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
AliciA PAterno PArsi • ArAsh PArsi • tomislAv PintAuer • lucio Gelmini • roberts W. hilts
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

COMPLETE SOLUTIONS MANUAL
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

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
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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 × 103 g)  (8.96 g cm–3 ) = 111.6 cm3 .
Next determine the radius using the relationship between volume of a sphere and radius.
V = 4
3
r3 = 111.6 cm3 = 4
3 (3.1416)r3 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/cm3
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
 

 
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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 = 21.33
 
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 1451.57 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
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/cm3
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%
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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.
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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 H2 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).
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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 105

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 
   
  
 
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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 1017
 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)  
 
24 3
2
(5.07 10 ) 1.8 10 0.16
= = 1.6
0.0980.065 + 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 O2 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 = 1.860 105

(d) 33,900 3.390 10 4
  (e) 6.321 10 4
 is correct (f) 5.0472 10 5.047 104 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.
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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 = 3.058 101
 
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

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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 = 127L mL
L mL (b) 15.8 1
1000 = 0.0158mL L
mL L
(c) 981 1
1000 = 0.9813
3
cm L
cm L (d)
3
3 6 3100 cm
2.65 m = 2.65 10 cm
1 m
 
  
 
28. (E) (a) 31000 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 = 29ft in
ft
cm
in
m
cm m  
(c) 1.42 453.6
1 = 644lb g
lb g (d) 248 0.4536
1 = 112lb kg
lb kg
(e)
3
34 qt 0.9464 dm
1.85 gal 7.00 dm
1 gal 1 qt
  
(f) 30.9464 L 1000 mL
3.72 qt = 3.52 10 mL
1 qt 1 L
  
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Chapter 1: Matter – Its Properties and Measurement
9
30. (M) (a)
2
2 6 21000 m
1.00 km = 1.00 10 m
1 km
 
  
 
(b)
3
3 6 3100 cm
1.00 m = 1.00 10 cm
1 m
 
  
 
(c)
2
2 6 25280 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 g10 g
m
m
æ ö æ ö÷ç ÷÷ çç ÷´ ´ =÷ çç ÷÷ ç ÷ç ç÷ è ø÷çè 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 = 3.2573
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.
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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.
1min1000 m 1 s
time = 1.45 km = 2.5 min
1 km 9.83 m 60 s
  
36. (M) (a) mass   25.0 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) 31000 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 cm1 in.
 
   
 
23
4 2
2
2.2 10 g 1 kg 100 cm
pressure = = 2.2 10 kg/m
1000 g 1 m1 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
r3 = 90.0  10–12 cm3
r3 = 2.15  10–11 cm3 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 °Ft t
   
high:  9 °F 9 °F
5 °C 5 °C(°F) (°C) 32 °F 50 °C 32 °F 122 °Ft t    
42. (E) high:    5 °C 5 °C
9 °F 9 °F(°C) = (°F) 32 °F = 118 °F 32 °F = 47.8 °C 48 °Ct t   
low:    5 °C 5 °C
9 °F 9 °F(°C) (°F) 32 °F 17 °F 32 °F 8.3 °Ct 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 °Ft 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 °Ct 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 = 3.96x – 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 = 0.444x – 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
 
  
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Solution Manual For General Chemistry: Principles and Modern Applications, 11th Edition

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