Instructor’s Solutions Manual
to accompany
Introduction
to Chemistry
A Conceptual Approach
Fourth Edition
Richard C. Bauer
Arizona State University
James P. Birk
Arizona State University
Pamela S. Marks
Arizona State University
,Chapter 1 – Matter and Energy
1.1 (a) mass; (b) chemical property; (c) mixture; (d) element; (e) energy; (f) physical property; (g) liquid; (h)
density; (i) homogeneous mixture; (j) solid state
1.2 (a) atom; (b) chemical change; (c) matter; (d) compound; (e) molecule; (f) physical change; (g) gas; (h)
potential energy; (i) hypothesis; (j) kinetic energy
1.3 When converting to scientific notation, count the number of places you need to move the decimal point.
Zeros to the left of the number are always dropped. For example, the number 0.002030 becomes
2.030 × 10–3 and the zeros to the left of 2030 are dropped. The zero to the right is only kept if it is
significant (covered later in this chapter). If the decimal point moves right, the exponent decreases. If the
decimal moves left the exponent increases.
(a) 2.95 × 104; (b) 8.2 × 10−5; (c) 6.5 × 108; (d) 1.00 × 10−2
1.4 When converting to scientific notation, count the number of places you need to move the decimal point.
Zeros to the left of the number are always dropped. For example, the number 0.002030 becomes
2.030 × 10- 3 and the zeros to the left of 2030 are dropped. The zero to the right is only kept if it is
significant (covered later in this chapter). If the decimal point moves right, the exponent decreases. If the
decimal moves left the exponent increases.
(a) 1.0 × 10−4; (b) 4.5 × 103; (c) 9.01 × 107; (d) 7.9 × 10−6
1.5 When converting from scientific notation to standard notation you may need to add place-holder zeros so
that the magnitude of the number is correct. For example, to get 1.86 × 10−5 into standard notation, you
need to increase the power by five, so the decimal moves to the left. In addition, you’ll need four
placeholder zeros to show the magnitude of the number.
(a) 0.0000186; (b) 10,000,000; (c) 453,000; (d) 0.0061
1.6 When converting from scientific notation to standard notation you may need to add place-holder zeros so
that the magnitude of the number is correct. For example, to get 1.86 × 10−5 into standard notation, you
need to increase the power by five, so the decimal moves to the left. In addition, you’ll need four
placeholder zeros to show the magnitude of the number.
(a) 8200; (b) 0.000002025; (c) 0.07; (d) 300000000
1.7 (a) 6.2 × 103; (b) 3.5 × 107; (c) 2.9 × 10−3; (d) 2.5 × 10−7; (e) 8.20 × 105; (f) 1.6 × 10−6
1.8 (a) 2.0 × 108; (b) 1.5 × 1014; (c) 3.0 × 10−10; (d) 8.5 × 10−6; (e) 8.56 × 105; (f) 1.26 × 108
1.9 Nonzero digits and zeros between nonzero digits are significant. Zeros at the end of a number and to the
right of the decimal are significant. Zeros to the left of the first nonzero digit and in exponentials (i.e.
× 103) are not significant. The number 0.0950 has three significant digits. The digits “950” are all
significant because (1) the 9 and 5 are nonzero, and (2) the zero is significant because it is at the end of the
number and to the right of the decimal.
(a) 3; (b) 2; (c) 4; (d) 2; (e) 3
1.10 Nonzero digits and zeros between nonzero digits are significant. Zeros at the end of a number and to the
right of the decimal are significant. Zeros to the left of the first nonzero digit and in exponentials (i.e.
× 103) are not significant. The number 0.04350 has four significant digits. The digits “4350” are all
significant because (1) the 4, 3, and 5 are nonzero, and (2) the zero is significant because it is at the end of
the number and to the right of the decimal.
(a) 3; (b) 4; (c) 4; (d) 4; (e) 2
1-1
,1.11 For operations involving multiplication, division, and powers, the answer will have the same number of
significant figures as the number with the fewest significant figures. For example, in part (c) the number
1.201 × 103 has four significant figures and the number 1.2 × 10−2 has two significant figures. The
calculated value is 14.412 which will be rounded to two significant figures, 14.
(a) 1.5; (b) 1.5; (c) 14; (d) 1.20
1.12 For operations involving multiplication, division, and powers, the answer will have the same number of
significant figures as the number with the fewest significant figures. For example, in part (a) the number
1.600 × 10−7 has four significant figures and the number 2.1 × 103 has two significant figures. The
calculated value is 3.36 × 103 which will be rounded to two significant figures, 3.4 × 10–4.
(a) 3.4 × 10–4; (b) 2.35; (c) 5.12; (d) 2.0
1.13 For operations involving addition and subtraction, the answer can only be as precise as the least precise
number. A number that has its last significant digit in the tenths place (one place past the decimal) has less
precision than a number that ends in the hundredths place (two places past the decimal). If you add these
two numbers together, you would have to round the answer to the tenths place. For example, in part (a)
1.6 + 1.15 gives a value of 2.75. This number will have to be rounded to the tenths place, 2.8.
(a) 2.8; (b) 0.28; (c) 2.8; (d) 0.049
1.14 For operations involving addition and subtraction, the answer can only be as precise as the least precise
number. A number that has its last significant digit in the tenths place (one place past the decimal) has less
precision than a number that ends in the ten thousandths place (four places past the decimal). If you add
these two numbers together, you would have to round the answer to the tenths place. For example, in part
(a) 87.5 + 1.3218 gives a value of 88.8218. This number will have to be rounded to the tenths place, 88.8.
(a) 88.8; (b) 12; (c) 0.22; (d) 1.80
1.15 When calculations involve multiple steps, the number of significant figures in subsequent steps requires us
to know the number of significant figures in the answers from the previous steps. We must keep track of
the last significant figure in the answer to each step. For example, in part (c) 0.35 m × 0.55 m gives a value
of 0.1925 m2. Following the rules of multiplication/division, this value should only be expressed to two
significant figures. However, to prevent rounding errors, we don’t round yet. We’ll make note that the
first step only has two significant figures by underlining the last significant digit, 0.1925 m2. In the second
step of the calculation we add this number to 25.2 m2. The value 25.3925 is obtained from the calculation.
Following the rules of addition/subtraction, the answer can only be as precise as the least precise number.
For this calculation, the number will have to be rounded to the tenths place, 25.4 m2.
(a)
( 20.90 kg − 12.90 kg ) =
(8.00 kg ) = 0.800 kg/L
10.00 L 10.00 L
45.82 g 0.64 g 45.82 g 0.64 g
− ÷ − 3
÷2
( 27 cm ) ( 0.6338498 cm )
(b) 2 =
( 3.0 cm ) ( 0.859 cm )
3 3 3
= 1.69704 g/cm − 1.00972 g/cm ÷ 2 = 0.68732 g/cm ÷ 2 = 0.34366 g/cm = 0.3 g/cm
3 3 3 3 3
(c) (0.35 m × 0.55 m) + 25.2 m2 = (0.1925 m2) + 25.2 m2 = 25.3925 m2 = 25.4 m2
1.16 When calculations involve multiple steps, the number of significant figures in subsequent steps requires us
to know the number of significant figures in the answers from the previous steps. We must keep track of
the last significant figure in the answer to each step. For example, in part (a) 0.25 m/s × 45.77 s gives a
value of 11.4425 m. Following the rules of multiplication/division, this value should only be expressed to
two significant figures. However, to prevent rounding errors, we don’t round yet. We’ll make note that the
1-2
, first step only has two significant figures by underlining the last significant digit, 11.4425 m. In the second
step of the calculation we add this number to 5.0 m. The value 16.4425 is obtained from the calculation.
Following the rules of addition/subtraction, the answer can only be as precise as the least precise number.
For this calculation, the number will have to be rounded to the ones place, 16 m.
(a) (0.25 m/s × 45.77 s) + 5.0 m = (11.4425 m) + 5.0 m = 16.4425 m = 16 m
2.523 lb 2.523 lb
(b) = = 0.63075 lb/gal = 0.63 lb/gal
( 62.9 gal − 58.9 gal ) ( 4.0 gal )
(c) (9.0 cm × 15.1 cm × 10.5 cm) + 75.7 cm3 = (1426.95 cm3) + 75.7 cm3 = 1502.65 cm3 = 1.5 × 103 cm3
1.17 (a) 1.21; (b) 0.204; (c) 1.84; (d) 42.2; (e) 0.00710
1.18 (a) 0.0205; (b) 1.36 × 104; (c) 13.5; (d) 16.2; (e) 1.00
1.19 When you are converting between a unit and the same base unit with a prefix (e.g. mm to m or visa versa)
you can find the conversion factors in Math Toolbox 1.3. Suppose you want to convert between
millimeters and meters. There are several ways you can do this. First, by definition milli is = 10–3, so
1 mm = 10–3 m. You might also already know that there are one thousand millimeters in a meter,
1000 mm = 1 m. Either conversion factor is correct. Next, you set up your calculation so that the
appropriate units cancel. The English-Metric conversions are also found in Math Toolbox 1.3.
−3
(a) Map: Length in mm
1 mm = 10 m
→ Length in m
Problem solution:
10−3 m
Length in m = 36 mm × = 0.036 m
1 mm
3
(b) Map: Mass in kg
1 kg = 10 g
→ Mass in g
Problem solution:
103 g
Mass in g = 357 kg × = 3.57 × 105 g
1 kg
−3
(c) Map: Volume in mL
1 mL = 10 L
→ Volume in L
Problem solution:
10−3 L
Volume in L = 76.50 mL × = 0.07650 L
1 mL
−2
(d) Map: Length in m
1 cm = 10 m
→ Length in cm
Problem solution:
cm
Length in cm = 0.0084670 m × −2 = 0.84670 cm
10 m
−9
(e) Map: Length in nm
1 nm = 10 m
→ Length in m
Problem solution:
10−9 m
Length in m = 597 nm × = 5.97 × 10−7 m
1 nm
(f) This is the first metric-English conversion, but the process is exactly the same. Note that the in-cm
conversion factor is exact, so it is not a factor in determining significant figures:
Map: Length in in →
1 in = 2.54 cm (exact)
Length in cm
Problem solution:
1-3
to accompany
Introduction
to Chemistry
A Conceptual Approach
Fourth Edition
Richard C. Bauer
Arizona State University
James P. Birk
Arizona State University
Pamela S. Marks
Arizona State University
,Chapter 1 – Matter and Energy
1.1 (a) mass; (b) chemical property; (c) mixture; (d) element; (e) energy; (f) physical property; (g) liquid; (h)
density; (i) homogeneous mixture; (j) solid state
1.2 (a) atom; (b) chemical change; (c) matter; (d) compound; (e) molecule; (f) physical change; (g) gas; (h)
potential energy; (i) hypothesis; (j) kinetic energy
1.3 When converting to scientific notation, count the number of places you need to move the decimal point.
Zeros to the left of the number are always dropped. For example, the number 0.002030 becomes
2.030 × 10–3 and the zeros to the left of 2030 are dropped. The zero to the right is only kept if it is
significant (covered later in this chapter). If the decimal point moves right, the exponent decreases. If the
decimal moves left the exponent increases.
(a) 2.95 × 104; (b) 8.2 × 10−5; (c) 6.5 × 108; (d) 1.00 × 10−2
1.4 When converting to scientific notation, count the number of places you need to move the decimal point.
Zeros to the left of the number are always dropped. For example, the number 0.002030 becomes
2.030 × 10- 3 and the zeros to the left of 2030 are dropped. The zero to the right is only kept if it is
significant (covered later in this chapter). If the decimal point moves right, the exponent decreases. If the
decimal moves left the exponent increases.
(a) 1.0 × 10−4; (b) 4.5 × 103; (c) 9.01 × 107; (d) 7.9 × 10−6
1.5 When converting from scientific notation to standard notation you may need to add place-holder zeros so
that the magnitude of the number is correct. For example, to get 1.86 × 10−5 into standard notation, you
need to increase the power by five, so the decimal moves to the left. In addition, you’ll need four
placeholder zeros to show the magnitude of the number.
(a) 0.0000186; (b) 10,000,000; (c) 453,000; (d) 0.0061
1.6 When converting from scientific notation to standard notation you may need to add place-holder zeros so
that the magnitude of the number is correct. For example, to get 1.86 × 10−5 into standard notation, you
need to increase the power by five, so the decimal moves to the left. In addition, you’ll need four
placeholder zeros to show the magnitude of the number.
(a) 8200; (b) 0.000002025; (c) 0.07; (d) 300000000
1.7 (a) 6.2 × 103; (b) 3.5 × 107; (c) 2.9 × 10−3; (d) 2.5 × 10−7; (e) 8.20 × 105; (f) 1.6 × 10−6
1.8 (a) 2.0 × 108; (b) 1.5 × 1014; (c) 3.0 × 10−10; (d) 8.5 × 10−6; (e) 8.56 × 105; (f) 1.26 × 108
1.9 Nonzero digits and zeros between nonzero digits are significant. Zeros at the end of a number and to the
right of the decimal are significant. Zeros to the left of the first nonzero digit and in exponentials (i.e.
× 103) are not significant. The number 0.0950 has three significant digits. The digits “950” are all
significant because (1) the 9 and 5 are nonzero, and (2) the zero is significant because it is at the end of the
number and to the right of the decimal.
(a) 3; (b) 2; (c) 4; (d) 2; (e) 3
1.10 Nonzero digits and zeros between nonzero digits are significant. Zeros at the end of a number and to the
right of the decimal are significant. Zeros to the left of the first nonzero digit and in exponentials (i.e.
× 103) are not significant. The number 0.04350 has four significant digits. The digits “4350” are all
significant because (1) the 4, 3, and 5 are nonzero, and (2) the zero is significant because it is at the end of
the number and to the right of the decimal.
(a) 3; (b) 4; (c) 4; (d) 4; (e) 2
1-1
,1.11 For operations involving multiplication, division, and powers, the answer will have the same number of
significant figures as the number with the fewest significant figures. For example, in part (c) the number
1.201 × 103 has four significant figures and the number 1.2 × 10−2 has two significant figures. The
calculated value is 14.412 which will be rounded to two significant figures, 14.
(a) 1.5; (b) 1.5; (c) 14; (d) 1.20
1.12 For operations involving multiplication, division, and powers, the answer will have the same number of
significant figures as the number with the fewest significant figures. For example, in part (a) the number
1.600 × 10−7 has four significant figures and the number 2.1 × 103 has two significant figures. The
calculated value is 3.36 × 103 which will be rounded to two significant figures, 3.4 × 10–4.
(a) 3.4 × 10–4; (b) 2.35; (c) 5.12; (d) 2.0
1.13 For operations involving addition and subtraction, the answer can only be as precise as the least precise
number. A number that has its last significant digit in the tenths place (one place past the decimal) has less
precision than a number that ends in the hundredths place (two places past the decimal). If you add these
two numbers together, you would have to round the answer to the tenths place. For example, in part (a)
1.6 + 1.15 gives a value of 2.75. This number will have to be rounded to the tenths place, 2.8.
(a) 2.8; (b) 0.28; (c) 2.8; (d) 0.049
1.14 For operations involving addition and subtraction, the answer can only be as precise as the least precise
number. A number that has its last significant digit in the tenths place (one place past the decimal) has less
precision than a number that ends in the ten thousandths place (four places past the decimal). If you add
these two numbers together, you would have to round the answer to the tenths place. For example, in part
(a) 87.5 + 1.3218 gives a value of 88.8218. This number will have to be rounded to the tenths place, 88.8.
(a) 88.8; (b) 12; (c) 0.22; (d) 1.80
1.15 When calculations involve multiple steps, the number of significant figures in subsequent steps requires us
to know the number of significant figures in the answers from the previous steps. We must keep track of
the last significant figure in the answer to each step. For example, in part (c) 0.35 m × 0.55 m gives a value
of 0.1925 m2. Following the rules of multiplication/division, this value should only be expressed to two
significant figures. However, to prevent rounding errors, we don’t round yet. We’ll make note that the
first step only has two significant figures by underlining the last significant digit, 0.1925 m2. In the second
step of the calculation we add this number to 25.2 m2. The value 25.3925 is obtained from the calculation.
Following the rules of addition/subtraction, the answer can only be as precise as the least precise number.
For this calculation, the number will have to be rounded to the tenths place, 25.4 m2.
(a)
( 20.90 kg − 12.90 kg ) =
(8.00 kg ) = 0.800 kg/L
10.00 L 10.00 L
45.82 g 0.64 g 45.82 g 0.64 g
− ÷ − 3
÷2
( 27 cm ) ( 0.6338498 cm )
(b) 2 =
( 3.0 cm ) ( 0.859 cm )
3 3 3
= 1.69704 g/cm − 1.00972 g/cm ÷ 2 = 0.68732 g/cm ÷ 2 = 0.34366 g/cm = 0.3 g/cm
3 3 3 3 3
(c) (0.35 m × 0.55 m) + 25.2 m2 = (0.1925 m2) + 25.2 m2 = 25.3925 m2 = 25.4 m2
1.16 When calculations involve multiple steps, the number of significant figures in subsequent steps requires us
to know the number of significant figures in the answers from the previous steps. We must keep track of
the last significant figure in the answer to each step. For example, in part (a) 0.25 m/s × 45.77 s gives a
value of 11.4425 m. Following the rules of multiplication/division, this value should only be expressed to
two significant figures. However, to prevent rounding errors, we don’t round yet. We’ll make note that the
1-2
, first step only has two significant figures by underlining the last significant digit, 11.4425 m. In the second
step of the calculation we add this number to 5.0 m. The value 16.4425 is obtained from the calculation.
Following the rules of addition/subtraction, the answer can only be as precise as the least precise number.
For this calculation, the number will have to be rounded to the ones place, 16 m.
(a) (0.25 m/s × 45.77 s) + 5.0 m = (11.4425 m) + 5.0 m = 16.4425 m = 16 m
2.523 lb 2.523 lb
(b) = = 0.63075 lb/gal = 0.63 lb/gal
( 62.9 gal − 58.9 gal ) ( 4.0 gal )
(c) (9.0 cm × 15.1 cm × 10.5 cm) + 75.7 cm3 = (1426.95 cm3) + 75.7 cm3 = 1502.65 cm3 = 1.5 × 103 cm3
1.17 (a) 1.21; (b) 0.204; (c) 1.84; (d) 42.2; (e) 0.00710
1.18 (a) 0.0205; (b) 1.36 × 104; (c) 13.5; (d) 16.2; (e) 1.00
1.19 When you are converting between a unit and the same base unit with a prefix (e.g. mm to m or visa versa)
you can find the conversion factors in Math Toolbox 1.3. Suppose you want to convert between
millimeters and meters. There are several ways you can do this. First, by definition milli is = 10–3, so
1 mm = 10–3 m. You might also already know that there are one thousand millimeters in a meter,
1000 mm = 1 m. Either conversion factor is correct. Next, you set up your calculation so that the
appropriate units cancel. The English-Metric conversions are also found in Math Toolbox 1.3.
−3
(a) Map: Length in mm
1 mm = 10 m
→ Length in m
Problem solution:
10−3 m
Length in m = 36 mm × = 0.036 m
1 mm
3
(b) Map: Mass in kg
1 kg = 10 g
→ Mass in g
Problem solution:
103 g
Mass in g = 357 kg × = 3.57 × 105 g
1 kg
−3
(c) Map: Volume in mL
1 mL = 10 L
→ Volume in L
Problem solution:
10−3 L
Volume in L = 76.50 mL × = 0.07650 L
1 mL
−2
(d) Map: Length in m
1 cm = 10 m
→ Length in cm
Problem solution:
cm
Length in cm = 0.0084670 m × −2 = 0.84670 cm
10 m
−9
(e) Map: Length in nm
1 nm = 10 m
→ Length in m
Problem solution:
10−9 m
Length in m = 597 nm × = 5.97 × 10−7 m
1 nm
(f) This is the first metric-English conversion, but the process is exactly the same. Note that the in-cm
conversion factor is exact, so it is not a factor in determining significant figures:
Map: Length in in →
1 in = 2.54 cm (exact)
Length in cm
Problem solution:
1-3
