Solutions Manual
Atkins and Jones’s Chemical Principles
The Quest for Insight 5th Edition
By
John Krenos,
Joseph Potenza,
Laurence Lavelle,
Yinfa Ma,
Carl Hoeger
( All Chapters Included - 100% Verified Solutions )
1
, FUNDAMENTALS
A.l (a) chemical; (b) physical; (c) physical
A.3 The temperature o f the injured camper, the evaporation and condensation
o f water are physical properties. The ignition o f propane is a chemical
change.
A.5. (a) Physical; (b) Chemical; (c) Chemical
A.7 (a) intensive; (b) intensive; (c) extensive; (d) extensive
( \
1 A'/log rah?
A.9 (a) 1000. grain x = 1 kilograin
1,000 grain
100 centibatman \
(b) 0.01 batman x = 1 centibatman
V 1batman )
1megamutchkin ] . ,,-
(c) lx l 0b mutchkin x -----^------------ = 1 megamutchkin1
0
106 mutchkin J
1 pint 1 quart 0.946 L 1000 mL
A .ll 1.00 cup x ( ^ ---- ) x ( - 1, ) x ( - ------ - ) x ( — —---- ) = 236 mL
2 cups 2 pints 1 quart 1L
. m
A .13 a- —
V
f 112.32 e mL 1
29.27 m L - 23.45 mL cm
- 19.3 g •cm'-'
SM-3
2
, m m
A. 15 d = — , rearranging gives V = —
V a
' 0.750 carat ^ ( 200 mg ig 1
l,3.51g-errf 3j 1carat J ^1000 mg J
= 0.0427 cm3
A .17 105.50 g -43.50 g=62.00 g=mHi0
m rr m wh,o 62.00 g
d=- = 62.00 cm
V d d,HO 0.9999 g •cm"1
\
96.75 g - 43.50 g
d,liquid = 0.8589 g- cm''
62.00 cm3
m . w m
A.19 d = — , rearrange gives V =—
V d
V = = 7.41 cm3
2.70 g.cm
Since the area is one centimeter, the thickness must be 7.41 cm
A.21 The result should have 3 significant figures because the number 3.25
has the least significant figures. The results should be 0.989
A.23 (a) 4.82 nm x ( 1Q0-^ — ) = 4,82 x 103 pm
lnm
(b) ( u (^ )x( ) x( ) = 30.5 mm3
min ImL 1cm 60s
(c) 1.88 ng x ) X ( - ^ ) - 1.88 x 10'12kg
10 ng 10 g
(d) ^ i lO W 2 .6 6 x 1 0-’ kg/m3
^ cm J ’ lOOOg lm 3
0.044g lOOOmg 1L
(e) ( )x( ° )x( j ) = 0.044 mg/cnf
L lg v 1000cm
SM-4
3
, m
A.25 (a) d = —
V
0.213^ 0-213g
1.100cm x 0.531cm x 0.212cm 0.1238cm-
- 1.72g •cm
This determination is more precise because the volume is not limited to
two significant figures as it is in part (b)
(b) d = -
V
41.003g-39.753g
20.37mL - 19.65mL
1.250g lmL ^
v0.72mL lcm3)
= 1.7g-cm“
A.27 (a) Formula: °X = 50 + 2°C
(b) "X = 50 + 2x22°C = 94"X
A.29 E k = —mv2
K 2
( lh ' 1000 in'!
= i(4 .2 kg)(14 km •I r 1)2
v3600 s j v 1km j
= 32 kg •nr -s~2
= 32 J
A.31 m=2.8 metric tons, v =100 km •hr'1, v, = 50 km •hr'
1
= —mv2
*• 2
SM-5
4
Atkins and Jones’s Chemical Principles
The Quest for Insight 5th Edition
By
John Krenos,
Joseph Potenza,
Laurence Lavelle,
Yinfa Ma,
Carl Hoeger
( All Chapters Included - 100% Verified Solutions )
1
, FUNDAMENTALS
A.l (a) chemical; (b) physical; (c) physical
A.3 The temperature o f the injured camper, the evaporation and condensation
o f water are physical properties. The ignition o f propane is a chemical
change.
A.5. (a) Physical; (b) Chemical; (c) Chemical
A.7 (a) intensive; (b) intensive; (c) extensive; (d) extensive
( \
1 A'/log rah?
A.9 (a) 1000. grain x = 1 kilograin
1,000 grain
100 centibatman \
(b) 0.01 batman x = 1 centibatman
V 1batman )
1megamutchkin ] . ,,-
(c) lx l 0b mutchkin x -----^------------ = 1 megamutchkin1
0
106 mutchkin J
1 pint 1 quart 0.946 L 1000 mL
A .ll 1.00 cup x ( ^ ---- ) x ( - 1, ) x ( - ------ - ) x ( — —---- ) = 236 mL
2 cups 2 pints 1 quart 1L
. m
A .13 a- —
V
f 112.32 e mL 1
29.27 m L - 23.45 mL cm
- 19.3 g •cm'-'
SM-3
2
, m m
A. 15 d = — , rearranging gives V = —
V a
' 0.750 carat ^ ( 200 mg ig 1
l,3.51g-errf 3j 1carat J ^1000 mg J
= 0.0427 cm3
A .17 105.50 g -43.50 g=62.00 g=mHi0
m rr m wh,o 62.00 g
d=- = 62.00 cm
V d d,HO 0.9999 g •cm"1
\
96.75 g - 43.50 g
d,liquid = 0.8589 g- cm''
62.00 cm3
m . w m
A.19 d = — , rearrange gives V =—
V d
V = = 7.41 cm3
2.70 g.cm
Since the area is one centimeter, the thickness must be 7.41 cm
A.21 The result should have 3 significant figures because the number 3.25
has the least significant figures. The results should be 0.989
A.23 (a) 4.82 nm x ( 1Q0-^ — ) = 4,82 x 103 pm
lnm
(b) ( u (^ )x( ) x( ) = 30.5 mm3
min ImL 1cm 60s
(c) 1.88 ng x ) X ( - ^ ) - 1.88 x 10'12kg
10 ng 10 g
(d) ^ i lO W 2 .6 6 x 1 0-’ kg/m3
^ cm J ’ lOOOg lm 3
0.044g lOOOmg 1L
(e) ( )x( ° )x( j ) = 0.044 mg/cnf
L lg v 1000cm
SM-4
3
, m
A.25 (a) d = —
V
0.213^ 0-213g
1.100cm x 0.531cm x 0.212cm 0.1238cm-
- 1.72g •cm
This determination is more precise because the volume is not limited to
two significant figures as it is in part (b)
(b) d = -
V
41.003g-39.753g
20.37mL - 19.65mL
1.250g lmL ^
v0.72mL lcm3)
= 1.7g-cm“
A.27 (a) Formula: °X = 50 + 2°C
(b) "X = 50 + 2x22°C = 94"X
A.29 E k = —mv2
K 2
( lh ' 1000 in'!
= i(4 .2 kg)(14 km •I r 1)2
v3600 s j v 1km j
= 32 kg •nr -s~2
= 32 J
A.31 m=2.8 metric tons, v =100 km •hr'1, v, = 50 km •hr'
1
= —mv2
*• 2
SM-5
4
