left_y=1960&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-13.jpg?height=226&width=1351&top_left_y=2234&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_645",
        "problem": "工业上可用氨水处理硫酸厂尾气中的 $\\mathrm{SO}_{2}$. 常温下, 向一定浓度的氨水中通入 $\\mathrm{SO}_{2}$ 气\n\n体, 溶液中 $\\operatorname{lgx}\\left[\\mathrm{x}=\\frac{\\mathrm{c}\\left(\\mathrm{SO}_{3}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)}\\right.$或 $\\frac{\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)}$ 与溶液 $\\mathrm{pH}$ 的关系如图所示。\n\n[图1]\n\n下列说法错误的是\nA: 曲线 II 代表 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)}$\nB: 当溶液的 $\\mathrm{pH}$ 由 7.2 降到 1.9 的过程中, 溶液中 $\\mathrm{HSO}_{3}^{-}$浓度呈增大趋势\nC: 当溶液中的 $\\mathrm{c}\\left(\\mathrm{SO}_{3}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$ 时, 溶液的 $\\mathrm{pH}$ 约为 4\nD: $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{3}$ 溶液中存在离子浓度关系: $\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)$ $+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n工业上可用氨水处理硫酸厂尾气中的 $\\mathrm{SO}_{2}$. 常温下, 向一定浓度的氨水中通入 $\\mathrm{SO}_{2}$ 气\n\n体, 溶液中 $\\operatorname{lgx}\\left[\\mathrm{x}=\\frac{\\mathrm{c}\\left(\\mathrm{SO}_{3}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)}\\right.$或 $\\frac{\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)}$ 与溶液 $\\mathrm{pH}$ 的关系如图所示。\n\n[图1]\n\n下列说法错误的是\n\nA: 曲线 II 代表 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)}$\nB: 当溶液的 $\\mathrm{pH}$ 由 7.2 降到 1.9 的过程中, 溶液中 $\\mathrm{HSO}_{3}^{-}$浓度呈增大趋势\nC: 当溶液中的 $\\mathrm{c}\\left(\\mathrm{SO}_{3}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$ 时, 溶液的 $\\mathrm{pH}$ 约为 4\nD: $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{3}$ 溶液中存在离子浓度关系: $\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}^{-}\\right)$ $+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-071.jpg?height=565&width=602&top_left_y=1442&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_352",
        "problem": "At what temperature is the following reaction at equilibrium when all substances are at standard pressure? Assume that entropies and enthalpies of reaction do not vary with temperature.\n\n$$\n\\mathrm{PCl}_{3}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{PCl}_{5}(g)\n$$\n\n| Substance | $\\Delta H^{\\mathrm{o}}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | $S^{\\circ}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ |\n| :---: | :---: | :---: |\n| $\\mathrm{PCl}_{3}(g)$ | -288.7 | 311.6 |\n| $\\mathrm{Cl}_{2}(g)$ | 0 | 223.1 |\n| $\\mathrm{PCl}_{5}(g)$ | -374.9 | 364.2 |\nA: $506 \\mathrm{~K}$\nB: $1640 \\mathrm{~K}$\nC: $1980 \\mathrm{~K}$\nD: $4260 \\mathrm{~K}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAt what temperature is the following reaction at equilibrium when all substances are at standard pressure? Assume that entropies and enthalpies of reaction do not vary with temperature.\n\n$$\n\\mathrm{PCl}_{3}(g)+\\mathrm{Cl}_{2}(g) \\rightleftharpoons \\mathrm{PCl}_{5}(g)\n$$\n\n| Substance | $\\Delta H^{\\mathrm{o}}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | $S^{\\circ}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ |\n| :---: | :---: | :---: |\n| $\\mathrm{PCl}_{3}(g)$ | -288.7 | 311.6 |\n| $\\mathrm{Cl}_{2}(g)$ | 0 | 223.1 |\n| $\\mathrm{PCl}_{5}(g)$ | -374.9 | 364.2 |\n\nA: $506 \\mathrm{~K}$\nB: $1640 \\mathrm{~K}$\nC: $1980 \\mathrm{~K}$\nD: $4260 \\mathrm{~K}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1172",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nTo prepare $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$, an oxidising agent must be used. One preparation uses $\\mathrm{HgCl}_{2}$. Mercury forms two chlorides: $\\mathrm{HgCl}_{2}$ and $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n\nGive the oxidation state of mercury in $\\mathrm{HgCl}_{2}$ and in $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nTo prepare $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$, an oxidising agent must be used. One preparation uses $\\mathrm{HgCl}_{2}$. Mercury forms two chlorides: $\\mathrm{HgCl}_{2}$ and $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n\nGive the oxidation state of mercury in $\\mathrm{HgCl}_{2}$ and in $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [$\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$, $\\mathrm{HgCl}_{2}$].\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "$\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$",
            "$\\mathrm{HgCl}_{2}$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_905",
        "problem": "等体积的两种一元酸“酸 1 ”和“酸 2 ”分别用等浓度的 $\\mathrm{NaOH}$ 溶液滴定, 滴定曲线如图所示。下列说法正确的是\n\n[图1]\nA: “酸 1 ”比“酸 2 ”的酸性弱\nB: “酸 2 ”的浓度为 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: “酸 2 ”的 $K_{a}$ 的数量级约为 $10^{-5}$ D．滴定“酸 1 ”和“酸 2 ”均可用甲基橙作指示剂\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n等体积的两种一元酸“酸 1 ”和“酸 2 ”分别用等浓度的 $\\mathrm{NaOH}$ 溶液滴定, 滴定曲线如图所示。下列说法正确的是\n\n[图1]\n\nA: “酸 1 ”比“酸 2 ”的酸性弱\nB: “酸 2 ”的浓度为 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: “酸 2 ”的 $K_{a}$ 的数量级约为 $10^{-5}$ D．滴定“酸 1 ”和“酸 2 ”均可用甲基橙作指示剂\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-026.jpg?height=431&width=642&top_left_y=1024&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1574",
        "problem": "At a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.Calculate the frequency in $\\mathrm{s}^{-1}$ of infrared photons that can be absorbed by HD molecule.\n\n(If you have been unable to calculate the value for $\\varepsilon_{\\mathrm{HD}}$ then use $8.000 \\cdot 10^{-20} \\mathrm{~J}$ for the calculation.)\n\nThe allowed electronic energies of $\\mathrm{H}$ atom are given by the expression:\n\n$$\nE=-\\frac{R_{H}}{\\mathrm{n}^{2}}, \\quad \\mathrm{n}=1,2, \\ldots \\text { where } R_{H}=13.5984 \\mathrm{eV}, \\text { and } 1 \\mathrm{eV}=1.602 \\cdot 10^{-19} \\mathrm{~J}\n$$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nAt a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.\n\nproblem:\nCalculate the frequency in $\\mathrm{s}^{-1}$ of infrared photons that can be absorbed by HD molecule.\n\n(If you have been unable to calculate the value for $\\varepsilon_{\\mathrm{HD}}$ then use $8.000 \\cdot 10^{-20} \\mathrm{~J}$ for the calculation.)\n\nThe allowed electronic energies of $\\mathrm{H}$ atom are given by the expression:\n\n$$\nE=-\\frac{R_{H}}{\\mathrm{n}^{2}}, \\quad \\mathrm{n}=1,2, \\ldots \\text { where } R_{H}=13.5984 \\mathrm{eV}, \\text { and } 1 \\mathrm{eV}=1.602 \\cdot 10^{-19} \\mathrm{~J}\n$$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of s, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "s"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_631",
        "problem": "下列图示与对应的叙述相符的是\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\n[图3]\n\n图 3\n\n[图4]\n\n图 4\n\n$2 \\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(\\mathrm{~g})$, 相同时间后测得 $\\mathrm{NO}_{2}$ 含量的曲线, 则该反应的 $\\triangle \\mathrm{H}<0$\nA: 图 1 表示某放热反应在无催化剂（a）和有催化剂（b）时反应的能量变化, 且加入催化剂改变反应的焓变\nB: 图 2 表示某明矾溶液中加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液, 沉淀的质量与加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液体积的关系，在加入 $20 \\mathrm{~mL} \\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液时铝离子恰好沉淀完全\nC: 图 3 表示在 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液中加水时溶液的导电性变化, 则 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液的 $\\mathrm{pH}: \\mathrm{a}>\\mathrm{b}$\nD: 图 4 表示等量 $\\mathrm{NO}_{2}$ 在容积相同的恒容密闭容器中, 不同温度下分别发生反应:\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列图示与对应的叙述相符的是\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\n[图3]\n\n图 3\n\n[图4]\n\n图 4\n\n$2 \\mathrm{NO}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{N}_{2} \\mathrm{O}_{4}(\\mathrm{~g})$, 相同时间后测得 $\\mathrm{NO}_{2}$ 含量的曲线, 则该反应的 $\\triangle \\mathrm{H}<0$\n\nA: 图 1 表示某放热反应在无催化剂（a）和有催化剂（b）时反应的能量变化, 且加入催化剂改变反应的焓变\nB: 图 2 表示某明矾溶液中加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液, 沉淀的质量与加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液体积的关系，在加入 $20 \\mathrm{~mL} \\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液时铝离子恰好沉淀完全\nC: 图 3 表示在 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液中加水时溶液的导电性变化, 则 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液的 $\\mathrm{pH}: \\mathrm{a}>\\mathrm{b}$\nD: 图 4 表示等量 $\\mathrm{NO}_{2}$ 在容积相同的恒容密闭容器中, 不同温度下分别发生反应:\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-008.jpg?height=383&width=288&top_left_y=1690&top_left_x=353",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-008.jpg?height=349&width=225&top_left_y=1710&top_left_x=764",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-008.jpg?height=304&width=284&top_left_y=1730&top_left_x=1046",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-008.jpg?height=291&width=297&top_left_y=1736&top_left_x=1391"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1093",
        "problem": "A typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nWhat is the purpose of these chemicals in the extraction solution?\nA: Make a very acidic solution\nB: Make a very neutral solution\nC: Make a very alkaline solution\nD: Make a very buffered solution\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nWhat is the purpose of these chemicals in the extraction solution?\n\nA: Make a very acidic solution\nB: Make a very neutral solution\nC: Make a very alkaline solution\nD: Make a very buffered solution\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-08.jpg?height=146&width=940&top_left_y=495&top_left_x=929"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_191",
        "problem": "Fluorine is the most electronegative element on the periodic table. As a result it always forms polar bonds with other non-metals. Despite this, which of the following fluorine containing compounds would be a nonpolar molecule?\nA: $\\mathrm{SF}_{4}$\nB: $\\mathrm{PF}_{3}$\nC: $\\mathrm{IF}_{5}$\nD: $\\mathrm{BrF}_{3}$\nE: $\\mathrm{XeF}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFluorine is the most electronegative element on the periodic table. As a result it always forms polar bonds with other non-metals. Despite this, which of the following fluorine containing compounds would be a nonpolar molecule?\n\nA: $\\mathrm{SF}_{4}$\nB: $\\mathrm{PF}_{3}$\nC: $\\mathrm{IF}_{5}$\nD: $\\mathrm{BrF}_{3}$\nE: $\\mathrm{XeF}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_450",
        "problem": "用下列实验装置进行的实验中，能达到相应实验目的的是\nA: 装置甲：防止铁钉生锈 [图1] 甲\nB: 装置乙：除去乙烯中混有的乙炔 [图2]\nC: 装置丙：实验室制取乙酸乙酯 [图3]\nD: 装置丁: 检验混合液中的 $\\mathrm{Fe}^{3+}$ 和 $\\mathrm{Cu}^{2+}$ 溶液 [图4]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n用下列实验装置进行的实验中，能达到相应实验目的的是\n\nA: 装置甲：防止铁钉生锈 [图1] 甲\nB: 装置乙：除去乙烯中混有的乙炔 [图2]\nC: 装置丙：实验室制取乙酸乙酯 [图3]\nD: 装置丁: 检验混合液中的 $\\mathrm{Fe}^{3+}$ 和 $\\mathrm{Cu}^{2+}$ 溶液 [图4]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-68.jpg?height=314&width=260&top_left_y=2219&top_left_x=864",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-69.jpg?height=385&width=277&top_left_y=176&top_left_x=958",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-69.jpg?height=357&width=471&top_left_y=598&top_left_x=947",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-69.jpg?height=334&width=191&top_left_y=961&top_left_x=1184"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_518",
        "problem": "科学家近日发明了一种无污染、无需净化高效产氢的可充电电池, 电池示意图如下。电极为金属锌和选择性催化材料, 实现了阴极析氢和 $\\mathrm{N}_{2} \\mathrm{H}_{4}$ 氧化两个独立的反应, 可稳定循环 600 次。下列说法错误的是\n\n[图1]\n\n$\\mathrm{KOH}$ 溶液阴离子交换膜\nA: 充电时, 阴极区溶液 $\\mathrm{pH}$ 减小\nB: 放电时, $\\mathrm{OH}^{-}$由 $\\mathrm{b}$ 极区向 $\\mathrm{a}$ 极区迁移\nC: 充电时, $b$ 极反应式为 $\\mathrm{N}_{2} \\mathrm{H}_{4}-4 \\mathrm{e}^{-}+4 \\mathrm{OH}^{-}=\\mathrm{N}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}$\nD: 产生 $1 \\mathrm{molH}_{2}, \\mathrm{a}$ 极质量增大 $65 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n科学家近日发明了一种无污染、无需净化高效产氢的可充电电池, 电池示意图如下。电极为金属锌和选择性催化材料, 实现了阴极析氢和 $\\mathrm{N}_{2} \\mathrm{H}_{4}$ 氧化两个独立的反应, 可稳定循环 600 次。下列说法错误的是\n\n[图1]\n\n$\\mathrm{KOH}$ 溶液阴离子交换膜\n\nA: 充电时, 阴极区溶液 $\\mathrm{pH}$ 减小\nB: 放电时, $\\mathrm{OH}^{-}$由 $\\mathrm{b}$ 极区向 $\\mathrm{a}$ 极区迁移\nC: 充电时, $b$ 极反应式为 $\\mathrm{N}_{2} \\mathrm{H}_{4}-4 \\mathrm{e}^{-}+4 \\mathrm{OH}^{-}=\\mathrm{N}_{2}+4 \\mathrm{H}_{2} \\mathrm{O}$\nD: 产生 $1 \\mathrm{molH}_{2}, \\mathrm{a}$ 极质量增大 $65 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-054.jpg?height=374&width=697&top_left_y=618&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1307",
        "problem": "Simple estimates of graphene properties\n\nGraphene is a two-dimensional, one atom thick carbon material (Fig. 1 a). Many layers of graphene stack together to form graphite (Fig. 1 b).\n[figure1]\n\nFig. 1. (a) The structure of graphene. Spheres are carbon atoms. They are arranged in hexagons. The area of one carbon hexagon is $5.16 \\cdot 10^{-20} \\mathrm{~m}^{2}$.\n\n(b) Crystal lattice of graphite. Three graphene layers are shown.\n\nSuch atomic structure was long considered to be unstable. However, in 2004 Andrey Geim and Konstantin Novoselov have reported production of the first samples of this unusual material. This groundbreaking invention was awarded by Nobel prize in 2010.\n\nExperimental studies of graphene are still restricted. Production of massive portions of the new substance still is a challenging synthetic problem. Many properties of graphene were estimated. Usually, there is not enough information for rigorous calculations, so we have to make assumptions and neglect unimportant factors. In this problem, you will estimate the adsorption properties of graphene.The single layer of nitrogen molecules adsorbed on the outer surface of graphite is shown in Fig. 2. Assume that the same arrangement of nitrogen molecules is formed on a graphene surface.\n\n[figure2]\n\nFig. 2. Nitrogen molecules $\\mathrm{N}_{2}$ (grey circles) on the outer surface of graphite\n\nHow many grams of nitrogen can be adsorbed on 1 gram of graphene assuming that the graphene layer is placed onto the surface of a solid support? Estimate the volume occupied by these nitrogen molecules after the complete desorption from $1 \\mathrm{~g}$ of graphene (pressure 1 bar, temperature $298 \\mathrm{~K}$ ).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nSimple estimates of graphene properties\n\nGraphene is a two-dimensional, one atom thick carbon material (Fig. 1 a). Many layers of graphene stack together to form graphite (Fig. 1 b).\n[figure1]\n\nFig. 1. (a) The structure of graphene. Spheres are carbon atoms. They are arranged in hexagons. The area of one carbon hexagon is $5.16 \\cdot 10^{-20} \\mathrm{~m}^{2}$.\n\n(b) Crystal lattice of graphite. Three graphene layers are shown.\n\nSuch atomic structure was long considered to be unstable. However, in 2004 Andrey Geim and Konstantin Novoselov have reported production of the first samples of this unusual material. This groundbreaking invention was awarded by Nobel prize in 2010.\n\nExperimental studies of graphene are still restricted. Production of massive portions of the new substance still is a challenging synthetic problem. Many properties of graphene were estimated. Usually, there is not enough information for rigorous calculations, so we have to make assumptions and neglect unimportant factors. In this problem, you will estimate the adsorption properties of graphene.\n\nproblem:\nThe single layer of nitrogen molecules adsorbed on the outer surface of graphite is shown in Fig. 2. Assume that the same arrangement of nitrogen molecules is formed on a graphene surface.\n\n[figure2]\n\nFig. 2. Nitrogen molecules $\\mathrm{N}_{2}$ (grey circles) on the outer surface of graphite\n\nHow many grams of nitrogen can be adsorbed on 1 gram of graphene assuming that the graphene layer is placed onto the surface of a solid support? Estimate the volume occupied by these nitrogen molecules after the complete desorption from $1 \\mathrm{~g}$ of graphene (pressure 1 bar, temperature $298 \\mathrm{~K}$ ).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{dm}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-244.jpg?height=540&width=1024&top_left_y=878&top_left_x=524",
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-245.jpg?height=534&width=700&top_left_y=567&top_left_x=701"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{dm}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_256",
        "problem": "Which of the following substances has the highest boiling point?\nA: bromine\nB: magnesium bromide\nC: phosphorus tribromide\nD: hydrogen bromide\nE: hydrogen\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following substances has the highest boiling point?\n\nA: bromine\nB: magnesium bromide\nC: phosphorus tribromide\nD: hydrogen bromide\nE: hydrogen\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_874",
        "problem": "用 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.0 \\mathrm{~mL} 0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液, 溶液 $\\mathrm{pH}$ 和温度随 $\\frac{\\mathrm{V}(\\mathrm{NaCH} \\text { 溶液 })}{\\mathrm{V}\\left(\\mathrm{H}_{2} A \\text { 溶液 }\\right)}$ 的变化曲线如图所示, 下列说法正确的\n\n[图1]\nA: 从 $W$ 至 Z 点, 一定存在 $K_{w}$ 不变, 且 $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(H^{+}\\right)=c\\left(O^{-}\\right)+2 c\\left(\\mathrm{~A}^{2-}\\right)+c(H A$ $\\left.{ }^{-}\\right)$\nB: 当 $\\frac{V(N a C-\\text { 容液 })}{V\\left(H_{2} A \\text { 溶液 }\\right)}=1$ 时, $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nC: $27^{\\circ} \\mathrm{C}, \\mathrm{H}_{2} \\mathrm{~A}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 2}=10^{-9.7}$\nD: $\\mathrm{Y}$ 点为第二反应终点\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用 $0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.0 \\mathrm{~mL} 0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~A}$ 溶液, 溶液 $\\mathrm{pH}$ 和温度随 $\\frac{\\mathrm{V}(\\mathrm{NaCH} \\text { 溶液 })}{\\mathrm{V}\\left(\\mathrm{H}_{2} A \\text { 溶液 }\\right)}$ 的变化曲线如图所示, 下列说法正确的\n\n[图1]\n\nA: 从 $W$ 至 Z 点, 一定存在 $K_{w}$ 不变, 且 $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(H^{+}\\right)=c\\left(O^{-}\\right)+2 c\\left(\\mathrm{~A}^{2-}\\right)+c(H A$ $\\left.{ }^{-}\\right)$\nB: 当 $\\frac{V(N a C-\\text { 容液 })}{V\\left(H_{2} A \\text { 溶液 }\\right)}=1$ 时, $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nC: $27^{\\circ} \\mathrm{C}, \\mathrm{H}_{2} \\mathrm{~A}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 2}=10^{-9.7}$\nD: $\\mathrm{Y}$ 点为第二反应终点\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-110.jpg?height=774&width=905&top_left_y=495&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_783",
        "problem": "羟胺的电离方程式为: $\\mathrm{NH}_{2} \\mathrm{OH}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{NH}_{3} \\mathrm{OH}^{+}+\\mathrm{OH}^{-}\\left(25^{\\circ} \\mathrm{C}\\right.$ 时, $\\mathrm{K}_{\\mathrm{b}}=9.0 \\times 10^{-}$ 9)。用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸滴定 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 羟胺溶液, 恒定 $25^{\\circ} \\mathrm{C}$ 时, 滴定过程中由水电离出来的 $\\mathrm{H}^{+}$浓度的负对数与盐酸体积的关系如图所示(已知： $\\lg 3=0.5$ )。下列说法试卷第 66 页，共 92 页\n\n[图1]\nA: 图中 $\\mathrm{V}_{1}>10$\nB: A 点对应溶液的 $\\mathrm{pH}=9.5$\nC: B、D 两点对应的溶液均为中性\nD: E 点对应溶液中存在: $c\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\mathrm{OH}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{2} \\mathrm{OH}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n羟胺的电离方程式为: $\\mathrm{NH}_{2} \\mathrm{OH}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{NH}_{3} \\mathrm{OH}^{+}+\\mathrm{OH}^{-}\\left(25^{\\circ} \\mathrm{C}\\right.$ 时, $\\mathrm{K}_{\\mathrm{b}}=9.0 \\times 10^{-}$ 9)。用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸滴定 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 羟胺溶液, 恒定 $25^{\\circ} \\mathrm{C}$ 时, 滴定过程中由水电离出来的 $\\mathrm{H}^{+}$浓度的负对数与盐酸体积的关系如图所示(已知： $\\lg 3=0.5$ )。下列说法试卷第 66 页，共 92 页\n\n[图1]\n\nA: 图中 $\\mathrm{V}_{1}>10$\nB: A 点对应溶液的 $\\mathrm{pH}=9.5$\nC: B、D 两点对应的溶液均为中性\nD: E 点对应溶液中存在: $c\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\mathrm{OH}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{2} \\mathrm{OH}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-67.jpg?height=248&width=442&top_left_y=270&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_283",
        "problem": "Calculate the amount (in mol) of $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ that reacts with $1 \\mathrm{~mol}$ of $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ under these conditions.\n$5.6311 \\mathrm{~g}$ of a potassium salt is dissolved in water and the solution made up to $100.0 \\mathrm{~mL}$.\n\nA $20.00 \\mathrm{~mL}$ sample of this solution is taken and $50.00 \\mathrm{~mL}$ of the same $0.1221 \\mathrm{~mol} \\mathrm{~L}^{-1}$ $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ as above is added. $\\mathrm{KB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ precipitates and is filtered off, then the excess $\\mathrm{B}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}{ }^{-}$is titrated with the same $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution as above, this time requiring $18.35 \\mathrm{~mL}$ for complete reaction.\n\nThe equations for the two reactions are as follows:\n\n- $\\mathrm{K}^{+}+\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4} \\rightarrow \\mathrm{KB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}(\\mathrm{~s})+\\mathrm{Na}^{+}$\n- $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}+2 \\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4} \\rightarrow \\mathrm{Hg}\\left(\\mathrm{B}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}\\right)_{2}+2 \\mathrm{NaNO}_{3}$ (but recall the nonstoichiometry of this reaction; the two reactants react in the ratio calculated above)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the amount (in mol) of $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ that reacts with $1 \\mathrm{~mol}$ of $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ under these conditions.\n$5.6311 \\mathrm{~g}$ of a potassium salt is dissolved in water and the solution made up to $100.0 \\mathrm{~mL}$.\n\nA $20.00 \\mathrm{~mL}$ sample of this solution is taken and $50.00 \\mathrm{~mL}$ of the same $0.1221 \\mathrm{~mol} \\mathrm{~L}^{-1}$ $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ as above is added. $\\mathrm{KB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ precipitates and is filtered off, then the excess $\\mathrm{B}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}{ }^{-}$is titrated with the same $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution as above, this time requiring $18.35 \\mathrm{~mL}$ for complete reaction.\n\nThe equations for the two reactions are as follows:\n\n- $\\mathrm{K}^{+}+\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4} \\rightarrow \\mathrm{KB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}(\\mathrm{~s})+\\mathrm{Na}^{+}$\n- $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}+2 \\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4} \\rightarrow \\mathrm{Hg}\\left(\\mathrm{B}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}\\right)_{2}+2 \\mathrm{NaNO}_{3}$ (but recall the nonstoichiometry of this reaction; the two reactants react in the ratio calculated above)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_989",
        "problem": "The average car in Canada uses $0.93 \\mathrm{~L}$ of gasoline to go $100 \\mathrm{~km}$. If it is assumed that gasoline is pure octane $\\left(\\mathrm{C}_{8} \\mathrm{H}_{18}\\right)$, with a density of $0.70 \\mathrm{~g} / \\mathrm{mL}$ and a molar mass of $114.2 \\mathrm{~g} / \\mathrm{mol}$, then how many moles of octane are consumed by driving $100 \\mathrm{~km}$ ?\nA: $\\quad 0.93 \\mathrm{~mol}$\nB: $5.7 \\mathrm{~mol}$\nC: $11 \\mathrm{~mol}$\nD: $5.7 \\times 10^{-4} \\mathrm{~mol}$\nE: $1.1 \\times 10^{-3} \\mathrm{~mol}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe average car in Canada uses $0.93 \\mathrm{~L}$ of gasoline to go $100 \\mathrm{~km}$. If it is assumed that gasoline is pure octane $\\left(\\mathrm{C}_{8} \\mathrm{H}_{18}\\right)$, with a density of $0.70 \\mathrm{~g} / \\mathrm{mL}$ and a molar mass of $114.2 \\mathrm{~g} / \\mathrm{mol}$, then how many moles of octane are consumed by driving $100 \\mathrm{~km}$ ?\n\nA: $\\quad 0.93 \\mathrm{~mol}$\nB: $5.7 \\mathrm{~mol}$\nC: $11 \\mathrm{~mol}$\nD: $5.7 \\times 10^{-4} \\mathrm{~mol}$\nE: $1.1 \\times 10^{-3} \\mathrm{~mol}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1469",
        "problem": "Phosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nPhosphoric acid is used as a fertiliser for agriculture. $1.00 \\times 10^{-3} \\mathrm{M}$ phosphoric acid is added to an aqueous soil suspension and the $\\mathrm{pH}$ is found to be 7.00 .\n\nDetermine the fractional concentrations of all the different phosphate species present in the solution. Assume that no component of the soil interacts with any phosphate species.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nPhosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nPhosphoric acid is used as a fertiliser for agriculture. $1.00 \\times 10^{-3} \\mathrm{M}$ phosphoric acid is added to an aqueous soil suspension and the $\\mathrm{pH}$ is found to be 7.00 .\n\nDetermine the fractional concentrations of all the different phosphate species present in the solution. Assume that no component of the soil interacts with any phosphate species.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [The fractional concentrations of  $\\mathrm{H}_{3} \\mathrm{PO}_{4} \\quad\\left(f_{0}\\right)$, The fractional concentrations of $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\left(f_{1}\\right)$, The fractional concentrations of $\\mathrm{HPO}_{4}^{2-} \\quad\\left(f_{2}\\right)$, The fractional concentrations of $\\mathrm{PO}_{4}^{3-} \\quad\\left(f_{3}\\right)$].\nTheir answer types are, in order, [numerical value, numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null,
            null,
            null
        ],
        "answer_sequence": [
            "The fractional concentrations of  $\\mathrm{H}_{3} \\mathrm{PO}_{4} \\quad\\left(f_{0}\\right)$",
            "The fractional concentrations of $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}\\left(f_{1}\\right)$",
            "The fractional concentrations of $\\mathrm{HPO}_{4}^{2-} \\quad\\left(f_{2}\\right)$",
            "The fractional concentrations of $\\mathrm{PO}_{4}^{3-} \\quad\\left(f_{3}\\right)$"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_440",
        "problem": "下列说法正确的是\nA: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right): \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=1: 10^{-2}$ 的溶液中 $\\mathrm{K}^{+} 、 \\mathrm{Ba}^{2+} 、 \\mathrm{ClO}^{-} 、 \\mathrm{CO}_{3}{ }^{2-}$ 一定能大量存在\nB: 室温下, 水电离出来的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=10^{-13} \\mathrm{~mol} / \\mathrm{L}$ 的溶液中 $\\mathrm{K}^{+} 、 \\mathrm{Cl}^{-} 、 \\mathrm{NO}_{3}{ }^{-} 、 \\mathrm{I}^{-}$一定能大量存在\nC: 相同温度下，等物质的量浓度的下列溶液： (1) $\\mathrm{H}_{2} \\mathrm{CO}_{3}(2) \\mathrm{Na}_{2} \\mathrm{CO}_{3}(3) \\mathrm{NaHCO}_{3}(4)\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$ 中 $\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2}\\right)$ 的大小关系为(2)>(4) $>$ (3) $>$ (1)\nD: 往 $0.1 \\mathrm{~mol} / \\mathrm{LCH}_{3} \\mathrm{COOH}$ 溶液中通入少量 $\\mathrm{HCl}$, 醋酸的电离平衡向逆反应方向移动, 且溶液中 $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right) / \\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$增大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列说法正确的是\n\nA: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right): \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=1: 10^{-2}$ 的溶液中 $\\mathrm{K}^{+} 、 \\mathrm{Ba}^{2+} 、 \\mathrm{ClO}^{-} 、 \\mathrm{CO}_{3}{ }^{2-}$ 一定能大量存在\nB: 室温下, 水电离出来的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=10^{-13} \\mathrm{~mol} / \\mathrm{L}$ 的溶液中 $\\mathrm{K}^{+} 、 \\mathrm{Cl}^{-} 、 \\mathrm{NO}_{3}{ }^{-} 、 \\mathrm{I}^{-}$一定能大量存在\nC: 相同温度下，等物质的量浓度的下列溶液： (1) $\\mathrm{H}_{2} \\mathrm{CO}_{3}(2) \\mathrm{Na}_{2} \\mathrm{CO}_{3}(3) \\mathrm{NaHCO}_{3}(4)\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$ 中 $\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2}\\right)$ 的大小关系为(2)>(4) $>$ (3) $>$ (1)\nD: 往 $0.1 \\mathrm{~mol} / \\mathrm{LCH}_{3} \\mathrm{COOH}$ 溶液中通入少量 $\\mathrm{HCl}$, 醋酸的电离平衡向逆反应方向移动, 且溶液中 $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right) / \\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$增大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_151",
        "problem": "Vitamin $\\mathrm{C}\\left(\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{6}\\right)$ supplements often come in $500 \\mathrm{mg}$ tablets. If a person consumes one $500 \\mathrm{mg}$ tablet of Vitamin C, how many moles of Vitamin $C$ would they be consuming?\nA: $2.84 \\times 10^{-3}$\nB: $8.81 \\times 10^{-2}$\nC: 1.00\nD: 2.84\nE: 88.1\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nVitamin $\\mathrm{C}\\left(\\mathrm{C}_{6} \\mathrm{H}_{8} \\mathrm{O}_{6}\\right)$ supplements often come in $500 \\mathrm{mg}$ tablets. If a person consumes one $500 \\mathrm{mg}$ tablet of Vitamin C, how many moles of Vitamin $C$ would they be consuming?\n\nA: $2.84 \\times 10^{-3}$\nB: $8.81 \\times 10^{-2}$\nC: 1.00\nD: 2.84\nE: 88.1\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_937",
        "problem": "向甲、乙、丙三个密闭容器中充入一定量的 $\\mathrm{A}$ 和 B, 发生反应:\n\n$\\mathrm{A}(\\mathrm{g})+\\mathrm{xB}(\\mathrm{g})-2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}=\\mathrm{a} \\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$\n\n各容器的反应温度、反应物起始量, 反应过程中 $\\mathrm{C}$ 的浓度随时间变化关系分别如下表和\n\n下图:\n\n| 容器 | 甲 | 乙 | 丙 |\n| :---: | :---: | :---: | :---: |\n| 容积 | $0.5 \\mathrm{~L}$ | $0.5 \\mathrm{~L}$ | $1.0 \\mathrm{~L}$ |\n| 温度/ ${ }^{\\circ} \\mathrm{C}$ | $\\mathrm{T}_{1}$ | $\\mathrm{~T}_{2}$ | $\\mathrm{~T}_{2}$ |\n| 反应物起始量 | $1.5 \\mathrm{~mol} \\mathrm{~A} \\quad 0.5 \\mathrm{~mol} \\mathrm{~B}$ | $1.5 \\mathrm{~mol} \\mathrm{~A} \\quad 0.5 \\mathrm{~mol} \\mathrm{~B}$ | $6.0 \\mathrm{~mol} \\mathrm{~A} \\quad 2.0 \\mathrm{~mol} \\mathrm{~B}$ |\n\n[图1]\nA: $10 \\mathrm{~min}$ 内甲容器中反应的平均速率 $\\mathrm{v}(\\mathrm{A})=0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 其他条件不变, 甲容器中在 $20 \\mathrm{~min}$ 后, 降低温度, 平衡向正反应方向移动\nC: 保持其他条件不变, 若起始时向乙容器中充入 $3.0 \\mathrm{~mol} \\mathrm{~A} 、 1.0 \\mathrm{~mol} \\mathrm{~B}$ 和 $2.0 \\mathrm{~mol}$ $\\mathrm{C}$, 则反应达到新平衡前 $\\mathrm{v}_{(\\text {(逆) }}>\\mathrm{v}_{\\text {(正) }}$\nD: $\\mathrm{T}_{2}{ }^{\\circ} \\mathrm{C}$, 向丙容器的平衡体系中再充入 $1.5 \\mathrm{~mol} \\mathrm{~A} 、 0.5 \\mathrm{~mol} \\mathrm{~B}$, 平衡时 $\\mathrm{C}$ 的体积分数大于 $25 \\%$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n向甲、乙、丙三个密闭容器中充入一定量的 $\\mathrm{A}$ 和 B, 发生反应:\n\n$\\mathrm{A}(\\mathrm{g})+\\mathrm{xB}(\\mathrm{g})-2 \\mathrm{C}(\\mathrm{g}) \\Delta \\mathrm{H}=\\mathrm{a} \\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$\n\n各容器的反应温度、反应物起始量, 反应过程中 $\\mathrm{C}$ 的浓度随时间变化关系分别如下表和\n\n下图:\n\n| 容器 | 甲 | 乙 | 丙 |\n| :---: | :---: | :---: | :---: |\n| 容积 | $0.5 \\mathrm{~L}$ | $0.5 \\mathrm{~L}$ | $1.0 \\mathrm{~L}$ |\n| 温度/ ${ }^{\\circ} \\mathrm{C}$ | $\\mathrm{T}_{1}$ | $\\mathrm{~T}_{2}$ | $\\mathrm{~T}_{2}$ |\n| 反应物起始量 | $1.5 \\mathrm{~mol} \\mathrm{~A} \\quad 0.5 \\mathrm{~mol} \\mathrm{~B}$ | $1.5 \\mathrm{~mol} \\mathrm{~A} \\quad 0.5 \\mathrm{~mol} \\mathrm{~B}$ | $6.0 \\mathrm{~mol} \\mathrm{~A} \\quad 2.0 \\mathrm{~mol} \\mathrm{~B}$ |\n\n[图1]\n\nA: $10 \\mathrm{~min}$ 内甲容器中反应的平均速率 $\\mathrm{v}(\\mathrm{A})=0.10 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 其他条件不变, 甲容器中在 $20 \\mathrm{~min}$ 后, 降低温度, 平衡向正反应方向移动\nC: 保持其他条件不变, 若起始时向乙容器中充入 $3.0 \\mathrm{~mol} \\mathrm{~A} 、 1.0 \\mathrm{~mol} \\mathrm{~B}$ 和 $2.0 \\mathrm{~mol}$ $\\mathrm{C}$, 则反应达到新平衡前 $\\mathrm{v}_{(\\text {(逆) }}>\\mathrm{v}_{\\text {(正) }}$\nD: $\\mathrm{T}_{2}{ }^{\\circ} \\mathrm{C}$, 向丙容器的平衡体系中再充入 $1.5 \\mathrm{~mol} \\mathrm{~A} 、 0.5 \\mathrm{~mol} \\mathrm{~B}$, 平衡时 $\\mathrm{C}$ 的体积分数大于 $25 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-107.jpg?height=302&width=774&top_left_y=1588&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_942",
        "problem": "利用 $\\mathrm{CH}_{4}$ 燃料电池电解制备 $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$ 并得到副产物 $\\mathrm{NaOH} 、 \\mathrm{H}_{2} 、 \\mathrm{Cl}_{2}$ 的装置如图所示。下列说法正确的是\n[图1]\nA: $\\mathrm{A} 、 \\mathrm{C}$ 膜均为阴离子交换膜, $\\mathrm{B}$ 膜为阳离子交换膜\nB: 若去掉 $\\mathrm{C}$ 膜, 产品室中有可能产生 $\\mathrm{CaHPO}_{4} 、 \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ 等沉淀\nC: $\\mathrm{a}$ 极电极反应: $\\mathrm{CH}_{4}+8 \\mathrm{e}^{-}+4 \\mathrm{O}^{2-}=\\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 标准状况下, a 极上通入 $1.12 \\mathrm{~L}$ 甲烷, 理论上阳极室溶液质量减少 $22.2 \\mathrm{~g}$ (忽略水分子的移动)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用 $\\mathrm{CH}_{4}$ 燃料电池电解制备 $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$ 并得到副产物 $\\mathrm{NaOH} 、 \\mathrm{H}_{2} 、 \\mathrm{Cl}_{2}$ 的装置如图所示。下列说法正确的是\n[图1]\n\nA: $\\mathrm{A} 、 \\mathrm{C}$ 膜均为阴离子交换膜, $\\mathrm{B}$ 膜为阳离子交换膜\nB: 若去掉 $\\mathrm{C}$ 膜, 产品室中有可能产生 $\\mathrm{CaHPO}_{4} 、 \\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$ 等沉淀\nC: $\\mathrm{a}$ 极电极反应: $\\mathrm{CH}_{4}+8 \\mathrm{e}^{-}+4 \\mathrm{O}^{2-}=\\mathrm{CO}_{2}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 标准状况下, a 极上通入 $1.12 \\mathrm{~L}$ 甲烷, 理论上阳极室溶液质量减少 $22.2 \\mathrm{~g}$ (忽略水分子的移动)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-49.jpg?height=374&width=1292&top_left_y=153&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_713",
        "problem": "近年来, 处理氮污染水已成为一个世界性课题。在金属 $\\mathrm{Pt} 、 \\mathrm{Cu}$ 和铱 (Ir) 的催化作用下, 密闭容器中的 $\\mathrm{H}_{2}$ 可高效转化酸性溶液中的硝态氮 $\\left(\\mathrm{NO}_{3}^{-}\\right)$, 其工作原理如图所示。\n\n[图1]\n\n下列说法正确的是\nA: 气体层中 Ir 颗粒表面的还原产物属于共价化合物\nB: 液体层中 $\\mathrm{NO}$ 在单原子铂表面发生的电极反应为: $2 \\mathrm{NO}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{N}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 若导电基体上的 Pt 颗粒增多，不利于降低溶液中的含氮量\nD: 液体层中 $\\mathrm{Pt}$ 的聚集状态不影响电极反应的选择性\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n近年来, 处理氮污染水已成为一个世界性课题。在金属 $\\mathrm{Pt} 、 \\mathrm{Cu}$ 和铱 (Ir) 的催化作用下, 密闭容器中的 $\\mathrm{H}_{2}$ 可高效转化酸性溶液中的硝态氮 $\\left(\\mathrm{NO}_{3}^{-}\\right)$, 其工作原理如图所示。\n\n[图1]\n\n下列说法正确的是\n\nA: 气体层中 Ir 颗粒表面的还原产物属于共价化合物\nB: 液体层中 $\\mathrm{NO}$ 在单原子铂表面发生的电极反应为: $2 \\mathrm{NO}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{N}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\mathrm{O}$\nC: 若导电基体上的 Pt 颗粒增多，不利于降低溶液中的含氮量\nD: 液体层中 $\\mathrm{Pt}$ 的聚集状态不影响电极反应的选择性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-71.jpg?height=479&width=1243&top_left_y=1645&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_855",
        "problem": "常温下, 现有 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液, $\\mathrm{pH}=7.8$ 。已知含氮(或含碳)各微粒的分布分数(平衡时, 各微粒浓度占总微粒浓度之和的分数)与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\nA: 当 $\\mathrm{pH}=9$ 时, 溶液中存在下列关系: $\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nB: $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液中存在下列关系: $\\frac{\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)}<1$\nC: 向 $\\mathrm{pH}=6.5$ 的上述溶液中逐滴滴加氢氧化钠溶液时, $\\mathrm{NH}_{4}{ }^{+}$和 $\\mathrm{HCO}_{3}-$ 浓度逐渐减小\nD: 分析可知, 常温下水解平衡常数 $\\left.\\mathrm{K}_{\\mathrm{h}}\\left(\\mathrm{HCO}_{3}\\right)^{-}\\right)$的数量级为 $10^{-7}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 现有 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液, $\\mathrm{pH}=7.8$ 。已知含氮(或含碳)各微粒的分布分数(平衡时, 各微粒浓度占总微粒浓度之和的分数)与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\n\nA: 当 $\\mathrm{pH}=9$ 时, 溶液中存在下列关系: $\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nB: $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液中存在下列关系: $\\frac{\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)}<1$\nC: 向 $\\mathrm{pH}=6.5$ 的上述溶液中逐滴滴加氢氧化钠溶液时, $\\mathrm{NH}_{4}{ }^{+}$和 $\\mathrm{HCO}_{3}-$ 浓度逐渐减小\nD: 分析可知, 常温下水解平衡常数 $\\left.\\mathrm{K}_{\\mathrm{h}}\\left(\\mathrm{HCO}_{3}\\right)^{-}\\right)$的数量级为 $10^{-7}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-66.jpg?height=317&width=1100&top_left_y=156&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_8",
        "problem": "Which molecule has the shortest carbon-nitrogen bond length?\nA: $\\mathrm{HCNO}$\nB: $\\mathrm{HNCO}$\nC: $\\mathrm{H}_{3} \\mathrm{CNO}$\nD: $\\mathrm{H}_{3} \\mathrm{CNO}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich molecule has the shortest carbon-nitrogen bond length?\n\nA: $\\mathrm{HCNO}$\nB: $\\mathrm{HNCO}$\nC: $\\mathrm{H}_{3} \\mathrm{CNO}$\nD: $\\mathrm{H}_{3} \\mathrm{CNO}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1554",
        "problem": "Type II electrodes that are made of a metal covered with a sparingly soluble salt of the metal are dipped into a soluble salt solution containing an anion of the sparingly soluble salt. The silver/silver chloride $(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$ and the calomel electrode $(\\mathrm{Hg}$, $\\left.\\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Cl}\\right)$ are examples of such electrodes. The standard emf of a cell built of those electrodes (-) Ag, $\\mathrm{AgCl}^{-} / \\mathrm{Cl}^{-} \\| \\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Hg}(+)$ is $E^{0}=0.0455 \\mathrm{~V}$ at $T=298 \\mathrm{~K}$. The temperature coefficient for this cell is $d E^{0} / d T=3.38 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}$.\n\nCalculate the solubility product of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ knowing that the standard potential of the calomel electrode is $E^{0}=0.798 \\mathrm{~V}$.\n\n$F=96487 \\mathrm{C} \\mathrm{mol}^{-1}, \\quad R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}, T=298 \\mathrm{~K}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nType II electrodes that are made of a metal covered with a sparingly soluble salt of the metal are dipped into a soluble salt solution containing an anion of the sparingly soluble salt. The silver/silver chloride $(\\mathrm{Ag}, \\mathrm{AgCl} / \\mathrm{Cl})$ and the calomel electrode $(\\mathrm{Hg}$, $\\left.\\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Cl}\\right)$ are examples of such electrodes. The standard emf of a cell built of those electrodes (-) Ag, $\\mathrm{AgCl}^{-} / \\mathrm{Cl}^{-} \\| \\mathrm{Hg}_{2} \\mathrm{Cl}_{2} / \\mathrm{Hg}(+)$ is $E^{0}=0.0455 \\mathrm{~V}$ at $T=298 \\mathrm{~K}$. The temperature coefficient for this cell is $d E^{0} / d T=3.38 \\times 10^{-4} \\mathrm{~V} \\mathrm{~K}^{-1}$.\n\nCalculate the solubility product of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ knowing that the standard potential of the calomel electrode is $E^{0}=0.798 \\mathrm{~V}$.\n\n$F=96487 \\mathrm{C} \\mathrm{mol}^{-1}, \\quad R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}, T=298 \\mathrm{~K}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1459",
        "problem": "If two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\\mathrm{H}_{2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\\mathrm{H}_{2}$ rapidly desorbs. This question examines two kinetic models for $\\mathrm{H}_{2}$ formation on the surface of a dust particle.\n\nIn both models, the rate constant for adsorption of $\\mathrm{H}$ atoms onto the surface of dust particles is $k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1}$. The typical number density of $\\mathrm{H}$ atoms (number of $\\mathrm{H}$ atoms per unit volume) in interstellar space is $[\\mathrm{H}]=10 \\mathrm{~cm}^{-3}$.\n\n[Note: In the following, you may treat numbers of surface-adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations. As a result, the units of the rate constants may be unfamiliar to you. Reaction rates have units of numbers of atoms or molecules per unit time.]\n\nThe $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$Reaction to form $\\mathrm{H}_{2}$ is assumed to be second order. On a dust particle the rate of removal of $\\mathrm{H}$ atoms by reaction is $k_{r} N^{2}$.\nWrite down an equation for the rate of change of $N$, including adsorption, desorption and reaction. Assuming steady state conditions, determine the value of $N$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIf two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\\mathrm{H}_{2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\\mathrm{H}_{2}$ rapidly desorbs. This question examines two kinetic models for $\\mathrm{H}_{2}$ formation on the surface of a dust particle.\n\nIn both models, the rate constant for adsorption of $\\mathrm{H}$ atoms onto the surface of dust particles is $k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1}$. The typical number density of $\\mathrm{H}$ atoms (number of $\\mathrm{H}$ atoms per unit volume) in interstellar space is $[\\mathrm{H}]=10 \\mathrm{~cm}^{-3}$.\n\n[Note: In the following, you may treat numbers of surface-adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations. As a result, the units of the rate constants may be unfamiliar to you. Reaction rates have units of numbers of atoms or molecules per unit time.]\n\nThe $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\nproblem:\nReaction to form $\\mathrm{H}_{2}$ is assumed to be second order. On a dust particle the rate of removal of $\\mathrm{H}$ atoms by reaction is $k_{r} N^{2}$.\nWrite down an equation for the rate of change of $N$, including adsorption, desorption and reaction. Assuming steady state conditions, determine the value of $N$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_67",
        "problem": "Nitrogen, $\\mathrm{N}_{2}$, has the following properties:\n\nnormal melting point: $63.2 \\mathrm{~K}$\n\nnormal boiling point: $77.4 \\mathrm{~K}$\n\ntriple point: $0.127 \\mathrm{~atm}, 63.1 \\mathrm{~K}$\n\ncritical point: $33.5 \\mathrm{~atm}, 126.0 \\mathrm{~K}$\n\nWhich statement about $\\mathrm{N}_{2}$ is correct?\nA: Liquid $\\mathrm{N}_{2}$ is denser than solid $\\mathrm{N}_{2}$.\nB: At sufficiently high pressure, $\\mathrm{N}_{2}$ can be liquefied at $150 \\mathrm{~K}$.\nC: Liquid $\\mathrm{N}_{2}$ and gaseous $\\mathrm{N}_{2}$ can coexist at $63.1 \\mathrm{~K}$ and $1 \\mathrm{~atm}$.\nD: If $\\mathrm{N}_{2}$ is heated from $60 \\mathrm{~K}$ to $70 \\mathrm{~K}$ at $0.100 \\mathrm{~atm}$, it sublimes.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNitrogen, $\\mathrm{N}_{2}$, has the following properties:\n\nnormal melting point: $63.2 \\mathrm{~K}$\n\nnormal boiling point: $77.4 \\mathrm{~K}$\n\ntriple point: $0.127 \\mathrm{~atm}, 63.1 \\mathrm{~K}$\n\ncritical point: $33.5 \\mathrm{~atm}, 126.0 \\mathrm{~K}$\n\nWhich statement about $\\mathrm{N}_{2}$ is correct?\n\nA: Liquid $\\mathrm{N}_{2}$ is denser than solid $\\mathrm{N}_{2}$.\nB: At sufficiently high pressure, $\\mathrm{N}_{2}$ can be liquefied at $150 \\mathrm{~K}$.\nC: Liquid $\\mathrm{N}_{2}$ and gaseous $\\mathrm{N}_{2}$ can coexist at $63.1 \\mathrm{~K}$ and $1 \\mathrm{~atm}$.\nD: If $\\mathrm{N}_{2}$ is heated from $60 \\mathrm{~K}$ to $70 \\mathrm{~K}$ at $0.100 \\mathrm{~atm}$, it sublimes.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_5",
        "problem": "Which of the following best describes the trend in the amount of energy released on addition of an electron to the gas-phase atoms (electron affinity) of the group 1 and group 2 elements in the same period?\nA: More energy is released by the group 1 elements because there is less electron-electron repulsion in the ions.\nB: More energy is released by the group 1 elements because the added electron enters an $s$ orbital rather than a $p$ orbital.\nC: More energy is released by the group 2 elements because they have more protons in the nucleus.\nD: More energy is released by the group 2 elements because addition of an electron to a group 1 element disrupts an unusually stable half-filled $s$ subshell.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following best describes the trend in the amount of energy released on addition of an electron to the gas-phase atoms (electron affinity) of the group 1 and group 2 elements in the same period?\n\nA: More energy is released by the group 1 elements because there is less electron-electron repulsion in the ions.\nB: More energy is released by the group 1 elements because the added electron enters an $s$ orbital rather than a $p$ orbital.\nC: More energy is released by the group 2 elements because they have more protons in the nucleus.\nD: More energy is released by the group 2 elements because addition of an electron to a group 1 element disrupts an unusually stable half-filled $s$ subshell.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_799",
        "problem": "周期表中有些元素有“对角线相似”现象(即周期表中处于对角线位置的两元素性质相似), 如 $\\mathrm{Li} 、 \\mathrm{Mg} ; \\mathrm{Be} 、 \\mathrm{Al} ; \\mathrm{B} 、 \\mathrm{Si}$ 等两两性质相似。现用熔融 $\\mathrm{LiCl}$ 电解的方法可得锂和氯气。若用已潮解的 $\\mathrm{LiCl}$ 加热蒸干并强热至熔融, 再用惰性电极电解, 结果得到金属锂和一种无色无味的气体。据此, 下列说法正确的是\nA: 无色气体为电解生成的锂与水反应放出的 $\\mathrm{H}_{2}$\nB: 电解前 $\\mathrm{LiCl}$ 在加热时已发生水解\nC: 电解时产生的无色气体是 $\\mathrm{O}_{2}$\nD: 无色气体是阳极放出的 $\\mathrm{Cl}_{2}$ 与水作用生成的 $\\mathrm{O}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n周期表中有些元素有“对角线相似”现象(即周期表中处于对角线位置的两元素性质相似), 如 $\\mathrm{Li} 、 \\mathrm{Mg} ; \\mathrm{Be} 、 \\mathrm{Al} ; \\mathrm{B} 、 \\mathrm{Si}$ 等两两性质相似。现用熔融 $\\mathrm{LiCl}$ 电解的方法可得锂和氯气。若用已潮解的 $\\mathrm{LiCl}$ 加热蒸干并强热至熔融, 再用惰性电极电解, 结果得到金属锂和一种无色无味的气体。据此, 下列说法正确的是\n\nA: 无色气体为电解生成的锂与水反应放出的 $\\mathrm{H}_{2}$\nB: 电解前 $\\mathrm{LiCl}$ 在加热时已发生水解\nC: 电解时产生的无色气体是 $\\mathrm{O}_{2}$\nD: 无色气体是阳极放出的 $\\mathrm{Cl}_{2}$ 与水作用生成的 $\\mathrm{O}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1264",
        "problem": "The molar solubility $s\\left(\\mathrm{~mol} \\mathrm{dm}{ }^{-3}\\right)$ of $\\mathrm{Th}\\left(\\mathrm{IO}_{3}\\right)_{4}$ as a function of the solubility product $K_{s p}$ of this sparingly soluble thorium salt is given by the equation:\nA: $s=\\left(K_{s p} / 128\\right)^{1 / 4}$\nB: $s=\\left(K_{s p} / 256\\right)^{1 / 5}$\nC: $s=256 K_{s p}{ }^{1 / 4}$\nD: $s=\\left(128 K_{s p}\\right)^{1 / 4}$\nE: $s=\\left(256 K_{s p}\\right)^{1 / 5}$\nF: $s=\\left(K_{s p} / 128\\right)^{1 / 5} / 2$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe molar solubility $s\\left(\\mathrm{~mol} \\mathrm{dm}{ }^{-3}\\right)$ of $\\mathrm{Th}\\left(\\mathrm{IO}_{3}\\right)_{4}$ as a function of the solubility product $K_{s p}$ of this sparingly soluble thorium salt is given by the equation:\n\nA: $s=\\left(K_{s p} / 128\\right)^{1 / 4}$\nB: $s=\\left(K_{s p} / 256\\right)^{1 / 5}$\nC: $s=256 K_{s p}{ }^{1 / 4}$\nD: $s=\\left(128 K_{s p}\\right)^{1 / 4}$\nE: $s=\\left(256 K_{s p}\\right)^{1 / 5}$\nF: $s=\\left(K_{s p} / 128\\right)^{1 / 5} / 2$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1515",
        "problem": "In Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nCalculate the solubility product constant for calcium carbonate from the data given above.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIn Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nCalculate the solubility product constant for calcium carbonate from the data given above.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-362.jpg?height=166&width=1354&top_left_y=1005&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_561",
        "problem": "用石墨电极电解 $100 \\mathrm{~mL}$ 的氯化钠溶液, 当电路中通过 $1 \\mathrm{~mole}-$ 时, 阴阳两极产生的气体体积比为 $5: 3$, 则原氯化钠溶液的物质的量浓度 $(\\mathrm{mol} / \\mathrm{L})$ 为 $(\\quad)$\nA: 4\nB: 2\nC: 1\nD: 5\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n用石墨电极电解 $100 \\mathrm{~mL}$ 的氯化钠溶液, 当电路中通过 $1 \\mathrm{~mole}-$ 时, 阴阳两极产生的气体体积比为 $5: 3$, 则原氯化钠溶液的物质的量浓度 $(\\mathrm{mol} / \\mathrm{L})$ 为 $(\\quad)$\n\nA: 4\nB: 2\nC: 1\nD: 5\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1497",
        "problem": "Potentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nWrite Nernst equations for redox systems of\n\na) solution I,\n\nb) solution II.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\n\nproblem:\nPotentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nWrite Nernst equations for redox systems of\n\na) solution I,\n\nb) solution II.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir answer types are, in order, [expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            null,
            null
        ],
        "answer_sequence": null,
        "type_sequence": [
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1473",
        "problem": "One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\nA common defect in this crystal is some copper atoms missing with the oxygen lattice unchanged. The composition of one such crystal sample was studied, and $0.2 \\%$ of all copper atoms were found to be in oxidation state +2 .What is $x$ in the empirical formula $\\mathrm{Cu}_{2-x} \\mathrm{O}$ of the crystal?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOne of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\nA common defect in this crystal is some copper atoms missing with the oxygen lattice unchanged. The composition of one such crystal sample was studied, and $0.2 \\%$ of all copper atoms were found to be in oxidation state +2 .\n\nproblem:\nWhat is $x$ in the empirical formula $\\mathrm{Cu}_{2-x} \\mathrm{O}$ of the crystal?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-139.jpg?height=410&width=1236&top_left_y=660&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_527",
        "problem": "$\\mathrm{NO}_{2} 、 \\mathrm{O}_{2}$ 和熔融 $\\mathrm{KNO}_{3}$ 可作燃料电池, 其原理如图所示。该电池在放电过程中石墨 I电极上生成氧化物 $Y, Y$ 可循环使用。下列说法正确的是\n\n[图1]\nA: 放电时, $\\mathrm{NO}_{3}^{-}$向石墨II电极迁移\nB: 电池总反应式为 $4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}=2 \\mathrm{~N}_{2} \\mathrm{O}_{5}$\nC: 石墨II附近发生的反应为 $2 \\mathrm{~N}_{2} \\mathrm{O}_{5}+\\mathrm{O}_{2}+4 \\mathrm{e}^{-}=4 \\mathrm{NO}_{3}^{-}$\nD: 当外电路通过 $4 \\mathrm{~mole}^{-}$，负极上共产生 $2 \\mathrm{molN}_{2} \\mathrm{O}_{5}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{NO}_{2} 、 \\mathrm{O}_{2}$ 和熔融 $\\mathrm{KNO}_{3}$ 可作燃料电池, 其原理如图所示。该电池在放电过程中石墨 I电极上生成氧化物 $Y, Y$ 可循环使用。下列说法正确的是\n\n[图1]\n\nA: 放电时, $\\mathrm{NO}_{3}^{-}$向石墨II电极迁移\nB: 电池总反应式为 $4 \\mathrm{NO}_{2}+\\mathrm{O}_{2}=2 \\mathrm{~N}_{2} \\mathrm{O}_{5}$\nC: 石墨II附近发生的反应为 $2 \\mathrm{~N}_{2} \\mathrm{O}_{5}+\\mathrm{O}_{2}+4 \\mathrm{e}^{-}=4 \\mathrm{NO}_{3}^{-}$\nD: 当外电路通过 $4 \\mathrm{~mole}^{-}$，负极上共产生 $2 \\mathrm{molN}_{2} \\mathrm{O}_{5}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-21.jpg?height=368&width=528&top_left_y=1866&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1192",
        "problem": "Determination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\n$5.624 \\mathrm{~g}$ of water were diluted by methanol up to a total volume of $1.000 \\mathrm{dm}^{3}$ (solution A); $22.45 \\mathrm{~cm}^{3}$ of this solution were used for titration of $15.00 \\mathrm{~cm}^{3}$ of a Fischer reagent (solution B).\n\nThen $25.00 \\mathrm{~cm}^{3}$ of methanol (of the same batch as used for the preparation of solution A) and $10.00 \\mathrm{~cm}^{3}$ of solution $\\mathbf{B}$ were mixed, and the mixture was titrated by the solution $\\mathbf{A} .10 .79 \\mathrm{~cm}^{3}$ of the latter solution were spent.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nDetermination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\n$5.624 \\mathrm{~g}$ of water were diluted by methanol up to a total volume of $1.000 \\mathrm{dm}^{3}$ (solution A); $22.45 \\mathrm{~cm}^{3}$ of this solution were used for titration of $15.00 \\mathrm{~cm}^{3}$ of a Fischer reagent (solution B).\n\nThen $25.00 \\mathrm{~cm}^{3}$ of methanol (of the same batch as used for the preparation of solution A) and $10.00 \\mathrm{~cm}^{3}$ of solution $\\mathbf{B}$ were mixed, and the mixture was titrated by the solution $\\mathbf{A} .10 .79 \\mathrm{~cm}^{3}$ of the latter solution were spent.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{mg} \\mathrm{cm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-651.jpg?height=213&width=900&top_left_y=776&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{mg} \\mathrm{cm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_400",
        "problem": "$\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}$ 是一种锂离子电池正极材料, 充放电过程中正极材料立方晶胞(示意图)的组成变化如图所示, 晶胞内未标出因放电产生的 0 价 $\\mathrm{Zn}$ 原子。下列说法错误的是\n\n[图1]\n\n$\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}$\n\n[图2]\n\n$\\mathrm{Li}_{\\mathrm{x}} \\mathrm{Zn}_{\\mathrm{y}} \\mathrm{S}$\n\n[图3]\n\n$\\mathrm{Li}_{2} \\mathrm{~S}$\nA: 每个 $\\mathrm{Li}_{\\mathrm{x}} \\mathrm{Zn}_{\\mathrm{y}} \\mathrm{S}$ 晶胞中包含的 $\\mathrm{Zn}^{2+}$ 个数为 1\nB: 每个 $\\mathrm{Li}_{2} \\mathrm{~S}$ 晶胞完全转化为 $\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}$ 晶胞, 电极反应式为 $\\mathrm{Li}_{2} \\mathrm{~S}-2 \\mathrm{e}^{-}+\\mathrm{mZn}=\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}+2 \\mathrm{Li}^{+}$\nC: 当 $\\mathrm{Li}_{x} \\mathrm{Zn}_{\\mathrm{y}} \\mathrm{S}$ 转化为 $\\mathrm{Li}_{2} \\mathrm{~S}$ 时, 每个晶胞有 $(2-\\mathrm{x})$ 个 $\\mathrm{Li}^{+}$迁移\nD: 若 $\\mathrm{Li}_{2} \\mathrm{~S}$ 的晶胞参数为 $\\mathrm{anm}$, 则 $\\mathrm{Li}^{+}$和 $\\mathrm{S}^{2}$-间的最短距离为 $\\frac{\\sqrt{3}}{4} \\mathrm{anm}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}$ 是一种锂离子电池正极材料, 充放电过程中正极材料立方晶胞(示意图)的组成变化如图所示, 晶胞内未标出因放电产生的 0 价 $\\mathrm{Zn}$ 原子。下列说法错误的是\n\n[图1]\n\n$\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}$\n\n[图2]\n\n$\\mathrm{Li}_{\\mathrm{x}} \\mathrm{Zn}_{\\mathrm{y}} \\mathrm{S}$\n\n[图3]\n\n$\\mathrm{Li}_{2} \\mathrm{~S}$\n\nA: 每个 $\\mathrm{Li}_{\\mathrm{x}} \\mathrm{Zn}_{\\mathrm{y}} \\mathrm{S}$ 晶胞中包含的 $\\mathrm{Zn}^{2+}$ 个数为 1\nB: 每个 $\\mathrm{Li}_{2} \\mathrm{~S}$ 晶胞完全转化为 $\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}$ 晶胞, 电极反应式为 $\\mathrm{Li}_{2} \\mathrm{~S}-2 \\mathrm{e}^{-}+\\mathrm{mZn}=\\mathrm{Zn}_{\\mathrm{m}} \\mathrm{S}+2 \\mathrm{Li}^{+}$\nC: 当 $\\mathrm{Li}_{x} \\mathrm{Zn}_{\\mathrm{y}} \\mathrm{S}$ 转化为 $\\mathrm{Li}_{2} \\mathrm{~S}$ 时, 每个晶胞有 $(2-\\mathrm{x})$ 个 $\\mathrm{Li}^{+}$迁移\nD: 若 $\\mathrm{Li}_{2} \\mathrm{~S}$ 的晶胞参数为 $\\mathrm{anm}$, 则 $\\mathrm{Li}^{+}$和 $\\mathrm{S}^{2}$-间的最短距离为 $\\frac{\\sqrt{3}}{4} \\mathrm{anm}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-038.jpg?height=297&width=308&top_left_y=545&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-038.jpg?height=243&width=649&top_left_y=598&top_left_x=675",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-038.jpg?height=309&width=328&top_left_y=528&top_left_x=1338",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-038.jpg?height=100&width=1380&top_left_y=2137&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1485",
        "problem": "Nitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample.\n\nCalculate the $\\mathrm{pH}$ of the solution which is titrated in 2.1 when $0 \\mathrm{~cm}^{3}, 9.65 \\mathrm{~cm}^{3}$, $19.30 \\mathrm{~cm}^{3}$ and $28.95 \\mathrm{~cm}^{3}$ of sodium hydroxide have been added. Disregard any volume change during the reaction of ammonia gas with hydrochloric acid. $K_{a}$ for ammonium ion is $5.7 \\times 10^{-10}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nNitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample.\n\nCalculate the $\\mathrm{pH}$ of the solution which is titrated in 2.1 when $0 \\mathrm{~cm}^{3}, 9.65 \\mathrm{~cm}^{3}$, $19.30 \\mathrm{~cm}^{3}$ and $28.95 \\mathrm{~cm}^{3}$ of sodium hydroxide have been added. Disregard any volume change during the reaction of ammonia gas with hydrochloric acid. $K_{a}$ for ammonium ion is $5.7 \\times 10^{-10}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_992",
        "problem": "Chlorine has two abundant stable isotopes, ${ }^{35} \\mathrm{Cl}$ and ${ }^{37} \\mathrm{Cl}$, with atomic masses of $34.97 \\mathrm{amu}$ and $36.96 \\mathrm{amu}$ respectively. What is the percent abundance of the heavier isotope?\nA: $78 \\%$\nB: $24 \\%$\nC: $64 \\%$\nD: $50 \\%$\nE: $36 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nChlorine has two abundant stable isotopes, ${ }^{35} \\mathrm{Cl}$ and ${ }^{37} \\mathrm{Cl}$, with atomic masses of $34.97 \\mathrm{amu}$ and $36.96 \\mathrm{amu}$ respectively. What is the percent abundance of the heavier isotope?\n\nA: $78 \\%$\nB: $24 \\%$\nC: $64 \\%$\nD: $50 \\%$\nE: $36 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1010",
        "problem": "What is the coefficient of $\\mathrm{O}_{2}$ when the following equation is balanced with the smallest whole-number coefficients?\n\n$\\_\\mathrm{Cr}_{2} \\mathrm{O}_{3}+\\ldots \\mathrm{KOH}+\\ldots \\mathrm{O}_{2} \\rightarrow \\mathrm{K}_{2} \\mathrm{CrO}_{4}+\\ldots \\mathrm{H}_{2} \\mathrm{O}$\nA: 2\nB: 3\nC: 4\nD: 5\nE: 6\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the coefficient of $\\mathrm{O}_{2}$ when the following equation is balanced with the smallest whole-number coefficients?\n\n$\\_\\mathrm{Cr}_{2} \\mathrm{O}_{3}+\\ldots \\mathrm{KOH}+\\ldots \\mathrm{O}_{2} \\rightarrow \\mathrm{K}_{2} \\mathrm{CrO}_{4}+\\ldots \\mathrm{H}_{2} \\mathrm{O}$\n\nA: 2\nB: 3\nC: 4\nD: 5\nE: 6\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_14",
        "problem": "The reaction between $\\mathbf{A}$ and $\\mathbf{B}$ involves an intermediate $\\mathbf{I}$ whose concentration remains very low throughout the reaction. The initial rates of a reaction as a function of [B] (at constant [A]) are shown below. Which statement about the mechanism is most consistent with the data?\n\n[figure1]\n\n[B]\nA: A initially reacts reversibly to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{B}$ in a process that is first-order in $\\mathbf{B}$.\nB: A initially reacts reversibly to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{B}$ in a process that is second-order in $\\mathbf{B}$.\nC: B initially reacts reversibly in a first-order process to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{A}$.\nD: $\\mathbf{B}$ initially reacts reversibly in a second-order process to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{A}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction between $\\mathbf{A}$ and $\\mathbf{B}$ involves an intermediate $\\mathbf{I}$ whose concentration remains very low throughout the reaction. The initial rates of a reaction as a function of [B] (at constant [A]) are shown below. Which statement about the mechanism is most consistent with the data?\n\n[figure1]\n\n[B]\n\nA: A initially reacts reversibly to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{B}$ in a process that is first-order in $\\mathbf{B}$.\nB: A initially reacts reversibly to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{B}$ in a process that is second-order in $\\mathbf{B}$.\nC: B initially reacts reversibly in a first-order process to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{A}$.\nD: $\\mathbf{B}$ initially reacts reversibly in a second-order process to form the intermediate $\\mathbf{I}$, which then reacts irreversibly with $\\mathbf{A}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-07.jpg?height=343&width=396&top_left_y=379&top_left_x=428"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_338",
        "problem": "The acid ionization constant for $\\mathrm{HNO}_{2}$ is $K_{a}=4.5 \\times 10^{-4}$ at $298 \\mathrm{~K}$. What is the $\\mathrm{pH}$ of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{2}(\\mathrm{aq})$ at $298 \\mathrm{~K}$ ? (Choose the closest value.)\nA: 1.00\nB: 2.17\nC: 1.67\nD: 3.23\nE: 6.53\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe acid ionization constant for $\\mathrm{HNO}_{2}$ is $K_{a}=4.5 \\times 10^{-4}$ at $298 \\mathrm{~K}$. What is the $\\mathrm{pH}$ of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{2}(\\mathrm{aq})$ at $298 \\mathrm{~K}$ ? (Choose the closest value.)\n\nA: 1.00\nB: 2.17\nC: 1.67\nD: 3.23\nE: 6.53\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_47",
        "problem": "$0.125 \\mathrm{~g}$ of strontium metal reacts with excess liquid ammonia to form strontium amide and hydrogen gas. How many moles of hydrogen are produced?\nA: $7.13 \\times 10^{-4} \\mathrm{~mol}$\nB: $1.43 \\times 10^{-3} \\mathrm{~mol}$\nC: $2.14 \\times 10^{-4} \\mathrm{~mol}$\nD: $2.85 \\times 10^{-3} \\mathrm{~mol}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n$0.125 \\mathrm{~g}$ of strontium metal reacts with excess liquid ammonia to form strontium amide and hydrogen gas. How many moles of hydrogen are produced?\n\nA: $7.13 \\times 10^{-4} \\mathrm{~mol}$\nB: $1.43 \\times 10^{-3} \\mathrm{~mol}$\nC: $2.14 \\times 10^{-4} \\mathrm{~mol}$\nD: $2.85 \\times 10^{-3} \\mathrm{~mol}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1516",
        "problem": "Prefix: $\\mathrm{pH}$ in the blood of a person changed from 7.40 to 7.00 due to lactic acid formed during physical activity. Let an aqueous solution having $\\mathrm{pH}=7.40$ and $\\left[\\mathrm{HCO}_{3}^{-}\\right]=$ 0.022 represent blood in the following calculation. How many moles of lactic acid have been added to $1.00 \\mathrm{dm}^{3}$ of this solution when its $\\mathrm{pH}$ has become 7.00 ?Lactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nBlood contains calcium. Determine the maximum concentration of \"free\" calcium ions in the solution $\\left(\\mathrm{pH}=7.40,\\left[\\mathrm{HCO}_{3}^{-}\\right]=0.022\\right)$ given in Prefix .",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nPrefix: $\\mathrm{pH}$ in the blood of a person changed from 7.40 to 7.00 due to lactic acid formed during physical activity. Let an aqueous solution having $\\mathrm{pH}=7.40$ and $\\left[\\mathrm{HCO}_{3}^{-}\\right]=$ 0.022 represent blood in the following calculation. How many moles of lactic acid have been added to $1.00 \\mathrm{dm}^{3}$ of this solution when its $\\mathrm{pH}$ has become 7.00 ?\n\nproblem:\nLactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nBlood contains calcium. Determine the maximum concentration of \"free\" calcium ions in the solution $\\left(\\mathrm{pH}=7.40,\\left[\\mathrm{HCO}_{3}^{-}\\right]=0.022\\right)$ given in Prefix .\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_556",
        "problem": "利用微生物电化学处理有机废水, 同时可淡化海水并获得酸碱。现以 $\\mathrm{NaCl}$ 溶液模拟海水、采用情性电极, 用如下图所示的装置处理有机废水 (以含有机酸 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液为例), 在直流电场作用下, 双极膜间的水解离成 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$。下列说法正确的是\n\n[图1]\nA: 膜 a 为阴离子交换膜, 膜 b 为阳离子交换膜\nB: 产品室生成的物质为盐酸\nC: 当阴极产生 $22.4 \\mathrm{~L}$ 气体时，理论上可除去模拟海水中 $11.7 \\mathrm{~g} \\mathrm{NaCl}$\nD: 阳极反应式为: $\\mathrm{CH}_{3} \\mathrm{COOH}+2 \\mathrm{H}_{2} \\mathrm{O}+8 \\mathrm{e}^{-}=2 \\mathrm{CO}_{2} \\uparrow+8 \\mathrm{H}^{+}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n利用微生物电化学处理有机废水, 同时可淡化海水并获得酸碱。现以 $\\mathrm{NaCl}$ 溶液模拟海水、采用情性电极, 用如下图所示的装置处理有机废水 (以含有机酸 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液为例), 在直流电场作用下, 双极膜间的水解离成 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$。下列说法正确的是\n\n[图1]\n\nA: 膜 a 为阴离子交换膜, 膜 b 为阳离子交换膜\nB: 产品室生成的物质为盐酸\nC: 当阴极产生 $22.4 \\mathrm{~L}$ 气体时，理论上可除去模拟海水中 $11.7 \\mathrm{~g} \\mathrm{NaCl}$\nD: 阳极反应式为: $\\mathrm{CH}_{3} \\mathrm{COOH}+2 \\mathrm{H}_{2} \\mathrm{O}+8 \\mathrm{e}^{-}=2 \\mathrm{CO}_{2} \\uparrow+8 \\mathrm{H}^{+}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-068.jpg?height=523&width=688&top_left_y=1189&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_994",
        "problem": "The ground state electronic configuration of a certain neutral atom is $[\\mathrm{Xe}] 6 s^{2} 4 f^{14} 5 d^{10} 6 p^{4}$. To which group of the periodic table does this atom belong?\nThis question was NOT marked. The electron configuration was mistakenly given as: $[X e] 6 s^{2} 5 f^{14} 6 d^{10} 6 p^{4}$\nA: group 1\nB: group 3\nC: group 6\nD: group 14\nE: group 16\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe ground state electronic configuration of a certain neutral atom is $[\\mathrm{Xe}] 6 s^{2} 4 f^{14} 5 d^{10} 6 p^{4}$. To which group of the periodic table does this atom belong?\nThis question was NOT marked. The electron configuration was mistakenly given as: $[X e] 6 s^{2} 5 f^{14} 6 d^{10} 6 p^{4}$\n\nA: group 1\nB: group 3\nC: group 6\nD: group 14\nE: group 16\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_921",
        "problem": "下列有关叙述正确的是\nA: $1 \\mathrm{~mol}$ 该有机物 [图1] $3 \\mathrm{~mol}$\nB: 若乙酸分子中的氧都是 ${ }^{18} \\mathrm{O}$, 乙醇分子中的氧都是 ${ }^{16} \\mathrm{O}$, 二者在浓 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ 作用下发生反应，一段时间后，分子中含有 ${ }^{18} \\mathrm{O}$ 的物质有 3 种\nC: 某中性有机物 $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{O}_{2}$ 在稀硫酸的作用下加热得到 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 两种有机物, $\\mathrm{Y}$ 经氧化可最终得到 $\\mathrm{X}$ ，则该中性有机物的结构可能有 4 种\nD: 已知 $\\|++\\xrightarrow{\\triangle}$, 那么要生成 [图2]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列有关叙述正确的是\n\nA: $1 \\mathrm{~mol}$ 该有机物 [图1] $3 \\mathrm{~mol}$\nB: 若乙酸分子中的氧都是 ${ }^{18} \\mathrm{O}$, 乙醇分子中的氧都是 ${ }^{16} \\mathrm{O}$, 二者在浓 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ 作用下发生反应，一段时间后，分子中含有 ${ }^{18} \\mathrm{O}$ 的物质有 3 种\nC: 某中性有机物 $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{O}_{2}$ 在稀硫酸的作用下加热得到 $\\mathrm{X}$ 和 $\\mathrm{Y}$ 两种有机物, $\\mathrm{Y}$ 经氧化可最终得到 $\\mathrm{X}$ ，则该中性有机物的结构可能有 4 种\nD: 已知 $\\|++\\xrightarrow{\\triangle}$, 那么要生成 [图2]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-67.jpg?height=186&width=997&top_left_y=1906&top_left_x=701",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-67.jpg?height=180&width=608&top_left_y=2531&top_left_x=1112"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_544",
        "problem": "二甲基砜是重要的有机合成中间体，可采用“成对间接电氧化”法合成。电解槽中某电极区反应过程如图所示, 其电解液显碱性。\n\n\n[图1]\n下列说法正确的是\nA: 电极反应式为 $\\mathrm{O}_{2}+2 \\mathrm{e}^{-}+2 \\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}_{2}$\nB: 离子交换膜为阴离子交换膜\nC: 反应前后 $\\mathrm{n}\\left(\\mathrm{WO}_{4}^{2-}\\right): \\mathrm{n}\\left(\\mathrm{WO}_{5}^{2-}\\right)$ 不变\nD: 消耗 $1 \\mathrm{~mol}$ 氧气, 理论上可得到 $1 \\mathrm{~mol}$ 二甲基砜\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n二甲基砜是重要的有机合成中间体，可采用“成对间接电氧化”法合成。电解槽中某电极区反应过程如图所示, 其电解液显碱性。\n\n\n[图1]\n下列说法正确的是\n\nA: 电极反应式为 $\\mathrm{O}_{2}+2 \\mathrm{e}^{-}+2 \\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}_{2}$\nB: 离子交换膜为阴离子交换膜\nC: 反应前后 $\\mathrm{n}\\left(\\mathrm{WO}_{4}^{2-}\\right): \\mathrm{n}\\left(\\mathrm{WO}_{5}^{2-}\\right)$ 不变\nD: 消耗 $1 \\mathrm{~mol}$ 氧气, 理论上可得到 $1 \\mathrm{~mol}$ 二甲基砜\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/B6n1xqbS/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1139",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nGive the name of the other by-product formed when $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ reacts with ethanoic anhydride.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nGive the name of the other by-product formed when $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ reacts with ethanoic anhydride.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1130",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nWhat would you expect the $\\mathrm{Cl}-\\mathrm{Sk}-\\mathrm{Cl}$ bond angle to be in a molecule of $\\mathrm{SkCl}_{2}$ if the bonding were covalent and the stuck-at-homium only formed bonds using its outermost valence $p$ orbitals?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nWhat would you expect the $\\mathrm{Cl}-\\mathrm{Sk}-\\mathrm{Cl}$ bond angle to be in a molecule of $\\mathrm{SkCl}_{2}$ if the bonding were covalent and the stuck-at-homium only formed bonds using its outermost valence $p$ orbitals?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $^{\\circ}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$^{\\circ}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_55",
        "problem": "Assuming that $\\Delta H^{\\circ}$ and $\\Delta S^{\\circ}$ do not vary significantly with temperature, at what temperature will graphite and diamond be in equilibrium at $1 \\mathrm{~atm}$ pressure?\n\n|  | $\\Delta H_{\\mathrm{o}}^{\\mathrm{e}}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | $S^{\\mathrm{o}}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{C}$ (s, graphite) | 0.0 | 5.9 |\n| $\\mathrm{C}$ (s, diamond) | 2.1 | 2.5 |\nA: $0.62 \\mathrm{~K}$\nB: $620 \\mathrm{~K}$\nC: $2400 \\mathrm{~K}$\nD: Graphite is more stable than diamond at all temperatures.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAssuming that $\\Delta H^{\\circ}$ and $\\Delta S^{\\circ}$ do not vary significantly with temperature, at what temperature will graphite and diamond be in equilibrium at $1 \\mathrm{~atm}$ pressure?\n\n|  | $\\Delta H_{\\mathrm{o}}^{\\mathrm{e}}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | $S^{\\mathrm{o}}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{C}$ (s, graphite) | 0.0 | 5.9 |\n| $\\mathrm{C}$ (s, diamond) | 2.1 | 2.5 |\n\nA: $0.62 \\mathrm{~K}$\nB: $620 \\mathrm{~K}$\nC: $2400 \\mathrm{~K}$\nD: Graphite is more stable than diamond at all temperatures.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_242",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount (in mol or mmol) of nitric acid that reacts with the $5.000 \\mathrm{~g}$ sample of this mineral.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amount (in mol or mmol) of nitric acid that reacts with the $5.000 \\mathrm{~g}$ sample of this mineral.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_620",
        "problem": "柠檬烯是一种食用香料, 其结构简式如图所示。下列有关柠檬烯的分析正确的是[图1]\nA: $1 \\mathrm{~mol}$ 柠檬烯可以和 $2 \\mathrm{molH}_{2}$ 发生加成反应\nB: 它的分子中所有原子一定在同一平面上\nC: 它和丁基苯( $\\mathrm{C}_{4} \\mathrm{H}_{9}$ )互为同分异构体\nD: 一定条件下, 它分别可以发生加成、取代、氧化、加聚等反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n柠檬烯是一种食用香料, 其结构简式如图所示。下列有关柠檬烯的分析正确的是[图1]\n\nA: $1 \\mathrm{~mol}$ 柠檬烯可以和 $2 \\mathrm{molH}_{2}$ 发生加成反应\nB: 它的分子中所有原子一定在同一平面上\nC: 它和丁基苯( $\\mathrm{C}_{4} \\mathrm{H}_{9}$ )互为同分异构体\nD: 一定条件下, 它分别可以发生加成、取代、氧化、加聚等反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/R0kJ4nw6/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1425",
        "problem": "Solutions containing $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and/or $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ are titrated with a strong base standard solution. Associate the contents of these solutions with the titration curves ( $\\mathrm{pH}$ vs. volume of titrant) shown in the figure:\n\n(For $\\mathrm{H}_{3} \\mathrm{PO}_{4}: p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\n\n[figure1]\n\nVolume of titrant $\\left(\\mathrm{cm}^{3}\\right)$\n\nThe sample contains both in a mole ratio $\\mathrm{H}_{3} \\mathrm{PO}_{4}: \\mathrm{NaH}_{2} \\mathrm{PO}_{4}=2: 1$:\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSolutions containing $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and/or $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ are titrated with a strong base standard solution. Associate the contents of these solutions with the titration curves ( $\\mathrm{pH}$ vs. volume of titrant) shown in the figure:\n\n(For $\\mathrm{H}_{3} \\mathrm{PO}_{4}: p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\n\n[figure1]\n\nVolume of titrant $\\left(\\mathrm{cm}^{3}\\right)$\n\nThe sample contains both in a mole ratio $\\mathrm{H}_{3} \\mathrm{PO}_{4}: \\mathrm{NaH}_{2} \\mathrm{PO}_{4}=2: 1$:\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-459.jpg?height=348&width=1413&top_left_y=934&top_left_x=273"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_689",
        "problem": "香子兰酸甲酯(甲)、甲醇、甲基丙烯酸甲酯(丙)、有机物(乙)、有机物(丁) 存在下列转化关系, 下列说法正确的是\n\n[图1]\nA: 丙的分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}_{2}$\nB: 乙 $\\rightarrow 丁$ 的反应类型为加成反应\nC: 乙、丁中的官能团种类均为 3 种\nD: 甲的同分异构体中含有苯环且取代基与甲完全相同有 9 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n香子兰酸甲酯(甲)、甲醇、甲基丙烯酸甲酯(丙)、有机物(乙)、有机物(丁) 存在下列转化关系, 下列说法正确的是\n\n[图1]\n\nA: 丙的分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{10} \\mathrm{O}_{2}$\nB: 乙 $\\rightarrow 丁$ 的反应类型为加成反应\nC: 乙、丁中的官能团种类均为 3 种\nD: 甲的同分异构体中含有苯环且取代基与甲完全相同有 9 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-63.jpg?height=283&width=1425&top_left_y=2043&top_left_x=353"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1267",
        "problem": "One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.What is the density of pure copper(I) oxide?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOne of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\n\nproblem:\nWhat is the density of pure copper(I) oxide?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~g} \\mathrm{~cm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-139.jpg?height=410&width=1236&top_left_y=660&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~g} \\mathrm{~cm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_682",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HA}$ 溶液, 体系中\n\n$\\lg c\\left(\\mathrm{~A}^{-}\\right) 、 \\lg c(\\mathrm{HA}) 、 \\mathrm{NaOH}$ 溶液的体积与溶液 $\\mathrm{pH}$ 的关系如图所示。下列说法错误的是\n\n[图1]\n\n## $\\lg c\\left(\\mathrm{~A}^{-}\\right)$或 $\\operatorname{g} c(\\mathrm{HA}) \\quad V[\\mathrm{NaOH}(\\mathrm{aq})] / \\mathrm{mL}$\nA: 图中曲线(1)表示 $\\lg c\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{pH}$ 的关系\nB: $25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaA}$ 溶液的 $\\mathrm{pH}$ 约为 9\nC: a 点溶液中, $c\\left(\\mathrm{~A}^{-}\\right)<c(\\mathrm{HA})$\nD: b 点时, $V \\approx 19.8 \\mathrm{~mL}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HA}$ 溶液, 体系中\n\n$\\lg c\\left(\\mathrm{~A}^{-}\\right) 、 \\lg c(\\mathrm{HA}) 、 \\mathrm{NaOH}$ 溶液的体积与溶液 $\\mathrm{pH}$ 的关系如图所示。下列说法错误的是\n\n[图1]\n\n## $\\lg c\\left(\\mathrm{~A}^{-}\\right)$或 $\\operatorname{g} c(\\mathrm{HA}) \\quad V[\\mathrm{NaOH}(\\mathrm{aq})] / \\mathrm{mL}$\n\nA: 图中曲线(1)表示 $\\lg c\\left(\\mathrm{~A}^{-}\\right)$与 $\\mathrm{pH}$ 的关系\nB: $25^{\\circ} \\mathrm{C}$ 时, $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaA}$ 溶液的 $\\mathrm{pH}$ 约为 9\nC: a 点溶液中, $c\\left(\\mathrm{~A}^{-}\\right)<c(\\mathrm{HA})$\nD: b 点时, $V \\approx 19.8 \\mathrm{~mL}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-032.jpg?height=614&width=714&top_left_y=167&top_left_x=408"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_78",
        "problem": "What is $\\Delta G^{\\circ}{ }_{\\mathrm{f}}$ of $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{~g})$ at $298 \\mathrm{~K}$ ?\n\n|  | $\\mathrm{C}_{6} \\mathrm{H}_{6}(l)$ | $\\mathrm{C}_{6} \\mathrm{H}_{6}(g)$ |\n| :--- | :--- | :--- |\n| $\\Delta H_{\\mathrm{f}}^{\\mathrm{o}}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | 49.0 | 82.9 |\n| $S^{\\mathrm{o}}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ | 173.3 | 269.0 |\n| $\\Delta G_{\\mathrm{f}}^{\\mathrm{o}}(298 \\mathrm{~K}), \\mathrm{kJ} \\mathrm{mol}^{-1}$ | 124.3 |  |\nA: $2.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $121.6 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $127.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $129.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is $\\Delta G^{\\circ}{ }_{\\mathrm{f}}$ of $\\mathrm{C}_{6} \\mathrm{H}_{6}(\\mathrm{~g})$ at $298 \\mathrm{~K}$ ?\n\n|  | $\\mathrm{C}_{6} \\mathrm{H}_{6}(l)$ | $\\mathrm{C}_{6} \\mathrm{H}_{6}(g)$ |\n| :--- | :--- | :--- |\n| $\\Delta H_{\\mathrm{f}}^{\\mathrm{o}}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | 49.0 | 82.9 |\n| $S^{\\mathrm{o}}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ | 173.3 | 269.0 |\n| $\\Delta G_{\\mathrm{f}}^{\\mathrm{o}}(298 \\mathrm{~K}), \\mathrm{kJ} \\mathrm{mol}^{-1}$ | 124.3 |  |\n\nA: $2.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $121.6 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $127.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $129.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1519",
        "problem": "Problem I:  Calculate the concentration of calcium ions in water in equilibrium with calcium carbonate in an atmosphere with a partial pressure of carbon dioxide of 1.00 bar.In Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nA $0.0150 \\mathrm{~m}$ solution of calcium hydroxide is saturated with carbon dioxide gas at a partial pressure of 1.00 bar. Calculate the concentration of calcium ions in the solution by considering the equilibrium equation given above in connection with problem I.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nProblem I:  Calculate the concentration of calcium ions in water in equilibrium with calcium carbonate in an atmosphere with a partial pressure of carbon dioxide of 1.00 bar.\n\nproblem:\nIn Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nA $0.0150 \\mathrm{~m}$ solution of calcium hydroxide is saturated with carbon dioxide gas at a partial pressure of 1.00 bar. Calculate the concentration of calcium ions in the solution by considering the equilibrium equation given above in connection with problem I.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-362.jpg?height=166&width=1354&top_left_y=1005&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_10",
        "problem": "A $0.10 \\mathrm{M}$ solution of glycolic acid is $3.8 \\%$ ionized. What is the percent ionization of a $0.50 \\mathrm{M}$ solution of glycolic acid?\nA: $1.7 \\%$\nB: $3.8 \\%$\nC: $8.5 \\%$\nD: $19.0 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $0.10 \\mathrm{M}$ solution of glycolic acid is $3.8 \\%$ ionized. What is the percent ionization of a $0.50 \\mathrm{M}$ solution of glycolic acid?\n\nA: $1.7 \\%$\nB: $3.8 \\%$\nC: $8.5 \\%$\nD: $19.0 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_390",
        "problem": "If elemental bromine is being formed according to the equation below at a rate of $0.056 \\mathrm{M} \\mathrm{s}^{-1}$, at what rate is bromide ion being consumed?\n\n$$\n\\begin{gathered}\n5 \\mathrm{Br}^{-}(a q)+\\mathrm{BrO}_{3}^{-}(a q)+6 \\mathrm{H}^{+}(a q) \\rightarrow \\\\\n3 \\mathrm{Br}_{2}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\nA: $0.019 \\mathrm{M} \\mathrm{s}^{-1}$\nB: $0.034 \\mathrm{M} \\mathrm{s}^{-1}$\nC: $0.056 \\mathrm{M} \\mathrm{s}^{-1}$\nD: $0.093 \\mathrm{M} \\mathrm{s}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf elemental bromine is being formed according to the equation below at a rate of $0.056 \\mathrm{M} \\mathrm{s}^{-1}$, at what rate is bromide ion being consumed?\n\n$$\n\\begin{gathered}\n5 \\mathrm{Br}^{-}(a q)+\\mathrm{BrO}_{3}^{-}(a q)+6 \\mathrm{H}^{+}(a q) \\rightarrow \\\\\n3 \\mathrm{Br}_{2}(a q)+3 \\mathrm{H}_{2} \\mathrm{O}(l)\n\\end{gathered}\n$$\n\nA: $0.019 \\mathrm{M} \\mathrm{s}^{-1}$\nB: $0.034 \\mathrm{M} \\mathrm{s}^{-1}$\nC: $0.056 \\mathrm{M} \\mathrm{s}^{-1}$\nD: $0.093 \\mathrm{M} \\mathrm{s}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1277",
        "problem": "Phosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nDetermine the $\\mathrm{pH}$ of the cola (ignoring the second and the third dissociation steps for phosphoric acid). Assume that the acidity of the cola arises only from phosphoric acid.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nPhosphoric acid is of a great importance in fertilizer industry. Besides, phosphoric acid and its various salts have a $\\mathrm{n}$ umber of applications in metal treatment, food, detergent and toothpaste industries.\n\nThe pK values of the three successive dissociations of phosphoric acid at $25^{\\circ} \\mathrm{C}$ are:\n\n$$\n\\begin{aligned}\n& p K_{1 a}=2.12 \\\\\n& p K_{2 a}=7.21 \\\\\n& p K_{3 a}=12.32\n\\end{aligned}\n$$\n\nDetermine the $\\mathrm{pH}$ of the cola (ignoring the second and the third dissociation steps for phosphoric acid). Assume that the acidity of the cola arises only from phosphoric acid.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_294",
        "problem": "In acidic solution, zinc metal reacts spontaneously with $\\mathrm{ReO}_{4}{ }^{-}$. The unbalanced chemical equation for the reaction is given below.\n\n$\\mathrm{Zn}(s)+\\mathrm{ReO}_{4}{ }^{-}(a q)+\\mathrm{H}^{+}(a q) \\rightarrow \\mathrm{Re}(s)+\\mathrm{Zn}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n\nWhat is the oxidation state of rhenium $(\\mathrm{Re})$ in $\\mathrm{ReO}_{4}^{-}$?\nA: 0\nB: +1\nC: +3\nD: +4\nE: +7\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn acidic solution, zinc metal reacts spontaneously with $\\mathrm{ReO}_{4}{ }^{-}$. The unbalanced chemical equation for the reaction is given below.\n\n$\\mathrm{Zn}(s)+\\mathrm{ReO}_{4}{ }^{-}(a q)+\\mathrm{H}^{+}(a q) \\rightarrow \\mathrm{Re}(s)+\\mathrm{Zn}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n\nWhat is the oxidation state of rhenium $(\\mathrm{Re})$ in $\\mathrm{ReO}_{4}^{-}$?\n\nA: 0\nB: +1\nC: +3\nD: +4\nE: +7\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_835",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{Na}_{2} \\mathrm{~S}$ 沉淀 $\\mathrm{Cu}^{2+} 、 \\mathrm{Zn}^{2+}$ 两种金属离子 $\\left(\\mathrm{M}^{2+}\\right)$, 所需 $\\mathrm{S}^{2-}$ 最低浓度的对数值 $\\operatorname{lgc}\\left(\\mathrm{S}^{2-}\\right)$ 与 $\\operatorname{lgc}\\left(\\mathrm{M}^{2+}\\right)$ 的关系如图所示, 下列说法不正确的是\n\n[图1]\nA: $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{K}(\\mathrm{CuS})$ 约为 $1 \\times 10^{-35}$\nB: 向 $100 \\mathrm{mLZn}^{2+} 、 \\mathrm{Cu}^{2+}$ 浓度均为 $10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的混合溶液 中逐滴加入 $10^{-4} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, $\\mathrm{Cu}^{2+}$ 先沉淀\nC: $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$\nD: 向 $\\mathrm{Cu}^{2+}$ 浓度为 $10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的工业废水中加入 $\\mathrm{ZnS}$ 粉末, 会有 $\\mathrm{CuS}$ 沉淀析出\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $\\mathrm{Na}_{2} \\mathrm{~S}$ 沉淀 $\\mathrm{Cu}^{2+} 、 \\mathrm{Zn}^{2+}$ 两种金属离子 $\\left(\\mathrm{M}^{2+}\\right)$, 所需 $\\mathrm{S}^{2-}$ 最低浓度的对数值 $\\operatorname{lgc}\\left(\\mathrm{S}^{2-}\\right)$ 与 $\\operatorname{lgc}\\left(\\mathrm{M}^{2+}\\right)$ 的关系如图所示, 下列说法不正确的是\n\n[图1]\n\nA: $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{K}(\\mathrm{CuS})$ 约为 $1 \\times 10^{-35}$\nB: 向 $100 \\mathrm{mLZn}^{2+} 、 \\mathrm{Cu}^{2+}$ 浓度均为 $10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的混合溶液 中逐滴加入 $10^{-4} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液, $\\mathrm{Cu}^{2+}$ 先沉淀\nC: $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)$\nD: 向 $\\mathrm{Cu}^{2+}$ 浓度为 $10^{-5} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的工业废水中加入 $\\mathrm{ZnS}$ 粉末, 会有 $\\mathrm{CuS}$ 沉淀析出\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-006.jpg?height=480&width=594&top_left_y=405&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_240",
        "problem": "Which of the following contains molecules with linear molecular geometry in the solid state?\nA: $\\mathrm{CO}_{2}$\nB: $\\mathrm{KCl}$\nC: MgO\nD: $\\mathrm{GeO}_{2}$\nE: $\\mathrm{MgCl}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following contains molecules with linear molecular geometry in the solid state?\n\nA: $\\mathrm{CO}_{2}$\nB: $\\mathrm{KCl}$\nC: MgO\nD: $\\mathrm{GeO}_{2}$\nE: $\\mathrm{MgCl}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_502",
        "problem": "钾离子电池因其低成本和较高的能量/功率密度, 引起了广泛关注。一种钾离子电池的总反应为 $x \\mathrm{KC}_{8}+\\mathrm{K}_{0.5-x} \\mathrm{MnO}_{2} \\stackrel{\\text { 充电 }}{\\stackrel{\\text { 放电 }}{\\rightleftharpoons}} 8 x \\mathrm{C}+\\mathrm{K}_{0.5} \\mathrm{MnO}_{2}$, 其工作原理如图所示。下列说法正确的是\n\n[图1]\nA: 该电池的交换膜为阳离子交换膜\nB: 充电时, a 极接电源的负极发生氧化反应\nC: 放电时, $\\mathrm{b}$ 极发生的反应为 $\\mathrm{K}_{0.5-x} \\mathrm{MnO}_{2}+x \\mathrm{~K}^{+}+x \\mathrm{e}^{-}=\\mathrm{K}_{0.5} \\mathrm{MnO}_{2}$\nD: 放电时, 理论上电路通过 $4 \\mathrm{~mole}^{-}, \\mathrm{a}$ 极质量减小 $39 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n钾离子电池因其低成本和较高的能量/功率密度, 引起了广泛关注。一种钾离子电池的总反应为 $x \\mathrm{KC}_{8}+\\mathrm{K}_{0.5-x} \\mathrm{MnO}_{2} \\stackrel{\\text { 充电 }}{\\stackrel{\\text { 放电 }}{\\rightleftharpoons}} 8 x \\mathrm{C}+\\mathrm{K}_{0.5} \\mathrm{MnO}_{2}$, 其工作原理如图所示。下列说法正确的是\n\n[图1]\n\nA: 该电池的交换膜为阳离子交换膜\nB: 充电时, a 极接电源的负极发生氧化反应\nC: 放电时, $\\mathrm{b}$ 极发生的反应为 $\\mathrm{K}_{0.5-x} \\mathrm{MnO}_{2}+x \\mathrm{~K}^{+}+x \\mathrm{e}^{-}=\\mathrm{K}_{0.5} \\mathrm{MnO}_{2}$\nD: 放电时, 理论上电路通过 $4 \\mathrm{~mole}^{-}, \\mathrm{a}$ 极质量减小 $39 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-27.jpg?height=523&width=582&top_left_y=158&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_594",
        "problem": "腺苷酸-AMP 的结构如图所示。下列说法不正确的是（）[图1]\nA: 分子式为 $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{5} \\mathrm{O}_{7} \\mathrm{P}$\nB: “磷酸结构”部分中心 $\\mathrm{P}$ 原子为 $\\mathrm{sp}^{3}$ 杂化\nC: “核糖结构”部分含有 5 个手性碳原子\nD: “腺嘌呤结构”中 5 个氮原子均可与过渡金属离子形成配位健\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n腺苷酸-AMP 的结构如图所示。下列说法不正确的是（）[图1]\n\nA: 分子式为 $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{~N}_{5} \\mathrm{O}_{7} \\mathrm{P}$\nB: “磷酸结构”部分中心 $\\mathrm{P}$ 原子为 $\\mathrm{sp}^{3}$ 杂化\nC: “核糖结构”部分含有 5 个手性碳原子\nD: “腺嘌呤结构”中 5 个氮原子均可与过渡金属离子形成配位健\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/bv7QZF2L/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_582",
        "problem": "光电化学装置可以将 $\\mathrm{CO}_{2}$ 还原为有机物, 实现碳资源的再生利用。图 1 是光电化学\n\n法实现 $\\mathrm{CO}_{2}$ 转化的 $\\mathrm{BiVO}_{4}|\\mathrm{NaCl}(\\mathrm{aq})| \\mathrm{Cu}$ 电解装置示意图。在不同电压条件下进行光照电解实验, 不同有机产物的法拉第效率（PE）如图 2 所示。\n\n[图1]\n\n图1 光电化学法转化装置示意图\n\n[图2]\n\n图2 有机产物的法拉第效率随电压变化图\n\n资料：I. 法拉第效率 $(\\mathrm{PE}) ; \\mathrm{PE}(\\mathrm{X})=n$ (生成 $\\mathrm{X}$ 的电子) $/ n$ (通过电极的电子)\n\nII. 选择性 $(\\mathrm{S}): \\mathrm{S}(\\mathrm{X})=n\\left(\\right.$ 生成 $\\mathrm{X}$ 的 $\\left.\\mathrm{CO}_{2}\\right) / n\\left(\\right.$ 发生反应的 $\\left.\\mathrm{CO}_{2}\\right) \\times 100 \\%$\n\n以下说法不正确的是\nA: 使用 $\\mathrm{BiVO}_{4}$ 光电极可以节约电能, 为阴极 $\\mathrm{CO}_{2}$ 的转化提供 $\\mathrm{H}+$\nB: 电解电压为 $0.75 \\mathrm{~V}$ 时, $\\mathrm{Cu}$ 电极只有 $\\mathrm{CO}_{2}$ 被还原\nC: 电解电压为 $0.9 \\mathrm{~V}$ 时, $\\mathrm{S}\\left(\\mathrm{CH}_{2} \\mathrm{O}\\right)=85 \\%$\nD: 电解电压为 $1.0 \\mathrm{~V}$ 时, 每生成 $1 \\mathrm{~mol} \\mathrm{CH}_{2} \\mathrm{O}$ 就有 $10 \\mathrm{~mol} \\mathrm{H}^{+}$移向甲室\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n光电化学装置可以将 $\\mathrm{CO}_{2}$ 还原为有机物, 实现碳资源的再生利用。图 1 是光电化学\n\n法实现 $\\mathrm{CO}_{2}$ 转化的 $\\mathrm{BiVO}_{4}|\\mathrm{NaCl}(\\mathrm{aq})| \\mathrm{Cu}$ 电解装置示意图。在不同电压条件下进行光照电解实验, 不同有机产物的法拉第效率（PE）如图 2 所示。\n\n[图1]\n\n图1 光电化学法转化装置示意图\n\n[图2]\n\n图2 有机产物的法拉第效率随电压变化图\n\n资料：I. 法拉第效率 $(\\mathrm{PE}) ; \\mathrm{PE}(\\mathrm{X})=n$ (生成 $\\mathrm{X}$ 的电子) $/ n$ (通过电极的电子)\n\nII. 选择性 $(\\mathrm{S}): \\mathrm{S}(\\mathrm{X})=n\\left(\\right.$ 生成 $\\mathrm{X}$ 的 $\\left.\\mathrm{CO}_{2}\\right) / n\\left(\\right.$ 发生反应的 $\\left.\\mathrm{CO}_{2}\\right) \\times 100 \\%$\n\n以下说法不正确的是\n\nA: 使用 $\\mathrm{BiVO}_{4}$ 光电极可以节约电能, 为阴极 $\\mathrm{CO}_{2}$ 的转化提供 $\\mathrm{H}+$\nB: 电解电压为 $0.75 \\mathrm{~V}$ 时, $\\mathrm{Cu}$ 电极只有 $\\mathrm{CO}_{2}$ 被还原\nC: 电解电压为 $0.9 \\mathrm{~V}$ 时, $\\mathrm{S}\\left(\\mathrm{CH}_{2} \\mathrm{O}\\right)=85 \\%$\nD: 电解电压为 $1.0 \\mathrm{~V}$ 时, 每生成 $1 \\mathrm{~mol} \\mathrm{CH}_{2} \\mathrm{O}$ 就有 $10 \\mathrm{~mol} \\mathrm{H}^{+}$移向甲室\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-01.jpg?height=480&width=651&top_left_y=819&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-01.jpg?height=565&width=794&top_left_y=777&top_left_x=994"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_720",
        "problem": "1. 有机物 X 完全燃烧的产物只有二氧化碳和水, 元素组成分析发现, 该物质中碳元素的质量分数为 $64.86 \\%$, 氢元素的质量分数为 $13.51 \\%$, 它的核磁共振氢谱有 5 组明显的吸收峰，且只含有两个甲基。下列关于有机物 X 的说法正确的是\nA: 分子式为 $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}$\nB: X只可能催化氧化成醛\nC: 与 X 互为官能团异构的同分异构体有 3 种\nD: X 可能消去生成两种结构的烯烃\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n1. 有机物 X 完全燃烧的产物只有二氧化碳和水, 元素组成分析发现, 该物质中碳元素的质量分数为 $64.86 \\%$, 氢元素的质量分数为 $13.51 \\%$, 它的核磁共振氢谱有 5 组明显的吸收峰，且只含有两个甲基。下列关于有机物 X 的说法正确的是\n\nA: 分子式为 $\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}$\nB: X只可能催化氧化成醛\nC: 与 X 互为官能团异构的同分异构体有 3 种\nD: X 可能消去生成两种结构的烯烃\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1363",
        "problem": "A feathery, greenish solid precipitate can be observed if chlorine gas is bubbled into water close to its freezing point. Similar precipitates form with other gases such as methane and noble gases. These materials are interesting because vast quantities of the so-called methane-hydrates are supposed to exist in nature (comparable in quantity with other natural gas deposits).\n\nThese precipitates all have related structures. The molecules of water just above its freezing point form a hydrogen-bonded structure. The gas molecules stabilize this framework by filling in the rather large cavities in the water structure forming clathrates. The crystals of chlorine and methane clathrates have the same structure. Their main characteristics are dodecahedra formed from 20 water molecules. The unit cell of the crystal can be thought as a body-centered cubic arrangement built from these dodecahedra which are almost spherical objects. The dodecahedra are connected via additional water molecules located on the faces of the unit cell. Two water molecules can be found on each face of the unit cell. The unit cell has an edge dimension of $1.182 \\mathrm{~nm}$. There are two types of cavities in this structure. One is the internal space in the dodecahedra (A). These are somewhat smaller than the other type of voids (B), of which there are 6 for each unit cell.\n\nIf all cavities contain a guest molecule, what is the ratio of the number of water to the number of guest molecules?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nA feathery, greenish solid precipitate can be observed if chlorine gas is bubbled into water close to its freezing point. Similar precipitates form with other gases such as methane and noble gases. These materials are interesting because vast quantities of the so-called methane-hydrates are supposed to exist in nature (comparable in quantity with other natural gas deposits).\n\nThese precipitates all have related structures. The molecules of water just above its freezing point form a hydrogen-bonded structure. The gas molecules stabilize this framework by filling in the rather large cavities in the water structure forming clathrates. The crystals of chlorine and methane clathrates have the same structure. Their main characteristics are dodecahedra formed from 20 water molecules. The unit cell of the crystal can be thought as a body-centered cubic arrangement built from these dodecahedra which are almost spherical objects. The dodecahedra are connected via additional water molecules located on the faces of the unit cell. Two water molecules can be found on each face of the unit cell. The unit cell has an edge dimension of $1.182 \\mathrm{~nm}$. There are two types of cavities in this structure. One is the internal space in the dodecahedra (A). These are somewhat smaller than the other type of voids (B), of which there are 6 for each unit cell.\n\nIf all cavities contain a guest molecule, what is the ratio of the number of water to the number of guest molecules?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_534",
        "problem": "氯化亚砜 $\\left(\\mathrm{SOCl}_{2}\\right)$ 是一种无色或淡黄色发烟液体, 遇水剧烈反应生成两种酸性气体,可用 $\\mathrm{SOCl}_{2}$ 和 $\\mathrm{FeCl}_{3} \\cdot \\mathrm{nH}_{2} \\mathrm{O}$ 制备无水 $\\mathrm{FeCl}_{3}$ 。由于该实验过程中可能会发生副反应使产品不纯, 为测定 $\\mathrm{n}$ 值, 设计如下实验方案。\n\n(1) 称取 $\\mathrm{m}_{1} \\mathrm{~g} \\mathrm{FeCl}_{3} \\cdot \\mathrm{HH}_{2} \\mathrm{O}$ 于试管中, 加入足量 $\\mathrm{SOCl}_{2}$, 充分反应后固体质量为 $\\mathrm{m}_{2} \\mathrm{~g}$ 。反应后的固体加入稀硫酸溶解, 并分成两等份。\n\n(2)向其中一份溶液中滴加试剂 X, 验证实验过程中发生了副反应。\n\n(3) 向另一份溶液中加入适量还原剂将 $\\mathrm{Fe}^{3+}$ 完全还原为 $\\mathrm{Fe}^{2+}$, 加入稀 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ 酸化后, 用新配制的 $\\mathrm{cmol} \\cdot \\mathrm{L}^{-1} \\mathrm{~K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 标准溶液滴定 $\\mathrm{Fe}^{2+}$, 达终点时消耗标准溶液 $\\mathrm{VmL}$ (滴定过程中 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 转化为 $\\mathrm{Cr}^{3+}, \\mathrm{Cl}^{-}$不反应)。\n\n$\\mathrm{FeCl}_{3} \\bullet \\mathrm{nH}_{2} \\mathrm{O}$ 中 $\\mathrm{n}$ 值计算正确的是\nA: $\\frac{162.5\\left(\\mathrm{~m}_{1}-\\mathrm{m}_{2}\\right)}{18 \\mathrm{~m}_{2}}$\nB: $\\frac{1000 \\mathrm{~m}_{1}}{108 \\mathrm{cV}}-\\frac{162.5}{18}$\nC: $\\frac{1000 \\mathrm{~m}_{1}}{216 \\mathrm{cV}}-\\frac{162.5}{18}$\nD: $\\frac{1000\\left(\\mathrm{~m}_{1}-\\mathrm{m}_{2}\\right)}{216 \\mathrm{cV}}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n氯化亚砜 $\\left(\\mathrm{SOCl}_{2}\\right)$ 是一种无色或淡黄色发烟液体, 遇水剧烈反应生成两种酸性气体,可用 $\\mathrm{SOCl}_{2}$ 和 $\\mathrm{FeCl}_{3} \\cdot \\mathrm{nH}_{2} \\mathrm{O}$ 制备无水 $\\mathrm{FeCl}_{3}$ 。由于该实验过程中可能会发生副反应使产品不纯, 为测定 $\\mathrm{n}$ 值, 设计如下实验方案。\n\n(1) 称取 $\\mathrm{m}_{1} \\mathrm{~g} \\mathrm{FeCl}_{3} \\cdot \\mathrm{HH}_{2} \\mathrm{O}$ 于试管中, 加入足量 $\\mathrm{SOCl}_{2}$, 充分反应后固体质量为 $\\mathrm{m}_{2} \\mathrm{~g}$ 。反应后的固体加入稀硫酸溶解, 并分成两等份。\n\n(2)向其中一份溶液中滴加试剂 X, 验证实验过程中发生了副反应。\n\n(3) 向另一份溶液中加入适量还原剂将 $\\mathrm{Fe}^{3+}$ 完全还原为 $\\mathrm{Fe}^{2+}$, 加入稀 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$ 酸化后, 用新配制的 $\\mathrm{cmol} \\cdot \\mathrm{L}^{-1} \\mathrm{~K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 标准溶液滴定 $\\mathrm{Fe}^{2+}$, 达终点时消耗标准溶液 $\\mathrm{VmL}$ (滴定过程中 $\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}$ 转化为 $\\mathrm{Cr}^{3+}, \\mathrm{Cl}^{-}$不反应)。\n\n$\\mathrm{FeCl}_{3} \\bullet \\mathrm{nH}_{2} \\mathrm{O}$ 中 $\\mathrm{n}$ 值计算正确的是\n\nA: $\\frac{162.5\\left(\\mathrm{~m}_{1}-\\mathrm{m}_{2}\\right)}{18 \\mathrm{~m}_{2}}$\nB: $\\frac{1000 \\mathrm{~m}_{1}}{108 \\mathrm{cV}}-\\frac{162.5}{18}$\nC: $\\frac{1000 \\mathrm{~m}_{1}}{216 \\mathrm{cV}}-\\frac{162.5}{18}$\nD: $\\frac{1000\\left(\\mathrm{~m}_{1}-\\mathrm{m}_{2}\\right)}{216 \\mathrm{cV}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1381",
        "problem": "Hydrogen is more energy-dense than carbon, by mass. Thus, historically there has been a move toward fuel with higher hydrogen content: coal $\\rightarrow$ oil $\\rightarrow$ natural gas $\\rightarrow$ hydrogen. Cost-effective production and safe storage of hydrogen are two major hurdles to the successful inauguration of a hydrogen economy.\n\nCalculate the ratio between heat generated when hydrogen is burned and heat generated when the same weight of carbon is burned. The difference comes to a large extent from the fact that the most abundant isotope of hydrogen has no neutron and hydrogen has no inner electron shell. $\\Delta H_{\\mathrm{f}}^{0}\\left[\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})\\right]=-286 \\mathrm{~kJ} / \\mathrm{mol}, \\Delta H_{\\mathrm{f}}{ }^{0}$ $\\left[\\mathrm{CO}_{2}(\\mathrm{~g})\\right]=-394 \\mathrm{~kJ} / \\mathrm{mol}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nHydrogen is more energy-dense than carbon, by mass. Thus, historically there has been a move toward fuel with higher hydrogen content: coal $\\rightarrow$ oil $\\rightarrow$ natural gas $\\rightarrow$ hydrogen. Cost-effective production and safe storage of hydrogen are two major hurdles to the successful inauguration of a hydrogen economy.\n\nCalculate the ratio between heat generated when hydrogen is burned and heat generated when the same weight of carbon is burned. The difference comes to a large extent from the fact that the most abundant isotope of hydrogen has no neutron and hydrogen has no inner electron shell. $\\Delta H_{\\mathrm{f}}^{0}\\left[\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})\\right]=-286 \\mathrm{~kJ} / \\mathrm{mol}, \\Delta H_{\\mathrm{f}}{ }^{0}$ $\\left[\\mathrm{CO}_{2}(\\mathrm{~g})\\right]=-394 \\mathrm{~kJ} / \\mathrm{mol}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1493",
        "problem": "Solutions containing $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and/or $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ are titrated with a strong base standard solution. Associate the contents of these solutions with the titration curves ( $\\mathrm{pH}$ vs. volume of titrant) shown in the figure:\n\n(For $\\mathrm{H}_{3} \\mathrm{PO}_{4}: p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\n\n[figure1]\n\nVolume of titrant $\\left(\\mathrm{cm}^{3}\\right)$\n\nThe sample contains $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ only:\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSolutions containing $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and/or $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ are titrated with a strong base standard solution. Associate the contents of these solutions with the titration curves ( $\\mathrm{pH}$ vs. volume of titrant) shown in the figure:\n\n(For $\\mathrm{H}_{3} \\mathrm{PO}_{4}: p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\n\n[figure1]\n\nVolume of titrant $\\left(\\mathrm{cm}^{3}\\right)$\n\nThe sample contains $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ only:\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-459.jpg?height=348&width=1413&top_left_y=934&top_left_x=273"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1000",
        "problem": "If equal volumes of $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solutions of $\\mathrm{NaOH}$ and $\\mathrm{HCl}$ are mixed, what is the $\\mathrm{pH}$ of the resulting solution at $298 \\mathrm{~K}$ ?\nA: 1\nB: 13\nC: 7\nD: 1.3\nE: 12.7\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf equal volumes of $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1}$ solutions of $\\mathrm{NaOH}$ and $\\mathrm{HCl}$ are mixed, what is the $\\mathrm{pH}$ of the resulting solution at $298 \\mathrm{~K}$ ?\n\nA: 1\nB: 13\nC: 7\nD: 1.3\nE: 12.7\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_708",
        "problem": "阴离子 $\\mathrm{PO}_{4}^{3-}$ 和二脲基分子能通过一种弱相互作用形成超分子阴离子配合物, 如下图所示(图中省略阴离子配合物中部分原子)。下列关于该阴离子配合物的说法正确的是\n\n[图1]\n\n二脲基分子\n\n[图2]\n\n阴离子配合物空间结构\nA: 所含元素电负性最大的是 $\\mathrm{N}$\nB: 二脲基分子中所有碳原子不可能共平面\nC: $\\mathrm{PO}_{4}^{3-}$ 离子与 $\\mathrm{SO}_{4}^{2-}$ 离子的空间结构不同\nD: 二脲基分子中 $\\mathrm{N}-\\mathrm{H}$ 的 $\\mathrm{H}$ 和 $\\mathrm{PO}_{4}^{3-}$ 离子的 $\\mathrm{O}$ 形成氢键\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n阴离子 $\\mathrm{PO}_{4}^{3-}$ 和二脲基分子能通过一种弱相互作用形成超分子阴离子配合物, 如下图所示(图中省略阴离子配合物中部分原子)。下列关于该阴离子配合物的说法正确的是\n\n[图1]\n\n二脲基分子\n\n[图2]\n\n阴离子配合物空间结构\n\nA: 所含元素电负性最大的是 $\\mathrm{N}$\nB: 二脲基分子中所有碳原子不可能共平面\nC: $\\mathrm{PO}_{4}^{3-}$ 离子与 $\\mathrm{SO}_{4}^{2-}$ 离子的空间结构不同\nD: 二脲基分子中 $\\mathrm{N}-\\mathrm{H}$ 的 $\\mathrm{H}$ 和 $\\mathrm{PO}_{4}^{3-}$ 离子的 $\\mathrm{O}$ 形成氢键\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-11.jpg?height=297&width=431&top_left_y=1325&top_left_x=344",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-11.jpg?height=354&width=379&top_left_y=1282&top_left_x=810"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_591",
        "problem": "某三元羧酸 $\\mathrm{H}_{3} \\mathrm{~A}$ 在表面活性剂、洗涤剂、润滑剂等方面具有重要的地位。常温时,向 $10 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中商入 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaOH}$ 溶液, $\\mathrm{H}_{3} \\mathrm{~A} 、 \\mathrm{H}_{2} \\mathrm{~A}^{-}$、 $\\mathrm{HA}^{2-} 、 \\mathrm{~A}^{3-}$ 的物质的量分数与溶液的 $\\mathrm{pH}$ 的关系如 图所示。下列说法错误的是\n\n[图1]\nA: 该温度下, $\\mathrm{H}_{3} \\mathrm{~A}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}=10^{-\\mathrm{a}}$\nB: 若 $\\mathrm{b}=7$, 加入 $20 \\mathrm{~mL} \\mathrm{NaOH}$ 溶液时, 溶液的 $\\mathrm{pH}<7$\nC: $\\mathrm{NaOH}$ 溶液的体积为 $30 \\mathrm{~mL}$ 时, 溶液中: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=3 \\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)$ $+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: 反应 $2 \\mathrm{H}_{2} \\mathrm{~A}^{-} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{~A}+\\mathrm{HA}^{2-}$ 在该温度下的平衡常数 $\\mathrm{K}=10^{\\mathrm{a}-\\mathrm{b}}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某三元羧酸 $\\mathrm{H}_{3} \\mathrm{~A}$ 在表面活性剂、洗涤剂、润滑剂等方面具有重要的地位。常温时,向 $10 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{3} \\mathrm{~A}$ 溶液中商入 $0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaOH}$ 溶液, $\\mathrm{H}_{3} \\mathrm{~A} 、 \\mathrm{H}_{2} \\mathrm{~A}^{-}$、 $\\mathrm{HA}^{2-} 、 \\mathrm{~A}^{3-}$ 的物质的量分数与溶液的 $\\mathrm{pH}$ 的关系如 图所示。下列说法错误的是\n\n[图1]\n\nA: 该温度下, $\\mathrm{H}_{3} \\mathrm{~A}$ 的电离常数 $\\mathrm{K}_{\\mathrm{a} 1}=10^{-\\mathrm{a}}$\nB: 若 $\\mathrm{b}=7$, 加入 $20 \\mathrm{~mL} \\mathrm{NaOH}$ 溶液时, 溶液的 $\\mathrm{pH}<7$\nC: $\\mathrm{NaOH}$ 溶液的体积为 $30 \\mathrm{~mL}$ 时, 溶液中: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=3 \\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~A}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{2-}\\right)$ $+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: 反应 $2 \\mathrm{H}_{2} \\mathrm{~A}^{-} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{~A}+\\mathrm{HA}^{2-}$ 在该温度下的平衡常数 $\\mathrm{K}=10^{\\mathrm{a}-\\mathrm{b}}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-32.jpg?height=415&width=802&top_left_y=158&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1249",
        "problem": "Nitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.Calculate the time elapsed to reduce the partial pressure of $\\mathrm{H}_{2}$ to the half of its initial value, if $\\mathrm{NO}$ with a pressure of $8.00 \\cdot 10^{2}$ torr and $\\mathrm{H}_{2}$ with 1.0 torr are mixed at $820^{\\circ} \\mathrm{C}$.\n\n(If you have been unable to calculate the value for the rate constant, you can use the value of $2 \\cdot 10^{-7}$ in appropriate unit.)\n\nA proposed mechanism for the reaction between $\\mathrm{NO}$ and $\\mathrm{H}_{2}$ is given below:\n\n$2 \\mathrm{NO}(g) \\xleftrightarrow[k_{-1}]{k_{1}} \\mathrm{~N}_{2} \\mathrm{O}_{2}(g)$\n\n$\\mathrm{N}_{2} \\mathrm{O}_{2}(g)+\\mathrm{H}_{2}(g)  \\stackrel{k_{2}}{\\longrightarrow} \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nNitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.\n\nproblem:\nCalculate the time elapsed to reduce the partial pressure of $\\mathrm{H}_{2}$ to the half of its initial value, if $\\mathrm{NO}$ with a pressure of $8.00 \\cdot 10^{2}$ torr and $\\mathrm{H}_{2}$ with 1.0 torr are mixed at $820^{\\circ} \\mathrm{C}$.\n\n(If you have been unable to calculate the value for the rate constant, you can use the value of $2 \\cdot 10^{-7}$ in appropriate unit.)\n\nA proposed mechanism for the reaction between $\\mathrm{NO}$ and $\\mathrm{H}_{2}$ is given below:\n\n$2 \\mathrm{NO}(g) \\xleftrightarrow[k_{-1}]{k_{1}} \\mathrm{~N}_{2} \\mathrm{O}_{2}(g)$\n\n$\\mathrm{N}_{2} \\mathrm{O}_{2}(g)+\\mathrm{H}_{2}(g)  \\stackrel{k_{2}}{\\longrightarrow} \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of s, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "s"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1233",
        "problem": "Contemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.Determine the ideal thermodynamic efficiency $(\\eta)$ of a fuel cell producing liquid water at $353.15 \\mathrm{~K}$. At this temperature, the enthalpy of formation of water is $\\Delta_{\\mathrm{f}} \\mathrm{H}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{I}\\right)=-281.64 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the corresponding reaction Gibbs energy change is $\\Delta_{\\mathrm{r}} G^{\\circ}=-225.85 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nContemporary means of transportation rely on burning fossil fuels, although the efficiency of real combustion engines is inherently limited and typically ranges between 20 and $40 \\%$.\n\nFuel cells represent a way to improve the engine efficiency for future vehicles. The engine efficiency can be improved by using hydrogen-based fuel cells.\n\nproblem:\nDetermine the ideal thermodynamic efficiency $(\\eta)$ of a fuel cell producing liquid water at $353.15 \\mathrm{~K}$. At this temperature, the enthalpy of formation of water is $\\Delta_{\\mathrm{f}} \\mathrm{H}^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}, \\mathrm{I}\\right)=-281.64 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the corresponding reaction Gibbs energy change is $\\Delta_{\\mathrm{r}} G^{\\circ}=-225.85 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1494",
        "problem": "One of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nHow many beta $\\left(\\beta^{-}\\right)$decays are there in this series? Show by calculation.\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOne of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nHow many beta $\\left(\\beta^{-}\\right)$decays are there in this series? Show by calculation.\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_75",
        "problem": "The graph below shows how the standard entropy of vaporization, $\\Delta S^{0}$ vap, varies with molar mass for straightchain alkanes $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{n} \\mathrm{CH}_{3}$.\n\n[figure1]\n\nWhich of the following graphs best shows how $\\Delta S^{\\circ}$ vap (solid squares, dashed line) varies with molar mass for straight-chain primary alcohols $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{n} \\mathrm{OH}$ ?\nA: [figure2]\nB: [figure3]\nC: [figure4]\nD: [figure5]\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe graph below shows how the standard entropy of vaporization, $\\Delta S^{0}$ vap, varies with molar mass for straightchain alkanes $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{n} \\mathrm{CH}_{3}$.\n\n[figure1]\n\nWhich of the following graphs best shows how $\\Delta S^{\\circ}$ vap (solid squares, dashed line) varies with molar mass for straight-chain primary alcohols $\\mathrm{CH}_{3}\\left(\\mathrm{CH}_{2}\\right)_{n} \\mathrm{OH}$ ?\n\nA: [figure2]\nB: [figure3]\nC: [figure4]\nD: [figure5]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-05.jpg?height=385&width=504&top_left_y=301&top_left_x=1342",
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-05.jpg?height=384&width=496&top_left_y=819&top_left_x=1259",
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-05.jpg?height=393&width=496&top_left_y=1208&top_left_x=1259",
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-05.jpg?height=390&width=483&top_left_y=1605&top_left_x=1271",
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-05.jpg?height=385&width=502&top_left_y=2003&top_left_x=1256"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_332",
        "problem": "Which of the following compounds is a solid at room temperature?\nA: $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$\nB: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\nC: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$\nD: $\\mathrm{C}_{8} \\mathrm{H}_{18}$\nE: $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OH}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following compounds is a solid at room temperature?\n\nA: $\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{C}-\\mathrm{H}$\nB: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{3}$\nC: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}$\nD: $\\mathrm{C}_{8} \\mathrm{H}_{18}$\nE: $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{OH}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_760",
        "problem": "常温下, 难溶盐 $\\mathrm{CuR} 、 \\mathrm{ZnR}$ 的沉淀溶解平衡曲线如图所示, 已知:\n\n$\\mathrm{ZnR}(\\mathrm{s})+\\mathrm{Cu}^{2+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{CuR}(\\mathrm{s})+\\mathrm{Zn}^{2+}(\\mathrm{aq})$ 热力学趋势很大。下列说法错误的是\n\n[图1]\nA: 常温下, $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnR}) / \\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuR})$ 约为 $1 \\times 10^{14}$\nB: 向 $\\mathrm{q}$ 点的溶液中加入少量对应金属的硝酸盐固体, 溶液组成可能变为 $\\mathrm{p}$ 点\nC: a 点对应的 CuR 溶解体系中, $v($ 溶解 $)>v($ 沉淀 $)$\nD: 向 $\\mathrm{c}\\left(\\mathrm{Zn}^{2+}\\right)=10 \\mathrm{c}\\left(\\mathrm{Cu}^{2+}\\right)$ 的混合溶液中加入 $\\mathrm{H}_{2} \\mathrm{R}$ 溶液, 首先析出 $\\mathrm{CuR}$ 沉淀\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 难溶盐 $\\mathrm{CuR} 、 \\mathrm{ZnR}$ 的沉淀溶解平衡曲线如图所示, 已知:\n\n$\\mathrm{ZnR}(\\mathrm{s})+\\mathrm{Cu}^{2+}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{CuR}(\\mathrm{s})+\\mathrm{Zn}^{2+}(\\mathrm{aq})$ 热力学趋势很大。下列说法错误的是\n\n[图1]\n\nA: 常温下, $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{ZnR}) / \\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuR})$ 约为 $1 \\times 10^{14}$\nB: 向 $\\mathrm{q}$ 点的溶液中加入少量对应金属的硝酸盐固体, 溶液组成可能变为 $\\mathrm{p}$ 点\nC: a 点对应的 CuR 溶解体系中, $v($ 溶解 $)>v($ 沉淀 $)$\nD: 向 $\\mathrm{c}\\left(\\mathrm{Zn}^{2+}\\right)=10 \\mathrm{c}\\left(\\mathrm{Cu}^{2+}\\right)$ 的混合溶液中加入 $\\mathrm{H}_{2} \\mathrm{R}$ 溶液, 首先析出 $\\mathrm{CuR}$ 沉淀\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-076.jpg?height=694&width=720&top_left_y=1755&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_721",
        "problem": "织物漂白剂亚氯酸钠 $\\left(\\mathrm{NaClO}_{2}\\right)$ 在溶液中可生成 $\\mathrm{ClO}_{2} 、 \\mathrm{HClO}_{2} 、 \\mathrm{ClO}_{2}{ }^{-} 、 \\mathrm{Cl}^{-}$等, 其中 $\\mathrm{HClO}_{2}$ 和 $\\mathrm{ClO}_{2}$ 都具有漂白作用, 但 $\\mathrm{ClO}_{2}$ 是有毒气体。 $25^{\\circ} \\mathrm{C}$ 时, 各组分含量随 $\\mathrm{pH}$ 变化情况如图所示 $\\left(\\mathrm{Cl}^{-}\\right.$没有画出)。下列说法错误的是\n\n[图1]\nA: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{HClO}_{2}$ 的电离平衡常数的数值 $\\mathrm{K}_{\\mathrm{a}}=10^{-6}$\nB: 使用该漂白剂的最佳 $\\mathrm{pH}$ 为 3.0\nC: $25^{\\circ} \\mathrm{C}$ 时, 等浓度的 $\\mathrm{HClO}_{2}$ 溶液和 $\\mathrm{NaClO}_{2}$ 溶液等体积混合后, 混合溶液中: $\\mathrm{c}\\left(\\mathrm{HClO}_{2}\\right)$ $+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{ClO}_{2}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 该温度下的 $\\mathrm{NaClO}_{2}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{ClO}_{2}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n织物漂白剂亚氯酸钠 $\\left(\\mathrm{NaClO}_{2}\\right)$ 在溶液中可生成 $\\mathrm{ClO}_{2} 、 \\mathrm{HClO}_{2} 、 \\mathrm{ClO}_{2}{ }^{-} 、 \\mathrm{Cl}^{-}$等, 其中 $\\mathrm{HClO}_{2}$ 和 $\\mathrm{ClO}_{2}$ 都具有漂白作用, 但 $\\mathrm{ClO}_{2}$ 是有毒气体。 $25^{\\circ} \\mathrm{C}$ 时, 各组分含量随 $\\mathrm{pH}$ 变化情况如图所示 $\\left(\\mathrm{Cl}^{-}\\right.$没有画出)。下列说法错误的是\n\n[图1]\n\nA: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{HClO}_{2}$ 的电离平衡常数的数值 $\\mathrm{K}_{\\mathrm{a}}=10^{-6}$\nB: 使用该漂白剂的最佳 $\\mathrm{pH}$ 为 3.0\nC: $25^{\\circ} \\mathrm{C}$ 时, 等浓度的 $\\mathrm{HClO}_{2}$ 溶液和 $\\mathrm{NaClO}_{2}$ 溶液等体积混合后, 混合溶液中: $\\mathrm{c}\\left(\\mathrm{HClO}_{2}\\right)$ $+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{ClO}_{2}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 该温度下的 $\\mathrm{NaClO}_{2}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{ClO}_{2}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-79.jpg?height=302&width=368&top_left_y=1251&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_179",
        "problem": "In the television series Breaking Bad, Walter White and Jesse Pinkman synthesize methamphetamine ( $N$-methyl-1-phenyl-2-propanamine) which is marketed as methamphetamine hydrochloride, or \"crystal meth\".\n\n[figure1]\n\n## methamphetamine\n\n hydrochlorideWhat is the molecular formula of methamphetamine hydrochloride?\nA: $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{ClN}$\nB: $\\mathrm{C}_{10} \\mathrm{H}_{10} \\mathrm{ClN}$\nC: $\\mathrm{C}_{10} \\mathrm{H}_{16} \\mathrm{CIN}$\nD: $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{ClN}$\nE: $\\mathrm{C}_{9} \\mathrm{H}_{16} \\mathrm{ClN}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the television series Breaking Bad, Walter White and Jesse Pinkman synthesize methamphetamine ( $N$-methyl-1-phenyl-2-propanamine) which is marketed as methamphetamine hydrochloride, or \"crystal meth\".\n\n[figure1]\n\n## methamphetamine\n\n hydrochlorideWhat is the molecular formula of methamphetamine hydrochloride?\n\nA: $\\mathrm{C}_{8} \\mathrm{H}_{16} \\mathrm{ClN}$\nB: $\\mathrm{C}_{10} \\mathrm{H}_{10} \\mathrm{ClN}$\nC: $\\mathrm{C}_{10} \\mathrm{H}_{16} \\mathrm{CIN}$\nD: $\\mathrm{C}_{10} \\mathrm{H}_{14} \\mathrm{ClN}$\nE: $\\mathrm{C}_{9} \\mathrm{H}_{16} \\mathrm{ClN}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b029a381ae33f0e16551g-2.jpg?height=214&width=351&top_left_y=377&top_left_x=435"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_670",
        "problem": "乙二酸为二元弱酸, 用 $\\mathrm{H}_{2} \\mathrm{~A}$ 表示。 $25^{\\circ} \\mathrm{C}$ 时, 向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NH}_{4} \\mathrm{HA}$ 溶液中滴加适量的盐酸或 $\\mathrm{NaOH}$ 溶液, 溶液中各含氮(或 $\\mathrm{A}$ )微粒的分布分数 $\\delta$ 与溶液 $\\mathrm{pH}$ 的关系如图所示(不考虑溶液中的 $\\mathrm{NH}_{3}$ 分子)。比如：溶液中 $\\mathrm{H}_{2} \\mathrm{~A}$ 的分布分数 $\\delta\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=$\n\n$$\n\\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}\n$$\n\n[图1]\n\n下列说法正确的是\nA: 向 $\\mathrm{NH}_{4} \\mathrm{HA}$ 溶液中滴加少量 $\\mathrm{NaOH}$ 溶液到 $\\mathrm{m}$ 点的过程中, $\\mathrm{HA}^{-}$反应的比例小于 $\\mathrm{NH}$\nB: $\\mathrm{f}$ 点时, 溶液中 $\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: $\\frac{\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}{\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}=10^{-3}$\nD: 在 $\\mathrm{d} 、 \\mathrm{n}$ 点溶液中均满足 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n乙二酸为二元弱酸, 用 $\\mathrm{H}_{2} \\mathrm{~A}$ 表示。 $25^{\\circ} \\mathrm{C}$ 时, 向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NH}_{4} \\mathrm{HA}$ 溶液中滴加适量的盐酸或 $\\mathrm{NaOH}$ 溶液, 溶液中各含氮(或 $\\mathrm{A}$ )微粒的分布分数 $\\delta$ 与溶液 $\\mathrm{pH}$ 的关系如图所示(不考虑溶液中的 $\\mathrm{NH}_{3}$ 分子)。比如：溶液中 $\\mathrm{H}_{2} \\mathrm{~A}$ 的分布分数 $\\delta\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=$\n\n$$\n\\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}\n$$\n\n[图1]\n\n下列说法正确的是\n\nA: 向 $\\mathrm{NH}_{4} \\mathrm{HA}$ 溶液中滴加少量 $\\mathrm{NaOH}$ 溶液到 $\\mathrm{m}$ 点的过程中, $\\mathrm{HA}^{-}$反应的比例小于 $\\mathrm{NH}$\nB: $\\mathrm{f}$ 点时, 溶液中 $\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: $\\frac{\\mathrm{K}_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}{\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}=10^{-3}$\nD: 在 $\\mathrm{d} 、 \\mathrm{n}$ 点溶液中均满足 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-081.jpg?height=454&width=693&top_left_y=898&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_647",
        "problem": "共价有机框架 (COF)是一类具有平面、多孔网状结构的材料, 其在催化、能量储存等领域有潜在的应用。一种 COF 材料的合成如图所示, 下列说法正确的是\nBDT\n[图1]\n\nHHTP COF\nA: $\\mathrm{BDT}$ 分子中 $\\mathrm{B} 、 \\mathrm{O} 、 \\mathrm{~S}$ 的杂化方式相同\nB: HHTP 的核磁共振氢谱有 2 组峰\nC: 合成 COF 的过程中发生了取代反应\nD: [图2]也可与 BDT 合成 COF 材料\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n共价有机框架 (COF)是一类具有平面、多孔网状结构的材料, 其在催化、能量储存等领域有潜在的应用。一种 COF 材料的合成如图所示, 下列说法正确的是\nBDT\n[图1]\n\nHHTP COF\n\nA: $\\mathrm{BDT}$ 分子中 $\\mathrm{B} 、 \\mathrm{O} 、 \\mathrm{~S}$ 的杂化方式相同\nB: HHTP 的核磁共振氢谱有 2 组峰\nC: 合成 COF 的过程中发生了取代反应\nD: [图2]也可与 BDT 合成 COF 材料\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-20.jpg?height=554&width=1454&top_left_y=1802&top_left_x=336",
            "https://i.postimg.cc/43w3TyRz/image.png",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-21.jpg?height=132&width=919&top_left_y=959&top_left_x=731"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_618",
        "problem": "化合物 $Y$ 具有增强免疫等功效，可由 $X$ 制得。下列有关 $X 、 Y$ 的说法正确的是\n[图1]\nA: 一定条件下 X 可发生氧化反应和消去反应\nB: $1 \\mathrm{molY}$ 最多能与 $4 \\mathrm{molNaOH}$ 反应\nC: $\\mathrm{X}$ 与足量 $\\mathrm{H}_{2}$ 反应后，产物分子中含有 8 个手性碳原子\nD: 等物质的量的 $\\mathrm{X} 、 \\mathrm{Y}$ 分别与足量 $\\mathrm{Br}_{2}$ 反应, 消耗 $\\mathrm{Br}_{2}$ 的物质的量相等\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化合物 $Y$ 具有增强免疫等功效，可由 $X$ 制得。下列有关 $X 、 Y$ 的说法正确的是\n[图1]\n\nA: 一定条件下 X 可发生氧化反应和消去反应\nB: $1 \\mathrm{molY}$ 最多能与 $4 \\mathrm{molNaOH}$ 反应\nC: $\\mathrm{X}$ 与足量 $\\mathrm{H}_{2}$ 反应后，产物分子中含有 8 个手性碳原子\nD: 等物质的量的 $\\mathrm{X} 、 \\mathrm{Y}$ 分别与足量 $\\mathrm{Br}_{2}$ 反应, 消耗 $\\mathrm{Br}_{2}$ 的物质的量相等\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-47.jpg?height=434&width=1418&top_left_y=1619&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1392",
        "problem": "At a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.Calculate the enthalpy change, $\\Delta H$, in $\\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$ for the following reaction at $0 \\mathrm{~K}$ :\n\n$\\mathrm{H}_{2}(g)+\\mathrm{D}_{2}(g) \\rightarrow 2 \\mathrm{HD}(g)$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nAt a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.\n\nproblem:\nCalculate the enthalpy change, $\\Delta H$, in $\\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$ for the following reaction at $0 \\mathrm{~K}$ :\n\n$\\mathrm{H}_{2}(g)+\\mathrm{D}_{2}(g) \\rightarrow 2 \\mathrm{HD}(g)$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_156",
        "problem": "Consider the following reaction, where the starting compound is treated with an unknown reagent over a catalytic surface to form the product:\n\n[figure1]\n\nWhich two terms can be used to describe this process?\nA: hydrogenation, elimination\nB: hydration, addition\nC: hydrogenation, substitution\nD: hydrogenation, addition\nE: hydration, substitution\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nConsider the following reaction, where the starting compound is treated with an unknown reagent over a catalytic surface to form the product:\n\n[figure1]\n\nWhich two terms can be used to describe this process?\n\nA: hydrogenation, elimination\nB: hydration, addition\nC: hydrogenation, substitution\nD: hydrogenation, addition\nE: hydration, substitution\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_343d5676bd7739089c6eg-3.jpg?height=151&width=687&top_left_y=914&top_left_x=1649"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_953",
        "problem": "在实验室中从废旧钴酸锂离子电池的正极材料(在铝笓上涂覆活性物质 $\\mathrm{LiCoO}_{2}$ ) 中回收钴、锂的操作流程如图:\n\n[图1]\n\n已知: 拆解废旧电池获取正极材料前, 先将其浸入 $\\mathrm{NaCl}$ 溶液中, 使电池短路而放电,此时溶液温度升高。下列说法错误的是\nA: 拆解废旧电池获取正极材过程中能量的主要转化方式为化学能 $\\rightarrow$ 热能 $\\rightarrow$ 电能\nB: “滤液”主要成分是 $\\mathrm{Na}\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$\nC: “酸浸”时消耗 $\\mathrm{LiCoO}_{2} 、 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3} 、 \\mathrm{H}_{2} \\mathrm{SO}_{4}$ 物质的量之比为 $6: 1: 3$\nD: 若用盐酸替代硫酸、 $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 溶液可同样达到“酸浸”的目的, 但产生 $\\mathrm{Cl}_{2}$ 污染环境\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n在实验室中从废旧钴酸锂离子电池的正极材料(在铝笓上涂覆活性物质 $\\mathrm{LiCoO}_{2}$ ) 中回收钴、锂的操作流程如图:\n\n[图1]\n\n已知: 拆解废旧电池获取正极材料前, 先将其浸入 $\\mathrm{NaCl}$ 溶液中, 使电池短路而放电,此时溶液温度升高。下列说法错误的是\n\nA: 拆解废旧电池获取正极材过程中能量的主要转化方式为化学能 $\\rightarrow$ 热能 $\\rightarrow$ 电能\nB: “滤液”主要成分是 $\\mathrm{Na}\\left[\\mathrm{Al}(\\mathrm{OH})_{4}\\right]$\nC: “酸浸”时消耗 $\\mathrm{LiCoO}_{2} 、 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3} 、 \\mathrm{H}_{2} \\mathrm{SO}_{4}$ 物质的量之比为 $6: 1: 3$\nD: 若用盐酸替代硫酸、 $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 溶液可同样达到“酸浸”的目的, 但产生 $\\mathrm{Cl}_{2}$ 污染环境\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-20.jpg?height=331&width=1464&top_left_y=154&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1016",
        "problem": "For which of the following reactions is the change in energy equal to the first ionization energy of oxygen?\nA: $\\mathrm{O}^{-}(g)+\\mathrm{e}^{-} \\rightarrow \\mathrm{O}^{2-}(g)$\nB: $\\mathrm{O}(g)+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{O}^{2-}(g)$\nC: $\\quad \\mathrm{O}(g) \\rightarrow \\mathrm{O}^{+}(g)+\\mathrm{e}^{-}$\nD: $\\mathrm{O}(g)+\\mathrm{e}^{-} \\rightarrow \\mathrm{O}^{-}(g)$\nE: $\\mathrm{O}(g) \\rightarrow \\mathrm{O}^{2+}(g)+2 \\mathrm{e}^{-}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor which of the following reactions is the change in energy equal to the first ionization energy of oxygen?\n\nA: $\\mathrm{O}^{-}(g)+\\mathrm{e}^{-} \\rightarrow \\mathrm{O}^{2-}(g)$\nB: $\\mathrm{O}(g)+2 \\mathrm{e}^{-} \\rightarrow \\mathrm{O}^{2-}(g)$\nC: $\\quad \\mathrm{O}(g) \\rightarrow \\mathrm{O}^{+}(g)+\\mathrm{e}^{-}$\nD: $\\mathrm{O}(g)+\\mathrm{e}^{-} \\rightarrow \\mathrm{O}^{-}(g)$\nE: $\\mathrm{O}(g) \\rightarrow \\mathrm{O}^{2+}(g)+2 \\mathrm{e}^{-}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_602",
        "problem": "表中列出了部分炔腈类化合物, 有关说法错误的是\n\n| 名称 | 氰基乙炔 | 氰基丁二炔 | $?$ | 氰基辛四炔 | 氰基癸五炔 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 化学式 | $\\mathrm{HC}_{3} \\mathrm{~N}$ | $\\mathrm{HC}_{5} \\mathrm{~N}$ | $\\mathrm{HC}$ |  |  |\n|  |  |  |  | $\\mathrm{HC}_{9} \\mathrm{~N}$ | $\\mathrm{HC}_{11} \\mathrm{~N}$ |\nA: 表中 $\\mathrm{HC}_{7} \\mathrm{~N}$ 的名称为氰基已三炔\nB: $\\mathrm{HC}_{9} \\mathrm{~N}$ 中所有原子均共线\nC: 以上 5 种物质互为同系物\nD: 可由乙炔和含氮化合物加聚制的 $\\mathrm{HC}_{\\mathrm{n}} \\mathrm{N}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n表中列出了部分炔腈类化合物, 有关说法错误的是\n\n| 名称 | 氰基乙炔 | 氰基丁二炔 | $?$ | 氰基辛四炔 | 氰基癸五炔 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| 化学式 | $\\mathrm{HC}_{3} \\mathrm{~N}$ | $\\mathrm{HC}_{5} \\mathrm{~N}$ | $\\mathrm{HC}$ |  |  |\n|  |  |  |  | $\\mathrm{HC}_{9} \\mathrm{~N}$ | $\\mathrm{HC}_{11} \\mathrm{~N}$ |\n\nA: 表中 $\\mathrm{HC}_{7} \\mathrm{~N}$ 的名称为氰基已三炔\nB: $\\mathrm{HC}_{9} \\mathrm{~N}$ 中所有原子均共线\nC: 以上 5 种物质互为同系物\nD: 可由乙炔和含氮化合物加聚制的 $\\mathrm{HC}_{\\mathrm{n}} \\mathrm{N}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_595",
        "problem": "化合物 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 。 $\\mathrm{X}$ 分子中只含一个苯环且苯环上只有一个取代基,其红外光谱和核磁共振氢谱如图。下列关于 X 的说法中不正确的是\n[图1]\nA: X 属于酯类化合物\nB: X 在碱性条件下的水解程度小于酸性条件\nC: 符合题中 X 分子结构特征的有机物只有 1 种\nD: $\\mathrm{X}$ 在一定条件下可与 $3 \\mathrm{~mol} \\mathrm{H}_{2}$ 发生加成反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化合物 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8} \\mathrm{O}_{2}$ 。 $\\mathrm{X}$ 分子中只含一个苯环且苯环上只有一个取代基,其红外光谱和核磁共振氢谱如图。下列关于 X 的说法中不正确的是\n[图1]\n\nA: X 属于酯类化合物\nB: X 在碱性条件下的水解程度小于酸性条件\nC: 符合题中 X 分子结构特征的有机物只有 1 种\nD: $\\mathrm{X}$ 在一定条件下可与 $3 \\mathrm{~mol} \\mathrm{H}_{2}$ 发生加成反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-12.jpg?height=374&width=1454&top_left_y=561&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_2",
        "problem": "Which experiment would be most appropriate for determining the number of components in a commercial nail polish remover?\nA: Gas chromatography\nB: Paper chromatography\nC: Boiling point determination\nD: Combustion analysis\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich experiment would be most appropriate for determining the number of components in a commercial nail polish remover?\n\nA: Gas chromatography\nB: Paper chromatography\nC: Boiling point determination\nD: Combustion analysis\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_250",
        "problem": "Which of the following molecules has the highest boiling point?\nA: $\\mathrm{CF}_{4}$\nB: $\\mathrm{CCl}_{4}$\nC: $\\mathrm{CBr}_{4}$\nD: $\\mathrm{Cl}_{4}$\nE: $\\mathrm{CH}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following molecules has the highest boiling point?\n\nA: $\\mathrm{CF}_{4}$\nB: $\\mathrm{CCl}_{4}$\nC: $\\mathrm{CBr}_{4}$\nD: $\\mathrm{Cl}_{4}$\nE: $\\mathrm{CH}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1391",
        "problem": "The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nCalculate the change in entropy of the universe.\n\nDoes your answer agree with the Second Law of Thermodynamics?\n\nTrue False\n\nThe pressure - temperature phase diagram of $\\mathrm{CO}_{2}$ is given below schematically. The diagram is not to scale.\n\n[figure1]\n\n$\\mathrm{t} /{ }^{\\circ} \\mathrm{C}$\n\nPhase diagram of $\\mathrm{CO}_{2}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nCalculate the change in entropy of the universe.\n\nDoes your answer agree with the Second Law of Thermodynamics?\n\nTrue False\n\nThe pressure - temperature phase diagram of $\\mathrm{CO}_{2}$ is given below schematically. The diagram is not to scale.\n\n[figure1]\n\n$\\mathrm{t} /{ }^{\\circ} \\mathrm{C}$\n\nPhase diagram of $\\mathrm{CO}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-383.jpg?height=646&width=923&top_left_y=428&top_left_x=435"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_263",
        "problem": "Calculate the molar mass of the original potassium salt, assuming that $1 \\mathrm{~mol}$ of the salt contains 1 mol of potassium ions.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the molar mass of the original potassium salt, assuming that $1 \\mathrm{~mol}$ of the salt contains 1 mol of potassium ions.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g/mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g/mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1455",
        "problem": "The undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nHA can be efficiently extracted only from acidic aqueous solutions.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nHA can be efficiently extracted only from acidic aqueous solutions.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-461.jpg?height=185&width=788&top_left_y=2209&top_left_x=634"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_221",
        "problem": "Estimate the absorbance of a solution with a potassium ion concentration of $15.4 \\mathrm{mg} \\mathrm{L}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nEstimate the absorbance of a solution with a potassium ion concentration of $15.4 \\mathrm{mg} \\mathrm{L}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_941",
        "problem": "常温下, 钠盐 $\\left(\\mathrm{Na}_{2} \\mathrm{XO}_{3}\\right)$ 溶液中微粒浓度的变化关系如图所示 $\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$。下列说法正确的是\n\n[图1]\nA: 曲线 $\\mathrm{N}$ 表示 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}\\right)}{\\mathrm{c}\\left(\\mathrm{HXO}_{3}^{-}\\right)}$与 $\\mathrm{pOH}$ 的变化关系\nB: 常温下, $\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}\\right)=10^{-10}$\nC: 当 $\\mathrm{pOH}=2$ 时, $\\mathrm{NaHXO}_{3}$ 溶液中: $\\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}\\right)}{\\mathrm{c}\\left(\\mathrm{XO}_{3}^{2-}\\right)}=10^{-8}$\nD: 向 $\\mathrm{Na}_{2} \\mathrm{XO}_{3}$ 溶液中滴加稀盐酸至中性时, 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{HXO}_{3}{ }^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{XO}_{3}{ }^{2}\\right.$ - )\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 钠盐 $\\left(\\mathrm{Na}_{2} \\mathrm{XO}_{3}\\right)$ 溶液中微粒浓度的变化关系如图所示 $\\left[\\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$。下列说法正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{N}$ 表示 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}\\right)}{\\mathrm{c}\\left(\\mathrm{HXO}_{3}^{-}\\right)}$与 $\\mathrm{pOH}$ 的变化关系\nB: 常温下, $\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}\\right)=10^{-10}$\nC: 当 $\\mathrm{pOH}=2$ 时, $\\mathrm{NaHXO}_{3}$ 溶液中: $\\frac{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}\\right)}{\\mathrm{c}\\left(\\mathrm{XO}_{3}^{2-}\\right)}=10^{-8}$\nD: 向 $\\mathrm{Na}_{2} \\mathrm{XO}_{3}$ 溶液中滴加稀盐酸至中性时, 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{HXO}_{3}{ }^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{XO}_{3}{ }^{2}\\right.$ - )\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-55.jpg?height=443&width=557&top_left_y=498&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_43",
        "problem": "The $K_{\\text {sp }}$ of $\\mathrm{AgCl}$ is $9.5 \\times 10^{-11}$ at $14{ }^{\\circ} \\mathrm{C}$ and is $7.8 \\times 10^{-10}$ at $42^{\\circ} \\mathrm{C}$. What is $\\Delta H^{\\circ}$ for the dissolution of $\\operatorname{AgCl}(s)$ ?\n\n$$\n\\operatorname{AgCl}(s) \\rightarrow \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)\n$$\nA: $-0.49 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $+0.37 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $+57 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $+220 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe $K_{\\text {sp }}$ of $\\mathrm{AgCl}$ is $9.5 \\times 10^{-11}$ at $14{ }^{\\circ} \\mathrm{C}$ and is $7.8 \\times 10^{-10}$ at $42^{\\circ} \\mathrm{C}$. What is $\\Delta H^{\\circ}$ for the dissolution of $\\operatorname{AgCl}(s)$ ?\n\n$$\n\\operatorname{AgCl}(s) \\rightarrow \\mathrm{Ag}^{+}(a q)+\\mathrm{Cl}^{-}(a q)\n$$\n\nA: $-0.49 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $+0.37 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $+57 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $+220 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_203",
        "problem": "Calculate the amount (in mol or mmol) of potassium iodide added.\nAqueous mixtures containing both silver and lead ions can be analysed by adding potassium iodide until both silver iodide (molar mass $234.8 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ) and lead iodide (molar mass $461.0 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ) have precipitated completely from solution, according to the following equations:\n\n$\\mathrm{Ag}^{+}+\\mathrm{I}^{-} \\rightarrow \\operatorname{AgI}(\\mathrm{s})$\n\n$\\mathrm{Pb}^{2+}+2 \\mathrm{I}^{-} \\rightarrow \\mathrm{PbI}_{2}(\\mathrm{~s})$\n\nThe concentration of the potassium iodide solution can be determined by analysing it for potassium content, as previously described.\n\nAn aqueous mixture containing both silver and lead ions reacts completely with $15.01 \\mathrm{~mL}$ of $0.2101 \\mathrm{~mol} \\mathrm{~L}^{-1}$ potassium iodide solution. The total mass of precipitate produced is $0.7328 \\mathrm{~g}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the amount (in mol or mmol) of potassium iodide added.\nAqueous mixtures containing both silver and lead ions can be analysed by adding potassium iodide until both silver iodide (molar mass $234.8 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ) and lead iodide (molar mass $461.0 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ) have precipitated completely from solution, according to the following equations:\n\n$\\mathrm{Ag}^{+}+\\mathrm{I}^{-} \\rightarrow \\operatorname{AgI}(\\mathrm{s})$\n\n$\\mathrm{Pb}^{2+}+2 \\mathrm{I}^{-} \\rightarrow \\mathrm{PbI}_{2}(\\mathrm{~s})$\n\nThe concentration of the potassium iodide solution can be determined by analysing it for potassium content, as previously described.\n\nAn aqueous mixture containing both silver and lead ions reacts completely with $15.01 \\mathrm{~mL}$ of $0.2101 \\mathrm{~mol} \\mathrm{~L}^{-1}$ potassium iodide solution. The total mass of precipitate produced is $0.7328 \\mathrm{~g}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mmol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mmol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1136",
        "problem": "This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nBalance the following equation for the preparation of potassium ferrate:\n\n[figure1]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question has more than one correct answer, you need to include them all.\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nBalance the following equation for the preparation of potassium ferrate:\n\n[figure1]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nTheir answer types are, in order, [numerical value, numerical value, numerical value, numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b1f97cb4cc7a1e9208fdg-7.jpg?height=63&width=1174&top_left_y=1439&top_left_x=561"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MA",
        "unit": [
            null,
            null,
            null,
            null,
            null,
            null
        ],
        "answer_sequence": null,
        "type_sequence": [
            "NV",
            "NV",
            "NV",
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1360",
        "problem": "Lactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nCalculate $\\mathrm{pH}$ in a $3.00 \\times 10^{-3} \\mathrm{M}$ solution of $\\mathrm{HL}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nLactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nCalculate $\\mathrm{pH}$ in a $3.00 \\times 10^{-3} \\mathrm{M}$ solution of $\\mathrm{HL}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_372",
        "problem": "What is the geometry of $\\mathrm{SF}_{4}$ ?\nA: Tetrahedral\nB: Square planar\nC: See-saw\nD: Trigonal monopyramidal\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the geometry of $\\mathrm{SF}_{4}$ ?\n\nA: Tetrahedral\nB: Square planar\nC: See-saw\nD: Trigonal monopyramidal\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_843",
        "problem": "$\\mathrm{t}^{\\circ} \\mathrm{C}$ 时, 配制一组 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{HCl}$或 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{NaOH}$ 的混合溶液, 溶液中部分微粒浓度的负对数 $(-\\operatorname{lgc})$ 与 $\\mathrm{pH}$ 关系如下图所示。下列说法正确的是（ ）\n\n[图1]\nA: $\\mathrm{pH}=6.3$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{pH}=7$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: $\\mathrm{pH}=\\mathrm{a}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, 反应 $\\mathrm{H}_{2} \\mathrm{CO}_{3}+\\mathrm{CO}_{3}^{2-} \\rightleftharpoons 2 \\mathrm{HCO}_{3}^{-}$的平衡常数为 $1.0 \\times 10^{4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{t}^{\\circ} \\mathrm{C}$ 时, 配制一组 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{HCl}$或 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{NaOH}$ 的混合溶液, 溶液中部分微粒浓度的负对数 $(-\\operatorname{lgc})$ 与 $\\mathrm{pH}$ 关系如下图所示。下列说法正确的是（ ）\n\n[图1]\n\nA: $\\mathrm{pH}=6.3$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{pH}=7$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: $\\mathrm{pH}=\\mathrm{a}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, 反应 $\\mathrm{H}_{2} \\mathrm{CO}_{3}+\\mathrm{CO}_{3}^{2-} \\rightleftharpoons 2 \\mathrm{HCO}_{3}^{-}$的平衡常数为 $1.0 \\times 10^{4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-097.jpg?height=494&width=805&top_left_y=176&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_683",
        "problem": "近期, 天津大学化学团队以 $\\mathrm{CO}_{2}$ 与辛胺为原料实现了甲酸和辛腈的高选择性合成,装置工作原理如图。下列说法正确的是\n\n[图1]\nA: $\\mathrm{Ni}_{2} \\mathrm{P}$ 电极与电源正极相连\nB: 辛胺转化为辛腈发生了还原反应\nC: $\\mathrm{In} / \\mathrm{In}_{2} \\mathrm{O}_{3-\\mathrm{x}}$ 电极上可能有副产物 $\\mathrm{H}_{2}$ 生成\nD: 在 $\\mathrm{In} / \\mathrm{In}_{2} \\mathrm{O}_{3-\\mathrm{x}}$ 电极上发生的反应为 $\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O}-2 \\mathrm{e}^{-}=\\mathrm{HCOO}^{-}+\\mathrm{OH}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n近期, 天津大学化学团队以 $\\mathrm{CO}_{2}$ 与辛胺为原料实现了甲酸和辛腈的高选择性合成,装置工作原理如图。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{Ni}_{2} \\mathrm{P}$ 电极与电源正极相连\nB: 辛胺转化为辛腈发生了还原反应\nC: $\\mathrm{In} / \\mathrm{In}_{2} \\mathrm{O}_{3-\\mathrm{x}}$ 电极上可能有副产物 $\\mathrm{H}_{2}$ 生成\nD: 在 $\\mathrm{In} / \\mathrm{In}_{2} \\mathrm{O}_{3-\\mathrm{x}}$ 电极上发生的反应为 $\\mathrm{CO}_{2}+\\mathrm{H}_{2} \\mathrm{O}-2 \\mathrm{e}^{-}=\\mathrm{HCOO}^{-}+\\mathrm{OH}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-78.jpg?height=557&width=1059&top_left_y=147&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1146",
        "problem": "This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nGiven that the following elements all form either molecules or ions by uniting with four oxygen atoms with the element in its maximum theoretical oxidation state, give the formula for each ion or molecule.\n\n| bromine $(\\mathrm{Br})$ | chromium $(\\mathrm{Cr})$ | arsenic $(\\mathrm{As})$ | germanium $(\\mathrm{Ge})$ |\n| :---: | :---: | :---: | ---: |\n| technetium $(\\mathrm{Tc})$ | thallium $(\\mathrm{TI})$ | iridium $(\\mathrm{Ir})$ | ruthenium $(\\mathrm{Ru})$ |",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nGiven that the following elements all form either molecules or ions by uniting with four oxygen atoms with the element in its maximum theoretical oxidation state, give the formula for each ion or molecule.\n\n| bromine $(\\mathrm{Br})$ | chromium $(\\mathrm{Cr})$ | arsenic $(\\mathrm{As})$ | germanium $(\\mathrm{Ge})$ |\n| :---: | :---: | :---: | ---: |\n| technetium $(\\mathrm{Tc})$ | thallium $(\\mathrm{TI})$ | iridium $(\\mathrm{Ir})$ | ruthenium $(\\mathrm{Ru})$ |\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [Br, Cr, As, Ge, Tc, TI, Ir, Ru].\nTheir answer types are, in order, [numerical value, numerical value, numerical value, numerical value, numerical value, numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null,
            null,
            null,
            null,
            null,
            null,
            null
        ],
        "answer_sequence": [
            "Br",
            "Cr",
            "As",
            "Ge",
            "Tc",
            "TI",
            "Ir",
            "Ru"
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV",
            "NV",
            "NV",
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_948",
        "problem": "在含单质碘的 $\\mathrm{KI}$ 溶液中存在可逆反应: $\\mathrm{I}_{2}(\\mathrm{aq})+\\mathrm{I}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{I}_{3}^{-}(\\mathrm{aq})$, 为测定该反应的平衡常数 $\\mathrm{K}$ 进行如下实验, 实验步骤如下:\n\n(1) 在装有 $400 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KI}$ 溶液的碘量瓶中加入足量 $\\mathrm{I}_{2}$, 充分搅拌溶解, 待过量的固体碘沉于瓶底后, 取 $42.5 \\mathrm{~mL}$ 上层清液, 用 $5 \\mathrm{mLCCl}_{4}$ 萃取, 充分振荡、静置、分液,得到 $42.5 \\mathrm{~mL}$ 萃取后的水溶液、 $5 \\mathrm{mLI}_{2} \\sim \\mathrm{CCl}_{4}$ 溶液。\n\n(2)取萃取后的 $5 \\mathrm{mLI}_{2} \\sim \\mathrm{CCl}_{4}$ 溶液于碘量瓶中, 加水充分振荡, 再加入质量分数为 $0.01 \\% \\mathrm{KI}$ 溶液, 充分振荡后, 静置 5 分钟, 注入 $4 \\mathrm{~mL} 0.2 \\%$ 的淀粉溶液, 用 $\\mathrm{cmol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 标准溶液滴定, 平行滴定 3 次, 平均消耗 $V_{1} \\mathrm{mLNa}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 溶液。\n\n(3)将萃取后的水溶液 $42.5 \\mathrm{~mL}$ 移入碘量瓶中, 注入 $4 \\mathrm{~mL} 0.2 \\%$ 的淀粉溶液, 用 $\\mathrm{cmol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 标准溶液滴定, 平行滴定 3 次, 平均消耗 $V_{2} \\mathrm{mLNa}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 溶液。\n\n已知: (1) $2 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}+\\mathrm{I}_{2}=2 \\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6}+2 \\mathrm{NaI}$; (2) $\\mathrm{I}^{-}$与 $\\mathrm{I}_{3}^{-}$难溶于 $\\mathrm{CCl}_{4}$; (3)达到溶解平衡后,\n$\\mathrm{I}_{2}$ 在 $\\mathrm{CCl}_{4}$ 层和水层中的分配比 $c\\left(\\mathrm{I}_{2}\\right)_{\\mathrm{CCl}_{4}}: c\\left(\\mathrm{I}_{2}\\right)_{\\mathrm{H}_{2} \\mathrm{O}}$ 为 $85: 1$\n\n[图1]\n\n碘量瓶示意图\n\n根据上述实验原理，下列关于实验误差分析的说法正确的是\nA: 步骤 (1)中萃取时若没有充分振荡, 则导致所测 $c\\left(\\mathrm{I}_{3}^{-}\\right)+c\\left(\\mathrm{I}_{2}\\right)$ 值偏小\nB: 步骤1)中吸取上层清液时, 不慎吸入碘固体, 则测得的 $\\mathrm{K}$ 偏大\nC: 步骤(2)中滴定前滴定管有气泡, 滴定后气泡消失, 则测得的 $\\mathrm{K}$ 偏大\nD: 步骤(3)中滴定终点时俯视读数, 则测得的 $\\mathrm{K}$ 偏小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在含单质碘的 $\\mathrm{KI}$ 溶液中存在可逆反应: $\\mathrm{I}_{2}(\\mathrm{aq})+\\mathrm{I}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{I}_{3}^{-}(\\mathrm{aq})$, 为测定该反应的平衡常数 $\\mathrm{K}$ 进行如下实验, 实验步骤如下:\n\n(1) 在装有 $400 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KI}$ 溶液的碘量瓶中加入足量 $\\mathrm{I}_{2}$, 充分搅拌溶解, 待过量的固体碘沉于瓶底后, 取 $42.5 \\mathrm{~mL}$ 上层清液, 用 $5 \\mathrm{mLCCl}_{4}$ 萃取, 充分振荡、静置、分液,得到 $42.5 \\mathrm{~mL}$ 萃取后的水溶液、 $5 \\mathrm{mLI}_{2} \\sim \\mathrm{CCl}_{4}$ 溶液。\n\n(2)取萃取后的 $5 \\mathrm{mLI}_{2} \\sim \\mathrm{CCl}_{4}$ 溶液于碘量瓶中, 加水充分振荡, 再加入质量分数为 $0.01 \\% \\mathrm{KI}$ 溶液, 充分振荡后, 静置 5 分钟, 注入 $4 \\mathrm{~mL} 0.2 \\%$ 的淀粉溶液, 用 $\\mathrm{cmol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 标准溶液滴定, 平行滴定 3 次, 平均消耗 $V_{1} \\mathrm{mLNa}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 溶液。\n\n(3)将萃取后的水溶液 $42.5 \\mathrm{~mL}$ 移入碘量瓶中, 注入 $4 \\mathrm{~mL} 0.2 \\%$ 的淀粉溶液, 用 $\\mathrm{cmol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 标准溶液滴定, 平行滴定 3 次, 平均消耗 $V_{2} \\mathrm{mLNa}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ 溶液。\n\n已知: (1) $2 \\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}+\\mathrm{I}_{2}=2 \\mathrm{Na}_{2} \\mathrm{~S}_{4} \\mathrm{O}_{6}+2 \\mathrm{NaI}$; (2) $\\mathrm{I}^{-}$与 $\\mathrm{I}_{3}^{-}$难溶于 $\\mathrm{CCl}_{4}$; (3)达到溶解平衡后,\n$\\mathrm{I}_{2}$ 在 $\\mathrm{CCl}_{4}$ 层和水层中的分配比 $c\\left(\\mathrm{I}_{2}\\right)_{\\mathrm{CCl}_{4}}: c\\left(\\mathrm{I}_{2}\\right)_{\\mathrm{H}_{2} \\mathrm{O}}$ 为 $85: 1$\n\n[图1]\n\n碘量瓶示意图\n\n根据上述实验原理，下列关于实验误差分析的说法正确的是\n\nA: 步骤 (1)中萃取时若没有充分振荡, 则导致所测 $c\\left(\\mathrm{I}_{3}^{-}\\right)+c\\left(\\mathrm{I}_{2}\\right)$ 值偏小\nB: 步骤1)中吸取上层清液时, 不慎吸入碘固体, 则测得的 $\\mathrm{K}$ 偏大\nC: 步骤(2)中滴定前滴定管有气泡, 滴定后气泡消失, 则测得的 $\\mathrm{K}$ 偏大\nD: 步骤(3)中滴定终点时俯视读数, 则测得的 $\\mathrm{K}$ 偏小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-062.jpg?height=371&width=194&top_left_y=297&top_left_x=451"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_11",
        "problem": "What is the correct equilibrium expression for the following reaction?\n\n$$\n8 \\mathrm{H}_{2} \\mathrm{~S}(a q)+4 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{S}_{8}(s)+8 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\nA: $$ K=\\frac{\\left[\\mathrm{S}_{8}\\right]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right] \\cdot P\\left(\\mathrm{O}_{2}\\right)} $$\nB: $K=\\frac{\\left[\\mathrm{S}_{8}\\right]\\left(8 \\cdot\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]\\right)^{8}}{\\left(8 \\cdot\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]\\right)^{8} \\cdot\\left(4 \\cdot P\\left(\\mathrm{O}_{2}\\right)\\right)^{4}}$\nC: $K=\\frac{\\left[\\mathrm{S}_{8}\\right]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]^{8}}{\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]^{8} \\cdot P\\left(\\mathrm{O}_{2}\\right)^{4}}$\nD: $K=\\frac{1}{\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]^{8} \\cdot P\\left(\\mathrm{O}_{2}\\right)^{4}}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the correct equilibrium expression for the following reaction?\n\n$$\n8 \\mathrm{H}_{2} \\mathrm{~S}(a q)+4 \\mathrm{O}_{2}(g) \\rightleftharpoons \\mathrm{S}_{8}(s)+8 \\mathrm{H}_{2} \\mathrm{O}(l)\n$$\n\nA: $$ K=\\frac{\\left[\\mathrm{S}_{8}\\right]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]}{\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right] \\cdot P\\left(\\mathrm{O}_{2}\\right)} $$\nB: $K=\\frac{\\left[\\mathrm{S}_{8}\\right]\\left(8 \\cdot\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]\\right)^{8}}{\\left(8 \\cdot\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]\\right)^{8} \\cdot\\left(4 \\cdot P\\left(\\mathrm{O}_{2}\\right)\\right)^{4}}$\nC: $K=\\frac{\\left[\\mathrm{S}_{8}\\right]\\left[\\mathrm{H}_{2} \\mathrm{O}\\right]^{8}}{\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]^{8} \\cdot P\\left(\\mathrm{O}_{2}\\right)^{4}}$\nD: $K=\\frac{1}{\\left[\\mathrm{H}_{2} \\mathrm{~S}\\right]^{8} \\cdot P\\left(\\mathrm{O}_{2}\\right)^{4}}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_255",
        "problem": "Calculate the amount (in mol) of phenylmagnesium bromide $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{MgBr}\\right)$ required to produce $12.73 \\mathrm{~g}$ of sodium tetraphenylborate (molar mass $342.2 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the amount (in mol) of phenylmagnesium bromide $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{MgBr}\\right)$ required to produce $12.73 \\mathrm{~g}$ of sodium tetraphenylborate (molar mass $342.2 \\mathrm{~g} \\mathrm{~mol}^{-1}$ ).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_158",
        "problem": "The process mass intensity (PMI) allows chemists to calculate how much material is used when generating a target amount of product in a chemical reaction. PMI is expressed as follows:\n\n$$\n\\mathrm{PMI}=[(\\text { mass of all input materials) } / \\text { (mass of desired product) }]\n$$\n\nTo synthesize moclobemide, an anti-depressant pharmaceutical, 0.00381 moles of 4-(2-aminoethyl)morpholine $\\left(\\mathrm{C}_{6} \\mathrm{H}_{14} \\mathrm{~N}_{2} \\mathrm{O}\\right)$ is dissolved in $20.0 \\mathrm{~mL}$ of triethylamine (density: $0.726 \\mathrm{~g} \\mathrm{~mL}^{-1}$ ) and 0.00384 moles of 4-chlorobenzoyl chloride $\\left(\\mathrm{C}_{7} \\mathrm{H}_{4} \\mathrm{Cl}_{2} \\mathrm{O}\\right)$ is added. After rapid stirring for 30 minutes, $10.0 \\mathrm{~mL}$ (density: $1.00 \\mathrm{~g} \\mathrm{~mL}^{-1}$ ) of water is added followed by $10.0 \\mathrm{~mL}$ of dichloromethane (density: $1.325 \\mathrm{~g} \\mathrm{~mL}^{-1}$ ) and then the mixture transferred to a separatory funnel. After extraction, the dichloromethane is dried with $5.0 \\mathrm{~g}$ of magnesium sulfate. At the end of the process, $0.826 \\mathrm{~g}$ of pure moclobemide is recovered.\n\nThe PMI for this reaction is:\nA: 43.8\nB: 47.1\nC: 49.3\nD: 53.2\nE: 55.9\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe process mass intensity (PMI) allows chemists to calculate how much material is used when generating a target amount of product in a chemical reaction. PMI is expressed as follows:\n\n$$\n\\mathrm{PMI}=[(\\text { mass of all input materials) } / \\text { (mass of desired product) }]\n$$\n\nTo synthesize moclobemide, an anti-depressant pharmaceutical, 0.00381 moles of 4-(2-aminoethyl)morpholine $\\left(\\mathrm{C}_{6} \\mathrm{H}_{14} \\mathrm{~N}_{2} \\mathrm{O}\\right)$ is dissolved in $20.0 \\mathrm{~mL}$ of triethylamine (density: $0.726 \\mathrm{~g} \\mathrm{~mL}^{-1}$ ) and 0.00384 moles of 4-chlorobenzoyl chloride $\\left(\\mathrm{C}_{7} \\mathrm{H}_{4} \\mathrm{Cl}_{2} \\mathrm{O}\\right)$ is added. After rapid stirring for 30 minutes, $10.0 \\mathrm{~mL}$ (density: $1.00 \\mathrm{~g} \\mathrm{~mL}^{-1}$ ) of water is added followed by $10.0 \\mathrm{~mL}$ of dichloromethane (density: $1.325 \\mathrm{~g} \\mathrm{~mL}^{-1}$ ) and then the mixture transferred to a separatory funnel. After extraction, the dichloromethane is dried with $5.0 \\mathrm{~g}$ of magnesium sulfate. At the end of the process, $0.826 \\mathrm{~g}$ of pure moclobemide is recovered.\n\nThe PMI for this reaction is:\n\nA: 43.8\nB: 47.1\nC: 49.3\nD: 53.2\nE: 55.9\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_970",
        "problem": "The chemical formula of barium perrhenate is $\\mathrm{Ba}\\left(\\mathrm{ReO}_{4}\\right)_{2}$. What is the charge on the perrhenate ion?\nA: +2\nB: +1\nC: 0\nD: -1\nE: -2\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe chemical formula of barium perrhenate is $\\mathrm{Ba}\\left(\\mathrm{ReO}_{4}\\right)_{2}$. What is the charge on the perrhenate ion?\n\nA: +2\nB: +1\nC: 0\nD: -1\nE: -2\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_260",
        "problem": "Calculate the concentration of chloride ions in the resulting solution when $50.0 \\mathrm{~mL}$ of $2.68 \\mathrm{~mol} \\mathrm{~L}^{-1}$ calcium chloride solution is mixed with $150 \\mathrm{~mL}$ of $1.13 \\mathrm{~mol} \\mathrm{~L}^{-1}$ silver nitrate solution.\nA: $0.49 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.67 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.85 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $1.34 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $\\quad 1.55 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nCalculate the concentration of chloride ions in the resulting solution when $50.0 \\mathrm{~mL}$ of $2.68 \\mathrm{~mol} \\mathrm{~L}^{-1}$ calcium chloride solution is mixed with $150 \\mathrm{~mL}$ of $1.13 \\mathrm{~mol} \\mathrm{~L}^{-1}$ silver nitrate solution.\n\nA: $0.49 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.67 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $0.85 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $1.34 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $\\quad 1.55 \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_402",
        "problem": "Cope 重排反应如图所示, $\\mathrm{K}\\left({ }^{\\mathrm{HO}}\\right.$ )经 Cope 重排反应可得到 $\\mathrm{M}$ 。已知: 羟基直接连在双键碳原子上不稳定会异构化成酮碳基或醛基。下列说法错误的是\n\n[图1]\nA: Y 中所有碳原子可能共平面\nB: $\\mathrm{M}$ 的结构简式为 [图2]\nC: 依据红外光谱可确证 $K 、 M$ 存在不同的官能团\nD: 含酚羟基且有手性碳原子的 $\\mathrm{M}$ 的同分异构体有 9 种(不考虑立体异构)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\nCope 重排反应如图所示, $\\mathrm{K}\\left({ }^{\\mathrm{HO}}\\right.$ )经 Cope 重排反应可得到 $\\mathrm{M}$ 。已知: 羟基直接连在双键碳原子上不稳定会异构化成酮碳基或醛基。下列说法错误的是\n\n[图1]\n\nA: Y 中所有碳原子可能共平面\nB: $\\mathrm{M}$ 的结构简式为 [图2]\nC: 依据红外光谱可确证 $K 、 M$ 存在不同的官能团\nD: 含酚羟基且有手性碳原子的 $\\mathrm{M}$ 的同分异构体有 9 种(不考虑立体异构)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-03.jpg?height=285&width=560&top_left_y=143&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-03.jpg?height=154&width=254&top_left_y=511&top_left_x=707",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-03.jpg?height=220&width=260&top_left_y=1186&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-03.jpg?height=225&width=403&top_left_y=1178&top_left_x=969",
            "https://i.postimg.cc/mD237Fg6/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_389",
        "problem": "Thallium-201, a radioactive isotope used to image the heart, has a half-life of $3.05 \\mathrm{~d}$. How long would it take for a sample of thallium-201 to decay to $18 \\%$ of its original activity?\nA: $4.4 \\mathrm{~d}$\nB: $6.1 \\mathrm{~d}$\nC: $7.5 \\mathrm{~d}$\nD: $17 \\mathrm{~d}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThallium-201, a radioactive isotope used to image the heart, has a half-life of $3.05 \\mathrm{~d}$. How long would it take for a sample of thallium-201 to decay to $18 \\%$ of its original activity?\n\nA: $4.4 \\mathrm{~d}$\nB: $6.1 \\mathrm{~d}$\nC: $7.5 \\mathrm{~d}$\nD: $17 \\mathrm{~d}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_309",
        "problem": "Which of the following ions, in its ground electronic state, does not have the same electronic configuration as a ground state Ar atom?\nA: $\\mathrm{P}^{3-}$\nB: $\\mathrm{Cl}^{-}$\nC: $\\mathrm{K}^{+}$\nD: $\\mathrm{Ca}^{2+}$\nE: $\\mathrm{Sc}^{2+}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following ions, in its ground electronic state, does not have the same electronic configuration as a ground state Ar atom?\n\nA: $\\mathrm{P}^{3-}$\nB: $\\mathrm{Cl}^{-}$\nC: $\\mathrm{K}^{+}$\nD: $\\mathrm{Ca}^{2+}$\nE: $\\mathrm{Sc}^{2+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_348",
        "problem": "Given the bond dissociation enthalpies (BDE) below, what is the approximate $\\Delta H^{\\circ}{ }_{\\mathrm{f}}$ for $\\mathrm{H}_{2} \\mathrm{O}(g)$ ?\n\n| Bond | BDE, $\\mathrm{kJ} \\mathrm{mol}^{-1}$ | Bond | BDE, $\\mathrm{kJ} \\mathrm{mol}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{H}-\\mathrm{H}$ | 432 | $\\mathrm{O}-\\mathrm{O}$ | 146 |\n| $\\mathrm{O}-\\mathrm{H}$ | 467 | $\\mathrm{O}=\\mathrm{O}$ | 495 |\nA: $-934 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-510 . \\mathrm{kJ} \\mathrm{mol}^{-1}$\nC: $-429 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-255 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGiven the bond dissociation enthalpies (BDE) below, what is the approximate $\\Delta H^{\\circ}{ }_{\\mathrm{f}}$ for $\\mathrm{H}_{2} \\mathrm{O}(g)$ ?\n\n| Bond | BDE, $\\mathrm{kJ} \\mathrm{mol}^{-1}$ | Bond | BDE, $\\mathrm{kJ} \\mathrm{mol}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n| $\\mathrm{H}-\\mathrm{H}$ | 432 | $\\mathrm{O}-\\mathrm{O}$ | 146 |\n| $\\mathrm{O}-\\mathrm{H}$ | 467 | $\\mathrm{O}=\\mathrm{O}$ | 495 |\n\nA: $-934 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-510 . \\mathrm{kJ} \\mathrm{mol}^{-1}$\nC: $-429 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-255 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1273",
        "problem": "Graphene is a two-dimensional, one atom thick carbon material (Fig. 1 a). Many layers of graphene stack together to form graphite (Fig. 1 b).\n[figure1]\n\nFig. 1. (a) The structure of graphene. Spheres are carbon atoms. They are arranged in hexagons. The area of one carbon hexagon is $5.16 \\cdot 10^{-20} \\mathrm{~m}^{2}$.\n\n(b) Crystal lattice of graphite. Three graphene layers are shown.\n\nSuch atomic structure was long considered to be unstable. However, in 2004 Andrey Geim and Konstantin Novoselov have reported production of the first samples of this unusual material. This groundbreaking invention was awarded by Nobel prize in 2010.\n\nExperimental studies of graphene are still restricted. Production of massive portions of the new substance still is a challenging synthetic problem. Many properties of graphene were estimated. Usually, there is not enough information for rigorous calculations, so we have to make assumptions and neglect unimportant factors. In this problem, you will estimate the adsorption properties of graphene.Estimate the specific surface of graphene open for adsorption in units $\\mathrm{m}^{2} \\mathrm{~g}^{-1}$. Consider that graphene plane is separated from any other solid or liquid substance.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nGraphene is a two-dimensional, one atom thick carbon material (Fig. 1 a). Many layers of graphene stack together to form graphite (Fig. 1 b).\n[figure1]\n\nFig. 1. (a) The structure of graphene. Spheres are carbon atoms. They are arranged in hexagons. The area of one carbon hexagon is $5.16 \\cdot 10^{-20} \\mathrm{~m}^{2}$.\n\n(b) Crystal lattice of graphite. Three graphene layers are shown.\n\nSuch atomic structure was long considered to be unstable. However, in 2004 Andrey Geim and Konstantin Novoselov have reported production of the first samples of this unusual material. This groundbreaking invention was awarded by Nobel prize in 2010.\n\nExperimental studies of graphene are still restricted. Production of massive portions of the new substance still is a challenging synthetic problem. Many properties of graphene were estimated. Usually, there is not enough information for rigorous calculations, so we have to make assumptions and neglect unimportant factors. In this problem, you will estimate the adsorption properties of graphene.\n\nproblem:\nEstimate the specific surface of graphene open for adsorption in units $\\mathrm{m}^{2} \\mathrm{~g}^{-1}$. Consider that graphene plane is separated from any other solid or liquid substance.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{m}^{2} \\mathrm{~g}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-244.jpg?height=540&width=1024&top_left_y=878&top_left_x=524"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{m}^{2} \\mathrm{~g}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_980",
        "problem": "How many protons are there in the nucleus of ${ }_{53}^{127}$ I?\nA: 7\nB: 53\nC: 74\nD: 127\nE: 180\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many protons are there in the nucleus of ${ }_{53}^{127}$ I?\n\nA: 7\nB: 53\nC: 74\nD: 127\nE: 180\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1569",
        "problem": "Many streams drain in areas where coal or metallic ores are mined. These streams have become acidic and contain high concentrations of dissolved iron and sulphate, due to sulphur-containing ores being exposed to the atmosphere or to oxygenated waters. The most common sulphur-containing mineral is pyrite, $\\mathrm{FeS}_{2}$, in which the oxidation state of iron is +2 . As the iron-rich streams mix with other waters, the dissolved iron precipitates as goethite, $\\mathrm{FeO}(\\mathrm{OH})$, which coats the stream bottom while the water remains acidic.\n\nThe concentration of iron as $\\mathrm{Fe}(\\mathrm{II})$ in a stream is $0.00835 \\mathrm{M}$. At a very narrow point in the stream it empties into a large pond, with a flow rate of $20.0 \\mathrm{I}$ each minute. The water in this stream is sufficiently aerated that $75 \\%$ of the $\\mathrm{Fe}(\\mathrm{II})$ is oxidized to $\\mathrm{Fe}(\\mathrm{III})$. The $\\mathrm{pH}$ of the pond is high enough ( $>7$ ) that the iron(III) precipitates immediately as $\\mathrm{Fe}(\\mathrm{OH})_{3}$ which on aging becomes $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$. What mass of $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ will be deposited on the bottom of the pond in two years?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nMany streams drain in areas where coal or metallic ores are mined. These streams have become acidic and contain high concentrations of dissolved iron and sulphate, due to sulphur-containing ores being exposed to the atmosphere or to oxygenated waters. The most common sulphur-containing mineral is pyrite, $\\mathrm{FeS}_{2}$, in which the oxidation state of iron is +2 . As the iron-rich streams mix with other waters, the dissolved iron precipitates as goethite, $\\mathrm{FeO}(\\mathrm{OH})$, which coats the stream bottom while the water remains acidic.\n\nThe concentration of iron as $\\mathrm{Fe}(\\mathrm{II})$ in a stream is $0.00835 \\mathrm{M}$. At a very narrow point in the stream it empties into a large pond, with a flow rate of $20.0 \\mathrm{I}$ each minute. The water in this stream is sufficiently aerated that $75 \\%$ of the $\\mathrm{Fe}(\\mathrm{II})$ is oxidized to $\\mathrm{Fe}(\\mathrm{III})$. The $\\mathrm{pH}$ of the pond is high enough ( $>7$ ) that the iron(III) precipitates immediately as $\\mathrm{Fe}(\\mathrm{OH})_{3}$ which on aging becomes $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$. What mass of $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$ will be deposited on the bottom of the pond in two years?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_881",
        "problem": "常温下, 将 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴加到 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 二元弱酸 $\\mathrm{H}_{2} \\mathrm{~A}$ 溶液中, 混合溶液的 $\\mathrm{pH}$ 随 $\\mathrm{NaOH}$ 溶液滴入量的关系如图所示。下列叙述不正确的是\n\n[图1]\nA: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right) \\approx 10^{-3}$\nB: $b$ 点时溶液中存在 $c\\left(H^{-}\\right)>c\\left(A^{2-}\\right)>c\\left(H_{2} A\\right)$\nC: $\\mathrm{c}$ 点时溶液中存在 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nD: 溶液中水的电离程度: $\\mathrm{c}>\\mathrm{b}>\\mathrm{a}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 将 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴加到 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 二元弱酸 $\\mathrm{H}_{2} \\mathrm{~A}$ 溶液中, 混合溶液的 $\\mathrm{pH}$ 随 $\\mathrm{NaOH}$ 溶液滴入量的关系如图所示。下列叙述不正确的是\n\n[图1]\n\nA: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right) \\approx 10^{-3}$\nB: $b$ 点时溶液中存在 $c\\left(H^{-}\\right)>c\\left(A^{2-}\\right)>c\\left(H_{2} A\\right)$\nC: $\\mathrm{c}$ 点时溶液中存在 $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nD: 溶液中水的电离程度: $\\mathrm{c}>\\mathrm{b}>\\mathrm{a}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-065.jpg?height=509&width=968&top_left_y=171&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_926",
        "problem": "在 $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{R}$ 及其钠盐的溶液中, $\\mathrm{H}_{2} \\mathrm{R} 、 \\mathrm{HR}^{-} 、 \\mathrm{R}^{2}$-分别在三者中所占的物质的量分数 $(\\alpha)$ 随溶液 $\\mathrm{pH}$ 变化关系如下图所示, 下列叙述错误的是\n\n[图1]\nA: $\\mathrm{H}_{2} \\mathrm{R}$ 是二元弱酸, 其 $\\mathrm{K}_{\\mathrm{a} 1}=1 \\times 10^{-2}$\nB: NaHR 在溶液中水解程度小于电离程度\nC: $\\mathrm{pH}<7.2$ 的溶液中一定存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 含 $\\mathrm{Na}_{2} \\mathrm{R}$ 与 $\\mathrm{NaHR}$ 各 $0.1 \\mathrm{~mol}$ 的混合溶液的 $\\mathrm{pH}=7.2$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在 $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{R}$ 及其钠盐的溶液中, $\\mathrm{H}_{2} \\mathrm{R} 、 \\mathrm{HR}^{-} 、 \\mathrm{R}^{2}$-分别在三者中所占的物质的量分数 $(\\alpha)$ 随溶液 $\\mathrm{pH}$ 变化关系如下图所示, 下列叙述错误的是\n\n[图1]\n\nA: $\\mathrm{H}_{2} \\mathrm{R}$ 是二元弱酸, 其 $\\mathrm{K}_{\\mathrm{a} 1}=1 \\times 10^{-2}$\nB: NaHR 在溶液中水解程度小于电离程度\nC: $\\mathrm{pH}<7.2$ 的溶液中一定存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HR}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{R}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 含 $\\mathrm{Na}_{2} \\mathrm{R}$ 与 $\\mathrm{NaHR}$ 各 $0.1 \\mathrm{~mol}$ 的混合溶液的 $\\mathrm{pH}=7.2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-84.jpg?height=320&width=985&top_left_y=1436&top_left_x=541"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_150",
        "problem": "The transition metal rhenium ( $\\mathrm{Re})$, is among the rarest elements in the earth's crust. The precursor to the pure metal, ammonium perrhenate, is produced during the refinement of molybdenum ores. The principal application of rhenium is as an alloy with nickel, used in jet engine components. Consider the unbalanced chemical equation for the hydrogen reduction of ammonium perrhenate:\n\n$$\n-\\mathrm{NH}_{4} \\mathrm{ReO}_{4}+{ }_{-} \\mathrm{H}_{2} \\longrightarrow-\\mathrm{Re}+{ }_{-} \\mathrm{H}_{2} \\mathrm{O}+{ }_{-} \\mathrm{NH}_{3}\n$$\n\nAssume the oxidation number of nitrogen in the reactants and products is unchanged. Determine the sum of the smallest integer coefficients for the stoichiometrically balanced chemical equation.\nA: 5\nB: 14\nC: 18\nD: 20\nE: 21\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe transition metal rhenium ( $\\mathrm{Re})$, is among the rarest elements in the earth's crust. The precursor to the pure metal, ammonium perrhenate, is produced during the refinement of molybdenum ores. The principal application of rhenium is as an alloy with nickel, used in jet engine components. Consider the unbalanced chemical equation for the hydrogen reduction of ammonium perrhenate:\n\n$$\n-\\mathrm{NH}_{4} \\mathrm{ReO}_{4}+{ }_{-} \\mathrm{H}_{2} \\longrightarrow-\\mathrm{Re}+{ }_{-} \\mathrm{H}_{2} \\mathrm{O}+{ }_{-} \\mathrm{NH}_{3}\n$$\n\nAssume the oxidation number of nitrogen in the reactants and products is unchanged. Determine the sum of the smallest integer coefficients for the stoichiometrically balanced chemical equation.\n\nA: 5\nB: 14\nC: 18\nD: 20\nE: 21\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1394",
        "problem": "In the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.A sample of $\\mathrm{YBCO}$ with $\\delta=0.25$ was subjected to $\\mathrm{X}$-ray diffraction using CuK $\\alpha$ radiation $(\\lambda=154.2 \\mathrm{pm})$. The lowest-angle diffraction peak was observed at $2 \\theta=7.450^{\\circ}$. Assuming that $a=b=(c / 3)$, calculate the values of $a$ and $c$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nIn the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.\n\nproblem:\nA sample of $\\mathrm{YBCO}$ with $\\delta=0.25$ was subjected to $\\mathrm{X}$-ray diffraction using CuK $\\alpha$ radiation $(\\lambda=154.2 \\mathrm{pm})$. The lowest-angle diffraction peak was observed at $2 \\theta=7.450^{\\circ}$. Assuming that $a=b=(c / 3)$, calculate the values of $a$ and $c$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the value of $a$, the value of $c$].\nTheir units are, in order, [pm, pm], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "pm",
            "pm"
        ],
        "answer_sequence": [
            "the value of $a$",
            "the value of $c$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_697",
        "problem": "有机物 $\\mathrm{X} \\rightarrow \\mathrm{Y}$ 的异构化反应如图所示。\n\n[图1]\n\n已知: 羟基直接连在双键上不稳定会异构化成羰基或醛基。下列说法错误的是\nA: $\\mathrm{X}$ 中的所有碳原子一定共平面\nB: 依据红外光谱可确定 X、Y 存在不同的官能团\nC: 含醛基且有手性碳原子的 $Y$ 的同分异构体有 5 种(不考虑立体异构)\nD: 类比上述反应, 一定条件下 [图2]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有机物 $\\mathrm{X} \\rightarrow \\mathrm{Y}$ 的异构化反应如图所示。\n\n[图1]\n\n已知: 羟基直接连在双键上不稳定会异构化成羰基或醛基。下列说法错误的是\n\nA: $\\mathrm{X}$ 中的所有碳原子一定共平面\nB: 依据红外光谱可确定 X、Y 存在不同的官能团\nC: 含醛基且有手性碳原子的 $Y$ 的同分异构体有 5 种(不考虑立体异构)\nD: 类比上述反应, 一定条件下 [图2]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-09.jpg?height=271&width=786&top_left_y=727&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-09.jpg?height=194&width=694&top_left_y=1331&top_left_x=887"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_377",
        "problem": "Addition of an electron to a gas-phase Si atom results in the release of more energy than addition of an electron to a gas-phase $\\mathrm{P}$ atom. What is the best explanation for their relative electron affinities?\nA: The electron added to Si experiences less electronelectron repulsion than the electron added to $\\mathrm{P}$.\nB: The electron added to Si enters a lower-energy subshell than the electron added to $\\mathrm{P}$.\nC: Si is more electronegative than $\\mathrm{P}$.\nD: Si is smaller than $\\mathrm{P}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAddition of an electron to a gas-phase Si atom results in the release of more energy than addition of an electron to a gas-phase $\\mathrm{P}$ atom. What is the best explanation for their relative electron affinities?\n\nA: The electron added to Si experiences less electronelectron repulsion than the electron added to $\\mathrm{P}$.\nB: The electron added to Si enters a lower-energy subshell than the electron added to $\\mathrm{P}$.\nC: Si is more electronegative than $\\mathrm{P}$.\nD: Si is smaller than $\\mathrm{P}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_752",
        "problem": "秦俑彩绘中含有难溶的铅白 $\\left(\\mathrm{PbCO}_{3}\\right)$ 和黄色的 $\\mathrm{PbI}_{2}$ 。常温下, $\\mathrm{PbCO}_{3}$ 和 $\\mathrm{PbI}_{2}$ 在不同\n\n的溶液中分别达到沉淀溶解平衡时 $\\mathrm{pM}$ 与 $\\mathrm{pR}$ 的关系如图所示, 其中 $\\mathrm{pM}$ 为阳离子浓度的负对数, $\\mathrm{pR}$ 为阴离子浓度的负对数。下列说法错误的是\n\n[图1]\nA: 完全沉淀废液中的 $\\mathrm{Pb}^{2+}, \\mathrm{I}^{-}$的效果不如 $\\mathrm{CO}_{3}^{2-}$\nB: $\\mathrm{z}$ 点, $\\mathrm{Q}=\\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{PbCO}_{3}\\right)$\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{PbI}_{2}\\right)$ 的数量级为 $10^{-10}$\nD: $\\mathrm{PbI}_{2}(\\mathrm{~s})$ 转化为 $\\mathrm{PbCO}_{3}(\\mathrm{~s})$ 的反应趋势很大\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n秦俑彩绘中含有难溶的铅白 $\\left(\\mathrm{PbCO}_{3}\\right)$ 和黄色的 $\\mathrm{PbI}_{2}$ 。常温下, $\\mathrm{PbCO}_{3}$ 和 $\\mathrm{PbI}_{2}$ 在不同\n\n的溶液中分别达到沉淀溶解平衡时 $\\mathrm{pM}$ 与 $\\mathrm{pR}$ 的关系如图所示, 其中 $\\mathrm{pM}$ 为阳离子浓度的负对数, $\\mathrm{pR}$ 为阴离子浓度的负对数。下列说法错误的是\n\n[图1]\n\nA: 完全沉淀废液中的 $\\mathrm{Pb}^{2+}, \\mathrm{I}^{-}$的效果不如 $\\mathrm{CO}_{3}^{2-}$\nB: $\\mathrm{z}$ 点, $\\mathrm{Q}=\\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{PbCO}_{3}\\right)$\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{PbI}_{2}\\right)$ 的数量级为 $10^{-10}$\nD: $\\mathrm{PbI}_{2}(\\mathrm{~s})$ 转化为 $\\mathrm{PbCO}_{3}(\\mathrm{~s})$ 的反应趋势很大\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-083.jpg?height=608&width=666&top_left_y=1963&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_945",
        "problem": "一种能将甘油 $\\left(\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}_{3}\\right)$ 和二氧化碳转化为甘油醛 $\\left(\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}_{3}\\right)$ 和合成气的电化学装置如图所示。下列说法正确的是\n\n[图1]\nA: 催化电极 $\\mathrm{a}$ 与电源负极相连\nB: 电解时催化电极 $\\mathrm{a}$ 附近的 $\\mathrm{pH}$ 减小\nC: 电解时 $\\mathrm{b}$ 极区 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液浓度升高\nD: 生成的甘油醛与合成气的物质的量相等\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种能将甘油 $\\left(\\mathrm{C}_{3} \\mathrm{H}_{8} \\mathrm{O}_{3}\\right)$ 和二氧化碳转化为甘油醛 $\\left(\\mathrm{C}_{3} \\mathrm{H}_{6} \\mathrm{O}_{3}\\right)$ 和合成气的电化学装置如图所示。下列说法正确的是\n\n[图1]\n\nA: 催化电极 $\\mathrm{a}$ 与电源负极相连\nB: 电解时催化电极 $\\mathrm{a}$ 附近的 $\\mathrm{pH}$ 减小\nC: 电解时 $\\mathrm{b}$ 极区 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液浓度升高\nD: 生成的甘油醛与合成气的物质的量相等\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-17.jpg?height=451&width=897&top_left_y=1228&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1194",
        "problem": "Application of kinetic studies in water treatment\n\nIndustrial waste is a major cause of water pollution and kinetic studies are carried out in a laboratory to design effluent treatment. 1,4-dioxane, more commonly known as dioxane $\\left(\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}\\right)$, an industrial solvent and by-product, is a significant water contaminant. It can be oxidised to hazard free chemicals using oxidants such as peroxodisulfate, ozone or hydrogen peroxide.\n\nThe data obtained in the kinetic study of oxidation of dioxane with potassium peroxodisulfate $\\left(\\mathrm{K}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}\\right)$ as oxidant and $\\mathrm{AgNO}_{3}$ as catalyst at $T=303.15 \\mathrm{~K}$ are given below. The reaction was monitored by the estimation of unreacted peroxodisulfate. The concentration of $\\mathrm{AgNO}_{3}$ used in this study was $1.00 \\cdot 10^{-3} \\mathrm{mmol} \\mathrm{dm}{ }^{-3}$.\n\n| Trial | Dioxane <br> $\\mathrm{mmol} \\cdot \\mathrm{dm}^{-3}$ | $\\mathrm{K}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}$ <br> $\\mathrm{mmol} \\cdot \\mathrm{dm}^{-3}$ | Initial rate <br> $\\mathrm{mmol} \\cdot \\mathrm{dm}^{-3} \\cdot \\mathrm{min}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.0100 | 2.50 | $1.661 \\cdot 10^{-2}$ |\n| 2 | 0.0100 | 5.10 | $3.380 \\cdot 10^{-2}$ |\n| 3 | 0.00500 | 13.8 | $9.200 \\cdot 10^{-2}$ |\n| 4 | 0.0110 | 13.8 | $9.201 \\cdot 10^{-2}$ |\n\nIn many countries the accepted maximum level of dioxane in drinking water is specified as $0.35 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nA water sample contains an initial dioxane concentration of $40.00 \\mu \\mathrm{gm}^{-3}$. Assume that $1 \\mathrm{~mol}$ dioxane requires $1 \\mathrm{~mol}$ of peroxodisulfate for oxidation. The concentration of $\\mathrm{AgNO}_{3}$ used in this study was $1.00 \\times 10^{-3} \\mathrm{mmol} \\cdot \\mathrm{dm}^{-3}$.Calculate the time in minutes the oxidation process has to continue in order to reach the accepted level of dioxane at $303.15 \\mathrm{~K}$ if the initial concentration of $\\mathrm{K}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}$ is $5.0 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~dm}^{-3}$. Assume that the rate law obtained from the data above is valid under these conditions.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nApplication of kinetic studies in water treatment\n\nIndustrial waste is a major cause of water pollution and kinetic studies are carried out in a laboratory to design effluent treatment. 1,4-dioxane, more commonly known as dioxane $\\left(\\mathrm{C}_{4} \\mathrm{H}_{8} \\mathrm{O}_{2}\\right)$, an industrial solvent and by-product, is a significant water contaminant. It can be oxidised to hazard free chemicals using oxidants such as peroxodisulfate, ozone or hydrogen peroxide.\n\nThe data obtained in the kinetic study of oxidation of dioxane with potassium peroxodisulfate $\\left(\\mathrm{K}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}\\right)$ as oxidant and $\\mathrm{AgNO}_{3}$ as catalyst at $T=303.15 \\mathrm{~K}$ are given below. The reaction was monitored by the estimation of unreacted peroxodisulfate. The concentration of $\\mathrm{AgNO}_{3}$ used in this study was $1.00 \\cdot 10^{-3} \\mathrm{mmol} \\mathrm{dm}{ }^{-3}$.\n\n| Trial | Dioxane <br> $\\mathrm{mmol} \\cdot \\mathrm{dm}^{-3}$ | $\\mathrm{K}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}$ <br> $\\mathrm{mmol} \\cdot \\mathrm{dm}^{-3}$ | Initial rate <br> $\\mathrm{mmol} \\cdot \\mathrm{dm}^{-3} \\cdot \\mathrm{min}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.0100 | 2.50 | $1.661 \\cdot 10^{-2}$ |\n| 2 | 0.0100 | 5.10 | $3.380 \\cdot 10^{-2}$ |\n| 3 | 0.00500 | 13.8 | $9.200 \\cdot 10^{-2}$ |\n| 4 | 0.0110 | 13.8 | $9.201 \\cdot 10^{-2}$ |\n\nIn many countries the accepted maximum level of dioxane in drinking water is specified as $0.35 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nA water sample contains an initial dioxane concentration of $40.00 \\mu \\mathrm{gm}^{-3}$. Assume that $1 \\mathrm{~mol}$ dioxane requires $1 \\mathrm{~mol}$ of peroxodisulfate for oxidation. The concentration of $\\mathrm{AgNO}_{3}$ used in this study was $1.00 \\times 10^{-3} \\mathrm{mmol} \\cdot \\mathrm{dm}^{-3}$.\n\nproblem:\nCalculate the time in minutes the oxidation process has to continue in order to reach the accepted level of dioxane at $303.15 \\mathrm{~K}$ if the initial concentration of $\\mathrm{K}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}$ is $5.0 \\times 10^{-6} \\mathrm{~mol} \\mathrm{~dm}^{-3}$. Assume that the rate law obtained from the data above is valid under these conditions.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of minutes, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "minutes"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_545",
        "problem": "药物 $Z$ 可用于治疗哮喘、系统性红斑狼疮等, 可由 $X$ (咖啡酸) 和 $Y(1,4$ 一环己二酮单乙二醇缩酮）为原料合成（如图）。下列说法中正确的是的\n\n[图1]\n\n(X)\n\n[图2]\n\n(Y)\n\n[图3]\n\n$(\\mathrm{Z})$\nA: $\\mathrm{X}$ 和 $\\mathrm{Y}$ 生成 $\\mathrm{Z}$ 的反应属于加成反应\nB: $\\mathrm{X}$ 中最多有 9 个碳原子共面\nC: $1 \\mathrm{molZ}$ 与 $\\mathrm{NaOH}$ 溶液反应, 最多消耗 $3 \\mathrm{molNaOH}$\nD: 可用 $\\mathrm{FeCl}_{3}$ 溶液检验此反应中是否有 $\\mathrm{Z}$ 生成\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n药物 $Z$ 可用于治疗哮喘、系统性红斑狼疮等, 可由 $X$ (咖啡酸) 和 $Y(1,4$ 一环己二酮单乙二醇缩酮）为原料合成（如图）。下列说法中正确的是的\n\n[图1]\n\n(X)\n\n[图2]\n\n(Y)\n\n[图3]\n\n$(\\mathrm{Z})$\n\nA: $\\mathrm{X}$ 和 $\\mathrm{Y}$ 生成 $\\mathrm{Z}$ 的反应属于加成反应\nB: $\\mathrm{X}$ 中最多有 9 个碳原子共面\nC: $1 \\mathrm{molZ}$ 与 $\\mathrm{NaOH}$ 溶液反应, 最多消耗 $3 \\mathrm{molNaOH}$\nD: 可用 $\\mathrm{FeCl}_{3}$ 溶液检验此反应中是否有 $\\mathrm{Z}$ 生成\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-62.jpg?height=174&width=505&top_left_y=1204&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-62.jpg?height=314&width=369&top_left_y=1145&top_left_x=935",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-62.jpg?height=271&width=471&top_left_y=1195&top_left_x=1324",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-62.jpg?height=263&width=828&top_left_y=2107&top_left_x=905"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1228",
        "problem": "A simple inorganic experiment\n\nCompound A which contains metal $\\mathbf{X}$ is a colourless crystalline solid and highly soluble in water. It is used as a reagent in analysis and gives in alkali media a binary compound B containing $6.9 \\%$ (mass) of oxygen. Under heating A decomposes with a mass loss of $36.5 \\%$.\n\nDetermine the metal $\\mathbf{X}$ and compounds $\\mathbf{A}$ and $\\mathbf{B}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nA simple inorganic experiment\n\nCompound A which contains metal $\\mathbf{X}$ is a colourless crystalline solid and highly soluble in water. It is used as a reagent in analysis and gives in alkali media a binary compound B containing $6.9 \\%$ (mass) of oxygen. Under heating A decomposes with a mass loss of $36.5 \\%$.\n\nDetermine the metal $\\mathbf{X}$ and compounds $\\mathbf{A}$ and $\\mathbf{B}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [X, A, B].\nTheir answer types are, in order, [expression, expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null,
            null
        ],
        "answer_sequence": [
            "X",
            "A",
            "B"
        ],
        "type_sequence": [
            "EX",
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_375",
        "problem": "The wavelength of one of the spectral lines of helium is $492 \\mathrm{~nm}$. What is the energy of a photon with this wavelength?\nA: $3.26 \\times 10^{-40} \\mathrm{~J}$\nB: $3.26 \\times 10^{-31} \\mathrm{~J}$\nC: $4.04 \\times 10^{-28} \\mathrm{~J}$\nD: $4.04 \\times 10^{-19} \\mathrm{~J}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe wavelength of one of the spectral lines of helium is $492 \\mathrm{~nm}$. What is the energy of a photon with this wavelength?\n\nA: $3.26 \\times 10^{-40} \\mathrm{~J}$\nB: $3.26 \\times 10^{-31} \\mathrm{~J}$\nC: $4.04 \\times 10^{-28} \\mathrm{~J}$\nD: $4.04 \\times 10^{-19} \\mathrm{~J}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_758",
        "problem": "$\\mathrm{CO}_{2}$ 在电流作用下能转化为 $\\mathrm{HCOO}^{-}$, 在不同的催化剂作用下的部分转化路径如下图所示。下列说法不正确的是\n\n[图1]\nA: 第(2)(5)步反应属于加成反应\nB: 第(1)(4)步反应需要电源提供电子\nC: 两种转化路径中原子均 $100 \\%$ 转化为产物\nD: 总反应可表示为: $\\mathrm{CO}_{2}+\\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{HCOO}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{CO}_{2}$ 在电流作用下能转化为 $\\mathrm{HCOO}^{-}$, 在不同的催化剂作用下的部分转化路径如下图所示。下列说法不正确的是\n\n[图1]\n\nA: 第(2)(5)步反应属于加成反应\nB: 第(1)(4)步反应需要电源提供电子\nC: 两种转化路径中原子均 $100 \\%$ 转化为产物\nD: 总反应可表示为: $\\mathrm{CO}_{2}+\\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{HCOO}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-095.jpg?height=963&width=1251&top_left_y=161&top_left_x=331"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_586",
        "problem": "一定温度下可逆反应: $\\mathrm{A}(\\mathrm{s})+2 \\mathrm{~B}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{C}(\\mathrm{g})+\\mathrm{D}(\\mathrm{g}) \\Delta H<0$ 。现将 $1 \\mathrm{~mol} \\mathrm{~A}$ 和 $2 \\mathrm{~mol}$ $\\mathrm{B}$ 加入甲容器中, 将 $4 \\mathrm{~mol} \\mathrm{C}$ 和 $2 \\mathrm{~mol} \\mathrm{D}$ 加入乙容器中, 此时控制活塞 $\\mathrm{P}$, 使乙的容积为甲的 2 倍, $t_{1}$ 时两容器内均达到平衡状态 (如图 1 所示, 隔板 $\\mathrm{K}$ 不能移动)。下列说法正确的是\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\n[图3]\n\n图 3\nA: 保持活塞位置不变, 升高温度, 达到新的平衡后, 甲、乙中 $\\mathrm{B}$ 的体积分数均增大\nB: 保持温度和活塞位置不变, 在甲中再加入 $1 \\mathrm{~mol} \\mathrm{~A}$ 和 $2 \\mathrm{~mol} \\mathrm{~B}$, 达到新的平衡后,甲中 $\\mathrm{C}$ 的浓度是乙中 $\\mathrm{C}$ 的浓度的 2 倍\nC: 保持温度和乙中的压强不变, $t_{2}$ 时分别向甲、乙中加入等质量的氦气后，甲、乙中反应速率变化情况分别如图 2 和图 3 所示 ( $t_{l}$ 前的反应速率变化已省略 $)$\nD: 保持温度不变, 移动活塞 P, 使乙的容积和甲相等, 达到新的平衡后, 乙中 C 的体积分数是甲中 $\\mathrm{C}$ 的体积分数的 2 倍\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一定温度下可逆反应: $\\mathrm{A}(\\mathrm{s})+2 \\mathrm{~B}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{C}(\\mathrm{g})+\\mathrm{D}(\\mathrm{g}) \\Delta H<0$ 。现将 $1 \\mathrm{~mol} \\mathrm{~A}$ 和 $2 \\mathrm{~mol}$ $\\mathrm{B}$ 加入甲容器中, 将 $4 \\mathrm{~mol} \\mathrm{C}$ 和 $2 \\mathrm{~mol} \\mathrm{D}$ 加入乙容器中, 此时控制活塞 $\\mathrm{P}$, 使乙的容积为甲的 2 倍, $t_{1}$ 时两容器内均达到平衡状态 (如图 1 所示, 隔板 $\\mathrm{K}$ 不能移动)。下列说法正确的是\n\n[图1]\n\n图 1\n\n[图2]\n\n图 2\n\n[图3]\n\n图 3\n\nA: 保持活塞位置不变, 升高温度, 达到新的平衡后, 甲、乙中 $\\mathrm{B}$ 的体积分数均增大\nB: 保持温度和活塞位置不变, 在甲中再加入 $1 \\mathrm{~mol} \\mathrm{~A}$ 和 $2 \\mathrm{~mol} \\mathrm{~B}$, 达到新的平衡后,甲中 $\\mathrm{C}$ 的浓度是乙中 $\\mathrm{C}$ 的浓度的 2 倍\nC: 保持温度和乙中的压强不变, $t_{2}$ 时分别向甲、乙中加入等质量的氦气后，甲、乙中反应速率变化情况分别如图 2 和图 3 所示 ( $t_{l}$ 前的反应速率变化已省略 $)$\nD: 保持温度不变, 移动活塞 P, 使乙的容积和甲相等, 达到新的平衡后, 乙中 C 的体积分数是甲中 $\\mathrm{C}$ 的体积分数的 2 倍\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-105.jpg?height=274&width=394&top_left_y=180&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-105.jpg?height=323&width=440&top_left_y=158&top_left_x=728",
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-105.jpg?height=326&width=511&top_left_y=157&top_left_x=1178"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1372",
        "problem": "At a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.\n\nThe total energy of $\\mathrm{H}_{2}$ molecule in its ground state is $-31.675 \\mathrm{eV}$, relative to the same reference as that of hydrogen atom.Calculate the electron affinity, EA, of $\\mathrm{H}_{2}{ }^{+}$ion in $\\mathrm{eV}$ if its dissociation energy is 2.650 $\\mathrm{eV}$. If you have been unable to calculate the value for the dissociation energy of $\\mathrm{H}_{2}$ then use $4.500 \\mathrm{eV}$ for the calculation.)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nAt a temperature of $0 \\mathrm{~K}$, the total energy of a gaseous diatomic molecule $A B$ is approximately given by:\n\n$E=E_{0}+E_{\\text {vib }}$ where $E_{0}$ is the electronic energy of the ground state, and $E_{\\text {vib }}$ is the vibrational energy.\n\nAllowed values of the vibrational energies are given by the expression:\n\n$$\nE_{\\mathrm{vib}}=\\left(\\nu+\\frac{1}{2}\\right) \\varepsilon \\quad v=0,1,2, \\ldots \\quad \\varepsilon=\\frac{h}{2 \\pi} \\sqrt{\\frac{k}{\\mu}} \\quad \\mu(\\mathrm{AB})=\\frac{m_{A} m_{B}}{m_{A}+m_{B}}\n$$\n\nwhere $h$ is the Planck's constant, $v$ is the vibrational quantum number, $k$ is the force constant, and $\\mu$ is the reduced mass of the molecule. At $0 \\mathrm{~K}$, it may be safely assumed that $v$ is zero, and $E_{0}$ and $k$ are independent of isotopic substitution in the molecule.\n\nDeuterium, $\\mathrm{D}$, is an isotope of hydrogen atom with mass number 2. For the $\\mathrm{H}_{2}$ molecule, $k$ is $575.11 \\mathrm{~N} \\mathrm{~m}^{-1}$, and the isotopic molar masses of $\\mathrm{H}$ and $\\mathrm{D}$ are 1.0078 and $2.0141 \\mathrm{~g} \\mathrm{~mol}^{-1}$, respectively.\n\nAt a temperature of $0 \\mathrm{~K}: \\varepsilon_{\\mathrm{H}_{2}}=1.1546 \\varepsilon_{\\mathrm{HD}}$ and $\\varepsilon_{\\mathrm{D}_{2}}=0.8167 \\varepsilon_{\\mathrm{HD}}$.\n\nThe total energy of $\\mathrm{H}_{2}$ molecule in its ground state is $-31.675 \\mathrm{eV}$, relative to the same reference as that of hydrogen atom.\n\nproblem:\nCalculate the electron affinity, EA, of $\\mathrm{H}_{2}{ }^{+}$ion in $\\mathrm{eV}$ if its dissociation energy is 2.650 $\\mathrm{eV}$. If you have been unable to calculate the value for the dissociation energy of $\\mathrm{H}_{2}$ then use $4.500 \\mathrm{eV}$ for the calculation.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of eV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "eV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_50",
        "problem": "What is $\\Delta G^{\\circ}$ (at $298 \\mathrm{~K}$ ) for $\\mathrm{ClF}_{3}(g)$ ?\n\n| Substance | $\\Delta H^{\\circ}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | $S^{\\circ}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ |\n| :---: | :---: | :---: |\n| $\\mathrm{F}_{2}(g)$ | 0 | 202.8 |\n| $\\mathrm{Cl}_{2}(g)$ | 0 | 223.1 |\n| $\\mathrm{ClF}_{3}(g)$ | -158.9 | 281.6 |\nA: $-78.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-83.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-118.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-242.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is $\\Delta G^{\\circ}$ (at $298 \\mathrm{~K}$ ) for $\\mathrm{ClF}_{3}(g)$ ?\n\n| Substance | $\\Delta H^{\\circ}, \\mathrm{kJ} \\mathrm{mol}^{-1}$ | $S^{\\circ}, \\mathrm{J} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$ |\n| :---: | :---: | :---: |\n| $\\mathrm{F}_{2}(g)$ | 0 | 202.8 |\n| $\\mathrm{Cl}_{2}(g)$ | 0 | 223.1 |\n| $\\mathrm{ClF}_{3}(g)$ | -158.9 | 281.6 |\n\nA: $-78.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-83.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $-118.9 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $-242.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_614",
        "problem": "室温下, 用 $0.010 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的氢氧化钠溶液滴定 $10 \\mathrm{~mL}$ 浓度为 $0.010 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 醋酸的滴定曲线如图所示。已知某碱 $\\mathrm{AOH}$ 的电离常数为 $1 \\times 10^{-6}$ 。下列说法错误的是\n\n$\\mathrm{pH}$\n\n[图1]\nA: $\\mathrm{CH}_{3} \\mathrm{COOA}$ 溶液中 $\\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{A}^{+}\\right)} \\approx 1$\nB: $\\mathrm{a}$ 点溶液中: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$\nC: 稀释 $\\mathrm{b}$ 点溶液时 $\\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{OCO}\\right)}$ 减小\nD: $\\mathrm{pH}=7$ 时溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 用 $0.010 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的氢氧化钠溶液滴定 $10 \\mathrm{~mL}$ 浓度为 $0.010 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 醋酸的滴定曲线如图所示。已知某碱 $\\mathrm{AOH}$ 的电离常数为 $1 \\times 10^{-6}$ 。下列说法错误的是\n\n$\\mathrm{pH}$\n\n[图1]\n\nA: $\\mathrm{CH}_{3} \\mathrm{COOA}$ 溶液中 $\\frac{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{A}^{+}\\right)} \\approx 1$\nB: $\\mathrm{a}$ 点溶液中: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$\nC: 稀释 $\\mathrm{b}$ 点溶液时 $\\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{OCO}\\right)}$ 减小\nD: $\\mathrm{pH}=7$ 时溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-103.jpg?height=421&width=1052&top_left_y=1460&top_left_x=385"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_313",
        "problem": "In acidic aqueous solution, zinc metal is oxidized to $\\mathrm{Zn}^{2+}$. The net ionic equation for the reaction is given below.\n\n$$\n\\mathrm{Zn}(\\mathrm{s})+2 \\mathrm{H}^{+}(\\mathrm{aq}) \\rightarrow \\mathrm{Zn}^{2+}(\\mathrm{aq})+\\mathrm{H}_{2}(g)\n$$\n\nIn an experiment, 5.0 grams of $\\mathrm{Zn}(\\mathrm{s})$ were added to $100 \\mathrm{~mL}$ of $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(a q)$. Which of the following changes to the procedure would not affect the initial rate of the reaction?\nA: warming the $\\mathrm{HCl}$ solution before adding the zinc\nB: using zinc powder instead of zinc granules\nC: using $50 \\mathrm{~mL}$ of $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(a q)$\nD: using $200 \\mathrm{~mL}$ of $0.50 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(a q)$\nE: using $100 \\mathrm{~mL}$ of $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn acidic aqueous solution, zinc metal is oxidized to $\\mathrm{Zn}^{2+}$. The net ionic equation for the reaction is given below.\n\n$$\n\\mathrm{Zn}(\\mathrm{s})+2 \\mathrm{H}^{+}(\\mathrm{aq}) \\rightarrow \\mathrm{Zn}^{2+}(\\mathrm{aq})+\\mathrm{H}_{2}(g)\n$$\n\nIn an experiment, 5.0 grams of $\\mathrm{Zn}(\\mathrm{s})$ were added to $100 \\mathrm{~mL}$ of $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(a q)$. Which of the following changes to the procedure would not affect the initial rate of the reaction?\n\nA: warming the $\\mathrm{HCl}$ solution before adding the zinc\nB: using zinc powder instead of zinc granules\nC: using $50 \\mathrm{~mL}$ of $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(a q)$\nD: using $200 \\mathrm{~mL}$ of $0.50 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}(a q)$\nE: using $100 \\mathrm{~mL}$ of $1.0 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{H}_{2} \\mathrm{SO}_{4}(\\mathrm{aq})$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_97",
        "problem": "The triple point of ethyne, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, is $192.4 \\mathrm{~K}$ and $1.2 \\mathrm{~atm}$. What may be concluded from this information?\nA: Solid ethyne sublimes rather than melting at $1 \\mathrm{~atm}$ pressure.\nB: Solid ethyne is less dense than liquid ethyne.\nC: Solid ethyne is not thermodynamically stable at $193.0 \\mathrm{~K}$.\nD: A sample of ethyne at $192.4 \\mathrm{~K}$ must have a pressure of $1.2 \\mathrm{~atm}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe triple point of ethyne, $\\mathrm{C}_{2} \\mathrm{H}_{2}$, is $192.4 \\mathrm{~K}$ and $1.2 \\mathrm{~atm}$. What may be concluded from this information?\n\nA: Solid ethyne sublimes rather than melting at $1 \\mathrm{~atm}$ pressure.\nB: Solid ethyne is less dense than liquid ethyne.\nC: Solid ethyne is not thermodynamically stable at $193.0 \\mathrm{~K}$.\nD: A sample of ethyne at $192.4 \\mathrm{~K}$ must have a pressure of $1.2 \\mathrm{~atm}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_790",
        "problem": "在缺氧的深层潮湿土壤中, 厌氧细菌会促进钢铁发生厌氧腐蚀, 其原理如图所示。为抑制腐蚀的发生, 通常将钢管、石墨电极分别与外接电源相连，使钢管表面形成致密的 $\\mathrm{Fe}_{3} \\mathrm{O}_{4}$ 薄膜。下列说法不正确的是\n\n[图1]\nA: 上述保护方法为阴极电保护法\nB: 腐蚀过程中存在 $\\mathrm{SO}_{4}^{2-}+4 \\mathrm{H}_{2} \\stackrel{\\text { 厌氧细菌 }}{=} \\mathrm{S}^{2-}+4 \\mathrm{H}_{2} \\mathrm{O}$\nC: 每生成 $1 \\mathrm{molFeS}$, 电路中转移的电子数为 $2 N_{\\mathrm{A}}$\nD: 通电后电子从钢管流向石墨电极, 再经潮湿的土壤回到钢管\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n在缺氧的深层潮湿土壤中, 厌氧细菌会促进钢铁发生厌氧腐蚀, 其原理如图所示。为抑制腐蚀的发生, 通常将钢管、石墨电极分别与外接电源相连，使钢管表面形成致密的 $\\mathrm{Fe}_{3} \\mathrm{O}_{4}$ 薄膜。下列说法不正确的是\n\n[图1]\n\nA: 上述保护方法为阴极电保护法\nB: 腐蚀过程中存在 $\\mathrm{SO}_{4}^{2-}+4 \\mathrm{H}_{2} \\stackrel{\\text { 厌氧细菌 }}{=} \\mathrm{S}^{2-}+4 \\mathrm{H}_{2} \\mathrm{O}$\nC: 每生成 $1 \\mathrm{molFeS}$, 电路中转移的电子数为 $2 N_{\\mathrm{A}}$\nD: 通电后电子从钢管流向石墨电极, 再经潮湿的土壤回到钢管\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-09.jpg?height=388&width=554&top_left_y=551&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1555",
        "problem": "The absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nAll solutions of $\\mathrm{HX}$ contain an excess of a strong acid. Only the undissociated $\\mathrm{HX}$ species absorb.\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nAll solutions of $\\mathrm{HX}$ contain an excess of a strong acid. Only the undissociated $\\mathrm{HX}$ species absorb.\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-463.jpg?height=454&width=777&top_left_y=321&top_left_x=545"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_975",
        "problem": "Which of the following particles is not a charged particle?\nA: $\\alpha$-particle\nB: $\\beta$-particle\nC: electron\nD: proton\nE: neutron\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following particles is not a charged particle?\n\nA: $\\alpha$-particle\nB: $\\beta$-particle\nC: electron\nD: proton\nE: neutron\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_702",
        "problem": "室温下, 在含 $\\mathrm{KSCN} 、 \\mathrm{KIO}_{3}$ 和 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 的溶液中滴加 $\\mathrm{AgNO}_{3}$ 溶液, 混合液中 $\\left.\\mathrm{pX}\\left\\{[\\mathrm{pX}=-\\lg c(\\mathrm{X})], \\mathrm{X}=\\mathrm{IO}_{3}^{-} 、 \\mathrm{SCN}^{-} 、 \\mathrm{CrO}_{4}^{2-}\\right\\} 、\\right]$ 和 $\\mathrm{pAg}\\left[\\mathrm{pAg}=-\\lg c\\left(\\mathrm{Ag}^{+}\\right)\\right]$的关系如图所示。已知: $K_{\\mathrm{sp}}(\\mathrm{AgSCN})<K_{\\mathrm{sp}}\\left(\\mathrm{AgIO}_{3}\\right)$ 。下列叙述正确的是\n\n[图1]\nA: 直线 $I$ 代表 $\\mathrm{pCrO}_{4}^{2-}$ 和 $\\mathrm{pAg}$ 的关系\nB: $K_{\\mathrm{sp}}\\left(\\mathrm{AgIO}_{3}\\right)=10^{-11.99}$\nC: 室温下, $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 的溶解度大于 $\\mathrm{AgSCN}$\nD: $\\mathrm{AgIO}_{3}+\\mathrm{SCN}^{-} \\rightleftharpoons \\mathrm{AgSCN}+\\mathrm{IO}_{3}^{-}$的平衡常数 $K$ 为 $10^{-4.49}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 在含 $\\mathrm{KSCN} 、 \\mathrm{KIO}_{3}$ 和 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 的溶液中滴加 $\\mathrm{AgNO}_{3}$ 溶液, 混合液中 $\\left.\\mathrm{pX}\\left\\{[\\mathrm{pX}=-\\lg c(\\mathrm{X})], \\mathrm{X}=\\mathrm{IO}_{3}^{-} 、 \\mathrm{SCN}^{-} 、 \\mathrm{CrO}_{4}^{2-}\\right\\} 、\\right]$ 和 $\\mathrm{pAg}\\left[\\mathrm{pAg}=-\\lg c\\left(\\mathrm{Ag}^{+}\\right)\\right]$的关系如图所示。已知: $K_{\\mathrm{sp}}(\\mathrm{AgSCN})<K_{\\mathrm{sp}}\\left(\\mathrm{AgIO}_{3}\\right)$ 。下列叙述正确的是\n\n[图1]\n\nA: 直线 $I$ 代表 $\\mathrm{pCrO}_{4}^{2-}$ 和 $\\mathrm{pAg}$ 的关系\nB: $K_{\\mathrm{sp}}\\left(\\mathrm{AgIO}_{3}\\right)=10^{-11.99}$\nC: 室温下, $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 的溶解度大于 $\\mathrm{AgSCN}$\nD: $\\mathrm{AgIO}_{3}+\\mathrm{SCN}^{-} \\rightleftharpoons \\mathrm{AgSCN}+\\mathrm{IO}_{3}^{-}$的平衡常数 $K$ 为 $10^{-4.49}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-010.jpg?height=597&width=728&top_left_y=587&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1205",
        "problem": "The label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\nCould it be possible that the dilute solution contained sulfuric acid?\nSulfuric acid: $\\quad \\mathrm{p} K_{\\mathrm{a} 2}=1.99$\n\n$\\square$ True $\\square$ False\n\nIf True, calculate the $\\mathrm{pH}$ (or at least try to estimate it) and show your work.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label.\n\nCould it be possible that the dilute solution contained sulfuric acid?\nSulfuric acid: $\\quad \\mathrm{p} K_{\\mathrm{a} 2}=1.99$\n\n$\\square$ True $\\square$ False\n\nIf True, calculate the $\\mathrm{pH}$ (or at least try to estimate it) and show your work.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_275",
        "problem": "One mole of any gas, when measured at $310.0 \\mathrm{~K}$ and $101.0 \\mathrm{kPa}$, has a volume of $25.52 \\mathrm{~L}$. Calculate the volume of $21.66 \\mathrm{~g}$ of nitrogen gas under these conditions.\nA: $16.51 \\mathrm{~L}$\nB: $\\quad 19.73 \\mathrm{~L}$\nC: $25.52 \\mathrm{~L}$\nD: $\\quad 33.01 \\mathrm{~L}$\nE: $\\quad 39.45 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOne mole of any gas, when measured at $310.0 \\mathrm{~K}$ and $101.0 \\mathrm{kPa}$, has a volume of $25.52 \\mathrm{~L}$. Calculate the volume of $21.66 \\mathrm{~g}$ of nitrogen gas under these conditions.\n\nA: $16.51 \\mathrm{~L}$\nB: $\\quad 19.73 \\mathrm{~L}$\nC: $25.52 \\mathrm{~L}$\nD: $\\quad 33.01 \\mathrm{~L}$\nE: $\\quad 39.45 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_129",
        "problem": "A system undergoes a reversible cyclic process and proceeds through a series of thermodynamic processes, exchanging heat and work with its surroundings and ultimately returning to its original state. Which one of the following statements is true? Assume that the surroundings are much, much larger than the system.\nA: $\\Delta S_{\\text {surroundings }}>0$\nB: $q>0$\nC: q $<0$\nD: $q=w=0$\nE: $\\Delta \\mathrm{S}>0$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA system undergoes a reversible cyclic process and proceeds through a series of thermodynamic processes, exchanging heat and work with its surroundings and ultimately returning to its original state. Which one of the following statements is true? Assume that the surroundings are much, much larger than the system.\n\nA: $\\Delta S_{\\text {surroundings }}>0$\nB: $q>0$\nC: q $<0$\nD: $q=w=0$\nE: $\\Delta \\mathrm{S}>0$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_215",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the amount (in mol or mmol) of ammonium ions in the $1.988 \\mathrm{~g}$ ammonium salt sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the amount (in mol or mmol) of ammonium ions in the $1.988 \\mathrm{~g}$ ammonium salt sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_870",
        "problem": "常温下, 已知 $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgCl})=1.8 \\times 10^{-10}, \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)=1.1 \\times 10^{-12}$ 以及 $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Cu}(\\mathrm{OH})_{2}\\right]=2 \\times 10^{-20}$下列说法正确的是\nA: 某 $\\mathrm{CuSO}_{4}$ 溶液里 $\\mathrm{c}\\left(\\mathrm{Cu}^{2+}\\right)=0.02 \\mathrm{~mol} / \\mathrm{L}$, 如要生成 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 沉淀, 应调整溶液的 $\\mathrm{pH}$ ，使之大于 5\nB: 要使 $0.2 \\mathrm{~mol} / \\mathrm{LCuSO}_{4}$ 溶液中 $\\mathrm{Cu}^{2+}$ 沉淀较为完全 $\\left(\\right.$ 使 $\\mathrm{Cu}^{2+}$ 浓度降至原来的 $10^{-5}$ ), 则应向溶液里加入 $\\mathrm{NaOH}$ 溶液, 使溶液 $\\mathrm{pH}$ 为 6\nC: 常温下 $\\mathrm{AgCl}$ 饱和溶液和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 饱和溶液的物质的量浓度比较: $\\mathrm{c}(\\mathrm{AgCl})>\\mathrm{c}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nD: 常温下, 在 $0.010 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{AgNO}_{3}$ 溶液中, $\\mathrm{AgCl}$ 与 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 分别能达到饱和时的物质的量浓度比较: $\\mathrm{c}(\\mathrm{AgCl})<\\mathrm{c}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 已知 $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{AgCl})=1.8 \\times 10^{-10}, \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)=1.1 \\times 10^{-12}$ 以及 $\\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Cu}(\\mathrm{OH})_{2}\\right]=2 \\times 10^{-20}$下列说法正确的是\n\nA: 某 $\\mathrm{CuSO}_{4}$ 溶液里 $\\mathrm{c}\\left(\\mathrm{Cu}^{2+}\\right)=0.02 \\mathrm{~mol} / \\mathrm{L}$, 如要生成 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 沉淀, 应调整溶液的 $\\mathrm{pH}$ ，使之大于 5\nB: 要使 $0.2 \\mathrm{~mol} / \\mathrm{LCuSO}_{4}$ 溶液中 $\\mathrm{Cu}^{2+}$ 沉淀较为完全 $\\left(\\right.$ 使 $\\mathrm{Cu}^{2+}$ 浓度降至原来的 $10^{-5}$ ), 则应向溶液里加入 $\\mathrm{NaOH}$ 溶液, 使溶液 $\\mathrm{pH}$ 为 6\nC: 常温下 $\\mathrm{AgCl}$ 饱和溶液和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 饱和溶液的物质的量浓度比较: $\\mathrm{c}(\\mathrm{AgCl})>\\mathrm{c}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\nD: 常温下, 在 $0.010 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{AgNO}_{3}$ 溶液中, $\\mathrm{AgCl}$ 与 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 分别能达到饱和时的物质的量浓度比较: $\\mathrm{c}(\\mathrm{AgCl})<\\mathrm{c}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1280",
        "problem": "Nitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.Calculate the initial rate of disappearance of NO, if $\\mathrm{NO}$ with a pressure of $2.00 \\cdot 10^{2}$ torr and $\\mathrm{H}_{2}$ with $1.00 \\cdot 10^{2}$ torr are mixed at $820{ }^{\\circ} \\mathrm{C}$.\n\n(If you have been unable to calculate the value for the rate constant, you can use the value of $2 \\cdot 10^{-7}$ in appropriate unit.)",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nNitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.\n\nproblem:\nCalculate the initial rate of disappearance of NO, if $\\mathrm{NO}$ with a pressure of $2.00 \\cdot 10^{2}$ torr and $\\mathrm{H}_{2}$ with $1.00 \\cdot 10^{2}$ torr are mixed at $820{ }^{\\circ} \\mathrm{C}$.\n\n(If you have been unable to calculate the value for the rate constant, you can use the value of $2 \\cdot 10^{-7}$ in appropriate unit.)\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $ torr $^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$ torr $^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1399",
        "problem": "One of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, $\\gamma=C_{p} / C_{V}$, which is exactly $5 / 3(1.67 \\pm 0.01)$ for a monoatomic gas. The ratio was derived from the measurement of speed of sound $v_{\\mathrm{s}}$ by using the following equation, where $f$ and $\\lambda$ are the frequency and wavelength of the sound, and $R, T$, and $M$ denote the molar gas constant, absolute temperature, and molar mass, respectively.\n\n$$\nv_{\\mathrm{s}}=f \\lambda=\\sqrt{\\frac{\\gamma R T}{M}}\n$$\n\nFor an unknown gas sample, the wavelength of the sound was measured to be $\\lambda=0.116 \\mathrm{~m}$ at a frequency of $f=3520 \\mathrm{~Hz}\\left(\\mathrm{~Hz}=\\mathrm{s}^{-1}\\right)$ and temperature of $15.0{ }^{\\circ} \\mathrm{C}$ and\nunder atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$. The density $\\rho$ of the gas for these conditions was measured to be $0.850 \\pm 0.005 \\mathrm{~kg} \\mathrm{~m}^{-3}$.Calculate the heat capacity ratio $\\gamma$ for this gas sample.\nA: $\\mathrm{HCl}$\nB: $\\mathrm{HF}$\nC: $\\mathrm{Ne}$\nD: $\\mathrm{Ar}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nOne of the strongest evidences for the monoatomicity of argon and helium is the ratio of the heat capacity under constant pressure to that at constant volume, $\\gamma=C_{p} / C_{V}$, which is exactly $5 / 3(1.67 \\pm 0.01)$ for a monoatomic gas. The ratio was derived from the measurement of speed of sound $v_{\\mathrm{s}}$ by using the following equation, where $f$ and $\\lambda$ are the frequency and wavelength of the sound, and $R, T$, and $M$ denote the molar gas constant, absolute temperature, and molar mass, respectively.\n\n$$\nv_{\\mathrm{s}}=f \\lambda=\\sqrt{\\frac{\\gamma R T}{M}}\n$$\n\nFor an unknown gas sample, the wavelength of the sound was measured to be $\\lambda=0.116 \\mathrm{~m}$ at a frequency of $f=3520 \\mathrm{~Hz}\\left(\\mathrm{~Hz}=\\mathrm{s}^{-1}\\right)$ and temperature of $15.0{ }^{\\circ} \\mathrm{C}$ and\nunder atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$. The density $\\rho$ of the gas for these conditions was measured to be $0.850 \\pm 0.005 \\mathrm{~kg} \\mathrm{~m}^{-3}$.\n\nproblem:\nCalculate the heat capacity ratio $\\gamma$ for this gas sample.\n\nA: $\\mathrm{HCl}$\nB: $\\mathrm{HF}$\nC: $\\mathrm{Ne}$\nD: $\\mathrm{Ar}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_174",
        "problem": "The first electron affinity (EA) for any element $X$ is represented by the equation: $X(g)+e^{-} \\rightarrow X^{-}(g)+E A_{1}$. Negative values for electron affinity indicate that energy is released when an atom gains an electron. Most highschool textbooks teach the trend that electron affinity increases across a period. According to the table below, which of the following statement(s) is/are true?\n\n| Element | $\\mathrm{Li}$ | $\\mathrm{Be}$ | $\\mathrm{B}$ | $\\mathrm{C}$ | $\\mathrm{N}$ | $\\mathrm{O}$ | $\\mathrm{F}$ | $\\mathrm{Ne}$ |\n| :--- | :---: | :--- | :---: | :---: | :---: | :---: | :---: | :---: |\n| EA (kJ mol-1) | -59.6 | 0 | -26.7 | -153.9 | -7 | -141 | -328 | 0 |\n\nI. The absolute value of the energy released when an atom gains an electron is always lower for metals than non-metals\n\nII. Metals cannot form anions\n\nIII. Neutral atoms with complete subshells do not release energy during anion formation\nA: I only\nB: II only\nC: I and III only\nD: III only\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe first electron affinity (EA) for any element $X$ is represented by the equation: $X(g)+e^{-} \\rightarrow X^{-}(g)+E A_{1}$. Negative values for electron affinity indicate that energy is released when an atom gains an electron. Most highschool textbooks teach the trend that electron affinity increases across a period. According to the table below, which of the following statement(s) is/are true?\n\n| Element | $\\mathrm{Li}$ | $\\mathrm{Be}$ | $\\mathrm{B}$ | $\\mathrm{C}$ | $\\mathrm{N}$ | $\\mathrm{O}$ | $\\mathrm{F}$ | $\\mathrm{Ne}$ |\n| :--- | :---: | :--- | :---: | :---: | :---: | :---: | :---: | :---: |\n| EA (kJ mol-1) | -59.6 | 0 | -26.7 | -153.9 | -7 | -141 | -328 | 0 |\n\nI. The absolute value of the energy released when an atom gains an electron is always lower for metals than non-metals\n\nII. Metals cannot form anions\n\nIII. Neutral atoms with complete subshells do not release energy during anion formation\n\nA: I only\nB: II only\nC: I and III only\nD: III only\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_472",
        "problem": "根据下列图示所得出的结论不正确的是\nA: 图甲是室温下 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水中滴加盐酸, 溶液中由水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$随加入盐酸体积的变化关系, 图中 $\\mathrm{b} 、 \\mathrm{~d}$ 两点溶液的 $\\mathrm{pH}$ 值均为 7 [图1] 甲\nB: 图乙是 $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$ 的平衡常数与反应温度的关系 曲线, 说明该反应的 $\\triangle H<0$ [图2] 乙\nC: 图丙是室温下用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 某一元酸 HX 的滴定曲线，该 滴定过程可以选择酚酞作为指示剂 [图3] 丙\nD: 图丁是室温下用 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 除去溶液中 $\\mathrm{Ba}^{2+}$ 达到沉淀溶解平衡时, 溶液中 $\\mathrm{c}\\left(\\mathrm{Ba}^{2+}\\right)$ 与 $\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)$ 的关系曲线, 说明 $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{BaSO}_{4}\\right)=1 \\times 10^{-10}$ [图4] 丁\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n根据下列图示所得出的结论不正确的是\n\nA: 图甲是室温下 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水中滴加盐酸, 溶液中由水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$随加入盐酸体积的变化关系, 图中 $\\mathrm{b} 、 \\mathrm{~d}$ 两点溶液的 $\\mathrm{pH}$ 值均为 7 [图1] 甲\nB: 图乙是 $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$ 的平衡常数与反应温度的关系 曲线, 说明该反应的 $\\triangle H<0$ [图2] 乙\nC: 图丙是室温下用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 某一元酸 HX 的滴定曲线，该 滴定过程可以选择酚酞作为指示剂 [图3] 丙\nD: 图丁是室温下用 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 除去溶液中 $\\mathrm{Ba}^{2+}$ 达到沉淀溶解平衡时, 溶液中 $\\mathrm{c}\\left(\\mathrm{Ba}^{2+}\\right)$ 与 $\\mathrm{c}\\left(\\mathrm{SO}_{4}{ }^{2-}\\right)$ 的关系曲线, 说明 $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{BaSO}_{4}\\right)=1 \\times 10^{-10}$ [图4] 丁\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-37.jpg?height=371&width=437&top_left_y=160&top_left_x=410",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-37.jpg?height=385&width=494&top_left_y=687&top_left_x=861",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-37.jpg?height=308&width=389&top_left_y=1325&top_left_x=408",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-37.jpg?height=348&width=423&top_left_y=1796&top_left_x=1299"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_833",
        "problem": "下列说法正确的是\nA: 1，4-二氧杂螺[2.2]丙烷的结构简式为 $\\mathrm{O}$, 二氯代物有 3 种\nB: 组成和结构可用 $\\mathrm{C}_{4} \\mathrm{H}_{9}-\\mathrm{O}-\\mathrm{C}_{2} \\mathrm{H}_{3} \\mathrm{Cl}_{2}$ 表示的有机物共有 12 种\nC: 化学式为 $\\mathrm{C}_{9} \\mathrm{H}_{18} \\mathrm{O}_{2}$ 且有芳香气味的有机物, 在酸性条件下加热水解产生相对分子质量相同的两种有机物, 则符合此条件的 $\\mathrm{C}_{9} \\mathrm{H}_{18} \\mathrm{O}_{2}$ 的结构有 16 种\nD: $\\mathrm{C}_{5} \\mathrm{H}_{10}$ 的同分异构体中属于烯烃的有 6 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列说法正确的是\n\nA: 1，4-二氧杂螺[2.2]丙烷的结构简式为 $\\mathrm{O}$, 二氯代物有 3 种\nB: 组成和结构可用 $\\mathrm{C}_{4} \\mathrm{H}_{9}-\\mathrm{O}-\\mathrm{C}_{2} \\mathrm{H}_{3} \\mathrm{Cl}_{2}$ 表示的有机物共有 12 种\nC: 化学式为 $\\mathrm{C}_{9} \\mathrm{H}_{18} \\mathrm{O}_{2}$ 且有芳香气味的有机物, 在酸性条件下加热水解产生相对分子质量相同的两种有机物, 则符合此条件的 $\\mathrm{C}_{9} \\mathrm{H}_{18} \\mathrm{O}_{2}$ 的结构有 16 种\nD: $\\mathrm{C}_{5} \\mathrm{H}_{10}$ 的同分异构体中属于烯烃的有 6 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_878",
        "problem": "难溶盐 $\\mathrm{CaF}_{2}$ 可溶于盐酸, 常温下, 用 $\\mathrm{HCl}$ 调节 $\\mathrm{CaF}_{2}$ 浊液的 $\\mathrm{pH}$, 测得体系中 $-\\lg c\\left(\\mathrm{~F}^{-}\\right)$\n\n或 $-\\lg c\\left(\\mathrm{Ca}^{2+}\\right)$ 与 $\\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HF})}$ 的关系如图所示。下列说法错误的是\n\n[图1]\nA: $\\mathrm{M}$ 代表 $-\\lg c\\left(\\mathrm{~F}^{-}\\right)$与 $\\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HF})}$ 的变化曲线\nB: 难溶盐 $\\mathrm{CaF}_{2}$ 的溶度积常数 $K_{\\mathrm{sp}}\\left(\\mathrm{CaF}_{2}\\right)=10^{-7.4}$\nC: $\\mathrm{Y}$ 点的溶液中存在 $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{Ca}^{2+}\\right)=c(\\mathrm{HF})$\nD: $\\mathrm{Z}$ 点的溶液中存在 $\\lg c\\left(\\mathrm{Ca}^{2+}\\right)-2 \\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HF})}=-4.2$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n难溶盐 $\\mathrm{CaF}_{2}$ 可溶于盐酸, 常温下, 用 $\\mathrm{HCl}$ 调节 $\\mathrm{CaF}_{2}$ 浊液的 $\\mathrm{pH}$, 测得体系中 $-\\lg c\\left(\\mathrm{~F}^{-}\\right)$\n\n或 $-\\lg c\\left(\\mathrm{Ca}^{2+}\\right)$ 与 $\\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HF})}$ 的关系如图所示。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{M}$ 代表 $-\\lg c\\left(\\mathrm{~F}^{-}\\right)$与 $\\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HF})}$ 的变化曲线\nB: 难溶盐 $\\mathrm{CaF}_{2}$ 的溶度积常数 $K_{\\mathrm{sp}}\\left(\\mathrm{CaF}_{2}\\right)=10^{-7.4}$\nC: $\\mathrm{Y}$ 点的溶液中存在 $c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{Ca}^{2+}\\right)=c(\\mathrm{HF})$\nD: $\\mathrm{Z}$ 点的溶液中存在 $\\lg c\\left(\\mathrm{Ca}^{2+}\\right)-2 \\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c(\\mathrm{HF})}=-4.2$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-024.jpg?height=680&width=717&top_left_y=171&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_441",
        "problem": "常温下, 向 $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2} 、 \\mathrm{~Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 和 $\\mathrm{HR}$ 的混合液中滴加 $\\mathrm{NaOH}$ 溶液, $\\mathrm{pM}$ 与 $\\mathrm{pH}$的关系如图所示。已知: $\\mathrm{pM}=-\\lg c(\\mathrm{M}), c(\\mathrm{M})$ 代表 $c\\left(\\mathrm{Co}^{2+}\\right) 、 c\\left(\\mathrm{~Pb}^{2+}\\right)$ 或 $\\frac{c\\left(\\mathrm{R}^{-}\\right)}{c(\\mathrm{HR})}$; $K_{\\mathrm{sp}}\\left[\\mathrm{Co}(\\mathrm{OH})_{2}\\right]>K_{\\mathrm{sp}}\\left[\\mathrm{Pb}(\\mathrm{OH})_{2}\\right]$; 当被沉淀的离子的物质的量浓度 $\\leq 1 \\times 10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 时,认为该离子已沉淀完全。下列叙述错误的是\n\n[图1]\nA: $\\mathrm{X} 、 \\mathrm{Z}$ 分别代表 $-\\lg c\\left(\\mathrm{~Pb}^{2+}\\right) 、-\\lg \\frac{c\\left(\\mathrm{R}^{-}\\right)}{c(\\mathrm{HR})}$ 与 $\\mathrm{pH}$ 的关系\nB: 图中 $\\mathrm{a}$ 点对应溶液的 $\\mathrm{pH}$ 为 6.5 , 此时溶液中 $c\\left(\\mathrm{R}^{-}\\right)<c(\\mathrm{HR})$\nC: 常温下, $\\mathrm{R}^{-}$的水解平衡常数 $K_{\\mathrm{h}}=1 \\times 10^{-9}$\nD: 向浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Co}^{2+}$ 和 $\\mathrm{Pb}^{2+}$ 的混合溶液中逐滴加入 $0.1 \\mathrm{~mol} / \\mathrm{LNaOH}$ 溶液,能通过沉淀的方式将两种离子分离\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $\\mathrm{Co}\\left(\\mathrm{NO}_{3}\\right)_{2} 、 \\mathrm{~Pb}\\left(\\mathrm{NO}_{3}\\right)_{2}$ 和 $\\mathrm{HR}$ 的混合液中滴加 $\\mathrm{NaOH}$ 溶液, $\\mathrm{pM}$ 与 $\\mathrm{pH}$的关系如图所示。已知: $\\mathrm{pM}=-\\lg c(\\mathrm{M}), c(\\mathrm{M})$ 代表 $c\\left(\\mathrm{Co}^{2+}\\right) 、 c\\left(\\mathrm{~Pb}^{2+}\\right)$ 或 $\\frac{c\\left(\\mathrm{R}^{-}\\right)}{c(\\mathrm{HR})}$; $K_{\\mathrm{sp}}\\left[\\mathrm{Co}(\\mathrm{OH})_{2}\\right]>K_{\\mathrm{sp}}\\left[\\mathrm{Pb}(\\mathrm{OH})_{2}\\right]$; 当被沉淀的离子的物质的量浓度 $\\leq 1 \\times 10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 时,认为该离子已沉淀完全。下列叙述错误的是\n\n[图1]\n\nA: $\\mathrm{X} 、 \\mathrm{Z}$ 分别代表 $-\\lg c\\left(\\mathrm{~Pb}^{2+}\\right) 、-\\lg \\frac{c\\left(\\mathrm{R}^{-}\\right)}{c(\\mathrm{HR})}$ 与 $\\mathrm{pH}$ 的关系\nB: 图中 $\\mathrm{a}$ 点对应溶液的 $\\mathrm{pH}$ 为 6.5 , 此时溶液中 $c\\left(\\mathrm{R}^{-}\\right)<c(\\mathrm{HR})$\nC: 常温下, $\\mathrm{R}^{-}$的水解平衡常数 $K_{\\mathrm{h}}=1 \\times 10^{-9}$\nD: 向浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Co}^{2+}$ 和 $\\mathrm{Pb}^{2+}$ 的混合溶液中逐滴加入 $0.1 \\mathrm{~mol} / \\mathrm{LNaOH}$ 溶液,能通过沉淀的方式将两种离子分离\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-030.jpg?height=682&width=856&top_left_y=173&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_495",
        "problem": "在 $25^{\\circ} \\mathrm{C}$ 时, 将 2 个铜电极插入到一定的 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 饱和溶液中, 通直流电电解并不断搅拌, 当阴极上收集到 a mol 气体时, 溶液中析出了 $\\mathrm{b} \\mathrm{mol}$ 的结晶水合物\n\n$\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ ，若保持温度不变，则所剩溶液中溶质的质量分数是\nA: $\\frac{b}{b+18 a}$\nB: $\\frac{b}{b+36 a}$\nC: $\\frac{71 b}{161 b+18 a}$\nD: $\\frac{71 b}{161 b+9 a}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在 $25^{\\circ} \\mathrm{C}$ 时, 将 2 个铜电极插入到一定的 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 饱和溶液中, 通直流电电解并不断搅拌, 当阴极上收集到 a mol 气体时, 溶液中析出了 $\\mathrm{b} \\mathrm{mol}$ 的结晶水合物\n\n$\\mathrm{Na}_{2} \\mathrm{SO}_{4} \\cdot 10 \\mathrm{H}_{2} \\mathrm{O}$ ，若保持温度不变，则所剩溶液中溶质的质量分数是\n\nA: $\\frac{b}{b+18 a}$\nB: $\\frac{b}{b+36 a}$\nC: $\\frac{71 b}{161 b+18 a}$\nD: $\\frac{71 b}{161 b+9 a}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_998",
        "problem": "Which group of elements has the greatest electron affinity?\nA: group 14\nB: group 15\nC: group 16\nD: group 17\nE: group 18\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich group of elements has the greatest electron affinity?\n\nA: group 14\nB: group 15\nC: group 16\nD: group 17\nE: group 18\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_264",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amounts (in mol or mmol) of $\\mathrm{Pb}^{2+}, \\mathrm{CO}_{3}{ }^{2-}, \\mathrm{OH}^{-}$present in the $5.000 \\mathrm{~g}$ sample of this mineral.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the chemical amounts (in mol or mmol) of $\\mathrm{Pb}^{2+}, \\mathrm{CO}_{3}{ }^{2-}, \\mathrm{OH}^{-}$present in the $5.000 \\mathrm{~g}$ sample of this mineral.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_703",
        "problem": "室温下, 向 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 的混合溶液中逐滴加入 $\\mathrm{BaCl}_{2}$ 溶液, 溶液中 $-1 \\mathrm{gc}\\left(\\mathrm{Ba}^{2+}\\right)$\n\n与 $1 \\mathrm{~g} \\frac{\\mathrm{d}\\left(\\mathrm{HCO}_{3}^{-}\\right)}{\\mathrm{d}\\left(\\mathrm{CO}_{3}^{2-}\\right)}$ 的关系如图所示 (已知: $\\mathrm{H}_{2} \\mathrm{CO}_{3} \\mathrm{~K}_{\\mathrm{a} 1}=4.2 \\times 10^{-7} 、 \\mathrm{~K}_{\\mathrm{a} 2}=5.6 \\times 10^{-11}$;\n$\\left.\\mathrm{BaCO}_{3} \\mathrm{~K}_{\\mathrm{sp}}=5 \\times 10^{-9}\\right)$ 。下列说法正确的是\n\n[图1]\nA: $\\mathrm{a}$ 对应溶液 $\\mathrm{pH}=6.5$\nB: $\\mathrm{b}$ 对应溶液的 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)=0.05 \\mathrm{~mol} / \\mathrm{L}$\nC: $\\mathrm{a} \\rightarrow \\mathrm{b}$ 的过程中, 溶液中 $\\frac{\\mathrm{d}\\left(\\mathrm{CO}_{3}^{2-}\\right) \\mathrm{d}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{d}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)}$ 一直增大\nD: $a$ 对应的溶液中一定存在: $d\\left(\\mathrm{Na}^{+}\\right)+2 d\\left(\\mathrm{Ba}^{2+}\\right)<3 \\mathrm{~d}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{d}\\left(\\mathrm{Cl}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 向 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 的混合溶液中逐滴加入 $\\mathrm{BaCl}_{2}$ 溶液, 溶液中 $-1 \\mathrm{gc}\\left(\\mathrm{Ba}^{2+}\\right)$\n\n与 $1 \\mathrm{~g} \\frac{\\mathrm{d}\\left(\\mathrm{HCO}_{3}^{-}\\right)}{\\mathrm{d}\\left(\\mathrm{CO}_{3}^{2-}\\right)}$ 的关系如图所示 (已知: $\\mathrm{H}_{2} \\mathrm{CO}_{3} \\mathrm{~K}_{\\mathrm{a} 1}=4.2 \\times 10^{-7} 、 \\mathrm{~K}_{\\mathrm{a} 2}=5.6 \\times 10^{-11}$;\n$\\left.\\mathrm{BaCO}_{3} \\mathrm{~K}_{\\mathrm{sp}}=5 \\times 10^{-9}\\right)$ 。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{a}$ 对应溶液 $\\mathrm{pH}=6.5$\nB: $\\mathrm{b}$ 对应溶液的 $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)=0.05 \\mathrm{~mol} / \\mathrm{L}$\nC: $\\mathrm{a} \\rightarrow \\mathrm{b}$ 的过程中, 溶液中 $\\frac{\\mathrm{d}\\left(\\mathrm{CO}_{3}^{2-}\\right) \\mathrm{d}\\left(\\mathrm{H}^{+}\\right)}{\\mathrm{d}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)}$ 一直增大\nD: $a$ 对应的溶液中一定存在: $d\\left(\\mathrm{Na}^{+}\\right)+2 d\\left(\\mathrm{Ba}^{2+}\\right)<3 \\mathrm{~d}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{d}\\left(\\mathrm{Cl}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-080.jpg?height=454&width=523&top_left_y=281&top_left_x=321"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_604",
        "problem": "室温下, 用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.0500 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液所得滴定曲线如图所示。下列关于溶液中微粒的物质的量浓度关系一定正确的是\n\n[图1]\nA: 在整个滴定过程中, 始终存在: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+$ $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nB: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=10.00 \\mathrm{~mL}$ 时: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nC: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=15.00 \\mathrm{~mL}$ 时: $3 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=30.00 \\mathrm{~mL}$ 时: $\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-$ $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n室温下, 用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.0500 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液所得滴定曲线如图所示。下列关于溶液中微粒的物质的量浓度关系一定正确的是\n\n[图1]\n\nA: 在整个滴定过程中, 始终存在: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+$ $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nB: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=10.00 \\mathrm{~mL}$ 时: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nC: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=15.00 \\mathrm{~mL}$ 时: $3 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=30.00 \\mathrm{~mL}$ 时: $\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-$ $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-093.jpg?height=460&width=711&top_left_y=1118&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_984",
        "problem": "Which of the following pairs of atomic symbols and elements is incorrect?\nA: $\\mathrm{Fe}$, iron\nB: $\\mathrm{Mg}$, magnesium\nC: $\\mathrm{Ca}$, calcium\nD: $\\mathrm{Br}$, boron\nE: Mn, manganese\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following pairs of atomic symbols and elements is incorrect?\n\nA: $\\mathrm{Fe}$, iron\nB: $\\mathrm{Mg}$, magnesium\nC: $\\mathrm{Ca}$, calcium\nD: $\\mathrm{Br}$, boron\nE: Mn, manganese\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_516",
        "problem": "有一种有机物的键线式酷似牛, 故称为牛式二烯炔醇。下列说法错误的是\n[图1]\nA: 牛式二烯炔醇在水中的溶解度比丙醇在水中的溶解度小\nB: 牛式二烯炔醇含有 1 种含氧官能团, 它属于不饱和醇类, 其分子式为 $\\mathrm{C}_{29} \\mathrm{H}_{45} \\mathrm{O}$\nC: 牛式二烯炔醇既不能发生消去反应, 也不能发生催化氧化反应生成醛类物质\nD: 牛式二烯炔醇既可以使溴水褪色, 又可以使酸性高锰酸钾溶液裉色, 二者裉色原理相同\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有一种有机物的键线式酷似牛, 故称为牛式二烯炔醇。下列说法错误的是\n[图1]\n\nA: 牛式二烯炔醇在水中的溶解度比丙醇在水中的溶解度小\nB: 牛式二烯炔醇含有 1 种含氧官能团, 它属于不饱和醇类, 其分子式为 $\\mathrm{C}_{29} \\mathrm{H}_{45} \\mathrm{O}$\nC: 牛式二烯炔醇既不能发生消去反应, 也不能发生催化氧化反应生成醛类物质\nD: 牛式二烯炔醇既可以使溴水褪色, 又可以使酸性高锰酸钾溶液裉色, 二者裉色原理相同\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-37.jpg?height=318&width=828&top_left_y=2102&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1346",
        "problem": "If two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\\mathrm{H}_{2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\\mathrm{H}_{2}$ rapidly desorbs. This question examines two kinetic models for $\\mathrm{H}_{2}$ formation on the surface of a dust particle.\n\nIn both models, the rate constant for adsorption of $\\mathrm{H}$ atoms onto the surface of dust particles is $k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1}$. The typical number density of $\\mathrm{H}$ atoms (number of $\\mathrm{H}$ atoms per unit volume) in interstellar space is $[\\mathrm{H}]=10 \\mathrm{~cm}^{-3}$.\n\n[Note: In the following, you may treat numbers of surface-adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations. As a result, the units of the rate constants may be unfamiliar to you. Reaction rates have units of numbers of atoms or molecules per unit time.]Calculate the rate at which $\\mathrm{H}$ atoms adsorb onto a dust particle. You may assume that this rate is constant throughout.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIf two atoms collide in interstellar space the energy of the resulting molecule is so great that it rapidly dissociates. Hydrogen atoms only react to give stable $\\mathrm{H}_{2}$ molecules on the surface of dust particles. The dust particles absorb most of the excess energy and the newly formed $\\mathrm{H}_{2}$ rapidly desorbs. This question examines two kinetic models for $\\mathrm{H}_{2}$ formation on the surface of a dust particle.\n\nIn both models, the rate constant for adsorption of $\\mathrm{H}$ atoms onto the surface of dust particles is $k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1}$. The typical number density of $\\mathrm{H}$ atoms (number of $\\mathrm{H}$ atoms per unit volume) in interstellar space is $[\\mathrm{H}]=10 \\mathrm{~cm}^{-3}$.\n\n[Note: In the following, you may treat numbers of surface-adsorbed atoms and number densities of gas-phase atoms in the same way as you would normally use concentrations in the rate equations. As a result, the units of the rate constants may be unfamiliar to you. Reaction rates have units of numbers of atoms or molecules per unit time.]\n\nproblem:\nCalculate the rate at which $\\mathrm{H}$ atoms adsorb onto a dust particle. You may assume that this rate is constant throughout.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~s}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~s}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_258",
        "problem": "Calculate the molar mass of potassium chloride (in $\\mathrm{g} \\mathrm{mol}^{-1}$ ).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the molar mass of potassium chloride (in $\\mathrm{g} \\mathrm{mol}^{-1}$ ).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g/mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g/mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_79",
        "problem": "Which of the following has the greatest molar solubility in water?\nA: $\\mathrm{ZnC}_{2} \\mathrm{O}_{4}\\left(K_{\\mathrm{sp}}=2.7 \\times 10^{-8}\\right)$\nB: $\\mathrm{BaCrO}_{4}\\left(K_{\\mathrm{sp}}=2.3 \\times 10^{-10}\\right)$\nC: $\\mathrm{CaF}_{2}\\left(K_{\\mathrm{sp}}=3.9 \\times 10^{-11}\\right)$\nD: $\\operatorname{AgBr}\\left(K_{\\mathrm{sp}}=5.0 \\times 10^{-13}\\right)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following has the greatest molar solubility in water?\n\nA: $\\mathrm{ZnC}_{2} \\mathrm{O}_{4}\\left(K_{\\mathrm{sp}}=2.7 \\times 10^{-8}\\right)$\nB: $\\mathrm{BaCrO}_{4}\\left(K_{\\mathrm{sp}}=2.3 \\times 10^{-10}\\right)$\nC: $\\mathrm{CaF}_{2}\\left(K_{\\mathrm{sp}}=3.9 \\times 10^{-11}\\right)$\nD: $\\operatorname{AgBr}\\left(K_{\\mathrm{sp}}=5.0 \\times 10^{-13}\\right)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_463",
        "problem": "Beckmann 重排是酮肟在一定条件下生成酰胺的反应, 机理中与羟基处于反位的基团迁移到缺电子的氮原子上，具体反应历程如图所示。\n\n[图1]\n\n已知： $\\mathrm{R} 、 \\mathrm{R}^{\\prime}$ 代表烷基。下列说法正确的是\nA: $\\mathrm{H}^{+}$在该反应过程中作催化剂\nB: 物质 I 中碳原子均为 $\\mathrm{sp}^{2}$ 杂化\nC: 物质 VII 能发生水解反应\nD: [图2]\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\nBeckmann 重排是酮肟在一定条件下生成酰胺的反应, 机理中与羟基处于反位的基团迁移到缺电子的氮原子上，具体反应历程如图所示。\n\n[图1]\n\n已知： $\\mathrm{R} 、 \\mathrm{R}^{\\prime}$ 代表烷基。下列说法正确的是\n\nA: $\\mathrm{H}^{+}$在该反应过程中作催化剂\nB: 物质 I 中碳原子均为 $\\mathrm{sp}^{2}$ 杂化\nC: 物质 VII 能发生水解反应\nD: [图2]\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-14.jpg?height=343&width=1425&top_left_y=1485&top_left_x=353",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-14.jpg?height=168&width=768&top_left_y=2269&top_left_x=473"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_714",
        "problem": "硅锰原电池是一种新型电池, 因其供电稳定, 储存量丰富而备受关注。硅锰原电池工作原理如图所示。下列说法正确的是\n\n[图1]\n\n质子交换膜\nA: 若将质子交换膜换成阴离子交换膜, 电解液换成 $\\mathrm{NaOH}$ 溶液, 电池的使用寿命会延长\nB: 放电过程中, 溶液中的 $\\mathrm{H}^{+}$会经质子交换膜由 $\\mathrm{Si} @ \\mathrm{C}$ 电极区移向 $\\mathrm{MnO}_{2}$ 电极区\nC: $\\mathrm{MnO}_{2}$ 电极上发生的电极反应式为 $\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 电路中每通过 $2 \\mathrm{~mol}$ 电子，负极材料质量会增加 $32 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n硅锰原电池是一种新型电池, 因其供电稳定, 储存量丰富而备受关注。硅锰原电池工作原理如图所示。下列说法正确的是\n\n[图1]\n\n质子交换膜\n\nA: 若将质子交换膜换成阴离子交换膜, 电解液换成 $\\mathrm{NaOH}$ 溶液, 电池的使用寿命会延长\nB: 放电过程中, 溶液中的 $\\mathrm{H}^{+}$会经质子交换膜由 $\\mathrm{Si} @ \\mathrm{C}$ 电极区移向 $\\mathrm{MnO}_{2}$ 电极区\nC: $\\mathrm{MnO}_{2}$ 电极上发生的电极反应式为 $\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$\nD: 电路中每通过 $2 \\mathrm{~mol}$ 电子，负极材料质量会增加 $32 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-14.jpg?height=337&width=411&top_left_y=771&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1412",
        "problem": "The complete electrolysis of $1 \\mathrm{~mol}$ of water requires the following amount of electric charge ( $\\mathrm{F}$ is the Faraday constant):\nA: $\\mathrm{F}$\nB: $(4 / 3) \\mathrm{F}$\nC: $(3 / 2) \\mathrm{F}$\nD: $2 \\mathrm{~F}$\nE: $3 \\mathrm{~F}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe complete electrolysis of $1 \\mathrm{~mol}$ of water requires the following amount of electric charge ( $\\mathrm{F}$ is the Faraday constant):\n\nA: $\\mathrm{F}$\nB: $(4 / 3) \\mathrm{F}$\nC: $(3 / 2) \\mathrm{F}$\nD: $2 \\mathrm{~F}$\nE: $3 \\mathrm{~F}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1081",
        "problem": "The electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\n\nCalculate the energy of an electron in a $2 p$ orbital in an excited hydrogen atom.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\n\nCalculate the energy of an electron in a $2 p$ orbital in an excited hydrogen atom.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~J}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_801f28df4b84ba4c94fag-09.jpg?height=568&width=571&top_left_y=327&top_left_x=1234"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~J}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_318",
        "problem": "When 1.50 grams of a compound containing only carbon, hydrogen, nitrogen and oxygen is burned completely in excess $\\mathrm{O}_{2}, 1.72 \\mathrm{~g} \\mathrm{CO}_{2}, 0.585 \\mathrm{~g} \\mathrm{NO}$ and $1.23 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}$ are produced. What is the empirical formula for the compound?\nA: $\\mathrm{C}_{2} \\mathrm{H}_{7} \\mathrm{O}_{2} \\mathrm{~N}$\nB: $\\mathrm{C}_{2} \\mathrm{H}_{14} \\mathrm{O}_{2} \\mathrm{~N}$\nC: $\\mathrm{CH}_{7} \\mathrm{ON}$\nD: $\\mathrm{C}_{2} \\mathrm{H}_{7} \\mathrm{ON}_{2}$\nE: $\\mathrm{CH}_{7} \\mathrm{O}_{2} \\mathrm{~N}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen 1.50 grams of a compound containing only carbon, hydrogen, nitrogen and oxygen is burned completely in excess $\\mathrm{O}_{2}, 1.72 \\mathrm{~g} \\mathrm{CO}_{2}, 0.585 \\mathrm{~g} \\mathrm{NO}$ and $1.23 \\mathrm{~g} \\mathrm{H}_{2} \\mathrm{O}$ are produced. What is the empirical formula for the compound?\n\nA: $\\mathrm{C}_{2} \\mathrm{H}_{7} \\mathrm{O}_{2} \\mathrm{~N}$\nB: $\\mathrm{C}_{2} \\mathrm{H}_{14} \\mathrm{O}_{2} \\mathrm{~N}$\nC: $\\mathrm{CH}_{7} \\mathrm{ON}$\nD: $\\mathrm{C}_{2} \\mathrm{H}_{7} \\mathrm{ON}_{2}$\nE: $\\mathrm{CH}_{7} \\mathrm{O}_{2} \\mathrm{~N}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_667",
        "problem": "糖 $\\mathrm{M}$ 能发生如下转化:\n\nM(\n\n[图1]\n\n已知: I. $\\underset{\\mathrm{OH}}{\\mathrm{RCH}-\\mathrm{CHR}^{\\prime}} \\xrightarrow[\\mathrm{H}_{2} \\mathrm{O}]{\\mathrm{H}_{5} \\mathrm{IO}_{6}} \\mathrm{RCHO}+\\mathrm{R}^{\\prime} \\mathrm{CHO}$\n\nII.\n\n[图2]\n\nIII. 分子中一个碳原子上同时连有两个羟基时不稳定, 易转变为羰基。\n\n下列说法正确的是\nA: $A . M \\rightarrow N 、 Q \\rightarrow K$ 两步转化均发生了氧化反应\nB: $\\mathrm{N} \\rightarrow \\mathrm{Q}$ 转化过程中, 另一种有机产物为丙醛\nC: $\\mathrm{N} 、 \\mathrm{Q} 、 \\mathrm{~K}$ 均能与新制的 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 悬浊液反应\nD: $1 \\mathrm{molM}$ 在 $\\mathrm{Cu}$ 催化下与足量 $\\mathrm{O}_{2}$ 反应消耗 $2.5 \\mathrm{molO}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n糖 $\\mathrm{M}$ 能发生如下转化:\n\nM(\n\n[图1]\n\n已知: I. $\\underset{\\mathrm{OH}}{\\mathrm{RCH}-\\mathrm{CHR}^{\\prime}} \\xrightarrow[\\mathrm{H}_{2} \\mathrm{O}]{\\mathrm{H}_{5} \\mathrm{IO}_{6}} \\mathrm{RCHO}+\\mathrm{R}^{\\prime} \\mathrm{CHO}$\n\nII.\n\n[图2]\n\nIII. 分子中一个碳原子上同时连有两个羟基时不稳定, 易转变为羰基。\n\n下列说法正确的是\n\nA: $A . M \\rightarrow N 、 Q \\rightarrow K$ 两步转化均发生了氧化反应\nB: $\\mathrm{N} \\rightarrow \\mathrm{Q}$ 转化过程中, 另一种有机产物为丙醛\nC: $\\mathrm{N} 、 \\mathrm{Q} 、 \\mathrm{~K}$ 均能与新制的 $\\mathrm{Cu}(\\mathrm{OH})_{2}$ 悬浊液反应\nD: $1 \\mathrm{molM}$ 在 $\\mathrm{Cu}$ 催化下与足量 $\\mathrm{O}_{2}$ 反应消耗 $2.5 \\mathrm{molO}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-10.jpg?height=183&width=1367&top_left_y=545&top_left_x=413",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-10.jpg?height=117&width=582&top_left_y=1012&top_left_x=403"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_995",
        "problem": "The Lewis structure (i.e. electron dot) structure for the $\\mathrm{HCN}$ molecule is given below.\n\n$$\n\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{N}:\n$$\n\nThe bond angle is nearest to\nA: $60^{\\circ}$\nB: $90^{\\circ}$\nC: $105^{\\circ}$\nD: $120^{\\circ}$\nE: $180^{\\circ}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe Lewis structure (i.e. electron dot) structure for the $\\mathrm{HCN}$ molecule is given below.\n\n$$\n\\mathrm{H}-\\mathrm{C} \\equiv \\mathrm{N}:\n$$\n\nThe bond angle is nearest to\n\nA: $60^{\\circ}$\nB: $90^{\\circ}$\nC: $105^{\\circ}$\nD: $120^{\\circ}$\nE: $180^{\\circ}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_154",
        "problem": "The following data were taken for the addition of solid barium fluoride to enough water to make $100.0 \\mathrm{~mL}$ of solution. What is the $\\mathrm{K}_{\\mathrm{sp}}$ of barium fluoride:\n\n$$\n\\mathrm{BaF}_{2}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Ba}^{2+}(\\mathrm{aq})+2 \\mathrm{~F}^{-}(\\mathrm{aq}) ?\n$$\n\n| Mass of solid <br> added (g) | Mass of solid <br> dissolved (g) | Mass of solid <br> undissolved (g) |\n| :---: | :---: | :---: |\n| 0.100 | 0.100 | 0 |\n| 0.200 | 0.200 | 0 |\n| 0.300 | 0.300 | 0 |\n| 0.400 | 0.319 | 0.081 |\nA: $1.30 \\times 10^{-1}$\nB: $3.25 \\times 10^{-2}$\nC: $2.41 \\times 10^{-5}$\nD: $6.03 \\times 10^{-6}$\nE: $6.03 \\times 10^{-9}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe following data were taken for the addition of solid barium fluoride to enough water to make $100.0 \\mathrm{~mL}$ of solution. What is the $\\mathrm{K}_{\\mathrm{sp}}$ of barium fluoride:\n\n$$\n\\mathrm{BaF}_{2}(\\mathrm{~s}) \\rightleftharpoons \\mathrm{Ba}^{2+}(\\mathrm{aq})+2 \\mathrm{~F}^{-}(\\mathrm{aq}) ?\n$$\n\n| Mass of solid <br> added (g) | Mass of solid <br> dissolved (g) | Mass of solid <br> undissolved (g) |\n| :---: | :---: | :---: |\n| 0.100 | 0.100 | 0 |\n| 0.200 | 0.200 | 0 |\n| 0.300 | 0.300 | 0 |\n| 0.400 | 0.319 | 0.081 |\n\nA: $1.30 \\times 10^{-1}$\nB: $3.25 \\times 10^{-2}$\nC: $2.41 \\times 10^{-5}$\nD: $6.03 \\times 10^{-6}$\nE: $6.03 \\times 10^{-9}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_434",
        "problem": "分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8}$ 的芳香烃 $\\mathrm{A}$, 一定条件下与氢气完全加成得到产物 $\\mathrm{B}$ 。下列说法正确的是\nA: 依据碳骨架分类, B 属于脂肪烃\nB: A 发生加聚反应生成的高分子化合物的结构简式为 [图1]\nC: B 的二氯代物有 23 种同分异构体(不包括立体异构)\nD: A 的同分异构体立方烷( [图2] )经硝化可得到六硝基立方烷，其可能的结构有 4 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n分子式为 $\\mathrm{C}_{8} \\mathrm{H}_{8}$ 的芳香烃 $\\mathrm{A}$, 一定条件下与氢气完全加成得到产物 $\\mathrm{B}$ 。下列说法正确的是\n\nA: 依据碳骨架分类, B 属于脂肪烃\nB: A 发生加聚反应生成的高分子化合物的结构简式为 [图1]\nC: B 的二氯代物有 23 种同分异构体(不包括立体异构)\nD: A 的同分异构体立方烷( [图2] )经硝化可得到六硝基立方烷，其可能的结构有 4 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-16.jpg?height=202&width=262&top_left_y=1972&top_left_x=1271",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-16.jpg?height=186&width=195&top_left_y=2283&top_left_x=859",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-16.jpg?height=100&width=206&top_left_y=2651&top_left_x=1116",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-17.jpg?height=209&width=512&top_left_y=478&top_left_x=515",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-17.jpg?height=364&width=1306&top_left_y=861&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_398",
        "problem": "Which of the following will increase $\\left[\\mathrm{Cl}^{-}\\right]$in a saturated solution of $\\mathrm{AgCl}$ in contact with excess solid silver chloride?\nI. Addition of $\\mathrm{AgCl}(s)$\nII. Addition of $\\mathrm{NH}_{3}(\\mathrm{aq})$\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following will increase $\\left[\\mathrm{Cl}^{-}\\right]$in a saturated solution of $\\mathrm{AgCl}$ in contact with excess solid silver chloride?\nI. Addition of $\\mathrm{AgCl}(s)$\nII. Addition of $\\mathrm{NH}_{3}(\\mathrm{aq})$\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_757",
        "problem": "一种以沸石笼为载体对 $\\mathrm{NO}$ 进行催化还原的原理如图所示。下列说法错误的是\n\n[图1]\nA: 反应(4)有极性键的断裂与生成\nB: 反应(2)(3)均为氧化还原反应\nC: 反应(5)的离子方程式为: $\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}^{2+}+\\mathrm{NO}=\\mathrm{N}_{2}+\\mathrm{NH}_{4}^{+}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}$\nD: 总反应还原 $1 \\mathrm{molNO}$ 消耗 $\\mathrm{O}_{2} 11.2 \\mathrm{~L}$ (标准状况)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种以沸石笼为载体对 $\\mathrm{NO}$ 进行催化还原的原理如图所示。下列说法错误的是\n\n[图1]\n\nA: 反应(4)有极性键的断裂与生成\nB: 反应(2)(3)均为氧化还原反应\nC: 反应(5)的离子方程式为: $\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{4}^{2+}+\\mathrm{NO}=\\mathrm{N}_{2}+\\mathrm{NH}_{4}^{+}+\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{Cu}\\left(\\mathrm{NH}_{3}\\right)_{2}^{+}$\nD: 总反应还原 $1 \\mathrm{molNO}$ 消耗 $\\mathrm{O}_{2} 11.2 \\mathrm{~L}$ (标准状况)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-048.jpg?height=842&width=1142&top_left_y=898&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_171",
        "problem": "One of the Twelve Principles of Green Chemistry is that \"synthetic methods should be designed to maximize the incorporation of all materials used in the process into the final product\". One way to consider this is to calculate the atom economy (AE) of a chemical reaction where $\\mathrm{AE}$ is defined as follows:\n\n$$\n\\text { atom economy }=\\frac{\\text { molecular mass of desired product }}{\\text { molecular mass of all reactants }} \\times 100 \\%\n$$\n\nThe atom economy of the reaction in the previous question (\\#16) is:\nA) $61.5 \\%$\nA: $61.5 \\%$\nB: $69.2 \\%$\nC: $100 \\%$\nD: $68.9 \\%$\nE: $74.4 \\%$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOne of the Twelve Principles of Green Chemistry is that \"synthetic methods should be designed to maximize the incorporation of all materials used in the process into the final product\". One way to consider this is to calculate the atom economy (AE) of a chemical reaction where $\\mathrm{AE}$ is defined as follows:\n\n$$\n\\text { atom economy }=\\frac{\\text { molecular mass of desired product }}{\\text { molecular mass of all reactants }} \\times 100 \\%\n$$\n\nThe atom economy of the reaction in the previous question (\\#16) is:\nA) $61.5 \\%$\n\nA: $61.5 \\%$\nB: $69.2 \\%$\nC: $100 \\%$\nD: $68.9 \\%$\nE: $74.4 \\%$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1077",
        "problem": "The energy needed to promote the electron in a hydrogen atom from the $1 \\mathrm{~s}$ orbital to the $2 p$ orbital is $=1.63 \\times 10^{-18} \\mathrm{~J}$.The electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\nWhen an electron returns from a higher energy orbital to a lower one, energy is given out as light (the cause of the familiar flame colours). The frequency of the light, $f$, (in $\\mathrm{Hz}$ ) is related to the energy of the transition, $\\Delta E$, by the equation:\n\n$$\n\\Delta E=h f \\quad\\left(\\text { where } h \\text { is Planck's constant }=6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}\\right) .\n$$\n\nCalculate the frequency of light for the electronic transition in a hydrogen atom from a $2 p$ orbital to the 1s orbital (the so-called Hydrogen Lyman- $\\alpha$ line).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe energy needed to promote the electron in a hydrogen atom from the $1 \\mathrm{~s}$ orbital to the $2 p$ orbital is $=1.63 \\times 10^{-18} \\mathrm{~J}$.\n\nproblem:\nThe electronic ground state (i.e. the lowest electronic state) of a hydrogen atom may be written $1 \\mathrm{~s}^{1}$ indicating that the single electron resides in the $1 \\mathrm{~s}$ orbital. If sufficient energy is given to the atom, the electron may be promoted from the $1 \\mathrm{~s}$ orbital to a higher energy orbital, such as the $2 p$ orbital or the $3 p$ orbital.\n\nThe energy of an electron in a hydrogen atom (or any ionized atom with nuclear charge $Z$ and with just one electron remaining) is given by the following equation:\n\n$$\nE_{n}=-R_{H} \\frac{Z^{2}}{n^{2}}\n$$\n\nThe energy of a free, ionized electron is zero; electrons in the atom have lower energy, hence the minus sign.\n\n[figure1]\n\nSupernova remnant E0102-72 as photographed by the UV / $x$ ray telescope Chandra. \n\nIn the equation, $Z$ is the number of protons in the nucleus ( $Z=1$ for hydrogen); $n$ is the principal quantum number ( $n=1$ for the 1 s orbital, 2 for the $2 s$ and $2 p$ orbitals, 3 for the 3s, 3 p and 3 d orbitals, etc.);\n\n$R_{H}$ is the Rydberg constant equal to the ionization energy of a hydrogen atom $\\left(R_{H}=2.179 \\times 10^{-18} \\mathrm{~J}\\right)$.\nWhen an electron returns from a higher energy orbital to a lower one, energy is given out as light (the cause of the familiar flame colours). The frequency of the light, $f$, (in $\\mathrm{Hz}$ ) is related to the energy of the transition, $\\Delta E$, by the equation:\n\n$$\n\\Delta E=h f \\quad\\left(\\text { where } h \\text { is Planck's constant }=6.626 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s}\\right) .\n$$\n\nCalculate the frequency of light for the electronic transition in a hydrogen atom from a $2 p$ orbital to the 1s orbital (the so-called Hydrogen Lyman- $\\alpha$ line).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of Hz, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_801f28df4b84ba4c94fag-09.jpg?height=568&width=571&top_left_y=327&top_left_x=1234"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "Hz"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_574",
        "problem": "TRAP 是一种温和的氧化剂, TRAP 试剂中的 $\\mathrm{RuO}_{4}^{-}$不会氧化碳碳双键, 可以将醇仅氧化至醛, 不会过度氧化为羧酸。TRAP 氧化醇的反应机理如下图, 下列说法正确的是\n\n[图1]\nA: 在上述 6 步反应中发生氧化还原反应的有 2 步\nB: 步骤(4)(5)中 $\\mathrm{NMO}$ 将 $\\mathrm{RuO}_{3}^{-}$还原, 生成 TRAP 试剂\nC: 步骤(6)的离子方程式为 $2 \\mathrm{RuO}_{3}^{-}=\\mathrm{RuO}_{2}+\\mathrm{RuO}_{4}^{2-}$\nD: 若 $\\mathrm{R}_{1}$ 为 $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}-, \\mathrm{R}_{2}$ 为- $\\mathrm{H}$, 则 TRAP 氧化该醇的主产物为 2-丁烯醛\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\nTRAP 是一种温和的氧化剂, TRAP 试剂中的 $\\mathrm{RuO}_{4}^{-}$不会氧化碳碳双键, 可以将醇仅氧化至醛, 不会过度氧化为羧酸。TRAP 氧化醇的反应机理如下图, 下列说法正确的是\n\n[图1]\n\nA: 在上述 6 步反应中发生氧化还原反应的有 2 步\nB: 步骤(4)(5)中 $\\mathrm{NMO}$ 将 $\\mathrm{RuO}_{3}^{-}$还原, 生成 TRAP 试剂\nC: 步骤(6)的离子方程式为 $2 \\mathrm{RuO}_{3}^{-}=\\mathrm{RuO}_{2}+\\mathrm{RuO}_{4}^{2-}$\nD: 若 $\\mathrm{R}_{1}$ 为 $\\mathrm{CH}_{3} \\mathrm{CH}=\\mathrm{CH}-, \\mathrm{R}_{2}$ 为- $\\mathrm{H}$, 则 TRAP 氧化该醇的主产物为 2-丁烯醛\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-40.jpg?height=706&width=845&top_left_y=415&top_left_x=320"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_886",
        "problem": "化合物乙是一种治疗神经类疾病的药物, 可由化合物甲经多步反应得到。下列有关化合物甲、乙的说法正确的是[图1]\nA: 甲分子中所有碳原子一定处于同一平面\nB: 乙中含有 2 个手性碳原子\nC: 用 $\\mathrm{NaHCO}_{3}$ 溶液或 $\\mathrm{FeCl}_{3}$ 溶液能鉴别化合物甲、乙\nD: 乙能与盐酸、 $\\mathrm{NaOH}$ 溶液反应, 且 $1 \\mathrm{~mol}$ 乙最多能与 $4 \\mathrm{~mol} \\mathrm{NaOH}$ 反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化合物乙是一种治疗神经类疾病的药物, 可由化合物甲经多步反应得到。下列有关化合物甲、乙的说法正确的是[图1]\n\nA: 甲分子中所有碳原子一定处于同一平面\nB: 乙中含有 2 个手性碳原子\nC: 用 $\\mathrm{NaHCO}_{3}$ 溶液或 $\\mathrm{FeCl}_{3}$ 溶液能鉴别化合物甲、乙\nD: 乙能与盐酸、 $\\mathrm{NaOH}$ 溶液反应, 且 $1 \\mathrm{~mol}$ 乙最多能与 $4 \\mathrm{~mol} \\mathrm{NaOH}$ 反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://i.postimg.cc/PNsnLHQK/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1334",
        "problem": "Oxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nCalculate the maximum power that the body can produce (assuming it is limited by oxygen transfer).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nCalculate the maximum power that the body can produce (assuming it is limited by oxygen transfer).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of W, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-418.jpg?height=823&width=1400&top_left_y=568&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "W"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_386",
        "problem": "Which best describes the bonding in $\\mathrm{Cu}(\\mathrm{s})$ ?\nA: The copper atoms are positively charged in a sea of delocalized electrons.\nB: The copper atoms are alternately positively and negatively charged.\nC: The copper atoms form covalent bonds to adjacent copper atoms.\nD: The copper atoms form hydrogen bonds to adjacent copper atoms.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich best describes the bonding in $\\mathrm{Cu}(\\mathrm{s})$ ?\n\nA: The copper atoms are positively charged in a sea of delocalized electrons.\nB: The copper atoms are alternately positively and negatively charged.\nC: The copper atoms form covalent bonds to adjacent copper atoms.\nD: The copper atoms form hydrogen bonds to adjacent copper atoms.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1483",
        "problem": "Oxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nThe special S-shaped uptake curve 1 is the result of subtle structural features of $\\mathrm{Hb}$. The deficient $\\mathrm{Hb}$ shown in curve 2 is not optimal because:\nA: The binding with $\\mathrm{O}_{2}$ is too weak.\nB: The binding with $\\mathrm{O}_{2}$ is too strong.\nC: The maximum oxygen capacity is too low.\nD: The deficiency is caused by carbon monoxide poisoning.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nThe special S-shaped uptake curve 1 is the result of subtle structural features of $\\mathrm{Hb}$. The deficient $\\mathrm{Hb}$ shown in curve 2 is not optimal because:\n\nA: The binding with $\\mathrm{O}_{2}$ is too weak.\nB: The binding with $\\mathrm{O}_{2}$ is too strong.\nC: The maximum oxygen capacity is too low.\nD: The deficiency is caused by carbon monoxide poisoning.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-418.jpg?height=823&width=1400&top_left_y=568&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1104",
        "problem": "To help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nCalculate the enthalpy of combustion of an isomer of octane.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nTo help tackle the causes of climate change, in September 2021 the UK government changed the standard petrol grade at fuel pumps from $E 5$ to E10 petrol. These E values refer to the percentage of ethanol in the ethanol-hydrocarbon fuel mixture. It has been argued that the change from E5 to E10 was one contributing factor to the petrol shortage in October 2021 as retailers attempted to remove their stock supplies of E5.\n\n[figure1]\n\nE values and octane numbers are both displayed on petrol pumps. There is only a small proportion of the straight chain isomer of octane in petrol as this tends to \"knock\" in the engine. Knocking is where the fuel ignites prematurely and this reduces engine efficiency. Branched chain isomers of octane knock much less and a lot of these are found in petrol. One major isomer is 2,2,4-trimethylpentane.\n\nBlending ethanol into fuel mixtures such as in E5 and E10 also reduces knocking. Ethanol is a biofuel and is often produced by fermenting sugar from crop plants. The plants capture carbon dioxide from the atmosphere and convert this into sugars such as glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$, through photosynthesis.\n\nThe anaerobic fermentation of glucose produces ethanol and carbon dioxide.\n\nOne of the characteristics of an effective fuel is the amount of energy it releases, or its enthalpy of combustion. One method to determine this is to use average bond enthalpies. Some average bond enthalpies are given below.\n\n| Bond | Average bond enthalpy / kJ mol ${ }^{-1}$ |\n| :---: | :---: |\n| C-C | 347 |\n| C-H | 413 |\n| O=O | 498 |\n| C-O | 358 |\n| C=O | 805 |\n| O-H | 464 |\n\nWhen calculated by this method, all the different isomers of octane have the same value.\n\nCalculate the enthalpy of combustion of an isomer of octane.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-02.jpg?height=459&width=923&top_left_y=356&top_left_x=949"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_991",
        "problem": "If the Kelvin temperature of a sample of ideal gas doubles (e.g. from $200 \\mathrm{~K}$ to $400 \\mathrm{~K}$ ), then the average kinetic energy of the molecules in the sample\nA: increases by a factor of $\\sqrt{2}$\nB: decreases by a factor of 2\nC: increases by a factor of 2\nD: increases by a factor of 4\nE: remains the same\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf the Kelvin temperature of a sample of ideal gas doubles (e.g. from $200 \\mathrm{~K}$ to $400 \\mathrm{~K}$ ), then the average kinetic energy of the molecules in the sample\n\nA: increases by a factor of $\\sqrt{2}$\nB: decreases by a factor of 2\nC: increases by a factor of 2\nD: increases by a factor of 4\nE: remains the same\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_961",
        "problem": "某固体 $\\mathrm{X}$, 可能含有 $\\mathrm{BaCl}_{2} 、 \\mathrm{NaHCO}_{3} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3} 、 \\mathrm{Cu} 、 \\mathrm{NaAlO}_{2}$ 中的一种或几种, 进行如下实验: (1)取一定量样品溶于水中, 得到固体 $\\mathrm{A}$ 和溶液 B; (2)向 $\\mathrm{A}$ 中加入足量稀盐酸, 得到澄清溶液 $\\mathrm{C}$ 。下列说法不正确的是\nA: 固体 $\\mathrm{A}$ 可能溶于 $\\mathrm{NaOH}$ 溶液\nB: 溶液 $\\mathrm{C}$ 能与 $\\mathrm{NaHCO}_{3}$ 溶液反应产生沉淀\nC: 向溶液 $\\mathrm{C}$ 中加入 $\\mathrm{KSCN}$ 溶液, 若不变色, 则混合物 $\\mathrm{X}$ 不含 $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$\nD: 向溶液 $\\mathrm{B}$ 中加入 $\\mathrm{NaOH}$ 溶液, 若出现白色沉淀, 则 $\\mathrm{X}$ 中必定有 $\\mathrm{BaCl}_{2}$ 和 $\\mathrm{NaHCO}_{3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某固体 $\\mathrm{X}$, 可能含有 $\\mathrm{BaCl}_{2} 、 \\mathrm{NaHCO}_{3} 、 \\mathrm{Fe}_{2} \\mathrm{O}_{3} 、 \\mathrm{Cu} 、 \\mathrm{NaAlO}_{2}$ 中的一种或几种, 进行如下实验: (1)取一定量样品溶于水中, 得到固体 $\\mathrm{A}$ 和溶液 B; (2)向 $\\mathrm{A}$ 中加入足量稀盐酸, 得到澄清溶液 $\\mathrm{C}$ 。下列说法不正确的是\n\nA: 固体 $\\mathrm{A}$ 可能溶于 $\\mathrm{NaOH}$ 溶液\nB: 溶液 $\\mathrm{C}$ 能与 $\\mathrm{NaHCO}_{3}$ 溶液反应产生沉淀\nC: 向溶液 $\\mathrm{C}$ 中加入 $\\mathrm{KSCN}$ 溶液, 若不变色, 则混合物 $\\mathrm{X}$ 不含 $\\mathrm{Fe}_{2} \\mathrm{O}_{3}$\nD: 向溶液 $\\mathrm{B}$ 中加入 $\\mathrm{NaOH}$ 溶液, 若出现白色沉淀, 则 $\\mathrm{X}$ 中必定有 $\\mathrm{BaCl}_{2}$ 和 $\\mathrm{NaHCO}_{3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1539",
        "problem": "Lactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nIn a saturated aqueous solution of $\\mathrm{CaCO}_{3}(\\mathrm{~s}) \\mathrm{pH}$ is measured to be 9.95. Calculate the solubility of calcium carbonate in water and show that the calculated value for the solubility product constant $K_{s p}$ is $5 \\times 10^{-9}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nLactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nIn a saturated aqueous solution of $\\mathrm{CaCO}_{3}(\\mathrm{~s}) \\mathrm{pH}$ is measured to be 9.95. Calculate the solubility of calcium carbonate in water and show that the calculated value for the solubility product constant $K_{s p}$ is $5 \\times 10^{-9}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1177",
        "problem": "This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nAn $8.0 \\mathrm{ml}$ stock solution of potassium ferrate was added to a buffered $\\mathrm{Mn}$ (II) solution as described above. The initial concentration of potassium ferrate in the resulting solution was 9.6 $\\mu \\mathrm{M}$. Calculate the concentration of the potassium ferrate stock solution in $\\mu \\mathrm{M}$.\n\nThe concentrations of the $\\mathrm{Mn}^{2+}$ ions that reacted with certain concentrations of the added potassium ferrate were measured and reported in the following graph:\n\n[figure1]\n\nKnowing the proportions in which the reactants combine, and the fact that iron(III) ions are formed, the researchers were able to work out the oxidation state of the manganese in the product.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nAn $8.0 \\mathrm{ml}$ stock solution of potassium ferrate was added to a buffered $\\mathrm{Mn}$ (II) solution as described above. The initial concentration of potassium ferrate in the resulting solution was 9.6 $\\mu \\mathrm{M}$. Calculate the concentration of the potassium ferrate stock solution in $\\mu \\mathrm{M}$.\n\nThe concentrations of the $\\mathrm{Mn}^{2+}$ ions that reacted with certain concentrations of the added potassium ferrate were measured and reported in the following graph:\n\n[figure1]\n\nKnowing the proportions in which the reactants combine, and the fact that iron(III) ions are formed, the researchers were able to work out the oxidation state of the manganese in the product.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of \\mathrm{~mol} \\mathrm{dm}^{-3}, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_b1f97cb4cc7a1e9208fdg-8.jpg?height=768&width=1031&top_left_y=750&top_left_x=564"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "\\mathrm{~mol} \\mathrm{dm}^{-3}"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_626",
        "problem": "1. 已知: $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Al}(\\mathrm{OH})_{3}$ 沉淀的生成与溶解的 $\\mathrm{pH}$ 如下表\n\n| 氢氧化物 | 溶液 $\\mathrm{pH}$ |  |  |  |\n| :--- | :---: | :---: | :---: | :---: |\n|  | 开始沉淀 | 沉淀完全 | 沉淀开始溶解 | 沉淀完全溶解 |\n| $\\mathrm{Fe}(\\mathrm{OH})_{3}$ | 2.3 | 3.4 | - | - |\n| $\\mathrm{Al}(\\mathrm{OH})_{3}$ | 3.3 | 5.2 | 7.8 | 12.8 |\n\n向 $\\mathrm{FeCl}_{3} 、 \\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$ 的混合溶液中逐滴加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液, 形成沉淀的情况如下图所示。\n\n以下推断错误的是\n\n[图1]\nA: $\\mathrm{AB}$ 段可能发生的反应是: $2 \\mathrm{Ba}^{2+}+3 \\mathrm{OH}^{-}+2 \\mathrm{SO}_{4}^{2-}+\\mathrm{Al}^{3+}+2 \\mathrm{Ba}^{2+}=2 \\mathrm{BaSO}_{4} \\downarrow+\\mathrm{Al}(\\mathrm{OH})_{3}$ $\\downarrow$\nB: $\\mathrm{C}$ 点对应的沉淀是: $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{BaSO}_{4}$\nC: $\\mathrm{OA}$ 段可能发生的反应是: $3 \\mathrm{Ba}^{2+}+6 \\mathrm{OH}^{-}+3 \\mathrm{SO}_{4}^{2-}+\\mathrm{Al}^{3+}+\\mathrm{Fe}^{3+}=3 \\mathrm{BaSO}_{4} \\downarrow+\\mathrm{Fe}(\\mathrm{OH})_{3} \\downarrow$ $+\\mathrm{Al}(\\mathrm{OH})_{3} \\downarrow$\nD: 据图计算原溶液中： $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n1. 已知: $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{Al}(\\mathrm{OH})_{3}$ 沉淀的生成与溶解的 $\\mathrm{pH}$ 如下表\n\n| 氢氧化物 | 溶液 $\\mathrm{pH}$ |  |  |  |\n| :--- | :---: | :---: | :---: | :---: |\n|  | 开始沉淀 | 沉淀完全 | 沉淀开始溶解 | 沉淀完全溶解 |\n| $\\mathrm{Fe}(\\mathrm{OH})_{3}$ | 2.3 | 3.4 | - | - |\n| $\\mathrm{Al}(\\mathrm{OH})_{3}$ | 3.3 | 5.2 | 7.8 | 12.8 |\n\n向 $\\mathrm{FeCl}_{3} 、 \\mathrm{Al}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3}$ 的混合溶液中逐滴加入 $\\mathrm{Ba}(\\mathrm{OH})_{2}$ 溶液, 形成沉淀的情况如下图所示。\n\n以下推断错误的是\n\n[图1]\n\nA: $\\mathrm{AB}$ 段可能发生的反应是: $2 \\mathrm{Ba}^{2+}+3 \\mathrm{OH}^{-}+2 \\mathrm{SO}_{4}^{2-}+\\mathrm{Al}^{3+}+2 \\mathrm{Ba}^{2+}=2 \\mathrm{BaSO}_{4} \\downarrow+\\mathrm{Al}(\\mathrm{OH})_{3}$ $\\downarrow$\nB: $\\mathrm{C}$ 点对应的沉淀是: $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 和 $\\mathrm{BaSO}_{4}$\nC: $\\mathrm{OA}$ 段可能发生的反应是: $3 \\mathrm{Ba}^{2+}+6 \\mathrm{OH}^{-}+3 \\mathrm{SO}_{4}^{2-}+\\mathrm{Al}^{3+}+\\mathrm{Fe}^{3+}=3 \\mathrm{BaSO}_{4} \\downarrow+\\mathrm{Fe}(\\mathrm{OH})_{3} \\downarrow$ $+\\mathrm{Al}(\\mathrm{OH})_{3} \\downarrow$\nD: 据图计算原溶液中： $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-001.jpg?height=442&width=531&top_left_y=1201&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_792",
        "problem": "为了探究镁铝在 $\\mathrm{NaOH}$ 溶液中的放电情况, 某研究小组设计了如图 1 所示的实验装\n\n置。反应过程中装置的电压变化如图 2 所示。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 镁电极始终为电池的负极\nB: $50 \\mathrm{~s}$ 后, 原电池中电子流动方向发生改变\nC: 判断原电池的正、负极既要依据金属活动性顺序, 又要考虑与电解质溶液有关\nD: 整个过程中正极的电极反应式均为: $2 \\mathrm{H}_{2} \\mathrm{O}-2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow+\\mathrm{OH}^{-}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n为了探究镁铝在 $\\mathrm{NaOH}$ 溶液中的放电情况, 某研究小组设计了如图 1 所示的实验装\n\n置。反应过程中装置的电压变化如图 2 所示。下列说法正确的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 镁电极始终为电池的负极\nB: $50 \\mathrm{~s}$ 后, 原电池中电子流动方向发生改变\nC: 判断原电池的正、负极既要依据金属活动性顺序, 又要考虑与电解质溶液有关\nD: 整个过程中正极的电极反应式均为: $2 \\mathrm{H}_{2} \\mathrm{O}-2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow+\\mathrm{OH}^{-}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-54.jpg?height=326&width=436&top_left_y=1182&top_left_x=336",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-54.jpg?height=291&width=545&top_left_y=1231&top_left_x=938"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_990",
        "problem": "After a large meal the $\\mathrm{pH}$ of your stomach drops to 1.78. What is $\\left[\\mathrm{H}^{+}\\right]$in your stomach after the meal?\nA: $1.66 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.250 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $1.78 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $1.83 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $6.03 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAfter a large meal the $\\mathrm{pH}$ of your stomach drops to 1.78. What is $\\left[\\mathrm{H}^{+}\\right]$in your stomach after the meal?\n\nA: $1.66 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nB: $\\quad 0.250 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nC: $1.78 \\mathrm{~mol} \\mathrm{~L}^{-1}$\nD: $1.83 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$\nE: $6.03 \\times 10^{-2} \\mathrm{~mol} \\mathrm{~L}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_823",
        "problem": "中科海钠公司研发的二次电池“钠离子电池”被欧盟评价为“锂电的性能、铅酸的价格”。其放电总反应为 $\\mathrm{Na}_{\\mathrm{x}} \\mathrm{C}_{\\mathrm{n}}+\\mathrm{Na}_{1-\\mathrm{x}} \\mathrm{MeO}_{2}=\\mathrm{nC}+\\mathrm{NaMeO}_{2}$, 工作原理如图所示。下列说法正确的是\n\n[图1]\nA: 放电时, 正极反应式: $\\mathrm{Na}_{1-\\mathrm{x}} \\mathrm{MeO}_{2}+\\mathrm{Xe}^{-}+\\mathrm{Na}^{+}=\\mathrm{NaMeO}_{2}$\nB: 充电时, 电流方向为 $\\mathrm{N} \\rightarrow$ 电解液 $\\rightarrow \\mathrm{M}$\nC: 放电时, $\\mathrm{M}$ 极每减少 $4.6 \\mathrm{~g}$, 电解液中将增加 $0.2 \\mathrm{molNa}^{+}$\nD: 充电时, 外电路每转移 $1 \\mathrm{~mole}$, 有 $1 \\mathrm{molNaMeO}_{2}$ 被氧化\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n中科海钠公司研发的二次电池“钠离子电池”被欧盟评价为“锂电的性能、铅酸的价格”。其放电总反应为 $\\mathrm{Na}_{\\mathrm{x}} \\mathrm{C}_{\\mathrm{n}}+\\mathrm{Na}_{1-\\mathrm{x}} \\mathrm{MeO}_{2}=\\mathrm{nC}+\\mathrm{NaMeO}_{2}$, 工作原理如图所示。下列说法正确的是\n\n[图1]\n\nA: 放电时, 正极反应式: $\\mathrm{Na}_{1-\\mathrm{x}} \\mathrm{MeO}_{2}+\\mathrm{Xe}^{-}+\\mathrm{Na}^{+}=\\mathrm{NaMeO}_{2}$\nB: 充电时, 电流方向为 $\\mathrm{N} \\rightarrow$ 电解液 $\\rightarrow \\mathrm{M}$\nC: 放电时, $\\mathrm{M}$ 极每减少 $4.6 \\mathrm{~g}$, 电解液中将增加 $0.2 \\mathrm{molNa}^{+}$\nD: 充电时, 外电路每转移 $1 \\mathrm{~mole}$, 有 $1 \\mathrm{molNaMeO}_{2}$ 被氧化\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-59.jpg?height=476&width=597&top_left_y=2172&top_left_x=387"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_906",
        "problem": "电化学脱硫是近几年国内外发展起来的一种新型脱硫方法, 某种电化学脱硫方法装置如下图所示, 不仅可以脱去烟气中的 $\\mathrm{SO}_{2}$ 还可以制得硫酸, 下列说法错误的是\n\n[图1]\nA: 该装置的能量是将化学能转化为电能\nB: 净化气的成分为二氧化碳和氧气\nC: 左端电极反应式为 $\\mathrm{O}_{2}+4 \\mathrm{e}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{OH}^{-}$\nD: 右端电极反应式为 $2 \\mathrm{SO}_{4}^{2-}-4 \\mathrm{e}^{-}=$ $2 \\mathrm{SO}_{3}+\\mathrm{O}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n电化学脱硫是近几年国内外发展起来的一种新型脱硫方法, 某种电化学脱硫方法装置如下图所示, 不仅可以脱去烟气中的 $\\mathrm{SO}_{2}$ 还可以制得硫酸, 下列说法错误的是\n\n[图1]\n\nA: 该装置的能量是将化学能转化为电能\nB: 净化气的成分为二氧化碳和氧气\nC: 左端电极反应式为 $\\mathrm{O}_{2}+4 \\mathrm{e}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{OH}^{-}$\nD: 右端电极反应式为 $2 \\mathrm{SO}_{4}^{2-}-4 \\mathrm{e}^{-}=$ $2 \\mathrm{SO}_{3}+\\mathrm{O}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-65.jpg?height=451&width=1424&top_left_y=211&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_908",
        "problem": "已知亚磷酸 $\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$ 为二元弱酸。 $25^{\\circ} \\mathrm{C}$ 时, 向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{H}_{3} \\mathrm{PO}_{3}$ 溶液中滴加同\n\n浓度 $\\mathrm{NaOH}$ 溶液, 混合溶液中的有关粒子浓度之比的对数与溶液 $\\mathrm{pH}$ 的关系如图所示,下列叙述正确的是\n\n[图1]\nA: $M 、 W$ 两点所在的直线表示 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)}$与 $\\mathrm{pH}$ 的关系\nB: 当 $\\mathrm{pH}=3$ 时, $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$\nC: $\\mathrm{N}$ 点: $3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{HPO}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}+\\mathrm{OH}^{-}$的平衡常数为 $10^{-12.57}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知亚磷酸 $\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$ 为二元弱酸。 $25^{\\circ} \\mathrm{C}$ 时, 向 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{H}_{3} \\mathrm{PO}_{3}$ 溶液中滴加同\n\n浓度 $\\mathrm{NaOH}$ 溶液, 混合溶液中的有关粒子浓度之比的对数与溶液 $\\mathrm{pH}$ 的关系如图所示,下列叙述正确的是\n\n[图1]\n\nA: $M 、 W$ 两点所在的直线表示 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)}$与 $\\mathrm{pH}$ 的关系\nB: 当 $\\mathrm{pH}=3$ 时, $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{HPO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{PO}_{3}\\right)$\nC: $\\mathrm{N}$ 点: $3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{HPO}_{3}^{2-}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{2} \\mathrm{PO}_{3}^{-}+\\mathrm{OH}^{-}$的平衡常数为 $10^{-12.57}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-086.jpg?height=454&width=1114&top_left_y=1526&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1384",
        "problem": "In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Later, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were measured by using a glass vessel with a known volume under atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$.\n\nTable 1. Mass of Chemical Nitrogen in the Vessel\n\n| From nitric oxide | $2.3001 \\mathrm{~g}$ |\n| :--- | :--- |\n| From nitrous oxide | $2.2990 \\mathrm{~g}$ |\n| From ammonium nitrite purified at a red heat | $2.2987 \\mathrm{~g}$ |\n| From urea | $2.2985 \\mathrm{~g}$ |\n| From ammonium nitrite purified in the cold | $2.2987 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 2 9 9 0} \\mathbf{~ g}$ |\n\nTable 2. Mass of Atmospheric Nitrogen in the Vessel\n\n| $\\mathrm{O}_{2}$ was removed by hot copper (1892) | $2.3103 \\mathrm{~g}$ |\n| :--- | :--- |\n| $\\mathrm{O}_{2}$ was removed by hot iron (1893) | $2.3100 \\mathrm{~g}$ |\n| $\\mathrm{O}_{2}$ was removed by ferrous hydrate (1894) | $2.3102 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 3 1 0 2} \\mathbf{~ g}$ |Estimate the mole fraction $x$ of argon in Rayleigh's atmospheric nitrogen, by assuming that argon and nitrogen were the only constituents. Use the mean masses of the atmospheric and chemical nitrogen for the calculation.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Later, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were measured by using a glass vessel with a known volume under atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$.\n\nTable 1. Mass of Chemical Nitrogen in the Vessel\n\n| From nitric oxide | $2.3001 \\mathrm{~g}$ |\n| :--- | :--- |\n| From nitrous oxide | $2.2990 \\mathrm{~g}$ |\n| From ammonium nitrite purified at a red heat | $2.2987 \\mathrm{~g}$ |\n| From urea | $2.2985 \\mathrm{~g}$ |\n| From ammonium nitrite purified in the cold | $2.2987 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 2 9 9 0} \\mathbf{~ g}$ |\n\nTable 2. Mass of Atmospheric Nitrogen in the Vessel\n\n| $\\mathrm{O}_{2}$ was removed by hot copper (1892) | $2.3103 \\mathrm{~g}$ |\n| :--- | :--- |\n| $\\mathrm{O}_{2}$ was removed by hot iron (1893) | $2.3100 \\mathrm{~g}$ |\n| $\\mathrm{O}_{2}$ was removed by ferrous hydrate (1894) | $2.3102 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 3 1 0 2} \\mathbf{~ g}$ |\n\nproblem:\nEstimate the mole fraction $x$ of argon in Rayleigh's atmospheric nitrogen, by assuming that argon and nitrogen were the only constituents. Use the mean masses of the atmospheric and chemical nitrogen for the calculation.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_496",
        "problem": "在 $25^{\\circ} \\mathrm{C}$ 时, 将 $1.0 \\mathrm{~L} \\mathrm{w} \\mathrm{mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液与 $0.1 \\mathrm{~mol} \\mathrm{NaOH}$ 固体混合, 充分反应。向混合液中加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 或 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 固体 (忽略体积和温度变化), 溶液 $\\mathrm{pH}$随加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 或 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 固体的物质的量的变化关系如图所示。下列叙述正确的是\n\n[图1]\nA: b 点混合液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\geq \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$\nB: 加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 过程中, $\\frac{\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right) \\times \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}$增大\nC: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{CH}_{3} \\mathrm{COOH}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{a}}=\\frac{0.2 \\times 10^{-7}}{\\mathrm{w}-0.1} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $a 、 b 、 c$ 对应的混合液中, 水的电离程度由大到小的顺序是 $c>a>b$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在 $25^{\\circ} \\mathrm{C}$ 时, 将 $1.0 \\mathrm{~L} \\mathrm{w} \\mathrm{mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液与 $0.1 \\mathrm{~mol} \\mathrm{NaOH}$ 固体混合, 充分反应。向混合液中加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 或 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 固体 (忽略体积和温度变化), 溶液 $\\mathrm{pH}$随加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 或 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 固体的物质的量的变化关系如图所示。下列叙述正确的是\n\n[图1]\n\nA: b 点混合液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\geq \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)$\nB: 加入 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 过程中, $\\frac{\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right) \\times \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)}$增大\nC: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{CH}_{3} \\mathrm{COOH}$ 的电离平衡常数 $\\mathrm{K}_{\\mathrm{a}}=\\frac{0.2 \\times 10^{-7}}{\\mathrm{w}-0.1} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $a 、 b 、 c$ 对应的混合液中, 水的电离程度由大到小的顺序是 $c>a>b$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-25.jpg?height=543&width=626&top_left_y=802&top_left_x=321"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_976",
        "problem": "1 Which of the following elements is not a metal?\nA: Se\nB: Sn\nC: $\\mathrm{Sr}$\nD: Sc\nE: Cs\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n1 Which of the following elements is not a metal?\n\nA: Se\nB: Sn\nC: $\\mathrm{Sr}$\nD: Sc\nE: Cs\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_524",
        "problem": "南京大学教授提出了一种高效的固碳新技术(如图)。利用锂一一二氧化碳电池, 在放电过程中二氧化碳转化为碳和碳酸锂, 在电池充电过程中通过选用合适的催化剂使碳酸锂单独被氧化分解, 而另一种放电产物碳留在电池内, 下列说法不正确的是\n\n[图1]\n\n聚合物电解质膜\nA: 该电池放电时, $\\mathrm{CO}_{2}$ 从放电进气口进入\nB: 该电池放电时的正极反应是 $3 \\mathrm{CO}_{2}+4 \\mathrm{Li}^{+}+4 \\mathrm{e}^{-}=2 \\mathrm{Li}_{2} \\mathrm{CO}_{3}+\\mathrm{C}$\nC: 该电池充电时 $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 转化为 $\\mathrm{Li}^{+}$\nD: 该电池放电时, 每转移 $2 \\mathrm{~mol}$ 电子, 可固定 $1 \\mathrm{molCO}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n南京大学教授提出了一种高效的固碳新技术(如图)。利用锂一一二氧化碳电池, 在放电过程中二氧化碳转化为碳和碳酸锂, 在电池充电过程中通过选用合适的催化剂使碳酸锂单独被氧化分解, 而另一种放电产物碳留在电池内, 下列说法不正确的是\n\n[图1]\n\n聚合物电解质膜\n\nA: 该电池放电时, $\\mathrm{CO}_{2}$ 从放电进气口进入\nB: 该电池放电时的正极反应是 $3 \\mathrm{CO}_{2}+4 \\mathrm{Li}^{+}+4 \\mathrm{e}^{-}=2 \\mathrm{Li}_{2} \\mathrm{CO}_{3}+\\mathrm{C}$\nC: 该电池充电时 $\\mathrm{Li}_{2} \\mathrm{CO}_{3}$ 转化为 $\\mathrm{Li}^{+}$\nD: 该电池放电时, 每转移 $2 \\mathrm{~mol}$ 电子, 可固定 $1 \\mathrm{molCO}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-112.jpg?height=565&width=942&top_left_y=1074&top_left_x=403"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_72",
        "problem": "Which of the following metals is most likely to be produced through electrolysis?\nA: Aluminum\nB: Gold\nC: Iron\nD: Mercury\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following metals is most likely to be produced through electrolysis?\n\nA: Aluminum\nB: Gold\nC: Iron\nD: Mercury\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_897",
        "problem": "聚合硫酸铁 $\\left[\\mathrm{Fe}(\\mathrm{OH}) \\mathrm{SO}_{4}\\right] n$ 能用作净水剂(絮凝剂), 可由绿矾 $\\left(\\mathrm{FeSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}\\right)$ 和 $\\mathrm{KClO}_{3}$在水溶液中反应得到。下列说法不正确的是\nA: $\\mathrm{KClO}_{3}$ 作氧化剂, 每生成 $1 \\mathrm{~mol}\\left[\\mathrm{Fe}(\\mathrm{OH}) \\mathrm{SO}_{4}\\right] n$ 消耗 $6 / n \\mathrm{~mol} \\mathrm{KClO}_{3}$\nB: 生成聚合硫酸铁后, 水溶液的 $\\mathrm{pH}$ 增大\nC: 聚合硫酸铁可在水中形成氢氧化铁胶体而净水\nD: 在相同条件下, $\\mathrm{Fe}^{3+}$ 比 $[\\mathrm{Fe}(\\mathrm{OH})]^{2+}$ 的水解能力更强\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n聚合硫酸铁 $\\left[\\mathrm{Fe}(\\mathrm{OH}) \\mathrm{SO}_{4}\\right] n$ 能用作净水剂(絮凝剂), 可由绿矾 $\\left(\\mathrm{FeSO}_{4} \\cdot 7 \\mathrm{H}_{2} \\mathrm{O}\\right)$ 和 $\\mathrm{KClO}_{3}$在水溶液中反应得到。下列说法不正确的是\n\nA: $\\mathrm{KClO}_{3}$ 作氧化剂, 每生成 $1 \\mathrm{~mol}\\left[\\mathrm{Fe}(\\mathrm{OH}) \\mathrm{SO}_{4}\\right] n$ 消耗 $6 / n \\mathrm{~mol} \\mathrm{KClO}_{3}$\nB: 生成聚合硫酸铁后, 水溶液的 $\\mathrm{pH}$ 增大\nC: 聚合硫酸铁可在水中形成氢氧化铁胶体而净水\nD: 在相同条件下, $\\mathrm{Fe}^{3+}$ 比 $[\\mathrm{Fe}(\\mathrm{OH})]^{2+}$ 的水解能力更强\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1487",
        "problem": "Propene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nPropene can be synthesized via a direct dehydrogenation of propane in the presence of a heterogeneous catalyst. However, such a reaction is not economically feasible due to the nature of the reaction itself. Provide a concise explanation to each of the questions below. Additional information: $H_{\\text {bond }}(\\mathrm{C}=\\mathrm{C})=1.77 \\mathrm{H}_{\\text {bond }}(\\mathrm{C}-\\mathrm{C}), \\mathrm{H}_{\\text {bond }}(\\mathrm{H}-\\mathrm{H})=$ $1.05 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})$, and $H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})=1.19 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$, where $H_{\\text {bond }}$ refers to average bond enthalpy of the indicated chemical bond.It is difficult to increase the amount of propene by increasing pressure at constant temperature. Which law or principle can best explain this phenomenon? Select your answer by marking \" $\\checkmark$ \" in one of the open circles.\nA: Boyle's law\nB: Charles' law\nC: Dalton's law\nD: Raoult's law\nE: Le Chatelier's principle\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nPropene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).\n\nPropene can be synthesized via a direct dehydrogenation of propane in the presence of a heterogeneous catalyst. However, such a reaction is not economically feasible due to the nature of the reaction itself. Provide a concise explanation to each of the questions below. Additional information: $H_{\\text {bond }}(\\mathrm{C}=\\mathrm{C})=1.77 \\mathrm{H}_{\\text {bond }}(\\mathrm{C}-\\mathrm{C}), \\mathrm{H}_{\\text {bond }}(\\mathrm{H}-\\mathrm{H})=$ $1.05 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})$, and $H_{\\text {bond }}(\\mathrm{C}-\\mathrm{H})=1.19 H_{\\text {bond }}(\\mathrm{C}-\\mathrm{C})$, where $H_{\\text {bond }}$ refers to average bond enthalpy of the indicated chemical bond.\n\nproblem:\nIt is difficult to increase the amount of propene by increasing pressure at constant temperature. Which law or principle can best explain this phenomenon? Select your answer by marking \" $\\checkmark$ \" in one of the open circles.\n\nA: Boyle's law\nB: Charles' law\nC: Dalton's law\nD: Raoult's law\nE: Le Chatelier's principle\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_633",
        "problem": "我国首创的海洋电池以铝板、铂网作电极, 海水为电解质溶液, 空气中的氧气与铝反应产生电流。电池总反应为: $4 \\mathrm{~A} 1+3 \\mathrm{O}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{Al}(\\mathrm{OH})_{3}$ 。则下列判断中, 正确的是\nA: 电池工作时, 铝板是正极\nB: 正极反应为: $\\mathrm{O}_{2}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=2 \\mathrm{OH}^{-}$\nC: 铂电极做成网状, 可增大与氧气的接触面积\nD: 该电池通常只需要更换铝板就可继续使用\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n我国首创的海洋电池以铝板、铂网作电极, 海水为电解质溶液, 空气中的氧气与铝反应产生电流。电池总反应为: $4 \\mathrm{~A} 1+3 \\mathrm{O}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}=4 \\mathrm{Al}(\\mathrm{OH})_{3}$ 。则下列判断中, 正确的是\n\nA: 电池工作时, 铝板是正极\nB: 正极反应为: $\\mathrm{O}_{2}+\\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=2 \\mathrm{OH}^{-}$\nC: 铂电极做成网状, 可增大与氧气的接触面积\nD: 该电池通常只需要更换铝板就可继续使用\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_342",
        "problem": "Which of the following is not a pair of isomers?\nA: ethyl benzene $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}-\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)$ and dimethyl benzene, $\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CH}_{3}\\right)_{2}$\nB: 1-propanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$ and 2-propanol $\\left(\\mathrm{CH}_{3} \\mathrm{CHOHCH}_{3}\\right)$\nC: ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)$ and dimethyl ether $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$\nD: 2-butanone $\\left(\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{CH}_{3}\\right)$ and 1-butanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$\nE: urea $\\left(\\mathrm{NH}_{2} \\mathrm{CONH}_{2}\\right)$ and ammonium cyanate $\\left(\\mathrm{NH}_{4} \\mathrm{CNO}\\right)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is not a pair of isomers?\n\nA: ethyl benzene $\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}-\\mathrm{C}_{2} \\mathrm{H}_{5}\\right)$ and dimethyl benzene, $\\mathrm{C}_{6} \\mathrm{H}_{4}\\left(\\mathrm{CH}_{3}\\right)_{2}$\nB: 1-propanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$ and 2-propanol $\\left(\\mathrm{CH}_{3} \\mathrm{CHOHCH}_{3}\\right)$\nC: ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{OH}\\right)$ and dimethyl ether $\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$\nD: 2-butanone $\\left(\\mathrm{CH}_{3} \\mathrm{COCH}_{2} \\mathrm{CH}_{3}\\right)$ and 1-butanol $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$\nE: urea $\\left(\\mathrm{NH}_{2} \\mathrm{CONH}_{2}\\right)$ and ammonium cyanate $\\left(\\mathrm{NH}_{4} \\mathrm{CNO}\\right)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_46",
        "problem": "Which alkyl halide reacts most rapidly with aqueous sodium hydroxide solution?\nA: $\\mathrm{CH}_{3} \\mathrm{Cl}$\nB: $\\mathrm{CH}_{3} \\mathrm{I}$\nC: $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{Cl}$\nD: $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{I}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich alkyl halide reacts most rapidly with aqueous sodium hydroxide solution?\n\nA: $\\mathrm{CH}_{3} \\mathrm{Cl}$\nB: $\\mathrm{CH}_{3} \\mathrm{I}$\nC: $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{Cl}$\nD: $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CCH}_{2} \\mathrm{I}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1137",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## Mendeleev's short-form table\n\nIn the second version of his table, Mendeleev updated the mass of indium and correctly positioned both it and thallium. He also arranged the Groups of similar elements vertically but there are too many elements within each Group; Mendeleev mixed elements from the transition metals with those from the main block of the periodic table. There is sense to this since the elements which Mendeleev places in a given Group form similar compounds - notably their oxides and their hydrides.\n\n[figure1]\n\nMendeleev's second periodic table from 1871\n\nSolid nitrogen $(\\mathrm{V})$ oxide has a different structure from the gaseous form and consists of linear cations, and trigonal-planar anions. Suggest the formula of the anion in solid nitrogen(V) oxide.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## Mendeleev's short-form table\n\nIn the second version of his table, Mendeleev updated the mass of indium and correctly positioned both it and thallium. He also arranged the Groups of similar elements vertically but there are too many elements within each Group; Mendeleev mixed elements from the transition metals with those from the main block of the periodic table. There is sense to this since the elements which Mendeleev places in a given Group form similar compounds - notably their oxides and their hydrides.\n\n[figure1]\n\nMendeleev's second periodic table from 1871\n\nSolid nitrogen $(\\mathrm{V})$ oxide has a different structure from the gaseous form and consists of linear cations, and trigonal-planar anions. Suggest the formula of the anion in solid nitrogen(V) oxide.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_bf14c3e936c3f7031d02g-05.jpg?height=742&width=1567&top_left_y=183&top_left_x=290"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1091",
        "problem": "The volume of the $\\mathrm{h}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$ is $3.74 \\times 10^{-23} \\mathrm{~cm}^{3}$.Boron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the density of $\\mathrm{h}-\\mathrm{BN}$ in $\\mathrm{g} \\mathrm{cm}^{-3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe volume of the $\\mathrm{h}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$ is $3.74 \\times 10^{-23} \\mathrm{~cm}^{3}$.\n\nproblem:\nBoron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the density of $\\mathrm{h}-\\mathrm{BN}$ in $\\mathrm{g} \\mathrm{cm}^{-3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~g} \\mathrm{~cm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=448&width=511&top_left_y=541&top_left_x=270",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=596&width=436&top_left_y=458&top_left_x=844",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=471&width=443&top_left_y=524&top_left_x=1389",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=603&width=305&top_left_y=1206&top_left_x=1315"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~g} \\mathrm{~cm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_476",
        "problem": "一定温度下, 在恒容密闭容器中发生如下反应: $2 \\mathrm{~A}(\\mathrm{~g})+\\mathrm{B}(\\mathrm{g}) \\rightleftharpoons 3 \\mathrm{C}(\\mathrm{g})$, 若反应开始时充入 $2 \\mathrm{~mol} \\mathrm{~A}$ 和 $2 \\mathrm{~mol} \\mathrm{~B}$, 达平衡后 $\\mathrm{A}$ 的体积分数为 $\\mathrm{a} \\%$ 。其他条件不变时, 若按下列\n四种配比作为起始物质, 平衡后 $\\mathrm{A}$ 的体积分数大于 $\\mathrm{a} \\%$ 的是\nA: $2.5 \\mathrm{~mol} \\mathrm{C}$\nB: $2 \\mathrm{~mol} \\mathrm{~A} 、 1 \\mathrm{~mol} \\mathrm{~B}$ 和 $10 \\mathrm{~mol} \\mathrm{He}$ (不参加反应)\nC: $1.5 \\mathrm{~mol} \\mathrm{~B}$ 和 $1.5 \\mathrm{~mol} \\mathrm{C}$\nD: $2 \\mathrm{~mol} \\mathrm{~A} 、 3 \\mathrm{~mol} \\mathrm{~B}$ 和 $3 \\mathrm{~mol} \\mathrm{C}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一定温度下, 在恒容密闭容器中发生如下反应: $2 \\mathrm{~A}(\\mathrm{~g})+\\mathrm{B}(\\mathrm{g}) \\rightleftharpoons 3 \\mathrm{C}(\\mathrm{g})$, 若反应开始时充入 $2 \\mathrm{~mol} \\mathrm{~A}$ 和 $2 \\mathrm{~mol} \\mathrm{~B}$, 达平衡后 $\\mathrm{A}$ 的体积分数为 $\\mathrm{a} \\%$ 。其他条件不变时, 若按下列\n四种配比作为起始物质, 平衡后 $\\mathrm{A}$ 的体积分数大于 $\\mathrm{a} \\%$ 的是\n\nA: $2.5 \\mathrm{~mol} \\mathrm{C}$\nB: $2 \\mathrm{~mol} \\mathrm{~A} 、 1 \\mathrm{~mol} \\mathrm{~B}$ 和 $10 \\mathrm{~mol} \\mathrm{He}$ (不参加反应)\nC: $1.5 \\mathrm{~mol} \\mathrm{~B}$ 和 $1.5 \\mathrm{~mol} \\mathrm{C}$\nD: $2 \\mathrm{~mol} \\mathrm{~A} 、 3 \\mathrm{~mol} \\mathrm{~B}$ 和 $3 \\mathrm{~mol} \\mathrm{C}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_103",
        "problem": "According to the Aufbau Principle in quantum mechanics, which of the following is the highest energy electron sublevel?\nA: $7 \\mathrm{~s}$\nB: $6 \\mathrm{p}$\nC: 6d\nD: 5f\nE: 5p\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAccording to the Aufbau Principle in quantum mechanics, which of the following is the highest energy electron sublevel?\n\nA: $7 \\mathrm{~s}$\nB: $6 \\mathrm{p}$\nC: 6d\nD: 5f\nE: 5p\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_128",
        "problem": "An unknown amino acid contains $9.5 \\%$ nitrogen by mass as determined by elemental analysis. Which of the following could be the unknown amino acid?\nA: arginine $\\mathrm{C}_{6} \\mathrm{H}_{14} \\mathrm{~N}_{4} \\mathrm{O}_{2}$\nB: histidine $\\mathrm{C}_{6} \\mathrm{H}_{9} \\mathrm{~N}_{3} \\mathrm{O}_{2}$\nC: cysteine $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{NO}_{2} \\mathrm{~S}$\nD: glycine $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{NO}_{2}$\nE: glutamic acid $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{NO}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn unknown amino acid contains $9.5 \\%$ nitrogen by mass as determined by elemental analysis. Which of the following could be the unknown amino acid?\n\nA: arginine $\\mathrm{C}_{6} \\mathrm{H}_{14} \\mathrm{~N}_{4} \\mathrm{O}_{2}$\nB: histidine $\\mathrm{C}_{6} \\mathrm{H}_{9} \\mathrm{~N}_{3} \\mathrm{O}_{2}$\nC: cysteine $\\mathrm{C}_{3} \\mathrm{H}_{7} \\mathrm{NO}_{2} \\mathrm{~S}$\nD: glycine $\\mathrm{C}_{2} \\mathrm{H}_{5} \\mathrm{NO}_{2}$\nE: glutamic acid $\\mathrm{C}_{5} \\mathrm{H}_{9} \\mathrm{NO}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1533",
        "problem": "The radioactive decay series of ${ }^{226} \\mathrm{Ra}$ is as follows:\n\n$$\n\\begin{aligned}\n& { }^{226} \\mathrm{Ra} \\xrightarrow{t}{ }^{222} \\mathrm{Rn} \\xrightarrow{3.825 \\mathrm{~d}}{ }^{218} \\mathrm{Po} \\xrightarrow{3.10 \\mathrm{~m}}{ }^{214} \\mathrm{~Pb} \\xrightarrow{26.8 \\mathrm{~m}}{ }^{214} \\mathrm{Bi} \\xrightarrow{19.9 \\mathrm{~m}} \\\\\n& { }^{214} \\mathrm{Po} \\xrightarrow{164.3 \\mu \\mathrm{s}} 210 \\mathrm{~Pb} \\xrightarrow{22.3 \\mathrm{y}} 210 \\mathrm{Bi} \\xrightarrow{5.013 \\mathrm{~d}} 210 \\mathrm{Po} \\xrightarrow{138.4 \\mathrm{~d}}{ }^{206 \\mathrm{~Pb}}\n\\end{aligned}\n$$\n\nThe times indicated are half-lives, the units are $y=$ years, $d=$ days, $m=$ minutes. The first decay, marked $t$ above, has a much longer half-life than the others.At the end of the 163 days the sample was found to contain $10.4 \\mathrm{~mm}^{3}$ of $\\mathrm{He}$, measured at $101325 \\mathrm{~Pa}$ and $273 \\mathrm{~K}$. Calculate the Avogadro constant from these data.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe radioactive decay series of ${ }^{226} \\mathrm{Ra}$ is as follows:\n\n$$\n\\begin{aligned}\n& { }^{226} \\mathrm{Ra} \\xrightarrow{t}{ }^{222} \\mathrm{Rn} \\xrightarrow{3.825 \\mathrm{~d}}{ }^{218} \\mathrm{Po} \\xrightarrow{3.10 \\mathrm{~m}}{ }^{214} \\mathrm{~Pb} \\xrightarrow{26.8 \\mathrm{~m}}{ }^{214} \\mathrm{Bi} \\xrightarrow{19.9 \\mathrm{~m}} \\\\\n& { }^{214} \\mathrm{Po} \\xrightarrow{164.3 \\mu \\mathrm{s}} 210 \\mathrm{~Pb} \\xrightarrow{22.3 \\mathrm{y}} 210 \\mathrm{Bi} \\xrightarrow{5.013 \\mathrm{~d}} 210 \\mathrm{Po} \\xrightarrow{138.4 \\mathrm{~d}}{ }^{206 \\mathrm{~Pb}}\n\\end{aligned}\n$$\n\nThe times indicated are half-lives, the units are $y=$ years, $d=$ days, $m=$ minutes. The first decay, marked $t$ above, has a much longer half-life than the others.\n\nproblem:\nAt the end of the 163 days the sample was found to contain $10.4 \\mathrm{~mm}^{3}$ of $\\mathrm{He}$, measured at $101325 \\mathrm{~Pa}$ and $273 \\mathrm{~K}$. Calculate the Avogadro constant from these data.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_573",
        "problem": "相同金属在其不同浓度盐溶液中可形成浓差电池。如下图所示装置是利用浓差电池电解 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液 ( $\\mathrm{a} 、 \\mathrm{~b}$ 电极均为石墨电极), 可以制得 $\\mathrm{O}_{2} 、 \\mathrm{H}_{2} 、 \\mathrm{H}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{NaOH}$ 。下列说法正确的是\n\n[图1]\nA: $\\mathrm{a}$ 电极的电极反应为: $2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow+2 \\mathrm{OH}^{-}$\nB: $\\mathrm{c}$ d 离子交换膜依次为阴离子交换膜和阳离子交换膜\nC: 电池放电过程中, $\\mathrm{Cu}(2)$ 电极上的电极反应为 $\\mathrm{Cu}-2 \\mathrm{e}^{-}=\\mathrm{Cu}^{2+}$\nD: 电池从开始工作到停止放电, 电解池阳极区理论上可生成 $1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n相同金属在其不同浓度盐溶液中可形成浓差电池。如下图所示装置是利用浓差电池电解 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液 ( $\\mathrm{a} 、 \\mathrm{~b}$ 电极均为石墨电极), 可以制得 $\\mathrm{O}_{2} 、 \\mathrm{H}_{2} 、 \\mathrm{H}_{2} \\mathrm{SO}_{4}$ 和 $\\mathrm{NaOH}$ 。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{a}$ 电极的电极反应为: $2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow+2 \\mathrm{OH}^{-}$\nB: $\\mathrm{c}$ d 离子交换膜依次为阴离子交换膜和阳离子交换膜\nC: 电池放电过程中, $\\mathrm{Cu}(2)$ 电极上的电极反应为 $\\mathrm{Cu}-2 \\mathrm{e}^{-}=\\mathrm{Cu}^{2+}$\nD: 电池从开始工作到停止放电, 电解池阳极区理论上可生成 $1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-62.jpg?height=534&width=1453&top_left_y=156&top_left_x=330"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_784",
        "problem": "工业上常用 $\\mathrm{H}_{2} \\mathrm{~S}$ 作沉淀剂除去废水中的 $\\mathrm{Zn}^{2+}$ 和 $\\mathrm{Mn}^{2+}$ 。通过调节溶液的 $\\mathrm{pH}$ 可使 $\\mathrm{Zn}^{2+}$和 $\\mathrm{Mn}^{2+}$ 逐一沉降, 处理过程中始终保持 $\\mathrm{H}_{2} \\mathrm{~S}$ 溶液为饱和状态即 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 体系中 $\\mathrm{S}^{2-} 、 \\mathrm{HS}^{-} 、 \\mathrm{Zn}^{2+} 、 \\mathrm{Mn}^{2+}$ 浓度 $(\\mathrm{mol} / \\mathrm{L})$ 的负对数 $\\mathrm{pM}$ 与 $\\mathrm{pH}$ 的关系如图所示。已知: $K_{\\mathrm{sp}}(\\mathrm{MnS})>K_{\\mathrm{sp}}(\\mathrm{ZnS})$ 。下列说法错误的是\n\n[图1]\nA: II表示 $-\\lg \\left(\\mathrm{S}^{2-}\\right)$ 与 $\\mathrm{pH}$ 的关系曲线\nB: $\\mathrm{Q}$ 点对应 $c\\left(\\mathrm{H}^{+}\\right)$的数量级为 $10^{-2}$\nC: $\\frac{K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)}{K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)}=10^{6.6}$\nD: 溶液中 $c\\left(\\mathrm{Zn}^{2+}\\right)$ 和 $c\\left(\\mathrm{Mn}^{2+}\\right)$ 均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}, \\mathrm{Zn}^{2+}$ 完全沉淀时溶液的最小 $\\mathrm{pH}$ 为 2 (金属离子的浓度 $\\leq 10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 可认为沉淀完全)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n工业上常用 $\\mathrm{H}_{2} \\mathrm{~S}$ 作沉淀剂除去废水中的 $\\mathrm{Zn}^{2+}$ 和 $\\mathrm{Mn}^{2+}$ 。通过调节溶液的 $\\mathrm{pH}$ 可使 $\\mathrm{Zn}^{2+}$和 $\\mathrm{Mn}^{2+}$ 逐一沉降, 处理过程中始终保持 $\\mathrm{H}_{2} \\mathrm{~S}$ 溶液为饱和状态即 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 体系中 $\\mathrm{S}^{2-} 、 \\mathrm{HS}^{-} 、 \\mathrm{Zn}^{2+} 、 \\mathrm{Mn}^{2+}$ 浓度 $(\\mathrm{mol} / \\mathrm{L})$ 的负对数 $\\mathrm{pM}$ 与 $\\mathrm{pH}$ 的关系如图所示。已知: $K_{\\mathrm{sp}}(\\mathrm{MnS})>K_{\\mathrm{sp}}(\\mathrm{ZnS})$ 。下列说法错误的是\n\n[图1]\n\nA: II表示 $-\\lg \\left(\\mathrm{S}^{2-}\\right)$ 与 $\\mathrm{pH}$ 的关系曲线\nB: $\\mathrm{Q}$ 点对应 $c\\left(\\mathrm{H}^{+}\\right)$的数量级为 $10^{-2}$\nC: $\\frac{K_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)}{K_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)}=10^{6.6}$\nD: 溶液中 $c\\left(\\mathrm{Zn}^{2+}\\right)$ 和 $c\\left(\\mathrm{Mn}^{2+}\\right)$ 均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}, \\mathrm{Zn}^{2+}$ 完全沉淀时溶液的最小 $\\mathrm{pH}$ 为 2 (金属离子的浓度 $\\leq 10^{-5} \\mathrm{~mol} / \\mathrm{L}$ 可认为沉淀完全)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-022.jpg?height=437&width=528&top_left_y=181&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_304",
        "problem": "If $1.00 \\mathrm{~L}$ of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{2}(\\mathrm{aq})$ is diluted with water to a final volume of $4.00 \\mathrm{~L}$, then which of the following statements regarding the new solution is true?\nA: The percent ionization of the acid decreases and the $\\mathrm{pH}$ remains the same.\nB: The percent ionization of the acid increases and the $\\mathrm{pH}$ decreases.\nC: The percent ionization of the acid increases and the $\\mathrm{pH}$ increases.\nD: The percent ionization of the acid decreases and the $\\mathrm{pH}$ decreases.\nE: The percent ionization of the acid increases and the $\\mathrm{pH}$ remains the same.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf $1.00 \\mathrm{~L}$ of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HNO}_{2}(\\mathrm{aq})$ is diluted with water to a final volume of $4.00 \\mathrm{~L}$, then which of the following statements regarding the new solution is true?\n\nA: The percent ionization of the acid decreases and the $\\mathrm{pH}$ remains the same.\nB: The percent ionization of the acid increases and the $\\mathrm{pH}$ decreases.\nC: The percent ionization of the acid increases and the $\\mathrm{pH}$ increases.\nD: The percent ionization of the acid decreases and the $\\mathrm{pH}$ decreases.\nE: The percent ionization of the acid increases and the $\\mathrm{pH}$ remains the same.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1491",
        "problem": "Rotational energy levels of diatomic molecules are well described by the formula $E_{\\mathrm{J}}=B \\mathrm{~J}(\\mathrm{~J}+1)$, where $\\mathrm{J}$ is the rotational quantum number of the molecule and $B$ its rotational constant. Constant $B$ is related to the reduced mass $\\mu$ and the bond length $R$ of the molecule through the equation\n\n$$\nB=\\frac{h^{2}}{8 \\pi^{2} \\mu R^{2}} .\n$$\n\nIn general, spectroscopic transitions appear at photon energies which are equal to the energy difference between appropriate states of a molecule ( $h \\nu=\\Delta E$ ). The observed rotational transitions occur between adjacent rotational levels, hence $\\Delta E=E_{J+1}-E_{J}=$ $2 B(\\mathrm{~J}+1)$. Consequently, successive rotational transitions that appear on the spectrum (such as the one shown here) follow the equation $h(\\Delta v)=2 B$.\n\nBy inspecting the spectrum provided, determine the following quantities for ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$ with appropriate units:\n\n$R$\n\n[figure1]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nRotational energy levels of diatomic molecules are well described by the formula $E_{\\mathrm{J}}=B \\mathrm{~J}(\\mathrm{~J}+1)$, where $\\mathrm{J}$ is the rotational quantum number of the molecule and $B$ its rotational constant. Constant $B$ is related to the reduced mass $\\mu$ and the bond length $R$ of the molecule through the equation\n\n$$\nB=\\frac{h^{2}}{8 \\pi^{2} \\mu R^{2}} .\n$$\n\nIn general, spectroscopic transitions appear at photon energies which are equal to the energy difference between appropriate states of a molecule ( $h \\nu=\\Delta E$ ). The observed rotational transitions occur between adjacent rotational levels, hence $\\Delta E=E_{J+1}-E_{J}=$ $2 B(\\mathrm{~J}+1)$. Consequently, successive rotational transitions that appear on the spectrum (such as the one shown here) follow the equation $h(\\Delta v)=2 B$.\n\nBy inspecting the spectrum provided, determine the following quantities for ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$ with appropriate units:\n\n$R$\n\n[figure1]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\AA$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-469.jpg?height=995&width=1525&top_left_y=1664&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\AA$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1498",
        "problem": "How many $\\mathrm{cm}^{3}$ of $1.00 \\mathrm{M} \\mathrm{NaOH}$ solution must be added to $100.0 \\mathrm{~cm}^{3}$ of $0.100 \\mathrm{M}$ $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ solution to obtain a phosphate buffer solution with $p H$ of about 7.2 ? (The $p K$ values for $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ are $p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\nA: $5.0 \\mathrm{~cm}^{3}$\nB: $10.0 \\mathrm{~cm}^{3}$\nC: $15.0 \\mathrm{~mL}$\nD: $20.0 \\mathrm{~mL}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many $\\mathrm{cm}^{3}$ of $1.00 \\mathrm{M} \\mathrm{NaOH}$ solution must be added to $100.0 \\mathrm{~cm}^{3}$ of $0.100 \\mathrm{M}$ $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ solution to obtain a phosphate buffer solution with $p H$ of about 7.2 ? (The $p K$ values for $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ are $p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\n\nA: $5.0 \\mathrm{~cm}^{3}$\nB: $10.0 \\mathrm{~cm}^{3}$\nC: $15.0 \\mathrm{~mL}$\nD: $20.0 \\mathrm{~mL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1364",
        "problem": "The unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nThe first order rate constant for the $\\mathrm{Cl} 2$ protein folding reaction can be determined by following the fluorescence intensity when a sample of unfolded protein is allowed to refold (typically the $\\mathrm{pH}$ of the solution is changed). The concentration of protein when a $1.0 \\mu \\mathrm{M}$ sample of unfolded $\\mathrm{Cl} 2$ was allowed to refold was measured at a temperature of $25{ }^{\\circ} \\mathrm{C}$ :\n\n| time / ms | 0 | 10 | 20 | 30 | 40 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| concentration $/ \\mu \\mathrm{M}$ | 1 | 0.64 | 0.36 | 0.23 | 0.14 |At $20^{\\circ} \\mathrm{C}$ the rate constant for the protein folding reaction is $33 \\mathrm{~s}^{-1}$. Calculate the activation energy for the protein folding reaction.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe unfolding reaction for many small proteins can be represented by the equilibrium:\n\nFolded $\\rightleftharpoons$ Unfolded\n\nYou may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature $T_{m}$ is defined as the temperature at which half of the molecules are unfolded and half are folded.\n\nThe first order rate constant for the $\\mathrm{Cl} 2$ protein folding reaction can be determined by following the fluorescence intensity when a sample of unfolded protein is allowed to refold (typically the $\\mathrm{pH}$ of the solution is changed). The concentration of protein when a $1.0 \\mu \\mathrm{M}$ sample of unfolded $\\mathrm{Cl} 2$ was allowed to refold was measured at a temperature of $25{ }^{\\circ} \\mathrm{C}$ :\n\n| time / ms | 0 | 10 | 20 | 30 | 40 |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| concentration $/ \\mu \\mathrm{M}$ | 1 | 0.64 | 0.36 | 0.23 | 0.14 |\n\nproblem:\nAt $20^{\\circ} \\mathrm{C}$ the rate constant for the protein folding reaction is $33 \\mathrm{~s}^{-1}$. Calculate the activation energy for the protein folding reaction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_290",
        "problem": "Which of the following groups of ions and atoms is comprised of species having exactly the same ground state electron configuration?\nA: $\\mathrm{Al}^{3+}, \\mathrm{O}^{2-}, \\mathrm{Ne}, \\mathrm{Cl}^{-}$\nB: $\\mathrm{Ca}, \\mathrm{Ti}^{2+}, \\mathrm{Cl}^{-}, \\mathrm{S}^{2-}$\nC: $\\mathrm{H}^{-}, \\mathrm{He}, \\mathrm{Li}^{2} \\mathrm{Be}^{2+}$\nD: $\\mathrm{Ne}, \\mathrm{Ar}, \\mathrm{Kr}, \\mathrm{Xe}$\nE: $\\mathrm{Ca}^{2+}, \\mathrm{Ti}^{4+}, \\mathrm{Cl}^{-}, \\mathrm{S}^{2-}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following groups of ions and atoms is comprised of species having exactly the same ground state electron configuration?\n\nA: $\\mathrm{Al}^{3+}, \\mathrm{O}^{2-}, \\mathrm{Ne}, \\mathrm{Cl}^{-}$\nB: $\\mathrm{Ca}, \\mathrm{Ti}^{2+}, \\mathrm{Cl}^{-}, \\mathrm{S}^{2-}$\nC: $\\mathrm{H}^{-}, \\mathrm{He}, \\mathrm{Li}^{2} \\mathrm{Be}^{2+}$\nD: $\\mathrm{Ne}, \\mathrm{Ar}, \\mathrm{Kr}, \\mathrm{Xe}$\nE: $\\mathrm{Ca}^{2+}, \\mathrm{Ti}^{4+}, \\mathrm{Cl}^{-}, \\mathrm{S}^{2-}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1423",
        "problem": "1. Calculate the concentration (in $\\mathrm{mol} \\mathrm{dm}^{-3}$ ) of $\\mathrm{CO}_{2}$ dissolved in distilled water equilibrated with the atmosphere in the year 2020.The concentration of carbon dioxide in the atmosphere has increased substantially during this century and is predicted to continue to increase. The $\\left[\\mathrm{CO}_{2}\\right]$ is expected to be about $440 \\mathrm{ppm}\\left(440 \\times 10^{-6} \\mathrm{~atm}\\right)$ in the year 2020 .\n\nCalculate the $\\mathrm{pH}$-value of the solution in 1.\n\nData:\n\nHenry's Law constant for $\\mathrm{CO}_{2}$ at $298 \\mathrm{~K}: 0.0343 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~atm}^{-1}$\n\nThermodynamic values, in $\\mathrm{kJ} / \\mathrm{mol}$ at $298 \\mathrm{~K}$ are:\n\n|  | $\\Delta_{f} G^{0}$ | $\\Delta_{f} H^{0}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CO}_{2}(\\mathrm{aq})$ | -386.2 | -412.9 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -237.2 | -285.8 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -587.1 | -691.2 |\n| $\\mathrm{H}^{+}(\\mathrm{aq})$ | 0.00 | 0.00 |",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n1. Calculate the concentration (in $\\mathrm{mol} \\mathrm{dm}^{-3}$ ) of $\\mathrm{CO}_{2}$ dissolved in distilled water equilibrated with the atmosphere in the year 2020.\n\nproblem:\nThe concentration of carbon dioxide in the atmosphere has increased substantially during this century and is predicted to continue to increase. The $\\left[\\mathrm{CO}_{2}\\right]$ is expected to be about $440 \\mathrm{ppm}\\left(440 \\times 10^{-6} \\mathrm{~atm}\\right)$ in the year 2020 .\n\nCalculate the $\\mathrm{pH}$-value of the solution in 1.\n\nData:\n\nHenry's Law constant for $\\mathrm{CO}_{2}$ at $298 \\mathrm{~K}: 0.0343 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~atm}^{-1}$\n\nThermodynamic values, in $\\mathrm{kJ} / \\mathrm{mol}$ at $298 \\mathrm{~K}$ are:\n\n|  | $\\Delta_{f} G^{0}$ | $\\Delta_{f} H^{0}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CO}_{2}(\\mathrm{aq})$ | -386.2 | -412.9 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -237.2 | -285.8 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -587.1 | -691.2 |\n| $\\mathrm{H}^{+}(\\mathrm{aq})$ | 0.00 | 0.00 |\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_194",
        "problem": "When you do extreme exercise, your body converts glucose to lactic acid $\\left(\\mathrm{HCH}_{3} \\mathrm{H}_{5} \\mathrm{O}_{3}\\right)$. Lactic acid has a $\\mathrm{K}_{\\mathrm{a}}=1.38 \\times 10^{-4}$. Buffer systems maintain the $\\mathrm{pH}$ of your blood at 7.4 during exercise. Without buffers, what would your blood $\\mathrm{pH}$ range be at equilibrium if $4.00 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$ of lactic acid dissociated according to the equation:\n\n$$\n\\mathrm{HCH}_{3} \\mathrm{H}_{5} \\mathrm{O}_{3}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{H}_{5} \\mathrm{O}_{3}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq})\n$$\nA: $2<\\mathrm{pH}<3$\nB: $3<\\mathrm{pH}<4$\nC: $4<\\mathrm{pH}<5$\nD: $5<\\mathrm{pH}<6$\nE: $6<\\mathrm{pH}<7$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen you do extreme exercise, your body converts glucose to lactic acid $\\left(\\mathrm{HCH}_{3} \\mathrm{H}_{5} \\mathrm{O}_{3}\\right)$. Lactic acid has a $\\mathrm{K}_{\\mathrm{a}}=1.38 \\times 10^{-4}$. Buffer systems maintain the $\\mathrm{pH}$ of your blood at 7.4 during exercise. Without buffers, what would your blood $\\mathrm{pH}$ range be at equilibrium if $4.00 \\times 10^{-3} \\mathrm{~mol} \\mathrm{~L}^{-1}$ of lactic acid dissociated according to the equation:\n\n$$\n\\mathrm{HCH}_{3} \\mathrm{H}_{5} \\mathrm{O}_{3}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{H}_{5} \\mathrm{O}_{3}^{-}(\\mathrm{aq})+\\mathrm{H}_{3} \\mathrm{O}^{+}(\\mathrm{aq})\n$$\n\nA: $2<\\mathrm{pH}<3$\nB: $3<\\mathrm{pH}<4$\nC: $4<\\mathrm{pH}<5$\nD: $5<\\mathrm{pH}<6$\nE: $6<\\mathrm{pH}<7$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_888",
        "problem": "$\\mathrm{t}^{\\circ} \\mathrm{C}$ 时, 配制一组 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{HCl}$\n\n或 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{NaOH}$ 的混合溶液, 溶液中部分微粒浓度的负对数 $(-\\operatorname{lgc})$ 与 $\\mathrm{pH}$ 关系如图所示。下列说法错误的是\n\n[图1]\nA: $\\mathrm{pH}=\\mathrm{a}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: $\\mathrm{pH}=7$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nC: $\\mathrm{pH}=10.3$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $25^{\\circ} \\mathrm{C}$ 时, 反应 $\\mathrm{H}_{2} \\mathrm{CO}_{3}+\\mathrm{CO}_{3}^{2-} \\rightleftharpoons 2 \\mathrm{HCO}_{3}^{-}$的平衡常数为 $1.0 \\times 10^{4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{t}^{\\circ} \\mathrm{C}$ 时, 配制一组 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{HCl}$\n\n或 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 与 $\\mathrm{NaOH}$ 的混合溶液, 溶液中部分微粒浓度的负对数 $(-\\operatorname{lgc})$ 与 $\\mathrm{pH}$ 关系如图所示。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{pH}=\\mathrm{a}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: $\\mathrm{pH}=7$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nC: $\\mathrm{pH}=10.3$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<1.000 \\times 10^{-3} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $25^{\\circ} \\mathrm{C}$ 时, 反应 $\\mathrm{H}_{2} \\mathrm{CO}_{3}+\\mathrm{CO}_{3}^{2-} \\rightleftharpoons 2 \\mathrm{HCO}_{3}^{-}$的平衡常数为 $1.0 \\times 10^{4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-078.jpg?height=517&width=851&top_left_y=1632&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_289",
        "problem": "Sodium sulfite $\\left(\\mathrm{Na}_{2} \\mathrm{SO}_{3}\\right)$ reacts with hydrochloric acid to produce sodium chloride, sulfur dioxide and water. What volume (in $\\mathrm{mL}$ ) of $10.1 \\mathrm{~mol} \\mathrm{~L}^{-1}$ hydrochloric acid is required for complete reaction with $32.6 \\mathrm{~g}$ of sodium sulfite?\nA: $12.8 \\mathrm{~mL}$\nB: $17.1 \\mathrm{~mL}$\nC: $25.6 \\mathrm{~mL}$\nD: $51.2 \\mathrm{~mL}$\nE: $102.4 \\mathrm{~mL}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSodium sulfite $\\left(\\mathrm{Na}_{2} \\mathrm{SO}_{3}\\right)$ reacts with hydrochloric acid to produce sodium chloride, sulfur dioxide and water. What volume (in $\\mathrm{mL}$ ) of $10.1 \\mathrm{~mol} \\mathrm{~L}^{-1}$ hydrochloric acid is required for complete reaction with $32.6 \\mathrm{~g}$ of sodium sulfite?\n\nA: $12.8 \\mathrm{~mL}$\nB: $17.1 \\mathrm{~mL}$\nC: $25.6 \\mathrm{~mL}$\nD: $51.2 \\mathrm{~mL}$\nE: $102.4 \\mathrm{~mL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1212",
        "problem": "In Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nCalculate the equilibrium concentration of hydrogen ions in a $0.0100 \\mathrm{~m}$ aqueous solution of sodium carbonate saturated with carbon dioxide at a carbon dioxide partial pressure of 1.00 bar. Ignore water dissociation effects.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIn Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.\n\nCarbon dioxide, $\\mathrm{CO}_{2}$, is a diprotic acid in aqueous solution. The $\\mathrm{p} K_{\\mathrm{a}}$-values at $0{ }^{\\circ} \\mathrm{C}$ are:\n\n[figure1]\n\nThe liquid volume change associated with dissolution of $\\mathrm{CO}_{2}$ may be neglected for all of the following problems. The temperature is to be taken as being $0{ }^{\\circ} \\mathrm{C}$.\n\nCalculate the equilibrium concentration of hydrogen ions in a $0.0100 \\mathrm{~m}$ aqueous solution of sodium carbonate saturated with carbon dioxide at a carbon dioxide partial pressure of 1.00 bar. Ignore water dissociation effects.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-362.jpg?height=166&width=1354&top_left_y=1005&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1067",
        "problem": "The 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nBicarbonate of soda (sodium hydrogencarbonate) is often used to treat ant stings.\n\nGiven that the density of methanoic acid is $1.2 \\mathrm{~g} \\mathrm{~cm}^{-3}$, the 'typical ant' injects $7.8 \\times 10^{-5}$ moles of methanoic acid. What mass of sodium hydrogen carbonate would be needed to neutralise completely the sting from this ant?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nBicarbonate of soda (sodium hydrogencarbonate) is often used to treat ant stings.\n\nGiven that the density of methanoic acid is $1.2 \\mathrm{~g} \\mathrm{~cm}^{-3}$, the 'typical ant' injects $7.8 \\times 10^{-5}$ moles of methanoic acid. What mass of sodium hydrogen carbonate would be needed to neutralise completely the sting from this ant?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_4408eb29f8e936fbb54dg-05.jpg?height=622&width=654&top_left_y=237&top_left_x=1089"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1230",
        "problem": "[figure1]\n\nFigure 2. Energy diagram of atomic orbitals of helium when an electron resides in the 1 s orbital.\n\nFigure 2 shows an energy diagram of the atomic orbitals of helium. The arrows indicate the \"allowed\" transitions according to the spectroscopic principle.Argon is also found in minerals such as malacon. Which equation explains the occurrence of argon in rocks among $[A]$ to $[D]$ below? Mark one.\nA: $ \\mathrm{ArF}_{2} \\rightarrow \\mathrm{Ar}+\\mathrm{F}_{2}$\nB: $ \\operatorname{ArXe} \\rightarrow \\mathrm{Ar}+\\mathrm{Xe}$\nC: $\\quad{ }^{40} \\mathrm{~K} \\rightarrow{ }^{40} \\mathrm{Ar}+\\varepsilon / \\beta^{+}$(electron capture / positron emission)\nD: $ \\quad{ }^{126} \\mathrm{I} \\rightarrow{ }^{126} \\mathrm{Ar}+\\beta^{-}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n[figure1]\n\nFigure 2. Energy diagram of atomic orbitals of helium when an electron resides in the 1 s orbital.\n\nFigure 2 shows an energy diagram of the atomic orbitals of helium. The arrows indicate the \"allowed\" transitions according to the spectroscopic principle.\n\nproblem:\nArgon is also found in minerals such as malacon. Which equation explains the occurrence of argon in rocks among $[A]$ to $[D]$ below? Mark one.\n\nA: $ \\mathrm{ArF}_{2} \\rightarrow \\mathrm{Ar}+\\mathrm{F}_{2}$\nB: $ \\operatorname{ArXe} \\rightarrow \\mathrm{Ar}+\\mathrm{Xe}$\nC: $\\quad{ }^{40} \\mathrm{~K} \\rightarrow{ }^{40} \\mathrm{Ar}+\\varepsilon / \\beta^{+}$(electron capture / positron emission)\nD: $ \\quad{ }^{126} \\mathrm{I} \\rightarrow{ }^{126} \\mathrm{Ar}+\\beta^{-}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-058.jpg?height=771&width=876&top_left_y=1816&top_left_x=196"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_895",
        "problem": "当某反应物的初始浓度为 $0.04 \\mathrm{~mol} \\cdot \\mathrm{dm}^{-3}$ 时, 反应的半衰期为 $400 \\mathrm{~s}$, 初始浓度为 0.024 $\\mathrm{mol} \\cdot \\mathrm{dm}^{-3}$ 时，半衰期为 $240 \\mathrm{~s}$, 则此反应为\nA: 零级反应\nB: 1.5 级反应\nC: 二级反应\nD: 一级反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n当某反应物的初始浓度为 $0.04 \\mathrm{~mol} \\cdot \\mathrm{dm}^{-3}$ 时, 反应的半衰期为 $400 \\mathrm{~s}$, 初始浓度为 0.024 $\\mathrm{mol} \\cdot \\mathrm{dm}^{-3}$ 时，半衰期为 $240 \\mathrm{~s}$, 则此反应为\n\nA: 零级反应\nB: 1.5 级反应\nC: 二级反应\nD: 一级反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_167",
        "problem": "The $1^{\\text {st }}$ and $3^{\\text {rd }}$ ionization energies of aluminum are $577.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and $2744.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ respectively. What data best matches the $2^{\\text {nd }}$ and $4^{\\text {th }}$ ionization energies of aluminum?\nA: | 700 | 3500 |\nB: | 700 | 5000 |\nC: | 700 | 11000 |\nD: | 1800 | 11000 |\nE: | 1800 | 5000 |\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe $1^{\\text {st }}$ and $3^{\\text {rd }}$ ionization energies of aluminum are $577.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and $2744.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ respectively. What data best matches the $2^{\\text {nd }}$ and $4^{\\text {th }}$ ionization energies of aluminum?\n\nA: | 700 | 3500 |\nB: | 700 | 5000 |\nC: | 700 | 11000 |\nD: | 1800 | 11000 |\nE: | 1800 | 5000 |\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_913",
        "problem": "室温下, $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=4.6 \\times 10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=5.6 \\times 10^{-11}, \\mathrm{~K}_{\\mathrm{a}}\\left(\\mathrm{HNO}_{2}\\right)=5.1 \\times 10^{-4}$ 。纯碱溶液可用于吸收 $\\mathrm{NO}_{2}$, 有关的化学反应为: $2 \\mathrm{NO}_{2}+\\mathrm{Na}_{2} \\mathrm{CO}_{3}=\\mathrm{NaNO}_{2}+\\mathrm{NaNO}_{3}+\\mathrm{CO}_{2}$ 。若溶液混合引起的体积变化可忽略, 室温时下列指定溶液中微粒物质的量浓度关系正确的是\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液和 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HNO}_{2}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}=\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{NO}_{2}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaNO}_{2}$ 溶液比较: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}(\\mathrm{CO}$ $\\left.{ }_{3}^{2-}\\right)<\\mathrm{c}\\left(\\mathrm{NO}_{2}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液吸收等量 $\\mathrm{NO}_{2}$, 无气体逸出: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)>\\mathrm{c}(\\mathrm{NO}$ $$ \\left.{ }_{2}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right) $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n室温下, $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=4.6 \\times 10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)=5.6 \\times 10^{-11}, \\mathrm{~K}_{\\mathrm{a}}\\left(\\mathrm{HNO}_{2}\\right)=5.1 \\times 10^{-4}$ 。纯碱溶液可用于吸收 $\\mathrm{NO}_{2}$, 有关的化学反应为: $2 \\mathrm{NO}_{2}+\\mathrm{Na}_{2} \\mathrm{CO}_{3}=\\mathrm{NaNO}_{2}+\\mathrm{NaNO}_{3}+\\mathrm{CO}_{2}$ 。若溶液混合引起的体积变化可忽略, 室温时下列指定溶液中微粒物质的量浓度关系正确的是\n\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液和 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HNO}_{2}$ 溶液等体积混合: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}=\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{NO}_{2}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaNO}_{2}$ 溶液比较: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}(\\mathrm{CO}$ $\\left.{ }_{3}^{2-}\\right)<\\mathrm{c}\\left(\\mathrm{NO}_{2}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液吸收等量 $\\mathrm{NO}_{2}$, 无气体逸出: $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NO}_{3}^{-}\\right)>\\mathrm{c}(\\mathrm{NO}$ $$ \\left.{ }_{2}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right) $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1302",
        "problem": "Bridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate $\\Delta H$ (in $\\mathrm{kJ}$ ) for transformation of $1.00 \\mathrm{~kg}$ of $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ to hemihydrate $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$. Is this reaction endothermic or is it exothermic?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate $\\Delta H$ (in $\\mathrm{kJ}$ ) for transformation of $1.00 \\mathrm{~kg}$ of $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ to hemihydrate $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$. Is this reaction endothermic or is it exothermic?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-345.jpg?height=951&width=1445&top_left_y=595&top_left_x=314"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1073",
        "problem": "Give half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{H}_{2} \\mathrm{O}_{2}$ to $\\mathrm{O}_{2}$\nA: oxidation\nB: reduction\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGive half-equations for the following reactions in aqueous acid solution. In each case, indicate whether the reaction is an oxidation or a reduction.\n\n$\\mathrm{H}_{2} \\mathrm{O}_{2}$ to $\\mathrm{O}_{2}$\n\nA: oxidation\nB: reduction\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_978",
        "problem": "The bubbles in boiling water are mostly\nA: $\\mathrm{He}$\nB: $\\mathrm{H}_{2} \\mathrm{O}$\nC: $\\mathrm{CO}_{2}$\nD: $\\mathrm{N}_{2}$\nE: $\\mathrm{O}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe bubbles in boiling water are mostly\n\nA: $\\mathrm{He}$\nB: $\\mathrm{H}_{2} \\mathrm{O}$\nC: $\\mathrm{CO}_{2}$\nD: $\\mathrm{N}_{2}$\nE: $\\mathrm{O}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_832",
        "problem": "随着金属有机化学的发展和锂离子电池的应用, 非水体系的电化学研究越来越深入。水溶液中, 以标准氢电极电势 $(\\mathrm{SHE})$ 为零点。在非水体系中, 通常选用二茂铁 $\\mathrm{Fe}\\left(\\mathrm{C}_{5} \\mathrm{H}_{5}\\right)_{2}$ (简写为 $\\mathrm{Fc}$ )与其氧化态 (表示为 $\\mathrm{Fc}^{+}$)组成的电对 $\\left(\\right.$表示为 $\\mathrm{Fc}^{+/ 0}$ ) 作为内标,\n\n$\\mathrm{E}^{\\ominus}\\left(\\mathrm{Fc}^{+/ 0}\\right)=0.400 \\mathrm{~V}$ (对 $\\mathrm{SHE}$ )。 $\\mathrm{Fc}^{+/ 0}$ 电对的应用, 有效解决了有机溶剂中物质的电极电势、\n\n酸碱解离常数等参数测定的问题。 $25^{\\circ} \\mathrm{C}$, 在水溶液 $(\\mathrm{aq})$ 和乙腈 $(\\mathrm{MeCN})$ 中得到如下数据:\n\n| 电极反应(solv 指溶剂) | $\\mathrm{E}^{\\ominus}(\\mathrm{aq})$ <br> 对 <br> SHE | $\\mathrm{E}^{\\ominus}(\\mathrm{MeCl}$ <br> 对 $\\mathrm{Fc}^{+/ 0}$ |\n| :---: | :---: | :---: |\n| (1) <br> $2 \\mathrm{H}^{+}($solv $)+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{H}_{2}(\\mathrm{~g})$ | 0 | -0.028 |\n| $(2)$ <br> $\\mathrm{O}_{2}(\\mathrm{~g})+4 \\mathrm{H}^{+}($solv $)+4 \\mathrm{e}^{-}$ | +1.23 | +1.21 |\n| $(3)$ <br> $\\mathrm{I}_{2}$ (solv) $+2 \\mathrm{e}^{-} \\rightleftharpoons 2 \\mathrm{I}^{-}(\\mathrm{solv})$ | +0.54 | -0.14 |\n\n\n| (4) |  |  |\n| :--- | :--- | :--- |\n| $\\mathrm{Fe}\\left(\\mathrm{Cp}^{*}\\right)_{2}^{+}($solv $)+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Fe}$ |  |  |\n| $\\mathrm{Cp}^{*}=\\mathrm{C}_{5}\\left(\\mathrm{CH}_{3}\\right)_{5}$ | - | -0.48 |\n|  |  |  |\n\n乙腈溶液中保持 $\\mathrm{O}_{2}(\\mathrm{~g})$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 均为标态, 反应(2)的电极电势随 $\\mathrm{H}^{+}$浓度发生变化:\n\n$\\mathrm{E} / \\mathrm{V}=1.21-0.0592 \\mathrm{pH}$ 。在 $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ 的乙腈溶液中, 测得反应(2)的电极电势为 $0.92 \\mathrm{~V}$ 。则下列计算结果中, 不正确的是\nA: 水溶液中, $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{solv})$ 的标准电池电动势 $\\mathrm{E}^{\\ominus}=1.23 \\mathrm{~V}$\nB: 乙腈溶液中, $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{solv})$ 的标准电池电动势 $\\mathrm{E}^{\\ominus}=1.238 \\mathrm{~V}$\nC: $\\mathrm{HCl}$ 在乙腈中的解离常数 $\\mathrm{K}_{\\mathrm{a}}=1.7 \\times 10^{-9}$\nD: 在 $25^{\\circ} \\mathrm{C}$ 的乙腈溶液中, $\\mathrm{I}_{2}$ 与 $\\mathrm{Fe}\\left(\\mathrm{Cp}^{*}\\right)_{2}$ 反应的平衡常数 $\\mathrm{K}^{\\ominus}=3.1 \\times 10^{-11}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n随着金属有机化学的发展和锂离子电池的应用, 非水体系的电化学研究越来越深入。水溶液中, 以标准氢电极电势 $(\\mathrm{SHE})$ 为零点。在非水体系中, 通常选用二茂铁 $\\mathrm{Fe}\\left(\\mathrm{C}_{5} \\mathrm{H}_{5}\\right)_{2}$ (简写为 $\\mathrm{Fc}$ )与其氧化态 (表示为 $\\mathrm{Fc}^{+}$)组成的电对 $\\left(\\right.$表示为 $\\mathrm{Fc}^{+/ 0}$ ) 作为内标,\n\n$\\mathrm{E}^{\\ominus}\\left(\\mathrm{Fc}^{+/ 0}\\right)=0.400 \\mathrm{~V}$ (对 $\\mathrm{SHE}$ )。 $\\mathrm{Fc}^{+/ 0}$ 电对的应用, 有效解决了有机溶剂中物质的电极电势、\n\n酸碱解离常数等参数测定的问题。 $25^{\\circ} \\mathrm{C}$, 在水溶液 $(\\mathrm{aq})$ 和乙腈 $(\\mathrm{MeCN})$ 中得到如下数据:\n\n| 电极反应(solv 指溶剂) | $\\mathrm{E}^{\\ominus}(\\mathrm{aq})$ <br> 对 <br> SHE | $\\mathrm{E}^{\\ominus}(\\mathrm{MeCl}$ <br> 对 $\\mathrm{Fc}^{+/ 0}$ |\n| :---: | :---: | :---: |\n| (1) <br> $2 \\mathrm{H}^{+}($solv $)+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{H}_{2}(\\mathrm{~g})$ | 0 | -0.028 |\n| $(2)$ <br> $\\mathrm{O}_{2}(\\mathrm{~g})+4 \\mathrm{H}^{+}($solv $)+4 \\mathrm{e}^{-}$ | +1.23 | +1.21 |\n| $(3)$ <br> $\\mathrm{I}_{2}$ (solv) $+2 \\mathrm{e}^{-} \\rightleftharpoons 2 \\mathrm{I}^{-}(\\mathrm{solv})$ | +0.54 | -0.14 |\n\n\n| (4) |  |  |\n| :--- | :--- | :--- |\n| $\\mathrm{Fe}\\left(\\mathrm{Cp}^{*}\\right)_{2}^{+}($solv $)+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Fe}$ |  |  |\n| $\\mathrm{Cp}^{*}=\\mathrm{C}_{5}\\left(\\mathrm{CH}_{3}\\right)_{5}$ | - | -0.48 |\n|  |  |  |\n\n乙腈溶液中保持 $\\mathrm{O}_{2}(\\mathrm{~g})$ 和 $\\mathrm{H}_{2} \\mathrm{O}$ 均为标态, 反应(2)的电极电势随 $\\mathrm{H}^{+}$浓度发生变化:\n\n$\\mathrm{E} / \\mathrm{V}=1.21-0.0592 \\mathrm{pH}$ 。在 $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ 的乙腈溶液中, 测得反应(2)的电极电势为 $0.92 \\mathrm{~V}$ 。则下列计算结果中, 不正确的是\n\nA: 水溶液中, $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{solv})$ 的标准电池电动势 $\\mathrm{E}^{\\ominus}=1.23 \\mathrm{~V}$\nB: 乙腈溶液中, $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{solv})$ 的标准电池电动势 $\\mathrm{E}^{\\ominus}=1.238 \\mathrm{~V}$\nC: $\\mathrm{HCl}$ 在乙腈中的解离常数 $\\mathrm{K}_{\\mathrm{a}}=1.7 \\times 10^{-9}$\nD: 在 $25^{\\circ} \\mathrm{C}$ 的乙腈溶液中, $\\mathrm{I}_{2}$ 与 $\\mathrm{Fe}\\left(\\mathrm{Cp}^{*}\\right)_{2}$ 反应的平衡常数 $\\mathrm{K}^{\\ominus}=3.1 \\times 10^{-11}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_796",
        "problem": "一定条件下向某密闭容器中加入 $0.3 \\mathrm{molA} 、 0.1 \\mathrm{molC}$ 和一定量的 $\\mathrm{B}$ 三种气体, 图 1 表示各物质浓度随时间的变化关系, 图 2 表示速率随时间的变化关系, $t_{2} 、 t_{3} 、 t_{4} 、 t_{5}$ 时刻各改变一种条件, 且改变的条件均不同。若 $\\mathrm{t}_{4}$ 时刻改变的条件是压强, 则下列说法错误的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\nA: 若 $\\mathrm{t}_{1}=15$, 则前 $15 \\mathrm{~s}$ 的平均反应速率 $\\mathrm{v}(\\mathrm{C})=0.004 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{s}^{-1}$\nB: 该反应的化学方程式为 $3 \\mathrm{~A}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{B}(\\mathrm{g})+2 \\mathrm{C}(\\mathrm{g})$\nC: $t_{2} 、 t_{3} 、 t_{5}$ 时刻改变的条件分别是升高温度、加入催化剂、增大反应物浓度\nD: B 的起始物质的量为 $0.04 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一定条件下向某密闭容器中加入 $0.3 \\mathrm{molA} 、 0.1 \\mathrm{molC}$ 和一定量的 $\\mathrm{B}$ 三种气体, 图 1 表示各物质浓度随时间的变化关系, 图 2 表示速率随时间的变化关系, $t_{2} 、 t_{3} 、 t_{4} 、 t_{5}$ 时刻各改变一种条件, 且改变的条件均不同。若 $\\mathrm{t}_{4}$ 时刻改变的条件是压强, 则下列说法错误的是\n\n[图1]\n\n图1\n\n[图2]\n\n图2\n\nA: 若 $\\mathrm{t}_{1}=15$, 则前 $15 \\mathrm{~s}$ 的平均反应速率 $\\mathrm{v}(\\mathrm{C})=0.004 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{s}^{-1}$\nB: 该反应的化学方程式为 $3 \\mathrm{~A}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{B}(\\mathrm{g})+2 \\mathrm{C}(\\mathrm{g})$\nC: $t_{2} 、 t_{3} 、 t_{5}$ 时刻改变的条件分别是升高温度、加入催化剂、增大反应物浓度\nD: B 的起始物质的量为 $0.04 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-031.jpg?height=415&width=671&top_left_y=158&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-031.jpg?height=394&width=551&top_left_y=177&top_left_x=1049"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_325",
        "problem": "The reaction below was studied using the method of initial rates.\n\n$$\n2 \\mathrm{HgCl}_{2}(\\mathrm{aq})+\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}(\\mathrm{aq}) \\rightarrow \\text { products }\n$$\n\nThe following data were recorded. (Rate refers to the initial rate of consumption of $\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$.)\n\n| Experiment | Initial $\\left[\\mathrm{HgCl}_{2}\\right]$ <br> (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ) | Initial $\\left.\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$ <br> (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ) | Rate <br> (in $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{hr}^{-1}$ ) |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.0836 | 0.202 | 0.260 |\n| 2 | 0.0836 | 0.404 | 1.04 |\n| 3 | 0.0334 | 0.404 | 0.416 |\n\nWhat is the rate law for the reaction?\nA: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]^{2}$\nB: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]^{2}\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$\nC: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$\nD: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]^{2}\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]^{2}$\nE: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]^{1 / 2}\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction below was studied using the method of initial rates.\n\n$$\n2 \\mathrm{HgCl}_{2}(\\mathrm{aq})+\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}(\\mathrm{aq}) \\rightarrow \\text { products }\n$$\n\nThe following data were recorded. (Rate refers to the initial rate of consumption of $\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}$.)\n\n| Experiment | Initial $\\left[\\mathrm{HgCl}_{2}\\right]$ <br> (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ) | Initial $\\left.\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$ <br> (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ) | Rate <br> (in $\\mathrm{mol} \\mathrm{L}^{-1} \\mathrm{hr}^{-1}$ ) |\n| :---: | :---: | :---: | :---: |\n| 1 | 0.0836 | 0.202 | 0.260 |\n| 2 | 0.0836 | 0.404 | 1.04 |\n| 3 | 0.0334 | 0.404 | 0.416 |\n\nWhat is the rate law for the reaction?\n\nA: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]^{2}$\nB: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]^{2}\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$\nC: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$\nD: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]^{2}\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]^{2}$\nE: Rate $=k\\left[\\mathrm{HgCl}_{2}\\right]^{1 / 2}\\left[\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right]$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_880",
        "problem": "已知氯酸钾和和亚硫酸氢钾可以发生氧化还原反应(无污染性气体产生), 且反应中被氧化被还原的元素个数都只有一个。若该反应的反应速率会随着 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$的提升而加快,\n下图为 $\\mathrm{ClO}_{3}^{-}$在单位时间内物质的量浓度变化表示的反应速率-时间图象, 下列说法正确的是\n\n[图1]\nA: 反应进行一段时间后速率下降可能是因为 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$的浓度下降导致。\nB: 纵坐标为 $\\mathrm{v}\\left(\\mathrm{H}^{+}\\right)$的 $\\mathrm{v}-\\mathrm{t}$ 曲线在通过平移后可与图中图象完全重合。\nC: 图中阴影部分的面积表示 $\\mathrm{t}_{1} \\sim \\mathrm{t}_{2}$ 时间内 $\\mathrm{ClO}_{3}^{-}$的物质的量的减少量。\nD: 系数为最简整数比时, 每摩尔反应转移 $6 \\mathrm{~mol} \\mathrm{e}$-并产生 $3 \\mathrm{~mol}$ 质子。\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知氯酸钾和和亚硫酸氢钾可以发生氧化还原反应(无污染性气体产生), 且反应中被氧化被还原的元素个数都只有一个。若该反应的反应速率会随着 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$的提升而加快,\n下图为 $\\mathrm{ClO}_{3}^{-}$在单位时间内物质的量浓度变化表示的反应速率-时间图象, 下列说法正确的是\n\n[图1]\n\nA: 反应进行一段时间后速率下降可能是因为 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$的浓度下降导致。\nB: 纵坐标为 $\\mathrm{v}\\left(\\mathrm{H}^{+}\\right)$的 $\\mathrm{v}-\\mathrm{t}$ 曲线在通过平移后可与图中图象完全重合。\nC: 图中阴影部分的面积表示 $\\mathrm{t}_{1} \\sim \\mathrm{t}_{2}$ 时间内 $\\mathrm{ClO}_{3}^{-}$的物质的量的减少量。\nD: 系数为最简整数比时, 每摩尔反应转移 $6 \\mathrm{~mol} \\mathrm{e}$-并产生 $3 \\mathrm{~mol}$ 质子。\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-054.jpg?height=437&width=588&top_left_y=364&top_left_x=354"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_529",
        "problem": "向 $1 \\mathrm{~L}$ 含 $0.01 \\mathrm{molNaAlO}_{2}$ 和 $0.02 \\mathrm{molNaOH}$ 的溶液中缓慢通入二氧化碳, 随 $n\\left(\\mathrm{CO}_{2}\\right)$ 增大，先后发生三个不同的反应。下列对应关系正确的是\n\n| 选项 | $n\\left(\\mathrm{CO}_{2}\\right) / \\mathrm{mol}$ | 溶液中离子的物质的量浓度 |\n| :--- | :--- | :--- |\n| $\\mathrm{A}$ | 0 | $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{AlO}_{2}^{-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$ |\n| $\\mathrm{B}$ | 0.01 | $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{AlO}_{2}^{-}\\right)>$ <br> $c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{CO}_{3}^{2-}\\right)$ |\n| $C$ | 0.015 | $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{CO}_{3}^{2-}\\right)>$ $c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{HCO}_{3}^{-}\\right)$ |\n| :--- | :--- | :--- |\n| $\\mathrm{D}$ | 0.03 | $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{HCO}_{3}^{-}\\right)+$ $c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)$ |\nA: A\nB: B\nC: $C$\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n向 $1 \\mathrm{~L}$ 含 $0.01 \\mathrm{molNaAlO}_{2}$ 和 $0.02 \\mathrm{molNaOH}$ 的溶液中缓慢通入二氧化碳, 随 $n\\left(\\mathrm{CO}_{2}\\right)$ 增大，先后发生三个不同的反应。下列对应关系正确的是\n\n| 选项 | $n\\left(\\mathrm{CO}_{2}\\right) / \\mathrm{mol}$ | 溶液中离子的物质的量浓度 |\n| :--- | :--- | :--- |\n| $\\mathrm{A}$ | 0 | $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{AlO}_{2}^{-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$ |\n| $\\mathrm{B}$ | 0.01 | $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{AlO}_{2}^{-}\\right)>$ <br> $c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{CO}_{3}^{2-}\\right)$ |\n| $C$ | 0.015 | $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{CO}_{3}^{2-}\\right)>$ $c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{HCO}_{3}^{-}\\right)$ |\n| :--- | :--- | :--- |\n| $\\mathrm{D}$ | 0.03 | $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{HCO}_{3}^{-}\\right)+$ $c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)$ |\n\nA: A\nB: B\nC: $C$\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_944",
        "problem": "测定溶液中乙二醛(含少量硝酸)含量的方法如下。\n\ni. 取 $\\mathrm{V} \\mathrm{mL}$ 待测溶液于雉形瓶中, 加入 2 滴酚酞溶液, 用浓度为 $0.20 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定至溶液恰好由无色变为粉红色，中和硝酸。\nii. 向 i 所得溶液加入过量 $0.50 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液 $\\mathrm{V}_{1} \\mathrm{~mL}$, 充全反应, 使乙二醛反应生成 $\\mathrm{CH}_{2} \\mathrm{OHCOONa}$ 。\n\niii. 用 $0.50 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 硫酸滴定 ii 中溶液至终点, 消耗硫酸体积为 $\\mathrm{V}_{2} \\mathrm{~mL}$ 。\n\n下列说法不正确的是\nA: 根据结构分析, pKa: $\\mathrm{CH}_{2} \\mathrm{OHCOOH}<\\mathrm{CH}_{3} \\mathrm{COOH}$\nB: 乙二醛反应生成 $\\mathrm{CH}_{2} \\mathrm{OHCOONa}$ 属于氧化还原反应\nC: 待测溶液中乙二醛浓度为 $\\left(0.5 \\mathrm{~V}_{1}-\\mathrm{V}_{2}\\right) / \\mathrm{V} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\nD: 若不进行步骤 $\\mathrm{i}$, 测得溶液中乙二醛浓度偏低\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n测定溶液中乙二醛(含少量硝酸)含量的方法如下。\n\ni. 取 $\\mathrm{V} \\mathrm{mL}$ 待测溶液于雉形瓶中, 加入 2 滴酚酞溶液, 用浓度为 $0.20 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定至溶液恰好由无色变为粉红色，中和硝酸。\nii. 向 i 所得溶液加入过量 $0.50 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液 $\\mathrm{V}_{1} \\mathrm{~mL}$, 充全反应, 使乙二醛反应生成 $\\mathrm{CH}_{2} \\mathrm{OHCOONa}$ 。\n\niii. 用 $0.50 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 硫酸滴定 ii 中溶液至终点, 消耗硫酸体积为 $\\mathrm{V}_{2} \\mathrm{~mL}$ 。\n\n下列说法不正确的是\n\nA: 根据结构分析, pKa: $\\mathrm{CH}_{2} \\mathrm{OHCOOH}<\\mathrm{CH}_{3} \\mathrm{COOH}$\nB: 乙二醛反应生成 $\\mathrm{CH}_{2} \\mathrm{OHCOONa}$ 属于氧化还原反应\nC: 待测溶液中乙二醛浓度为 $\\left(0.5 \\mathrm{~V}_{1}-\\mathrm{V}_{2}\\right) / \\mathrm{V} \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\nD: 若不进行步骤 $\\mathrm{i}$, 测得溶液中乙二醛浓度偏低\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1393",
        "problem": "The $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\nModel B attempts to analyse the probability that the dust particles carry 0,1 or $2 \\mathrm{H}$ atoms. The three states are linked by the following reaction scheme. The assumption is made that no more than 2 atoms may be adsorbed simultaneously.\n\n[figure1]\n\n$x_{0}, x_{1}$ and $x_{2}$ are the fractions of dust particles existing in state 0,1 or 2 , respectively. These fractions may be treated in the same way as concentrations in the following kinetic analysis. For a system in state $m$ with fraction $x_{m}$, the rates of the three possible processes are\n\nAdsorption $(m \\rightarrow m+1):$ rate $=k_{\\mathrm{a}}[\\mathrm{H}] x_{m}$\n\nDesorption $(m \\rightarrow m-1):$ rate $=k_{\\mathrm{d}} m x_{m}$\n\nReaction $(m \\rightarrow m-2)$ : rate $=1 / 2 k_{\\mathrm{r}} m(m-1) x_{m}$Assuming steady-state conditions, use the above rate equations to find expressions for the ratios $x_{2} / x_{1}$ and $x_{1} / x_{0}$, and evaluate these ratios.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe $\\mathrm{H}$ atoms are mobile on the surface. When they meet they react to form $\\mathrm{H}_{2}$, which then desorbs. The two kinetic models under consideration differ in the way the reaction is modelled, but share the same rate constants $k_{a}, k_{d}$, and $k_{r}$, for adsorption, desorption, and bimolecular reaction, as given below.\n\n$$\n\\begin{aligned}\n& k_{a}=1.4 \\cdot 10^{-5} \\mathrm{~cm}^{3} \\mathrm{~s}^{-1} \\\\\n& k_{d}=1.9 \\cdot 10^{-3} \\mathrm{~s}^{-1} \\\\\n& k_{r}=5.1 \\cdot 10^{4} \\mathrm{~s}^{-1}\n\\end{aligned}\n$$\n\nModel B attempts to analyse the probability that the dust particles carry 0,1 or $2 \\mathrm{H}$ atoms. The three states are linked by the following reaction scheme. The assumption is made that no more than 2 atoms may be adsorbed simultaneously.\n\n[figure1]\n\n$x_{0}, x_{1}$ and $x_{2}$ are the fractions of dust particles existing in state 0,1 or 2 , respectively. These fractions may be treated in the same way as concentrations in the following kinetic analysis. For a system in state $m$ with fraction $x_{m}$, the rates of the three possible processes are\n\nAdsorption $(m \\rightarrow m+1):$ rate $=k_{\\mathrm{a}}[\\mathrm{H}] x_{m}$\n\nDesorption $(m \\rightarrow m-1):$ rate $=k_{\\mathrm{d}} m x_{m}$\n\nReaction $(m \\rightarrow m-2)$ : rate $=1 / 2 k_{\\mathrm{r}} m(m-1) x_{m}$\n\nproblem:\nAssuming steady-state conditions, use the above rate equations to find expressions for the ratios $x_{2} / x_{1}$ and $x_{1} / x_{0}$, and evaluate these ratios.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-015.jpg?height=263&width=357&top_left_y=1205&top_left_x=861"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1026",
        "problem": "The 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nWhen you are bitten by an ant it does not inject you with all of its methanoic acid but keeps a little in reserve. Assuming a 'typical ant' injects $80 \\%$ of its methanoic acid, the total volume of pure methanoic acid contained in a 'typical ant' is $3.75 \\times 10^{-3} \\mathrm{~cm}^{3}$. Then how many 'typical ant' ants would have to be distilled to produce $1.0 \\mathrm{dm}^{3}$ of pure methanoic acid?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nWhen you are bitten by an ant it does not inject you with all of its methanoic acid but keeps a little in reserve. Assuming a 'typical ant' injects $80 \\%$ of its methanoic acid, the total volume of pure methanoic acid contained in a 'typical ant' is $3.75 \\times 10^{-3} \\mathrm{~cm}^{3}$. Then how many 'typical ant' ants would have to be distilled to produce $1.0 \\mathrm{dm}^{3}$ of pure methanoic acid?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_4408eb29f8e936fbb54dg-05.jpg?height=622&width=654&top_left_y=237&top_left_x=1089"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_25",
        "problem": "Weighed samples of iron are added in increments to a given amount of liquid bromine and allowed to react completely. The reaction produces a single product that is isolated and weighed. Which statement best describes the experiments summarized by the graph?\n\n[figure1]\nA: When $1.00 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, $\\mathrm{Fe}$ is the limiting reactant.\nB: When $2.00 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, $10.6 \\mathrm{~g}$ of $\\mathrm{FeBr}_{2}$ is formed.\nC: When $2.50 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, both reactants are used completely.\nD: When $3.50 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, there is an excess of $\\mathrm{Br}_{2}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWeighed samples of iron are added in increments to a given amount of liquid bromine and allowed to react completely. The reaction produces a single product that is isolated and weighed. Which statement best describes the experiments summarized by the graph?\n\n[figure1]\n\nA: When $1.00 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, $\\mathrm{Fe}$ is the limiting reactant.\nB: When $2.00 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, $10.6 \\mathrm{~g}$ of $\\mathrm{FeBr}_{2}$ is formed.\nC: When $2.50 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, both reactants are used completely.\nD: When $3.50 \\mathrm{~g}$ of $\\mathrm{Fe}$ is added, there is an excess of $\\mathrm{Br}_{2}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_f219c4a43c56eb2776d8g-03.jpg?height=428&width=485&top_left_y=1342&top_left_x=386"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1120",
        "problem": "A typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nDuring the test a person swabs their nose/throat and places it in the extraction solution. The resulting solution is called the test solution. A test sample of a few drops $\\left(0.10 \\mathrm{~cm}^{3}\\right)$ of the test solution are placed on the sample pad.\n\nIf the person taking the test has COVID-19, there will typically be $7.1 \\times 10^{6}$ virus particles per $\\mathrm{cm}^{3}$ in the test solution. Each virus particle has approximately 20 spike proteins on its surface.\n\nWhat is the concentration of spike proteins, [SP], in the test sample placed on the pad in mol dm ${ }^{-3}$ ?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nA typical lateral flow test works by a sample liquid flowing along a paper medium (from left-to-right in the picture). Lateral flow tests for COVID-19 use the strong binding between spike proteins on the surface of virus particles and antibodies to detect coronavirus. The red/pink colour in many lateral flow tests is from gold nanoparticles coated with antibodies.\n\n[figure1]\n\nTo use a particular lateral flow test, a swab is taken and placed in $1.0 \\mathrm{~cm}^{3}$ extraction solution containing $7.3 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$ and $4.6 \\mathrm{mmol} \\mathrm{dm}^{-3} \\mathrm{KH}_{2} \\mathrm{PO}_{4}$ at pH 7.4 at $25^{\\circ} \\mathrm{C}$.\n\nDuring the test a person swabs their nose/throat and places it in the extraction solution. The resulting solution is called the test solution. A test sample of a few drops $\\left(0.10 \\mathrm{~cm}^{3}\\right)$ of the test solution are placed on the sample pad.\n\nIf the person taking the test has COVID-19, there will typically be $7.1 \\times 10^{6}$ virus particles per $\\mathrm{cm}^{3}$ in the test solution. Each virus particle has approximately 20 spike proteins on its surface.\n\nWhat is the concentration of spike proteins, [SP], in the test sample placed on the pad in mol dm ${ }^{-3}$ ?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_88f0e5267ac93b183637g-08.jpg?height=146&width=940&top_left_y=495&top_left_x=929"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_731",
        "problem": "电化学脱硫在金属冶炼和废水处理中均有应用。一种电化学脱硫工作原理示意图如图所示。该装置工作时, 下列说法正确的是\n\n[图1]\nA: a 为直流电源负极\nB: 阴极区溶液 $\\mathrm{pH}$ 减小\nC: $\\mathrm{Mn}^{2+} 、 \\mathrm{Mn}^{3+}$ 之间转化可加快电子转移\nD: 导线中流过 $4.5 \\mathrm{~mole}$-同时阳极区溶液质量增加 $44 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n电化学脱硫在金属冶炼和废水处理中均有应用。一种电化学脱硫工作原理示意图如图所示。该装置工作时, 下列说法正确的是\n\n[图1]\n\nA: a 为直流电源负极\nB: 阴极区溶液 $\\mathrm{pH}$ 减小\nC: $\\mathrm{Mn}^{2+} 、 \\mathrm{Mn}^{3+}$ 之间转化可加快电子转移\nD: 导线中流过 $4.5 \\mathrm{~mole}$-同时阳极区溶液质量增加 $44 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-77.jpg?height=426&width=877&top_left_y=161&top_left_x=361"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_884",
        "problem": "$25^{\\circ} \\mathrm{C}$ 将浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{HA}$ 溶液和 $\\mathrm{BOH}$ 溶液按体积分别为 $\\mathrm{V}_{\\mathrm{a}}$ 和 $\\mathrm{V}_{\\mathrm{b}}$ 混合, 保\n\n持 $V_{a}+V_{b}=100 m L$, 且生成的 $B A$ 可溶于水。已知 $V_{a} 、 V_{b}$ 与混合液 $p H$ 关系如图。下列说法正确的是\n\n$\\mathrm{V} / \\mathrm{mL}$\n\n[图1]\nA: 电离平衡常数 $\\mathrm{K}(\\mathrm{HA})>\\mathrm{K}(\\mathrm{BOH})$\nB: 水的电离程度: $\\mathrm{Z}>\\mathrm{Y}>\\mathrm{X}$\nC: $x$ 点存在: $2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}(\\mathrm{HA})=\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)+3 \\mathrm{c}(\\mathrm{BOH})+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $\\mathrm{z}$ 点存在: $\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)>\\mathrm{c}(\\mathrm{HA})>\\mathrm{c}(\\mathrm{BOH})>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 将浓度均为 $0.1 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{HA}$ 溶液和 $\\mathrm{BOH}$ 溶液按体积分别为 $\\mathrm{V}_{\\mathrm{a}}$ 和 $\\mathrm{V}_{\\mathrm{b}}$ 混合, 保\n\n持 $V_{a}+V_{b}=100 m L$, 且生成的 $B A$ 可溶于水。已知 $V_{a} 、 V_{b}$ 与混合液 $p H$ 关系如图。下列说法正确的是\n\n$\\mathrm{V} / \\mathrm{mL}$\n\n[图1]\n\nA: 电离平衡常数 $\\mathrm{K}(\\mathrm{HA})>\\mathrm{K}(\\mathrm{BOH})$\nB: 水的电离程度: $\\mathrm{Z}>\\mathrm{Y}>\\mathrm{X}$\nC: $x$ 点存在: $2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+2 \\mathrm{c}(\\mathrm{HA})=\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)+3 \\mathrm{c}(\\mathrm{BOH})+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: $\\mathrm{z}$ 点存在: $\\mathrm{c}\\left(\\mathrm{B}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{-}\\right)>\\mathrm{c}(\\mathrm{HA})>\\mathrm{c}(\\mathrm{BOH})>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-10.jpg?height=417&width=731&top_left_y=817&top_left_x=363"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_349",
        "problem": "An element is a solid at room temperature but soft enough to be cut with an ordinary knife. When placed in water, the element reacts violently. What element is it?\nA: $\\mathrm{Na}$\nB: $\\mathrm{Mg}$\nC: $\\mathrm{Cu}$\nD: $\\mathrm{Hg}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn element is a solid at room temperature but soft enough to be cut with an ordinary knife. When placed in water, the element reacts violently. What element is it?\n\nA: $\\mathrm{Na}$\nB: $\\mathrm{Mg}$\nC: $\\mathrm{Cu}$\nD: $\\mathrm{Hg}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1450",
        "problem": "A less common agent for iodizing salt is potassium iodide, which cannot be easily measured by iodometric titration.\n\nOne possible method for analyzing iodide in the presence of chloride is potentiometric titration. However, this method is not very precise in the presence of large amounts of chloride. In this method, a silver wire is immersed in the solution (containing iodide and chloride) to be analyzed and silver ion is gradually added to the solution. The potential of the silver wire is measured relative to a reference electrode consisting of a silver wire in a solution of $\\mathrm{AgNO}_{3}\\left(c=1.000\\right.$ mol dm$\\left.{ }^{-3}\\right)$. The measured potentials are negative and the absolute values of these potentials are reported. The solution to be analysed has a volume of $1.000 \\mathrm{dm}^{3}$ (which you may assume does not change as silver ion is added) and $t=25.0^{\\circ} \\mathrm{C}$.\n\nThe results of this experiment are governed by three equilibria: the solubility of $\\mathrm{Agl}(\\mathrm{s})\\left[K_{\\mathrm{spl}}\\right]$ and $\\mathrm{AgCl}(\\mathrm{s})\\left[K_{\\mathrm{spCl}}\\right]$ and the formation of $\\mathrm{AgCl}_{2}{ }^{-}(\\mathrm{aq})\\left[K_{\\mathrm{f}}\\right]$. (Iodide also forms complex ions with silver but this may be neglected at the very low concentrations of iodide present in this experiment).\n\n$$\n\\begin{array}{ll}\n\\mathrm{Agl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{I}^{-}(\\mathrm{aq}) & K_{\\mathrm{spl}} \\\\\n\\mathrm{AgCl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) & K_{\\mathrm{spCl}} \\\\\n\\mathrm{Ag}^{+}(\\mathrm{aq})+2 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{AgCl}_{2}^{-}(\\mathrm{aq}) & K_{\\mathrm{f}}\n\\end{array}\n$$\n\nBelow are shown the results of two experiments measuring the observed potential as a function of added number of moles of silver ion. Experiment $\\mathbf{A}$ (solid circles) was carried out with $1.000 \\mathrm{dm}^{3}$ of iodide solution with concentration $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and containing no chloride ion. Experiment $\\mathbf{B}$ (open circles) was done using $1.000 \\mathrm{dm}^{3}$ of solution with a concentration of iodide $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and that of chloride equal to $1.00 \\times 10^{-1} \\mathrm{~mol} \\mathrm{dm}^{-3}$.\n\n[figure1]\n\n| $\\mu \\mathrm{mol} \\mathrm{Ag}^{+}$ <br> added | $\\|E\\|, \\mathrm{V}$ <br> exp. A | $\\|E\\|, \\mathrm{V}$ <br> exp. B |\n| :--- | :--- | :--- |\n| 1.00 | 0.637 | 0.637 |\n| 3.00 | 0.631 | 0.631 |\n| 5.00 | 0.622 | 0.622 |\n| 7.00 | 0.609 | 0.610 |\n| 9.00 | 0.581 | 0.584 |\n| 10.0 | 0.468 | 0.558 |\n| 11.0 | 0.355 | 0.531 |\n| 12.0 | 0.337 | 0.517 |\n| 13.0 | 0.327 | 0.517 |\n| 15.0 | 0.313 | 0.517 |Select an appropriate data from the experiments and use it to calculate the solubility product of Agl, ($\\left.K_{\\mathrm{spl} .}\\right)$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nA less common agent for iodizing salt is potassium iodide, which cannot be easily measured by iodometric titration.\n\nOne possible method for analyzing iodide in the presence of chloride is potentiometric titration. However, this method is not very precise in the presence of large amounts of chloride. In this method, a silver wire is immersed in the solution (containing iodide and chloride) to be analyzed and silver ion is gradually added to the solution. The potential of the silver wire is measured relative to a reference electrode consisting of a silver wire in a solution of $\\mathrm{AgNO}_{3}\\left(c=1.000\\right.$ mol dm$\\left.{ }^{-3}\\right)$. The measured potentials are negative and the absolute values of these potentials are reported. The solution to be analysed has a volume of $1.000 \\mathrm{dm}^{3}$ (which you may assume does not change as silver ion is added) and $t=25.0^{\\circ} \\mathrm{C}$.\n\nThe results of this experiment are governed by three equilibria: the solubility of $\\mathrm{Agl}(\\mathrm{s})\\left[K_{\\mathrm{spl}}\\right]$ and $\\mathrm{AgCl}(\\mathrm{s})\\left[K_{\\mathrm{spCl}}\\right]$ and the formation of $\\mathrm{AgCl}_{2}{ }^{-}(\\mathrm{aq})\\left[K_{\\mathrm{f}}\\right]$. (Iodide also forms complex ions with silver but this may be neglected at the very low concentrations of iodide present in this experiment).\n\n$$\n\\begin{array}{ll}\n\\mathrm{Agl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{I}^{-}(\\mathrm{aq}) & K_{\\mathrm{spl}} \\\\\n\\mathrm{AgCl}(\\mathrm{s}) \\rightleftharpoons \\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{Cl}^{-}(\\mathrm{aq}) & K_{\\mathrm{spCl}} \\\\\n\\mathrm{Ag}^{+}(\\mathrm{aq})+2 \\mathrm{Cl}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{AgCl}_{2}^{-}(\\mathrm{aq}) & K_{\\mathrm{f}}\n\\end{array}\n$$\n\nBelow are shown the results of two experiments measuring the observed potential as a function of added number of moles of silver ion. Experiment $\\mathbf{A}$ (solid circles) was carried out with $1.000 \\mathrm{dm}^{3}$ of iodide solution with concentration $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and containing no chloride ion. Experiment $\\mathbf{B}$ (open circles) was done using $1.000 \\mathrm{dm}^{3}$ of solution with a concentration of iodide $1.00 \\times 10^{-5} \\mathrm{~mol} \\mathrm{dm}^{-3}$ and that of chloride equal to $1.00 \\times 10^{-1} \\mathrm{~mol} \\mathrm{dm}^{-3}$.\n\n[figure1]\n\n| $\\mu \\mathrm{mol} \\mathrm{Ag}^{+}$ <br> added | $\\|E\\|, \\mathrm{V}$ <br> exp. A | $\\|E\\|, \\mathrm{V}$ <br> exp. B |\n| :--- | :--- | :--- |\n| 1.00 | 0.637 | 0.637 |\n| 3.00 | 0.631 | 0.631 |\n| 5.00 | 0.622 | 0.622 |\n| 7.00 | 0.609 | 0.610 |\n| 9.00 | 0.581 | 0.584 |\n| 10.0 | 0.468 | 0.558 |\n| 11.0 | 0.355 | 0.531 |\n| 12.0 | 0.337 | 0.517 |\n| 13.0 | 0.327 | 0.517 |\n| 15.0 | 0.313 | 0.517 |\n\nproblem:\nSelect an appropriate data from the experiments and use it to calculate the solubility product of Agl, ($\\left.K_{\\mathrm{spl} .}\\right)$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-145.jpg?height=968&width=1051&top_left_y=1361&top_left_x=180"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_488",
        "problem": "以食盐等为原料制备六水合高氯酸铜 $\\left[\\mathrm{Cu}\\left(\\mathrm{ClO}_{4}\\right)_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}\\right]$ 的一种工艺流程如下:\n\n[图1]\n\n下列说法正确的是\nA: “电解I”时阳极可用不锈钢材质\nB: “歧化反应”的产物之一为 $\\mathrm{NaClO}_{4}$\nC: “操作 $\\mathrm{a}$ ”是过滤\nD: “反应II”的离子方程式为 $\\mathrm{Cu}_{2}(\\mathrm{OH})_{2} \\mathrm{CO}_{3}+4 \\mathrm{H}^{+}=2 \\mathrm{Cu}^{2+}+\\mathrm{CO}_{2} \\uparrow+3 \\mathrm{H}_{2} \\mathrm{O}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n以食盐等为原料制备六水合高氯酸铜 $\\left[\\mathrm{Cu}\\left(\\mathrm{ClO}_{4}\\right)_{2} \\cdot 6 \\mathrm{H}_{2} \\mathrm{O}\\right]$ 的一种工艺流程如下:\n\n[图1]\n\n下列说法正确的是\n\nA: “电解I”时阳极可用不锈钢材质\nB: “歧化反应”的产物之一为 $\\mathrm{NaClO}_{4}$\nC: “操作 $\\mathrm{a}$ ”是过滤\nD: “反应II”的离子方程式为 $\\mathrm{Cu}_{2}(\\mathrm{OH})_{2} \\mathrm{CO}_{3}+4 \\mathrm{H}^{+}=2 \\mathrm{Cu}^{2+}+\\mathrm{CO}_{2} \\uparrow+3 \\mathrm{H}_{2} \\mathrm{O}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-79.jpg?height=414&width=1468&top_left_y=975&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_205",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the mass (in g) of the KHP solution required to react completely with $4.359 \\mathrm{~g}$ of the sodium hydroxide solution",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the mass (in g) of the KHP solution required to react completely with $4.359 \\mathrm{~g}$ of the sodium hydroxide solution\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_382",
        "problem": "How many hydroxide ions are in $2.5 \\mathrm{~mol} \\mathrm{Mg}(\\mathrm{OH})_{2}$ ?\nA: $3.0 \\times 10^{23}$\nB: $6.0 \\times 10^{23}$\nC: $1.5 \\times 10^{24}$\nD: $3.0 \\times 10^{24}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many hydroxide ions are in $2.5 \\mathrm{~mol} \\mathrm{Mg}(\\mathrm{OH})_{2}$ ?\n\nA: $3.0 \\times 10^{23}$\nB: $6.0 \\times 10^{23}$\nC: $1.5 \\times 10^{24}$\nD: $3.0 \\times 10^{24}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1054",
        "problem": "This question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nThe density of liquid hydrogen is $0.071 \\mathrm{~g} \\mathrm{~cm}^{-3}$.\n\nSpaceX's aim to colonise Mars leads to the choice of methane as an alternative to hydrogen. Methane can be formed using Mars' natural resources via the Sabatier process as the atmosphere of the planet is made up of $95.3 \\%$ carbon dioxide by volume. In the Sabatier process methane and water are formed from the reaction of carbon dioxide and hydrogen gas.\n\nSpaceX has recently developed a new engine, the Raptor, that uses liquid methane and LOX. Energy is required to turn them into the gaseous phase before they react. The enthalpy change of vaporisation of methane is $+8.2 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the enthalpy change of vaporisation of oxygen is $+6.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. The enthalpy change of combustion of methane, shown below, is $-890.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\n$$\n\\mathrm{CH}_{4(\\mathrm{~g})}+2 \\mathrm{O}_{2(\\mathrm{~g})} \\rightarrow \\mathrm{CO}_{2(\\mathrm{~g})}+2 \\mathrm{H}_{2} \\mathrm{O}_{(\\mathrm{g})}\n$$\n\nCalculate the enthalpy change of reaction, in $\\mathrm{kJ}$, when one mole of liquid methane reacts with liquid oxygen to form gaseous carbon dioxide and gaseous water.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nThe density of liquid hydrogen is $0.071 \\mathrm{~g} \\mathrm{~cm}^{-3}$.\n\nSpaceX's aim to colonise Mars leads to the choice of methane as an alternative to hydrogen. Methane can be formed using Mars' natural resources via the Sabatier process as the atmosphere of the planet is made up of $95.3 \\%$ carbon dioxide by volume. In the Sabatier process methane and water are formed from the reaction of carbon dioxide and hydrogen gas.\n\nSpaceX has recently developed a new engine, the Raptor, that uses liquid methane and LOX. Energy is required to turn them into the gaseous phase before they react. The enthalpy change of vaporisation of methane is $+8.2 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the enthalpy change of vaporisation of oxygen is $+6.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. The enthalpy change of combustion of methane, shown below, is $-890.8 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\n$$\n\\mathrm{CH}_{4(\\mathrm{~g})}+2 \\mathrm{O}_{2(\\mathrm{~g})} \\rightarrow \\mathrm{CO}_{2(\\mathrm{~g})}+2 \\mathrm{H}_{2} \\mathrm{O}_{(\\mathrm{g})}\n$$\n\nCalculate the enthalpy change of reaction, in $\\mathrm{kJ}$, when one mole of liquid methane reacts with liquid oxygen to form gaseous carbon dioxide and gaseous water.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kJ}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-02.jpg?height=434&width=619&top_left_y=317&top_left_x=1181",
            "https://cdn.mathpix.com/cropped/2024_03_14_1046d64c5e1013f7935dg-02.jpg?height=291&width=805&top_left_y=2056&top_left_x=614"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kJ}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_135",
        "problem": "A reaction $A B+C$, undergoes a single displacement reaction. Which of the following would be a product of the reaction if $A$ is a group 2 metal and $C$ is a group 1 metal?\nA: $\\mathrm{CA}_{2}$\nB: B\nC: CB\nD: $\\mathrm{A}_{2}$\nE: $\\mathrm{C}_{2} \\mathrm{~B}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA reaction $A B+C$, undergoes a single displacement reaction. Which of the following would be a product of the reaction if $A$ is a group 2 metal and $C$ is a group 1 metal?\n\nA: $\\mathrm{CA}_{2}$\nB: B\nC: CB\nD: $\\mathrm{A}_{2}$\nE: $\\mathrm{C}_{2} \\mathrm{~B}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_306",
        "problem": "Consider the compounds $\\mathrm{HF}, \\mathrm{HCl}, \\mathrm{HBr}$ and $\\mathrm{HI}$. Of these compounds, which one has the highest boiling point and which one is the strongest acid in water?\nA: HF has the highest boiling point and is the strongest acid\nB: $\\mathrm{HI}$ has the highest boiling point and is the strongest acid\nC: $\\mathrm{HF}$ has the highest boiling point and $\\mathrm{HI}$ is the strongest acid\nD: $\\mathrm{HI}$ has the highest boiling point and $\\mathrm{HF}$ is the strongest acid\nE: $\\mathrm{HI}$ has the highest boiling point and $\\mathrm{HCl}$ is the strongest acid\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nConsider the compounds $\\mathrm{HF}, \\mathrm{HCl}, \\mathrm{HBr}$ and $\\mathrm{HI}$. Of these compounds, which one has the highest boiling point and which one is the strongest acid in water?\n\nA: HF has the highest boiling point and is the strongest acid\nB: $\\mathrm{HI}$ has the highest boiling point and is the strongest acid\nC: $\\mathrm{HF}$ has the highest boiling point and $\\mathrm{HI}$ is the strongest acid\nD: $\\mathrm{HI}$ has the highest boiling point and $\\mathrm{HF}$ is the strongest acid\nE: $\\mathrm{HI}$ has the highest boiling point and $\\mathrm{HCl}$ is the strongest acid\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1559",
        "problem": "When the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\nWhat is the molar concentration of chloride in the sample?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nWhen the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\nWhat is the molar concentration of chloride in the sample?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1227",
        "problem": "The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nCalculate $\\Delta S_{\\text {sur }}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nCalculate $\\Delta S_{\\text {sur }}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{JK}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{JK}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_817",
        "problem": "$300^{\\circ} \\mathrm{C}$ 时, 将气体 $\\mathrm{X}$ 和气体 $\\mathrm{Y}$ 各 $0.16 \\mathrm{~mol}$ 充入 $10 \\mathrm{~L}$ 恒容密闭容器中, 发生反应 $\\mathrm{X}(\\mathrm{g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{Z}(\\mathrm{g}) \\Delta H<0$, 一段时间后达到平衡。反应过程中测定的数据如下表:下列说法正确的是（ ）\n\n| $\\mathrm{t} / \\mathrm{min}$ | 2 | 4 | 7 | 9 |\n| :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{n}(\\mathrm{Y}) / \\mathrm{mol}$ | 0.12 | 0.11 | 0.10 | 0.10 |\nA: 当 $\\mathrm{v}_{\\text {逆 }}(\\mathrm{X})=2 \\mathrm{v}_{\\text {正 }}(\\mathrm{Z})$ 可以说明反应达平衡\nB: 反应前 $2 \\mathrm{~min}$ 的平均速率 $\\mathrm{v}(\\mathrm{Z})=4.0 \\times 10^{-3} \\mathrm{~mol}^{\\circ} \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nC: 其他条件不变, 再充入 $0.2 \\mathrm{~mol} \\mathrm{Z}$, 平衡时 $\\mathrm{X}$ 的体积分数增大\nD: 该反应在 $350^{\\circ} \\mathrm{C}$ 时的平衡常数小于 1.44\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$300^{\\circ} \\mathrm{C}$ 时, 将气体 $\\mathrm{X}$ 和气体 $\\mathrm{Y}$ 各 $0.16 \\mathrm{~mol}$ 充入 $10 \\mathrm{~L}$ 恒容密闭容器中, 发生反应 $\\mathrm{X}(\\mathrm{g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{Z}(\\mathrm{g}) \\Delta H<0$, 一段时间后达到平衡。反应过程中测定的数据如下表:下列说法正确的是（ ）\n\n| $\\mathrm{t} / \\mathrm{min}$ | 2 | 4 | 7 | 9 |\n| :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{n}(\\mathrm{Y}) / \\mathrm{mol}$ | 0.12 | 0.11 | 0.10 | 0.10 |\n\nA: 当 $\\mathrm{v}_{\\text {逆 }}(\\mathrm{X})=2 \\mathrm{v}_{\\text {正 }}(\\mathrm{Z})$ 可以说明反应达平衡\nB: 反应前 $2 \\mathrm{~min}$ 的平均速率 $\\mathrm{v}(\\mathrm{Z})=4.0 \\times 10^{-3} \\mathrm{~mol}^{\\circ} \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nC: 其他条件不变, 再充入 $0.2 \\mathrm{~mol} \\mathrm{Z}$, 平衡时 $\\mathrm{X}$ 的体积分数增大\nD: 该反应在 $350^{\\circ} \\mathrm{C}$ 时的平衡常数小于 1.44\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1441",
        "problem": "Nitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.Express the experimentally determined rate constant $k$ in terms of $k_{1}, k_{-1}$ and $k_{2}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an equation.\nHere is some context information for this question, which might assist you in solving it:\nNitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.\n\nproblem:\nExpress the experimentally determined rate constant $k$ in terms of $k_{1}, k_{-1}$ and $k_{2}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an equation, e.g. ANSWER=\\frac{x^2}{4}+\\frac{y^2}{2}=1",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EQ",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_119",
        "problem": "Hypophosphatemia is a condition of abnormally low phosphate concentration in the human blood stream, occurring in $2 \\%$ of hospitalized patients. The treatment includes administration of intravenous phosphate buffer to increase blood phosphate concentration. However, since phosphoric acid is a weak acid, care needs to be taken to maintain blood $\\mathrm{pH}$ at 7.4. Given the data below, and assuming similar concentrations of both species of the acid-base conjugate pair, determine the most effective buffer combination to achieve a $\\mathrm{pH}$ of 7.4.\n\n$\\mathrm{H}_{3} \\mathrm{PO}_{4}^{--} ; K_{\\mathrm{a}}=7.2 \\times 10^{-3} \\quad \\mathrm{H}_{2} \\mathrm{PO}_{4}^{-} ; K_{\\mathrm{a}}=6.3 \\times 10^{-8} \\quad \\mathrm{HPO}_{4}^{2-} ; K_{\\mathrm{a}}=4.2 \\times 10^{-13}$\nA: $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\nB: $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and $\\mathrm{HPO}_{4}{ }^{2-}$\nC: $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$and $\\mathrm{HPO}_{4}{ }^{2-}$\nD: $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$and $\\mathrm{PO}_{4}^{3-}$\nE: $\\mathrm{HPO}_{4}^{2-}$ and $\\mathrm{PO}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHypophosphatemia is a condition of abnormally low phosphate concentration in the human blood stream, occurring in $2 \\%$ of hospitalized patients. The treatment includes administration of intravenous phosphate buffer to increase blood phosphate concentration. However, since phosphoric acid is a weak acid, care needs to be taken to maintain blood $\\mathrm{pH}$ at 7.4. Given the data below, and assuming similar concentrations of both species of the acid-base conjugate pair, determine the most effective buffer combination to achieve a $\\mathrm{pH}$ of 7.4.\n\n$\\mathrm{H}_{3} \\mathrm{PO}_{4}^{--} ; K_{\\mathrm{a}}=7.2 \\times 10^{-3} \\quad \\mathrm{H}_{2} \\mathrm{PO}_{4}^{-} ; K_{\\mathrm{a}}=6.3 \\times 10^{-8} \\quad \\mathrm{HPO}_{4}^{2-} ; K_{\\mathrm{a}}=4.2 \\times 10^{-13}$\n\nA: $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$\nB: $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and $\\mathrm{HPO}_{4}{ }^{2-}$\nC: $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$and $\\mathrm{HPO}_{4}{ }^{2-}$\nD: $\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}$and $\\mathrm{PO}_{4}^{3-}$\nE: $\\mathrm{HPO}_{4}^{2-}$ and $\\mathrm{PO}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1383",
        "problem": "The smallest diffraction angle of a monochromatic beam of X-rays in a certain experiment is $11.5^{\\circ}$. Based on this we must expect a beam of X-rays diffracted at:\nA: 22.0 degrees\nB: 22.5 degrees\nC: 23.0 degrees\nD: 23.5 degrees\nE: 24.0 degrees\nF: 24.5 degrees\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe smallest diffraction angle of a monochromatic beam of X-rays in a certain experiment is $11.5^{\\circ}$. Based on this we must expect a beam of X-rays diffracted at:\n\nA: 22.0 degrees\nB: 22.5 degrees\nC: 23.0 degrees\nD: 23.5 degrees\nE: 24.0 degrees\nF: 24.5 degrees\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1170",
        "problem": "Two equations for the production of copper from copper(II) oxide and carbon:\n\n$$\n\\begin{aligned}\n\\mathrm{CuO}(\\mathrm{s})+\\mathrm{C}(\\mathrm{s}) & \\longrightarrow \\mathrm{CO}(\\mathrm{g})+\\mathrm{Cu}(\\mathrm{s}) \\\\\n2 \\mathrm{CuO}(\\mathrm{s})+\\mathrm{C}(\\mathrm{s}) & \\longrightarrow \\mathrm{CO}_{2}(\\mathrm{~g})+2 \\mathrm{Cu}(\\mathrm{s})\n\\end{aligned}\n$$This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nWhat is the minimum theoretical mass of carbon needed to produce $100 \\mathrm{~g}$ of copper?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nTwo equations for the production of copper from copper(II) oxide and carbon:\n\n$$\n\\begin{aligned}\n\\mathrm{CuO}(\\mathrm{s})+\\mathrm{C}(\\mathrm{s}) & \\longrightarrow \\mathrm{CO}(\\mathrm{g})+\\mathrm{Cu}(\\mathrm{s}) \\\\\n2 \\mathrm{CuO}(\\mathrm{s})+\\mathrm{C}(\\mathrm{s}) & \\longrightarrow \\mathrm{CO}_{2}(\\mathrm{~g})+2 \\mathrm{Cu}(\\mathrm{s})\n\\end{aligned}\n$$\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nWhat is the minimum theoretical mass of carbon needed to produce $100 \\mathrm{~g}$ of copper?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_22",
        "problem": "A student has $10 \\mathrm{~mL}$ of a solution that might contain any or all of the following cations at $0.01 \\mathrm{M}$ concentrations: $\\mathrm{Mn}^{2+}, \\mathrm{Ba}^{2+}, \\mathrm{Ag}^{+}$, and $\\mathrm{Cu}^{2+}$. Addition of $10 \\mathrm{~mL}$ of $1 \\mathrm{M}$ $\\mathrm{HCl}$ causes a precipitate to form. After the precipitate is filtered off, $1 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{SO}_{4}$ is added to the filtrate and another precipitate forms. What is the second precipitate?\nA: $\\mathrm{MnSO}_{4}$\nB: $\\mathrm{BaSO}_{4}$\nC: $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\nD: A mixture of $\\mathrm{BaSO}_{4}$ and $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student has $10 \\mathrm{~mL}$ of a solution that might contain any or all of the following cations at $0.01 \\mathrm{M}$ concentrations: $\\mathrm{Mn}^{2+}, \\mathrm{Ba}^{2+}, \\mathrm{Ag}^{+}$, and $\\mathrm{Cu}^{2+}$. Addition of $10 \\mathrm{~mL}$ of $1 \\mathrm{M}$ $\\mathrm{HCl}$ causes a precipitate to form. After the precipitate is filtered off, $1 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{SO}_{4}$ is added to the filtrate and another precipitate forms. What is the second precipitate?\n\nA: $\\mathrm{MnSO}_{4}$\nB: $\\mathrm{BaSO}_{4}$\nC: $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\nD: A mixture of $\\mathrm{BaSO}_{4}$ and $\\mathrm{Ag}_{2} \\mathrm{SO}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_800",
        "problem": "某反应 $\\mathrm{A}(\\mathrm{g})+\\mathrm{B}(\\mathrm{g}) \\rightarrow \\mathrm{C}(\\mathrm{g})+\\mathrm{D}(\\mathrm{g})$ 的速率方程为 $\\mathrm{v}=\\mathrm{kc}^{\\mathrm{m}}(\\mathrm{A}) \\cdot \\mathrm{c}^{\\mathrm{n}}(\\mathrm{B})$, 其中 $\\mathrm{k}$ 是反应速率常数, 受温度、催化剂的影响。其半衰期(当剩余反应物恰好是起始的一半时所需的时间)为 $\\frac{0.8}{\\mathrm{k}}$ 。改变反应物浓度时, 反应的瞬时速率如表所示;\n\n| $\\mathrm{c}(\\mathrm{A}) /(\\mathrm{mol} / \\mathrm{L})$ | 0.25 | 0.50 | 1.00 | 0.50 | 1.00 | $\\mathrm{c}_{1}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{c}(\\mathrm{B}) /(\\mathrm{mol} / \\mathrm{L})$ | 0.050 | 0.050 | 0.100 | 0.100 | 0.200 | $\\mathrm{c}_{2}$ |\n| $\\mathrm{v} / 10^{-3} \\mathrm{~mol} /(\\mathrm{L} \\cdot \\mathrm{min})$ | 1.6 | 3.2 | $\\mathrm{v}_{1}$ | 3.2 | $\\mathrm{v}_{2}$ | 4.8 |\n\n下列说法不正确的是\nA: 上述表格中的 $\\mathrm{c}_{1}=0.75 、 \\mathrm{v}_{1}=6.4$\nB: 该反应的速率常数 $\\mathrm{k}=6.4 \\times 10^{-4} \\mathrm{~min}^{-1}$\nC: 在过量的 B 存在时，反应掉 $75 \\%$ 的 A 所需的时间是 $250 \\mathrm{~min}$ D．升温、加入催化剂，均可使 $\\mathrm{k}$ 增大导致反应的瞬时速率加快\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某反应 $\\mathrm{A}(\\mathrm{g})+\\mathrm{B}(\\mathrm{g}) \\rightarrow \\mathrm{C}(\\mathrm{g})+\\mathrm{D}(\\mathrm{g})$ 的速率方程为 $\\mathrm{v}=\\mathrm{kc}^{\\mathrm{m}}(\\mathrm{A}) \\cdot \\mathrm{c}^{\\mathrm{n}}(\\mathrm{B})$, 其中 $\\mathrm{k}$ 是反应速率常数, 受温度、催化剂的影响。其半衰期(当剩余反应物恰好是起始的一半时所需的时间)为 $\\frac{0.8}{\\mathrm{k}}$ 。改变反应物浓度时, 反应的瞬时速率如表所示;\n\n| $\\mathrm{c}(\\mathrm{A}) /(\\mathrm{mol} / \\mathrm{L})$ | 0.25 | 0.50 | 1.00 | 0.50 | 1.00 | $\\mathrm{c}_{1}$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{c}(\\mathrm{B}) /(\\mathrm{mol} / \\mathrm{L})$ | 0.050 | 0.050 | 0.100 | 0.100 | 0.200 | $\\mathrm{c}_{2}$ |\n| $\\mathrm{v} / 10^{-3} \\mathrm{~mol} /(\\mathrm{L} \\cdot \\mathrm{min})$ | 1.6 | 3.2 | $\\mathrm{v}_{1}$ | 3.2 | $\\mathrm{v}_{2}$ | 4.8 |\n\n下列说法不正确的是\n\nA: 上述表格中的 $\\mathrm{c}_{1}=0.75 、 \\mathrm{v}_{1}=6.4$\nB: 该反应的速率常数 $\\mathrm{k}=6.4 \\times 10^{-4} \\mathrm{~min}^{-1}$\nC: 在过量的 B 存在时，反应掉 $75 \\%$ 的 A 所需的时间是 $250 \\mathrm{~min}$ D．升温、加入催化剂，均可使 $\\mathrm{k}$ 增大导致反应的瞬时速率加快\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_947",
        "problem": "中国科学院将分子 $\\mathrm{I}_{2}$ 引入电解质中调整充电和放电反应途径, 研制出了高功率可充电 $\\mathrm{LiSOCl}_{2}$ 电池, 工作原理如图所示, 已知 $\\mathrm{SOCl}_{2}$ 可与水发生反应。下列有关说法正确的是\n\n[图1]\nA: 该电池既可选用含水电解液, 也可选无水电解液\nB: 放电时, $\\mathrm{SOCl}_{2}$ 最终被氧化为 $\\mathrm{S}$\nC: 充电时, 阴极反应式: $2 \\mathrm{LiCl}+\\mathrm{I}_{2}+2 \\mathrm{e}^{-}=2 \\mathrm{ICl}+2 \\mathrm{Li}^{+}$\nD: 放电时, 每产生 $11.2 \\mathrm{~L}$ (标准状况下) $\\mathrm{SO}_{2}$ 时, 电路中转移 $2 \\mathrm{~mol}$ 电子\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n中国科学院将分子 $\\mathrm{I}_{2}$ 引入电解质中调整充电和放电反应途径, 研制出了高功率可充电 $\\mathrm{LiSOCl}_{2}$ 电池, 工作原理如图所示, 已知 $\\mathrm{SOCl}_{2}$ 可与水发生反应。下列有关说法正确的是\n\n[图1]\n\nA: 该电池既可选用含水电解液, 也可选无水电解液\nB: 放电时, $\\mathrm{SOCl}_{2}$ 最终被氧化为 $\\mathrm{S}$\nC: 充电时, 阴极反应式: $2 \\mathrm{LiCl}+\\mathrm{I}_{2}+2 \\mathrm{e}^{-}=2 \\mathrm{ICl}+2 \\mathrm{Li}^{+}$\nD: 放电时, 每产生 $11.2 \\mathrm{~L}$ (标准状况下) $\\mathrm{SO}_{2}$ 时, 电路中转移 $2 \\mathrm{~mol}$ 电子\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-009.jpg?height=477&width=591&top_left_y=938&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_340",
        "problem": "What is the $\\mathrm{pH}$ of $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HClO}_{2}(\\mathrm{aq})$ ?\n\n$$\nK_{\\mathrm{a}}=1.1 \\times 10^{-2} \\text { for } \\mathrm{HClO}_{2}\n$$\nA: 1.98\nB: 5.11\nC: 1.55\nD: 2.52\nE: 1.00\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the $\\mathrm{pH}$ of $0.10 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HClO}_{2}(\\mathrm{aq})$ ?\n\n$$\nK_{\\mathrm{a}}=1.1 \\times 10^{-2} \\text { for } \\mathrm{HClO}_{2}\n$$\n\nA: 1.98\nB: 5.11\nC: 1.55\nD: 2.52\nE: 1.00\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1290",
        "problem": "Metals of the iron subgroup are effective catalysts of hydrogenation of $\\mathrm{CO}$ (FischerTropsch reaction)\n\n$$\n\\mathrm{CO}+3 \\mathrm{H}_{2} \\quad \\mathrm{Fe}, \\mathrm{Co} \\rightarrow \\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\nCatalyst (e.g. cobalt) is often used in the form of solid nanoparticles that have a spherical structure (fig.1). The reduction in size of the catalyst increases catalytic activity significantly. The unwanted side-reaction however involves the oxidation of the catalyst:\n\n$$\n\\mathrm{Co}(\\mathrm{s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftarrows \\mathrm{CoO}(\\mathrm{s})+\\mathrm{H}_{2}(\\mathrm{~g})\n$$\n\nSolid cobalt oxide (bulk) is formed in the reaction vessel. This causes an irreversible loss of the catalyst's mass. Solid cobalt oxide can also be deposited on the surface of $\\mathrm{Co}(\\mathrm{s})$. In this case the new spherical layer is formed around the surface of the catalyst (see figure 2) and the catalytic activity drops.\n\n[figure1]\n\nFig. 1\n\n[figure2]\n\nFig. 2\n\nLet us see how formation of nanoparticles affects the equilibrium of reaction (1).\n\n$$\nG^{0}(r)=G^{0}(\\text { bulk })+\\frac{2 \\sigma}{r} V\n$$\n\nCalculate the standard Gibbs energy $\\Delta_{\\mathrm{r}} G^{0}(1)$ and the equilibrium constant for the reaction (1) at $T=500 \\mathrm{~K}$.\n\nReference data:\n\n| Substance | $\\rho, \\mathrm{g} \\mathrm{cm}^{-3}$ | $\\Delta_{\\mathrm{f}} G_{500}^{0}, \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{Co}(\\mathrm{s})$ | 8.90 |  |\n| $\\mathrm{CoO}(\\mathrm{s})$ | 5.68 | -198.4 |\n| $\\mathrm{H}_{2} \\mathrm{O}$ (gas) |  | -219.1 |",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nMetals of the iron subgroup are effective catalysts of hydrogenation of $\\mathrm{CO}$ (FischerTropsch reaction)\n\n$$\n\\mathrm{CO}+3 \\mathrm{H}_{2} \\quad \\mathrm{Fe}, \\mathrm{Co} \\rightarrow \\mathrm{CH}_{4}+\\mathrm{H}_{2} \\mathrm{O}\n$$\n\nCatalyst (e.g. cobalt) is often used in the form of solid nanoparticles that have a spherical structure (fig.1). The reduction in size of the catalyst increases catalytic activity significantly. The unwanted side-reaction however involves the oxidation of the catalyst:\n\n$$\n\\mathrm{Co}(\\mathrm{s})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\rightleftarrows \\mathrm{CoO}(\\mathrm{s})+\\mathrm{H}_{2}(\\mathrm{~g})\n$$\n\nSolid cobalt oxide (bulk) is formed in the reaction vessel. This causes an irreversible loss of the catalyst's mass. Solid cobalt oxide can also be deposited on the surface of $\\mathrm{Co}(\\mathrm{s})$. In this case the new spherical layer is formed around the surface of the catalyst (see figure 2) and the catalytic activity drops.\n\n[figure1]\n\nFig. 1\n\n[figure2]\n\nFig. 2\n\nLet us see how formation of nanoparticles affects the equilibrium of reaction (1).\n\n$$\nG^{0}(r)=G^{0}(\\text { bulk })+\\frac{2 \\sigma}{r} V\n$$\n\nCalculate the standard Gibbs energy $\\Delta_{\\mathrm{r}} G^{0}(1)$ and the equilibrium constant for the reaction (1) at $T=500 \\mathrm{~K}$.\n\nReference data:\n\n| Substance | $\\rho, \\mathrm{g} \\mathrm{cm}^{-3}$ | $\\Delta_{\\mathrm{f}} G_{500}^{0}, \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{Co}(\\mathrm{s})$ | 8.90 |  |\n| $\\mathrm{CoO}(\\mathrm{s})$ | 5.68 | -198.4 |\n| $\\mathrm{H}_{2} \\mathrm{O}$ (gas) |  | -219.1 |\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-640.jpg?height=411&width=348&top_left_y=1231&top_left_x=223",
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-640.jpg?height=409&width=985&top_left_y=1189&top_left_x=821"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_362",
        "problem": "Which aqueous solution has the highest boiling point?\nA: $1.0 \\mathrm{~m}$ acetic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$\nB: $1.0 \\mathrm{~m}$ sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$\nC: $1.0 \\mathrm{~m}$ phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$\nD: $1.0 \\mathrm{~m}$ glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich aqueous solution has the highest boiling point?\n\nA: $1.0 \\mathrm{~m}$ acetic acid, $\\mathrm{CH}_{3} \\mathrm{COOH}$\nB: $1.0 \\mathrm{~m}$ sulfuric acid, $\\mathrm{H}_{2} \\mathrm{SO}_{4}$\nC: $1.0 \\mathrm{~m}$ phosphoric acid, $\\mathrm{H}_{3} \\mathrm{PO}_{4}$\nD: $1.0 \\mathrm{~m}$ glucose, $\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_523",
        "problem": "某传感器可利用电化学原理测定混合气体中 $\\mathrm{O}_{2}$ 的含量, 工作示意图如下。单位时间内, 通过传感器的待测气体为 $\\mathrm{pL}$ (标准状况), 某电极质量增加了 $\\mathrm{q} g$ 。下列说法正确的是\n\n[图1]\nA: $\\mathrm{Pb}$ 电极上的电极反应式为: $2 \\mathrm{~Pb}-4 \\mathrm{e}^{-}+4 \\mathrm{OH}^{-}=2 \\mathrm{PbO}+2 \\mathrm{H}_{2} \\mathrm{O}$\nB: 反应过程中转移的电子数为 $\\mathrm{qN}_{\\mathrm{A}} / 48$\nC: 反应过程中正极区溶液的 $\\mathrm{pH}$ 减小\nD: 待测气体中 $\\mathrm{O}_{2}$ 的体积分数为 $7 \\mathrm{q} / 60 \\mathrm{p}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某传感器可利用电化学原理测定混合气体中 $\\mathrm{O}_{2}$ 的含量, 工作示意图如下。单位时间内, 通过传感器的待测气体为 $\\mathrm{pL}$ (标准状况), 某电极质量增加了 $\\mathrm{q} g$ 。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{Pb}$ 电极上的电极反应式为: $2 \\mathrm{~Pb}-4 \\mathrm{e}^{-}+4 \\mathrm{OH}^{-}=2 \\mathrm{PbO}+2 \\mathrm{H}_{2} \\mathrm{O}$\nB: 反应过程中转移的电子数为 $\\mathrm{qN}_{\\mathrm{A}} / 48$\nC: 反应过程中正极区溶液的 $\\mathrm{pH}$ 减小\nD: 待测气体中 $\\mathrm{O}_{2}$ 的体积分数为 $7 \\mathrm{q} / 60 \\mathrm{p}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-32.jpg?height=585&width=494&top_left_y=1786&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_834",
        "problem": "某酸 $\\mathrm{H}_{3} \\mathrm{XO}_{3}$ 溶液中含 $\\mathrm{X}$ 粒子的物质的量分数与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\n\n## $\\mathrm{O}$\n\n[图2]\nA: 在 $\\mathrm{H}_{3} \\mathrm{XO}_{3}$ 溶液中滴加少量 $\\mathrm{NaOH}$ 溶液, 发生反应的离子方程式为 $\\mathrm{H}_{3} \\mathrm{XO}_{3}+3 \\mathrm{OH}^{-}=\\mathrm{XO}_{3}^{3-}+3 \\mathrm{H}_{2} \\mathrm{O}$\nB: $\\mathrm{NaH}_{2} \\mathrm{XO}_{3}$ 溶液中有 $c\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}^{-}\\right)>c\\left(\\mathrm{H}_{3} \\mathrm{XO}_{3}\\right)>c\\left(\\mathrm{HXO}_{3}^{2-}\\right)$\nC: $\\mathrm{Na}_{2} \\mathrm{HXO}_{3}$ 溶液中有 $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{HXO}_{3}^{2-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}^{-}\\right)+c\\left(\\mathrm{XO}_{3}^{3-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某酸 $\\mathrm{H}_{3} \\mathrm{XO}_{3}$ 溶液中含 $\\mathrm{X}$ 粒子的物质的量分数与 $\\mathrm{pH}$ 的关系如图所示。下列说法正确的是\n\n[图1]\n\n## $\\mathrm{O}$\n\n[图2]\n\nA: 在 $\\mathrm{H}_{3} \\mathrm{XO}_{3}$ 溶液中滴加少量 $\\mathrm{NaOH}$ 溶液, 发生反应的离子方程式为 $\\mathrm{H}_{3} \\mathrm{XO}_{3}+3 \\mathrm{OH}^{-}=\\mathrm{XO}_{3}^{3-}+3 \\mathrm{H}_{2} \\mathrm{O}$\nB: $\\mathrm{NaH}_{2} \\mathrm{XO}_{3}$ 溶液中有 $c\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}^{-}\\right)>c\\left(\\mathrm{H}_{3} \\mathrm{XO}_{3}\\right)>c\\left(\\mathrm{HXO}_{3}^{2-}\\right)$\nC: $\\mathrm{Na}_{2} \\mathrm{HXO}_{3}$ 溶液中有 $c\\left(\\mathrm{Na}^{+}\\right)+c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{HXO}_{3}^{2-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{XO}_{3}^{-}\\right)+c\\left(\\mathrm{XO}_{3}^{3-}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-006.jpg?height=733&width=965&top_left_y=1164&top_left_x=337",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-006.jpg?height=294&width=680&top_left_y=1995&top_left_x=791",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-007.jpg?height=243&width=1220&top_left_y=330&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_590",
        "problem": "下列有关消除反应原料、试剂、产物和反应类型都正确的是\n\n[图1]\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列有关消除反应原料、试剂、产物和反应类型都正确的是\n\n[图1]\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-50.jpg?height=1322&width=1445&top_left_y=1315&top_left_x=311"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1579",
        "problem": "This result is due to the presence of products that are not expected to be found in the residue. Give two of them that under these experimental conditions can plausibly account for the data.\n\nTraditionally, in industry the analysis and the yield are expressed as percentage of oxide. The phosphorous content is expressed as if it were $\\mathrm{P}_{2} \\mathrm{O}_{5}$.\n\nIf $n_{2}$ is the amount of a soluble product obtained, $n_{1}$ the amount of a substance added as acid, $n_{0}$ the amount of apatite added, the yield is:\n\n$$\n\\begin{aligned}\n& r_{\\exp }=\\frac{n_{2}}{n_{1}+n_{0}} 100 \\\\\n& m_{2}=0.144 \\mathrm{~g} \\text { of residue is obtained on the filter. }\n\\end{aligned}\n$$The elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nThe experimental yield is over $100 \\%$. Calculate a value of $r$ nearer to the real yield.\n\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 ; \\mathrm{S}: 32.\n$$\n\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThis result is due to the presence of products that are not expected to be found in the residue. Give two of them that under these experimental conditions can plausibly account for the data.\n\nTraditionally, in industry the analysis and the yield are expressed as percentage of oxide. The phosphorous content is expressed as if it were $\\mathrm{P}_{2} \\mathrm{O}_{5}$.\n\nIf $n_{2}$ is the amount of a soluble product obtained, $n_{1}$ the amount of a substance added as acid, $n_{0}$ the amount of apatite added, the yield is:\n\n$$\n\\begin{aligned}\n& r_{\\exp }=\\frac{n_{2}}{n_{1}+n_{0}} 100 \\\\\n& m_{2}=0.144 \\mathrm{~g} \\text { of residue is obtained on the filter. }\n\\end{aligned}\n$$\n\nproblem:\nThe elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nThe experimental yield is over $100 \\%$. Calculate a value of $r$ nearer to the real yield.\n\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 ; \\mathrm{S}: 32.\n$$\n\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1216",
        "problem": "5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.\n\n[figure1]\n\nUshabti figurines from Egyptian pharaoh tomb covered with Egyptian blue and a Chinese blue soap dinspenser sold at Alibaba\n\nThe ancient method of preparation for these pigments can easily by reproduced in a modern laboratory.\n\nWhen considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.\n\nTo make Egyptian blue, one should heat $10.0 \\mathrm{~g}$ of mineral $\\mathrm{A}$ with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ and $9.05 \\mathrm{~g}$ of mineral $\\mathrm{B}$ at $800-900^{\\circ} \\mathrm{C}$ for a prolonged time. A volume of $16.7 \\mathrm{dm}^{3}$ of a mixture of two gaseous products are released (the volume is measured at $850^{\\circ} \\mathrm{C}$ and $1.013 \\times 10^{5} \\mathrm{~Pa}$ (1.013 bar) pressure. In result, $34.0 \\mathrm{~g}$ of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to $0^{\\circ} \\mathrm{C}$, the gaseous volume reduces to $3.04 \\mathrm{dm}^{3}$.Calculate the mass of the gaseous mixture formed upon heating of $\\mathbf{A}$ with $\\mathbf{B}$ and $\\mathrm{SiO}_{2}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.\n\n[figure1]\n\nUshabti figurines from Egyptian pharaoh tomb covered with Egyptian blue and a Chinese blue soap dinspenser sold at Alibaba\n\nThe ancient method of preparation for these pigments can easily by reproduced in a modern laboratory.\n\nWhen considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.\n\nTo make Egyptian blue, one should heat $10.0 \\mathrm{~g}$ of mineral $\\mathrm{A}$ with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ and $9.05 \\mathrm{~g}$ of mineral $\\mathrm{B}$ at $800-900^{\\circ} \\mathrm{C}$ for a prolonged time. A volume of $16.7 \\mathrm{dm}^{3}$ of a mixture of two gaseous products are released (the volume is measured at $850^{\\circ} \\mathrm{C}$ and $1.013 \\times 10^{5} \\mathrm{~Pa}$ (1.013 bar) pressure. In result, $34.0 \\mathrm{~g}$ of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to $0^{\\circ} \\mathrm{C}$, the gaseous volume reduces to $3.04 \\mathrm{dm}^{3}$.\n\nproblem:\nCalculate the mass of the gaseous mixture formed upon heating of $\\mathbf{A}$ with $\\mathbf{B}$ and $\\mathrm{SiO}_{2}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-159.jpg?height=611&width=1357&top_left_y=705&top_left_x=378"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_32",
        "problem": "What is the standard reduction potential for $\\mathrm{Cr}^{2+}(a q)$ to $\\mathrm{Cr}(s)$ ?\n\n| Half-Reaction | $E^{\\circ}, \\mathrm{V}$ |\n| :---: | :---: |\n| $\\mathrm{Cr}^{3+}(a q)+3 e^{-} \\rightarrow \\mathrm{Cr}(s)$ | -0.73 |\n| $\\mathrm{Cr}^{3+}(a q)+e^{-} \\rightarrow \\mathrm{Cr}^{2+}(a q)$ | -0.50 |\nA: $-0.23 \\mathrm{~V}$\nB: $-0.85 \\mathrm{~V}$\nC: $-1.23 \\mathrm{~V}$\nD: $-1.69 \\mathrm{~V}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the standard reduction potential for $\\mathrm{Cr}^{2+}(a q)$ to $\\mathrm{Cr}(s)$ ?\n\n| Half-Reaction | $E^{\\circ}, \\mathrm{V}$ |\n| :---: | :---: |\n| $\\mathrm{Cr}^{3+}(a q)+3 e^{-} \\rightarrow \\mathrm{Cr}(s)$ | -0.73 |\n| $\\mathrm{Cr}^{3+}(a q)+e^{-} \\rightarrow \\mathrm{Cr}^{2+}(a q)$ | -0.50 |\n\nA: $-0.23 \\mathrm{~V}$\nB: $-0.85 \\mathrm{~V}$\nC: $-1.23 \\mathrm{~V}$\nD: $-1.69 \\mathrm{~V}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1401",
        "problem": "In 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Later, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were measured by using a glass vessel with a known volume under atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$.\n\nTable 1. Mass of Chemical Nitrogen in the Vessel\n\n| From nitric oxide | $2.3001 \\mathrm{~g}$ |\n| :--- | :--- |\n| From nitrous oxide | $2.2990 \\mathrm{~g}$ |\n| From ammonium nitrite purified at a red heat | $2.2987 \\mathrm{~g}$ |\n| From urea | $2.2985 \\mathrm{~g}$ |\n| From ammonium nitrite purified in the cold | $2.2987 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 2 9 9 0} \\mathbf{~ g}$ |\n\nTable 2. Mass of Atmospheric Nitrogen in the Vessel\n\n| $\\mathrm{O}_{2}$ was removed by hot copper (1892) | $2.3103 \\mathrm{~g}$ |\n| :--- | :--- |\n| $\\mathrm{O}_{2}$ was removed by hot iron (1893) | $2.3100 \\mathrm{~g}$ |\n| $\\mathrm{O}_{2}$ was removed by ferrous hydrate (1894) | $2.3102 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 3 1 0 2} \\mathbf{~ g}$ |Calculate the volume $V\\left[\\mathrm{~m}^{3}\\right]$ of the vessel used by Rayleigh from the mean mass of chemical nitrogen, which must have been pure nitrogen. Assume that the measurements were carried out at a temperature of $15.0^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn 1894, Lord Rayleigh reported that the mass of chemically prepared nitrogen was different from that of nitrogen extracted from the atmosphere, as shown in Tables 1 and 2. Later, this difference was attributed to the presence of argon in atmospheric nitrogen. The masses of gases were measured by using a glass vessel with a known volume under atmospheric pressure $\\left(1.013 \\cdot 10^{5} \\mathrm{~Pa}\\right)$.\n\nTable 1. Mass of Chemical Nitrogen in the Vessel\n\n| From nitric oxide | $2.3001 \\mathrm{~g}$ |\n| :--- | :--- |\n| From nitrous oxide | $2.2990 \\mathrm{~g}$ |\n| From ammonium nitrite purified at a red heat | $2.2987 \\mathrm{~g}$ |\n| From urea | $2.2985 \\mathrm{~g}$ |\n| From ammonium nitrite purified in the cold | $2.2987 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 2 9 9 0} \\mathbf{~ g}$ |\n\nTable 2. Mass of Atmospheric Nitrogen in the Vessel\n\n| $\\mathrm{O}_{2}$ was removed by hot copper (1892) | $2.3103 \\mathrm{~g}$ |\n| :--- | :--- |\n| $\\mathrm{O}_{2}$ was removed by hot iron (1893) | $2.3100 \\mathrm{~g}$ |\n| $\\mathrm{O}_{2}$ was removed by ferrous hydrate (1894) | $2.3102 \\mathrm{~g}$ |\n| Mean | $\\mathbf{2 . 3 1 0 2} \\mathbf{~ g}$ |\n\nproblem:\nCalculate the volume $V\\left[\\mathrm{~m}^{3}\\right]$ of the vessel used by Rayleigh from the mean mass of chemical nitrogen, which must have been pure nitrogen. Assume that the measurements were carried out at a temperature of $15.0^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{m}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{m}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_513",
        "problem": "乙醇、液氨、水都可以发生自偶电离, 如 $\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{OH}^{-}, \\mathrm{NH}_{3}+\\mathrm{NH}_{3} \\rightleftharpoons \\mathrm{NH}$ ${ }_{4}^{+}+\\mathrm{NH}_{2}^{-}$, 则下列叙述正确的是\nA: 乙醇的电离方程式为: $2 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{2}^{+}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}^{-}$\nB: 乙醇的电离方程式为: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}^{-}+\\mathrm{H}^{+}$\nC: 若液氨的离子积常数为: $1.0 \\times 10^{-28}$, 则液氨浓度为 $1.0 \\times 10^{-14} \\mathrm{~mol} / \\mathrm{L}$\nD: 若可用与 $\\mathrm{pH}$ 相当的定义来规定 $\\mathrm{pOH} 、 \\mathrm{pNH}_{2}$ 等, 则乙醇中与 $\\mathrm{pH}$ 相当的为 $\\operatorname{lgc}\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{2}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n乙醇、液氨、水都可以发生自偶电离, 如 $\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\mathrm{O} \\rightleftharpoons \\mathrm{H}_{3} \\mathrm{O}^{+}+\\mathrm{OH}^{-}, \\mathrm{NH}_{3}+\\mathrm{NH}_{3} \\rightleftharpoons \\mathrm{NH}$ ${ }_{4}^{+}+\\mathrm{NH}_{2}^{-}$, 则下列叙述正确的是\n\nA: 乙醇的电离方程式为: $2 \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{2}^{+}+\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}^{-}$\nB: 乙醇的电离方程式为: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH} \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{O}^{-}+\\mathrm{H}^{+}$\nC: 若液氨的离子积常数为: $1.0 \\times 10^{-28}$, 则液氨浓度为 $1.0 \\times 10^{-14} \\mathrm{~mol} / \\mathrm{L}$\nD: 若可用与 $\\mathrm{pH}$ 相当的定义来规定 $\\mathrm{pOH} 、 \\mathrm{pNH}_{2}$ 等, 则乙醇中与 $\\mathrm{pH}$ 相当的为 $\\operatorname{lgc}\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}_{2}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_24",
        "problem": "Which statement about the properties of barium chloride and mercury(II) chloride is correct?\nA: $\\mathrm{BaCl}_{2}$ has a higher melting point than $\\mathrm{HgCl}_{2}$.\nB: $\\mathrm{BaCl}_{2}$ has a higher solubility in nonpolar solvents than $\\mathrm{HgCl}_{2}$.\nC: $\\mathrm{BaCl}_{2}$ has a higher vapor pressure than $\\mathrm{HgCl}_{2}$.\nD: Molten $\\mathrm{BaCl}_{2}$ has a lower electrical conductivity than molten $\\mathrm{HgCl}_{2}$.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich statement about the properties of barium chloride and mercury(II) chloride is correct?\n\nA: $\\mathrm{BaCl}_{2}$ has a higher melting point than $\\mathrm{HgCl}_{2}$.\nB: $\\mathrm{BaCl}_{2}$ has a higher solubility in nonpolar solvents than $\\mathrm{HgCl}_{2}$.\nC: $\\mathrm{BaCl}_{2}$ has a higher vapor pressure than $\\mathrm{HgCl}_{2}$.\nD: Molten $\\mathrm{BaCl}_{2}$ has a lower electrical conductivity than molten $\\mathrm{HgCl}_{2}$.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_555",
        "problem": "室温下, 甘氨酸在水溶液中主要以 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-} 、 \\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}$ 和 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}$三种微粒形式存在, $0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 甘氨酸溶液中 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO} 、 \\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}$ 和 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}$-的浓度对数值与 $\\mathrm{pH}$ 的关系如图所示。下列说法不正确的是\n\n[图1]\nA: 曲线(3)表示 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}^{-}$\nB: 甘氨酸溶液显弱酸性\nC: $\\mathrm{pH}=7$ 时, $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)$\nD: $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COONa}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温下, 甘氨酸在水溶液中主要以 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-} 、 \\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}$ 和 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}$三种微粒形式存在, $0.001 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 甘氨酸溶液中 $\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO} 、 \\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}$ 和 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}$-的浓度对数值与 $\\mathrm{pH}$ 的关系如图所示。下列说法不正确的是\n\n[图1]\n\nA: 曲线(3)表示 $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}^{-}$\nB: 甘氨酸溶液显弱酸性\nC: $\\mathrm{pH}=7$ 时, $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COOH}\\right)$\nD: $\\mathrm{H}_{2} \\mathrm{NCH}_{2} \\mathrm{COONa}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-080.jpg?height=485&width=942&top_left_y=1345&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_146",
        "problem": "Removing all lone pairs of electrons on the central atom of $\\mathrm{ClF}_{3}$ would change the geometry\nA: from trigonal pyramidal to trigonal planar\nB: from $T$-shaped to trigonal planar\nC: from trigonal bipyramidal to trigonal pyramidal\nD: from trigonal bipyramidal to trigonal planar\nE: minimally. The shape would remain trigonal planar.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRemoving all lone pairs of electrons on the central atom of $\\mathrm{ClF}_{3}$ would change the geometry\n\nA: from trigonal pyramidal to trigonal planar\nB: from $T$-shaped to trigonal planar\nC: from trigonal bipyramidal to trigonal pyramidal\nD: from trigonal bipyramidal to trigonal planar\nE: minimally. The shape would remain trigonal planar.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_803",
        "problem": "常温下, 向一定浓度邻苯二甲酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{~A}\\right.$ 表示 $)$ 溶液中通入 $\\mathrm{HCl}$ 气体, 保持溶液体积和温度不变, 测得 $-\\lg \\mathrm{X}$ 与 $\\mathrm{pOH}\\left[\\mathrm{X}\\right.$ 为 $\\left.\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right) 、 \\mathrm{c}\\left(\\mathrm{A}^{2-}\\right) 、 \\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)} ; \\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$的变化关系如图所示。下列说法不正确的是\n\n[图1]\nA: 曲线 $\\mathrm{L}_{1}$ 表示 $-\\operatorname{lgc}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\nB: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=1 \\times 10^{-3}$\nC: 水的电离程度: $a>b>c$\nD: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向一定浓度邻苯二甲酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{~A}\\right.$ 表示 $)$ 溶液中通入 $\\mathrm{HCl}$ 气体, 保持溶液体积和温度不变, 测得 $-\\lg \\mathrm{X}$ 与 $\\mathrm{pOH}\\left[\\mathrm{X}\\right.$ 为 $\\left.\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right) 、 \\mathrm{c}\\left(\\mathrm{A}^{2-}\\right) 、 \\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)} ; \\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right]$的变化关系如图所示。下列说法不正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{L}_{1}$ 表示 $-\\operatorname{lgc}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\nB: $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=1 \\times 10^{-3}$\nC: 水的电离程度: $a>b>c$\nD: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-001.jpg?height=380&width=697&top_left_y=804&top_left_x=354"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_478",
        "problem": "利用 $\\mathrm{H}_{2}$ 和 $\\mathrm{CO}$ 反应生成 $\\mathrm{CH}_{4}$ 的过程中主要涉及的反应如下:\n\n反应I $\\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g})=\\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta H_{1}=-206.2 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n反应II $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})=\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta H_{2}=-41.2 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n$\\left[\\mathrm{CH}_{4}\\right.$ 的产率 $\\frac{\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)}{\\mathrm{n}(\\mathrm{CO})} \\times 100 \\%, \\mathrm{CH}_{4}$ 的选择性 $\\left.\\frac{\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)_{\\text {生成 }}}{\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)_{\\text {生成 }}+\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)_{\\text {生成 }}} \\times 100 \\%\\right]$ 。保持温度\n\n一定, 在固定容积的密闭容器中进行上述反应, 平衡时 $\\mathrm{CH}_{4}$ 和 $\\mathrm{CO}_{2}$ 的产率及 $\\mathrm{CO}$ 和 $\\mathrm{H}_{2}$\n\n的转化率随 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 的变化情况如图所示。下列说法不正确的是\n\n[图1]\nA: 当容器内气体总压不变时, 反应II达到平衡状态\nB: 曲线 $\\mathrm{c}$ 表示 $\\mathrm{CH}_{4}$ 的产率随 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 的变化\nC: $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}=0.5$, 反应达平衡时, $\\mathrm{CH}_{4}$ 的选择性为 $50 \\%$\nD: 随着 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 增大, $\\mathrm{CO}_{2}$ 的选择性先增大后减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n利用 $\\mathrm{H}_{2}$ 和 $\\mathrm{CO}$ 反应生成 $\\mathrm{CH}_{4}$ 的过程中主要涉及的反应如下:\n\n反应I $\\mathrm{CO}(\\mathrm{g})+3 \\mathrm{H}_{2}(\\mathrm{~g})=\\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta H_{1}=-206.2 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n反应II $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})=\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\quad \\Delta H_{2}=-41.2 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n$\\left[\\mathrm{CH}_{4}\\right.$ 的产率 $\\frac{\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)}{\\mathrm{n}(\\mathrm{CO})} \\times 100 \\%, \\mathrm{CH}_{4}$ 的选择性 $\\left.\\frac{\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)_{\\text {生成 }}}{\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)_{\\text {生成 }}+\\mathrm{n}\\left(\\mathrm{CH}_{4}\\right)_{\\text {生成 }}} \\times 100 \\%\\right]$ 。保持温度\n\n一定, 在固定容积的密闭容器中进行上述反应, 平衡时 $\\mathrm{CH}_{4}$ 和 $\\mathrm{CO}_{2}$ 的产率及 $\\mathrm{CO}$ 和 $\\mathrm{H}_{2}$\n\n的转化率随 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 的变化情况如图所示。下列说法不正确的是\n\n[图1]\n\nA: 当容器内气体总压不变时, 反应II达到平衡状态\nB: 曲线 $\\mathrm{c}$ 表示 $\\mathrm{CH}_{4}$ 的产率随 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 的变化\nC: $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}=0.5$, 反应达平衡时, $\\mathrm{CH}_{4}$ 的选择性为 $50 \\%$\nD: 随着 $\\frac{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}{\\mathrm{n}(\\mathrm{CO})}$ 增大, $\\mathrm{CO}_{2}$ 的选择性先增大后减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-098.jpg?height=477&width=571&top_left_y=915&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_303",
        "problem": "A compound of carbon, hydrogen and oxygen is found to be $52.13 \\%$ carbon by mass, $13.13 \\%$ hydrogen by mass, and $34.74 \\%$ oxygen by mass. What is the simplest formula of the compound?\n$\\mathrm{H}, 1.008 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nC, $12.01 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nO, $16.00 \\mathrm{~g} \\mathrm{~mol}^{-1}$\nA: $\\mathrm{C}_{5} \\mathrm{H}_{8} \\mathrm{O}$\nB: $\\mathrm{C}_{3} \\mathrm{H}_{4} \\mathrm{O}_{3}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}$\nD: $\\mathrm{CH}_{2} \\mathrm{O}_{2}$\nE: $\\mathrm{CHO}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA compound of carbon, hydrogen and oxygen is found to be $52.13 \\%$ carbon by mass, $13.13 \\%$ hydrogen by mass, and $34.74 \\%$ oxygen by mass. What is the simplest formula of the compound?\n$\\mathrm{H}, 1.008 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nC, $12.01 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nO, $16.00 \\mathrm{~g} \\mathrm{~mol}^{-1}$\n\nA: $\\mathrm{C}_{5} \\mathrm{H}_{8} \\mathrm{O}$\nB: $\\mathrm{C}_{3} \\mathrm{H}_{4} \\mathrm{O}_{3}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}$\nD: $\\mathrm{CH}_{2} \\mathrm{O}_{2}$\nE: $\\mathrm{CHO}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_629",
        "problem": "某种离子型铁的氧化物晶胞如图所示, 它由 $\\mathrm{A} 、 \\mathrm{~B}$ 单元组成。若通过 $\\mathrm{Li}^{+}$嵌入或脱嵌晶胞的棱心和体心, 可将该晶体设计为某锂电池的正极材料。有关说法错误的是\n[图1]\n\n已知: 脱嵌率 $=\\frac{\\text { 一个晶胞中脱嵌出的 } \\mathrm{Li}^{+} \\text {数 }}{\\text { 一个晶胞中嵌入 } \\mathrm{Li}^{+} \\text {的最大值 }} \\times 100 \\%$\nA: $\\mathrm{Li}^{+}$脱嵌过程为该电池放电过程\nB: 充电时, 该锂电池的正极反应为 $\\mathrm{LiFe}_{6} \\mathrm{O}_{8}-\\mathrm{xe}^{-}=\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{Fe}_{6} \\mathrm{O}_{8}+\\mathrm{xLi}^{+}$\nC: 若该正极材料中 $n\\left(\\mathrm{Fe}^{2+}\\right): n\\left(\\mathrm{Fe}^{3+}\\right)=5: 7$, 则脱嵌率为 $50 \\%$\nD: $\\mathrm{B}_{1}$ 中 $\\mathrm{M}$ 原子分数坐标为 $(0,0,0)$, 则 $\\mathrm{A}_{3}$ 中 $\\mathrm{Q}$ 原子分数坐标为 $\\left(\\frac{7}{8}, \\frac{7}{8}, \\frac{7}{8}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某种离子型铁的氧化物晶胞如图所示, 它由 $\\mathrm{A} 、 \\mathrm{~B}$ 单元组成。若通过 $\\mathrm{Li}^{+}$嵌入或脱嵌晶胞的棱心和体心, 可将该晶体设计为某锂电池的正极材料。有关说法错误的是\n[图1]\n\n已知: 脱嵌率 $=\\frac{\\text { 一个晶胞中脱嵌出的 } \\mathrm{Li}^{+} \\text {数 }}{\\text { 一个晶胞中嵌入 } \\mathrm{Li}^{+} \\text {的最大值 }} \\times 100 \\%$\n\nA: $\\mathrm{Li}^{+}$脱嵌过程为该电池放电过程\nB: 充电时, 该锂电池的正极反应为 $\\mathrm{LiFe}_{6} \\mathrm{O}_{8}-\\mathrm{xe}^{-}=\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{Fe}_{6} \\mathrm{O}_{8}+\\mathrm{xLi}^{+}$\nC: 若该正极材料中 $n\\left(\\mathrm{Fe}^{2+}\\right): n\\left(\\mathrm{Fe}^{3+}\\right)=5: 7$, 则脱嵌率为 $50 \\%$\nD: $\\mathrm{B}_{1}$ 中 $\\mathrm{M}$ 原子分数坐标为 $(0,0,0)$, 则 $\\mathrm{A}_{3}$ 中 $\\mathrm{Q}$ 原子分数坐标为 $\\left(\\frac{7}{8}, \\frac{7}{8}, \\frac{7}{8}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-037.jpg?height=376&width=1444&top_left_y=914&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_117",
        "problem": "The first ionization energy of phosphorus is lower than that of:\nA: chlorine\nB: silicon\nC: sodium\nD: barium\nE: bismuth\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe first ionization energy of phosphorus is lower than that of:\n\nA: chlorine\nB: silicon\nC: sodium\nD: barium\nE: bismuth\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_122",
        "problem": "A common method for cleaning up an acid spill is to spread sodium carbonate over the spill to neutralize it. If $50.0 \\mathrm{~mL}$ of $0.75 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ spilled on the countertop, what is the minimum amount of sodium carbonate required to neutralize the spill?\nA: $1.6 \\mathrm{~g}$\nB: $1.9 \\mathrm{~g}$\nC: $2.0 \\mathrm{~g}$\nD: $3.1 \\mathrm{~g}$\nE: $4.0 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA common method for cleaning up an acid spill is to spread sodium carbonate over the spill to neutralize it. If $50.0 \\mathrm{~mL}$ of $0.75 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{HCl}$ spilled on the countertop, what is the minimum amount of sodium carbonate required to neutralize the spill?\n\nA: $1.6 \\mathrm{~g}$\nB: $1.9 \\mathrm{~g}$\nC: $2.0 \\mathrm{~g}$\nD: $3.1 \\mathrm{~g}$\nE: $4.0 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_920",
        "problem": "据 2022 年 1 月统计, 我国光伏发电并网装机容量突破 3 亿千瓦, 连续七年稳居全球首位。已知四甲基氢氧化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}\\right]$ 常用作电子工业清洗剂, 以四甲基氯化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NCl}\\right]$ 为原料, 采用电渗析法合成 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}\\right]$, 工作原理如图。下列说法正确的是\n\n[图1]\nA: 光伏并网发电装置中 $\\mathrm{N}$ 型半导体为正极\nB: $\\mathrm{c} 、 \\mathrm{e}$ 为阳离子交换膜, $\\mathrm{d}$ 为阴离子交换膜\nC: 保持电流恒定, 温度越高合成四甲基氢氧化铵的速率越快\nD: 制备 $182 \\mathrm{~g}$ 四甲基氢氧化铵, 两极共产生 $33.6 \\mathrm{~L}$ 气体(标准状况)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n据 2022 年 1 月统计, 我国光伏发电并网装机容量突破 3 亿千瓦, 连续七年稳居全球首位。已知四甲基氢氧化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}\\right]$ 常用作电子工业清洗剂, 以四甲基氯化铵 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NCl}\\right]$ 为原料, 采用电渗析法合成 $\\left[\\left(\\mathrm{CH}_{3}\\right)_{4} \\mathrm{NOH}\\right]$, 工作原理如图。下列说法正确的是\n\n[图1]\n\nA: 光伏并网发电装置中 $\\mathrm{N}$ 型半导体为正极\nB: $\\mathrm{c} 、 \\mathrm{e}$ 为阳离子交换膜, $\\mathrm{d}$ 为阴离子交换膜\nC: 保持电流恒定, 温度越高合成四甲基氢氧化铵的速率越快\nD: 制备 $182 \\mathrm{~g}$ 四甲基氢氧化铵, 两极共产生 $33.6 \\mathrm{~L}$ 气体(标准状况)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-29.jpg?height=449&width=1333&top_left_y=141&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_6",
        "problem": "A gas-phase hydrogen atom absorbs a photon of visible light and then emits a photon of ultraviolet light. What may be concluded about its initial and its final values of the principal quantum number $n$ ?\nA: Initially $n=1$ and finally $n=2$\nB: Initially $n=2$ and finally $n=1$\nC: Initially $n=2$ and finally $n=4$\nD: This scenario is impossible because ultraviolet light is more energetic than visible light.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA gas-phase hydrogen atom absorbs a photon of visible light and then emits a photon of ultraviolet light. What may be concluded about its initial and its final values of the principal quantum number $n$ ?\n\nA: Initially $n=1$ and finally $n=2$\nB: Initially $n=2$ and finally $n=1$\nC: Initially $n=2$ and finally $n=4$\nD: This scenario is impossible because ultraviolet light is more energetic than visible light.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_588",
        "problem": "有机物 $\\mathrm{X}$ 是一种重要的有机合成中间体。为研究 $\\mathrm{X}$ 的组成与结构, 进行了如下实验:\n\n(1)有机物 X 的质谱图如图所示。\n\n[图1]\n\n(2)将 $10.0 \\mathrm{~g} \\mathrm{X}$ 在足量 $\\mathrm{O}_{2}$ 中充分燃烧, 并将其产物依次通过足量的无水 $\\mathrm{CaCl}_{2}$ 和 $\\mathrm{KOH}$ 浓溶液, 发现无水 $\\mathrm{CaCl}_{2}$ 增重 $7.2 \\mathrm{~g}, \\mathrm{KOH}$ 浓溶液增重 $22.0 \\mathrm{~g}$ 。\n\n(3)经红外光谱测定, 有机物 $\\mathrm{X}$ 中含有醛基; 有机物 $\\mathrm{X}$ 的核磁共振氢谱图上有 2 个吸收峰, 峰面积之比为 $3: 1$ 。\n\n下列有关说法中正确的是\nA: 该有机物 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{8} \\mathrm{O}_{2}$\nB: 该有机物 $\\mathrm{X}$ 的相对分子质量为 55\nC: 该有机物 $\\mathrm{X}$ 的结构简式为 $\\mathrm{CHO}\\left(\\mathrm{CH}_{2}\\right)_{3} \\mathrm{CHO}$\nD: 与 X 官能团种类、数量完全相同的同分异构体还有 3 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n有机物 $\\mathrm{X}$ 是一种重要的有机合成中间体。为研究 $\\mathrm{X}$ 的组成与结构, 进行了如下实验:\n\n(1)有机物 X 的质谱图如图所示。\n\n[图1]\n\n(2)将 $10.0 \\mathrm{~g} \\mathrm{X}$ 在足量 $\\mathrm{O}_{2}$ 中充分燃烧, 并将其产物依次通过足量的无水 $\\mathrm{CaCl}_{2}$ 和 $\\mathrm{KOH}$ 浓溶液, 发现无水 $\\mathrm{CaCl}_{2}$ 增重 $7.2 \\mathrm{~g}, \\mathrm{KOH}$ 浓溶液增重 $22.0 \\mathrm{~g}$ 。\n\n(3)经红外光谱测定, 有机物 $\\mathrm{X}$ 中含有醛基; 有机物 $\\mathrm{X}$ 的核磁共振氢谱图上有 2 个吸收峰, 峰面积之比为 $3: 1$ 。\n\n下列有关说法中正确的是\n\nA: 该有机物 $\\mathrm{X}$ 的分子式为 $\\mathrm{C}_{5} \\mathrm{H}_{8} \\mathrm{O}_{2}$\nB: 该有机物 $\\mathrm{X}$ 的相对分子质量为 55\nC: 该有机物 $\\mathrm{X}$ 的结构简式为 $\\mathrm{CHO}\\left(\\mathrm{CH}_{2}\\right)_{3} \\mathrm{CHO}$\nD: 与 X 官能团种类、数量完全相同的同分异构体还有 3 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-61.jpg?height=391&width=728&top_left_y=1095&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1247",
        "problem": "Oxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nCalculate how much oxygen (in $\\mathrm{mol} \\mathrm{s}^{-1}$ ) can be deposited in tissue by blood with normal $\\mathrm{Hb}(1)$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOxygen is of vital importance for all of us. Oxygen enters the body via the lungs and is transported to the tissues in our body by blood. There it can deliver energy by the oxidation of sugars:\n\n$$\n\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}+6 \\mathrm{O}_{2} \\rightarrow 6 \\mathrm{CO}_{2}+6 \\mathrm{H}_{2} \\mathrm{O}\n$$\n\nThis reaction releases $400 \\mathrm{~kJ}$ of energy per mol of oxygen. $\\mathrm{O}_{2}$ uptake by blood is at four heme $(\\mathrm{Hm}$ ) groups in the protein hemoglobin $(\\mathrm{Hb})$.\n\nFree $\\mathrm{Hm}$ consists of an $\\mathrm{Fe}^{2+}$ ion attached to four $\\mathrm{N}$ atoms of a porphyrin ${ }^{2-}$ ligand. Oxygen can bind at the coordination site of $\\mathrm{Fe}^{2+}$ giving a $\\mathrm{HmO}_{2}$ complex. Carbon monoxide can be complexed similarly, giving a $\\mathrm{Hm} \\cdot \\mathrm{CO}$ complex. $\\mathrm{CO}$ is a poison as it binds more strongly to $\\mathrm{Hm}$ than $\\mathrm{O}_{2}$ does. The equilibrium constant $K_{1}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{CO} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{CO}   \\qquad(1)\n$$\n\nis 10000 times larger than the equilibrium constant $K_{2}$ for the reaction:\n\n$$\n\\mathrm{Hm}+\\mathrm{O}_{2} \\rightleftharpoons \\mathrm{Hm} \\cdot \\mathrm{O}_{2} \\qquad(2)\n$$\n\nEach $\\mathrm{Hb}$ molecule can take up four molecules of $\\mathrm{O}_{2}$. Blood in contact with $\\mathrm{O}_{2}$ absorbs a fraction of this amount, depending on the oxygen pressure, as shown in Figure 1 (curve 1). Also shown are the curves (2) and (3) for blood with two kinds of deficient $\\mathrm{Hb}$. These occur in patients with certain hereditary diseases.\n\nRelevant data: $\\mathrm{O}_{2}$ pressure in lungs is $15 \\mathrm{kPa}$; in the muscles it is $2 \\mathrm{kPa}$. The maximum[^13]flow of blood through heart and lungs is $4 \\times 10^{-4} \\mathrm{~m}^{3} \\mathrm{~s}^{-1}$. The red cells in blood occupy $40 \\%$ of the blood volume; inside the cells the concentration of $\\mathrm{Hb}$ is $340 \\mathrm{~kg} \\mathrm{~m}^{-3}$; $\\mathrm{Hb}$ has a molar mass of $64 \\mathrm{~kg} \\mathrm{~mol}^{-1} . R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1} . T=298 \\mathrm{~K}$.\n\n[figure1]\n\nFigure 1\n\nCalculate how much oxygen (in $\\mathrm{mol} \\mathrm{s}^{-1}$ ) can be deposited in tissue by blood with normal $\\mathrm{Hb}(1)$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{~s}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-418.jpg?height=823&width=1400&top_left_y=568&top_left_x=288"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{~s}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_267",
        "problem": "When the given amounts of each reagent are mixed together, which of the following will release the largest mass of $\\mathrm{CO}_{2}$ ?\nA: $\\quad 0.3 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\nB: $\\quad 0.2 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.3 \\mathrm{~mol} \\mathrm{HCl}$\nC: $0.2 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.2 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\nD: $0.3 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.3 \\mathrm{~mol} \\mathrm{HCl}$\nE: $\\quad 0.1 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.3 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\nF: $\\quad 0.1 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.2 \\mathrm{~mol} \\mathrm{HCl}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen the given amounts of each reagent are mixed together, which of the following will release the largest mass of $\\mathrm{CO}_{2}$ ?\n\nA: $\\quad 0.3 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.1 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\nB: $\\quad 0.2 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.3 \\mathrm{~mol} \\mathrm{HCl}$\nC: $0.2 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.2 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\nD: $0.3 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.3 \\mathrm{~mol} \\mathrm{HCl}$\nE: $\\quad 0.1 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.3 \\mathrm{~mol} \\mathrm{H}_{2} \\mathrm{SO}_{4}$\nF: $\\quad 0.1 \\mathrm{~mol} \\mathrm{CuCO}_{3}$ and $0.2 \\mathrm{~mol} \\mathrm{HCl}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1405",
        "problem": "The electronic energy levels $E_{n}^{H}$ of a hydrogen atom are given by the equation\n\n$$\nE_{n}^{\\mathrm{H}}=-\\frac{R y}{n^{2}} \\quad(n=1,2,3 \\cdots)\n$$\n\nHere $n$ is a principal quantum number, and $R y$ is a constant with dimensions of energy. The energy from $n=1$ to $n=2$ of the hydrogen atom is $10.2 \\mathrm{eV}$.Calculate the ionization energy $E_{\\mathrm{B}}(\\mathrm{eV})$ of the hydrogen atom to the first decimal place.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe electronic energy levels $E_{n}^{H}$ of a hydrogen atom are given by the equation\n\n$$\nE_{n}^{\\mathrm{H}}=-\\frac{R y}{n^{2}} \\quad(n=1,2,3 \\cdots)\n$$\n\nHere $n$ is a principal quantum number, and $R y$ is a constant with dimensions of energy. The energy from $n=1$ to $n=2$ of the hydrogen atom is $10.2 \\mathrm{eV}$.\n\nproblem:\nCalculate the ionization energy $E_{\\mathrm{B}}(\\mathrm{eV})$ of the hydrogen atom to the first decimal place.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of eV, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "eV"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1492",
        "problem": "Nitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.Experimental rate law: $R=k\\left(p_{\\mathrm{NO}}\\right)^{2} p_{\\mathrm{H}_{2}}$$. Calculate the rate constant.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nNitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.\n\nproblem:\nExperimental rate law: $R=k\\left(p_{\\mathrm{NO}}\\right)^{2} p_{\\mathrm{H}_{2}}$$. Calculate the rate constant.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $ torr $^{-2} \\mathrm{~s}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$ torr $^{-2} \\mathrm{~s}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_455",
        "problem": "含氯苯的废水可通过加入适量乙酸钠, 设计成微生物电池将氯苯转化为苯而除去,其原理如图所示。下列叙述正确的是\n\n[图1]\nA: 氯苯被氧化生成苯\nB: $\\mathrm{N}$ 极为电池的负极\nC: $\\mathrm{M}$ 极的电极反应式为 $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{Cl}+\\mathrm{e}^{-}=\\mathrm{C}_{6} \\mathrm{H}_{6}+\\mathrm{Cl}^{-}$\nD: 每生成 $1 \\mathrm{molCO}_{2}$, 由 $\\mathrm{N}$ 极区进入 $\\mathrm{M}$ 极区的 $\\mathrm{H}^{+}$为 $4 \\mathrm{~mol}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n含氯苯的废水可通过加入适量乙酸钠, 设计成微生物电池将氯苯转化为苯而除去,其原理如图所示。下列叙述正确的是\n\n[图1]\n\nA: 氯苯被氧化生成苯\nB: $\\mathrm{N}$ 极为电池的负极\nC: $\\mathrm{M}$ 极的电极反应式为 $\\mathrm{C}_{6} \\mathrm{H}_{5} \\mathrm{Cl}+\\mathrm{e}^{-}=\\mathrm{C}_{6} \\mathrm{H}_{6}+\\mathrm{Cl}^{-}$\nD: 每生成 $1 \\mathrm{molCO}_{2}$, 由 $\\mathrm{N}$ 极区进入 $\\mathrm{M}$ 极区的 $\\mathrm{H}^{+}$为 $4 \\mathrm{~mol}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-35.jpg?height=422&width=862&top_left_y=1705&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1140",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## Mendeleev's short-form table\n\nIn the second version of his table, Mendeleev updated the mass of indium and correctly positioned both it and thallium. He also arranged the Groups of similar elements vertically but there are too many elements within each Group; Mendeleev mixed elements from the transition metals with those from the main block of the periodic table. There is sense to this since the elements which Mendeleev places in a given Group form similar compounds - notably their oxides and their hydrides.\n\n[figure1]\n\nMendeleev's second periodic table from 1871\n\nThe element from Mendeleev's Group V, i.e., nitrogen, forms oxides with the elements in the +5 oxidation state.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## Mendeleev's short-form table\n\nIn the second version of his table, Mendeleev updated the mass of indium and correctly positioned both it and thallium. He also arranged the Groups of similar elements vertically but there are too many elements within each Group; Mendeleev mixed elements from the transition metals with those from the main block of the periodic table. There is sense to this since the elements which Mendeleev places in a given Group form similar compounds - notably their oxides and their hydrides.\n\n[figure1]\n\nMendeleev's second periodic table from 1871\n\nThe element from Mendeleev's Group V, i.e., nitrogen, forms oxides with the elements in the +5 oxidation state.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_bf14c3e936c3f7031d02g-05.jpg?height=742&width=1567&top_left_y=183&top_left_x=290"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_208",
        "problem": "What is the total number of valence electrons in the $\\mathrm{PO}_{2}{ }^{3-}$ ion?\nA: 15\nB: 17\nC: 20\nD: 30\nE: 34\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the total number of valence electrons in the $\\mathrm{PO}_{2}{ }^{3-}$ ion?\n\nA: 15\nB: 17\nC: 20\nD: 30\nE: 34\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1474",
        "problem": "$\\mathbf{H}$ is stable to acids, but is hydrolysed in alkali. A $0.7934 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) was heated with excess aqueous sodium hydroxide. Cobalt(III) oxide was formed and ammonia gas given off. The ammonia produced was distilled off and absorbed into $50.0 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c_{\\mathrm{HCl}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$. The residual $\\mathrm{HCl}$ required $24.8 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{KOH}$ solution $\\left(c_{\\mathrm{KOH}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to be neutralized.\n\nThe remaining suspension of cobalt(III) oxide was allowed to cool, approximately $1 \\mathrm{~g}$ of potassium iodide was added, and then the mixture was acidified with aqueous $\\mathrm{HCl}$. The liberated iodine was then titrated with aqueous solution of sodium thiosulfate $(c=0.200$ mol $\\mathrm{dm}^{-3}$ ) and required $21.0 \\mathrm{~cm}^{3}$ for complete reaction.Calculate the percentage, by mass, of cobalt in $\\mathbf{H}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n$\\mathbf{H}$ is stable to acids, but is hydrolysed in alkali. A $0.7934 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) was heated with excess aqueous sodium hydroxide. Cobalt(III) oxide was formed and ammonia gas given off. The ammonia produced was distilled off and absorbed into $50.0 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c_{\\mathrm{HCl}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$. The residual $\\mathrm{HCl}$ required $24.8 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{KOH}$ solution $\\left(c_{\\mathrm{KOH}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to be neutralized.\n\nThe remaining suspension of cobalt(III) oxide was allowed to cool, approximately $1 \\mathrm{~g}$ of potassium iodide was added, and then the mixture was acidified with aqueous $\\mathrm{HCl}$. The liberated iodine was then titrated with aqueous solution of sodium thiosulfate $(c=0.200$ mol $\\mathrm{dm}^{-3}$ ) and required $21.0 \\mathrm{~cm}^{3}$ for complete reaction.\n\nproblem:\nCalculate the percentage, by mass, of cobalt in $\\mathbf{H}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_887",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液 $(\\mathrm{pH}=7.8$ ) 中通入 $\\mathrm{HCl}$ 或加入 $\\mathrm{NaOH}$ 调节 $\\mathrm{pH}$, 不考虑溶液体积变化且过程中无气体逸出。含碳(或氮)微粒的分布分数 $\\delta$ [如: $\\left.\\delta\\left(\\mathrm{HCO}_{3}^{-}=\\frac{c\\left(\\mathrm{HCO}_{3}^{-}\\right)}{c_{\\text {总 }}(\\text { 含碳微粒 })}\\right)\\right]$ 与 $\\mathrm{pH}$ 关系如图。下列说法错误的是\n\n[图1]\nA: $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液在 $\\mathrm{pH}=7.8$ 时存在: $c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nB: 曲线 $\\mathrm{M}$ 表示 $\\mathrm{NH}_{4}^{+}$的分布分数随 $\\mathrm{pH}$ 变化\nC: $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的 $\\mathrm{K}_{\\mathrm{a} 2}=10^{-10.3}$\nD: 该体系在 $\\mathrm{pH}=8.3$ 时, 溶液中 $c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=\\frac{1}{11} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 向 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液 $(\\mathrm{pH}=7.8$ ) 中通入 $\\mathrm{HCl}$ 或加入 $\\mathrm{NaOH}$ 调节 $\\mathrm{pH}$, 不考虑溶液体积变化且过程中无气体逸出。含碳(或氮)微粒的分布分数 $\\delta$ [如: $\\left.\\delta\\left(\\mathrm{HCO}_{3}^{-}=\\frac{c\\left(\\mathrm{HCO}_{3}^{-}\\right)}{c_{\\text {总 }}(\\text { 含碳微粒 })}\\right)\\right]$ 与 $\\mathrm{pH}$ 关系如图。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{NH}_{4} \\mathrm{HCO}_{3}$ 溶液在 $\\mathrm{pH}=7.8$ 时存在: $c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nB: 曲线 $\\mathrm{M}$ 表示 $\\mathrm{NH}_{4}^{+}$的分布分数随 $\\mathrm{pH}$ 变化\nC: $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的 $\\mathrm{K}_{\\mathrm{a} 2}=10^{-10.3}$\nD: 该体系在 $\\mathrm{pH}=8.3$ 时, 溶液中 $c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=\\frac{1}{11} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-029.jpg?height=517&width=417&top_left_y=869&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_918",
        "problem": "学习了酸碱中和滴定实验后, 常温下, 某同学在两个相同的特制容器中分别加入 $20 \\mathrm{~mL} 0.4 / \\mathrm{mol}_{2} \\mathrm{~L}_{2} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液和 $40 \\mathrm{~mL} 0.2 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液，再分别用 $0.4 \\mathrm{~mol} / \\mathrm{L}$ 盐酸滴定, 利用 $\\mathrm{pH}$ 计和压力传感器检测, 得到如图曲线。下列说法正确的是\n\n[图1]\nA: 水的电离程度: $e$ 点 $<f$ 点\nB: 当滴加盐酸的体积为 $V_{1} \\mathrm{~mL}$ 时 ( $\\mathrm{a}$ 点、 $\\mathrm{b}$ 点), 所发生的反应用离子方程式表示为 $\\mathrm{HCO}$ ${ }_{3}^{-}+\\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2} \\uparrow$\nC: 若用与盐酸等物质的量浓度的醋酸溶液滴定, 则曲线甲中点位置不变\nD: $\\mathrm{c}(20,8.3)$ 点处, $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\left(0.2+10^{-5.7}\\right) \\mathrm{mol} / \\mathrm{L}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n学习了酸碱中和滴定实验后, 常温下, 某同学在两个相同的特制容器中分别加入 $20 \\mathrm{~mL} 0.4 / \\mathrm{mol}_{2} \\mathrm{~L}_{2} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液和 $40 \\mathrm{~mL} 0.2 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液，再分别用 $0.4 \\mathrm{~mol} / \\mathrm{L}$ 盐酸滴定, 利用 $\\mathrm{pH}$ 计和压力传感器检测, 得到如图曲线。下列说法正确的是\n\n[图1]\n\nA: 水的电离程度: $e$ 点 $<f$ 点\nB: 当滴加盐酸的体积为 $V_{1} \\mathrm{~mL}$ 时 ( $\\mathrm{a}$ 点、 $\\mathrm{b}$ 点), 所发生的反应用离子方程式表示为 $\\mathrm{HCO}$ ${ }_{3}^{-}+\\mathrm{H}^{+}=\\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{2} \\uparrow$\nC: 若用与盐酸等物质的量浓度的醋酸溶液滴定, 则曲线甲中点位置不变\nD: $\\mathrm{c}(20,8.3)$ 点处, $\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\left(0.2+10^{-5.7}\\right) \\mathrm{mol} / \\mathrm{L}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-012.jpg?height=554&width=957&top_left_y=1987&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_268",
        "problem": "Rubidium forms a number of oxides, one of which contains $78.08 \\%$ rubidium by mass. What is the empirical formula of this oxide?\nA: $\\mathrm{Rb}_{6} \\mathrm{O}$\nB: $\\mathrm{RbO}$\nC: $\\mathrm{RbO}_{2}$\nD: $\\mathrm{Rb}_{2} \\mathrm{O}$\nE: $\\mathrm{Rb}_{2} \\mathrm{O}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nRubidium forms a number of oxides, one of which contains $78.08 \\%$ rubidium by mass. What is the empirical formula of this oxide?\n\nA: $\\mathrm{Rb}_{6} \\mathrm{O}$\nB: $\\mathrm{RbO}$\nC: $\\mathrm{RbO}_{2}$\nD: $\\mathrm{Rb}_{2} \\mathrm{O}$\nE: $\\mathrm{Rb}_{2} \\mathrm{O}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_837",
        "problem": "常温下, 向 $1 \\mathrm{LpH}=10$ 的 $\\mathrm{NaOH}$ 溶液中持续通入 $\\mathrm{CO}_{2}$ 。通入的 $\\mathrm{CO}_{2}$ 的体积 $(\\mathrm{V})$ 与溶液中水电离出的 $\\mathrm{OH}^{-}$离子浓度(c)的关系如图所示。下列叙述不正确的是( )\n\n[图1]\nA: $b$ 点溶液中: $c\\left(\\mathrm{OH}^{-}\\right)=1 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{a}$ 点溶液中: 水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=1 \\times 10^{-10} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nD: $\\mathrm{d}$ 点溶液中: 溶液是 $\\mathrm{NaHCO}_{3}$ 和 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的混合液\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $1 \\mathrm{LpH}=10$ 的 $\\mathrm{NaOH}$ 溶液中持续通入 $\\mathrm{CO}_{2}$ 。通入的 $\\mathrm{CO}_{2}$ 的体积 $(\\mathrm{V})$ 与溶液中水电离出的 $\\mathrm{OH}^{-}$离子浓度(c)的关系如图所示。下列叙述不正确的是( )\n\n[图1]\n\nA: $b$ 点溶液中: $c\\left(\\mathrm{OH}^{-}\\right)=1 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $\\mathrm{a}$ 点溶液中: 水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=1 \\times 10^{-10} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nC: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)$\nD: $\\mathrm{d}$ 点溶液中: 溶液是 $\\mathrm{NaHCO}_{3}$ 和 $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的混合液\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-69.jpg?height=437&width=674&top_left_y=884&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1258",
        "problem": "The elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nFrom what mass of apatite should one start if all the reactions are stoichiometric?\n\nStarting with $m_{0}$ of obtained apatite, $m_{1}=5.49 \\mathrm{~g}$ of residue are obtained.\n\n\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 \\text {; }  \\mathrm{S}: 32\n$$\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe elemental phosphorus is present in the nature as phosphate in a complex mineral apatite. This mineral contains, in addition to phosphate, silica and the following ions: $\\mathrm{Ca}^{2+}, \\mathrm{CO}_{3}^{2-}, \\mathrm{SO}_{4}^{2-}, \\mathrm{SiO}_{3}^{2-}$, and $\\mathrm{F}^{-}$.\n\nLet us assume that this mineral is a mixture of tricalcium phosphate, $\\mathrm{Ca}_{3}\\left(\\mathrm{PO}_{4}\\right)_{2}$, calcium sulphate, calcium fluoride, calcium carbonate and silica.\n\nFor uses as fertilizer the calcium bis(dihydrogenphosphate), $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, which is soluble in water, has been prepared. For this purpose, apatite is treated with a mixture of phosphoric and sulphuric acid. At the same time this operation eliminates the majority of impurities.\n\nThe elemental analysis of an apatite gave the following results in which, except of fluorine, the elemental composition is expressed as if the elements were in the form of oxides:\n\n|  | $\\mathrm{CaO}$ | $\\mathrm{P}_{2} \\mathrm{O}_{5}$ | $\\mathrm{SiO}_{2}$ | $\\mathrm{~F}$ | $\\mathrm{SO}_{3}$ | $\\mathrm{CO}_{2}$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $\\%$ by mass | 47.3 | 28.4 | 3.4 | 3.4 | 3.5 | 6.1 |\n\nOperation 1 - A sample of $m_{0}$ of this mineral is treated with $50.0 \\mathrm{~cm}^{3}$ of a solution containing $0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}$ phosphoric and $0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ sulphuric acids. The mixture is completely dehydrated by heating up to about $70{ }^{\\circ} \\mathrm{C}$ avoiding temperature rising above $90{ }^{\\circ} \\mathrm{C}$. This operation is carried out under the hood since toxic gaseous substances are emitted. The dry residue is ground and weighed; $m_{1}$ is the mass of the residue obtained.\n\nIn these conditions only dihydrogenphosphate, $\\mathrm{Ca}\\left(\\mathrm{H}_{2} \\mathrm{PO}_{4}\\right)_{2}$, is formed while silica and silicate do not react.\n\nOperation 2 - $1.00 \\mathrm{~g}$ of this residue is treated with $50.0 \\mathrm{~cm}^{3}$ of water at $40{ }^{\\circ} \\mathrm{C}$, then filtered, dried and weighed. The mass of the residue obtained is $m_{2}$. This new residue is mainly containing gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, whose solubility can be considered as constant between $20^{\\circ} \\mathrm{C}$ and $50{ }^{\\circ} \\mathrm{C}$ and is equal to $2.3 \\mathrm{~g} \\mathrm{dm}^{-3}$.\n\nFrom what mass of apatite should one start if all the reactions are stoichiometric?\n\nStarting with $m_{0}$ of obtained apatite, $m_{1}=5.49 \\mathrm{~g}$ of residue are obtained.\n\n\n\n$$\n\\text { Relative atomic masses of } \\mathrm{P}: 31 ; \\mathrm{Ca}: 40 ; \\mathrm{O}: 16 ; \\mathrm{H}: 1 ; \\mathrm{F}: 19 ; \\mathrm{C}: 12 ; \\mathrm{Si}: 28 \\text {; }  \\mathrm{S}: 32\n$$\n\nValues of $p K: \\quad \\frac{\\mathrm{HSO}_{4}^{-}}{\\mathrm{SO}_{4}^{2-}}=2 \\quad \\frac{\\mathrm{HF}}{\\mathrm{F}^{-}}=3 \\quad \\frac{\\mathrm{H}_{3} \\mathrm{PO}_{4}}{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}=2 \\quad \\frac{\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}}{\\mathrm{HPO}_{4}^{2-}}=7 \\quad \\frac{\\mathrm{HPO}_{4}^{2-}}{\\mathrm{PO}_{4}^{3-}}=12$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1259",
        "problem": "Werner was also the first person to separate the enantiomers of an octahedral compound, $\\mathbf{H}$, which contained no carbon atoms. The compound, $\\mathbf{H}$, is composed of only cobalt, ammonia, chloride and an oxygen species which could be either $\\mathrm{H}_{2} \\mathrm{O}$, or $\\mathrm{HO}^{-}$or $\\mathrm{O}^{2-}$. The compound contains octahedrally coordinated cobalt ions. All of the chloride is easily removed from the compound by titration with aqueous silver nitrate. A $0.2872 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) required $22.8 \\mathrm{~cm}^{3}$ of a silver nitrate solution ( $c=0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) to exchange all of the chloride.\n\nCalculate the percentage, by mass, of chloride in $\\mathbf{H}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nWerner was also the first person to separate the enantiomers of an octahedral compound, $\\mathbf{H}$, which contained no carbon atoms. The compound, $\\mathbf{H}$, is composed of only cobalt, ammonia, chloride and an oxygen species which could be either $\\mathrm{H}_{2} \\mathrm{O}$, or $\\mathrm{HO}^{-}$or $\\mathrm{O}^{2-}$. The compound contains octahedrally coordinated cobalt ions. All of the chloride is easily removed from the compound by titration with aqueous silver nitrate. A $0.2872 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) required $22.8 \\mathrm{~cm}^{3}$ of a silver nitrate solution ( $c=0.100 \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) to exchange all of the chloride.\n\nCalculate the percentage, by mass, of chloride in $\\mathbf{H}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_844",
        "problem": "下列说法正确的是\nA: 铁表面镀铜时, 铁与电源的正极相连, 铜与电源的负极相连\nB: $1 \\mathrm{molCl}_{2}$ 与足量的铁完全反应, 转移的电子数约为 $3 \\times 6.02 \\times 10^{23}$\nC: $3 \\mathrm{C}(\\mathrm{s})+\\mathrm{CaO}(\\mathrm{s})=\\mathrm{CnC}_{2}(\\mathrm{~s})+\\mathrm{CO}(\\mathrm{g})$ 在常温下不能自发进行. 说明该反应的 $\\Delta \\mathrm{H}>0$\nD: 室温下, $\\mathrm{pH}=3$ 的盐酸溶液与 $\\mathrm{pH}=11$ 的氨水等体积混合, 溶液 $\\mathrm{pH}>7$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列说法正确的是\n\nA: 铁表面镀铜时, 铁与电源的正极相连, 铜与电源的负极相连\nB: $1 \\mathrm{molCl}_{2}$ 与足量的铁完全反应, 转移的电子数约为 $3 \\times 6.02 \\times 10^{23}$\nC: $3 \\mathrm{C}(\\mathrm{s})+\\mathrm{CaO}(\\mathrm{s})=\\mathrm{CnC}_{2}(\\mathrm{~s})+\\mathrm{CO}(\\mathrm{g})$ 在常温下不能自发进行. 说明该反应的 $\\Delta \\mathrm{H}>0$\nD: 室温下, $\\mathrm{pH}=3$ 的盐酸溶液与 $\\mathrm{pH}=11$ 的氨水等体积混合, 溶液 $\\mathrm{pH}>7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_743",
        "problem": "$\\mathrm{KHC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ (四草酸钾, 记作 PT)是一种分析试剂。室温时, $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 的 $p K_{a 1} 、 p K_{a 2}$ 分别为 1.23、4.19 $\\left(p K_{a}=-\\lg K_{a}\\right)$ 。下列指定溶液中微粒物质的量浓度关系错误的是\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 溶液中: $c\\left(\\mathrm{~K}^{+}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 中滴加 $\\mathrm{NaOH}$ 至溶液 $\\mathrm{pH}=4.19: \\quad c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{~K}^{+}\\right)<c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)-$ $c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 中滴加 $\\mathrm{NaOH}$ 至溶液呈中性: $c\\left(\\mathrm{~K}^{+}\\right)>c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 与 $0.3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合: $c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{~K}^{+}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+$ $c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)+c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{KHC}_{2} \\mathrm{O}_{4} \\cdot \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$ (四草酸钾, 记作 PT)是一种分析试剂。室温时, $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 的 $p K_{a 1} 、 p K_{a 2}$ 分别为 1.23、4.19 $\\left(p K_{a}=-\\lg K_{a}\\right)$ 。下列指定溶液中微粒物质的量浓度关系错误的是\n\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 溶液中: $c\\left(\\mathrm{~K}^{+}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 中滴加 $\\mathrm{NaOH}$ 至溶液 $\\mathrm{pH}=4.19: \\quad c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{~K}^{+}\\right)<c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)-$ $c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 中滴加 $\\mathrm{NaOH}$ 至溶液呈中性: $c\\left(\\mathrm{~K}^{+}\\right)>c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{PT}$ 与 $0.3 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合: $c\\left(\\mathrm{Na}^{+}\\right)-c\\left(\\mathrm{~K}^{+}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+$ $c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)+c\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_427",
        "problem": "赖氨酸 $\\left[\\mathrm{H}_{3} \\mathrm{~N}^{+}\\left(\\mathrm{CH}_{2}\\right)_{4} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{COO}\\right.$, 用 $\\mathrm{HR}$ 表示 $]$ 是人体必需氨基酸, 其盐酸盐 $\\left(\\mathrm{H}_{3} \\mathrm{RCl}_{2}\\right)$在水溶液中存在如下平衡: $\\mathrm{H}_{3} \\mathrm{R}^{2+} \\stackrel{\\mathrm{K}_{1}}{\\rightleftharpoons} \\mathrm{H}_{2} \\mathrm{R}^{+} \\stackrel{\\mathrm{K}_{2} \\rightleftharpoons}{\\rightleftharpoons} \\mathrm{HR} \\stackrel{\\mathrm{K}_{3}}{\\rightleftharpoons} \\mathrm{R}^{-}$。向一定浓度的 $\\mathrm{H}_{3} \\mathrm{RCl}_{2}$\n溶液中滴加 $\\mathrm{NaOH}$ 溶液, 溶液中 $\\mathrm{H}_{3} \\mathrm{R}^{2+} 、 \\mathrm{H}_{2} \\mathrm{R}^{+} 、 \\mathrm{HR}$ 和 $\\mathrm{R}$ 的分布系数 $\\delta(\\mathrm{x})$ 随 $\\mathrm{pH}$ 变化如图所示。已知 $\\delta(\\mathrm{x})=\\frac{\\mathrm{c}(\\mathrm{x})}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{R}^{2+}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{+}\\right)+\\mathrm{c}(\\mathrm{HR})+\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)}$, 下列表述正确的是\n\n[图1]\nA: $\\frac{\\mathrm{K}_{2}}{\\mathrm{~K}_{1}}>\\frac{\\mathrm{K}_{3}}{\\mathrm{~K}_{2}}$\nB: $M$ 点, $c\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)=2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: $\\mathrm{O}$ 点, $\\mathrm{pH}=\\frac{-\\lg \\mathrm{K}_{2}-\\lg \\mathrm{K}_{3}}{2}$\nD: P 点, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}(\\mathrm{Cl}-)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n赖氨酸 $\\left[\\mathrm{H}_{3} \\mathrm{~N}^{+}\\left(\\mathrm{CH}_{2}\\right)_{4} \\mathrm{CH}\\left(\\mathrm{NH}_{2}\\right) \\mathrm{COO}\\right.$, 用 $\\mathrm{HR}$ 表示 $]$ 是人体必需氨基酸, 其盐酸盐 $\\left(\\mathrm{H}_{3} \\mathrm{RCl}_{2}\\right)$在水溶液中存在如下平衡: $\\mathrm{H}_{3} \\mathrm{R}^{2+} \\stackrel{\\mathrm{K}_{1}}{\\rightleftharpoons} \\mathrm{H}_{2} \\mathrm{R}^{+} \\stackrel{\\mathrm{K}_{2} \\rightleftharpoons}{\\rightleftharpoons} \\mathrm{HR} \\stackrel{\\mathrm{K}_{3}}{\\rightleftharpoons} \\mathrm{R}^{-}$。向一定浓度的 $\\mathrm{H}_{3} \\mathrm{RCl}_{2}$\n溶液中滴加 $\\mathrm{NaOH}$ 溶液, 溶液中 $\\mathrm{H}_{3} \\mathrm{R}^{2+} 、 \\mathrm{H}_{2} \\mathrm{R}^{+} 、 \\mathrm{HR}$ 和 $\\mathrm{R}$ 的分布系数 $\\delta(\\mathrm{x})$ 随 $\\mathrm{pH}$ 变化如图所示。已知 $\\delta(\\mathrm{x})=\\frac{\\mathrm{c}(\\mathrm{x})}{\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{R}^{2+}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{+}\\right)+\\mathrm{c}(\\mathrm{HR})+\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)}$, 下列表述正确的是\n\n[图1]\n\nA: $\\frac{\\mathrm{K}_{2}}{\\mathrm{~K}_{1}}>\\frac{\\mathrm{K}_{3}}{\\mathrm{~K}_{2}}$\nB: $M$ 点, $c\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)=2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{R}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nC: $\\mathrm{O}$ 点, $\\mathrm{pH}=\\frac{-\\lg \\mathrm{K}_{2}-\\lg \\mathrm{K}_{3}}{2}$\nD: P 点, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}(\\mathrm{Cl}-)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-102.jpg?height=626&width=848&top_left_y=358&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1422",
        "problem": "Which of the following acids is the strongest?\nA: perchloric acid, $\\mathrm{HClO}_{4}$\nB: chloric acid, $\\mathrm{HClO}_{3}$\nC: chlorous acid, $\\mathrm{HClO}_{2}$\nD: hypochlorous, $\\mathrm{HClO}$ (e) All of them are equally strong because they all contain chlorine\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following acids is the strongest?\n\nA: perchloric acid, $\\mathrm{HClO}_{4}$\nB: chloric acid, $\\mathrm{HClO}_{3}$\nC: chlorous acid, $\\mathrm{HClO}_{2}$\nD: hypochlorous, $\\mathrm{HClO}$ (e) All of them are equally strong because they all contain chlorine\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1131",
        "problem": "This question is about periodic acid\n\nSugars are important as fuels for our cells and are also being developed as biofuels for manufacturing. An enormous variety of sugars exists in nature. In particular, sugars have many structural isomers, each of which has different properties. Periodic acid is a useful reagent for helping to determine which isomer of sugar is present in a sample.\n\n\"Please note it's pronounced purrr-I-odic acid!\"\n\nPeriodic acid exists in two forms: orthoperiodic acid, $\\mathrm{H}_{5} \\mathrm{IO}_{6}$, and metaperiodic acid, $\\mathrm{HIO}_{4}$. Metaperiodic acid can be formed by heating orthoperiodic acid.\n\nGive the oxidation state of iodine in periodic acid.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about periodic acid\n\nSugars are important as fuels for our cells and are also being developed as biofuels for manufacturing. An enormous variety of sugars exists in nature. In particular, sugars have many structural isomers, each of which has different properties. Periodic acid is a useful reagent for helping to determine which isomer of sugar is present in a sample.\n\n\"Please note it's pronounced purrr-I-odic acid!\"\n\nPeriodic acid exists in two forms: orthoperiodic acid, $\\mathrm{H}_{5} \\mathrm{IO}_{6}$, and metaperiodic acid, $\\mathrm{HIO}_{4}$. Metaperiodic acid can be formed by heating orthoperiodic acid.\n\nGive the oxidation state of iodine in periodic acid.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_742",
        "problem": "微生物脱盐池是一种高效、经济的能源装置, 利用微生物处理有机废水获得电能,同时可实现海水淡化。现以氯化钠溶液模拟海水, 采用惰性电极, 用下图的装置处理有机废水(以含 $\\mathrm{CH}_{3} \\mathrm{COO}^{-}$的溶液为例), 下列说法正确的是\n\n[图1]\nA: 该电池可在高温环境下工作\nB: $\\mathrm{Y}$ 为阳离子选择性交换膜\nC: 负极反应为: $\\mathrm{CH}_{3} \\mathrm{COO}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}-8 \\mathrm{e}^{-}=2 \\mathrm{CO}_{2} \\uparrow+7 \\mathrm{H}^{+}$\nD: 每消耗 $2.24 \\mathrm{~L}$ (标准状况)的空气, 有 $0.4 \\mathrm{~mol}$ 电子转移\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n微生物脱盐池是一种高效、经济的能源装置, 利用微生物处理有机废水获得电能,同时可实现海水淡化。现以氯化钠溶液模拟海水, 采用惰性电极, 用下图的装置处理有机废水(以含 $\\mathrm{CH}_{3} \\mathrm{COO}^{-}$的溶液为例), 下列说法正确的是\n\n[图1]\n\nA: 该电池可在高温环境下工作\nB: $\\mathrm{Y}$ 为阳离子选择性交换膜\nC: 负极反应为: $\\mathrm{CH}_{3} \\mathrm{COO}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}-8 \\mathrm{e}^{-}=2 \\mathrm{CO}_{2} \\uparrow+7 \\mathrm{H}^{+}$\nD: 每消耗 $2.24 \\mathrm{~L}$ (标准状况)的空气, 有 $0.4 \\mathrm{~mol}$ 电子转移\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-33.jpg?height=694&width=1082&top_left_y=2080&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_828",
        "problem": "已知 $X 、 M$ 都是中学常见元素, 下列对两个离子反应通式的推断中, 其中正确的是 （甲） $\\mathrm{XO}_{3}{ }^{\\mathrm{n}-}+\\mathrm{X}^{\\mathrm{n}-}+\\mathrm{H}^{+} \\rightarrow \\mathrm{X}$ 单质 $+\\mathrm{H}_{2} \\mathrm{O}$ (未配平);（乙） $\\mathrm{M}^{\\mathrm{m}+}+\\mathrm{mOH}^{-}=\\mathrm{M}(\\mathrm{OH})_{\\mathrm{m} \\downarrow} \\downarrow$\n\n(1) 若 $n=1$, 则 $\\mathrm{XO}_{3}{ }^{n-}$ 中 $X$ 元素为 +5 价, $X$ 位于周期表第 $V A$ 族\n\n(2)若 $n=2$, 则 $X$ 最高价氧化物的水化物可能与它的氢化物反应\n\n(3)若 $\\mathrm{m}=1$, 则 $\\mathrm{M}\\left(\\mathrm{NO}_{3}\\right)_{\\mathrm{m}}$ 溶液和氨水互滴时的现象可能不同\n\n(4) 若 $m=2$, 则在空气中蒸干、灼烧 $\\mathrm{MSO}_{4}$ 溶液一定能得到 $\\mathrm{MSO}_{4}$\n\n(5)若 $\\mathrm{m}=3$, 则 $\\mathrm{MCl}_{3}$ 与足量氢氧化钠溶液反应一定生成 $\\mathrm{M}(\\mathrm{OH})_{\\mathrm{m}}$\nA: (2)(3)\nB: (1) (3)\nC: (4) 5\nD: (2)(4)\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知 $X 、 M$ 都是中学常见元素, 下列对两个离子反应通式的推断中, 其中正确的是 （甲） $\\mathrm{XO}_{3}{ }^{\\mathrm{n}-}+\\mathrm{X}^{\\mathrm{n}-}+\\mathrm{H}^{+} \\rightarrow \\mathrm{X}$ 单质 $+\\mathrm{H}_{2} \\mathrm{O}$ (未配平);（乙） $\\mathrm{M}^{\\mathrm{m}+}+\\mathrm{mOH}^{-}=\\mathrm{M}(\\mathrm{OH})_{\\mathrm{m} \\downarrow} \\downarrow$\n\n(1) 若 $n=1$, 则 $\\mathrm{XO}_{3}{ }^{n-}$ 中 $X$ 元素为 +5 价, $X$ 位于周期表第 $V A$ 族\n\n(2)若 $n=2$, 则 $X$ 最高价氧化物的水化物可能与它的氢化物反应\n\n(3)若 $\\mathrm{m}=1$, 则 $\\mathrm{M}\\left(\\mathrm{NO}_{3}\\right)_{\\mathrm{m}}$ 溶液和氨水互滴时的现象可能不同\n\n(4) 若 $m=2$, 则在空气中蒸干、灼烧 $\\mathrm{MSO}_{4}$ 溶液一定能得到 $\\mathrm{MSO}_{4}$\n\n(5)若 $\\mathrm{m}=3$, 则 $\\mathrm{MCl}_{3}$ 与足量氢氧化钠溶液反应一定生成 $\\mathrm{M}(\\mathrm{OH})_{\\mathrm{m}}$\n\nA: (2)(3)\nB: (1) (3)\nC: (4) 5\nD: (2)(4)\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_49",
        "problem": "A student wishes to determine the molar mass of a pure solid organic compound. Which measurement would be most useful?\nA: Melting point of the solid\nB: Melting point depression of a mixture of the solid with 1,4-dichlorobenzene\nC: Heat of combustion\nD: Solubility in water\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student wishes to determine the molar mass of a pure solid organic compound. Which measurement would be most useful?\n\nA: Melting point of the solid\nB: Melting point depression of a mixture of the solid with 1,4-dichlorobenzene\nC: Heat of combustion\nD: Solubility in water\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1118",
        "problem": "The volume of this bubble in $\\mathrm{m}^{3}$ is $1.77 \\times 10^{-6} \\mathrm{~m}^{3}$.Former Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nMany varities of Swiss cheese, such as Emmental, are famous for the holes or 'eyes' that appear in the cheese. To produce the holes another species of bacteria, Propionibacterium freudenreichii is important. This bacterium carries out the reaction of lactic acid to propanoic acid, ethanoic acid, carbon dioxide and water. The production of carbon dioxide causes the bubbles to appear.\n\n[figure3]\n\nAssume during fermentation at $21^{\\circ} \\mathrm{C}$, a spherical bubble of diameter $1.5 \\mathrm{~cm}$ appears in the cheese.\n\nAssuming the bubble is pure $\\mathrm{CO}_{2}$ at atmospheric pressure, $p_{\\text {atm }}=101325 \\mathrm{~Pa}$, calculate the mass of lactic acid which was fermented by the bacteria to produce this bubble. For the calculation assume that $\\mathrm{CO}_{2}$ obeys the ideal gas law,\n\n$$\np V=n R T\n$$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe volume of this bubble in $\\mathrm{m}^{3}$ is $1.77 \\times 10^{-6} \\mathrm{~m}^{3}$.\n\nproblem:\nFormer Prime Minister Liz Truss once said that \"We import two thirds of our cheese. That is a disgrace.\"\n\nWhen we started writing this paper in summer 2022, we thought a question on cheese would be timely, given that the 2023 International Chemistry Olympiad is being held in Switzerland (a country with many famous cheeses) and that Liz Truss would be the Prime Minister at the time of Round 1.\n\nWhile there are many differences in the process of cheese manufacture, the conversion of lactose to lactic acid during fermentation is a key chemical process wherever the cheese is from.\n\n[figure1]\n\n[figure2]\n\nMany varities of Swiss cheese, such as Emmental, are famous for the holes or 'eyes' that appear in the cheese. To produce the holes another species of bacteria, Propionibacterium freudenreichii is important. This bacterium carries out the reaction of lactic acid to propanoic acid, ethanoic acid, carbon dioxide and water. The production of carbon dioxide causes the bubbles to appear.\n\n[figure3]\n\nAssume during fermentation at $21^{\\circ} \\mathrm{C}$, a spherical bubble of diameter $1.5 \\mathrm{~cm}$ appears in the cheese.\n\nAssuming the bubble is pure $\\mathrm{CO}_{2}$ at atmospheric pressure, $p_{\\text {atm }}=101325 \\mathrm{~Pa}$, calculate the mass of lactic acid which was fermented by the bacteria to produce this bubble. For the calculation assume that $\\mathrm{CO}_{2}$ obeys the ideal gas law,\n\n$$\np V=n R T\n$$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~g}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-13.jpg?height=616&width=699&top_left_y=323&top_left_x=1178",
            "https://i.postimg.cc/Vs42tGMJ/5.png",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-14.jpg?height=528&width=785&top_left_y=747&top_left_x=1047"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~g}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1152",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nwhich is reduced during this reaction?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nwhich is reduced during this reaction?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_609",
        "problem": "已知常温下 $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $\\mathrm{K}_{\\mathrm{a} 1}=2.7 \\times 10^{-6}, \\mathrm{~K}_{\\mathrm{a} 2}=6.3 \\times 10^{-10}, \\mathrm{HClO}$ 的 $\\mathrm{Ka}=3 \\times 10^{-8}$ 。下列说法正确的是\nA: $\\mathrm{NaHA}$ 溶液中存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 向 $\\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中通入少量 $\\mathrm{Cl}_{2}$ 的离子方程式: $\\mathrm{Cl}_{2}+2 \\mathrm{~A}^{2-}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{Cl}^{-}+\\mathrm{ClO}^{-}+2 \\mathrm{HA}^{-}$\nC: $\\mathrm{NaClO}$ 溶液中水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$大于等浓度的 $\\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 等浓度、等体积的 $\\mathrm{NaHA}$ 和 $\\mathrm{NaClO}$ 的混合溶液中存在: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{ClO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}(\\mathrm{HClO})+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知常温下 $\\mathrm{H}_{2} \\mathrm{~A}$ 的 $\\mathrm{K}_{\\mathrm{a} 1}=2.7 \\times 10^{-6}, \\mathrm{~K}_{\\mathrm{a} 2}=6.3 \\times 10^{-10}, \\mathrm{HClO}$ 的 $\\mathrm{Ka}=3 \\times 10^{-8}$ 。下列说法正确的是\n\nA: $\\mathrm{NaHA}$ 溶液中存在: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 向 $\\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中通入少量 $\\mathrm{Cl}_{2}$ 的离子方程式: $\\mathrm{Cl}_{2}+2 \\mathrm{~A}^{2-}+\\mathrm{H}_{2} \\mathrm{O}=\\mathrm{Cl}^{-}+\\mathrm{ClO}^{-}+2 \\mathrm{HA}^{-}$\nC: $\\mathrm{NaClO}$ 溶液中水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$大于等浓度的 $\\mathrm{Na}_{2} \\mathrm{~A}$ 溶液中水电离出的 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 等浓度、等体积的 $\\mathrm{NaHA}$ 和 $\\mathrm{NaClO}$ 的混合溶液中存在: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{ClO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}(\\mathrm{HClO})+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_856",
        "problem": "在容积相同的 4 个密闭容器中, 按不同方式投入反应物, 保持恒温、恒容, 测得反应达到平衡时的有关数据如下, 已知: $\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-92.4 \\mathrm{~kJ} / \\mathrm{mol}$\n\n| 容器 | 甲 | 乙 | 丙 | 丁 |\n| :---: | :---: | :---: | :---: | :---: |\n| 反应温度 $\\mathrm{T} /{ }^{\\circ} \\mathrm{C}$ | 400 | 400 | 400 | 500 |\n| 反应物投入量 | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ | $2 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $4 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ |\n| $\\mathrm{NH}_{3}$ 的浓度 $(\\mathrm{mol} / \\mathrm{L})$ | $c_{1}$ | $c_{2}$ | $c_{3}$ | $c_{4}$ |\n| 反应的能量变化 | 放出 $a \\mathrm{~kJ}$ | 吸收 $b \\mathrm{~kJ}$ | 吸收 $c \\mathrm{~kJ}$ | 放出 $d \\mathrm{~kJ}$ |\n| 体系压强 $(\\mathrm{Pa})$ | $p_{1}$ | $p_{2}$ | $p_{3}$ | $p_{4}$ |\n| 反应物转化率 | $\\alpha_{1}$ | $\\alpha_{2}$ | $\\alpha_{3}$ | $\\alpha_{4}$ |\n| 平衡常数 $K$ | $K_{1}$ | $K_{2}$ | $K_{3}$ | $K_{4}$ |\n\n下列说法正确的是\nA: $2 \\mathrm{p}_{2}<\\mathrm{p}_{3}, \\mathrm{a}+\\mathrm{b}=92.4$\nB: $2 \\mathrm{c}_{1}>\\mathrm{c}_{3}, \\alpha_{1}+\\alpha_{3}<1$\nC: $\\mathrm{K}_{3}>\\mathrm{K}_{4}, \\alpha_{2}+\\alpha_{4}<1$\nD: $\\frac{c}{2}+\\mathrm{d}<92.4,2 \\mathrm{p}_{4}>\\mathrm{p}_{3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n在容积相同的 4 个密闭容器中, 按不同方式投入反应物, 保持恒温、恒容, 测得反应达到平衡时的有关数据如下, 已知: $\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-92.4 \\mathrm{~kJ} / \\mathrm{mol}$\n\n| 容器 | 甲 | 乙 | 丙 | 丁 |\n| :---: | :---: | :---: | :---: | :---: |\n| 反应温度 $\\mathrm{T} /{ }^{\\circ} \\mathrm{C}$ | 400 | 400 | 400 | 500 |\n| 反应物投入量 | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ | $2 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $4 \\mathrm{~mol} \\mathrm{NH}_{3}$ | $1 \\mathrm{~mol} \\mathrm{~N}_{2} 、 3 \\mathrm{~mol} \\mathrm{H}_{2}$ |\n| $\\mathrm{NH}_{3}$ 的浓度 $(\\mathrm{mol} / \\mathrm{L})$ | $c_{1}$ | $c_{2}$ | $c_{3}$ | $c_{4}$ |\n| 反应的能量变化 | 放出 $a \\mathrm{~kJ}$ | 吸收 $b \\mathrm{~kJ}$ | 吸收 $c \\mathrm{~kJ}$ | 放出 $d \\mathrm{~kJ}$ |\n| 体系压强 $(\\mathrm{Pa})$ | $p_{1}$ | $p_{2}$ | $p_{3}$ | $p_{4}$ |\n| 反应物转化率 | $\\alpha_{1}$ | $\\alpha_{2}$ | $\\alpha_{3}$ | $\\alpha_{4}$ |\n| 平衡常数 $K$ | $K_{1}$ | $K_{2}$ | $K_{3}$ | $K_{4}$ |\n\n下列说法正确的是\n\nA: $2 \\mathrm{p}_{2}<\\mathrm{p}_{3}, \\mathrm{a}+\\mathrm{b}=92.4$\nB: $2 \\mathrm{c}_{1}>\\mathrm{c}_{3}, \\alpha_{1}+\\alpha_{3}<1$\nC: $\\mathrm{K}_{3}>\\mathrm{K}_{4}, \\alpha_{2}+\\alpha_{4}<1$\nD: $\\frac{c}{2}+\\mathrm{d}<92.4,2 \\mathrm{p}_{4}>\\mathrm{p}_{3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-096.jpg?height=877&width=1068&top_left_y=150&top_left_x=320"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_999",
        "problem": "The formula of a compound is $\\mathrm{X}_{2} \\mathrm{O}$. Which of the following is $X$ least likely to be?\nA: barium $(\\mathrm{Ba})$\nB: $\\operatorname{sodium}(\\mathrm{Na})$\nC: cesium (Cs)\nD: hydrogen $(\\mathrm{H})$\nE: copper $(\\mathrm{Cu})$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe formula of a compound is $\\mathrm{X}_{2} \\mathrm{O}$. Which of the following is $X$ least likely to be?\n\nA: barium $(\\mathrm{Ba})$\nB: $\\operatorname{sodium}(\\mathrm{Na})$\nC: cesium (Cs)\nD: hydrogen $(\\mathrm{H})$\nE: copper $(\\mathrm{Cu})$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_663",
        "problem": "下列实验事实不能证明苯分子中不存在独立的碳碳双键的是\n\n[图1]\nA: 用扫描隧道显微镜获得的苯分子图像(如图)\nB: 苯可与氢气在一定条件下发生加成反应\nC: 苯不能使酸性高锰酸钾溶液裉色\nD: 环已烷 (1)、环已烯(1)和苯(1)的标准燃烧热分别为 $-3916 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} 、-3747 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$ 和 $-3265 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列实验事实不能证明苯分子中不存在独立的碳碳双键的是\n\n[图1]\n\nA: 用扫描隧道显微镜获得的苯分子图像(如图)\nB: 苯可与氢气在一定条件下发生加成反应\nC: 苯不能使酸性高锰酸钾溶液裉色\nD: 环已烷 (1)、环已烯(1)和苯(1)的标准燃烧热分别为 $-3916 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} 、-3747 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$ 和 $-3265 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-01.jpg?height=211&width=317&top_left_y=563&top_left_x=321"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_92",
        "problem": "A sample of $0.040 \\mathrm{~mol}$ hypochlorite ion is treated with varying amounts of $1.0 \\mathrm{M}$ aqueous $\\mathrm{H}_{2} \\mathrm{O}_{2}$. Which graph represents the amount of $\\mathrm{O}_{2}(g)$ that is evolved according to the chemical reaction shown?\n\n$$\n\\mathrm{OCl}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\rightarrow \\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(g)\n$$\nA: [figure1] mL $1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\nB: [figure2] mL $1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\nC: [figure3] $\\mathrm{mL} 1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\nD: [figure4] mL $1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA sample of $0.040 \\mathrm{~mol}$ hypochlorite ion is treated with varying amounts of $1.0 \\mathrm{M}$ aqueous $\\mathrm{H}_{2} \\mathrm{O}_{2}$. Which graph represents the amount of $\\mathrm{O}_{2}(g)$ that is evolved according to the chemical reaction shown?\n\n$$\n\\mathrm{OCl}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}_{2}(a q) \\rightarrow \\mathrm{Cl}^{-}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)+\\mathrm{O}_{2}(g)\n$$\n\nA: [figure1] mL $1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\nB: [figure2] mL $1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\nC: [figure3] $\\mathrm{mL} 1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\nD: [figure4] mL $1.0 \\mathrm{M} \\mathrm{H}_{2} \\mathrm{O}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=252&width=327&top_left_y=1804&top_left_x=281",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=293&width=325&top_left_y=2168&top_left_x=282",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=241&width=331&top_left_y=1809&top_left_x=707",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-03.jpg?height=261&width=333&top_left_y=2168&top_left_x=706"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_190",
        "problem": "A closed $600.0 \\mathrm{~mL}$ flask contains solid mercuric oxide and air initially at $21.0^{\\circ} \\mathrm{C}$ and $101.3 \\mathrm{kPa}$. When heated, mercuric oxide decomposes completely according to the reaction:\n\n$$\n2 \\mathrm{HgO}(\\mathrm{s}) \\rightarrow 2 \\mathrm{Hg}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g})\n$$\n\nAfter heating, the flask is at a temperature of $75.2^{\\circ} \\mathrm{C}$ and has a pressure of $205.5 \\mathrm{kPa}$. What mass of mercury metal is in the flask when the reaction is complete?\nA: $7.11 \\mathrm{~g}$\nB: $4.33 \\mathrm{~g}$\nC: $3.56 \\mathrm{~g}$\nD: $17.1 \\mathrm{~g}$\nE: $8.66 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA closed $600.0 \\mathrm{~mL}$ flask contains solid mercuric oxide and air initially at $21.0^{\\circ} \\mathrm{C}$ and $101.3 \\mathrm{kPa}$. When heated, mercuric oxide decomposes completely according to the reaction:\n\n$$\n2 \\mathrm{HgO}(\\mathrm{s}) \\rightarrow 2 \\mathrm{Hg}(\\mathrm{s})+\\mathrm{O}_{2}(\\mathrm{~g})\n$$\n\nAfter heating, the flask is at a temperature of $75.2^{\\circ} \\mathrm{C}$ and has a pressure of $205.5 \\mathrm{kPa}$. What mass of mercury metal is in the flask when the reaction is complete?\n\nA: $7.11 \\mathrm{~g}$\nB: $4.33 \\mathrm{~g}$\nC: $3.56 \\mathrm{~g}$\nD: $17.1 \\mathrm{~g}$\nE: $8.66 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_519",
        "problem": "常温下, 向 $25 \\mathrm{~mL} 0.12 \\mathrm{~mol} / \\mathrm{L} \\mathrm{AgNO}_{3}$ 溶液中逐滴入一定浓度的氨水, 先出现沉淀,继续滴加氨水至沉淀溶解。该过程中加入氨水的体积 $\\mathrm{V}$ 与溶液中 $\\lg \\left[\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) / \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)\\right]$的关系如图所示。已知 $\\mathrm{e}$ 点对应的溶液迅速由浑浊变得澄清, 且此时溶液中的 $\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$与 $\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)$ 均约为 $2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L}$ 。下列叙述正确的是\n\n[图1]\nA: a 点对应溶液中存在四种离子, 其中 $\\mathrm{Ag}^{+}$浓度最大\nB: b 点对应溶液中: $c\\left(\\mathrm{Ag}^{+}\\right)+c\\left\\{\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right\\}=c\\left(\\mathrm{NO}_{3}^{-}\\right)$\nC: 与葡萄糖发生银镜反应, 最好选择 cd 段溶液\nD: 由 $\\mathrm{e}$ 点可知, 反应 $\\mathrm{Ag}^{+}+2 \\mathrm{NH}_{3}=\\mathrm{Ag}\\left[\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$的平衡常数的数量级为 $10^{7}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $25 \\mathrm{~mL} 0.12 \\mathrm{~mol} / \\mathrm{L} \\mathrm{AgNO}_{3}$ 溶液中逐滴入一定浓度的氨水, 先出现沉淀,继续滴加氨水至沉淀溶解。该过程中加入氨水的体积 $\\mathrm{V}$ 与溶液中 $\\lg \\left[\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) / \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)\\right]$的关系如图所示。已知 $\\mathrm{e}$ 点对应的溶液迅速由浑浊变得澄清, 且此时溶液中的 $\\mathrm{c}\\left(\\mathrm{Ag}^{+}\\right)$与 $\\mathrm{c}\\left(\\mathrm{NH}_{3}\\right)$ 均约为 $2 \\times 10^{-3} \\mathrm{~mol} / \\mathrm{L}$ 。下列叙述正确的是\n\n[图1]\n\nA: a 点对应溶液中存在四种离子, 其中 $\\mathrm{Ag}^{+}$浓度最大\nB: b 点对应溶液中: $c\\left(\\mathrm{Ag}^{+}\\right)+c\\left\\{\\left[\\mathrm{Ag}\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}\\right\\}=c\\left(\\mathrm{NO}_{3}^{-}\\right)$\nC: 与葡萄糖发生银镜反应, 最好选择 cd 段溶液\nD: 由 $\\mathrm{e}$ 点可知, 反应 $\\mathrm{Ag}^{+}+2 \\mathrm{NH}_{3}=\\mathrm{Ag}\\left[\\left(\\mathrm{NH}_{3}\\right)_{2}\\right]^{+}$的平衡常数的数量级为 $10^{7}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-041.jpg?height=634&width=1108&top_left_y=163&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_277",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the amount of pure KHP (in $\\mathrm{mol}$ or $\\mathrm{mmol}$ ) required to react completely with $4.359 \\mathrm{~g}$ of the sodium hydroxide solution.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the amount of pure KHP (in $\\mathrm{mol}$ or $\\mathrm{mmol}$ ) required to react completely with $4.359 \\mathrm{~g}$ of the sodium hydroxide solution.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mmol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mmol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_700",
        "problem": "2022 年北京大学课题组发现 $[4+2)]$ 反应并对反应机理进行研究, 该 $[4+2]$ 反应是通过在环丁酮 2-位引入乙烯基取代作为导向基诱导铑对环丁酮的碳-碳键活化，同时酮碳基可以与路易斯酸配位，进一步活化环丁酮的碳-碳键。这一协同催化模式选择性切断羰基与季碳中心之间的碳-碳键, 不仅成功地构建六元环系, 还成功拓展到烷基及多种电子效应的芳基上, 大部分底物都具有中等到良好的收率，对氰基，酮羰基，酯基及酰胺等官能团具有良好的兼容性，机理如下图，下列表述错误的是\n\n[图1]\nA: 生成 TS-OA 发生了氧化加成, 这一步被认为是反应的决速步\nB: 其中 $\\mathrm{ZnCl}_{2}$ 被认为具有促进 $\\mathrm{C}-\\mathrm{C}$ 的活化和原位生成活性阳离子铑的作用\nC: 由 TS-OA 生成 B 有配位键的形成\nD: 该反应为合成 4 元环物质提供一种新的 C-C 耦合方式\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n2022 年北京大学课题组发现 $[4+2)]$ 反应并对反应机理进行研究, 该 $[4+2]$ 反应是通过在环丁酮 2-位引入乙烯基取代作为导向基诱导铑对环丁酮的碳-碳键活化，同时酮碳基可以与路易斯酸配位，进一步活化环丁酮的碳-碳键。这一协同催化模式选择性切断羰基与季碳中心之间的碳-碳键, 不仅成功地构建六元环系, 还成功拓展到烷基及多种电子效应的芳基上, 大部分底物都具有中等到良好的收率，对氰基，酮羰基，酯基及酰胺等官能团具有良好的兼容性，机理如下图，下列表述错误的是\n\n[图1]\n\nA: 生成 TS-OA 发生了氧化加成, 这一步被认为是反应的决速步\nB: 其中 $\\mathrm{ZnCl}_{2}$ 被认为具有促进 $\\mathrm{C}-\\mathrm{C}$ 的活化和原位生成活性阳离子铑的作用\nC: 由 TS-OA 生成 B 有配位键的形成\nD: 该反应为合成 4 元环物质提供一种新的 C-C 耦合方式\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-100.jpg?height=1462&width=1019&top_left_y=157&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_76",
        "problem": "The triple point of bromine is $265.9 \\mathrm{~K}$ and 0.058 bar, and the standard entropy of vaporization of bromine is $89.0 \\mathrm{~J}$ $\\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$. What is the standard enthalpy of vaporization of bromine?\nA: $17.4 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $23.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $26.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $30.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe triple point of bromine is $265.9 \\mathrm{~K}$ and 0.058 bar, and the standard entropy of vaporization of bromine is $89.0 \\mathrm{~J}$ $\\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$. What is the standard enthalpy of vaporization of bromine?\n\nA: $17.4 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $23.7 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $26.5 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $30.0 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_65",
        "problem": "In which molecule are there two distinct sets of sulfurfluorine bond lengths?\nA: $\\mathrm{SF}_{2}$\nB: $\\mathrm{SOF}_{2}$\nC: $\\mathrm{SF}_{4}$\nD: $\\mathrm{SF}_{6}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn which molecule are there two distinct sets of sulfurfluorine bond lengths?\n\nA: $\\mathrm{SF}_{2}$\nB: $\\mathrm{SOF}_{2}$\nC: $\\mathrm{SF}_{4}$\nD: $\\mathrm{SF}_{6}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_360",
        "problem": "${ }^{52} \\mathrm{Mn}$ undergoes radioactive decay to give ${ }^{52} \\mathrm{Cr}$ by what decay mode?\nA: Alpha emission\nB: Beta emission\nC: Positron emission\nD: Gamma emission\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n${ }^{52} \\mathrm{Mn}$ undergoes radioactive decay to give ${ }^{52} \\mathrm{Cr}$ by what decay mode?\n\nA: Alpha emission\nB: Beta emission\nC: Positron emission\nD: Gamma emission\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1504",
        "problem": "Clathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.Determine $n$ (not necessarily integer) in the formula of methane hydrate, $\\mathrm{CH}_{4} \\cdot n \\mathrm{H}_{2} \\mathrm{O}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nClathrate gun\n\nThe only gun that is able to kill all living people in one shot\n\nOn the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. \n\n[figure1]\n\nIt is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.\n\nUpon decomposition of $1.00 \\mathrm{~g}$ of a methane hydrate with a fixed composition at $25{ }^{\\circ} \\mathrm{C}$ and atmospheric ( $101.3 \\mathrm{kPa}$ ) pressure, $205 \\mathrm{~cm}^{3}$ of methane is released.\n\nproblem:\nDetermine $n$ (not necessarily integer) in the formula of methane hydrate, $\\mathrm{CH}_{4} \\cdot n \\mathrm{H}_{2} \\mathrm{O}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-222.jpg?height=425&width=434&top_left_y=1255&top_left_x=1459"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_977",
        "problem": "What volume of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NaOH}(\\mathrm{aq})$ is required to neutralize $0.245 \\mathrm{~L}$ of $0.200 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{H}_{3} \\mathrm{PO}_{4}(\\mathrm{aq})$ ?\nA: $0.490 \\mathrm{~L}$\nB: $\\quad 0.500 \\mathrm{~L}$\nC: $\\quad 1.47 \\mathrm{~L}$\nD: $2.30 \\mathrm{~L}$\nE: $\\quad 1.47 \\mathrm{~mL}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat volume of $0.100 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NaOH}(\\mathrm{aq})$ is required to neutralize $0.245 \\mathrm{~L}$ of $0.200 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{H}_{3} \\mathrm{PO}_{4}(\\mathrm{aq})$ ?\n\nA: $0.490 \\mathrm{~L}$\nB: $\\quad 0.500 \\mathrm{~L}$\nC: $\\quad 1.47 \\mathrm{~L}$\nD: $2.30 \\mathrm{~L}$\nE: $\\quad 1.47 \\mathrm{~mL}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_367",
        "problem": "A $0.1 \\mathrm{M}$ solution of which salt is the most basic?\nA: $\\mathrm{NaNO}_{3}$\nB: $\\mathrm{NaClO}_{4}$\nC: $\\mathrm{NaHSO}_{4}$\nD: $\\mathrm{NaHCO}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $0.1 \\mathrm{M}$ solution of which salt is the most basic?\n\nA: $\\mathrm{NaNO}_{3}$\nB: $\\mathrm{NaClO}_{4}$\nC: $\\mathrm{NaHSO}_{4}$\nD: $\\mathrm{NaHCO}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_912",
        "problem": "常温下将 $\\mathrm{NaOH}$ 溶液滴加到已二酸 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 溶液中, 混合溶液的 $\\mathrm{pH}$ 与离子浓度变化的关系如图所示。下列叙述错误的是\n\n[图1]\nA: $\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 的数量级为 $10^{-6}$\nB: 曲线 $\\mathrm{N}$ 表示 $\\mathrm{pH}$ 与 $\\lg \\frac{c\\left(H X^{-}\\right)}{c\\left(H_{2} X\\right)}$ 的变化关系\nC: $\\mathrm{NaHX}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 当混合溶液呈中性时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下将 $\\mathrm{NaOH}$ 溶液滴加到已二酸 $\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 溶液中, 混合溶液的 $\\mathrm{pH}$ 与离子浓度变化的关系如图所示。下列叙述错误的是\n\n[图1]\n\nA: $\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{X}\\right)$ 的数量级为 $10^{-6}$\nB: 曲线 $\\mathrm{N}$ 表示 $\\mathrm{pH}$ 与 $\\lg \\frac{c\\left(H X^{-}\\right)}{c\\left(H_{2} X\\right)}$ 的变化关系\nC: $\\mathrm{NaHX}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nD: 当混合溶液呈中性时, $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HX}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{X}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-82.jpg?height=648&width=813&top_left_y=1943&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1396",
        "problem": "Distribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of $\\mathrm{PO}_{4}{ }_{4}^{3-}$ and $\\mathrm{SiO}_{4}{ }^{4-}$ in alkaline extract\n\nBoth phosphate and silicate ions can react with molybdate in alkaline solution, producing the yellow molybdophosphate and molybdatosilicate. Further reduction with ascorbic acid produces intense color molybdenum blue compounds. Both complexes exhibit maximum absorption at $800 \\mathrm{~nm}$. Addition of tartaric acid helps preventing interference from silicate in the determination of phosphate.\n\nTwo series of phosphate standard are treated with and without tartaric acid whereas a series of silicate standard is not treated with tartaric acid. Linear equations obtained from those calibration curves are as follows:\n\n| Conditions | Linear equations |\n| :---: | :---: |\n| Phosphate with and without tartaric acid | $y=6720 \\mathrm{x}_{1}$ |\n| Silicate without tartaric acid | $\\mathrm{y}=868 \\mathrm{x}_{2}$ |\n\n$\\mathrm{y}$ is absorbance at $800 \\mathrm{~nm}$,\n\n$\\mathrm{x}_{1}$ is concentration of phosphate as $\\mathrm{mol} \\mathrm{dm}^{-3}$,\n\n$\\mathrm{x}_{2}$ is concentration of silicate as $\\mathrm{mol} \\mathrm{dm}^{-3}$\n\nAbsorbance at $800 \\mathrm{~nm}$ of the alkaline fraction of the soil extract after treated with and without tartaric acid are 0.267 and 0.510 , respectively.Calculate the phosphate concentration in the alkaline soil extract in $\\mathrm{mol} \\mathrm{~dm}^{-3}$ and calculate the corresponding phosphorous in $\\mathrm{mg} \\mathrm{~dm}^{-3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nDistribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of $\\mathrm{PO}_{4}{ }_{4}^{3-}$ and $\\mathrm{SiO}_{4}{ }^{4-}$ in alkaline extract\n\nBoth phosphate and silicate ions can react with molybdate in alkaline solution, producing the yellow molybdophosphate and molybdatosilicate. Further reduction with ascorbic acid produces intense color molybdenum blue compounds. Both complexes exhibit maximum absorption at $800 \\mathrm{~nm}$. Addition of tartaric acid helps preventing interference from silicate in the determination of phosphate.\n\nTwo series of phosphate standard are treated with and without tartaric acid whereas a series of silicate standard is not treated with tartaric acid. Linear equations obtained from those calibration curves are as follows:\n\n| Conditions | Linear equations |\n| :---: | :---: |\n| Phosphate with and without tartaric acid | $y=6720 \\mathrm{x}_{1}$ |\n| Silicate without tartaric acid | $\\mathrm{y}=868 \\mathrm{x}_{2}$ |\n\n$\\mathrm{y}$ is absorbance at $800 \\mathrm{~nm}$,\n\n$\\mathrm{x}_{1}$ is concentration of phosphate as $\\mathrm{mol} \\mathrm{dm}^{-3}$,\n\n$\\mathrm{x}_{2}$ is concentration of silicate as $\\mathrm{mol} \\mathrm{dm}^{-3}$\n\nAbsorbance at $800 \\mathrm{~nm}$ of the alkaline fraction of the soil extract after treated with and without tartaric acid are 0.267 and 0.510 , respectively.\n\nproblem:\nCalculate the phosphate concentration in the alkaline soil extract in $\\mathrm{mol} \\mathrm{~dm}^{-3}$ and calculate the corresponding phosphorous in $\\mathrm{mg} \\mathrm{~dm}^{-3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{mg} \\mathrm{~dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{mg} \\mathrm{~dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1080",
        "problem": "Polonium is a radioactive group VI element, discovered in 1898 by Marie Curie. It occurs naturally in trace amounts in some uranium ores but is now made by neutron irradiation of ${ }^{209} \\mathrm{Bi}$. This produces short-lived ${ }^{210} \\mathrm{Bi}$ which decays to polonium by the emission of a betaparticle (an electron):\n\n$$\n{ }_{83}^{209} \\mathrm{Bi}+{ }_{0}^{1} n \\rightarrow{ }_{83}^{210} \\mathrm{Bi}+\\gamma \\quad{ }_{83}^{210} \\mathrm{Bi} \\rightarrow{ }_{84}^{210} \\mathrm{Po}+{ }_{-1}^{0} \\beta\n$$\n\nPolonium-210 has a half life of 138 days and decays by emitting an alpha particle (a helium nucleus).\n\n[figure1]\n\nDue to its very short half life and the impedance of the alpha-particles it emits, metallic polonium and its compounds are self heating; $1 \\mathrm{~g}$ of metal produces $141 \\mathrm{~W}$. This led to its use in Radioisotope Heater Units (RHUs) to keep satellites warm and functioning in space, and in Radioisotope Thermal Generators (RTGs) to produce electrical power. More recently, plutonium-238 has been used instead of polonium. ${ }^{238} \\mathrm{Pu}$ has a much longer half-life but produces less power $\\left(0.56 \\mathrm{~W} \\mathrm{~g}^{-1}\\right)$.\n\nWhat will be the power output of ${ }^{210} \\mathrm{Po}$ after 1 year?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nPolonium is a radioactive group VI element, discovered in 1898 by Marie Curie. It occurs naturally in trace amounts in some uranium ores but is now made by neutron irradiation of ${ }^{209} \\mathrm{Bi}$. This produces short-lived ${ }^{210} \\mathrm{Bi}$ which decays to polonium by the emission of a betaparticle (an electron):\n\n$$\n{ }_{83}^{209} \\mathrm{Bi}+{ }_{0}^{1} n \\rightarrow{ }_{83}^{210} \\mathrm{Bi}+\\gamma \\quad{ }_{83}^{210} \\mathrm{Bi} \\rightarrow{ }_{84}^{210} \\mathrm{Po}+{ }_{-1}^{0} \\beta\n$$\n\nPolonium-210 has a half life of 138 days and decays by emitting an alpha particle (a helium nucleus).\n\n[figure1]\n\nDue to its very short half life and the impedance of the alpha-particles it emits, metallic polonium and its compounds are self heating; $1 \\mathrm{~g}$ of metal produces $141 \\mathrm{~W}$. This led to its use in Radioisotope Heater Units (RHUs) to keep satellites warm and functioning in space, and in Radioisotope Thermal Generators (RTGs) to produce electrical power. More recently, plutonium-238 has been used instead of polonium. ${ }^{238} \\mathrm{Pu}$ has a much longer half-life but produces less power $\\left(0.56 \\mathrm{~W} \\mathrm{~g}^{-1}\\right)$.\n\nWhat will be the power output of ${ }^{210} \\mathrm{Po}$ after 1 year?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{W} \\mathrm{g}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_56fb624b7c8173291b3eg-10.jpg?height=500&width=648&top_left_y=367&top_left_x=1132"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{W} \\mathrm{g}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_253",
        "problem": "Estimate the potassium ion concentration (in $\\mathrm{mg} \\mathrm{L}^{-1}$ ) of a solution with an absorbance of 0.68 . Give your answer to 1 decimal place.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nEstimate the potassium ion concentration (in $\\mathrm{mg} \\mathrm{L}^{-1}$ ) of a solution with an absorbance of 0.68 . Give your answer to 1 decimal place.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg/l, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg/l"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1175",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nHow many protons  does an iodide ion contain",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n## the tellurium-iodine problem\n\nAlthough the elements in Mendeleev's table are primarily arranged by atomic mass, this was not the case with tellurium (Te) and iodine. Mendeleev realised that the chemical properties of the elements meant that tellurium had to come before iodine, but the atomic masses did not support this order. He marked the mass of tellurium with a question mark to highlight its suspicious value.\n\nThe modern value for the relative atomic mass of tellurium is one of the least precise: $127.60 \\pm$ 0.03 . The reason for the uncertainty is that naturally occurring tellurium is a mix of 8 different isotopes whose proportions can vary depending on the sample. In contrast, naturally occurring iodine consists of a single isotope - iodine 127 - and so its relative mass is known to a high precision: $126.904472 \\pm 0.000003$.\n\nHow many protons  does an iodide ion contain\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1017",
        "problem": "Polonium-210 $\\left({ }^{210} \\mathrm{Po}\\right)$ is radioactive, extremely toxic, and it decays according to the chemical equation below. What is the missing product in the equation?\n\n$$\n{ }^{210} \\mathrm{Po} \\rightarrow ?+{ }_{2}^{4} \\mathrm{He}\n$$\nA: ${ }^{214} \\mathrm{Po}$\nB: ${ }^{212} \\mathrm{TI}$\nC: ${ }^{206} \\mathrm{~Pb}$\nD: ${ }^{214} \\mathrm{Rn}$\nE: ${ }^{210} \\mathrm{Po}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPolonium-210 $\\left({ }^{210} \\mathrm{Po}\\right)$ is radioactive, extremely toxic, and it decays according to the chemical equation below. What is the missing product in the equation?\n\n$$\n{ }^{210} \\mathrm{Po} \\rightarrow ?+{ }_{2}^{4} \\mathrm{He}\n$$\n\nA: ${ }^{214} \\mathrm{Po}$\nB: ${ }^{212} \\mathrm{TI}$\nC: ${ }^{206} \\mathrm{~Pb}$\nD: ${ }^{214} \\mathrm{Rn}$\nE: ${ }^{210} \\mathrm{Po}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1182",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nIt is predicted that in order to complete the next row of the periodic table starting with electrons in the 8s shell, it will also be necessary to fill elements in the first row of the g-block.\n\n How many $5 \\mathrm{~g}$ orbitals will need to be filled?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nIt is predicted that in order to complete the next row of the periodic table starting with electrons in the 8s shell, it will also be necessary to fill elements in the first row of the g-block.\n\n How many $5 \\mathrm{~g}$ orbitals will need to be filled?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_655",
        "problem": "下列溶液中粒子浓度关系一定正确的是 ( )\nA: 常温下, 氨水与硫酸混合后, 溶液的 $\\mathrm{pH}=7: c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{SO}_{4}^{2-}\\right)$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中: $c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{HCO}_{3}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{Na}^{+}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$ 溶液与 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水等体积混合: $2 c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=c\\left(\\mathrm{NH}_{4}^{+}\\right)+2 c\\left(\\mathrm{OH}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的硫化钠溶液中: $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{HS}^{-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列溶液中粒子浓度关系一定正确的是 ( )\n\nA: 常温下, 氨水与硫酸混合后, 溶液的 $\\mathrm{pH}=7: c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{SO}_{4}^{2-}\\right)$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{NaHCO}_{3}$ 溶液中: $c\\left(\\mathrm{OH}^{-}\\right)+c\\left(\\mathrm{HCO}_{3}^{-}\\right)+c\\left(\\mathrm{CO}_{3}^{2-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{Na}^{+}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCl}$ 溶液与 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 氨水等体积混合: $2 c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=c\\left(\\mathrm{NH}_{4}^{+}\\right)+2 c\\left(\\mathrm{OH}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的硫化钠溶液中: $c\\left(\\mathrm{OH}^{-}\\right)=c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{HS}^{-}\\right)+c\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1532",
        "problem": "The transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nTo determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at $298 \\mathrm{~K}$ are shown in the tables below. (Use the grids if you like.)\n\n| $[\\mathrm{ArCl}]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.1 | 0.2 | 0.4 | 0.6 |\n| :--- | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $1.88 \\times 10^{-5}$ | $4.13 \\times 10^{-5}$ | $9.42 \\times 10^{-5}$ | $1.50 \\times 10^{-4}$ |\n\n[figure1]\n\n| $\\left[\\mathrm{NiLL}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | $6 \\times 10^{-3}$ | $9 \\times 10^{-3}$ | $1.2 \\times 10^{-2}$ | $1.5 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $6.01 \\times 10^{-5}$ | $7.80 \\times 10^{-5}$ | $1.10 \\times 10^{-4}$ |\n\n\n| $\\left[\\mathrm{L}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.06 | 0.09 | 0.12 | 0.15 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $5.8 \\times 10^{-5}$ | $4.3 \\times 10^{-5}$ | $3.4 \\times 10^{-5}$ | $2.8 \\times 10^{-5}$ |Determine the order with respect to $\\mathrm{[NiLL']}$ assuming it is integer.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nTo determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at $298 \\mathrm{~K}$ are shown in the tables below. (Use the grids if you like.)\n\n| $[\\mathrm{ArCl}]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.1 | 0.2 | 0.4 | 0.6 |\n| :--- | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $1.88 \\times 10^{-5}$ | $4.13 \\times 10^{-5}$ | $9.42 \\times 10^{-5}$ | $1.50 \\times 10^{-4}$ |\n\n[figure1]\n\n| $\\left[\\mathrm{NiLL}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | $6 \\times 10^{-3}$ | $9 \\times 10^{-3}$ | $1.2 \\times 10^{-2}$ | $1.5 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $6.01 \\times 10^{-5}$ | $7.80 \\times 10^{-5}$ | $1.10 \\times 10^{-4}$ |\n\n\n| $\\left[\\mathrm{L}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.06 | 0.09 | 0.12 | 0.15 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $5.8 \\times 10^{-5}$ | $4.3 \\times 10^{-5}$ | $3.4 \\times 10^{-5}$ | $2.8 \\times 10^{-5}$ |\n\nproblem:\nDetermine the order with respect to $\\mathrm{[NiLL']}$ assuming it is integer.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-032.jpg?height=834&width=902&top_left_y=1553&top_left_x=583"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1176",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nOne compound that may be produced by this method is methylstuck-at-homium trichloride. One mole of this reacts with four moles of sodium hydroxide to produce the salt sodium methylstuck-athomanoate (a salt of methylstuck-at-homanoic acid) and two other by-products.\n\nGive the equation for the reaction of methylstuck-at-homium trichloride with sodium hydroxide.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nOne compound that may be produced by this method is methylstuck-at-homium trichloride. One mole of this reacts with four moles of sodium hydroxide to produce the salt sodium methylstuck-athomanoate (a salt of methylstuck-at-homanoic acid) and two other by-products.\n\nGive the equation for the reaction of methylstuck-at-homium trichloride with sodium hydroxide.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1309",
        "problem": "5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.\n\n[figure1]\n\nUshabti figurines from Egyptian pharaoh tomb covered with Egyptian blue and a Chinese blue soap dinspenser sold at Alibaba\n\nThe ancient method of preparation for these pigments can easily by reproduced in a modern laboratory.\n\nWhen considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.\n\nTo make Egyptian blue, one should heat $10.0 \\mathrm{~g}$ of mineral $\\mathrm{A}$ with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ and $9.05 \\mathrm{~g}$ of mineral $\\mathrm{B}$ at $800-900^{\\circ} \\mathrm{C}$ for a prolonged time. A volume of $16.7 \\mathrm{dm}^{3}$ of a mixture of two gaseous products are released (the volume is measured at $850^{\\circ} \\mathrm{C}$ and $1.013 \\times 10^{5} \\mathrm{~Pa}$ (1.013 bar) pressure. In result, $34.0 \\mathrm{~g}$ of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to $0^{\\circ} \\mathrm{C}$, the gaseous volume reduces to $3.04 \\mathrm{dm}^{3}$.\n\nWhen $10.0 \\mathrm{~g}$ of mineral $\\mathbf{A}$ is heated with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ in the absence of $\\mathbf{B}$, it forms $8.34 \\mathrm{dm}^{3}$ of gaseous products at a temperature of $850^{\\circ} \\mathrm{C}$ and pressure of $1.013 \\cdot 10^{5} \\mathrm{~Pa}(=1.013$ bar). Mineral A contains only one metal.Calculate the molar mass of mineral B. Hint: it is an ionic solid insoluble in water and containing no water of crystallization.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\n5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.\n\n[figure1]\n\nUshabti figurines from Egyptian pharaoh tomb covered with Egyptian blue and a Chinese blue soap dinspenser sold at Alibaba\n\nThe ancient method of preparation for these pigments can easily by reproduced in a modern laboratory.\n\nWhen considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.\n\nTo make Egyptian blue, one should heat $10.0 \\mathrm{~g}$ of mineral $\\mathrm{A}$ with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ and $9.05 \\mathrm{~g}$ of mineral $\\mathrm{B}$ at $800-900^{\\circ} \\mathrm{C}$ for a prolonged time. A volume of $16.7 \\mathrm{dm}^{3}$ of a mixture of two gaseous products are released (the volume is measured at $850^{\\circ} \\mathrm{C}$ and $1.013 \\times 10^{5} \\mathrm{~Pa}$ (1.013 bar) pressure. In result, $34.0 \\mathrm{~g}$ of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to $0^{\\circ} \\mathrm{C}$, the gaseous volume reduces to $3.04 \\mathrm{dm}^{3}$.\n\nWhen $10.0 \\mathrm{~g}$ of mineral $\\mathbf{A}$ is heated with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ in the absence of $\\mathbf{B}$, it forms $8.34 \\mathrm{dm}^{3}$ of gaseous products at a temperature of $850^{\\circ} \\mathrm{C}$ and pressure of $1.013 \\cdot 10^{5} \\mathrm{~Pa}(=1.013$ bar). Mineral A contains only one metal.\n\nproblem:\nCalculate the molar mass of mineral B. Hint: it is an ionic solid insoluble in water and containing no water of crystallization.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-159.jpg?height=611&width=1357&top_left_y=705&top_left_x=378"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_559",
        "problem": "将一定量的有机物充分燃烧后的产物通入足量的石灰水中完全吸收, 经过滤得沉淀 $20 \\mathrm{~g}$ ，滤液质量比原石灰水减少 $5.8 \\mathrm{~g}$, 该有机物可能是\nA: 甲烷\nB: 乙烷\nC: 丙烷\nD: 乙醇 $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n将一定量的有机物充分燃烧后的产物通入足量的石灰水中完全吸收, 经过滤得沉淀 $20 \\mathrm{~g}$ ，滤液质量比原石灰水减少 $5.8 \\mathrm{~g}$, 该有机物可能是\n\nA: 甲烷\nB: 乙烷\nC: 丙烷\nD: 乙醇 $\\left(\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1144",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nTo prepare $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$, an oxidising agent must be used. One preparation uses $\\mathrm{HgCl}_{2}$. Mercury forms two chlorides: $\\mathrm{HgCl}_{2}$ and $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n\n $\\mathrm{SkCl}_{4}$ smokes in air because it reacts with water vapour. In the simplest equation for this reaction, an oxide of stuck-at-homium is formed and one other product. Give an equation for this reaction.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nTo prepare $\\mathrm{SkCl}_{4}$ from $\\mathrm{SkCl}_{2}$, an oxidising agent must be used. One preparation uses $\\mathrm{HgCl}_{2}$. Mercury forms two chlorides: $\\mathrm{HgCl}_{2}$ and $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$.\n\n $\\mathrm{SkCl}_{4}$ smokes in air because it reacts with water vapour. In the simplest equation for this reaction, an oxide of stuck-at-homium is formed and one other product. Give an equation for this reaction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_187",
        "problem": "To increase strength and hardness when forging knives and blades, hot steel can be quenched by rapidly cooling in water. A $454 \\mathrm{~g}$ steel blade is heated to a uniform temperature, and then quenched in $2000 \\mathrm{~mL}$ of $25.0^{\\circ} \\mathrm{C}$ water. If the steel blade loses $173.7 \\mathrm{~kJ}$ of heat during the quenching process, what is the final temperature of the water? The specific heat capacity of water is $4.18 \\mathrm{~J} \\mathrm{~g}^{-1}{ }^{\\circ} \\mathrm{C}^{-1}$. Assume no water evaporates during the quenching process.\nA: $16.9^{\\circ} \\mathrm{C}$\nB: $20.7^{\\circ} \\mathrm{C}$\nC: $41.9^{\\circ} \\mathrm{C}$\nD: $45.8^{\\circ} \\mathrm{C}$\nE: $91.5^{\\circ} \\mathrm{C}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nTo increase strength and hardness when forging knives and blades, hot steel can be quenched by rapidly cooling in water. A $454 \\mathrm{~g}$ steel blade is heated to a uniform temperature, and then quenched in $2000 \\mathrm{~mL}$ of $25.0^{\\circ} \\mathrm{C}$ water. If the steel blade loses $173.7 \\mathrm{~kJ}$ of heat during the quenching process, what is the final temperature of the water? The specific heat capacity of water is $4.18 \\mathrm{~J} \\mathrm{~g}^{-1}{ }^{\\circ} \\mathrm{C}^{-1}$. Assume no water evaporates during the quenching process.\n\nA: $16.9^{\\circ} \\mathrm{C}$\nB: $20.7^{\\circ} \\mathrm{C}$\nC: $41.9^{\\circ} \\mathrm{C}$\nD: $45.8^{\\circ} \\mathrm{C}$\nE: $91.5^{\\circ} \\mathrm{C}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_243",
        "problem": "Which of the following is both an empirical formula and a molecular formula?\nA: $\\mathrm{C}_{3} \\mathrm{~F}_{6}$\nB: $\\mathrm{C}_{3} \\mathrm{~F}_{8}$\nC: $\\mathrm{C}_{4} \\mathrm{~F}_{6}$\nD: $\\mathrm{C}_{4} \\mathrm{~F}_{8}$\nE: $\\mathrm{C}_{4} \\mathrm{~F}_{10}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is both an empirical formula and a molecular formula?\n\nA: $\\mathrm{C}_{3} \\mathrm{~F}_{6}$\nB: $\\mathrm{C}_{3} \\mathrm{~F}_{8}$\nC: $\\mathrm{C}_{4} \\mathrm{~F}_{6}$\nD: $\\mathrm{C}_{4} \\mathrm{~F}_{8}$\nE: $\\mathrm{C}_{4} \\mathrm{~F}_{10}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1125",
        "problem": "This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nAssuming the reaction takes place with the reactants in the whole-number ratio above, how many electrons are formally transferred during the reaction?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nAssuming the reaction takes place with the reactants in the whole-number ratio above, how many electrons are formally transferred during the reaction?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_224",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the molar mass of molecule $\\mathbf{1 .}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the molar mass of molecule $\\mathbf{1 .}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g/mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g/mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1132",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhen Mendeleev created his first table indium had only just been discovered and its chemical properties had not been fully studied. The error in the mass arose from the fact that the correct formula for indium oxide was not known.\n\nIn one experiment to determine the atomic mass of indium, $0.5135 \\mathrm{~g}$ of indium metal was converted to $0.6243 \\mathrm{~g}$ of the oxide. Using these data, and the modern relative atomic mass of oxygen, calculate the apparent atomic mass of indium assuming:\n\nthe formula for indium oxide is $\\mathrm{InO}_{2}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nWhen Mendeleev created his first table indium had only just been discovered and its chemical properties had not been fully studied. The error in the mass arose from the fact that the correct formula for indium oxide was not known.\n\nIn one experiment to determine the atomic mass of indium, $0.5135 \\mathrm{~g}$ of indium metal was converted to $0.6243 \\mathrm{~g}$ of the oxide. Using these data, and the modern relative atomic mass of oxygen, calculate the apparent atomic mass of indium assuming:\n\nthe formula for indium oxide is $\\mathrm{InO}_{2}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1402",
        "problem": "Regarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe transmittance is inversely proportional to the logarithm of absorbance.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nRegarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe transmittance is inversely proportional to the logarithm of absorbance.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_621",
        "problem": "工业上可利用氨水吸收 $\\mathrm{SO}_{2}$ 和 $\\mathrm{NO}_{2}$, 原理如图所示。已知: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的 $K_{\\mathrm{b}}=10^{-4.7}, \\mathrm{H}_{2} \\mathrm{SO}_{3}$ 的 $K_{\\mathrm{a}_{1}}=10^{-1.8}, K_{\\mathrm{a}_{2}}=10^{-7.0}$ 。下列说法正确的是\n\n[图1]\nA: 反应 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\mathrm{SO}_{3} \\rightleftharpoons \\mathrm{NH}_{4}^{+}+\\mathrm{HSO}_{3}^{-}+\\mathrm{H}_{2} \\mathrm{O}$ 的平衡常数 $K=10^{-6.5}$\nB: 向氨水中通入 $\\mathrm{SO}_{2}$ 至 $\\mathrm{pH}=7: c\\left(\\mathrm{HSO}_{3}^{-}\\right)>c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)$\nC: 测得 $\\mathrm{NH}_{4} \\mathrm{HSO}_{3}$ 溶液 $\\mathrm{pH}=4.1: \\frac{c\\left(\\mathrm{NH}_{4}^{+}\\right)}{c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}>\\frac{c\\left(\\mathrm{HSO}_{3}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)}$\nD: $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶液中: $c\\left(\\mathrm{H}^{+}\\right)=2 c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n工业上可利用氨水吸收 $\\mathrm{SO}_{2}$ 和 $\\mathrm{NO}_{2}$, 原理如图所示。已知: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 的 $K_{\\mathrm{b}}=10^{-4.7}, \\mathrm{H}_{2} \\mathrm{SO}_{3}$ 的 $K_{\\mathrm{a}_{1}}=10^{-1.8}, K_{\\mathrm{a}_{2}}=10^{-7.0}$ 。下列说法正确的是\n\n[图1]\n\nA: 反应 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\mathrm{SO}_{3} \\rightleftharpoons \\mathrm{NH}_{4}^{+}+\\mathrm{HSO}_{3}^{-}+\\mathrm{H}_{2} \\mathrm{O}$ 的平衡常数 $K=10^{-6.5}$\nB: 向氨水中通入 $\\mathrm{SO}_{2}$ 至 $\\mathrm{pH}=7: c\\left(\\mathrm{HSO}_{3}^{-}\\right)>c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{H}^{+}\\right)=c\\left(\\mathrm{OH}^{-}\\right)$\nC: 测得 $\\mathrm{NH}_{4} \\mathrm{HSO}_{3}$ 溶液 $\\mathrm{pH}=4.1: \\frac{c\\left(\\mathrm{NH}_{4}^{+}\\right)}{c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)}>\\frac{c\\left(\\mathrm{HSO}_{3}^{-}\\right)}{c\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)}$\nD: $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ 溶液中: $c\\left(\\mathrm{H}^{+}\\right)=2 c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)+c\\left(\\mathrm{OH}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-024.jpg?height=160&width=865&top_left_y=405&top_left_x=344"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_343",
        "problem": "In acidic solution, zinc metal reacts spontaneously with $\\mathrm{ReO}_{4}{ }^{-}$. The unbalanced chemical equation for the reaction is given below.\n\n$\\mathrm{Zn}(s)+\\mathrm{ReO}_{4}{ }^{-}(a q)+\\mathrm{H}^{+}(a q) \\rightarrow \\mathrm{Re}(s)+\\mathrm{Zn}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n\nWhat is the coefficient of zinc $(\\mathrm{Zn})$ when the equation above for the reaction is balanced using the smallest whole number coefficients?\nA: 1\nB: 2\nC: 7\nD: 16\nE: none of the above\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn acidic solution, zinc metal reacts spontaneously with $\\mathrm{ReO}_{4}{ }^{-}$. The unbalanced chemical equation for the reaction is given below.\n\n$\\mathrm{Zn}(s)+\\mathrm{ReO}_{4}{ }^{-}(a q)+\\mathrm{H}^{+}(a q) \\rightarrow \\mathrm{Re}(s)+\\mathrm{Zn}^{2+}(a q)+\\mathrm{H}_{2} \\mathrm{O}(l)$\n\nWhat is the coefficient of zinc $(\\mathrm{Zn})$ when the equation above for the reaction is balanced using the smallest whole number coefficients?\n\nA: 1\nB: 2\nC: 7\nD: 16\nE: none of the above\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1041",
        "problem": "This question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nThe density of liquid hydrogen is $0.071 \\mathrm{~g} \\mathrm{~cm}^{-3}$.\n\nCalculate the number of moles of hydrogen molecules in $1 \\mathrm{dm}^{3}$ of liquid hydrogen.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about rocket fuels\n\nThe NASA Artemis uses liquid oxygen (LOX) and liquid hydrogen fuel sources. These fuels ensure the rocket mass is low and alongside their large enthalpy change of combustion enable a rocket to overcome gravity.\n\nIn 2022 there were a number of failed launch attempts due to a leak in the hydrogen fuel.\n\n[figure1]\n\nIn the rocket engine the fuel components are first vaporised before reacting to form water.\n\nThe bond enthalpy of $\\mathrm{H}-\\mathrm{H}$ is $432 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$ and the average bond enthalpy of $\\mathrm{O}-\\mathrm{H}$ is $460 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$. Assume the enthalpy change for the reaction in part (a) is $-241 \\mathrm{~kJ}$ per mole of hydrogen gas.\n\nThe density of liquid hydrogen is $0.071 \\mathrm{~g} \\mathrm{~cm}^{-3}$.\n\nCalculate the number of moles of hydrogen molecules in $1 \\mathrm{dm}^{3}$ of liquid hydrogen.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-02.jpg?height=434&width=619&top_left_y=317&top_left_x=1181"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_252",
        "problem": "Calculate the chemical amount (in mol or mmol) of potassium in the original $5.6311 \\mathrm{~g}$ of potassium salt.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the chemical amount (in mol or mmol) of potassium in the original $5.6311 \\mathrm{~g}$ of potassium salt.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mmol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mmol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_95",
        "problem": "An unknown metal $M$ reacts with aqueous sulfuric acid to form hydrogen gas and the divalent metal ion $\\mathrm{M}^{2+}(a q)$. A student collects the hydrogen over water and calculates the molar mass of $M$ based on the volume of hydrogen collected. Which error will result in a value of the calculated molar mass that is higher than the true molar mass?\nA: The concentration of sulfuric acid is $2.0 \\mathrm{M}$ rather than $1.0 \\mathrm{M}$ as recommended by the procedure.\nB: Some of the metal sample has oxidized to the metal oxide.\nC: The barometric pressure is assumed to be 760 torr when in fact it is 740 torr.\nD: The vapor pressure of water is not included in the calculation.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAn unknown metal $M$ reacts with aqueous sulfuric acid to form hydrogen gas and the divalent metal ion $\\mathrm{M}^{2+}(a q)$. A student collects the hydrogen over water and calculates the molar mass of $M$ based on the volume of hydrogen collected. Which error will result in a value of the calculated molar mass that is higher than the true molar mass?\n\nA: The concentration of sulfuric acid is $2.0 \\mathrm{M}$ rather than $1.0 \\mathrm{M}$ as recommended by the procedure.\nB: Some of the metal sample has oxidized to the metal oxide.\nC: The barometric pressure is assumed to be 760 torr when in fact it is 740 torr.\nD: The vapor pressure of water is not included in the calculation.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_248",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the molar mass of KHP (in $\\mathrm{g} \\mathrm{mol}^{-1}$ )\n$20.58 \\mathrm{~g}$ of $\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is dissolved in water, giving a solution with a mass of $118.48 \\mathrm{~g}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nIn the absence of volumetric glassware, it is possible to use only mass measurements to determine the composition of solutions.\n\n$\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is an acid commonly used in such determinations.\nCalculate the molar mass of KHP (in $\\mathrm{g} \\mathrm{mol}^{-1}$ )\n$20.58 \\mathrm{~g}$ of $\\mathrm{KHP}\\left(\\mathrm{KC}_{8} \\mathrm{H}_{5} \\mathrm{O}_{4}\\right)$ is dissolved in water, giving a solution with a mass of $118.48 \\mathrm{~g}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g/mol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g/mol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_808",
        "problem": "一定温度下, 在三个容积相同的恒容密闭容器中按不同方式投入反应物, 发生反应 $2 \\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{SO}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-196.0 \\mathrm{~kJ} . \\mathrm{mol}^{-1}$, 测得反应的相关数据如下, 下列说法正确的是 ( )\n\n|  | 容器 1 | 容器 2 | 器 3 |\n| :--- | :--- | :--- | :--- |\n| 应温度 $\\mathrm{T} / \\mathrm{K}$ | 700 | 700 | 800 |\n| 反应物投入量 | $2 \\mathrm{~mol} \\mathrm{SO}_{2} 、 1 \\mathrm{~mol} \\mathrm{O}_{2}$ | $4 \\mathrm{~mol} \\mathrm{SO}_{3}$ | $2 \\mathrm{~mol} \\mathrm{SO}_{3}$ |\n| 平衡 $\\mathrm{v}$ 正 $\\left(\\mathrm{SO}_{2}\\right) / \\mathrm{mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{S}^{-1}$ | $\\mathrm{v}_{1}$ | $\\mathrm{v}_{2}$ | $\\mathrm{v}_{3}$ |\n| 平衡 $\\mathrm{c}\\left(\\mathrm{SO}_{3}\\right) / \\mathrm{mol}^{-1} \\mathrm{~L}^{-1}$ | $\\mathrm{c}_{1}$ | $\\mathrm{c}_{2}$ | $\\mathrm{c}_{3}$ |\n| 平衡体系总压强 $\\mathrm{p} / \\mathrm{Pa}$ | $\\mathrm{p}_{1}$ | $\\mathrm{p}_{2}$ | $\\mathrm{p}_{3}$ |\n| 反应能量变化 | 放出 $\\mathrm{a} \\mathrm{KJ}$ | 吸收 $\\mathrm{b} \\mathrm{KJ}$ | 吸收 $\\mathrm{c} \\mathrm{KJ}^{2}$ |\n| 物质的平衡转化率 $\\alpha$ | $\\alpha_{1}\\left(\\mathrm{SO}_{2}\\right)$ | $\\alpha_{2}\\left(\\mathrm{SO}_{3}\\right)$ | $\\alpha_{3}\\left(\\mathrm{SO}_{2}\\right)$ |\n| 平衡常数 $\\mathrm{K}$ | $\\mathrm{K}_{1}$ | $\\mathrm{~K}_{2}$ | $\\mathrm{~K}_{3}$ |\nA: $\\mathrm{v}_{1}<\\mathrm{v}_{2}, \\mathrm{c}_{2}<2 \\mathrm{c}_{1}$\nB: $\\mathrm{K}_{1}>\\mathrm{K}_{3}, \\mathrm{p}_{2}>2 \\mathrm{p}_{3}$\nC: $\\mathrm{v}_{1}<\\mathrm{v}_{3}, \\alpha_{1}\\left(\\mathrm{SO}_{2}\\right)<\\alpha_{2}\\left(\\mathrm{SO}_{2}\\right)$\nD: $\\mathrm{c}_{2}>2 \\mathrm{c}_{3}, \\quad \\alpha_{2}\\left(\\mathrm{SO}_{3}\\right)+\\alpha_{3}\\left(\\mathrm{SO}_{2}\\right)<1$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n一定温度下, 在三个容积相同的恒容密闭容器中按不同方式投入反应物, 发生反应 $2 \\mathrm{SO}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{SO}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-196.0 \\mathrm{~kJ} . \\mathrm{mol}^{-1}$, 测得反应的相关数据如下, 下列说法正确的是 ( )\n\n|  | 容器 1 | 容器 2 | 器 3 |\n| :--- | :--- | :--- | :--- |\n| 应温度 $\\mathrm{T} / \\mathrm{K}$ | 700 | 700 | 800 |\n| 反应物投入量 | $2 \\mathrm{~mol} \\mathrm{SO}_{2} 、 1 \\mathrm{~mol} \\mathrm{O}_{2}$ | $4 \\mathrm{~mol} \\mathrm{SO}_{3}$ | $2 \\mathrm{~mol} \\mathrm{SO}_{3}$ |\n| 平衡 $\\mathrm{v}$ 正 $\\left(\\mathrm{SO}_{2}\\right) / \\mathrm{mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{S}^{-1}$ | $\\mathrm{v}_{1}$ | $\\mathrm{v}_{2}$ | $\\mathrm{v}_{3}$ |\n| 平衡 $\\mathrm{c}\\left(\\mathrm{SO}_{3}\\right) / \\mathrm{mol}^{-1} \\mathrm{~L}^{-1}$ | $\\mathrm{c}_{1}$ | $\\mathrm{c}_{2}$ | $\\mathrm{c}_{3}$ |\n| 平衡体系总压强 $\\mathrm{p} / \\mathrm{Pa}$ | $\\mathrm{p}_{1}$ | $\\mathrm{p}_{2}$ | $\\mathrm{p}_{3}$ |\n| 反应能量变化 | 放出 $\\mathrm{a} \\mathrm{KJ}$ | 吸收 $\\mathrm{b} \\mathrm{KJ}$ | 吸收 $\\mathrm{c} \\mathrm{KJ}^{2}$ |\n| 物质的平衡转化率 $\\alpha$ | $\\alpha_{1}\\left(\\mathrm{SO}_{2}\\right)$ | $\\alpha_{2}\\left(\\mathrm{SO}_{3}\\right)$ | $\\alpha_{3}\\left(\\mathrm{SO}_{2}\\right)$ |\n| 平衡常数 $\\mathrm{K}$ | $\\mathrm{K}_{1}$ | $\\mathrm{~K}_{2}$ | $\\mathrm{~K}_{3}$ |\n\nA: $\\mathrm{v}_{1}<\\mathrm{v}_{2}, \\mathrm{c}_{2}<2 \\mathrm{c}_{1}$\nB: $\\mathrm{K}_{1}>\\mathrm{K}_{3}, \\mathrm{p}_{2}>2 \\mathrm{p}_{3}$\nC: $\\mathrm{v}_{1}<\\mathrm{v}_{3}, \\alpha_{1}\\left(\\mathrm{SO}_{2}\\right)<\\alpha_{2}\\left(\\mathrm{SO}_{2}\\right)$\nD: $\\mathrm{c}_{2}>2 \\mathrm{c}_{3}, \\quad \\alpha_{2}\\left(\\mathrm{SO}_{3}\\right)+\\alpha_{3}\\left(\\mathrm{SO}_{2}\\right)<1$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1219",
        "problem": "The transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nTo determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at $298 \\mathrm{~K}$ are shown in the tables below. (Use the grids if you like.)\n\n| $[\\mathrm{ArCl}]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.1 | 0.2 | 0.4 | 0.6 |\n| :--- | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $1.88 \\times 10^{-5}$ | $4.13 \\times 10^{-5}$ | $9.42 \\times 10^{-5}$ | $1.50 \\times 10^{-4}$ |\n\n[figure1]\n\n| $\\left[\\mathrm{NiLL}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | $6 \\times 10^{-3}$ | $9 \\times 10^{-3}$ | $1.2 \\times 10^{-2}$ | $1.5 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $6.01 \\times 10^{-5}$ | $7.80 \\times 10^{-5}$ | $1.10 \\times 10^{-4}$ |\n\n\n| $\\left[\\mathrm{L}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.06 | 0.09 | 0.12 | 0.15 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $5.8 \\times 10^{-5}$ | $4.3 \\times 10^{-5}$ | $3.4 \\times 10^{-5}$ | $2.8 \\times 10^{-5}$ |Determine the order with respect to $\\mathrm{[L']}$ assuming it is integer.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nThe transition-metal-catalyzed amination of aryl halides has become one of the most powerful methods to synthesize arylamines. The overall reaction for the nickelcatalyzed amination of aryl chloride in basic conditions is:\n\n$$\n\\mathrm{ArCl}+\\mathrm{RNH}_{2} \\xrightarrow[\\mathrm{NaO} \\text { 'Bu, solvent }]{\\mathrm{NiLL'}} \\mathrm{Ar}-\\mathrm{NHR}+\\mathrm{HCl}\n$$\n\nin which NiLL' is the nickel complex catalyst. The reaction goes through several steps in which the catalyst, reactants, and solvent may be involved in elementary steps.\n\nTo determine the reaction order with respect to each reactant, the dependence of the initial rate of the reaction on the concentrations of each reagent was carried out with all other reagents present in large excess. Some kinetic data at $298 \\mathrm{~K}$ are shown in the tables below. (Use the grids if you like.)\n\n| $[\\mathrm{ArCl}]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.1 | 0.2 | 0.4 | 0.6 |\n| :--- | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $1.88 \\times 10^{-5}$ | $4.13 \\times 10^{-5}$ | $9.42 \\times 10^{-5}$ | $1.50 \\times 10^{-4}$ |\n\n[figure1]\n\n| $\\left[\\mathrm{NiLL}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | $6 \\times 10^{-3}$ | $9 \\times 10^{-3}$ | $1.2 \\times 10^{-2}$ | $1.5 \\times 10^{-2}$ |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $4.12 \\times 10^{-5}$ | $6.01 \\times 10^{-5}$ | $7.80 \\times 10^{-5}$ | $1.10 \\times 10^{-4}$ |\n\n\n| $\\left[\\mathrm{L}^{\\prime}\\right]$ <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3}\\right)$ | 0.06 | 0.09 | 0.12 | 0.15 |\n| :---: | :---: | :---: | :---: | :---: |\n| Initial rate <br> $\\left(\\mathrm{mol} \\mathrm{dm}^{-3} \\mathrm{~s}^{-1}\\right)$ | $5.8 \\times 10^{-5}$ | $4.3 \\times 10^{-5}$ | $3.4 \\times 10^{-5}$ | $2.8 \\times 10^{-5}$ |\n\nproblem:\nDetermine the order with respect to $\\mathrm{[L']}$ assuming it is integer.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-032.jpg?height=834&width=902&top_left_y=1553&top_left_x=583"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1297",
        "problem": "IONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\nCalculate the composition of the solution (in $\\mathrm{mol} \\mathrm{dm}^{-3}$ ) obtained on dissolving $1.00 \\times 10^{-2} \\mathrm{~mol}$ of copper(I) in $1.0 \\mathrm{dm}^{3}$ of water.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nIONIC SOLUTIONS - AQUEOUS SOLUTIONS OF COPPER SALTS\n\nThis part is about the acidity of the hydrated $\\mathrm{Cu}^{2+}$ ion and the precipitation of the hydroxide.\n\nConsider a $1.00 \\times 10^{-2} \\mathrm{~mol} \\mathrm{dm}^{-3}$ solution of copper(II) nitrate. The $\\mathrm{pH}$ of this solution is 4.65.\n\nDisproportionation of copper(I) ions\n\nThe $\\mathrm{Cu}^{+}$ion is involved in two redox couples:\n\nCouple 1: $\\quad \\mathrm{Cu}^{+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}$\n\nStandard electrode potential $E_{1}^{0}=+0.52 \\mathrm{~V}$\n\nCouple 2: $\\mathrm{Cu}^{2+}+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}^{+}$\n\nStandard electrode potential $E_{2}^{0}=+0.16 \\mathrm{~V}$\nCalculate the composition of the solution (in $\\mathrm{mol} \\mathrm{dm}^{-3}$ ) obtained on dissolving $1.00 \\times 10^{-2} \\mathrm{~mol}$ of copper(I) in $1.0 \\mathrm{dm}^{3}$ of water.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [The quantity of $\\left[\\mathrm{Cu}^{+}\\right]$, The quantity of $\\left[\\mathrm{Cu}^{2+}\\right]$].\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "The quantity of $\\left[\\mathrm{Cu}^{+}\\right]$",
            "The quantity of $\\left[\\mathrm{Cu}^{2+}\\right]$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_431",
        "problem": "近日, 我国学者在 Science 报道了一种氯离子介导的电化学合成方法, 能将乙烯高效清洁、选择性地转化为环氧乙烷, 电化学反应的过程如图所示。在电解结束后, 将阴、阳极电解液输出混合, 便可反应生成环氧乙烷。下列叙述错误的是\n\n[图1]\n\n阴离子交换膜\n\n[图2]\nA: 电解过程中 $\\mathrm{Cl}^{-}$透过交换膜向右侧移动\nB: 工作过程中 $\\mathrm{Ni}$ 电极附近的电极反应式为 $2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=2 \\mathrm{OH}^{-}+\\mathrm{H}_{2} \\uparrow$\nC: 每生成 $1 \\mathrm{~mol}$ 环氧乙烷, 理论上电路中转移电子数为 $2 \\mathrm{~N}_{\\mathrm{A}}$\nD: 电解结束溶液混合后 $\\mathrm{KCl}$ 的浓度比电解前的小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n近日, 我国学者在 Science 报道了一种氯离子介导的电化学合成方法, 能将乙烯高效清洁、选择性地转化为环氧乙烷, 电化学反应的过程如图所示。在电解结束后, 将阴、阳极电解液输出混合, 便可反应生成环氧乙烷。下列叙述错误的是\n\n[图1]\n\n阴离子交换膜\n\n[图2]\n\nA: 电解过程中 $\\mathrm{Cl}^{-}$透过交换膜向右侧移动\nB: 工作过程中 $\\mathrm{Ni}$ 电极附近的电极反应式为 $2 \\mathrm{H}_{2} \\mathrm{O}+2 \\mathrm{e}^{-}=2 \\mathrm{OH}^{-}+\\mathrm{H}_{2} \\uparrow$\nC: 每生成 $1 \\mathrm{~mol}$ 环氧乙烷, 理论上电路中转移电子数为 $2 \\mathrm{~N}_{\\mathrm{A}}$\nD: 电解结束溶液混合后 $\\mathrm{KCl}$ 的浓度比电解前的小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-75.jpg?height=405&width=717&top_left_y=183&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-75.jpg?height=154&width=942&top_left_y=640&top_left_x=360"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_173",
        "problem": "The molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. Lauryl alcohol $\\left(\\mathrm{C}_{12} \\mathrm{H}_{26} \\mathrm{O}\\right)$ is prepared from coconut oil and is used to make sodium lauryl sulfate, a synthetic detergent. What is the molality of a solution of $17.1 \\mathrm{~g}$ lauryl alcohol dissolved in 3.21 moles of ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}\\right)$ ?\nA: $0.310 \\mathrm{~m}$\nB: $0.621 \\mathrm{~m}$\nC: $0.842 \\mathrm{~m}$\nD: $1.41 \\mathrm{~m}$\nE: $2.52 \\mathrm{~m}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe molality (m) of a solution is defined as the number of moles of solute per kilogram of solvent. Lauryl alcohol $\\left(\\mathrm{C}_{12} \\mathrm{H}_{26} \\mathrm{O}\\right)$ is prepared from coconut oil and is used to make sodium lauryl sulfate, a synthetic detergent. What is the molality of a solution of $17.1 \\mathrm{~g}$ lauryl alcohol dissolved in 3.21 moles of ethanol $\\left(\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{O}\\right)$ ?\n\nA: $0.310 \\mathrm{~m}$\nB: $0.621 \\mathrm{~m}$\nC: $0.842 \\mathrm{~m}$\nD: $1.41 \\mathrm{~m}$\nE: $2.52 \\mathrm{~m}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_265",
        "problem": "A $1.620 \\mathrm{~g}$ of $\\mathrm{XF}_{6}$ can be produced from $1.000 \\mathrm{~g}$ of element $\\mathbf{X}$.\n\nWhich of the following could be element $\\mathbf{X}$ ?\nA: W\nB: Se\nC: Mo\nD: $\\mathrm{Rh}$\nE: $\\mathrm{U}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $1.620 \\mathrm{~g}$ of $\\mathrm{XF}_{6}$ can be produced from $1.000 \\mathrm{~g}$ of element $\\mathbf{X}$.\n\nWhich of the following could be element $\\mathbf{X}$ ?\n\nA: W\nB: Se\nC: Mo\nD: $\\mathrm{Rh}$\nE: $\\mathrm{U}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_564",
        "problem": "为完成下列各组实验, 所选玻璃仪器和试剂均准确、完整的是(不考虑存放试剂的容器)\n\n| 选 | 实验目的 | 玻璃仪器 | 试剂 |\n| :---: | :---: | :---: | :---: |\n| A | 实验室制备乙烯 <br> 气体 | 圆底烧瓶、酒精灯、导 <br> 管 | 乙醇、浓硫酸 |\n| B | 验证 1-溴丁烷发 <br> 生消去反应 | 酒精灯、圆底烧瓶、导 <br> 管、试管 | 1-滔丁烷、乙醇、氢氧化钠、酸性高 <br> 锰酸钾溶液 |\n| C | 硫酸四氨合铜晶 <br> 体的形成 | 胶头滴管、试管、玻璃 <br> 棒 | 硫酸铜溶液、氨水、乙醇 |\n| D | 验证纤维素的水 <br> 解产物 | 酒精灯、试管、玻璃棒、 <br> 胶头滴管 | 脱脂棉、浓硫酸、蒸馏水、氢氧化钠 <br> 溶液、硫酸铜溶液 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n为完成下列各组实验, 所选玻璃仪器和试剂均准确、完整的是(不考虑存放试剂的容器)\n\n| 选 | 实验目的 | 玻璃仪器 | 试剂 |\n| :---: | :---: | :---: | :---: |\n| A | 实验室制备乙烯 <br> 气体 | 圆底烧瓶、酒精灯、导 <br> 管 | 乙醇、浓硫酸 |\n| B | 验证 1-溴丁烷发 <br> 生消去反应 | 酒精灯、圆底烧瓶、导 <br> 管、试管 | 1-滔丁烷、乙醇、氢氧化钠、酸性高 <br> 锰酸钾溶液 |\n| C | 硫酸四氨合铜晶 <br> 体的形成 | 胶头滴管、试管、玻璃 <br> 棒 | 硫酸铜溶液、氨水、乙醇 |\n| D | 验证纤维素的水 <br> 解产物 | 酒精灯、试管、玻璃棒、 <br> 胶头滴管 | 脱脂棉、浓硫酸、蒸馏水、氢氧化钠 <br> 溶液、硫酸铜溶液 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_436",
        "problem": "室温下, 用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.0500 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液所得滴定曲线如图所示。下列关于溶液中微粒的物质的量浓度关系一定正确的是\n\n[图1]\nA: 在整个滴定过程中, 始终存在: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nB: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=10.00 \\mathrm{~mL}$ 时: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nC: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=15.00 \\mathrm{~mL}$ 时: $3 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=30.00 \\mathrm{~mL}$ 时： $\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n室温下, 用 $0.1000 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液滴定 $20.00 \\mathrm{~mL} 0.0500 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液所得滴定曲线如图所示。下列关于溶液中微粒的物质的量浓度关系一定正确的是\n\n[图1]\n\nA: 在整个滴定过程中, 始终存在: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nB: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=10.00 \\mathrm{~mL}$ 时: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nC: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=15.00 \\mathrm{~mL}$ 时: $3 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nD: 当 $\\mathrm{V}[\\mathrm{NaOH}(\\mathrm{aq})]=30.00 \\mathrm{~mL}$ 时： $\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)=\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-083.jpg?height=488&width=743&top_left_y=1886&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1272",
        "problem": "Determination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\n\n\nA known amount of water $(21.537 \\mathrm{~g})$ was placed into a $1.000 \\mathrm{dm}^{3}$ volumetric flask which was filled by methanol up to the mark. For titration of $10.00 \\mathrm{~cm}^{3}$ of the obtained solution, $22.70 \\mathrm{~cm}^{3}$ of Fischer reagent solution were needed, whereas 2.20 $\\mathrm{cm}^{3}$ of iodine were used for titration of $25.00 \\mathrm{~cm}^{3}$ of methanol.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nDetermination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\n\n\nA known amount of water $(21.537 \\mathrm{~g})$ was placed into a $1.000 \\mathrm{dm}^{3}$ volumetric flask which was filled by methanol up to the mark. For titration of $10.00 \\mathrm{~cm}^{3}$ of the obtained solution, $22.70 \\mathrm{~cm}^{3}$ of Fischer reagent solution were needed, whereas 2.20 $\\mathrm{cm}^{3}$ of iodine were used for titration of $25.00 \\mathrm{~cm}^{3}$ of methanol.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{mg} \\mathrm{cm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-651.jpg?height=213&width=900&top_left_y=776&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{mg} \\mathrm{cm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_678",
        "problem": "$\\operatorname{AgCl}\\left(\\mathrm{K}_{\\mathrm{sp}}=1.6 \\times 10^{-10}\\right)$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ (砖红色) 都是难溶电解质, 以\n\n$\\mathrm{pAg}\\left[\\mathrm{pAg}=-\\operatorname{lgc}\\left(\\mathrm{Ag}^{+}\\right)\\right]$对 $\\mathrm{pCl}$ 和 $\\mathrm{pCrO}_{4}$ 作图的沉淀平衡曲线如图所示。下列说法不正确的是\n\n[图1]\n\n氯化银和铭酸银沉淀的平衡曲线\nA: 阴影区域 $\\mathrm{AgCl}$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 都将溶解\nB: (2) 线是 $\\mathrm{AgCl}$ 的沉淀平衡曲线\nC: 向含有 $\\mathrm{AgCl}(\\mathrm{s})$ 的 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KCl}$ 溶液中加入 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$, 白色固体逐渐变为砖红色\nD: $2 \\mathrm{pAg}+\\mathrm{pCrO}_{4}=11.72$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\operatorname{AgCl}\\left(\\mathrm{K}_{\\mathrm{sp}}=1.6 \\times 10^{-10}\\right)$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ (砖红色) 都是难溶电解质, 以\n\n$\\mathrm{pAg}\\left[\\mathrm{pAg}=-\\operatorname{lgc}\\left(\\mathrm{Ag}^{+}\\right)\\right]$对 $\\mathrm{pCl}$ 和 $\\mathrm{pCrO}_{4}$ 作图的沉淀平衡曲线如图所示。下列说法不正确的是\n\n[图1]\n\n氯化银和铭酸银沉淀的平衡曲线\n\nA: 阴影区域 $\\mathrm{AgCl}$ 和 $\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}$ 都将溶解\nB: (2) 线是 $\\mathrm{AgCl}$ 的沉淀平衡曲线\nC: 向含有 $\\mathrm{AgCl}(\\mathrm{s})$ 的 $1.0 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KCl}$ 溶液中加入 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$, 白色固体逐渐变为砖红色\nD: $2 \\mathrm{pAg}+\\mathrm{pCrO}_{4}=11.72$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-109.jpg?height=477&width=1262&top_left_y=1692&top_left_x=357"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_230",
        "problem": "Biological tissue samples are often stained with dyes, which are coloured organic salts.\n\n- Basic dyes consist of a coloured cation and a colourless anion.\n- Acidic dyes consist of a coloured anion and a colourless cation.\n\nWhich of the following dyes are acidic? Select all that apply.\nA: $\\mathrm{C}_{21} \\mathrm{H}_{22} \\mathrm{~N}_{3} \\mathrm{Cl}$\nB: $\\mathrm{C}_{25} \\mathrm{H}_{33} \\mathrm{~N}_{2} \\mathrm{O}_{2} \\mathrm{Cl}$\nC: $\\mathrm{C}_{16} \\mathrm{H}_{17} \\mathrm{~N}_{2} \\mathrm{ClS}$\nD: $\\mathrm{C}_{19} \\mathrm{H}_{17} \\mathrm{~N}_{2} \\mathrm{NaO}_{5} \\mathrm{~S}$\nE: $\\mathrm{C}_{33} \\mathrm{H}_{43} \\mathrm{~N}_{3} \\mathrm{Na}_{2} \\mathrm{O}_{8} \\mathrm{~S}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nBiological tissue samples are often stained with dyes, which are coloured organic salts.\n\n- Basic dyes consist of a coloured cation and a colourless anion.\n- Acidic dyes consist of a coloured anion and a colourless cation.\n\nWhich of the following dyes are acidic? Select all that apply.\n\nA: $\\mathrm{C}_{21} \\mathrm{H}_{22} \\mathrm{~N}_{3} \\mathrm{Cl}$\nB: $\\mathrm{C}_{25} \\mathrm{H}_{33} \\mathrm{~N}_{2} \\mathrm{O}_{2} \\mathrm{Cl}$\nC: $\\mathrm{C}_{16} \\mathrm{H}_{17} \\mathrm{~N}_{2} \\mathrm{ClS}$\nD: $\\mathrm{C}_{19} \\mathrm{H}_{17} \\mathrm{~N}_{2} \\mathrm{NaO}_{5} \\mathrm{~S}$\nE: $\\mathrm{C}_{33} \\mathrm{H}_{43} \\mathrm{~N}_{3} \\mathrm{Na}_{2} \\mathrm{O}_{8} \\mathrm{~S}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_138",
        "problem": "One component of gastric (stomach) fluid is hydrochloric acid ( $\\mathrm{HCl})$. Baking soda $\\left(\\mathrm{NaHCO}_{3}\\right)$ will neutralize $\\mathrm{HCl}$ according to the reaction: $\\mathrm{NaHCO}_{3}(\\mathrm{aq})+\\mathrm{HCl}(\\mathrm{aq}) \\rightarrow \\mathrm{NaCl}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+\\mathrm{CO}_{2}(\\mathrm{~g})$\n\nHow many grams of $\\mathrm{CO}_{2}$ are produced when $300 \\mathrm{~mL}$ of $1.22 \\mathrm{~mol} \\mathrm{~L}^{-1}$ $\\mathrm{HCl}$ react with $175 \\mathrm{~mL}$ of $1.55 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NaHCO}_{3}$ ?\nA: $9.89 \\mathrm{~g}$\nB: $10.5 \\mathrm{~g}$\nC: $11.9 \\mathrm{~g}$\nD: $14.4 \\mathrm{~g}$\nE: $16.1 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOne component of gastric (stomach) fluid is hydrochloric acid ( $\\mathrm{HCl})$. Baking soda $\\left(\\mathrm{NaHCO}_{3}\\right)$ will neutralize $\\mathrm{HCl}$ according to the reaction: $\\mathrm{NaHCO}_{3}(\\mathrm{aq})+\\mathrm{HCl}(\\mathrm{aq}) \\rightarrow \\mathrm{NaCl}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+\\mathrm{CO}_{2}(\\mathrm{~g})$\n\nHow many grams of $\\mathrm{CO}_{2}$ are produced when $300 \\mathrm{~mL}$ of $1.22 \\mathrm{~mol} \\mathrm{~L}^{-1}$ $\\mathrm{HCl}$ react with $175 \\mathrm{~mL}$ of $1.55 \\mathrm{~mol} \\mathrm{~L}^{-1} \\mathrm{NaHCO}_{3}$ ?\n\nA: $9.89 \\mathrm{~g}$\nB: $10.5 \\mathrm{~g}$\nC: $11.9 \\mathrm{~g}$\nD: $14.4 \\mathrm{~g}$\nE: $16.1 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_107",
        "problem": "Penicillamine is an important organic compound used in the treatment of rheumatoid arthritis. One molecule of penicillamine contains a single sulfur atom and the weight percentage of sulfur in penicillamine is $21.49 \\%$. What is the molecular weight of penicillamine in $\\mathrm{g} \\mathrm{mol}^{-1}$ ?\nA: 85.40\nB: 101.3\nC: 125.2\nD: 137.6\nE: 149.2\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nPenicillamine is an important organic compound used in the treatment of rheumatoid arthritis. One molecule of penicillamine contains a single sulfur atom and the weight percentage of sulfur in penicillamine is $21.49 \\%$. What is the molecular weight of penicillamine in $\\mathrm{g} \\mathrm{mol}^{-1}$ ?\n\nA: 85.40\nB: 101.3\nC: 125.2\nD: 137.6\nE: 149.2\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_973",
        "problem": "Which of the following types of radiation has the highest energy per photon?\nA: radio waves\nB: ultraviolet radiation\nC: infrared radiation\nD: x-rays\nE: purple laser light\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following types of radiation has the highest energy per photon?\n\nA: radio waves\nB: ultraviolet radiation\nC: infrared radiation\nD: x-rays\nE: purple laser light\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1263",
        "problem": "5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.\n\n[figure1]\n\nUshabti figurines from Egyptian pharaoh tomb covered with Egyptian blue and a Chinese blue soap dinspenser sold at Alibaba\n\nThe ancient method of preparation for these pigments can easily by reproduced in a modern laboratory.\n\nWhen considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.\n\nTo make Egyptian blue, one should heat $10.0 \\mathrm{~g}$ of mineral $\\mathrm{A}$ with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ and $9.05 \\mathrm{~g}$ of mineral $\\mathrm{B}$ at $800-900^{\\circ} \\mathrm{C}$ for a prolonged time. A volume of $16.7 \\mathrm{dm}^{3}$ of a mixture of two gaseous products are released (the volume is measured at $850^{\\circ} \\mathrm{C}$ and $1.013 \\times 10^{5} \\mathrm{~Pa}$ (1.013 bar) pressure. In result, $34.0 \\mathrm{~g}$ of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to $0^{\\circ} \\mathrm{C}$, the gaseous volume reduces to $3.04 \\mathrm{dm}^{3}$.\n\nIn order to obtain Chinese blue, one should take $17.8 \\mathrm{~g}$ of mineral $\\mathbf{C}$ instead of mineral $\\mathbf{B}$ (keeping the amounts of mineral $\\mathbf{A}$ and $\\mathrm{SiO}_{2}$ same as for Egyptian blue), and run the reaction at higher temperatures. Besides the pigment, the same gaseous products in the same quantities are formed as in the preparation of Egyptian blue.\n\nElemental analysis of some samples of Chinese blue shows traces of sulfur. This led to a conclusion that those were synthesized using another common mineral instead of $\\mathbf{C}$.Could the temperature of synthesis of Chinese blue be decreased if this mineral is used instead of $\\mathbf{C}$ ?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\nHere is some context information for this question, which might assist you in solving it:\n5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.\n\n[figure1]\n\nUshabti figurines from Egyptian pharaoh tomb covered with Egyptian blue and a Chinese blue soap dinspenser sold at Alibaba\n\nThe ancient method of preparation for these pigments can easily by reproduced in a modern laboratory.\n\nWhen considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.\n\nTo make Egyptian blue, one should heat $10.0 \\mathrm{~g}$ of mineral $\\mathrm{A}$ with $21.7 \\mathrm{~g}$ of $\\mathrm{SiO}_{2}$ and $9.05 \\mathrm{~g}$ of mineral $\\mathrm{B}$ at $800-900^{\\circ} \\mathrm{C}$ for a prolonged time. A volume of $16.7 \\mathrm{dm}^{3}$ of a mixture of two gaseous products are released (the volume is measured at $850^{\\circ} \\mathrm{C}$ and $1.013 \\times 10^{5} \\mathrm{~Pa}$ (1.013 bar) pressure. In result, $34.0 \\mathrm{~g}$ of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to $0^{\\circ} \\mathrm{C}$, the gaseous volume reduces to $3.04 \\mathrm{dm}^{3}$.\n\nIn order to obtain Chinese blue, one should take $17.8 \\mathrm{~g}$ of mineral $\\mathbf{C}$ instead of mineral $\\mathbf{B}$ (keeping the amounts of mineral $\\mathbf{A}$ and $\\mathrm{SiO}_{2}$ same as for Egyptian blue), and run the reaction at higher temperatures. Besides the pigment, the same gaseous products in the same quantities are formed as in the preparation of Egyptian blue.\n\nElemental analysis of some samples of Chinese blue shows traces of sulfur. This led to a conclusion that those were synthesized using another common mineral instead of $\\mathbf{C}$.\n\nproblem:\nCould the temperature of synthesis of Chinese blue be decreased if this mineral is used instead of $\\mathbf{C}$ ?\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-159.jpg?height=611&width=1357&top_left_y=705&top_left_x=378"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1308",
        "problem": "Diatoms, microscopic organisms, are an abundant food source in the oceans producing carbohydrates from carbon dioxide and water by photosynthesis:\n\nEighteen percent of a $9.1 \\times 10^{4} \\mathrm{~kg}$ whale's mass is carbon. Carbon can be returned to the atmosphere as $\\mathrm{CO}_{2}$ and then removed from the atmosphere by weathering of rocks containing calcium silicate.\n$\\mathrm{CaSiO}_{3}(\\mathrm{~s})+2 \\mathrm{CO}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightarrow \\mathrm{Ca}^{2+}(\\mathrm{aq})+2 \\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})+\\mathrm{H}_{4} \\mathrm{SiO}_{4}(\\mathrm{aq})$\n\nWhat are the maximum number of grams of $\\mathrm{CaSiO}_{3}$ that can be weathered by the carbon dioxide produced from the decomposition of 1000 blue whales, the number estimated to die annually?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nDiatoms, microscopic organisms, are an abundant food source in the oceans producing carbohydrates from carbon dioxide and water by photosynthesis:\n\nEighteen percent of a $9.1 \\times 10^{4} \\mathrm{~kg}$ whale's mass is carbon. Carbon can be returned to the atmosphere as $\\mathrm{CO}_{2}$ and then removed from the atmosphere by weathering of rocks containing calcium silicate.\n$\\mathrm{CaSiO}_{3}(\\mathrm{~s})+2 \\mathrm{CO}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightarrow \\mathrm{Ca}^{2+}(\\mathrm{aq})+2 \\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})+\\mathrm{H}_{4} \\mathrm{SiO}_{4}(\\mathrm{aq})$\n\nWhat are the maximum number of grams of $\\mathrm{CaSiO}_{3}$ that can be weathered by the carbon dioxide produced from the decomposition of 1000 blue whales, the number estimated to die annually?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1056",
        "problem": "The 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nBicarbonate of soda (sodium hydrogencarbonate) is often used to treat ant stings.\n\nGiven that the density of methanoic acid is $1.2 \\mathrm{~g} \\mathrm{~cm}^{-3}$, how many moles of methanoic acid does the 'typical ant' inject?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe 'simplest' carboxylic acid is called methanoic acid and has formula $\\mathrm{HCOOH}$. It occurs naturally in ants and used to be prepared by distilling them! This gave rise to the earlier name for methanoic acid - formic acid - after the Latin word formica for ant.\n\nWhen an ant bites, it injects a solution containing $50 \\%$ by volume of methanoic acid. A typical ant may inject around $6.0 \\times 10^{-3} \\mathrm{~cm}^{3}$ of this solution.\n\n[figure1]\n\nA Formica rufa worker ant, just after biting the photographer!\n\nBicarbonate of soda (sodium hydrogencarbonate) is often used to treat ant stings.\n\nGiven that the density of methanoic acid is $1.2 \\mathrm{~g} \\mathrm{~cm}^{-3}$, how many moles of methanoic acid does the 'typical ant' inject?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of moles, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_4408eb29f8e936fbb54dg-05.jpg?height=622&width=654&top_left_y=237&top_left_x=1089"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "moles"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_31",
        "problem": "Given\n\n$\\begin{array}{ll}\\mathrm{O}(g)+e^{-}(g) \\rightarrow \\mathrm{O}^{-}(g) & \\Delta H^{0}=-142 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\ \\mathrm{O}(g)+2 e^{-}(g) \\rightarrow \\mathrm{O}^{2-}(g) & \\Delta H^{0}=+702 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\\end{array}$\n\nWhat is $\\Delta H^{\\circ}$ for the following reaction?\n\n$$\n\\mathrm{O}^{-}(g)+e^{-}(g) \\rightarrow \\mathrm{O}^{2-}(g)\n$$\nA: $-844 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-560 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $+560 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $+844 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nGiven\n\n$\\begin{array}{ll}\\mathrm{O}(g)+e^{-}(g) \\rightarrow \\mathrm{O}^{-}(g) & \\Delta H^{0}=-142 \\mathrm{~kJ} \\mathrm{~mol}^{-1} \\\\ \\mathrm{O}(g)+2 e^{-}(g) \\rightarrow \\mathrm{O}^{2-}(g) & \\Delta H^{0}=+702 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\\end{array}$\n\nWhat is $\\Delta H^{\\circ}$ for the following reaction?\n\n$$\n\\mathrm{O}^{-}(g)+e^{-}(g) \\rightarrow \\mathrm{O}^{2-}(g)\n$$\n\nA: $-844 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nB: $-560 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nC: $+560 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\nD: $+844 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_235",
        "problem": "Basic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the mass of carbon dioxide released when $10.00 \\mathrm{~g}$ of this mineral reacts with excess nitric acid.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBasic carbonate minerals are found widely in nature and, as their name suggests, contain carbonate ions $\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$ together with other basic ions such as hydroxide $\\left(\\mathrm{OH}^{-}\\right)$or oxide $\\left(\\mathrm{O}^{2-}\\right)$ and various metal cations.\n\nCarbonate content can be determined by measuring the mass of carbon dioxide given off when the mineral is treated with excess nitric acid.\nCalculate the mass of carbon dioxide released when $10.00 \\mathrm{~g}$ of this mineral reacts with excess nitric acid.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_985",
        "problem": "How many litres of gaseous methane $\\left(\\mathrm{CH}_{4}\\right)$ must be burned in oxygen to produce enough $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$ to fill a 3.0-L balloon? Assume that $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$ are the only combustion products and that the temperature and pressure remain constant.\nA: $1.0 \\mathrm{~L}$\nB: $\\quad 1.5 \\mathrm{~L}$\nC: $2.0 \\mathrm{~L}$\nD: $2.5 \\mathrm{~L}$\nE: $3.0 \\mathrm{~L}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many litres of gaseous methane $\\left(\\mathrm{CH}_{4}\\right)$ must be burned in oxygen to produce enough $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$ to fill a 3.0-L balloon? Assume that $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$ are the only combustion products and that the temperature and pressure remain constant.\n\nA: $1.0 \\mathrm{~L}$\nB: $\\quad 1.5 \\mathrm{~L}$\nC: $2.0 \\mathrm{~L}$\nD: $2.5 \\mathrm{~L}$\nE: $3.0 \\mathrm{~L}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_322",
        "problem": "| Half-Reaction | $E^{0}$ |\n| :--- | ---: |\n| $\\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Ag}(\\mathrm{s})$ | $+0.80 \\mathrm{~V}$ |\n| $\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+4 \\mathrm{e}^{-} \\rightleftharpoons 4 \\mathrm{OH}^{-}(\\mathrm{aq})$ | $+0.40 \\mathrm{~V}$ |\n| $\\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}(\\mathrm{s})$ | $+0.34 \\mathrm{~V}$ |\n| $2 \\mathrm{H}^{+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{H}(g)$ | $0.0 \\mathrm{~V}$ |\n| $\\mathrm{Sn}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Sn}(\\mathrm{s})$ | $-0.14 \\mathrm{~V}$ |\n| $\\mathrm{Ni}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Ni}(\\mathrm{s})$ | $-0.25 \\mathrm{~V}$ |\n| $\\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Fe}(\\mathrm{s})$ | $-0.41 \\mathrm{~V}$ |\n| $\\mathrm{Cr}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cr}(\\mathrm{s})$ | $-0.74 \\mathrm{~V}$ |\n| $\\mathrm{Zn}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Zn}(\\mathrm{s})$ | $-0.76 \\mathrm{~V}$ |\n| $2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{H}(g)+2 \\mathrm{OH}^{-}(\\mathrm{aq})$ | $-0.83 \\mathrm{~V}$ |\n| $\\mathrm{Al}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Al}(\\mathrm{s})$ | $-1.66 \\mathrm{~V}$ |\n\nThe metal $\\mathrm{X}$ dissolves in $\\mathrm{HCl}(\\mathrm{aq})$ but does not react in pure water, even its powdered form. It is a better reducing agent than $\\mathrm{Ni}(s)$. It forms an oxide with the formula $\\mathrm{X}_{2} \\mathrm{O}_{3}$. What is $\\mathrm{X}$ ?\nA: silver, $\\mathrm{Ag}$\nB: copper, $\\mathrm{Cu}$\nC: zinc, $\\mathrm{Zn}$\nD: aluminum, $\\mathrm{Al}$\nE: chromium, $\\mathrm{Cr}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n| Half-Reaction | $E^{0}$ |\n| :--- | ---: |\n| $\\mathrm{Ag}^{+}(\\mathrm{aq})+\\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Ag}(\\mathrm{s})$ | $+0.80 \\mathrm{~V}$ |\n| $\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+4 \\mathrm{e}^{-} \\rightleftharpoons 4 \\mathrm{OH}^{-}(\\mathrm{aq})$ | $+0.40 \\mathrm{~V}$ |\n| $\\mathrm{Cu}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cu}(\\mathrm{s})$ | $+0.34 \\mathrm{~V}$ |\n| $2 \\mathrm{H}^{+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{H}(g)$ | $0.0 \\mathrm{~V}$ |\n| $\\mathrm{Sn}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Sn}(\\mathrm{s})$ | $-0.14 \\mathrm{~V}$ |\n| $\\mathrm{Ni}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Ni}(\\mathrm{s})$ | $-0.25 \\mathrm{~V}$ |\n| $\\mathrm{Fe}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Fe}(\\mathrm{s})$ | $-0.41 \\mathrm{~V}$ |\n| $\\mathrm{Cr}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Cr}(\\mathrm{s})$ | $-0.74 \\mathrm{~V}$ |\n| $\\mathrm{Zn}^{2+}(\\mathrm{aq})+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Zn}(\\mathrm{s})$ | $-0.76 \\mathrm{~V}$ |\n| $2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{H}(g)+2 \\mathrm{OH}^{-}(\\mathrm{aq})$ | $-0.83 \\mathrm{~V}$ |\n| $\\mathrm{Al}^{3+}(\\mathrm{aq})+3 \\mathrm{e}^{-} \\rightleftharpoons \\mathrm{Al}(\\mathrm{s})$ | $-1.66 \\mathrm{~V}$ |\n\nThe metal $\\mathrm{X}$ dissolves in $\\mathrm{HCl}(\\mathrm{aq})$ but does not react in pure water, even its powdered form. It is a better reducing agent than $\\mathrm{Ni}(s)$. It forms an oxide with the formula $\\mathrm{X}_{2} \\mathrm{O}_{3}$. What is $\\mathrm{X}$ ?\n\nA: silver, $\\mathrm{Ag}$\nB: copper, $\\mathrm{Cu}$\nC: zinc, $\\mathrm{Zn}$\nD: aluminum, $\\mathrm{Al}$\nE: chromium, $\\mathrm{Cr}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_768",
        "problem": "常温下, $\\mathrm{p} K_{\\mathrm{sp}}(\\mathrm{CuS})=35.2, \\mathrm{p} K_{\\mathrm{sp}}(\\mathrm{FeS})=17.2\\left(\\mathrm{p} K_{\\mathrm{sp}}=-\\lg K_{\\mathrm{sp}}\\right)$ 。解平衡曲线如图所示(图中 $\\mathrm{R}$ 表示 $\\mathrm{Cu}$ 或 $\\mathrm{Fe}$ ), 下列说法错误的是\n\n[图1]\nA: 当 $c\\left(\\mathrm{~S}^{2}\\right)=10^{-4.9} \\mathrm{~mol} / \\mathrm{L}$ 时, 两溶液中 $c\\left(\\mathrm{Fe}^{2+}\\right)$ 与 $c\\left(\\mathrm{Cu}^{2+}\\right)$ 之比等于 $10^{18}$\nB: 曲线I表示 $\\mathrm{FeS}$\nC: 当 $c\\left(\\mathrm{Fe}^{2^{+}}\\right)=c\\left(\\mathrm{Cu}^{2+}\\right)$ 的混合溶液中滴入 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液时, 先生成 $\\mathrm{CuS}$ 沉淀\nD: $\\mathrm{M}$ 点的 $\\mathrm{RS}$ 饱和溶液中加入 $\\mathrm{Na}_{2} \\mathrm{~S}$, 溶液组成由 $\\mathrm{M}$ 点向 $\\mathrm{P}$ 点移动\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, $\\mathrm{p} K_{\\mathrm{sp}}(\\mathrm{CuS})=35.2, \\mathrm{p} K_{\\mathrm{sp}}(\\mathrm{FeS})=17.2\\left(\\mathrm{p} K_{\\mathrm{sp}}=-\\lg K_{\\mathrm{sp}}\\right)$ 。解平衡曲线如图所示(图中 $\\mathrm{R}$ 表示 $\\mathrm{Cu}$ 或 $\\mathrm{Fe}$ ), 下列说法错误的是\n\n[图1]\n\nA: 当 $c\\left(\\mathrm{~S}^{2}\\right)=10^{-4.9} \\mathrm{~mol} / \\mathrm{L}$ 时, 两溶液中 $c\\left(\\mathrm{Fe}^{2+}\\right)$ 与 $c\\left(\\mathrm{Cu}^{2+}\\right)$ 之比等于 $10^{18}$\nB: 曲线I表示 $\\mathrm{FeS}$\nC: 当 $c\\left(\\mathrm{Fe}^{2^{+}}\\right)=c\\left(\\mathrm{Cu}^{2+}\\right)$ 的混合溶液中滴入 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液时, 先生成 $\\mathrm{CuS}$ 沉淀\nD: $\\mathrm{M}$ 点的 $\\mathrm{RS}$ 饱和溶液中加入 $\\mathrm{Na}_{2} \\mathrm{~S}$, 溶液组成由 $\\mathrm{M}$ 点向 $\\mathrm{P}$ 点移动\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-032.jpg?height=605&width=625&top_left_y=337&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1236",
        "problem": "$\\mathbf{H}$ is stable to acids, but is hydrolysed in alkali. A $0.7934 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) was heated with excess aqueous sodium hydroxide. Cobalt(III) oxide was formed and ammonia gas given off. The ammonia produced was distilled off and absorbed into $50.0 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c_{\\mathrm{HCl}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$. The residual $\\mathrm{HCl}$ required $24.8 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{KOH}$ solution $\\left(c_{\\mathrm{KOH}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to be neutralized.\n\nThe remaining suspension of cobalt(III) oxide was allowed to cool, approximately $1 \\mathrm{~g}$ of potassium iodide was added, and then the mixture was acidified with aqueous $\\mathrm{HCl}$. The liberated iodine was then titrated with aqueous solution of sodium thiosulfate $(c=0.200$ mol $\\mathrm{dm}^{-3}$ ) and required $21.0 \\mathrm{~cm}^{3}$ for complete reaction.Calculate the identity of the oxygen species contained in $\\mathbf{H}$. Show your working.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\nHere is some context information for this question, which might assist you in solving it:\n$\\mathbf{H}$ is stable to acids, but is hydrolysed in alkali. A $0.7934 \\mathrm{~g}$ sample of $\\mathbf{H}$ (containing no water of crystallization) was heated with excess aqueous sodium hydroxide. Cobalt(III) oxide was formed and ammonia gas given off. The ammonia produced was distilled off and absorbed into $50.0 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c_{\\mathrm{HCl}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$. The residual $\\mathrm{HCl}$ required $24.8 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{KOH}$ solution $\\left(c_{\\mathrm{KOH}}=0.500 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right.$ ) to be neutralized.\n\nThe remaining suspension of cobalt(III) oxide was allowed to cool, approximately $1 \\mathrm{~g}$ of potassium iodide was added, and then the mixture was acidified with aqueous $\\mathrm{HCl}$. The liberated iodine was then titrated with aqueous solution of sodium thiosulfate $(c=0.200$ mol $\\mathrm{dm}^{-3}$ ) and required $21.0 \\mathrm{~cm}^{3}$ for complete reaction.\n\nproblem:\nCalculate the identity of the oxygen species contained in $\\mathbf{H}$. Show your working.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_826",
        "problem": "已知: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaSO}_{4}\\right)=4.0 \\times 10^{-6} 、 \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)=4.0 \\times 10^{-9} 、 \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{PbCO}_{3}\\right)=8.4 \\times 10^{-14}$,三种盐的沉淀溶解平衡曲线如图所示, $\\mathrm{pM}=-\\operatorname{lgc}$ (阴离子)、 $\\mathrm{pN}=-\\operatorname{lgc}$ (阳离子)。下列说法错误的是\n\n[图1]\nA: a 线是 $\\mathrm{CaSO}_{4}$ 沉淀溶解平衡曲线\nB: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向 $0.001 \\mathrm{~mol} \\mathrm{CaCO} 3$ 沉淀中加入 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液(反应后溶液为 $1 \\mathrm{~L}$ ), 使 $\\mathrm{CaCO}_{3}$沉淀会转化为 $\\mathrm{CaSO}_{4}$ 沉淀, 所需 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 的物质的量至少为 $1.001 \\mathrm{~mol}$\nC: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向 $10 \\mathrm{~mL}$ 水中加入 $\\mathrm{CaCO}_{3}$ 和 $\\mathrm{PbCO}_{3}$ 至二者均饱和, 溶液中 $\\frac{\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)}{\\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right)}=$ $\\frac{\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)}{\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{PbCO}_{3}\\right)}$\nD: 若 $\\mathrm{d}$ 点表示 $\\mathrm{CaCO}_{3}$ 的饱和溶液加入 $\\mathrm{CaSO}_{4}$ 饱和溶液等体积混合则: $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)<$ $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaSO}_{4}\\right)=4.0 \\times 10^{-6} 、 \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)=4.0 \\times 10^{-9} 、 \\mathrm{~K}_{\\mathrm{sp}}\\left(\\mathrm{PbCO}_{3}\\right)=8.4 \\times 10^{-14}$,三种盐的沉淀溶解平衡曲线如图所示, $\\mathrm{pM}=-\\operatorname{lgc}$ (阴离子)、 $\\mathrm{pN}=-\\operatorname{lgc}$ (阳离子)。下列说法错误的是\n\n[图1]\n\nA: a 线是 $\\mathrm{CaSO}_{4}$ 沉淀溶解平衡曲线\nB: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向 $0.001 \\mathrm{~mol} \\mathrm{CaCO} 3$ 沉淀中加入 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 溶液(反应后溶液为 $1 \\mathrm{~L}$ ), 使 $\\mathrm{CaCO}_{3}$沉淀会转化为 $\\mathrm{CaSO}_{4}$ 沉淀, 所需 $\\mathrm{Na}_{2} \\mathrm{SO}_{4}$ 的物质的量至少为 $1.001 \\mathrm{~mol}$\nC: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 向 $10 \\mathrm{~mL}$ 水中加入 $\\mathrm{CaCO}_{3}$ 和 $\\mathrm{PbCO}_{3}$ 至二者均饱和, 溶液中 $\\frac{\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)}{\\mathrm{c}\\left(\\mathrm{Pb}^{2+}\\right)}=$ $\\frac{\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{CaCO}_{3}\\right)}{\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{PbCO}_{3}\\right)}$\nD: 若 $\\mathrm{d}$ 点表示 $\\mathrm{CaCO}_{3}$ 的饱和溶液加入 $\\mathrm{CaSO}_{4}$ 饱和溶液等体积混合则: $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)<$ $\\mathrm{c}\\left(\\mathrm{Ca}^{2+}\\right)<\\mathrm{c}\\left(\\mathrm{SO}_{4}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-033.jpg?height=440&width=525&top_left_y=174&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_207",
        "problem": "Which of the following compounds has a trigonal pyramidal geometry?\nA: $\\mathrm{NCl}_{3}$\nB: $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\nC: $\\mathrm{COCl}_{2}$\nD: $\\mathrm{CH}_{4}$\nE: $\\mathrm{BCl}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following compounds has a trigonal pyramidal geometry?\n\nA: $\\mathrm{NCl}_{3}$\nB: $\\mathrm{CH}_{2} \\mathrm{Cl}_{2}$\nC: $\\mathrm{COCl}_{2}$\nD: $\\mathrm{CH}_{4}$\nE: $\\mathrm{BCl}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_124",
        "problem": "A $2.0 \\mathrm{~L}$ balloon initially holds $3.0 \\mathrm{~mol}$ of helium. When $3.0 \\mathrm{~mol}$ of helium are added, the volume of the balloon increases to $4.0 \\mathrm{~L}$ and the temperature remains unchanged. Which expression correctly describes the final pressure of the system in terms of the initial pressure, P? Assume Ideal Gas behaviour.\nA: $P$\nB: $2 \\mathrm{P}$\nC: $3 \\mathrm{P}$\nD: $4 \\mathrm{P}$\nE: $6 \\mathrm{P}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA $2.0 \\mathrm{~L}$ balloon initially holds $3.0 \\mathrm{~mol}$ of helium. When $3.0 \\mathrm{~mol}$ of helium are added, the volume of the balloon increases to $4.0 \\mathrm{~L}$ and the temperature remains unchanged. Which expression correctly describes the final pressure of the system in terms of the initial pressure, P? Assume Ideal Gas behaviour.\n\nA: $P$\nB: $2 \\mathrm{P}$\nC: $3 \\mathrm{P}$\nD: $4 \\mathrm{P}$\nE: $6 \\mathrm{P}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1162",
        "problem": "This question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nIn this reaction, which element is oxidised, and which element is reduced?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThis question is about the chlorides of a mythical metal\n\nThe mythical metal stuck-at-homium, symbol Sk, forms just two chlorides $-\\mathrm{SkCl}_{2}$ which is a white solid, and $\\mathrm{SkCl}_{4}$, a colourless volatile oily liquid which fumes on contact with air. The preparation of $\\mathrm{SkCl}_{4}$ was first described over four hundred years ago, but its formation may well have been described earlier and may have been one of the winged dragons referred to by the alchemists.\n\nAqueous $\\mathrm{SkCl}_{2}$ is easily prepared by dissolving metallic stuck-at-homium in dilute hydrochloric acid. On evaporation, the dihydrate is produced, $\\mathrm{SkCl}_{2} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$. This can be dehydrated to give anhydrous $\\mathrm{SkCl}_{2}$ using ethanoic anhydride, $\\left(\\mathrm{CH}_{3} \\mathrm{CO}\\right)_{2} \\mathrm{O}$. The structure of solid anhydrous $\\mathrm{SkCl}_{2}$ consists of chains of $\\mathrm{SkCl}_{2}$ units in which there are two different $\\mathrm{Sk}-\\mathrm{Cl}$ bond lengths and two different chlorine environments. The anhydrous solid melts and then boils at a temperature less than $700^{\\circ} \\mathrm{C}$.\n\nIn this reaction, which element is oxidised, and which element is reduced?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [Oxidised, Reduced].\nTheir answer types are, in order, [expression, expression].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "Oxidised",
            "Reduced"
        ],
        "type_sequence": [
            "EX",
            "EX"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_344",
        "problem": "At a certain temperature, the equilibrium constant for the reaction below is $K_{p}=0.100$.\n\n$$\n\\mathrm{P}_{4}(g) \\rightleftharpoons 2 \\mathrm{P}_{2}(g)\n$$\n\nIn an experiment, some $\\mathrm{P}_{4}$ gas was added to an empty reaction vessel and then the vessel was quickly sealed. The total pressure at equilibrium was $1.00 \\mathrm{~atm}$. What was the initial pressure of $\\mathrm{P}_{4}$ used in this experiment?\nA: $1.00 \\mathrm{~atm}$\nB: $\\quad 0.730 \\mathrm{~atm}$\nC: $\\quad 0.752 \\mathrm{~atm}$\nD: $0.865 \\mathrm{~atm}$\nE: $\\quad 0.667 \\mathrm{~atm}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAt a certain temperature, the equilibrium constant for the reaction below is $K_{p}=0.100$.\n\n$$\n\\mathrm{P}_{4}(g) \\rightleftharpoons 2 \\mathrm{P}_{2}(g)\n$$\n\nIn an experiment, some $\\mathrm{P}_{4}$ gas was added to an empty reaction vessel and then the vessel was quickly sealed. The total pressure at equilibrium was $1.00 \\mathrm{~atm}$. What was the initial pressure of $\\mathrm{P}_{4}$ used in this experiment?\n\nA: $1.00 \\mathrm{~atm}$\nB: $\\quad 0.730 \\mathrm{~atm}$\nC: $\\quad 0.752 \\mathrm{~atm}$\nD: $0.865 \\mathrm{~atm}$\nE: $\\quad 0.667 \\mathrm{~atm}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_407",
        "problem": "已知: $25^{\\circ} \\mathrm{C}, \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 电离平衡常数 $K=1.76 \\times 10^{-5}{ }_{\\circ} 25^{\\circ} \\mathrm{C}$, 向 $1 \\mathrm{~L} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 某一元酸 $\\mathrm{HR}$溶液中逐渐通入氨气, 若溶液温度和体积保持不变, 所得混合溶液的 $\\mathrm{pH}$ 与 $\\lg \\frac{c\\left(R^{-}\\right)}{c(H R)}$ 变化的关系如图所示。下列叙述正确的是\n\n[图1]\nA: 由图可推知: $25^{\\circ} \\mathrm{C}, 0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaR}$ 溶液的 $\\mathrm{pH}$ 约为 10\nB: 当通入 $0.1 \\mathrm{~mol} \\mathrm{NH}_{3}$ 时, 所得溶液中: $c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{R}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nC: $\\mathrm{pH}=7$ 时, 所得溶液中: $c(\\mathrm{HR})>c\\left(\\mathrm{R}^{-}\\right)=c\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$\nD: $\\mathrm{pH}=10$ 时, 所得溶液中: $c\\left(\\mathrm{R}^{-}\\right)>c(\\mathrm{HR}), c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知: $25^{\\circ} \\mathrm{C}, \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$ 电离平衡常数 $K=1.76 \\times 10^{-5}{ }_{\\circ} 25^{\\circ} \\mathrm{C}$, 向 $1 \\mathrm{~L} 0.1 \\mathrm{~mol} / \\mathrm{L}$ 某一元酸 $\\mathrm{HR}$溶液中逐渐通入氨气, 若溶液温度和体积保持不变, 所得混合溶液的 $\\mathrm{pH}$ 与 $\\lg \\frac{c\\left(R^{-}\\right)}{c(H R)}$ 变化的关系如图所示。下列叙述正确的是\n\n[图1]\n\nA: 由图可推知: $25^{\\circ} \\mathrm{C}, 0.1 \\mathrm{~mol} / \\mathrm{L} \\mathrm{NaR}$ 溶液的 $\\mathrm{pH}$ 约为 10\nB: 当通入 $0.1 \\mathrm{~mol} \\mathrm{NH}_{3}$ 时, 所得溶液中: $c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{R}^{-}\\right)>c\\left(\\mathrm{OH}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)$\nC: $\\mathrm{pH}=7$ 时, 所得溶液中: $c(\\mathrm{HR})>c\\left(\\mathrm{R}^{-}\\right)=c\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$\nD: $\\mathrm{pH}=10$ 时, 所得溶液中: $c\\left(\\mathrm{R}^{-}\\right)>c(\\mathrm{HR}), c\\left(\\mathrm{NH}_{4}^{+}\\right)>c\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-65.jpg?height=334&width=582&top_left_y=184&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_257",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the concentration of the sodium hydroxide solution (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ).\nThe ammonium ion content of a salt can be determined using the following procedure. A $1.988 \\mathrm{~g}$ sample of an ammonium salt is placed in a flask and heated with $50.00 \\mathrm{~mL}$ of $0.5493 \\mathrm{~mol} \\mathrm{~L}^{-1}$ potassium hydroxide solution (a known excess). The ammonium and hydroxide ions react to produce water and ammonia, which is expelled from the flask by evaporation:\n\n$\\mathrm{NH}_{4}^{+}(\\mathrm{aq})+\\mathrm{OH}^{-}(\\mathrm{aq}) \\rightarrow \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+\\mathrm{NH}_{3}(\\mathrm{~g})$\n\nThe potassium hydroxide remaining in the flask after all of the ammonia is expelled is determined with $0.1032 \\mathrm{~mol} \\mathrm{~L}^{-1}$ hydrochloric acid, $23.89 \\mathrm{~mL}$ of which is required for complete reaction.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the concentration of the sodium hydroxide solution (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ).\nThe ammonium ion content of a salt can be determined using the following procedure. A $1.988 \\mathrm{~g}$ sample of an ammonium salt is placed in a flask and heated with $50.00 \\mathrm{~mL}$ of $0.5493 \\mathrm{~mol} \\mathrm{~L}^{-1}$ potassium hydroxide solution (a known excess). The ammonium and hydroxide ions react to produce water and ammonia, which is expelled from the flask by evaporation:\n\n$\\mathrm{NH}_{4}^{+}(\\mathrm{aq})+\\mathrm{OH}^{-}(\\mathrm{aq}) \\rightarrow \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})+\\mathrm{NH}_{3}(\\mathrm{~g})$\n\nThe potassium hydroxide remaining in the flask after all of the ammonia is expelled is determined with $0.1032 \\mathrm{~mol} \\mathrm{~L}^{-1}$ hydrochloric acid, $23.89 \\mathrm{~mL}$ of which is required for complete reaction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol/L, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol/L"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_727",
        "problem": "常温下, 下列溶液中有关微粒的物质的量浓度关系正确的是\nA: $\\mathrm{pH}=8$ 的 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=9.9 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{SO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}{ }^{-}\\right)+\\mathrm{c}$ $\\left(\\mathrm{OH}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$ $+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液中通入 $\\mathrm{HCl}$ 至溶液 $\\mathrm{pH}=7: \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}$ $\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 下列溶液中有关微粒的物质的量浓度关系正确的是\n\nA: $\\mathrm{pH}=8$ 的 $\\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=9.9 \\times 10^{-7} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{SO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{SO}_{3}{ }^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HSO}_{3}{ }^{-}\\right)+\\mathrm{c}$ $\\left(\\mathrm{OH}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$ $+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液中通入 $\\mathrm{HCl}$ 至溶液 $\\mathrm{pH}=7: \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}$ $\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_898",
        "problem": "30. 某温度下, 分别向 $20 \\mathrm{~mL}$ 浓度均为 $\\mathrm{xmol} / \\mathrm{L}$ 的 $\\mathrm{NaCl}$ 和 $\\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ 溶液中滴加\n\n$0.1 \\mathrm{~mol} / \\mathrm{LAgNO}_{3}$ 溶液, 滴加过程中 $-\\operatorname{lgc}\\left(\\mathrm{Cl}^{-}\\right)$和 $-\\operatorname{lgc}\\left(\\mathrm{CrO}_{4}^{2-}\\right)$ 与 $\\mathrm{AgNO}_{3}$ 溶液的体积关系如图所示。下列说法不正确的是\n\n[图1]\nA: $\\mathrm{x}=0.1$\nB: 曲线 $\\mathrm{I}$ 代表 $\\mathrm{NaCl}$ 溶液\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$ 约为 $4 \\times 10^{-12}$\nD: $\\mathrm{y}=9$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n30. 某温度下, 分别向 $20 \\mathrm{~mL}$ 浓度均为 $\\mathrm{xmol} / \\mathrm{L}$ 的 $\\mathrm{NaCl}$ 和 $\\mathrm{Na}_{2} \\mathrm{CrO}_{4}$ 溶液中滴加\n\n$0.1 \\mathrm{~mol} / \\mathrm{LAgNO}_{3}$ 溶液, 滴加过程中 $-\\operatorname{lgc}\\left(\\mathrm{Cl}^{-}\\right)$和 $-\\operatorname{lgc}\\left(\\mathrm{CrO}_{4}^{2-}\\right)$ 与 $\\mathrm{AgNO}_{3}$ 溶液的体积关系如图所示。下列说法不正确的是\n\n[图1]\n\nA: $\\mathrm{x}=0.1$\nB: 曲线 $\\mathrm{I}$ 代表 $\\mathrm{NaCl}$ 溶液\nC: $\\mathrm{K}_{\\mathrm{sp}}\\left(\\mathrm{Ag}_{2} \\mathrm{CrO}_{4}\\right)$ 约为 $4 \\times 10^{-12}$\nD: $\\mathrm{y}=9$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-30.jpg?height=508&width=554&top_left_y=174&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_956",
        "problem": "研究表明, 利用微电流电解产生羟基自由基 $(\\cdot \\mathrm{OH})$ 的强氧化性可使藻细胞光系统受损，丧失光合作用能力，达到除藻效果，其原理如下图。下列说法错误的是\n\n[图1]\nA: 铂电极 A 为阴极\nB: A 电极反应为 $4 \\mathrm{OH}^{-}-4 \\mathrm{e}^{-}=2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2} \\uparrow$\nC: 氧化性顺序： $\\mathrm{OH}>\\mathrm{H}_{2} \\mathrm{O}_{2}>\\mathrm{O}_{2}$\nD: 电路中通过 $0.15 \\mathrm{~mol}$ 电子, $\\mathrm{B}$ 电极生成 $0.05 \\mathrm{~mol} \\cdot \\mathrm{OH}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n研究表明, 利用微电流电解产生羟基自由基 $(\\cdot \\mathrm{OH})$ 的强氧化性可使藻细胞光系统受损，丧失光合作用能力，达到除藻效果，其原理如下图。下列说法错误的是\n\n[图1]\n\nA: 铂电极 A 为阴极\nB: A 电极反应为 $4 \\mathrm{OH}^{-}-4 \\mathrm{e}^{-}=2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{O}_{2} \\uparrow$\nC: 氧化性顺序： $\\mathrm{OH}>\\mathrm{H}_{2} \\mathrm{O}_{2}>\\mathrm{O}_{2}$\nD: 电路中通过 $0.15 \\mathrm{~mol}$ 电子, $\\mathrm{B}$ 电极生成 $0.05 \\mathrm{~mol} \\cdot \\mathrm{OH}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-07.jpg?height=486&width=1243&top_left_y=1730&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1409",
        "problem": "In the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.\n\nTwo identical samples of YBCO with an unknown value of $\\delta$ were prepared. The first sample was dissolved in $5 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$, evolving $\\mathrm{O}_{2}$. After boiling to expel gases, cooling, and addition of $10 \\mathrm{~cm}^{3}$ of $\\mathrm{KI}$ solution ( $c=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) under Ar, titration with thiosulfate to the starch endpoint required $1.542 \\cdot 10^{-4}$ mol thiosulfate. The second sample of YBCO was added under Ar directly to $7 \\mathrm{~cm}^{3}$ of a solution in which $c(\\mathrm{KI})=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}$ and $\\mathrm{c}(\\mathrm{HCl})=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$. Titration of this solution required $1.696 \\cdot 10^{-4} \\mathrm{~mol}$ thiosulfate to reach the endpoint.Calculate the amount of substance of $\\mathrm{Cu}$ (in mol) in each of these samples of YBCO.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nIn the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of $90 \\mathrm{~K}$. One such material contains yttrium, barium, copper and oxygen and is called \"YBCO\". It has a nominal composition of $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7}$, but its actual composition is variable according to the formula $\\mathrm{YBa}_{2} \\mathrm{Cu}_{3} \\mathrm{O}_{7-\\delta}$ $(0<\\delta<0.5)$.\n\nTwo identical samples of YBCO with an unknown value of $\\delta$ were prepared. The first sample was dissolved in $5 \\mathrm{~cm}^{3}$ of aqueous $\\mathrm{HCl}\\left(c=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}\\right)$, evolving $\\mathrm{O}_{2}$. After boiling to expel gases, cooling, and addition of $10 \\mathrm{~cm}^{3}$ of $\\mathrm{KI}$ solution ( $c=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$ ) under Ar, titration with thiosulfate to the starch endpoint required $1.542 \\cdot 10^{-4}$ mol thiosulfate. The second sample of YBCO was added under Ar directly to $7 \\mathrm{~cm}^{3}$ of a solution in which $c(\\mathrm{KI})=1.0 \\mathrm{~mol} \\mathrm{dm}^{-3}$ and $\\mathrm{c}(\\mathrm{HCl})=0.7 \\mathrm{~mol} \\mathrm{dm}^{-3}$. Titration of this solution required $1.696 \\cdot 10^{-4} \\mathrm{~mol}$ thiosulfate to reach the endpoint.\n\nproblem:\nCalculate the amount of substance of $\\mathrm{Cu}$ (in mol) in each of these samples of YBCO.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_865",
        "problem": "某同学研究 $\\mathrm{FeSO}_{4}$ 溶液和 $\\mathrm{AgNO}_{3}$ 溶液的反应, 设计如下对比实验。\n\n[图1]\n\n下列说法正确的是\nA: II中电流表指针向左偏转的原因是 $\\mathrm{Fe}^{2+}$ 氧化了银电极\nB: II中若将银电极换成石墨电极, 电流表指针可能不再向左偏转\nC: 对比I、II可知, I中 $\\mathrm{NO}_{3}-$ 氧化了 $\\mathrm{Fe}^{2+}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某同学研究 $\\mathrm{FeSO}_{4}$ 溶液和 $\\mathrm{AgNO}_{3}$ 溶液的反应, 设计如下对比实验。\n\n[图1]\n\n下列说法正确的是\n\nA: II中电流表指针向左偏转的原因是 $\\mathrm{Fe}^{2+}$ 氧化了银电极\nB: II中若将银电极换成石墨电极, 电流表指针可能不再向左偏转\nC: 对比I、II可知, I中 $\\mathrm{NO}_{3}-$ 氧化了 $\\mathrm{Fe}^{2+}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-26.jpg?height=894&width=1445&top_left_y=1255&top_left_x=311"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_453",
        "problem": "甘氨酸在水溶液中主要以偶极离子 $\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)$的形式存在, 它是两性电解质,有两个可解离基团, 解离方程如下:\n\n[图1]\n\n$$\n\\mathrm{R}^{+} \\quad \\mathrm{R}^{+-} \\quad \\mathrm{R}^{+-}\n$$\n\n当调节溶液的 $\\mathrm{pH}$ 使甘氨酸所带正负电荷正好相等时, 甘氨酸所带的净电荷为零, 在电场中不发生移动现象, 此时溶液的 $\\mathrm{pH}$ 叫等电点。下列有关说法中不正确的是\nA: 甘氨酸晶体中含有阴离子和阳离子\nB: 等电点时, $R^{+-} 、 R^{+} 、 R^{-}$的数量关系是 $R^{+-=}=R^{+}=R^{-}$\nC: 已知甘氨酸的等电点为 5.97, 说明甘氨酸中羧基更易解离\nD: $\\mathrm{pH}$ 小于 5.97 的甘氨酸溶液中, 在电场中将向阴极移动\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n甘氨酸在水溶液中主要以偶极离子 $\\left(\\mathrm{H}_{3} \\mathrm{~N}^{+} \\mathrm{CH}_{2} \\mathrm{COO}^{-}\\right)$的形式存在, 它是两性电解质,有两个可解离基团, 解离方程如下:\n\n[图1]\n\n$$\n\\mathrm{R}^{+} \\quad \\mathrm{R}^{+-} \\quad \\mathrm{R}^{+-}\n$$\n\n当调节溶液的 $\\mathrm{pH}$ 使甘氨酸所带正负电荷正好相等时, 甘氨酸所带的净电荷为零, 在电场中不发生移动现象, 此时溶液的 $\\mathrm{pH}$ 叫等电点。下列有关说法中不正确的是\n\nA: 甘氨酸晶体中含有阴离子和阳离子\nB: 等电点时, $R^{+-} 、 R^{+} 、 R^{-}$的数量关系是 $R^{+-=}=R^{+}=R^{-}$\nC: 已知甘氨酸的等电点为 5.97, 说明甘氨酸中羧基更易解离\nD: $\\mathrm{pH}$ 小于 5.97 的甘氨酸溶液中, 在电场中将向阴极移动\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-50.jpg?height=74&width=985&top_left_y=477&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1429",
        "problem": "Determination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\nCalculate the composition of the hydrate $\\mathrm{Fe}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3} \\cdot \\mathrm{xH}_{2} \\mathrm{O}(\\mathrm{x}=$ integer $)$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nDetermination of water by the classical Fischer method involves titration of a sample solution (or suspension) in methanol by a methanolic iodine solution, containing also an excess of $\\mathrm{SO}_{2}$ and pyridine $\\left(\\mathrm{C}_{5} \\mathrm{H}_{5} \\mathrm{~N}\\right.$, Py) - Fischer reagent. The following reactions occur during the titration:\n\n[figure1]\n\nlodine content is usually expressed in $\\mathrm{mg}$ of water reacting with $1 \\mathrm{~cm}^{3}$ of the titrant solution (hereunder $\\mathrm{T}, \\mathrm{mg} \\mathrm{cm}^{-3}$ ), which equals the mass of water $(\\mathrm{mg}$ ) reacting with 1.00 $\\mathrm{cm}^{3}$ of the iodine solution. $T$ is determined experimentally by titration of a sample with a known water content. The sample may be, for example, a hydrated compound or a standard solution of water in methanol. In the latter case it should be taken into account that methanol itself can contain certain amount of water.\n\nCalculate the composition of the hydrate $\\mathrm{Fe}_{2}\\left(\\mathrm{SO}_{4}\\right)_{3} \\cdot \\mathrm{xH}_{2} \\mathrm{O}(\\mathrm{x}=$ integer $)$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-651.jpg?height=213&width=900&top_left_y=776&top_left_x=286"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_769",
        "problem": "二氧化氯 $\\left(\\mathrm{ClO}_{2}\\right)$ 是一种黄绿色、易溶于水的气体, 常用于污染物的处理。工业上通过情性电极电解氯化铵和盐酸的方法制备 $\\mathrm{ClO}_{2}$ 的原理如图所示。下列说法正确的是\n\n[图1]\nA: a 极与电源的负极连接, $Y$ 溶液是稀盐酸\nB: $\\mathrm{a}$ 极上发生的反应为 $\\mathrm{NH}_{4}^{+}+6 \\mathrm{e}^{-}+3 \\mathrm{Cl}^{-}=\\mathrm{NCl}_{3}+4 \\mathrm{H}^{+}$\nC: 二氧化氯发生器内发生的氧化还原反应中, 氧化剂与还原剂的物质的量之比为 $1: 6$\nD: 当 $0.6 \\mathrm{~mol}$ 阴离子通过离子交换膜时, 理论上二氧化氯发生器中产生标准状况下 $2.24 \\mathrm{LNH}_{3}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n二氧化氯 $\\left(\\mathrm{ClO}_{2}\\right)$ 是一种黄绿色、易溶于水的气体, 常用于污染物的处理。工业上通过情性电极电解氯化铵和盐酸的方法制备 $\\mathrm{ClO}_{2}$ 的原理如图所示。下列说法正确的是\n\n[图1]\n\nA: a 极与电源的负极连接, $Y$ 溶液是稀盐酸\nB: $\\mathrm{a}$ 极上发生的反应为 $\\mathrm{NH}_{4}^{+}+6 \\mathrm{e}^{-}+3 \\mathrm{Cl}^{-}=\\mathrm{NCl}_{3}+4 \\mathrm{H}^{+}$\nC: 二氧化氯发生器内发生的氧化还原反应中, 氧化剂与还原剂的物质的量之比为 $1: 6$\nD: 当 $0.6 \\mathrm{~mol}$ 阴离子通过离子交换膜时, 理论上二氧化氯发生器中产生标准状况下 $2.24 \\mathrm{LNH}_{3}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-30.jpg?height=459&width=1108&top_left_y=1798&top_left_x=331"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1385",
        "problem": "One of the first accurate determinations of the Avogadro constant was carried out by studying the vertical distribution under gravity of colloidal particles suspended in water. In one such experiment, particles with radius $2.12 \\cdot 10^{-7} \\mathrm{~m}$ and density $1.206 \\cdot 10^{3} \\mathrm{~kg} \\mathrm{~m}^{-3}$ were suspended in a tube of water at $15^{\\circ} \\mathrm{C}$. After a llowing sufficient time to equilibrate, the mean numbers of particles per unit volume observed at four heights from the bottom of the tube were:\n\n| height $/ 10^{-6} \\mathrm{~m}$ | 5 | 35 | 65 | 95 |\n| :--- | :---: | :---: | :---: | :---: |\n| mean number <br> per unit volume | 4.00 | 1.88 | 0.90 | 0.48 |Assuming the particles to be spherical, calculate:",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nOne of the first accurate determinations of the Avogadro constant was carried out by studying the vertical distribution under gravity of colloidal particles suspended in water. In one such experiment, particles with radius $2.12 \\cdot 10^{-7} \\mathrm{~m}$ and density $1.206 \\cdot 10^{3} \\mathrm{~kg} \\mathrm{~m}^{-3}$ were suspended in a tube of water at $15^{\\circ} \\mathrm{C}$. After a llowing sufficient time to equilibrate, the mean numbers of particles per unit volume observed at four heights from the bottom of the tube were:\n\n| height $/ 10^{-6} \\mathrm{~m}$ | 5 | 35 | 65 | 95 |\n| :--- | :---: | :---: | :---: | :---: |\n| mean number <br> per unit volume | 4.00 | 1.88 | 0.90 | 0.48 |\n\nproblem:\nAssuming the particles to be spherical, calculate:\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [ the mass, $m$, of a particle, the mass, $m_{\\mathrm{H}_{2} \\mathrm{O}}$, of the water it displaces, the effective mass, $m^{\\star}$, of the particle in water accounting for buoyancy (i.e. taking account of the upthrust due to the displaced volume of water). Take the density of water to be $999 \\mathrm{~kg} \\mathrm{~m}^{-3}$.].\nTheir units are, in order, [$\\mathrm{~kg}$, $\\mathrm{~kg}$, $\\mathrm{~kg}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "$\\mathrm{~kg}$",
            "$\\mathrm{~kg}$",
            "$\\mathrm{~kg}$"
        ],
        "answer_sequence": [
            " the mass, $m$, of a particle",
            "the mass, $m_{\\mathrm{H}_{2} \\mathrm{O}}$, of the water it displaces",
            "the effective mass, $m^{\\star}$, of the particle in water accounting for buoyancy (i.e. taking account of the upthrust due to the displaced volume of water). Take the density of water to be $999 \\mathrm{~kg} \\mathrm{~m}^{-3}$."
        ],
        "type_sequence": [
            "NV",
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_123",
        "problem": "The melting point of $\\mathrm{CaS}$ is higher than that of $\\mathrm{KCl}$. Explanations for this observation include which of the following?\nI. $\\quad \\mathrm{Ca}^{2+}$ is more positively charged than $\\mathrm{K}^{+}$.\n\nII. $\\quad \\mathrm{S}^{2-}$ is more negatively charged than $\\mathrm{Cl}^{-}$.\n\nIII. The $\\mathrm{K}^{+}$ion is smaller than the $\\mathrm{Ca}^{2+}$ ion.\nA: II only\nB: I and II only\nC: I and III only\nD: II and III only\nE: I, II, and III\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe melting point of $\\mathrm{CaS}$ is higher than that of $\\mathrm{KCl}$. Explanations for this observation include which of the following?\nI. $\\quad \\mathrm{Ca}^{2+}$ is more positively charged than $\\mathrm{K}^{+}$.\n\nII. $\\quad \\mathrm{S}^{2-}$ is more negatively charged than $\\mathrm{Cl}^{-}$.\n\nIII. The $\\mathrm{K}^{+}$ion is smaller than the $\\mathrm{Ca}^{2+}$ ion.\n\nA: II only\nB: I and II only\nC: I and III only\nD: II and III only\nE: I, II, and III\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1558",
        "problem": "${ }^{32} \\mathrm{P}$ labelled phosphorus pentachloride (half-life $t_{1 / 2}=14.3$ days) is used to study the electrophilic attack of $\\mathrm{PCl}_{4}{ }^{+}$cation on nitrogen or on oxygen.\n\n[figure1]\n\nThe reaction is carried out in $\\mathrm{CCl}_{4}$ and the solvent and product IV distilled off. Samples of III (remaining in the distillation flask), of IV (in the distillate) and samples of the starting material II are hydrolyzed by heating with a strong sodium hydroxide solution. The phosphate ions formed are precipitated as ammonium magnesium phosphate. Purified samples of the three precipitates are then dissolved by known volumes of water and the radioactivity measured.\n\nCalculate the solubility for $\\mathrm{Mg}\\left(\\mathrm{NH}_{4}\\right) \\mathrm{PO}_{4}$ at $\\mathrm{pH}$ equal to 10 under idealized conditions (activity coefficients can be neglected).",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\n${ }^{32} \\mathrm{P}$ labelled phosphorus pentachloride (half-life $t_{1 / 2}=14.3$ days) is used to study the electrophilic attack of $\\mathrm{PCl}_{4}{ }^{+}$cation on nitrogen or on oxygen.\n\n[figure1]\n\nThe reaction is carried out in $\\mathrm{CCl}_{4}$ and the solvent and product IV distilled off. Samples of III (remaining in the distillation flask), of IV (in the distillate) and samples of the starting material II are hydrolyzed by heating with a strong sodium hydroxide solution. The phosphate ions formed are precipitated as ammonium magnesium phosphate. Purified samples of the three precipitates are then dissolved by known volumes of water and the radioactivity measured.\n\nCalculate the solubility for $\\mathrm{Mg}\\left(\\mathrm{NH}_{4}\\right) \\mathrm{PO}_{4}$ at $\\mathrm{pH}$ equal to 10 under idealized conditions (activity coefficients can be neglected).\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol} \\mathrm{dm}^{-3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-013.jpg?height=297&width=1659&top_left_y=568&top_left_x=198"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol} \\mathrm{dm}^{-3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1283",
        "problem": "Diatoms, microscopic organisms, are an abundant food source in the oceans producing carbohydrates from carbon dioxide and water by photosynthesis:\n\nThree percent of the mass of a $9.1 \\times 10^{4} \\mathrm{~kg}$ adult whale is nitrogen. When a $9.1 \\times 10^{4}$ $\\mathrm{kg}$ blue whale dies, what is the maximum mass of $\\mathrm{NH}_{4}^{+}$that can become available for other marine organisms?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nDiatoms, microscopic organisms, are an abundant food source in the oceans producing carbohydrates from carbon dioxide and water by photosynthesis:\n\nThree percent of the mass of a $9.1 \\times 10^{4} \\mathrm{~kg}$ adult whale is nitrogen. When a $9.1 \\times 10^{4}$ $\\mathrm{kg}$ blue whale dies, what is the maximum mass of $\\mathrm{NH}_{4}^{+}$that can become available for other marine organisms?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~kg} \\mathrm{NH}_{4}^{+}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~kg} \\mathrm{NH}_{4}^{+}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_381",
        "problem": "How many total orbitals are there with principal quantum number $n=4$ ?\nA: 1\nB: 4\nC: 9\nD: 16\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many total orbitals are there with principal quantum number $n=4$ ?\n\nA: 1\nB: 4\nC: 9\nD: 16\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_57",
        "problem": "Which aqueous solution has the highest normal boiling point?\nA: $0.10 \\mathrm{M} \\mathrm{Na}_{3} \\mathrm{PO}_{4}$\nB: $0.30 \\mathrm{M} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$\nC: $0.50 \\mathrm{M} \\mathrm{NaH}_{2} \\mathrm{PO}_{4}$\nD: $0.70 \\mathrm{M} \\mathrm{H}_{3} \\mathrm{PO}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich aqueous solution has the highest normal boiling point?\n\nA: $0.10 \\mathrm{M} \\mathrm{Na}_{3} \\mathrm{PO}_{4}$\nB: $0.30 \\mathrm{M} \\mathrm{Na}_{2} \\mathrm{HPO}_{4}$\nC: $0.50 \\mathrm{M} \\mathrm{NaH}_{2} \\mathrm{PO}_{4}$\nD: $0.70 \\mathrm{M} \\mathrm{H}_{3} \\mathrm{PO}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1022",
        "problem": "How many moles of water are there in $1.80 \\mathrm{~L}$ of $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})$ at $1.00 \\mathrm{~atm}$ and $298 \\mathrm{~K}$ ? The density of water is $1.00 \\mathrm{~g} / \\mathrm{mL}$ at $1.00 \\mathrm{~atm}$ and $298 \\mathrm{~K}$.\nA: $1.00 \\mathrm{~mol}$\nB: $\\quad 0.0736 \\mathrm{~mol}$\nC: $55.6 \\mathrm{~mol}$\nD: $1.00 \\times 10^{2} \\mathrm{~mol}$\nE: $13.6 \\mathrm{~mol}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many moles of water are there in $1.80 \\mathrm{~L}$ of $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{I})$ at $1.00 \\mathrm{~atm}$ and $298 \\mathrm{~K}$ ? The density of water is $1.00 \\mathrm{~g} / \\mathrm{mL}$ at $1.00 \\mathrm{~atm}$ and $298 \\mathrm{~K}$.\n\nA: $1.00 \\mathrm{~mol}$\nB: $\\quad 0.0736 \\mathrm{~mol}$\nC: $55.6 \\mathrm{~mol}$\nD: $1.00 \\times 10^{2} \\mathrm{~mol}$\nE: $13.6 \\mathrm{~mol}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1149",
        "problem": "This question is about benzene and its isomers\n\nWorking out the structure of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, was one of the great achievements of $19^{\\text {th }}$ century chemists. The carcinogenic hydrocarbon was first discovered by Michael Faraday at the Royal Institution in 1825 during the repeated fractional distillation of some oil residues.\n\nFaraday analysed the pure benzene by passing its vapour over heated copper(II) oxide to produce carbon dioxide, water and copper. He collected and measured the volume of liquid water and gaseous $\\mathrm{CO}_{2}$ produced from a given mass of benzene.\n\nFaraday confirmed his analysis by a second method - exploding a mixture of benzene vapour with excess oxygen gas and measuring the volume of carbon dioxide produced by absorbing it with potassium hydroxide solution. He took 7.0 'volumes' of the mixture (the precise units are not important) and found that 2.1 volumes of carbon dioxide were produced.\n\nTaking the half-life to be exactly 2 days, how many days would it take a sample of Dewarbenzene to convert to $99 \\%$ normal benzene?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about benzene and its isomers\n\nWorking out the structure of benzene, $\\mathrm{C}_{6} \\mathrm{H}_{6}$, was one of the great achievements of $19^{\\text {th }}$ century chemists. The carcinogenic hydrocarbon was first discovered by Michael Faraday at the Royal Institution in 1825 during the repeated fractional distillation of some oil residues.\n\nFaraday analysed the pure benzene by passing its vapour over heated copper(II) oxide to produce carbon dioxide, water and copper. He collected and measured the volume of liquid water and gaseous $\\mathrm{CO}_{2}$ produced from a given mass of benzene.\n\nFaraday confirmed his analysis by a second method - exploding a mixture of benzene vapour with excess oxygen gas and measuring the volume of carbon dioxide produced by absorbing it with potassium hydroxide solution. He took 7.0 'volumes' of the mixture (the precise units are not important) and found that 2.1 volumes of carbon dioxide were produced.\n\nTaking the half-life to be exactly 2 days, how many days would it take a sample of Dewarbenzene to convert to $99 \\%$ normal benzene?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_547",
        "problem": "常温下, 将一定量稀硫酸滴入高铁酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{FeO}_{4}\\right)$ 溶液中, 溶液中含铁微粒 $\\mathrm{FeO}_{4}^{2-}$ 、 $\\mathrm{HFeO}_{4}^{-} 、 \\mathrm{H}_{2} \\mathrm{FeO}_{4} 、 \\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}$的物质的量分数 $\\delta(\\mathrm{X})$ 随 $\\mathrm{pOH}$ 的变化如图,\n\n$$\n\\left[\\delta(\\mathrm{X})=\\frac{\\mathrm{c}(\\mathrm{X})}{\\mathrm{c}\\left(\\mathrm{FeO}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HFeO}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{FeO}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}\\right)}, \\quad \\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right] \\text {。下列说法正 }\n$$\n\n确的是\n\n[图1]\nA: 曲线 $\\mathrm{I}$ 表示 $\\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}$的变化曲线\nB: $a 、 b 、 c$ 三点水的电离程度相等\nC: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{pH}=7$ 的溶液中存在: $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HFeO}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{FeO}_{4}^{2-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{FeO}_{4} \\rightleftharpoons \\mathrm{FeO}_{4}^{2-}+2 \\mathrm{H}^{+}$的平衡常数 $\\mathrm{K}=10^{-4.8}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 将一定量稀硫酸滴入高铁酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{FeO}_{4}\\right)$ 溶液中, 溶液中含铁微粒 $\\mathrm{FeO}_{4}^{2-}$ 、 $\\mathrm{HFeO}_{4}^{-} 、 \\mathrm{H}_{2} \\mathrm{FeO}_{4} 、 \\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}$的物质的量分数 $\\delta(\\mathrm{X})$ 随 $\\mathrm{pOH}$ 的变化如图,\n\n$$\n\\left[\\delta(\\mathrm{X})=\\frac{\\mathrm{c}(\\mathrm{X})}{\\mathrm{c}\\left(\\mathrm{FeO}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{HFeO}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{FeO}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}\\right)}, \\quad \\mathrm{pOH}=-\\operatorname{lgc}\\left(\\mathrm{OH}^{-}\\right)\\right] \\text {。下列说法正 }\n$$\n\n确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{I}$ 表示 $\\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}$的变化曲线\nB: $a 、 b 、 c$ 三点水的电离程度相等\nC: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{pH}=7$ 的溶液中存在: $\\mathrm{c}\\left(\\mathrm{H}_{3} \\mathrm{FeO}_{4}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HFeO}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{FeO}_{4}^{2-}\\right)$\nD: $25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{FeO}_{4} \\rightleftharpoons \\mathrm{FeO}_{4}^{2-}+2 \\mathrm{H}^{+}$的平衡常数 $\\mathrm{K}=10^{-4.8}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-047.jpg?height=448&width=757&top_left_y=1529&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_366",
        "problem": "What is the concentration of a solution of $\\mathrm{K}_{2} \\mathrm{CO}_{3}$ that has $\\mathrm{pH}=11.90$ ? (For $\\mathrm{H}_{2} \\mathrm{CO}_{3}, K_{\\mathrm{a} 1}=4.2 \\times 10^{-7}, K_{\\mathrm{a} 2}=4.8 \\times$ $10^{-11}$.)\nA: $3.0 \\times 10^{-1} \\mathrm{M}$\nB: $2.6 \\times 10^{-2} \\mathrm{M}$\nC: $7.9 \\times 10^{-3} \\mathrm{M}$\nD: $1.3 \\times 10^{-12} \\mathrm{M}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the concentration of a solution of $\\mathrm{K}_{2} \\mathrm{CO}_{3}$ that has $\\mathrm{pH}=11.90$ ? (For $\\mathrm{H}_{2} \\mathrm{CO}_{3}, K_{\\mathrm{a} 1}=4.2 \\times 10^{-7}, K_{\\mathrm{a} 2}=4.8 \\times$ $10^{-11}$.)\n\nA: $3.0 \\times 10^{-1} \\mathrm{M}$\nB: $2.6 \\times 10^{-2} \\mathrm{M}$\nC: $7.9 \\times 10^{-3} \\mathrm{M}$\nD: $1.3 \\times 10^{-12} \\mathrm{M}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1404",
        "problem": "Regarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe absorbance is proportional to the concentration of the absorbing compound.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nRegarding Beer's law, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe absorbance is proportional to the concentration of the absorbing compound.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1330",
        "problem": "Methanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |Calculate $K_p$ for the reaction at $298 \\mathrm{~K}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nMethanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |\n\nproblem:\nCalculate $K_p$ for the reaction at $298 \\mathrm{~K}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_500",
        "problem": "$\\mathrm{H}_{2} \\mathrm{~S}$ 为二元弱酸。 $20^{\\circ} \\mathrm{C}$ 时, 向 $0.100 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中缓慢通入 $\\mathrm{HCl}$ 气体 (忽略溶液体积的变化及 $\\mathrm{H}_{2} \\mathrm{~S}$ 的挥发)。下列指定溶液中微粒的物质的量浓度关系一定正确的是\nA: 通入 $\\mathrm{HCl}$ 气体之前: $\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: $\\mathrm{pH}=7$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nC: $\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)$ 的碱性溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)>0.100 \\mathrm{~mol} / \\mathrm{L}+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nD: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=0.100 \\mathrm{~mol} / \\mathrm{L}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)-\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{H}_{2} \\mathrm{~S}$ 为二元弱酸。 $20^{\\circ} \\mathrm{C}$ 时, 向 $0.100 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{Na}_{2} \\mathrm{~S}$ 溶液中缓慢通入 $\\mathrm{HCl}$ 气体 (忽略溶液体积的变化及 $\\mathrm{H}_{2} \\mathrm{~S}$ 的挥发)。下列指定溶液中微粒的物质的量浓度关系一定正确的是\n\nA: 通入 $\\mathrm{HCl}$ 气体之前: $\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nB: $\\mathrm{pH}=7$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nC: $\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)$ 的碱性溶液中: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{HS}^{-}\\right)>0.100 \\mathrm{~mol} / \\mathrm{L}+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)$\nD: $\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)=0.100 \\mathrm{~mol} / \\mathrm{L}$ 的溶液中: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)-\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_651",
        "problem": "将等物质的量浓度的 $\\mathrm{CuSO}_{4}$ 和 $\\mathrm{NaCl}$ 等体积混合后, 用石墨电极进行电解, 电解过程中，溶液 $\\mathrm{pH}$ 随时间 $\\mathrm{t}$ 变化的曲线如图所示，则下列说法正确的是\n\n[图1]\nA: 阴极先析出 $\\mathrm{Cl}_{2}$, 后析出 $\\mathrm{O}_{2}$, 阳极先产生 $\\mathrm{Cu}$, 后析出 $\\mathrm{H}_{2}$\nB: $\\mathrm{AB}$ 段阳极只产生 $\\mathrm{Cl}_{2}$, 阴极只产生 $\\mathrm{Cu}$\nC: $\\mathrm{BC}$ 段表示在阴极上是 $\\mathrm{H}^{+}$放电产生了 $\\mathrm{H}_{2}$\nD: CD 段相当于电解水\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n将等物质的量浓度的 $\\mathrm{CuSO}_{4}$ 和 $\\mathrm{NaCl}$ 等体积混合后, 用石墨电极进行电解, 电解过程中，溶液 $\\mathrm{pH}$ 随时间 $\\mathrm{t}$ 变化的曲线如图所示，则下列说法正确的是\n\n[图1]\n\nA: 阴极先析出 $\\mathrm{Cl}_{2}$, 后析出 $\\mathrm{O}_{2}$, 阳极先产生 $\\mathrm{Cu}$, 后析出 $\\mathrm{H}_{2}$\nB: $\\mathrm{AB}$ 段阳极只产生 $\\mathrm{Cl}_{2}$, 阴极只产生 $\\mathrm{Cu}$\nC: $\\mathrm{BC}$ 段表示在阴极上是 $\\mathrm{H}^{+}$放电产生了 $\\mathrm{H}_{2}$\nD: CD 段相当于电解水\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-76.jpg?height=300&width=463&top_left_y=164&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_40",
        "problem": "One mole of which hydrocarbon requires $8 \\mathrm{~mol} \\mathrm{O}_{2}$ to achieve complete combustion to give carbon dioxide and water?\nA: $\\mathrm{C}_{3} \\mathrm{H}_{8}$\nB: $\\mathrm{C}_{4} \\mathrm{H}_{10}$\nC: $\\mathrm{C}_{5} \\mathrm{H}_{10}$\nD: $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nOne mole of which hydrocarbon requires $8 \\mathrm{~mol} \\mathrm{O}_{2}$ to achieve complete combustion to give carbon dioxide and water?\n\nA: $\\mathrm{C}_{3} \\mathrm{H}_{8}$\nB: $\\mathrm{C}_{4} \\mathrm{H}_{10}$\nC: $\\mathrm{C}_{5} \\mathrm{H}_{10}$\nD: $\\mathrm{C}_{5} \\mathrm{H}_{12}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1166",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidation states of elements in Groups 2 and 12 are +2, what are the valence electrons of barium?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nGiven the maximum oxidation states of elements in Groups 2 and 12 are +2, what are the valence electrons of barium?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_902",
        "problem": "某小组同学设计实验探究镁与铵盐溶液的反应, 记录如下:\n\n| 实验 <br> 编号 | $(1)$ | $(2)$ | $(3$ | $(4)$ |\n| :--- | :--- | :--- | :--- | :--- |\n| 溶液 | $1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ | $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ | $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ | $\\mathrm{H}_{2} \\mathrm{O}$ |\n\n\n| 种类 |  | $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ |  |  |\n| :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{m}(\\mathrm{Mg})$ | 0.48 | 0.48 | 0.48 | 0.48 |\n| $\\mathrm{V}($ 溶 <br> 液 $)$ <br> $/ \\mathrm{mL}$ | 100 | 100 | 100 | 100 |\n| 实验 <br> 现象 | 白色沉淀 体产生, 并产生 | 有气体产生, 并产生 | 有气体产生, 并产生 | 几乎看不 |\n| $6 \\mathrm{r}$ 时 |  | 白沉淀 | 到现象 |  |\n| $\\mathrm{V}\\left(\\mathrm{H}_{2}\\right)$ | 433 | 255 | 347 |  |\n\n根据上述实验所得结论正确的是\nA: 实验(2)中发生的反应为 $\\mathrm{Mg}+\\mathrm{NH}_{4}^{+}+2 \\mathrm{H}_{2} \\mathrm{O}=\\mathrm{Mg}^{2+}+2 \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\uparrow$\nB: 由实验 1 ) (3)可得, 溶液中 $c\\left(\\mathrm{NH}_{4}^{+}\\right)$越大, 反应速率越快\nC: 由实验 1)、(2)可得, 溶液中阴离子的种类对产生 $\\mathrm{H}_{2}$ 的速率有影响\nD: 由实验 (1) (4)可得, 溶液的 $\\mathrm{pH}$ 越小, 产生 $\\mathrm{H}_{2}$ 的速率越快\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某小组同学设计实验探究镁与铵盐溶液的反应, 记录如下:\n\n| 实验 <br> 编号 | $(1)$ | $(2)$ | $(3$ | $(4)$ |\n| :--- | :--- | :--- | :--- | :--- |\n| 溶液 | $1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ | $0.5 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ | $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ | $\\mathrm{H}_{2} \\mathrm{O}$ |\n\n\n| 种类 |  | $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{SO}_{4}$ |  |  |\n| :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{m}(\\mathrm{Mg})$ | 0.48 | 0.48 | 0.48 | 0.48 |\n| $\\mathrm{V}($ 溶 <br> 液 $)$ <br> $/ \\mathrm{mL}$ | 100 | 100 | 100 | 100 |\n| 实验 <br> 现象 | 白色沉淀 体产生, 并产生 | 有气体产生, 并产生 | 有气体产生, 并产生 | 几乎看不 |\n| $6 \\mathrm{r}$ 时 |  | 白沉淀 | 到现象 |  |\n| $\\mathrm{V}\\left(\\mathrm{H}_{2}\\right)$ | 433 | 255 | 347 |  |\n\n根据上述实验所得结论正确的是\n\nA: 实验(2)中发生的反应为 $\\mathrm{Mg}+\\mathrm{NH}_{4}^{+}+2 \\mathrm{H}_{2} \\mathrm{O}=\\mathrm{Mg}^{2+}+2 \\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{H}_{2} \\uparrow$\nB: 由实验 1 ) (3)可得, 溶液中 $c\\left(\\mathrm{NH}_{4}^{+}\\right)$越大, 反应速率越快\nC: 由实验 1)、(2)可得, 溶液中阴离子的种类对产生 $\\mathrm{H}_{2}$ 的速率有影响\nD: 由实验 (1) (4)可得, 溶液的 $\\mathrm{pH}$ 越小, 产生 $\\mathrm{H}_{2}$ 的速率越快\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_775",
        "problem": "某化学兴趣小组对电解质溶液作如下的归纳总结 (均在常温下)。其中不正确的是\nA: $\\mathrm{pH}=3$ 的强酸溶液 $1 \\mathrm{~mL}$, 加水稀释至 $100 \\mathrm{~mL}$ 后, 溶液 $\\mathrm{pH}$ 升高 2 个单位\nB: $1 \\mathrm{~L} 0.50 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液与 $2 \\mathrm{~L} 0.25 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液含 $\\mathrm{NH}_{4}+$ 物质的量后者大\nC: $\\mathrm{pH}=8.3$ 的 $\\mathrm{NaHCO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nD: $\\mathrm{pH}=4$ 、浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{CH}_{3} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{COONa}$ 混合溶液中: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right.$ $\\left.{ }^{-}\\right)-\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=2 \\times\\left(10^{-4}-10^{-10}\\right) \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某化学兴趣小组对电解质溶液作如下的归纳总结 (均在常温下)。其中不正确的是\n\nA: $\\mathrm{pH}=3$ 的强酸溶液 $1 \\mathrm{~mL}$, 加水稀释至 $100 \\mathrm{~mL}$ 后, 溶液 $\\mathrm{pH}$ 升高 2 个单位\nB: $1 \\mathrm{~L} 0.50 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液与 $2 \\mathrm{~L} 0.25 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液含 $\\mathrm{NH}_{4}+$ 物质的量后者大\nC: $\\mathrm{pH}=8.3$ 的 $\\mathrm{NaHCO}_{3}$ 溶液: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}{ }^{-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}{ }^{2-}\\right)$\nD: $\\mathrm{pH}=4$ 、浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{CH}_{3} \\mathrm{COOH} 、 \\mathrm{CH}_{3} \\mathrm{COONa}$ 混合溶液中: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right.$ $\\left.{ }^{-}\\right)-\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=2 \\times\\left(10^{-4}-10^{-10}\\right) \\mathrm{mol} \\cdot \\mathrm{L}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_218",
        "problem": "Which of the following elements has the highest $7^{\\text {th }}$ ionisation energy?\nA: $\\mathrm{Mg}$\nB: $\\mathrm{Cl}$\nC: $\\mathrm{P}$\nD: $\\mathrm{S}$\nE: $\\mathrm{Al}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following elements has the highest $7^{\\text {th }}$ ionisation energy?\n\nA: $\\mathrm{Mg}$\nB: $\\mathrm{Cl}$\nC: $\\mathrm{P}$\nD: $\\mathrm{S}$\nE: $\\mathrm{Al}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1009",
        "problem": "Which group of atoms and ions contain the same number of electrons?\nA: $\\mathrm{F}, \\mathrm{Ne}, \\mathrm{Na}$\nB: $\\mathrm{O}^{2-}, \\mathrm{S}^{2-}, \\mathrm{Se}^{2-}$\nC: Mg, Al, Si\nD: $\\mathrm{Ca}^{2+}, \\mathrm{Fe}^{3+}, \\mathrm{Zn}^{2+}$\nE: $\\mathrm{Cl}, \\mathrm{Ar}, \\mathrm{K}^{+}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich group of atoms and ions contain the same number of electrons?\n\nA: $\\mathrm{F}, \\mathrm{Ne}, \\mathrm{Na}$\nB: $\\mathrm{O}^{2-}, \\mathrm{S}^{2-}, \\mathrm{Se}^{2-}$\nC: Mg, Al, Si\nD: $\\mathrm{Ca}^{2+}, \\mathrm{Fe}^{3+}, \\mathrm{Zn}^{2+}$\nE: $\\mathrm{Cl}, \\mathrm{Ar}, \\mathrm{K}^{+}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_409",
        "problem": "《环境科学》刊发了我国科研部门采用零价铁活化过硫酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}\\right.$, 其中 $\\mathrm{S}$ 为 +6 价) 去除废水中的正五价砷 $[\\operatorname{As}(\\mathrm{V})]$ 的研究成果, 其反应机制模型如图所示。设阿伏加德罗常数的值为 $\\mathrm{N}_{\\mathrm{A}}, \\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]=2.7 \\times 10^{-39}$ 。下列叙述正确的是\n\n[图1]\nA: $1 \\mathrm{~mol}$ 过硫酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}\\right)$ 含 $2 \\mathrm{~N}_{\\mathrm{A}}$ 个过氧键\nB: 若 $56 \\mathrm{gFe}$ 参加反应, 共有 $\\mathrm{N}_{\\mathrm{A}}$ 个 $\\mathrm{S}_{2} \\mathrm{O}_{8}^{2-}$ 被还原\nC: 室温下, 中间产物 $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 溶于水所得饱和溶液中 $\\mathrm{c}\\left(\\mathrm{Fe}^{3+}\\right)$ 为 $2.7 \\times 10^{-18} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{pH}$ 越小, 越有利于去除废水中的正五价砷\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n《环境科学》刊发了我国科研部门采用零价铁活化过硫酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}\\right.$, 其中 $\\mathrm{S}$ 为 +6 价) 去除废水中的正五价砷 $[\\operatorname{As}(\\mathrm{V})]$ 的研究成果, 其反应机制模型如图所示。设阿伏加德罗常数的值为 $\\mathrm{N}_{\\mathrm{A}}, \\mathrm{K}_{\\mathrm{sp}}\\left[\\mathrm{Fe}(\\mathrm{OH})_{3}\\right]=2.7 \\times 10^{-39}$ 。下列叙述正确的是\n\n[图1]\n\nA: $1 \\mathrm{~mol}$ 过硫酸钠 $\\left(\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{8}\\right)$ 含 $2 \\mathrm{~N}_{\\mathrm{A}}$ 个过氧键\nB: 若 $56 \\mathrm{gFe}$ 参加反应, 共有 $\\mathrm{N}_{\\mathrm{A}}$ 个 $\\mathrm{S}_{2} \\mathrm{O}_{8}^{2-}$ 被还原\nC: 室温下, 中间产物 $\\mathrm{Fe}(\\mathrm{OH})_{3}$ 溶于水所得饱和溶液中 $\\mathrm{c}\\left(\\mathrm{Fe}^{3+}\\right)$ 为 $2.7 \\times 10^{-18} \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{pH}$ 越小, 越有利于去除废水中的正五价砷\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-38.jpg?height=594&width=939&top_left_y=1779&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1443",
        "problem": "The concentration of carbon dioxide in the atmosphere has increased substantially during this century and is predicted to continue to increase. The $\\left[\\mathrm{CO}_{2}\\right]$ is expected to be about $440 \\mathrm{ppm}\\left(440 \\times 10^{-6} \\mathrm{~atm}\\right)$ in the year 2020 .\n\nCalculate the enthalpy of reaction between $\\mathrm{CO}_{2}(\\mathrm{aq})$ and $\\mathrm{H}_{2} \\mathrm{O}$.\n\nData:\n\nHenry's Law constant for $\\mathrm{CO}_{2}$ at $298 \\mathrm{~K}: 0.0343 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~atm}^{-1}$\n\nThermodynamic values, in $\\mathrm{kJ} / \\mathrm{mol}$ at $298 \\mathrm{~K}$ are:\n\n|  | $\\Delta_{f} G^{0}$ | $\\Delta_{f} H^{0}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CO}_{2}(\\mathrm{aq})$ | -386.2 | -412.9 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -237.2 | -285.8 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -587.1 | -691.2 |\n| $\\mathrm{H}^{+}(\\mathrm{aq})$ | 0.00 | 0.00 |",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe concentration of carbon dioxide in the atmosphere has increased substantially during this century and is predicted to continue to increase. The $\\left[\\mathrm{CO}_{2}\\right]$ is expected to be about $440 \\mathrm{ppm}\\left(440 \\times 10^{-6} \\mathrm{~atm}\\right)$ in the year 2020 .\n\nCalculate the enthalpy of reaction between $\\mathrm{CO}_{2}(\\mathrm{aq})$ and $\\mathrm{H}_{2} \\mathrm{O}$.\n\nData:\n\nHenry's Law constant for $\\mathrm{CO}_{2}$ at $298 \\mathrm{~K}: 0.0343 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1} \\mathrm{~atm}^{-1}$\n\nThermodynamic values, in $\\mathrm{kJ} / \\mathrm{mol}$ at $298 \\mathrm{~K}$ are:\n\n|  | $\\Delta_{f} G^{0}$ | $\\Delta_{f} H^{0}$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CO}_{2}(\\mathrm{aq})$ | -386.2 | -412.9 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$ | -237.2 | -285.8 |\n| $\\mathrm{HCO}_{3}{ }^{-}(\\mathrm{aq})$ | -587.1 | -691.2 |\n| $\\mathrm{H}^{+}(\\mathrm{aq})$ | 0.00 | 0.00 |\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $ \\mathrm{~kJ} \\mathrm{~mol}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$ \\mathrm{~kJ} \\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_86",
        "problem": "The normal boiling point of $\\mathrm{CH}_{2} \\mathrm{Br}_{2}$ is $97^{\\circ} \\mathrm{C}$ while the normal boiling point of $\\mathrm{CHBr}_{3}$ is $149^{\\circ} \\mathrm{C}$. Differences in which interactions are most responsible for this difference in boiling points?\nA: London dispersion forces\nB: Dipole-dipole interactions\nC: Hydrogen bonding\nD: Carbon-bromine covalent bonding\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe normal boiling point of $\\mathrm{CH}_{2} \\mathrm{Br}_{2}$ is $97^{\\circ} \\mathrm{C}$ while the normal boiling point of $\\mathrm{CHBr}_{3}$ is $149^{\\circ} \\mathrm{C}$. Differences in which interactions are most responsible for this difference in boiling points?\n\nA: London dispersion forces\nB: Dipole-dipole interactions\nC: Hydrogen bonding\nD: Carbon-bromine covalent bonding\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_968",
        "problem": "Which of the following is an ionic solid?\nA: $\\mathrm{N}_{2} \\mathrm{O}$\nB: $\\mathrm{HCl}$\nC: $\\mathrm{LiCl}$\nD: $\\mathrm{CO}_{2}$\nE: $\\mathrm{CH}_{4}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following is an ionic solid?\n\nA: $\\mathrm{N}_{2} \\mathrm{O}$\nB: $\\mathrm{HCl}$\nC: $\\mathrm{LiCl}$\nD: $\\mathrm{CO}_{2}$\nE: $\\mathrm{CH}_{4}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_909",
        "problem": "常温下, 向 $1 \\mathrm{~L} 1.0 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaClO}$ 溶液中缓慢通入 $\\mathrm{SO}_{2}$ 气体, 使其充分吸收, 溶液 $\\mathrm{pH}$ 与通入 $\\mathrm{SO}_{2}$ 物质的量关系如图所示(忽略溶液体积的变化和 $\\mathrm{NaClO} 、 \\mathrm{HClO}$ 的分解)。下列说法错误的是\n\n[图1]\nA: 常温下, $\\mathrm{HClO}$ 电离平衡常数的数量级为 $10^{-8}$\nB: $\\mathrm{a}$ 点溶液中存在 $4 c\\left(\\mathrm{Cl}^{-}\\right)=c(\\mathrm{HClO})+c\\left(\\mathrm{ClO}^{-}\\right)$\nC: $\\mathrm{b}$ 点溶液中存在 $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{SO}_{4}^{2-}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)$\nD: $\\mathrm{c}$ 点溶液中 $c\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{~mol} / \\mathrm{L}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $1 \\mathrm{~L} 1.0 \\mathrm{~mol} / \\mathrm{L}$ 的 $\\mathrm{NaClO}$ 溶液中缓慢通入 $\\mathrm{SO}_{2}$ 气体, 使其充分吸收, 溶液 $\\mathrm{pH}$ 与通入 $\\mathrm{SO}_{2}$ 物质的量关系如图所示(忽略溶液体积的变化和 $\\mathrm{NaClO} 、 \\mathrm{HClO}$ 的分解)。下列说法错误的是\n\n[图1]\n\nA: 常温下, $\\mathrm{HClO}$ 电离平衡常数的数量级为 $10^{-8}$\nB: $\\mathrm{a}$ 点溶液中存在 $4 c\\left(\\mathrm{Cl}^{-}\\right)=c(\\mathrm{HClO})+c\\left(\\mathrm{ClO}^{-}\\right)$\nC: $\\mathrm{b}$ 点溶液中存在 $c\\left(\\mathrm{Na}^{+}\\right)>c\\left(\\mathrm{SO}_{4}^{2-}\\right)>c\\left(\\mathrm{Cl}^{-}\\right)>c\\left(\\mathrm{H}^{+}\\right)>c\\left(\\mathrm{OH}^{-}\\right)$\nD: $\\mathrm{c}$ 点溶液中 $c\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{~mol} / \\mathrm{L}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-20.jpg?height=512&width=813&top_left_y=652&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_393",
        "problem": "How many unpaired electrons are in the superoxide ion, $\\mathrm{O}_{2}^{-}$?\nA: Zero\nB: One\nC: Two\nD: Three\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many unpaired electrons are in the superoxide ion, $\\mathrm{O}_{2}^{-}$?\n\nA: Zero\nB: One\nC: Two\nD: Three\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_68",
        "problem": "Which halogen has the greatest first ionization energy?\nA: $\\mathrm{F}$\nB: $\\mathrm{Cl}$\nC: $\\mathrm{Br}$\nD: I\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich halogen has the greatest first ionization energy?\n\nA: $\\mathrm{F}$\nB: $\\mathrm{Cl}$\nC: $\\mathrm{Br}$\nD: I\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_132",
        "problem": "Artificial photosynthesis involves splitting water with solar energy. This clean energy reaction is: $2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightarrow 2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})$\n\nWhich of the following statements about the splitting of water is incorrect:\nA: The oxidation state of hydrogen in water is +1\nB: The oxidation state of hydrogen gas is 0\nC: Water is the oxidizing and reducing agent\nD: The oxidation state of oxygen in water is - 1\nE: The oxygen atoms are oxidized in this process\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nArtificial photosynthesis involves splitting water with solar energy. This clean energy reaction is: $2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightarrow 2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})$\n\nWhich of the following statements about the splitting of water is incorrect:\n\nA: The oxidation state of hydrogen in water is +1\nB: The oxidation state of hydrogen gas is 0\nC: Water is the oxidizing and reducing agent\nD: The oxidation state of oxygen in water is - 1\nE: The oxygen atoms are oxidized in this process\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_603",
        "problem": "常温下, 下列说法正确的是\nA: $\\mathrm{pH}$ 加和为 12 的强酸与弱酸混合, $\\mathrm{pH}$ 可能为 6\nB: $\\mathrm{pH}$ 加和为 12 的强酸与弱碱等体积混合, 溶液一定呈碱性\nC: $\\mathrm{pH}$ 加和为 12 的弱酸与强碱混合, 若成中性, 则 $\\mathrm{v}$ (酸) $>\\mathrm{v}$ (碱)\nD: $\\mathrm{pH}$ 加和为 12 的强酸与强碱混合, 若成中性, 则两溶液体积比为 100: 1\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 下列说法正确的是\n\nA: $\\mathrm{pH}$ 加和为 12 的强酸与弱酸混合, $\\mathrm{pH}$ 可能为 6\nB: $\\mathrm{pH}$ 加和为 12 的强酸与弱碱等体积混合, 溶液一定呈碱性\nC: $\\mathrm{pH}$ 加和为 12 的弱酸与强碱混合, 若成中性, 则 $\\mathrm{v}$ (酸) $>\\mathrm{v}$ (碱)\nD: $\\mathrm{pH}$ 加和为 12 的强酸与强碱混合, 若成中性, 则两溶液体积比为 100: 1\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_106",
        "problem": "Another way to make diamonds in the lab is by converting graphite to diamond. At $2000 \\mathrm{~K}$ and $200000 \\mathrm{~atm}$ :\n\n$\\mathrm{C}$ (graphite) $\\rightleftharpoons \\mathrm{C}$ (diamond); $\\Delta \\mathrm{G}=-10 \\mathrm{~kJ} \\mathrm{~mol}^{-1}, \\Delta \\mathrm{S}=-10 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$\n\nWhich of the following is a good approximation of the equilibrium temperature for the reaction at 200000 atm if enthalpy and entropy are assumed to be temperature independent?\nA: $\\mathrm{T}_{\\text {eq }}=2000 \\mathrm{~K}$\nB: $\\mathrm{T}_{\\mathrm{eq}}=3000 \\mathrm{~K}$\nC: $\\mathrm{T}_{\\text {eq }}=1000 \\mathrm{~K}$\nD: $\\mathrm{T}_{\\text {eq }}=4000 \\mathrm{~K}$\nE: $\\mathrm{T}_{\\mathrm{eq}}=100 \\mathrm{~K}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAnother way to make diamonds in the lab is by converting graphite to diamond. At $2000 \\mathrm{~K}$ and $200000 \\mathrm{~atm}$ :\n\n$\\mathrm{C}$ (graphite) $\\rightleftharpoons \\mathrm{C}$ (diamond); $\\Delta \\mathrm{G}=-10 \\mathrm{~kJ} \\mathrm{~mol}^{-1}, \\Delta \\mathrm{S}=-10 \\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}$\n\nWhich of the following is a good approximation of the equilibrium temperature for the reaction at 200000 atm if enthalpy and entropy are assumed to be temperature independent?\n\nA: $\\mathrm{T}_{\\text {eq }}=2000 \\mathrm{~K}$\nB: $\\mathrm{T}_{\\mathrm{eq}}=3000 \\mathrm{~K}$\nC: $\\mathrm{T}_{\\text {eq }}=1000 \\mathrm{~K}$\nD: $\\mathrm{T}_{\\text {eq }}=4000 \\mathrm{~K}$\nE: $\\mathrm{T}_{\\mathrm{eq}}=100 \\mathrm{~K}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_608",
        "problem": "某密闭容器中发生如下反应: $X(\\mathrm{~g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{Z}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}<0$ 。如图表示该反应的速率 $(v)$ 随时间 $(t)$ 变化的关系， $t_{2} 、 t_{3} 、 t_{5}$ 时刻外界条件有所改变，但都没有改变各物质的初始加入量。下列说法中正确的是\n\n[图1]\nA: $\\mathrm{t}_{2}$ 时一定加入了催化剂\nB: $\\mathrm{t}_{3}-\\mathrm{t}_{4}$ 平衡正向移动\nC: $\\mathrm{t}_{5}$ 时可能增大了压强\nD: $\\mathrm{t}_{6}$ 之后转化率最低\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某密闭容器中发生如下反应: $X(\\mathrm{~g})+\\mathrm{Y}(\\mathrm{g}) \\rightleftharpoons 2 \\mathrm{Z}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}<0$ 。如图表示该反应的速率 $(v)$ 随时间 $(t)$ 变化的关系， $t_{2} 、 t_{3} 、 t_{5}$ 时刻外界条件有所改变，但都没有改变各物质的初始加入量。下列说法中正确的是\n\n[图1]\n\nA: $\\mathrm{t}_{2}$ 时一定加入了催化剂\nB: $\\mathrm{t}_{3}-\\mathrm{t}_{4}$ 平衡正向移动\nC: $\\mathrm{t}_{5}$ 时可能增大了压强\nD: $\\mathrm{t}_{6}$ 之后转化率最低\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-035.jpg?height=386&width=765&top_left_y=641&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_397",
        "problem": "The reaction of tert-butyl bromide with azide ion in aqueous solution is proposed to proceed by the following mechanism:\n\n$$\n\\begin{aligned}\n& \\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}(a q) \\underset{k_{-1}}{k_{1}}\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}^{+}(a q)+\\mathrm{Br}^{-}(a q) \\\\\n& \\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}^{+}(a q)+\\mathrm{N}_{3}^{-}(a q) \\xrightarrow{k_{2}}\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CN}_{3}(a q)\n\\end{aligned}\n$$\n\nAssuming that $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}^{+}(a q)$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants, what is the rate law for this reaction?\nA: Rate $=k_{1}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]$\nB: Rate $=k_{2}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]\\left[\\mathrm{N}_{3}^{-}\\right]$\nC: $\\quad$ Rate $=\\frac{k_{1} k_{2}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]\\left[\\mathrm{N}_{3}^{-}\\right]}{k_{-1}\\left[\\mathrm{Br}^{-}\\right]}$\nD: Rate $=\\frac{k_{1} k_{2}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]\\left[\\mathrm{N}_{3}^{-}\\right]}{k_{-1}\\left[\\mathrm{Br}^{-}\\right]+k_{2}\\left[\\mathrm{~N}_{3}{ }^{-}\\right]}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe reaction of tert-butyl bromide with azide ion in aqueous solution is proposed to proceed by the following mechanism:\n\n$$\n\\begin{aligned}\n& \\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}(a q) \\underset{k_{-1}}{k_{1}}\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}^{+}(a q)+\\mathrm{Br}^{-}(a q) \\\\\n& \\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}^{+}(a q)+\\mathrm{N}_{3}^{-}(a q) \\xrightarrow{k_{2}}\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CN}_{3}(a q)\n\\end{aligned}\n$$\n\nAssuming that $\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{C}^{+}(a q)$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants, what is the rate law for this reaction?\n\nA: Rate $=k_{1}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]$\nB: Rate $=k_{2}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]\\left[\\mathrm{N}_{3}^{-}\\right]$\nC: $\\quad$ Rate $=\\frac{k_{1} k_{2}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]\\left[\\mathrm{N}_{3}^{-}\\right]}{k_{-1}\\left[\\mathrm{Br}^{-}\\right]}$\nD: Rate $=\\frac{k_{1} k_{2}\\left[\\left(\\mathrm{CH}_{3}\\right)_{3} \\mathrm{CBr}\\right]\\left[\\mathrm{N}_{3}^{-}\\right]}{k_{-1}\\left[\\mathrm{Br}^{-}\\right]+k_{2}\\left[\\mathrm{~N}_{3}{ }^{-}\\right]}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1331",
        "problem": "Bridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate the equilibrium pressure (in bar) of water vapour in a closed vessel containing $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ at $25^{\\circ} \\mathrm{C}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBridge between Denmark and Sweden\n\n[figure1]\n\nOn July 1, 2000, the combined tunnel and bridge connecting Denmark and Sweden was officially opened. It consists of a tunnel from Copenhagen to an artificial island, and a bridge from the island to Malmö in Sweden. The major construction materials employed are concrete and steel. This problem deals with chemical reactions relating to production and degradation of such materials.\n\nConcrete is produced from a mixture of cement, water, sand and small stones. Cement consists primarily of calcium silicates and calcium aluminates formed by heating and grinding of clay and limestone. In the later steps of cement production a small amount of gypsum, $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}$, is added to improve subsequent hardening of the concrete. The use of elevated temperatures during the final production may lead to formation of unwanted hemihydrate, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}$. Consider the following reaction:\n\n$$\n\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s}) \\rightarrow \\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})+1 \\frac{1}{2} \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})\n$$\n\nThe following thermodynamic data apply at $25^{\\circ} \\mathrm{C}$, standard pressure: 1.00 bar:\n\n| Compound | $\\Delta_{\\dot{f}} \\mathrm{H}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $\\mathrm{S}\\left(\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :--- | :---: | :---: |\n| $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -2021.0 | 194.0 |\n| $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ | -1575.0 | 130.5 |\n| $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ | -241.8 | 188.6 |\n\nGas constant: $R=8.314 \\mathrm{~J} \\mathrm{~mol}^{-1} \\mathrm{~K}^{-1}=0.08314 \\mathrm{~L} \\mathrm{bar} \\mathrm{mol}^{-1} \\mathrm{~K}^{-1}$\n\n$$\n0^{\\circ} \\mathrm{C}=273.15 \\mathrm{~K} .\n$$\n\nCalculate the equilibrium pressure (in bar) of water vapour in a closed vessel containing $\\mathrm{CaSO}_{4} \\cdot 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$, $\\mathrm{CaSO}_{4} \\cdot 1 / 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{s})$ and $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ at $25^{\\circ} \\mathrm{C}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of bar, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-345.jpg?height=951&width=1445&top_left_y=595&top_left_x=314"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "bar"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1573",
        "problem": "The absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\n\nPure aqueous solutions of $\\mathrm{HX}$ were used. Both $\\mathrm{HX}$ and $\\mathrm{X}^{-}$absorb. Measurements were obtained at a wavelength where the molar absorptivities of $X^{-}$and $H X$ are equal and different than zero.\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\n\nPure aqueous solutions of $\\mathrm{HX}$ were used. Both $\\mathrm{HX}$ and $\\mathrm{X}^{-}$absorb. Measurements were obtained at a wavelength where the molar absorptivities of $X^{-}$and $H X$ are equal and different than zero.\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-463.jpg?height=454&width=777&top_left_y=321&top_left_x=545"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_54",
        "problem": "Which of the following properties is not typical of metallic solids?\nA: High vapor pressure\nB: High coordination number of atoms in the lattice\nC: High electrical conductivity\nD: High thermal conductivity\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following properties is not typical of metallic solids?\n\nA: High vapor pressure\nB: High coordination number of atoms in the lattice\nC: High electrical conductivity\nD: High thermal conductivity\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_893",
        "problem": "已知室温下 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=10^{-12.9}$ 。通过下列实验探究含硫化合物的性质。\n\n实验 1 ：测得 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~S}$ 溶液 $\\mathrm{pH}=4.1$\n\n实验 2: 向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHS}$ 溶液中逐滴加入 $5 \\mathrm{~mL}$ 水, 用 $\\mathrm{pH}$ 计监测过程中 $\\mathrm{pH}$ 变化\n\n实验 3：向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~S}$ 溶液中逐滴加入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, 直至 $\\mathrm{pH}=7$\n\n实验 4: 向 $5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~S}$ 中滴加 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{MnSO}_{4}$ 溶液, 产生粉色沉淀, 再加几滴 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CuSO}_{4}$ 溶液, 产生黑色沉淀下列说法不正确的是\nA: 由实验 1 可知: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~S}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)<\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 实验 2 加水过程中, 监测结果为溶液的 $\\mathrm{pH}$ 不断减小\nC: 实验 3 所得溶液中存在: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 由实验 4 可知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{MnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n已知室温下 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=10^{-7}, \\mathrm{~K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)=10^{-12.9}$ 。通过下列实验探究含硫化合物的性质。\n\n实验 1 ：测得 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~S}$ 溶液 $\\mathrm{pH}=4.1$\n\n实验 2: 向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHS}$ 溶液中逐滴加入 $5 \\mathrm{~mL}$ 水, 用 $\\mathrm{pH}$ 计监测过程中 $\\mathrm{pH}$ 变化\n\n实验 3：向 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~S}$ 溶液中逐滴加入 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液, 直至 $\\mathrm{pH}=7$\n\n实验 4: 向 $5 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{~S}$ 中滴加 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{MnSO}_{4}$ 溶液, 产生粉色沉淀, 再加几滴 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CuSO}_{4}$ 溶液, 产生黑色沉淀下列说法不正确的是\n\nA: 由实验 1 可知: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{~S}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)<\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: 实验 2 加水过程中, 监测结果为溶液的 $\\mathrm{pH}$ 不断减小\nC: 实验 3 所得溶液中存在: $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~S}\\right)+\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{S}^{2-}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: 由实验 4 可知: $\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{MnS})>\\mathrm{K}_{\\mathrm{sp}}(\\mathrm{CuS})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_679",
        "problem": "以秸秆为原料合成 PEF 树脂的路线如图所示。下列说法错误的是\n[图1]\nA: 异构化时，官能团由羰基变为醛基\nB: PEF 树脂可降解, 单体 $\\mathrm{a}$ 为乙二醇\nC: 5-HMF 和 FDCA 中所有碳原子可能共平面\nD: $5-\\mathrm{HMF}$ 的同分异构体中含苯环结构的有 2 种\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n以秸秆为原料合成 PEF 树脂的路线如图所示。下列说法错误的是\n[图1]\n\nA: 异构化时，官能团由羰基变为醛基\nB: PEF 树脂可降解, 单体 $\\mathrm{a}$ 为乙二醇\nC: 5-HMF 和 FDCA 中所有碳原子可能共平面\nD: $5-\\mathrm{HMF}$ 的同分异构体中含苯环结构的有 2 种\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-05.jpg?height=348&width=1098&top_left_y=840&top_left_x=336",
            "https://i.postimg.cc/fybN45Tj/image.png"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1565",
        "problem": "The absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nPure aqueous solutions of $H X$ were used. Only the undissociated species $H X$ absorb.\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe absorbance of solutions of the weak acid HX were obtained. Associate the expected form of the resulting working curve with those shown in figure, under the following conditions:\n\n[figure1]\n\nTotal concentration of $\\mathrm{HX}$\n\nPure aqueous solutions of $H X$ were used. Only the undissociated species $H X$ absorb.\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-463.jpg?height=454&width=777&top_left_y=321&top_left_x=545"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_916",
        "problem": "碳酸二甲酯 $\\left[\\left(\\mathrm{CH}_{3} \\mathrm{O}\\right)_{2} \\mathrm{CO}\\right]$ 被誉为当今有机合成的“新基石”, 一种电化学合成它的工作原理如图所示:\n\n[图1]\n\n下列说法正确的是\nA: B 为电源的正极\nB: 左室电极反应: $2 \\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{CO}-2 \\mathrm{e}^{-}=\\left(\\mathrm{CH}_{3} \\mathrm{O}\\right)_{2} \\mathrm{CO}+2 \\mathrm{H}^{+}$\nC: 若参加反应的 $\\mathrm{O}_{2}$ 为 $1.12 \\mathrm{~L}$ (标准状况), 则理论上可制得 $9 \\mathrm{~g}\\left(\\mathrm{CH}_{3} \\mathrm{O}\\right)_{2} \\mathrm{CO}$\nD: 若反应转移的电子为 $2 \\mathrm{~mol}$, 则左室与右室溶液的质量差为 $4 \\mathrm{~g}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n碳酸二甲酯 $\\left[\\left(\\mathrm{CH}_{3} \\mathrm{O}\\right)_{2} \\mathrm{CO}\\right]$ 被誉为当今有机合成的“新基石”, 一种电化学合成它的工作原理如图所示:\n\n[图1]\n\n下列说法正确的是\n\nA: B 为电源的正极\nB: 左室电极反应: $2 \\mathrm{CH}_{3} \\mathrm{OH}+\\mathrm{CO}-2 \\mathrm{e}^{-}=\\left(\\mathrm{CH}_{3} \\mathrm{O}\\right)_{2} \\mathrm{CO}+2 \\mathrm{H}^{+}$\nC: 若参加反应的 $\\mathrm{O}_{2}$ 为 $1.12 \\mathrm{~L}$ (标准状况), 则理论上可制得 $9 \\mathrm{~g}\\left(\\mathrm{CH}_{3} \\mathrm{O}\\right)_{2} \\mathrm{CO}$\nD: 若反应转移的电子为 $2 \\mathrm{~mol}$, 则左室与右室溶液的质量差为 $4 \\mathrm{~g}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-075.jpg?height=513&width=1022&top_left_y=2065&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_751",
        "problem": "常温下, 向 $1 \\mathrm{~L} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 一元酸 $\\mathrm{HR}$ 溶液中逐渐通入氨气 $\\left[\\right.$ 已知常温下 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$电离平衡常数 $\\mathrm{K}=1.76 \\times 10^{-5} \\mathrm{~J}$, 使溶液温度和体积保持不变, 混合溶液的 $\\mathrm{pH}$ 与离子浓度变化的关系如图所示。下列叙述正确的是\n\n[图1]\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HR}$ 溶液的 $\\mathrm{pH}$ 为 5\nB: $\\mathrm{HR}$ 为弱酸, 常温时随着氨气的通入, $\\mathrm{c}\\left(\\mathrm{R}^{-}\\right) /\\left[\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right) . \\mathrm{c}(\\mathrm{HR})\\right]$ 逐渐增大\nC: 当通入 $0.1 \\mathrm{~mol} \\mathrm{NH}_{3}$ 时, $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}>\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)\\right.$\nD: 当 $\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)=\\mathrm{c}(\\mathrm{HR})$ 时 溶液必为中性\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向 $1 \\mathrm{~L} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 一元酸 $\\mathrm{HR}$ 溶液中逐渐通入氨气 $\\left[\\right.$ 已知常温下 $\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}$电离平衡常数 $\\mathrm{K}=1.76 \\times 10^{-5} \\mathrm{~J}$, 使溶液温度和体积保持不变, 混合溶液的 $\\mathrm{pH}$ 与离子浓度变化的关系如图所示。下列叙述正确的是\n\n[图1]\n\nA: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HR}$ 溶液的 $\\mathrm{pH}$ 为 5\nB: $\\mathrm{HR}$ 为弱酸, 常温时随着氨气的通入, $\\mathrm{c}\\left(\\mathrm{R}^{-}\\right) /\\left[\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right) . \\mathrm{c}(\\mathrm{HR})\\right]$ 逐渐增大\nC: 当通入 $0.1 \\mathrm{~mol} \\mathrm{NH}_{3}$ 时, $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}>\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)\\right.$\nD: 当 $\\mathrm{c}\\left(\\mathrm{R}^{-}\\right)=\\mathrm{c}(\\mathrm{HR})$ 时 溶液必为中性\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-34.jpg?height=542&width=893&top_left_y=1805&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_972",
        "problem": "The average mass of a solid copper penny is $2.63 \\mathrm{~g}$. What is the mass of one mole of pennies?\nA: $1.58 \\times 10^{24} \\mathrm{~g}$\nB: $6.02 \\times 10^{23} \\mathrm{~g}$\nC: $6.36 \\times 10^{23} \\mathrm{~g}$\nD: $63.6 \\mathrm{~g}$\nE: $\\quad 1.58 \\times 10^{23} \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe average mass of a solid copper penny is $2.63 \\mathrm{~g}$. What is the mass of one mole of pennies?\n\nA: $1.58 \\times 10^{24} \\mathrm{~g}$\nB: $6.02 \\times 10^{23} \\mathrm{~g}$\nC: $6.36 \\times 10^{23} \\mathrm{~g}$\nD: $63.6 \\mathrm{~g}$\nE: $\\quad 1.58 \\times 10^{23} \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1426",
        "problem": "Distribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of total phosphate $\\left(\\mathrm{PO}_{4}{ }_{4}^{3-}\\right)$ and silicate $\\left(\\mathrm{SiO}_{4}^{4-}\\right)$\n\nA 5.00 gram of soil sample is digested to give a final volume of $50.0 \\mathrm{~cm}^{3}$ digesting solution which dissolves total phosphorus and silicon. The extract is analyzed for the total concentrations of phosphorus and silicon. The concentrations of phosphorus and silicon are found to be $5.16 \\mathrm{mg} \\mathrm{dm}^{-3}$ and $5.35 \\mathrm{mg} \\mathrm{dm}^{-3}$, respectively.Determine the mass of $\\mathrm{PO}_{4}{ }^{3-}$ in $\\mathrm{mg}$ per $1.00 \\mathrm{~g}$ of soil.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nDistribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of total phosphate $\\left(\\mathrm{PO}_{4}{ }_{4}^{3-}\\right)$ and silicate $\\left(\\mathrm{SiO}_{4}^{4-}\\right)$\n\nA 5.00 gram of soil sample is digested to give a final volume of $50.0 \\mathrm{~cm}^{3}$ digesting solution which dissolves total phosphorus and silicon. The extract is analyzed for the total concentrations of phosphorus and silicon. The concentrations of phosphorus and silicon are found to be $5.16 \\mathrm{mg} \\mathrm{dm}^{-3}$ and $5.35 \\mathrm{mg} \\mathrm{dm}^{-3}$, respectively.\n\nproblem:\nDetermine the mass of $\\mathrm{PO}_{4}{ }^{3-}$ in $\\mathrm{mg}$ per $1.00 \\mathrm{~g}$ of soil.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_546",
        "problem": "恒温恒容时, 向体积为 $2.0 \\mathrm{~L}$ 密闭容器中充入 $1.0 \\mathrm{molPCl}$, 反应 $\\mathrm{PCl}_{5}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{PCl}_{3}(\\mathrm{~g})+$ $\\mathrm{Cl}_{2}(\\mathrm{~g})$ 经一段时间后达到平衡。反应过程中测定的部分数据见下表:\n\n| $\\mathrm{t} / \\mathrm{s}$ | 0 | 50 | 150 | 250 | 350 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{n}\\left(\\mathrm{PCl}_{3}\\right) / \\mathrm{mol}$ | 0 | 0.16 | 0.19 | 0.20 | 0.20 |\n\n下列说法正确的是\nA: 反应在前 $50 \\mathrm{~s}$ 的平均速率为 $\\mathrm{v}\\left(\\mathrm{PCl}_{3}\\right)=0.0016 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{s}^{-1}$\nB: 保持其他条件不变, 升高温度, 平衡时, $\\mathrm{c}\\left(\\mathrm{PCl}_{3}\\right)=0.12 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 则反应的 $\\triangle \\mathrm{H}<0$\nC: 相同温度下, 起始时向容器中充入 $1.0 \\mathrm{~mol} \\mathrm{PCl}_{5} 、 0.20 \\mathrm{~mol} \\mathrm{PCl}_{3}$ 和 $0.20 \\mathrm{molCl}_{2}$, 达到平衡前 $\\mathrm{v}($ 正 $)>\\mathrm{v}($ 逆 $)$\nD: 相同温度下, 起始时向容器中充入 $2.0 \\mathrm{~mol} \\mathrm{PCl}_{3} 、 2.0 \\mathrm{~mol} \\mathrm{Cl}_{2}$, 达到平衡时, $\\mathrm{PCl}_{3}$的转化率小于 $80 \\%$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n恒温恒容时, 向体积为 $2.0 \\mathrm{~L}$ 密闭容器中充入 $1.0 \\mathrm{molPCl}$, 反应 $\\mathrm{PCl}_{5}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{PCl}_{3}(\\mathrm{~g})+$ $\\mathrm{Cl}_{2}(\\mathrm{~g})$ 经一段时间后达到平衡。反应过程中测定的部分数据见下表:\n\n| $\\mathrm{t} / \\mathrm{s}$ | 0 | 50 | 150 | 250 | 350 |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| $\\mathrm{n}\\left(\\mathrm{PCl}_{3}\\right) / \\mathrm{mol}$ | 0 | 0.16 | 0.19 | 0.20 | 0.20 |\n\n下列说法正确的是\n\nA: 反应在前 $50 \\mathrm{~s}$ 的平均速率为 $\\mathrm{v}\\left(\\mathrm{PCl}_{3}\\right)=0.0016 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{s}^{-1}$\nB: 保持其他条件不变, 升高温度, 平衡时, $\\mathrm{c}\\left(\\mathrm{PCl}_{3}\\right)=0.12 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$, 则反应的 $\\triangle \\mathrm{H}<0$\nC: 相同温度下, 起始时向容器中充入 $1.0 \\mathrm{~mol} \\mathrm{PCl}_{5} 、 0.20 \\mathrm{~mol} \\mathrm{PCl}_{3}$ 和 $0.20 \\mathrm{molCl}_{2}$, 达到平衡前 $\\mathrm{v}($ 正 $)>\\mathrm{v}($ 逆 $)$\nD: 相同温度下, 起始时向容器中充入 $2.0 \\mathrm{~mol} \\mathrm{PCl}_{3} 、 2.0 \\mathrm{~mol} \\mathrm{Cl}_{2}$, 达到平衡时, $\\mathrm{PCl}_{3}$的转化率小于 $80 \\%$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_474",
        "problem": "2022 年 7 月 10 日正式上市的比亚迪“汉”汽车, 配置磷酸铁锂“刀片电池”, 进而解决磷酸铁锂电池能量密度低的问题。“刀片电池”放电时的总反应:\n\n$\\mathrm{Li}_{\\mathrm{x}} \\mathrm{C}_{6}+\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}=6 \\mathrm{C}+\\mathrm{LiFePO}_{4}$, 工作原理如图所示, 下列说法错误的是\n\n[图1]\nA: 放电时, 负极反应式为 $\\mathrm{Li}_{\\mathrm{x}} \\mathrm{C}_{6}-\\mathrm{xe}^{-}=\\mathrm{xLi}^{+}+6 \\mathrm{C}$\nB: 用充电桩给汽车电池充电的过程中, 阴极质量不变\nC: 充电时的铝箔与电源正极相连\nD: 放电时，电子由铜箔通过隔膜流向铝箔。 $\\mathrm{Li}^{+}$移向正极，使正极质量增加\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n2022 年 7 月 10 日正式上市的比亚迪“汉”汽车, 配置磷酸铁锂“刀片电池”, 进而解决磷酸铁锂电池能量密度低的问题。“刀片电池”放电时的总反应:\n\n$\\mathrm{Li}_{\\mathrm{x}} \\mathrm{C}_{6}+\\mathrm{Li}_{1-\\mathrm{x}} \\mathrm{FePO}_{4}=6 \\mathrm{C}+\\mathrm{LiFePO}_{4}$, 工作原理如图所示, 下列说法错误的是\n\n[图1]\n\nA: 放电时, 负极反应式为 $\\mathrm{Li}_{\\mathrm{x}} \\mathrm{C}_{6}-\\mathrm{xe}^{-}=\\mathrm{xLi}^{+}+6 \\mathrm{C}$\nB: 用充电桩给汽车电池充电的过程中, 阴极质量不变\nC: 充电时的铝箔与电源正极相连\nD: 放电时，电子由铜箔通过隔膜流向铝箔。 $\\mathrm{Li}^{+}$移向正极，使正极质量增加\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-26.jpg?height=454&width=722&top_left_y=895&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_356",
        "problem": "Which piece of equipment would give the most precise delivery of $25.0 \\mathrm{~mL}$ of a solution?\nA: 25-mL graduated cylinder\nB: 25-mL syringe\nC: 25-mL beaker\nD: 25-mL volumetric pipet\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich piece of equipment would give the most precise delivery of $25.0 \\mathrm{~mL}$ of a solution?\n\nA: 25-mL graduated cylinder\nB: 25-mL syringe\nC: 25-mL beaker\nD: 25-mL volumetric pipet\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_53",
        "problem": "A student standardizes a solution of aqueous $\\mathrm{NaOH}$ against a measured mass of solid potassium hydrogen phthalate. She then uses this $\\mathrm{NaOH}$ solution to titrate a measured mass of an unknown monocarboxylic acid to its phenolphthalein endpoint to determine its molar mass.\n\nWhich errors will lead to a value of the molar mass that is too high?\n\nI. The potassium hydrogen phthalate is partially hydrated.\n\nII. The $\\mathrm{NaOH}$ solution is allowed to stand after being standardized and absorbs some carbon dioxide from the air.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student standardizes a solution of aqueous $\\mathrm{NaOH}$ against a measured mass of solid potassium hydrogen phthalate. She then uses this $\\mathrm{NaOH}$ solution to titrate a measured mass of an unknown monocarboxylic acid to its phenolphthalein endpoint to determine its molar mass.\n\nWhich errors will lead to a value of the molar mass that is too high?\n\nI. The potassium hydrogen phthalate is partially hydrated.\n\nII. The $\\mathrm{NaOH}$ solution is allowed to stand after being standardized and absorbs some carbon dioxide from the air.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_308",
        "problem": "Which of the following molecules is linear?\nA: $\\mathrm{H}_{2} \\mathrm{O}$\nB: $\\mathrm{O}_{3}$\nC: $\\mathrm{NH}_{3}$\nD: HCN\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following molecules is linear?\n\nA: $\\mathrm{H}_{2} \\mathrm{O}$\nB: $\\mathrm{O}_{3}$\nC: $\\mathrm{NH}_{3}$\nD: HCN\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1251",
        "problem": "When the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\n| Buffer | systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid |\n| :--- | :---: |\n|1. $\\quad 0.10 \\mathrm{M}$ lactic acid / $0.10 \\mathrm{M}$ sodium lactate | $1.4 \\times 10^{-4}$ |\n|2. $\\quad 0.10 \\mathrm{M}$ acetic acid / $0.10 \\mathrm{M}$ sodium acetate | $1.8 \\times 10^{-5}$ |\n|3. $\\quad 0.10 \\mathrm{M}$ sodium dihydrogen phosphate / $0.10 \\mathrm{M}$ sodium hydrogen phosphate | $6.2 \\times 10^{-8}$ |\n|4. $\\quad 0.10 \\mathrm{M}$ ammonium chloride / $0.10 \\mathrm{M}$ ammonia | $5.6 \\times 10^{-10}$ |\nA: Buffer: $\\quad 0.10 \\mathrm{M}$ lactic acid / $0.10 \\mathrm{M}$ sodium lactate; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $1.4 \\times 10^{-4}$\nB: Buffer: $\\quad 0.10 \\mathrm{M}$ acetic acid / $0.10 \\mathrm{M}$ sodium acetate; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $1.8 \\times 10^{-5}$\nC: Buffer: $\\quad 0.10 \\mathrm{M}$ sodium dihydrogen phosphate / $0.10 \\mathrm{M}$ sodium hydrogen phosphate; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $6.2 \\times 10^{-8}$\nD: Buffer: $\\quad 0.10 \\mathrm{M}$ ammonium chloride / $0.10 \\mathrm{M}$ ammonia; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $5.6 \\times 10^{-10}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhen the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is $8 \\mathrm{ppm}\\left(8 \\mathrm{mg} \\mathrm{dm}^{-3}\\right)$.\n\nAfter one week of heavy rain, the following analysis is done on water from the bay. To a $50.00 \\mathrm{~cm}^{3}$ sample of bay water a few drops of a $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ solution are added. The sample is then titrated with $16.16 \\mathrm{~cm}^{3}$ of a $0.00164 \\mathrm{M} \\mathrm{AgNO}_{3}$ solution. After $\\mathrm{AgNO}_{3}$ solution has been added to the sample a bright red-orange precipitate forms.\n\n| Buffer | systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid |\n| :--- | :---: |\n|1. $\\quad 0.10 \\mathrm{M}$ lactic acid / $0.10 \\mathrm{M}$ sodium lactate | $1.4 \\times 10^{-4}$ |\n|2. $\\quad 0.10 \\mathrm{M}$ acetic acid / $0.10 \\mathrm{M}$ sodium acetate | $1.8 \\times 10^{-5}$ |\n|3. $\\quad 0.10 \\mathrm{M}$ sodium dihydrogen phosphate / $0.10 \\mathrm{M}$ sodium hydrogen phosphate | $6.2 \\times 10^{-8}$ |\n|4. $\\quad 0.10 \\mathrm{M}$ ammonium chloride / $0.10 \\mathrm{M}$ ammonia | $5.6 \\times 10^{-10}$ |\n\nA: Buffer: $\\quad 0.10 \\mathrm{M}$ lactic acid / $0.10 \\mathrm{M}$ sodium lactate; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $1.4 \\times 10^{-4}$\nB: Buffer: $\\quad 0.10 \\mathrm{M}$ acetic acid / $0.10 \\mathrm{M}$ sodium acetate; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $1.8 \\times 10^{-5}$\nC: Buffer: $\\quad 0.10 \\mathrm{M}$ sodium dihydrogen phosphate / $0.10 \\mathrm{M}$ sodium hydrogen phosphate; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $6.2 \\times 10^{-8}$\nD: Buffer: $\\quad 0.10 \\mathrm{M}$ ammonium chloride / $0.10 \\mathrm{M}$ ammonia; systems $\\mathrm{K}_{\\mathrm{a}}$ of weak acid: $5.6 \\times 10^{-10}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_328",
        "problem": "Which of the following reagents could be used to separate the metal ions in an aqueous mixture of $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}$ and $\\mathrm{AgNO}_{3}$ ?\nA: $\\mathrm{NH}_{3}$\nB: $\\mathrm{KOH}$\nC: $\\mathrm{N a C l}$\nD: $\\mathrm{HNO}_{3}$\nE: $\\mathrm{CaCO}_{3}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following reagents could be used to separate the metal ions in an aqueous mixture of $\\mathrm{Fe}\\left(\\mathrm{NO}_{3}\\right)_{3}$ and $\\mathrm{AgNO}_{3}$ ?\n\nA: $\\mathrm{NH}_{3}$\nB: $\\mathrm{KOH}$\nC: $\\mathrm{N a C l}$\nD: $\\mathrm{HNO}_{3}$\nE: $\\mathrm{CaCO}_{3}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_780",
        "problem": "已知: $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 的 $\\mathrm{pK}_{\\mathrm{a} 1}=1.22, \\mathrm{pK}_{\\mathrm{a} 2}=4.19, \\mathrm{CH}_{3} \\mathrm{COOH}$ 的 $\\mathrm{pK}_{\\mathrm{a}}=4.76$ (电离常数 $\\mathrm{K}$ 的负对数 $-1 \\mathrm{gK}=\\mathrm{pK})$ 。下列说法正确的是\nA: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{HC}_{2} \\mathrm{O}_{4}$ 和 $\\mathrm{CH}_{3} \\mathrm{COONH}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$前者小于后者\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KHC}_{2} \\mathrm{O}_{4}$ 溶液滴加氨水至中性: $\\left(\\mathrm{NH}_{4}{ }^{+}\\right)<\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{~K}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液滴加盐酸至 $\\mathrm{pH}=1.22: \\quad \\mathrm{C}\\left(\\mathrm{H}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)-3 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液中滴加 $\\mathrm{KHC}_{2} \\mathrm{O}_{4}$ 溶液至 $\\mathrm{PH}=4.76$; $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}-\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n已知: $25^{\\circ} \\mathrm{C}$ 时 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 的 $\\mathrm{pK}_{\\mathrm{a} 1}=1.22, \\mathrm{pK}_{\\mathrm{a} 2}=4.19, \\mathrm{CH}_{3} \\mathrm{COOH}$ 的 $\\mathrm{pK}_{\\mathrm{a}}=4.76$ (电离常数 $\\mathrm{K}$ 的负对数 $-1 \\mathrm{gK}=\\mathrm{pK})$ 。下列说法正确的是\n\nA: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NH}_{4} \\mathrm{HC}_{2} \\mathrm{O}_{4}$ 和 $\\mathrm{CH}_{3} \\mathrm{COONH}_{4}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$前者小于后者\nB: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{KHC}_{2} \\mathrm{O}_{4}$ 溶液滴加氨水至中性: $\\left(\\mathrm{NH}_{4}{ }^{+}\\right)<\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{~K}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液滴加盐酸至 $\\mathrm{pH}=1.22: \\quad \\mathrm{C}\\left(\\mathrm{H}^{+}\\right)-\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)-3 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 溶液中滴加 $\\mathrm{KHC}_{2} \\mathrm{O}_{4}$ 溶液至 $\\mathrm{PH}=4.76$; $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}-\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_9",
        "problem": "What is the oxidation state of nitrogen in magnesium nitride?\nA: -3\nB: -2\nC: +3\nD: +5\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the oxidation state of nitrogen in magnesium nitride?\n\nA: -3\nB: -2\nC: +3\nD: +5\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_33",
        "problem": "Which graph best describes the radial wavefunction of a $2 p$ orbital?\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich graph best describes the radial wavefunction of a $2 p$ orbital?\n\nA: [figure1]\nB: [figure2]\nC: [figure3]\nD: [figure4]\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-07.jpg?height=201&width=363&top_left_y=1759&top_left_x=1247",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-07.jpg?height=214&width=361&top_left_y=2059&top_left_x=1248",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-07.jpg?height=238&width=371&top_left_y=1727&top_left_x=1663",
            "https://cdn.mathpix.com/cropped/2024_03_06_49f2b155cb809a9525fcg-07.jpg?height=233&width=369&top_left_y=2011&top_left_x=1667"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_543",
        "problem": "镓 (Ga)的化学性质与铝相似, 电解精炼法提纯镓的原理如图。已知三种金属活动顺序为 $\\mathrm{Zn}>\\mathrm{Ga}>\\mathrm{Fe}$ 。下列说法错误的是\n\n[图1]\nA: 阳极泥的主要成分是铁和铜\nB: 若电压过高, 阴极可能会产生 $\\mathrm{H}_{2}$ 导致电解效率下降\nC: 电子流向为 $\\mathrm{N}$ 极 $\\rightarrow$ 粗 $\\mathrm{Ga} \\rightarrow \\mathrm{NaOH}(\\mathrm{aq}) \\rightarrow$ 高纯 $\\mathrm{Ga} \\rightarrow \\mathrm{M}$ 极\nD: 阴极反应为 $\\mathrm{Ga}^{3+}+3 \\mathrm{e}^{-}=\\mathrm{Ga}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n镓 (Ga)的化学性质与铝相似, 电解精炼法提纯镓的原理如图。已知三种金属活动顺序为 $\\mathrm{Zn}>\\mathrm{Ga}>\\mathrm{Fe}$ 。下列说法错误的是\n\n[图1]\n\nA: 阳极泥的主要成分是铁和铜\nB: 若电压过高, 阴极可能会产生 $\\mathrm{H}_{2}$ 导致电解效率下降\nC: 电子流向为 $\\mathrm{N}$ 极 $\\rightarrow$ 粗 $\\mathrm{Ga} \\rightarrow \\mathrm{NaOH}(\\mathrm{aq}) \\rightarrow$ 高纯 $\\mathrm{Ga} \\rightarrow \\mathrm{M}$ 极\nD: 阴极反应为 $\\mathrm{Ga}^{3+}+3 \\mathrm{e}^{-}=\\mathrm{Ga}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-72.jpg?height=522&width=785&top_left_y=1004&top_left_x=336"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_202",
        "problem": "Calculate the chemical amount (in mol or mmol) of $\\mathrm{KB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ precipitated.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the chemical amount (in mol or mmol) of $\\mathrm{KB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ precipitated.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mmol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mmol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_750",
        "problem": "某饱和一元醛和一元酮的混合物共 $3.0 \\mathrm{~g}$, 跟足量的银氨溶液反应可还原出 $16.2 \\mathrm{~g}$银。下列说法正确的是\nA: 混合物中一定有甲醛\nB: 混合物中可能含乙醛\nC: 醛与酮的质量比为 $3: 5$\nD: 该酮为丙酮\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n某饱和一元醛和一元酮的混合物共 $3.0 \\mathrm{~g}$, 跟足量的银氨溶液反应可还原出 $16.2 \\mathrm{~g}$银。下列说法正确的是\n\nA: 混合物中一定有甲醛\nB: 混合物中可能含乙醛\nC: 醛与酮的质量比为 $3: 5$\nD: 该酮为丙酮\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_148",
        "problem": "If the total pressure increases in each of the reaction mixtures below, for which reaction would the product yield remain unchanged at equilibrium?\nA: $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\leftrightharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$\nB: $2 \\mathrm{NO}(\\mathrm{g})+\\mathrm{Cl}_{2}(\\mathrm{~g}) \\leftrightharpoons 2 \\mathrm{NOCl}(\\mathrm{g})$\nC: $2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\leftrightharpoons 2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{S}_{2}(\\mathrm{~g})$\nD: $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\nE: $3 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{CO}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIf the total pressure increases in each of the reaction mixtures below, for which reaction would the product yield remain unchanged at equilibrium?\n\nA: $\\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\leftrightharpoons \\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g})$\nB: $2 \\mathrm{NO}(\\mathrm{g})+\\mathrm{Cl}_{2}(\\mathrm{~g}) \\leftrightharpoons 2 \\mathrm{NOCl}(\\mathrm{g})$\nC: $2 \\mathrm{H}_{2} \\mathrm{~S}(\\mathrm{~g}) \\leftrightharpoons 2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{S}_{2}(\\mathrm{~g})$\nD: $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\nE: $3 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{CO}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CH}_{4}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_563",
        "problem": "常温下, $K_{\\mathrm{a}}(\\mathrm{HCOOH})=1.77 \\times 10^{-4}, \\mathrm{~K}_{\\mathrm{a}}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=1.75 \\times 10^{-5}$,\n\n$K_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=1.76 \\times 10^{-5}$, 下列说法正确的是\nA: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOONa}$ 和 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中阳离子的物质的量浓度之和:前者小于后者\nB: 用相同浓度的 $\\mathrm{NaOH}$ 溶液分别滴定等体积 $\\mathrm{pH}$ 均为 3 的 $\\mathrm{HCOOH}$ 和 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液至终点, 消耗 $\\mathrm{NaOH}$ 溶液的体积相等\nC: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCOOH}$ 溶液与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合后的溶液中: $\\mathrm{c}\\left(\\mathrm{HCOO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}(\\mathrm{HCOOH})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸等体积混合后的溶液 $(\\mathrm{pH}<7)$ 中: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, $K_{\\mathrm{a}}(\\mathrm{HCOOH})=1.77 \\times 10^{-4}, \\mathrm{~K}_{\\mathrm{a}}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)=1.75 \\times 10^{-5}$,\n\n$K_{\\mathrm{b}}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=1.76 \\times 10^{-5}$, 下列说法正确的是\n\nA: 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{HCOONa}$ 和 $\\mathrm{NH}_{4} \\mathrm{Cl}$ 溶液中阳离子的物质的量浓度之和:前者小于后者\nB: 用相同浓度的 $\\mathrm{NaOH}$ 溶液分别滴定等体积 $\\mathrm{pH}$ 均为 3 的 $\\mathrm{HCOOH}$ 和 $\\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液至终点, 消耗 $\\mathrm{NaOH}$ 溶液的体积相等\nC: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HCOOH}$ 溶液与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合后的溶液中: $\\mathrm{c}\\left(\\mathrm{HCOO}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}(\\mathrm{HCOOH})+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COONa}$ 与 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 盐酸等体积混合后的溶液 $(\\mathrm{pH}<7)$ 中: $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{Cl}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1124",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nUsing the modern atomic mass of indium, calculate the maximum mass of indium oxide that could be formed from $1.00 \\mathrm{~g}$ of indium metal.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  the atomic mass of indium\n\nOne error is the mass of the element indium (In). To determine the atomic mass of indium, a known quantity of the metal was dissolved in acid, a solution of sodium hydroxide was added to precipitate indium hydroxide and then this was heated to form indium oxide, $\\ln _{2} \\mathrm{O}_{3}$.\n\nUsing the modern atomic mass of indium, calculate the maximum mass of indium oxide that could be formed from $1.00 \\mathrm{~g}$ of indium metal.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1311",
        "problem": "Nitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.Select the schematic energy diagram that is consistent with the proposed reaction mechanism and experimental rate law.\nA: ![]([figure1])\nB: ![]([figure2])\nC: ![]([figure3])\nD: ![]([figure4])\nE: ![]([figure5])\nF: ![]([figure6])\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\nNitrogen oxides, common pollutants in the ambient air, are primarily nitric oxide, $\\mathrm{NO}$, and nitrogen dioxide, $\\mathrm{NO}_{2}$. Atmospheric nitric oxide is produced mainly during thunderstorms and in the internal combustion engines. At high temperatures NO reacts with $\\mathrm{H}_{2}$ to produce nitrous oxide, $\\mathrm{N}_{2} \\mathrm{O}$, a greenhouse gas.\n\n$$\n2 \\mathrm{NO}(g)+\\mathrm{H}_{2}(g) \\rightarrow \\mathrm{N}_{2} \\mathrm{O}(g)+\\mathrm{H}_{2} \\mathrm{O}(g)\n$$\n\nTo study the kinetics of this reaction at $820^{\\circ} \\mathrm{C}$, i nitial rates for the formation of $\\mathrm{N}_{2} \\mathrm{O}$ were measured using various initial partial pressures of $\\mathrm{NO}$ and $\\mathrm{H}_{2}$.\n\n| Exp. | Initial pressure, torr |  | Initial rate of production of <br> $\\mathrm{N}_{2} \\mathrm{O}$, torr $\\mathrm{s}^{-1}$ |\n| :---: | :---: | :---: | :---: |\n|  | $p_{\\mathrm{NO}}$ | $p_{\\mathrm{H}_{2}}$ |  |\n| 1 | 120.0 | 60.0 | $8.66 \\cdot 10^{-2}$ |\n| 2 | 60.0 | 60.0 |  $2.17 \\cdot 10^{-2}$  |\n| 3 | 60.0 | 180.0 |$6.62 \\cdot 10^{-2}$  |\n\nThroughout this problem do not use concentrations. Use units of pressure (torr) and time in seconds.\n\nproblem:\nSelect the schematic energy diagram that is consistent with the proposed reaction mechanism and experimental rate law.\n\nA: ![]([figure1])\nB: ![]([figure2])\nC: ![]([figure3])\nD: ![]([figure4])\nE: ![]([figure5])\nF: ![]([figure6])\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E, F].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-113.jpg?height=520&width=417&top_left_y=1967&top_left_x=343",
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-113.jpg?height=523&width=422&top_left_y=1966&top_left_x=834",
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-113.jpg?height=520&width=406&top_left_y=1967&top_left_x=1339",
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-114.jpg?height=525&width=408&top_left_y=360&top_left_x=330",
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-114.jpg?height=525&width=422&top_left_y=357&top_left_x=834",
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-114.jpg?height=519&width=420&top_left_y=360&top_left_x=1338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_241",
        "problem": "Calculate the mass (in g) of lead(II) iodide produced.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the mass (in g) of lead(II) iodide produced.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of g, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "g"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1453",
        "problem": "$10.0 \\mathrm{~cm}^{3}$ of $0.50 \\mathrm{M} \\mathrm{HCl}$ and $10.0 \\mathrm{~cm}^{3}$ of $0.50 \\mathrm{M} \\mathrm{NaOH}$ solutions, both at the same temperature, are mixed in a calorimeter. A temperature increase of $\\Delta T$ is recorded. Estimate the temperature increase if $5.0 \\mathrm{~cm}^{3}$ of $0.50 \\mathrm{M} \\mathrm{NaOH}$ were used instead of 10.0 $\\mathrm{cm}^{3}$. Thermal I osses are negligible and the specific heats of both solutions are taken as equal.\nA: $(1 / 2) \\Delta T$\nB: $(2 / 3) \\Delta T$\nC: $(3 / 4) \\Delta T$\nD: $\\Delta T$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\n$10.0 \\mathrm{~cm}^{3}$ of $0.50 \\mathrm{M} \\mathrm{HCl}$ and $10.0 \\mathrm{~cm}^{3}$ of $0.50 \\mathrm{M} \\mathrm{NaOH}$ solutions, both at the same temperature, are mixed in a calorimeter. A temperature increase of $\\Delta T$ is recorded. Estimate the temperature increase if $5.0 \\mathrm{~cm}^{3}$ of $0.50 \\mathrm{M} \\mathrm{NaOH}$ were used instead of 10.0 $\\mathrm{cm}^{3}$. Thermal I osses are negligible and the specific heats of both solutions are taken as equal.\n\nA: $(1 / 2) \\Delta T$\nB: $(2 / 3) \\Delta T$\nC: $(3 / 4) \\Delta T$\nD: $\\Delta T$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_735",
        "problem": "化合物 $Y$ 具有保肝、抗炎、增强免疫等功效，可由 $X$ 制得。下列有关化合物 $X 、 Y$的说法正确的是\n[图1]\nA: 一定条件下 X 可发生氧化、取代、消去反应\nB: $1 \\mathrm{~mol} \\mathrm{Y}$ 最多能与 $4 \\mathrm{~mol} \\mathrm{NaOH}$ 反应\nC: $\\mathrm{X}$ 与足量 $\\mathrm{H}_{2}$ 反应后，每个产物分子中含有 8 个手性碳原子\nD: 等物质的量的 $\\mathrm{X} 、 \\mathrm{Y}$ 分别与足量 $\\mathrm{Br}_{2}$ 反应, 最多消耗 $\\mathrm{Br}_{2}$ 的物质的量相等\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n化合物 $Y$ 具有保肝、抗炎、增强免疫等功效，可由 $X$ 制得。下列有关化合物 $X 、 Y$的说法正确的是\n[图1]\n\nA: 一定条件下 X 可发生氧化、取代、消去反应\nB: $1 \\mathrm{~mol} \\mathrm{Y}$ 最多能与 $4 \\mathrm{~mol} \\mathrm{NaOH}$ 反应\nC: $\\mathrm{X}$ 与足量 $\\mathrm{H}_{2}$ 反应后，每个产物分子中含有 8 个手性碳原子\nD: 等物质的量的 $\\mathrm{X} 、 \\mathrm{Y}$ 分别与足量 $\\mathrm{Br}_{2}$ 反应, 最多消耗 $\\mathrm{Br}_{2}$ 的物质的量相等\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-73.jpg?height=422&width=1464&top_left_y=650&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-73.jpg?height=311&width=874&top_left_y=1895&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_717",
        "problem": "使用 SNCR 脱硝技术的原理是 $4 \\mathrm{NO}(\\mathrm{g})+4 \\mathrm{NH}_{3}(\\mathrm{~g})=4 \\mathrm{~N}_{2}(\\mathrm{~g})+6 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, 下图是其在密闭体系中研究反应条件对烟气脱硝效率的实验结果。下列说法正确的是\n[图1]\nA: 从图 1 判断，该反应的正反应方向是放热反应\nB: 从图 2 判断, 减少氨气的浓度有助于提高 $\\mathrm{NO}$ 的转化率\nC: 从图 1 判断, 脱硝的最佳温度约为 $925^{\\circ} \\mathrm{C}$\nD: 从图 2 判断, 综合考虑脱硝效率和运行成本最佳氨氮摩尔比应为 2.5\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n使用 SNCR 脱硝技术的原理是 $4 \\mathrm{NO}(\\mathrm{g})+4 \\mathrm{NH}_{3}(\\mathrm{~g})=4 \\mathrm{~N}_{2}(\\mathrm{~g})+6 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, 下图是其在密闭体系中研究反应条件对烟气脱硝效率的实验结果。下列说法正确的是\n[图1]\n\nA: 从图 1 判断，该反应的正反应方向是放热反应\nB: 从图 2 判断, 减少氨气的浓度有助于提高 $\\mathrm{NO}$ 的转化率\nC: 从图 1 判断, 脱硝的最佳温度约为 $925^{\\circ} \\mathrm{C}$\nD: 从图 2 判断, 综合考虑脱硝效率和运行成本最佳氨氮摩尔比应为 2.5\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-74.jpg?height=392&width=1122&top_left_y=580&top_left_x=358"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_761",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 某混合溶液中 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)=0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 由水电离出的 $\\mathrm{c}_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$的对数 $\\operatorname{lgc}_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}, \\lg \\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}$的关系如图所示。下列说法不正确的是\n\n[图1]\nA: 曲线 $\\mathrm{L}_{2}$ 表示 $\\operatorname{lgc}_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}$ 的变化关系\nB: X 点时存在 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)<\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)<\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nC: $\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=10^{-8}$\nD: $\\mathrm{Y}$ 点时溶液的 $\\mathrm{pH}=7$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 某混合溶液中 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)+\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)=0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 由水电离出的 $\\mathrm{c}_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$的对数 $\\operatorname{lgc}_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}, \\lg \\frac{\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}$的关系如图所示。下列说法不正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{L}_{2}$ 表示 $\\operatorname{lgc}_{\\text {水 }}\\left(\\mathrm{H}^{+}\\right)$与 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)}{\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)}$ 的变化关系\nB: X 点时存在 $\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)<\\mathrm{c}\\left(\\mathrm{HA}^{-}\\right)<\\mathrm{c}\\left(\\mathrm{A}^{2-}\\right)$\nC: $\\mathrm{K}_{\\mathrm{a} 2}\\left(\\mathrm{H}_{2} \\mathrm{~A}\\right)=10^{-8}$\nD: $\\mathrm{Y}$ 点时溶液的 $\\mathrm{pH}=7$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-058.jpg?height=383&width=571&top_left_y=168&top_left_x=343"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_435",
        "problem": "在恒压密闭容器中, 充入起始量一定的 $\\mathrm{CO}_{2}$ 和 $\\mathrm{H}_{2}$, 主要发生下列反应:\n\n$\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}=41.2 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n$2 \\mathrm{CO}_{2}(\\mathrm{~g})+6 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}=-122.5 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n达平衡时, $\\mathrm{CO}_{2}$ 转化率和 $\\mathrm{CO}$ 的选择性 $\\left(\\mathrm{CO}\\right.$ 的选择性 $=\\frac{\\mathrm{n}_{\\text {生成 }}(\\mathrm{CO})}{\\mathrm{n}_{\\text {反立 }}\\left(\\mathrm{CO}_{2}\\right)}$ 随温度的变化如图所示, 下列说法不正确的是。\n\n[图1]\nA: 图中曲线(1表示平衡时 $\\mathrm{CO}$ 的选择性随温度的变化\nB: 温度一定, 通过增大压强能提高 $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ 的平衡产率\nC: 一定温度下, 增大 $\\frac{\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)}{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}$, 一定能降低 $\\mathrm{CO}_{2}$ 平衡转化率\nD: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 起始投入 $2 \\mathrm{~mol} \\mathrm{CO}_{2} 、 6 \\mathrm{~mol} \\mathrm{H}_{2}$, 达平衡时生成 $$ \\mathrm{n}\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)=\\left(2 \\mathrm{a} \\%-2 \\mathrm{a}^{2} \\times 10^{-4}\\right) \\mathrm{mol} $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n在恒压密闭容器中, 充入起始量一定的 $\\mathrm{CO}_{2}$ 和 $\\mathrm{H}_{2}$, 主要发生下列反应:\n\n$\\mathrm{CO}_{2}(\\mathrm{~g})+\\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CO}(\\mathrm{g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}=41.2 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n$2 \\mathrm{CO}_{2}(\\mathrm{~g})+6 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\quad \\Delta \\mathrm{H}=-122.5 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\n\n达平衡时, $\\mathrm{CO}_{2}$ 转化率和 $\\mathrm{CO}$ 的选择性 $\\left(\\mathrm{CO}\\right.$ 的选择性 $=\\frac{\\mathrm{n}_{\\text {生成 }}(\\mathrm{CO})}{\\mathrm{n}_{\\text {反立 }}\\left(\\mathrm{CO}_{2}\\right)}$ 随温度的变化如图所示, 下列说法不正确的是。\n\n[图1]\n\nA: 图中曲线(1表示平衡时 $\\mathrm{CO}$ 的选择性随温度的变化\nB: 温度一定, 通过增大压强能提高 $\\mathrm{CH}_{3} \\mathrm{CH}_{3}$ 的平衡产率\nC: 一定温度下, 增大 $\\frac{\\mathrm{n}\\left(\\mathrm{CO}_{2}\\right)}{\\mathrm{n}\\left(\\mathrm{H}_{2}\\right)}$, 一定能降低 $\\mathrm{CO}_{2}$ 平衡转化率\nD: $\\mathrm{T}^{\\circ} \\mathrm{C}$ 时, 起始投入 $2 \\mathrm{~mol} \\mathrm{CO}_{2} 、 6 \\mathrm{~mol} \\mathrm{H}_{2}$, 达平衡时生成 $$ \\mathrm{n}\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)=\\left(2 \\mathrm{a} \\%-2 \\mathrm{a}^{2} \\times 10^{-4}\\right) \\mathrm{mol} $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-057.jpg?height=529&width=711&top_left_y=158&top_left_x=341",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-058.jpg?height=228&width=959&top_left_y=443&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1352",
        "problem": "The energy of stable states of the hydrogen atom is given by: $E_{n}=-2.18 \\times 10^{-18} / \\mathrm{n}^{2}[\\mathrm{~J}]$ where $\\mathrm{n}$ denotes the principal quantum number.\n\nCalculate the de Broglie wavelength of the electrons emitted from a copper crystal when irradiated by photons from the first line and the sixth line of the Lyman series.\n\n$h=6.6256 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s} ; \\quad m_{\\mathrm{e}}=9.1091 \\times 10^{-31} \\mathrm{~kg} ; \\quad c=2.99792 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}$",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nThe energy of stable states of the hydrogen atom is given by: $E_{n}=-2.18 \\times 10^{-18} / \\mathrm{n}^{2}[\\mathrm{~J}]$ where $\\mathrm{n}$ denotes the principal quantum number.\n\nCalculate the de Broglie wavelength of the electrons emitted from a copper crystal when irradiated by photons from the first line and the sixth line of the Lyman series.\n\n$h=6.6256 \\times 10^{-34} \\mathrm{~J} \\mathrm{~s} ; \\quad m_{\\mathrm{e}}=9.1091 \\times 10^{-31} \\mathrm{~kg} ; \\quad c=2.99792 \\times 10^{8} \\mathrm{~m} \\mathrm{~s}^{-1}$\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [$\\Delta E_{2 \\rightarrow 1}$, $\\Delta E_{7 \\rightarrow 1}$].\nTheir units are, in order, [$\\AA $, $\\AA $], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "$\\AA $",
            "$\\AA $"
        ],
        "answer_sequence": [
            "$\\Delta E_{2 \\rightarrow 1}$",
            "$\\Delta E_{7 \\rightarrow 1}$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_477",
        "problem": "高分子有机物聚吡咯(PPy)是一种性能优异的光敏型半导体, 其制成的纳米管在紫\n\n外光照射、关闭周期内会发生如下反应:\n[图1]\n电流。某科研组使用 PPy 构建了一种浓差电池, 用来提取天然水中的氢能, 其构造如图所示。下列叙述正确的是\n\n[图2]\nA: $a$ 为负极, $b$ 为正极\nB: $\\mathrm{b}$ 极电极方程式为 $2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow$\nC: C. 纳米管道中的离子电流由 PPy 阳离子、 $\\mathrm{H}^{+} 、 \\mathrm{Na}^{+} 、 \\mathrm{Cl}^{-}$的定向移动形成\nD: 照射一段时间后关闭光源，纳米管道中仍能存在微弱电流\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n高分子有机物聚吡咯(PPy)是一种性能优异的光敏型半导体, 其制成的纳米管在紫\n\n外光照射、关闭周期内会发生如下反应:\n[图1]\n电流。某科研组使用 PPy 构建了一种浓差电池, 用来提取天然水中的氢能, 其构造如图所示。下列叙述正确的是\n\n[图2]\n\nA: $a$ 为负极, $b$ 为正极\nB: $\\mathrm{b}$ 极电极方程式为 $2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=\\mathrm{H}_{2} \\uparrow$\nC: C. 纳米管道中的离子电流由 PPy 阳离子、 $\\mathrm{H}^{+} 、 \\mathrm{Na}^{+} 、 \\mathrm{Cl}^{-}$的定向移动形成\nD: 照射一段时间后关闭光源，纳米管道中仍能存在微弱电流\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-084.jpg?height=410&width=1376&top_left_y=1412&top_left_x=340",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-084.jpg?height=306&width=534&top_left_y=2006&top_left_x=338",
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-085.jpg?height=192&width=774&top_left_y=732&top_left_x=948"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_596",
        "problem": "下列溶液中微粒的物质的量浓度关系正确的是\nA: 常温下 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的下列溶液中: (1) $\\mathrm{NH}_{4} \\mathrm{Al}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 、 (2) $\\mathrm{NH}_{4} \\mathrm{Cl} 、(3)$ $\\mathrm{CH}_{3} \\mathrm{COONH}_{4}, \\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$由大到小的顺序为(2)>(1)>(3)\nB: 常温下 $0.4 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液和 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合后溶液显酸性, 则溶液中粒子浓度由大到小的顺序为 $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}(\\mathrm{Na}$ $\\left.{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液与 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 溶液等体积混合所得溶液中: $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的氨水与 $0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液等体积混合所得溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)$ $+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列溶液中微粒的物质的量浓度关系正确的是\n\nA: 常温下 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的下列溶液中: (1) $\\mathrm{NH}_{4} \\mathrm{Al}\\left(\\mathrm{SO}_{4}\\right)_{2}$ 、 (2) $\\mathrm{NH}_{4} \\mathrm{Cl} 、(3)$ $\\mathrm{CH}_{3} \\mathrm{COONH}_{4}, \\mathrm{c}\\left(\\mathrm{NH}_{4}{ }^{+}\\right)$由大到小的顺序为(2)>(1)>(3)\nB: 常温下 $0.4 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{CH}_{3} \\mathrm{COOH}$ 溶液和 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液等体积混合后溶液显酸性, 则溶液中粒子浓度由大到小的顺序为 $\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COO}^{-}\\right)>\\mathrm{c}(\\mathrm{Na}$ $\\left.{ }^{+}\\right)>\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{COOH}\\right)>\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nC: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液与 $0.2 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaHCO}_{3}$ 溶液等体积混合所得溶液中: $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+3 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}^{+}\\right)$\nD: $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的氨水与 $0.05 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液等体积混合所得溶液中: $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)$ $+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=2 \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1074",
        "problem": "When ammonium dichromate( $\\mathrm{VI})$ is added gradually to molten ammonium thiocyanate, Reinecke's salt is formed. It has the formula $\\mathrm{NH}_{4}\\left[\\mathrm{Cr}(\\mathrm{SCN})_{x}\\left(\\mathrm{NH}_{3}\\right)_{y}\\right]$ and the following composition by mass:\n\n$$\n\\begin{array}{ll}\n\\mathrm{Cr} & 15.5 \\% \\\\\n\\mathrm{~S} & 38.15 \\% \\\\\n\\mathrm{~N} & 29.2 \\%\n\\end{array}\n$$\n\nCalculate the values of $x$ and $y$ in the above formula.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nWhen ammonium dichromate( $\\mathrm{VI})$ is added gradually to molten ammonium thiocyanate, Reinecke's salt is formed. It has the formula $\\mathrm{NH}_{4}\\left[\\mathrm{Cr}(\\mathrm{SCN})_{x}\\left(\\mathrm{NH}_{3}\\right)_{y}\\right]$ and the following composition by mass:\n\n$$\n\\begin{array}{ll}\n\\mathrm{Cr} & 15.5 \\% \\\\\n\\mathrm{~S} & 38.15 \\% \\\\\n\\mathrm{~N} & 29.2 \\%\n\\end{array}\n$$\n\nCalculate the values of $x$ and $y$ in the above formula.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [the value of $x$, the value of $y$].\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "the value of $x$",
            "the value of $y$"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_327",
        "problem": "Which of the following has the highest melting point?\nA: $\\mathrm{I}_{2}(\\mathrm{~s})$\nB: $\\mathrm{C}_{60}(\\mathrm{~s})$\nC: $\\mathrm{NaCl}(\\mathrm{s})$\nD: $\\operatorname{LiF}(\\mathrm{s})$\nE: Xe(s)\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following has the highest melting point?\n\nA: $\\mathrm{I}_{2}(\\mathrm{~s})$\nB: $\\mathrm{C}_{60}(\\mathrm{~s})$\nC: $\\mathrm{NaCl}(\\mathrm{s})$\nD: $\\operatorname{LiF}(\\mathrm{s})$\nE: Xe(s)\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_287",
        "problem": "If the concentration of the $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ solution is actually $0.1221 \\mathrm{~mol} \\mathrm{~L}^{-1}$, calculate the chemical amount (in mol or mmol) of $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ added.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nIf the concentration of the $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ solution is actually $0.1221 \\mathrm{~mol} \\mathrm{~L}^{-1}$, calculate the chemical amount (in mol or mmol) of $\\mathrm{NaB}\\left(\\mathrm{C}_{6} \\mathrm{H}_{5}\\right)_{4}$ added.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mmol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mmol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_692",
        "problem": "两种气态烃以任意比例混合, $105^{\\circ} \\mathrm{C}$ 时, $1 \\mathrm{~L}$ 该混合烃与 $9 \\mathrm{~L} \\mathrm{O}_{2}$ 混合, 充分燃烧后恢复到原状况, 所得气体体积仍为 $10 \\mathrm{~L}$ 。下列各组混合烃中不符合此条件的是( )\nA: $\\mathrm{CH}_{4} 、 \\mathrm{C}_{2} \\mathrm{H}_{4}$\nB: $\\mathrm{CH}_{4} 、 \\mathrm{C}_{3} \\mathrm{H}_{6}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{4} 、 \\mathrm{C}_{3} \\mathrm{H}_{4}$\nD: $\\mathrm{CH}_{4} 、 \\mathrm{C}_{3} \\mathrm{H}_{6}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n两种气态烃以任意比例混合, $105^{\\circ} \\mathrm{C}$ 时, $1 \\mathrm{~L}$ 该混合烃与 $9 \\mathrm{~L} \\mathrm{O}_{2}$ 混合, 充分燃烧后恢复到原状况, 所得气体体积仍为 $10 \\mathrm{~L}$ 。下列各组混合烃中不符合此条件的是( )\n\nA: $\\mathrm{CH}_{4} 、 \\mathrm{C}_{2} \\mathrm{H}_{4}$\nB: $\\mathrm{CH}_{4} 、 \\mathrm{C}_{3} \\mathrm{H}_{6}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{4} 、 \\mathrm{C}_{3} \\mathrm{H}_{4}$\nD: $\\mathrm{CH}_{4} 、 \\mathrm{C}_{3} \\mathrm{H}_{6}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1201",
        "problem": "Which one of the following equations must be used for the exact calculation of $\\left[\\mathrm{H}^{+}\\right]$of an aqueous $\\mathrm{HCl}$ solution at any concentration $c_{\\mathrm{HCl}}$ ? $\\left(K_{w}=1 \\times 10^{-14}\\right)$.\nA: $\\left[\\mathrm{H}^{+}\\right]=c_{\\mathrm{HCl}}$\nB: $\\left[\\mathrm{H}^{+}\\right]=c_{\\mathrm{HCl}}+K_{w} /\\left[\\mathrm{H}^{+}\\right]$\nC: $\\left[\\mathrm{H}^{+}\\right]=C_{\\mathrm{HCl}}+K_{w}$\nD: $\\left[\\mathrm{H}^{+}\\right]=c_{\\mathrm{HCl}}-K_{w} /\\left[\\mathrm{H}^{+}\\right]$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich one of the following equations must be used for the exact calculation of $\\left[\\mathrm{H}^{+}\\right]$of an aqueous $\\mathrm{HCl}$ solution at any concentration $c_{\\mathrm{HCl}}$ ? $\\left(K_{w}=1 \\times 10^{-14}\\right)$.\n\nA: $\\left[\\mathrm{H}^{+}\\right]=c_{\\mathrm{HCl}}$\nB: $\\left[\\mathrm{H}^{+}\\right]=c_{\\mathrm{HCl}}+K_{w} /\\left[\\mathrm{H}^{+}\\right]$\nC: $\\left[\\mathrm{H}^{+}\\right]=C_{\\mathrm{HCl}}+K_{w}$\nD: $\\left[\\mathrm{H}^{+}\\right]=c_{\\mathrm{HCl}}-K_{w} /\\left[\\mathrm{H}^{+}\\right]$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_471",
        "problem": "辅酶 $\\mathrm{Q}_{10}$ 具有预防动脉硬化的功效, 其结构简式如下。下列有关辅酶 $\\mathrm{Q}_{10}$ 的说法正确的是\n\n[图1]\nA: 分子式为 $\\mathrm{C}_{60} \\mathrm{H}_{90} \\mathrm{O}_{4}$\nB: 分子中含有 14 个甲基\nC: 分子中的四个氧原子一定在同一平面\nD: 可发生加成反应，不能发生取代反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n辅酶 $\\mathrm{Q}_{10}$ 具有预防动脉硬化的功效, 其结构简式如下。下列有关辅酶 $\\mathrm{Q}_{10}$ 的说法正确的是\n\n[图1]\n\nA: 分子式为 $\\mathrm{C}_{60} \\mathrm{H}_{90} \\mathrm{O}_{4}$\nB: 分子中含有 14 个甲基\nC: 分子中的四个氧原子一定在同一平面\nD: 可发生加成反应，不能发生取代反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-25.jpg?height=294&width=525&top_left_y=1435&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1466",
        "problem": "Beach sand mineral, monazite, is a rich source of thorium, available in large quantities in the state of Kerala in India. A typical monazite sample contains about $9 \\%$ $\\mathrm{ThO}_{2}$ and $0.35 \\% \\mathrm{U}_{3} \\mathrm{O}_{8} \\cdot{ }^{208} \\mathrm{~Pb}$ a ${ }^{206} \\mathrm{~Pb}$ are the stable end-products in the radioactive decay series of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$, respectively. All the lead $(\\mathrm{Pb})$ found in monazite is of radiogenic origin.\n\nThe isotopic atom ratio ${ }^{208} \\mathrm{~Pb} /{ }^{232} \\mathrm{Th}$, measured mass spectrometrically, in a monazite sample was found to be 0.104 . The half-lives of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U} 1.41 \\times 10^{10}$ years and $4.47 \\times 10^{9}$ years, respectively. Assume that ${ }^{208} \\mathrm{~Pb},{ }^{206} \\mathrm{~Pb},{ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$ remained entirely in the monazite sample since the formation of monazite mineral.\n\nCalculate the age (time elapsed since its formation) of the monazite sample.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBeach sand mineral, monazite, is a rich source of thorium, available in large quantities in the state of Kerala in India. A typical monazite sample contains about $9 \\%$ $\\mathrm{ThO}_{2}$ and $0.35 \\% \\mathrm{U}_{3} \\mathrm{O}_{8} \\cdot{ }^{208} \\mathrm{~Pb}$ a ${ }^{206} \\mathrm{~Pb}$ are the stable end-products in the radioactive decay series of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$, respectively. All the lead $(\\mathrm{Pb})$ found in monazite is of radiogenic origin.\n\nThe isotopic atom ratio ${ }^{208} \\mathrm{~Pb} /{ }^{232} \\mathrm{Th}$, measured mass spectrometrically, in a monazite sample was found to be 0.104 . The half-lives of ${ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U} 1.41 \\times 10^{10}$ years and $4.47 \\times 10^{9}$ years, respectively. Assume that ${ }^{208} \\mathrm{~Pb},{ }^{206} \\mathrm{~Pb},{ }^{232} \\mathrm{Th}$ and ${ }^{238} \\mathrm{U}$ remained entirely in the monazite sample since the formation of monazite mineral.\n\nCalculate the age (time elapsed since its formation) of the monazite sample.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of years, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "years"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1189",
        "problem": "The unit cell is the smallest repeating unit in a crystal structure. The unit cell of a gold crystal is found by X-ray diffraction to have the face-centred cubic unit structure (i.e. where the centre of an atom is located at each corner of a cube and in the middle of each face). The side of the unit cell is found to be $0.408 \\mathrm{~nm}$.Hence calculate the mass of a gold atom and the Avogadro constant, given that the relative atomic mass of $\\mathrm{Au}$ is 196.97.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\nHere is some context information for this question, which might assist you in solving it:\nThe unit cell is the smallest repeating unit in a crystal structure. The unit cell of a gold crystal is found by X-ray diffraction to have the face-centred cubic unit structure (i.e. where the centre of an atom is located at each corner of a cube and in the middle of each face). The side of the unit cell is found to be $0.408 \\mathrm{~nm}$.\n\nproblem:\nHence calculate the mass of a gold atom and the Avogadro constant, given that the relative atomic mass of $\\mathrm{Au}$ is 196.97.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [ the mass of a gold atom, the Avogadro constant].\nTheir units are, in order, [kg, $\\mathrm{~mol}^{-1}$], but units shouldn't be included in your concluded answer.\nTheir answer types are, in order, [numerical value, numerical value].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            "kg",
            "$\\mathrm{~mol}^{-1}$"
        ],
        "answer_sequence": [
            " the mass of a gold atom",
            "the Avogadro constant"
        ],
        "type_sequence": [
            "NV",
            "NV"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_83",
        "problem": "What is the concentration of calcium ions in $350 . \\mathrm{mL}$ of an aqueous solution containing $7.50 \\mathrm{~g} \\mathrm{CaCl}_{2}$ ?\nA: $0.0676 \\mathrm{M}$\nB: $0.193 \\mathrm{M}$\nC: $0.284 \\mathrm{M}$\nD: $0.535 \\mathrm{M}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the concentration of calcium ions in $350 . \\mathrm{mL}$ of an aqueous solution containing $7.50 \\mathrm{~g} \\mathrm{CaCl}_{2}$ ?\n\nA: $0.0676 \\mathrm{M}$\nB: $0.193 \\mathrm{M}$\nC: $0.284 \\mathrm{M}$\nD: $0.535 \\mathrm{M}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_753",
        "problem": "实验小组利用传感器探究 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 的性质。\n\n[图1]\n\n下列分析不正确的是 ( )\nA: (1)与(2)的实验数据基本相同, 说明(2)中的 $\\mathrm{OH}^{-}$未参与该反应\nB: 加入试剂体积相同时, (2)所得沉淀质量等于(3)所得沉淀质量\nC: b 点对应溶液中水的电离程度小于 $\\mathrm{c}$ 点对应溶液中水的电离程度\nD: 从起始到 $\\mathrm{a}$ 点过程中反应的离子方程式为: $$ \\mathrm{Ca}^{2+}+2 \\mathrm{OH}^{-}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{CaCO}_{3} \\downarrow+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{3}^{2-} $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n实验小组利用传感器探究 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 和 $\\mathrm{NaHCO}_{3}$ 的性质。\n\n[图1]\n\n下列分析不正确的是 ( )\n\nA: (1)与(2)的实验数据基本相同, 说明(2)中的 $\\mathrm{OH}^{-}$未参与该反应\nB: 加入试剂体积相同时, (2)所得沉淀质量等于(3)所得沉淀质量\nC: b 点对应溶液中水的电离程度小于 $\\mathrm{c}$ 点对应溶液中水的电离程度\nD: 从起始到 $\\mathrm{a}$ 点过程中反应的离子方程式为: $$ \\mathrm{Ca}^{2+}+2 \\mathrm{OH}^{-}+2 \\mathrm{HCO}_{3}^{-}=\\mathrm{CaCO}_{3} \\downarrow+2 \\mathrm{H}_{2} \\mathrm{O}+\\mathrm{CO}_{3}^{2-} $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-066.jpg?height=976&width=1445&top_left_y=577&top_left_x=311"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1179",
        "problem": "This question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nCalculate the percentage decrease in mass when copper(II) oxide is reduced to copper metal.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about oxidation, reduction and ferrets\n\nThe term 'oxidation' was first used to suggest processes when oxygen was introduced into a substance, but later came, more generally, to be associated with the loss of electrons from species.\n\nThe term 'reduction' was initially used when metal oxides were 'reduced' to the metal since the overall mass was found to decrease in the process.\n\n'Oxidation states' are the hypothetical charges that atoms would have if the bonds in the compounds were totally ionic with no covalent contribution; this assumption is never true in real compounds.\n\nCalculate the percentage decrease in mass when copper(II) oxide is reduced to copper metal.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of percentage, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "percentage"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_359",
        "problem": "Which of the following are true for a spontaneous process in a system at constant temperature and pressure?\nI. $\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surr }}>0$\nII. $\\Delta G_{\\text {sys }}<0$\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich of the following are true for a spontaneous process in a system at constant temperature and pressure?\nI. $\\Delta S_{\\text {sys }}+\\Delta S_{\\text {surr }}>0$\nII. $\\Delta G_{\\text {sys }}<0$\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_355",
        "problem": "A student determines the acetic acid concentration of a sample of distilled vinegar by titration of $25.00 \\mathrm{~mL}$ of the vinegar with standardized sodium hydroxide solution using phenolphthalein as an indicator. Which error will give an acetic acid content for the vinegar that is too low?\nA: Some of the vinegar is spilled when being transferred from the volumetric flask to the titration flask.\nB: The $\\mathrm{NaOH}$ solution is allowed to stand for a prolonged period after standardization and absorbs carbon dioxide from the air.\nC: The endpoint is recorded when the solution turns dark red instead of faint pink.\nD: The vinegar is diluted with distilled water in the titration flask before the $\\mathrm{NaOH}$ solution is added.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA student determines the acetic acid concentration of a sample of distilled vinegar by titration of $25.00 \\mathrm{~mL}$ of the vinegar with standardized sodium hydroxide solution using phenolphthalein as an indicator. Which error will give an acetic acid content for the vinegar that is too low?\n\nA: Some of the vinegar is spilled when being transferred from the volumetric flask to the titration flask.\nB: The $\\mathrm{NaOH}$ solution is allowed to stand for a prolonged period after standardization and absorbs carbon dioxide from the air.\nC: The endpoint is recorded when the solution turns dark red instead of faint pink.\nD: The vinegar is diluted with distilled water in the titration flask before the $\\mathrm{NaOH}$ solution is added.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1287",
        "problem": "One of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nHow much energy in MeV is released in the complete chain?\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nOne of naturally occurring radioactive decay series begins with ${ }_{90}^{232}$ Th and ends with a stable ${ }_{82}^{208} \\mathrm{~Pb}$.\n\nHow much energy in MeV is released in the complete chain?\n\nThe necessary atomic masses are :\n\n${ }_{2}^{4} \\mathrm{He}=4.00260 \\mathrm{u}, \\quad{ }_{82}^{208} \\mathrm{~Pb}=207.97664 \\mathrm{u}, \\quad{ }_{90}^{232} \\mathrm{Th}=232.03805 \\mathrm{u} ;$ and $1 \\mathrm{u}=931.5 \\mathrm{MeV}$\n\n$1 \\mathrm{MeV}=1.602 \\times 10^{-13} \\mathrm{~J}$\n\n$N_{A}=6.022 \\times 10^{23} \\mathrm{~mol}^{-1}$\n\nThe molar volume of an ideal gas at $0^{\\circ} \\mathrm{C}$ and $1 \\mathrm{~atm}$ is $22.4 \\mathrm{dm}^{3} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{MeV}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{MeV}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1254",
        "problem": "Solutions containing $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and/or $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ are titrated with a strong base standard solution. Associate the contents of these solutions with the titration curves ( $\\mathrm{pH}$ vs. volume of titrant) shown in the figure:\n\n(For $\\mathrm{H}_{3} \\mathrm{PO}_{4}: p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\n\n[figure1]\n\nVolume of titrant $\\left(\\mathrm{cm}^{3}\\right)$\n\nThe sample contains both in a mole ratio $\\mathrm{H}_{3} \\mathrm{PO}_{4}: \\mathrm{NaH}_{2} \\mathrm{PO}_{4}=1: 1$:\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nSolutions containing $\\mathrm{H}_{3} \\mathrm{PO}_{4}$ and/or $\\mathrm{NaH}_{2} \\mathrm{PO}_{4}$ are titrated with a strong base standard solution. Associate the contents of these solutions with the titration curves ( $\\mathrm{pH}$ vs. volume of titrant) shown in the figure:\n\n(For $\\mathrm{H}_{3} \\mathrm{PO}_{4}: p K_{1}=2.1, p K_{2}=7.2, p K_{3}=12.0$ )\n\n[figure1]\n\nVolume of titrant $\\left(\\mathrm{cm}^{3}\\right)$\n\nThe sample contains both in a mole ratio $\\mathrm{H}_{3} \\mathrm{PO}_{4}: \\mathrm{NaH}_{2} \\mathrm{PO}_{4}=1: 1$:\n\nA: Curve A\nB: Curve B\nC: Curve C\nD: Curve D\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-459.jpg?height=348&width=1413&top_left_y=934&top_left_x=273"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_666",
        "problem": "一种将燃料电池与电解池组合制备 $\\mathrm{KMnO}_{4}$ 的装置如图所示 ( 电极甲、乙、丙、丁均为惰性电极）。该装置工作时，下列说法不正确的是\n\n[图1]\nA: 甲为正极, 丙为阴极\nB: 丁极的电极反应式为 $\\mathrm{MnO}_{4}^{2-}-\\mathrm{e}^{-}=\\mathrm{MnO}_{4}^{-}$\nC: $\\mathrm{KOH}$ 溶液的质量分数: $\\mathrm{c} \\%>\\mathrm{a} \\%>b \\%$\nD: 标准状况下, 乙电极上每消耗 $22.4 \\mathrm{~L}$ 气体时, 理论上有 $4 \\mathrm{molK}^{+}$移入阴极区\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n一种将燃料电池与电解池组合制备 $\\mathrm{KMnO}_{4}$ 的装置如图所示 ( 电极甲、乙、丙、丁均为惰性电极）。该装置工作时，下列说法不正确的是\n\n[图1]\n\nA: 甲为正极, 丙为阴极\nB: 丁极的电极反应式为 $\\mathrm{MnO}_{4}^{2-}-\\mathrm{e}^{-}=\\mathrm{MnO}_{4}^{-}$\nC: $\\mathrm{KOH}$ 溶液的质量分数: $\\mathrm{c} \\%>\\mathrm{a} \\%>b \\%$\nD: 标准状况下, 乙电极上每消耗 $22.4 \\mathrm{~L}$ 气体时, 理论上有 $4 \\mathrm{molK}^{+}$移入阴极区\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-44.jpg?height=514&width=1456&top_left_y=1682&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1417",
        "problem": "Lactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nCalculate the value of the equilibrium constant for the reaction between lactic acid and hydrogen carbonate.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is an expression.\n\nproblem:\nLactic acid is formed in the muscles during intense activity (anaerobic metabolism). In the blood, lactic acid is neutralized by reaction with hydrogen carbonate. This will be illustrated by the following calculations:\n\nLactic acid written as HL is monoprotic, and the acid dissociation constant is $K_{\\mathrm{HL}}=1.4 \\times 10^{-4}$.\n\nThe acid dissociation constants for carbonic acid are: $K_{\\mathrm{a} 1}=4.5 \\times 10^{-7}$ and $K_{\\mathrm{a} 2}=$ $4.7 \\times 10^{-11}$. All carbon dioxide remains dissolved during the reactions.\n\nCalculate the value of the equilibrium constant for the reaction between lactic acid and hydrogen carbonate.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is an expression without equals signs, e.g. ANSWER=\\frac{1}{2} g t^2",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "EX",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_238",
        "problem": "Which of the following molecules contain more than 3 lone pairs of valence electrons? Select all that apply.\nA: $\\mathrm{H}_{2}$\nB: $\\mathrm{CH}_{4}$\nC: $\\mathrm{Cl}_{2} \\mathrm{O}$\nD: $\\mathrm{CO}_{2}$\nE: $\\mathrm{HCl}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (more than one correct answer).\n\nproblem:\nWhich of the following molecules contain more than 3 lone pairs of valence electrons? Select all that apply.\n\nA: $\\mathrm{H}_{2}$\nB: $\\mathrm{CH}_{4}$\nC: $\\mathrm{Cl}_{2} \\mathrm{O}$\nD: $\\mathrm{CO}_{2}$\nE: $\\mathrm{HCl}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be two or more of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_438",
        "problem": "下列方案设计、现象和结论都正确的是\n\n| 目的 | 方案设计 | 现象和结论 |\n| :---: | :---: | :---: |\n| $\\left\\{\\begin{array}{l}\\text { 鉴别同为白色粉 } \\\\ \\text { 末状晶体的尿素 } \\\\ {\\left[\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}\\right] \\text { 和氯 }} \\\\ \\text { 化铵 }\\left(\\mathrm{NH}_{4} \\mathrm{Cl}\\right)\\end{array}\\right.$ | 分别取少量晶体于试管中, 加入 <br> 足量浓 $\\mathrm{NaOH}$ 溶液加热, 在试管 <br> 口放置湿润的红色石荵试纸, 观 <br> 察试纸是否变色 | 若试纸不变蓝色, 说明该晶体 <br> 为尿素, 若试纸变蓝色, 说明 <br> 该晶体为氯化铵 |\n| 检验钠盐试剂柜 <br> 中某瓶标签残破 <br> 的纯净白色晶体 <br> 是不是 $\\mathrm{NaHCO}_{3}$ | 取少量晶体于试管中, 加热, 观 <br> 察现象; 冷却后加入足量盐酸, <br> 观察是否产生无色无味气体, 并 <br> 通入澄清石灰水 | 若加热时试管口出现小水珠, <br> 加盐酸后产生无色无味气体, <br> 且能使石灰水变浑浊, 则该试 <br> 剂为 $\\mathrm{NaHCO}_{3}$ |\n| $\\left\\{\\begin{array}{l}\\text { 探究人体汗液中 } \\\\ \\text { 是否含有氨基酸 }\\end{array}\\right.$ | 取 $1 \\mathrm{~mL}$ 汗液, 加入 5 滴 $0.1 \\%$ 茚三 <br> 酮溶液加热至沸腾, 观察现象 | 若溶液呈现紫色, 则汗液中可 <br> 能含有氨基酸, 但也可能是含 <br> 有蛋白质 |\n| [图1] | 在试管甲中加入 $1 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot L^{-1}$ <br> 高锰酸钾溶液和 $2 \\mathrm{~mL}$ 蒸馏水, 试 <br> 管乙中加入 $2 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 高锰 <br> 酸钾溶液和 $1 \\mathrm{~mL}$ 蒸馏水, 再同时 <br> 向两试管加入 $1 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 草 <br> 酸溶液, 观察现象 | 若试管乙中溶液紫色先褪去, <br> 则反应物浓度增大时, 反应速 <br> 率增大 |\nA: A\nB: B\nC: C\nD: D\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列方案设计、现象和结论都正确的是\n\n| 目的 | 方案设计 | 现象和结论 |\n| :---: | :---: | :---: |\n| $\\left\\{\\begin{array}{l}\\text { 鉴别同为白色粉 } \\\\ \\text { 末状晶体的尿素 } \\\\ {\\left[\\mathrm{CO}\\left(\\mathrm{NH}_{2}\\right)_{2}\\right] \\text { 和氯 }} \\\\ \\text { 化铵 }\\left(\\mathrm{NH}_{4} \\mathrm{Cl}\\right)\\end{array}\\right.$ | 分别取少量晶体于试管中, 加入 <br> 足量浓 $\\mathrm{NaOH}$ 溶液加热, 在试管 <br> 口放置湿润的红色石荵试纸, 观 <br> 察试纸是否变色 | 若试纸不变蓝色, 说明该晶体 <br> 为尿素, 若试纸变蓝色, 说明 <br> 该晶体为氯化铵 |\n| 检验钠盐试剂柜 <br> 中某瓶标签残破 <br> 的纯净白色晶体 <br> 是不是 $\\mathrm{NaHCO}_{3}$ | 取少量晶体于试管中, 加热, 观 <br> 察现象; 冷却后加入足量盐酸, <br> 观察是否产生无色无味气体, 并 <br> 通入澄清石灰水 | 若加热时试管口出现小水珠, <br> 加盐酸后产生无色无味气体, <br> 且能使石灰水变浑浊, 则该试 <br> 剂为 $\\mathrm{NaHCO}_{3}$ |\n| $\\left\\{\\begin{array}{l}\\text { 探究人体汗液中 } \\\\ \\text { 是否含有氨基酸 }\\end{array}\\right.$ | 取 $1 \\mathrm{~mL}$ 汗液, 加入 5 滴 $0.1 \\%$ 茚三 <br> 酮溶液加热至沸腾, 观察现象 | 若溶液呈现紫色, 则汗液中可 <br> 能含有氨基酸, 但也可能是含 <br> 有蛋白质 |\n| [图1] | 在试管甲中加入 $1 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot L^{-1}$ <br> 高锰酸钾溶液和 $2 \\mathrm{~mL}$ 蒸馏水, 试 <br> 管乙中加入 $2 \\mathrm{~mL} 0.01 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 高锰 <br> 酸钾溶液和 $1 \\mathrm{~mL}$ 蒸馏水, 再同时 <br> 向两试管加入 $1 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 草 <br> 酸溶液, 观察现象 | 若试管乙中溶液紫色先褪去, <br> 则反应物浓度增大时, 反应速 <br> 率增大 |\n\nA: A\nB: B\nC: C\nD: D\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-097.jpg?height=470&width=309&top_left_y=1426&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_701",
        "problem": "二甲醚是一种清洁能源，用水煤气制取二甲醚的原理如下：\n\nI. $\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})$\n\nII. $2 \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\n$500 \\mathrm{~K}$ 时, 在 $2 \\mathrm{~L}$ 恒容密闭容器中充入 $4 \\mathrm{~mol} \\mathrm{CO}$ 和 $8 \\mathrm{~mol} \\mathrm{H}_{2}, 4 \\mathrm{~min}$ 达到平衡, 平衡时 $\\mathrm{CO}$的转化率为 $80 \\%$, 且 $2 \\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$, 则下列说法不正确的是\nA: $0 \\sim 4 \\mathrm{~min}$, 反应 $\\mathrm{I}$ 的 $\\mathrm{v}\\left(\\mathrm{H}_{2}\\right)=0.8 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 反应 II 中 $\\mathrm{CH}_{3} \\mathrm{OH}$ 的转化率为 $80 \\%$\nC: 反应 $\\mathrm{I}$ 的平衡常数 $\\mathrm{K}=6.25$\nD: 容器总压强保持不变, 可证明反应体系到达平衡状态\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n二甲醚是一种清洁能源，用水煤气制取二甲醚的原理如下：\n\nI. $\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g})$\n\nII. $2 \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$\n\n$500 \\mathrm{~K}$ 时, 在 $2 \\mathrm{~L}$ 恒容密闭容器中充入 $4 \\mathrm{~mol} \\mathrm{CO}$ 和 $8 \\mathrm{~mol} \\mathrm{H}_{2}, 4 \\mathrm{~min}$ 达到平衡, 平衡时 $\\mathrm{CO}$的转化率为 $80 \\%$, 且 $2 \\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{OH}\\right)=\\mathrm{c}\\left(\\mathrm{CH}_{3} \\mathrm{OCH}_{3}\\right)$, 则下列说法不正确的是\n\nA: $0 \\sim 4 \\mathrm{~min}$, 反应 $\\mathrm{I}$ 的 $\\mathrm{v}\\left(\\mathrm{H}_{2}\\right)=0.8 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\cdot \\mathrm{min}^{-1}$\nB: 反应 II 中 $\\mathrm{CH}_{3} \\mathrm{OH}$ 的转化率为 $80 \\%$\nC: 反应 $\\mathrm{I}$ 的平衡常数 $\\mathrm{K}=6.25$\nD: 容器总压强保持不变, 可证明反应体系到达平衡状态\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1561",
        "problem": "For sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.A fuel cell (see figure) is made up of three segments sandwiched together: the anode, the electrolyte, and the cathode. Hydrogen is used as fuel and oxygen as oxidant. Two chemical\n\n[figure1]\nreactions occur at the interfaces of the three different segments:\n\n$\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+4 e^{-} \\rightarrow 4 \\mathrm{OH}^{-}(a q)$\n$\\mathrm{H}_{2}(g)+2 \\mathrm{OH}^{-}(a q) \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 e^{-}$\n\nThe net result of the two reactions is\n\n$2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n\nThe hydrogen for the fuel cell is supplied from the hydrolysis of sodium borohydride.\n\nCalculate the standard potential for the cathode half reaction if the standard reduction potential for the anode half reaction is $-0.83 \\mathrm{~V}$ and $\\Delta_{f} G^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(I)\\right)$ is $-237 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nFor sustainable energy, hydrogen appears to be the best energy carrier. The most efficient way of using hydrogen is generation of electrical energy in a fuel cell. However, storing hydrogen in large quantities is a challenge in fuel cell applications. Among the chemical hydrides considered as solid hydrogen storage materials, sodium borohydride $\\left(\\mathrm{NaBH}_{4}\\right)$, being non-toxic, stable and environmentally benign, appears to be the most promising one. The hydrolysis of sodium borohydride that releases $\\mathrm{H}_{2}$ gas is a slow reaction at ambient temperature and, therefore, needs to be catalyzed.\n\n$$\n\\mathrm{NaBH}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(I)  \\xrightarrow{catalyst} \\mathrm{Na}^{+}(\\mathrm{aq})+\\mathrm{BO}_{2}^{-}(\\mathrm{aq})+4 \\mathrm{H}_{2}(g)\n$$\n\nColloidal ruthenium(0) nanoclusters are the most active catalysts in this hydrolysis even at room temperature and lead to a complete $\\mathrm{H}_{2}$ release from sodium borohydride. Kinetic studies show that the catalytic hydrolysis of $\\mathrm{NaBH}_{4}$ is a first order reaction with respect to the catalyst, but a zero order with respect to the substrate. The rate of hydrogen production per mole of ruthenium is $92 \\mathrm{~mol} \\mathrm{H}_{2} \\cdot(\\mathrm{mol} \\mathrm{Ru})^{-1} \\cdot \\min ^{-1}$ at $25^{\\circ} \\mathrm{C}$.\n\nproblem:\nA fuel cell (see figure) is made up of three segments sandwiched together: the anode, the electrolyte, and the cathode. Hydrogen is used as fuel and oxygen as oxidant. Two chemical\n\n[figure1]\nreactions occur at the interfaces of the three different segments:\n\n$\\mathrm{O}_{2}(g)+2 \\mathrm{H}_{2} \\mathrm{O}(l)+4 e^{-} \\rightarrow 4 \\mathrm{OH}^{-}(a q)$\n$\\mathrm{H}_{2}(g)+2 \\mathrm{OH}^{-}(a q) \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)+2 e^{-}$\n\nThe net result of the two reactions is\n\n$2 \\mathrm{H}_{2}(g)+\\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(l)$\n\nThe hydrogen for the fuel cell is supplied from the hydrolysis of sodium borohydride.\n\nCalculate the standard potential for the cathode half reaction if the standard reduction potential for the anode half reaction is $-0.83 \\mathrm{~V}$ and $\\Delta_{f} G^{\\circ}\\left(\\mathrm{H}_{2} \\mathrm{O}(I)\\right)$ is $-237 \\mathrm{~kJ} \\mathrm{~mol}^{-1}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of V, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d067913db7e87f3acbf3g-125.jpg?height=602&width=717&top_left_y=1869&top_left_x=1109"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "V"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_879",
        "problem": "某装置中模拟 $\\mathrm{CO}_{2}$ 捕获和转化, 先进行反应 1 CaO $+\\mathrm{CO}_{2}=\\mathrm{CaCO}_{3}$, 反应完毕后向装置中以恒定流速通入恒定组成的 $\\mathrm{N}_{2} 、 \\mathrm{CH}_{4}$ 混合气, 进行反应(2)\n\n$\\mathrm{CaCO}_{3}+\\mathrm{CH}_{4} \\stackrel{\\text { 催化剂 }}{=} \\mathrm{CaO}+2 \\mathrm{CO}+2 \\mathrm{H}_{2}$, 单位时间流出气体各组分的物质的量随反应时\n\n间变化如图所示。第二阶段反应中始终未检测到 $\\mathrm{CO}_{2}$ 。下列说法错误的是\n\n[图1]\nA: $\\mathrm{t}_{1}$ 后催化剂催化效果逐渐降低\nB: $\\mathrm{t}_{1} \\sim \\mathrm{t}_{3}$, 存在生成 $\\mathrm{H}_{2}$ 的副反应\nC: $\\mathrm{t}_{3}$ 时催化剂上开始积碳\nD: $t_{2}$ 时, 副反应生成 $H_{2}$ 的速率小于反应(2)生成 $H_{2}$ 速率\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某装置中模拟 $\\mathrm{CO}_{2}$ 捕获和转化, 先进行反应 1 CaO $+\\mathrm{CO}_{2}=\\mathrm{CaCO}_{3}$, 反应完毕后向装置中以恒定流速通入恒定组成的 $\\mathrm{N}_{2} 、 \\mathrm{CH}_{4}$ 混合气, 进行反应(2)\n\n$\\mathrm{CaCO}_{3}+\\mathrm{CH}_{4} \\stackrel{\\text { 催化剂 }}{=} \\mathrm{CaO}+2 \\mathrm{CO}+2 \\mathrm{H}_{2}$, 单位时间流出气体各组分的物质的量随反应时\n\n间变化如图所示。第二阶段反应中始终未检测到 $\\mathrm{CO}_{2}$ 。下列说法错误的是\n\n[图1]\n\nA: $\\mathrm{t}_{1}$ 后催化剂催化效果逐渐降低\nB: $\\mathrm{t}_{1} \\sim \\mathrm{t}_{3}$, 存在生成 $\\mathrm{H}_{2}$ 的副反应\nC: $\\mathrm{t}_{3}$ 时催化剂上开始积碳\nD: $t_{2}$ 时, 副反应生成 $H_{2}$ 的速率小于反应(2)生成 $H_{2}$ 速率\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-034.jpg?height=448&width=714&top_left_y=741&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_734",
        "problem": "某无色溶液中可能含有以下离子中的一种或几种 $\\mathrm{Na}^{+} 、 \\mathrm{Ag}^{+} 、 \\mathrm{Fe}^{3+} 、 \\mathrm{NH}_{4}^{+} 、 \\mathrm{Al}^{3+} 、 \\mathrm{CO}$\n\n${ }_{3}^{2-} 、 \\mathrm{AlO}_{2}^{-} 、 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-} 、 \\mathrm{SO}_{4}^{2-}$ 。现取该溶液进行有关实验, 实验结果如图所示:\n\n[图1]\n\n下列说法不正确的是\nA: 淡黄色沉淀甲不可能为 $\\mathrm{AgBr}$\nB: 气体甲可能是混合气体\nC: 综合上述信息可以确定肯定存在的离子有: $\\mathrm{Na}^{+} 、 \\mathrm{AlO}_{2}^{-} 、 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}$\nD: 由溶液甲生成气体乙的途径只有: $\\mathrm{Al}^{3+}+3 \\mathrm{HCO}_{3}^{-}=\\mathrm{Al}(\\mathrm{OH})_{3} \\downarrow+3 \\mathrm{CO}_{2} \\uparrow$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n某无色溶液中可能含有以下离子中的一种或几种 $\\mathrm{Na}^{+} 、 \\mathrm{Ag}^{+} 、 \\mathrm{Fe}^{3+} 、 \\mathrm{NH}_{4}^{+} 、 \\mathrm{Al}^{3+} 、 \\mathrm{CO}$\n\n${ }_{3}^{2-} 、 \\mathrm{AlO}_{2}^{-} 、 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-} 、 \\mathrm{SO}_{4}^{2-}$ 。现取该溶液进行有关实验, 实验结果如图所示:\n\n[图1]\n\n下列说法不正确的是\n\nA: 淡黄色沉淀甲不可能为 $\\mathrm{AgBr}$\nB: 气体甲可能是混合气体\nC: 综合上述信息可以确定肯定存在的离子有: $\\mathrm{Na}^{+} 、 \\mathrm{AlO}_{2}^{-} 、 \\mathrm{~S}_{2} \\mathrm{O}_{3}^{2-}$\nD: 由溶液甲生成气体乙的途径只有: $\\mathrm{Al}^{3+}+3 \\mathrm{HCO}_{3}^{-}=\\mathrm{Al}(\\mathrm{OH})_{3} \\downarrow+3 \\mathrm{CO}_{2} \\uparrow$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-050.jpg?height=305&width=1150&top_left_y=1435&top_left_x=333"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_938",
        "problem": "制备新型锂离子电池电极材料磷酸亚铁锂 $\\left(\\mathrm{LiFePO}_{4}\\right)$ 的方法如下。\n\n方法一: 将碳酸锂、乙酸亚铁 $\\left[\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)_{2} \\mathrm{Fe}\\right]$ 、磷酸二氢铵按一定比例混合、充分研磨后，在 $800^{\\circ} \\mathrm{C}$ 左右煅烧制得产品。\n\n方法二: 将一定浓度的磷酸二氢铵、氯化锂混合溶液作为电解液, 以铁棒为阳极, 石墨为阴极, 电解析出磷酸亚铁锂沉淀。沉淀经过滤、洗涤、干燥, 在 $800^{\\circ} \\mathrm{C}$ 左右煅烧制得产品。下列说法错误的是\nA: 上述两种方法制备磷酸亚铁锂的过程都必须在情性气体氛围中进行\nB: 方法一所得固体产品中混有 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$, 同时生成的乙酸以气体逸出\nC: 方法二阴极反应式为 $\\mathrm{Fe}+\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}+\\mathrm{Li}^{+}-2 \\mathrm{e}^{-}=\\mathrm{LiFePO}_{4}+2 \\mathrm{H}^{+}$\nD: 锂离子电池中常用有机聚合物作为正负极之间锂离子迁移的介质\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n制备新型锂离子电池电极材料磷酸亚铁锂 $\\left(\\mathrm{LiFePO}_{4}\\right)$ 的方法如下。\n\n方法一: 将碳酸锂、乙酸亚铁 $\\left[\\left(\\mathrm{CH}_{3} \\mathrm{COO}\\right)_{2} \\mathrm{Fe}\\right]$ 、磷酸二氢铵按一定比例混合、充分研磨后，在 $800^{\\circ} \\mathrm{C}$ 左右煅烧制得产品。\n\n方法二: 将一定浓度的磷酸二氢铵、氯化锂混合溶液作为电解液, 以铁棒为阳极, 石墨为阴极, 电解析出磷酸亚铁锂沉淀。沉淀经过滤、洗涤、干燥, 在 $800^{\\circ} \\mathrm{C}$ 左右煅烧制得产品。下列说法错误的是\n\nA: 上述两种方法制备磷酸亚铁锂的过程都必须在情性气体氛围中进行\nB: 方法一所得固体产品中混有 $\\left(\\mathrm{NH}_{4}\\right)_{2} \\mathrm{CO}_{3}$, 同时生成的乙酸以气体逸出\nC: 方法二阴极反应式为 $\\mathrm{Fe}+\\mathrm{H}_{2} \\mathrm{PO}_{4}^{-}+\\mathrm{Li}^{+}-2 \\mathrm{e}^{-}=\\mathrm{LiFePO}_{4}+2 \\mathrm{H}^{+}$\nD: 锂离子电池中常用有机聚合物作为正负极之间锂离子迁移的介质\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_341",
        "problem": "According to the reaction profile below, what is $\\Delta H$ for the reaction $4 \\mathrm{HBr}(g)+\\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g)+2 \\mathrm{Br}_{2}(g)$ ?\n\n[figure1]\n\nReaction progress\nA: $276 \\mathrm{~kJ}$\nB: $-276 \\mathrm{~kJ}$\nC: $434 \\mathrm{~kJ}$\nD: $-434 \\mathrm{~kJ}$\nE: $158 \\mathrm{~kJ}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAccording to the reaction profile below, what is $\\Delta H$ for the reaction $4 \\mathrm{HBr}(g)+\\mathrm{O}_{2}(g) \\rightarrow 2 \\mathrm{H}_{2} \\mathrm{O}(g)+2 \\mathrm{Br}_{2}(g)$ ?\n\n[figure1]\n\nReaction progress\n\nA: $276 \\mathrm{~kJ}$\nB: $-276 \\mathrm{~kJ}$\nC: $434 \\mathrm{~kJ}$\nD: $-434 \\mathrm{~kJ}$\nE: $158 \\mathrm{~kJ}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_92a3e606264c5ee1789ag-5.jpg?height=461&width=786&top_left_y=325&top_left_x=1119"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_739",
        "problem": "利用电渗析法再生钠碱循环脱硫中的吸收液, 并获取高浓度的 $\\mathrm{SO}_{2}$, 工作原理如图所示。已知双极膜在电流作用下可将水解离, 膜两侧分别得到 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$。下列说法正确的是\n\n[图1]\n\n图1 钠碱循环脱硫\n\n[图2]\n\n图2 电渗析法再生吸收液\nA: 双极膜右侧得到的是 $\\mathrm{OH}$\nB: 再生吸收液从 $\\mathrm{N}$ 室流出\nC: 相同条件下，阳极和阴极产生的气体体积比为 2:1\nD: $\\mathrm{M}$ 室中 $\\mathrm{SO}_{3}^{2-} 、 \\mathrm{HSO}_{3}^{-}$分别与 $\\mathrm{H}^{+}$发生反应可得高浓度 $\\mathrm{SO}_{2}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n利用电渗析法再生钠碱循环脱硫中的吸收液, 并获取高浓度的 $\\mathrm{SO}_{2}$, 工作原理如图所示。已知双极膜在电流作用下可将水解离, 膜两侧分别得到 $\\mathrm{H}^{+}$和 $\\mathrm{OH}^{-}$。下列说法正确的是\n\n[图1]\n\n图1 钠碱循环脱硫\n\n[图2]\n\n图2 电渗析法再生吸收液\n\nA: 双极膜右侧得到的是 $\\mathrm{OH}$\nB: 再生吸收液从 $\\mathrm{N}$ 室流出\nC: 相同条件下，阳极和阴极产生的气体体积比为 2:1\nD: $\\mathrm{M}$ 室中 $\\mathrm{SO}_{3}^{2-} 、 \\mathrm{HSO}_{3}^{-}$分别与 $\\mathrm{H}^{+}$发生反应可得高浓度 $\\mathrm{SO}_{2}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-50.jpg?height=340&width=627&top_left_y=227&top_left_x=360",
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-50.jpg?height=369&width=714&top_left_y=221&top_left_x=1068"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1286",
        "problem": "Kinetic isotope effect (KIE) is a phenomenon associated with a change in rate constant of the reaction when one of the atoms is replaced by its isotope. KIE can be used to confirm whether a particular bond to hydrogen is broken in the reaction. Harmonic oscillator model is used to estimate the difference in the rate between $\\mathrm{C}-\\mathrm{H}$ and ${ }^{1} \\mathrm{C}-\\mathrm{D}$ bond activation ( $\\mathrm{D}-$ deuterium).\n\nThe vibrational frequency $(v)$ represented by harmonic oscillator model is\n\n$$\nv=\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{\\mu}}\n$$\n\nwhere $k$ is the force constant and $\\mu$ is the reduced mass.\n\nThe vibrational energies of the molecule are given by\n\n$$\nE_{n}=\\left(n+\\frac{1}{2}\\right) h v\n$$\n\nwhere $n$ is vibrational quantum number with possible values of $0,1,2, \\ldots$ The energy of the lowest vibrational energy level ( $\\mathrm{E}_{n}$ at $n=0$ ) is called zero-point vibrational energy (ZPE).Calculate the reduced mass of $\\mathrm{C}-\\mathrm{H}\\left(\\mu_{\\mathrm{CH}}\\right)$  in atomic mass unit. Assume that the mass of deuterium is twice that of hydrogen.\n\nNote: If a student is unable to calculate the values for $\\mu_{\\text { }}$ and $\\mu_{\\mathrm{CD}}$ he/she can use $\\mu_{C H}=1.008$ and $\\mu_{C D}=2.016$ for the subsequent parts of the question. The given values are not necessarily be close to the correct values.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nKinetic isotope effect (KIE) is a phenomenon associated with a change in rate constant of the reaction when one of the atoms is replaced by its isotope. KIE can be used to confirm whether a particular bond to hydrogen is broken in the reaction. Harmonic oscillator model is used to estimate the difference in the rate between $\\mathrm{C}-\\mathrm{H}$ and ${ }^{1} \\mathrm{C}-\\mathrm{D}$ bond activation ( $\\mathrm{D}-$ deuterium).\n\nThe vibrational frequency $(v)$ represented by harmonic oscillator model is\n\n$$\nv=\\frac{1}{2 \\pi} \\sqrt{\\frac{k}{\\mu}}\n$$\n\nwhere $k$ is the force constant and $\\mu$ is the reduced mass.\n\nThe vibrational energies of the molecule are given by\n\n$$\nE_{n}=\\left(n+\\frac{1}{2}\\right) h v\n$$\n\nwhere $n$ is vibrational quantum number with possible values of $0,1,2, \\ldots$ The energy of the lowest vibrational energy level ( $\\mathrm{E}_{n}$ at $n=0$ ) is called zero-point vibrational energy (ZPE).\n\nproblem:\nCalculate the reduced mass of $\\mathrm{C}-\\mathrm{H}\\left(\\mu_{\\mathrm{CH}}\\right)$  in atomic mass unit. Assume that the mass of deuterium is twice that of hydrogen.\n\nNote: If a student is unable to calculate the values for $\\mu_{\\text { }}$ and $\\mu_{\\mathrm{CD}}$ he/she can use $\\mu_{C H}=1.008$ and $\\mu_{C D}=2.016$ for the subsequent parts of the question. The given values are not necessarily be close to the correct values.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of amu, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "amu"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_69",
        "problem": "For the galvanic cell\n\n$$\n\\mathrm{Zn}(s)\\left|\\mathrm{Zn}^{2+}(a q) \\| \\mathrm{Cu}^{2+}(a q)\\right| \\mathrm{Cu}(s)\n$$\n\nthe standard cell potential is $1.10 \\mathrm{~V}$. What is the potential at $25^{\\circ} \\mathrm{C}$ if the concentrations of $\\mathrm{Zn}^{2+}$ and $\\mathrm{Cu}^{2+}$ are $2.5 \\mathrm{M}$ and $0.10 \\mathrm{M}$ respectively?\nA: $1.14 \\mathrm{~V}$\nB: $1.10 \\mathrm{~V}$\nC: $1.06 \\mathrm{~V}$\nD: $1.02 \\mathrm{~V}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nFor the galvanic cell\n\n$$\n\\mathrm{Zn}(s)\\left|\\mathrm{Zn}^{2+}(a q) \\| \\mathrm{Cu}^{2+}(a q)\\right| \\mathrm{Cu}(s)\n$$\n\nthe standard cell potential is $1.10 \\mathrm{~V}$. What is the potential at $25^{\\circ} \\mathrm{C}$ if the concentrations of $\\mathrm{Zn}^{2+}$ and $\\mathrm{Cu}^{2+}$ are $2.5 \\mathrm{M}$ and $0.10 \\mathrm{M}$ respectively?\n\nA: $1.14 \\mathrm{~V}$\nB: $1.10 \\mathrm{~V}$\nC: $1.06 \\mathrm{~V}$\nD: $1.02 \\mathrm{~V}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_489",
        "problem": "$\\mathrm{Mg} / \\mathrm{H}_{2} \\mathrm{O}_{2}$ 酸性燃料电池采用海水作电解质(加入一定量的酸), 下列说法正确的是\nA: 电池总反应为: $\\mathrm{Mg}+\\mathrm{H}_{2} \\mathrm{O}_{2}=\\mathrm{Mg}(\\mathrm{OH})_{2}$\nB: 正极发生的电极反应为: $\\mathrm{H}_{2} \\mathrm{O}_{2}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=2 \\mathrm{H}_{2} \\mathrm{O}$\nC: 工作时, 正极周围海水的 $\\mathrm{pH}$ 增大\nD: 电池工作时, 溶液中的 $\\mathrm{H}^{+}$向负极移动\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$\\mathrm{Mg} / \\mathrm{H}_{2} \\mathrm{O}_{2}$ 酸性燃料电池采用海水作电解质(加入一定量的酸), 下列说法正确的是\n\nA: 电池总反应为: $\\mathrm{Mg}+\\mathrm{H}_{2} \\mathrm{O}_{2}=\\mathrm{Mg}(\\mathrm{OH})_{2}$\nB: 正极发生的电极反应为: $\\mathrm{H}_{2} \\mathrm{O}_{2}+2 \\mathrm{H}^{+}+2 \\mathrm{e}^{-}=2 \\mathrm{H}_{2} \\mathrm{O}$\nC: 工作时, 正极周围海水的 $\\mathrm{pH}$ 增大\nD: 电池工作时, 溶液中的 $\\mathrm{H}^{+}$向负极移动\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_369",
        "problem": "The normal boiling points of molecular fluorine, chlorine, bromine, and iodine increase in that order. Which of the following statements accounts for this increase?\nA: The chemical reactivity decreases in that order.\nB: The London dispersion forces increase in that order.\nC: The dipole-dipole forces increase in that order.\nD: The hydrogen bonding increases in that order.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe normal boiling points of molecular fluorine, chlorine, bromine, and iodine increase in that order. Which of the following statements accounts for this increase?\n\nA: The chemical reactivity decreases in that order.\nB: The London dispersion forces increase in that order.\nC: The dipole-dipole forces increase in that order.\nD: The hydrogen bonding increases in that order.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_420",
        "problem": "$\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中存在多个平衡。在考虑平衡: $(1) \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons 2 \\mathrm{HCrO}_{4}^{-}(\\mathrm{aq})$ $\\mathrm{K}_{1}=3.0 \\times 10^{-2}\\left(25^{\\circ} \\mathrm{C}\\right)$; (2) $\\mathrm{HCrO}_{4}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) \\mathrm{K}_{2}=3.3 \\times 10^{-7}\\left(25^{\\circ} \\mathrm{C}\\right)$ 的条件下, $25^{\\circ} \\mathrm{C}$ 时, 向 $0.10 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中加入 $\\mathrm{NaOH}$, 溶液中 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)}$, 随 $\\mathrm{pH}$ 的变化关系如图所示, 溶液体积变化可忽略。下列说法中正确的是\n\n[图1]\nA: 已知 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)=1.29 \\times 10^{-2}$, 由(2)可知向 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 溶液中通 $\\mathrm{SO}_{2}$ 可制得 $\\mathrm{KHCrO}_{4}$\nB: $\\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}^{2}\\left(\\mathrm{CrO}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)}$ 的值随溶液 $\\mathrm{pH}$ 的增大逐渐减小\nC: 当溶液 $\\mathrm{pH}=9$ 时, 溶液中的 $\\mathrm{HCrO}_{4}^{-}$平衡浓度约为 $6 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L}$\nD: 当溶液 $\\mathrm{pH}=7$ 时, $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCrO}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$\\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中存在多个平衡。在考虑平衡: $(1) \\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\rightleftharpoons 2 \\mathrm{HCrO}_{4}^{-}(\\mathrm{aq})$ $\\mathrm{K}_{1}=3.0 \\times 10^{-2}\\left(25^{\\circ} \\mathrm{C}\\right)$; (2) $\\mathrm{HCrO}_{4}^{-}(\\mathrm{aq}) \\rightleftharpoons \\mathrm{CrO}_{4}^{2-}(\\mathrm{aq})+\\mathrm{H}^{+}(\\mathrm{aq}) \\mathrm{K}_{2}=3.3 \\times 10^{-7}\\left(25^{\\circ} \\mathrm{C}\\right)$ 的条件下, $25^{\\circ} \\mathrm{C}$ 时, 向 $0.10 \\mathrm{~mol} / \\mathrm{L} \\mathrm{K}_{2} \\mathrm{Cr}_{2} \\mathrm{O}_{7}$ 溶液中加入 $\\mathrm{NaOH}$, 溶液中 $\\lg \\frac{\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)}$, 随 $\\mathrm{pH}$ 的变化关系如图所示, 溶液体积变化可忽略。下列说法中正确的是\n\n[图1]\n\nA: 已知 $\\mathrm{K}_{\\mathrm{a} 1}\\left(\\mathrm{H}_{2} \\mathrm{SO}_{3}\\right)=1.29 \\times 10^{-2}$, 由(2)可知向 $\\mathrm{K}_{2} \\mathrm{CrO}_{4}$ 溶液中通 $\\mathrm{SO}_{2}$ 可制得 $\\mathrm{KHCrO}_{4}$\nB: $\\frac{\\mathrm{c}\\left(\\mathrm{H}^{+}\\right) \\cdot \\mathrm{c}^{2}\\left(\\mathrm{CrO}_{4}^{2-}\\right)}{\\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)}$ 的值随溶液 $\\mathrm{pH}$ 的增大逐渐减小\nC: 当溶液 $\\mathrm{pH}=9$ 时, 溶液中的 $\\mathrm{HCrO}_{4}^{-}$平衡浓度约为 $6 \\times 10^{-4} \\mathrm{~mol} / \\mathrm{L}$\nD: 当溶液 $\\mathrm{pH}=7$ 时, $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCrO}_{4}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CrO}_{4}^{2-}\\right)+2 \\mathrm{c}\\left(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-017.jpg?height=503&width=531&top_left_y=1276&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_806",
        "problem": "下列说法正确的是\nA: 熔点：正戊烷>2，2-二甲基戊烷>2，3-二甲基丁烷>丙烷\nB: 密度: $\\mathrm{CCl}_{4}>\\mathrm{H}_{2} \\mathrm{O}>$ 苯\nC: 同质量的物质燃烧耗 $\\mathrm{O}_{2}$ 量：甲烷>乙烷>乙烯>乙炔\nD: 同物质的量物质燃烧耗 $\\mathrm{O}_{2}$ 量：环已烷>苯>苯甲酸\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n下列说法正确的是\n\nA: 熔点：正戊烷>2，2-二甲基戊烷>2，3-二甲基丁烷>丙烷\nB: 密度: $\\mathrm{CCl}_{4}>\\mathrm{H}_{2} \\mathrm{O}>$ 苯\nC: 同质量的物质燃烧耗 $\\mathrm{O}_{2}$ 量：甲烷>乙烷>乙烯>乙炔\nD: 同物质的量物质燃烧耗 $\\mathrm{O}_{2}$ 量：环已烷>苯>苯甲酸\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1110",
        "problem": "Boron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the volume of the $\\mathrm{h}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nBoron nitride has several crystalline forms. This includes a cubic form (c-BN), a hexagonal layered structure similar to graphite ( $\\mathrm{h}-\\mathrm{BN}$ ), and a wurtzite structure ( $\\mathrm{w}-\\mathrm{BN}$ ). The different forms can be interconverted by changing the pressure and temperature. Diagrams of the three different structures are shown below.\n\n[figure1]\n\nUnit cell of c-BN\n\n[figure2]\n\nLayer structure of $\\mathrm{h}-\\mathrm{BN}$\n\n[figure3]\n\nUnit cell of $w-B N$\n\nThe unit cell of $\\mathrm{c}-\\mathrm{BN}$ is cubic, with cell parameters of $\\mathrm{a}=3.63 \\AA$.\n\nThe unit cell of $h-B N$ is a right regular hexagonal prism, with cell parameters of $\\mathrm{b}=1.47 \\AA$ and $\\mathrm{c}=6.66 \\AA$.\n\nThe unit cell of $\\mathrm{w}-\\mathrm{BN}$ is a right rhombic prism, with cell parameters of $d=2.54 \\AA$ and $\\mathrm{e}=3.63 \\AA$ and interior angles of $60^{\\circ}$ and $120^{\\circ}$.\n\n[figure4]\n\nUnit cell of h-BN\n\nYou may find it helpful to refer to the physical constants and formulae page for useful equations for the next part of this question.\n\nCalculate the volume of the $\\mathrm{h}-\\mathrm{BN}$ unit cell in $\\mathrm{cm}^{3}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~cm}^{3}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=448&width=511&top_left_y=541&top_left_x=270",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=596&width=436&top_left_y=458&top_left_x=844",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=471&width=443&top_left_y=524&top_left_x=1389",
            "https://cdn.mathpix.com/cropped/2024_03_14_5fa6a064d09014135c8ag-06.jpg?height=603&width=305&top_left_y=1206&top_left_x=1315"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~cm}^{3}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1361",
        "problem": "Rotational energy levels of diatomic molecules are well described by the formula $E_{\\mathrm{J}}=B \\mathrm{~J}(\\mathrm{~J}+1)$, where $\\mathrm{J}$ is the rotational quantum number of the molecule and $B$ its rotational constant. Constant $B$ is related to the reduced mass $\\mu$ and the bond length $R$ of the molecule through the equation\n\n$$\nB=\\frac{h^{2}}{8 \\pi^{2} \\mu R^{2}} .\n$$\n\nIn general, spectroscopic transitions appear at photon energies which are equal to the energy difference between appropriate states of a molecule ( $h \\nu=\\Delta E$ ). The observed rotational transitions occur between adjacent rotational levels, hence $\\Delta E=E_{J+1}-E_{J}=$ $2 B(\\mathrm{~J}+1)$. Consequently, successive rotational transitions that appear on the spectrum (such as the one shown here) follow the equation $h(\\Delta v)=2 B$.\n\nBy inspecting the spectrum provided, determine the following quantities for ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$ with appropriate units:\n\n$\\Delta v$\n\n[figure1]",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nRotational energy levels of diatomic molecules are well described by the formula $E_{\\mathrm{J}}=B \\mathrm{~J}(\\mathrm{~J}+1)$, where $\\mathrm{J}$ is the rotational quantum number of the molecule and $B$ its rotational constant. Constant $B$ is related to the reduced mass $\\mu$ and the bond length $R$ of the molecule through the equation\n\n$$\nB=\\frac{h^{2}}{8 \\pi^{2} \\mu R^{2}} .\n$$\n\nIn general, spectroscopic transitions appear at photon energies which are equal to the energy difference between appropriate states of a molecule ( $h \\nu=\\Delta E$ ). The observed rotational transitions occur between adjacent rotational levels, hence $\\Delta E=E_{J+1}-E_{J}=$ $2 B(\\mathrm{~J}+1)$. Consequently, successive rotational transitions that appear on the spectrum (such as the one shown here) follow the equation $h(\\Delta v)=2 B$.\n\nBy inspecting the spectrum provided, determine the following quantities for ${ }^{12} \\mathrm{C}^{16} \\mathrm{O}$ with appropriate units:\n\n$\\Delta v$\n\n[figure1]\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{GHz}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-469.jpg?height=995&width=1525&top_left_y=1664&top_left_x=268"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{GHz}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_216",
        "problem": "Calculate the mass of potassium (in $\\mathrm{mg}$ ) present in $5.10 \\mathrm{mg}$ of potassium chloride. Give your answer to two decimal places.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the mass of potassium (in $\\mathrm{mg}$ ) present in $5.10 \\mathrm{mg}$ of potassium chloride. Give your answer to two decimal places.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1368",
        "problem": "The undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe distribution constant $\\left(K_{D}\\right)$ of the acid HA depends on the $\\mathrm{pH}$ of the aqueous phase.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a True or False question.\n\nproblem:\nThe undissociated form of a weak organic acid HA can be extracted from the aqueous phase by a water-immiscible organic solvent according to the scheme:\n\n[figure1]\n\nRegarding this extraction, are the following statements correct $(\\mathrm{True})$ or not $(\\mathrm{False})$ ?\n\nThe distribution constant $\\left(K_{D}\\right)$ of the acid HA depends on the $\\mathrm{pH}$ of the aqueous phase.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be either \"True\" or \"False\".",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_9a7e6af5bdf769e8e1dfg-461.jpg?height=185&width=788&top_left_y=2209&top_left_x=634"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "TF",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_648",
        "problem": "室温时, 体积为 $1 \\mathrm{~mL} 、$ 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的两种常见物质 $\\mathrm{X}(\\mathrm{OH})_{\\mathrm{n}} 、 \\mathrm{~K}_{2} \\mathrm{AO}_{3}$ 的溶液分别加水稀释至体积为 $\\mathrm{VmL}, \\mathrm{pH}$ 随 $\\lg$ 的变化情况如下图所示。下列叙述错误的是\n\n[图1]\nA: $n=1$\nB: $\\mathrm{H}_{2} \\mathrm{AO}_{3}$ 的二级电离常数 $\\mathrm{K}_{\\mathrm{a} 2}$ 约为 $1.0 \\times 10^{-10.2}$\nC: $\\mathrm{pH}=10$ 的两种溶液中水的电离程度相同\nD: 升高温度, $\\mathrm{K}_{2} \\mathrm{AO}_{3}$ 溶液 $\\mathrm{pH}$ 增大, $\\mathrm{X}(\\mathrm{OH})_{\\mathrm{n}}$ 溶液 $\\mathrm{pH}$ 减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n室温时, 体积为 $1 \\mathrm{~mL} 、$ 浓度均为 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的两种常见物质 $\\mathrm{X}(\\mathrm{OH})_{\\mathrm{n}} 、 \\mathrm{~K}_{2} \\mathrm{AO}_{3}$ 的溶液分别加水稀释至体积为 $\\mathrm{VmL}, \\mathrm{pH}$ 随 $\\lg$ 的变化情况如下图所示。下列叙述错误的是\n\n[图1]\n\nA: $n=1$\nB: $\\mathrm{H}_{2} \\mathrm{AO}_{3}$ 的二级电离常数 $\\mathrm{K}_{\\mathrm{a} 2}$ 约为 $1.0 \\times 10^{-10.2}$\nC: $\\mathrm{pH}=10$ 的两种溶液中水的电离程度相同\nD: 升高温度, $\\mathrm{K}_{2} \\mathrm{AO}_{3}$ 溶液 $\\mathrm{pH}$ 增大, $\\mathrm{X}(\\mathrm{OH})_{\\mathrm{n}}$ 溶液 $\\mathrm{pH}$ 减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-70.jpg?height=511&width=488&top_left_y=164&top_left_x=356"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_481",
        "problem": "丙二酸 $\\left.(\\mathrm{HOOCCHH})_{2} \\mathrm{COOH}\\right)$ 是二元弱酸。 $298 \\mathrm{~K}$ 时, 向一定浓度的 $\\mathrm{K}_{2} \\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}$ 溶液中滴加盐酸, 混合溶液中含碳粒子浓度 $\\mathrm{pX}$ 与 $\\mathrm{pH}$ 的关系如图所示。已知: $\\mathrm{pX}=-\\operatorname{lgX}$, $\\mathrm{X}=\\frac{c\\left(\\mathrm{H}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}^{-}\\right)}{c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)}$ 或 $\\left.\\frac{c\\left(\\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}^{2-}\\right)}{c\\left(\\mathrm{H}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}^{-}\\right)}\\right)$下列叙述正确的是\n\n[图1]\nA: 直线 $\\mathrm{L}_{2}$ 表示 $\\mathrm{pH}$ 与 $\\mathrm{p}\\left[\\frac{c\\left(\\mathrm{H}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}^{-}\\right)}{c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)}\\right]$ 的关系\nB: $\\mathrm{K}_{2} \\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}$ 稀溶液与等体积、等浓度的盐酸混合后, $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)<c\\left(\\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}^{2-}\\right)$\nC: 当 $c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}^{2-}\\right)$ 时, 溶液 $\\mathrm{pH}=3.26$\nD: $\\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c\\left(\\mathrm{OH}^{-}\\right)}=0$ 的溶液中的溶质可能是 $\\mathrm{KH}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}$ 和 $\\mathrm{K}_{2} \\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n丙二酸 $\\left.(\\mathrm{HOOCCHH})_{2} \\mathrm{COOH}\\right)$ 是二元弱酸。 $298 \\mathrm{~K}$ 时, 向一定浓度的 $\\mathrm{K}_{2} \\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}$ 溶液中滴加盐酸, 混合溶液中含碳粒子浓度 $\\mathrm{pX}$ 与 $\\mathrm{pH}$ 的关系如图所示。已知: $\\mathrm{pX}=-\\operatorname{lgX}$, $\\mathrm{X}=\\frac{c\\left(\\mathrm{H}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}^{-}\\right)}{c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)}$ 或 $\\left.\\frac{c\\left(\\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}^{2-}\\right)}{c\\left(\\mathrm{H}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}^{-}\\right)}\\right)$下列叙述正确的是\n\n[图1]\n\nA: 直线 $\\mathrm{L}_{2}$ 表示 $\\mathrm{pH}$ 与 $\\mathrm{p}\\left[\\frac{c\\left(\\mathrm{H}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}^{-}\\right)}{c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)}\\right]$ 的关系\nB: $\\mathrm{K}_{2} \\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}$ 稀溶液与等体积、等浓度的盐酸混合后, $c\\left(\\mathrm{H}^{+}\\right)+c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)<c\\left(\\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}^{2-}\\right)$\nC: 当 $c\\left(\\mathrm{H}_{4} \\mathrm{C}_{3} \\mathrm{O}_{4}\\right)=c\\left(\\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}^{2-}\\right)$ 时, 溶液 $\\mathrm{pH}=3.26$\nD: $\\lg \\frac{c\\left(\\mathrm{H}^{+}\\right)}{c\\left(\\mathrm{OH}^{-}\\right)}=0$ 的溶液中的溶质可能是 $\\mathrm{KH}_{3} \\mathrm{C}_{3} \\mathrm{O}_{4}$ 和 $\\mathrm{K}_{2} \\mathrm{H}_{2} \\mathrm{C}_{3} \\mathrm{O}_{4}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-088.jpg?height=579&width=851&top_left_y=1161&top_left_x=337"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1538",
        "problem": "The production of zinc from zinc sulphide proceeds in two stages: the roasting of zinc sulphide in the air and the reduction of the zinc oxide formed by carbon monoxide. In this problem we will consider the roasting of zinc sulphide.\n\nThis operation consists in burning zinc sulphide in the air. The equation of the reaction taking place is as follows:\n\n$$\n\\mathrm{ZnS}(\\mathrm{s})+3 / 2 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{ZnO}(\\mathrm{s})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\Delta_{r} H_{1350}^{0}=-448.98 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\n\nIndustrially this reaction is carried out at $1350 \\mathrm{~K}$.\nSuppose that the zinc containing mineral contains only zinc sulphide, ZnS.\n\nAssuming that the gangue does not react during the roasting, calculate the minimum $\\mathrm{ZnS}$ content of the mineral for which the reaction would be self-sustaining at $1350 \\mathrm{~K}$ despite the presence of silica. Give the answer is grams of ZnS per hundred grams of zinc blend.\n\nData:\n\nStandard molar heat capacities averaged over the temperature range considered (in $\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}$ ):\n\n| ZnS (solid): | 58.05 | $\\mathrm{ZnO}$ (solid): | 51.64 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{SO}_{2}$ (gas): | 51.10 | $\\mathrm{O}_{2}$ (gas): | 34.24 |\n| $\\mathrm{~N}_{2}$ (gas): | 30.65 | $\\mathrm{SiO}_{2}$ (solid): | 72.50 |\n| Molar masses (in $\\mathrm{g} \\mathrm{mol}^{-1}$ ): | $\\mathrm{ZnS}: 97.5$ | $\\mathrm{SiO}_{2}: 60.1$ |  |",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe production of zinc from zinc sulphide proceeds in two stages: the roasting of zinc sulphide in the air and the reduction of the zinc oxide formed by carbon monoxide. In this problem we will consider the roasting of zinc sulphide.\n\nThis operation consists in burning zinc sulphide in the air. The equation of the reaction taking place is as follows:\n\n$$\n\\mathrm{ZnS}(\\mathrm{s})+3 / 2 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{ZnO}(\\mathrm{s})+\\mathrm{SO}_{2}(\\mathrm{~g}) \\quad \\Delta_{r} H_{1350}^{0}=-448.98 \\mathrm{~kJ} \\mathrm{~mol}^{-1}\n$$\n\nIndustrially this reaction is carried out at $1350 \\mathrm{~K}$.\nSuppose that the zinc containing mineral contains only zinc sulphide, ZnS.\n\nAssuming that the gangue does not react during the roasting, calculate the minimum $\\mathrm{ZnS}$ content of the mineral for which the reaction would be self-sustaining at $1350 \\mathrm{~K}$ despite the presence of silica. Give the answer is grams of ZnS per hundred grams of zinc blend.\n\nData:\n\nStandard molar heat capacities averaged over the temperature range considered (in $\\mathrm{J} \\mathrm{K}^{-1} \\mathrm{~mol}^{-1}$ ):\n\n| ZnS (solid): | 58.05 | $\\mathrm{ZnO}$ (solid): | 51.64 |\n| :--- | :--- | :--- | :--- |\n| $\\mathrm{SO}_{2}$ (gas): | 51.10 | $\\mathrm{O}_{2}$ (gas): | 34.24 |\n| $\\mathrm{~N}_{2}$ (gas): | 30.65 | $\\mathrm{SiO}_{2}$ (solid): | 72.50 |\n| Molar masses (in $\\mathrm{g} \\mathrm{mol}^{-1}$ ): | $\\mathrm{ZnS}: 97.5$ | $\\mathrm{SiO}_{2}: 60.1$ |  |\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of %, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "%"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_291",
        "problem": "In the laboratory, one must never dip a stirring rod into a reagent bottle. This is because\nA: the bottle may tip over\nB: the stirring rod might break\nC: the rod might puncture the bottle\nD: the contents of the bottle may become contaminated\nE: reagent can creep up the rod and come in contact with one's hand\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nIn the laboratory, one must never dip a stirring rod into a reagent bottle. This is because\n\nA: the bottle may tip over\nB: the stirring rod might break\nC: the rod might puncture the bottle\nD: the contents of the bottle may become contaminated\nE: reagent can creep up the rod and come in contact with one's hand\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_548",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KOH}$ 溶液滴定 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液, 所得滴定曲线如图所示。忽略混合时溶液体积的变化, 下列有关各点溶液中粒子浓度关系正确的是\n\n[图1]\nA: a 点溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nB: b 点溶液中: $c\\left(\\mathrm{~K}^{+}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{--}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nC: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)-\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{d}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 用 $0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{KOH}$ 溶液滴定 $10 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$ 的 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液, 所得滴定曲线如图所示。忽略混合时溶液体积的变化, 下列有关各点溶液中粒子浓度关系正确的是\n\n[图1]\n\nA: a 点溶液中: $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nB: b 点溶液中: $c\\left(\\mathrm{~K}^{+}\\right)>c\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{--}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)>\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$\nC: $\\mathrm{c}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)-\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\nD: $\\mathrm{d}$ 点溶液中: $\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)-\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)+\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}^{-}\\right)$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-091.jpg?height=300&width=480&top_left_y=164&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_63",
        "problem": "The volume of a sample of gaseous $\\mathrm{SF}_{6}$ would increase by more than a factor of two under which of these conditions?\n\nI. It is heated from $300.0 \\mathrm{~K}$ to $600.0 \\mathrm{~K}$ at a pressure of $1.00 \\mathrm{~atm}$.\n\nII. The pressure is decreased from $2.00 \\mathrm{~atm}$ to 1.00 atm at $300 \\mathrm{~K}$.\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe volume of a sample of gaseous $\\mathrm{SF}_{6}$ would increase by more than a factor of two under which of these conditions?\n\nI. It is heated from $300.0 \\mathrm{~K}$ to $600.0 \\mathrm{~K}$ at a pressure of $1.00 \\mathrm{~atm}$.\n\nII. The pressure is decreased from $2.00 \\mathrm{~atm}$ to 1.00 atm at $300 \\mathrm{~K}$.\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_201",
        "problem": "Estimate the absorbance of a solution made by dissolving $4.5 \\mathrm{mg}$ of $\\mathrm{KCl}$ in water and adjusting the solution volume to $0.500 \\mathrm{~L}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nEstimate the absorbance of a solution made by dissolving $4.5 \\mathrm{mg}$ of $\\mathrm{KCl}$ in water and adjusting the solution volume to $0.500 \\mathrm{~L}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_863",
        "problem": "下列热化学方程式中, 正确的是\nA: 甲烷的燃烧热为 $890.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$, 则甲烷燃烧的热化学方程式可表示为 $\\mathrm{CH}_{4}(\\mathrm{~g})+2 \\mathrm{O}_{2}(\\mathrm{~g})=\\mathrm{CO}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\Delta \\mathrm{H}=-890.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\nB: 在 $101 \\mathrm{kPa}$ 时, $2 \\mathrm{~g} \\mathrm{H}_{2}$ 完全燃烧生成液态水, 放出 $285.8 \\mathrm{~kJ}$ 热量, 氢气燃烧的热化学方程式表示为 $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})=2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\Delta \\mathrm{H}=-571.6 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} \\mathrm{l}$\nC: $\\mathrm{HCl}$ 和 $\\mathrm{NaOH}$ 反应的中和热 $\\Delta \\mathrm{H}=-57.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$, 则 $\\mathrm{CH}_{3} \\mathrm{COOH}(\\mathrm{aq})+\\mathrm{NaOH}(\\mathrm{aq})=\\mathrm{CH}_{3} \\mathrm{COONa}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\Delta \\mathrm{H}=-57.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\nD: $500^{\\circ} \\mathrm{C} 、 30 \\mathrm{MPa}$ 下, 将 $0.5 \\mathrm{~mol} \\mathrm{~N}_{2}(\\mathrm{~g})$ 和 $1.5 \\mathrm{~mol} \\mathrm{H}_{2}(\\mathrm{~g})$ 置于密闭容器中充分反应生成 $\\mathrm{NH}_{3}(\\mathrm{~g})$ 放热 $19.3 \\mathrm{~kJ}$, 其热化学方程式为 $\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-38.6$ $\\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n下列热化学方程式中, 正确的是\n\nA: 甲烷的燃烧热为 $890.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$, 则甲烷燃烧的热化学方程式可表示为 $\\mathrm{CH}_{4}(\\mathrm{~g})+2 \\mathrm{O}_{2}(\\mathrm{~g})=\\mathrm{CO}_{2}(\\mathrm{~g})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g}) \\Delta \\mathrm{H}=-890.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\nB: 在 $101 \\mathrm{kPa}$ 时, $2 \\mathrm{~g} \\mathrm{H}_{2}$ 完全燃烧生成液态水, 放出 $285.8 \\mathrm{~kJ}$ 热量, 氢气燃烧的热化学方程式表示为 $2 \\mathrm{H}_{2}(\\mathrm{~g})+\\mathrm{O}_{2}(\\mathrm{~g})=2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\Delta \\mathrm{H}=-571.6 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1} \\mathrm{l}$\nC: $\\mathrm{HCl}$ 和 $\\mathrm{NaOH}$ 反应的中和热 $\\Delta \\mathrm{H}=-57.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$, 则 $\\mathrm{CH}_{3} \\mathrm{COOH}(\\mathrm{aq})+\\mathrm{NaOH}(\\mathrm{aq})=\\mathrm{CH}_{3} \\mathrm{COONa}(\\mathrm{aq})+\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l}) \\Delta \\mathrm{H}=-57.3 \\mathrm{~kJ} \\cdot \\mathrm{mol}^{-1}$\nD: $500^{\\circ} \\mathrm{C} 、 30 \\mathrm{MPa}$ 下, 将 $0.5 \\mathrm{~mol} \\mathrm{~N}_{2}(\\mathrm{~g})$ 和 $1.5 \\mathrm{~mol} \\mathrm{H}_{2}(\\mathrm{~g})$ 置于密闭容器中充分反应生成 $\\mathrm{NH}_{3}(\\mathrm{~g})$ 放热 $19.3 \\mathrm{~kJ}$, 其热化学方程式为 $\\mathrm{N}_{2}(\\mathrm{~g})+3 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons 2 \\mathrm{NH}_{3}(\\mathrm{~g}) \\Delta \\mathrm{H}=-38.6$ $\\mathrm{kJ} \\cdot \\mathrm{mol}^{-1}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_232",
        "problem": "Calculate the chemical amount (in mol or mmol) of $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ added.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nCalculate the chemical amount (in mol or mmol) of $\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ added.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mmol, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mmol"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_934",
        "problem": "$0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液分别滴入 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HX}$ 溶液与 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n$\\mathrm{HCl}$ 溶液中, 其 $\\mathrm{pH}$ 随滴入 $\\mathrm{NaOH}$ 溶液体积变化的图像如图所示。下列说法正确的是\n\n[图1]\nA: $b$ 点: $c(X) \\cdot c\\left(\\mathrm{OH}^{-}\\right)=10^{-12} \\mathrm{~mol}^{2} \\cdot \\mathrm{L}^{-2}$\nB: $c$ 点: $c\\left(X^{-}\\right)-c(H X)=c\\left(\\mathrm{OH}^{-}\\right)-c\\left(\\mathrm{H}^{+}\\right)$\nC: $a 、 d$ 点溶液混合后为酸性\nD: 水的电离程度: $d>c>b>a$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{NaOH}$ 溶液分别滴入 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{HX}$ 溶液与 $20 \\mathrm{~mL} 0.1 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1}$\n$\\mathrm{HCl}$ 溶液中, 其 $\\mathrm{pH}$ 随滴入 $\\mathrm{NaOH}$ 溶液体积变化的图像如图所示。下列说法正确的是\n\n[图1]\n\nA: $b$ 点: $c(X) \\cdot c\\left(\\mathrm{OH}^{-}\\right)=10^{-12} \\mathrm{~mol}^{2} \\cdot \\mathrm{L}^{-2}$\nB: $c$ 点: $c\\left(X^{-}\\right)-c(H X)=c\\left(\\mathrm{OH}^{-}\\right)-c\\left(\\mathrm{H}^{+}\\right)$\nC: $a 、 d$ 点溶液混合后为酸性\nD: 水的电离程度: $d>c>b>a$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-095.jpg?height=731&width=917&top_left_y=248&top_left_x=321"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1008",
        "problem": "What volume of $\\mathrm{CO}_{2}$ is produced when you burn exactly 1.0 litre of gaseous propane $\\left(\\mathrm{C}_{3} \\mathrm{H}_{8}\\right)$ in the presence of excess oxygen in your backyard barbecue? Assume $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$ are the only combustion products and that the pressure and temperature remain constant.\nA: 1.0\nB: 1.5\nC: 2.0\nD: 2.5\nE: 3.0\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat volume of $\\mathrm{CO}_{2}$ is produced when you burn exactly 1.0 litre of gaseous propane $\\left(\\mathrm{C}_{3} \\mathrm{H}_{8}\\right)$ in the presence of excess oxygen in your backyard barbecue? Assume $\\mathrm{H}_{2} \\mathrm{O}$ and $\\mathrm{CO}_{2}$ are the only combustion products and that the pressure and temperature remain constant.\n\nA: 1.0\nB: 1.5\nC: 2.0\nD: 2.5\nE: 3.0\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_213",
        "problem": "How many hydrogen atoms are there in $0.2 \\mathrm{~mol}$ of ammonium sulfate?\nA: $6.0 \\times 10^{22}$\nB: $1.2 \\times 10^{23}$\nC: $2.4 \\times 10^{23}$\nD: $4.8 \\times 10^{23}$\nE: $9.6 \\times 10^{23}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nHow many hydrogen atoms are there in $0.2 \\mathrm{~mol}$ of ammonium sulfate?\n\nA: $6.0 \\times 10^{22}$\nB: $1.2 \\times 10^{23}$\nC: $2.4 \\times 10^{23}$\nD: $4.8 \\times 10^{23}$\nE: $9.6 \\times 10^{23}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_399",
        "problem": "Which is formed at the cathode during the electrolysis of aqueous $\\mathrm{AgF}$ ?\nA: $\\operatorname{Ag}(s)$\nB: $\\mathrm{H}_{2}(g)$\nC: $\\mathrm{O}_{2}(g)$\nD: $\\mathrm{F}_{2}(g)$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhich is formed at the cathode during the electrolysis of aqueous $\\mathrm{AgF}$ ?\n\nA: $\\operatorname{Ag}(s)$\nB: $\\mathrm{H}_{2}(g)$\nC: $\\mathrm{O}_{2}(g)$\nD: $\\mathrm{F}_{2}(g)$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_741",
        "problem": "常温下, 用 $\\mathrm{NaOH}$ 调节不同浓度的二元酸 $\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$ 的酸碱性, 在其 $\\mathrm{pH}$ 为 10 时溶液中 $\\mathrm{pHCO}_{3}^{-}$与 $\\mathrm{pH}_{2} \\mathrm{CO}_{3}$ 或 $\\mathrm{pCO}_{3}^{2-}$ 的关系如图所示 $[\\mathrm{pX}=-\\operatorname{lgc}(\\mathrm{X})]$ 。下列说法错误的是\n\n[图1]\nA: II 代表 $\\mathrm{pCO}_{3}^{2-}$\nB: $\\mathrm{NaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: 相同浓度的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液和 $\\mathrm{NaHCO}_{3}$ 溶液的 $\\mathrm{pH}$ : 前者大于后者\nD: 碳酸的第一步电离常数为 $1 \\times 10^{-7}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 用 $\\mathrm{NaOH}$ 调节不同浓度的二元酸 $\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$ 的酸碱性, 在其 $\\mathrm{pH}$ 为 10 时溶液中 $\\mathrm{pHCO}_{3}^{-}$与 $\\mathrm{pH}_{2} \\mathrm{CO}_{3}$ 或 $\\mathrm{pCO}_{3}^{2-}$ 的关系如图所示 $[\\mathrm{pX}=-\\operatorname{lgc}(\\mathrm{X})]$ 。下列说法错误的是\n\n[图1]\n\nA: II 代表 $\\mathrm{pCO}_{3}^{2-}$\nB: $\\mathrm{NaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)<\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+2 \\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nC: 相同浓度的 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 溶液和 $\\mathrm{NaHCO}_{3}$ 溶液的 $\\mathrm{pH}$ : 前者大于后者\nD: 碳酸的第一步电离常数为 $1 \\times 10^{-7}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_1f04a2abf501b4d5eb93g-029.jpg?height=540&width=799&top_left_y=1106&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_501",
        "problem": "将 $\\mathrm{CO}_{2}$ 转化为二甲醚的反应原理为 $2 \\mathrm{CO}_{2}(\\mathrm{~g})+6 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, 一定条件下, 现有两个体积均为 $1.0 \\mathrm{~L}$ 的恒容密闭容器甲和乙, 在甲中充入 $0.1 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 0.18 $\\mathrm{mol} \\mathrm{H}_{2}$, 在乙中充入 $0.2 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 $0.36 \\mathrm{~mol} \\mathrm{H}_{2}$, 发生上述反应并达到平衡。该反应中 $\\mathrm{CO}_{2}$的平衡转化率随温度的变化曲线如图所示。下列说法正确的是\n\n[图1]\nA: 曲线 $\\mathrm{X}$ 表示的是容器甲中 $\\mathrm{CO}_{2}$ 的平衡转化率随温度的变化\nB: 体系的总压强 $\\mathrm{P}$ 总: $\\mathrm{P}$ 总 $($ 状态 $\\mathrm{II})<2 \\mathrm{P}$ 总 $($ 状态III)\nC: 将状态I对应的容器升温到 $\\mathrm{T}_{2} \\mathrm{~K}$, 可变成状态 II\nD: $\\mathrm{T}_{2} \\mathrm{~K}$ 时, 向空的容器甲中充入 $\\mathrm{CO}_{2}(\\mathrm{~g}) 、 \\mathrm{H}_{2}(\\mathrm{~g}) 、 \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})$ 和 $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ 各 1 mol,反应将向正方向进行\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n将 $\\mathrm{CO}_{2}$ 转化为二甲醚的反应原理为 $2 \\mathrm{CO}_{2}(\\mathrm{~g})+6 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightleftharpoons \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})+3 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$, 一定条件下, 现有两个体积均为 $1.0 \\mathrm{~L}$ 的恒容密闭容器甲和乙, 在甲中充入 $0.1 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 0.18 $\\mathrm{mol} \\mathrm{H}_{2}$, 在乙中充入 $0.2 \\mathrm{~mol} \\mathrm{CO}_{2}$ 和 $0.36 \\mathrm{~mol} \\mathrm{H}_{2}$, 发生上述反应并达到平衡。该反应中 $\\mathrm{CO}_{2}$的平衡转化率随温度的变化曲线如图所示。下列说法正确的是\n\n[图1]\n\nA: 曲线 $\\mathrm{X}$ 表示的是容器甲中 $\\mathrm{CO}_{2}$ 的平衡转化率随温度的变化\nB: 体系的总压强 $\\mathrm{P}$ 总: $\\mathrm{P}$ 总 $($ 状态 $\\mathrm{II})<2 \\mathrm{P}$ 总 $($ 状态III)\nC: 将状态I对应的容器升温到 $\\mathrm{T}_{2} \\mathrm{~K}$, 可变成状态 II\nD: $\\mathrm{T}_{2} \\mathrm{~K}$ 时, 向空的容器甲中充入 $\\mathrm{CO}_{2}(\\mathrm{~g}) 、 \\mathrm{H}_{2}(\\mathrm{~g}) 、 \\mathrm{CH}_{3} \\mathrm{OCH}_{3}(\\mathrm{~g})$ 和 $\\mathrm{H}_{2} \\mathrm{O}(\\mathrm{g})$ 各 1 mol,反应将向正方向进行\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_94f971e8ac2d3727b8beg-089.jpg?height=417&width=506&top_left_y=2307&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1581",
        "problem": "Palindromic sequences are an interesting class of DNA. In a palindromic double-stranded DNA (dsDNA) species, the sequence of one strand read in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction matches the $5^{\\prime} \\rightarrow 3^{\\prime}$ reading on the complementary strand. Hence, a palindromic dsDNA consists of two identical strands that are complementary to each other. An example is the so-called\n\nDrew-Dickerson dodecanucleotide (1):\n\n[figure1]How many different palindromic double-stranded DNA undecanucleotides (i.e., dsDNA species with eleven base pairs) exist?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nPalindromic sequences are an interesting class of DNA. In a palindromic double-stranded DNA (dsDNA) species, the sequence of one strand read in the $5^{\\prime} \\rightarrow 3^{\\prime}$ direction matches the $5^{\\prime} \\rightarrow 3^{\\prime}$ reading on the complementary strand. Hence, a palindromic dsDNA consists of two identical strands that are complementary to each other. An example is the so-called\n\nDrew-Dickerson dodecanucleotide (1):\n\n[figure1]\n\nproblem:\nHow many different palindromic double-stranded DNA undecanucleotides (i.e., dsDNA species with eleven base pairs) exist?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-260.jpg?height=140&width=440&top_left_y=1729&top_left_x=814"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1316",
        "problem": "The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nCalculate $\\Delta S_{\\text {sys }}$ assuming $\\mathrm{CO}_{2}$ to be an ideal gas.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThe second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nCalculate $\\Delta S_{\\text {sys }}$ assuming $\\mathrm{CO}_{2}$ to be an ideal gas.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{JK}^{-1}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{JK}^{-1}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1300",
        "problem": "$Problem I: \\mathrm{CO}_{2}$ gas, initially at a pressure of 4.0 bar and temperature of $10.0{ }^{\\circ} \\mathrm{C}$ is cooled at constant pressure. In this process A: it goes first to the liquid phase and then to the solid phase. B: it goes to the solid phase without going through the liquid phase.The second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nStarting with the same pressure and temperature as above (in problem I), $\\mathrm{CO}_{2}$ is compressed isothermatically. In this process,\n\nA: it goes first to the liquid phase and then to the solid phase.\n\nB: it goes to the solid phase without going through the liquid phase.\nA: it goes first to the liquid phase and then to the solid phase.\nB: it goes to the solid phase without going through the liquid phase.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\nHere is some context information for this question, which might assist you in solving it:\n$Problem I: \\mathrm{CO}_{2}$ gas, initially at a pressure of 4.0 bar and temperature of $10.0{ }^{\\circ} \\mathrm{C}$ is cooled at constant pressure. In this process A: it goes first to the liquid phase and then to the solid phase. B: it goes to the solid phase without going through the liquid phase.\n\nproblem:\nThe second law of thermodynamics is a fundamental law of science. In this problem we consider the thermodynamics of an ideal gas, phase transitions and chemical equilibrium.\n\n3.00 mol of $\\mathrm{CO}_{2}$ gas expands isothermically (in thermal contact with the surroundings; temperature $=15^{\\circ} \\mathrm{C}$ ) against a fixed external pressure of 1.00 bar. The initial and final volumes of the gas are $10.0 \\mathrm{dm}^{3}$ and $30.0 \\mathrm{dm}^{3}$, respectively.\n\nStarting with the same pressure and temperature as above (in problem I), $\\mathrm{CO}_{2}$ is compressed isothermatically. In this process,\n\nA: it goes first to the liquid phase and then to the solid phase.\n\nB: it goes to the solid phase without going through the liquid phase.\n\nA: it goes first to the liquid phase and then to the solid phase.\nB: it goes to the solid phase without going through the liquid phase.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_857",
        "problem": "氯霉素(D)的部分合成路线如下:\n\n[图1]\n\n下列说法正确的是\nA: $\\mathrm{A} . \\mathrm{A} \\rightarrow \\mathrm{B}$ 的反应过程中, 每生成 1 个 $\\mathrm{B}$ 分子, $\\mathrm{A}$ 中断裂 2 个碳氧 $\\sigma$ 键\nB: B 的含苯环同分异构体中最多有 8 种不同化学环境的氢原子\nC: B 转化为 $\\mathrm{C}$ 的过程中有配位键的形成\nD: 氯霉素 D 可发生还原、消去反应等, 不能发生缩聚反应\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n氯霉素(D)的部分合成路线如下:\n\n[图1]\n\n下列说法正确的是\n\nA: $\\mathrm{A} . \\mathrm{A} \\rightarrow \\mathrm{B}$ 的反应过程中, 每生成 1 个 $\\mathrm{B}$ 分子, $\\mathrm{A}$ 中断裂 2 个碳氧 $\\sigma$ 键\nB: B 的含苯环同分异构体中最多有 8 种不同化学环境的氢原子\nC: B 转化为 $\\mathrm{C}$ 的过程中有配位键的形成\nD: 氯霉素 D 可发生还原、消去反应等, 不能发生缩聚反应\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-07.jpg?height=183&width=1211&top_left_y=1936&top_left_x=340"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_542",
        "problem": "常温下, 向某浓度的 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中逐滴加入已知浓度的 $\\mathrm{NaOH}$ 溶液, 若 $\\mathrm{pC}$ 表示溶液中溶质微粒的物质的量浓度的负对数, 则所得溶液中 $\\mathrm{pC}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right), \\mathrm{pC}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)$、 $\\mathrm{pC}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$ 与溶液 $\\mathrm{pH}$ 的变化关系如图所示。已知: $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} \\rightleftharpoons \\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}+\\mathrm{H}^{+} \\mathrm{K}_{\\mathrm{a} 1} ; \\mathrm{HC}_{2} \\mathrm{O}_{4}$ $-\\rightleftharpoons \\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}+\\mathrm{H}^{+} \\mathrm{K}_{\\mathrm{a} 2}$ 。则下列说法正确的是\n\n[图1]\nA: 当 $\\mathrm{pH}=3$ 时, 溶液中 $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)<\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nB: $\\mathrm{pH}$ 由 3 增大到 5.3 的过程中, 水的电离程度逐渐减小\nC: 常温下, $\\mathrm{K}_{\\mathrm{a} 2}=10^{-5.3}$\nD: 常下随着 $\\mathrm{pH}$ 的增大: $\\mathrm{c}^{2}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right) /\\left[\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right) \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)\\right]$ 的值先增大后减小\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n常温下, 向某浓度的 $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}$ 溶液中逐滴加入已知浓度的 $\\mathrm{NaOH}$ 溶液, 若 $\\mathrm{pC}$ 表示溶液中溶质微粒的物质的量浓度的负对数, 则所得溶液中 $\\mathrm{pC}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right), \\mathrm{pC}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)$、 $\\mathrm{pC}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)$ 与溶液 $\\mathrm{pH}$ 的变化关系如图所示。已知: $\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4} \\rightleftharpoons \\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}+\\mathrm{H}^{+} \\mathrm{K}_{\\mathrm{a} 1} ; \\mathrm{HC}_{2} \\mathrm{O}_{4}$ $-\\rightleftharpoons \\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}+\\mathrm{H}^{+} \\mathrm{K}_{\\mathrm{a} 2}$ 。则下列说法正确的是\n\n[图1]\n\nA: 当 $\\mathrm{pH}=3$ 时, 溶液中 $\\mathrm{c}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right)<\\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)=\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right)$\nB: $\\mathrm{pH}$ 由 3 增大到 5.3 的过程中, 水的电离程度逐渐减小\nC: 常温下, $\\mathrm{K}_{\\mathrm{a} 2}=10^{-5.3}$\nD: 常下随着 $\\mathrm{pH}$ 的增大: $\\mathrm{c}^{2}\\left(\\mathrm{HC}_{2} \\mathrm{O}_{4}{ }^{-}\\right) /\\left[\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}\\right) \\mathrm{c}\\left(\\mathrm{C}_{2} \\mathrm{O}_{4}{ }^{2-}\\right)\\right]$ 的值先增大后减小\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-54.jpg?height=539&width=805&top_left_y=804&top_left_x=363",
            "https://cdn.mathpix.com/cropped/2024_03_31_8494b19d3cbb27e021b1g-54.jpg?height=54&width=1071&top_left_y=2497&top_left_x=338"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1005",
        "problem": "Aluminum dissolves in acidic solution according to the chemical equation below.\n\n$$\n2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\rightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)\n$$\n\nHow many grams of aluminum $\\left(27 \\mathrm{~g} \\mathrm{~mol}^{-1}\\right)$ are required to produce $0.50 \\mathrm{~mol} \\mathrm{H}_{2}$ ?\nA: $20 \\mathrm{~g}$\nB: $9.0 \\mathrm{~g}$\nC: $\\quad 14 \\mathrm{~g}$\nD: $27 \\mathrm{~g}$\nE: $\\quad 0.24 \\mathrm{~g}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nAluminum dissolves in acidic solution according to the chemical equation below.\n\n$$\n2 \\mathrm{Al}(s)+6 \\mathrm{HCl}(a q) \\rightarrow 2 \\mathrm{AlCl}_{3}(a q)+3 \\mathrm{H}_{2}(g)\n$$\n\nHow many grams of aluminum $\\left(27 \\mathrm{~g} \\mathrm{~mol}^{-1}\\right)$ are required to produce $0.50 \\mathrm{~mol} \\mathrm{H}_{2}$ ?\n\nA: $20 \\mathrm{~g}$\nB: $9.0 \\mathrm{~g}$\nC: $\\quad 14 \\mathrm{~g}$\nD: $27 \\mathrm{~g}$\nE: $\\quad 0.24 \\mathrm{~g}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1564",
        "problem": "Methanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |Calculate $\\Delta H^{0}$ for the reaction at $298 \\mathrm{~K}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nMethanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:\n\n$$\n\\mathrm{CO}(\\mathrm{g})+2 \\mathrm{H}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{CH}_{3} \\mathrm{OH}(\\mathrm{g}) .\n$$\n\nThe standard enthalpy of formation $\\left(\\Delta H f^{0}\\right)$ and the absolute entropy $\\left(S^{\\circ}\\right)$ for each of the three gases at room temperature ( $298 \\mathrm{~K})$ and at a standard pressure of 1 bar are given as follows.\n\n| $\\mathrm{Gas}$ | $\\Delta H_{f}^{\\mathrm{o}}\\left(\\mathrm{kJ} \\mathrm{mol}^{-1}\\right)$ | $S^{o}\\left(\\mathrm{~J} \\mathrm{~K}^{-1} \\mathrm{~mol}^{-1}\\right)$ |\n| :---: | :---: | :---: |\n| $\\mathrm{CO}(g)$ | -111 | 198 |\n| $\\mathrm{H}_{2}(g)$ | 0 | 131 |\n| $\\mathrm{CH}_{3} \\mathrm{OH}(g)$ | -201 | 240 |\n\nproblem:\nCalculate $\\Delta H^{0}$ for the reaction at $298 \\mathrm{~K}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of kJ, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "kJ"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1534",
        "problem": "Potentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nWhat is the absolute range of variation for the following physical values\n\na) transmittance $T$;\n\nb) absorbance $A$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis question involves multiple quantities to be determined.\n\nproblem:\nPotentiometric and spectrophotometric methods are widely used for the determination of equilibrium concentrations and equilibrium constants in solution. Both methods are frequently used in combination to achieve simultaneous determination of several species.\n\nSolution I contains a mixture of $\\mathrm{FeCl}_{2}(\\mathrm{aq})$ and $\\mathrm{FeCl}_{3}(\\mathrm{aq})$, and solution II contains a mixture of $\\mathrm{K}_{4} \\mathrm{Fe}(\\mathrm{CN})_{6}$ and $\\mathrm{K}_{3} \\mathrm{Fe}(\\mathrm{CN})_{6}$. The concentrations of iron-containing species satisfy the relations $\\left[\\mathrm{Fe}^{2+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}\\right]_{11}$ and $\\left[\\mathrm{Fe}^{3+}\\right]_{1}=\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]_{1 \\mid}$. The potential of platinum electrode immersed into the solution I is $0.652 \\mathrm{~V}$, while the potential of platinum electrode immersed into solution II is $0.242 \\mathrm{~V}$. The transmittance of the solution II measured relative to the solution I at $420 \\mathrm{~nm}$ is $10.7 \\%$ (optical pathlength $I=5.02 \\mathrm{~mm}$ ). The complexes $\\mathrm{Fe}(\\mathrm{CN})_{6}^{4-}, \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+}$, and $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ do not absorb light at $420 \\mathrm{~nm}$.\n\nMolar absorption at this wavelength $\\varepsilon\\left(\\left[\\mathrm{Fe}(\\mathrm{CN})_{6}^{3-}\\right]\\right)=1100 \\mathrm{M}^{-1} \\mathrm{~cm}^{-1}$.\n\nStandard redox potential for $\\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{3+} / \\mathrm{Fe}\\left(\\mathrm{H}_{2} \\mathrm{O}\\right)_{6}^{2+}$ is $0.771 \\mathrm{~V}$.\n\nThe factor before the logarithm in the Nernst equation is 0.0590 .\n\nWhat is the absolute range of variation for the following physical values\n\na) transmittance $T$;\n\nb) absorbance $A$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nYour final quantities should be output in the following order: [transmittance $T$, absorbance $A$].\nTheir answer types are, in order, [range interval, range interval].\nPlease end your response with: \"The final answers are \\boxed{ANSWER}\", where ANSWER should be the sequence of your final answers, separated by commas, for example: 5, 7, 2.5",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "MPV",
        "unit": [
            null,
            null
        ],
        "answer_sequence": [
            "transmittance $T$",
            "absorbance $A$"
        ],
        "type_sequence": [
            "IN",
            "IN"
        ],
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_687",
        "problem": "将废旧锌锰电池进行回收处理以实现资源的再生利用, 初步处理后的废料中含有 $\\mathrm{MnO}_{2} 、 \\mathrm{MnOOH} \\mathrm{Zn}(\\mathrm{OH})_{2}$ 、及 $\\mathrm{Fe}$ 等, 用该废料制备 $\\mathrm{Zn}$ 和 $\\mathrm{MnO}_{2}$ 的一种工艺流程如下: (已知: $\\mathrm{Mn}$ 的金属活动性强于 $\\mathrm{Fe}, \\mathrm{Mn}^{2+}$ 在酸性条件下比较稳定, $\\mathrm{pH}$ 大于 5.5 时易被氧化)。下列说法正确的是\n\n[图1]\nA: $\\mathrm{Mn}$ 原子的基态电子排布图为 $[\\mathrm{Ar}] 3 \\mathrm{~d}^{5} 4 \\mathrm{~s}^{2}$\nB: “酸浸”步骤中的离子反应方程式为 $\\mathrm{Fe}+\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}=\\mathrm{Fe}^{2+}+\\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$\nC: “净化”时通入 $\\mathrm{O}_{2}$ 是为了氧化 $\\mathrm{Fe}^{2+}$, 加入 $\\mathrm{MnCO}_{3}$ 时必须控制 $\\mathrm{pH}$ 小于 5.5\nD: “电解”时的阳极反应式为 $\\mathrm{Mn}^{2+}-2 \\mathrm{e}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}=\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n将废旧锌锰电池进行回收处理以实现资源的再生利用, 初步处理后的废料中含有 $\\mathrm{MnO}_{2} 、 \\mathrm{MnOOH} \\mathrm{Zn}(\\mathrm{OH})_{2}$ 、及 $\\mathrm{Fe}$ 等, 用该废料制备 $\\mathrm{Zn}$ 和 $\\mathrm{MnO}_{2}$ 的一种工艺流程如下: (已知: $\\mathrm{Mn}$ 的金属活动性强于 $\\mathrm{Fe}, \\mathrm{Mn}^{2+}$ 在酸性条件下比较稳定, $\\mathrm{pH}$ 大于 5.5 时易被氧化)。下列说法正确的是\n\n[图1]\n\nA: $\\mathrm{Mn}$ 原子的基态电子排布图为 $[\\mathrm{Ar}] 3 \\mathrm{~d}^{5} 4 \\mathrm{~s}^{2}$\nB: “酸浸”步骤中的离子反应方程式为 $\\mathrm{Fe}+\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}=\\mathrm{Fe}^{2+}+\\mathrm{Mn}^{2+}+2 \\mathrm{H}_{2} \\mathrm{O}$\nC: “净化”时通入 $\\mathrm{O}_{2}$ 是为了氧化 $\\mathrm{Fe}^{2+}$, 加入 $\\mathrm{MnCO}_{3}$ 时必须控制 $\\mathrm{pH}$ 小于 5.5\nD: “电解”时的阳极反应式为 $\\mathrm{Mn}^{2+}-2 \\mathrm{e}^{-}+2 \\mathrm{H}_{2} \\mathrm{O}=\\mathrm{MnO}_{2}+4 \\mathrm{H}^{+}$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-11.jpg?height=301&width=1331&top_left_y=155&top_left_x=334"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_380",
        "problem": "Molecules from which class of biopolymers can react with water, in the presence of suitable enzymes, to form smaller examples of that class of biopolymers?\nI. Proteins\nII. Polysaccharides\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nMolecules from which class of biopolymers can react with water, in the presence of suitable enzymes, to form smaller examples of that class of biopolymers?\nI. Proteins\nII. Polysaccharides\n\nA: I only\nB: II only\nC: Both I and II\nD: Neither I nor II\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1366",
        "problem": "One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.Calculate the smallest $\\mathrm{O}-\\mathrm{O}distance in the structure?",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nOne of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.\n[figure1]\n\nThe two figures above depict the cubic unit cell of the $\\mathrm{Cu}_{2} \\mathrm{O}$ crystal. The lattice constant of the structure is $427.0 \\mathrm{pm}$.\n\nproblem:\nCalculate the smallest $\\mathrm{O}-\\mathrm{O}distance in the structure?\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of pm, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_14_d4566978659c41b7e32eg-139.jpg?height=410&width=1236&top_left_y=660&top_left_x=296"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "pm"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1351",
        "problem": "The molar mass of glucose $\\left(\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\\right)$ is $180 \\mathrm{~g} \\mathrm{~mol}^{-1}$ and $N_{A}$ is the Avogadro constant. Which one of the following statements is not correct?\nA: An aqueous $0.5 \\mathrm{M}$ solution of glucose is prepared by dissolving $90 \\mathrm{~g}$ of glucose to give $1000 \\mathrm{~cm}^{3}$ of solution.\nB: $1.00 \\mathrm{mmol}$ amount of glucose has a mass of $180 \\mathrm{mg}$.\nC: $0.0100 \\mathrm{~mol}$ of glucose comprises of $0.0100 \\times 24 \\times \\mathrm{N}_{A}$ atoms.\nD: $90.0 \\mathrm{~g}$ glucose contain $3 \\times \\mathrm{N}_{\\mathrm{A}}$ atoms of carbon.\nE: $100 \\mathrm{~cm}^{3}$ of a $0.10 \\mathrm{M}$ solution contain $18 \\mathrm{~g}$ of glucose.\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nThe molar mass of glucose $\\left(\\mathrm{C}_{6} \\mathrm{H}_{12} \\mathrm{O}_{6}\\right)$ is $180 \\mathrm{~g} \\mathrm{~mol}^{-1}$ and $N_{A}$ is the Avogadro constant. Which one of the following statements is not correct?\n\nA: An aqueous $0.5 \\mathrm{M}$ solution of glucose is prepared by dissolving $90 \\mathrm{~g}$ of glucose to give $1000 \\mathrm{~cm}^{3}$ of solution.\nB: $1.00 \\mathrm{mmol}$ amount of glucose has a mass of $180 \\mathrm{mg}$.\nC: $0.0100 \\mathrm{~mol}$ of glucose comprises of $0.0100 \\times 24 \\times \\mathrm{N}_{A}$ atoms.\nD: $90.0 \\mathrm{~g}$ glucose contain $3 \\times \\mathrm{N}_{\\mathrm{A}}$ atoms of carbon.\nE: $100 \\mathrm{~cm}^{3}$ of a $0.10 \\mathrm{M}$ solution contain $18 \\mathrm{~g}$ of glucose.\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1321",
        "problem": "Many streams drain in areas where coal or metallic ores are mined. These streams have become acidic and contain high concentrations of dissolved iron and sulphate, due to sulphur-containing ores being exposed to the atmosphere or to oxygenated waters. The most common sulphur-containing mineral is pyrite, $\\mathrm{FeS}_{2}$, in which the oxidation state of iron is +2 . As the iron-rich streams mix with other waters, the dissolved iron precipitates as goethite, $\\mathrm{FeO}(\\mathrm{OH})$, which coats the stream bottom while the water remains acidic.\n\nCalculate how many moles of pyrite would be required to bring $1.0 \\mathrm{dm}^{3}$ of pure water to a $\\mathrm{pH}$ of 3.0 if the pyrite was completely converted into $\\mathrm{FeO}(\\mathrm{OH})$ and $\\mathrm{H}^{+}$ions. Neglect the formation of $\\mathrm{HSO}_{4}{ }^{-}$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nMany streams drain in areas where coal or metallic ores are mined. These streams have become acidic and contain high concentrations of dissolved iron and sulphate, due to sulphur-containing ores being exposed to the atmosphere or to oxygenated waters. The most common sulphur-containing mineral is pyrite, $\\mathrm{FeS}_{2}$, in which the oxidation state of iron is +2 . As the iron-rich streams mix with other waters, the dissolved iron precipitates as goethite, $\\mathrm{FeO}(\\mathrm{OH})$, which coats the stream bottom while the water remains acidic.\n\nCalculate how many moles of pyrite would be required to bring $1.0 \\mathrm{dm}^{3}$ of pure water to a $\\mathrm{pH}$ of 3.0 if the pyrite was completely converted into $\\mathrm{FeO}(\\mathrm{OH})$ and $\\mathrm{H}^{+}$ions. Neglect the formation of $\\mathrm{HSO}_{4}{ }^{-}$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of $\\mathrm{~mol}$, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "$\\mathrm{~mol}$"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_854",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离常数: $\\mathrm{K}_{1}=4.5 \\times 10^{-7}, \\mathrm{~K}_{2}=4.7 \\times 10^{-11}$, 下列说法中正确的是\nA: $0.1 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液中: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nC: $\\mathrm{NaHCO}_{3}$ 和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 混合液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: $0.5 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$ 之比小于 5\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, $\\mathrm{H}_{2} \\mathrm{CO}_{3}$ 的电离常数: $\\mathrm{K}_{1}=4.5 \\times 10^{-7}, \\mathrm{~K}_{2}=4.7 \\times 10^{-11}$, 下列说法中正确的是\n\nA: $0.1 \\mathrm{~mol} / \\mathrm{LNaHCO}_{3}$ 溶液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nB: $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液中: $2 \\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{H}_{2} \\mathrm{CO}_{3}\\right)$\nC: $\\mathrm{NaHCO}_{3}$ 和 $\\mathrm{Na}_{2} \\mathrm{CO}_{3}$ 混合液中: $\\mathrm{c}\\left(\\mathrm{Na}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)>\\mathrm{c}\\left(\\mathrm{HCO}_{3}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)+\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$\nD: $0.5 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液与 $0.1 \\mathrm{~mol} / \\mathrm{LNa}_{2} \\mathrm{CO}_{3}$ 溶液中 $\\mathrm{c}\\left(\\mathrm{CO}_{3}^{2-}\\right)$ 之比小于 5\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_505",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 向 $20.0 \\mathrm{~mL} 0.100 \\mathrm{~mol} / \\mathrm{L} \\mathrm{HR}$ 溶液中滴加 $0.100 \\mathrm{~mol} / \\mathrm{L} \\mathrm{MOH}$ 溶液, 混合溶液的 $\\mathrm{pH}$ 与加入 $\\mathrm{MOH}$ 溶液体积的变化关系如图所示。下列说法错误的是\n\n[图1]\nA: HR 的电离常数的数量级为 $10^{-6}$\nB: $0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{MOH}$ 溶液的电离度大于 $1 \\%$\nC: $\\mathrm{P}$ 点溶液中水电离出的 $\\mathrm{H}^{+}$浓度一定大于 $10^{-7} \\mathrm{~mol} / \\mathrm{L}$\nD: $\\mathrm{Q}$ 点溶液中存在 $c\\left(\\mathrm{M}^{+}\\right)>c\\left(\\mathrm{R}^{-}\\right)>c(\\mathrm{MOH})>c(\\mathrm{HR})$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 向 $20.0 \\mathrm{~mL} 0.100 \\mathrm{~mol} / \\mathrm{L} \\mathrm{HR}$ 溶液中滴加 $0.100 \\mathrm{~mol} / \\mathrm{L} \\mathrm{MOH}$ 溶液, 混合溶液的 $\\mathrm{pH}$ 与加入 $\\mathrm{MOH}$ 溶液体积的变化关系如图所示。下列说法错误的是\n\n[图1]\n\nA: HR 的电离常数的数量级为 $10^{-6}$\nB: $0.100 \\mathrm{~mol} \\cdot \\mathrm{L}^{-1} \\mathrm{MOH}$ 溶液的电离度大于 $1 \\%$\nC: $\\mathrm{P}$ 点溶液中水电离出的 $\\mathrm{H}^{+}$浓度一定大于 $10^{-7} \\mathrm{~mol} / \\mathrm{L}$\nD: $\\mathrm{Q}$ 点溶液中存在 $c\\left(\\mathrm{M}^{+}\\right)>c\\left(\\mathrm{R}^{-}\\right)>c(\\mathrm{MOH})>c(\\mathrm{HR})$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-089.jpg?height=385&width=574&top_left_y=2149&top_left_x=318"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_365",
        "problem": "What is the $K_{\\mathrm{sp}}$ of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ at $298 \\mathrm{~K}$ ?\n\n| Half-Reaction | $E^{\\mathrm{o}}, \\mathrm{V}$ (at $\\left.298 \\mathrm{~K}\\right)$ |\n| :---: | :---: |\n| $\\mathrm{Hg}_{2}{ }^{2+}+2 e^{-} \\rightarrow 2 \\mathrm{Hg}(l)$ | +0.80 |\n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(\\mathrm{~s})+2 e^{-} \\rightarrow 2 \\mathrm{Hg}(l)+2 \\mathrm{Cl}^{-}(a q)$ | +0.31 |\nA: $2.6 \\times 10^{-17}$\nB: $3.3 \\times 10^{-11}$\nC: $5.1 \\times 10^{-9}$\nD: $5.7 \\times 10^{-6}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nWhat is the $K_{\\mathrm{sp}}$ of $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}$ at $298 \\mathrm{~K}$ ?\n\n| Half-Reaction | $E^{\\mathrm{o}}, \\mathrm{V}$ (at $\\left.298 \\mathrm{~K}\\right)$ |\n| :---: | :---: |\n| $\\mathrm{Hg}_{2}{ }^{2+}+2 e^{-} \\rightarrow 2 \\mathrm{Hg}(l)$ | +0.80 |\n| $\\mathrm{Hg}_{2} \\mathrm{Cl}_{2}(\\mathrm{~s})+2 e^{-} \\rightarrow 2 \\mathrm{Hg}(l)+2 \\mathrm{Cl}^{-}(a q)$ | +0.31 |\n\nA: $2.6 \\times 10^{-17}$\nB: $3.3 \\times 10^{-11}$\nC: $5.1 \\times 10^{-9}$\nD: $5.7 \\times 10^{-6}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_48",
        "problem": "A diamagnetic compound is $71.4 \\%$ arsenic by mass. What is its molecular formula?\nA: $\\mathrm{CH}_{3} \\mathrm{AsN}$\nB: $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{As}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{As}_{2} \\mathrm{~N}_{2}$\nD: $\\mathrm{C}_{4} \\mathrm{H}_{12} \\mathrm{As}_{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nA diamagnetic compound is $71.4 \\%$ arsenic by mass. What is its molecular formula?\n\nA: $\\mathrm{CH}_{3} \\mathrm{AsN}$\nB: $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{As}$\nC: $\\mathrm{C}_{2} \\mathrm{H}_{6} \\mathrm{As}_{2} \\mathrm{~N}_{2}$\nD: $\\mathrm{C}_{4} \\mathrm{H}_{12} \\mathrm{As}_{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_1156",
        "problem": "This question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nIt is predicted that in order to complete the next row of the periodic table starting with electrons in the 8s shell, it will also be necessary to fill elements in the first row of the g-block.\n\nAssuming no orbitals from any shells in the $9^{\\text {th }}$ period are occupied, predict the atomic number of the element beneath oganesson $(\\mathrm{Og})$.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThis question is about the Periodic Table.\n\n2019 marks the $150^{\\text {th }}$ anniversary of the publication of Dmitri Mendeleev's first periodic table, shown on the right. Unlike our modern tables, in this version the Groups of elements are arranged in horizontal rows.\n\nSince the structure of the atom was completely unknown at the time, there was no concept of atomic number and the elements were ordered by their atomic masses. The numbers shown in the table are the atomic masses as known at the time. Before the 1860s, atomic masses were not accurate enough for all the trends and patterns in any arrangement of the elements to be discovered.\n\nMendeleev's table is pretty impressive, but there are a number of errors...\nОІЫТ'Б СПСТЕМЫ ЭЛЕМЕНТОВ'Ь,\n\nОСНОВАННОӤ НА ИХ'Ъ АТОМНОМ'Ъ ВТСТ И ХНМИЧЕСКОМ'b СХОДСТВЪ.\n\n$$\n\\mathrm{Ti}=50 \\quad \\mathrm{Zr}=90 \\quad ?=180 .\n$$\n\n$\\mathrm{V}=-51 \\quad \\mathrm{Nb}=94 \\quad \\mathrm{Ta}=182$\n\n$\\mathrm{Cr}=52 \\quad \\mathrm{M}=-96 \\quad \\mathrm{~W}=186$.\n\n$\\mathrm{Mn}-55 \\mathrm{Rh}=104,4 \\mathrm{Pl}=197$,\n\nFe $=56 \\quad \\mathrm{R} \\|=104,4 \\quad \\mathrm{Ir}-198$.\n\n$\\mathrm{Ni}=\\mathrm{Co}=59 \\quad \\mathrm{P}=106, \\mathrm{i}$ Os -199.\n\n$\\mathrm{H}=1 \\quad \\mathrm{Cu}=63,4 \\quad \\mathrm{Ag}=108 \\quad \\mathrm{Hg}=200$. $\\mathrm{Be}-9,4 \\mathrm{Mg}=24 \\quad \\mathrm{Zn}=65,2 \\mathrm{Cd}=112$\n\n$\\mathrm{B}=11 \\quad \\mathrm{Al}=27,4 \\quad ?=68 . \\mathrm{Ur}=116 \\quad \\mathrm{~A} \\|=197$ ?\n\n$\\mathrm{C}=12 \\quad \\mathrm{Si}=28 \\quad ?=70 \\quad \\mathrm{Sn}=118$\n\n$\\mathrm{N}=14 \\quad \\mathrm{P}=31 \\quad \\mathrm{As}=75 \\quad \\mathrm{Sb}=122 \\quad \\mathrm{Bi}=210$ ?\n\n$0=16 \\quad \\mathrm{~S}=32 \\mathrm{Se}=79,4 \\mathrm{Te}=128$ ?\n\n$\\mathrm{F}=19 \\quad \\mathrm{Cl}=35,5 \\mathrm{Br} \\cdots 80 \\quad \\mathrm{I}=127$\n\n$\\mathrm{Li}=7 \\mathrm{Na}=23 \\quad \\mathrm{~K}=39 \\mathrm{Rb}=85,4 \\quad \\mathrm{Cs}-133 \\quad \\mathrm{Tl}=204$.\n\nCa $=40 \\quad \\mathrm{Sr} \\Rightarrow 57, ; \\quad \\mathrm{Ba}=137 \\quad \\mathrm{~Pb}=207$.\n\n$?=45 \\mathrm{Ce}=92$\n\n?Er $=56 \\quad \\mathrm{La}=94$\n\n? $\\mathrm{Yl}=60 \\quad \\mathrm{Di}=95$\n\n? $\\mathbf{n}=75, \\mathbf{T} \\mathbf{T h}=118 ?$\n\n##  electron configurations\n\nWe now know that elements in the same Group of the periodic table have similar properties because they have the same number of electrons in their outermost (valence) shell. In Mendeleev's second table, elements in the same Group have the same number of valence electrons, even though some elements are from the main block of the periodic table, and some from the transition metals.\n\nOur modern periodic table is based on the electronic configurations of the elements.\n\nFor elements from Groups 1 and 2, the valence electrons are in $s$ orbitals whereas for elements in Groups 13-18, they are in $s$ and $p$ orbitals. In both cases the orbitals have principal quantum numbers the same as the period number.\n\nFor elements in Groups 3-12, the electrons are filling d orbitals in shells with principal quantum numbers one less that the period number. The valence electrons are the s electrons and perhaps some / all of the d electrons.\n\nFor the lanthanoids and actinoids, the electrons are filling $f$ orbitals in shells with principal quantum numbers two less than the period number.\n\nIt is predicted that in order to complete the next row of the periodic table starting with electrons in the 8s shell, it will also be necessary to fill elements in the first row of the g-block.\n\nAssuming no orbitals from any shells in the $9^{\\text {th }}$ period are occupied, predict the atomic number of the element beneath oganesson $(\\mathrm{Og})$.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_581",
        "problem": "合成药物异搏定路线中某一步骤如下:\n\n[图1]\n\n下列说法错误的是\nA: 物质 X 中所有原子可能在同一平面内\nB: 物质 X 在空气中易被氧化\nC: 等物质的量的 $\\mathrm{X} 、 \\mathrm{Y}$ 分别与 $\\mathrm{NaOH}$ 反应, 最多消耗 $\\mathrm{NaOH}$ 的物质的量之比为 $1: 2$\nD: 等物质的量的 $X 、 Z$ 分别与 $\\mathrm{H}_{2}$ 加成, 最多消耗 $\\mathrm{H}_{2}$ 的物质的量之比为 4: 6\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n合成药物异搏定路线中某一步骤如下:\n\n[图1]\n\n下列说法错误的是\n\nA: 物质 X 中所有原子可能在同一平面内\nB: 物质 X 在空气中易被氧化\nC: 等物质的量的 $\\mathrm{X} 、 \\mathrm{Y}$ 分别与 $\\mathrm{NaOH}$ 反应, 最多消耗 $\\mathrm{NaOH}$ 的物质的量之比为 $1: 2$\nD: 等物质的量的 $X 、 Z$ 分别与 $\\mathrm{H}_{2}$ 加成, 最多消耗 $\\mathrm{H}_{2}$ 的物质的量之比为 4: 6\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_d1b519b524f279de7793g-67.jpg?height=263&width=1328&top_left_y=157&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_1470",
        "problem": "Distribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of total phosphate $\\left(\\mathrm{PO}_{4}{ }_{4}^{3-}\\right)$ and silicate $\\left(\\mathrm{SiO}_{4}^{4-}\\right)$\n\nA 5.00 gram of soil sample is digested to give a final volume of $50.0 \\mathrm{~cm}^{3}$ digesting solution which dissolves total phosphorus and silicon. The extract is analyzed for the total concentrations of phosphorus and silicon. The concentrations of phosphorus and silicon are found to be $5.16 \\mathrm{mg} \\mathrm{dm}^{-3}$ and $5.35 \\mathrm{mg} \\mathrm{dm}^{-3}$, respectively.Determine the mass of $\\mathrm{PO}_{4}{ }^{3-}$ in $\\mathrm{mg}$ per $1.00 \\mathrm{~g}$ of soil.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\nHere is some context information for this question, which might assist you in solving it:\nDistribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:\n\nDetermination of total phosphate $\\left(\\mathrm{PO}_{4}{ }_{4}^{3-}\\right)$ and silicate $\\left(\\mathrm{SiO}_{4}^{4-}\\right)$\n\nA 5.00 gram of soil sample is digested to give a final volume of $50.0 \\mathrm{~cm}^{3}$ digesting solution which dissolves total phosphorus and silicon. The extract is analyzed for the total concentrations of phosphorus and silicon. The concentrations of phosphorus and silicon are found to be $5.16 \\mathrm{mg} \\mathrm{dm}^{-3}$ and $5.35 \\mathrm{mg} \\mathrm{dm}^{-3}$, respectively.\n\nproblem:\nDetermine the mass of $\\mathrm{PO}_{4}{ }^{3-}$ in $\\mathrm{mg}$ per $1.00 \\mathrm{~g}$ of soil.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mg, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mg"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_273",
        "problem": "Thermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the concentration of the oxalic acid solution (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ).\nOxalic acid reacts with sodium hydroxide to produce sodium oxalate and water, according to the following chemical equation:\n\n$\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}($ aq $)+2 \\mathrm{NaOH}($ aq $) \\rightarrow \\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$\n\n$20.00 \\mathrm{~mL}$ of the oxalic acid solution above requires $18.57 \\mathrm{~mL}$ of a sodium hydroxide solution for complete reaction.",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThe answer to this question is a numerical value.\n\nproblem:\nThermogravimetric analysis involves measuring the mass of a sample as it is heated.\n\nCompound $\\mathbf{A}$ decomposes with increasing temperature to give a sequence of compounds $\\mathbf{B}$, $\\mathbf{C}$ and $\\mathbf{D}$, all of which contain calcium. At each stage, a small molecule is also given off (denoted molecules 1, 2 and 3 respectively), which results in the mass of each successive compound being smaller.\n\n[figure1]\n\nThe mass of each compound, expressed as a percentage of the original mass of $\\mathbf{A}$, is recorded in the table below.\n\n| Compound | Percentage of original <br> mass of A remaining |\n| :---: | :---: |\n| A | 100.0 |\n| B | 87.67 |\n| C | 68.50 |\n| D | 38.38 |\n\nIt is known that compound $\\mathbf{B}$ is calcium oxalate, $\\mathrm{CaC}_{2} \\mathrm{O}_{4}$.\nCalculate the concentration of the oxalic acid solution (in $\\mathrm{mol} \\mathrm{L}^{-1}$ ).\nOxalic acid reacts with sodium hydroxide to produce sodium oxalate and water, according to the following chemical equation:\n\n$\\mathrm{H}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}($ aq $)+2 \\mathrm{NaOH}($ aq $) \\rightarrow \\mathrm{Na}_{2} \\mathrm{C}_{2} \\mathrm{O}_{4}(\\mathrm{aq})+2 \\mathrm{H}_{2} \\mathrm{O}(\\mathrm{l})$\n\n$20.00 \\mathrm{~mL}$ of the oxalic acid solution above requires $18.57 \\mathrm{~mL}$ of a sodium hydroxide solution for complete reaction.\n\nAll mathematical formulas and symbols you output should be represented with LaTeX!\nYou can solve it step by step.\nRemember, your answer should be calculated in the unit of mol/L, but when concluding your final answer, do not include the unit.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER is the numerical value without any units.",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_06_b203bdbc8bac24f6fc7cg-09.jpg?height=322&width=948&top_left_y=570&top_left_x=425"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "NV",
        "unit": [
            "mol/L"
        ],
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_133",
        "problem": "Use the information provided below to calculate the enthalpy of reaction when one mole of chlorine trifluoride gas decomposes into one mole of chlorine monofluoride gas and one mole of gaseous fluorine.\n\n$$\n\\begin{array}{ll}\n\\left.2 \\mathrm{ClF}^{\\mathrm{g}}\\right)+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{Cl}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{F}_{2} \\mathrm{O}(\\mathrm{g}) & \\Delta \\mathrm{H}=+167.4 \\mathrm{~kJ} \\\\\n2 \\mathrm{ClF}_{3}(\\mathrm{~g})+2 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{Cl}_{2} \\mathrm{O}(\\mathrm{g})+3 \\mathrm{~F}_{2} \\mathrm{O}(\\mathrm{g}) & \\Delta \\mathrm{H}=+341.4 \\mathrm{~kJ} \\\\\n3 \\mathrm{~F}_{2} \\mathrm{O}(\\mathrm{g}) \\rightarrow 3 \\mathrm{~F}_{2}(\\mathrm{~g})+\\frac{3}{2} \\mathrm{O}_{2}(\\mathrm{~g}) & \\Delta \\mathrm{H}=+65.1 \\mathrm{~kJ}\n\\end{array}\n$$\nA: $-217.4 \\mathrm{~kJ}$\nB: $+25.0 \\mathrm{~kJ}$\nC: $+68.4 \\mathrm{~kJ}$\nD: $+108.7 \\mathrm{~kJ}$\nE: $573.9 \\mathrm{~kJ}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nUse the information provided below to calculate the enthalpy of reaction when one mole of chlorine trifluoride gas decomposes into one mole of chlorine monofluoride gas and one mole of gaseous fluorine.\n\n$$\n\\begin{array}{ll}\n\\left.2 \\mathrm{ClF}^{\\mathrm{g}}\\right)+\\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{Cl}_{2} \\mathrm{O}(\\mathrm{g})+\\mathrm{F}_{2} \\mathrm{O}(\\mathrm{g}) & \\Delta \\mathrm{H}=+167.4 \\mathrm{~kJ} \\\\\n2 \\mathrm{ClF}_{3}(\\mathrm{~g})+2 \\mathrm{O}_{2}(\\mathrm{~g}) \\rightarrow \\mathrm{Cl}_{2} \\mathrm{O}(\\mathrm{g})+3 \\mathrm{~F}_{2} \\mathrm{O}(\\mathrm{g}) & \\Delta \\mathrm{H}=+341.4 \\mathrm{~kJ} \\\\\n3 \\mathrm{~F}_{2} \\mathrm{O}(\\mathrm{g}) \\rightarrow 3 \\mathrm{~F}_{2}(\\mathrm{~g})+\\frac{3}{2} \\mathrm{O}_{2}(\\mathrm{~g}) & \\Delta \\mathrm{H}=+65.1 \\mathrm{~kJ}\n\\end{array}\n$$\n\nA: $-217.4 \\mathrm{~kJ}$\nB: $+25.0 \\mathrm{~kJ}$\nC: $+68.4 \\mathrm{~kJ}$\nD: $+108.7 \\mathrm{~kJ}$\nE: $573.9 \\mathrm{~kJ}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D, E].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_82",
        "problem": "Nitric oxide is formed at high temperature in the presence of oxygen and nitrogen. A proposed mechanism for its formation is shown:\n\n$$\n\\begin{array}{ll}\n\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{O}(g) & \\text { fast, unfavorable } \\\\\n\\mathrm{N}_{2}(g)+\\mathrm{O}(g) \\rightarrow \\mathrm{NO}(g)+\\mathrm{N}(g) & \\text { slow } \\\\\n\\mathrm{N}(g)+\\mathrm{O}(g) \\rightarrow \\mathrm{NO}(g) & \\text { very fast }\n\\end{array}\n$$\n\nWhat rate law is predicted by this mechanism?\nA: Rate $=k\\left[\\mathrm{O}_{2}\\right]$\nB: Rate $=k\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]$\nC: Rate $=k\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]^{1 / 2}$\nD: Rate $=k\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]^{2}$\n",
        "prompt": "You are participating in an international Chemistry competition and need to solve the following question.\nThis is a multiple choice question (only one correct answer).\n\nproblem:\nNitric oxide is formed at high temperature in the presence of oxygen and nitrogen. A proposed mechanism for its formation is shown:\n\n$$\n\\begin{array}{ll}\n\\mathrm{O}_{2}(g) \\rightleftharpoons 2 \\mathrm{O}(g) & \\text { fast, unfavorable } \\\\\n\\mathrm{N}_{2}(g)+\\mathrm{O}(g) \\rightarrow \\mathrm{NO}(g)+\\mathrm{N}(g) & \\text { slow } \\\\\n\\mathrm{N}(g)+\\mathrm{O}(g) \\rightarrow \\mathrm{NO}(g) & \\text { very fast }\n\\end{array}\n$$\n\nWhat rate law is predicted by this mechanism?\n\nA: Rate $=k\\left[\\mathrm{O}_{2}\\right]$\nB: Rate $=k\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]$\nC: Rate $=k\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]^{1 / 2}$\nD: Rate $=k\\left[\\mathrm{~N}_{2}\\right]\\left[\\mathrm{O}_{2}\\right]^{2}$\n\nYou can solve it step by step.\nPlease end your response with: \"The final answer is $\\boxed{ANSWER}$\", where ANSWER should be one of the options: [A, B, C, D].",
        "figure_urls": null,
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "EN",
        "modality": "text-only"
    },
    {
        "id": "Chemistry_615",
        "problem": "钠碱脱硫液 $\\left(\\mathrm{NaOH}+\\mathrm{Na}_{2} \\mathrm{SO}_{3}\\right)$ 吸收一定量 $\\mathrm{SO}_{2}$ 气体后, 可通过以下装置实现再生。下列说法正确的是\n\n[图1]\nA: 电极 $\\mathrm{a}$ 为负极\nB: $\\mathrm{m}$ 应为阳离子交换膜\nC: 出液 2 仍为 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, 与进液的浓度相同\nD: 出液 1 可使澳水裉色\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个多选题（有多个正确答案）。\n\n问题:\n钠碱脱硫液 $\\left(\\mathrm{NaOH}+\\mathrm{Na}_{2} \\mathrm{SO}_{3}\\right)$ 吸收一定量 $\\mathrm{SO}_{2}$ 气体后, 可通过以下装置实现再生。下列说法正确的是\n\n[图1]\n\nA: 电极 $\\mathrm{a}$ 为负极\nB: $\\mathrm{m}$ 应为阳离子交换膜\nC: 出液 2 仍为 $\\mathrm{H}_{2} \\mathrm{SO}_{4}$, 与进液的浓度相同\nD: 出液 1 可使澳水裉色\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为两个或更多的选项：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_8e75292c6204fb2a9e3ag-28.jpg?height=643&width=940&top_left_y=158&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "MC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    },
    {
        "id": "Chemistry_894",
        "problem": "$25^{\\circ} \\mathrm{C}$ 时, 向 $100 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{LNH}_{4} \\mathrm{HRO}_{4}$ 溶液中滴加 $0.1 \\mathrm{~mol} / \\mathrm{LKOH}$ 溶液, 溶液 $\\mathrm{pH}$ 与\n\n$\\mathrm{KOH}$ 溶液体积的关系曲线如图所示 (已知: $\\mathrm{NH}_{4} \\mathrm{HRO}_{4}=\\mathrm{NH}_{4}^{+}+\\mathrm{RO}_{4}^{2-}+\\mathrm{H}^{+}$), 对图中 $a$ 、\n\n$\\mathrm{b} 、 \\mathrm{c} 、 d$ 四个点分析中, 不正确的是\n\n[图1]\nA: $a$ 点水的电离程度最大, 且 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $b$ 点 $c\\left(\\mathrm{~K}^{+}\\right)+c\\left(\\mathrm{NH}_{4}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{RO}_{4}^{2-}\\right)$\nC: $\\mathrm{c}$ 点 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)$\nD: $d$ 点 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{RO}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$, 且 $$ \\mathrm{c}\\left(\\mathrm{RO}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right) $$\n",
        "prompt": "你正在参加一个国际化学竞赛，并需要解决以下问题。\n这是一个单选题（只有一个正确答案）。\n\n问题:\n$25^{\\circ} \\mathrm{C}$ 时, 向 $100 \\mathrm{~mL} 0.1 \\mathrm{~mol} / \\mathrm{LNH}_{4} \\mathrm{HRO}_{4}$ 溶液中滴加 $0.1 \\mathrm{~mol} / \\mathrm{LKOH}$ 溶液, 溶液 $\\mathrm{pH}$ 与\n\n$\\mathrm{KOH}$ 溶液体积的关系曲线如图所示 (已知: $\\mathrm{NH}_{4} \\mathrm{HRO}_{4}=\\mathrm{NH}_{4}^{+}+\\mathrm{RO}_{4}^{2-}+\\mathrm{H}^{+}$), 对图中 $a$ 、\n\n$\\mathrm{b} 、 \\mathrm{c} 、 d$ 四个点分析中, 不正确的是\n\n[图1]\n\nA: $a$ 点水的电离程度最大, 且 $\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)=\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$\nB: $b$ 点 $c\\left(\\mathrm{~K}^{+}\\right)+c\\left(\\mathrm{NH}_{4}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{RO}_{4}^{2-}\\right)$\nC: $\\mathrm{c}$ 点 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{NH}_{3} \\cdot \\mathrm{H}_{2} \\mathrm{O}\\right)=\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)$\nD: $d$ 点 $\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{K}^{+}\\right)+\\mathrm{c}\\left(\\mathrm{H}^{+}\\right)=2 \\mathrm{c}\\left(\\mathrm{RO}_{4}^{2-}\\right)+\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)$, 且 $$ \\mathrm{c}\\left(\\mathrm{RO}_{4}^{2-}\\right)>\\mathrm{c}\\left(\\mathrm{OH}^{-}\\right)>\\mathrm{c}\\left(\\mathrm{NH}_{4}^{+}\\right) $$\n\n你可以一步一步来解决这个问题，并输出详细的解答过程。\n你需要在输出的最后用以下格式总结答案：“最终答案是$\\boxed{ANSWER}$”，其中ANSWER应为以下选项之一：[A, B, C, D]",
        "figure_urls": [
            "https://cdn.mathpix.com/cropped/2024_03_31_9ae9edb59ad378207fffg-106.jpg?height=406&width=706&top_left_y=822&top_left_x=341"
        ],
        "answer": null,
        "solution": null,
        "answer_type": "SC",
        "unit": null,
        "answer_sequence": null,
        "type_sequence": null,
        "test_cases": null,
        "subject": "Chemistry",
        "language": "ZH",
        "modality": "multi-modal"
    }
]