2
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.21
density =
1.22
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
586 g
=
= 3.12 g/mL
volume
188 mL
density =
mass
volume
Solution:
mass = density × volume
mass of ethanol =
1.23
? °C = (°F − 32°F) ×
0.798 g
× 17.4 mL = 13.9 g
1 mL
5°C
9°F
5°C
= 35°C
9°F
5°C
= − 11°C
(12°F − 32°F) ×
9 °F
5°C
(102°F − 32°F) ×
= 39°C
9°F
5°C
= 1011°C
(1852°F − 32°F) ×
9°F
9°F ⎞
⎛
⎜ °C × 5°C ⎟ + 32°F
⎝
⎠
(a)
? °C = (95°F − 32°F) ×
(b)
? °C =
(c)
? °C =
(d)
? °C =
(e)
? °F =
9°F ⎞
⎛
? °F = ⎜ −273.15 °C ×
+ 32°F = − 459.67°F
°C ⎟⎠
5
⎝
1.24
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°F − 32°F) ×
(b)
5°C
= 41°C
9°F
Conversion from Celsius to Fahrenheit.
9°F ⎞
⎛
? °F = ⎜ °C ×
+ 32°F
5°C ⎟⎠
⎝
9°F ⎞
⎛
? °F = ⎜ −11.5 °C ×
+ 32°F = 11.3 °F
5
°C ⎟⎠
⎝
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(c)
Conversion from Celsius to Fahrenheit.
9°F ⎞
⎛
? °F = ⎜ °C ×
+ 32°F
5
°C ⎟⎠
⎝
9°F ⎞
⎛
? °F = ⎜ 6.3 × 103 °C ×
+ 32°F = 1.1 × 104 °F
5°C ⎟⎠
⎝
(d)
Conversion from Fahrenheit to Celsius.
? °C = (°F − 32°F) ×
5°C
9°F
? °C = (451°F − 32°F) ×
1.25
1.26
K = (°C + 273°C)
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)
2
2
1K
1°C
°C = K − 273 = 77 K − 273 = −196°C
K = (°C + 273°C)
(b)
°C = 4.2 K − 273 = −269°C
(c)
°C = 601 K − 273 = 328°C
1.29
(a)
2.7 × 10
1.30
(a)
10
−2
−8
3.56 × 10
(b)
2
(c)
10
−8
(a)
(b)
4
(d)
−2
= 0.0152
indicates that the decimal point must be moved 8 places to the left.
7.78 × 10
1.31
4.7764 × 10
indicates that the decimal point must be moved two places to the left.
1.52 × 10
(b)
5°C
= 233°C
9°F
−8
−1
= 0.0000000778
145.75 + (2.3 × 10 ) = 145.75 + 0.23 = 1.4598 × 10
79500
2.5 × 102
=
7.95 × 104
2.5 × 102
−3
2
= 3.2 × 102
−4
−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
4
6
10
−3
9.6 × 10
−2
3
4
1.32
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(a) Addition using scientific notation.
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.
3
n
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.
−3
9.5 × 10
−3
+ 8.5 × 10
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
n
10 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)
= 1.14 × 10
+2 + 8
= 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
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
5
Subtract the N parts of the numbers, keeping the exponent, n, the same.
5
8.5 × 10
5
− 9.0 × 10
−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
n
10 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
3
13 × 10 = 1.3 × 10
1.33
(a)
(e)
four
three
1.34
(a)
(e)
one
two or three
1.35
(a)
10.6 m
1.36
(a)
Division
(b)
(f)
two
one
(b)
(f)
(b)
three
one
0.79 g
(c)
(c)
(g)
five
one
(c)
(g)
three
one or two
(d)
(h)
two, three, or four
two
(d)
four
2
16.5 cm
Strategy: The number of significant figures in the answer is determined by the original number having the
smallest number of significant figures.
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
(Why are there no units?)
6
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(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
(c)
−3
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.
1.37
1.38
1 dm
= 226 dm
0.1 m
(a)
? dm = 22.6 m ×
(b)
? kg = 25.4 mg ×
(c)
? L = 556 mL ×
(d)
?
g
cm 3
=
10.6 kg
1 m3
0.001 g
1 kg
×
= 2.54 × 10−5 kg
1 mg
1000 g
1 × 10−3 L
= 0.556 L
1 mL
3
×
1000 g ⎛ 1 × 10−2 m ⎞
×⎜
⎟ = 0.0106 g/cm 3
⎜ 1 cm ⎟
1 kg
⎝
⎠
(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
−3
grams to milligrams (1 mg = 1 × 10 g). Arrange the appropriate conversion factors so that pounds and
grams cancel, and the unit milligrams is obtained in your answer.
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
Solution: The sequence of conversions is
days → hours → minutes → seconds
Using the following conversion factors,
24 h
1 day
60 min
1h
60 s
1 min
we can write
? s = 365.24 day ×
24 h 60 min
60 s
×
×
= 3.1557 × 107 s
1 day
1h
1 min
Check: Does your answer seem reasonable? Should there be a very large number of seconds in 1 year?
1.609 km 1000 m
1s
1 min
×
×
×
= 8.3 min
8
1 mi
1 km
60 s
3.00 × 10 m
1.41
(93 × 106 mi) ×
1.42
(a)
? in/s =
(b)
? m/min =
(c)
? km/h =
1.43
6.0 ft ×
168 lb ×
1 mi
5280 ft 12 in 1 min
×
×
×
= 81 in/s
13 min
1 mi
1 ft
60 s
1 mi
1609 m
×
= 1.2 × 102 m/min
13 min
1 mi
1 mi
1609 m
1 km
60 min
×
×
×
= 7.4 km/h
13 min
1 mi
1000 m
1h
1m
= 1.8 m
3.28 ft
453.6 g
1 kg
×
= 76.2 kg
1 lb
1000 g
55 mi 1.609 km
×
= 88 km/h
1h
1 mi
1.44
? km/h =
1.45
62 m
1 mi
3600 s
×
×
= 1.4 × 102 mph
1s
1609 m
1h
1.46
0.62 ppm Pb =
0.62 g Pb
1 × 106 g blood
0.62 g Pb
6.0 × 103 g of blood ×
= 3.7 × 10−3 g Pb
6
1 × 10 g blood
9
10
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.47
(a)
1.42 yr ×
365 day 24 h 3600 s 3.00 × 108 m
1 mi
×
×
×
×
= 8.35 × 1012 mi
1 yr
1 day
1h
1s
1609 m
(b)
32.4 yd ×
36 in 2.54 cm
×
= 2.96 × 103 cm
1 yd
1 in
(c)
3.0 × 1010 cm
1 in
1 ft
×
×
= 9.8 × 108 ft/s
1s
2.54 cm 12 in
(a)
? m = 185 nm ×
(b)
? s = (4.5 × 109 yr) ×
(c)
⎛ 0.01 m ⎞
−5 3
? m 3 = 71.2 cm3 × ⎜
⎟ = 7.12 × 10 m
1
cm
⎝
⎠
(d)
⎛ 1 cm ⎞
1L
? L = 88.6 m × ⎜
= 8.86 × 104 L
⎟ ×
⎜ 1 × 10−2 m ⎟ 1000 cm3
⎝
⎠
1.48
1 × 10−9 m
= 1.85 × 10−7 m
1 nm
365 day 24 h 3600 s
×
×
= 1.4 × 1017 s
1 yr
1 day
1h
3
3
3
2.70 g
3
density =
1.50
density =
1.51
Substance
(a) water
(b) carbon
(c) iron
(d) hydrogen gas
(e) sucrose
(f) table salt
(g) mercury
(h) gold
(i) air
1.52
See Section 1.6 of your text for a discussion of these terms.
1 cm3
×
⎛ 1 cm ⎞
1 kg
3
3
×⎜
⎟ = 2.70 × 10 kg/m
1000 g ⎝ 0.01 m ⎠
1.49
0.625 g
1L
1 mL
×
×
= 6.25 × 10−4 g/cm 3
1L
1000 mL 1 cm3
Qualitative Statement
colorless liquid
black solid (graphite)
rusts easily
colorless gas
tastes sweet
tastes salty
liquid at room temperature
a precious metal
a mixture of gases
Quantitative Statement
freezes at 0°C
3
density = 2.26 g/cm
3
density = 7.86 g/cm
melts at −255.3°C
at 0°C, 179 g of sucrose dissolves in 100 g of H2O
melts at 801°C
boils at 357°C
3
density = 19.3 g/cm
contains 20% oxygen by volume
(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.
(c)
Physical property. The color of the hemoglobin can be observed and measured without changing its
composition or identity.
16
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
(b)
1.76
4
62 kg = 6.2 × 10 g
O:
C:
H:
1.77
|0.864 g − 0.837 g|
× 100% = 3.1%
0.864 g
4
4
4
(6.2 × 10 g)(0.65) = 4.0 × 10 g O
4
4
(6.2 × 10 g)(0.18) = 1.1 × 10 g C
4
3
(6.2 × 10 g)(0.10) = 6.2 × 10 g H
3
N: (6.2 × 10 g)(0.03) = 2 × 10 g N
4
2
Ca: (6.2 × 10 g)(0.016) = 9.9 × 10 g Ca
4
2
P: (6.2 × 10 g)(0.012) = 7.4 × 10 g P
3 minutes 44.39 seconds = 224.39 seconds
Time to run 1500 meters is:
1500 m ×
1.78
1 mi
224.39 s
×
= 209.19 s = 3 min 29.19 s
1609 m
1 mi
2
2
? °C = (7.3 × 10 − 273) K = 4.6 × 10 °C
9°F ⎞
⎛
? °F = ⎜ (4.6 × 102 °C) ×
+ 32°F = 8.6 × 102 °F
5°C ⎟⎠
⎝
1.79
? g Cu = (5.11 × 103 kg ore) ×
1.80
(8.0 × 104 tons Au) ×
1.81
? g Au =
34.63% Cu 1000 g
×
= 1.77 × 106 g Cu
100% ore
1 kg
2000 lb Au 16 oz Au
$350
×
×
= $9.0 × 1011 or 900 billion dollars
1 ton Au
1 lb Au
1 oz Au
4.0 × 10−12 g Au
1 mL
×
× (1.5 × 1021 L seawater) = 6.0 × 1012 g Au
1 mL seawater
0.001 L
value of gold = (6.0 × 1012 g Au) ×
1 lb
16 oz $350
×
×
= $7.4 × 1013
453.6 g
1 lb
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
= 5.4 × 1022 Fe atoms
1.0 g Fe
1.82
? Fe atoms = 4.9 g Fe ×
1.83
mass of Earth's crust = (5.9 × 1021 tons) ×
0.50% crust
= 2.95 × 1019 tons
100% Earth
mass of silicon in crust = (2.95 × 1019 tons crust) ×
27.2% Si
2000 lb
1 kg
×
×
= 7.3 × 1021 kg Si
100% crust
1 ton
2.205 lb
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.84
17
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
−10
1.3 × 10 m long. Let n be the number of times we can cut the Cu wire in half. We can write:
n
⎛1⎞
−10
m
⎜ 2 ⎟ × 0.1 m = 1.3 × 10
⎝ ⎠
n
⎛1⎞
−9
⎜ 2 ⎟ = 1.3 × 10 m
⎝ ⎠
Taking the log of both sides of the equation:
⎛1⎞
n log ⎜ ⎟ = log(1.3 × 10−9 )
⎝2⎠
n = 30 times
5000 mi 1 gal gas 9.5 kg CO2
×
×
= 9.5 × 1010 kg CO 2
1 car
20 mi
1 gal gas
1.85
(40 × 106 cars) ×
1.86
Volume = area × thickness.
From the density, we can calculate the volume of the Al foil.
Volume =
mass
3.636 g
=
= 1.3472 cm3
density
2.699 g / cm3
2
2
Convert the unit of area from ft to cm .
2
2
⎛ 12 in ⎞ ⎛ 2.54 cm ⎞
2
1.000 ft × ⎜
⎟ ×⎜
⎟ = 929.03 cm
⎝ 1 ft ⎠ ⎝ 1 in ⎠
2
thickness =
volume
1.3472 cm3
=
= 1.450 × 10−3 cm = 1.450 × 10−2 mm
2
area
929.03 cm
1.87
(a)
(b)
1.88
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.
homogeneous
heterogeneous. The air will contain particulate matter, clouds, etc. This mixture is not homogeneous.
(2.0 × 104 gallons H 2 O) ×
3.79 L
1 mL
1g
×
×
= 7.58 × 107 g H 2 O
1 gallon 0.001 L 1 mL
Next, let’s calculate the mass of chlorine that needs to be added to the pool.
(7.58 × 107 g H 2 O) ×
1 g chlorine
1 × 106 g H 2 O
= 75.8 g chlorine
18
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
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 ×
100% soln
1 mL soln
×
= 1.3 × 103 mL of chlorine solution
6% chlorine
1 g soln
1 yr
1.89
(2.0 × 1022 J) ×
1.90
We assume that the thickness of the oil layer is equivalent to the length of one oil molecule. We can
calculate the thickness of the oil layer from the volume and surface area.
1.8 × 1020 J
= 1.1 × 102 yr
2
⎛ 1 cm ⎞
5
2
40 m 2 × ⎜
⎟ = 4.0 × 10 cm
0.01
m
⎝
⎠
3
0.10 mL = 0.10 cm
Volume = surface area × thickness
thickness =
volume
0.10 cm3
=
= 2.5 × 10−7 cm
5
2
surface area
4.0 × 10 cm
Converting to nm:
(2.5 × 10−7 cm) ×
1.91
0.01 m
1 nm
×
= 2.5 nm
1 cm
1 × 10−9 m
The mass of water used by 50,000 people in 1 year is:
50, 000 people ×
150 gal water
3.79 L 1000 mL 1.0 g H 2 O 365 days
×
×
×
×
= 1.04 × 1013 g H 2 O/yr
1 person each day
1 gal.
1L
1 mL H 2 O
1 yr
A concentration of 1 ppm of fluorine is needed. In other words, 1 g of fluorine is needed per million grams
of water. NaF is 45.0% fluorine by mass. The amount of NaF needed per year in kg is:
(1.04 × 1013 g H 2 O) ×
1g F
6
10 g H 2 O
×
100% NaF
1 kg
×
= 2.3 × 104 kg NaF
45% F
1000 g
An average person uses 150 gallons of water per day. This is equal to 569 L of water. If only 6 L of water is
used for drinking and cooking, 563 L of water is used for purposes in which NaF is not necessary. Therefore
the amount of NaF wasted is:
563 L
× 100% = 99%
569 L
1.92
3
3
(a)
⎛ 1 ft ⎞ ⎛ 1 in ⎞ 1 cm3
1 mL
×⎜
×
= $3.06 × 10−3 /L
⎟ ×⎜
⎟ ×
3
12
in
2.54
cm
1
mL
0.001
L
15.0 ft
⎝
⎠ ⎝
⎠
(b)
2.1 L water ×
$1.30
0.304 ft 3 gas
$1.30
×
= $0.055 = 5.5¢
1 L water
15.0 ft 3
CHAPTER 1: CHEMISTRY--THE STUDY OF CHANGE
1.96
21
We want to calculate the mass of the cylinder, which can be calculated from its volume and density. The
2
volume of a cylinder is πr l. The density of the alloy can be calculated using the mass percentages of each
element and the given densities of each element.
The volume of the cylinder is:
2
V = πr l
2
V = π(6.44 cm) (44.37 cm)
3
V = 5781 cm
The density of the cylinder is:
3
3
3
density = (0.7942)(8.94 g/cm ) + (0.2058)(7.31 g/cm ) = 8.605 g/cm
Now, we can calculate the mass of the cylinder.
mass = density × volume
3
3
4
mass = (8.605 g/cm )(5781 cm ) = 4.97 × 10 g
The assumption made in the calculation is that the alloy must be homogeneous in composition.
1.97
It would be more difficult to prove that the unknown substance is an element. Most compounds would
decompose on heating, making them easy to identify. For example, see Figure 4.13(a) of the text. On
heating, the compound HgO decomposes to elemental mercury (Hg) and oxygen gas (O2).
1.98
The density of the mixed solution should be based on the percentage of each liquid and its density. Because
the solid object is suspended in the mixed solution, it should have the same density as this solution. The
density of the mixed solution is:
(0.4137)(2.0514 g/mL) + (0.5863)(2.6678 g/mL) = 2.413 g/mL
As discussed, the density of the object should have the same density as the mixed solution (2.413 g/mL).
Yes, this procedure can be used in general to determine the densities of solids. This procedure is called the
flotation method. It is based on the assumptions that the liquids are totally miscible and that the volumes of
the liquids are additive.
1.99
Gently heat the liquid to see if any solid remains after the liquid evaporates. Also, collect the vapor and then
compare the densities of the condensed liquid with the original liquid. The composition of a mixed liquid
would change with evaporation along with its density.
1.100
When the carbon dioxide gas is released, the mass of the solution will decrease. If we know the starting mass
of the solution and the mass of solution after the reaction is complete (given in the problem), we can calculate
the mass of carbon dioxide produced. Then, using the density of carbon dioxide, we can calculate the
volume of carbon dioxide released.
1.140 g
Mass of hydrochloric acid = 40.00 mL ×
= 45.60 g
1 mL
Mass of solution before reaction = 45.60 g + 1.328 g = 46.928 g