Chemistry Reference
In-Depth Information
(e)
The heat of vaporization per gram is the heat of part (d)
divided by the number of grams, but changed in sign be-
cause the heat of vaporization is heat
added to the system:
H
2
O
100
1.36
1
kJ
0.60 g
2.3 kJ/g
50
2
O
2
(g) £ CO(g)
14.92
C(s)
¢
H
110 kJ
0
H
2
Te
C(s)
O
2
(g) £ CO
2
(g)
¢
H
393 kJ
H
2
Se
50
(H
2
O)
Let
x
number of moles of CO formed
H
2
S
100
x
(
110 kJ)
(1.75
x
)(
393 kJ)
546 kJ
0
10
20
30
40
50
60
110
x
687.
8
393
x
546
283
x
141.
8
Atomic number
x
0.500
8
mol CO
1.75
x
1.25 mol CO
2
0.500
8
mol CO takes 0.250 mol
1.25 mol
O
2
CO
2
takes 1.25 mol
O
2
15 Solutions
15.1
1.50 mol
O
2
required
(a)
The first solution is unsaturated because it could hold
2.00 g additional of the solute at the temperature
specified.
(b)
The second solution is saturated.
(c)
The third solution is obviously saturated because it is in
contact with excess solid.
14.94
This problem may be thought of as involving limiting quan-
tities, in that the heat available can be compared with the
heat required for a certain purpose. The heat required to raise
the temperature to
0°C
is given by
(35.0 g) (2.089 J/g
#
°C) (15°C)
Heat
mc
¢
t
11
0
0 J
1.1
0
kJ
15.2
(a)
KCl is more soluble at
(b)
10°C.
KNO
3
is more soluble at
40°C.
More heat than that is available. After the ice has been raised
to the melting point, there will still be available
15.3
The names differ by only one letter. The symbol for molality
is lowercase
m;
for molarity, it is capital
M.
Kilograms rather
than liters are involved in molality. A quantity of solvent
rather than a quantity of solution appears in the denominator
of the expression for molality.
2.10 kJ
1.1
0
kJ
1.0
0
kJ
The ice will start to melt. The heat required to melt all the ice
is
(35.0 g) (335 J/g)
11,700 J
11.7 kJ
15.4
They are similar in that they both are ratios of each com-
ponent to the total and in that the total is specified in
each case (100% for percent composition and 1 mol for
mole fraction). They differ in the units used to calculate
them—moles for mole fraction and masses for percent
composition.
Because not that much heat is still available, not all the ice
will melt. At the end, there will be a mixture of water and ice,
so the temperature will be
0°C.
14.99
The sudden expansion of liquid does work pushing
back the atmosphere and overcoming intermolecular forces
in the liquid. The energy to do that work comes from the
molecules themselves, so the average energy of the mole-
cules is lowered. The
CO
2
15.5
(a)
A change of temperature will change the volume of the
solution, and thus change its molarity.
(b)
No
CO
2
condenses to a solid because of
this loss of energy.
a
1 mol NaCl
55.85 g NaCl
15.6
12.0 g NaCl
b
0.214
9
mol NaCl
14.104
1
2
CO(g)
O
2
(g)
CO
2
(g)
3
0.214
9
mol NaCl
0.125 kg H
2
O
1.72 m
25
C
25
C
1
2
4
15.7
The percentage of water is
100.0% total
25.0% formaldehyde
75.0% water
1
2
75
C
CO(g)
O
2
(g)
CO
2
(g)
15.8
By definition, the sum of the mole fractions must be 1.000;
therefore,
H
75
C
H
1
mc
t
(28.0 g)(1.04 J/g·
C)(
50
C)
14
6
0 J
H
2
mc
t
(32.0 g)(0.922 J/g·
C)(
50
C)
73
8
J
H
3
H
combustion
283 kJ
H
4
mc
t
(44.0 g)(0.852 J/g·
C)(
50
C)
18
7
0 J
H
H
1
H
2
H
3
H
4
(
1.4
6
kJ)
(
0.73
8
kJ)
(
283 kJ)
(1.8
7
kJ)
283 kJ
14.106
A plot of these data shows that water should boil at about
75
C. Because it actually boils 175
C higher than
that, the effects of hydrogen bonding are obviously very
important.
X
alcohol
1.000
0.930
0.070
1
2
15.9
By definition, the sum of the mole fractions must be 1.000;
therefore,
X
CH
2
O
1.000
0.911
0.0450
0.044
a
1 mol KCl
74.6 g KCl
15.10
43.0 g KCl
b
0.576
4
mol KCl
a
1 mol H
2
O
18.0 g H
2
O
275 g H
2
O
b
15.2
8
mol H
2
O
0.576
4
0.576
4
15.2
8
X
KCl
0.0364
