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Solution Manual For General Chemistry: Principles and Modern Applications, 11th Edition

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    Petrucci Herring Madura Bissonnette P r i n c i P l e s a n d M o d e r n a P P l i c a t i o n s General Chemistry e l e v e n t h e d i t i o n Alici A P A terno P A rsi • Ar A sh P A rsi • t omisl A v Pint A uer • l ucio Gelmini • r oberts W. h ilts c o M P l e t e s o l u t i o n s M a n u a l 05/02/16 11:01 am

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    C OMPLETE S OLUTIONS M ANUAL Alicia Paterno Parsi Arash Parsi Tomislav Pintauer Duquesne University Duquesne University Duquesne University Lucio Gelmini Robert W. Hilts Grant MacEwan College Grant MacEwan College General Chemistry Principles and Modern Applications Eleventh Edition Ralph H. Petrucci California State University, San Bernardino F. Geoffrey Herring University of British Columbia Jeffry D. Madura Duquesne University Carey Bissonnette University of Waterloo A01_PETR5044_11_CSM_FM.pdf 1 1/29/16 10:02:33 PM

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    Contents Preface.............................................................................................................................................. iv Chapter 1 Matter—Its Properties and Measurement ..................................................................1 Chapter 2 Atoms and the Atomic Theory .................................................................................27 Chapter 3 Chemical Compounds ..............................................................................................60 Chapter 4 Chemical Reactions ...............................................................................................110 Chapter 5 Introduction to Reactions in Aqueous Solutions ...................................................160 Chapter 6 Gases ......................................................................................................................204 Chapter 7 Thermochemistry ...................................................................................................256 Chapter 8 Electrons in Atoms.................................................................................................300 Chapter 9 The Periodic Table and Some Atomic Properties ..................................................346 Chapter 10 Chemical Bonding I: Basic Concepts ....................................................................376 Chapter 11 Chemical Bonding II: Valence Bond and Molecular Orbital Theories .................444 Chapter 12 Intermolecular Forces: Liquids and Solids ............................................................501 Chapter 13 Spontaneous Change: Entropy and Gibbs Energy .................................................556 Chapter 14 Solutions and Their Physical Properties ................................................................610 Chapter 15 Principles of Chemical Equilibrium.......................................................................661 Chapter 16 Acids and Bases .....................................................................................................717 Chapter 17 Additional Aspects of Acid–Base Equilibria .........................................................786 Chapter 18 Solubility and Complex-Ion Equilibria ..................................................................887 Chapter 19 Electrochemistry ....................................................................................................943 Chapter 20 Chemical Kinetics ................................................................................................1011 Chapter 21 Chemistry of the Main-Group Elements I: Groups 1, 2, 13, and 14 ....................1065 Chapter 22 Chemistry of the Main-Group Elements II: Groups 18, 17, 16, 15, and Hydrogen .......................................................................................................1097 Chapter 23 The Transition Elements ......................................................................................1138 Chapter 24 Complex Ions and Coordination Compounds ......................................................1166 Chapter 25 Nuclear Chemistry ...............................................................................................1201 Chapter 26 Structures of Organic Compounds .......................................................................1226 Chapter 27 Reactions of Organic Compounds .......................................................................1269 Chapter 28 Chemistry of the Living State ..............................................................................1306 A01_PETR5044_11_CSM_FM.pdf 3 1/29/16 10:02:35 PM

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    1 CHAPTER 1 MATTER—ITS PROPERTIES AND MEASUREMENT PRACTICE EXAMPLES 1A (E) Convert the Fahrenheit temperature to Celsius and compare.     5 °C 5 °C 9 °F 9 °F °C °F 32 °F 350 °F 32 °F 177 °C.      1B (E) We convert the Fahrenheit temperature to Celsius.     5 °C 5 °C 9 °F 9 °F °C °F 32 °F 15 °F 32 °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. 291.4 g 108.6 g density = = 1.46 g/mL 125 mL  2B (E) First determine the volume required. V = (1.000 × 10 3 g)  (8.96 g cm –3 ) = 111.6 cm 3 . Next determine the radius using the relationship between volume of a sphere and radius. V = 4 3  r 3 = 111.6 cm 3 = 4 3 (3.1416) r 3 r = 3 111.6 3 4(3.1416)  = 2.987 cm  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. mass 28.4 g rock density = = = 2.76 g/mL volume 44.1 mL rock & water 33.8 mL water  = 2.76 g/cm 3 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. 1000 mL 0.71 g gasohol 10 g ethanol 1 kg ethanol ethanol mass = 25 L gasohol 1 L 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. 100.0 g rubbing alcohol rubbing alcohol mass 15.0 g (isopropyl alcohol) 70.0 g (isopropyl alcohol) 21.43 g rubbing alcohol 21.4 g rubbing alcohol density 0.857 g/mL 25.0 mL      M01_PETR5044_11_CSM_C01.pdf 1 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 2 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 0.000456 6.422 10 = 21.3 3   5B (M) For this calculation, the value 1.3 10 3   has the least precision (two significant figures), thus the final answer must also be quoted to two significant figures. 8.21 10 1.3 10 0.00236 4.071 10 = 1.1 10 4 3 2 6         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 exponential notation.     3 2 1.302 10 + 952.7 1302 + 952.7 2255 = = = 15.6 157 12.22 145 1.57 10 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 V Oil = m / d = 76.2 g / 0.926 g/mL = 82.3 mL = V Mg-Al d Mg-Al = 211.5 g / 82.3 mL = 2.57 g/cm 3 Now, since the density is a linear function of the composition, d Mg-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 = 0.96 x + 1.74 x = 0.86 = mass % of Al Mass % of Mg = 1 – 0.86 = 0.14, 14% M01_PETR5044_11_CSM_C01.pdf 2 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 3 B (M) Stepwise approach: Mass of seawater = d  V = 1.027 g/mL × 1500 mL = 1540.5 g 2.67 g NaCl 39.34 g Na 1540.5 g seawater 16.18 g Na 100 g seawater 100 g NaCl    Then, convert mass of Na to atoms of Na 23 26 1 kg Na 1 Na atom 16.18 g Na 4.239 10 Na atoms 1000 g Na 3.817 10 kg Na       Conversion Pathway: 26 2.67 g NaCl 39.34 g Na 1 kg Na 1 Na atom 1540.5 g seawater 100 g seawater 100 g NaCl 1000 g Na 3.8175 10 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 grassroots 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. M01_PETR5044_11_CSM_C01.pdf 3 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 4 6. (E) For a theory to be considered 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. 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) 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, CO 2 (g), and H 2 O(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. 8. (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.) 9. (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). M01_PETR5044_11_CSM_C01.pdf 4 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 5 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) 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. 11. (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). 12. (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) 3 8950. = 8.950 10 (4 sig. fig.)  (b) 4 10, 700. = 1.0700 10 (5 sig. fig.)  (c) 0.0240 = 2.40 10 2   (d) 0.0047 = 4.7 10 3   (e) 938.3 = 9.383 10 2  (f) 275,482 = 2.75482 10 5  14. (E) (a) 3 3 5.12 10 = 0.00512    (b) 5 5 8.05 10 = 0.0000805    (c) 4 4 291.1 10 = 0.02911    (d) 2 2 72.1 10 = 0.721    15. (E) (a) 34,000 centimeters / second = 3.4 10 4  cm/s (b) 3 6378 km 6.378 10 = ´ km (c) (trillionth = 1  10 –12 ) hence, 74  10 –12 m or 7.4  10 –11 m (d) 3 2 3 5 3 3 (2.2 10 ) (4.7 10 ) 2.7 10 4.6 10 5.8 10 5.8 10            M01_PETR5044_11_CSM_C01.pdf 5 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 6 16. (E) (a) 173 thousand trillion watts W = 173,000,000,000,000,000 = 1.73 10 17  W (b) one ten-millionth of a meter = 7 1 10, 000, 000 m = 1 10 m    (c) (trillionth = 1  10 –12 ) hence, 142  10 –12 m or 1.42  10 –10 m (d)     2 4 3 2 (5.07 10 ) 1.8 10 0.16 = = 1.6 0.098 0.065 + 3.3 10       Significant Figures 17. (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 O 2 is an estimated value derived from experimental data, which contains some inherent error. 18. (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. 19. (E) Each of the following is expressed to four significant figures. (a) 3984.6 3985  (b) 422.04 422.0  (c) 186,000 = 1.860 10 5  (d) 33,900 3.390 10 4   (e) 6.321 10 4  is correct (f) 5.0472 10 5.047 10 4 4      20. (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. M01_PETR5044_11_CSM_C01.pdf 6 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 7 (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. 21. (E) (a) 4 0.406 0.0023 = 9.3 10    (b) 0.1357 16.80 0.096 = 2.2 10 1     (c) 0.458 + 0.12 0.037 = 5.4 10 1    (d) 32.18 + 0.055 1.652 = 3.058 10 1   22. (M) (a) 320 24.9 0.080 = 3.2 10 2.49 10 8.0 10 = 1.0 10 2 1 2 5        (b) 432.7 6.5 0.002300 62 0.103 = 4.327 10 6.5 2.300 10 6.2 10 1.03 10 = 1.0 2 3 1 1             (c) 1 1 1 1 32.44 + 4.9 0.304 3.244 10 + 4.9 3.04 10 = = 4.47 10 82.94 8.294 10         (d) 8.002 + 0.3040 13.4 0.066 +1.02 = 8.002 + 3.040 10 1.34 10 6.6 10 +1.02 = 5.79 10 1 1 2 1          23. (M) (a) 4 2.44 10  (b) 3 1.5 10  (c) 40.0 (d) 3 2.131 10  (e) 3 4.8 10   24. (M) (a) 1 7.5 10  (b) 12 6.3 10  (c) 3 4.6 10  (d) 1 1.058 10   (e) 3 4.2 10   25. (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. 24 h 1 h 1 h 9 days = 216.000 h 3min = 0.050 h 44 s = 0.012 h 1 d 60 min 3600 s    total time = 216.000 h + 0.050 h + 0.012 h = 216.062 h 25, 012 mi 1.609344 km average speed = = 186.30 km/h 216.062 h 1 mi  M01_PETR5044_11_CSM_C01.pdf 7 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 8 (b) First compute the mass of fuel remaining mass gal qt gal L qt mL L g mL lb g lb = 14 4 1 0.9464 1 1000 1 0.70 1 1 453.6 = 82      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”) is not to the nearest pound. 0.4536 kg mass of fuel used = (9000 lb 82 lb) 4045 kg 1 lb 25, 012 mi 1.609344 km fuel consumption = = 9.95 km/kg or ~10 km/kg 4045 kg 1 mi     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 18 3 3 10 ft  ) would have more realistically reflected the precision with which the proved reserve was known. Units of Measurement 27. (E) (a) 0.127 1000 1 = 127 L mL L mL  (b) 15.8 1 1000 = 0.0158 mL L mL L  (c) 981 1 1000 = 0.981 3 3 cm L cm L  (d) 3 3 6 3 100 cm 2.65 m = 2.65 10 cm 1 m         28. (E) (a) 3 1000 g 2.35 kg = 2.35 10 g 1 kg   (b) 1 kg 792 g = 0.792 kg 1000 g  (c) 1 cm 3869 mm = 386.9 cm 10 mm  (d) 10 mm 0.043 cm = 0.43 mm 1 cm  29. (E) (a) 2.54 cm 68.4 in. = 174 cm 1 in.  (b) 94 12 . 1 2.54 1 . 1 100 = 29 ft in ft cm in m cm m    (c) 1.42 453.6 1 = 644 lb g lb g  (d) 248 0.4536 1 = 112 lb kg lb kg  (e) 3 3 4 qt 0.9464 dm 1.85 gal 7.00 dm 1 gal 1 qt    (f) 3 0.9464 L 1000 mL 3.72 qt = 3.52 10 mL 1 qt 1 L    M01_PETR5044_11_CSM_C01.pdf 8 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 9 30. (M) (a) 2 2 6 2 1000 m 1.00 km = 1.00 10 m 1 km         (b) 3 3 6 3 100 cm 1.00 m = 1.00 10 cm 1 m         (c) 2 2 6 2 5280 ft 12 in. 2.54 cm 1 m 1.00 mi = 2.59 10 m 1 mi 1 ft 1 in. 100 cm            31. (E) Express both masses in the same units for comparison. 3 1 g 10 mg 3245 g 3.245 mg, 6 1 g 10 g m m æ ö æ ö ÷ ç ÷ ÷ ç ç ÷ ´ ´ = ÷ ç ç ÷ ÷ ç ÷ ç ç ÷ è ø ÷ ç è ø which is larger than 0.00515 mg. 32. (E) Express both masses in the same units for comparison. 1000 g 0.000475 kg = 0.475 1 kg  g, which is smaller than 3257 1 10 = 3.257 3 mg g mg  g. 33 . (E) Conversion pathway approach: height hands in hand cm in m cm = 15 4 . 1 2.54 1 . 1 100 = 1.5    m Stepwise approach: 4 in. 15 hands 60 in. 1 hand 2.54 cm 60 in. 152.4 cm 1 in. 1 m 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 chain 1 furlong 1 mi 5280 ft 12 in. 2.54 cm 1.00 link = 20.1 cm 100 links 10 chains 8 furlongs 1 mi 1 ft 1 in.       35 . (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. time = 100.0 m = 10. s 100 yd 36 in. 1 m    The final answer can only be quoted to a maximum of two significant figures. M01_PETR5044_11_CSM_C01.pdf 9 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 10 (b) We need to convert yards to meters. 100 yd 36 in. 2.54 cm 1 m speed = = 9.83 m/s 9.3 s 1 yd 1 in. 100 cm    (c) The speed is used as a conversion factor. 1min 1000 m 1 s time = 1.45 km = 2.5 min 1 km 9.83 m 60 s    36. (M) (a) mass   2 5.0 gr 1.0 g 1000 mg mg = 2 tablets = 6.7 10 1 tablet 15 gr 1 g     mg (b) dosage 2 6.7 10 mg 1 lb 1000 g rate = = 9.2 161 lb 453.6 g 1 kg    mg aspirin/kg body weight (c) 3 1000 g 2 tablets 1 day time = 1.0 kg = 1.5 10 days 1 kg 0.67 g 2 tablets     37. (D) 2 2 2 100 m 100 cm 1 in. 1 ft 1 mi 640 acres 1 hectare = 1 hm 1 hm 1 m 2.54 cm 12 in. 5280 ft 1 mi             1 hectare = 2.47 acres 38. (D) Here we must convert pounds per cubic inch into grams per cubic centimeter: density for metallic iron = 3 0.284 lb 1 in.  454 g 1 lb  3 3 (1 in.) (2.54 cm) = 7.87 3 g cm 39 . (D) 2 3 2 2 32 lb 453.6 g 1 in. pressure = = 2.2 10 g/cm 1 lb 2.54 cm 1 in.          2 3 4 2 2 2.2 10 g 1 kg 100 cm pressure = = 2.2 10 kg/m 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 = 4/3  r 3 = 90.0  10 –12 cm 3 r 3 = 2.15  10 –11 cm 3 and r = 2.78  10 –4 cm Thus, the diameter is 2  r = 2  2.78  10 –4 cm 10 mm 1 cm  = 5.56  10 –3 mm M01_PETR5044_11_CSM_C01.pdf 10 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 11 Temperature Scales 41. (E) low:   9 F 9 °F 5 °C 5 °C (°F) (°C) + 32 °F 10 °C + 32 °F 14 °F t t      high:   9 °F 9 °F 5 °C 5 °C (°F) (°C) 32 °F 50 °C 32 °F 122 °F t t      42. (E) high:     5 °C 5 °C 9 °F 9 °F (°C) = (°F) 32 °F = 118 °F 32 °F = 47.8 °C 48 °C t t    low:     5 °C 5 °C 9 °F 9 °F (°C) (°F) 32 °F 17 °F 32 °F 8.3 °C t t       43. (M) Let us determine the Fahrenheit equivalent of absolute zero.   9 °F 9 °F 5 °C 5 °C (°F) (°C) 32 °F 273.15 °C + 32 °F 459.7 °F t t       A temperature of 465 °F  cannot be achieved because it is below absolute zero. 44. (M) Determine the Celsius temperature that corresponds to the highest Fahrenheit temperature, 240 °F.     5 °C 5 °C 9 °F 9 °F (°C) (°F) 32 °F 240 °F 32 °F 116 °C t t      Because 116 °C is above the range of the thermometer, this thermometer cannot be used in this candy-making assignment. 45 . (D) (a) From the data provided we can write down the following relationship: –38.9  C = 0  M and 356.9  C = 100  M. To find the mathematical relationship between these two scales, we can treat each relationship as a point on a two-dimensional Cartesian graph: Therefore, the equation for the line is y = 3.96 x – 38.9. The algebraic relationship between the two temperature scales is t (  C) = 3.96(  M) – 38.9 or rearranging, t (  M) = ( C) 38.9 3.96 t   Alternatively, note that the change in temperature in °C corresponding to a change of 100 °M is [356.9 – (–38.9)] = 395.8 °C, hence, (100 °M/395.8 °C) = 1 °M/3.96 °C. This factor must be multiplied by the number of degrees Celsius above zero on the M scale. This number of degrees is t (°C) + 38.9, which leads to the general equation t (°M) = [ t (°C) + 38.9]/3.96. The boiling point of water is 100  C, corresponding to t (  M) = 100 38.9 3.96  = 35.1  M (b) t (°M) = 273.15 38.9 3.96   = –59.2  M would be the absolute zero on this scale. M01_PETR5044_11_CSM_C01.pdf 11 1/29/16 10:03:14 PM

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    Chapter 1: Matter – Its Properties and Measurement 12 46. (D) (a) From the data provided we can write down the following relationship: –77.75 = 0  A and –33.35  C = 100  A. To find the mathematical relationship between these two scales, we can treat each relationship as a point on a two- dimensional Cartesian graph. Therefore, the equation for the line is y = 0.444 x – 77.75 The algebraic relationship between the two temperature scales is t (  C) = 0.444(  A) – 77.75 or rearranging t (  A) = ( C) 77.75 0.444 t   The boiling point of water (100  C) corresponds to t (  A) = 100 77.75 0.444  = 400  A (b) t (  A) = 273.15 77.75 0.444   = –440  A Density 47. (E) butyric acid mass 2088 g 1 L density 0.958 g/mL volume 2.18 L 1000 mL     48. (E) chloroform mass 22.54 kg 1 L 1000 g density 1.48 g/mL volume 15.2 L 1000 mL 1 kg      49. (M) The mass of acetone is the difference in masses between empty and filled masses. Conversion pathway approach: 437.5 lb 75.0 lb 453.6 g 1 gal 1 L density 0.790 g/mL 55.0 gal 1 lb 3.785 L 1000 mL       Stepwise approach: 5 437.5 lb 75.0 lb 362.5 lb 453.6 g 362.5 lb 1.644 10 g 1 lb      M01_PETR5044_11_CSM_C01.pdf 12 1/29/16 10:03:14 PM

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