GeneralChemistry10ed_solutions - CHAPTER 1 MATTERITS PROPERTIES AND MEASUREMENT PRACTICE EXAMPLES 1A(E Convert the Fahrenheit temperature to Celsius and

GeneralChemistry10ed_solutions - CHAPTER 1 MATTERITS...

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Unformatted text preview: CHAPTER 1 MATTER—ITS PROPERTIES AND MEASUREMENT PRACTICE EXAMPLES 1A (E) Convert the Fahrenheit temperature to Celsius and compare. C = F 32 F 5 C = 350 F 32 F 5 C = 177 C . 9 F 9 F 1B (E) We convert the Fahrenheit temperature to Celsius. C = F 32 F 5 C = 15 F 32 F 5 C = 26 C . The antifreeze only protects to 9 F 9 F 22 C and thus it will not offer protection to temperatures as low as 15 F = 26.1 C . 2A (E) The mass is the difference between the mass of the full and empty flask. density = 291.4 g 108.6 g = 1.46 g/mL 125 mL 2B (E) First determine the volume required. V = (1.000 × 103 g) (8.96 g cm-3) = 111.6 cm3. Next determine the radius using the relationship between volume of a sphere and radius. 4 4 111.6 3 r= 3 = 2.987 cm V = r3 = 111.6 cm3 = (3.1416)r3 3 3 4(3.1416) 3A (E) The volume of the stone is the difference between the level in the graduated cylinder with the stone present and with it absent. density = mass 28.4 g rock = = 2.76 g/mL = 2.76 g/cm3 volume 44.1 mL rock &water 33.8 mL water 3B (E) The water level will remain unchanged. The mass of the ice cube displaces the same mass of liquid water. A 10.0 g ice cube will displace 10.0 g of water. When the ice cube melts, it simply replaces the displaced water, leaving the liquid level unchanged. 4A (E) The mass of ethanol can be found using dimensional analysis. ethanol mass = 25 L gasohol 1000 mL 0.71 g gasohol 10 g ethanol 1 kg ethanol 1L 1 mL gasohol 100 g gasohol 1000 g ethanol = 1.8 kg ethanol 4B (E) We use the mass percent to determine the mass of the 25.0 mL sample. rubbing alcohol mass = 15.0 g (2-propanol) rubbing alcohol density = 100.0 g rubbing alcohol = 21.43 g rubbing alcohol 70.0 g (2-propanol) 21.4 g = 0.857 g/mL 25.0 mL 1 Chapter 1: Matter – Its Properties and Measurement 5A (M) For this calculation, the value 0.000456 has the least precision (three significant figures), thus the final answer must also be quoted to three significant figures. 62.356 = 21.3 0.000456 6.422 103 5B (M) For this calculation, the value 1.3 103 has the least precision (two significant figures), thus the final answer must also be quoted to two significant figures. 8.21 104 1.3 103 = 1.1 106 2 0.00236 4.071 10 6A (M) The number in the calculation that has the least precision is 102.1 (+0.1), thus the final answer must be quoted to just one decimal place. 0.236 +128.55 102.1 = 26.7 6B (M) This is easier to visualize if the numbers are not in scientific notation. 1.302 103 + 952.7 = 1302 + 952.7 = 2255 = 15.6 1.57 102 12.22 157 12.22 145 INTEGRATIVE EXAMPLE A (D) Stepwise Approach: First, determine the density of the alloy by the oil displacement. Mass of oil displaced = Mass of alloy in air – Mass of alloy in oil = 211.5 g – 135.3 g = 76.2 g VOil = m / D = 76.2 g / 0.926 g/mL = 82.3 mL = VMg-Al DMg-Al = 211.5 g / 82.3 mL = 2.57 g/cc Now, since the density is a linear function of the composition, DMg-Al = mx + b, where x is the mass fraction of Mg, and b is the y-intercept. Substituting 0 for x (no Al in the alloy), everything is Mg and the equation becomes: 1.74 = m · 0 + b. Therefore, b = 1.74 Assuming 1 for x (100% by weight Al): 2.70 = (m × 1) + 1.74, therefore, m = 0.96 Therefore, for an alloy: 2.57 = 0.96x + 1.74 x = 0.86 = mass % of Al Mass % of Mg = 1 – 0.86 = 0.14, 14% 2 Chapter 1: Matter – Its Properties and Measurement B (M) Stepwise approach: Mass of seawater = D • V = 1.027 g/mL × 1500 mL = 1540.5 g 1540.5 g seawater 2.67 g NaCl 39.34 g Na = 16.18 g Na 100 g seawater 100 g NaCl Then, convert mass of Na to atoms of Na 16.18 g Na 1 kg Na 1 Na atom = 4.239 1023 Na atoms 1000 g Na 3.817 1026 kg Na Conversion Pathway: 1540.5 g seawater 2.67 g NaCl 39.34 g Na 1 kg Na 1 Na atom 100 g seawater 100 g NaCl 1000 g Na 3.8175 1026 kg Na EXERCISES The Scientific Method 1. (E) One theory is preferred over another if it can correctly predict a wider range of phenomena and if it has fewer assumptions. 2. (E) No. The greater the number of experiments that conform to the predictions of the law, the more confidence we have in the law. There is no point at which the law is ever verified with absolute certainty. 3. (E) For a given set of conditions, a cause, is expected to produce a certain result or effect. Although these cause-and-effect relationships may be difficult to unravel at times (“God is subtle”), they nevertheless do exist (“He is not malicious”). 4. (E) As opposed to scientific laws, legislative laws are voted on by people and thus are subject to the whims and desires of the electorate. Legislative laws can be revoked by a grass roots majority, whereas scientific laws can only be modified if they do not account for experimental observations. As well, legislative laws are imposed on people, who are expected to modify their behaviors, whereas, scientific laws cannot be imposed on nature, nor will nature change to suit a particular scientific law that is proposed. 5. (E) The experiment should be carefully set up so as to create a controlled situation in which one can make careful observations after altering the experimental parameters, preferably one at a time. The results must be reproducible (to within experimental error) and, as more and more experiments are conducted, a pattern should begin to emerge, from which a comparison to the current theory can be made. 3 Chapter 1: Matter – Its Properties and Measurement 6. (E) For a theory to be considered as plausible, it must, first and foremost, agree with and/or predict the results from controlled experiments. It should also involve the fewest number of assumptions (i.e., follow Occam’s Razor). The best theories predict new phenomena that are subsequently observed after the appropriate experiments have been performed. Properties and Classification of Matter 7. (E) When an object displays a physical property it retains its basic chemical identity. By contrast, the display of a chemical property is accompanied by a change in composition. (a) (b) Chemical: The paper is converted to ash, CO2(g), and H2O(g) along with the evolution of considerable energy. (c) Chemical: The green patina is the result of the combination of water, oxygen, and carbon dioxide with the copper in the bronze to produce basic copper carbonate. (d) 8. Physical: The iron nail is not changed in any significant way when it is attracted to a magnet. Its basic chemical identity is unchanged. Physical: Neither the block of wood nor the water has changed its identity. (E) When an object displays a physical property it retains its basic chemical identity. By contrast, the display of a chemical property is accompanied by a change in composition. (a) (b) Physical: The marble slab is not changed into another substance by feeling it. (c) Physical: The sapphire retains its identity as it displays its color. (d) 9. Chemical: The change in the color of the apple indicates that a new substance (oxidized apple) has formed by reaction with air. Chemical: After firing, the properties of the clay have changed from soft and pliable to rigid and brittle. New substances have formed. (Many of the changes involve driving off water and slightly melting the silicates that remain. These molten substances cool and harden when removed from the kiln.) (E) (a) Homogeneous mixture: Air is a mixture of nitrogen, oxygen, argon, and traces of other gases. By “fresh,” we mean no particles of smoke, pollen, etc., are present. Such species would produce a heterogeneous mixture. (b) Heterogeneous mixture: A silver plated spoon has a surface coating of the element silver and an underlying baser metal (typically iron). This would make the coated spoon a heterogeneous mixture. (c) Heterogeneous mixture: Garlic salt is simply garlic powder mixed with table salt. Pieces of garlic can be distinguished from those of salt by careful examination. (d) Substance: Ice is simply solid water (assuming no air bubbles). 4 Chapter 1: Matter – Its Properties and Measurement 10. (E) (a) Heterogeneous mixture: We can clearly see air pockets within the solid matrix. On close examination, we can distinguish different kinds of solids by their colors. (b) (c) Substance: This is assuming that no gases or organic chemicals are dissolved in the water. (d) 11. Homogeneous mixture: Modern inks are solutions of dyes in water. Older inks often were heterogeneous mixtures: suspensions of particles of carbon black (soot) in water. Heterogeneous mixture: The pieces of orange pulp can be seen through a microscope. Most “cloudy” liquids are heterogeneous mixtures; the small particles impede the transmission of light. (E) (a) If a magnet is drawn through the mixture, the iron filings will be attracted to the magnet and the wood will be left behind. (b) (c) Olive oil will float to the top of a container and can be separated from water, which is more dense. It would be best to use something with a narrow opening that has the ability to drain off the water layer at the bottom (i.e., buret). (d) 12. When the glass-sucrose mixture is mixed with water, the sucrose will dissolve, whereas the glass will not. The water can then be boiled off to produce pure sucrose. The gold flakes will settle to the bottom if the mixture is left undisturbed. The water then can be decanted (i.e., carefully poured off). (E) (a) Physical: This is simply a mixture of sand and sugar (i.e., not chemically bonded). (b) Chemical: Oxygen needs to be removed from the iron oxide. (c) Physical: Seawater is a solution of various substances dissolved in water. (d) Physical: The water-sand slurry is simply a heterogeneous mixture. Exponential Arithmetic 13. (E) (a) 8950. = 8.950 103 (4 sig. fig.) (b) 10, 700. = 1.0700 104 (5 sig. fig.) (d) 14. (E) (a) 3.21 102 = 0.0321 (c) 15. 0.0047 = 4.7 103 (e) 121.9 105 = 0.001219 (c) 0.0240 = 2.40 102 938.3 = 9.383 102 (f) 275,482 = 2.75482 105 (b) 5.08 104 = 0.000508 (d) 16.2 102 = 0.162 (E) (a) 34,000 centimeters / second = 3.4 104 cm/s (b) (c) (d) six thousand three hundred seventy eight kilometers = 6378 km = 6.378 103 km (trillionth = 1 10-12) hence, 74 10-12 m or 7.4 10-11 m (2.2 103 ) (4.7 102 ) 2.7 103 4.6 105 3 3 5.8 10 5.8 10 5 Chapter 1: Matter – Its Properties and Measurement 16. (E) (a) 173 thousand trillion watts = 173,000,000,000,000,000 W = 1.73 1017 W (b) one ten millionth of a meter = 1 10, 000, 000 m = 1 107 m (c) (trillionth = 1 10-12) hence, 142 10-12 m or 1.42 10-10 m (d) (5.07 104 ) 1.8 103 0.065 + 3.3 10 2 2 = 0.16 = 1.6 0.098 Significant Figures 17. (E) (a) An exact number—500 sheets in a ream of paper. (b) (c) Measured quantity: The distance between any pair of planetary bodies can only be determined through certain astronomical measurements, which are subject to error. (d) 18. Pouring the milk into the bottle is a process that is subject to error; there can be slightly more or slightly less than one liter of milk in the bottle. This is a measured quantity. Measured quantity: the internuclear separation quoted for O2 is an estimated value derived from experimental data, which contains some inherent error. (E) (a) The number of pages in the text is determined by counting; the result is an exact number. (b) (c) Measured quantity: The area is determined by calculations based on measurements. These measurements are subject to error. (d) 19. An exact number. Although the number of days can vary from one month to another (say, from January to February), the month of January always has 31 days. Measured quantity: Average internuclear distance for adjacent atoms in a gold medal is an estimated value derived from X-ray diffraction data, which contain some inherent error. (E) Each of the following is expressed to four significant figures. (a) (d) 20. 3984.6 3985 (b) 422.04 422.0 33,900 3.390 104 (e) 6.321 104 is correct (c) 186,000 = 1.860 105 (f) 5.0472 104 5.047 104 (E) (a) 450 has two or three significant figures; trailing zeros left of the decimal are indeterminate, if no decimal point is present. (b) 98.6 has three significant figures; non-zero digits are significant. (c) 0.0033 has two significant digits; leading zeros are not significant. (d) 902.10 has five significant digits; trailing zeros to the right of the decimal point are significant, as are zeros flanked by non-zero digits. (e) 0.02173 has four significant digits; leading zeros are not significant. 6 Chapter 1: Matter – Its Properties and Measurement (f) (g) 7.02 has three significant figures; zeros flanked by non-zero digits are significant. (h) 21. 7000 can have anywhere from one to four significant figures; trailing zeros left of the decimal are indeterminate, if no decimal point is shown. 67,000,000 can have anywhere from two to eight significant figures; there is no way to determine which, if any, of the zeros are significant, without the presence of a decimal point. (E) (a) 0.406 0.0023 = 9.3 104 (c) 22. (b) (d) 0.458 + 0.12 0.037 = 5.4 101 0.1357 16.80 0.096 = 2.2 101 32.18 + 0.055 1.652 = 3.058 101 320 24.9 3.2 102 2.49 101 = = 1.0 105 (M) (a) 2 0.080 8.0 10 (b) (c) 32.44 + 4.9 0.304 3.244 101 + 4.9 3.04 101 = = 4.47 101 1 82.94 8.294 10 (d) 23. 432.7 6.5 0.002300 4.327 102 6.5 2.300 103 = = 1.0 6.2 101 1.03 101 62 0.103 8.002 + 0.3040 8.002 + 3.040 101 = = 5.79 101 2 1 13.4 0.066 +1.02 1.34 10 6.6 10 +1.02 (M) (a) 2.44 104 (b) 1.5 103 2.131103 (e) 4.8 10-3 (M) (a) 7.5 101 (b) 6.3 1012 1.058 101 (e) 4.2 10-3 (quadratic equation solution) (d) 24. (d) 25. (c) 40.0 (c) 4.6 103 (M) (a) The average speed is obtained by dividing the distance traveled (in miles) by the elapsed time (in hours). First, we need to obtain the elapsed time, in hours. 9 days 24 h 1h 1h = 216.000 h 3min = 0.050 h 44 s = 0.012 h 1d 60 min 3600 s total time = 216.000 h + 0.050 h + 0.012 h = 216.062 h average speed = 25, 012 mi 1.609344 km = 186.30 km/h 216.062 h 1 mi 7 Chapter 1: Matter – Its Properties and Measurement (b) First compute the mass of fuel remaining mass = 14 gal 4 qt 0.9464 L 1000 mL 0.70 g 1 lb = 82 lb 1 gal 1 qt 1L 1 mL 453.6 g Next determine the mass of fuel used, and then finally, the fuel consumption. Notice that the initial quantity of fuel is not known precisely, perhaps at best to the nearest 10 lb, certainly (“nearly 9000 lb”) not to the nearest pound. 0.4536 kg 4045 kg 1 lb 25, 012 mi 1.609344 km fuel consumption = = 9.95 km/kg or ~10 km/kg 4045 kg 1 mi mass of fuel used = (9000 lb 82 lb) 26. (M) If the proved reserve truly was an estimate, rather than an actual measurement, it would have been difficult to estimate it to the nearest trillion cubic feet. A statement such as 2,911,000 trillion cubic feet (or even 3 1018 ft 3 ) would have more realistically reflected the precision with which the proved reserve was known. Units of Measurement 27. (E) (a) 0.127 L (c) 28. 29. (d) 1.42 lb (e) 1.85 gal× 100 cm 6 3 2.65 m = 2.65 10 cm 1m 3 1000 g = 1.55 103 g (b) 1 kg 642 g 1 cm = 289.6 cm 10 mm (d) 0.086 cm 2.54 cm = 174 cm 1 in. (b) 94 ft (d) 248 lb 0.4536 kg = 112 kg 1 lb (f) 3.72 qt 0.9464 L 1000 mL = 3.52 103 mL 1 qt 1L 2896 mm (c) 15.8 mL 3 1L 981 cm = 0.981 L 1000 cm3 (E) (a) 68.4 in. 1L = 0.0158 L 1000 mL (b) 3 (E) (a) 1.55 kg (c) 1000 mL = 127 mL 1L 453.6 g = 644 g 1 lb 4 qt 1 gal × 0.9464 dm 1 qt 1 kg = 0.642 kg 1000 g 12 in. 2.54 cm 1m = 29 m 1 ft 1 in. 100 cm 3 =7.00 dm 3 8 10 mm = 0.86 mm 1 cm Chapter 1: Matter – Its Properties and Measurement 2 30. 1000 m 6 2 (M) (a) 1.00 km 2 = 1.00 10 m 1 km 3 (b) 100 cm 6 3 1.00 m = 1.00 10 cm 1m (c) 1m 5280 ft 12 in. 2.54 cm 6 2 1.00 mi 2 = 2.59 10 m 1 ft 1 in. 100 cm 1 mi 3 2 31. (E) Express both masses in the same units for comparison. 1g 103 mg 3245 g 6 = 3.245 mg , which is larger than 0.00515 mg. 10 g 1 g 32. (E) Express both masses in the same units for comparison. 0.000475 kg which is smaller than 3257 mg 33. 1000 g = 0.475 g, 1 kg 1g = 3.257 g. 103 mg (E) Conversion pathway approach: 4 in. 2.54 cm 1m = 1.5 m height = 15 hands 1 hand 1 in. 100 cm Stepwise approach: 4 in. 15 hands 60 in. 1 hand 2.54 cm 60 in. 152.4 cm 1 in. 1m 152.4 cm = 1.524 m = 1.5 m 100 cm 34. (M) A mile is defined as being 5280 ft in length. We must use this conversion factor to find the length of a link in inches. 1.00 link 35. 1 chain 1 furlong 1 mile 5280 ft 12 in. 2.54 cm = 20.1 cm 100 links 10 chains 8 furlongs 1 mi 1 ft 1 in. (M) (a) We use the speed as a conversion factor, but need to convert yards into meters. 9.3 s 1 yd 39.37 in. = 10. s 100 yd 36 in. 1m The final answer can only be quoted to a maximum of two significant figures. time = 100.0 m 9 Chapter 1: Matter – Its Properties and Measurement (b) We need to convert yards to meters. speed = (c) 36. 100 yd 36 in. 2.54 cm 1m = 9.83 m/s 9.3 s 1 yd 1 in. 100 cm The speed is used as a conversion factor. 1min 1000 m 1s time = 1.45 km = 2.5 min 1 km 9.83 m 60 s (M) (a) mass mg = 2 tablets 5.0 gr 1.0 g 1000 mg = 6.7 102 mg 1 tablet 15 gr 1g (b) 6.7 102 mg 1 lb 1000 g dosage rate = = 9.5 mg aspirin/kg body weight 155 lb 453.6 g 1 kg (c) time = 1.0 kg 1000 g 2 tablets 1 day = 1.5 103 days 1 kg 0.67 g 2 tablets 2 37. 1 in. 1 ft 1 mi 640 acres 100 m 100 cm (D) 1 hectare = 1 hm 1m 2.54 cm 12 in. 5280 ft 1 mi 2 1 hm 2 1 hectare = 2.47 acres 38. (D) Here we must convert pounds per cubic inch into grams per cubic centimeter: density for metallic iron = 0.284 lb 454 g (1 in.)3 g = 7.87 3 3 1 in. 1 lb cm3 (2.54 cm) 2 39. 32 lb 453.6 g 1 in. 3 2 (D) pressure = = 2.2 10 g/cm 1 lb 2.54 cm 1 in.2 2 2.2 103 g 1 kg 100 cm 4 2 pressure = = 2.2 10 kg/m 2 1000 g 1 m 1 cm 40. (D) First we will calculate the radius for a typical red blood cell using the equation for the volume of a sphere. V = 4/3r3 = 90.0 10-12 cm3 r3 = 2.15 10-11 cm3 and r = 2.78 10-4 cm 10 mm Thus, the diameter is 2 r = 2 2.78 10-4 cm = 5.56 10-3 mm 1 cm 10 Chapter 1: Matter – Its Properties and Measurement Stepwise approach: 437.5 lb - 75.0 lb = 362.5 lb 453.6 g 1.644 105g 362.5 lb 1 lb 3.785 L 208 L 55.0 gal 1 gal 1000 mL 2.08 105 mL 208 L 1L 1.644 105g = 0.790 g/mL 2.08 105 mL 50. (M) Density is a conversion factor. b g volume = 283.2 g filled 121.3 g empty 1 mL = 102 mL 1.59 g 1000 mL 0.9867 g antifreeze 8.50 g acetone 1L 1 mL antifreeze 100.0 g antifreeze 1 kg = 0.629 kg acetone 1000 g 51. (M) acetone mass = 7.50 L antifreeze 52. (M) solution mass = 1.00 kg sucrose 1000 g sucrose 100.00 g solution = 9.95 103 g solution 1 kg sucrose 10.05 g sucrose 53. (M) fertilizer mass = 225 g nitrogen 54. 1 kg N 100 kg fertilizer = 1.07 kg fertilizer 1000 g N 21 kg N (M) 1000 mL vinegar 1.006 g vinegar 5.4 g acetic acid × × 1L 1 mL vinegar 100 g vinegar macetic acid = 54.3 g acetic acid macetic acid =1.00 L vinegar× 55. (M) The calculated volume of the iron block is converted to its mass by using the provided density. mass = 52.8 cm 6.74 cm 3.73 cm 7.86 56. g = 1.04 104 g iron cm3 (D) The calculated volume of the steel cylinder is converted to its mass by using the provided density. mass = V(density) = πr 2 h(d) = 3.14159 1.88 cm 18.35 cm × 7.75 2 13 g =1.58 × 103 g steel 3 cm Chapter 1: Matter – Its Properties and Measurement 57. (M) We start by determining the mass of each item. (1) mass of iron bar = 81.5 cm 2.1 cm 1.6 cm 7.86 g/cm3 = 2.2 103 g iron (2) 100 cm 3 3 mass of Al foil = 12.12 m 3.62 m 0.003 cm 2.70 g Al/cm = 4 10 g Al 1m (3) 1000 cm3 mass of water = 4.051 L 0.998 g / cm3 = 4.04 103 g water 1L 2 In order of increasing mass, the items are: iron bar aluminum foil water. Please bear in mind, however, that, strictly speaking, the rules for significant figures do not allow us to distinguish between the masses of aluminum and water. 58. (M) Total volume of 125 pieces of shot V = 8.9 mL 8.4 mL = 0.5 mL; 59. mass 0.5 mL 1cm3 8.92 g = = 0.04 g/shot shot 125 shot 1mL 1 cm3 (D) First determine the volume of the aluminum foil, then its area, and finally its thickness. 1 cm3 2 volume = 2.568 g = 0.951 cm3 ; area = 22.86 cm = 522.6 cm 2 2.70 g thickness = volume 0.951 cm3 10 mm = = 1.82 102 mm 2 area 1 cm 522.6 cm 60. (D) The vertical piece of steel has a volume = 12.78 cm 1.35 cm 2.75 cm = 47.4 cm3 The horizontal piece of stee...
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  • Fall '09
  • Hempstead
  • Chemistry, Mass, Orders of magnitude, Bromoform, g Na, kg Na

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