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1
Chemistry in Our Lives
1.1

a. Chemistry is the science of the composition and properties of matter.
b. A chemist is a scientist who studies the composition and changes of matter.
c. A chemical is a substance that is used in or produced by a chemical process.

1.2

Your friends may give a variety of definitions, most of which will probably not agree with the dictionary definitions.

1.3

Many chemicals are listed on a vitamin bottle such as: vitamin A, vitamin B3, vitamin B12, vitamin
C, folic acid, etc.

1.4

Many chemicals are listed on a cereal box such as: vitamin A, vitamin B6, vitamin B12, vitamin C,
folic acid, sugar, salt, iron, etc.

1.5

Typical items found in a medicine cabinet and chemicals they contain:
Antacid tablets: calcium carbonate, cellulose, starch, stearic acid, silicon dioxide
Mouthwash: water, alcohol, glycerol, sodium benzoate, benzoic acid
Cough suppressant: menthol, beta-carotene, sucrose, glucose

1.6

Typical items found in cleansers are: water, ammonia, sodium silicate, and sodium phosphate.

1.7

No. All of these ingredients are chemicals.

1.8

No. All of these ingredients are chemicals.

1.9

An advantage of a pesticide is that it protects crops from damage by various insects. Some
disadvantages are that a pesticide can destroy beneficial insects, be retained in a crop that is
eventually eaten by animals or humans, or pollute ground water.

1.10 Some advantages of eating sugar is that it gives energy and tastes sweet. Some disadvantages are
that sugar can lead to obesity and tooth decay.
1.11 a.
b.
c.
d.


A hypothesis proposes a possible explanation for a natural phenomenon.
An experiment is a procedure that tests the validity of a hypothesis.
A theory is a hypothesis that has been validated many times by many scientists.
An observation is a description or measurement of a natural phenomenon.

1.12 a.
b.
c.
d.

hypothesis
observation
experiment
theory

1.13 (1)
(2)
(3)
(4)
(5)
(6)

observation
hypothesis
experiment
observation
observation
theory

1.14 (1) observation

(2) hypothesis
(3) experiment

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Chapter 1

(4) experiment
(5) observation
(6) hypothesis or theory
1.15 There are several things a student can do to be successful in chemistry, including forming a study
group, going to lecture, working sample problems and study checks, working problems and
checking answers, reading the assignment ahead of class, going to the instructor’s office hours,
and keeping a problem notebook.
1.16 There are many things that make it difficult to learn chemistry, including not going to lecture
regularly, not working problems and study checks, not reading the assignment ahead of class, not
going to the instructor’s office hours, and waiting until the night before an exam to study.
1.17 a. Form a study group.
c. Visit the professor during office hours.
e. Become an active learner.
1.18 a, c, d, and e.
1.19 Yes. Sherlock’s investigation includes observations (gathering data), formulating a hypothesis,
testing the hypothesis, and modifying it until the hypothesis is validated.
1.20 Sherlock meant that you should not propose a theory until you have data from experiments and
observations.

1.21 a. Determination of a melting point with a thermometer is an observation.
b. Describing a reason for the extinction of dinosaurs is a hypothesis or theory.
c. Measuring the speed of a race is an observation.
1.22 a. observation
b. observation
c. hypothesis or theory
1.23 A hypothesis, which is a possible explanation for an observation, can be tested with experiments.
1.24 Experimentation is used to test and verify a hypothesis.
1.25 b. Another hypothesis needs to be written when experimental results do not support the previous
hypothesis.
c. More experiments are needed for a new hypothesis.
1.26 b. Many experiments by many scientists validate the hypothesis.
1.27 A successful study plan would include:
b. Working the sample problems throughout the chapter.
c. Going to the instructor’s office during office hours.
1.28 A successful study plan would include:
b. Forming a study group and discussing problems with others.
c. Working problems in a notebook for reference before exams.
1.29 a.
b.
c.
d.

(1)
(2)
(3)
(2)

observation
hypothesis

experiment
hypothesis

1.30 a.
b.
c.
d.

(1)
(2)
(1)
(1)

observation
hypothesis
observation
observation

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2
Measurements
2.1

a. meter, length

d. second, time

b. gram, mass
e. Celsius, temperature

c. liter, volume

2.2

a. liter, volume
d. gram, mass

b. meter, length
e. kelvin, temperature

c. kilogram, mass

2.3

a. meter; both
d. second, both

b. kilogram, both
e. Celsius, metric

c. foot, neither

2.4

a. cubic meter, both

d. liter, metric

b. kelvin, SI
e. gram, metric

c. Fahrenheit; neither

2.5

a. gram; metric
d. meter; both

b. liter; metric
e. second; both

c. Celsius; metric

2.6

a. kelvin, SI
d. meter, both

b. kilogram, both
e. cubic meter, both

c. liter, metric

2.7

a.

b.
c.
d.
e.
f.

2.8

a. 1.8 ϫ 108 g
e. 2.4 ϫ 10Ϫ2 s

2.9

a.
b.
c.
d.

Move the decimal point left four places to give 5.5 ϫ 104 m.
Move the decimal point left two places to give 4.8 ϫ 102 g.
Move the decimal point right six places to give 5 ϫ 10Ϫ6 cm.
Move the decimal point right four places to give 1.4 ϫ 10Ϫ4 s.
Move the decimal point right three places to give 7.85 ϫ 10Ϫ3 L.
Move the decimal point left six places to give 6.7 ϫ 105 kg.
c. 7.5 ϫ 105 g

d. 1.5 ϫ 10Ϫ1 m

The value 7.2 ϫ 103, which is also 72 ϫ 102, is greater than 8.2 ϫ 102.
The value 3.2 ϫ 10Ϫ2, which is also 320 ϫ 10Ϫ4, is greater than 4.5 ϫ 10Ϫ4.

The value 1 ϫ 104 or 10 000 is greater than 1 ϫ 10Ϫ4, or 0.0001.
The value 6.8 ϫ 10Ϫ2 or 0.068 is greater than 0.00052.

2.10 a. 5.5 ϫ 10Ϫ9
2.11 a.
b.
c.
d.

b. 6 ϫ 10Ϫ5 m
f. 1.5 ϫ 103 m3

b. 3.4 ϫ 102

c. 5 ϫ 10Ϫ8

d. 4 ϫ 10Ϫ10

The standard number is 1.2 times the power of 104, or 10 000, which gives 12 000.
The standard number is 8.25 times the power of 10Ϫ2, or 0.01, which gives 0.0825.
The standard number is 4 times the power of 106, or 1 000 000, which gives 4 000 000.
The standard number is 5 times the power of 10Ϫ3, or 0.001, which gives 0.005.

2.12 a. 0.000 036

b. 87 500

c. 0.03

d. 212 000


2.13 a. The estimated digit is the last digit reported in a measurement. In 8.6 m, the 6 in the first
decimal (tenths) place was estimated and has some uncertainty.
b. The estimated digit is the 5 in the second decimal (hundredths) place.
c. The estimated digit would be the 0 in the first decimal (tenths) place.
2.14 a. The estimated digit is the last digit reported in a measurement. In 125.04 g, the 4 in the
hundredths place is estimated and has some uncertainty.
b. The estimated digit is the 7 in the third decimal (thousandths) place.
c. The estimated digit would be the 8 in the first decimal (tenths) place.

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Chapter 2

2.15 Measured numbers are obtained using some kind of measuring tool. Exact numbers are numbers
obtained by counting or from a definition in the metric or the U.S measuring systems.
a. measured
b. exact
c. exact
d. measured
2.16 a. exact

b. measured

c. measured


d. measured

2.17 Measured numbers are obtained using some kind of measuring tool. Exact numbers are numbers
obtained by counting or from a definition in the metric or the U.S. measuring systems.
a. The value 6 oz of meat is obtained by measurement, whereas 3 hamburgers is a counted/
exact number.
b. None; both 1 table and 4 chairs are counted/exact numbers.
c. Both 0.75 lb and 350 g are obtained by measurements.
d. None; the values in a definition are exact numbers.
2.18 a. 5 pizzas
2.19 a.
b.
c.
d.
e.

c. 3 onions

d. 5 cars

Zeros preceding significant digits are not significant.
Zeros between significant digits are significant.
Zeros after significant digits in a decimal number are significant.
Zeros in the coefficient of a number written in scientific notation are significant.
Zeros in a number with no decimal point are considered as placeholders only and not significant.

2.20 a. significant
e. significant
2.21 a.

b.
c.
d.
e.
f.

b. 6 nickels

b. significant

c. not significant

d. not significant

All five numbers are significant figures.
Only the two nonzero numbers are significant; the preceding zeros are placeholders.
Only the two nonzero numbers are significant; the zeros that follow are placeholders.
All three numbers in the coefficient of a number written in scientific notation are significant.
All four numbers including the last zero in a decimal number are significant.
All three numbers including the zeros that follow a nonzero digit in a decimal number are
significant.

2.22 a. 4 SF
e. 3 SF

b. 6 SF
f. 2 SF

c. 3 SF


d. 3 SF

2.23 Both measurements in c have 2 significant figures and both measurements in d have 4 significant
figures.
2.24 In a and b both pairs have three significant figures. In d both pairs have two significant figures.
2.25 a. 5 000 is the same as 5 ϫ 1000, which is written in scientific notation as 5 ϫ 103.
b. 30 000 is the same as 3 ϫ 10 000, which is written in scientific notation as 3 ϫ 104.
c. 100 000 is the same as 1 ϫ 100 000, which is written in scientific notation as 1 ϫ 105.
1
d. 0.000 25 is the same as 2.5 ϫ
, which is written in scientific notation as 2.5 ϫ 10Ϫ4.
10 000
2.26 a. 5.1 ϫ 106 g

b. 2.6 ϫ 104 s

c. 4.0 ϫ 104 m

d. 8.2 ϫ 10Ϫ4 kg

2.27 Calculators carry out mathematical computations and display without regard to significant figures.
Our task is to round the calculator’s answer to the number of significant figures or digits allowed
by the values of the original data.
2.28 The number in the calculator display does not show the correct number of significant figures.
Thus, a significant zero must be added.
2.29 To round a number, determine how many significant figures are kept and drop all remaining digits.
There is no change in the retained figures if the first digit dropped is 0 to 4. However, if the first
digit dropped is 5 to 9, raise the last retained digit by 1.

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Measurements

To round 1.854, drop the 4 and keep 1.85.
To round 184.2038, drop 2038 and keep 184.
To round 0.004 738 265, drop 8265 and increase the retained digits by 1, or 0.004 74.
To round 8807 to three significant figures, drop 7 and increase the retained digits to 8810,
keeping a zero in the ones place as a placeholder. In scientific notation: 8.81 ϫ 103.
e. To round 1.832 149, drop 2149 and keep 1.83 as the rounded value.
a.
b.
c.
d.

2.30 a. 1.9
2.31 a.
b.
c.
d.

b. 180

c. 0.0047

d. 8800


e. 1.8

Drop 55 and increase the last digit by 1, which gives 56.9 m.
Drop 25, and keep remaining digits as 0.00228 g.
Drop 27, keep remaining digits, and add two zeros as placeholders, 11 500 s (1.15 ϫ 104 s).
Add a significant zero to give three significant figures, 8.10 L.

2.32 a. 3.3 m

b. 1.9 ϫ 102 g

c. 0.0023 m

d. 2.0 L

2.33 a. Because the value of 0.034 has 2 SFs, the answer 1.6 can have only 2 SFs.
b. The measurement 5 has 1 SF, which allows 1 SF in the answer (0.01).
c. The measurement 1.25 has 3 SFs, which allows 3 SFs in the answer (27.6):
34.56
ϭ 27.6
1.25
d. The measurement 25 has 2 SFs, which allows 2 SFs in the answer (3.5):
(0.2465)(25)
ϭ 3.5
1.78
e. The measurement 2.8 ϫ 104 has 2 SFs, which allows 2 SFs in the answer (0.14):
(2.8 ϫ 104)(5.05 ϫ 10Ϫ6) ϭ 0.14 (1.4 ϫ 10Ϫ1)
f. The measurement 8 ϫ 103 has 1 SF, which allows 1 SF in the answer (0.8):
(3.45 ϫ 10Ϫ2)(1.8 ϫ 105)

ϭ 8 ϫ 10Ϫ1 (0.8)
(8 ϫ 103)
2.34 a. 7 ϫ 104
d. 0.0055

b. 0.005
e. 6 ϫ 106

c. 15
f. 8.58

2.35 The answer of addition/subtraction problems has the same number of places as the measurement
with the largest place.
a.
45.48 cm
2 decimal places
+ 8.057 cm
3 decimal places
53.54 cm
2 decimal places
b.

23.45 g
104.1 g
+ 0.025 g
127.6 g

2 decimal places
1 decimal place
3 decimal places

1 decimal place

c.

145.675 mL
Ϫ 24.2 mL
121.5 mL

3 decimal places
1 decimal place
1 decimal place

d.

1.08 L
Ϫ 0.585 L
0.50 L

2 decimal places
3 decimal places
2 decimal places

2.36 a.
b.
c.
d.

5.08 g ϩ 25.1 g ϭ 30.2 g
85.66 cm ϩ 104.10 cm ϩ 0.025 cm ϭ 189.79 cm
24.568 mL Ϫ 14.25 mL ϭ 10.32 mL

0.2654 L Ϫ 0.2585 L ϭ 0.0069 L

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Chapter 2

2.37 The km/h markings indicate how many kilometers (how much distance) will be traversed in
1 hour’s time if the speed is held constant. The mph markings indicate the same distance traversed
but measured in miles during the 1 hour of travel.
2.38 On the speedometer, 80 kph is about 50 mph. You are not exceeding the 55 mph speed limit if
your speedometer reads 80 km/hr (kph).
2.39 Because the prefix kilo means one thousand times, a kilogram is equal to 1000 grams.
2.40 Because the prefix centi means one hundredth, a centimeter is one hundredth of a meter.
2.41 a. mg
e. ␮L

b. dL
f. ns

c. km

2.42 a. centimeter
d. gigameter

b. kilogram

e. microgram

c. deciliter
f. picogram

2.43 a. 0.01
d. 0.1

b. 1000
e. 1 000 000

c. 0.001
f. 10Ϫ9

2.44 a. 1 decigram
d. 1 centigram

b. 1 microgram
e. 1 milligram

c. 1 kilogram
f. 1 picogram

2.45 a. 100 cm

b. 1000 m

c. 0.001 m

d. 1000 mL


2.46 a. 1 kg ϭ 1000 g

b. 1 mL ϭ 0.001 L

c. 1 g ϭ 0.001 kg

d. 1 g ϭ 1000 mg

2.47 a.
b.
c.
d.

d. kg

A kilogram, which is 1000 g, is larger than a milligram (0.001 g).
A milliliter, which is 10Ϫ3 L, is larger than a microliter (10Ϫ6 L).
A km, which is 103 (1000) m, is larger than a cm (10Ϫ2 m or 0.01 m).
A kL, which is 103 (1000) L, is larger than a dL (10Ϫ1 L or 0.1 L).

2.48 a. mg

c. ␮m

b. mm

2.49 Because a conversion factor can be inverted to give a second factor.

d. mL

1m
100 cm
and
.
100 cm
1m

2.50 Verify that the units cancel when the conversion factors are applied.
2.51 The numerator and denominator are from the equality: 1 kg ϭ 1000 g
2.52 1 m ϭ 100 cm
2.53 a. 1 yd ϭ 3 ft,
b. 1 L ϭ 1000 mL,
c. 1 min ϭ 60 s,
d. 1 dL ϭ 100 mL,
2.54 a. 1 gal ϭ 4 qt,
b. 1 m ϭ 1000 mm,
c. 1 week ϭ 7 days,
d. $1 ϭ 4 quarters,

6

1 yd
3 ft
1L
1000 mL
1 min
60 s
1 dL
100 mL
1 gal

4 qt
1m
1000 mm
1 week
7 days
$1
4 quarters

and
and
and
and
and
and
and
and

3 ft
1 yd
1000 mL
1L
60 s
1 min
100 mL
1 dL
4 qt
1 gal
1000
1m
7 days

1 week
4 quarters
$1


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Measurements

2.55 The equalities between the metric prefixes can be written as two conversion factors.
1m
100 cm
a. 1 m ϭ 100 cm,
and
100 cm
1m
1g
1000 mg
b. 1 g ϭ 1000 mg,
and
1000 mg
1g
1L
1000 mL
c. 1 L ϭ 1000 mL,
and
1000 mL
1L

6
1 kg
10 mg
d. 1 kg ϭ 106 mg,
and
1 kg
106 mg
e. (1 m)3 ϭ (100 cm)3,
2.56 a. 1 in. ϭ 2.54 cm,
b. 1 kg ϭ 2.205 lb,
c. 1 lb ϭ 453.6 g,
d. 1.057 qt ϭ 1 L,
e. 1 in.2 ϭ (2.54)2 cm2,

(100 cm)3
(1 m)3
1 in.
2.54 cm
1 kg
2.205 lb
1 lb
453.6 g
1.057 qt
1L
(2.54)2 cm2
1 in.2

and
and
and

and
and
and

(1 m)3
(100 cm)3
2.54 cm
1 in.
2.205 lb
1 kg
453.6 g
1 lb
1L
1.057 qt
1 in.2
(2.54)2 cm2

2.57 An equality stated in a problem can be written as two conversion factors, which are true only for
that problem.
3.5 m
1s
a. 3.5 m ϭ 1 s,
and
1s
3.5 m
0.65 g
1 mL
b. 1 mL ϭ 0.65 g,
and
1 mL

0.65 g
46.0 km
1.0 gal
c. 1.0 gal ϭ 46.0 km,
and
1.0 gal
46.0 km
d. Percent means parts silver per 100 parts sterling silver. Using grams (g) as the mass unit,
100 g sterling ϭ 93 g silver,
e. ppb indicates ␮g/kg,
2.58 a. 32 mi ϭ 1 gal,
b. 20 drops ϭ 1 mL,
c. ppm indicates mg/kg,
d. 58 g gold ϭ 100 g jewelry,
e. $3.19 ϭ 1 gal,

93 g silver
100 g sterling
29 ␮g
1 kg
32 mi
1 gal
20 drops
1 mL
32 mg
1 kg
58 g gold
100 g jewelry
$3.19
1 gal


and
and
and
and
and
and
and

100 g sterling
93 g silver
1 kg
29 ␮g
1 gal
32 mi
1 mL
20 drops
1 kg
32 mg
100 g jewelry
58 g gold
1 gal
$3.19

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Chapter 2

2.59 When using a conversion factor, you are trying to cancel existing units and arrive at a new
(desired) unit. The conversion factor must be set up to give unit cancellation.
2.60 The new (desired) unit should be in the numerator of the conversion factor.
2.61 a. Plan: cm : m
1m
ϭ 1.75 m
175 cm ϫ
100 cm
b. Plan: mL : L
1L
ϭ 5.5 L
5500 mL ϫ
1000 mL
c. Plan: kg : g
1000 g
0.0055 kg ϫ
ϭ 5.5 g
1 kg
d. Plan: cm3 : m3
1 m3
350 cm3 ϫ
ϭ 3.5 ϫ 10Ϫ4 m3
(100)3 cm3
2.62 a. Plan: mg : g
1g
ϭ 0.8 g
1000 mg

b. Plan: dL : mL
100 mL
ϭ 85 mL
0.85 dL ϫ
1 dL
c. Plan: mg : g
1g
2840 mg ϫ
ϭ 2.84 g
1000 mg
d. Plan: m2 : km2
1 km2
150 000 m2 ϫ
ϭ 0.15 km2
(1000)2 m2
800 mg ϫ

2.63 a. Plan: qt : mL
1L
1000 mL
ϫ
ϭ 710. mL
0.750 qt ϫ
1.057 qt
1L
b. Plan: stone : lb : kg
14 lb
1 kg
ϫ
ϭ 74.9 kg

11.8 stones ϫ
1 stone
2.205 lb
c. Plan: in. : cm : mm
2.54 cm
10 mm
ϫ
ϭ 495 mm
19.5 in. ϫ
1 in.
1 cm
d. Plan: ␮m : m : cm : in.
1m
100 cm
1 in.
0.50 ␮m ϫ
ϫ
ϫ
ϭ 2.0 ϫ 10Ϫ5 in.
6
10 ␮m
1m
2.54 cm
1 lb
453.6 g
ϫ
ϭ 110 g
16 oz
1 lb
1 qt

1L
1000 mL
b. 5.0 pt ϫ
ϫ
ϫ
ϭ 2400 mL
2 pt
1.057 qt
1L
1 km
c. 120 000 mi ϫ
ϭ 190 000 km
0.6214 mi

2.64 a. 4.0 oz ϫ

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Measurements

1.057 qt
1 gal
ϫ
ϭ 12.2 gal
1L

4 qt
18.5 gal Ϫ 12.2 gal ϭ 6.3 gal

d. 46.0 L ϫ

2.65 a. Plan: ft : in. : cm : m
12 in.
2.54 cm
1m
78.0 ft ϫ
ϫ
ϫ
ϭ 23.8 m (length)
1 ft
1 in.
100 cm
b. Plan: ft : in. : cm : m : m2
12 in.
2.54 cm
1m
27.0 ft ϫ
ϫ
ϫ
ϭ 8.23 m (width)
1ft
1 in.
100 cm
Area ϭ 23.8 m ϫ 8.23 m ϭ 196 m2
c. Plan: m : km : hr : min : s
1 km

1 hr
60 min
60 s
ϫ
ϫ
ϫ
ϭ 0.463 s
23.8 m ϫ
1000 m
185 km
1 hr
1 min
d. Plan: m2 : cm2 : in.2 : ft2 : gal : qt : L
(100 cm)2
(1 in.)2
(1 ft)2
1 gal
4 qt
1L
196 m2 ϫ
ϫ
ϫ
ϫ
ϫ
ϫ
ϭ 53.2 L
2
2
2
2

1m
(2.54 cm)
(12 in.)
150 ft
1 gal
1.057 qt
2.66 a. 91.4 m

b. 41 m

c. 4500 m2

d. 2.7 s

2.67 Each of the following require a percent factor from the problem information.
a. Plan: g crust : g oxygen (percent equality: 100.0 g crust ϭ 46.7 g oxygen)
46.7 g oxygen
ϭ 152 g oxygen
325 g crust ϫ
100.0 g crust
b. Plan: g crust : g magnesium (percent equality: 100.0 g crust ϭ 2.1 g magnesium)
2.1 g magnesium
ϭ 0.026 g magnesium
1.25 g crust ϫ
100.0 g crust
c. Plan: oz : lb : g : g nitrogen (percent equality: 100.0 g fertilizer ϭ 15 g nitrogen)
1 lb
453.6 g
15 g nitrogen
ϫ

ϫ
ϭ 43 g nitrogen
10.0 oz fertilizer ϫ
16 oz
1 lb
100.0 g fertilizer
d. Plan: kg pecans : kg choc. bars : lb (percent equality: 100.0 kg bars ϭ 22.0 kg pecans)
100 kg choc. bars
2.205 lb
5.0 kg pecans ϫ
ϫ
ϭ 50. lb of chocolate bars
22.0 kg pecans
1 kg
2.68 a. 0.045 kg

b. 2530 g

c. 29 g fiber/cake

d. 4.0 oz

2.69 Because the density of aluminum is 2.70 g/cm3, silver is 10.5 g/cm3, and lead is 11.3 g/cm3, we
can identify the unknown metal by calculating its density as follows:
217 g metal
ϭ 11.3 g/cm3
The metal is lead.
19.2 cm3 metal
2.70 The volume of a cube, 2.0 cm on each edge, is calculated as follows:
(2.0 cm)3 ϫ 1 mL/1 cm 3 ϭ 8.0 mL

A cube will displace its volume when submerged in water, so the final volume reading in each
graduated cylinder is: 40.0 mL water ϩ 8.0 mL metal ϭ 48.0 mL total volume
2.71 Density is the mass of a substance divided by its volume. The densities of solids and liquids are
usually stated in g/mL or g/cm3.
Mass (grams)
Density ϭ
Volume (mL)

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Chapter 2

a.
b.
c.

d.
e.

2.72 a.
b.
c.
d.
e.


24.0 g
ϭ 1.20 g/mL
20.0 mL
0.870 g
453.6 g
0.250 lb
ϭ
ϫ
130.3 mL
1 lb
mL
Volume of gem: 34.5 mL total Ϫ 20.0 mL water ϭ 14.5 mL
45.0 g
Density of gem:
ϭ 3.10 g/mL
14.5 mL
485.6 g
ϭ 4.26 g/cm3
114 cm3
1 qt
1L
1000 mL
ϫ
ϫ
ϭ 47.3 mL
0.100 pt ϫ
2 pt
1.057 qt
1L
Mass of syrup ϭ 182.48 g Ϫ 115.25 g ϭ 67.23 g

67.23 g
ϭ 1.42 g/mL
Density of syrup ϭ
47.3 mL
1220 g/3500 mL ϭ 0.35 g/mL
155 g/125 mL
ϭ 1.24 g/mL
5.025 g/5.00 mL ϭ 1.01 g/mL
275 g/207 cm3 ϭ 1.33 g/cm3
140 g/10 000 mL ϭ 0.014 g/mL

2.73 a. 1.50 kg alcohol ϫ

1000 g
1 mL
1L
ϫ
ϫ
ϭ 1.91 L
1 kg alcohol
0.785 g
1000 mL

13.6 g
ϭ 88 g
1 mL
7.8 g
1 lb
16 oz
c. 225 mL ϫ

ϫ
ϫ
ϭ 62 oz
1 mL
453.6 g
1 lb
1 mL
8.92 g
d. 74.1 cm3 ϫ
ϫ
ϭ 661 g
3
1 cm
1 mL
4 qt
1000 mL
0.66 g
1 kg
e. 12.0 gal ϫ
ϫ
ϫ
ϫ
ϭ 30. kg
1 gal
1.057 qt
1 mL
1000 g
b. 6.5 mL ϫ

1 mL

ϭ 3.39 mL silver metal
10.5 g
18.0 mL water ϩ 3.39 mL silver ϭ 21.4 mL total volume
1 mL
ϭ 0.61 mL mercury metal
8.3 g ϫ
13.6 g
4 qt
1L
1000 mL
1.0 g
1 lb
35 gal ϫ
ϫ
ϫ
ϫ
ϫ
ϭ 290 lb
1 gal
1.057 qt
1L
1 mL
453.6 g
82.1 mL
0.904 kg

2.74 a. 35.6 g ϫ

b.
c.

d.
e.
2.75 a.
b.
c.
d.

The number of legs is a counted number; it is exact.
The height is measured with a ruler or tape measure; it is a measured number.
The number of chairs is a counted number; it is exact.
The area is measured with a ruler or tape measure; it is a measured number.

2.76 a. 61.5ЊC

b. 53.80ЊC

c. 4.9ЊC

2.77 a. length ϭ 6.96 cm; width ϭ 4.75 cm
b. length ϭ 69.6 mm; width ϭ 47.5 mm
c. There are three significant figures in the length measurement.

10


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Measurements


d. There are three significant figures in the width measurement.
e. 33.3 cm2
f. Since there are three significant figures in the width and length measurements, there are three
significant figures in the area.
2.78 a. Length is 3.7 cm; the 7 is the estimated digit.
b. Length is 2.50 cm; the 0 is the estimated digit.
c. Length is 4.10 cm; the 0 is the estimated digit.
2.79 The volume of the object is: 23.1 mL Ϫ 18.5 mL ϭ 4.6 mL.
8.24 g
The mass is 8.24 g and the density is:
ϭ 1.8 g/mL.
4.6 mL
2.80 a.
b.
c.
d.

This is cube 3, since it has sunk to the bottom.
This is cube 4, since it is floating about one-third out of the water.
This is cube 1, since it is floating about one-half out of the water.
This is cube 2, since it is floating just at the surface of the water.

2.81 A is vegetable oil, B is water, and C is mercury.
2.82 A. would be gold; it has the highest density and the smallest volume.
B. would be silver; its density is intermediate and the volume is intermediate.
C. would be aluminum; it has the lowest density and the largest volume.
2.83 a. Drop 8 and increase retained digits by 1 to give 0.000 0126 L (1.26 ϫ 10Ϫ5 L).
b. Drop 8 and increase retained digits by 1 to give 3.53 ϫ 102 kg.
c. Drop 111, keep retained digits, and add three zeros as placeholders to give 125 000 m3,

or 1.25 ϫ 105 m3.
d. Drop 03 and keep retained digits to give 58.7 m.
e. Add 2 significant zeros to give 3.00 ϫ 10Ϫ3 s.
f. Drop 26 and keep retained digits, 0.0108 g.
2.84 265 g
2.85 Plan: ft : in. : cm : m : min
12 in.
2.54 cm
1m
1 min
7500 ft ϫ
ϫ
ϫ
ϫ
ϭ 42 min
1 ft
1 in.
100 cm
55.0 m
2.86 a. 22 kg salmon ϩ 5.5 kg crab ϩ 3.48 kg oysters ϭ 31 kg seafood
2.205 lb
b. 31 kg seafood (total) ϫ
ϭ 68 lb
1 kg
2.87 Plan: lb : g : onions
453.6 g onions
1 onion
ϫ
ϭ 16 onions
4.0 lb onions ϫ

1 lb onions
115 g onions
Because the number of onions is a counting number, the value for onions, 15.8, is rounded to a
whole number, 16.
2.88 $1420 ϫ

1 kg
1 lb
ϫ
ϭ 4 ϫ 102 kg
$1.75
2.205 lb

2.89 a. Plan: oz : crackers
6 crackers
ϭ 96 crackers
8.0 oz ϫ
0.50 oz
b. Plan: crackers : servings : g : lb : oz
1 serving
4 g fat
1 lb
16 oz
10 crackers ϫ
ϫ
ϫ
ϫ
ϭ 0.2 oz fat
6 crackers
1 serving

453.6 g
1 lb

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Chapter 2

c. Plan: boxes : oz : servings : mg : g
8.0 oz
1 serving
140 mg sodium
1g
50 boxes ϫ
ϫ
ϫ
ϫ
ϭ 110 g sodium
1 box
0.50 oz
1 serving
1000 mg
2.90 75 000 mL ϫ

1L
1.057 qt

1 gal
ϫ
ϫ
ϭ 20. gal
1000 mL
1L
4 qt

2.91 Plan: lb : kg : pesos : dollar : cents
1 kg
48 peses
1 dollar
100 cents
0.45 lb ϫ
ϫ
ϫ
ϫ
ϭ 91 cents
2.205 lb
1 kg
10.8 peses
1 dollar
Because the calculation is for a counted number of cents, the value 90.7 is rounded to 91.
15.0 oz protein
1 lb
453.6 g
ϫ
ϫ
ϭ 34 g protein
100.0 oz burger

16 oz
1 lb
Yes, the hamburger contains 34 g of protein, which is 10 grams more than she is allowed. To stay
within her diet, Celeste could have only a 5.6-oz burger, as shown by the following calculation:
1 lb
16 oz
100.0-oz burger
ϫ
ϫ
ϭ 5.6-oz burger
24 g protein ϫ
453.6 g
1 lb
15.0 oz protein
This one burger would use her entire day’s allowance of protein!

2.92 8.0 oz burger ϫ

2.93 Plan: tubes : oz : lb : g : kg sunscreen : kg benzyl salicylate
4.0 oz
1 lb
453.6 g
1 kg
2.50 kg
ϫ
ϫ
ϫ
ϫ
ϭ 0.92 kg
325 tubes ϫ

1 tube
16 oz
1 lb
1000 g
100 kg sunscreen
2.94 442.5 mL total Ϫ 325.2 mL water ϭ 117.3 mL object
3.15 oz object
1 lb
453.6 g
ϫ
ϫ
ϭ 0.761 g/mL
117.3 mL object
16 oz
1 lb
2.95 This problem has two units. Convert g to mg, and convert L in the denominator to dL.
1.85 g
1000 mg
1L
ϫ
ϫ
ϭ 185 mg/dL
1L
1g
10 dL
2.96 9.60 L
2.97 The difference between the initial volume of the water and its volume with the lead object will
give us the volume of the object. 285 mL total Ϫ 215 mL water ϭ 70 mL lead
Using the density of lead, we can convert mL to the mass in grams of the lead object.
11.3 g lead

70. mL lead ϫ
ϭ 790 g lead
1 mL lead
1 cm3 iron
1 mL
ϫ
ϭ 1.91 mL iron
7.86 g iron
1 cm3
1 cm3 lead
1 mL
20.0 g lead ϫ
ϫ
ϭ 1.77 mL lead
11.3 g lead
1 cm3
155 mL water ϩ 1.91 mL iron ϩ 1.77 mL lead ϭ 159 mL total volume

2.98 15.0 g iron ϫ

2.99 Plan: L gas : mL gas : g gas : g oil : mL oil : cm3 oil
1.00 L gas ϫ

1000 mL gas
0.66 g gas
1 g oil
1 mL oil
1 cm3
ϫ
ϫ

ϫ
ϫ
ϭ 720 cm3 oil
1 L gas
1 mL gas
1 g gas
0.92 g oil
1 mL

2.100 Plan: kg : g : mL : L : qt
1000 g
1 mL alcohol
1L
1.057 qt
1.50 kg alcohol ϫ
ϫ
ϫ
ϫ
ϭ 2.02 qt alcohol
1 kg
0.785 g alcohol
1000 mL
1L

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Measurements

2.101 a. Plan: kg mass : kg fat : lb (percent equality: 100.0 kg mass ϭ 3.0 kg fat)
3.0 kg fat
2.205 lb
ϫ
ϭ 3.0 lb fat
45 kg body weight ϫ
100.0 kg body mass
1 kg
b. Plan: L fat : mL : g : lb
1000 mL
0.94 g fat
1 lb
ϫ
ϫ
ϭ 6.2 lb fat
3.0 L fat ϫ
1L
1 mL fat
453.6 g
2.102 5.77 kg
2.103 Plan: cm3 : g : g silver : lb : oz (percent equality: 100 g sterling ϭ 92.5 g silver)
10.3 g
92.5 g silver
1 lb
16 oz
27.0 cm3 ϫ
ϫ

ϫ
ϫ
ϭ 9.07 oz pure silver
1 cm3
100 g
453.6 g
1 lb
2.104 Plan: kg : lb body mass : lb water
2.205 lb
55 lb water
65 kg body mass ϫ
ϫ
ϭ 79 lb water
1 kg
100. lb body mass
2.105 Since the balance can measure mass to 0.001 g, the mass should be given to 0.001 g; you should
record the mass of the object as 34.075 g.
2.106 The student who reports 5.8 cm is not reading to the nearest mm. The others are estimating
differently.
2.107 6.4 gal
2.108 a. 79 cups

b. 314 cans

c. 157 tablets

2.109 3.8 ϫ 102 g aluminum
2.110 a. 69 m3

b. 6.9 ϫ 104 kg


2.111 a. 43 g

b. 3.9 g copper

c. 2.8 cm3

2.112 0.203 mm
2.113 75.7 mL ϩ 4.8 mL (silver) ϩ 2.6 mL (gold) ϭ 82.9 mL

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3
Matter and Energy
3.1

A pure substance has a definite composition. A mixture has a variable composition.
a. pure substance
b. mixture
c. pure substance

d. pure substance

3.2

a. mixture
c. mixture

3.3

Elements are the simplest type of pure substance. Compounds contain two or more elements in the
same ratio.
a. element
b. compound
c. element
d. compound

3.4

a. element
c. compound

3.5

A homogeneous mixture has a uniform composition. A heterogeneous mixture does not have a uniform composition throughout the mixture.
a. heterogeneous
b. homogeneous
c. homogeneous
d. heterogeneous

3.6


a. homogeneous
c. homogeneous

3.7

a.
b.
c.
d.

3.8

a. physical property
c. chemical property

3.9

a.
b.
c.
d.
e.

b. element
d. element

b. heterogeneous
d. heterogeneous


Color is a physical property.
The ability of hydrogen to react with oxygen is a chemical property.
The temperature at which a substance freezes is a physical property.
The change in milk left in a warm room is a chemical property.
b. chemical property
d. physical property

The change of water from gas to liquid is a physical change.
The reaction of cesium with water is a chemical change.
The melting of gold from solid to liquid is a physical change.
The change in shape is a physical change.
Dissolving sugar in water only separates the sugar molecules: it is a physical change.

3.10 a. physical change
c. physical change
e. physical change
3.11 a.
b.
c.
d.
e.

b. pure substance
d. pure substance

b. chemical change
d. chemical change

The reactivity of a substance is a chemical property.
The state of a substance is a physical property.

The color of a substance is a physical property.
The reaction with hydrogen is a chemical property.
The temperature at which a substance melts is a physical property.

3.12 a. physical property
c. physical property
e. physical property

b. chemical property
d. chemical property

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Chapter 3

3.13 The Fahrenheit temperature scale is still used in the United States. A normal body temperature is
98.6ЊF on this scale. To convert her temperature to the equivalent reading on the Celsius scale, the
following calculation must be performed:
(99.8ЊF Ϫ 32)
ϭ 37.7ЊC
1.8
Because a normal body temperature is 37.0ЊC, her temperature of 37.7ЊC would be a mild fever.
3.14 Because Mexico uses the Celsius temperature scale, he is accustomed to setting the oven’s temperature in Celsius degrees. I would advise him that ovens in the United States are calibrated in
Fahrenheit degrees and that we would need to determine the equivalent of 175ЊC on the Fahrenheit
scale, as follows:

1.8 (175ЊC) ϩ 32 ϭ 347ЊF
We would set the oven to 350°F and watch the cooking time carefully.
3.15 In temperature calculations, the number of significant figures is determined by the measured temperature. The values of 32 and 1.8 are exact.
a. 1.8 (37.0ЊC) ϩ 32 ϭ 66.6 ϩ 32 ϭ 98.6ЊF
33.3
(65.3ЊF Ϫ 32Њ)
b.
ϭ
ϭ 18.5ЊC
1.8
1.8
c. Ϫ27ЊC ϩ 273 ϭ 246 K
d. 62ЊC ϩ 273 ϭ 335 K
82
(114ЊF Ϫ 32)
e.
ϭ
ϭ 46ЊC
1.8
1.8
40
(72ЊF Ϫ 32)
f.
ϭ
ϭ 22ЊC; 22ЊC ϩ 273 ϭ 295 K
1.8
1.8
3.16 a. 1.8 (25ЊC) ϩ 32
b. 1.8 (155ЊC) ϩ 32
Ϫ57

Ϫ25ЊF Ϫ 32
c.
ϭ
1.8
1.8
d. 224 K Ϫ 273
e. 545 K Ϫ 273
f. 875 K Ϫ 273

ϭ 45 ϩ 32 ϭ 77ЊF
ϭ 279 ϩ 32 ϭ 311ЊF
ϭ Ϫ32ЊC
ϭ Ϫ49ЊC
ϭ 272ЊC
ϭ 602ЊC; 1.8 (602ЊC) ϩ 32 ϭ 1080 ϩ 32 ϭ 1110ЊF

74
(106ЊF Ϫ 32)
ϭ
ϭ 41ЊC
1.8
1.8
(103ЊF Ϫ 32)
71
b.
ϭ
ϭ 39ЊC
1.8
1.8
No, there is no need to phone the doctor. The child’s temperature is less than 40.0ЊC.


3.17 a.

145ЊF Ϫ 32
113
ϭ
ϭ 62.8ЊC (1.8 is exact)
1.8
1.8
b. 1.8 (20.6ЊC) ϩ 32 ϭ 37.1 ϩ 32 ϭ 69.1ЊF (32 is exact)

3.18 a.

3.19 At the top of the hill, the energy of the car is in the form of potential energy. As it descends, potential energy is converted to kinetic energy. When the car reaches the bottom, all its energy is kinetic energy.
3.20 As the elevator moves to the top of the ramp, the skier’s potential energy increases. As the skier
goes down the ramp (ski jump), potential energy is converted to kinetic energy.
3.21 a. potential; stored energy
c. potential; stored energy

16

b. kinetic; energy of motion
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Matter and Energy


3.22 a. potential

b. potential

c. kinetic

d. kinetic

3.23 a. Using a hair dryer converts electrical energy into heat energy (the air is warmed) and
mechanical energy.
b. Using a fan converts electrical energy into mechanical energy.
c. Burning gasoline converts chemical energy into mechanical energy as the car is propelled
down the road and heat energy as the engine warms up.
d. Radiant energy is converted into heat energy by the solar water heater.
3.24 a.
b.
c.
d.

Electrical energy becomes radiant and heat energies when the filament glows.
Burning gas converts chemical energy into radiant and heat energies.
Electrical energy is converted into heat energy.
The chemical energy of the log is converted into radiant and heat energies.

3.25 a. (2) light bulb
b. (3) hydroelectric power plant
c. (1) heater
3.26 a. (3) ceiling fan
b. (2) automobile engineering plant

c. (1) battery
1 kcal
ϭ 3.5 kcal
1000 cal
1 cal
ϫ
ϭ 99.2 cal
4.184 J
4.184 J
ϫ
ϭ 120 J
1 cal
1 cal
1000 J
ϫ
ϫ
ϭ 1100 cal
1 kJ
4.184 J

3.27 a. 3500 cal ϫ
b. 415 J
c. 28 cal
d. 4.5 kJ
3.28 a. 8100 cal
c. 10.7 kJ

b. 0.325 kJ
d. 10 500 J


3.29 Copper, which has the lowest specific heat, would reach the highest temperature.
3.30 Because the final temperature of B is lower, that means that B has a higher specific heat.
3.31 SH ϭ

J
m ϫ ⌬⌻

a. ⌬T ϭ 83.6ЊC Ϫ 24.2ЊC ϭ 59.4ЊC
b. ⌬T ϭ 57.9ЊC Ϫ 35.0ЊC ϭ 22.9ЊC
3.32 a. 0.227 J/g ЊC

312 J
ϭ 0.389 J/g ЊC
13.5 g ϫ 59.4ЊC
345 J
SH ϭ
ϭ 0.313 J/g ЊC
48.2 g ϫ 22.9ЊC
SH ϭ

b. 0.514 J/g ЊC

3.33 The heat required is given by the relationship: Heat (q) ϭ m ϫ ⌬T ϫ SH.
If heat is added, q is ϩ, if heat is lost q is Ϫ.
a. q ϭ m ϫ ⌬T ϫ SH ϭ 25.0 g ϫ (25.7ЊC Ϫ 12.5ЊC) ϫ 4.184 J/g ЊC ϭ
1 cal
ϭ ϩ330 cal
25.0 g ϫ 13.2ЊC ϫ 4.184 J/g ЊC ϭ ϩ1380 J; ϩ1380 J ϫ
4.184 J
b. q ϭ m ϫ ⌬T ϫ SH ϭ 38.0 g ϫ (246ЊC Ϫ 122ЊC) ϫ 0.385 J/g ЊC ϭ

1 cal
38.0 g ϫ 124ЊC ϫ 0.385 J/g ЊC ϭ ϩ1810 J; ϩ1810 J ϫ
ϭ ϩ434 cal
4.184 J

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Chapter 3

c. q ϭ m ϫ ⌬T ϫ SH ϭ 15.0 g ϫ (Ϫ42.0ЊC Ϫ 65.0ЊC) ϫ 2.46 J/g ЊC ϭ
1 cal
ϭ Ϫ904 cal
15.0 g ϫ Ϫ107ЊC ϫ 2.46 J/g ЊC ϭ Ϫ3780 J; Ϫ3780 J ϫ
4.184 J
d. q ϭ m ϫ ⌬T ϫ SH ϭ 125 g ϫ (55ЊC Ϫ 118ЊC) ϫ 0.450 J/g ЊC ϭ
1 cal
125 g ϫ Ϫ63ЊC ϫ 0.450 J/g ЊC ϭ Ϫ3500 J; Ϫ3500 J ϫ
ϭ Ϫ850 cal
4.184 J
3.34 a. ϩ1.30 ϫ 103 J, ϩ311 cal
b. Ϫ2.65 ϫ 104 J, Ϫ6.32 ϫ 103 cal
c. ϩ383 J, ϩ91.6 cal
d. Ϫ246 J, Ϫ58.8 cal
3.35 a. m ϭ
b. m ϭ

ϭ
c. m ϭ
ϭ
d. m ϭ

q
225 J
225 J
ϭ
ϭ
ϭ 54.5 g
⌬T ϫ SH
(47.0ЊC Ϫ 15.0ЊC) ϫ 0.129 J/g ЊC
32.0ЊC ϫ 0.129 J/g ЊC
Ϫ8.40 kJ
1000 J
q
ϭ
ϫ
⌬T ϫ SH
(82.0ЊC Ϫ 168.0ЊC) ϫ 0.45029 J/g ЊC
1 kJ
Ϫ8400 J
ϭ 217 g
Ϫ86.0ЊC ϫ 0.450 J/g ЊC
8.80 kJ
1000 J
q
ϭ
ϫ

⌬T ϫ SH
(26.8ЊC Ϫ 12.5ЊC) ϫ 0.897 J/g ЊC
1 kJ
8800 J
ϭ 686 g
14.3ЊC ϫ 0.897 J/g ЊC
q
Ϫ14 200 J
Ϫ14 200 J
ϭ
ϫ
ϭ 190 g
⌬T ϫ SH
(42ЊC Ϫ 185ЊC) ϫ 0.523 J/g ЊC
Ϫ143ЊC ϫ 0.523 J/g ЊC

3.36 a. 26.6 g
3.37 a. ⌬T ϭ
b. ⌬T ϭ
c. ⌬T ϭ
d. ⌬T ϭ

b. 113 g
m
m
m
m

3.38 a. 55.3ЊC


q
ϫ SH
q
ϫ SH
q
ϫ SH
q
ϫ SH

c. 15.1 g

d. 31.6 g

1580 J
ϭ 176ЊC
20.0 g ϫ 0.450 J/g ЊC
7.10 kJ
1000 J
ϭ
ϫ
ϭ 11.3ЊC
150.0 g ϫ 4.184 J/g ЊC
1 kJ
7680 J
ϭ
ϭ 700 ЊC
85.0 g ϫ 0.129 J/g ЊC
6.75 kJ
1000 J
ϭ 351ЊC

ϭ
ϫ
50.0 g ϫ 0.385 J/g ЊC
1 kJ
ϭ

b. 121ЊC

c. 2.02ЊC

d. 57.4ЊC

4.184 J
1 kJ
1 kcal
ϫ 10.5ЊC ϫ
ϭ 22.2 kJ ϫ
ϭ 5.30 kcal
g ЊC
1000 J
4.184 kJ
4.184 J
1 kJ
1 kcal
b. 4980 g ϫ
ϫ 42ЊC ϫ
ϭ 871 kJ ϫ
ϭ 208 kcal
g ЊC
1000 J

4.184 kJ
4.184 J
1 kJ
1 kcal
c. 1250 g ϫ
ϫ 25ЊC ϫ
ϭ 132 kJ ϫ
ϭ 31.6 kcal
g ЊC
1000 J
4.184 kJ

3.39 a. 505 g ϫ

3.40 To calculate heat, we need the mass of the water, its specific heat, and the temperature change.
Then divide the heat energy absorbed by the water (which is equivalent to the energy released by
the combustion of fuel) by the grams of octane burned, as follows:
4.184 J
1 kJ
1
ϭ 48.1 kJ/g octane
ϫ 16.1ЊC ϫ
ϫ
357 g ϫ
g ЊC
1000 J
0.500 g octane
48.1 kJ/g
ϭ 11.5 kcal/g
4.184 kJ/kcal


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Matter and Energy

3.41 a. Because the orange juice contains both carbohydrate and protein, two calculations will be
needed:
17 kJ
26 g carbohydrate ϫ
ϭ 442 kJ (not rounded)
g carbohydrate
17 kJ
ϭ 34 kJ (not rounded)
2 g protein ϫ
g protein
Total: 442 kJ ϩ 34 kJ ϭ 476 kJ (not rounded) ϭ 480 kJ (rounded to tens)
1 kcal
ϭ 110 kcal
480 kJ ϫ
4.184 kJ
17 kJ
b. 18 g carbohydrate ϫ
ϭ 310 kJ
g carbohydrate
4 kcal

ϭ 70 kcal
18 g carbohydrate ϫ
g carbohydrate
38 kJ
c. 14 g fat ϫ
ϭ 530 kJ
g fat
9 kcal
ϭ 130 kcal
14 g fat ϫ
g fat
d. Three calculations are needed:
17 kJ
ϭ 1156 kJ (not rounded)
g carbohydrate
17 kJ
ϫ
ϭ 2550 kJ (not rounded)
150 g protein
g protein
38 kJ
5.0 g fat
ϫ
ϭ 190 kJ
g fat
Total: 1156 kJ ϩ 2550 kJ ϩ 190 kJ ϭ 3896 kJ (not rounded)
ϭ 3900 kJ (rounded to tens place)
1 kcal
ϭ 932 kcal ϭ 930 kcal (rounded)
3900 kJ ϫ

4.184 kJ
68 g carbohydrate ϫ

3.42 a. Three calculations are needed:
17 kJ
ϭ 102 kJ
g carbohydrate
38 kJ
ϫ
ϭ 608 kJ
16 g fat
g fat
17 kJ kcal
ϫ
ϭ 119 kJ
7 g protein
g protein
Total: 102 kJ ϩ 608 kJ ϩ 119 kJ
ϭ 829 kj ϭ 830 kJ
1 kcal
830 kJ ϫ
ϭ 198 kcal ϭ 200 kcal
4.184 kJ
All values are rounded to the tens place.
38 kJ
b. 7 g fat
ϫ
ϭ 266 kJ
g fat
17 kJ

ϭ 153 kJ
9 g carbohydrate ϫ
g carbohydrate
17 kJ
2 g protein
ϫ
ϭ 34 kJ
g protein
6 g carbohydrate ϫ

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Chapter 3

266 kJ ϩ 153 kJ ϩ 34 kJ ϭ 453 kJ ϭ 450 kJ ϫ

1 kcal
ϭ 107 ϭ 110 kcal
4.184 kJ

All values are rounded to the tens place.
c. With just carbohydrate present, only a single calculation is required.
17 kJ
ϭ 595 ϭ 600 kJ
35 g carbohydrate ϫ

g carbohydrate
1 kcal
ϭ 143 kcal ϭ 140 kcal
600 kJ ϫ
4.184 kJ
17 kJ
ϭ 272 kJ
d. 16 g carbohydrate ϫ
g carbohydrate
17 kJ
ϫ
ϭ 153 kJ
9 g protein
g protein
38 kJ
12 g fat ϫ
ϭ 456 kJ
g fat
Total ϭ 881 kJ ϭ 880 kJ
1 kcal
880 kJ ϫ
ϭ 210 kcal
4.184 kJ
3.43 a. compound

b. mixture

c. element

3.44 a is a homogeneous mixture because it has the same properties throughout.

3.45 b and c are heterogeneous mixtures because they are not the same throughout the mixture.
3.46 a. homogeneous
d. homogeneous

b. heterogeneous
e. heterogeneous

c. heterogeneous
f. homogeneous

3.47 gold 250 J or 59 cal; aluminum 240 J or 58 cal; silver 250 J or 59 cal
The heat needed for 10.0 cm3 samples of the metals is almost the same.
3.48 a. cheeseburger (31 g ϩ 34) ϫ 4 kcal/g ϭ 260 kcal ϩ (29 ϫ 9 ϭ 260 kcal)
Total 520 kcal
french fries (3 g ϩ 29 g) ϫ 4 kcal/g ϭ 130 kcal ϩ (11 ϫ 9 ϭ 100 kcal)
Total 230 kcal
chocolate shake (11 g ϩ 60. g) ϫ 4 kcal/g ϭ 280 kcal ϩ (9 ϫ 9 ϭ 80 kcal)
Total 360 kcal
kcal for total meal ϭ 1090 kcal
sleeping 60 kcal/hr 1090 kcal ϫ 1 hr/60 kcal ϭ 18 hr sleeping
b. running 750 kcal/hr 1090 kcal ϫ 1 hr/750 kcal ϭ 1.5 hr running
3.49 a. element
d. element

b. compound
e. mixture

c. mixture

3.50 a. heterogeneous

d. heterogeneous

b. homogeneous
e. homogeneous

c. homogeneous

3.51 a. Appearance is a physical property.
b. The melting point of gold is a physical property.
c. The ability of gold to conduct electricity is a physical property.
d. The ability of gold to form a new substance with sulfur is a chemical property.
3.52 a. physical property
c. physical property
3.53 a.
b.
c.
d.

20

b. chemical property
d. physical property

Plant growth is a chemical change.
A change of state from liquid to solid is a physical change.
Chopping wood into smaller pieces is a physical change.
Burning wood, which forms new substances, is a chemical change.


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Matter and Energy

3.54 a. physical change
c. chemical change

b. physical change
d. chemical change

3.55 a. Tablets are solid.
d. Air is a mixture of gases.

b. Helium in a balloon is a gas.
e. Charcoal is a solid.

c. Milk is a liquid.

3.56 a. solid
d. gas

b. liquid
e. liquid

c. solid

3.57 a. 56.7ЊC, 330 K
3.58 a. 331 K, 136ЊF


b. Ϫ56.5ЊC, 216 K
b. 184K, Ϫ129ЊF

3.59 Ϫ26ЊC, 247 K
3.60 115.7ЊF, 319.5 K
3.61 3500 kcal
3.62 This question requires three calculations to obtain the total kcal.
4 kcal
ϭ 50 kcal due to carbohydrate
12 g carbohydrate ϫ
1 g carbohydrate
9 kcal
ϫ
ϭ 80 kcal due to fat
9 g fat
1 g fat
4 kcal
ϫ
ϭ 40 kcal due to protein
9 g protein
1 g protein
Total: 50 kcal ϩ 80 kcal ϩ 40 kcal ϭ 170 kcal
3.63 Water has a higher specific heat than sand, which means that a large amount of energy is required
to cause a significant temperature change. Even a small amount of energy will cause a significant
temperature change in the sand.
3.64 883 g ϫ

4.184 J
1 kJ
ϫ 23ЊC ϫ

ϭ 85 kJ
g ЊC
1000 J

3.65 725 g ϫ

4.18 J
1 kJ
ϫ 28ЊC ϫ
ϭ 85 kJ
g ЊC
1000 J

0.385 J
ϭ Ϫ472 J
g ЊC
472 J
1 g ЊC
mwater ϭ
ϫ
ϭ 5.13 g H2O
22ЊC
4.184 J

3.66 25.0 g ϫ (Ϫ49ЊC) ϫ

3.67 50.0 g water ϫ 9.4ЊC ϫ
SHmetal ϭ
3.68


4.183 J
ϭ 1970 J
g ЊC

Ϫ1970 J
ϭ 1.1 J/g ЊC
25.0 g (Ϫ70.6ЊC)

18.9 kJ
1000 J
1 cal
1 kcal
ϫ
ϫ
ϫ
ϭ 9.0 kcal/g oil
0.50 g oil
1 kJ
4.18 J
1000 cal
1000 J
ϭ 8 400 000 J
1 kJ
Heat ϭ m ϫ ⌬T ϫ SH
8 400 000 J
Heat
ϭ
ϭ 40ЊC
⌬T ϭ
m ϫ SH

50 000 g (4.184 J/g ЊC)
Final T ϭ 20ЊC ϩ 40ЊC ϭ 60ЊC

3.69 8400 kJ ϫ

21


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Chapter 3

3.70 1.5 gal ϫ

4 qt
1L
1000 mL 0.66 g
11.5 kcal
ϫ
ϫ
ϫ
ϭ 4.3 ϫ 104 kcal
1 gal
1.057 qt
1L
1 mL
1g


3.71 The heat lost by the water is equal to the heat gained by the copper, for both: q ϭ m ϫ ⌬T ϫ SH.
For water: q ϭ m ϫ ⌬T ϫ SH ϭ 50.0 g ϫ (24.0ЊC Ϫ 16.0ЊC) ϫ 4.184 J/g ЊC ϭ
50.0 g ϫ 8.0ЊC ϫ 4.184 J/g ЊC ϭ 1670 J (not rounded)
1670 J
1670 J
q
For copper: m ϭ
ϭ
ϭ
⌬T ϫ SH
(86.0ЊC Ϫ 24.0ЊC) ϫ 0.385 J/g ЊC
62.0ЊC ϫ 0.385 J/g ЊC
ϭ 70 g
3.72 The meal contains a total of 21 g protein, a total of 38 g fat, and a total of 124 g carbohydrate. The
total caloric content of the meal must be determined to answer the question.
4 kcal
ϭ 84 kcal due to protein
1 g protein
9 kcal
ϫ
ϭ 340 kcal due to fat
38 g fat
1 g fat
4 kcal
124 g carbohydrate ϫ
ϭ 500 kcal due to carbohydrate
1 g carbohydrate
Total: 84 kcal ϩ 340 kcal ϩ 500. kcal ϭ (924) ϭ 920 kcal
Using Table 3.11, 1.8 hr of swimming are needed to “burn off ” the caloric content of the meal.
1 hr

ϭ 1.8 hr
920 kcal ϫ
500 kcal
21 g protein

ϫ

3.73 a. 45 g protein, 140 g carbohydrate, 53 g fat
b. 71 g protein, 210 g carbohydrate, 84 g fat
c. 98 g protein, 290 g carbohydrate, 120 g fat
3.74 q (water) ϭ 85.0 g ϫ 32ЊC ϫ 4.184 J/g ЊC ϭ 11 400 J
Ϫ11 400 J
SHmetal ϭ
ϭ 0.397 J/g ЊC ϭ 0.40 J/g ЊC
(Ϫ239ЊC)(125 g)
3.75 a. 26.5 g
c. 176ЊC

b. 54.5 g
d. 0.27 cal/g ЊC

Answers to Combining Ideas Chapters 1–3
CI 1 a. five significant figures
b. 294.10 troy ounce ϫ

31.1035 g
ϭ 9147.5 g gold
1 troy ounce

1 kg

ϭ 9.1475 kg gold
1000 g
1 cm3
c. 9147.5 g ϫ
ϭ 474 cm3
19.3 g
2.54 cm
d. 0.0035 in. ϫ
ϭ 0.0089 cm
1 in.
474 cm3
ϭ 5.3 ϫ 104 cm2
0.0089 cm
1 m2
ϭ 5.3 m2
5.3 ϫ 104 cm2 ϫ
(102 cm)2
e. (1064ЊC ϫ 1.8) ϩ 32 ϭ 1947ЊF; 1064ЊC ϩ 273 ϭ 1337 K
9147.5 g ϫ

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Matter and Energy

f. T ϭ (85 Ϫ 32)ր1.8 ϭ 29ЊC

⌬T ϭ 29 Ϫ 22 ϭ 7ЊC

T ϭ (72 Ϫ 32)/1.8 ϭ 22ЊC

Heat ϭ m ϫ SH ϫ ⌬T ϭ 9147.5 g ϫ
CI 2 a.
b.
c.
d.
e.

0.129 J
1 kJ
ϫ 7ЊC ϫ
ϭ 8 kJ
g ЊC
1000 J

330 km
$15.50
4.6 hr (4 hr 40 min)
1.6 ϫ 104 g
6.6 ϫ 105 kJ

453.6 g
1 cm3
ϫ
ϭ 14 cm3
1 lb
7.86 g

0.25 lb
453.6 g
1 cm3
b. 30 nails ϫ
ϫ
ϫ
ϭ 5.8 cm3 ϭ 5.8 mL
75 nails
1 lb
7.86 g
17.6 mL water ϩ 5.8 mL nails ϭ 23.4 mL new volume level
(15.6ЊF Ϫ 32)
c.
ϭ Ϫ9.11ЊC
1.8
(125.1ЊF Ϫ 32)
ϭ 51.7ЊC ⌬T ϭ 51.7 Ϫ (Ϫ9.11) ϭ 60.8ЊC
1.8
453.6 g
0.450 J
Heat ϭ 0.25 lb ϫ
ϫ 60.8ЊC ϫ
ϭ 3100 J or 3.1 ϫ 103 J
1 lb
g ЊC
453.6 g
0.450 J
4.184 J
d. 0.25 lb ϫ
ϫ (55 Ϫ Tf)ЊC ϫ

ϭ 325 g H2O ϫ (Tf Ϫ 4.0)ЊC ϫ
1 lb
g ЊC
g ЊC
2810 J Ϫ 51Tf J ϭ 1360 Tf J Ϫ 5440 J
2810 J ϩ 5440 J ϭ 51 Tf J ϩ 1360 Tf J
8250 J ϭ 1410 J Tf
8250 J
Tf ϭ
ϭ 5.9ЊC
1410 J

CI 3 a. 0.25 lb ϫ

CI 4 a.
b.
c.
d.

185 kcal
774 kJ
161 kJ
110 min

(12 in.)2
(2.54 cm)3
1 mL
1L
ϫ
28

in.
ϫ
ϫ
ϫ
ϭ 1.7 ϫ 103 L
2
3
3
1 ft
1 in.
1 cm
1000 mL
1.0 kg
b. 1.7 ϫ 103 L ϫ
ϭ 1.7 ϫ 103 kg
1L
1ЊC
c. 105ЊF Ϫ 62ЊF ϭ 43ЊF 43Њ F ϫ
ϭ 24ЊC ϭ ⌬T
1.8ЊF
4.184 kJ
ϭ 1.7 ϫ 105 kJ
1.7 ϫ 103 kg ϫ 24ЊC ϫ
kg ЊC
1 min
1 hr
d. 1.7 ϫ 105 kJ ϫ
ϫ
ϭ 0.47 hr
6000 kJ

60 min

CI 5 a. 25 ft2 ϫ

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