SOLUTIONS - CHAPTER 1 Problems 1) Identify each of the following as a physical property or a chemical property a) When calcium carbonate is heated, it releases carbon dioxide and forms calcium oxide chemical, as it deals with the conversion of calcium carbonate into reaction products upon the application of heat b) The normal melting point of lithium metal is 179 C physical, as it does not involve conversion of lithium into other substances c) The density of copper metal is 8.92 g/cm 3 physical, as it does not involve the conversion of copper into other substances d) In the presence of oxygen, wood will burn chemical, as it deals with the reaction of wood with oxygen by combustion e) Radon is a radioactive gas physical, though this is a close call. Radioactivity converts one element into another element, but does not do so by a chemical reaction. 2) SI units are important in chemistry. Give the SI units for the following, and indicate whether the unit is fundamental or derived. a) length - m, meters (fundamental) b) volume - m 3, cubic meters (derived) c) mass - kg, kilograms (fundamental) d) time - s, seconds (fundamental) e) velocity - m/s, meters per second (derived) 3) What is the difference between mass and weight? Mass is a measure of the quantity or amount of matter possessed by an object, while weight is a measure of the force of attraction between an object and a large body, such as a moon, planet, or star, due to gravity. The mass of an object is independent of where it is measured, while the weight of an object depends on where the measurement is carried out. 4) (Burdge, 1.26) a) Normally the human body can endure a temperature of 105. ºF for only a short period of time without permanent damage to the brain and other vital organs. What is this temperature in degrees Celsius. ºC = 5 /9 [ ºF - 32. ] = 5 /9 [ 105. - 32. ] = 41. ºC b) Ethylene glycol is a liquid organic compound that is used as an antifreeze in car radiators. It freezes at - 11.5 ºC. Calculate its freezing temperature in degrees Fahrenheit. ºF = 9 /5 ºC + 32 = 9 /5 (- 11.5) + 32 = 11.3 ºF 1
c) The temperature at the surface of the sun is about 6300. ºC. What is this temperature in degrees Fahrenheit? ºF = 9 /5 ºC + 32 = 9 /5 (6300) + 32 = 11400 ºF (Note that there is some ambiguity as to how many significant figures should be given in the answer). 5) a) How many microliters ( L) are in 1 L? # L = 1. L 10 6 L = 10 6 L. 1 L Note that this is an exact conversion factor. b) How many microliters are in 24.6 ml? # L = 20.4 ml 1 L 10 6 L = 20400 L = 2.04 x 10 4 L 1000 ml 1 L It is best in this problem to give the final answer in scientific notation so it is clear the answer has three significant figures. 6) (Burdge, 1.30) Convert the following temperatures into degrees Celsius: a) 77. K, the boiling point of liquid nitrogen ºC = K - 273.15 = 77-273.15 = - 196. ºC b) 4.2 K, the boiling point of liquid helium ºC = K - 273.15 = 4.2-273.15 = - 268.9 ºC c) 601. K, the melting point of lead ºC = K - 273.15 = 601. - 273.15 = 328. ºC 7) (Burdge, 1.36) Express the following numbers in scientific notation a) 0.000000027 2.7 x 10-8 b) 356 3.56 x 10 2 c) 47,764 4.7764 x 10 4 d) 0.096 9.6 x 10-2 2
8) (Burdge, 1.35) Distinguish between the terms accuracy and precision. In general, explain why a precise measurement does not always guarantee an accurate result. Precision refers to how close a series of measurements of the same thing are to one another (and so is a measure of the reproducibility or degree of scatter in the measurements). Accuracy refers to how close a measurement or average of several measurements is to the true value of the thing being measured. If there is bias (systematic error) in a measurement the measurement might be precise (all the results are close together) but not accurate (the average result is far from the true value). An example would be a measurement of mass using an analytical balance that has not been correctly zeroed, and so has a systematic error of 1.5 g in the masses it displays. The results from successive measurements of the mass of the same object might be close together (precise) but they would be off from the true value for mass by 1.5 g because of the failure to correctly zero the balance. 9) (Burdge, 1.40) Determine the number of significant figures in each of the following measurements: a) 4867 mi 4 - all digits are nonzero and so significant b) 56 ml 2 - all digits are nonzero and so significant c) 60,104 tons 5 - the two zeros that appear are between nonzero digits, and so all digits are significant d) 2900 2, 3, or 4 - there is no decimal point, and so no way to tell whether the trailing zeros are significant or placeholders e) 40.2 g/cm 3 3 - the zero appears between two nonzero digits, and so all digits are significant f) 0.0000003 cm 1 - the leading zeros are not significant g) 0.7 min 1 - the leading zero is not significant h) 46 amu 2 - all digits are nonzero and so significant 10) A 25.00 ml sample of liquid benzene has a mass of 21.962 g at room temperature. What is the density of benzene? D = mass So D = 21.962 g = 0.8785 g/ml = 0.8785 g/cm 3 volume 25.00 ml Note that I have converted the final answer into units of g/cm 3 (using the fact that 1 ml = 1 cm 3, an exact conversion factor) because g/cm 3 is the common unit used for the density of a solid or a liquid. 3
11) Carry out the following conversions. a) How many grams of meat are there in a quarter-pound hamburger (0.25 lb)? # grams = 0.25 lb 453.6 g = 110 g = 1.1 x 10 2 g 1. lb It is best to give the answer in scientific notation to make clear that it has 2 significant figures. b) The height of the Sears tower in Chicago is 1454 ft. What is the height in meters? # meters = 1454 ft 12 in 2.54 cm 1m = 443.2 m 1 ft 1 in 100 cm c) The land area of Australia is 2,941,526 mi 2. What is the area in units of m 2? area = 2941526 mi 2 [(5280 ft/mi) x (12 in/ft) x (2.54 cm/in) x (1 m/100 cm)] 2 = 7.618517 x 10 12 m 2 Since area has units of mi 2, the conversion factors all have to be squared as well. Note that since all the conversion factors between miles and meters are exact, the number of significant figures in the answer is 7. If you used the direct conversion factor between miles and kilometers given in the back cover of the book, which is approximate, you would get the incorrect number of significant figures in your answer! 12) Express the results of the following calculations to the correct number of significant figures. a) (4.884) (2.05) = 10.0 (3 sig figs, since 2.05 has 3 sig figs) b) 94.61/3.7 = 26. (2 sig figs, since 3.7 has 2 sig figs) c) 3.7/94.61 = 0.039 (2 sig figs, since 3.7 has 2 sig figs) d) 5502.3 + 24. + 0.01 = 5526 (sig figs determined by the number 24.) e) (86.3 + 1.42) - 0.09 = 87.6 (sig figs determined by the number 86.3) f) (5.7) (2.31) = 13. (2 sig figs, since 5.7 has 2 sig figs) 13) (Burdge, 1.50) Carry out the following conversions a) 22.6 m to decimeters #decimeters = 22.6 m 10 dm = 226. dm 1 m 4
b) 25.4 mg to kilograms # kilograms = 25.4 mg 1g 1 kg = 0.0000254 kg = 2.54 x 10-5 kg 1000 mg 1000 g c) 556 ml to liters # liters = 556 ml 1 L = 0.556 L 1000 ml d) 10.6 kg/m 3 to g/cm 3 # g = 10.6 kg 1000 g 1 m 1 m 1 m = 0.0106 g = 1.06 x 10-2 g cm 3 m 3 1 kg 100 cm 100 cm 100 cm cm 3 cm 3 Note that I could have done the volume conversion using 1 m 3 = 10 6 cm 3, but I wanted to show explicitly how you need to use the 1 m = 100 cm conversion three times when going from m 3 to cm 3. 14) (Burdge, 1.56) How many minutes does it take light from the sun to reach Earth? (The distance from the sun to Earth is 93 million miles; the speed of light is 3.00 x 10 8 m/s) While there is a direct conversion from miles to meters, I am going to do this calculation the long way to better show dimensional analysis in action. I have broken the calculation into two parts. I first find the distance between the Earth and the sun in meters, and then use the speed of light to convert from distance to time. # meters = 93 x 10 6 miles 5280 ft 12 in 2.54 cm 1 m = 1.497 x 10 11 m 1 mile 1 ft 1 in 100 cm # minutes = 1.497 x 10 11 m 1 s 1 min = 8.3 min 3.00 x 10 8 m 60 s The number of significant figures in the final result is two, because 93 million miles has two significant figures. Note that I round to the correct number of significant figures at the end of the calculation, and not in the intermediate steps, to avoid roundoff error. 15) The density of chloroform, a common organic solvent, is D = 1.4832 g/ml at 20 C. How many milliliters of chloroform would you need to have a mass of 112.5 g of chloroform? # ml = 112.5 g 1 ml = 75.85 ml 1.4832 g 5
16) (Burdge, 1.62) Carry out the following conversions: a) 1.42 km to miles # miles = 1.42 km 1000 m 100 cm 1 in 1 ft 1 mile = 0.882 mi 1 km 1 m 2.54 cm 12 in 5280 ft There is a direct conversion factor between miles and kilometers, but I have chosen this longer conversion to show how a multistep conversion can be carried out b) 32.4 yards to centimeters # cm = 32.4 yd 36 in 2.54 cm = 2.96 x 10 3 cm 1 yd 1 in I have written the result in scientific notation to make clear that the final answer has three significant figures. c) 3.0 x 10 10 cm/s to ft/s # ft = 3.0 x 10 10 cm 1 in 1 ft = 9.8 x 10 8 ft/s s s 2.54 cm 12 in 17) An irregularly shaped piece of silicon weighing 8.763 g is dropped into a graduated cylinder containing water. The initial level of water in the cylinder is 25.12 ml. After the piece of silicon is dropped into the graduated cylinder the level of water is 28.88 ml. Based on this information, what is the density of silicon? The volume occupied by the piece of silicon is V = 28.88 ml - 25.12 ml = 3.76 ml D = mass So D = 8.763 g = 2.33 g/ml = 2.33 g/cm 3 volume 3.76mL 18) A 1.0 ounce (oz) piece of chocolate contains 15 mg of caffeine. A 6.0 ounce regular cup of coffee contains 105 mg of caffeine. How much chocolate would you have to consume to get as much caffeine as you would get from 2 cups of coffee? g caffeine (from coffee) = 2 cups 105 mg = 210 mg = 2.10 x 10 2 mg 1 cup So oz chocolate = 2.10 x 10 2 mg caffeine 1.0 oz chocolate = 14.0 oz chocolate 15 mg caffeine 6
19) (Burdge, 1.72) In determining the density of a rectangular metal bar, a student made the following measurements: length = 8.53 cm; width = 2.4 cm; height = 1.0 cm; mass = 52.7064 g. Calculate the density of the metal to the correct number of significant figures. volume = (length) x (width) x (height) = (8.53 cm) (2.4 cm) (1.0 cm) = 20.47 cm 3 density = mass = 52.7064 g = 2.575 g/cm 3 = 2.6 g/cm 3 volume 20.47 cm 3 There are two significant figures in the final result because both width and height are given to two significant figures. 20) (Burdge, 1.76) The speed of sound in air at room temperature is about 343 m/s. Calculate this speed in miles per hour (1 mi = 1609. m). We will first convert to miles/s, and then to miles/hr. # mile = 343 m 100 cm 1 in 1 ft 1 mile = 0.2131 mile/s s 1 s 1 m 2.54 cm 12 in 5280 ft # mile = 0.2131 mile 60 s 60 min = 767 mile/hr hr s 1 min 1 hr Since the speed m/s was given to three significant figures, the final result is also given to three significant figures. 21) (Burdge, 1.96) A sheet of aluminum (Al) foil has a total area of 1.000 ft 2 and a mass of 3.636 g. What is the thickness of the foil in mm? (density of Al = 2.699 g/cm 3 )? We first convert the area from ft 2 to cm 2. Area = 1.000 ft 2 12 in 12 in 2.54 cm 2.54 cm = 929.03 cm 2 1 ft 1 ft 1 in 1 in The volume of the aluminum foil can be found from the mass of the foil and the density of aluminum. Volume = 3.636 g 1 cm 3 = 1.3472 cm 3 2.699 g But Volume = (Area) (thickness), and so thickness = Volume = 1.3472 cm 3 = 0.001450 cm = 0.01450 mm = 1.450 x 10-2 mm Area 929.03 cm 2 7
22) (Burdge, 1.108) Bronze is an alloy made of copper (Cu) and tin (Sn). Calculate the mass of a bronze cylinder of radius 6.44 cm and length 44.37 cm. The composition of the bronze cylinder is 79.42 percent Cu and 20.58 percent Sn, and the densities of Cu and Sn are 8.94 g/cm 3 and 7.31 g/cm 3, respectively. What assumptions must you make in this calculation? We first find the volume of the brass cylinder. For a cylinder Volume = r 2 h = (6.44 cm) 2 (44.37 cm) = 5781.1 cm 3 If we assume that the volume of the alloy is additive (that is, that the volume of the alloy is equal to the volume of the copper + the volume of the tin used to make the alloy) we can then find the volume of a 1.0000 g sample of the alloy. A 1.000 g sample will contain 0.7942 g Cu and 0.2058 g Sn, and so Volume of 1.000 g = (0.7942 g Cu) 1 cm 3 + (0.2058 g Sn) 1 cm 3 = 0.11699 cm 3 8.94 g 7.31 g 1 gm We can use this as a conversion factor to find the total mass of the cylinder Mass cylinder = 5781.1 cm 3 1 gm = 4.94 x 10 4 g = 49.4 kg 0.11699 cm 3 As with some of the other problems, we have waited until the end of the calculation to round our results to the correct number of significant figures to avoid roundoff error. 8