CHAPTER 1
CHEMISTRY: THE STUDY OF CHANGE
Problem Categories
Biological: 1.26, 1.50, 1.71, 1.72, 1.80, 1.86, 1.96, 1.97, 1.105, 1.114.
Conceptual: 1.3, 1.4, 1.17, 1.18, 1.11, 1.12, 1.56, 1.64, 1.91, 1.94, 1.101, 1.103, 1.117.
Environmental: 1.72, 1.89, 1.91, 1.98, 1.109, 1.112.
Industrial: 1.53, 1.57, 1.83.
Difficulty Level
Easy: 1.3, 1.9, 1.10, 1.11, 1.12, 1.17, 1.23, 1.24, 1.25, 1.26, 1.27, 1.28, 1.31, 1.32, 1.33, 1.34, 1.35, 1.36, 1.56, 1.57,
1.66, 1.79, 1.82, 1.86, 1.91.
Medium: 1.4, 1.18, 1.37, 1.38, 1.39, 1.40, 1.41, 1.42, 1.43, 1.44, 1.45, 1.46, 1.47, 1.48, 1.49, 1.50, 1.51, 1.52, 1.53,
1.54, 1.55, 1.58, 1.59, 1.60, 1.61, 1.62, 1.63, 1.64, 1.65, 1.72, 1.73, 1.74, 1.75, 1.76, 1.77, 1.78, 1.80, 1.81, 1.83, 1.84,
1.85, 1.87, 1.93, 1.96, 1.97, 1.98.
Difficult: 1.67, 1.68, 1.69, 1.70, 1.71, 1.88, 1.89, 1.90, 1.92, 1.94, 1.95, 1.99, 1.100, 1.101, 1.102, 1.103, 1.104, 1.105,
1.106.

1.3

(a)

Quantitative. This statement clearly involves a measurable distance.

(b)

Qualitative. This is a value judgment. There is no numerical scale of measurement for artistic
excellence.

(c)

Qualitative. If the numerical values for the densities of ice and water were given, it would be a
quantitative statement.

(d)

Qualitative. Another value judgment.

(e)

Qualitative. Even though numbers are involved, they are not the result of measurement.

1.4

(a)

hypothesis

1.9

Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum;

(b)

law

(c)

theory

Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon.

1.10


1.11

(a)

Cs

(b)

Ge

(c)

Ga

(d)

Sr

(e)

U

(f)

Se

(g)

Ne


(h)

Cd

(a)

element

(b)

compound

(c)

element

(d)

compound


2

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

1.12

(a)


homogeneous mixture

(b)

element

(c)

compound

(d)

homogeneous mixture

(e)

heterogeneous mixture

(f)

heterogeneous mixture

(g)

element

(a)

Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are


1.17

changed.
(b)

Chemical property. The fertilizer is consumed by the growing plants; it is turned into vegetable matter
(different composition).

(c)

Physical property. The measurement of the boiling point of water does not change its identity or
composition.

(d)

Physical property. The measurement of the densities of lead and aluminum does not change their
composition.

(e)

Chemical property. When uranium undergoes nuclear decay, the products are chemically different
substances.

1.18

(a)

Physical change. The helium isn’t changed in any way by leaking out of the balloon.

(b)


Chemical change in the battery.

(c)

Physical change. The orange juice concentrate can be regenerated by evaporation of the water.

(d)

Chemical change. Photosynthesis changes water, carbon dioxide, etc., into complex organic matter.

(e)

Physical change. The salt can be recovered unchanged by evaporation.

586 g
mass
=
= 3.12 g/mL
volume
188 mL

1.23

density =

1.24

Strategy: We are given the density and volume of a liquid and asked to calculate the mass of the liquid.


Rearrange the density equation, Equation (1.1) of the text, to solve for mass.

density =

mass
volume

Solution:

mass = density ´ volume
mass of methanol =

0.7918 g
´ 89.9 mL = 71.2 g
1 mL

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1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

3

Full file at />5C
1.25

? C = (F - 32F) ´


(a)
(b)
(c)
(d)
(e)

9F

5C
= 35C
9F
5C
? C = (12 - 32)F ´
= - 11C
9F
5C
? C = (102 - 32)F ´
= 39C
9F
5C
? C = (1852 - 32)F ´
= 1011C
9F
æ
9F ö÷
÷ + 32F
? F = ççC ´
5C ø÷÷
èç

? C = (95 - 32)F ´

æ
9F ö÷
÷ + 32F = - 459.67F
? F = çç-273.15 C ´
5C ÷÷ø
èç
1.26

Strategy: Find the appropriate equations for converting between Fahrenheit and Celsius and between
Celsius and Fahrenheit given in Section 1.7 of the text. Substitute the temperature values given in the
problem into the appropriate equation.

(a)

Conversion from Fahrenheit to Celsius.

? C = (F - 32F) ´

5C
9F

? C = (105 - 32)F ´
(b)

5C
= 41C
9F


Conversion from Celsius to Fahrenheit.

æ
9F ö÷
÷ + 32F
? F = ççC ´
çè
5C ø÷÷
æ
9F ÷ö
÷ + 32F = 11.3 F
? F = çç-11.5 C ´
çè
5C ÷÷ø
(c)

Conversion from Celsius to Fahrenheit.

æ
9F ö÷
÷ + 32F
? F = ççC ´
5C ø÷÷
èç
æ
9F ÷ö
÷÷ + 32F = 1.1 ´ 10 4 F
? F = çç6.3 ´ 103 C ´
çè
5C ÷ø

(d)

Conversion from Fahrenheit to Celsius.

? C = (F - 32F) ´

5C
9F

? C = (451 - 32)F ´

5C
= 233C
9F

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4

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

1.27

K = (C + 273C)

1.28

1K

1C

(a)

K = 113C + 273C = 386 K

(b)

K = 37C + 273C = 3.10 ´ 10 K

(c)

K = 357C + 273C = 6.30 ´ 10 K

(a)

K = (C + 273C)

2

2

1K
1C

C = K - 273 = 77 K - 273 = -196C

(b)

C = 4.2 K - 273 = -269C


(c)

C = 601 K - 273 = 328C

1.31

(a)

2.7 ´ 10

1.32

(a)

10

-2

-8

2

3.56 ´ 10

(b)

(c)

-2


-8

10

-8

(a)

(b)

1.34

= 0.0152

indicates that the decimal point must be moved 8 places to the left.
7.78 ´ 10

1.33

-2

9.6 ´ 10

(d)

indicates that the decimal point must be moved two places to the left.
1.52 ´ 10

(b)


4

4.7764 ´ 10

= 0.0000000778

-1

2

145.75 + (2.3 ´ 10 ) = 145.75 + 0.23 = 1.4598 ´ 10
79500
2.5 ´ 10

2

=

7.95 ´ 10 4
2.5 ´ 10

2

-3

= 3.2 × 10 2
-4

-3


-3

-3

(c)

(7.0 ´ 10 ) - (8.0 ´ 10 ) = (7.0 ´ 10 ) - (0.80 ´ 10 ) = 6.2 ´ 10

(d)

(1.0 ´ 10 ) ´ (9.9 ´ 10 ) = 9.9 ´ 10

(a)

Addition using scientific notation.

4

6

10

n

Strategy: Let’s express scientific notation as N ´ 10 . When adding numbers using scientific notation, we

must write each quantity with the same exponent, n. We can then add the N parts of the numbers, keeping the
exponent, n, the same.
Solution: Write each quantity with the same exponent, n.

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

n

3

5

3

Let’s write 0.0095 in such a way that n = -3. We have decreased 10 by 10 , so we must increase N by 10 .
Move the decimal point 3 places to the right.
0.0095 = 9.5 ´ 10

-3

Add the N parts of the numbers, keeping the exponent, n, the same.
9.5 ´ 10

-3

+ 8.5 ´ 10

-3
-3


18.0 ´ 10

The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10
n
to express N between 1 and 10 (1.8), we must increase 10 by a factor of 10. The exponent, n, is increased by
1 from -3 to -2.
18.0 ´ 10
(b)

-3

= 1.8 ´ 10

-2

Division using scientific notation.
n

Strategy: Let’s express scientific notation as N ´ 10 . When dividing numbers using scientific notation,
divide the N parts of the numbers in the usual way. To come up with the correct exponent, n, we subtract the
exponents.
Solution: Make sure that all numbers are expressed in scientific notation.

653 = 6.53 ´ 10

2

Divide the N parts of the numbers in the usual way.
6.53 ¸ 5.75 = 1.14
Subtract the exponents, n.


1.14 ´ 10
(c)

+ 2 - (- 8)

+2+8

= 1.14 ´ 10

= 1.14 ´ 10

10

Subtraction using scientific notation.
n

Strategy: Let’s express scientific notation as N ´ 10 . When subtracting numbers using scientific notation,
we must write each quantity with the same exponent, n. We can then subtract the N parts of the numbers,
keeping the exponent, n, the same.
Solution: Write each quantity with the same exponent, n.

Let’s write 850,000 in such a way that n = 5. This means to move the decimal point five places to the left.
850,000 = 8.5 ´ 10

5

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6

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

Subtract the N parts of the numbers, keeping the exponent, n, the same.
8.5 ´ 10

5

- 9.0 ´ 10

5

-0.5 ´ 10

5

The usual practice is to express N as a number between 1 and 10. Since we must increase N by a factor of 10
n

to express N between 1 and 10 (5), we must decrease 10 by a factor of 10. The exponent, n, is decreased by
1 from 5 to 4.
5

-0.5 ´ 10 = -5 ´ 10

(d)

4


Multiplication using scientific notation.
n

Strategy: Let’s express scientific notation as N ´ 10 . When multiplying numbers using scientific

notation, multiply the N parts of the numbers in the usual way. To come up with the correct exponent, n, we
add the exponents.
Solution: Multiply the N parts of the numbers in the usual way.

3.6 ´ 3.6 = 13
Add the exponents, n.

13 ´ 10

- 4 + (+ 6)

= 13 ´ 10

2

The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10
n

to express N between 1 and 10 (1.3), we must increase 10 by a factor of 10. The exponent, n, is increased by
1 from 2 to 3.
2

13 ´ 10 = 1.3 ´ 10


3

(a)

four

(b)

two

(c)

five

(d)

two, three, or four

(e)

three

(f)

one

(g)

one


(h)

two

(a)

one

(b)

three

(c)

three

(e)

two or three

(f)

one

(g)

one or two

1.37


(a)

10.6 m

1.38

(a) Division

1.35

1.36

(b)

0.79 g

(c)

2

16.5 cm

(d)

(d)

four

6


3

1 × 10 g/cm

Strategy: The number of significant figures in the answer is determined by the original number having the

smallest number of significant figures.
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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

7

Solution:

7.310 km
= 1.283
5.70 km
The 3 (bolded) is a nonsignificant digit because the original number 5.70 only has three significant digits.
Therefore, the answer has only three significant digits.
The correct answer rounded off to the correct number of significant figures is:
1.28
(b)

(Why are there no units?)

Subtraction


Strategy: The number of significant figures to the right of the decimal point in the answer is determined by

the lowest number of digits to the right of the decimal point in any of the original numbers.
Solution: Writing both numbers in decimal notation, we have

0.00326 mg
- 0.0000788 mg

0.0031812 mg
The bolded numbers are nonsignificant digits because the number 0.00326 has five digits to the right of the
decimal point. Therefore, we carry five digits to the right of the decimal point in our answer.
The correct answer rounded off to the correct number of significant figures is:
-3

0.00318 mg = 3.18 ´ 10

(c)

mg

Addition

Strategy: The number of significant figures to the right of the decimal point in the answer is determined by

the lowest number of digits to the right of the decimal point in any of the original numbers.
Solution: Writing both numbers with exponents = +7, we have
7

7


7

(0.402 ´ 10 dm) + (7.74 ´ 10 dm) = 8.14 ´ 10 dm
7

Since 7.74 ´ 10 has only two digits to the right of the decimal point, two digits are carried to the right of the
decimal point in the final answer.
(d)

Subtraction, addition, and division

Strategy: For subtraction and addition, the number of significant figures to the right of the decimal point in

that part of the calculation is determined by the lowest number of digits to the right of the decimal point in
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.


8

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

any of the original numbers. For the division part of the calculation, the number of significant figures in the
answer is determined by the number having the smallest number of significant figures. First, perform the
subtraction and addition parts to the correct number of significant figures, and then perform the division.
Solution:

(7.8 m - 0.34 m)
(1.15 s + 0.82 s)


1.39

=

7.5 m
1.97 s

= 3.8 m/s

Calculating the mean for each set of data, we find:
Student A: 87.6 mL
Student B: 87.1 mL
Student C: 87.8 mL
From these calculations, we can conclude that the volume measurements made by Student B were the most
accurate of the three students. The precision in the measurements made by both students B and C are fairly
high, while the measurements made by student A are less precise. In summary:
Student A: neither accurate nor precise
Student B: both accurate and precise
Student C: precise, but not accurate

1.40

Calculating the mean for each set of data, we find:
Tailor X: 31.5 in
Tailor Y: 32.6 in
Tailor Z: 32.1 in
From these calculations, we can conclude that the seam measurements made by Tailor Z were the most
accurate of the three tailors. The precision in the measurements made by both tailors X and Z are fairly high,
while the measurements made by tailor Y are less precise. In summary:
Tailor X: most precise

Tailor Y: least accurate and least precise
Tailor Z: most accurate

1.41

1 dm

(a)

? dm = 22.6 m ´

(b)

? kg = 25.4 mg ´

(c)

? L = 556 mL ´

0.1 m

= 226 dm

0.001g
1mg

´

1 ´ 10-3 L
1 mL


1kg
1000 g

= 2.54 ´ 10 5 kg

= 0.556 L

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

(d)

1.42

?

g
cm 3

=

9

æ1 ´ 10-2 m ÷ö3
ç
÷÷÷ = 0.0106 g / cm 3

´
´ ççç
1 kg
çç 1 cm ÷÷÷
è
ø

10.6 kg

1000 g

1 m3

(a)
Strategy: The problem may be stated as

? mg = 242 lb
A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This
relationship will allow conversion from pounds to grams. A metric conversion is then needed to convert
grams to milligrams (1 mg = 1 ´ 10

-3

g). Arrange the appropriate conversion factors so that pounds and

grams cancel, and the unit milligrams is obtained in your answer.
Solution: The sequence of conversions is

lb  grams  mg
Using the following conversion factors,

453.6 g

1 mg

1 lb

1 ´ 10-3 g

we obtain the answer in one step:
? mg = 242 lb ´

1 mg
453.6 g
´
= 1.10 ´ 108 mg
-3
1 lb
1 ´ 10 g

Check: Does your answer seem reasonable? Should 242 lb be equivalent to 110 million mg? How many

mg are in 1 lb? There are 453,600 mg in 1 lb.
(b)
Strategy: The problem may be stated as
3

3

? m = 68.3 cm
-2


Recall that 1 cm = 1 ´ 10

3

3

m. We need to set up a conversion factor to convert from cm to m .

Solution: We need the following conversion factor so that centimeters cancel and we end up with meters.
1 ´ 10-2 m
1 cm

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10

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

Since this conversion factor deals with length and we want volume, it must therefore be cubed to give
1 ´ 10-2 m

´

1 cm

1 ´ 10-2 m
1 cm


´

1 ´ 10-2 m
1 cm

æ1 ´ 10-2 m ö÷3
ç
÷÷
= ççç
÷÷
1
cm
ççè
ø÷÷

We can write
?m

æ1 ´ 10-2 m ö÷3
ç
÷÷
-5 3
= 68.3 cm ´ ççç
÷÷ = 6.83 ´ 10 m
çç 1 cm
è
ø÷÷
3


3

3

Check: We know that 1 cm = 1 ´ 10
-6

1 ´ 10

-6

3

1

3

m . We started with 6.83 ´ 10 cm . Multiplying this quantity by

-5

gives 6.83 ´ 10 .

(c)
Strategy: The problem may be stated as
3

? L = 7.2 m

3


3

3

In Chapter 1 of the text, a conversion is given between liters and cm (1 L = 1000 cm ). If we can convert m
3

to cm , we can then convert to liters. Recall that 1 cm = 1 ´ 10

-2

m. We need to set up two conversion

3

3

3

factors to convert from m to L. Arrange the appropriate conversion factors so that m and cm cancel, and
the unit liters is obtained in your answer.
Solution: The sequence of conversions is
3

3

m  cm  L
Using the following conversion factors,
æ 1 cm ö÷3

çç
÷÷
çç
÷
çç1 ´ 10-2 m ÷÷÷
è
ø

1L
1000 cm 3

the answer is obtained in one step:
æ
ç
1 cm
? L = 7.2 m3 ´ ççç
çç1 ´ 10-2 m
è

ö÷3
1L
÷÷÷ ´
= 7.2 ´ 10 3 L
÷÷
3
1000
cm
ø÷
3


3

3

Check: From the above conversion factors you can show that 1 m = 1 ´ 10 L. Therefore, 7 m would
3

equal 7 ´ 10 L, which is close to the answer.

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

11

(d)
Strategy: The problem may be stated as

? lb = 28.3 mg
A relationship between pounds and grams is given on the end sheet of your text (1 lb = 453.6 g). This
relationship will allow conversion from grams to pounds. If we can convert from mg to grams, we can then
-6

convert from grams to pounds. Recall that 1 mg = 1 ´ 10

g. Arrange the appropriate conversion factors so

that mg and grams cancel, and the unit pounds is obtained in your answer.

Solution: The sequence of conversions is
mg  g  lb

Using the following conversion factors,
1 ´ 10-6 g

1 lb

1 mg

453.6 g

we can write
? lb = 28.3 mg ´

1 ´ 10-6 g
1 mg

´

1 lb
= 6.24 ´ 10-8 lb
453.6 g

Check: Does the answer seem reasonable? What number does the prefix m represent? Should 28.3 mg be a
very small mass?

1255 m
1.43


1.44

1s

´

1 mi
1609 m

´

3600 s
1h

= 2808 mi/ h

Strategy: The problem may be stated as

? s = 365.24 days
You should know conversion factors that will allow you to convert between days and hours, between hours
and minutes, and between minutes and seconds. Make sure to arrange the conversion factors so that days,
hours, and minutes cancel, leaving units of seconds for the answer.
Solution: The sequence of conversions is

days  hours  minutes  seconds
Using the following conversion factors,
24 h

60 min


60 s

1 day

1h

1 min

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12

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

we can write
? s = 365.24 days ´

24 h
1 day

´

60 min
60 s
´
= 3.1557 ´ 107 s
1h
1 min


Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year?

1.45

(93 ´ 106 mi) ´

1.46

(a)

? in/s =

(b)

? m/min =

(c)

? km/h =

1.47

6.0 ft ´

168 lb ´

1.48

1.49


1.50

? mph =

62 m
1s

´

1.609 km
1 mi

1000 m
1 km

1s

´

8

3.00 ´ 10 m

´

1 min
60 s

= 8.3 min


1 mi
5280 ft
12 in
1 min
´
´
´
= 118 in/s
8.92 min
1 mi
1 ft
60 s
1 mi
1609 m
´
= 1.80 ´ 10 2 m/min
8.92 min
1 mi

1 mi
1609 m
1 km
60 min
´
´
´
= 10.8 km/h
8.92 min
1 mi

1000 m
1h

1m
= 1.8 m
3.28 ft
453.6 g
1 lb

´

1 kg
1000 g

= 76.2 kg

286 km
1 mi
´
= 178 mph
1h
1.609 km

1 mi
1609 m

0.62 ppm Pb =

´


3600 s
1h

(a)

1.42 yr ´

(b)

32.4 yd ´

= 1.4 ´ 102 mph

0.62 g Pb
1 ´ 106 g blood

6.0 ´ 103 g of blood ´

1.51

´

0.62 g Pb
1 ´ 10 g blood

365 days
1 yr

36 in
1 yd


6

´

´

24 h
1 day

2.54 cm
1 in

´

= 3.7 ´ 10-3 g Pb

3600 s
1h

´

3.00 ´ 108 m
1s

´

1 mi
1609 m


= 8.35 ´ 1012 mi

= 2.96 ´ 10 3 cm

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1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

(c)

1.52

3.0 ´ 1010 cm
1s

1 in

´

2.54 cm

1 ft
12 in

1 lb


(a)

? lbs = 70 kg ´

(b)

? s = 14 ´ 109 yr ´

(c)

? m = 90 in ´

(d)

æ
? L = 88.6 m ´ ççç
çè

365 d
1 yr

1 in

´

= 9.8 ´ 108 ft /s

= 1.5 ´ 102 lbs

0.4536 kg


2.54 cm

3

´

´

24 h
1d

´

3600 s
1h

= 4.4 ´ 1017 s

1 ´ 10-2 m
= 2.3 m
1 cm

3
1L
÷÷ö ´
= 8.86 ´ 10 4 L
÷÷
3
-2

1 ´ 10 m ø
1000 cm

1 cm

1.53

æ 1 cm ÷ö3
ç
´
´ çç
density =
÷÷÷ = 2.70 ´ 10 3 kg/m 3
3
ç
1000 g
çè 0.01 m ÷ø
1 cm

1.54

density =

1.55

Substance

1.56

2.70 g


13

1 kg

0.625 g
1L
1 mL
´
´
= 6.25 ´ 10-4 g/cm 3
1L
1000 mL
1 cm 3

Qualitative Statement

Quantitative Statement

(a)

water

colorless liquid

freezes at 0C

(b)

carbon


black solid (graphite)

density = 2.26 g/cm

(c)

iron

rusts easily

density = 7.86 g/cm

(d)

hydrogen gas

colorless gas

melts at -255.3C

(e)

sucrose

tastes sweet

at 0C, 179 g of sucrose dissolves in 100 g of H2O

(f)


table salt

tastes salty

melts at 801C

(g)

mercury

liquid at room temperature

boils at 357C

(h)

gold

a precious metal

density = 19.3 g/cm

(i)

air

a mixture of gases

contains 20% oxygen by volume


3
3

3

See Section 1.6 of your text for a discussion of these terms.
(a)

Chemical property. Iron has changed its composition and identity by chemically combining with
oxygen and water.

(b)

Chemical property. The water reacts with chemicals in the air (such as sulfur dioxide) to produce acids,
thus changing the composition and identity of the water.

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14

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

(c)

Physical property. The color of the hemoglobin can be observed and measured without changing its
composition or identity.


(d)

Physical property. The evaporation of water does not change its chemical properties. Evaporation is a
change in matter from the liquid state to the gaseous state.

(e)

Chemical property. The carbon dioxide is chemically converted into other molecules.
1 ton

1.57

(95.0 ´ 109 lb of sulfuric acid) ´

1.58

Volume of rectangular bar = length ´ width ´ height
density =

1.59

3

2.0 ´ 10 lb

= 4.75 ´ 10 7 tons of sulfuric acid

m
52.7064 g
=

= 2.6 g/cm 3
V
(8.53 cm)(2.4 cm)(1.0 cm)

mass = density ´ volume
(a)

3

mass = (19.3 g/cm ) ´ [

4
3
4
p(10.0 cm) ] = 8.08 ´ 10 g
3

3
æ
1 cm ÷ö
ç
÷÷ = 1.4 ´ 10-6 g
(b) mass = (21.4 g/cm 3 ) ´ çç0.040 mm ´
çç
10 mm ÷÷ø
è

(c)
1.60


mass = (0.798 g/mL) (50.0 mL) = 39.9 g

You are asked to solve for the inner diameter of the bottle. If we can calculate the volume that the cooking
2
oil occupies, we can calculate the radius of the cylinder. The volume of the cylinder is, Vcylinder = pr h (r is
the inner radius of the cylinder, and h is the height of the cylinder). The cylinder diameter is 2r.
volume of oil filling bottle =

volume of oil filling bottle =

mass of oil
density of oil
1360 g
= 1.43 ´ 103 mL = 1.43 ´ 103 cm3
0.953 g/mL

Next, solve for the radius of the cylinder.
2

Volume of cylinder = pr h
r=

r=

volume
p´h

1.43 ´ 103 cm3
p ´ 21.5 cm


= 4.60 cm

The inner diameter of the bottle equals 2r.
Bottle diameter = 2r = 2(4.60 cm) = 9.20 cm
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1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

1.61

15

From the mass of the water and its density, we can calculate the volume that the water occupies. The volume
that the water occupies is equal to the volume of the flask.
volume =

mass
density

Mass of water = 87.39 g - 56.12 g = 31.27 g

Volume of the flask =

31.27 g
mass
= 31.35 cm 3

=
3
density
0.9976 g/cm

1.62

343 m
1 mi
3600 s
´
´
= 767 mph
1s
1609 m
1h

1.63

The volume of silver is equal to the volume of water it displaces.
3

Volume of silver = 260.5 mL - 242.0 mL = 18.5 mL = 18.5 cm
density =

1.64

194.3 g
18.5 cm 3


= 10.5 g/cm 3

In order to work this problem, you need to understand the physical principles involved in the experiment in
Problem 1.61. The volume of the water displaced must equal the volume of the piece of silver. If the silver
did not sink, would you have been able to determine the volume of the piece of silver?
The liquid must be less dense than the ice in order for the ice to sink. The temperature of the experiment must
be maintained at or below 0°C to prevent the ice from melting.

1.65

The volume of a sphere is:

V =

3
4 3
4 æç 48.6 cm ÷÷ö
4
3
pr = p çç
÷ = 6.01 ´ 10 cm
3
3 çè 2 ÷÷ø

density =

1.66

Volume =


6.852 ´ 105 g
mass
3
=
= 11.4 g/cm
4
3
volume
6.01 ´ 10 cm

mass
density

Volume occupied by Li =

1.20 ´ 103 g
0.53 g / cm

3

= 2.3 ´ 10 3 cm 3

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16

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE


1.67

For the Fahrenheit thermometer, we must convert the possible error of 0.1°F to °C.

? C = 0.1F ´

5C
= 0.056C
9F

The percent error is the amount of uncertainty in a measurement divided by the value of the measurement,
converted to percent by multiplication by 100.
Percent error =

known error in a measurement
value of the measurement

´ 100%

For the Fahrenheit thermometer,

percent error =

0.056C
´ 100% = 0.1%
38.9C

For the Celsius thermometer,

percent error =


0.1C
´ 100% = 0.3%
38.9C

Which thermometer is more accurate?
1.68

To work this problem, we need to convert from cubic feet to L. Some tables will have a conversion factor of
3
28.3 L = 1 ft , but we can also calculate it using the dimensional analysis method described in Section 1.9 of
the text.
First, converting from cubic feet to liters:

(

æ12 in ÷ö3
÷ ´
5.0 ´ 107 ft 3 ´ çç
çè 1 ft ø÷÷

)

1 ´ 10-3 L
æ
ö3
çç 2.54 cm ÷÷ ´ 1 mL ´
= 1.42 ´ 109 L
çè 1 in ø÷÷
3

1
mL
1 cm

The mass of vanillin (in g) is:
2.0 ´ 10-11 g vanillin
1L

(

)

´ 1.42 ´ 109 L = 2.84 ´ 10-2 g vanillin

The cost is:
= $0.064 = 6.4 c
(2.84 ´ 10-2 g vanillin) ´ 50 g$112
vanillin

1.69

æ
9F ö÷
÷ + 32F
? F = ççC ´
çè
5C ø÷÷
Let temperature = t

9

t + 32F
5
9
t - t = 32F
5
4
- t = 32F
5

t =

t = - 40F = - 40C
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1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

1.70

17

There are 78.3 + 117.3 = 195.6 Celsius degrees between 0°S and 100°S. We can write this as a unit factor.

æ
ö
çç195.6C ÷÷
çè 100S ø÷÷

Set up the equation like a Celsius to Fahrenheit conversion. We need to subtract 117.3C, because the zero
point on the new scale is 117.3C lower than the zero point on the Celsius scale.

æ195.6C ö÷
÷ (? S ) - 117.3C
? C = çç
èç 100S ÷÷ø

1.71

Solving for ? °S gives:

æ 100S ÷ö
÷
? S = (? C + 117.3C) çç
çè195.6C ÷÷ø

For 25°C we have:

æ 100S ÷ö
÷ = 73S
? S = (25 + 117.3)C çç
çè195.6C ÷÷ø

The key to solving this problem is to realize that all the oxygen needed must come from the 4% difference
(20%-16%) between inhaled and exhaled air.
The 240 mL of pure oxygen/min requirement comes from the 4% of inhaled air that is oxygen.
240 mL of pure oxygen/min = (0.04)(volume of inhaled air/min)

Volume of inhaled air/min =


240 mL of oxygen/min
= 6000 mL of inhaled air/min
0.04

Since there are 12 breaths per min,
volume of air/breath =

1.72

(a)

(b)

1.73

6000 mL of inhaled air
1 min
8.6 ´ 103 L of air
1 day

´

´

1 min
6000 mL of inhaled air
´
= 5 ´ 10 2 mL/breath
1 min

12 breaths

0.001 L
1 mL

´

60 min
1h

2.1 ´ 10-6 L CO
1 L of air

´

24 h
1 day

= 8.6 ´ 103 L of air/day

= 0.018 L CO/day

Twenty-five grams of the least dense metal (solid A) will occupy the greatest volume of the three metals, and
25.0 g of the most dense metal (solid B) will occupy the least volume.
We can calculate the volume occupied by each metal and then add the volume of water (20.0 mL) to find the
total volume occupied by the metal and water.

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18

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

25.0 g A ´

1 mL

= 8.6 mL
2.9 g A
Total volume = 8.6 mL + 20.0 mL = 28.6 mL

Solid A:

1 mL

25.0 g B ´

Solid B:

= 3.0 mL

8.3 g B

Total volume = 3.0 mL + 20.0 mL = 23.0 mL

25.0 g C ´

Solid C:


1 mL
3.3 g C

= 7.6 mL

Total volume = 7.6 mL + 20.0 mL = 27.6 mL

Therefore, we have: (a) solid C, (b) solid B, and (c) solid A.

1.74

The diameter of the basketball can be calculated from its circumference. We can then use the diameter of a
ball as a conversion factor to determine the number of basketballs needed to circle the equator.
Circumference = 2πr
29.6 in
circumference
=
= 9.42 in
p
p
The circumference of the Earth is 2πr = (2 × π × 6400 km) = 40212 km
d = 2r =

40212 km ´

1000 m
1km

´


1 cm
-2

1´10

´
/m

1 in
1 ball
´
= 168,000,000.balls
2.54 cm 9.42 in

We round up to an integer number of basketballs with 3 significant figures.

1.75

Assume that the crucible is platinum. Let’s calculate the volume of the crucible and then compare that to the
volume of water that the crucible displaces.
volume =

mass
density

Volume of crucible =

860.2 g
21.45 g/cm 3


Volume of water displaced =

= 40.10 cm 3

(860.2 - 820.2) g
0.9986 g/cm 3

= 40.1 cm 3

The volumes are the same (within experimental error), so the crucible is made of platinum.

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1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

1.76

19

Volume = surface area ´ depth
3

2

Recall that 1 L = 1 dm . Let’s convert the surface area to units of dm and the depth to units of dm.


(

8

surface area = 1.8 ´ 10 km

(

)

depth = 3.9 ´ 103 m ´

2

æ1000 m ÷ö2
æ 1 dm ÷ö2
çç
ç
÷
÷÷ = 1.8 ´ 1016 dm 2
´ç
÷ ´ çç
çç 1 km ÷÷
çç 0.1 m ÷÷
è
ø
è
ø


)

1 dm
0.1 m

= 3.9 ´ 10 4 dm
16

2

4

20

3

20

Volume = surface area ´ depth = (1.8 ´ 10 dm ) (3.9 ´ 10 dm) = 7.0 ´ 10 dm = 7.0 ´ 10 L

1.77

31.103 g Au

(a)

2.41 troy oz Au ´

(b)


1 troy oz = 31.103 g

= 75.0 g Au

1 troy oz Au

? g in 1 oz = 1 oz ´

1 lb
16 oz

´

453.6 g
1 lb

= 28.35 g

A troy ounce is heavier than an ounce.

1.78

Volume of sphere =

4 3
πr
3

3
4 æç15 cm ö÷÷

3
3
ç
Volume = p ç
÷ = 1.77 ´ 10 cm
3 çè 2 ø÷÷

(

)

mass = volume ´ density = 1.77 ´ 103 cm 3 ´

4.0 ´ 101 kg Os ´

1.79

(a)

(b)

1.80

2.205 lb
1 kg

0.864 g - 0.837 g
0.864 g

1 cm 3


´

1 kg
1000 g

= 4.0 ´ 101 kg Os

= 88 lb Os

0.798 g/mL - 0.802 g/mL
0.798 g/mL

22.57 g Os

´ 100% = 0.5%

´ 100% = 3.1%

4

62 kg = 6.2 ´ 10 g
O:

4

4

(6.2 ´ 10 g)(0.65) = 4.0 ´ 10 g O


N:

4

3

(6.2 ´ 10 g)(0.03) = 2 ´ 10 g N

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20

1.81

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

4

4

4

3

C:

(6.2 ´ 10 g)(0.18) = 1.1 ´ 10 g C


H:

(6.2 ´ 10 g)(0.10) = 6.2 ´ 10 g H

4

2

4

2

Ca: (6.2 ´ 10 g)(0.016) = 9.9 ´ 10 g Ca
P:

(6.2 ´ 10 g)(0.012) = 7.4 ´ 10 g P

3 minutes 43.13 seconds = 223.13 seconds

Time to run 1500 meters is:
1500 m ´

1.82

1 mi
1609 m

´

2


223.13 s
1 mi

= 208.01 s = 3 min 28.01 s

2

? C = (7.3 ´ 10 - 273) K = 4.6 ´ 10 C
æ
9o F ö÷÷
ç
o
2
? F = çç(4.6 ´ 102 o C) ´
÷÷ + 32 F = 8.6 ´ 10  F
ççè
5o C ø÷

1.83

? g Cu = (5.11 ´ 103 kg ore) ´

1.84

(8.0 ´ 104 tons Au) ´

1.85

? g Au =


34.63% Cu
100% ore

2000 lb Au
1 ton Au

4.0 ´ 10-12 g Au
1 mL seawater

´

´

1000 g

1 lb Au

0.001 L

= 1.77 ´ 106 g Cu

1 kg

16 oz Au

1 mL

value of gold = (6.0 ´ 1012 g Au) ´


´

´

$948
= $2.4 ´ 1012 or 2.4 trillion dollars
1 oz Au

´ (1.5 ´ 1021 L seawater) = 6.0 ´ 1012 g Au

1 lb
453.6 g

´

16 oz
1 lb

´

$948
= $2.0 ´ 1014
1 oz

No one has become rich mining gold from the ocean, because the cost of recovering the gold would outweigh
the price of the gold.

1.1 ´ 1022 Fe atoms

1.86


? Fe atoms = 4.9 g Fe

1.87

mass of Earth's crust = (5.9 ´ 1021 tons) ´

1.0 g Fe

= 5.4 ´ 10 22 Fe atoms

0.50% crust
100% Earth

mass of silicon in crust = (2.95 ´ 1019 tons crust) ´

= 2.95 ´ 1019 tons
27.2% Si

100% crust

´

2000 lb
1 ton

´

1 kg
2.205 lb


= 7.3 ´ 1021 kg Si

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1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

1.88

21

10 cm = 0.1 m. We need to find the number of times the 0.1 m wire must be cut in half until the piece left is
equal to the diameter of a Cu atom, which is (2)(1.3 ´ 10- m). Let n be the number of times we can cut the
10

Cu wire in half. We can write:
æ 1 ö÷n
çç ÷ ´ 0.1 m = 2.6 ´ 1010 m
èç 2 ø÷÷
æ 1 ÷ön
çç ÷ = 2.6 ´ 10-9 m
çè 2 ÷÷ø

Taking the log of both sides of the equation:

æ1ö

nlog çç ÷÷÷ = log(2.6 ´ 10-9 )
çè 2 ø÷
n = 29 times

5000 mi

1.89

(250 ´ 106 cars) ´

1.90

Volume = area ´ thickness.

1 car

´

1 gal gas
20 mi

´

9.5 kg CO2
= 5.9 ´ 1011 kg CO 2
1 gal gas

From the density, we can calculate the volume of the Al foil.
Volume =


3.636 g
mass
=
= 1.3472 cm 3
density
2.699 g / cm 3
2

2

Convert the unit of area from ft to cm .
æ12 in ÷ö2
æ 2.54 cm ÷ö2
çç
ç
÷
÷÷ = 929.03 cm 2
1.000 ft ´ç
÷ ´ çç
çç 1 ft ÷÷
çç 1 in ÷÷
è
ø
è
ø
2

thickness =

1.91


1.3472 cm3
volume
=
= 1.450 × 10-3 cm = 1.450 ´ 10-2 mm
2
area
929.03 cm

(a)

homogeneous

(b)

heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous.

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22

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

1.92

First, let’s calculate the mass (in g) of water in the pool. We perform this conversion because we know there
is 1 g of chlorine needed per million grams of water.


(2.0 ´ 104 gollons H2O) ´ 1 gallon ´ 0.001 L ´ 1 mL = 7.58 ´ 107 g H2O
3.79 L

1 mL

1g

Next, let’s calculate the mass of chlorine that needs to be added to the pool.

(7.58 ´ 107 g H2 O) ´ 1 ´ 106 g H O
1 g chlorine

= 75.8 g chlorine

2

The chlorine solution is only 6 percent chlorine by mass. We can now calculate the volume of chlorine
solution that must be added to the pool.
75.8 g chlorine ´

1.93

100% soln
6% chlorine

´

1 mL soln
1 g soln


= 1.3 ´ 10 3 mL of chlorine solution

The volume of the cylinder is:
2

2

3

V = πr h = π(0.25 cm) (10 cm) = 2.0 cm

The number of Al atoms in the cylinder is:
2.0 cm 3 ´

2.70 g
1 cm

1.94

(a)

3

´

1 Al atom
-23

4.48 ´ 10


= 1.2 ´ 10 23 Al atoms
g

The volume of the pycnometer can be calculated by determining the mass of water that the pycnometer
holds and then using the density to convert to volume.
(43.1195 - 32.0764) g ´

(b)

0.99820 g

= 11.063 mL

Using the volume of the pycnometer from part (a), we can calculate the density of ethanol.
(40.8051 - 32.0764) g
11.063 mL

(c)

1 mL

= 0.78900 g/ mL

From the volume of water added and the volume of the pycnometer, we can calculate the volume of the
zinc granules by difference. Then, we can calculate the density of zinc.
volume of water = (62.7728 - 32.0764 - 22.8476) g ´

1 mL
0.99820 g


= 7.8630 mL

volume of zinc granules = 11.063 mL - 7.8630 mL = 3.200 mL
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1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

density of zinc =

1.95

22.8476 g
3.200 mL

23

= 7.140 g/mL

Let the fraction of gold = x, and the fraction of sand = (1 – x). We set up an equation to solve for x.
3

3

3

(x)(19.3 g/cm ) + (1 – x) (2.95 g/cm ) = 4.17 g/cm

19.3x – 2.95x + 2.95 = 4.17
x = 0.0746

Converting to a percentage, the mixture contains 7.46% gold.

1.96

First, convert 10 mm to units of cm.
10 mm ´

1 ´ 10-4 cm
1 mm

= 1.0 ´ 10-3 cm

Now, substitute into the given equation to solve for time.
t =

(1.0 ´ 10-3 cm)2
x2
=
= 0.88 s
2D
2(5.7 ´ 10-7 cm 2 /s)

It takes 0.88 seconds for a glucose molecule to diffuse 10 mm.

1.97

11


The mass of a human brain is about 1 kg (1000 g) and contains about 10 cells. The mass of a brain cell is:
1000 g
11

1 ´ 10 cells

= 1 ´ 10-8 g/cell

Assuming that each cell is completely filled with water (density = 1 g/mL), we can calculate the volume of
each cell. Then, assuming the cell to be cubic, we can solve for the length of one side of such a cell.
1 ´ 10-8 g
1 cell

Vcube = a
a = (V)

´

1 mL
1g

´

1 cm 3
1 mL

= 1 ´ 10-8 cm 3 /cell

3


1/3

= (1 × 10

-8

3 1/3

cm )

= 0.002 cm
11

Next, the height of a single cell is a, 0.002 cm. If 10 cells are spread out in a thin layer a single cell thick,
11

the surface area can be calculated from the volume of 10 cells and the height of a single cell.
V = surface area × height

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24

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE

11


The volume of 10 brain cells is:
1000 g ´

1 mL
1g

´

1 cm 3
1 mL

= 1000 cm 3

The surface area is:
æ1 ´ 10-2 m ÷ö2
1000 cm3
V
÷÷
5
2 çç
1 2
Surface area =
=
= 5 ´ 10 cm ´ çç
÷ = 5 ´ 10 m
height
0.002 cm
çç 1 cm ÷÷÷
è
ø


1.98

(a)

A concentration of CO of 800 ppm in air would mean that there are 800 parts by volume of CO per
1 million parts by volume of air. Using a volume unit of liters, 800 ppm CO means that there are 800 L
of CO per 1 million liters of air. The volume in liters occupied by CO in the room is:
æ 1 cm ÷ö3
1L
ç
÷÷
17.6 m ´ 8.80 m ´ 2.64 m = 409 m3 ´ ççç
= 4.09 ´ 105 L air
÷÷ ´
2
3
çç1 ´ 10 m ÷÷
1000
cm
è
ø
5

4.09 ´ 10 L air ´

(b)

3


1 m3

´

1 ´ 10-3 g
1 mg

-3

1 mg = 1 × 10
120 mg
1 dL

1.99

1 ´ 106 L air

= 327 L CO

1 mg = 1 × 10- g and 1 L = 1000 cm . We convert mg/m to g/L:
0.050 mg

(c)

8.00 ´ 102 L CO

´

3


3

æ1 ´ 10-2 m ÷ö3 1000 cm3
ç
÷÷
´ ççç
= 5.0 ´ 10-8 g/L
÷÷ ´
1
L
çç 1 cm
÷
÷ø
è
-2

mg and 1 mL = 1 × 10

1 mg
1 ´ 10-3 mg

´

1 ´ 10-2 dL
1 mL

dL. We convert mg/dL to mg/mL:

= 1.20 ´ 10 3 μg /mL


This problem is similar in concept to a limiting reagent problem. We need sets of coins with 3 quarters,
1 nickel, and 2 dimes. First, we need to find the total number of each type of coin.
Number of quarters = (33.871 ´ 103 g) ´
Number of nickels = (10.432 ´ 103 g) ´
Number of dimes = (7.990 ´ 103 g) ´

1 quarter
5.645 g
1 nickel
4.967 g

1 dime
2.316 g

= 6000 quarters

= 2100 nickels

= 3450 dimes

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.


1: CHEMISTRY: THE STUDY OF CHANGE
Solution Manual for Chemistry 13th Edition ByCHAPTER
Chang

25


Next, we need to find which coin limits the number of sets that can be assembled. For each set of coins, we
need 2 dimes for every 1 nickel.
2100 nickels ´

2 dimes
1 nickel

= 4200 dimes

We do not have enough dimes.
For each set of coins, we need 2 dimes for every 3 quarters.
6000 quarters ´

2 dimes
3 quarters

= 4000 dimes

Again, we do not have enough dimes, and therefore the number of dimes is our “limiting reagent”.
If we need 2 dimes per set, the number of sets that can be assembled is:
3450 dimes ´

1 set
2 dimes

= 1725 sets

The mass of each set is:
æ
5.645 g ÷ö æç

4.967 g ÷ö æç
2.316 g ÷ö
çç
÷÷ + ç1 nickel ´
÷÷ + ç2 dimes ´
÷÷ = 26.534 g/set
´
3
quarters
çç
ç
ç
1 quarter ÷÷ø ççè
1 nickel ÷÷ø ççè
1 dime ÷÷ø
çè

Finally, the total mass of 1725 sets of coins is:
1725 sets ´

1.100

26.534 g
1 set

= 4.577 ´ 104 g

We wish to calculate the density and radius of the ball bearing. For both calculations, we need the volume of
the ball bearing. The data from the first experiment can be used to calculate the density of the mineral oil. In
the second experiment, the density of the mineral oil can then be used to determine what part of the 40.00 mL

volume is due to the mineral oil and what part is due to the ball bearing. Once the volume of the ball bearing
is determined, we can calculate its density and radius.
From experiment one:
Mass of oil = 159.446 g - 124.966 g = 34.480 g

Density of oil =

34.480 g
40.00 mL

= 0.8620 g/mL

From the second experiment:
Mass of oil = 50.952 g - 18.713 g = 32.239 g

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any
manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.