General chemistry 10th edition complete solutions manual petrucci

CHAPTER MATTER—ITS PROPERTIES AND MEASUREMENT PRACTICE EXAMPLES 1A (E) Convert the Fahrenheit temperature to Celsius and compare C = F 32 F 59 CF = 350 F 32 F 59 CF = 177 C 1B (E) We convert the Fahrenheit temperature to Celsius C = F 32 F 59 CF = 15 F 32 F C F = 26 C The antifreeze only protects to 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 4 111.6 r = 111.6 cm3 = (3.1416)r3 r= = 2.987 cm V= 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 kg ethanol 1L 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 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 10 has the least precision (two significant figures), thus the final answer must also be quoted to two significant figures 8.21 104 1.3 10 = 1.1 106 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 = = = 15.6 157 12.22 145 1.57 102 12.22 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 for x (no Al in the alloy), everything is Mg and the equation becomes: 1.74 = m · + b Therefore, b = 1.74 Assuming 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 = – 0.86 = 0.14, 14% Chapter 1: Matter – Its Properties and Measurement B (M) Stepwise approach: Mass of seawater = D V = 1.027 g/mL ì 1500 mL = 1540.5 g 1540.5 g seawater 2.67 g NaCl 100 g seawater 39.34 g Na = 16.18 g Na 100 g NaCl Then, convert mass of Na to atoms of Na 16.18 g Na kg Na 1000 g Na Na atom = 4.239 1023 Na atoms 3.817 10 26 kg Na Conversion Pathway: 1540.5 g seawater 2.67 g NaCl 100 g seawater 39.34 g Na 100 g NaCl kg Na 1000 g Na Na atom 3.8175 10 26 kg Na EXERCISES The Scientific Method (E) One theory is preferred over another if it can correctly predict a wider range of phenomena and if it has fewer assumptions (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 (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 exist (“He is not malicious”) (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 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 (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 Chapter 1: Matter – Its Properties and Measurement (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 (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) Physical: The iron nail is not changed in any significant way when it is attracted to a magnet Its basic chemical identity is unchanged (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) 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) Chemical: The change in the color of the apple indicates that a new substance (oxidized apple) has formed by reaction with air (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) 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) Chapter 1: Matter – Its Properties and Measurement 10 11 12 (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) 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 (c) Substance: This is assuming that no gases or organic chemicals are dissolved in the water (d) 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) 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 (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) 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) (E) (a) 3.21 10 = 0.0321 (c) 15 0.0047 = 4.7 10 121.9 10 = 0.001219 (c) 0.0240 = 2.40 10 938.3 = 9.383 102 (f) 275,482 = 2.75482 105 (b) 5.08 10 = 0.000508 (d) 16.2 10 = 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 = 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 5.8 10 5.8 10 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 = 10, 000, 000 m = 10 (c) (trillionth = (d) (5.07 104 ) 10-12) hence, 142 1.8 10 0.065 + 3.3 10 = 10-12 m or 1.42 m 10-10 m 0.16 = 1.6 0.098 Significant Figures 17 18 19 20 (E) (a) An exact number—500 sheets in a ream of paper (b) 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 (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) 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) 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 (c) Measured quantity: The area is determined by calculations based on measurements These measurements are subject to error (d) 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) 3984.6 3985 (b) 422.04 422.0 (d) 33,900 3.390 104 (e) 6.321 104 is correct (c) 186,000 = 1.860 105 (f) 5.0472 10 5.047 10 (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 Chapter 1: Matter – Its Properties and Measurement 21 (f) 7000 can have anywhere from one to four significant figures; trailing zeros left of the decimal are indeterminate, if no decimal point is shown (g) 7.02 has three significant figures; zeros flanked by non-zero digits are significant (h) 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 10 (c) 22 23 0.1357 16.80 0.096 = 2.2 10 (d) 32.18 + 0.055 1.652 = 3.058 101 (b) 432.7 6.5 0.002300 4.327 102 6.5 2.300 10 = 62 0.103 6.2 101 1.03 10 (c) 32.44 + 4.9 0.304 3.244 101 + 4.9 3.04 10 = 82.94 8.294 101 (d) 8.002 + 0.3040 8.002 + 3.040 10 = = 5.79 10 13.4 0.066 +1.02 1.34 101 6.6 10 +1.02 (M) (a) 2.44 104 (b) 1.5 103 2.131 103 (e) 4.8 (M) (a) 7.5 101 (b) 6.3 1012 (e) 4.2 (d) 25 0.458 + 0.12 0.037 = 5.4 10 (b) 320 24.9 3.2 102 2.49 101 = = 1.0 105 (M) (a) 0.080 8.0 10 (d) 24 1.058 10 = 1.0 = 4.47 10 1 (c) 40.0 (c) 4.6 103 10-3 10-3 (quadratic equation solution) (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 days 24 h 1h 1h = 216.000 h 3min = 0.050 h 44 s = 0.012 h 1d 60 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 mi Chapter 1: Matter – Its Properties and Measurement (b) First compute the mass of fuel remaining qt gal mass = 14 gal 0.9464 L 1000 mL qt 1L 0.70 g mL lb = 82 lb 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 lb 25, 012 mi 1.609344 km fuel consumption = = 9.95 km/kg or ~10 km/kg 4045 kg 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 1018 ft ) would have more realistically reflected the precision with which the proved reserve was known Units of Measurement 27 (c) 28 29 1000 mL = 127 mL 1L (E) (a) 0.127 L 981 cm (b) 1L = 0.981 L 1000 cm3 (d) 1L = 0.0158 L 1000 mL 15.8 mL 2.65 m 100 cm 1m (E) (a) 1.55 kg 1000 g = 1.55 103 g (b) kg 642 g (c) cm = 289.6 cm 10 mm (d) 0.086 cm 2.54 cm = 174 cm in (b) 94 ft (d) 248 lb (f) 3.72 qt 2896 mm (E) (a) 68.4 in (c) 1.42 lb (e) 1.85 gal× 453.6 g = 644 g lb qt gal × 0.9464 dm qt =7.00 dm = 2.65 106 cm3 kg = 0.642 kg 1000 g 10 mm = 0.86 mm cm 12 in 2.54 cm 1m = 29 m ft in 100 cm 0.4536 kg = 112 kg lb 0.9464 L 1000 mL qt 1L = 3.52 103 mL Chapter 27: Reactions of Organic Compounds Electrophilic Aromatic Substitution 1283 Chapter 27: Reactions of Organic Compounds 1284 Chapter 27: Reactions of Organic Compounds 1285 Chapter 27: Reactions of Organic Compounds Reactions of Alkanes 1286 Chapter 27: Reactions of Organic Compounds (b) Whereas the flourine radical is so reactive, its attack on alkane is purely statistical There are six times as many 1° hydrogens and 2° hydrogens in the reactant molecule Polymerization Reactions 45 (M) The reason lies in the statistics During polymerization reaction, polymers of different chain-lengths are formed As a result, we can only speak of average molecular weight 1287 Chapter 27: Reactions of Organic Compounds 46 (M) Dacron is the polymer formed in the condensation of a carboxylic acid and alcohol It is called a polyester because it contains ester linkages along polymer chains To calculate the percent of oxygen by mass, we consider the smallest repeating unit in Dacron %0= × × 16.00 64 × 100% = × 100% = 33.3% 10 × 12.00 + × 16.00 + × 1.008 192.064 48 (M) No polymer formation is expected because ethyl alcohol does not contain two functional groups capable of undergoing a condensation reaction With an —OH group on each of its three carbons, glycerol is capable of polymerizing with terephthalic acid Synthesis of Organic Compounds 1288 Chapter 27: Reactions of Organic Compounds INTEGRATIVE AND ADVANCED EXERCISES 1289 Chapter 27: Reactions of Organic Compounds 1290 Chapter 27: Reactions of Organic Compounds 1291 Chapter 27: Reactions of Organic Compounds 1292 Chapter 27: Reactions of Organic Compounds 1293 Chapter 27: Reactions of Organic Compounds 1294 Chapter 27: Reactions of Organic Compounds 1295 Chapter 27: Reactions of Organic Compounds To calculate the yield of each product, find the number of hydrogens in the starting alkane that give rise to the product under consideration and multiply by the relative reactivity corresponding to the type of hydrogen Product 1-chloro-2-methylbutane 1-chloro-3-methylbutane 2-chloro-3-methylbutane 2-chloro-2-methylbutane Relative yield 6×1=6 3×1=3 2×3=6 × 4.3 = 4.3 Total = 19.3 1296 Absolute yield 6/19.3 = 0.31 = 31% 3/19.3 = 0.16 = 16% 6/19.3 = 0.31 = 31% 4/19.3 = 0.21 = 21% Chapter 27: Reactions of Organic Compounds 1297 ... 102, c = 7.987 1 0-6 , d = 1.6880 1 0-2 (simplicity) (b - 2ct)(1 dt) - (a bt - ct )d f (t) = (quotient (1 dt) rule) f (t) = b bdt - 2ct - 2cdt - ad - bdt cdt (1 dt) = b - 2ct - cdt - ad (1 dt) = (max)... C (F 32) 59 (a) F C - 49 C (C C = -4 5 49 - 32) 95 = 95 (C - 81) C = 59 C - 95 (81) C = 59 C - 45 Hence: C = -1 01.25 When it is ~ -1 01 C, the temperature in Fahrenheit is -1 50 F (49 lower) (b)... expressed to four significant figures 68 (M) 1.543 = 1.5794 – 1.836 × 1 0-3 (t-15) 1.543 - 1.5794 = – 1.836 × 1 0-3 (t-15) = 0.0364 0.0364 (t-15) = t = 19.8 + 15 = 34.8 C = 19.8 C 1.836 10 69 (D) volume needed