Chapter
stringclasses 18
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|---|---|---|
1 | 205-208 | 14
Sn2+ + 2e– ¾® Sn
–0 14
AgI + e– ¾® Ag + I–
–0 15
Ni2+ + 2e– ¾® Ni
–0 23
V3+ + e– ¾® V2+
–0 |
1 | 206-209 | 14
AgI + e– ¾® Ag + I–
–0 15
Ni2+ + 2e– ¾® Ni
–0 23
V3+ + e– ¾® V2+
–0 26
Co2+ + 2e– ¾® Co
–0 |
1 | 207-210 | 15
Ni2+ + 2e– ¾® Ni
–0 23
V3+ + e– ¾® V2+
–0 26
Co2+ + 2e– ¾® Co
–0 28
In3+ + 3e– ¾® In
–0 |
1 | 208-211 | 23
V3+ + e– ¾® V2+
–0 26
Co2+ + 2e– ¾® Co
–0 28
In3+ + 3e– ¾® In
–0 34
Tl+ + e– ¾® Tl
–0 |
1 | 209-212 | 26
Co2+ + 2e– ¾® Co
–0 28
In3+ + 3e– ¾® In
–0 34
Tl+ + e– ¾® Tl
–0 34
PbSO4 + 2e– ¾® Pb + SO2–
4
–0 |
1 | 210-213 | 28
In3+ + 3e– ¾® In
–0 34
Tl+ + e– ¾® Tl
–0 34
PbSO4 + 2e– ¾® Pb + SO2–
4
–0 36
Ti3+ + e– ¾® Ti2+
–0 |
1 | 211-214 | 34
Tl+ + e– ¾® Tl
–0 34
PbSO4 + 2e– ¾® Pb + SO2–
4
–0 36
Ti3+ + e– ¾® Ti2+
–0 37
Cd2+ + 2e– ¾® Cd
–0 |
1 | 212-215 | 34
PbSO4 + 2e– ¾® Pb + SO2–
4
–0 36
Ti3+ + e– ¾® Ti2+
–0 37
Cd2+ + 2e– ¾® Cd
–0 40
In2+ + e– ¾® In+
–0 |
1 | 213-216 | 36
Ti3+ + e– ¾® Ti2+
–0 37
Cd2+ + 2e– ¾® Cd
–0 40
In2+ + e– ¾® In+
–0 40
Cr3+ + e– ¾® Cr2+
–0 |
1 | 214-217 | 37
Cd2+ + 2e– ¾® Cd
–0 40
In2+ + e– ¾® In+
–0 40
Cr3+ + e– ¾® Cr2+
–0 41
Fe2+ + 2e– ¾® Fe
–0 |
1 | 215-218 | 40
In2+ + e– ¾® In+
–0 40
Cr3+ + e– ¾® Cr2+
–0 41
Fe2+ + 2e– ¾® Fe
–0 44
In3+ + 2e– ¾® In+
–0 |
1 | 216-219 | 40
Cr3+ + e– ¾® Cr2+
–0 41
Fe2+ + 2e– ¾® Fe
–0 44
In3+ + 2e– ¾® In+
–0 44
S + 2e– ¾® S2–
–0 |
1 | 217-220 | 41
Fe2+ + 2e– ¾® Fe
–0 44
In3+ + 2e– ¾® In+
–0 44
S + 2e– ¾® S2–
–0 48
In3+ + e– ¾® In2+
–0 |
1 | 218-221 | 44
In3+ + 2e– ¾® In+
–0 44
S + 2e– ¾® S2–
–0 48
In3+ + e– ¾® In2+
–0 49
U4+ + e– ¾® U3+
–0 |
1 | 219-222 | 44
S + 2e– ¾® S2–
–0 48
In3+ + e– ¾® In2+
–0 49
U4+ + e– ¾® U3+
–0 61
Cr3+ + 3e– ¾® Cr
–0 |
1 | 220-223 | 48
In3+ + e– ¾® In2+
–0 49
U4+ + e– ¾® U3+
–0 61
Cr3+ + 3e– ¾® Cr
–0 74
Zn2+ + 2e– ¾® Zn
–0 |
1 | 221-224 | 49
U4+ + e– ¾® U3+
–0 61
Cr3+ + 3e– ¾® Cr
–0 74
Zn2+ + 2e– ¾® Zn
–0 76
(continued)
Standard potentials at 298 K in electrochemical order
APPENDIX III
Rationalised 2023-24
144
Chemistry
Reduction half-reaction
Eo/V
Cd(OH)2 + 2e– ¾® Cd + 2OH–
–0 |
1 | 222-225 | 61
Cr3+ + 3e– ¾® Cr
–0 74
Zn2+ + 2e– ¾® Zn
–0 76
(continued)
Standard potentials at 298 K in electrochemical order
APPENDIX III
Rationalised 2023-24
144
Chemistry
Reduction half-reaction
Eo/V
Cd(OH)2 + 2e– ¾® Cd + 2OH–
–0 81
2H2O + 2e– ¾® H2 + 2OH–
–0 |
1 | 223-226 | 74
Zn2+ + 2e– ¾® Zn
–0 76
(continued)
Standard potentials at 298 K in electrochemical order
APPENDIX III
Rationalised 2023-24
144
Chemistry
Reduction half-reaction
Eo/V
Cd(OH)2 + 2e– ¾® Cd + 2OH–
–0 81
2H2O + 2e– ¾® H2 + 2OH–
–0 83
Cr2+ + 2e– ¾® Cr
–0 |
1 | 224-227 | 76
(continued)
Standard potentials at 298 K in electrochemical order
APPENDIX III
Rationalised 2023-24
144
Chemistry
Reduction half-reaction
Eo/V
Cd(OH)2 + 2e– ¾® Cd + 2OH–
–0 81
2H2O + 2e– ¾® H2 + 2OH–
–0 83
Cr2+ + 2e– ¾® Cr
–0 91
Mn2+ + 2e– ¾® Mn
–1 |
1 | 225-228 | 81
2H2O + 2e– ¾® H2 + 2OH–
–0 83
Cr2+ + 2e– ¾® Cr
–0 91
Mn2+ + 2e– ¾® Mn
–1 18
V2+ + 2e– ¾® V
–1 |
1 | 226-229 | 83
Cr2+ + 2e– ¾® Cr
–0 91
Mn2+ + 2e– ¾® Mn
–1 18
V2+ + 2e– ¾® V
–1 19
Ti2+ + 2e– ¾® Ti
–1 |
1 | 227-230 | 91
Mn2+ + 2e– ¾® Mn
–1 18
V2+ + 2e– ¾® V
–1 19
Ti2+ + 2e– ¾® Ti
–1 63
Al3+ + 3e– ¾® Al
–1 |
1 | 228-231 | 18
V2+ + 2e– ¾® V
–1 19
Ti2+ + 2e– ¾® Ti
–1 63
Al3+ + 3e– ¾® Al
–1 66
U3+ + 3e– ¾® U
–1 |
1 | 229-232 | 19
Ti2+ + 2e– ¾® Ti
–1 63
Al3+ + 3e– ¾® Al
–1 66
U3+ + 3e– ¾® U
–1 79
Sc3+ + 3e– ¾® Sc
–2 |
1 | 230-233 | 63
Al3+ + 3e– ¾® Al
–1 66
U3+ + 3e– ¾® U
–1 79
Sc3+ + 3e– ¾® Sc
–2 09
Mg2+ + 2e– ¾® Mg
–2 |
1 | 231-234 | 66
U3+ + 3e– ¾® U
–1 79
Sc3+ + 3e– ¾® Sc
–2 09
Mg2+ + 2e– ¾® Mg
–2 36
Ce3+ + 3e– ¾® Ce
–2 |
1 | 232-235 | 79
Sc3+ + 3e– ¾® Sc
–2 09
Mg2+ + 2e– ¾® Mg
–2 36
Ce3+ + 3e– ¾® Ce
–2 48
Reduction half-reaction
Eo/V
La3+ + 3e– ¾® La
–2 |
1 | 233-236 | 09
Mg2+ + 2e– ¾® Mg
–2 36
Ce3+ + 3e– ¾® Ce
–2 48
Reduction half-reaction
Eo/V
La3+ + 3e– ¾® La
–2 52
Na+ + e– ¾® Na
–2 |
1 | 234-237 | 36
Ce3+ + 3e– ¾® Ce
–2 48
Reduction half-reaction
Eo/V
La3+ + 3e– ¾® La
–2 52
Na+ + e– ¾® Na
–2 71
Ca2+ + 2e– ¾® Ca
–2 |
1 | 235-238 | 48
Reduction half-reaction
Eo/V
La3+ + 3e– ¾® La
–2 52
Na+ + e– ¾® Na
–2 71
Ca2+ + 2e– ¾® Ca
–2 87
Sr2+ + 2e– ¾® Sr
–2 |
1 | 236-239 | 52
Na+ + e– ¾® Na
–2 71
Ca2+ + 2e– ¾® Ca
–2 87
Sr2+ + 2e– ¾® Sr
–2 89
Ba2+ + 2e– ¾® Ba
–2 |
1 | 237-240 | 71
Ca2+ + 2e– ¾® Ca
–2 87
Sr2+ + 2e– ¾® Sr
–2 89
Ba2+ + 2e– ¾® Ba
–2 91
Ra2+ + 2e– ¾® Ra
–2 |
1 | 238-241 | 87
Sr2+ + 2e– ¾® Sr
–2 89
Ba2+ + 2e– ¾® Ba
–2 91
Ra2+ + 2e– ¾® Ra
–2 92
Cs+ + e– ¾® Cs
–2 |
1 | 239-242 | 89
Ba2+ + 2e– ¾® Ba
–2 91
Ra2+ + 2e– ¾® Ra
–2 92
Cs+ + e– ¾® Cs
–2 92
Rb+ + e– ¾® Rb
–2 |
1 | 240-243 | 91
Ra2+ + 2e– ¾® Ra
–2 92
Cs+ + e– ¾® Cs
–2 92
Rb+ + e– ¾® Rb
–2 93
K+ +e– ¾® K
–2 |
1 | 241-244 | 92
Cs+ + e– ¾® Cs
–2 92
Rb+ + e– ¾® Rb
–2 93
K+ +e– ¾® K
–2 93
Li+ + e– ¾® Li
–3 |
1 | 242-245 | 92
Rb+ + e– ¾® Rb
–2 93
K+ +e– ¾® K
–2 93
Li+ + e– ¾® Li
–3 05
APPENDIX III CONTINUED
Rationalised 2023-24
145
Appendix
Sometimes, a numerical expression may involve multiplication, division or rational powers of large
numbers |
1 | 243-246 | 93
K+ +e– ¾® K
–2 93
Li+ + e– ¾® Li
–3 05
APPENDIX III CONTINUED
Rationalised 2023-24
145
Appendix
Sometimes, a numerical expression may involve multiplication, division or rational powers of large
numbers For such calculations, logarithms are very useful |
1 | 244-247 | 93
Li+ + e– ¾® Li
–3 05
APPENDIX III CONTINUED
Rationalised 2023-24
145
Appendix
Sometimes, a numerical expression may involve multiplication, division or rational powers of large
numbers For such calculations, logarithms are very useful They help us in making difficult calculations
easy |
1 | 245-248 | 05
APPENDIX III CONTINUED
Rationalised 2023-24
145
Appendix
Sometimes, a numerical expression may involve multiplication, division or rational powers of large
numbers For such calculations, logarithms are very useful They help us in making difficult calculations
easy In Chemistry, logarithm values are required in solving problems of chemical kinetics, thermodynamics,
electrochemistry, etc |
1 | 246-249 | For such calculations, logarithms are very useful They help us in making difficult calculations
easy In Chemistry, logarithm values are required in solving problems of chemical kinetics, thermodynamics,
electrochemistry, etc We shall first introduce this concept, and discuss the laws, which will have to be
followed in working with logarithms, and then apply this technique to a number of problems to show
how it makes difficult calculations simple |
1 | 247-250 | They help us in making difficult calculations
easy In Chemistry, logarithm values are required in solving problems of chemical kinetics, thermodynamics,
electrochemistry, etc We shall first introduce this concept, and discuss the laws, which will have to be
followed in working with logarithms, and then apply this technique to a number of problems to show
how it makes difficult calculations simple We know that
23 = 8, 32 = 9, 53 = 125, 70 = 1
In general, for a positive real number a, and a rational number m, let am = b,
where b is a real number |
1 | 248-251 | In Chemistry, logarithm values are required in solving problems of chemical kinetics, thermodynamics,
electrochemistry, etc We shall first introduce this concept, and discuss the laws, which will have to be
followed in working with logarithms, and then apply this technique to a number of problems to show
how it makes difficult calculations simple We know that
23 = 8, 32 = 9, 53 = 125, 70 = 1
In general, for a positive real number a, and a rational number m, let am = b,
where b is a real number In other words
the mth power of base a is b |
1 | 249-252 | We shall first introduce this concept, and discuss the laws, which will have to be
followed in working with logarithms, and then apply this technique to a number of problems to show
how it makes difficult calculations simple We know that
23 = 8, 32 = 9, 53 = 125, 70 = 1
In general, for a positive real number a, and a rational number m, let am = b,
where b is a real number In other words
the mth power of base a is b Another way of stating the same fact is
logarithm of b to base a is m |
1 | 250-253 | We know that
23 = 8, 32 = 9, 53 = 125, 70 = 1
In general, for a positive real number a, and a rational number m, let am = b,
where b is a real number In other words
the mth power of base a is b Another way of stating the same fact is
logarithm of b to base a is m If for a positive real number a, a ¹ 1
am = b,
we say that m is the logarithm of b to the base a |
1 | 251-254 | In other words
the mth power of base a is b Another way of stating the same fact is
logarithm of b to base a is m If for a positive real number a, a ¹ 1
am = b,
we say that m is the logarithm of b to the base a We write this as
logab
m ,
“log” being the abbreviation of the word “logarithm” |
1 | 252-255 | Another way of stating the same fact is
logarithm of b to base a is m If for a positive real number a, a ¹ 1
am = b,
we say that m is the logarithm of b to the base a We write this as
logab
m ,
“log” being the abbreviation of the word “logarithm” Thus, we have
=
=
=
=
=
=
=
=
3
2
2
3
3
5
0
7
log
8
3,
Since2
8
log
9
2,
Since3
9
log 125
3,
Since5
125
log 1
0,
Since7
1
Laws of Logarithms
In the following discussion, we shall take logarithms to any base a, (a > 0 and a ¹ 1)
First Law: loga
(mn) = logam + logan
Proof: Suppose that logam = x and logan = y
Then ax= m, ay = n
Hence mn = ax |
1 | 253-256 | If for a positive real number a, a ¹ 1
am = b,
we say that m is the logarithm of b to the base a We write this as
logab
m ,
“log” being the abbreviation of the word “logarithm” Thus, we have
=
=
=
=
=
=
=
=
3
2
2
3
3
5
0
7
log
8
3,
Since2
8
log
9
2,
Since3
9
log 125
3,
Since5
125
log 1
0,
Since7
1
Laws of Logarithms
In the following discussion, we shall take logarithms to any base a, (a > 0 and a ¹ 1)
First Law: loga
(mn) = logam + logan
Proof: Suppose that logam = x and logan = y
Then ax= m, ay = n
Hence mn = ax ay = ax+y
It now follows from the definition of logarithms that
loga (mn) = x + y = loga m – loga n
Second Law: loga
m
n
= loga m – logan
Proof: Let logam = x, logan = y
Logarithms
APPENDIX IV
Rationalised 2023-24
146
Chemistry
Then ax = m, ay = n
Hence
x
x y
ay
m
a
n
a
Therefore
a
a
a
m
log
x
y
log
m
log
n
n
Third Law : loga(mn) = n logam
Proof : As before, if logam = x, then ax = m
Then
n
n
x
nx
m
a
a
giving loga(mn) = nx = n loga m
Thus according to First Law: “the log of the product of two numbers is equal to the sum of their logs |
1 | 254-257 | We write this as
logab
m ,
“log” being the abbreviation of the word “logarithm” Thus, we have
=
=
=
=
=
=
=
=
3
2
2
3
3
5
0
7
log
8
3,
Since2
8
log
9
2,
Since3
9
log 125
3,
Since5
125
log 1
0,
Since7
1
Laws of Logarithms
In the following discussion, we shall take logarithms to any base a, (a > 0 and a ¹ 1)
First Law: loga
(mn) = logam + logan
Proof: Suppose that logam = x and logan = y
Then ax= m, ay = n
Hence mn = ax ay = ax+y
It now follows from the definition of logarithms that
loga (mn) = x + y = loga m – loga n
Second Law: loga
m
n
= loga m – logan
Proof: Let logam = x, logan = y
Logarithms
APPENDIX IV
Rationalised 2023-24
146
Chemistry
Then ax = m, ay = n
Hence
x
x y
ay
m
a
n
a
Therefore
a
a
a
m
log
x
y
log
m
log
n
n
Third Law : loga(mn) = n logam
Proof : As before, if logam = x, then ax = m
Then
n
n
x
nx
m
a
a
giving loga(mn) = nx = n loga m
Thus according to First Law: “the log of the product of two numbers is equal to the sum of their logs Similarly, the Second Law says: the log of the ratio of two numbers is the difference of their logs |
1 | 255-258 | Thus, we have
=
=
=
=
=
=
=
=
3
2
2
3
3
5
0
7
log
8
3,
Since2
8
log
9
2,
Since3
9
log 125
3,
Since5
125
log 1
0,
Since7
1
Laws of Logarithms
In the following discussion, we shall take logarithms to any base a, (a > 0 and a ¹ 1)
First Law: loga
(mn) = logam + logan
Proof: Suppose that logam = x and logan = y
Then ax= m, ay = n
Hence mn = ax ay = ax+y
It now follows from the definition of logarithms that
loga (mn) = x + y = loga m – loga n
Second Law: loga
m
n
= loga m – logan
Proof: Let logam = x, logan = y
Logarithms
APPENDIX IV
Rationalised 2023-24
146
Chemistry
Then ax = m, ay = n
Hence
x
x y
ay
m
a
n
a
Therefore
a
a
a
m
log
x
y
log
m
log
n
n
Third Law : loga(mn) = n logam
Proof : As before, if logam = x, then ax = m
Then
n
n
x
nx
m
a
a
giving loga(mn) = nx = n loga m
Thus according to First Law: “the log of the product of two numbers is equal to the sum of their logs Similarly, the Second Law says: the log of the ratio of two numbers is the difference of their logs Thus,
the use of these laws converts a problem of multiplication/division into a problem of addition/subtraction,
which are far easier to perform than multiplication/division |
1 | 256-259 | ay = ax+y
It now follows from the definition of logarithms that
loga (mn) = x + y = loga m – loga n
Second Law: loga
m
n
= loga m – logan
Proof: Let logam = x, logan = y
Logarithms
APPENDIX IV
Rationalised 2023-24
146
Chemistry
Then ax = m, ay = n
Hence
x
x y
ay
m
a
n
a
Therefore
a
a
a
m
log
x
y
log
m
log
n
n
Third Law : loga(mn) = n logam
Proof : As before, if logam = x, then ax = m
Then
n
n
x
nx
m
a
a
giving loga(mn) = nx = n loga m
Thus according to First Law: “the log of the product of two numbers is equal to the sum of their logs Similarly, the Second Law says: the log of the ratio of two numbers is the difference of their logs Thus,
the use of these laws converts a problem of multiplication/division into a problem of addition/subtraction,
which are far easier to perform than multiplication/division That is why logarithms are so useful in
all numerical computations |
1 | 257-260 | Similarly, the Second Law says: the log of the ratio of two numbers is the difference of their logs Thus,
the use of these laws converts a problem of multiplication/division into a problem of addition/subtraction,
which are far easier to perform than multiplication/division That is why logarithms are so useful in
all numerical computations Logarithms to Base 10
Because number 10 is the base of writing numbers, it is very convenient to use logarithms to the base
10 |
1 | 258-261 | Thus,
the use of these laws converts a problem of multiplication/division into a problem of addition/subtraction,
which are far easier to perform than multiplication/division That is why logarithms are so useful in
all numerical computations Logarithms to Base 10
Because number 10 is the base of writing numbers, it is very convenient to use logarithms to the base
10 Some examples are:
log10
10 = 1,
since 101 = 10
log10 100 = 2,
since 102 = 100
log10 10000 = 4,
since 104 = 10000
log10 0 |
1 | 259-262 | That is why logarithms are so useful in
all numerical computations Logarithms to Base 10
Because number 10 is the base of writing numbers, it is very convenient to use logarithms to the base
10 Some examples are:
log10
10 = 1,
since 101 = 10
log10 100 = 2,
since 102 = 100
log10 10000 = 4,
since 104 = 10000
log10 0 01 = –2,
since 10–2 = 0 |
1 | 260-263 | Logarithms to Base 10
Because number 10 is the base of writing numbers, it is very convenient to use logarithms to the base
10 Some examples are:
log10
10 = 1,
since 101 = 10
log10 100 = 2,
since 102 = 100
log10 10000 = 4,
since 104 = 10000
log10 0 01 = –2,
since 10–2 = 0 01
log10 0 |
1 | 261-264 | Some examples are:
log10
10 = 1,
since 101 = 10
log10 100 = 2,
since 102 = 100
log10 10000 = 4,
since 104 = 10000
log10 0 01 = –2,
since 10–2 = 0 01
log10 0 001 = –3,
since 10–3
= 0 |
1 | 262-265 | 01 = –2,
since 10–2 = 0 01
log10 0 001 = –3,
since 10–3
= 0 001
and log101 = 0
since 100 = 1
The above results indicate that if n is an integral power of 10, i |
1 | 263-266 | 01
log10 0 001 = –3,
since 10–3
= 0 001
and log101 = 0
since 100 = 1
The above results indicate that if n is an integral power of 10, i e |
1 | 264-267 | 001 = –3,
since 10–3
= 0 001
and log101 = 0
since 100 = 1
The above results indicate that if n is an integral power of 10, i e , 1 followed by several zeros or
1 preceded by several zeros immediately to the right of the decimal point, then log n can be easily found |
1 | 265-268 | 001
and log101 = 0
since 100 = 1
The above results indicate that if n is an integral power of 10, i e , 1 followed by several zeros or
1 preceded by several zeros immediately to the right of the decimal point, then log n can be easily found If n is not an integral power of 10, then it is not easy to calculate log n |
1 | 266-269 | e , 1 followed by several zeros or
1 preceded by several zeros immediately to the right of the decimal point, then log n can be easily found If n is not an integral power of 10, then it is not easy to calculate log n But mathematicians have
made tables from which we can read off approximate value of the logarithm of any positive number
between 1 and 10 |
1 | 267-270 | , 1 followed by several zeros or
1 preceded by several zeros immediately to the right of the decimal point, then log n can be easily found If n is not an integral power of 10, then it is not easy to calculate log n But mathematicians have
made tables from which we can read off approximate value of the logarithm of any positive number
between 1 and 10 And these are sufficient for us to calculate the logarithm of any number expressed
in decimal form |
1 | 268-271 | If n is not an integral power of 10, then it is not easy to calculate log n But mathematicians have
made tables from which we can read off approximate value of the logarithm of any positive number
between 1 and 10 And these are sufficient for us to calculate the logarithm of any number expressed
in decimal form For this purpose, we always express the given decimal as the product of an integral
power of 10 and a number between 1 and 10 |
1 | 269-272 | But mathematicians have
made tables from which we can read off approximate value of the logarithm of any positive number
between 1 and 10 And these are sufficient for us to calculate the logarithm of any number expressed
in decimal form For this purpose, we always express the given decimal as the product of an integral
power of 10 and a number between 1 and 10 Standard Form of Decimal
We can express any number in decimal form, as the product of (i) an integral power of 10, and (ii)
a number between 1 and 10 |
1 | 270-273 | And these are sufficient for us to calculate the logarithm of any number expressed
in decimal form For this purpose, we always express the given decimal as the product of an integral
power of 10 and a number between 1 and 10 Standard Form of Decimal
We can express any number in decimal form, as the product of (i) an integral power of 10, and (ii)
a number between 1 and 10 Here are some examples:
(i) 25 |
1 | 271-274 | For this purpose, we always express the given decimal as the product of an integral
power of 10 and a number between 1 and 10 Standard Form of Decimal
We can express any number in decimal form, as the product of (i) an integral power of 10, and (ii)
a number between 1 and 10 Here are some examples:
(i) 25 2 lies between 10 and 100
25 |
1 | 272-275 | Standard Form of Decimal
We can express any number in decimal form, as the product of (i) an integral power of 10, and (ii)
a number between 1 and 10 Here are some examples:
(i) 25 2 lies between 10 and 100
25 2 =
1
25 |
1 | 273-276 | Here are some examples:
(i) 25 2 lies between 10 and 100
25 2 =
1
25 2
10
2 |
1 | 274-277 | 2 lies between 10 and 100
25 2 =
1
25 2
10
2 52
10
10
(ii) 1038 |
1 | 275-278 | 2 =
1
25 2
10
2 52
10
10
(ii) 1038 4 lies between 1000 and 10000 |
1 | 276-279 | 2
10
2 52
10
10
(ii) 1038 4 lies between 1000 and 10000 3
3
1 0 3 8 |
1 | 277-280 | 52
10
10
(ii) 1038 4 lies between 1000 and 10000 3
3
1 0 3 8 4
1 0 3 8 |
1 | 278-281 | 4 lies between 1000 and 10000 3
3
1 0 3 8 4
1 0 3 8 4
1 0
1 |
1 | 279-282 | 3
3
1 0 3 8 4
1 0 3 8 4
1 0
1 0 3 8 4
1 0
1 0 0 0
(iii) 0 |
1 | 280-283 | 4
1 0 3 8 4
1 0
1 0 3 8 4
1 0
1 0 0 0
(iii) 0 005 lies between 0 |
1 | 281-284 | 4
1 0
1 0 3 8 4
1 0
1 0 0 0
(iii) 0 005 lies between 0 001 and 0 |
1 | 282-285 | 0 3 8 4
1 0
1 0 0 0
(iii) 0 005 lies between 0 001 and 0 01
∴ 0 |
1 | 283-286 | 005 lies between 0 001 and 0 01
∴ 0 005 = (0 |
1 | 284-287 | 001 and 0 01
∴ 0 005 = (0 005 × 1000) × 10
–3 = 5 |
1 | 285-288 | 01
∴ 0 005 = (0 005 × 1000) × 10
–3 = 5 0 × 10
–3
(iv) 0 |
1 | 286-289 | 005 = (0 005 × 1000) × 10
–3 = 5 0 × 10
–3
(iv) 0 00025 lies between 0 |
1 | 287-290 | 005 × 1000) × 10
–3 = 5 0 × 10
–3
(iv) 0 00025 lies between 0 0001 and 0 |
1 | 288-291 | 0 × 10
–3
(iv) 0 00025 lies between 0 0001 and 0 001
∴ 0 |
1 | 289-292 | 00025 lies between 0 0001 and 0 001
∴ 0 00025 = (0 |
1 | 290-293 | 0001 and 0 001
∴ 0 00025 = (0 00025 × 10000) × 10
–4 = 2 |
1 | 291-294 | 001
∴ 0 00025 = (0 00025 × 10000) × 10
–4 = 2 5 × 10
–4
Rationalised 2023-24
147
Appendix
In each case, we divide or multiply the decimal by a power of 10, to bring one non-zero digit to the left
of the decimal point, and do the reverse operation by the same power of 10, indicated separately |
1 | 292-295 | 00025 = (0 00025 × 10000) × 10
–4 = 2 5 × 10
–4
Rationalised 2023-24
147
Appendix
In each case, we divide or multiply the decimal by a power of 10, to bring one non-zero digit to the left
of the decimal point, and do the reverse operation by the same power of 10, indicated separately Thus, any positive decimal can be written in the form
n = m × 10
p
where p is an integer (positive, zero or negative) and 1< m < 10 |
1 | 293-296 | 00025 × 10000) × 10
–4 = 2 5 × 10
–4
Rationalised 2023-24
147
Appendix
In each case, we divide or multiply the decimal by a power of 10, to bring one non-zero digit to the left
of the decimal point, and do the reverse operation by the same power of 10, indicated separately Thus, any positive decimal can be written in the form
n = m × 10
p
where p is an integer (positive, zero or negative) and 1< m < 10 This is called the “standard form of n |
1 | 294-297 | 5 × 10
–4
Rationalised 2023-24
147
Appendix
In each case, we divide or multiply the decimal by a power of 10, to bring one non-zero digit to the left
of the decimal point, and do the reverse operation by the same power of 10, indicated separately Thus, any positive decimal can be written in the form
n = m × 10
p
where p is an integer (positive, zero or negative) and 1< m < 10 This is called the “standard form of n ”
Working Rule
1 |
1 | 295-298 | Thus, any positive decimal can be written in the form
n = m × 10
p
where p is an integer (positive, zero or negative) and 1< m < 10 This is called the “standard form of n ”
Working Rule
1 Move the decimal point to the left, or to the right, as may be necessary, to bring one non-zero digit
2 |
1 | 296-299 | This is called the “standard form of n ”
Working Rule
1 Move the decimal point to the left, or to the right, as may be necessary, to bring one non-zero digit
2 to the left of decimal point |
1 | 297-300 | ”
Working Rule
1 Move the decimal point to the left, or to the right, as may be necessary, to bring one non-zero digit
2 to the left of decimal point (i) If you move p places to the left, multiply by 10
(ii) If you move p places to the right, multiply by 10p |
1 | 298-301 | Move the decimal point to the left, or to the right, as may be necessary, to bring one non-zero digit
2 to the left of decimal point (i) If you move p places to the left, multiply by 10
(ii) If you move p places to the right, multiply by 10p –p |
1 | 299-302 | to the left of decimal point (i) If you move p places to the left, multiply by 10
(ii) If you move p places to the right, multiply by 10p –p (iii) If you do not move the decimal point at all, multiply by 10
0 |
1 | 300-303 | (i) If you move p places to the left, multiply by 10
(ii) If you move p places to the right, multiply by 10p –p (iii) If you do not move the decimal point at all, multiply by 10
0 (iv) Write the new decimal obtained by the power of 10 (of step 2) to obtain the standard form of
the given decimal |
1 | 301-304 | –p (iii) If you do not move the decimal point at all, multiply by 10
0 (iv) Write the new decimal obtained by the power of 10 (of step 2) to obtain the standard form of
the given decimal Characteristic and Mantissa
Consider the standard form of n
n = m ×10
p, where 1 < m < 10
Taking logarithms to the base 10 and using the laws of logarithms
log n = log m + log 10
p
= log m + p log 10
= p + log m
Here p is an integer and as 1 < m < 10, so 0 < log m < 1, i |
1 | 302-305 | (iii) If you do not move the decimal point at all, multiply by 10
0 (iv) Write the new decimal obtained by the power of 10 (of step 2) to obtain the standard form of
the given decimal Characteristic and Mantissa
Consider the standard form of n
n = m ×10
p, where 1 < m < 10
Taking logarithms to the base 10 and using the laws of logarithms
log n = log m + log 10
p
= log m + p log 10
= p + log m
Here p is an integer and as 1 < m < 10, so 0 < log m < 1, i e |
1 | 303-306 | (iv) Write the new decimal obtained by the power of 10 (of step 2) to obtain the standard form of
the given decimal Characteristic and Mantissa
Consider the standard form of n
n = m ×10
p, where 1 < m < 10
Taking logarithms to the base 10 and using the laws of logarithms
log n = log m + log 10
p
= log m + p log 10
= p + log m
Here p is an integer and as 1 < m < 10, so 0 < log m < 1, i e , m lies between 0 and 1 |
1 | 304-307 | Characteristic and Mantissa
Consider the standard form of n
n = m ×10
p, where 1 < m < 10
Taking logarithms to the base 10 and using the laws of logarithms
log n = log m + log 10
p
= log m + p log 10
= p + log m
Here p is an integer and as 1 < m < 10, so 0 < log m < 1, i e , m lies between 0 and 1 When log
n has been expressed as p + log m, where p is an integer and 0 log m < 1, we say that p is the
“characteristic” of log n and that log m is the “mantissa of log n |
Subsets and Splits