Solution and Answer Guide: Ebbing, General Chemistry, 12e, 2026, 9798214194400;
Chapter 1: Chemistry and Measurement
Solution and Answer Guide
EBBING, GENERAL CHEMISTRY, 12E, 2026, 9798214194400;
CHAPTER 1: CHEMISTRY AND MEASUREMENT
TABLE OF CONTENTS
Solutions to Chapter Integrated Exercises .........................................................................1
Answers to Chapter Integrated Concept Checks .............................................................. 3
Answers to End of Chapter Self-Assessment and Review Questions ............................. 5
Answers to End of Chapter Concept Explorations ............................................................ 8
Answers to End of Chapter Conceptual Problems ............................................................ 9
Solutions to End of Chapter Practice Problems ...............................................................12
Solutions to End of Chapter General Problems ............................................................... 23
Solutions to End of Chapter Strategy Problems .............................................................. 34
Solutions to End of Chapter Capstone Problems ............................................................ 37
SOLUTIONS TO CHAPTER INTEGRATED EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given
first with one nonsignificant figure, and the last significant figure is underlined. The final
answer is then rounded to the correct number of significant figures. In multistep problems,
intermediate answers are given with at least one nonsignificant figure; however, only the final
answer has been rounded off.
1.1.
Solution:
a. The factor 9.1 has the fewest significant figures, so the answer should be reported
to two significant figures.
5.61 × 7.891
= 4.86 = 4.9
9.1
b. The number with the least number of decimal places is 8.91. Therefore, round the
answer to two decimal places.
8.91 − 6.435 = 2.475 = 2.48
c. The number with the least number of decimal places is 6.81. Therefore, round the
answer to two decimal places.
6.81 − 6.730 = 0.080 = 0.08
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 1
accessible website, in whole or in part.
, Solution and Answer Guide: Ebbing, General Chemistry, 12e, 2026, 9798214194400;
Chapter 1: Chemistry and Measurement
d. You first do the subtraction within parentheses. In this step, the number with the
least number of decimal places is 6.81, so the result of the subtraction has two
decimal places. The least significant figure for this step is underlined.
38.91 × (6.81 − 6.730) = 38.91 × 0.080
Next, perform the multiplication. In this step, the factor 0.080 has the fewest
significant figures, so round the answer to one significant figure.
38.91 × 0.080 = 3.11 = 3
1.2.
Solution:
a. 1.84 x 10−9 m = 1.84 nm
b. 5.67 x 10−12 s = 5.67 ps
c. 7.85 x 10−3 g = 7.85 mg
d. 9.7 x 103 m = 9.7 km
e. 0.000732 s = 0.732 ms, or 732 µs
f. 0.000000000154 m = 0.154 nm, or 154 pm
1.3.
Solution:
a. Substituting, we find that
5°C 5°C
TC = × (TF − 32°F) = × (102.5°F − 32°F) = 39.167°C
9°F 9°F
= 39.2°C
b. Substituting, we find that
1 K 1 K
TK = Tc × + 273.15 K = −78°C × + 273.15 K = 195.15 K
1°C 1°C
= 195 K
1.4.
Solution:
Recall that density equals mass divided by volume. You substitute 159 g for the mass
and 20.2 g/cm3 for the volume.
m 159 g
d= = = 7.871 g/cm3 = 7.87 g/cm3
V 20.2 cm3
The density of the metal equals that of iron.
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 2
accessible website, in whole or in part.
, Solution and Answer Guide: Ebbing, General Chemistry, 12e, 2026, 9798214194400;
Chapter 1: Chemistry and Measurement
1.5.
Solution:
Rearrange the formula defining the density to obtain the volume.
m
V=
d
Substitute 30.3 g for the mass and 0.789 g/cm3 for the density.
30.3 g
V= = 38.40 cm3 = 38.4 cm3
0.789 g/cm3
1.6.
Solution:
Since one pm = 10−12 m, and the prefix milli- means 10−3, you can write
10−12 m 1 mm
121 pm × × = 1.21 × 10−7 mm
1 pm 10−3 m
1.7.
Solution:
3 3
10−10 m 1 dm
67.6 Å ×
3
× −1 = 6.76 × 10 dm
−26 3
1 Å 10 m
1.8.
Solution:
From the definitions, you obtain the following conversion factors:
36 in 2.54 cm 10-2 m
1= 1= 1=
1 yd 1 in 1 cm
The conversion factor for yards to meters is as follows:
36 in 2.54 cm 10-2 m
1.000 yd × × × = 0.9144 m (exact)
1 yd 1 in 1 cm
Finally,
0.9144 m
3.54 yd × = 3.237 m = 3.24 m
1 yd
ANSWERS TO CHAPTER INTEGRATED CONCEPT CHECKS
1.1.
Solution:
Box A contains a collection of identical units; therefore, it must represent an element.
Box B contains a compound because a compound is the chemical combination of two
or more elements (two elements in this case). Box C contains a mixture because it is
made up of two different substances.
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 3
accessible website, in whole or in part.
, Solution and Answer Guide: Ebbing, General Chemistry, 12e, 2026, 9798214194400;
Chapter 1: Chemistry and Measurement
1.2.
Solution:
a. For a person who weighs less than 100 pounds, two significant figures are typically
used, although one significant figure is possible (for example, 60 pounds). For a
person who weighs 100 pounds or more, three significant figures are typically used
to report the weight (given to the whole pound), although people often round to
the nearest unit of 10, which may result in reporting the weight with two significant
figures (for example, 170 pounds).
b. Assuming a weight of 165 pounds, rounded to two significant figures this would be
reported as 1.7 x 102 pounds.
c. For example, 165 lb weighed on a scale that can measure in 100-lb increments would
be 200 lb. Using the conversion factor 1 lb = 0.4536 kg, 165 lb is equivalent to 74.8 kg.
Thus, on a scale that can measure in 50-kg increments, 165 lb would be 50 kg.
1.3.
Solution:
a. If your leg is approximately 32 inches long, this would be equivalent to 0.81 m, 8.1
dm, or 81 cm.
b. One story is approximately 10 feet, so three stories is 30 feet. This would be
equivalent to approximately 9 m.
c. Normal body temperature is 98.6°F, or 37.0°C. Thus, if your body temperature were
39°C (102°F), you would feel as if you had a moderate fever.
d. Room temperature is approximately 72°F, or 22°C. Thus, if you were sitting in a
room at 23°C (73°F), you would be comfortable in a short-sleeve shirt.
1.4.
Solution:
Gold is a very unreactive substance, so comparing physical properties is probably your
best option. However, color is a physical property you cannot rely on in this case to get
your answer.
One experiment you could perform is to determine the densities of the metal and the
chunk of gold. You could measure the mass of the nugget on a balance and the volume
of the nugget by water displacement. Using this information, you could calculate the
density of the nugget. Repeat the experiment and calculations for the sample of gold.
If the nugget is gold, the two densities should be equal and be 19.3 g/cm3.
Also, you could determine the melting points of the metal and the chunk of pure gold.
The two melting points should be the same (1338 K) if the metal is gold.
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 4
accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
Solution and Answer Guide
EBBING, GENERAL CHEMISTRY, 12E, 2026, 9798214194400;
CHAPTER 1: CHEMISTRY AND MEASUREMENT
TABLE OF CONTENTS
Solutions to Chapter Integrated Exercises .........................................................................1
Answers to Chapter Integrated Concept Checks .............................................................. 3
Answers to End of Chapter Self-Assessment and Review Questions ............................. 5
Answers to End of Chapter Concept Explorations ............................................................ 8
Answers to End of Chapter Conceptual Problems ............................................................ 9
Solutions to End of Chapter Practice Problems ...............................................................12
Solutions to End of Chapter General Problems ............................................................... 23
Solutions to End of Chapter Strategy Problems .............................................................. 34
Solutions to End of Chapter Capstone Problems ............................................................ 37
SOLUTIONS TO CHAPTER INTEGRATED EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given
first with one nonsignificant figure, and the last significant figure is underlined. The final
answer is then rounded to the correct number of significant figures. In multistep problems,
intermediate answers are given with at least one nonsignificant figure; however, only the final
answer has been rounded off.
1.1.
Solution:
a. The factor 9.1 has the fewest significant figures, so the answer should be reported
to two significant figures.
5.61 × 7.891
= 4.86 = 4.9
9.1
b. The number with the least number of decimal places is 8.91. Therefore, round the
answer to two decimal places.
8.91 − 6.435 = 2.475 = 2.48
c. The number with the least number of decimal places is 6.81. Therefore, round the
answer to two decimal places.
6.81 − 6.730 = 0.080 = 0.08
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 1
accessible website, in whole or in part.
, Solution and Answer Guide: Ebbing, General Chemistry, 12e, 2026, 9798214194400;
Chapter 1: Chemistry and Measurement
d. You first do the subtraction within parentheses. In this step, the number with the
least number of decimal places is 6.81, so the result of the subtraction has two
decimal places. The least significant figure for this step is underlined.
38.91 × (6.81 − 6.730) = 38.91 × 0.080
Next, perform the multiplication. In this step, the factor 0.080 has the fewest
significant figures, so round the answer to one significant figure.
38.91 × 0.080 = 3.11 = 3
1.2.
Solution:
a. 1.84 x 10−9 m = 1.84 nm
b. 5.67 x 10−12 s = 5.67 ps
c. 7.85 x 10−3 g = 7.85 mg
d. 9.7 x 103 m = 9.7 km
e. 0.000732 s = 0.732 ms, or 732 µs
f. 0.000000000154 m = 0.154 nm, or 154 pm
1.3.
Solution:
a. Substituting, we find that
5°C 5°C
TC = × (TF − 32°F) = × (102.5°F − 32°F) = 39.167°C
9°F 9°F
= 39.2°C
b. Substituting, we find that
1 K 1 K
TK = Tc × + 273.15 K = −78°C × + 273.15 K = 195.15 K
1°C 1°C
= 195 K
1.4.
Solution:
Recall that density equals mass divided by volume. You substitute 159 g for the mass
and 20.2 g/cm3 for the volume.
m 159 g
d= = = 7.871 g/cm3 = 7.87 g/cm3
V 20.2 cm3
The density of the metal equals that of iron.
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 2
accessible website, in whole or in part.
, Solution and Answer Guide: Ebbing, General Chemistry, 12e, 2026, 9798214194400;
Chapter 1: Chemistry and Measurement
1.5.
Solution:
Rearrange the formula defining the density to obtain the volume.
m
V=
d
Substitute 30.3 g for the mass and 0.789 g/cm3 for the density.
30.3 g
V= = 38.40 cm3 = 38.4 cm3
0.789 g/cm3
1.6.
Solution:
Since one pm = 10−12 m, and the prefix milli- means 10−3, you can write
10−12 m 1 mm
121 pm × × = 1.21 × 10−7 mm
1 pm 10−3 m
1.7.
Solution:
3 3
10−10 m 1 dm
67.6 Å ×
3
× −1 = 6.76 × 10 dm
−26 3
1 Å 10 m
1.8.
Solution:
From the definitions, you obtain the following conversion factors:
36 in 2.54 cm 10-2 m
1= 1= 1=
1 yd 1 in 1 cm
The conversion factor for yards to meters is as follows:
36 in 2.54 cm 10-2 m
1.000 yd × × × = 0.9144 m (exact)
1 yd 1 in 1 cm
Finally,
0.9144 m
3.54 yd × = 3.237 m = 3.24 m
1 yd
ANSWERS TO CHAPTER INTEGRATED CONCEPT CHECKS
1.1.
Solution:
Box A contains a collection of identical units; therefore, it must represent an element.
Box B contains a compound because a compound is the chemical combination of two
or more elements (two elements in this case). Box C contains a mixture because it is
made up of two different substances.
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 3
accessible website, in whole or in part.
, Solution and Answer Guide: Ebbing, General Chemistry, 12e, 2026, 9798214194400;
Chapter 1: Chemistry and Measurement
1.2.
Solution:
a. For a person who weighs less than 100 pounds, two significant figures are typically
used, although one significant figure is possible (for example, 60 pounds). For a
person who weighs 100 pounds or more, three significant figures are typically used
to report the weight (given to the whole pound), although people often round to
the nearest unit of 10, which may result in reporting the weight with two significant
figures (for example, 170 pounds).
b. Assuming a weight of 165 pounds, rounded to two significant figures this would be
reported as 1.7 x 102 pounds.
c. For example, 165 lb weighed on a scale that can measure in 100-lb increments would
be 200 lb. Using the conversion factor 1 lb = 0.4536 kg, 165 lb is equivalent to 74.8 kg.
Thus, on a scale that can measure in 50-kg increments, 165 lb would be 50 kg.
1.3.
Solution:
a. If your leg is approximately 32 inches long, this would be equivalent to 0.81 m, 8.1
dm, or 81 cm.
b. One story is approximately 10 feet, so three stories is 30 feet. This would be
equivalent to approximately 9 m.
c. Normal body temperature is 98.6°F, or 37.0°C. Thus, if your body temperature were
39°C (102°F), you would feel as if you had a moderate fever.
d. Room temperature is approximately 72°F, or 22°C. Thus, if you were sitting in a
room at 23°C (73°F), you would be comfortable in a short-sleeve shirt.
1.4.
Solution:
Gold is a very unreactive substance, so comparing physical properties is probably your
best option. However, color is a physical property you cannot rely on in this case to get
your answer.
One experiment you could perform is to determine the densities of the metal and the
chunk of gold. You could measure the mass of the nugget on a balance and the volume
of the nugget by water displacement. Using this information, you could calculate the
density of the nugget. Repeat the experiment and calculations for the sample of gold.
If the nugget is gold, the two densities should be equal and be 19.3 g/cm3.
Also, you could determine the melting points of the metal and the chunk of pure gold.
The two melting points should be the same (1338 K) if the metal is gold.
© 2026 Cengage Learning, Inc. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly 4
accessible website, in whole or in part.
