Chapter01rw-final

Chapter01rw-final - 1 1.1 (a) (b) (c) (d) 1.2 Introduction:...

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Unformatted text preview: 1 1.1 (a) (b) (c) (d) 1.2 Introduction: Matter and Measurement Visualizing Concepts Pure elements contain only one kind of atom. Atoms can be present singly or as tightly bound groups called molecules. Compounds contain two or more kinds of atoms bound tightly into molecules. Mixtures contain more than one kind of atom and/or molecule, not bound into discrete particles. pure element: i, v mixture of elements: vi pure compound: iv mixture of an element and a compound: ii, iii After a physical change, the identities of the substances involved are the same as their identity before the change. That is, molecules retain their original composition. During a chemical change, at least one new substance is produced; rearrangement of atoms into new molecules occurs. The diagram represents a chemical change, because the molecules after the change are different than the molecules before the change. 1.3 (a) (f) time volume (b) density (g) (c) length (d) area (e) temperature temperature 1.4 Density is the ratio of mass to volume. For a sphere, size is like volume both are determined by the radius of the sphere. For spheres of the same size or volume, the denominator of the density relationship is the same. The denser the sphere, the heavier it is. A list from lightest to heaviest is in order of increasing density and mass. The aluminum sphere (density = 2.70 g/cm3) is lightest, then nickel (density = 8.90 g/cm3), then silver (density = 10.409 g/cm3). Measurements (darts) that are close to each other are precise. Measurements that are close to the “true value” (the bull’s eye) are accurate. (a) (b) Figure ii represents data that are both accurate and precise. The darts are close to the bull’s eye and each other. Figure i represents data that are precise but inaccurate. The darts are near each other but their center point (average value) is far from the bull’s eye. 1.5 1 1 Matter and Measurement (c) 1.6 (a) Solutions to Exercises Figure iii represents data that are imprecise but their average value is accurate. The darts are far from each other, but their average value, or geometric center point, is close to the bull’s eye. 7.5 cm. There are two significant figures in this measurement; the number of cm can be read precisely, but there is some estimating (uncertainty) required to read tenths of a centimeter. Listing two significant figures is consistent with the convention that measured quantities are reported so that there is uncertainty in only the last digit. 140°F. The temperature can be read to the nearest 50°F and estimated to the nearest 5–10°F. Since there is uncertainty in the tens digit, the measurement has two significant figures. (b) 1.7 The determined age of the artifact, 1,900 years, has two significant figures. There is uncertainty in the hundreds place, indicating that the minimum uncertainty in age is 100 years. The 20-year period since the age was determined is not significant relative to the determined age. (a) Volume = length × width × height. Because the operation is multiplication, the dimension with fewest significant figures (sig figs) determines the number of sig figs in the result. The dimension “2.0 cm” has 2 sig figs, so the volume is reported with 2 sig figs. Density = mass/volume. Because the operation is division, again the datum with fewer significant figures determines the number of sig figs in the result. While mass has 4 sig figs, volume [from (a)] has 2 sig figs, so density is also reported to 2 sig figs. 1.8 (b) 1.9 In order to cancel units, the conversion factor must have the unit being canceled opposite the starting position. For example, if the unit cm starts in the numerator, then the conversion factor must have cm in its denominator. However, if the unit cm starts in the denominator, the conversion factor must have cm in the numerator. Ideally, this will lead to the desired units in the appropriate location, numerator or denominator. However, the inverse of the answer can be taken when necessary. Given: mi/hr Find: km/s Given mi/hr use 1 ⎯ ⎯ ⎯⎯ ⎯km⎯ → 0.62 mi use 1 hr ⎯⎯ ⎯ ⎯ ⎯→ 60 min ⎯use 1⎯ ⎯ → ⎯ ⎯ hr⎯ 60 min Find km/s 1.10 km/hr km/min Classification and Properties of Matter 1.11 (a) (b) (c) (d) heterogeneous mixture homogeneous mixture (If there are undissolved particles, such as sand or decaying plants, the mixture is heterogeneous.) pure substance homogeneous mixture 2 1 Matter and Measurement 1.12 (a) (b) (c) (d) 1.13 (a) (g) 1.14 (a) (g) 1.15 A(s) homogeneous mixture heterogeneous mixture (particles in liquid) pure substance heterogeneous mixture S nickel C calcium (b) Mg (h) sodium (b) N (h) helium B(s) + C(g) (c) K (d) Cl Solutions to Exercises (e) Cu (f) fluorine (i) aluminum (c) Br (i) lead (j) silicon (e) Fe (f) phosphorus (d) Zn (j) silver When solid carbon is burned in excess oxygen gas, the two elements combine to form a gaseous compound, carbon dioxide. Clearly substance C is this compound. Since C is produced when A is heated in the absence of oxygen (from air), both the carbon and oxygen in C must have been present in A originally. A is, therefore, a compound composed of two or more elements chemically combined. Without more information on the chemical or physical properties of B, we cannot determine absolutely whether it is an element or a compound. However, few if any elements exist as white solids, so B is probably also a compound. 1.16 Before modern instrumentation, the classification of a pure substance as an element was determined by whether it could be broken down into component elements. Scientists subjected the substance to all known chemical means of decomposition, and if the results were negative, the substance was an element. Classification by negative results was somewhat ambiguous, since an effective decomposition technique might exist but not yet have been discovered. Physical properties: silvery white (color); lustrous; melting point = 649°C; boiling point = 1105°C; density at 20°C = 1.738 g/cm 3 ; pounded into sheets (malleable); drawn into wires (ductile); good conductor. Chemical properties: burns in air to give intense white light; reacts with Cl 2 to produce brittle white solid. Physical properties: silver-grey (color); melting point = 420°C; hardness = 2.5 Mohs; density = 7.13 g/cm 3 at 25°C. Chemical properties: metal; reacts with sulfuric acid to produce hydrogen gas; reacts slowly with oxygen at elevated temperatures to produce ZnO. (a) (a) (b) (c) (d) chemical chemical physical physical (The production of H 2 O is a chemical change, but its condensation is a physical change.) physical (The production of soot is a chemical change, but its deposition is a physical change.) (b) physical (c) physical (d) chemical (e) chemical 1.17 1.18 1.19 1.20 3 1 Matter and Measurement 1.21 (a) Solutions to Exercises Take advantage of the different water solubilities of the two solids. Add water to dissolve the sugar; filter this mixture, collecting the sand on the filter paper and the sugar water in the flask. Evaporate the water from the flask to recover solid sugar. Either the melting-point difference or magnetism difference between iron and sulfur can be used to separate these two elements. Heat the mixture until the sulfur melts, then decant (pour off) the liquid sulfur. Or use a magnet to attract the iron particles, leaving the solid sulfur behind. (b) 1.22 Take advantage of differences in physical properties to separate the components of a mixture. First heat the liquid to 100°C to evaporate the water. This is conveniently done in a distillation apparatus (Figure 1.13) so that the water can be collected. After the water is completely evaporated and if there is a residue, measure the physical properties of the residue such as color, density, and melting point. Compare the observed properties of the residue to those of table salt, NaCl. If the properties match, the colorless liquid contained table salt. If the properties don’t match, the liquid contained a different dissolved solid. If there is no residue, no dissolved solid is present. Units and Measurement 1.23 (a) (f) 1.24 (a) (b) (c) (d) 1 × 10 – 1 1 × 10 3 (b) 1 × 10 – 2 (c) (h) 1 × 10 – 1 5 1 × 10 – 3 (d) 1 × 10 – 6 (i) 1 × 10 – 12 (e) 1 × 10 6 (g) 1 × 10 – 9 6.35 × 10 − 2 L × 6.5 × 10 −6 s × 9.5 × 10 − 4 m × 1 mL = 63.5 mL 1 × 10 − 3 L 1 μs = 6.5 μs = 0.95 mm = 4.23 mm 3 1 × 10 −3 L 1 mL 1 μL 1 × 10 −6 L 1 × 10 −6 s 1 mm 1 × 10 − 3 m (1 × 10 4.23 × 10 −9 m 3 × 1 3 mm 3 −3 3 )m × 3 4.23 mm 3 × (10 −1 ) 3 cm 3 1 mm 3 3 1 mL 1 cm × 3 × × = 4.23 μL (e) 12.5 × 10 −8 kg × 1 × 10 3 g 1 kg 1 mg 1 × 10 − 3 g = 0.125 mg (125 μg) (f) 3.5 × 10 −10 g × 1 ng 1 × 10 −9 g 1 × 10 −15 s 1 fs = 0.35 ng (g) 1.25 (a) 6.54 × 10 9 fs × × 1 μs 1 × 10 −6 s = 6.54 μs °C = 5/9 (°F – 32°); 5/9 (62 – 32) = 17°C 4 1 Matter and Measurement (b) (c) (d) (e) 1.26 (a) (b) (c) °F = 9/5 (°C) + 32°; 9/5 (216.7) + 32 = 422.1°F K = °C + 273.15; 233°C + 273.15 = 506 K 315 K – 273 = 42°C; 9/5 (42°C) + 32 = 108°F Solutions to Exercises °C = 5/9 (°F – 32°); 5/9 (2500 – 32) = 1371°C; 1371°C + 273 = 1644 K (assuming 2500 C has 4 sig figs) °C = 5/9 (87°F – 32°) = 31°C K = 25°C + 273 = 298 K; °F = 9/5 (25°C) + 32 = 77°F °C = 5/9 (175°F – 32°) = 79.444 = 79.4°C K = °C + 273.15 = 79.444°C + 273.15 = 352.6 K (d) °F = 9/5 (755°C) + 32 = 1391°F; K = 755°C + 273.15 = 1028 K (It could be argued that the result of 9/5 (755) has 3 sig figs, so the final Fahrenheit temperature should have 3 sig figs, 1390°F.) (e) melting point = –248.6°C + 273.15 = 24.6 K boiling point = –246.1°C + 273.15 = 27.1 K density = 39.73 g mass = = 1.59 g/mL or 1.59 g/cm 3 volume 25.0 mL 1.27 (a) (The units cm 3 and mL will be used interchangeably in this manual.) Carbon tetrachloride, 1.59 g/mL, is more dense than water, 1.00 g/mL; carbon tetrachloride will sink rather than float on water. (b) (c) 1.28 (a) 75.00 cm 3 × 21.45 g cm 3 = 1.609 × 10 3 g (1.609 kg) 87.50 g × 1 cm = 50.3452 = 50.35 cm 3 = 50.35 mL 1.738 g 3 volume = length 3 (cm 3 ); density = mass/volume (g/cm 3 ) volume = (1.500) 3 cm 3 = 3.375 cm 3 density = 76.31 g 3.375 cm 3 = 22.61 g/cm 3 osmium (b) 125.0 mL × 4.51 g 1 cm 3 × = 563.75 = 564 g titanium 1 mL 1 cm 3 0.8787 g 1 mL × = 131.8 g benzene 1 mL 1 × 10 − 3 L (c) 0.1500 L × 1.29 (a) density = 38.5 g 45 mL = 0.86 g/mL The substance is probably toluene, density = 0.866 g/mL. 5 1 Matter and Measurement (b) (c) 1.30 (a) 45.0 g × 1 mL = 40.4 mL ethylene glycol 1.114 g Solutions to Exercises ( 5.00) 3 cm 3 × 21.95 g 25.0 mL 8.90 g 1 cm 3 = 1.11 × 10 3 g (1.11 kg) nickel = 0.878 g/mL The tabulated value has four significant figures, while the experimental value has three. The tabulated value rounded to three figures is 0.879. The values agree within one in the last significant figure of the experimental value; the two results agree. The liquid could be benzene. (b) (c) 15.0 g × 1 mL = 19.3 mL cyclohexan e 0.7781 g r = d/2 = 5.0 cm/2 = 2.5 cm V = 4/3 π r 3 = 4/3 × π × (2.5) 3 cm 3 = 65 cm 3 = 7.4 × 10 2 g cm 3 (The answer has two significant figures because the diameter had only two figures.) 65.4498 cm 3 × 11.34 g Note: This is the first exercise where “intermediate rounding” occurs. In this manual, when a solution is given in steps, the intermediate result will be rounded to the correct number of significant figures. However, the unrounded number will be used in subsequent calculations. The final answer will appear with the correct number of significant figures. That is, calculators need not be cleared and new numbers entered in the middle of a calculation sequence. This may result in a small discrepancy in the last significant digit between student-calculated answers and those given in the manual. These variations occur in any analysis of numerical data. For example, in this exercise the volume of the sphere, 65.4498 cm 3 , is rounded to 65 cm 3 , but 65.4498 is retained in the subsequent calculation of mass, 7.4 × 10 2 g. In this case, 65 cm 3 × 11.34 g/cm 3 also yields 7.4 × 10 2 g. In other exercises, the correctly rounded results of the two methods may not be identical. 1.31 thickness = volume/area volume = 200 mg × 1 × 10 −3 g 1 mg 12 2 in 2 × × 1 cm 3 = 0.01035 = 0.0104 cm 3 19.32 g 2.54 2 cm 2 area = 2.4 ft × 1.0 ft × thickness = 0.01035 cm 3 2 ,230 cm 2 1 nm = 2.23 × 10 3 = 2.2 × 10 3 cm 2 1 ft 2 in 2 1 × 10 −2 m × = 4.6 × 10 −8 m 1 cm = 46 nm thick 4.6 × 10 −8 m × 1 × 10 −9 m 6 1 Matter and Measurement 1.32 Calculate the volume of the rod: 2.17 kg × 1000 g 1 kg × 1 cm 3 = 931.3 = 931 cm 3 2.33 g 2 Solutions to Exercises 4V ⎛d⎞ ⎛4V⎞ ;d =⎜ V = π r 2 h; d = 2r, r = d/2; V = π ⎜ ⎟ h ; d 2 = ⎟ πh ⎝ πh⎠ ⎝2⎠ 1/ 2 ⎛ 4 (931.3) cm 3 d=⎜ ⎜ π (16.8) cm ⎝ ⎞ ⎟ ⎟ ⎠ 1/ 2 = 8.401 = 8.40 cm Uncertainty in Measurement 1.33 Exact: (c), (d), and (f) (All others depend on measurements and standards that have margins of error, e.g., the length of a week as defined by the earth’s rotation.) Exact: (b), (e) (The number of students is exact on any given day.) (a) (a) (a) (d) 1.38 1.39 (a) (a) (b) (c) (d) 1.40 (a) (b) 3 4 (b) 2 (b) 3 (c) 5 (c) 4 (d) 3 (d) 5 (e) 5 (e) 6 (c) 8.543 × 10 – 3 1.34 1.35 1.36 1.37 1.025 × 10 2 2.579 × 10 – 4 7.93 × 10 3 mi (b) 6.570 × 10 5 (e) –3.572 × 10 – 2 (b) 4.001 × 10 4 km 12.0550 + 9.05 = 21.105 = 21.11 (For addition and subtraction, the minimum number of decimal places, here two, determines decimal places in the result.) 257.2 – 19.789 = 237.4 (6.21 × 10 3 )(0.1050) = 652 (For multiplication and division, the minimum number of significant figures, here three, determines sig figs in the result.) 0.0577/0.753 = 7.66 × 10 – 2 [320.5 – 6104.5/2.3] = –2.3 × 10 3 (The intermediate result has two significant figures, so only the thousand and hundred places in the answer are significant.) [285.3 × 10 5 – 0.01200 × 10 5 ] × 2.8954 = 8.260 × 10 7 (Since subtraction depends on decimal places, both numbers must have the same exponent to determine decimal places/sig figs. The intermediate result has 1 decimal place and 4 sig figs, so the answer has 4 sig figs.) (0.0045 × 20,000.0) 2 sig figs /0 dec pl 863 3 sig figs + (2813 × 12) = 3.4 × 10 4 2 sig figs /first 2 digits (3.45 × 108)] (3 sig figs /0 dec pl) [0 dec pl/3 sig figs] – = 7.62 × 10 5 = 3 sig figs (c) (d) × × [1255 7 1 Matter and Measurement Dimensional Analysis 1.41 Solutions to Exercises In each conversion factor, the old unit appears in the denominator, so it cancels, and the new unit appears in the numerator. (a) mm → nm : mg → kg : 1 × 10 −3 m 1 nm × = 1 × 106 nm / mm −9 1 mm 1 × 10 m × 1 kg 1000 g 1 cm 1 × 10 −2 (b) 1 × 10 −3 g 1 mg 1000 m 1 km × = 1 × 10 −6 kg/mg (c) km → ft : m × 1 in 1 ft × = 3.28 × 103 km/ft 2.54 cm 12 in (d) 1.42 in 3 → cm 3 : (2.54) 3 cm 3 1 3 in 3 = 16.4 cm3/in3 In each conversion factor, the old unit appears in the denominator, so it cancels, and the new unit appears in the numerator. (a) μm → mm : 1 × 10 −6 m 1 mm × = 1 × 10 − 3 mm / μm 1 μm 1 × 10 − 3 m 1 × 10 −3 s 1 ms × 1 ns 1 × 10 −9 s = 1 × 10 6 ns / ms (b) (c) (d) ms → ns : mi → km: 1.6093 km/mi ft 3 → L : (12) 3 in 3 1 ft 3 × (2.54) 3 cm 3 1 in 3 × 1L 1000 cm 3 = 28.3 L/ft 3 1.43 (a) (b) (c) (d) (e) (f) 0.076 L × 1000 mL = 76 mL 1L 1 nm 1 × 10 −9 m 1 × 10 −9 s 1 ns = 50. nm = 6.88 × 10 − 4 s 5.0 × 10 −8 m × 6.88 × 10 5 ns × 0.50 lb × 1.55 kg m3 5.850 gal hr × 453.6 g 1 lb 1 kg × = 226.8 = 2.3 × 10 2 g × 1m3 (10) 3 dm 3 × The data and the result have 2 sig figs. 1000 g 1 dm 3 = 1.55 g/L 1L 3.7854 L 1 hr 1 min × × = 6.151 × 10 − 3 L/s 1 gal 60 min 60 s Estimated answer: 6 × 4 = 24; 24/60 = 0.4; 0.4/60 = 0.0066 = 7 × 10 – 3. This agrees with the calculated answer of 6.151 × 10 – 3 L/s. 8 1 Matter and Measurement 1.44 (a) (b) 1454 ft × 1 yd 3 ft × Solutions to Exercises 2.998 × 10 8 m 1 km 60 s 60 min × × × = 1.079 × 10 9 km/hr s 1000 m 1 min 1 hr 1m = 443.18 = 443.2 m 1.0936 yd 1 3 dm 3 (1 × 10 ) m × −1 3 3 (c) 3,666 ,500 m 3 × × 1L 1 dm 3 = 3.6665 × 10 9 L 1 × 10 −3 g 1 mg (d) 232 mg cholesterol 100 mL blood 5.00 days × 0.0550 mi × 1 mL 1 × 10 − 3 L × 5.2 L × = 12 g cholesterol 1.45 (a) (b) (c) 24 hr 60 min 60 s × × = 4.32 × 10 5 s 1 day 1 hr 1 min 1.6093 km 1000 m × = 88.5 m mi 1 km 1 gal $1.89 $0.499 × = gal 3.7854 L L 1 × 10 −2 m 0.510 in 2.54 cm 1 km 1 ms 60 s 60 min km × × × × × × = 46.6 −3 ms 1 in 1 cm 1000 m 1 min 1 hr hr 1 × 10 s (d) Estimate: 0.5 × 2.5 = 1.25; 1.25 × 0.01 ≈ 0.01; 0.01 × 60 × 60 ≈ 36 km/hr (e) 22.50 gal min × 3.7854 L 1 min × = 1.41953 = 1.420 L/s gal 60 s Estimate: 20 × 4 = 80; 80/60 ≈ 1.3 L/s (f) 0.02500 ft 3 × 12 3 in 3 1 ft 3 × 2.54 3 cm 3 1 in 3 = 707.9 cm 3 Estimate: 10 3 = 1000; 3 3 = 27; 1000 × 27 = 27,000; 27,000/0.04 ≈ 700 cm 3 1.46 (a) (b) 0.105 in × 0.650 qt × 1 × 10 −2 m 1 mm 2.54 cm × × = 2.667 = 2.67 mm cm in 1 × 10 − 3 m 1L 1 mL × = 614.94 = 615 mL 1.057 qt 1 × 10 −3 L (c) 8.75 μ m 1 × 10 −6 m 1 km 60 s 60 min × × × × = 3.15 × 10 − 5 km/hr 3 s 1 μm 1 min 1 hr 1 × 10 m 1.955 m 3 × (1.0936) 3 yd 3 1m 3 (d) (e) (f) = 2.55695 = 2.557 yd 3 $3.99 2.205 lb × = 8.798 = $8.80/kg lb 1 kg 8.75 lb ft 3 × 453.59 g 1 lb × 1 ft 3 12 3 in 3 × 1 in 3 2.54 3 cm 3 × 1 cm 3 = 0.140 g/mL 1 mL 9 1 Matter and Measurement 1.47 (a) 31 gal × 4 qt 1 gal × 1L = 1.2 × 10 2 L 1.057 qt Solutions to Exercises Estimate: (30 × 4)/1 ≈ 120 L (b) 6 mg × 1 kg × 150 lb = 4 × 10 2 mg kg (body) 2.205 lb Estimate: 6/2 = 3; 3 × 150 = 450 mg (c) 254 mi 1.609 km 1 gal 1.057 qt 9.64 km × × × = 11.2 gal 1 mi 4 qt 1L L Estimate: 250/10 = 25; 1.6/4 = 0.4; 25 × 0.4 × 1 ≈ 10 km/L (d) 50 cups 1 lb × 1 qt 1L 1000 mL 1 lb 26 mL × × × = 4 cups 1.057 qt 1L 453.6 g g Estimate: 50/4 = 12; 1000/500 = 2; (12 × 2)/1 ≈ 24 mL/g 1.48 (a) 1486 mi × charge 1 km × = 10.6 charges 0.62137 mi 225 km Since charges are integral events, 11 charges are required. (b) 14 m 1 km 1 mi 60 s 60 min × × × × = 31 mi/hr 3 s 1 hr 1 × 10 m 1.6093 km 1 min 450 in 3 × ( 2.54) 3 cm 3 1 in 3 × 1 mL 1 cm 3 × × 1 × 10 −3 L = 7.37 L 1 mL (c) (d) 1.49 2.4 × 10 5 barrels × 1L × = 3.8 × 10 7 L 1 barrel 1 gal 1.057 qt 42 gal 4 qt 12.5 ft × 15.5 ft × 8.0 ft = 1580 = 1.6 × 10 3 ft 3 (2 sig figs) 1550 ft 3 × (1 yd) 3 (3 ft) 3 × (1 m) 3 (1.0936) yd 3 3 × 10 3 dm 3 1m 3 × 1L 1 dm 3 × 1.19 g L × 1 kg 1000 g = 52 kg air Estimate: 1550/30 = 50; (50 × 1)/1 ≈ 50 kg 1.50 9.0 ft × 14.5 ft × 18.8 ft = 2453.4 = 2.5 × 10 3 ft 3 2453.4 ft 3 × (1 yd) 3 ( 3 ft) 3 × (1 m) 3 (1.094 yd) 3 × 48 μg CO 1 × 10 −6 g × = 3.3 × 10 − 3 g CO 1 μg 1m3 1.51 Select a common unit for comparison, in this case the cm. 1 in ≈ 2.5 cm, 1 m = 100 cm 57 cm = 57 cm 14 in ≈ 35 cm 1.1 m = 110 cm The order of length from shortest to longest is 14-in shoe < 57-cm string < 1.1-m pipe. 10 1 Matter and Measurement 1.52 1 kg > 2 lb, 1 L ≈ 1 qt 5 lb potatoes < 2.5 kg 5 kg sugar = 5 kg Solutions to Exercises Select a common unit for comparison, in this case the kg. 1 gal = 4 qt ≈ 4 L. 1 mL H 2 O = 1 g H 2 O. 1 L = 1000 g, 4 L = 4000 g = 4 kg The order of mass from lightest to heaviest is 5 lb potatoes < 1 gal water < 5 kg sugar. 1.53 (a) (b) 26.73 g total × $25.00 × 0.90 g Ag 1 g total × 1 tr oz $1.18 × = $0.91 31.1 g Ag 1 tr oz 31.1 g 1 g total 1 tr oz 1 coin × × × = 2.4 coins $13.25 1 tr oz 0.90 g Ag 26.73 g Since coins come in integer numbers, 3 coins are required. 1.54 A wire is a very long, thin cylinder of volume, V = π r 2 h, where h is the length of the wire and π r 2 is the cross-sectional area of the wire. Strategy: 1) Calculate total volume of copper in cm 3 from mass and density 2) h (length in cm) = 3) Change cm → ft 150 lb Cu × 453.6 g 1 lb Cu × 1 cm 3 = 7610.7 = 7.61 × 10 3 cm 3 8.94 g V π r2 r = d/2 = 8.25 mm × h= V πr 2 1 cm 1 × = 0.4125 = 0.413 cm 10 mm 2 2 = 7610.7 cm 3 π(0.4125) cm 2 = 1.4237 × 10 4 = 1.42 × 10 4 cm 1 in 1 ft × = 467 ft 2.54 cm 12 in (too difficult to estimate) 1.4237 × 10 4 cm × Additional Exercises 1.55 1.56 Composition is the contents of a substance, the kinds of elements that are present and their relative amounts. Structure is the arrangement of these contents. (a) A gold coin is probably a solid solution. Pure gold (element 79) is too soft and too valuable to be used for coinage, so other metals are added. However, the simple term “gold coin” does not give a specific indication of the other metals in the mixture. A cup of coffee is a solution if there are no suspended solids (coffee grounds). It is a heterogeneous mixture if there are grounds. If cream or sugar is added, the homogeneity of the mixture depends on how thoroughly the components are mixed. 11 1 Matter and Measurement (b) Solutions to Exercises A wood plank is a heterogeneous mixture of various cellulose components. The different domains in the mixture are visible as wood grain or knots. The ambiguity in each of these examples is that the name of the substance does not provide a complete description of the material. We must rely on mental images, and these vary from person to person. A hypothesis is a possible explanation for certain phenomena based on preliminary experimental data. A theory may be more general, and has a significant body of experimental evidence to support it; a theory has withstood the test of experimentation. A scientific law is a summary or statement of natural behavior; it tells how matter behaves. A theory is an explanation of natural behavior; it attempts to explain why matter behaves the way it does. 1.57 (a) (b) 1.58 Any sample of vitamin C has the same relative amount of carbon and oxygen; the ratio of oxygen to carbon in the isolated sample is the same as the ratio in synthesized vitamin C. 2.00 g O 1.50 g C = xgO 6.35 g C ; x= (2.00 g O) (6.35 g C) 1.50 g C = 8.47 g O This calculation assumes the law of constant composition. 1.59 (a) I. (22.52 + 22.48 + 22.54)/3 = 22.51 II. (22.64 + 22.58 + 22.62)/3 = 22.61 Based on the average, set I is more accurate. That is, it is closer to the true value of 22.52%. (b) Average deviation = ∑ | value − average | /3 I. | 22.52 – 22.51 | + |22.48 – 22.51 | + |22.54 – 22.51 |/3 = 0.02 II. | 22.64 – 22.61 | + |22.58 – 22.61 | + |22.62 – 22.61 |/3 = 0.02 The two sets display the same precision, even though set I is more accurate. 1.60 (a) Inappropriate. The circulation of a widely-read publication like National Geographic would vary over a year’s time, and could simply not be counted to the nearest single subscriber. Probably about four significant figures would be appropriate. Inappropriate. In a county with 5 million people, the population fluctuates with moves, births, and deaths. The population cannot be known precisely to the nearest person, even over the course of a day. There would be uncertainty in at least the tens, probably the hundreds place in the population. Appropriate. The percentage has three significant figures. In a population as large as the United States, the number of people named Brown can surely be counted by census data or otherwise to a precision of three significant figures. (b) (c) 12 1 Matter and Measurement 1.61 (a) (e) 1.62 (a) volume time m (b) s2 Solutions to Exercises (c) volume (g) temperature × m= × kg − m 2 s2 2 1 kg − m = s s3 (b) area (f) kg − m s2 (d) density length (c) kg − m s2 (d) kg - m s2 × 1 m2 = kg m - s2 (e) kg − m 2 s2 1.63 (a) (b) 2.4 × 10 5 mi × 2.4 × 10 5 mi × 1.609 km 1000 m × = 3.9 × 10 8 m 1 mi 1 km 1.609 km 1 hr 60 min 60 s × × × = 4.0 × 10 6 s 1 mi 350 km 1 hr 1 min 1.64 (a) (b) (c) (d) 575 ft × 1 quarter 12 in 2.54 cm 10 mm × × × = 1.1307 × 10 5 = 1.13 × 10 5 quarters 1 ft 1 in 1 cm 1.55 mm 5.67 g 1 quarter = 6.41 × 10 5 g (641 kg) 1.1307 × 10 5 quarters × 1.1307 × 10 5 quarters × 1 dollar = $28 ,268 = $2.83 × 10 4 4 quarters $8.7 × 10 12 × 1 stack = 3.1 × 10 8 stacks (approximately 310 million stacks) $28,268 1.65 (a) (1.094 yd) 3 $1950 1 acre 3 ft (1 m) 3 (1 dm) 3 × × × × × = acre − ft 1 yd 1L 4840 yd 2 (1 m) 3 (10 dm) 3 $1.583 × 10 – 3/L or 0.1583 ¢/L (b) 1.66 (0.158 ¢/L to 3 sig figs) 1 year $2.671 $2.67 1 acre − ft $1950 × × × 1 household = = day day 2 households − year 365 days acre - ft There are 209.1 degrees between the freezing and boiling points on the Celsius (C) scale and 100 degrees on the glycol (G) scale. Also, –11.5°C = 0°G. By analogy with °F and °C, ºG = 100 209.1 (ºC + 11.5) or ºC = ( ºG) − 11.5 209.1 100 These equations correctly relate the freezing point (and boiling point) of ethylene glycol on the two scales. f.p. of H 2 O : ºG = 100 (0ºC + 11.5) = 5.50ºG 209.1 13 1 Matter and Measurement 1.67 The most dense liquid, Hg, will sink; the least dense, cyclohexane, will float; H 2 O will be in the middle. Solutions to Exercises 1.68 Density is the ratio of mass and volume. For substances with different densities, the greater the density the smaller the volume of substance that will contain a certain mass. Since volume is directly related to diameter (V = 4/3 π r 3 = 1/6 π d 3 ), the more dense the substance, the smaller the diameter of a ball that contains a certain mass. The order of the sphere sizes (diameters) is the reverse order of densities: Pb < Ag < Al. The mass of water in the bottle does not change with temperature, but the density (ratio of mass to volume) does. That is, the amount of volume occupied by a certain mass of water changes with temperature. Calculate the mass of water in the bottle at 25°C, and then the volume occupied by this mass at −10°C. (a) 25°C: 1.50 L H2O × 1000 cm 3 1L × 0.997 g H 2 O 1 cm 3 1 cm 3 0.917 g H 2 O × 1.69 = 1.4955 × 103 = 1.50 × 103 g H2O −10°C: 1.4955 × 103 g H2O × 1L 1000 cm 3 = 1.6309 = 1.63 L (b) If the soft-drink bottle is completely filled with 1.50 L of water, the 1.63 L of ice cannot be contained in the bottle. The extra volume of ice will push through any opening in the bottle, or crack the bottle to create an opening. 1.70 mass of toluene = 58.58 g – 32.65 g = 25.93 g volume of toluene = 25.93 g × 1 mL = 30.0116 = 30.0 mL 0.864 g volume of solid = 50.00 mL – 30.0116 mL = 19.9884 = 20.0 mL density of solid = 1.71 (a) (b) 32.65 g 19.9884 mL = 1.63 g/mL density = (35.66 g – 14.23 g)/4.59 cm 3 = 4.67 g/cm 3 34.5 kg × 1000 g 1 mL 1L × × = 2.54 L 1 kg 13.6 g 1000 mL 14 1 Matter and Measurement (c) 1.011 × 10 5 cm 3 × 19.3 g cm 3 = 1.95 × 10 6 g Solutions to Exercises V = 4/3 π r 3 = 4/3 π (28.9 cm) 3 = 1.0111 × 10 5 = 1.01 × 10 5 = 1.01 × 10 5 cm 3 The sphere weighs 1950 kg or 4300 pounds. The student is unlikely to be able to carry the sphere. 1.72 0.500 L battery acid × 640 g battery acid × 1.28 g 1000 mL × = 640 g battery acid L mL 38.1 g sulfuric acid = 243.84 = 244 g sulfuric acid 100 g battery acid × 1 in 3 (2.54) cm 3 3 1.73 (a) 40 lb peat 14 × 20 × 30 in 3 × 453.6 g 1 lb = 0.13 g/cm 3 peat 1 gal 1.057 qt 453.6 g 40 lb soil 1 × 10 −3 L 1 mL × × × × × = 2.5 g/cm 3 soil 3 4 qt 1L 1.9 gal 1 mL 1 lb 1 cm No. Volume must be specified in order to compare mass. The densities tell us that a certain volume of peat moss is “lighter” (weighs less) than the same volume of top soil. (b) 1 bag peat = 14 × 20 × 30 = 8.4 × 10 3 in 3 10. ft × 20. ft × 2.0 in × 57,600 in 3 × 1 bag 8.4 × 10 3 in 3 12 2 in 2 ft 2 = 57 ,600 = 5.8 × 10 4 in 3 peat needed = 6.9 bags needed (Buy 7 bags of peat.) 1.74 (a) Calculate the volume of the coin. It is a cylinder, V = π r 2 h, r = 2 π d2 h d ⎛d⎞ , V = π⎜ ⎟ h = 2 4 ⎝2⎠ Then use density of pure gold to calculate mass. V=π × (2.2) 2 cm 2 1 cm × 3.0 mm × = 1.140 = 1.1 cm 3 4 10 mm 19.3 g gold = 22.01 = 22 g pure gold 1 cm 3 1 tr oz $640 = $452.93 = 4.5 × 102 ($450) × 31.1 g tr oz 1.140 cm3 × (b) 22.01 g gold × 1.75 8.0 oz × 84 cm 3 50 ft 2 453.6 g 1 lb 1 cm 3 × × = 84.00 = 84 cm 3 16 oz lb 2.70 g × 1 2 ft 2 12 in 2 2 × 1 2 in 2 2.54 cm 2 2 × 10 mm = 0.018 mm 1 cm 15 1 Matter and Measurement 1.76 11.86 g ethanol × Solutions to Exercises 1 cm 3 = 15.0317 = 15.03 cm 3 , volume of cylinder 0.789 g ethanol 1/ 2 ⎡ 15.0317 cm 3 ⎤ V = πr 2 h; r = (V/πh) 1/2 = ⎢ ⎥ ⎣ π × 15.0 cm ⎦ d = 2r = 1.13 cm = 0.5648 = 0.565 cm 1.77 (a) Let x = mass of Au in jewelry 9.85 - x = mass of Ag in jewelry The total volume of jewelry = volume of Au + volume of Ag 0.675 cm 3 = x g × 0.675 = 1 cm 3 1 cm 3 + (9.85 − x)g × 19.3 g 10.5 g (To solve, multiply both sides by (19.3) (10.5)) 9.85 − x x + 19.3 10.5 0.675 (19.3)(10.5) = 10.5 x + (9.85 – x)(19.3) 136.79 = 10.5 x + 190.105 – 19.3 x –53.315 = –8.8 x x = 6.06 g Au; 9.85 g total – 6.06 g Au = 3.79 g Ag mass % Au = 6.06 g Au 9.85 g jewelry × 100 = 61.5% Au (b) 1.78 24 carats × 0.615 = 15 carat gold A solution can be separated into components by physical means, so separation would be attempted. If the liquid is a solution, the solute could be a solid or a liquid; these two kinds of solutions would be separated differently. Therefore, divide the liquid into several samples and do different tests on each. Try evaporating the solvent from one sample. If a solid remains, the liquid is a solution and the solute is a solid. If the result is negative, try distilling a sample to see if two or more liquids with different boiling points are present. If this result is negative, the liquid is probably a pure substance, but negative results are never entirely conclusive. We might not have tried the appropriate separation technique. The separation is successful if two distinct spots are seen on the paper. To quantify the characteristics of the separation, calculate a reference value for each spot that is distance travelled by spot distance travelled by solvent 1.79 If the values for the two spots are fairly different, the separation is successful. (One could measure the distance between the spots, but this would depend on the length of paper used and be different for each experiment. The values suggested above are independent of the length of paper.) 16 1 Matter and Measurement 1.80 The densities are: hexane – 0.6603 g/cm 3 benzene – 0.87654 g/cm 3 methylene iodide (methane, diiodo) – 3.3254 g/cm 3 Solutions to Exercises carbon tetrachloride (methane, tetrachloro) – 1.5940 g/cm 3 Only methylene iodide will separate the two granular solids. The undesirable solid (2.04 g/cm 3 ) is less dense than methylene iodide and will float; the desired material is more dense than methylene iodide and will sink. The other three liquids are less dense than both solids and will not produce separation. 1.81 (a) Density is the ratio of mass and volume; density = mass/volume. Propylene glycol (PG) is C3H6(OH)2 density = (b) 51.80 g = 1.036 = 1.04 g/mL 50.0 mL Use density of ethylene glycol (EG) to calculate the mass of EG in 1.00 gal. Then use the relationship 62 g EG = 76 g PG to calculate the mass of PG required to provide the same protection as 1.00 gal of EG. 1.00 gal EG × 3.7854 L 1000 mL 1.12 g EG × × = 4.2396 × 103 = 4.24 × 103 g EG 1 gal 1L mL 76 g PG = 5.1970 × 103 = 5.2 × 103 g PG required 62 g EG 4.2396 × 103 g EG × (c) Use the mass of PG required to produce the same protection as 1.00 gal of EG calculated in part (b) and the density of PG calculated in part (a) to calculate the volume of PG required to provide this protection. 5.1970 × 10 3 g PG × 1 mL 1L 1 gal × × = 1.325 = 1.3 gal PG 1.036 g 1000 mL 3.7854 L 1.82 Study (a) is likely to be both precise and accurate, because the errors are carefully controlled. The secondary weight standard will be resistant to chemical and physical changes, the balance is carefully calibrated, and weighings are likely to be made by the same person. The relatively large number of measurements is likely to minimize the effect of random errors on the average value. The accuracy and precision of study (b) depend on the veracity of the participants’ responses, which cannot be carefully controlled. It also depends on the definition of “comparable lifestyle.” The percentages are not precise, because the broad definition of lifestyle leads to a range of results (scatter). The relatively large number of participants improves the precision and accuracy. In general, controlling errors and maximizing the number of data points in a study improves precision and accuracy. 17 ...
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