Issuu

1 Chemistry in Our Lives Learning Goals • • • •

Define the term chemistry, and identify chemicals. Describe the activities that are part of the scientific method. Identify strategies that are effective for learning. Develop a study plan for learning chemistry. Review math concepts used in chemistry: place values, positive and negative numbers, percentages, solving equations, and interpreting graphs. • Write a number in scientific notation.

Chapter Outline Chapter Opener: Forensic Scientist Chemistry and Chemicals Scientific Method: Thinking Like a Scientist Chemistry Link to Health: Early Chemist: Paracelsus Studying and Learning Chemistry Key Math Skills for Chemistry Writing Numbers in Scientific Notation Clinical Update: Forensic Evidence Helps Solve the Crime

1.1 1.2 1.3 1.4 1.5

Key Math Skills • • • • • •

Identifying Place Values (1.4) Using Positive and Negative Numbers in Calculations (1.4) Calculating Percentages (1.4) Solving Equations (1.4) Interpreting Graphs (1.4) Writing Numbers in Scientific Notation (1.5)

Answers and Solutions to Text Problems 1.1

a. Chemistry is the study of the composition, structure, properties, and reactions of matter. b. A chemical is a substance that has the same composition and properties wherever it is found.

1.2

Your friends may give a variety of definitions, most of which will probably not agree with the dictionary definitions.

1.3

Many chemicals are listed on a bottle of multivitamins such as vitamin A, vitamin B3, vitamin B12, vitamin C, and folic acid.

1.4

Many chemicals are listed on a cereal box such as vitamin A, vitamin B6, vitamin B12, vitamin C, folic acid, sugar, salt, and iron.

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Chapter 1

1.5

Typical items found in a medicine cabinet and some of the chemicals they contain are as follows: Antacid tablets: calcium carbonate, cellulose, starch, stearic acid, silicon dioxide Mouthwash: water, alcohol, thymol, glycerol, sodium benzoate, benzoic acid Cough suppressant: menthol, beta-carotene, sucrose, glucose

1.6

Typical chemicals found in dishwashing products are: water, sodium lauryl sulfate, sodium laureth sulfate, dimethyl amine oxide, sodium chloride, phenoxyethanol.

1.7

a. observation d. observation

b. hypothesis e. observation

c. experiment f. conclusion

1.8

a. observation d. experiment

b. hypothesis e. observation

c. experiment f. hypothesis

1.9

a. observation

b. hypothesis

c. experiment

d. experiment

1.10 a. hypothesis

b. observation

c. experiment

d. conclusion

1.11 There are several things you can do that will help you successfully learn chemistry, including attending class regularly, forming a study group, reading the assigned pages in the textbook before class, answering the Engage questions as you read new material in the textbook, trying to solve the Sample Problems first before reading the provided Solutions, working the Study Checks and Practice Problems and checking the Answers, self-testing during and after reading each Section, retesting on new information a few days later, going to the instructor’s office hours, and keeping a problem notebook. 1.12 Many things make it difficult to learn chemistry, including not going to class regularly, not working Sample Problems and Study Checks, not reading the assignment ahead of class, not going to the instructor’s office hours, and waiting until the night before an exam to study. 1.13 Ways you can enhance your learning of chemistry include: a. forming a study group. c. asking yourself questions while reading the text. e. answering the Engage questions. 1.14 Ways you can enhance your learning of chemistry include: a. studying different topics at the same time. c. attending review sessions. d. working the problems again after a few days. e. keeping a problem notebook. 1.15 a. The bolded 8 is in the thousandths place. b. The bolded 6 is in the ones place. c. The bolded 6 is in the hundreds place. 1.16 a. The bolded 5 is in the tenths place. b. The bolded 7 is in the tens place. c. The bolded 0 is in the hundredths place. 1.17 a. 15 - (-8) = 15 + 8 = 23 b. -8 + (-22) = -30 c. 4 * (-2) + 6 = -8 + 6 = -2 1.18 a. -11 - (-9) = -11 + 9 = -2 b. 34 + (-55) = -21 -56 = -7 c. 8 1.19 a. The graph shows the relationship between the temperature of a cup of tea and time. b. The vertical axis measures temperature, in °C.

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Chemistry in Our Lives

c. The values on the vertical axis range from 20 °C to 80 °C. d. As time increases, the temperature decreases. 1.20 a. The horizontal axis measures time, in minutes. b. The values on the horizontal axis range from 0 min to 100 min. c. After 20 min, the temperature of the tea is about 56 °C. d. About 38 min were required for the tea to reach a temperature of 45 °C. 4a + 4 = 40 1.21 a. 4a + 4 - 4 = 40 - 4 4a = 36 4a 36 = 4 4 a = 9 a b. = 7 6 a 6 a b = 6(7) 6 a = 42 1.22 a. 2b + 7 = b + 10 2b + 7 - 7 = b + 10 - 7 2b = b + 3 2b - b = b - b + 3 b = 3 b. 3b - 4 = 24 - b 3b - 4 + 4 = 24 - b + 4 3b = 28 - b 3b + b = 28 - b + b 4b = 28 4b 28 = 4 4 b = 7 21 flu shots * 100, = 84, received flu shots 25 patients b. total grams of alloy = 56 g silver + 22 g copper = 78 g of alloy 56 g silver * 100, = 72, silver 78 g alloy c. total number of coins = 11 nickels + 5 quarters + 7 dimes = 23 coins 7 dimes * 100, = 30, dimes 23 coins

1.23 a.

22 boys * 100, = 63, boys 35 babies b. total grams of alloy = 67 g gold + 35 g zinc = 102 g of alloy 35 g zinc * 100, = 34, zinc 102 g alloy c. total number of coins = 15 pennies + 14 dimes + 6 quarters = 35 coins 15 pennies * 100, = 43, pennies 35 coins

1.24 a.

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Chapter 1

c. The values on the vertical axis range from 20 °C to 40 °C. d. As time increases, the temperature decreases. 1.26 a. The horizontal axis measures time, in hours, since death. b. The values on the horizontal axis range from 0 h to 25 h. c. About 15 hours were needed to reach a body temperature of 28 °C. d. Since it takes about 5 hours to reach a body temperature of 34 °C, the time of death is estimated to be (9 p.m. – 5 h =) 4 p.m. 1.27 a. Move the decimal point four places to the left to give 5.5 * 104. b. Move the decimal point two places to the left to give 4.8 * 102. c. Move the decimal point six places to the right to give 5 * 10-6. d. Move the decimal point four places to the right to give 1.4 * 10-4. e. Move the decimal point three places to the right to give 7.2 * 10-3. f. Move the decimal point five places to the left to give 6.7 * 105. 1.28 a. Move the decimal point eight places to the left to give 1.8 * 108. b. Move the decimal point five places to the right to give 6 * 10-5. c. Move the decimal point two places to the left to give 7.5 * 102. d. Move the decimal point one place to the right to give 1.5 * 10-1. e. Move the decimal point two places to the right to give 2.4 * 10-2. f. Move the decimal point three places to the left to give 1.5 * 103. 1.29 a. 7.2 * 103, which is also 7200, is larger than 8.2 * 102 or 820. b. 3.2 * 10-2, which is also 0.032, is larger than 4.5 * 10-4 or 0.000 45. c. 1 * 104, which is also 10 000, is larger than 1 * 10-4 or 0.0001. d. 6.8 * 10-2, which is also 0.068, is larger than 0.000 52. 1.30 a. 5.5 * 10-9, which is also 0.000 000 005 5, is smaller than 4.9 * 10-3 or 0.0049. b. 3.4 * 102, which is also 340, is smaller than 1250. c. 0.000 000 4 is smaller than 5.0 * 102 or 500. d. 2.50 * 102, which is also 250, is smaller than 4 * 105 or 400 000. 1.31 a. hypothesis

b. conclusion

c. experiment

d. observation

1.32 a. hypothesis

b. experiment

c. observation

d. conclusion

1.33

120 g ethylene glycol * 100, = 27, ethylene glycol 450 g liquid

1.34

1.5 g ethylene glycol * 100, = 0.15, ethylene glycol 1000 g body mass

1.35 No. All of these ingredients are chemicals. 1.36 No. All of these ingredients are chemicals. 1.37 Yes. Sherlock’s investigation includes making observations (gathering data), formulating a hypothesis, testing the hypothesis, and modifying it until one of the hypotheses is validated. 1.38 Holmes stresses the important first step of the scientific method: making observations and collecting data. 1.39 a. When two negative numbers are added, the answer has a negative sign. b. When a positive and negative number are multiplied, the answer has a negative sign. 1.40 a. When a negative number is subtracted from a positive number, the answer has a positive sign. b. When two negative numbers are divided, the answer has a positive sign. 1.41 a. Describing the appearance of a patient is an observation. b. Formulating a reason for the extinction of dinosaurs is a hypothesis. c. Measuring a patient’s blood pressure is an observation.

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Chemistry in Our Lives

1.42 a. Measuring the composition of a sample is an observation. b. Recording a change in a sample is an observation. c. Formulating a reason as to why a phenomenon has happened is a hypothesis. 1.43 If experimental results do not support your hypothesis, you should: b. modify your hypothesis. c. do more experiments. 1.44 A hypothesis is confirmed when: b. many experiments validate the hypothesis. 1.45 A successful study plan would include: b. working the Sample Problems as you go through a chapter. c. self-testing. 1.46 A successful study plan would include: b. forming a study group and discussing the problems together. c. working problems in a notebook for easy reference. 1.47 a. 4 * (-8) = -32 b. -12 - 48 = -12 + (-48) = -60 -168 c. = 42 -4 1.48 a. -95 - (-11) = -95 + 11 = -84 152 b. = -8 -19 c. 4 - 56 = 4 + (-56) = -52 1.49 total number of gumdrops = 16 orange + 8 yellow + 16 black = 40 gumdrops 8 yellow gumdrops a. * 100, = 20, yellow gumdrops 40 total gumdrops 16 black gumdrops b. * 100, = 40, black gumdrops 40 total gumdrops 1.50 total number of students = 12 As + 18 Bs + 20 Cs = 50 students 18 Bs * 100, = 36, Bs a. 50 total students 20 Cs * 100, = 40, Cs b. 50 total students 1.51 a. Move the decimal point five places to the left to give 1.2 * 105. b. Move the decimal point seven places to the right to give 3.4 * 10-7. c. Move the decimal point two places to the right to give 6.6 * 10-2. d. Move the decimal point three places to the left to give 2.7 * 103. 1.52 a. Move the decimal point three places to the right to give 4.2 * 10-3. b. Move the decimal point two places to the left to give 3.1 * 102. c. Move the decimal point eight places to the left to give 8.9 * 108. d. Move the decimal point eight places to the right to give 5.6 * 10-8. 1.53 a. observation

b. hypothesis

c. conclusion

1.54 a. observation

b. experiment

c. conclusion

1.55 a. Self-testing allows you to check on what you understand. b. Forming a study group can motivate you to study, fill in gaps, and correct misunderstandings by teaching and learning together. c. Reading the assignment before class prepares you to learn new material. Copyright © 2019 Pearson Education, Inc.

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Chapter 1

1.56 a. Studying only the night before an exam does not allow you time to make connections between new and previously learned information for longer lasting memory and more efficient retrieval of the information. b. Not going to class does not allow you to interact with your professor and peers. Class can give you another perspective on the material; the professor may go over examples or applications not covered in the textbook, and student questions and discussion may elaborate on the material or provide new insights. c. Not practicing the problems in the text does not allow you to apply problem solving to the new concepts you are learning. 1.57 a. observation

b. hypothesis

c. experiment

d. conclusion

1.58 a. observation

b. hypothesis

c. observation

d. observation

2x + 5 = 41 1.59 a. 2x + 5 - 5 = 41 - 5 2x = 36 2x 36 = 2 2 x = 18 5x = 40 b. 3 5x 3 a b = 3(40) 3 5x = 120 5x 120 = 5 5 x = 24 1.60 a. 3z - (-6) = 12 3z + 6 = 12 3z + 6 - 6 = 12 - 6 3z = 6 3z 6 = 3 3 z = 2 4z b. = -8 -12 4z -12 a b = -12(-8) -12 4z = 96 4z 96 = 4 4 z = 24 1.61 a. The graph shows the relationship between the solubility of carbon dioxide in water and temperature. b. The vertical axis measures the solubility of carbon dioxide in water (g CO2/100 g water). c. The values on the vertical axis range from 0 to 0.35 g CO2/100 g water. d. As temperature increases, the solubility of carbon dioxide in water decreases. 1.62 a. The horizontal axis measures temperature, in °C. b. The values on the horizontal axis range from 0 °C to 60 °C. c. At 25 °C, the solubility of carbon dioxide in water is about 0.17 g CO2/100 g water. d. Carbon dioxide has a solubility of 0.20 g CO2/100 g water at a temperature of about 16 °C.

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2 Chemistry and Measurements Learning Goals • Write the names and abbreviations for the metric and SI units used in measurements of volume, length, mass, temperature, and time. • Identify a number as measured or exact; determine the number of significant figures in a measured number. • Give the correct number of significant figures for a calculated answer. • Use the numerical values of prefixes to write a metric equality. • Write a conversion factor for two units that describe the same quantity. • Use conversion factors to change from one unit to another. • Calculate the density of a substance; use the density to calculate the mass or volume of a substance.

Chapter Outline Chapter Opener: Registered Nurse Units of Measurement Measured Numbers and Significant Figures Significant Figures in Calculations Prefixes and Equalities Writing Conversion Factors Problem Solving Using Unit Conversion Chemistry Link to Health: Toxicology and Risk–Benefit Assessment Density Chemistry Link to Health: Bone Density Clinical Update: Greg’s Visit with His Doctor

2.1 2.2 2.3 2.4 2.5 2.6 2.7

Key Math Skill • Rounding Off (2.3)

Core Chemistry Skills • • • • • •

Counting Significant Figures (2.2) Using Significant Figures in Calculations (2.3) Using Prefixes (2.4) Writing Conversion Factors from Equalities (2.5) Using Conversion Factors (2.6) Using Density as a Conversion Factor (2.7)

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Chapter 2

Answers and Solutions to Text Problems 2.1

a. The abbreviation for the unit gram is g. b. The abbreviation for the unit degree Celsius is °C. c. The abbreviation for the unit liter is L. d. The abbreviation for the unit pound is lb. e. The abbreviation for the unit second is s.

2.2

a. The abbreviation for the unit kilogram is kg. b. The abbreviation for the unit kelvin is K. c. The abbreviation for the unit quart is qt. d. The abbreviation for the unit meter is m. e. The abbreviation for the unit centimeter is cm.

2.3

a. A liter is a unit of volume. b. A centimeter is a unit of length. c. A kilometer is a unit of length. d. A second is a unit of time.

2.4

a. A kilometer is a unit of length. b. A kilogram is a unit of mass. c. A degree Celsius is a unit of temperature. d. A liter is a unit of volume.

2.5

a. The unit is a meter, which is a unit of length. b. The unit is a gram, which is a unit of mass. c. The unit is a milliliter, which is a unit of volume. d. The unit is a second, which is a unit of time. e. The unit is a degree Celsius, which is a unit of temperature.

2.6

a. The unit is a liter, which is a unit of volume. b. The unit is a centimeter, which is a unit of length. c. The unit is a kilogram, which is a unit of mass. d. The unit is an hour, which is a unit of time. e. The unit is a kelvin, which is a unit of temperature.

2.7

a. The unit is a second, which is a unit of time. b. The unit is a kilogram, which is a unit of mass. c. The unit is a gram, which is a unit of mass. d. The unit is a degree Celsius, which is a unit of temperature.

2.8

a. The unit is a degree Celsius, which is a unit of temperature. b. The unit is a second, which is a unit of time. c. The unit is a milliliter, which is a unit of volume. d. The unit is a gram, which is a unit of mass.

2.9

a. All five numbers are significant figures (5 SFs). b. Only the two nonzero numbers are significant (2 SFs); the preceding zeros are placeholders. c. Only the two nonzero numbers are significant (2 SFs); the zeros that follow are placeholders. d. All three numbers in the coefficient of a number written in scientific notation are significant (3 SFs). e. All four numbers to the right of the decimal point, including the last zero in the decimal number, are significant (4 SFs). f. All three numbers, including the zeros at the end of a decimal number, are significant (3 SFs).

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Chemistry and Measurements

2.10 a. All four numbers, including the last zero, in a decimal number, are significant (4 SFs). b. All six numbers are significant figures (6 SFs). c. All three numbers, including the zeros at the end of a decimal number, are significant (3 SFs). d. All three numbers are significant figures (3 SFs). e. There are three significant figures (3 SFs), since the zero between the two nonzero numbers is significant; the zeros that follow are placeholders only. f. Both of the numbers in the coefficient of a number written in scientific notation are significant (2 SFs). 2.11 Both measurements in part b have three significant figures, and both measurements in part c have two significant figures. 2.12 Both measurements in part a have three significant figures. 2.13 a. Zeros at the beginning of a decimal number are not significant. b. Zeros between nonzero digits are significant. c. Zeros at the end of a decimal number are significant. d. Zeros in the coefficient of a number written in scientific notation are significant. e. Zeros used as placeholders in a large number without a decimal point are not significant. 2.14 a. Zeros between nonzero digits are significant. b. Zeros at the end of a decimal number are significant. c. Zeros at the beginning of a decimal number are not significant. d. Zeros used as placeholders in a large number without a decimal point are not significant. e. Zeros in the coefficient of a number written in scientific notation are significant. 2.15 a. 5000 L is the same as 5 * 1000 L, which is written in scientific notation as 5.0 * 103 L with two significant figures. b. 30 000 g is the same as 3 * 10 000 g, which is written in scientific notation as 3.0 * 104 g with two significant figures. c. 100 000 m is the same as 1 * 100 000 m, which is written in scientific notation as 1.0 * 105 m with two significant figures. 1 d. 0.000 25 cm is the same as 2.5 * cm, which is written in scientific notation as 10 000 2.5 * 10-4 cm. 2.16 a. 5 100 000 g is the same as 5.1 * 1 000 000 g, which is written in scientific notation as 5.1 * 106 g. b. 26 000 s is the same as 2.6 * 10 000 s, which is written in scientific notation as 2.6 * 104 s. c. 40 000 m is the same as 4 * 10 000 m, which is written in scientific notation as 4.0 * 104 m with two significant figures. 1 d. 0.000 820 kg is the same as 8.2 * kg, which is written in scientific notation as 10 000 -4 8.2 * 10 kg. 2.17 Measured numbers are obtained using some type of measuring device. Exact numbers are numbers obtained by counting items or using a definition that compares two units in the same measuring system. a. The value 67.5 kg is a measured number; measurement of mass requires a measuring device. b. The value 2 tablets is obtained by counting, making it an exact number. c. The values in the metric definition 1 L = 1000 mL are exact numbers. d. The value 1720 km is a measured number; measurement of distance requires a measuring device.

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Chapter 2

2.18 Measured numbers are obtained using some type of measuring device. Exact numbers are numbers obtained by counting items or using a definition that compares two units in the same measuring system. a. The value 31 students is obtained by counting, making it an exact number. b. The value 1.20 * 108 yr is a measured number; measurement of time requires a measuring device. c. The value 104 kg is a measured number; measurement of mass requires a measuring device. d. The value 184 mg/dL is a measured number; measurement of mass and volume requires measuring devices. 2.19 Measured numbers are obtained using some type of measuring device. Exact numbers are numbers obtained by counting items or using a definition that compares two units in the same measuring system. a. 6 oz of hamburger meat is a measured number (3 hamburgers is a counted/exact number). b. Neither are measured numbers (both 1 table and 4 chairs are counted/exact numbers). c. Both 0.75 lb of grapes and 350 g of butter are measured numbers. d. Neither are measured numbers (the values in a definition are exact numbers). 2.20 Measured numbers are obtained using some type of measuring device. Exact numbers are numbers obtained by counting items or using a definition that compares two units in the same measuring system. a. 5 pizzas is a counted/exact number (50.0 g of cheese is a measured number). b. 6 nickels is a counted/exact number (16 g of nickel is a measured number). c. 3 onions is a counted/exact number (3 lb of onions is a measured number). d. 5 cars is a counted/exact number (5 miles is a measured number). 2.21 a. 1.607 kg is a measured number and has 4 SFs. b. 130 mcg is a measured number and has 2 SFs. c. 4.02 * 106 red blood cells is a measured number and has 3 SFs. d. 23 babies is a counted/exact number. 2.22 a. 103.5 °F is a measured number and has 4 SFs. b. 21 tablets is a counted/exact number. c. 0.46 s is a measured number and has 2 SFs. d. 1.20 * 1010 neurons is a measured number and has 3 SFs. 2.23 a. 1.85 kg; the last digit is dropped since it is 4 or less. b. 88.2 L; since the fourth digit is 4 or less, the last three digits are dropped. c. 0.004 74 cm; since the fourth significant digit (the first digit to be dropped) is 5 or greater, the last retained digit is increased by 1 when the last four digits are dropped. d. 8810 m; since the fourth significant digit (the first digit to be dropped) is 5 or greater, the last retained digit is increased by 1 when the last digit is dropped (a nonsignificant zero is added at the end as a placeholder). e. 1.83 * 105 s; since the fourth digit is 4 or less, the last digit is dropped. The * 105 is retained so that the magnitude of the answer is not changed. 2.24 a. 1.9 kg; since the third significant digit (the first digit to be dropped) is 5 or greater, the last retained digit is increased by 1 when the last two digits are dropped. b. 88 L; since the third significant digit is 4 or less, the last four digits are dropped. c. 0.0047 cm; since the third significant digit is 4 or less, the last five digits are dropped. d. 8800 m; since the third digit is 4 or less, the last digit is changed to a zero as a placeholder. e. 1.8 * 105 s; since the third digit is 4 or less, the last two digits are dropped. The * 105 is retained so that the magnitude of the answer is not changed.

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Chemistry and Measurements

2.25 a. To round off 56.855 m to three significant figures, drop the final digits 55 and increase the last retained digit by 1 to give 56.9 m. b. To round off 0.002 282 g to three significant figures, drop the final digit 2 to give 0.002 28 g. c. To round off 11 527 s to three significant figures, drop the final digits 27 and add two zeros as placeholders to give 11 500 s (1.15 * 104 s). d. To express 8.1 L to three significant figures, add a significant zero to give 8.10 L. 2.26 a. To round off 3.2805 m to two significant figures, drop the final digits 805 and increase the last retained digit by 1 to give 3.3 m. b. To round off 1.855 * 102 g to two significant figures, drop the final digits 55 and increase the last retained digit by 1 to give 1.9 * 102 g. c. To round off 0.002 341 mL to two significant figures, drop the final digits 41 to give 0.0023 mL. d. To express 2 L to two significant figures, add a significant zero to give 2.0 L. 2.27 a. 45.7 * 0.034 = 1.6 Two significant figures are allowed since 0.034 has 2 SFs. b. 0.002 78 * 5 = 0.01 One significant figure is allowed since 5 has 1 SF. 34.56 c. = 27.6 Three significant figures are allowed since 1.25 has 3 SFs. 1.25 (0.2465)(25) d. = 3.5 Two significant figures are allowed since 25 has 2 SFs. 1.78 e. (2.8 * 104)(5.05 * 10-6) = 0.14 or 1.4 * 10-1 Two significant figures are allowed since 2.8 * 104 has 2 SFs. (3.45 * 10-2)(1.8 * 105) f. = 0.8 or 8 * 10-1 One significant figure is allowed since (8 * 103) 8 * 103 has 1 SF. 2.28 a. 400 * 185 = 7 * 104 One significant figure is allowed since 400 has 1 SF. 2.40 b. = 0.005 or 5 * 10-3 One significant figure is allowed since 4 has 1 SF. (4)(125) c. 0.825 * 3.6 * 5.1 = 15 Two significant figures are allowed since 3.6 and 5.1 both have 2 SFs. (3.5)(0.261) = 0.0055 or 5.5 * 10-3 Two significant figures are allowed since 3.5 has 2 SFs. d. (8.24)(20.0) (5 * 10-5)(1.05 * 104) e. = 6 * 106 One significant figure is allowed since 5 * 10-5 (8.24 * 10-8) has 1 SF. (4.25 * 102)(2.56 * 10-3) f. = 8.58 Three significant figures are allowed since 56.5 (2.245 * 10-3)(56.5) (and others) has 3 SFs. 2.29 a. 45.48 + 8.057 = 53.54 Two decimal places are allowed since 45.48 has two decimal places. b. 23.45 + 104.1 + 0.025 = 127.6 One decimal place is allowed since 104.1 has one decimal place. c. 145.675 - 24.2 = 121.5 One decimal place is allowed since 24.2 has one decimal place. d. 1.08 - 0.585 = 0.50 Two decimal places are allowed since 1.08 has two decimal places. e. 2300 + 196.11 = 2500 Answer is rounded off to the hundreds place since 2300 is expressed to the hundreds place. f. 145.111 - 22.9 + 34.49 = 156.7 One decimal place is allowed since 22.9 has one decimal place.

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Chapter 2

2.30 a. 5.08 + 25.1 = 30.2 One decimal place is allowed since 25.1 has one decimal place. b. 85.66 + 104.10 + 0.025 = 189.79 Two decimal places are allowed since 85.66 and 104.10 have two decimal places. c. 24.568 - 14.25 = 10.32 Two decimal places are allowed since 14.25 has two decimal places. d. 0.2654 - 0.2585 = 0.0069 Four decimal places are allowed since both numbers have four decimal places. e. 66.77 + 17 - 0.33 = 83 Answer is rounded off to the ones place since 17 is expressed to the ones place. f. 460 - 33.77 = 430 Answer is rounded off to the tens place since 460 is expressed to the tens place. 2.31 a. mg

b. dL

c. km

d. pg

2.32 a. Gg

b. Mm

c. mL or mcL

d. fs

2.33 a. centiliter

b. kilogram

c. millisecond

d. petameter

2.34 a. deciliter

b. terasecond

c. microgram

d. picometer

2.35 a. 0.01 c. 0.001 (or 1 * 10-3)

b. 1 000 000 000 000 (or 1 * 1012) d. 0.1

2.36 a. 1 000 000 000 (or 1 * 109) c. 1 000 000 (or 1 * 106)

b. 0.000 001 (or 1 * 10-6) d. 0.000 000 001 (or 1 * 10-9)

2.37 a. decigram

b. microgram

c. kilogram

d. centigram

2.38 a. gigameter

b. megameter

c. millimeter

d. picometer

2.39 a. 1 m = 100 cm c. 1 mm = 0.001 m

b. 1 m = 1 * 109 nm d. 1 L = 1000 mL

2.40 a. 1 Mg = 1 * 106 g c. 1 g = 0.001 kg

b. 1 mL = 1000 mL d. 1 g = 1000 mg

2.41 a. kilogram, since 103 g is greater than 10-3 g b. milliliter, since 10-3 L is greater than 10-6 L c. km, since 103 m is greater than 100 m d. kL, since 103 L is greater than 10-1 L e. nanometer, since 10-9 m is greater than 10-12 m 2.42 a. mg, since 10-3 g is smaller than 100 g b. nanometer, since 10-9 m is smaller than 10-2 m c. micrometer, since 10-6 m is smaller than 10-3 m d. mL, since 10-3 L is smaller than 10-1 L e. centigram, since 10-2 g is smaller than 106 g

1m 100 cm and 100 cm 1m 2.44 You can check that you have written the correct conversion factors for an equality by looking at the number and units in each; the information on one side of the equality appears in the numerator of a conversion factor while the information on the other side of the equality appears in the denominator. To generate a second conversion factor, we invert the first conversion factor so the numbers and units in the numerator and denominator switch places. 2.43 A conversion factor can be inverted to give a second conversion factor:

100 cm 1m and 1m 100 cm 1 * 109 ng 1g and b. 1 g = 1 * 109 ng; 1g 1 * 109 ng

2.45 a. 1 m = 100 cm;

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Chemistry and Measurements

1000 L 1 kL and 1 kL 1000 L 1000 ms 1s and d. 1 s = 1000 ms; 1s 1000 ms 100 mm 1 dm e. 1 dm = 100 mm; and 1 dm 100 mm 2.54 cm 1 in. 2.46 a. 1 in. = 2.54 cm; and 1 in. 2.54 cm 0.621 mi 1 km b. 1 km = 0.621 mi; and 1 km 0.621 mi 454 g 1 lb and c. 1 lb = 454 g; 1 lb 454 g 10 dL 1L d. 1 L = 10 dL; and 1L 10 dL 12 1 * 10 pg 1g and e. 1 g = 1 * 1012 pg; 1g 1 * 1012 pg c. 1 kL = 1000 L;

1 yd 3 ft and ; the 1 yd and 3 ft are both exact (U.S. definition). 1 yd 3 ft 1 kg 2.20 lb b. 1 kg = 2.20 lb; and ; the 2.20 lb is measured: it has 3 SFs; the 1 kg is exact. 1 kg 2.20 lb 1 gal gasoline 27 mi c. 1 gal of gasoline = 27 mi; and ; the 27 mi is measured: 1 gal gasoline 27 mi it has 2 SFs; the 1 gal is exact. 93 g silver 100 g sterling d. 100 g of sterling = 93 g of silver; and ; the 93 g is measured: 100 g sterling 93 g silver it has 2 SFs; the 100 g is exact. 60 sec 1 min and ; the 1 min and 60 sec are both exact. e. 1 min = 60 sec; 1 min 60 sec 2.47 a. 1 yd = 3 ft;

2.48 a. 1 L = 1.06 qt;

1.06 qt 1L and ; the 1.06 qt is measured: it has 3 SFs; the 1 L is exact. 1L 1.06 qt

b. 1 lb of oranges = +1.29;

1 lb oranges +1.29 and ; the $1.29 is measured: it has 1 lb oranges +1.29

3 SFs; the 1 lb is exact. 7 days 1 week c. 1 week = 7 days; and ; the 1 week and 7 days are both exact. 1 week 7 days

100 mL 1 dL and ; the 1 dL and 100 mL are both exact 1 dL 100 mL (metric definition). 75 g gold 100 g gold ring and ; the 75 g is e. 100 g of gold ring = 75 g of gold; 100 g gold ring 75 g gold measured: it has 2 SFs; the 100 g is exact. d. 1 dL = 100 mL;

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Chapter 2

3.5 m 1s and ; the 3.5 m is measured: it has 2 SFs; the 1 s is exact. 1s 3.5 m 3.5 g potassium 1 day b. 1 day = 3.5 g of potassium; and ; the 3.5 g is measured: 1 day 3.5 g potassium it has 2 SFs; the 1 day is exact. 1 L gasoline 26.0 km c. 1 L of gasoline = 26.0 km; and ; the 26.0 km is measured: 1 L gasoline 26.0 km it has 3 SFs; the 1 L is exact. 29 mcg pesticide 1 kg plums d. 1 kg of plums = 29 mcg of pesticide; and ; the 29 mcg 1 kg plums 29 mcg pesticide is measured: it has 2 SFs; the 1 kg is exact. 28.2 g silicon 100 g crust e. 100 g of crust = 28.2 g of silicon; and ; the 28.2 g is measured: 100 g crust 28.2 g silicon it has 3 SFs; the 100 g is exact. 2.49 a. 1 s = 3.5 m;

150 mcg iodine 1 day and ; the 150 mcg is measured: 1 day 150 mcg iodine it has 2 SFs; the 1 day is exact. 32 mg nitrate 1 kg water and ; the 32 mg is measured: b. 1 kg of water = 32 mg of nitrate; 1 kg water 32 mg nitrate it has 2 SFs; the 1 kg is exact. 58 g gold 100 g jewelry c. 100 g of jewelry = 58 g of gold; and ; the 58 g is measured: 100 g jewelry 58 g gold it has 2 SFs; the 100 g is exact. +1.65 1 L milk and ; the $1.65 is measured: it has 3 SFs; the 1 L d. 1 L of milk = +1.65; 1 L milk +1.65 is exact. 1000 kg 1 metric ton and ; the 1 metric ton and e. 1 metric ton = 1000 kg; 1 metric ton 1000 kg 1000 kg are both exact (metric definition). 2.50 a. 1 day = 150 mcg of iodine;

630 mg calcium 1 tablet and ; the 630 mg is 1 tablet 630 mg calcium measured: it has 2 SFs; the 1 tablet is exact. 60 mg vitamin C 1 day and ; the 60 mg is b. 1 day = 60 mg of vitamin C; 1 day 60 mg vitamin C measured: it has 1 SF; the 1 day is exact. 50 mg atenolol 1 tablet and ; the 50 mg is c. 1 tablet = 50 mg of atenolol; 1 tablet 50 mg atenolol measured: it has 1 SF; the 1 tablet is exact. 81 mg aspirin 1 tablet d. 1 tablet = 81 mg of aspirin; and ; the 81 mg is measured: 1 tablet 81 mg aspirin it has 2 SFs; the 1 tablet is exact. 2.51 a. 1 tablet = 630 mg of calcium;

10 mg furosemide 1 mL and ; the 10 mg is 1 mL 10 mg furosemide measured: it has 1 SF; the 1 mL is exact. 70. mcg selenium 1 day and ; the 70. mcg is b. 1 day = 70. mcg of selenium; 1 day 70. mcg selenium measured: it has 2 SFs; the 1 day is exact. 2.52 a. 1 mL = 10 mg of furosemide;

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Chemistry and Measurements

85 mL saline solution 1 hour and ; 1 hour 85 mL saline solution the 85 mL is measured: it has 2 SFs; the 1 hour is exact. 360 mg omega@3 fatty acids and d. 1 capsule = 360 mg of omega-3 fatty acids; 1 capsule 1 capsule ; the 360 mg is measured: it has 2 SFs; the 1 capsule is exact. 360 mg omega@3 fatty acids c. 1 hour = 85 mL of saline solution;

10 mg Atarax 5 mL syrup and 5 mL syrup 10 mg Atarax 0.25 g Lanoxin 1 tablet b. 1 tablet = 0.25 g of Lanoxin; and 1 tablet 0.25 g Lanoxin 300 mg Motrin 1 tablet c. 1 tablet = 300 mg of Motrin; and 1 tablet 300 mg Motrin

2.53 a. 5 mL of syrup = 10 mg of Atarax;

2.5 mg Coumadin 1 tablet and 1 tablet 2.5 mg Coumadin 100 mg Clozapine 1 tablet b. 1 tablet = 100 mg of Clozapine; and 1 tablet 100 mg Clozapine 1.5 g Cefuroxime 1 mL solution c. 1 mL of solution = 1.5 g of Cefuroxime; and 1 mL solution 1.5 g Cefuroxime

2.54 a. 1 tablet = 2.5 mg of Coumadin;

2.55 a. Given

44.2 mL Need liters 1L Plan mL S L 1000 mL 1L Set-up 44.2 mL * = 0.0442 L (3 SFs) 1000 mL b. Given 8.65 m Need nanometers 1 * 109 nm Plan m S nm 1m 1 * 109 nm Set-up 8.65 m * = 8.65 * 109 nm (3 SFs) 1m c. Given 5.2 * 108 g Need megagrams 1 Mg

Plan

g S Mg

2.56 a. Given

4.82 * 10-5 L Need

1 * 106 g 1 Mg Set-up 5.2 * 108 g * = 5.2 * 102 Mg (2 SFs) 1 * 106 g d. Given 0.72 ks Need milliseconds 1000 s 1000 ms Plan ks S s S ms 1 ks 1s 1000 s 1000 ms Set-up 0.72 ks * * = 7.2 * 105 ms (2 SFs) 1 ks 1s picoliters

Plan Set-up

1 * 10 pL 1L 1 * 1012 pL 4.82 * 10-5 L * = 4.82 * 107 pL (3 SFs) 1L

L S pL

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Chapter 2

b. Given 575.2 dm Need kilometers 1m 1 km Plan dm S m S km 10 dm 1000 m 1m 1 km * = 0.057 52 km (4 SFs) Set-up 575.2 dm * 10 dm 1000 m c. Given 5 * 10-4 kg Need micrograms 1000 g 1 * 106 mg Plan kg S g S mg 1 kg 1g 1000 g 1 * 106 mg * = 5 * 105 mg (1 SF) Set-up 5 * 10-4 kg * 1 kg 1g d. Given 6.4 * 1010 ps Need seconds 1s Plan ps S s 1 * 1012 ps 1s Set-up 6.4 * 1010 ps * = 0.064 s (2 SFs) 1 * 1012 ps 2.57 a. Given

3.428 lb Need kilograms 1 kg Plan lb S kg 2.20 lb 1 kg = 1.56 kg (3 SFs) Set-up 3.428 lb * 2.20 lb b. Given 1.6 m Need inches 39.4 in. Plan m S in. 1m 39.4 in. = 63 in. (2 SFs) Set-up 1.6 m * 1m c. Given 4.2 L Need quarts 1.06 qt Plan L S qt 1L 1.06 qt = 4.5 qt (2 SFs) Set-up 4.2 L * 1L d. Given 0.672 ft Need millimeters 12 in. 2.54 cm 10 mm Plan ft S in. S cm S mm 1 ft 1 in. 1 cm 12 in. 2.54 cm 10 mm * * = 205 mm (3 SFs) Set-up 0.672 ft * 1 ft 1 in. 1 cm 2.58 a. Given 0.21 lb Need grams 454 g Plan lb S g 1 lb 454 g = 95 g (2 SFs) Set-up 0.21 lb * 1 lb b. Given 11.6 in. Need centimeters 2.54 cm Plan in. S cm 1 in. 2.54 cm = 29.5 cm (3 SFs) Set-up 11.6 in. * 1 in.

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Chemistry and Measurements

c. Given 0.15 qt Need milliliters 946 mL Plan qt S mL 1 qt 946 mL = 140 mL (2 SFs) Set-up 0.15 qt * 1 qt d. Given 35.41 kg Need pounds 2.20 lb Plan kg S lb 1 kg 2.20 lb = 77.9 lb (3 SFs) Set-up 35.41 kg * 1 kg 2.59 a. Given 175 cm Need meters 1m Plan cm S m 100 cm 1m = 1.75 m (3 SFs) Set-up 175 cm * 100 cm b. Given 5000 mL Need liters 1L Plan mL S L 1000 mL 1L = 5 L (1 SF) Set-up 5000 mL * 1000 mL c. Given 0.0055 kg Need grams 1000 g Plan kg S g 1 kg 1000 g Set-up 0.0055 kg * = 5.5 g (2 SFs) 1 kg d. Given 3500 cm3 Need liters 1 mL 1L Plan cm3 S mL S L 3 1000 mL 1 cm 1 mL 1L Set-up 3500 cm3 * * = 3.5 L (2 SFs) 3 1000 mL 1 cm 2.60 a. Given 800 mg Need grams 1g Plan mg S g 1000 mg 1g = 0.8 g (1 SF) Set-up 800 mg * 1000 mg b. Given 3.2 dL Need milliliters 100 mL Plan dL S mL 1 dL 100 mL = 320 mL (2 SFs) Set-up 3.2 dL * 1 dL c. Given 2840 mg Need grams 1g Plan mg S g 1000 mg 1g = 2.84 g (3 SFs) Set-up 2840 mg * 1000 mg

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Chapter 2

d. Given

0.29 kg Need grams 1000 g Plan kg S g 1 kg 1000 g Set-up 0.29 kg * = 290 g (2 SFs) 1 kg 2.61 a. Given

0.500 qt Need milliliters 946 mL Plan qt S mL 1 qt 946 mL = 473 mL (3 SFs) Set-up 0.500 qt * 1 qt b. Given 175 lb Need kilograms 1 kg Plan lb S kg 2.20 lb 1 kg Set-up 175 lb * = 79.5 kg (3 SFs) 2.20 lb c. Given 74 kg body mass, 15, body fat Need pounds of body fat Plan kg of body mass S kg of body fat S lb of body fat (percent equality: 100 kg of body mass = 15 g of body fat) 15 kg body fat 2.20 lb body fat 100 kg body mass 1 kg body fat 15 kg body fat 2.20 lb body fat Set-up 74 kg body mass * * 100 kg body mass 1 kg body fat = 24 lb of body fat (2 SFs) d. Given 10.0 oz of fertilizer, 15, nitrogen Need grams of nitrogen Plan oz of fertilizer S lb of fertilizer S g of fertilizer S g of nitrogen (percent equality: 100 g of fertilizer = 15 g of nitrogen) 1 lb 454 g 15 g nitrogen 16 oz 1 lb 100 g fertilizer Set-up

454 g fertilizer 15 g nitrogen 1 lb fertilizer * * 16 oz fertilizer 1 lb fertilizer 100 g fertilizer = 43 g of nitrogen (2 SFs)

10.0 oz fertilizer *

2.62 a. Given 0.750 L of wine, 12, alcohol Need milliliters of alcohol Plan L of wine S mL of wine S mL of alcohol (percent equality: 100 mL of wine = 12 mL of alcohol) 1000 mL wine 12 mL alcohol 1 L wine 100 mL wine 1000 mL wine 12 mL alcohol Set-up 0.750 L wine * * = 90. mL of alcohol (2 SFs) 1 L wine 100 mL wine b. Given 1 high-fiber muffin, 51, fiber, 6 muffins = 12 oz Need grams of fiber Plan number of muffins S oz S lb S g of muffin S g of fiber (percent equality: 100 g of fiber muffin = 51 g of fiber) 454 g 51 g fiber 12 oz 1 lb 6 muffins 16 oz 1 lb 100 g muffin 454 g 51 g fiber 12 oz 1 lb * * * = 29 g of fiber (2 SFs) 6 muffins 16 oz 1 lb 100 g muffin c. Given 1.43 kg of peanut butter, 8.0, of peanut butter/sandwich Need ounces of peanut butter/sandwich Set-up

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1 muffin *

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Chemistry and Measurements

g of peanut butter (pb) S lb of pb S oz of pb S oz of pb/sandwich k (percent equality: 100 oz of peanut butter = 8.0 oz of pb/sandwich) 2.20 lb 16 oz 8.0 oz pb/sandwich 1 kg 1 lb 100 oz pb 2.20 lb pb 16 oz pb 8.0 oz pb/sandwich Set-up 1.43 kg peanut butter * * * 1 kg pb 1 lb pb 100 oz pb = 4.0 oz of peanut butter/sandwich (2 SFs) d. Given 5.0 kg of pecans, 22.0, pecans Need pounds of chocolate bars Plan kg of pecans S kg of bars S lb of bars 100 kg choc. bars 2.20 lb (percent equality: 100 kg of bars = 22.0 kg of pecans) 22.0 kg pecans 1 kg 100 kg choc. bars 2.20 lb Set-up 5.0 kg pecans * * = 50. lb of chocolate bars (2 SFs) 22.0 kg pecans 1 kg Plan

2.63 a. Given

250 L of water Need gallons of water 1.06 qt 1 gal Plan L S qt S gal 1L 4 qt 1.06 qt 1 gal Set-up 250 L * * = 66 gal (2 SFs) 1L 4 qt b. Given 0.024 g of sulfa drug, 8-mg tablets Need number of tablets Plan g of sulfa drug S mg of sulfa drug S number of tablets 1000 mg 1 tablet 1g 8 mg sulfa drug 1000 mg 1 tablet Set-up 0.024 g sulfa drug * * = 3 tablets (1 SF) 1g 8 mg sulfa drug c. Given 34-lb child, 115 mg of ampicillin/kg of body mass Need milligrams of ampicillin Plan lb of body mass S kg of body mass S mg of ampicillin 1 kg 115 mg ampicillin 2.20 lb 1 kg body mass 1 kg body mass 115 mg ampicillin Set-up 34 lb body mass * * 2.20 lb body mass 1 kg body mass = 1800 mg of ampicillin (2 SFs) d. Given 4.0 oz of ointment Need grams of ointment 1 lb 454 g Plan oz S lb S g 16 oz 1 lb 454 g 1 lb * = 110 g of ointment (2 SFs) Set-up 4.0 oz * 16 oz 1 lb

2.64 a. Given 1 day, 1.0 g of tetracycline/6 h, 500-mg tablets Need number of tablets Plan days S hours S g of tetracycline S mg of tetracycline S number of tablets 1.0 g tetracycline 1000 mg 24 h 1 tablet 1 day 6h 1g 500 mg tetracycline Set-up 1.0 g tetracycline 1000 mg 24 h 1 tablet 1 day * * * * = 8 tablets (1 SF) 1 day 6h 1g 500 mg tetracycline b. Given 180-lb patient, 5.00 mg of medication/kg of body mass Need milligrams of medication Plan lb of body mass S kg of body mass S mg of medication 1 kg 5.00 mg medication 2.20 lb 1 kg body mass Copyright © 2019 Pearson Education, Inc.

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Chapter 2

1 kg body mass 5.00 mg medication * 2.20 lb body mass 1 kg body mass = 410 mg of medication (2 SFs) c. Given 0.50 mg of atropine, 0.10 mg of atropine/mL of solution Need milliliters of solution 1 mL solution Plan mg of atropine S mL of solution 0.10 mg atropine 1 mL solution Set-up 0.50 mg atropine * = 5.0 mL of solution (2 SFs) 0.10 mg atropine d. Given 5.0 pt of plasma Need milliliters of plasma 473 mL Plan pt S mL 1 pt 473 mL Set-up 5.0 pt * = 2400 mL of plasma (2 SFs) 1 pt Set-up

180 lb body mass *

2.65 a. Given 500. mL of IV saline solution, 80. mL/h Need infusion time in hours 1h Plan mL of IV saline solution S hours 80. mL saline solution 1h Set-up 500. mL saline solution * = 6.3 h (2 SFs) 80. mL saline solution b. Given 72.6-lb child, 1.5 mg of Medrol/kg of body mass, 20. mg of Medrol/mL of solution Need milliliters of Medrol solution Plan lb of body mass S kg of body mass S mg of Medrol S mL of solution 1 kg 1.5 mg Medrol 1 mL solution 2.20 lb 1 kg body mass 20. mg Medrol 1 kg body mass 1.5 mg Medrol 1 mL solution Set-up 72.6 lb body mass * * * 2.20 lb body mass 1 kg body mass 20. mg Medrol = 2.5 mL of Medrol solution (2 SFs) 2.66 a. Given o rdered dose 12.5 mg of promethazine, stock 25 mg of promethazine/mL of solution Need milliliters of promethazine solution 1 mL solution Plan mg of promethazine S mL of stock solution 25 mg promethazine 1 mL solution Set-up 12.5 mg promethazine * = 0.50 mL of stock solution (2 SFs) 25 mg promethazine b. Given 67-lb child, 25 mg of ampicillin/kg of body mass, 250 mg of ampicillin/capsule Need number of capsules Plan lb of body mass S kg of body mass S mg of ampicillin S number of capsules 1 kg body mass 25 mg ampicillin 1 capsule 2.20 lb body mass 1 kg body mass 250 mg ampicillin 1 kg body mass 25 mg ampicillin 1 capsule Set-up 67 lb body mass * * * 2.20 lb body mass 1 kg body mass 250 mg ampicillin = 3.0 capsules of ampicillin (2 SFs) mass (grams) 2.67 Density is the mass of a substance divided by its volume. Density = volume (mL) 3 The densities of solids and liquids are usually stated in g/mL or g/cm , so in some problems the units will need to be converted. mass (grams) 24.0 g = = 1.20 g/mL (3 SFs) a. Density = volume (mL) 20.0 mL b. Given 0.250 lb of butter, 130.3 mL Need density (g/mL)

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Chemistry and Measurements

454 g lb S g, then calculate density 1 lb 454 g Set-up 0.250 lb * = 113.5 g (3 SFs allowed) 1 lb 113.5 g mass 6 Density = = = 0.871 g/mL (3 SFs) volume 130.3 mL c. Given 12.00 mL initial volume, 13.45 mL final volume, 4.50 g Need density (g/mL) Plan calculate volume by difference, then calculate density Set-up volume of gem: 13.45 mL total - 12.00 mL water = 1.45 mL 4.50 g mass 6 Density = = = 3.10 g/mL (3 SFs) volume 1.45 mL 3.85 g mass d. Density = = = 1.28 g/mL (3 SFs) volume 3.00 mL Plan

155 g mass = = 1.24 g/mL (3 SFs) volume 125 mL 2.68 lb, 3.5 L Need density (g/mL)

2.68 a. Density = b. Given

454 g 1000 mL 1 lb 1L 454 g 1000 mL Set-up 2.68 lb * = 1220 g (3 SFs) and 3.5 L * = 3500 mL (2 SFs) 1 lb 1L 1220 g mass 6 Density = = = 0.35 g/mL (2 SFs) volume 3500 mL 4.004 g mass c. Density = = = 1.001 g/mL (4 SFs) volume 4.000 mL d. Given 1.65 lb, 170 mL Need density (g/mL) 454 g Plan lb S g, then calculate density 1 lb 454 g Set-up 1.65 lb * = 749 g (3 SFs) 1 lb 749 g mass 6 Density = = = 4.4 g/mL (2 SFs) volume 170 mL

Plan

lb S g

and

L S mL, then calculate density

2.69 a. Given 514.1 g, 114 cm3 Need density (g/mL) Plan convert volume cm3 S mL, then calculate density 1 mL Set-up 114 cm3 * = 114 mL 1 cm3 514.1 g mass 6 Density = = = 4.51 g/mL (3 SFs) volume 114 mL b. Given 0.100 pt, 115.25 g initial, 182.48 g final Need density (g/mL) Plan pt S mL, then calculate mass by difference, then calculate density 473 mL 1 pt 473 mL Set-up 0.100 pt * = 47.3 mL (3 SFs) 1 pt mass of syrup = 182.48 g - 115.25 g = 67.23 g 67.23 g mass 6 Density = = = 1.42 g/mL (3 SFs) volume 47.3 mL

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Chapter 2

c. Given 8.51 kg, 3.15 L Need Plan

kg S g

and

density (g/mL)

L S mL, then calculate density

1000 g 1 kg

1000 mL 1L

1000 g = 8510 g (3 SFs) and 1 kg 1000 mL 3.15 L * = 3150 mL (3 SFs) 1L 8510 g mass 6 Density = = = 2.70 g/mL (3 SFs) volume 3150 mL 2.70 a. Given 275 g, 207 cm3 Need density (g/mL) Plan convert volume cm3 S mL, then calculate density 1 mL Set-up 207 cm3 * = 207 mL 1 cm3 275 g mass = = 1.33 g/mL (3 SFs) 6 Density = volume 207 mL b. Given 0.104 kg, 14.3 cm3 Need density (g/mL) 1000 g 1 mL Plan kg S g and cm3 S mL, then calculate density 1 kg 1 cm3 1000 g 1 mL Set-up 0.104 kg * = 104 g (3 SFs) and 14.3 cm3 * = 14.3 mL (3 SFs) 1 kg 1 cm3 104 g mass 6 Density = = = 7.27 g/mL (3 SFs) volume 14.3 mL 43.5 g mass = = 0.791 g/mL (3 SFs) c. Density = volume 55.0 mL Set-up

8.51 kg *

2.71 In these problems, the density is used as a conversion factor. a. Given 1.50 kg of ethanol Need liters of ethanol 1000 g 1 mL 1L Plan kg S g S mL S L 1 kg 0.79 g 1000 mL 1000 g 1 mL 1L Set-up 1.50 kg alcohol * * * 1 kg alcohol 0.79 g 1000 mL = 1.9 L of ethanol (2 SFs) b. Given 6.5 mL of mercury Need grams of mercury 13.6 g Plan mL S g 1 mL 13.6 g = 88 g of mercury (2 SFs) Set-up 6.5 mL * 1 mL c. Given 225 cm3 of silver Need ounces of silver 1 mL 10.5 g 1 lb 16 oz Plan cm3 S mL S g S lb S oz 3 3 454 g 1 lb 1 cm 1 cm 10.5 g 1 mL 1 lb 16 oz * * * = 83.3 oz of silver (3 SFs) Set-up 225 cm3 * 3 1 mL 454 g 1 lb 1 cm 2.72 a. Given 18.0 mL of water, 35.6 g of silver Need final volume (mL) 1 mL Plan g S mL, then calculate the final volume using addition 10.5 g 1 mL = 3.39 mL of silver (3 SFs) Set-up 35.6 g * 10.5 g 18.0 mL water + 3.39 mL silver = 21.4 mL total volume (rounded to tenths)

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Chemistry and Measurements

b. Given

8.3 g of mercury Need milliliters of mercury 1 mL Plan g S mL 13.6 g 1 mL = 0.61 mL of mercury (2 SFs) Set-up 8.3 g * 13.6 g c. Given 35 gal of water Need kilograms of water 4 qt 946 mL 1.00 g 1 kg Plan gal S qt S mL S g S kg 1 gal 1 qt 1 mL 1000 g 4 qt 1.00 g 1 kg 946 mL Set-up 35 gal * * * * = 130 kg of water (2 SFs) 1 gal 1 qt 1 mL 1000 g 2.73 a. Given 74.1 cm3 of copper Need grams of copper 1 mL 8.92 g Plan cm3 S mL S g 1 cm3 1 mL 8.92 g 1 mL Set-up 74.1 cm3 * * = 661 g of copper (3 SFs) 1 mL 1 cm3 b. Given

12.0 gal of gasoline Need

kilograms of gasoline 4 qt 946 mL 0.74 g 1 kg Plan gal S qt S mL S g S kg 1 gal 1 qt 1 mL 1000 g 4 qt 0.74 g 1 kg 946 mL Set-up 12.0 gal * * * * = 34 kg of gasoline (2 SFs) 1 gal 1 qt 1 mL 1000 g c. Given 27 g of ice Need cubic centimeters of ice 1 mL 1 cm3 Plan g S mL S cm3 0.92 g 1 mL 1 mL 1 cm3 Set-up 27 g * * = 29 cm3 (2 SFs) 0.92 g 1 mL 2.74 a. Given

1.2 kg of olive oil Need milliliters of olive oil 1000 g 1 mL Plan kg S g S mL 1 kg 0.92 g 1000 g 1 mL Set-up 1.2 kg * * = 1.3 * 103 mL (2 SFs) 1 kg 0.92 g b. Given 115 cm3 of iron Need kilograms of iron 1 kg 1 mL 7.86 g Plan cm3 S mL S g S kg 3 1 mL 1000 g 1 cm 7.86 g 1 kg 1 mL Set-up 115 cm3 * * * = 0.904 kg of iron (3 SFs) 1 mL 1000 g 1 cm3 c. Given 7.3 L of helium Need grams of helium 0.179 g Plan L S g 1L 0.179 g Set-up 7.3 L * = 1.3 g (2 SFs) 1L 2.75 Because the density of aluminum is 2.70 g/cm3, silver is 10.5 g/cm3, and lead is 11.3 g/cm3, we can identify the unknown metal by calculating its density as follows: Density =

217 g mass of metal = = 11.3 g/cm3 (3 SFs) volume of metal 19.2 cm3

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Chapter 2

2.76 The volume of a cube, 2.0 cm on each edge, is calculated as follows: 1 mL Volume = (length)3 = (2.0 cm)3 * = 8.0 mL (2 SFs) 1 cm3 (Both cubes have the same volume, but their masses differ.) A cube will displace its volume when submerged in water, so the final volume reading in each graduated cylinder is: 40.0 mL water + 8.0 mL metal = 48.0 mL total volume density of substance 1.030 g/mL = = 1.03 (3 SFs) density of water 1.00 g/mL mass of glucose solution 20.6 g b. Density = = = 1.03 g/mL (3 SFs) volume of glucose solution 20.0 mL density of substance c. Specific gravity = density of water 6 Density of substance = specific gravity * density of water = 0.92 * 1.00 g/mL = 0.92 g/mL Mass of substance = volume of substance * density of substance = 750 mL * 0.92 g/mL = 690 g (2 SFs) density of substance d. Specific gravity = density of water 6 Density of substance = specific gravity * density of water = 0.850 * 1.00 g/mL = 0.850 g/mL 325 g mass of solution 6 Volume of solution = = = 382 mL (3 SFs) density of solution 0.850 g/mL 2.77 a. Specific gravity =

density of substance 1.02 g/mL = = 1.02 (3 SFs) density of water 1.00 g/mL 190 mg 1g mass of VLDL b. Density = = * = 0.95 g/mL (2 SFs) volume of VLDL 0.200 mL 1000 mg density of substance c. Specific gravity = density of water 6 Density of substance = specific gravity * density of water = 0.86 * 1.00 g/mL = 0.86 g/mL Mass of substance = volume of substance * density of substance 1000 mL = 2.15 L * 0.86 g/mL * = 1800 g (2 SFs) 1L mass of urine sample 5.025 g d. Density = = = 1.005 g/mL volume of urine sample 5.000 mL density of sample 1.005 g/mL Specific gravity = = = 1.01 (3 SFs) density of water 1.00 g/mL Since the specific gravity of the urine sample falls within the normal range for urine (1.003 to 1.030), these results do not indicate that the patient has type 2 diabetes.

2.78 a. Specific gravity =

42 mcg iron 1 dL blood and 1 dL blood 42 mcg iron b. Given 8.0-mL blood sample, 42 mcg of iron/dL Need micrograms of iron 42 mcg iron 1 dL blood Plan mL of blood S dL of blood S mcg of iron 100 mL blood 1 dL blood 42 mcg iron 1 dL blood Set-up 8.0 mL blood * * = 3.4 mcg of iron (2 SFs) 100 mL blood 1 dL blood

2.79 a. 1 dL of blood = 42 mcg of iron;

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Chemistry and Measurements

50 mg iron 1 tablet and 1 tablet 50 mg iron b. Given 1 tablet twice daily, 50 mg of iron/tablet Need grams of iron per week Plan days S tablets S mg of iron S g of iron 7 days 2 tablets 50 mg iron 1 g iron 1 week 1 day 1 tablet 1000 mg iron 7 days 50 mg iron 1 g iron 2 tablets Set-up * * * = 0.7 g of iron/week (1 SF) 1 week 1 day 1 tablet 1000 mg iron

2.80 a. 1 tablet = 50 mg of iron;

2.81 Both measurements in part c have two significant figures, and both measurements in part d have four significant figures. 2.82 Both measurements in part a have three significant figures, and in parts b and c, both pairs have two significant figures. 2.83 a. The number of legs is a counted number; it is exact. b. The height is measured with a ruler or tape measure; it is a measured number. c. The number of chairs is a counted number; it is exact. d. The area is calculated from the length and width, which are measured with a ruler or tape measure; it is a measured number. 2.84 a. Length is 3.7 cm (2 SFs); the 7 is the estimated digit. b. Length is 2.50 cm (3 SFs); the 0 is the estimated digit. c. Length is 4.10 cm (3 SFs); the 0 is the estimated digit. 2.85 61.5 °C 2.86 4.8 °C 2.54 cm = 97.5 cm (3 SFs) 1 in. 2.54 cm b. length = 24.2 in. * = 61.5 cm (3 SFs) 1 in. c. There are three significant figures in the length measurement. d. Area = length * width = 97.5 cm * 61.5 cm = 6.00 * 103 cm2 (3 SFs)

2.87 a. length = 38.4 in. *

2.54 cm = 17.8 cm (3 SFs) 1 in. 2.54 cm b. width = 6.00 in. * = 15.2 cm (3 SFs) 1 in. c. There are three significant figures in the width measurement. d. Volume = length * width * height = 17.8 cm * 15.2 cm * 10.2 cm = 2.76 * 103 cm3 (3 SFs) 2.89 a. Diagram 3; a cube that has a greater density than the water will sink to the bottom. b. Diagram 4; a cube with a density of 0.80 g/mL will be about four-fifths submerged in the water. c. Diagram 1; a cube with a density that is one-half the density of water will be one-half submerged in the water. d. Diagram 2; a cube with the same density as water will float just at the surface of the water.

2.88 a. length = 7.00 in. *

2.90 Given 18.5 mL initial volume, 23.1 mL final volume, 8.24 g mass Need density (g/mL) Plan calculate volume by difference, then calculate density Set-up The volume of the object is 23.1 mL - 18.5 mL = 4.6 mL 8.24 g mass 6 Density = = = 1.8 g/mL (2 SFs) volume 4.6 mL

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Chapter 2

2.91 Since all three solids have a mass of 10.0 g, the one with the smallest volume must have the highest density; the one with the largest volume will have the lowest density. A would be gold; it has the highest density (19.3 g/mL) and the smallest volume. B would be silver; its density is intermediate (10.5 g/mL) and the volume is intermediate. C would be aluminum; it has the lowest density (2.70 g/mL) and the largest volume. 2.92 The liquid with the highest density will be at the bottom of the cylinder; the liquid with the lowest density will be at the top of the cylinder: A is vegetable oil (D = 0.92 g/mL), B is water (D = 1.00 g/mL), C is mercury (D = 13.6 g/mL) 2.93 The green cube has the same volume as the gray cube. However, the green cube has a larger mass on the scale, which means that its mass/volume ratio is larger. Thus, the density of the green cube is higher than the density of the gray cube. 2.94 The green cube has the same mass as the gray cube. However, the green cube has a larger volume, which means that its mass/volume ratio is smaller. Thus, the density of the green cube is lower than the density of the gray cube. 2.95 a. To round off 0.000 012 58 L to three significant figures, drop the final digit 8 and increase the last retained digit by 1 to give 0.000 012 6 L or 1.26 * 10-5 L. b. To round off 3.528 * 102 kg to three significant figures, drop the final digit 8 and increase the last retained digit by 1 to give 353 kg (3.53 * 102 kg). c. To express 125 111 m to three significant figures, drop the final digits 111 and add three zeros as placeholders to give 125 000 m (or 1.25 * 105 m). d. To express 34.9673 s to three significant figures, drop the final digits 673 and increase the last retained digit by 1 to give 35.0 s. 2.96 a. To express 58.703 mL to three significant figures, drop the final digits 03 to give 58.7 mL. b. To express 3 * 10-3 s to three significant figures, add two significant zeros to give 3.00 * 10-3 s. c. To express 0.010 826 g to three significant figures, drop the final digits 26 to give 0.0108 g or 1.08 * 10-2 g. d. To round off 1.7484 * 103 ms to three significant figures, drop the final digits 84 and increase the last retained digit by 1 to give 1.75 * 103 ms. 2.97 a. The total mass is the sum of the individual components of the dessert. 137.25 g + 84 g + 43.7 g = 265 g. No places to the right of the decimal point are allowed since the mass of the fudge sauce (84 g) has no digits to the right of the decimal point. b. Given grams of dessert from part a Need pounds of dessert 1 lb Plan g S lb 454 g 1 lb = 0.584 lb of dessert (3 SFs) Set-up 265 g dessert (total) * 454 g 2.98 a. The total mass is the sum of the individual components of the order. 22 kg salmon + 5.5 kg crab + 3.48 kg oysters = 31 kg of seafood. No places to the right of the decimal point are allowed since the mass of the salmon (22 kg) has no digits to the right of the decimal point. b. Given kilograms of seafood from part a Need pounds of seafood 2.20 lb Plan kg S lb 1 kg 2.20 lb = 68 lb of seafood (2 SFs) Set-up 31 kg seafood (total) * 1 kg

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Chemistry and Measurements

2.99 Given 1.95 euros/kg of grapes, $1.14/euro Need cost in dollars per pound 1 kg 1.95 euros +1.14 Plan euros/kg S euros/lb S +/lb 1 kg grapes 2.20 lb 1 euro 1 kg 1.95 euros +1.14 Set-up * * = +1.01/lb of grapes (3 SFs) 1 kg grapes 2.20 lb 1 euro 2.100 Given 0.45 lb of avocado, 48 pesos/kg, 18 pesos/dollar Need cost in cents 1 kg 48 pesos 1 dollar 100 cents Plan lb S kg S pesos S dollars S cents 2.20 lb 1 kg 18 pesos 1 dollar 1 kg 48 pesos 1 dollar 100 cents Set-up 0.45 lb * * * * = 55 cents (2 SFs) 2.20 lb 1 kg 18 pesos 1 dollar 2.101 Given

4.0 lb of onions

Need

2.102 Given

$1420, $1.75/lb Need

2.103 Given

7500 ft Need

number of onions 454 g 1 onion Plan lb S g S number of onions 1 lb 115 g 454 g 1 onion Set-up 4.0 lb onions * * = 16 onions (2 SFs) 1 lb 115 g kilograms of potatoes 1 kg 1 lb Plan + S lb of potatoes S kg of potatoes +1.75 2.20 lb 1 kg 1 lb Set-up +1420 * * = 369 kg of potatoes (3 SFs) +1.75 2.20 lb

Plan Set-up

minutes

12 in. 2.54 cm 1m 1 min 1 ft 1 in. 100 cm 55.0 m 12 in. 2.54 cm 1m 1 min 7500 ft * * * * = 42 min (2 SFs) 1 ft 1 in. 100 cm 55.0 m

ft S in. S cm S m S min

2.104 Given

1700 km Need hours 0.621 mi 1h Plan km S mi S h 1 km 63 mi 0.621 mi 1h * = 17 h (2 SFs) Set-up 1700 km * 1 km 63 mi

2.105 Given 215 mL initial, 285 mL final volume, density of lead 11.3 g/mL Need grams of lead 11.3 g Plan calculate the volume by difference and mL S g 1 mL Set-up The difference between the initial volume of the water and its volume with the lead object will give us the volume of the lead object: 285 mL total - 215 mL water = 70. mL of lead, then 11.3 g lead 70. mL lead * = 790 g of lead (2 SFs) 1 mL lead 2.106 Given 155 mL initially, 15.0 g of iron, 20.0 g of lead Need final volume (mL) Plan for both iron and lead g S mL then calculate the final volume using addition 1 mL iron 1 mL lead 7.86 g iron 11.3 g lead

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Chapter 2

1 mL iron = 1.91 mL of iron (3 SFs) 7.86 g iron 1 mL lead 20.0 g lead * = 1.77 mL of lead (3 SFs) 11.3 g lead 6 final volume = 155 mL water + 1.91 mL iron + 1.77 mL lead = 159 mL (no places to the right of the decimal point are allowed) Set-up 15.0 g iron *

2.107 Given 1.2 kg of gasoline Need milliliters of gasoline 1000 g 1 mL Plan kg S g S mL 1 mL 0.74 g 1000 g 1 mL Set-up 1.2 kg * * = 1600 mL (1.6 * 103 mL) of gasoline (2 SFs) 1 kg 0.74 g 2.108 Given 3.40 kg of ethanol Need quarts of ethanol 1000 g 1 mL 1 qt Plan kg S g S mL S qt 1 kg 0.79 g 946 mL 1000 g 1 qt 1 mL Set-up 3.40 kg * * * = 4.5 qt of ethanol (2 SFs) 1 kg 0.79 g 946 mL 2.109 a. Given

8.0 oz Need

number of crackers 6 crackers Plan oz S number of crackers 0.50 oz 6 crackers Set-up 8.0 oz * = 96 crackers (2 SFs) 0.50 oz

b. Given 10 crackers, 4 g of fat/serving Need ounces of fat Plan number of crackers S servings S g of fat S lb of fat S oz of fat 1 serving 4 g fat 1 lb 16 oz 6 crackers 1 serving 454 g 1 lb 1 serving 4 g fat 1 lb 16 oz Set-up 10 crackers * * * * = 0.2 oz of fat (1 SF) 6 crackers 1 serving 454 g 1 lb c. Given 2.4 g of sodium, 140 mg of sodium/serving Need number of servings 100 mg sodium 1 serving Plan mg of sodium S servings 1 g sodium 140 mg sodium 1000 mg sodium 1 serving Set-up 2.4 g sodium * * = 17 servings (2 SFs) 1 g sodium 140 mg sodium 2.110 Given 75 000 mL of water Need gallons of water 1 qt 1 gal Plan mL S qt S gal 946 mL 4 qt 1 qt 1 gal Set-up 75 000 mL * * = 20. gal of water (2 SFs) 946 mL 4 qt 2.111 Given 10 days, 4 tablets/day, 250-mg tablets Need ounces of amoxicillin Plan days S tablets S mg of amoxicillin S g S lb S oz of amoxicillin 1g 4 tablets 250 mg amoxicillin 1 lb 16 oz 1 day 1 tablet 1000 mg 454 g 1 lb 250 mg amoxicillin 1g 4 tablets 1 lb 16 oz Set-up 10 days * * * * * 1 day 1 tablet 1000 mg 454 g 1 lb = 0.35 oz of amoxicillin (2 SFs)

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Chemistry and Measurements

2.112 Given 1.2 oz of protein Need Plan Set-up

grams of protein

oz of protein S lb of protein S g of protein

1 lb 16 oz

454 g 1 lb

454 g 1 lb * = 34 g of protein (2 SFs) 16 oz 1 lb Yes, she has exceeded her protein limit for the day since she consumed 34 g of protein, which is 10 grams more than she is allowed.

1.2 oz protein *

2.113 Given 5.0 mL of elixir, 30. mg of phenobarbital/7.5 mL Need milligrams of phenobarbital 30. mg phenobarbital Plan mL of elixir S mg of phenobarbital 7.5 mL elixir 30. mg phenobarbital Set-up 5.0 mL elixir * = 20. mg of phenobarbital (2 SFs) 7.5 mL elixir 2.114 Given

2.0 mg of morphine, 10. mg of morphine/mL Need milliliters of solution 1 mL solution Plan mg of morphine S mL of solution 10. mg morphine 1 mL solution Set-up 2.0 mg morphine * = 0.20 mL of solution (2 SFs) 10. mg morphine

2.115 Because the balance can measure mass to 0.001 g, the mass should be reported to 0.001 g. You should record the mass of the object as 31.075 g. 2.116 Any measurement contains some estimation. The three different students were estimating the final digit, and they came up with different estimates. The student who reported 5.8 cm should have given the value of 5.80 cm. 2.117 Given

3.0-h trip Need

gallons of gasoline 55 mi 1 km 1L Plan h S mi S km S L S qt S gal 1h 0.621 mi 11 km 1.06 qt 1 gal 55 mi 1 km 1L Set-up 3.0 h * * * * * 1h 0.621 mi 11 km 1L 4 qt = 6.4 gal of gasoline (2 SFs)

1.06 qt 1L

1 gal 4 qt

2.118 Given 325 tubes of sunscreen, 4.0 oz/tube, 2.50, benzyl salicylate Need kilograms of benzyl salicylate Plan tubes S oz of sunscreen S lb of sunscreen S kg of sunscreen S kg of benzyl salicylate 1 kg 2.50 kg benzyl salicylate 4.0 oz sunscreen 1 lb 1 tube 16 oz 2.20 lb 100 kg sunscreen 1 kg 2.50 kg benzyl salicylate 4.0 oz sunscreen 1 lb * * * Set-up 325 tubes * 1 tube 16 oz 2.20 lb 100 kg sunscreen = 0.92 kg of benzyl salicylate (2 SFs) 2.119 Given 1.50 L of gasoline Need milliliters of olive oil Plan L of gasoline S mL of gasoline S g of gasoline S g of olive oil S mL of olive oil (equality from question: 1 g of olive oil = 1 g of gasoline) 1000 mL gasoline 0.74 g gasoline 1 mL olive oil 1 L gasoline 1 mL gasoline 0.92 g olive oil 0.74 g gasoline 1000 mL * = 1110 g of gasoline Set-up 1.50 L gasoline * 1L 1 mL gasoline 1 mL olive oil 1110 g olive oil * = 1200 mL (1.2 * 103 mL) of olive oil (2 SFs) 0.92 g olive oil Copyright © 2019 Pearson Education, Inc.

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Chapter 2

2.120 Given 75.5 mL initially, 50.0 g of silver, 50.0 g of gold Need final volume (mL) Plan for both silver and gold g S mL then calculate the final volume using addition 1 mL 1 mL 10.5 g silver 19.3 g gold 1 mL silver = 4.76 mL of silver Set-up 50.0 g silver * 10.5 g silver 1 mL gold 50.0 g gold * = 2.59 mL of gold 19.3 g gold 6 final volume = 75.5 mL water + 4.76 mL silver + 2.59 mL gold = 82.9 mL (one place to right of decimal) 2.121 a. Given 65 kg of body mass, 3.0, fat Need pounds of fat Plan kg of body mass S kg of fat S lb of fat (percent equality: 100 kg of body mass = 3.0 kg of fat) 3.0 kg fat 2.20 lb fat 100 kg body mass 1 kg fat 3.0 kg fat 2.20 lb fat Set-up 65 kg body mass * * = 4.3 lb of fat (2 SFs) 100 kg body mass 1 kg fat b. Given 3.0 L of fat Need pounds of fat 1000 mL 0.909 g 1 lb Plan L S mL S g S lb 1L 1 mL 454 g 0.909 g 1000 mL 1 lb Set-up 3.0 L * * * = 6.0 lb of fat (2 SFs) 1L 1 mL 454 g 2.122 Given 180 bottles Need kilograms of ethanol Plan bottles S pt S mL S g of mouthwash S g of ethanol S kg of ethanol 1.06 pt 473 mL 0.876 g 21.6 g ethanol 1 kg ethanol 1 bottle 1 pt 1 mL 100 g mouthwash 1000 g ethanol 1.06 pt 0.876 g 21.6 g ethanol 473 mL Set-up 180 bottles * * * * 1 bottle 1 pt 1 mL 100 g mouthwash 1 kg ethanol * = 17.1 kg of ethanol (3 SFs) 1000 g ethanol

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3 Matter and Energy Learning Goals • • • • • • •

Classify examples of matter as pure substances or mixtures. Identify the states and the physical and chemical properties of matter. Given a temperature, calculate the corresponding temperature on another scale. Identify energy as potential or kinetic; convert between units of energy. Use the energy values to calculate the kilocalories (kcal) or kilojoules (kJ) for a food. Use specific heat to calculate heat loss or gain. Describe the changes of state between solids, liquids, and gases; calculate the energy released or absorbed.

Chapter Outline Chapter Opener: Dietitian Classification of Matter Chemistry Link to Health: Breathing Mixtures States and Properties of Matter Temperature Chemistry Link to Health: Variation in Body Temperature Energy Energy and Nutrition Chemistry Link to Health: Losing and Gaining Weight Specific Heat Changes of State Chemistry Link to Health: Steam Burns Clinical Update: A Diet and Exercise Program

3.1 3.2 3.3 3.4 3.5 3.6 3.7

Core Chemistry Skills • • • • •

Identifying Physical and Chemical Changes (3.2) Converting between Temperature Scales (3.3) Using Energy Units (3.4) Using the Heat Equation (3.6) Calculating Heat for Change of State (3.7)

Answers and Solutions to Text Problems 3.1

Elements are the simplest type of pure substance, containing only one type of atom. Compounds contain two or more elements chemically combined in a specific proportion. a. A silicon chip is an element since it contains only one type of atom (Si). b. Hydrogen peroxide (H2O2) is a compound since it contains two elements (H, O) that are chemically combined. c. Oxygen gas (O2) is an element since it contains only one type of atom (O). Copyright © 2019 Pearson Education, Inc.

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Chapter 3

d. Rust (Fe 2O3) is a compound since it contains two elements (Fe, O) that are chemically combined. e. Methane (CH4) in natural gas is a compound since it contains two elements (C, H) that are c hemically combined. 3.2

Elements are the simplest type of pure substance, containing only one type of atom. Compounds contain two or more elements chemically combined in a specific proportion. a. Helium gas is an element since it contains only one type of atom (He). b. Sulfur is an element since it contains only one type of atom (S). c. Sugar (C 12H22O11) is a compound since it contains three elements (C, H, O) that are chemically combined. d. Mercury in a thermometer is an element since it contains only one type of atom (Hg). e. Lye (NaOH) is a compound since it contains three elements (Na, O, H) that are chemically combined.

3.3

A pure substance is matter that has a fixed or definite composition: either elements or compounds. In a mixture, two or more substances are physically mixed but not chemically combined. a. Baking soda is composed of one type of matter (NaHCO3), which makes it a pure substance. b. A blueberry muffin is composed of several substances mixed together, which makes it a mixture. c. Ice is composed of one type of matter (H2O), which makes it a pure substance. d. Zinc is composed of one type of matter (Zn), which makes it a pure substance. e. Trimix is a physical mixture of oxygen, nitrogen, and helium gases, which makes it a mixture.

3.4

A pure substance is matter that has a fixed or definite composition: either elements or compounds. In a mixture, two or more substances are physically mixed but not chemically combined. a. A soft drink is composed of several substances mixed together (e.g., water, sugar, carbon dioxide), which makes it a mixture. b. Propane is composed of one type of matter (C 3H8), which makes it a pure substance. c. A cheese sandwich is composed of several substances physically but not chemically combined, which makes it a mixture. d. An iron nail is composed of one type of matter (Fe), which makes it a pure substance. e. Salt substitute is composed of one type of matter (KCl), which makes it a pure substance.

3.5

A homogeneous mixture has a uniform composition; a heterogeneous mixture does not have a uniform composition throughout the mixture. a. Vegetable soup is a heterogeneous mixture since it has chunks of vegetables. b. Tea is a homogeneous mixture since it has a uniform composition. c. Fruit salad is a heterogeneous mixture since it has chunks of fruit. d. Tea with ice and lemon slices is a heterogeneous mixture since it has chunks of ice and lemon.

3.6

A homogeneous mixture has a uniform composition; a heterogeneous mixture does not have a uniform composition throughout the mixture. a. Nonfat milk is a homogeneous mixture since it has a uniform composition. b. Chocolate-chip ice cream is a heterogeneous mixture since it has chunks of chocolate. c. A peanut butter sandwich is a heterogeneous mixture since it has layers of bread and peanut butter. d. Cranberry juice is a homogeneous mixture since it has a uniform composition.

3.7

a. A gas has no definite volume or shape. b. In a gas, the particles do not interact with each other. c. In a solid, the particles are held in a rigid structure.

3.8

a. A liquid has a definite volume but takes the shape of its container. b. In a gas, the particles are very far apart. c. A gas occupies the entire volume of the container.

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Matter and Energy

3.9

A physical property is a characteristic of a substance—such as color, shape, odor, luster, size, melting point, and density—that can be observed without affecting the identity of the substance. A chemical property is a characteristic that indicates the ability of a substance to change into a new substance. a. Color and physical state are physical properties. b. The ability to react with oxygen is a chemical property. c. The temperature of a substance is a physical property. d. The ability to corrode iron is a chemical property. e. Burning butane gas in oxygen forms new substances, which makes it a chemical property.

3.10 A physical property is a characteristic of a substance—such as color, shape, odor, luster, size, melting point, and density—that can be observed without affecting the identity of the substance. A chemical property is a characteristic that indicates the ability of a substance to change into a new substance. a. Color and physical state are physical properties. b. The browning of apples is a chemical reaction and is thus a chemical property. c. The ability of phosphorus to ignite in air is a chemical property. d. The physical state of a substance is a physical property. e. Compressing propane gas to liquid propane does not involve formation of new substances, which makes it a physical property. 3.11 When matter undergoes a physical change, its state or appearance changes, but its composition remains the same. When a chemical change occurs, the original substance is converted into a new substance, which has different physical and chemical properties. a. Water vapor condensing is a physical change since the physical form of the water changes, but the composition of the substance does not. b. Cesium metal reacting is a chemical change since new substances form. c. Gold melting is a physical change since the physical state changes, but not the composition of the substance. d. Cutting a puzzle is a physical change since the size and shape change, but not the composition of the substance. e. Grating cheese is a physical change since the size and shape change, but not the composition of the substance. 3.12 When matter undergoes a physical change, its state or appearance changes, but its composition remains the same. When a chemical change occurs, the original substance is converted into a new substance, which has different physical and chemical properties. a. Rolling pie dough is a physical change since it changes the shape and size of the dough, but not the composition of the substance. b. Silver tarnishing is a chemical change since new substances form. c. Cutting a tree is a physical change since the size and shape changes, but not the composition of the substance. d. Digesting food is a chemical change since new substances form. e. Melting a chocolate bar is a physical change since the shape of the bar changes, but no new substances are formed. 3.13 a. The high reactivity of fluorine is a chemical property since it allows for the formation of new substances. b. The physical state of fluorine is a physical property. c. The color of fluorine is a physical property. d. The reactivity of fluorine with hydrogen is a chemical property since it allows for the formation of new substances. e. The melting point of fluorine is a physical property. Copyright © 2019 Pearson Education, Inc.

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Chapter 3

3.14 a. The melting point of zirconium is a physical property. b. The lack of reactivity of zirconium is a chemical property since it makes the formation of new substances unlikely. c. The color of zirconium is a physical property. d. The reactivity of zirconium is a chemical property since it allows for the formation of new substances. e. The appearance of zirconium is a physical property. 3.15 The Fahrenheit temperature scale is still used in the United States. A normal body temperature is 98.6 °F on this scale. To convert her 99.8 °F temperature to the equivalent reading on the Celsius scale, the following calculation must be performed: TC =

(99.8 - 32) 67.8 = = 37.7 °C (3 SFs) (1.8 and 32 are exact numbers) 1.8 1.8

Because a normal body temperature is 37.0 on the Celsius scale, her temperature of 37.7 °C would indicate a mild fever. 3.16 The Celsius temperature scale is used in Mexico. Typical baking temperatures are higher than 175 °F, generally 325 to 350 °F. To convert this temperature to the equivalent reading on the Fahrenheit scale that is still used in the United States, the following calculation must be performed: TF = 1.8(175) + 32 = 315 + 32 = 347 °F (3 SFs) (1.8 and 32 are exact numbers) 3.17 To convert Celsius to Fahrenheit: TF = 1.8(TC) + 32 (1.8 and 32 are exact numbers) TF - 32 (1.8 and 32 are exact numbers) To convert Fahrenheit to Celsius: TC = 1.8 To convert Celsius to Kelvin: TK = TC + 273 To convert Kelvin to Celsius: TC = TK - 273 a. TF = 1.8(TC) + 32 = 1.8(37.0) + 32 = 66.6 + 32 = 98.6 °F (65.3 - 32) TF - 32 33.3 = = = 18.5 °C b. TC = 1.8 1.8 1.8 c. TK = TC + 273 = -27 + 273 = 246 K d. TK = TC + 273 = 62 + 273 = 335 K (114 - 32) TF - 32 82 = = = 46 °C e. TC = 1.8 1.8 1.8 3.18 To convert Celsius to Fahrenheit: TF = 1.8(TC) + 32 (1.8 and 32 are exact numbers) T - 32 To convert Fahrenheit to Celsius: TC = F (1.8 and 32 are exact numbers) 1.8 To convert Celsius to Kelvin: TK = TC + 273 To convert Kelvin to Celsius: TC = TK - 273 a. TF = 1.8(TC) + 32 = 1.8(25) + 32 = 45 + 32 = 77 °F b. TF = 1.8(TC) + 32 = 1.8(155) + 32 = 279 + 32 = 311 °F (-25 - 32) T - 32 -57 c. TC = F = = = -32 °C 1.8 1.8 1.8 d. TC = TK - 273 = 224 - 273 = -49 °C e. TK = TC + 273 = 145 + 273 = 418 K (106 - 32) TF - 32 74 = = = 41 °C 1.8 1.8 1.8 (103 - 32) T - 32 71 = = = 39 °C b. TC = F 1.8 1.8 1.8 No, there is no need to phone the doctor. The child’s temperature is less than 40.0 °C. 3.19 a. TC =

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Matter and Energy

(145 - 32) TF - 32 113 = = = 62.8 °C 1.8 1.8 1.8 b. TF = 1.8(TC) + 32 = 1.8(20.6) + 32 = 37.1 + 32 = 69.1 °F

3.20 a. TC =

3.21 As the roller-coaster car climbs to the top of the ramp, its kinetic energy is converted to potential energy. At the top, the car has its maximum potential energy. As the car descends, potential energy is converted into kinetic energy. At the bottom, all of its energy is kinetic. 3.22 As the elevator moves to the top of the jump, the skier’s potential energy is increasing. As the skier goes down the ramp, potential energy is being converted into kinetic energy. 3.23 a. Water at the top of the waterfall has potential energy. b. Kicking a ball gives it kinetic energy. c. The energy in a lump of coal is potential energy. d. A skier at the top of a hill has potential energy. 3.24 a. The energy in your food is potential energy. b. A tightly wound spring has potential energy. c. A car speeding down the freeway has kinetic energy. d. The movements that occur during an earthquake involve kinetic energy. 3.25 a. Given

3500 cal Need kilocalories 1 kcal Plan cal S kcal 1000 cal 1 kcal Set-up 3500 cal * = 3.5 kcal (2 SFs) 1000 cal b. Given

415 J Need calories 1 cal Plan J S cal 4.184 J 1 cal Set-up 415 J * = 99.2 cal (3 SFs) 4.184 J c. Given

28 cal Need joules 4.184 J Plan cal S J 1 cal 4.184 J Set-up 28 cal * = 120 J (2 SFs) 1 cal d. Given

4.5 kJ Need

calories 1000 J 1 cal Plan kJ S J S cal 1 kJ 4.184 J 1000 J 1 cal Set-up 4.5 kJ * * = 1100 cal (2 SFs) 1 kJ 4.184 J 3.26 a. Given

8.1 kcal Need calories 1000 cal Plan kcal S cal 1 kcal 1000 cal Set-up 8.1 kcal * = 8100 cal (2 SFs) 1 kcal b. Given

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Chapter 3

c. Given

2550 cal Need kilojoules 4.184 J 1 kJ Plan cal S J S kJ 1 cal 1000 J 4.184 J 1 kJ * = 10.7 kJ (3 SFs) Set-up 2550 cal * 1 cal 1000 J d. Given

2.50 kcal Need joules 1000 cal 4.184 J Plan kcal S cal S J 1 kcal 1 cal 1000 cal 4.184 J * = 10 500 J (3 SFs) Set-up 2.50 kcal * 1 kcal 1 cal 3.27 a. Given 3.0 h, 270 kJ/h Need joules 1000 J Plan h S kJ S J 1 kJ 270 kJ 1000 J * = 8.1 * 105 J (2 SFs) Set-up 3.0 h * 1.0 h 1 kJ b. Given 3.0 h, 270 kJ/h Need

kilocalories 1000 J 1 cal 1 kcal Plan h S kJ S J S cal S kcal 1 kJ 4.184 J 1000 cal 270 kJ 1000 J 1 cal 1 kcal * * * = 190 kcal (2 SFs) Set-up 3.0 h * 1.0 h 1 kJ 4.184 J 1000 cal

3.28 a. Given

750 kcal Need

b. Given

750 kcal Need

joules 1000 cal 4.184 J Plan kcal S cal S J 1 kcal 1 cal 1000 cal 4.184 J * = 3.1 * 106 J (2 SFs) Set-up 750 kcal * 1 kcal 1 cal kilojoules 1000 cal 4.184 J 1 kJ Plan kcal S cal S J S kJ 1 kcal 1 cal 1000 J 1000 cal 4.184 J 1 kJ * * = 3100 kJ (2 SFs) Set-up 750 kcal * 1 kcal 1 cal 1000 J

3.29 a. Given

125 kJ Need

b. Given

870. kJ Need

3.30 a. Given

131 kJ Need

kilocalories 1000 J 1 cal 1 kcal Plan kJ S J S cal S kcal 1 kJ 4.184 J 1000 cal 1000 J 1 cal 1 kcal * * = 29.9 kcal (3 SFs) Set-up 125 kJ * 1 kJ 4.184 J 1000 cal kilocalories 1000 J 1 cal 1 kcal Plan kJ S J S cal S kcal 1 kJ 4.184 J 1000 cal 1000 J 1 cal 1 kcal * * = 208 kcal (3 SFs) Set-up 870. kJ * 1 kJ 4.184 J 1000 cal kilocalories 1000 J 1 cal 1 kcal Plan kJ S J S cal S kcal 1 kJ 4.184 J 1000 cal 1000 J 1 cal 1 kcal * * = 31.3 kcal (3 SFs) Set-up 131 kJ * 1 kJ 4.184 J 1000 cal

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Matter and Energy

b. Given

23.4 kJ Need

kilocalories 1000 J 1 cal 1 kcal Plan kJ S J S cal S kcal 1 kJ 4.184 J 1000 cal 1000 J 1 cal 1 kcal * * = 5.59 kcal (3 SFs) Set-up 23.4 kJ * 1 kJ 4.184 J 1000 cal

3.31 a. Given one cup of orange juice that contains 26 g of carbohydrate, 2 g of protein, and no fat Need total energy in kilojoules Energy (rounded off to the tens place)

Food Type

Mass

Energy Value

Carbohydrate

26 g

17 kJ 1g

440 kJ

Protein

17 kJ 1g

30 kJ

470 kJ

Total energy content

b. Given one apple that provides 72 kcal of energy and contains no fat or protein Need grams of carbohydrate 1 g carbohydrate Plan kcal S g of carbohydrate 4 kcal 1 g carbohydrate Set-up 72 kcal * = 18 g of carbohydrate (2 SFs) 4 kcal c. Given one tablespoon of vegetable oil that contains 14 g of fat, and no carbohydrate or protein Need total energy in kilocalories Food Type

Mass

Fat

14 g

Energy (rounded off to the tens place)

Energy Value

9 kcal 1g

130 kcal

d. Given one avocado that provides 410 kcal in total, and contains 13 g of carbohydrate and 5 g of protein Need grams of fat Energy Value

Energy 1 rounded off to the tens place 2

Food Type

Mass

Carbohydrate

13 g

4 kcal 1g

Fat

9 kcal 1g

? kcal

Protein

4 kcal 1g

20 kcal

410 kcal

Total energy content

50 kcal

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Chapter 3

3.32 a. Given two tablespoons of peanut butter that contains 6 g of carbohydrate, 16 g of fat, and 7 g of protein Need total energy in kilojoules Food Type

Mass

Energy (rounded off to the tens place)

Energy Value

Carbohydrate

17 kJ 1g

100 kJ

Fat

16 g

38 kJ 1g

610 kJ

Protein

17 kJ 1g

120 kJ

830 kJ

Total energy content

b. Given one cup of soup that provides 110 kcal in total and contains 9 g of carbohydrate and 6 g of fat Need grams of protein Food Type

Mass

Carbohydrate

Fat

Protein

Energy (rounded off to the tens place)

Energy Value

4 kcal 1g 9 kcal 1g 4 kcal 1g

40 kcal

50 kcal

? kcal

Total energy content = 110 kcal 6 Energy from protein = 110 kcal - (40 kcal + 50 kcal) = 20 kcal 1 g protein = 5 g of protein (1 SF) 4 kcal c. Given one can of cola that contains 40. g of carbohydrate and no fat or protein Need total energy in kilocalories 6 20 kcal *

Food Type

Mass

Energy (rounded off to the tens place)

Energy Value

4 kcal = 160 kcal 1g d. Given a diet that consists of 68 g of carbohydrate, 9 g of fat, and 150 g of protein Need total energy in kilocalories *

Carbohydrate

40. g

Food Type

Mass

Carbohydrate

68 g

Fat

9 g

Protein

150 g

4 kcal 1g 9 kcal 1g 4 kcal 1g

Total energy content

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Energy (rounded off to the tens place)

Energy Value

270 kcal

80 kcal

600 kcal

950 kcal

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Matter and Energy

3.33 Given one cup of clam chowder that contains 16 g of carbohydrate, 12 g of fat, and 9 g of protein Need total energy in kilocalories and kilojoules Energy (rounded off to the tens place)

Food Type

Mass

Energy Value

Carbohydrate

16 g

4 kcal (or 17 kJ) 1g

60 kcal (or 270 kJ)

Fat

12 g

9 kcal (or 38 kJ) 1g

110 kcal (or 460 kJ)

Protein

4 kcal (or 17 kJ) 1g

40 kcal (or 150 kJ)

Total energy content

210 kcal (or 880 kJ)

3.34 Using the energy totals from Table 3.8: Meal

Energy

Chicken, no skin, 3 oz 110 kcal Broccoli, 3 oz 30 kcal Apple, 1 medium 60 kcal Milk, nonfat, 1 cup 90 kcal Total = 290 kcal 3.35 3.2 L glucose solution *

5.0 g glucose 1000 mL solution 4 kcal * * 1 L solution 100. mL solution 1 g glucose

= 640 kcal (rounded off to the tens place) 3.36 Given a high-protein diet contains 70.0 g of carbohydrate, 5.0 g of fat, and 150 g of protein Need total energy in kilocalories and kilojoules Energy (rounded off to the tens place)

Food Type

Mass

Energy Value

Carbohydrate

70.0 g

4 kcal (or 17 kJ) 1g

280 kcal (or 1190 kJ)

Fat

5.0 g

9 kcal (or 38 kJ) 1g

50 kcal (or 190 kJ)

Protein

150 g

4 kcal (or 17 kJ) 1g

600 kcal (or 2550 kJ)

Total energy content

930 kcal (or 3930 kJ)

3.37 Copper, which has the lowest specific heat of the samples, would reach the highest temperature. 3.38 The specific heat of A must be less than the specific heat of B because the amount of temperature change (∆T ) depends on the amount of heat supplied (or removed, if cooling), heat the amount of material present (m), and its specific heat (SH ): ∆T = . m * SH If the same amount of heat is applied to each substance and equal masses of A and B are used, the only variable that remains is the specific heat for each material. A substance with a low specific heat will produce a larger temperature change than a substance with a high specific heat. Copyright © 2019 Pearson Education, Inc.

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Chapter 3

3.39 a. Given SHwater = 1.00 cal/g °C; m = 8.5 g; ∆T = 36 °C - 15 °C = 21 °C Need heat in calories Plan Heat = m * ∆T * SH 1.00 cal Set-up Heat = 8.5 g * 21 °C * = 180 cal (2 SFs) g °C b. Given SHwater = 4.184 J/g °C; m = 25 g; ∆T = 86 °C - 61 °C = 25 °C Need heat in joules Plan Heat = m * ∆T * SH 4.184 J Set-up Heat = 25 g * 25 °C * = 2600 J (2 SFs) g °C c. Given SHwater = 1.00 cal/g °C; m = 150 g; ∆T = 77 °C - 15 °C = 62 °C Need heat in kilocalories 1 kcal Plan Heat = m * ∆T * SH then cal S kcal 1000 cal 1.00 cal 1 kcal Set-up Heat = 150 g * 62 °C * * = 9.3 kcal (2 SFs) g °C 1000 cal d. Given SHcopper = 0.385 J/g °C; m = 175 g; ∆T = 188 °C - 28 °C = 160. °C Need heat in kilojoules 1 kJ Plan Heat = m * ∆T * SH then J S kJ 1000 J 0.385 J 1 kJ Set-up Heat = 175 g * 160. °C * * = 10.8 kJ (3 SFs) g °C 1000 J 3.40 a. Given SHwater = 1.00 cal/g °C; m = 85 g; ∆T = 45 °C - 25 °C = 20 °C Need heat in calories Plan Heat = m * ∆T * SH 1.00 cal Set-up Heat = 85 g * 20 °C * = 1700 cal (2 SFs) g °C b. Given SHwater = 4.184 J/g °C; m = 75 g; ∆T = 66 °C - 22 °C = 44 °C Need heat in joules Plan Heat = m * ∆T * SH 4.184 J Set-up Heat = 75 g * 44 °C * = 14 000 J (2 SFs) g °C c. Given SHwater = 1.00 cal/g °C; m = 5.0 kg; ∆T = 28 °C - 22 °C = 6 °C Need heat in kilocalories 1000 g 1 kcal Plan kg S g ; then Heat = m * ∆T * SH; then cal S kcal 1 kg 1000 cal 1000 g 1.00 cal 1 kcal Set-up Heat = 5.0 kg * * 6 °C * * = 30 kcal (1 SF) 1 kg g °C 1000 cal d. Given SHgold = 0.129 J/g °C; m = 224 g; ∆T = 185 °C - 18 °C = 167 °C Need heat in kilojoules 1 kJ Plan Heat = m * ∆T * SH then J S kJ 1000 J 0.129 J 1 kJ Set-up Heat = 224 g * 167 °C * * = 4.83 kJ (3 SFs) g °C 1000 J 3.41 a. Given SHwater = 4.184 J/g °C = 1.00 cal/g °C; m = 25.0 g; ∆T = 25.7 °C - 12.5 °C = 13.2 °C Need heat in joules and calories Plan Heat = m * ∆T * SH

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