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CHAPTER 13
PROPERTIES OF LIQUIDS
SOLUTIONS TO REVIEW QUESTIONS
1.
The potential energy is greater in the liquid water than in the ice. The heat necessary to
melt the ice increases the potential energy of the liquid, thus allowing the molecules
greater freedom of motion. The potential energy of steam (gas) is greater than that of
liquid water.
2.
At 0°C, all three substances, H 2S, H2Se, and H2Te, are gases, because they all have
boiling points below 0°C.
3.
Liquids are made of particles that are close together, they are not compressible and they
have definite volume. Solids also exhibit similar properties.
4.
Liquids take the shape of the container they are in. Gases also exhibit this property.
5.
The pressure of the atmosphere must be 1.00 atmosphere, otherwise the water would be
boiling at some other temperature.
6.
–
O
H
H
+
7.
Melting point, boiling point, heat of fusion, heat of vaporization, density, and crystal
structure in the solid state are some of the physical properties of water that would be very
different, if the molecules were linear and nonpolar instead of bent and highly polar. For
example, the boiling point, melting point, heat of fusion and heat of vaporization would
be lower because linear molecules have no dipole moment and the attraction among
molecules would be much less.
8.
Prefixes preceding the word hydrate are used in naming hydrates, indicating the number
of molecules of water present in the formulas. The prefixes used are:
mono = 1
hexa = 6
9.
di = 2
hepta = 7
tri = 3
octa = 8
tetra = 4
nona = 9
penta = 5
deca = 10
The distillation setup in Figure 13.13 would be satisfactory for separating salt and water,
but not for separating ethyl alcohol and water. In the first case, the water is easily
vaporized, the salt is not, so the water boils off and condenses to a pure liquid. In the
second case, both substances are easily vaporized, so both would vaporize (though not to
and equal degree) and the condensed liquid would contain both substances.
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10.
The thermometer would be at about 70°C. The liquid is boiling, which means its vapor
pressure equals the confining pressure. From Table 13.1, we find that ethyl alcohol has a
vapor pressure of 543 torr at 70°C.
11.
The water in both containers would have the same vapor pressure, for it is a function of
the temperature of the liquid.
12.
Vapor pressure is the pressure exerted by a vapor when it is in equilibrium with its liquid.
The liquid molecules on the surface and the gas molecules above the liquid are the
important players in vapor pressure. One liter of water will evaporate at the same rate as
100 mL of water if they are at the same temperature and have the same surface area. The
liquid and vapor in both cases will come to equilibrium in the same amount of time and
they will have the same vapor pressure. Vapor pressure is a characteristic of the type of
liquid, not the number of molecules of a liquid.
13.
In Figure 13.1, it would be case (b) in which the atmosphere would reach saturation. The
vapor pressure of water is the same in both (a) and (b), but since (a) is an open container
the vapor escapes into the atmosphere and doesn’t reach saturation.
14.
If ethyl ether and ethyl alcohol were both placed in a closed container, (a) both substances
would be present in the vapor, for both are volatile liquids; (b) ethyl ether would have
more molecules in the vapor because it has a higher vapor pressure at a given
temperature.
15.
The vapor pressure observed in (c) would remain unchanged. The presence of more water
in (b) does not change the magnitude of the vapor pressure of the water. The temperature
of the water determines the magnitude of the vapor pressure.
16.
At 30 torr, H2O would boil at approximately 29°C, ethyl alcohol at 14°C, and ethyl ether
at some temperature below 0°C.
17.
(a)
(b)
(c)
18.
Based on Figure 13.7:
(a) Line BC is horizontal because the temperature remains constant during the entire
process of melting. The energy input is absorbed in changing from the solid to the
liquid state.
(b) During BC, both solid and liquid phases are present.
(c) The line DE represents the change from liquid water to steam (vapor) at the boiling
temperature of water.
At a pressure of 500 torr, water boils at 88°C.
The normal boiling point of ethyl alcohol is 78°C.
At a pressure of 0.50 atm (380 torr), ethyl ether boils at 16°C.
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19.
Physical properties of water:
(a) melting point, 0°C
(b) boiling point, 100°C (at 1 atm pressure)
(c) colorless
(d) odorless
(e) tasteless
(f) heat of fusion, 335 J> g (80 cal> g)
(g) heat of vaporization, 2.26 kJ> g (540 cal> g)
(h) density = 1.0 g>mL (at 4°C)
(i) specific heat = 4.184 J>g°C
20.
For water, to have its maximum density, the temperature must be 4°C, and the pressure
sufficient to keep it liquid. d = 1.0 g>mL
21.
Apply heat to an ice-water mixture, the heat energy is absorbed to melt the ice (heat of
fusion), rather than warm the water, so the temperature remains constant until all the ice
has melted.
22.
Ice at 0°C contains less heat energy than water at 0°C. Heat must be added to convert ice
to water, so the water will contain that much additional heat energy.
23.
Ice floats in water because it is less dense than water. The density of ice at 0°C is
0.915 g> mL. Liquid water, however, has a density of 1.00 g> mL. Ice will sink in ethyl
alcohol, which has a density of 0.789 g> mL.
24.
The heat of vaporization of water would be lower if water molecules were linear instead
of bent. If linear, the molecules of water would be nonpolar. The relatively high heat of
vaporization of water is a result of the molecule being highly polar and having strong
dipole-dipole and hydrogen bonding attraction for other water molecules.
25.
Ethyl alcohol exhibits hydrogen bonding; ethyl ether does not. This is indicated by the
high heat of vaporization of ethyl alcohol, even though its molar mass is much less than
the molar mass of ethyl ether.
26.
A linear water molecule, being nonpolar, would exhibit less hydrogen bonding than the
highly polar, bent, water molecule. The polar molecule has a greater intermolecular
attractive force than a nonpolar molecule.
27.
Water, at 80°C, will have fewer hydrogen bonds than water at 40°C. At the higher
temperature, the molecules of water are moving faster than at the lower temperature. This
results in less hydrogen bonding at the higher temperature.
28.
H2NCH2CH2NH2 has two polar NH2 groups. It should, therefore, show more hydrogen
bonding and a higher boiling point (117°C) versus 49°C for CH3CH2CH2NH2.
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29.
Rubbing alcohol feels cold when applied to the skin, because the evaporation of the
alcohol absorbs heat from the skin. The alcohol has a fairly high vapor pressure (low
boiling point) and evaporates quite rapidly. This produces the cooling effect.
30.
(a)
(b)
Order of increasing rate of evaporation: mercury, acetic acid, water, toluene,
benzene, carbon tetrachloride, methyl alcohol, bromine.
Highest boiling point is mercury. Lowest boiling point is bromine.
31.
Water boils when its vapor pressure equals the prevailing atmospheric pressure over the
water. In order for water to boil at 50°C, the pressure over the water would need to be
reduced to a point equal to the vapor pressure of the water (92.5 torr).
32.
In a pressure cooker, the temperature at which the water boils increases above its normal
boiling point, because the water vapor (steam) formed by boiling cannot escape. This
results in an increased pressure over water and, consequently, an increased boiling
temperature.
33.
Vapor pressure varies with temperature. The temperature at which the vapor pressure of a
liquid equals the prevailing pressure is the boiling point of the liquid.
34.
As temperature increases, molecular velocities increase. At higher molecular velocities, it
becomes easier for molecules to break away from the attractive forces in the liquid.
35.
Water has a relatively high boiling point because there is a high attraction between
molecules due to hydrogen bonding.
36.
Ammonia would have a higher vapor pressure than SO2 at -40°C because it has a lower
boiling point (NH3 is more volatile than SO2).
37.
As the temperature of a liquid increases, the kinetic energy of the molecules as well as the
vapor pressure of the liquid increases. When the vapor pressure of the liquid equals the
external pressure, boiling begins with many of the molecules having enough energy to
escape from the liquid. Bubbles of vapor are formed throughout the liquid and these
bubbles rise to the surface, escaping as boiling continues.
38.
HF has a higher boiling point than HCl because of the strong hydrogen bonding in HF
(F is the most electronegative element). Neither F2 nor Cl 2 will have hydrogen bonding,
so the compound, F2 , with the lower molar mass, has the lower boiling point.
39.
The boiling liquid remains at constant temperature because the added heat energy is being
used to convert the liquid to a gas, i.e., to supply the heat of vaporization for the liquid at
its boiling point.
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40.
34.6°C, the boiling point of ethyl ether. (See Table 13.2)
41.
If the lake is in an area where the temperature is below freezing for part of the year, the
expected temperature would be 4°C at the bottom of the lake. This is because the surface
water would cool to 4°C (maximum density) and sink.
42.
The formation of hydrogen and oxygen from water is an endothermic reaction, due to the
following evidence:
(a) Energy must continually be provided to the system for the reaction to proceed. The
reaction will cease when the energy source is removed.
(b) The reverse reaction, burning hydrogen in oxygen, releases energy as heat.
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CHAPTER 13
SOLUTIONS TO EXERCISES
1.
Acid anhydride: [HClO4 , Cl 2O7]
2.
Acid anhydride: [H 2SO3 , SO2]
3.
Basic anhydrides: [LiOH, Li 2O]
[NaOH, Na 2O]
[Mg(OH)2 , MgO]
4.
Basic anhydrides: [KOH, K 2O]
[Ba(OH)2 , BaO]
[Ca(OH)2 , CaO]
5.
(a)
Ba(OH)2
(b)
2 CH 3OH + 3O2 ¡ 2 CO2 + 4 H 2O
(c)
2 Rb + 2 H 2O ¡ 2 RbOH + H 2
(d)
SnCl 2 # 2 H 2O
(e)
HNO3 + NaOH ¡ NaNO3 + H 2O
(f)
CO2 + H 2O ¡ H 2CO3
(a)
Li 2O + H 2O ¡ 2 LiOH
(b)
2 KOH
(c)
Ba + 2 H 2O ¡ Ba(OH)2 + H 2
(d)
Cl 2 + H 2O ¡ HCl + HClO
(e)
SO3 + H 2O ¡ H 2SO4
(f)
H 2SO3 + 2 KOH ¡ K 2SO3 + 2 H 2O
7.
(a)
(b)
(c)
barium bromide dihydrate
aluminum chloride hexahydrate
iron(III) phosphate tetrahydrate
8.
(a)
(b)
(c)
magnesium ammonium phosphate hexahydrate
iron(II) sulfate heptahydrate
tin(IV) chloride pentahydrate
9.
Deionized water is water from which the ions have been removed.
(a) Hard water contains dissolved calcium and magnesium salts.
(b) Soft water is free of ions that cause hardness (Ca2+ and Mg 2+) but it may contain
other ions such as Na+ and K +.
6.
¢
¢
[H 3PO4 , P2O5]
[H 2CO3 , CO2]
[H 2SO4 , SO3]
[HNO3 , N2O5]
" BaO + H O
2
¢
" SnCl + 2 H O
2
2
" K O + H O
2
2
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10.
Deionized water is water from which the ions have been removed.
(a) Distilled water has been vaporized by boiling and recondensed. It is free of
nonvolatile impurities, but may still contain any volatile impurities that were
initially present in the water.
(b) Natural waters are generally not pure, but contain dissolved minerals and suspended
matter, and can even contain harmful bacteria.
11.
Hydrogen bonding occurs in molecules containing hydrogen which is bonded to either
fluorine, oxygen, or nitrogen.
(a) C2H6 will not hydrogen bond; hydrogen is not bonded to either fluorine, oxygen, or
nitrogen.
(b) NH3 will hydrogen bond; hydrogen is bonded to nitrogen.
(c) H2O will hydrogen bond; hydrogen is bonded to oxygen.
(d) HI will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or
nitrogen.
(e) C2H5OH will hydrogen bond; one of the hydrogen is bonded to oxygen.
12.
Hydrogen bonding occurs in molecules containing hydrogen which is bonded to either
fluorine, oxygen, or nitrogen.
(a) HF will hydrogen bond; hydrogen is bonded to fluorine.
(b) CH4 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or
nitrogen.
(c) H2O2 will hydrogen bond; hydrogen is bonded to oxygen.
(d) CH3OH will hydrogen bond; one of the hydrogens is bonded to oxygen.
(e) H2 will not hydrogen bond; hydrogen is not bonded to fluorine, oxygen, or
nitrogen.
13.
NH3
H
H
H N
H
H N
H
H2O
H
H
O
H
O
H
C2H5OH
H
H C
H
C
H
H
O
H O
H
H
C
H
H
C H
H
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14.
HF
F H
F H
H2O2
H
H O
O
O H
O
H
CH3OH
H
H
H C
H
H
C
O
H
H O
H
15.
Water spreading out on a glass plate is an example of adhesive forces. The water
molecules have stronger attractive forces toward the glass surface than they do to other
water molecules.
16.
Water forming beaded droplets is an example of cohesive forces. The water molecules
have stronger attractive forces for other water molecules that they do for the surface.
17.
(100. g CoCl 2 # 6 H 2O)a
18.
(100. g FeI 2 # 4 H 2O)a
19.
(100. g CoCl 2 # 6 H 2O)a
20.
(100. g FeI 2 # 4 H 2O)a
21.
Assume 1 mol of the compound which contains seven moles of water.
% H 2O = ¢
22.
1 mol
b = 0.420 mol CoCl 2 # 6 H 2O
238.0 g
1 mol
b = 0.262 mol FeI 2 # 4 H 2O
381.7 g
6 mol H 2O
1 mol
b¢
≤ = 2.52 mol H 2O
238.0 g 1 mol CoCl 2 # 6 H 2O
4 mol H 2O
1 mol
b¢
≤ = 1.05 mol H 2O
381.7 g 1 mol FeI 2 # 4 H 2O
172118.02 g2
g H 2O
11002
=
a
b11002 = 51.17% H 2O
≤
g MgSO4 # 7 H 2O
246.5 g
Assume 1 mol of the hydrate which contains 18 mol of water.
% H 2O =
1182118.02 g2
g H 2O
=
a
b11002 = 48.67% H 2O
g Al 2(SO4)3 # 18 H 2O
666.5 g
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23.
Assume 100. g of the compound.
10.14221100. g2 = 14.2 g H 2O
10.85821100. g2 = 85.8 g Pb(C2H 3O2)2
1 mol
(14.2 g H 2O)a
b = 0.788 mol H 2O
18.02 g
1 mol
(85.8 g Pb(C2H 3O2)2)a
b = 0.264 mol Pb(C2H 3O2)2
325.3 g
In the formula for the hydrate, there is one mole of Pb(C2H 3O2)2 , so divide each of the
moles calculated by 0.264.
0.264 mol Pb(C2H 3O2)2
= 1 Pb(C2H 3O2)2
0.264 mol
0.788 mol H 2O
= 2.98 H 2O
0.264 mol
Therefore, the formula is Pb(C2H 3O2)2 # 3 H 2O.
24.
25.0 g hydrate - 16.9 g FePO4 = 8.1 g H 2O driven off
1 mol
0.45
(8.1 g H 2O)a
b = 0.45 mol H 2O
= 4.0
18.02 g
0.112
1 mol
b = 0.112 mol FePO4
150.8 g
The formula is FePO4 # 4 H 2O.
(16.9 g FePO4)a
25.
(a)
(b)
0.112
= 1.00
0.112
Energy (E a) to heat the water from 20.°C ¡ 100.°C
4.184 J
E a = (m)(specific heat)(¢T) = (120. g)a
b (80.°C) = 4.0 * 104 J
g°C
Energy (E b) to convert water to steam: heat of vaporization = 2.26 * 103 J>g
E b = (m)(heat of vaporization) = (120. g)(2.26 * 103 J>g) = 2.71 * 105 J
E total = E a + E b = (4.0 * 104 J) + (2.71 * 105 J) = 3.11 * 105 J
26.
(a)
(b)
Energy (E a) to cool the water from 24°C ¡ 0°C
4.184 J
E a = (m)(specific heat)(¢T) = (126 g)a
b (24°C) = 1.3 * 104 J
g°C
Energy (E b) to convert water to ice
E b = (m)(heat of fusion) = (126 g)(335 J>g) = 4.22 * 104 J
E total = E a + E b = (1.3 * 104 J) + (4.22 * 104 J) = 5.5 * 104 J
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27.
Energy released in cooling the water: 25°C to 0°C
E = (m)(specific heat)(¢T) = (300. g)a
4.184 J
b(25°C) = 3.1 * 104 J
g°C
Energy required to melt the ice
E = (m)(heat of fusion) = (100. g)(335 J>g) = 3.35 * 104 J
Less energy is released in cooling the water than is required to melt the ice. Ice will
remain and the water will be at 0°C.
28.
Energy to heat the water = energy to condense the steam
4.184 J
(300. g)a
b(100.°C - 25°C) = (m)(2259 J>g)
g°C
m = 42 g (grams of steam required to heat the water to 100.°C) 42 g of steam are
required to heat 300. g of water to 100.°C. Since only 35 g of steam are added to the
system, the final temperature will be less than 100.°C. Not sufficient steam.
29.
Energy lost by warm water = energy gained by the ice
x = final temperature
mass(H 2O) = (1.5 L H 2O)a
(1500 g)a
1000 mL 1.0 g
ba
b = 1500 g
L
mL
J
4.184 J
4.184 J
b(75°C - x) = (75 g)a 335 b + (75 g)a
b (x - 0°C)
g
g°C
g°C
(4.707 * 105 J) - (6276x J>°C) = 2.51 * 104 J + 313.8x J>°C
4.46 * 105 J = 6.5898x 103x J/°C
x = 68°C
30.
E = (m)(heat of fusion)
(500. g)(335 J>g) = 167,000 J needed to melt the ice
9560 J 6 167,500 J
Since 167,500 J are required to melt all the ice, and only 9560 J are available, the system
will be at 0°C. It will be a mixture of ice and water.
31.
(a)
2 Na + 2 H 2O ¡ 2 NaOH + H 2
(1.00 g Na)a
2 mol H 2O 18.02 g
1 mol
b¢
b = 0.784 g H 2O
≤a
22.99 g
2 mol Na
mol
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(b)
MgO + H 2O ¡ Mg(OH)2
(1.00 g MgO)a
(c)
N2O5 + H 2O ¡ 2 HNO3
(1.00 g N2O5)a
32.
(a)
2 mol H 2O 18.02 g
b = 18.0 g H 2O
≤a
2 mol K
mol
Ca + 2 H 2O ¡ Ca(OH)2 + H 2
(1.00 mol Ca) ¢
(c)
18.02 g
1 mol H 2O
1 mol
b¢
b = 0.167 g H 2O
≤a
108.0 g 1 mol N2O5
mol
2 K + 2 H 2O ¡ 2 KOH + H 2
(1.00 mol K) ¢
(b)
18.02 g
1 mol H 2O
1 mol
b¢
b = 0.447 g H 2O
≤a
40.31 g 1 mol MgO
mol
2 mol H 2O 18.02 g
b = 36.0 g H 2O
≤a
1 mol Ca
mol
SO3 + H 2O ¡ H 2SO4
(1.00 mol SO3) ¢
1 mol H 2O 18.02 g
b = 18.0 g H 2O
≤a
1 mol SO3
mol
33.
Water forms droplets because of surface tension, or the desire for a droplet of water to
minimize its ratio of surface area to volume. The molecules of water inside the drop are
attracted to other water molecules all around them, but on the surface of the droplet the
water molecules feel an inward attraction only. This inward attraction of surface water
molecules for internal water molecules is what holds the droplets together and minimizes
their surface area.
34.
Steam molecules will cause a more severe burn. Steam molecules contain more energy at
100°C than water molecules at 100°C due to the energy absorbed during the vaporization
stage (heat of vaporization).
35.
The alcohol has a higher vapor pressure than water and thus evaporates faster than
water. When the alcohol evaporates it absorbs energy from the water, cooling the water.
Eventually the water will lose enough energy to change from a liquid to a solid
(freeze).
36.
When one leaves the swimming pool, water starts to evaporate from the skin of the body.
Part of the energy needed for evaporation is absorbed from the skin, resulting in the cool
feeling.
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37.
(a)
100
T (°C)
80
60
40
From 0°C to 40.°C solid X warms until
at 40.°C it begins to melt. The
temperature remains at 40.0°C until all
of X is melted. After that, liquid X will
warm steadily to 65°C where it will boil
and remain at 65°C until all of the
liquid becomes vapor. Beyond 65°C, the
vapor will warm steadily until 100°C.
20
0
Joules
(b)
= (60. g)(3.5 J>g°C)(40.°C)
Joules needed (0°C to 40°C)
= (60. g)(80. J>g)
Joules needed at 40°C
Joules needed (40°C to 65°C) = (60. g)(3.5 J>g°C)(25°C)
= (60. g)(190 J>g)
Joules needed at 65°C
Joules needed (65°C to 100°C) = (60. g)(3.5 J>g°C)(35°C)
Total Joules needed
(each step rounded to two significant figures)
=
=
=
=
=
8400 J
4800 J
5300 J
11,000 J
7400 L
37,000 J
38.
During phase changes (ice melting to liquid water or liquid water evaporating to steam),
all the heat energy is used to cause the phase change. Once the phase change is complete
the heat energy is once again used to increase the temperature of the substance.
39.
As the temperature of a liquid increases, the molecules gain kinetic energy thereby
increasing their escaping tendency (vapor pressure).
40.
Since boiling occurs when vapor pressure equals atmospheric pressure, the graph in
Figure 13.4 indicates that water will boil at about 75°C at 270 torr pressure.
41.
CuSO4 (anhydrous) is gray white. When exposed to moisture, it turns bright blue forming
CuSO4 # 5 H 2O. The color change is an indicator of moisture in the environment.
42.
MgSO4 # 7 H 2O
43.
Soap can soften hard water by forming a precipitate with, and thus removing, the calcium
and magnesium ions. This precipitate is a greasy scum and is very objectionable, so it is a
poor way to soften water.
Na 2HPO4 # 12 H 2O
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44.
Chlorine is commonly used to destroy bacteria in water. Ozone and ultraviolet radiation
are also used in some places.
45.
Ozone, O3
46.
When organic pollutants in water are oxidized by dissolved oxygen, there may not be
sufficient dissolved oxygen to sustain marine life, such as fish. Most marine life forms
depend on dissolved oxygen for cellular respiration.
47.
Liquids that are stored in ceramic containers should never be drunk, for they are likely to
have dissolved some of the lead from the ceramic. If the ceramic is glazed, the liquid is
less apt to dissolve lead from the ceramic.
48.
Na 2 zeolite(s) + Mg 2+(aq) ¡ Mg zeolite(s) + 2 Na+(aq)
49.
Softening of hard water using sodium carbonate:
CaCl 2(aq) + Na 2CO3(aq) ¡ CaCO3(s) + 2 NaCl(aq)
50.
Heating Curve for H2S
Temperature (°C)
–50
Liquid S gas (phase change)
–60
liquid
–70
–80
–90
gas
Solid S liquid
(phase change)
solid
–100
Heat added ¡
51.
(a)
Melt ice: Ea = (m)(heat of fusion) = (225 g)(335 J>g) = 7.54 * 104 J
(b)
Warm the water: E b = (m)(specific heat)(¢T)
= (225 g)a
(c)
4.184 J
b(100.°C) = 9.41 * 104 J
g°C
Vaporize the water:
Ec = (m)(heat of vaporization) = (225 g)(2259 J>g) = 5.08 * 105 J
Etotal = Ea + Eb + Ec = 6.78 * 105 J
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52.
The heat of vaporization of water is 2.26 kJ> g.
(2.26 kJ>g)a
18.02 g
b = 40.7 kJ>mol
mol
0.096 cal
b(150. - 20.0°C)
g°C
= 3.1 * 103 cal (3.1 kcal)
53.
E = (m)(specific heat)(¢T) = (250. g)a
54.
Energy liberated when steam at 100.0°C condenses to water at 100.0°C
(50.0 mol steam) a
18.02 g 2.26 kJ 1000 J
ba
b = 2.04 * 106 J
ba
g
mol
kJ
Energy liberated in cooling water from 100.0°C to 30.0°C
(50.0 mol H 2O)a
18.02 g 4.184 J
ba
b (100.0°C - 30.0°C) = 2.64 * 105 J
mol
g°C
Total energy liberated = 2.04 * 106 J + 2.64 * 105 J = 2.30 * 106 J
55.
Energy to warm the ice from -10.0°C to 0°C
(100. g)a
2.01 J
b(10.0°C) = 2010 J
g°C
Energy to melt the ice at 0°C
(100. g)(335 J>g) = 33,500 J
Energy to heat the water from 0°C to 20.0°C
(100. g)a
4.184 J
b(20.0°C) = 8370 J
g°C
E total = 2010 J + 33,500 J + 8370 J = 4.39 * 104 J = 43.9 kJ
56.
2 H 2O(l) ¡ 2 H 2(g) + O2(g)
The conversion is L O2 ¡ mol O2 ¡ mol H 2O ¡ g H 2O
2 mol H 2O 18.02 g
1 mol
(25.0 L O2)a
ba
ba
b = 40.2 g H 2O
22.4 L
1 mol O2
mol
57.
The conversion is
molecules
molecules
molecules
molecules
mol
¡
¡
¡
¡
s
day
day
hr
min
1.00 mol H 2O 6.022 * 1023 molecules 1.00 day
1 hr
1 min
ba
ba
b
¢
≤¢
≤a
day
mol
24 hr
60 min
60 s
= 6.97 * 1018 molecules>s
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58.
Liquid water has a density of 1.00 g> mL.
d =
m
V
V =
18.02 g
m
=
= 18.0 mL
d
1.00 g>mL
(volume of 1 mole)
1.00 mole of water vapor at STP has a volume of 22.4 L (gas)
59.
2 H 2(g) + O2(g) ¡ 2 H 2O(g)
(a)
(b)
60.
1 mL O2
≤ = 40.0 mL O2 react with 80.0 mL of H 2
2 mL H 2
Since 60.0 mL of O2 are available, some oxygen remains unreacted.
(80.0 mL H 2) ¢
60.0 mL - 40.0 mL = 20.0 mL O2 unreacted.
Energy absorbed by the student when steam at 100.°C changes to water at 100.°C
2.26 kJ
(1.5 g steam)a
b = 3.4 kJ (3.4 * 103 J)
g
Energy absorbed when water cools from 100.°C to 20.0°C
E = (m)(specific heat)(¢t)
E = (1.5 g)a
4.184 J
b(100.°C - 20.0°C) = 5.0 * 102 J
g°C
E total = 3.4 * 103 J + 5.0 * 102 J = 3.9 * 103 J
61.
(a)
(b)
won’t hydrogen bond – no hydrogen in the molecule
will hydrogen bond
CH 3 ¬ O ¬ H––––O ¬ CH 3
ƒ
H
(c)
won’t hydrogen bond; no hydrogen covalently bonded to a strongly electronegative
element
will hydrogen bond
H ¬ O––––H
ƒ
ƒ
H H
won’t hydrogen bond; hydrogen must be attached to N, O, or F
(d)
(e)
62.
During the fusion (melting) of a substance the temperature remains constant so a
temperature factor is not needed.
63.
Energy needed to heat Cu to its melting point:
E = (50.0 g)(0.385 J>g°C)(1083°C - 25.0°C) = 2.04 * 104 J
Energy needed to melt the Cu.
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E = (50.0 g)(134 J>g) = 6.70 * 103 J
E total = 2.04 * 104 J + 6.70 * 103 J = 2.71 * 104 J
64.
A lake freezes from the top down because the density of water as a solid is lower than
that of the liquid, causing the ice to float on the top of the liquid. Since the liquid remains
below the solid, marine, and aquatic life continues.
65.
Heat lost by warm water = heat gained by ice
m = grams of ice to lower temperature of water to 0.0°C.
4.184 J
b(45°C - 0.0°C) = (m)(335 J>g)
g°C
68 g = m (grams of ice melted)
68 g of ice melted. Therefore, 150. g - 68 g = 82 g ice remain.
(120. g)a
66.
Mass solution - mass H 2O = mass H 2SO4
(122 mL)(1.26 g>mL) - (100. mL)(1.00 g>mL) = 54 g H 2SO4 added
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