CHAPTER ONE
CHEMICAL FOUNDATIONS
For Review
1.
a. Law versus theory: A law is a concise statement or equation that summarizes observed
behavior. A theory is a set of hypotheses that gives an overall explanation of some
phenomenon. A law summarizes what happens; a theory (or model) attempts to explain
why it happens.
b. Theory versus experiment: A theory is an explanation of why things behave the way they
do, while an experiment is the process of observing that behavior. Theories attempt to
explain the results of experiments and are, in turn, tested by further experiments.
c. Qualitative versus quantitative: A qualitative observation only describes a quality while a
quantitative observation attaches a number to the observation. Examples: Qualitative
observations: The water was hot to the touch. Mercury was found in the drinking water.
Quantitative observations: The temperature of the water was 62C. The concentration of
mercury in the drinking water was 1.5 ppm.
d. Hypothesis versus theory: Both are explanations of experimental observation. A theory is
a set of hypotheses that has been tested over time and found to still be valid, with
(perhaps) some modifications.
2.
No, it is useful whenever a systematic approach of observation and hypothesis testing can be
used.
3.
a. No.
b. Yes
c. Yes
Only statements b and c can be determined from experiment.
4.
Volume readings are estimated to one decimal place past the markings on the glassware. The
assumed uncertainty is ±1 in the estimated digit. For glassware a, the volume would be
estimated to the tenths place since the markings are to the ones place. A sample reading
would be 4.2 with an uncertainty of ±0.1. This reading has two significant figures. For
glassware b, 10.52 ±0.01 would be a sample reading and the uncertainty; this reading has four
significant figures. For glassware c, 18 ±1 would be a sample reading and the uncertainty,
with the reading having two significant figures.
5.
Accuracy: How close a measurement or series of measurements are to an accepted or true
value.
Precision: How close a series of measurements of the same thing are to each other. The
results, average = 14.91  0.03%, are precise (close to each other) but are not accurate (not
close to the true value.
1
2
CHAPTER 1
CHEMICAL FOUNDATIONS
6.
In both sets of rules, the lease precise number determines the number of significant figures in
the final result. For multiplication/division, the number of significant figures in the result is
the same as the number of significant figures in the least precise number used in the
calculation. For addition/subtraction, the result has the same number of decimal places as the
least precise number used in the calculation (not necessarily the number with the fewest
significant figures).
7.
Consider gold with a density of 19.32 g/cm3. The two conversion factors are:
19.32 g
1 cm3
or
1 cm3
19.32 g
Use the first form when converting from the volume of gold in cm3 to the mass of gold and
use the second form when converting from mass of gold to volume of gold. When using
conversion factors, concentrate on the units crossing off.
8.
To convert from Celsius to Kelvin, a constant number of 273 is added to the Celsius
temperature. Because of this, T(C) = T(K). When converting from Fahrenheit to Celsius,
one conversion that must occur is to multiply the Fahrenheit temperature by a factor less than
one (5/9). Therefore, the Fahrenheit scale is more expansive than the Celsius scale, and 1F
would correspond to a smaller temperature change than 1C or 1K.
9.
Chemical changes involve the making and breaking of chemical bonds. Physical changes do
not. The identity of a substance changes after a chemical change, but not after a physical
change.
10.
Many techniques of chemical analysis require relatively pure samples. Thus, a separation step
often is necessary to remove materials that will interfere with the analytical measurement.
Questions
17
A law summarizes what happens, e.g., law of conservation of mass in a chemical reaction or
the ideal gas law, PV = nRT. A theory (model) is an attempt to explain why something
happens. Dalton’s atomic theory explains why mass is conserved in a chemical reaction. The
kinetic molecular theory explains why pressure and volume are inversely related at constant
temperature and moles of gas present as well as explaining the other mathematical
relationships summarized in PV = nRT.
18.
The fundamental steps are:
1. making observations
2. formulating hypotheses
3. performing experiments to test the hypotheses
The key to the scientific method is performing experiments to test hypotheses. If after the test
of time, the hypotheses seem to account satisfactorily for some aspect of natural behavior,
then the set of tested hypotheses turn into a theory (model). However, scientists continue to
perform experiments to refine or replace existing theories.
CHAPTER 1
19.
CHEMICAL FOUNDATIONS
3
A qualitative observation expresses what makes something what it is; it does not involve a
number, e.g., the air we breathe is a mixture of gases, ice is less dense than water, rotten milk
stinks.
The SI units are mass in kilograms, length in meters, and volume in the derived units of m3.
The assumed uncertainty in a number is 1 in the last significant figure of the number. The
precision of an instrument is related to the number of significant figures associated with an
experimental reading on that instrument. Different instruments for measuring mass, length, or
volume have varying degrees of precision. Some instruments only give a few significant
figures for a measurement while others will give more significant figures.
20.
Precision: reproducibility; Accuracy: the agreement of a measurement with the true value.
a. imprecise and inaccurate data: 12.32 cm, 9.63 cm, 11.98 cm, 13.34 cm
b. precise but inaccurate data: 8.76 cm, 8.79 cm, 8.72 cm, 8.75 cm
c. precise and accurate data: 10.60 cm, 10.65 cm, 10.63 cm, 10.64 cm
Data can be imprecise if the measuring device is imprecise as well as if the user of the
measuring device has poor skills. Data can be inaccurate due to a systematic error in the
measuring device or with the user. For example, a balance may read all masses as weighing
0.2500 g too high or the user of a graduated cylinder may read all measurements 0.05 mL too
low.
A set of measurements which are imprecise implies that all the numbers are not close to each
other. If the numbers aren’t reproducible, then all of the numbers can’t be very close to the
true value. Some say that if the average of imprecise data gives the true value, then it is
accurate data; a better description is that the data takers are extremely lucky.
21.
Significant figures are the digits we associate with a number. They contain all of the certain
digits and the first uncertain digit (the first estimated digit). What follows is one thousand
indicated to varying numbers of significant figures: 1000 or 1 × 103 (1 S.F.); 1.0 × 103 (2
S.F.); 1.00 × 103 (3 S.F.); 1000. or 1.000 × 103 (4 S.F.).
To perform the calculation, the addition/subtraction significant figure rule is applied to 1.5 
1.0. The result of this is the one significant figure answer of 0.5. Next, the
multiplication/division rule is applied to 0.5/0.50. A one significant figure number divided by
a two significant figure number yields an answer with one significant figure (answer = 1).
22.
The volume per mass is the reciprocal of the density (1/density). The volume per mass
conversion factor has units of cm3/g and is useful when converting from the mass of an object
to its volume in cm3.
23.
Straight line equation: y = mx + b where m is the slope of the line and b is the y-intercept. For
the TF vs. TC plot:
TF = (9/5)TC + 32
y= m x + b
The slope of the plot is 1.8 (= 9/5) and the y-intercept is 32F.
4
CHAPTER 1
CHEMICAL FOUNDATIONS
For the TC vs. TK plot:
TC = TK  273
y= mx +b
The slope of the plot is one and the y-intercept is 273C.
24.
a. coffee; saltwater; the air we breathe (N2 + O2 + others); brass (Cu + Zn)
b. book; human being; tree; desk
c. sodium chloride (NaCl); water (H2O); glucose (C6H12O6); carbon dioxide (CO2)
d. nitrogen (N2); oxygen (O2); copper (Cu); zinc (Zn)
e. boiling water; freezing water; melting a popsicle; dry ice subliming
f.
Elecrolysis of molten sodium chloride to produce sodium and chlorine gas; the explosive
reaction between oxygen and hydrogen to produce water; photosynthesis which converts
H2O and CO2 into C6H12O6 and O2; the combustion of gasoline in our car to produce CO2
and H2O
Exercises
Significant Figures and Unit Conversions
25.
a.
exact
c. exact
26.
27.
28.
29.
b. inexact
d. inexact (π has an infinite number of decimal places.)
a. one significant figure (S.F.). The implied uncertainty is 1000 pages. More significant
figures should be added if a more precise number is known.
b. two S.F.
c. four S.F.
d. two S.F.
e. infinite number of S.F. (exact number)
a. 6.07 × 10 15 ; 3 S.F.
b. 0.003840; 4 S.F.
c. 17.00; 4 S.F.
d. 8 × 108; 1 S.F.
e. 463.8052; 7 S.F.
f.
g. 301; 3 S.F.
h. 300.; 3 S.F.
a. 100; 1 S.F.
b. 1.0 × 102; 2 S.F.
c. 1.00 × 103; 3 S.F.
d. 100.; 3 S.F.
e. 0.0048; 2 S.F.
f.
g. 4.80 × 10 3 ; 3 S.F.
h. 4.800 × 10 3 ; 4 S.F.
f.
one S.F.
300; 1 S.F.
0.00480; 3 S.F.
When rounding, the last significant figure stays the same if the number after this significant
figure is less than 5 and increases by one if the number is greater than or equal to 5.
CHAPTER 1
30.
31.
CHEMICAL FOUNDATIONS
a. 3.42 × 10 4
a. 5 × 102
b. 1.034 × 104
b. 4.8 × 102
5
c. 1.7992 × 101
c. 4.80 × 102
d. 3.37 × 105
d. 4.800 × 102
For addition and/or subtraction, the result has the same number of decimal places as the
number in the calculation with the fewest decimal places. When the result is rounded to the
correct number of significant figures, the last significant figure stays the same if the number
after this significant figure is less than 5 and increases by one if the number is greater than or
equal to 5. The underline shows the last significant figure in the intermediate answers.
a. 212.2 + 26.7 + 402.09 = 640.99 = 641.0
b. 1.0028 + 0.221 + 0.10337 = 1.32717 = 1.327
c. 52.331 + 26.01  0.9981 = 77.3429 = 77.34
d. 2.01 × 102 + 3.014 × 103 = 2.01 × 102 + 30.14 × 102 = 32.15 × 102 = 3215
When the exponents are different, it is easiest to apply the addition/subtraction rule when
all numbers are based on the same power of 10.
e. 7.255  6.8350 = 0.42 = 0.420 (first uncertain digit is in the third decimal place).
32.
For multiplication and/or division, the result has the same number of significant figures as the
number in the calculation with the fewest significant figures.
a.
0.102  0.0821  273
 2.2635  2.26
1.01
b. 0.14 × 6.022 × 1023 = 8.431 × 1022 = 8.4 × 1022; Since 0.14 only has two significant
figures, the result should only have two significant figures.
c. 4.0 × 104 × 5.021 × 10 3 × 7.34993 × 102 = 1.476 × 105 = 1.5 × 105
d.
33.
2.00  10 6
 6.6667  1012  6.67  1012
3.00  10 7
a. Here, apply the multiplication/division rule first; then apply the addition/subtraction rule
to arrive at the one decimal place answer. We will generally round off at intermediate
steps in order to show the correct number of significant figures. However, you should
round off at the end of all the mathematical operations in order to avoid round-off error.
The best way to do calculations is to keep track of the correct number of significant
figures during intermediate steps, but round off at the end. For this problem, we
underlined the last significant figure in the intermediate steps.
2.526 0.470 80.705
= 0.8148 + 0.7544 + 186.558 = 188.1


3.1
0.623 0.4326
b. Here, the mathematical operation requires that we apply the addition/subtraction rule
first, then apply the multiplication/division rule.
6
CHAPTER 1
CHEMICAL FOUNDATIONS
6.404  2.91 6.404  2.91
= 12

18.7  17.1
1.6
c. 6.071 × 10 5  8.2 × 10 6  0.521 × 10 4 = 60.71 × 10 6  8.2 × 10 6  52.1 × 10 6
= 0.41 × 10 6 = 4 × 10 7
d.
3.8  10 12  4.0  10 13
38  10 13  4.0  10 13
42  10 13


 6.3  10  26
4  1012  6.3  1013
4  1012  63  1012
67  1012
e.
9.5  4.1  2.8  3.175 19.575
= 4.89 = 4.9

4
4
Uncertainty appears in the first decimal place. The average of several numbers can only
be as precise as the least precise number. Averages can be exceptions to the significant
figure rules.
f.
34.
8.925  8.905
0.020
× 100 = 0.22
 100 
8.925
8.925
a. 6.022 × 1023 × 1.05 × 102 = 6.32 × 1025
b.
6.6262  10 34  2.998  10 8
 7.82  10 17
9
2.54  10
c. 1.285 × 10 2 + 1.24 × 10 3 + 1.879 × 10 1
= 0.1285 × 10 1 + 0.0124 × 10 1 + 1.879 × 10 1 = 2.020 × 10 1
When the exponents are different, it is easiest to apply the addition/subtraction rule when
all numbers are based on the same power of 10.
d. 1.285  10 2 1.24  10 3 = 1.285  10 2 0.124  10 2 = 1.161  10 2
35.
e.
(1.00866  1.00728 )
0.00138

 2.29  10 27
23
6.02205  10
6.02205  10 23
f.
9.875  10 2  9.795  10 2
0.080  10 2

100

 100  8.1  10 1
2
2
9.875  10
9.875  10
g.
9.42  10 2  8.234  10 2  1.625  10 3 0.942  10 3  0.824  10 3  1.625  10 3

3
3
= 1.130 × 103
a. 8.43 cm ×
1m
1000 mm

 84.3 mm
100 cm
m
c. 294.5 nm 
b. 2.41 × 102 cm ×
1m
100 cm

 2.945  10 5 cm
9
m
1  10 nm
1m
= 2.41 m
100 cm
CHAPTER 1
CHEMICAL FOUNDATIONS
d. 1.445 ×104 m ×
f.
36.
1 km
= 14.45 km
1000 m
1  1012 m 1  10 9 nm

 6.50  10 23 nm
Tm
m
1g
1 kg

 25  10 18 kg  2.5  10 17 kg
15
1  10 fg 1000 g
c. 25 fg 
d. 8.0 dm3 ×
e. 1 mL 
37.
1000 mm
= 2.353 × 105 mm
m
1  1012 g
1 kg

 1  10 9 kg
Tg
1000 g
b. 6.50 × 102 Tm 
1 μg 
e. 235.3 m ×
1m
1  10 6 μm

 0.9033 μm
m
1  10 9 nm
903.3 nm 
a. 1 Tg 
f.
7
1L
dm 3
= 8.0 L (1 L = 1 dm3 = 1000 cm3 = 1000 mL)
1L
1  10 6 μL

 1  10 3 μL
1 000 mL
L
1g
1  1012 pg

 1  10 6 pg
6
g
1  10 μg
a. Appropriate conversion factors are found in Appendix 6. In general, the number of
significant figures we use in the conversion factors will be one more than the number of
significant figures from the numbers given in the problem. This is usually sufficient to
avoid round-off error.
3.91 kg ×
16 oz
1 lb
= 8.62 lb; 0.62 lb ×
= 9.9 oz
0.4536 kg
lb
Baby’s weight = 8 lb and 9.9 oz or to the nearest ounce, 8 lb and 10. oz.
51.4 cm ×
1 in
= 20.2 in ≈ 20 1/4 in = baby’s height
2.54 cm
b. 25,000 mi ×
1.61 km
1000 m
= 4.0 × 104 km; 4.0 × 104 km ×
= 4.0 × 107 m
mi
km

1m  
1m 
   2.1 dm 
 = 1.2 × 10 2 m3
c. V = 1 × w × h = 1.0 m ×  5.6 cm 
100
cm
10
dm

 

8
CHAPTER 1
CHEMICAL FOUNDATIONS
3
1L
 1 0 dm 
1.2 × 10 2 m3 × 
= 12 L
 
dm 3
 m 
3
1000 cm3  1 in 
 = 730 in3; 730 in3 ×
12 L ×
 
L
2
.
54
cm


38.
39.
a. 908 oz ×
1 lb 0.4536 kg
= 25.7 kg

16 oz
lb
b. 12.8 L ×
1 qt
1 gal
= 3.38 gal

0.9463 L 4 qt
c. 125 mL ×
1L
1 qt
= 0.132 qt

1000 mL 0.9463 L
d. 2.89 gal ×
4 qt
1L
1000 mL
= 1.09 × 104 mL


1 gal 1 .057 qt
1L
e. 4.48 lb ×
453 .6 g
= 2.03 × 103 g
1 lb
f.
1L
1.06 qt
= 0.58 qt

1000 mL
L
550 mL ×
a. 1.25 mi ×
3
 1 ft 

 = 0.42 ft3
12
in


8 furlongs
40 rods
= 10.0 furlongs; 10.0 furlongs ×
= 4.00 × 102 rods
furlong
mi
4.00 × 102 rods ×
2.01 × 103 m ×
5.5 yd 36 in 2.54 cm
1m
= 2.01 × 103 m



rod
yd
in
100 cm
1 km
= 2.01 km
1000 m
b. Let's assume we know this distance to ± 1 yard. First convert 26 miles to yards.
26 mi ×
5280 ft 1 yd
= 45,760. yd

mi
3 ft
26 mi + 385 yd = 45,760. yd + 385 yd = 46,145 yards
46,145 yard ×
1 rod
1 furlong
= 8390.0 rods; 8390.0 rods ×
= 209.75 furlongs
5.5 yd
40 rods
46,145 yard ×
36 in 2.54 cm
1m
1 km
= 42,195 m; 42,195 m ×
= 42.195 km


yd
in
100 cm
1000 m
CHAPTER 1
CHEMICAL FOUNDATIONS
9
2
40.
10,000 m 2  1 km 
 = 1  10 2 km 2
a. 1 ha ×
 
ha
 1000 m 
160 rod 2  5.5 yd 36 in 2.54 cm
1m
b. 5.5 acre ×
 



acre
yd
in
100 cm
 rod
2

 = 2.2 × 104 m2

2
 1 km 
 = 0.022 km2
2.2 × 10 m ×
= 2.2 ha; 2.2 × 10 m × 
4
2
1000
m
1  10 m


4
2
1 ha
4
2
c. Area of lot = 120 ft × 75 ft = 9.0 × 103 ft2
 1 yd 1 rod
9.0 × 10 ft × 

 3 ft 5.5 yd
3
2
2

1 acre
$6,500
$31,000
 
= 0.21 acre;

2
0.21 acre
acre
 160 rod
We can use our result from (b) to get the conversion factor between acres and ha (5.5 acre
= 2.2 ha.). Thus, 1 ha = 2.5 acre.
1 ha
$6,500
$77,000
= 0.084 ha; The price is:

2.5 acre
0.084 ha
ha
0.21 acre ×
41.
a. 1 troy lb ×
12 troy oz 20 pw 24 grains 0.0648 g 1 kg
= 0.373 kg




troy lb
troy oz
pw
grain
1000 g
1 troy lb = 0.373 kg ×
b. 1 troy oz ×
2.205 lb
= 0.822 lb
kg
20 pw 24 grains 0.0648 g
= 31.1 g


troy oz
pw
grain
1 troy oz = 31.1 g ×
1 carat
= 156 carats
0.200 g
c. 1 troy lb = 0.373 kg; 0.373 kg ×
42.
a. 1 grain ap ×
1000 g 1 cm 3

= 19.3 cm3
kg
19.3 g
1 scruple
1 dram ap 3.888 g
= 0.06480 g


20 grain ap 3 scruples dram ap
From the previous question, we are given that 1 grain troy = 0.0648 g = 1 grain ap. So,
the two are the same.
b. 1 oz ap ×
8 dram ap 3.888 g 1 oz troy *
= 1.00 oz troy


oz ap
dram ap
31.1 g
*See Exercise 41b.
10
CHAPTER 1
c. 5.00 × 102 mg ×
0.386 scruple ×
d. 1 scruple ×
43.
CHEMICAL FOUNDATIONS
1g
1 dram ap 3 scruples
= 0.386 scruple


1000 mg
3.888 g
dram ap
20 grains ap
= 7.72 grains ap
scruple
1 dram ap 3.888 g
= 1.296 g

3 scruples dram ap

3.00  108 m  1.094 yd 60 s 60 min
1 knot
 
warp 1.71 =  5.00 




s
m
min
hr
2000 yd / hr


= 2.95 × 109 knots

3.00  108 m 
1 km
1 mi
60 s 60 min
 5.00 
 
= 3.36 × 109 mi/hr





s
1000
m
1
.
609
km
min
hr


44.
100. m
100 . m 1 km
60 s 60 min
= 10.2 m/s;
= 36.8 km/hr



9.77 s
9.77 s 1000 m min
hr
100. m 1.0936 yd 3 ft
33.6 ft
1 mi
60 s 60 min
= 33.6 ft/s;
= 22.9 mi/hr





9.77 s
m
yd
s
5280 ft min
hr
1.00 × 102 yd ×
45.
1m
9.77 s
= 8.93 s

1.0936 yd 100. m
65 km 0.6214 mi
= 40.4 = 40. mi/hr

hr
km
To the correct number of significant figures, 65 km/hr does not violate a 40. mi/hr speed
limit.
46.
112 km ×
0.6214 mi 1 hr
= 1.1 hr = 1 hr and 6 min

km
65 mi
112 km ×
0.6214 mi 1 gal
3.785 L
= 9.4 L of gasoline


km
28 mi
gal
47.
$17.25 U.S. $1.00 Canadian
1 gal
= $0.68 Canadian/L


8.21 gal
$0.82 U.S.
3.7854 L
48.
1.5 teaspoons ×
80. mg acet
= 240 mg acetaminophen
0.50 teaspoon
CHAPTER 1
CHEMICAL FOUNDATIONS
11
240 mg acet
1 lb
= 22 mg acetaminophen/kg

24 lb
0.454 kg
240 mg acet
1 lb
= 15 mg acetaminophen/kg

35 lb
0.454 kg
The range is from 15 mg to 22 mg acetaminophen per kg of body weight.
Temperature
49.
a. TC =
b. TC =
c. TC =
d. TC =
50.
5
9
5
9
5
9
5
9
(TF  32) =
5
9
(459C  32) = 273C; TK = TC + 273 = 273C + 273 = 0 K
(40.F  32) = 40.C; TK = 40.C + 273 = 233 K
(68F  32) = 20.C; TK = 20.C + 273 = 293 K
(7 × 107F  32) = 4 × 107C; TK = 4 × 107C + 273 = 4 × 107 K
96.1°F ± 0.2°F; First, convert 96.1°F to °C. TC =
5
5
(TF - 32) =
(96.1 - 32) = 35.6°C
9
9
A change in temperature of 9°F is equal to a change in temperature of 5°C.
uncertainty is:
5 C
= ± 0.1°C. Thus, 96.1 ± 0.2°F = 35.6 ± 0.1°C
9 F
9
9
a. TF =
× TC + 32 = × 39.2°C + 32 = 102.6°F (Note: 32 is exact.)
5
5
± 0.2°F ×
51.
TK = TC + 273.2 = 39.2 + 273.2 = 312.4 K
b. TF =
9
× (25) + 32 = 13°F; TK = 25 + 273 = 248 K
5
c.
9
× (273) + 32 = -459°F; TK = 273 + 273 = 0 K
5
TF =
9
× 801 + 32 = 1470°F; TK = 801 + 273 = 1074 K
5
a. TC = TK  273 = 233  273 = -40.°C
d. TF =
52.
TF =
9
9
× TC + 32 = × (40.) + 32 = 40.°F
5
5
b. TC = 4  273 = 269°C; TF =
9
× (269) + 32 = 452°F
5
c. TC = 298  273 = 25°C; TF =
9
× 25 + 32 = 77°F
5
So the
12
CHAPTER 1
d. TC = 3680  273 = 3410°C; TF =
CHEMICAL FOUNDATIONS
9
× 3410 + 32 = 6170°F
5
Density
3
53.
453.6 g
 2.54 cm 
= 1.6 × 105 g; V = 1.2 × 104 in3  
 = 2.0 × 105 cm3
in
lb


5
mass
1  10 g

 0.80 g / cm3
density =
5
3
volume
2.0  10 cm
mass = 350 lb 
Because the material has a density less than water, it will float in water.
54.
V
4 3 4
2.0 g
= 3.8 g/cm3
π r   3.14  (0.50 cm)3  0.52 cm3 ; d 
3
3
0.52 cm3
The ball will sink.
55.
V
density =
56.
3
4 3 4
1000 m 100 cm 

33
3
π r   3.14   7.0  10 5 km 

  1.4  10 cm
3
3
km
m


mass

volume
2  10 36 kg 
1.4  10 33
1000 g
kg
= 1.4 × 106 g/cm3 = 1 × 106 g/cm3
cm 3
V = l × w × h = 2.9 cm × 3.5 cm × 10.0 cm = 1.0 × 102 cm3
d = density =
615.0 g
1.0  10 cm
2
3

6.2 g
cm3
0.200 g 1 cm 3

= 0.28 cm3
carat
3.51 g
57.
5.0 carat 
58.
2.8 mL 
59.
V = 21.6 mL  12.7 mL = 8.9 mL; density =
60.
5.25 g 
1 cm 3 3.51 g
1 carat


= 49 carats
3
mL
cm
0.200 g
1 cm 3
10.5 g
33.42 g
= 3.8 g/mL = 3.8 g/cm3
8.9 mL
= 0.500 cm3 = 0.500 mL
The volume in the cylinder will rise to 11.7 mL (11.2 mL + 0.500 mL = 11.7 mL).
61.
a. Both have the same mass of 1.0 kg.
b. 1.0 mL of mercury; Mercury has a greater density than water. Note: 1 mL = 1 cm3
1.0 mL 
13.6 g
0.998 g
= 14 g of mercury; 1.0 mL 
= 1.0 g of water
mL
mL
CHAPTER 1
CHEMICAL FOUNDATIONS
13
c. Same; Both represent 19.3 g of substance.
19.3 mL 
0.9982 g
19.32 g
= 19.3 g of water; 1.00 mL 
= 19.3 g of gold
mL
mL
d. 1.0 L of benzene (880 g vs 670 g)
75 mL 
8.96 g
1000 mL 0.880 g
= 670 g of copper; 1.0 L 
= 880 g of benzene

L
mL
mL
2
62.
 5280 ft 
Volume of lake = 100 mi × 
 × 20 ft = 6 × 1010 ft3
 mi 
2
3
 12 in 2.54 cm  1 mL 0.4 μg
6 × 1010 ft3 × 
= 7 × 1014 μg mercury


 
in 
mL
cm3
 ft
7 × 1014 μg × 
63.
1g

1 kg
10 μg 10 3 g
6
= 7 × 105 kg of mercury
a. 1.0 kg feather; Feathers are less dense than lead.
b. 100 g water; Water is less dense than gold. c. Same; Both volumes are 1.0 L.
64.
1 cm3
= 3.0 × 105 cm3 [H2(g) = hydrogen gas]
0.000084 g
a. H2(g): V = 25.0 g ×
b. H2O(l): V = 25.0 g ×
c. Fe(s): V = 25.0 g ×
1 cm3
= 25.0 cm3 [H2O(l) = water]
0.9982 g
1 cm3
= 3.18 cm3 [Fe(s) = iron]
7.87 g
Notice the huge volume of the gaseous H2 sample as compared to the liquid and solid
samples. The same mass of gas occupies a volume that is over 10,000 times larger than the
liquid sample. Gases are indeed mostly empty space.
65.
V = 1.00 × 103 g ×
1 cm 3
= 44.3 cm3
22.57 g
44.3 cm3 = 1 × w × h = 4.00 cm × 4.00 cm × h, h = 2.77 cm
66.
V = 22 g ×
1 cm 3
= 2.5 cm3; V = π r2 × l where l = length of the wire
8.96 g
2
 1 cm 
 0.25 mm 
 × l, l = 5.1 × 103 cm = 170 ft
2.5 cm3 = π × 
 × 
2
10
mm




2
14
CHAPTER 1
CHEMICAL FOUNDATIONS
Classification and Separation of Matter
67.
A gas has molecules that are very far apart from each other while a solid or liquid has
molecules that are very close together. An element has the same type of atom, whereas a
compound contains two or more different elements. Picture i represents an element that
exists as two atoms bonded together (like H2 or O2 or N2). Picture iv represents a compound
(like CO, NO, or HF). Pictures iii and iv contain representations of elements that exist as
individual atoms (like Ar, Ne, or He).
a. Picture iv represents a gaseous compound. Note that pictures ii and iii also contain a
gaseous compound, but they also both have a gaseous element present.
b. Picture vi represents a mixture of two gaseous elements.
c. Picture v represents a solid element.
d. Pictures ii and iii both represent a mixture of a gaseous element and a gaseous compound.
68.
Solid: rigid; has a fixed volume and shape; slightly compressible
Liquid: definite volume but no specific shape; assumes shape of the container; slightly
compressible
Gas: no fixed volume or shape; easily compressible
Pure substance: has constant composition; can be composed of either compounds or elements
Element: substances that cannot be decomposed into simpler substances by chemical or
physical means.
Compound: a substance that can be broken down into simpler substances (elements) by
chemical processes.
Homogeneous mixture: a mixture of pure substances that has visibly indistinguishable parts.
Heterogeneous mixture: a mixture of pure substances that has visibly distinguishable parts.
Solution: a homogeneous mixture; can be a solid, liquid or gas
Chemical change: a given substance becomes a new substance or substances with different
properties and different composition.
Physical change: changes the form (g, l, or s) of a substance but does no change the chemical
composition of the substance.
69.
Homogeneous: Having visibly indistinguishable parts (the same throughout).
Heterogeneous: Having visibly distinguishable parts (not uniform throughout).
a. heterogeneous (Due to hinges, handles, locks, etc.)
CHAPTER 1
CHEMICAL FOUNDATIONS
15
b. homogeneous (hopefully; If you live in a heavily polluted area, air may be
heterogeneous.)
70.
c. homogeneous
d. homogeneous (hopefully, if not polluted)
e. heterogeneous
f.
a.
pure
b. mixture
c. mixture
d.
pure
f.
pure
g. mixture
h. mixture
i.
pure
heterogeneous
e. mixture (copper and zinc)
Iron and uranium are elements. Water and table salt are compounds. Water is H2O and table
salt is NaCl. Compounds are composed of two or more elements.
71.
A physical change is a change in the state of a substance (solid, liquid and gas are the three
states of matter); a physical change does not change the chemical composition of the
substance. A chemical change is a change in which a given substance is converted into
another substance having a different formula (composition).
a. Vaporization refers to a liquid converting to a gas, so this is a physical change. The
formula (composition) of the moth ball does not change.
b. This is a chemical change since hydrofluoric acid (HF) is reacting with glass (SiO2) to
form new compounds which wash away.
c. This is a physical change since all that is happening is the conversion of liquid alcohol to
gaseous alcohol. The alcohol formula (C2H5OH) does not change.
d. This is a chemical change since the acid is reacting with cotton to form new compounds.
72.
a. Distillation separates components of a mixture, so the orange liquid is a mixture (has an
average color of the yellow liquid and the red solid). Distillation utilizes boiling point
differences to separate out the components of a mixture. Distillation is a physical change
because the components of the mixture do not become different compounds or elements.
b. Decomposition is a type of chemical reaction. The crystalline solid is a compound, and
decomposition is a chemical change where new substances are formed.
c. Tea is a mixture of tea compounds dissolved in water. The process of mixing sugar into
tea is a physical change. Sugar doesn’t react with the tea compounds, it just makes the
solution sweeter.
Additional Exercises
1 capsule
= 24 capsules
0.65 g
73.
15.6 g ×
74.
126 gal ×
4 qt
1L
= 477 L

gal 1.057 qt
16
75.
CHAPTER 1
CHEMICAL FOUNDATIONS
100 cm  
100 cm 

Total volume =  200 . m 
   300 . m 
 × 4.0 cm = 2.4 × 109 cm3
m  
m 

2
2

2.54 cm 
 12 in 
 2.54 cm   
2
Vol. of topsoil covered by 1 bag = 10. ft  
  
   1.0 in 

in 
 ft 
 in   

= 2.4 × 104 cm3
2.4 × 109 cm3 
76.
1 bag
2.4  10 cm
4
3
= 1.0 × 105 bags topsoil
a. No; If the volumes were the same, then the gold idol would have a much greater mass
because gold is much more dense than sand.
b. Mass = 1.0 L 
1 000 cm3 1 9.32 g
1 kg


= 19.32 kg (= 42.59 lb)
3
L
1000
g
cm
It wouldn't be easy to play catch with the idol because it would have a mass of over 40
pounds.
77.
18.5 cm ×
10.0 o F
= 35.2F increase; Tfinal = 98.6 + 35.2 = 133.8F
5.25 cm
Tc = 5/9 (133.8 – 32) = 56.56C
78.
massbenzene = 58.80 g  25.00 g = 33.80 g; Vbenzene = 33.80 g 
Vsolid = 50.0 cm3  38.4 cm3 = 11.6 cm3; density =
79.
1 cm 3
= 38.4 cm3
0.880 g
25.00 g
= 2.16 g/cm3
11.6 cm3
a. Volume × density = mass; the orange block is more dense. Because mass (orange) >
mass (blue) and because volume (orange) < volume (blue), the density of the orange
block must be greater to account for the larger mass of the orange block.
b. Which block is more dense cannot be determined. Because mass (orange) > mass (blue)
and because volume (orange) > volume (blue), the density of the orange block may or
may not be larger than the blue block. If the blue block is more dense, its density cannot
be so large that its mass is larger than the orange block’s mass.
c. The blue block is more dense. Because mass (blue ) = mass (orange) and because
volume (blue) < volume (orange), the density of the blue block must be larger in order to
equate the masses.
d. The blue block is more dense. Because mass (blue) > mass (orange) and because the
volumes are equal, the density of the blue block must be larger in order to give the blue
block the larger mass.
CHAPTER 1
80.
CHEMICAL FOUNDATIONS
17
4π r 3 4π  c

Circumference = c = 2π r; V =

3
3  2π
Largest density =
5.25 oz
(9.00 in )
=
3
5.25 oz
=
3
12.3 in
3

c3
  2
6π

0.427 oz
in 3
6π 2
Smallest density =
Maximum range is:
81.
5.00 oz
(9.25 in ) 3
6π 2
=
5.00 oz
13.4 in
(0.373  0.427 ) oz
V = Vfinal - Vinitial; d =
in 3
=
0.73 oz
in 3
or 0.40 ± 0.03 oz/in3 (Uncertainty in 2nd decimal
place.)
28.90 g
9.8 cm  6.4 cm
3
3
3

28.90 g
3.4 cm3
= 8.5 g/cm3
dmax =
massmax
; We get Vmin from 9.7 cm3  6.5 cm3 = 3.2 cm3.
Vmin
dmax =
28.87 g
8.0 g
massmin
28.93 g
9.0 g
; dmin =
=


3
3
3
3
Vmax
3.2 cm
cm
9.9 cm  6.3 cm
cm3
The density is: 8.5 ± 0.5 g/cm3.
Challenge Problems
82.
In general, glassware is estimated to one place past the markings.
a.
128.7 mL glassware
130
b.
18 mL glassware
30
c. 23.45 mL glassware
24
129
20
128
127
read to tenth’s place
10
read to one’s place
23
read to two decimal places
128.7 + 18 + 23.45 = 170.15 = 170. (Due to 18, the sum would be only known to the
ones place.)
18
83.
84.
CHAPTER 1
a.
2.70  2.64
× 100 = 2%
2.70
c.
1.000  0.9981
0.002
× 100 =
× 100 = 0.2%
1.000
1.000
b.
CHEMICAL FOUNDATIONS
| 16.12  16.48 |
× 100 = 2.2%
16.12
a. At some point in 1982, the composition of the metal used in minting pennies was
changed because the mass changed during this year (assuming the volume of the pennies
were constant).
b. It should be expressed as 3.08 ± 0.05 g. The uncertainty in the second decimal place will
swamp any effect of the next decimal places.
85.
Heavy pennies (old): mean mass = 3.08 ± 0.05 g
(2.467  2.545  2.518)
= 2.51 ± 0.04 g
3
Light pennies (new): mean mass =
Because we are assuming that volume is additive, let’s calculate the volume of 100. g of each
type of penny then calculate the density of the alloy. For 100. g of the old pennies, 95 g will
be Cu (copper) and 5 g will be Zn (zinc).
V = 95 g Cu ×
1 cm3
1 cm3
 5 g Zn 
= 11.3 cm3 (carrying one extra sig. fig.)
8.96 g
7.14 g
Density of old pennies =
100 . g
11.3 cm
3
= 8.8 g/cm3
For 100. g of new pennies, 97.6 g will be Zn and 2.4 g will be Cu.
V = 2.4 g Cu ×
1 cm 3
1 cm 3
+ 97.6 g Zn ×
= 13.94 cm3 (carrying one extra sig. fig.)
8.96 g
7.14 g
Density of new pennies =
100 . g
3
= 7.17 g/cm3
13.94 cm
mass
d=
; Because the volume of both types of pennies are assumed equal, then:
volume
3
d new mass new 7.17 g / cm


= 0.81
3
d old
mass old
8.8 g / cm
mass new 2.51 g
The calculated average mass ratio is:
= 0.815

mass old 3.08 g
To the first two decimal places, the ratios are the same. If the assumptions are correct, then
we can reasonably conclude that the difference in mass is accounted for by the difference in
alloy used.
CHAPTER 1
86.
CHEMICAL FOUNDATIONS
a.
100oA
115oC
100oA
160oC
0oA
-45oC
19
A change in temperature of 160°C equals a
change in temperature of 100°A.
So,
160C
is our unit conversion for a
100A
degree change in temperature.
At the freezing point: 0°A = -45°C
Combining the two pieces of information:
TA = (TC + 45°C) ×
b. TC = (TF - 32) ×
TF  32 =
c. TC = TA ×
100A
5A
8C
= (TC + 45°C) ×
or TC = TA ×
 45°C
160C
8C
5A
5
8
5
; TC = TA ×
 45 = (TF - 32) ×
9
5
9
9
72
72F
8


 81, TF = TA ×
 49F
 TA   45 = TA 
5
5
25
25A


8
8
3  Tc
 45 and TC = TA; So, TC = TC ×  45,
= 45, TC = 75°C = 75°A
5
5
5
8A
72F
45°C = 93°C; TF = 86°A ×
 49°F = 199°F = 2.0 × 102°F
25A
5C
5A
e. TA = (45°C + 45°C) ×
= 56°A
8C
d. TC = 86°A ×
87.
Let x = mass of copper and y = mass of silver.
105.0 g = x + y and 10.12 mL =
y
x
; Solving:

8.96 10.5
x
105.0  x 


10.12 
 × 8.96 × 10.5, 952.1 = 10.5 x + 940.8  8.96 x
8.96
10.5 

(carrying 1 extra significant figure)
11.3 = 1.54 x, x = 7.3 g; mass %Cu =
88.
7.3 g
× 100 = 7.0% Cu
105.0 g
a.
2 compounds
compound and element (diatomic)
20
CHAPTER 1
CHEMICAL FOUNDATIONS
b.
gas element (monoatomic)
atoms/molecules far apart;
random order; takes volume
of container
89.
liquid element
atoms/molecules close
together; somewhat
ordered arrangement;
takes volume of container
solid element
atoms/molecules
close together;
ordered arrangement;
has its own volume
a. One possibility is that rope B is not attached to anything and rope A and rope C are
connected via a pair of pulleys and/or gears.
b. Try to pull rope B out of the box. Measure the distance moved by C for a given
movement of A. Hold either A or C firmly while pulling on the other.
90.
The bubbles of gas is air in the sand that is escaping; methanol and sand are not reacting.
We will assume that the mass of trapped air is insignificant.
mass of dry sand = 37.3488 g  22.8317 g = 14.5171 g
mass of methanol = 45.2613 g  37.3488 g = 7.9125 g
Volume of sand particles (air absent) = volume of sand and methanol  volume of methanol
Volume of sand particles (air absent) = 17.6 mL  10.00 mL = 7.6 mL
density of dry sand (air present) =
14.5171 g
= 1.45 g/mL
10.0 mL
7.9125 g
= 0.7913 g/mL
10.00 mL
14.5171 g
density of sand particles (air absent) =
= 1.9 g/mL
7.6 mL
density of methanol =
Integrative Problems
91.
2.97  108 persons  0.0100 = 2.97  106 persons contributing
$4.75  10 8
$160 .
20 nickels
= $160./person;
= 3.20 × 103 nickels/person

6
2.97  10 persons
person
$1
$160 .
1 pound sterling
= 85.6 pounds sterling/person

person
$1.869
CHAPTER 1
92.
CHEMICAL FOUNDATIONS
22610 kg 1000 g
1 m3


= 22.61 g/cm3
kg
m3
1  10 6 cm3
volume of block = 10.0 cm  8.0 cm  9.0 cm = 720 cm3;
93.
21
At 200.0 F: TC =
At 100.0 F: TC =
5
(200.0 F 32 F) = 93.33 C; TK = 93.33 + 273.15 = 366.48 K
9
5
9
22 .61 g
× 720 cm3 = 1.6 × 104 g
cm3
(100.0 F 32 F) = 73.33 C; TK = 73.33 C + 273.15
= 199.82 K
T(C) = [93.33 C (73.33 C)] = 166.66 C; T(K) = [366.48 K 199.82 K]
= 166.66 K
The “300 Club” name only works for the Fahrenheit scale; it does not hold true for the
Celsius and Kelvin scales.
Marathon Problem
3
94.
 2.54 cm 
a. Vgold = r2h = 3.14 × (0.25/2 in)2 × 1.5 in × 
 = 1.2 cm3
in


23.1984 g
dgold (at 86 F) =
= 19 g/cm3
1.2 cm3
b. Calculate the density of the liquid at 86F, then determine the density at 40.F.
massliquid = 79.16 g 73.47 g = 5.69 g
volumefinal = 8.5 cm3= Vgold + Vliquid, Vliquid = 8.5 cm3  1.2 cm3 = 7.3 cm3
dliquid (at 86 F) =
5.69 g
7.3 cm3
= 0.78 g/cm3
The density will increase by 1.0 % for every 10. C drop in temperature. The temperature
drop is 86  40. = 46 F. Because 1F is equivalent to 5/9C, the temperature drop in
C equals 46(5/9) = 26C. Because there is a 1% increase in density for every 10.C
drop in temperature, there will be a 2.6% increase in density for the 26C temperature
drop.
densityliquid (at 40.C) = 1.026 × 0.78 g/cm3 = 0.80 g/cm3