Download CHAPTER 1 CHEMICAL FOUNDATIONS 1 CHAPTER ONE

CHAPTER ONE
CHEMICAL FOUNDATIONS
.
2.
No, it is useful whenever a systematic approach of observation and hypothesis testing can be
used.
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.
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).
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.
Questions
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
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CHAPTER 1
CHEMICAL FOUNDATIONS
Exercises
Significant Figures and Unit Conversions
26.
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)
f.
30.
a. 5 × 102
b. 4.8 × 102
d. 4.800 × 102
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.
c. 4.80 × 102
one S.F.
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.
34.
2.00  10 6
 6.6667  1012  6.67  1012
7
3.00  10
a. 6.022 × 1023 × 1.05 × 102 = 6.32 × 1025
b.
6.6262  10 34  2.998  10 8
 7.82  10 17
2.54  10 9
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
e.
(1.00866  1.00728 )
0.00138

 2.29  10 27
23
23
6.02205  10
6.02205  10
CHAPTER 1
38.
42.
CHEMICAL FOUNDATIONS
3
f.
9.875  10 2  9.795  10 2
0.080  10 2

100

 100  8.1  10 1
9.875  10 2
9.875  10 2
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. 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 grain ap ×
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
c. 5.00 × 102 mg ×
0.386 scruple ×
d. 1 scruple ×
*See Exercise 41b.
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 km
1 mi
60 s 60 min
 5.00 
 



= 3.36 × 109 mi/hr


s
1000
m
1
.
609
km
min
hr


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CHAPTER 1
CHEMICAL FOUNDATIONS
Temperature
50.
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:
± 0.2°F ×
58.
2.8 mL 
60.
5.25 g 
So the
5 C
= ± 0.1°C. Thus, 96.1 ± 0.2°F = 35.6 ± 0.1°C
9 F
1 cm 3 3.51 g
1 carat


= 49 carats
3
mL
cm
0.200 g
1 cm 3
10.5 g
= 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).
Classification and Separation of Matter
70.
a.
pure
b. mixture
c. mixture
d.
pure
f.
pure
g. mixture
h. mixture
i.
pure
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.
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.