CONCEPTUAL EXAMPLE 1.1

1. CONCEPTUAL EXAMPLE 1.1 Risk–Benefit Analysis Solution The heroin would provide the benefit of pain relief, but its use for such purposes has been judged to be too risky by the U.S. Food and Drug Administration. The DQ is low. The heroin would provide the benefit of pain relief. The risk of addiction in a dying person is irrelevant. Heroin is used this way in Great Britain, but it is banned for any purpose in the United States. The DQ is uncertain. (Both answers involve judgments that are not clearly scientific; people can differ in their assessments of each.) Exercise 1.1A Exercise 1.1B Chloramphenicol is a powerful antibacterial drug that often destroys bacteria unaffected by other drugs. It is highly dangerous to some individuals, however, causing fatal aplastic anemia in about 1 in 30,000 people. Do a risk–benefit analysis for the use of chloramphenicol in (a) sick farm animals, from which people might consume milk or meat with residues of the drug, and (b) a person with Rocky Mountain spotted fever faced with a high probability of death or permanent disability. A four-year clinical trial (6800 participants) of the drug tamoxifen in healthy women at high risk showed a 49% lower rate of breast cancer than a group of 6800 women taking a placebo. Tamoxifen has a low rate of serious side effects, including potentially fatal blood clots, uterine cancer, hot flashes, loss of libido, and cataracts. Do a risk–benefit analysis for the use of tamoxifen in (a) all women and (b) women at high risk. Some medical doctors think heroin is more effective for the relief of severe pain than other medicines. However, heroin is highly addictive and often renders the user unable to function in society. Do a risk–benefit analysis for the use of heroin in treating the pain of (a) a young athlete’s broken leg and (b) a terminally ill cancer patient.

2. CONCEPTUAL EXAMPLE 1.1 Risk–Benefit Analysis continued Exercise 1.1C Rotavirus is the most common cause of severe diarrhea among children. This virus causes hospitalization of approximately 55,000 children each year in the United States and the death of over 600,000 children annually worldwide. A vaccine can prevent most cases. There is a strong association between the vaccine and bowel obstruction in some infants. Do a risk–benefit analysis for the use of the vaccine in (a) the United States and (b) worldwide.

3. CONCEPTUAL EXAMPLE 1.2 Mass and Weight Solution The person’s mass would be the same (62.5 kg) as on Earth; the quantity of matter has not changed. The person would weigh only 0.376 x 124 lb = 46.6 lb; the force of attraction between planet and person is only 0.376 times that on Earth. Exercise 1.2A Exercise 1.2B At the surface of Venus, the force of gravity is 0.903 times that on Earth’s surface. (a) What would be the mass of a standard 1.00-kg object on Venus? (b) A man who weighs 198 lb on Earth would weigh how much on the surface of Venus? On Jupiter, at the boundary between the gaseous atmosphere and the liquid that makes up the bulk of the planet, the force of gravity is 2.34 times that on Earth. (a) What would be the mass of a 52.5-kg woman at that location on Jupiter? (b) A man who weighs 212 lb on Earth would weigh how much on Jupiter? On the planet Mercury gravity is 0.376 times that on Earth. (a) What would be the mass on Mercury of a person who has a mass of 62.5 kilograms (kg) on Earth? (b) What would be the weight on Mercury of a person who weighs 124 pounds (lb) on Earth?

4. CONCEPTUAL EXAMPLE 1.3 Chemical Change and Physical Change Solution We examine each change and determine whether there has been a change in composition or structure. In other words, we ask “Have new substances that are chemically different been created?” If so, the change is chemical; if not, it is physical. Physical change: The composition of the hair is not changed by cutting. Chemical change: The compositions of curds and whey are different from the composition of the milk. Physical change: Liquid water and invisible water vapor formed when liquid water boils have the same composition; the water merely changes from a liquid to a gas. Chemical change: New substances, hydrogen and oxygen, are formed. Exercise 1.3A Which of the following events involve chemical changes and which involve physical changes? Gasoline vaporizes from an open container. A piece of magnesium metal burns in air to form a white powder called magnesium oxide. A dull knife is sharpened with a whetstone. Which of the following events involve chemical changes and which involve physical changes? Your hair is cut. Lemon juice converts milk to curds and whey. Water boils. Water is broken down into hydrogen gas and oxygen gas.

5. CONCEPTUAL EXAMPLE 1.3 Chemical Change and Physical Change continued Exercise 1.3B Which of the following events involve chemical changes and which involve physical changes? A steel wrench left out in the rain becomes rusty. A stick of butter melts. A wooden log is burned. A piece of wood is ground up into sawdust.

6. CONCEPTUAL EXAMPLE 1.4 Elements and Compounds Solution C, Ca, and In represent elements (each is a single symbol). HI, BN, and HBr are composed of two symbols each and represent compounds. Exercise 1.4A Exercise 1.4B Which of the following represent elements and which represent compounds? He CuO No NO KI Os How many different elements are represented in the entire list of Exercise 1.4A? Which of the following represent elements and which represent compounds? C Ca HI BN In HBr

7. EXAMPLE 1.5 Prefixes and Powers of Ten Solution Our goal is to replace each power of ten with the appropriate prefix from Table 1.5. For example, 10–3 = 0.001, corresponding to milli(unit). (It doesn’t matter what the unit is; here we are dealing only with the prefixes.) 10–3 corresponds to the prefix milli-; 2.89 mg; that is, 10–3x 9 (unit) = milli(unit). 103 corresponds to the prefix kilo-; 4.30 km; that is, 103x (unit) = kilo(unit). Exercise 1.5 Convert each of the following measurements to a unit that replaces the power of ten by a prefix. 7.24 x 103 g b. 4.29 x 10–6 m c. 7.91 x 10-3 s d. 2.29 x 10–2 g e. 7.90 x 106 m Convert each of the following measurements to a unit that replaces the power of ten by a prefix. a. 2.89 x 10–3 g b. 4.30 x 103 m

8. Solution • Our goal is to find the power of ten that relates the given unit to the SI base unit. That is, centi(base unit) = 10-2x (base unit) • 4.12 centimeter = 4.12 x 10-2 m • b. To change microsecond to the base unit second, we replace the prefix micro- by 10-6. • To obtain an answer in the conventional exponential form, we also need to replace the coefficient • 947 by 9.47 x 102. The result of these two changes is • 947 ms = 947 x 10-6 s = 9.47 x 102x 10-6 s = 9.47 x 10-4 s • To change nanometer to the base unit meter, we replace the prefix nano- by 10–9. The answer in exponential form is 3.17 x 10-9 m. Exercise 1.6A Exercise 1.6B Use exponential notation to express each of the following measurements in terms of an SI base unit. a. 7.45 nm b. 5.25 ms d. 1.415 km e. 2.06 mm Use exponential notation to express each of the following measurements in terms of an SI base unit. a. 284 nm b. 119 ms d. 754 km e. 6.19 x 106 mm EXAMPLE 1.6 5 Prefixes and Powers of Ten Use exponential notation to express each of the following measurements in terms of an SI base unit. a. 4.12 cm b. 947 ms c. 3.17 nm

9. EXAMPLE 1.7 5 Math, Length, Area, Volume Solution In customary units, a two-year-old child should weigh about 20–25 lb. Knowing that 1 kg is about 2.2 lb, the only reasonable answer is 10 kg. In customary units, a two-year-old child should be about 30–36 in. tall. Knowing that 1 cm is a little less than 0.5 in., the answer must be a little more than twice 30–36, or a bit more than 70–72 cm. Thus the only reasonable answer 85 cm. Exercise 1.7A Exercise 1.7B Without doing a detailed calculation, determine which of the following is a reasonable area for the front cover of your textbook. 500 mm2 50 cm2 500 cm2 50 m2 Without doing a detailed calculation, determine which of the following is a reasonable volume for your textbook. 1600 mm3 16 cm3 1600 cm3 1.6 m3 Without doing a detailed calculation, determine which of the following is a reasonable (a) mass (weight) and (b) height for a typical two-year-old child. Mass: 10 mg 10 g 10 kg 100 g Height: 85 mm 85 cm 850 cm 8.5 m

10. EXAMPLE 1.8 5 Unit Conversions Solution a. We start with the given quantity 1.83 kg and use the equivalence 1 kg = 1000 g to form a conversion factor (Appendix A) that allows us to cancel the unit kg and end with the unit g. b. Here we start with the given quantity 729 mL and use the equivalences 106mL = 1 L and 103 mL = 1 L to form conversion factors that allows us to cancel the unit mL and end with the unit mL. 1000 g 1.83 kg x x = 1830 g 1 kg Exercise 1.8 Convert (a) 7.5 m to millimeters, (b) 2056 mL to liters, (c) 2.06 g to micrograms, and (d) 0.738 cm to millimeters. 1 L 1000 mL x 729 mL x = 0.729 mL 106 mL 1 L Convert (a) 1.83 kg to grams and (b) 729 mL to milliliters.

11. EXAMPLE 1.9 5 Mass, Volume, and Density Solution From Table 1.6 we see that the density of gold (19.3 g/cm3) is greater than that of copper (8.94 g/cm3). It takes a larger block of copper to have a mass of 50.00 g than of gold. A 50.0-g block of copper has a greater volume than a 50.0-g block of gold. From Table 1.6 we see that the density of ethyl alcohol (0.789 g/mL) is greater than that of hexane (0.660 g/mL). Because it is more dense (that is, it has more mass in each unit of volume), 225 mL of ethyl alcohol has a greater mass than does 225 mL of hexane. Exercise 1.9A Exercise 1.9B Without doing a detailed calculation, determine which has the greater volume, a 500.0-g block of ice or a 500.0-g block of magnesium. Wood does not dissolve in water and will float on water if its density is less than that of water. Padouk (d = 0.86 g/cm3) and ebony (d = 1.2g/cm3) are tropical woods. Which will float on water and which will sink in water? Answer the following without doing a detailed calculation. (a) Which has the greater volume, a 50.0-g block of copper or a 50.0-g block of gold? (b) Which has the greater mass, 225 mL of ethyl alcohol or 225 mL of hexane?

12. EXAMPLE 1.10 5 Density from Mass and Volume Solution The given quantities are m = 156 g and V = 20.0 cm3 We can use the equation that defines density. mass (m) m 156 g Exercise 1.10A Exercise 1.10B = 7.80 g/cm3 = d = V 20.0 cm3 What is the density, in grams per cubic centimeter, of a metal alloy if a cube that measures 2.00 cm on an edge has a mass of 80.2g? What is the density, in grams per milliliter, of salt solution if 52.5 mL has a mass of 58.5 g? density (d) volume (V) What is the density of iron if 156 g of iron occupies a volume of 20.0 cm3?

13. EXAMPLE 1.11 5 Mass from Density and Volume Solution We can express density as a ratio, 0.703 g/1.00 mL, and use it as a conversion factor. We also need the factor 1000 mL/1 L to convert to milliliters. 0.703 g 1000 mL m = dxV = = 703 g x 1 L x 1 mL 1 L Exercise 1.11A Exercise 1.11B What is the mass of 200.0 mL of glycerol, which has a density of 1.264 g/mL at 20 °C? What is the mass of a lead brick that is 2.50 cm x 3.50 cm x 8.25 cm on an edge? (The density of lead at 20 °C is 11.34 g/cm3.) What is the mass of 1.00 L of gasoline if its density is 0.703 g/mL?

14. EXAMPLE 1.12 5 Volume from Mass and Density Solution Here we can express the inverse of density as a ratio, 1.00 mL/13.6 g, and use it as a conversion factor. 1 mL V = 461 g x = 34.1 mL 13.534 g Exercise 1.12A Exercise 1.12B What volume is occupied by a 845-g piece of magnesium? (See Table 1.6.) What volume is occupied by 225 g of hexane? (See Table 1.6.) What volume is occupied by 461 g of mercury? (See Table 1.6)

15. EXAMPLE 1.13 5 Temperature Conversions Solution K = °C + 273.15 K = 36 + 273.15 = 309 K Exercise 1.13A Exercise 1.13B What is the boiling point of water (100 °C) expressed in kelvins? Express a temperature of –78 °C in kelvins. Ether boils at 36 °C. What is the boiling point of ether on the Kelvin scale?

16. EXAMPLE 1.14 5 Energy Conversions Solution Exercise 1.14A Exercise 1.14B A European woman on average consumes food with an energy content of 7525 kJ per day. What is her daily intake in kilocalories (food calories)? It takes about 12 kJ of energy to melt a single ice cube. How many kilocalories does it take to melt three ice cubes? 1000 cal 4.184 J 1 kJ 10.3 kcal x x x = 43.1 kJ 1 kcal 1 cal 1000 J When 1.00 g of gasoline burns, it yields about 10.3 kcal of energy. What is the quantity of energy in kilojoules (kJ)?