Chapter 1–1
Chapter 1 Matter and Measurement
Solutions to In-Chapter Problems
1.1
Natural materials are isolated from natural sources, whereas synthetic materials are produced by
chemists in the laboratory. Natural: water, rock, tree. Synthetic: tennis ball, CD, plastic bag.
1.2
Naturally occurring: ice, blood. Synthetic: gloves, mask, plastic syringe, stainless steel needle.
1.3
Physical properties can be observed or measured without changing the composition of the
material (a and d). Chemical properties determine how a substance can be converted to another
substance by chemical reactions (b, c, and e).
1.4
This represents a chemical change because the “particles” on the left are different from the
particles on the right. For example, on the left side there are particles containing only two red
balls, while on the right there are none of these.
1.5
Representation (a) is a pure substance since each particle contains one red and two gray spheres.
Representation (b) is a mixture since some of the particles are only red, and some are red and
black.
1.6
A pure substance is composed of a single component and has a constant composition regardless
of the sample size (d). A mixture is composed of more than one component. The composition of
a mixture can vary depending on the sample (a, b, and c).
1.7
An element is a pure substance that cannot be broken down into simpler substances by a chemical
reaction (a). A compound is a pure substance formed by combining two or more elements
together (b, c, and d).
1.8
Use Table 1.2 to determine the prefix for each unit.
a. a million liters = megaliter
c. a hundredth of a gram = centigram
b. a thousandth of a second = millisecond
d. a tenth of a liter = deciliter
1.9
One nanometer = 0.000 000 001 m (one billionth of a meter); therefore, 1 m = 1,000,000,000 nm.
1.10
One microgram = 0.000 001 g (one millionth of a gram); therefore, 1 g = 1,000,000 µg.
1.11
Use Table 1.2 to determine which quantity is larger.
a. 3 cL
c. 5 km
d. 2 mL
b. 1 µg
1.12
All nonzero digits are significant. A zero is significant only if it occurs between two nonzero
digits, or at the end of a number with a decimal point. The significant figures are in bold.
a. 23.45
4 significant figures
c. 230
2 significant figures
e. 0.202
3 significant figures
g. 1,245,006
7 significant figures
b. 23.057
5 significant figures
d. 231.0
4 significant figures
f. 0.003 60
3 significant figures
h. 1,200,000
2 significant figures
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Matter and Measurement 1–2
1.13
All nonzero digits are significant. A zero is significant only if it occurs between two nonzero
digits, or at the end of a number with a decimal point. The significant figures are in bold.
a. 10,040
4 significant figures
c. 1,004.00
6 significant figures
e. 1.0040
5 significant figures
g. 0.001 004
4 significant figures
b. 10,040.
5 significant figures
d. 1.004
4 significant figures
f. 0.1004
4 significant figures
h. 0.010 040 0
6 significant figures
1.14
A zero is significant only if it occurs between two nonzero digits, or at the end of a number with a
decimal point.
a. 0.003 04
No Yes
1.15
b. 26,045
Yes
c. 1,000,034
Yes
d. 0.304 00
No Yes
When the number to be rounded off is 4 or less, it and all other digits to the right are dropped.
When the number is 5 or greater, 1 is added to the digit to its left.
a. 1.2735
Since this number is 7 (5 or greater),
round the 2 to its left up by one.
1.3
c. 3,836.9
Since this number is 3 (4 or less), drop
it and all numbers to its right.
3,800
b. 0.002 536 22
Since this number is 3 (4 or less), drop
it and all numbers to its right.
0.0025
1.16
The answers must have the same number of significant figures as the original number with the
fewest number of significant figures.
a. 10.70 × 3.5 = 37.45
Since 3.5 has only 2 significant figures, round
the answer to give it two significant figures.
37
b. 0.206 ÷ 25,993 = 0.000 007 93
Since 0.206 has 3 significant figures, the answer
has the appropriate number of significant figures.
1.17
c. 1,300 ÷ 41.2 = 31.553 398
Since 1,300 has only 2 significant figures, round
the answer to give it 2 significant figures.
32
d. 120.5 × 26 = 3,133
Since 26 has only 2 significant figures, round
the answer to give it 2 significant figures.
3,100
The answers must have the same number of decimal places as the original number with the fewest
decimal places.
a. 27.8 cm + 0.246 cm = 28.046 cm
Since 27.8 has one digit after the decimal point,
round the answer to one digit after the decimal
point.
28.0 cm
b. 102.66 mL + 0.857 mL + 24.0 mL =
127.517 mL
Since 24.0 has one digit after the decimal point,
round the answer to one digit after the decimal
point.
127.5 mL
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Chapter 1–3
c. 54.6 mg – 25 mg = 29.6 mg
Since 25 has zero digits after the decimal point,
round the answer to the nearest whole number.
30. mg
1.18
d. 2.35 s – 0.266 s = 2.084 s
Since 2.35 has 2 digits after the decimal point,
round the answer to 2 digits after the decimal
point.
2.08 s
To write a number in scientific notation:
[1] Move the decimal point to give a number between 1 and 10.
[2] Multiply the result by 10x, where x is the number of places the decimal point was moved.
0.000 098 g/dL
= 9.8 x 10–5 g/dL
the number of places the decimal
point was moved to the right
Move the decimal point five places to the right.
1.19
1.20
To write a number in scientific notation:
[1] Move the decimal point to give a number between 1 and 10.
[2] Multiply the result by 10x, where x is the number of places the decimal point was moved.
a. 93,200 = 9.32 × 104
The decimal point was moved 4
places to the left.
c. 6,780,000 = 6.78 × 106
The decimal point was moved 6
places to the left.
b. 0.000 725 = 7.25 × 10–4
The decimal point was moved 4
places to the right.
d. 0.000 030 = 3.0 × 10–5
The decimal point was moved 5
places to the right.
e. 4,520,000,000,000 =
4.52 × 1012
The decimal point was moved
12 places to the left.
f. 0.000 000 000 028 =
2.8 × 10–11
The decimal point was moved
11 places to the right.
The exponent in 10x tells how many places to move the decimal point in the coefficient to
generate a standard number. The decimal point goes to the right when x is positive and to the left
when x is negative.
Answer:
6.02 x 1021
6.020000000000000000000
6,020,000,000,000,000,000,000
Move the decimal point to the right 21 places.
1.21
The exponent in 10x tells how many places to move the decimal point in the coefficient to
generate a standard number. The decimal point goes to the right when x is positive and to the left
when x is negative.
a. 6.5 × 103 = 6,500
The decimal point was moved 3
places to the right.
b. 3.26 × 10–5 = 0.000 032 6
The decimal point was moved 5
places to the left.
c. 3.780 × 10–2 = 0.037 80
The decimal point was moved 2
places to the left.
d. 1.04 × 108 = 104,000,000
The decimal point was moved 8
places to the right.
e. 2.221 × 106 = 2,221,000
The decimal point was moved
6 placed to the right.
f. 4.5 × 10–10 =
0.000 000 000 45
The decimal point was moved
10 places to the left.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Matter and Measurement 1–4
1.22
Use the equalities in Tables 1.3 and 1.4 to write a fraction that shows the relationship between the
two units.
0.621 mi
1 km
1 km
0.621 mi
c. 454 g
1 lb
1 lb
454 g
b. 1000 mm
1m
1m
1000 mm
d. 1000 µg
1 mg
1 mg
1000 µg
a.
1.23
To convert 4,120 km to miles:
[1] Identify the original quantity and the desired quantity, including units.
[2] Write out the conversion factor(s) needed to solve the problem.
[3] Set up and solve the problem.
[4] Write the answer using the correct number of significant figures and check by estimation.
[1]
4,120 km
? mi
original quantity
[2]
desired quantity
1 km
Two possible
conversion factors:
or
0.621 mi
0.621 mi
1 km
Choose this factor to cancel
the unwanted unit, km.
conversion factor
[3]
x
4,120 km
0.621 mi
=
1 km
original quantity
2,558.52 mi
desired quantity
The number of km (unwanted unit) cancels.
[4] The initial number has three significant figures, so the final answer is rounded to 2,560 mi.
1.24
Use conversion factors to solve the problems.
a.
b.
c.
d.
1.25
10 dL
1L
28.3 g
40.0 oz x
1 oz
2.54 cm
32 in.
x
1 in.
10 mm
10 cm x
1 cm
25 L
x
=
250 dL
=
1,132 g = 1,130 g rounded to 3 significant figures
=
81.28 cm = 81 cm rounded to 2 significant figures
=
100 mm
Use conversion factors to solve the problems.
Kilometers do not cancel.
a. 6,250 ft
1 mi
x
5,280 ft
Feet cancel.
x
1 km
0.621 mi
=
1.91 km
Miles cancel.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 1–5
Liters do not cancel.
b. 3 cups
1 qt
x
2.54 cm
x
1 ft
=
140 cm
1 in.
Feet cancel.
1.26
0.7 L
Quarts cancel.
Centimeters do not cancel.
12 in.
x
=
1.06 qt
Cups cancel.
c. 4.5 ft
1L
x
4 cups
Inches cancel.
Convert mL to tsp.
1 tsp
7.5 mL x
=
5.0 mL
1.5 tsp
1.27
0.100 mg
x
1000 µg
1 tablet
x
1 mg
=
4 tablets
25 µg
1.28
1.29
5 mL
100 mg
x
160 mg
=
8 mL Children's Motrin
Convert from °C to °F and K using the formulas listed in Section 1.9.
°F =
=
1.8(oC) + 32
K =
1.8(28.5) + 32 = 83.3 °F
=
oC
+ 273
28.5 + 273
= 302 K
1.30
a. °F =
1.8(oC) + 32
=
1.8(20.) + 32
b. oC
=
o
F –
c. °C =
=
= 68 °F
1.8
=
1.31
=
1.8
= 25 °C
1.8(oC) + 32
=
1.8(25) + 32
=
150. – 32
298 – 273
°F =
d. K =
32
K – 273
= 77 °F
°C + 273
75 + 273
= 348 K
66 oC
To convert volume (mL) to mass (g), multiply the volume by the density (g/mL).
10.0 mL
x
0.713 g
mL
=
7.13 g
Milliliters cancel.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Matter and Measurement 1–6
1.32
a. 100. g
x
1 mL
0.92 g
=
110 mL
1L
b. 110 mL x
1000 mL
= 0.11 L
Grams cancel.
1.33
a. Given equal volumes of A and B, the mass of A will be larger.
b. Given equal masses of A and B, B will have the larger volume.
1.34
a.
b.
specific gravity
2.3 =
=
density of a substance (g/mL)
density of water (g/mL)
density of a substance (g/mL)
=
0.80 g/mL
1 g/mL
=
0.80
density = 2.3 g/mL
1 g/mL
Solutions to End-of-Chapter Problems
1.35
Representation (a) is a pure element since each particle contains only gray spheres.
Representation (b) is a pure compound since each particle contains gray and black spheres.
Representation (c) is a mixture, because some of the particles are only gray, and some are gray
and black. Representation (d) is a mixture, because some of the particles are gray, and some are
blue.
1.36
Representation (d) illustrates a mixture of two elements. Representation (c) illustrates a mixture
of compound and an element.
1.37
An element is a pure substance that cannot be broken down into simpler substances by a chemical
reaction. A compound is a pure substance formed by combining two or more elements.
1.38
A compound is a pure substance formed by chemically combining two or more elements in a
specific ratio by mass. The composition of a compound does not vary from sample to sample. A
mixture is formed by physically combining two or more substances. The composition of a
mixture can vary depending on the sample.
1.39
Phase
Solid
Liquid
Gas
a. Volume
Definite
Definite
Not fixed
b. Shape
Definite
Assumes shape of container
None
c. Organization
Very organized
Less organized
Disorganized
d. Particle Proximity
Very close
Close
Far apart
1.40
Physical properties are those that can be observed or measured without changing the composition
of the material. Chemical properties are those that determine how a substance can be converted to
another substance.
1.41
A chemical change converts one substance to another substance by a chemical reaction (b). A
physical change can be observed or measured without changing the composition of the material
(a and c).
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Chapter 1–7
1.42
A chemical change converts one substance to another by a chemical reaction (c). A physical
change can be observed or measured without changing the composition of the material (a and b).
1.43
This is a physical change since the compound CO2 is unchanged in this transition. The same
“particles” exist at the beginning and end of the process.
1.44
This is a chemical change since the nitrogen molecules (blue spheres) and hydrogen molecules
(gray spheres) have been converted to ammonia molecules (one blue sphere and three gray
spheres).
1.45
a and b. The temperature on the Fahrenheit thermometer is 76.5 °F, which has three significant
figures.
c. oC
=
oF
–
32
=
76.5 – 32
1.8
1.8
1.46
= 24.7 oC
a and b. The length of the crayon is 4.5 cm, which has two significant figures.
c.
4.5 cm
x
1m
100 cm
=
4.5 x 10–2 m
1.47
An exact number results from counting objects or is part of a definition, such as having 20 people
in a class. An inexact number results from a measurement or observation and contains some
uncertainty, such as the distance from the earth to the sun, 9.3 × 107 miles.
1.48
a. 10 (cloves of garlic) is an exact number and two (tablespoons of oil) is an inexact number.
b. five (puppies) is an exact number and 10 (lb) is an inexact number.
c. four (bicycles) is an exact number and 250 (mi) is an inexact number.
d. 4 (cm) is an inexact number and 12 (stitches) is an exact number.
1.49
Compare the measurements using Table 1.2. (< means less than; > means greater than.)
a. 5 mL < 5 dL
1.50
c. 5 cm > 5 mm
d. 10 Ms > 10 ms
Compare the measurements using Table 1.2. (< means less than; > means greater than.)
a. 10 km > 10 m
1.51
b. 10 mg > 10 µg
b. 10 L > 10 mL
c. 10 g > 10 µg
d. 10 cm > 10 mm
All nonzero digits are significant. A zero is significant only if it occurs between two nonzero
digits, or at the end of a number with a decimal point. The significant figures are in bold.
a. 16.00
4 significant figures
d. 1,600,000
2 significant figures
g. 1.060 × 1010
4 significant figures
b. 160
2 significant figures
c. 0.001 60
3 significant figures
e. 1.06
3 significant figures
f. 0.1600
4 significant figures
h. 1.6 × 10–6
2 significant figures
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Matter and Measurement 1–8
1.52
1.53
All nonzero digits are significant. A zero is significant only if it occurs between two nonzero
digits, or at the end of a number with a decimal point. The significant figures are in bold.
a. 160.
3 significant figures
d. 1.60
3 significant figures
g. 1.600 × 10–10
4 significant figures
b. 160.0
4 significant figures
c. 0.000 16
2 significant figures
e. 1,600.
4 significant figures
f. 1.060
4 significant figures
h. 1.6 × 106
2 significant figures
When the number to be rounded off is 4 or less, it and all other digits to the right are dropped.
When the number is 5 or greater, 1 is added to the digit to its left.
a. 25,401 = 25,400
b. 1,248,486 = 1,250,000
1.54
e. 195.371 = 195
f. 196.814 = 197
When the number to be rounded off is 4 or less, it and all other digits to the right are dropped.
When the number is 5 or greater, 1 is added to the digit to its left.
a. 25,401 = 2.540 × 104
b. 1,248,486 = 1,248,000
1.55
c. 0.001 265 982 = 0.001 27
d. 0.123 456 = 0.123
c. 0.001 265 982 = 0.001 266
d. 0.123 456 = 0.123 4
e. 195.371 = 195.4
f. 196.814 = 196.8
The answers in problems with multiplication and division must have the same number of decimal
places as the original number with the fewest decimal places. The answers in problems with
addition and subtraction must have the same number of digits after the decimal point as the
original number with the fewest digits after the decimal point.
a. 53.6 × 0.41 = 21.976
Since 0.41 has two significant
figures, round the answer to
two significant figures.
22
c. 65.2 ÷ 12 = 5.43333
Since 12 has two significant
figures, round the answer to
two significant figures.
5.4
e. 694.2 × 0.2 = 138.84
Since 0.2 has one significant
figure, round the answer to one
significant figure.
100
b. 25.825 – 3.86 = 21.965
Since 3.86 has two digits after
the decimal point, round the
answer to two digits after the
decimal point.
21.97
d. 41.0 + 9.135 = 50.135
Since 41.0 has one digit after
the decimal point, round the
answer to one digit after the
decimal point.
50.1
f. 1,045 – 1.26 = 1,043.74
Since 1,045 has no digits after
the decimal point, round the
answer to the closest whole
number.
1,044
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 1–9
1.56
1.57
The answers in problems with multiplication and division must have the same number of decimal
places as the original number with the fewest decimal places. The answers in problems with
addition and subtraction must have the same number of digits after the decimal point as the
original number with the fewest digits after the decimal point.
a. 49,682 × 0.80 = 39,745.6
Since 0.80 has two significant
figures, round the answer to
two significant figures.
40,000
c. 1,000 ÷ 2.34 = 427.3504
Since 1,000 has one significant
figure, round the answer to one
significant figure.
400
e. 25,000 ÷ 0.4356 = 57,392.10
Since 25,000 has two
significant figure, round the
answer to two significant
figures.
57,000
b. 66.815 + 2.82 = 69.6350
Since 2.82 has two digits after
the decimal point, round the
answer to two digits after the
decimal point.
69.64
d. 21 – 0.88 = 20.1200
Since 21 has no digits after the
decimal point, round the answer
to the closest whole number.
20.
f. 21.5381 + 26.55 = 48.0881
Since 26.55 has two digits after
the decimal point, round the
answer to two digits after the
decimal point.
48.09
To write a number in scientific notation:
[1] Move the decimal point to give a number between 1 and 10.
[2] Multiply the result by 10x, where x is the number of places the decimal point was moved.
a. 1,234 g = 1.234 × 103 g
The decimal point was moved 3 places to the
left.
b. 0.000 016 2 m = 1.62 × 10–5 m
The decimal point was moved 5 places to the
right.
c. 5,244,000 L = 5.244 × 106 L
The decimal point was moved 6 places to the
left.
1.58
d. 0.005 62 g = 5.62 × 10–3 g
The decimal point was moved 3 places to the
right.
e. 44,000 km = 4.4 × 104 km
The decimal point was moved 4 places to the
left.
To write a number in scientific notation:
[1] Move the decimal point to give a number between 1 and 10.
[2] Multiply the result by 10x, where x is the number of places the decimal point was moved.
a. 0.001 25 m = 1.25 × 10–3 m
The decimal point was moved 3 places to the
right.
b. 8,100,000,000 lb = 8.1 × 109 lb
The decimal point was moved 9 places to the
left.
c. 54,235.6 m = 5.42356 × 104 m
The decimal point was moved 4 places to the
left.
d. 0.000 001 899 L = 1.899 × 10–6 L
The decimal point was moved 6 places to the
right.
e. 4,440 s = 4.440 × 103 s
The decimal point was moved 3 places to the
left.
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Matter and Measurement 1–10
1.59
The exponent in 10x tells how many places to move the decimal point in the coefficient to
generate a standard number. The decimal point goes to the right when x is positive and to the left
when x is negative.
a. 3.4 × 108 = 340,000,000
The decimal point was moved 8 places to the
right.
b. 5.822 × 10–5 = 0.000 058 22
The decimal point was moved 5 places to the
left.
1.60
The exponent in 10x tells how many places to move the decimal point in the coefficient to
generate a standard number. The decimal point goes to the right when x is positive and to the left
when x is negative.
a. 4.02 × 1010 = 40,200,000,000
The decimal point was moved 10 places to the
right.
b. 2.46 × 10–3 = 0.002 46
The decimal point was moved 3 places to the
left.
1.61
c. 6.86 × 109 = 6,860,000,000
The decimal point was moved 9 places to the
right.
d. 1.00 × 10–7 = 0.000 000 100
The decimal point was moved 7 places to the
left.
Compare the two numbers. The number in bold is larger.
a. 4.44 × 103 or 4.8 × 102
b. 5.6 × 10–6 or 5.6 × 10–5
1.62
c. 3 × 102 = 300
The decimal point was moved 2 places to the
right.
d. 6.86 × 10–8 = 0.000 000 068 6
The decimal point was moved 8 places to the
left.
c. 1.3 × 108 or 52,300,000
d. 9.8 × 10–4 or 0.000 089
Compare the three numbers and order from smallest to largest.
a. 7 × 104 < 5.06 × 106 < 2.5 × 108
b. 8.6 × 10–6 < 2.5 × 10–4 < 6.3 × 10–2
1.63
Write the number in scientific notation.
a. 0.000 400 g of folate = 4.00 × 10–4 g
The decimal point was moved 4 places to the
right.
b. 0.002 g of copper = 2 × 10–3 g
The decimal point was moved 3 places to the
right.
1.64
c. 0.000 080 g of vitamin K = 8.0 × 10–5 g
The decimal point was moved 5 places to the
right.
d. 3,400 mg of chloride = 3.4 × 103 mg
The decimal point was moved 3 places to the
left.
Use conversion factors to solve the problem and write the answer in scientific notation.
a. 0.40 µm x
1m
1x
106
µm
= 4.0 x 10–7 m
b. 4.0 x 10–7 m x
100 cm
1m
x
1 in
2.54 cm
= 1.6 x 10–5 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.
Chapter 1–11
1.65
1.66
Use conversion factors to solve the problems.
a.
1.5 kg
x
b.
1.5 kg
x
1 kg
2.20 lb
1,500 g x
1 oz
28.3 g
= 53 oz
= 3.3 lb
1 kg
1 gr
65 mg
= 5.0 gr
Use conversion factors to solve the problems.
1000 mg
=
1g
a.
300 g
b.
2L
c.
5.0 cm
x
1m
=
100 cm
d.
300 g
x
1 oz
28.3 g
e.
2 ft x
f.
1.68
c.
= 1,500 g
Use conversion factors to solve the problem.
325 mg x
1.67
1000 g
x
1,000,000 µL
1L
x
x
3.5 yd
12 in.
1 ft
x
3 ft
1 yd
x
300,000 mg
=
=
2,000,000 µL
0.050 m
10.60 oz = 10 oz rounded
to one significant figure
1m
=
39.4 in.
0.6091 m = 0.6 m rounded
to one significant figure
12 in.
1m
= 3.198 m = 3.2 m rounded
x
1 ft
to two significant figures
39.4 in.
Use conversion factors to solve the problems.
a.
25 µL x
b. 35 kg
d.
1000 g
300 mL
e. 3 cups x
x
1 x 106 µL
x
c. 2.36 mL
1000 mL
1L
=
2.5 x 10–2 mL
1L
=
1 kg
x
1L
=
1000 mL
x
1 qt
=
946 mL
3.5 x 104 g
2.36 x 10–3 L
0.317 qt = 0.3 qt rounded
to one significant figure
1L
1 qt
x
1.06 qt
4 cups
=
0.7075 L = 0.7 L rounded
to one significant figure
© 2013 by 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.
Matter and Measurement 1–12
f.
1.69
1 kg
2,000 lb
x
1 ton
2.20 lb
2.5 tons x
Use conversion factors to solve the problems.
a.
234 lb
x
1 kg
2.20 lb
=
b.
50. in.
x
2.54 cm
1 in.
=
1 qt
2 pints
c. 3.0 pints x
106.36 kg = 106 kg rounded
to three significant figures
127 cm = 130 cm rounded
to two significant figures
1L
= 1.415 L = 1.4 L rounded
1.06 qt
to two significant figures
x
1.8(oC) + 32
d. °F =
=
1.70
1.8(37.7) + 32 = 99.9 °F
Use conversion factors to solve the problems.
a. 53.2 kg
x
2.20 lb
1 kg
b. 90. cm
x
1 in.
=
2.54 cm
c. 5.0 qt
d. oC
x
–
117.04 lb = 117 lb rounded
to three significant figures
35.433 in. = 35. in. rounded
to two significant figures
1000 mL =
4,716.98 mL = 4,700 mL rounded
x
1L
to two significant figures
1L
1.06 qt
oF
=
=
32
103.5 – 32
=
1.8
1.8
1.71
= 39.7 oC
Use conversion factors to solve the problems.
a. 1 qt x
x
b. 1 L
1.72
= 2,273 kg = 2300 kg rounded
to two significant figures
946 mL
1 qt
1.06 qt
946 mL rounded to three significant figures
=
32 fl oz
1 qt
x
1L
=
33.92 fl oz = 33.9 fl oz rounded to three significant figures
Use conversion factors to solve the problem.
26 mi x
385 yd x
1 km
=
0.621 mi
3 ft
1 yd
x
41.87 km = 42 km rounded
to two significant figures
1 mi
5280 ft
x
1 km
0.621 mi
=
0.3523 km = 0.352 km rounded
to three significant figures
Total distance = 41.87 km + 0.352 km = 42.222 km = 42.22 km rounded to 2 digits
after the decimal
1.73
Convert from °C to °F and K using the formulas listed in Section 1.9.
a.
°F =
1.8(oC) + 32
=
1.8(53) + 32
K =
=
127 °F
=
oC
+ 273
53 + 273
= 326 K
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manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.
Chapter 1–13
b.
oC
=
oF
–
K =
32
1.8
=
1.74
1.8
=
C + 273
177 oC
1.8(oC) + 32
°F =
b.
K =
=
1.8(–183) + 32 = –297 °F
=
1.8(oC) + 32
°F =
K =
1.8(–162) + 32 = –292 °F
=
=
oC
+ 273
–183 + 273
oC
= 90. K
+ 273
–162 + 273
=
oC
=
oF
–
32
b.
oC
=
1.8
=
oF
–
32
1.8
10 – 32
= –12 °C < –10 °C
higher
temperature
10 °F
1.8
=
higher temperature
–50 – 32
1.8
= –45 °C > – 50 °C
–50 °F
Convert the temperatures to a common unit to compare. Order the values from lowest to highest.
a.
oC
=
oF
–
32
oC
= K – 273
1.8
=
=
0 – 32
= –18
1.8
0 – 273
b.
oC
=
oF
–
= –273 oC
32
0K
oC
= K - 273
1.8
=
100 – 32
1.8
0 K < 0 oF < 0 oC
oC
0 oF
= 100 – 273 = –173 oC
100 K < 100 oF < 100 oC
= 82 oC
100 oF
1.77
111 K
Convert the temperatures to a common unit to compare.
a.
1.76
= 450. K
177 + 273
Convert from °C to °F and K using the formulas listed in Section 1.9.
a.
1.75
=
350. – 32
o
100 K
Density is the mass per unit volume, usually reported in g/mL or g/cc. Specific gravity is the
ratio of the density of a substance to the density of water and has no units.
1.78
20.0 g
x
1 cc
1.56 g
=
12.8 g
1.79
122 g
= 1.01 g/mL
121 mL
© 2013 by 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.
Matter and Measurement 1–14
1.80
Calculate the volume of water using the density of water; use that volume to calculate density of
saline.
24.5 g
1 mL
x
=
25.6 g
24.5 mL
1.00 g
= 1.04 g/mL
24.5 mL
1.81
1 qt x
946 mL
1.03 g
x
1 qt
1 mL
x
1 kg
1000 g
=
0.974 38 kg = 0.974 kg
1.82
1 gallon x
1.83
1.84
0.66 g
4 qt
946 mL
x
x
1 gallon
1 qt
1 mL
1 kg
1000 g
x
=
2.497 kg = 2 kg
The density of a substance determines whether it floats or sinks in a liquid. The less dense liquid
is the upper layer. The density of water is 1.0 g/mL.
a. heptane
(0.684 g/mL < 1.0 g/mL)
c. water
(1.0 g/mL < 1.49 g/mL)
b. olive oil
(0.92 g/mL < 1.0 g/mL)
d. water
(1.0 g/mL < 1.59 g/mL)
The density of a substance determines whether it floats or sinks in a liquid. The less dense liquid
is the upper layer. The density of water is 1.0 g/cc.
a. aluminum will sink
(1.70 g/cc > 1.0 g/cc)
c. Styrofoam will float
(0.100 g/cc < 1.00 g/cc)
b. lead will sink
(11.34g/cc > 1.0 g/cc)
d. maple wood will float
(0.74g/mL < 1.0 g/cc)
1.85
a.
b.
specific gravity
=
density of mercury (g/mL)
13.6 g/mL
=
density of water (g/mL)
density of ethanol (g/mL)
0.789 =
density = 0.789 g/mL
1 g/mL
= 13.6
1 g/mL
1.86
Specific gravity is a unitless quantity because it compares the density of a substance with the
density of water. As a result, the units in the numerator and the units in the denominator cancel.
1.87
Use conversion factors to solve the problems.
a.
186 mg
dL
x
1g
1000 mg
= 0.186 g/dL
b.
186 mg
dL
x
10 dL
1L
= 1,860 mg/L
© 2013 by 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.
Chapter 1–15
1.88
Use conversion factors to solve the problems.
a. 15.5 g
1000 mg
x
4
= 1.55 x 10 mg
1g
b. 15.5 g
x
1,000,000 µg
7
= 1.55 x 10 µg
1g
1.89
1.5 g x
1000 mg
1g
1 tablet
500 mg
x
= 3 tablets
1.90
1.05 g
x 1000 mL x
1L
1.8 L
1 kg
x
1000 g
1 mL
= 1.89 kg
70.7 kg + 1.89 kg = 72.59 kg; 72.59 kg - 69.3 kg = 3.29 kg = 3.3 kg sweat lost
2.20 lb
3.3 kg x
=
7.3 lb
1 kg
1.91
3.5 g x
1000 mg
1g
1 banana
451 mg
x
= 7.8 bananas (8 bananas)
1.92
13.0 oz x
250 mg
1 oz
1g
1000 mg
x
= 3.25 g of sodium
3.25 g – 2.4 g = 0.8 g more sodium than recommended daily value
1.93
a.
20 mL
1 dose
x
$10.00
300. mL
$0.67
dose
=
b.
2 tablespoons = 30. mL
30 mL
1 dose
x
$10.00
300. mL
=
$1.00
dose
1.94
a. 1.93 mg x
1g
1000 mg
b. 20 cigarettes x
1.93 x 10–3 g x 1,000,000 µg
1g
= 1.93 x 10-3 g
1.93 mg
= 38.6 mg
cigarette
= 1.93 x 103 µg
38.6 mg > 21mg ; the smoker will get less
nicotine from the patch
1.95
2 tablets x 325 mg/tablet = 650. mg
0.510 kg x
1000 g
1 kg
x 1000 mg
1g
x
1 dose
650. mg
= 784.6 = 784 full doses
© 2013 by 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.
Matter and Measurement 1–16
1.96
2 tsp x
5.0 mL
= 10. mL
1 tsp
10. mL
x
400. mg
5.0 mL
x
1g
1000 mg
= 0.80 g
10. mL
x
40. mg
5.0 mL
x
1g
1000 mg
= 0.080 g
The daily dose contains 0.80 g of both aluminum hydroxide and
magnesium hydroxide and 0.080 g of simethicone.
1.97
4 times
2.0 g
x
x 1000 mg
1g
time
1 tablet
500. mg
x
= 16 tablets
1.98
a.
b.
20. min x
90. mL
x
1h
60 min
x
1h
x
150 mL
150 mL
c.
60 min
= 36 min
1h
1h
600. mL x
= 50. mL
1h
= 4.0 hours
150 mL
d.
2.0 g
1000 mg
1g
x
mL
x
x
1h
150 mL
90. mg
x
60 min
1h
= 8.9 min
1.99
2.0 mg
1 kg
x
1 kg
110 lb = 1.0 x 102 mg
x
2.20 lb
1.100
Number of doses = 3 x 7 = 21
21 x 28 kg x
10. mg
kg
x
1g
1000 mg
= 5.9 g
1.101
42 lb x
1 kg
2.20 lb
x
10 mg
1 kg
x
1 tablet
80 mg
= 2.4 = 2 tablets
© 2013 by 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.
Chapter 1–17
1.102
Dose in pill form = 160 mg + 4(80. mg) = 480 mg
a.
40. kg
x 3.2 mg
kg
+
4 x
40. kg
x 1.6mg
kg
= 384 mg
Pill form will give a 40. kg individual a larger dose.
b.
100. kg x 3.2 mg
kg
+
100. kg x 1.6mg
kg
4 x
= 960 mg
Injection form will give a 100. kg individual a larger dose.
1.103
1.5 tsp
x
5.0 mL
1 tsp
x
100. mg
x
5 mL
1g
1000 mg
= 0.15 g
1.104
200 lb x
1 kg
2.20 lb
x
10 µg
1 kg
x
1 mg
1000 µg
= 0.909 mg = 0.9 mg
© 2013 by 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.