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CHAPTER 4
PROPERTIES OF MATTER
SOLUTIONS TO REVIEW QUESTIONS
1.
Gaseous state: 393 K ¼ 120.0 C; boiling point of acetic acid is 118.0 C.
2.
Liquid state: melting point of chlorine is –101.6 C and boiling point of chlorine is 34.6 C.
3.
Small bubbles appear at each electrode, and a gas collects above each electrode. The system now contains
water and two different gases.
4.
A new substance is always formed during a chemical change, but never formed during physical changes.
5.
Potential energy is the energy of position. By the position of an object, it has the potential of movement to
a lower energy state. Kinetic energy is the energy matter possesses due to its motion.
6. (a)
(b)
118:0 C þ 273:15 ¼ 391:2 K
ð118:0 CÞð1:8Þ þ 32 ¼ 244:4 F
7. (a)
(b)
ð16:7 CÞð1:8Þ þ 32 ¼ 62:1 F
16:7 C þ 273:15 ¼ 289:9 K
8.
Boiling water is a physical change so the chemical make-up of the water does not change. Therefore, the
bubbles that you see in boiling water are . . . water vapor!
9.
Dissolving salt in water is a physical change. If the salt-water solution is boiled the water will evaporate
away leaving the salt behind.
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- Chapter 4 -
SOLUTIONS TO EXERCISES
1. (a)
(b)
(c)
(d)
physical
chemical
physical
physical
(e)
(f)
(g)
(h)
chemical
physical
chemical
chemical
2. (a)
(b)
(c)
(d)
physical
physical
chemical
physical
(e)
(f)
(g)
(h)
physical
physical
chemical
chemical
3. Although the appearance of the platinum wire changed during the heating, the original appearance was
restored when the wire cooled. No change in the composition of the platinum could be detected.
4. A copper wire, like the platinum wire, changes to a glowing red color when heated (physical change).
Upon cooling, the original appearance of the copper wire is not restored, but a new substance, black
copper(II) oxide appears (chemical change).
5. Reactants:
Product:
copper, oxygen
copper(II) oxide
6. Reactant:
Product:
water
hydrogen, oxygen
7. (a)
(b)
(c)
chemical
physical
chemical
(d)
(e)
(f)
chemical
chemical
physical
8. (a)
(b)
(c)
physical
chemical
chemical
(d)
(e)
(f)
physical
chemical
chemical
9. (a)
(b)
(c)
kinetic energy
kinetic energy
potential energy
(d)
(e)
potential energy
kinetic energy
10. (a)
(b)
(c)
kinetic energy
potential energy
potential energy
(d)
(e)
kinetic energy
kinetic energy
11. The kinetic energy is converted to thermal energy (heat), chiefly in the brake system, and eventually
dissipated into the atmosphere.
12. The transformation of kinetic energy to thermal energy (heat) is responsible for the fiery reentry of a
space vehicle.
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- Chapter 4 13. (a)
(b)
(c)
þ
þ
(d)
(e)
þ
14. (a)
(b)
(c)
þ
þ
(d)
(e)
15.
heat ¼ ðmÞðspecific heatÞðDtÞ
¼ ð125 gÞð0:900 J=g CÞð95:5 C 19:0 CÞ
¼ 8:61 103 J
16.
heat ¼ ðmÞðspecific heatÞðDtÞ
¼ ð65 gÞð0:128 J=g CÞð98:5 C 22:0 CÞ
¼ 6:4 102 J
17.
heat ¼ ðmÞðspecific heatÞðDtÞ
¼ ð25:0 gÞð2:138 J=g CÞð78:5 C 22:5 CÞ
¼ 2:99 103 J
18.
heat ¼ ðmÞðspecific heatÞðDtÞ
¼ ð35:0 gÞð2:604 J=g CÞð82:4 C 21:2 CÞ
¼ 5:58 103 J
19. heat ¼ ðmÞðspecific heatÞðDtÞ; change kJ to J
specific heat ¼
heat
2:50 103 J
¼
¼ 0:230 J=g C
mðDtÞ ð135 gÞð100:0 C 19:5 CÞ
20. heat ¼ ðmÞðspecific heatÞðDtÞ; change kJ to J
specific heat ¼
heat
1:075 104 J
¼
¼ 0:128 J=g C
mðDtÞ ð275 gÞð327:5 C 21:2 CÞ
21. heat lost by copper ¼ heat gained by water x ¼ final temperature
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
ð155:0gÞð0:385 J=g CÞð150:0 C xÞ ¼ ð250:0gÞð4:184 J=g CÞðx 19:8 CÞ
8950 J 59:675x J= C ¼ 1046 x J= C 20710 J
8950 J þ 20710 J ¼ 1046 x J= C þ 59:675x J= C
29660 J ¼ 1106 x J= C
26:8 C ¼ x
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- Chapter 4 22. heat lost by copper ¼ heat gained by water x ¼ final temperature
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
ð225:0gÞð0:900 J=g CÞð125:5 C xÞ ¼ ð500:0gÞð4:184 J=g CÞðx 22:5 CÞ
25410 J 202:5x J= C ¼ 2090 x J= C 47070 J
25410 J þ 47070 J ¼ 2090 x J= C þ 202:5x J= C
72480 J ¼ 2290 x J= C
31:7 C ¼ x
23. (a)
heat lost by metal ¼ heat gained by water
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
ð110:0 gÞðspecific heatÞð92:0 C 24:2 CÞ ¼ ð75:0 gÞð4:184 J=g CÞð24:2 C 21:0 CÞ
ðspecific heatÞ 7458 g C ¼ 1004:16 J
specific heat ¼ 0:13 J=g C
(b)
24. (a)
Iron has a specific heat of 0.47 J/g C and lead has a specific heat of 0.13 J/g C. The metal could
possibly be lead, but further tests would be needed to determine this. The metal cannot be iron.
heat lost by metal ¼ heat gained by water
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
ð40:0 gÞðspecific heatÞð62:0 C 21:0 CÞ ¼ ð85:0 gÞð4:184 J=g CÞð21:0 C 19:2 CÞ
specific heat 1640 g C ¼ 640:152 J
specific heat ¼ 0:39 J=g C
(b)
Gold has a specific heat of 0.13 J/g C. Therefore, the metal could not be pure gold. It is possible
that the metal is copper, since the specific heat of copper is 0.385 J/g C. Further tests would be
needed to determine the identity of the metal.
25. When wood burns carbon dioxide is released into the air. Therefore, the mass of the wood before burning
will not equal the mass of the wood after burning. However, the law of conservation of mass still holds
true. Some of the original mass of the wood is converted to gases that are released into the atmosphere.
26. The product is a compound. The tin reacted with oxygen from the air, producing an oxide of tin.
27. heat ¼ ðmÞðspecific heatÞðDtÞ
mass ¼
heat
1:69 104 J
¼
¼ 52:5 g H2 O
ðspecific heatÞðDtÞ ð4:184 J=g CÞð100:0 C 23:0 CÞ
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- Chapter 4 28. heat ¼ ðmÞðspecific heatÞðDtÞ
heat
mass ¼
; convert grams to pounds
ðspecific heatÞðDtÞ
¼
29.
3:25 104 J
¼ 238 g gold
ð0:131 J=g CÞð1064:4 C 23:2 CÞ
1 lb
ð238 gÞ
¼ 0:525 lb gold
453:6 g
heat ¼ ðmÞðspecific heatÞðDtÞ
change kJ to J
4
4:00 10 J ¼ ð500:0 gÞð4:184 J=g CÞðx 10:0 CÞ
x ¼ final temperature
4:00 104 J ¼ 2092x J= C 20920 J
60,920 J ¼ 2092x J= C
60,920 C ¼ 2092x
29:1 C ¼ x
30.
heat lost by iron ¼ heat gained by water x ¼ mass of iron in grams
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
xð0:473 J=g CÞð125:0 C 25:6 CÞ ð375 gÞð4:184 J=g CÞð25:6 C 19:8 CÞ
47:0x J=g ¼ 9:1 103 J
x¼
31.
9:1 103 J
¼ 190 g Fe
47:0 J=g
heat lost by copper ¼ heat gained by water x ¼ mass of copper in grams
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
xð0:385 J=g CÞð275:1 C 29:7 CÞ ¼ ð272 gÞð4:184 J=g CÞð29:7 C 21:0 CÞ
94:5x J=g ¼ 9:9 103 J
x¼
32. (a)
9:9 103 J
¼ 1:0102 g Cu
94:5 J=g
heat ¼ ðmÞðspecific heatÞðDtÞ
ð100:0 gÞð0:0921 cal=g CÞð100:0 C 10:0 CÞ ¼ 829 cal to heat Cu
(b)
let x ¼ temperature of Al after adding 829 cal
829 cal ¼ ð100:0 gÞð0:215 cal=g CÞðx 10:0 CÞ
829 cal ¼ ð21:5 cal= CÞx 215 cal
x ¼ 48:6 C ðfinal temperature for aluminumÞ
Therefore the copper gets hotter since it ended up at 100.0 C.
Note: You can figure this out without calculation if you consider the specific heats of the metals.
Since the specific heat of copper is much less than aluminum the copper heats more easily.
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- Chapter 4 33.
heat lost by iron ¼ heat gained by water
x ¼ initial temperature of iron
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
ð500:0 gÞð0:473 J=g CÞðx 90:0 CÞ ¼ ð400:0 gÞð4:184 J=g CÞð90:0 C 10:0 CÞ
J
237 x 2:13 104 J ¼ 1:34 105 J
C
J
237 x ¼ 1:55 105 J C
C
x¼
1:55 105 J C
237 J
x ¼ 654 C
34.
heat lost by metal ¼ heat gained by water
x ¼ specific heat of metal
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
ð20:0 gÞðxÞð203: C 29:0 CÞ ¼ ð100:0 gÞð4:184 J=g CÞð29:0 C 25:0 CÞ
ð3480 g CÞx ¼ 1674 J
x ¼ 0:48 J
35.
heat lost ¼ heat gained
x ¼ final temperature
ðmÞðspecific heatÞðDtÞ ¼ ðmÞðspecific heatÞðDtÞ
ðspecific heats are the sameÞ
Let x ¼ final temperature
J
J
ð10:0 gÞ ð50:0 C xÞ ¼ ð50:0 gÞ ðx 10:0 CÞ
gC
gC
500:g C 10:0 gx ¼ 50:0 gx 500 g C
1:00 103 g C ¼ 60:0 gx
x¼
1:00 103 g C
¼ 16:7 C
60:0 g
36. Specific heats for the metals are Fe: 0.473 J/g C; Cu: 0.385 J/g C; Al: 0.900 J/g C. The metal with the lowest
specific heat will warm most quickly, therefore, the copper pan heats fastest, and fries the egg fastest.
37. In order for the water to boil both the pan and water must reach 100.0 C. Specific heat for copper is
0.385 J/g C.
J
J
ð300:0 gÞ 0:385 ð100: C 25 CÞ þ ð800:0 gÞ 4:184 ð100: C 25 CÞ
gC
gC
¼ 8:7 103 J þ 2:5 105 J
¼ 2:6 105 J needed
to heat the pan and water to 100 C
1s
ð2:6 105 JÞ
¼ 414 s ¼ 6:9 min ¼ 6 min þ 54 s
628 cal
The water will boil at 6:06 and 54 s p.m.
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- Chapter 4 38. Heat is transferred from the molecules of water on the surface to the air above them. As you blow you
move the warmed air molecules away from the surface replacing them with cooler ones which are
warmed by the coffee and cool it in a repeating cycle. Inserting a spoon into hot coffee cools the coffee by
heat transfer as well. Heat is transferred from the coffee to the spoon lowering the temperature of the
coffee and raising the temperature of the spoon.
39. The potatoes will cook at the same rate whether the water boils vigorously or slowly. Once the boiling
point is reached the water temperature remains constant. The energy available is the same so the cooking
time should be equal.
40. ð250 mLÞð0:04Þ ¼ 10 mL fat
ð10 mLÞð0:8 g=mLÞ ¼ 8 g fat in a glass of milk
41. mercury þ sulfur ! compound
The mercury and sulfur react to form a compound since the properties of the product are different from
the properties of either reactant.
g ¼ 1:36 103 g mercury
ð100:0 mLÞ 13:6
mL
mercury þ sulfur ! compound
1360 g þ 100:0 g
1460 g
This supports the Law of Conservation of Matter since the mass of the product is equal to the mass of the
reactants.
42. According to the law of conservation of energy the amount of potential energy the ball has initially
should equal the amount of kinetic energy it has at the bottom of the hill. If the hill that the ball rolled
down was frictionless then the ball should role up the other side until it has reached the same level as
where it started. However, no hill is really frictionless. Therefore, some of the ball’s potential energy is
converted to kinetic energy of motion and some is lost by way of heat friction between the ball and the
surface of the hill.
43. A chemical change has occurred. Hydrogen molecules and oxygen molecules have combined to form
water molecules.
44. (a)
(b)
As the water is heated, molecules that exist in the liquid state are changed into molecules in the
gas gaseous state.
A physical change has occurred. No new substance was formed during the heating process. Only a
change in state occurred.
45. Upon heating the substance, the appearance of the bright blue solid changes to black and a brownish
orange gas is released. A chemical change has occurred.
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