Solution Manual For Chemistry, 13th Edition

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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 (b) law (c) theory
1.9 Li, lithium; F, fluorine; P, phosphorus; Cu, copper; As, arsenic; Zn, zinc; Cl, chlorine; Pt, platinum;
Mg, magnesium; U, uranium; Al, aluminum; Si, silicon; Ne, neon.
1.10 (a) Cs (b) Ge (c) Ga (d) Sr
(e) U (f) Se (g) Ne (h) Cd
1.11 (a) element (b) compound (c) element (d) compound

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE2
1.12 (a) homogeneous mixture (b) element (c) compound
(d) homogeneous mixture (e) heterogeneous mixture (f) heterogeneous mixture
(g) element
1.17 (a) Chemical property. Oxygen gas is consumed in a combustion reaction; its composition and identity are
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.
1.23 586 gmass
volume 188 mL
= = =density 3.12 g/mL
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.
= mass
density volume
Solution:
mass = density ´ volume
= ´ =
0.7918 g 89.9 mL
1 mL
mass of methanol 71.2 g

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 3
1.25 
 =  -  ´ 
5 C
? C ( F 32 F) 9 F
(a) 
 = -  ´ =

5 C
? C (95 32) F 9 F 35 C
(b) 
 = -  ´ = -

5 C
? C (12 32) F 9 F 11 C
(c) 
 = -  ´ =

5 C
? C (102 32) F 9 F 39 C
(d) 
 = -  ´ =

5 C
? C (1852 32) F 9 F 1011 C
(e) 9 F
? F C 32 F
5 C
æ ö ÷ç ÷ =  ´ + ç ÷ç ÷ç è ø
æ ö ÷ç ÷ = -  ´ +  = -ç ÷ç ÷ç è ø
9 F
? F 273.15 C 32 F
5 C 459.67 F
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.

 =  -  ´ 
5 C
? C ( F 32 F) 9 F

= -  ´ =

5 C
(105 32) F 9 F
? C 41 C 
(b) Conversion from Celsius to Fahrenheit.
æ ö ÷ç ÷ =  ´ + ç ÷ç ÷ç è ø
9 F
? F C 32 F
5 C
æ ö ÷ç ÷= -  ´ +  =ç ÷ç ÷ç è ø
9 F
11.5 C 32 F
5 C
? F 11.3 F 
(c) Conversion from Celsius to Fahrenheit.
æ ö ÷ç ÷ =  ´ + ç ÷ç ÷ç è ø
9 F
? F C 32 F
5 C
æ ö ÷ç ÷= ´  ´ +  =ç ÷ç ÷ç è ø
3 9 F
6.3 10 C 32 F
5 C
4
? F 1.1 10 F ´ 
(d) Conversion from Fahrenheit to Celsius.

 =  -  ´ 
5 C
? C ( F 32 F) 9 F

= -  ´ =

5 C
(451 32) F 9 F
? C 233 C 

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE4
1.27 =  +  
1 K
K ( C 273 C) 1 C
(a) K = 113C + 273C = 386 K
(b) K = 37C + 273C = 3.10 ´ 10 2 K
(c) K = 357C + 273C = 6.30 ´ 10 2 K
1.28 (a) =  +  
1 K
K ( C 273 C) 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 ´ 10-8 (b) 3.56 ´ 10 2 (c) 4.7764 ´ 10 4 (d) 9.6 ´ 10-2
1.32 (a) 10-2 indicates that the decimal point must be moved two places to the left.
1.52 ´ 10-2 = 0.0152
(b) 10-8 indicates that the decimal point must be moved 8 places to the left.
7.78 ´ 10-8 = 0.0000000778
1.33 (a) 145.75 + (2.3 ´ 10-1
) = 145.75 + 0.23 = 1.4598 ´ 10 2
(b)
4
2 2
7.95 1079500
2.5 10 2.5 10
´
= =
´ ´
2
3.2 × 10
(c) (7.0 ´ 10-3
) - (8.0 ´ 10-4
) = (7.0 ´ 10-3
) - (0.80 ´ 10-3
) = 6.2 ´ 10-3
(d) (1.0 ´ 10 4) ´ (9.9 ´ 10 6
) = 9.9 ´ 10 10
1.34 (a) Addition using scientific notation.
Strategy: Let’s express scientific notation as N ´ 10n. 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.

CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 5
Let’s write 0.0095 in such a way that n = -3. We have decreased 10n by 10 3, so we must increase N by 10 3
.
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
18.0 ´ 10-3
The usual practice is to express N as a number between 1 and 10. Since we must decrease N by a factor of 10
to express N between 1 and 10 (1.8), we must increase 10n by a factor of 10. The exponent, n, is increased by
1 from -3 to -2.
18.0 ´ 10-3 = 1.8 ´ 10-2
(b) Division using scientific notation.
Strategy: Let’s express scientific notation as N ´ 10n. 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+ 2 - (- 8) = 1.14 ´ 10+ 2 + 8 = 1.14 ´ 10 10
(c) Subtraction using scientific notation.
Strategy: Let’s express scientific notation as N ´ 10n. 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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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE6
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
to express N between 1 and 10 (5), we must decrease 10n by a factor of 10. The exponent, n, is decreased by
1 from 5 to 4.
-0.5 ´ 10 5 = -5 ´ 10 4
(d) Multiplication using scientific notation.
Strategy: Let’s express scientific notation as N ´ 10n. 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
to express N between 1 and 10 (1.3), we must increase 10n by a factor of 10. The exponent, n, is increased by
1 from 2 to 3.
13 ´ 10 2 = 1.3 ´ 10 3
1.35 (a) four (b) two (c) five (d) two, three, or four
(e) three (f) one (g) one (h) two
1.36 (a) one (b) three (c) three (d) four
(e) two or three (f) one (g) one or two
1.37 (a) 10.6 m (b) 0.79 g (c) 16.5 cm
2 (d) 1 × 10 6 g/cm
3
1.38 (a) Division
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
5.70 km = 1.283
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 (Why are there no units?)
(b) 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:
0.00318 mg = 3.18 ´ 10-3 mg
(c) 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
(0.402 ´ 10 7 dm) + (7.74 ´ 10 7 dm) = 8.14 ´ 10 7 dm
Since 7.74 ´ 10 7 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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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE8
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) 7.5 m
(1.15 s 0.82 s) 1.97 s
- = =
+ 3.8 m/s
1.39 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 (a) = ´ =
1 dm
22.6 m 0.1 m
? dm 226 dm
(b) 0.001g 1kg
25.4 mg 1mg 1000 g
= ´ ´ = 5
? kg 2.54 10 kg
-
´
(c)
3
1 10 L
556 mL 1 mL
-
´
= ´ =? L 0.556 L

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 9
(d)
3
2
3
10.6 kg 1000 g 1 10 m
1 kg 1 cm1 m
-æ ö÷´ç ÷ç ÷ç= ´ ´ =÷ç ÷ç ÷÷çè ø
3
3
g
? 0.0106 g cm
cm
/
1.42 (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 lb 3
1 mg
1 10 g-
´
we obtain the answer in one step:
-
= ´ ´ =
´ 3
1 mg453.6 g
242 lb 1 lb 1 10 g
8
? mg 1.10 10 mg´
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
? m
3 = 68.3 cm
3
Recall that 1 cm = 1 ´ 10-2 m. We need to set up a conversion factor to convert from cm 3 to m
3.
Solution: We need the following conversion factor so that centimeters cancel and we end up with meters.
-
´ 2
1 10 m
1 cm

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE10
Since this conversion factor deals with length and we want volume, it must therefore be cubed to give
- - - -æ ö÷´ ´ ´ ´ç ÷ç ÷ç´ ´ = ÷ç ÷ç ÷÷çè ø
3
2 2 2 2
1 10 m 1 10 m 1 10 m 1 10 m
1 cm 1 cm 1 cm 1 cm
We can write
-æ ö÷´ç ÷ç ÷ç= ´ =÷ç ÷ç ÷÷çè ø
3
2
3 1 10 m
68.3 cm 1 cm
3 5 3
? m 6.83 10 m-
´
Check: We know that 1 cm
3 = 1 ´ 10-6 m
3. We started with 6.83 ´ 10 1 cm
3. Multiplying this quantity by
1 ´ 10-6 gives 6.83 ´ 10-5
.
(c)
Strategy: The problem may be stated as
? L = 7.2 m
3
In Chapter 1 of the text, a conversion is given between liters and cm
3 (1 L = 1000 cm
3
). If we can convert m
3
to cm
3, we can then convert to liters. Recall that 1 cm = 1 ´ 10-2 m. We need to set up two conversion
factors to convert from m
3 to L. Arrange the appropriate conversion factors so that m
3 and cm
3 cancel, and
the unit liters is obtained in your answer.
Solution: The sequence of conversions is
m
3  cm
3  L
Using the following conversion factors,
-
æ ö÷ç ÷ç ÷ç ÷ç ÷ç ÷´ ÷çè ø
3
2
1 cm
1 10 m 3
1 L
1000 cm
the answer is obtained in one step:
-
æ ö÷ç ÷ç ÷ç= ´ ´ =÷ç ÷ç ÷´ ÷çè ø
3
3
2 3
1 cm 1 L
7.2 m
1 10 m 1000 cm
3
? L 7.2 10 L´
Check: From the above conversion factors you can show that 1 m
3 = 1 ´ 10 3 L. Therefore, 7 m
3 would
equal 7 ´ 10 3 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
convert from grams to pounds. Recall that 1 mg = 1 ´ 10-6 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,
-
´
m
6
1 10 g
1 g
1 lb
453.6 g
we can write
-
´
= m ´ ´ =
m
6
1 10 g 1 lb
28.3 g 1 g 453.6 g
8
? lb 6.24 10 lb-
´
Check: Does the answer seem reasonable? What number does the prefix m represent? Should 28.3 mg be a
very small mass?
1.43 1255 m 1 mi 3600 s
1 s 1609 m 1 h
´ ´ = 2808 mi h/
1.44 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
1 day
60 min
1 h
60 s
1 min

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE12
we can write
24 h 60 min 60 s
365.24 days 1 day 1 h 1 min
= ´ ´ ´ = ´ 7
? s 3.1557 10 s
Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year?
1.45 ´ ´ ´ ´ ´ =
´
6
8
1.609 km 1000 m 1 s 1 min
(93 10 mi) 1 mi 1 km 60 s3.00 10 m
8.3 min
1.46 (a) ? in/s = ´ ´ ´
1 mi 5280 ft 12 in 1 min
8.92 min 1 mi 1 ft 60 s = 118 in/s
(b) ? m/min = ´
1 mi 1609 m
8.92 min 1 mi = 1.80 ´ 2
10 m/min
(c) ? km/h = ´ ´ ´
1 mi 1609 m 1 km 60 min
8.92 min 1 mi 1000 m 1 h = 10.8 km/h
1.47 ´ =
1 m
6.0 ft 3.28 ft 1.8 m
´ ´ =
453.6 g 1 kg
168 lb 1 lb 1000 g 76.2 kg
1.48 ? mph = ´
286 km 1 mi
1 h 1.609 km = 178 mph
1.49 ´ ´ =
62 m 1 mi 3600 s
1 s 1609 m 1 h
2
1.4 10 mph´
1.50 =
´ 6
0.62 g Pb
0.62 ppm Pb
1 10 g blood
´ ´ = ´
´
3
6
0.62 g Pb
6.0 10 g of blood
1 10 g blood
3
3.7 10 g Pb-
1.51 (a)
8
365 days 24 h 3600 s 3.00 10 m 1 mi
1.42 yr 1 yr 1 day 1 h 1 s 1609 m
´
´ ´ ´ ´ ´ = 12
8.35 10 mi´
(b) ´ ´
36 in 2.54 cm
32.4 yd 1 yd 1 in
3
2.96 10 cm= ´

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 13
(c)
10
3.0 10 cm 1 in 1 ft
1 s 2.54 cm 12 in
´ ´ ´ = 8
9.8 10 ft /s´
1.52 (a) = ´ =
1 lb
70 kg 0.4536 kg
2
? lbs 1.5 10 lbs´
(b) = ´ ´ ´ ´ =9 365 d 24 h 3600 s
14 10 yr 1 yr 1 d 1 h
17
? s 4.4 10 s´
(c)
-
´
= ´ ´
22.54 cm 1 10 m
90 in 1 in 1 cm
? m 2.3 m=
(d) -
æ ö÷ç ÷= ´ ´ç ÷ç ÷çè ø´
3
3
2 3
1 cm 1 L
88.6 m
1 10 m 1000 cm
4
? L 8.86 10 L= ´
1.53
3
3
2.70 g 1 kg 1 cm
1000 g 0.01 m1 cm
æ ö÷ç ÷ç= ´ ´ =÷ç ÷ç ÷÷çè ø
3 3
density 2.70 10 kg m´ /
1.54 3
0.625 g 1 L 1 mL
1 L 1000 mL 1 cm
-
= ´ ´ = 4 3
density 6.25 10 g cm´ /
1.55 Substance Qualitative Statement Quantitative Statement
(a) water colorless liquid freezes at 0C
(b) carbon black solid (graphite) density = 2.26 g/cm
3
(c) iron rusts easily density = 7.86 g/cm
3
(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 H 2O
(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
3
(i) air a mixture of gases contains 20% oxygen by volume
1.56 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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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE14
(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.
1.57 ´ ´ =
´
9
3
1 ton
(95.0 10 lb of sulfuric acid)
2.0 10 lb
7
4.75 10 tons of sulfuric acid´
1.58 Volume of rectangular bar = length ´ width ´ height
52.7064 g
= (8.53 cm)(2.4 cm)(1.0 cm)
= = 3
density 2.6 g cm
m
V /
1.59 mass = density ´ volume
(a) mass = (19.3 g/cm
3
) ´ [ 4
3 p(10.0 cm) 3] = 8.08 ´ 10
4 g
(b) -
æ ö÷ç ÷ç= ´ ´ =÷ç ÷ç ÷÷çè ø
3
3 1 cm
(21.4 g/cm ) 0.040 mm 10 mm
6
mass 1.4 10 g´
(c) mass = (0.798 g/mL) (50.0 mL) = 39.9 g
1.60 You are asked to solve for the inner diameter of the bottle. If we can calculate the volume that the cooking
oil occupies, we can calculate the radius of the cylinder. The volume of the cylinder is, Vcylinder = pr2h (r is
the inner radius of the cylinder, and h is the height of the cylinder). The cylinder diameter is 2r.
= mass of oil
volume of oil filling bottle density of oil
= = ´ = ´3 3 31360 g
volume of oil filling bottle 1.43 10 mL 1.43 10 cm
0.953 g/mL
Next, solve for the radius of the cylinder.
Volume of cylinder = pr2h
volume
= p ´
r h
´
= =
p ´
3 3
1.43 10 cm 4.60 cm
21.5 cm
r
The inner diameter of the bottle equals 2r.
Bottle diameter = 2r = 2(4.60 cm) = 9.20 cm

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CHAPTER 1: CHEMISTRY: THE STUDY OF CHANGE 15
1.61 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.
= mass
volume density
Mass of water = 87.39 g - 56.12 g = 31.27 g
= =
3
31.27 gmass =
density 0.9976 g/cm
3
Volume of the flask 31.35 cm
1.62 ´ ´ =
343 m 1 mi 3600 s
1 s 1609 m 1 h 767 mph
1.63 The volume of silver is equal to the volume of water it displaces.
Volume of silver = 260.5 mL - 242.0 mL = 18.5 mL = 18.5 cm
3
3
194.3 g
18.5 cm
= = 3
density 10.5 g cm/
1.64 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:
æ ö÷ç ÷ç= p = p = ´÷ç ÷ç ÷çè ø
3
3 4 3
48.6 cm4 4 6.01 10 cm
3 3 2
V r
´
= = =
´
5
4 3
6.852 10 gmass
volume 6.01 10 cm
density 11.4 g/cm
3
1.66 = mass
Volume density
´
= =
3
3
1.20 10 g
0.53 g / cm
3 3
Volume occupied by Li 2.3 10 cm´

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Solution Manual For Chemistry, 13th Edition

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