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CHAPTER 1
Chemistry and Measurement
โ SOLUTIONS TO EXERCISES
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
1.1.
From the law of conservation of mass,
Mass of wood + mass of air = mass of ash + mass of gases
Substituting, you obtain
1.85 grams + 9.45 grams = 0.28 grams + mass of gases
or,
Mass of gases = (1.85 + 9.45 โ 0.28) grams = 11.02 grams
Thus, the mass of gases in the vessel at the end of the experiment is 11.02 grams.
1.2.
Physical properties: soft, silvery-colored metal; melts at 64ยฐC.
Chemical properties: reacts vigorously with water, with oxygen, and with chlorine.
1.3.
a.
The factor 9.1 has the fewest significant figures, so the answer should be reported to two
significant figures.
5.61 ๏ด 7.891
= 4.86 = 4.9
9.1
b.
The number with the least number of decimal places is 8.91. Therefore, round the answer to
two decimal places.
8.91 โ 6.435 = 2.475 = 2.48
c.
The number with the least number of decimal places is 6.81. Therefore, round the answer to
two decimal places.
6.81 โ 6.730 = 0.080 = 0.08
d.
You first do the subtraction within parentheses. In this step, the number with the least
number of decimal places is 6.81, so the result of the subtraction has two decimal places.
The least significant figure for this step is underlined.
38.91 ๏ด (6.81 โ 6.730) = 38.91 ๏ด 0.080
Next, perform the multiplication. In this step, the factor 0.080 has the fewest significant
figures, so round the answer to one significant figure.
38.91 ๏ด 0.080 = 3.11 = 3
1
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2
Chapter 1: Chemistry and Measurement
1.4.
a.
1.84 x 10โ9 m = 1.84 nm
b.
5.67 x 10โ12 s = 5.67 ps
c.
7.85 x 10โ3 g = 7.85 mg
d.
9.7 x 103 m = 9.7 km
e.
0.000732 s = 0.732 ms, or 732 ยตs
f.
0.000000000154 m = 0.154 nm, or 154 pm
a.
Substituting, we find that
1.5.
tC =
5๏ฐC
5๏ฐC
๏ด (tF โ 32ยฐF) =
๏ด (102.5ยฐF โ 32ยฐF) = 39.167ยฐC
9๏ฐF
9๏ฐF
= 39.2ยฐC
b.
Substituting, we find that
1K๏ถ
1K๏ถ
๏ฆ
๏ฆ
TK = ๏ง tc ๏ด
๏ท + 273.15 K = ๏ง ๏ญ78๏ฐC ๏ด
๏ท + 273.15 K = 195.15 K
1๏ฐC ๏ธ
1๏ฐC ๏ธ
๏จ
๏จ
= 195 K
1.6.
Recall that density equals mass divided by volume. You substitute 159 g for the mass and 20.2
g/cm3 for the volume.
d=
m
159 g
=
= 7.871 g/cm3 = 7.87 g/cm3
V
20.2 cm3
The density of the metal equals that of iron.
1.7.
Rearrange the formula defining the density to obtain the volume.
V=
m
d
Substitute 30.3 g for the mass and 0.789 g/cm3 for the density.
V=
1.8.
30.3 g
= 38.40 cm3 = 38.4 cm3
0.789 g/cm3
Since one pm = 10โ12 m, and the prefix milli- means 10โ3, you can write
121 pm ๏ด
1 mm
10๏ญ12 m
๏ด ๏ญ3
= 1.21 ๏ด 10โ7 mm
1 pm
10 m
3
3
1.9.
๏ฆ 10๏ญ10 m ๏ถ
๏ฆ 1 dm ๏ถ
โ26
3
67.6 ร
3 ๏ด ๏ง
๏ท ๏ด ๏ง ๏ญ1
๏ท = 6.76 ๏ด 10 dm
๏จ 10 m ๏ธ
๏จ 1ร
๏ธ
1.10.
From the definitions, you obtain the following conversion factors:
1=
36 in
1 yd
1=
2.54 cm
1 in
1=
10-2 m
1 cm
The conversion factor for yards to meters is as follows:
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Chapter 1: Chemistry and Measurement
1.000 yd x
3
2.54 cm 10-2 m
36 in
x
x
= 0.9144 m (exact)
1 in
1 cm
1 yd
Finally,
3.54 yd x
0.9144 m
= 3.237 m = 3.24 m
1 yd
โ ANSWERS TO CONCEPT CHECKS
1.1.
Box A contains a collection of identical units; therefore, it must represent an element. Box B
contains a compound because a compound is the chemical combination of two or more elements
(two elements in this case). Box C contains a mixture because it is made up of two different
substances.
1.2.
a.
For a person who weighs less than 100 pounds, two significant figures are typically used,
although one significant figure is possible (for example, 60 pounds). For a person who
weighs 100 pounds or more, three significant figures are typically used to report the weight
(given to the whole pound), although people often round to the nearest unit of 10, which
may result in reporting the weight with two significant figures (for example, 170 pounds).
b.
Assuming a weight of 165 pounds, rounded to two significant figures this would be
reported as 1.7 x 102 pounds.
c.
For example, 165 lb weighed on a scale that can measure in 100-lb increments would be
200 lb. Using the conversion factor 1 lb = 0.4536 kg, 165 lb is equivalent to 74.8 kg. Thus,
on a scale that can measure in 50-kg increments, 165 lb would be 50 kg.
a.
If your leg is approximately 32 inches long, this would be equivalent to 0.81 m, 8.1 dm, or
81 cm.
b.
One story is approximately 10 feet, so three stories is 30 feet. This would be equivalent to
approximately 9 m.
c.
Normal body temperature is 98.6ยฐF, or 37.0ยฐC. Thus, if your body temperature were 39ยฐC
(102ยฐF), you would feel as if you had a moderate fever.
d.
Room temperature is approximately 72ยฐF, or 22ยฐC. Thus, if you were sitting in a room at
23ยฐC (73ยฐF), you would be comfortable in a short-sleeve shirt.
1.3.
1.4.
Gold is a very unreactive substance, so comparing physical properties is probably your best
option. However, color is a physical property you cannot rely on in this case to get your answer.
One experiment you could perform is to determine the densities of the metal and the chunk of
gold. You could measure the mass of the nugget on a balance and the volume of the nugget by
water displacement. Using this information, you could calculate the density of the nugget. Repeat
the experiment and calculations for the sample of gold. If the nugget is gold, the two densities
should be equal and be 19.3 g/cm3.
Also, you could determine the melting points of the metal and the chunk of pure gold. The two
melting points should be the same (1338 K) if the metal is gold.
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4
Chapter 1: Chemistry and Measurement
โ ANSWERS TO SELF-ASSESSMENT AND REVIEW QUESTIONS
1.1.
One area of technology that chemistry has changed is the characteristics of materials. The liquidcrystal displays (LCDs) in devices such as watches, cell phones, computer monitors, and
televisions are materials made of molecules designed by chemists. Electronics and
communications have been transformed by the development of optical fibers to replace copper
wires. In biology, chemistry has changed the way scientists view life. Biochemists have found
that all forms of life share many of the same molecules and molecular processes.
1.2.
An experiment is an observation of natural phenomena carried out in a controlled manner so that
the results can be duplicated and rational conclusions obtained. A theory is a tested explanation of
basic natural phenomena. They are related in that a theory is based on the results of many
experiments and is fruitful in suggesting other, new experiments. Also, an experiment can
disprove a theory but can never prove it absolutely. A hypothesis is a tentative explanation of
some regularity of nature.
1.3.
Rosenberg conducted controlled experiments and noted a basic relationship that could be stated
as a hypothesisโthat is, that certain platinum compounds inhibit cell division. This led him to do
new experiments on the anticancer activity of these compounds.
1.4.
Matter is the general term for the material things around us. It is whatever occupies space and can
be perceived by our senses. Mass is the quantity of matter in a material. The difference between
mass and weight is that mass remains the same wherever it is measured, but weight is
proportional to the mass of the object divided by the square of the distance between the center of
mass of the object and that of the earth.
1.5.
The law of conservation of mass states that the total mass remains constant during a chemical
change (chemical reaction). To demonstrate this law, place a sample of wood in a sealed vessel
with air, and weigh it. Heat the vessel to burn the wood, and weigh the vessel after the
experiment. The weight before the experiment and that after it should be the same.
1.6.
Mercury metal, which is a liquid, reacts with oxygen gas to form solid mercury(II) oxide. The
color changes from that of metallic mercury (silvery) to a color that varies from red to yellow
depending on the particle size of the oxide.
1.7.
Gases are easily compressible and fluid. Liquids are relatively incompressible and fluid. Solids
are relatively incompressible and rigid.
1.8.
An example of a substance is the element sodium. Among its physical properties: It is a solid, and
it melts at 98ยฐC. Among its chemical properties: It reacts vigorously with water, and it burns in
chlorine gas to form sodium chloride.
1.9.
An example of an element: sodium; of a compound: sodium chloride, or table salt; of a
heterogeneous mixture: salt and sugar; of a homogeneous mixture: sodium chloride dissolved in
water to form a solution.
1.10.
A glass of bubbling carbonated beverage with ice cubes contains three phases: gas, liquid, and
solid.
1.11.
A compound may be decomposed by chemical reactions into elements. An element cannot be
decomposed by any chemical reaction. Thus, a compound cannot also be an element in any case.
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Chapter 1: Chemistry and Measurement
5
1.12.
The precision refers to the closeness of the set of values obtained from identical measurements of
a quantity. The number of digits reported for the value of a measured or calculated quantity
(significant figures) indicates the precision of the value.
1.13.
Multiplication and division rule: In performing the calculation 100.0 x 0.0634 รท 25.31, the
calculator display shows 0.2504938. We would report the answer as 0.250 because the factor
0.0634 has the least number of significant figures (three).
Addition and subtraction rule: In performing the calculation 184.2 + 2.324, the calculator display
shows 186.524. Because the quantity 184.2 has the least number of decimal places (one), the
answer is reported as 186.5.
1.14.
An exact number is a number that arises when you count items or sometimes when you define a
unit. For example, a foot is defined to be 12 inches. A measured number is the result of a
comparison of a physical quantity with a fixed standard of measurement. For example, a steel rod
measures 9.12 centimeters, or 9.12 times the standard centimeter unit of measurement.
1.15.
For a given unit, the SI system uses prefixes to obtain units of different sizes. Units for all other
possible quantities are obtained by deriving them from any of the seven base units. You do this by
using the base units in equations that define other physical quantities.
1.16.
An absolute temperature scale is a scale in which the lowest temperature that can be attained
theoretically is zero. Degrees Celsius and kelvins have units of equal size and are related by the
formula
1ยฐC
tC = (TK โ 273.15 K) ๏ด
1K
1.17.
The density of an object is its mass per unit volume. Because the density is characteristic of a
substance, it can be helpful in identifying it. Density can also be useful in determining whether a
substance is pure. It also provides a useful relationship between mass and volume.
1.18.
Units should be carried along because (1) the units for the answers will come out in the
calculations, and (2), if you make an error in arranging factors in the calculation, this will become
apparent because the final units will be nonsense.
1.19.
The answer is c, three significant figures.
1.20.
The answer is a, 4.43 x 102 mm.
1.21.
The answer is e, 75 mL.
1.22.
The answer is c, 0.23 mg.
โ ANSWERS TO CONCEPT EXPLORATIONS
1.23.
a.
First, check the physical appearance of each sample. Check the particles that make up each
sample for consistency and hardness. Also, note any odor. Then perform on each sample
some experiments to measure physical properties such as melting point, density, and
solubility in water. Compare all of these results and see if they match.
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6
1.24.
Chapter 1: Chemistry and Measurement
b.
It is easier to prove that the compounds were different by finding one physical property that
is different, say different melting points. To prove the two compounds were the same
would require showing that every physical property was the same.
c.
Of the properties listed in part a, the melting point would be most convincing. It is not
difficult to measure, and it is relatively accurate. The density of a powder is not as easy to
determine as the melting point, and solubility is not reliable enough on its own.
d.
No. Since neither solution reached a saturation point, there is not enough information to tell
if there was a difference in behavior. Many white powders dissolve in water. Their
chemical compositions are not the same.
Part 1
a.
3 g + 1.4 g + 3.3 g = 7.7 g = 8 g
b.
First, 3 g + 1.4 g = 4.4 g = 4 g. Then, 4 g + 3.3 g = 7.3 g = 7 g.
c.
Yes, the answer in part a is more accurate. When you round off intermediate steps, you
accumulate small errors and your answer is not as accurate.
d.
The answer 29 g is correct.
e.
This answer is incorrect. It should be 3 x 101 with only one significant figure in the answer.
The student probably applied the rule for addition (instead of for multiplication) after the
first step.
f.
The answer 28.5 g is correct.
g.
Donโt round off intermediate answers. Indicate the round-off position after each step by
underlining the least significant digit.
Part 2
a.
The calculated answer is incorrect. It should be 11 cm3. The answer given has too many
significant figures. There is also a small round off error due to using a rounded-off value
for the density.
b.
This is a better answer. It is reported with the correct number of significant figures (three).
It can be improved by using all of the digits given for the density.
c.
V=
d.
There was no rounding off of intermediate steps; all the factors are as accurate as possible.
1 cm3
10 ball bearings
1.234 g
๏ด
๏ด
= 3.90889 = 3.909 cm3
3.1569 g
1
1 ball bearing
โ ANSWERS TO CONCEPTUAL PROBLEMS
1.25.
1.26.
a.
Two phases: liquid and solid.
b.
Three phases: liquid water, solid quartz, and solid seashells.
If the material is a pure compound, all samples should have the same melting point, the same
color, and the same elemental composition. If it is a mixture, these properties should differ
depending on the composition.
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Chapter 1: Chemistry and Measurement
1.27.
7
a.
You need to establish two points on the thermometer with known (defined) temperaturesโ
for example, the freezing point (0๏ฐC) and boiling point (100๏ฐC) of water. You could first
immerse the thermometer in an ice-water bath and mark the level at this point as 0๏ฐC. Then,
immerse the thermometer in boiling water, and mark the level at this point as 100๏ฐC. As
long as the two points are far enough apart to obtain readings of the desired accuracy, the
thermometer can be used in experiments.
b.
You could make 19 evenly spaced marks on the thermometer between the two original
points, each representing a difference of 5ยฐC. You may divide the space between the two
original points into fewer spaces as long as you can read the thermometer to obtain the
desired accuracy.
1.28.
a.
1.29.
a.
b.
c.
To answer this question, you need to develop an equation that converts between ๏ฐF and
๏ฐYS. To do so, you need to recognize that one degree on the Your Scale does not
correspond to one degree on the Fahrenheit scale and that โ100๏ฐF corresponds to 0๏ฐ on
Your Scale (different โzeroโ points). As stated in the problem, in the desired range of 100
Your Scale degrees, there are 120 Fahrenheit degrees. Therefore, the relationship can be
expressed as 120๏ฐF = 100๏ฐYS, since it covers the same temperature range. Now you need
to โscaleโ the two systems so that they correctly convert from one scale to the other. You
could set up an equation with the known data points and then employ the information from
the relationship above.
For example, to construct the conversion between ๏ฐYS and ๏ฐF, you could perform the
following steps:
Step 1:
๏ฐF = ๏ฐYS
Not a true statement, but one you would like to make true.
Step 2:
ยฐF = ยฐYS ๏ด
120ยฐF
100ยฐYS
This equation takes into account the difference in the size between the temperature unit on
the two scales but will not give you the correct answer because it doesnโt take into account
the different zero points.
Step 3: By subtracting 100๏ฐF from your equation from Step 2, you now have the complete
equation that converts between ๏ฐF and ๏ฐYS.
๏ฐF = (๏ฐYS ๏ด
b.
120ยฐF
) โ 100๏ฐF
100ยฐYS
Using the relationship from part a, 66ยฐYS is equivalent to
(66๏ฐYS ๏ด
120ยฐF
) โ 100๏ฐF = โ20.8๏ฐF = โ21๏ฐF
100ยฐYS
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8
Chapter 1: Chemistry and Measurement
1.30.
Some physical properties you could measure are density, hardness, color, and conductivity.
Chemical properties of sodium would include reaction with air, reaction with water, reaction with
chlorine, reaction with acids, bases, etc.
1.31.
The empty boxes are identical, so they donโt contribute to any mass or density difference. In
packing the boxes with wooden cubes, it is simple to show that 125 wooden cubes will fit
neatly into a box with no significant pockets of air, or void spaces, remaining in the packed
box. Alternatively, as you will learn later in discussions on crystal structure, packing the box
with wooden spheres depends on how the spheres pack. In terms of space utilization, the least
efficient packing pattern involves spheres packed directly on top of each other; in other
words, spheres would pack directly centered into positions where the wooden cubes would
be, so 125 spheres would go in the box. A more efficient packing pattern results if the
spheres occupy “valley” spots made available between every set of 3 spheres on adjacent
layers; in such cases, more spheres than cubes would pack into the box. Regardless
of the number of spheres packed into a box, there will still exist significant void space
between the packed spheres that is filled with air which is much less dense than wood. As a
result, the box containing the spheres will be less efficiently packed with actual wood. You
can thus conclude that the box containing the cubes must have a greater mass of wood;
hence, it must have a greater density.
1.32.
a.
Since the bead is less dense than any of the liquids in the container, the bead will float on
top of all the liquids.
b.
First, determine the density of the plastic bead. Since density is mass divided by volume,
you get
d=
m
3.92 ๏ด 10๏ญ2 g
=
= 0.911 g/mL = 0.91 g/mL
V
0.043 mL
Thus, the glass bead will pass through the top three layers and float on the ethylene glycol
layer, which is more dense.
1.33.
1.34.
c.
Since the bead sinks all the way to the bottom, it must be more dense than 1.114 g/mL.
a.
A paper clip has a mass of about 1 g.
b.
Answers will vary depending on your particular sample. Keeping in mind that the SI unit
for mass is kg, the approximate weights for the items presented in the problem are as
follows: a grain of sand, 1 ๏ด 10โ5 kg; a paper clip, 1 x 10โ3 kg; a nickel, 5 ๏ด 10โ3 kg; a 5.0gallon bucket of water, 2.0 ๏ด 101 kg; a brick, 3 kg; a car, 1 ๏ด 103 kg.
When taking measurements, never throw away meaningful information even if there is some
uncertainty in the final digit. In this case, you are certain that the nail is between 5 and 6 cm. The
uncertain, yet still important, digit is between the 5 and 6 cm measurements. You can estimate
with reasonable precision that it is about 0.7 cm from the 5 cm mark, so an acceptable answer
would be 5.7 cm. Another person might argue that the length of the nail is closer to 5.8 cm, which
is also acceptable given the precision of the ruler. In any case, an answer of 5.7 or 5.8 should
provide useful information about the length of the nail. If you were to report the length of the nail
as 6 cm, you would be discarding potentially useful length information provided by the
measuring instrument. If a higher degree of measurement precision were needed (more significant
figures), you would need to switch to a more precise rulerโfor example, one that had mm
markings.
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Chapter 1: Chemistry and Measurement
1.35.
a.
The number of significant figures in this answer follows the rules for multiplication and
division. Here, the measurement with the fewest significant figures is the reported volume
0.310 m3, which has three. Therefore, the answer will have three significant figures. Since
Volume = L x W x H, you can rearrange and solve for one of the measurements, say the
length.
L=
b.
9
V
0.310 m3
=
= 0.83496 m = 0.835 m
(0.7120 m) (0.52145 m)
W ๏ด H
The number of significant figures in this answer follows the rules for addition and
subtraction. The measurement with the least number of decimal places is the result 1.509
m, which has three. Therefore, the answer will have three decimal places. Since the result is
the sum of the three measurements, the third length is obtained by subtracting the other two
measurements from the total.
Length = 1.509 m โ 0.7120 m โ 0.52145 m = 0.27555 m = 0.276 m
1.36.
The mass of something (how heavy it is) depends on how much of the item, material, substance,
or collection of things you have. The density of something is the mass of a specific amount
(volume) of an item, material, substance, or collection of things. You could use 1 kg of feathers
and 1 kg of water to illustrate that they have the same mass yet have very different volumes;
therefore, they have different densities.
โ SOLUTIONS TO PRACTICE PROBLEMS
Note on significant figures: If the final answer to a solution needs to be rounded off, it is given first with
one nonsignificant figure, and the last significant figure is underlined. The final answer is then rounded to
the correct number of significant figures. In multistep problems, intermediate answers are given with at
least one nonsignificant figure; however, only the final answer has been rounded off.
1.37.
By the law of conservation of mass:
Mass of sodium carbonate + mass of acetic acid solution = mass of contents of reaction vessel +
mass of carbon dioxide
Plugging in gives
15.5 g + 19.7 g = 28.7 g + mass of carbon dioxide
Mass of carbon dioxide = 15.5 g + 19.7 g โ 28.7 g = 6.5 g
1.38.
By the law of conservation of mass:
Mass of iron + mass of acid = mass of contents of beaker + mass of hydrogen
Plugging in gives
5.6 g + 15.0 = 20.4 g + mass of hydrogen
Mass of hydrogen = 5.6 g + 15.0 g โ 20.4 g = 0.2 g
1.39.
By the law of conservation of mass:
Mass of zinc + mass of sulfur = mass of zinc sulfide
Rearranging and plugging in give
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10
Chapter 1: Chemistry and Measurement
Mass of zinc sulfide = 65.4 g + 32.1 g = 97.5 g
For the second part, let x = mass of zinc sulfide that could be produced. By the law of
conservation of mass:
36.9 g + mass of sulfur = x
Write a proportion that relates the mass of zinc reacted to the mass of zinc sulfide formed, which
should be the same for both cases.
mass zinc
36.9 g
65.4 g
=
=
x
mass zinc sulfide
97.5 g
Solving gives x = 55.01 g = 55.0 g
1.40.
By the law of conservation of mass:
Mass of aluminum + mass of bromine = mass of aluminum bromide
Plugging in and solving give
27.0 g + Mass of bromine = 266.7 g
Mass of bromine = 266.7 g โ 27.0 g = 239.7 g
For the second part, let x = mass of bromine that reacts. By the law of conservation of mass:
18.1 g + x = mass of aluminum bromide
Write a proportion that relates the mass of aluminum reacted to the mass of bromine reacted,
which should be the same for both cases.
mass aluminum
18.1 g
27.0 g
=
=
mass bromine
x
239.7 g
Solving gives x = 160.7 g = 161 g
1.41.
a. Solid
b. Liquid
c. Gas
d. Solid
1.42.
a. Solid
b. Solid
c. Solid
d. Liquid
1.43.
a.
Physical change
b.
Physical change
c.
Chemical change
d.
Physical change
a.
Physical change
b.
Chemical change
c.
Chemical change
d.
Physical change
1.44.
1.45.
Physical change: Liquid mercury is cooled to solid mercury.
Chemical changes: (1) Solid mercury oxide forms liquid mercury metal and gaseous oxygen; (2)
glowing wood and oxygen form burning wood (form ash and gaseous products).
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Chapter 1: Chemistry and Measurement
1.46.
11
Physical changes: (1) Solid iodine is heated to gaseous iodine; (2) gaseous iodine is cooled to
form solid iodine.
Chemical change: Solid iodine and zinc metal are ignited to form a white powder.
1.47.
1.48.
1.49.
a.
Physical property
b.
Chemical property
c.
Physical property
d.
Physical property
e.
Chemical property
a.
Physical property
b.
Chemical property
c.
Physical property
d.
Chemical property
e.
Physical property
Physical properties: (1) Iodine is solid; (2) the solid has lustrous blue-black crystals;
(3) the crystals vaporize readily to a violet-colored gas.
Chemical properties: (1) Iodine combines with many metals, such as with aluminum to give
aluminum iodide.
1.50.
Physical properties: (1) is a solid; (2) has an orange-red color; (3) has a density of
11.1 g/cm3; (4) is insoluble in water.
Chemical property: Mercury(II) oxide decomposes when heated to give mercury and oxygen.
1.51.
1.52.
1.53.
a.
Physical process
b.
Chemical reaction
c.
Physical process
d.
Chemical reaction
e.
Physical process
a.
Chemical reaction
b.
Physical process
c.
Physical process
d.
Physical process
e.
Chemical reaction
a.
Solution
b.
Substance
c.
Substance
d.
Heterogeneous mixture
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12
Chapter 1: Chemistry and Measurement
1.54.
a.
Homogeneous mixture, if fresh; heterogeneous mixture, if spoiled
b.
Substance
c.
Solution
d.
Substance
a.
A pure substance with two phases present, liquid and gas.
b.
A mixture with two phases present, solid and liquid.
c.
A pure substance with two phases present, solid and liquid.
d.
A mixture with two phases present, solid and solid.
a.
A mixture with two phases present, solid and liquid.
b.
A mixture with two phases present, solid and liquid.
c.
A mixture with two phases present, solid and solid.
d.
A pure substance with two phases present, liquid and gas.
a.
four
b.
three
c.
four
d.
five
e.
three
f.
four
a.
five
b.
four
c.
two
d.
four
e.
three
f.
four
1.55.
1.56.
1.57.
1.58.
1.59.
40,000 km = 4.0 x 104 km
1.60.
150,000,000 km = 1.50 ๏ด 108 km
1.61.
a.
8.71 ๏ด 0.0301
= 8.457 = 8.5
0.031
b.
0.71 + 89.3 = 90.01 = 90.0
c.
934 ๏ด 0.00435 + 107 = 4.0629 + 107 = 111.06 = 111
d.
(847.89 โ 847.73) ๏ด 14673 = 0.16 ๏ด 14673 = 2347 = 2.3 ๏ด 103
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
1.62.
1.63.
a.
8.71 ๏ด 0.57
= 0.8456 = 0.85
5.871
b.
8.937 โ 8.930 = 0.007
c.
8.937 + 8.930 = 17.867
d.
0.00015 ๏ด 54.6 + 1.002 = 0.00819 + 1.002 = 1.0101 = 1.010
13
The volume of the first sphere is
V1 = (4/3)๏ฐr3 = (4/3)๏ฐ ๏ด (4.52 cm)3 = 386.82 cm3
The volume of the second sphere is
V2 = (4/3)๏ฐr3 = (4/3)๏ฐ ๏ด (4.72 cm)3 = 440.47 cm3
The difference in volume is
V2 โ V1 = 440.47 cm3 โ 386.82 cm3 = 53.65 cm3 = 54 cm3
1.64.
The length of the cylinder between the two marks is
l = 3.50 cm โ 3.20 cm = 0.30 cm
The volume of iron contained between the marks is
V = ๏ฐr2l = ๏ฐ ๏ด (1.500 cm)2 ๏ด 0.30 cm = 2.12 cm3 = 2.1 cm3
1.65.
1.66.
1.67.
1.68.
a.
5.89 ๏ด 10โ12 s = 5.89 ps
b.
0.2010 m = 2.01 dm
c.
2.560 ๏ด 10โ9 g = 2.560 ng
d.
6.05 ๏ด 103 m = 6.05 km
a.
4.851 ๏ด 10โ6 g = 4.851 ยตg
b.
3.16 ๏ด 10โ2 m = 3.16 cm
c.
2.591 ๏ด 10โ9 s = 2.591 ns
d.
8.93 ๏ด 10โ12 g = 8.93 pg
a.
6.15 ps = 6.15 ๏ด 10โ12 s
b.
3.781 ยตm = 3.781 ๏ด 10โ6 m
c.
1.546 ร
= 1.546 ๏ด 10โ10 m
d.
9.7 mg = 9.7 ๏ด 10โ3 g
a.
6.20 km = 6.20 ๏ด 103 m
b.
1.98 ns = 1.98 ๏ด 10โ9 s
c.
2.54 cm = 2.54 ๏ด 10โ2 m
d.
5.23 ยตg = 5.23 ๏ด 10โ6 g
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14
Chapter 1: Chemistry and Measurement
1.69.
a.
tC =
5ยฐC
5ยฐC
๏ด (tF โ 32ยฐF) =
๏ด (68ยฐF โ 32ยฐF) = 20.0ยฐC = 20.ยฐC
9ยฐF
9ยฐF
b.
tC =
5ยฐC
5ยฐC
๏ด (tF โ 32ยฐF) =
๏ด (โ23ยฐF โ 32ยฐF) = โ30.56ยฐC = โ31ยฐC
9ยฐF
9ยฐF
c.
tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (26ยฐC ๏ด
) + 32ยฐF = 78.8ยฐF = 79ยฐF
5ยฐC
5ยฐC
d.
tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (โ81ยฐC ๏ด
) + 32ยฐF = โ113.8ยฐF = โ114ยฐF
5ยฐC
5ยฐC
a.
tC =
5ยฐC
5ยฐC
๏ด (tF โ 32ยฐF) =
๏ด (51ยฐF โ 32ยฐF) = 10.556๏ฐC = 11๏ฐC
9ยฐF
9ยฐF
b.
tC =
5ยฐC
5ยฐC
๏ด (tF โ 32ยฐF) =
๏ด (โ11ยฐF โ 32ยฐF) = โ23.9ยฐC = โ24ยฐC
9ยฐF
9ยฐF
c.
tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (โ41ยฐC ๏ด
) + 32ยฐF = โ41.8ยฐF = โ42ยฐF
5ยฐC
5ยฐC
d.
tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (22ยฐC ๏ด
) + 32ยฐF = 71.6ยฐF = 72ยฐF
5ยฐC
5ยฐC
1.70.
1.71.
tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (โ20.0ยฐC ๏ด
) + 32ยฐF = โ4.0ยฐF = โ4.0ยฐF
5ยฐC
5ยฐC
1.72.
tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (โ222.7ยฐC ๏ด
) + 32ยฐF = โ368.86ยฐF = โ368.9ยฐF
5ยฐC
5ยฐC
1.73.
d=
m
12.4 g
=
= 7.560 g/cm3 = 7.56 g/cm3
1.64 cm3
V
1.74.
d=
m
23.6 g
=
= 0.7867 g/mL = 0.787 g/mL
V
30.0 mL
1.75.
First, determine the density of the liquid.
d=
m
6.71 g
=
= 0.7894 = 0.79 g/mL
8.5 mL
V
The density is closest to ethanol (0.789 g/cm3).
1.76.
First, determine the density of the mineral sample.
d=
m
5.94 g
=
= 8.137 = 8.1 g/cm3
3
0.73 cm
V
The density is closest to cinnabar (8.10 g/cm3).
1.77.
The mass of platinum is obtained as follows.
Mass = d ๏ด V = 21.4 g/cm3 ๏ด 5.9 cm3 = 126 g = 1.3 ๏ด 102 g
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
1.78.
15
The mass of gasoline is obtained as follows.
Mass = d ๏ด V = 0.70 g/mL ๏ด 43.8 mL = 30.66 g = 31 g
1.79.
The volume of ethanol is obtained as follows. Recall that 1 mL = 1 cm3.
m
19.8 g
=
= 25.09 cm3 = 25.1 cm3 = 25.1 mL
d
0.789 g/cm3
Volume =
1.80.
The volume of bromine is obtained as follows.
m
88.5 g
=
= 28.54 mL = 28.5 mL
d
3.10 g/mL
Volume =
1.81.
Since 1 kg = 103 g, and 1 mg = 10โ3 g, you can write
0.450 kg x
1.82.
Since 1 mg = 10โ3 g, and 1 ยตg = 10โ6 g, you can write
611 mg x
1.83.
1 cm
10๏ญ9 m
๏ด ๏ญ2
= 5.55 ๏ด 10โ5 cm
10 m
1 nm
Since 1 ร
= 10โ10 m, and 1 mm = 10โ3 m, you can write
0.96 ร
๏ด
1.85.
10-3 g
1 ฮผg
๏ด -6 = 6.11 ๏ด 105 ยตg
1 mg
10 g
Since 1 nm = 10โ9 m, and 1 cm = 10โ2 m, you can write
555 nm ๏ด
1.84.
103 g
1 mg
๏ด ๏ญ3 = 4.50 ๏ด 105 mg
1 kg
10 g
1 mm
10๏ญ10 m
๏ด ๏ญ3
= 9.6 ๏ด 10โ8 mm
1ร
10 m
Since 1 km = 103 m, you can write
3
๏ฆ 103 m ๏ถ
17
3
3.73 ๏ด 108 km3 ๏ด ๏ง
๏ท = 3.73 ๏ด 10 m
1
km
๏จ
๏ธ
Now, 1 dm = 10โ1 m. Also, note that 1 dm3 = 1 L. Therefore, you can write
3
๏ฆ 1 dm ๏ถ
3.73 ๏ด 1017 m3 ๏ด ๏ง -1 ๏ท = 3.73 ๏ด 1020 dm3 = 3.73 ๏ด 1020 L
๏จ 10 m ๏ธ
1.86.
1 ยตm = 10โ6 m, and 1 dm = 10โ1 m. Also, note that 1 dm3 = 1 L. Therefore, you can write
3
3
๏ฆ 10-6 m ๏ถ
๏ฆ 1 dm ๏ถ
โ15
3
โ15
1.3 ยตm ๏ด ๏ง
๏ท ๏ด ๏ง -1 ๏ท = 1.3 ๏ด 10 dm = 1.3 ๏ด 10 L
10
m
1
ฮผm
๏จ
๏ธ
๏จ
๏ธ
3
1.87.
3.58 short ton x
2000 lb
16 oz
1g
๏ด
๏ด
= 3.248 ๏ด 106 g = 3.25 ๏ด 106 g
1 short ton
1 lb
0.03527 oz
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16
Chapter 1: Chemistry and Measurement
1.88.
3.15 Btu ๏ด
1.89.
2425 fathoms ๏ด
1.90.
1.3 x 1010 barrels ๏ด
1.91.
1L
๏ฆ 2.54 cm ๏ถ
(24.2 in.) ๏ด (15.9 in.) ๏ด (14.8 in.) ๏ด ๏ง
= 93.32 L = 93.3 L
๏ท ๏ด
1
in.
1000
cm3
๏จ
๏ธ
1.92.
33 worms
๏ฆ 1000 m ๏ถ
(1.00 km) ๏ด (2.0 km) ๏ด (1 m) ๏ด ๏ง
= 6.60 ๏ด 107 = 6.6 ๏ด 107 worms
๏ท ๏ด
1 m3
๏จ 1 km ๏ธ
252.0 cal
4.184 J
๏ด
= 3321 J = 3.32 ๏ด 103 J
1 Btu
1 cal
6 ft
12 in.
2.54 ๏ด 10๏ญ2 m
๏ด
๏ด
= 4434.8 m = 4.435 ๏ด 103 m
1 fathom
1 ft
1 in.
42 gal
9.46 x 10-4 m3
4 qt
๏ด
๏ด
= 2.066 ๏ด 109 m3 = 2.1 x 109 m3
1 qt
1 barrel 1 gal
3
2
โ SOLUTIONS TO GENERAL PROBLEMS
1.93.
From the law of conservation of mass,
Mass of sodium + mass of water = mass of hydrogen + mass of solution
Substituting, you obtain
19.70 g + 126.22 g = mass of hydrogen + 145.06 g
or,
Mass of hydrogen = 19.70 g + 126.22 g โ 145.06 g = 0.86 g
Thus, the mass of hydrogen produced was 0.86 g.
1.94.
From the law of conservation of mass,
Mass of tablet + mass of acid solution = mass of carbon dioxide + mass of solution
Substituting, you obtain
0.853 g + 56.519 g = mass of carbon dioxide + 57.152 g
Mass of carbon dioxide = 0.853 g + 56.519 g โ 57.152 g = 0.220 g
Thus, the mass of carbon dioxide produced was 0.220 g.
1.95.
From the law of conservation of mass,
Mass of aluminum + mass of iron(III) oxide = mass of iron +
mass of aluminum oxide + mass of unreacted iron(III) oxide
5.40 g + 18.50 g = 11.17 g + 10.20 g + mass of iron(III) oxide unreacted
Mass of iron(III) oxide unreacted = 5.40 g + 18.50 g โ 11.17 g โ 10.20 g = 2.53 g
Thus, the mass of unreacted iron(III) oxide is 2.53 g.
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Chapter 1: Chemistry and Measurement
1.96.
17
From the law of conservation of mass,
Mass of sodium bromide + mass of chlorine reacted = mass of bromine +
mass of sodium chloride
20.6 g + mass of chlorine reacted = 16.0 g + 11.7 g
Mass of chlorine reacted = 16.0 g + 11.7 g โ 20.6 g = 7.1 g
Thus, the mass of chlorine that reacted is 7.1 g.
1.97.
50.90 g + 5.680 g + 53.3 g = 109.88 g = 109.9 g total
1.98.
66.5 g + 58.2 g + 5.279 g = 129.979 g = 130.0 g total
1.99.
a. Chemical
b. Physical
c. Physical
d. Chemical
1.100. a. Physical
b. Chemical
c. Physical
d. Chemical
1.101. Compounds always contain the same proportions of the elements by mass. Thus, if we let X be
the proportion of iron in a sample, we can calculate the proportion of iron in each sample as
follows.
Sample A:
X=
mass of iron
1.094 g
=
= 0.72068 = 0.7207
mass of sample 1.518 g
Sample B:
X=
mass of iron
1.449 g
=
= 0.70476 = 0.7048
mass of sample
2.056 g
Sample C:
X=
mass of iron
1.335 g
=
= 0.71276 = 0.7128
mass of sample 1.873 g
Since each sample has a different proportion of iron by mass, the material is not a compound.
1.102. Compounds always contain the same proportions of the elements by mass. Thus, if we let X be
the proportion of mercury in a sample, we can calculate the proportion of mercury in each sample
as follows.
Sample A:
X=
mass of mercury
0.9641 g
=
= 0.92612 = 0.9261
mass of sample
1.0410 g
Sample B:
X=
mass of mercury 1.4293 g
=
= 0.92607 = 0.9261
mass of sample
1.5434 g
Sample C:
X=
mass of mercury 1.1283 g
=
= 0.92612 = 0.9261
mass of sample
1.2183 g
Since each sample has the same proportion of mercury by mass, the data are consistent with the
hypothesis that the material is a compound.
1.103. V = (edge)3 = (39.3 cm)3 = 6.069 ๏ด 104 cm3 = 6.07 ๏ด 104 cm3
1.104. V = ๏ฐr2l = ๏ฐ ๏ด (2.56 cm)2 ๏ด 56.32 cm = 1159 cm3 = 1.16 ๏ด 103 cm3
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18
Chapter 1: Chemistry and Measurement
1.105. V = LWH = 47.8 in. ๏ด 12.5 in. ๏ด 19.5 in. ๏ด
1 gal
= 50.43 gal = 50.4 gal
231 in 3
1.106. The volume in cubic inches is
V = (4/3)๏ฐr3 = (4/3)๏ฐ ๏ด (175.0 in.)3 = 2.24492 ๏ด 107 in3 = 2.245 ๏ด 107 in3
The volume in imperial gallons is
V = 2.24492 x 107 in3 ๏ด
1 gal
= 8.09275 ๏ด 104 gal = 8.093 ๏ด 104 gal
277.4 in 3
1.107. The volume of the first sphere is given by
V1 = (4/3)๏ฐr3 = (4/3)๏ฐ ๏ด (5.61 cm)3 = 739.5 cm3
The volume of the second sphere is given by
V2 = (4/3)๏ฐr3 = (4/3)๏ฐ ๏ด (5.85 cm)3 = 838.6 cm3
The difference in volume between the two spheres is given by
V = V2 โ V1 = 838.6 cm3 โ 739.5 cm3 = 9.91 ๏ด 101 = 9.9 ๏ด 101 cm3
1.108. The surface area of the first circle is given by
S1 = ๏ฐr2 = ๏ฐ ๏ด (7.98 cm)2 = 200.0 cm2
The surface area of the second circle is given by
S2 = ๏ฐr2 = ๏ฐ ๏ด (8.50 cm)2 = 226.9 cm2
The difference in surface area between the two circles is
Difference = S2 โ S1 = 226.9 cm2 โ 200.0 cm2 = 26.9 cm2 = 27 cm2
1.109. a.
56.1- 51.1
= 7.59 ๏ด 10โ1 = 7.6 ๏ด 10โ1
6.58
b.
56.1 + 51.1
= 1.629 ๏ด 101 = 1.63 ๏ด 101
6.58
c.
(9.1 + 8.6) ๏ด 26.91 = 4.763 ๏ด 102 = 4.76 ๏ด 102
d.
0.0065 ๏ด 3.21 + 0.0911 = 1.119 ๏ด 10โ1 = 1.12 ๏ด 10โ1
1.110. a.
9.345 ๏ญ 9.005
= 0.03465 = 0.0347
9.811
b.
9.345 + 9.005
= 1.8703 = 1.870
9.811
c.
(7.50 + 7.53) ๏ด 3.71 = 55.761 = 55.8
d.
0.71 ๏ด 0.36 + 17.36 = 17.6156 = 17.62
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
1.111. a.
9.12 cg
b.
66 pm
c.
7.1 ยตm
d.
56 nm
1.112. a.
1.86 cg
b.
77 pm
c.
6.5 nm
d.
0.85 ยตm
1.113. a.
1.07 ๏ด 10โ12 s
b.
5.8 ๏ด 10โ6 m
c.
3.19 ๏ด 10โ7 m
d.
1.53 ๏ด 10โ2 s
1.114. a.
6.6 ๏ด 10โ3 K
b.
2.75 ๏ด 10โ10 m
c.
2.21 ๏ด 10โ2 s
d.
4.5 ๏ด 10โ5 m
1.115. tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (3410ยฐC ๏ด
) + 32ยฐF = 6170ยฐF = 6170ยฐF
5ยฐC
5ยฐC
1.116. tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (1677ยฐC ๏ด
) + 32ยฐF = 3050.6ยฐF = 3051ยฐF
5ยฐC
5ยฐC
1.117. tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (825ยฐC ๏ด
) + 32ยฐF = 1517ยฐF = 1.52 x 103ยฐF
5ยฐC
5ยฐC
1.118. tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (50ยฐC ๏ด
) + 32ยฐF = 122ยฐF = 122ยฐF (more likely 120 ยฐF)
5ยฐC
5ยฐC
19
1.119. The temperature in kelvins is
1K
1K
) + 273.15 K = (29.8ยฐC ๏ด
) + 273.15 K = 302.95 K
1ยฐC
1ยฐC
= 303.0 K
TK = (tC ๏ด
The temperature in degrees Fahrenheit is
tF = (tC ๏ด
9ยฐF
9ยฐF
) + 32ยฐF = (29.8ยฐC ๏ด
) + 32ยฐF = 85.64ยฐF = 85.6ยฐF
5ยฐC
5ยฐC
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20
Chapter 1: Chemistry and Measurement
1.120. The temperature in kelvins is
1K
1K
TK = (tC ๏ด
) + 273.15 K = (โ38.9ยฐC ๏ด
) + 273.15 K = 234.25 K
1ยฐC
1ยฐC
= 234.3 K
The temperature in degrees Fahrenheit is
9ยฐF
9ยฐF
tF = (tC ๏ด
) + 32ยฐF = (โ38.9ยฐC ๏ด
) + 32ยฐF = โ38.02ยฐF = โ38.0ยฐF
5ยฐC
5ยฐC
1.121. The temperature in degrees Celsius is
tC =
5ยฐC
5ยฐC
๏ด (tF โ 32ยฐF) =
๏ด (1666ยฐF โ 32ยฐF) = 907.77ยฐC = 907.8ยฐC
9ยฐF
9ยฐF
The temperature in kelvins is
1K
1K
) + 273.15 K = (907.77ยฐC ๏ด
) + 273.15 K = 1180.92 K
1ยฐC
1ยฐC
= 1180.9 K
TK = (tC ๏ด
1.122. The temperature in degrees Celsius is
tC =
5ยฐC
5ยฐC
๏ด (tF โ 32ยฐF) =
๏ด (236ยฐF โ 32ยฐF) = 113.3ยฐC = 113ยฐC
9ยฐF
9ยฐF
The temperature in kelvins is
TK = (tC ๏ด
= 386 K
1K
1K
) + 273.15 K = (113.3ยฐC ๏ด
) + 273.15 K = 386.4 K
1ยฐC
1ยฐC
3
1.123. Density =
2.70 g
1 kg
๏ฆ 1 cm ๏ถ
๏ด 3 ๏ด ๏ง ๏ญ2 ๏ท = 2.70 ๏ด 103 kg/m3
3
1 cm
10 g
๏จ 10 m ๏ธ
1.124. Density =
5.96 g
1 kg
๏ฆ 1 cm ๏ถ
๏ด 3 ๏ด ๏ง ๏ญ2 ๏ท = 5.96 ๏ด 103 kg/m3
3
1 cm
10 g
๏จ 10 m ๏ธ
3
1.125. The volume of the quartz is 67.1 mL โ 52.2 mL = 14.9 mL. Then, the density is
Density =
mass
39.8 g
=
= 2.671 g/mL = 2.67 g/mL = 2.67 g/cm3
volume 14.9 mL
1.126. First, determine the volume of water in the flask. The mass of water is obtained as follows. 109.3
g โ 70.7 g = 38.6 g. Now, using the density (0.997 g/cm3),
Volume =
mass
38.6 g
=
= 38.716 cm3 = 38.716 mL
density
0.997 g/ cm3
Now, the volume of the ore is 53.2 mL โ 38.716 mL = 14.48 mL, or 14.48 cm3. Therefore, the
density of the hematite ore is
Density =
70.7 g
mass
=
= 4.881 g/cm3 = 4.88 g/cm3
volume 14.48 cm3
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
21
1.127. First, determine the density of the liquid sample.
mass
22.3 g
=
= 1.486 g/mL = 1.49 g/mL = 1.49 g/cm3
volume 15.0 mL
Density =
This density is closest to that of chloroform (1.489 g/cm3), so the unknown liquid is chloroform.
1.128. First, determine the density of the calcite sample.
mass
35.6 g
=
= 2.759 g/cm3 = 2.76 g/cm3
volume 12.9 cm3
Density =
Since a substance will float on the liquids with greater densities, calcite will float on
tetrabromoethane (2.96 g/cm3) and methylene iodode (3.33 g/cm3).
1.129. First, determine the volume of the cube of platinum.
V = (edge)3 = (4.40 cm)3 = 85.18 cm3
Now, use the density to determine the mass of the platinum.
Mass = d x V = 21.4 g/cm3 ๏ด 85.18 cm3 = 1822.9 g = 1.82 ๏ด 103 g
1.130. First, determine the volume of the cylinder of silicone.
V = ๏ฐr2l = ๏ฐ x (4.00 cm)2 ๏ด 12.40 cm = 623.29 cm3
Now, use the density to determine the mass of the silicon.
Mass = d ๏ด V = 2.33 g/cm3 x 623.29 cm3 = 1452.2 g = 1.45 ๏ด 103 g
1.131. Volume =
mass
35.00 g
=
= 33.238 mL = 33.24 mL
density 1.053 g/ mL
1.132. First, convert kilograms to grams (1 kg = 103 g). Thus, 0.070 kg = 70 g. Then
mass
70 g
=
= 77.61 mL
density
0.902 g/ mL
Volume =
Finally, convert the volume to liters (1000 mL = 1 L).
Volume = 77.61 mL ๏ด
1L
= 0.07761 L = 0.078 L
1000 mL
1.133. a.
8.45 kg ๏ด
103 g
1 ฮผg
๏ด ๏ญ6 = 8.45 ๏ด 109 ๏ญg
1 kg
10 g
b.
318 ยตs ๏ด
1 ms
10๏ญ6 s
๏ด ๏ญ3 = 3.18 ๏ด 10โ1 ms
1 ๏ญs
10 s
c.
93 km ๏ด
1 nm
103 m
๏ด ๏ญ9
= 9.3 ๏ด 1013 nm
10 m
1 km
d.
37.1 mm ๏ด
1 cm
10๏ญ3 m
๏ด ๏ญ2
= 3.71 cm
10 m
1 mm
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
22
Chapter 1: Chemistry and Measurement
1 ๏ญm
10๏ญ10 m
๏ด ๏ญ6
= 1.27 ๏ด 10โ2 ยตm
10 m
1ร
1.134. a.
127 ร
๏ด
b.
21.0 kg ๏ด
103 g
1 mg
๏ด ๏ญ3 = 2.10 ๏ด 107 mg
1 kg
10 g
c.
1.09 cm x
1 mm
10๏ญ2 m
๏ด ๏ญ3
= 10.9 mm
10 m
1 cm
d.
4.6 ns ๏ด
1.135. a.
5.91 kg ๏ด
103 g
1 mg
๏ด ๏ญ3 = 5.91 ๏ด 106 mg
1 kg
10 g
b.
753 mg ๏ด
10๏ญ3 g
1 ฮผg
๏ด ๏ญ6 = 7.53 ๏ด 105 ยตg
1 mg
10 g
c.
90.1 MHz ๏ด
d.
498 mJ ๏ด
1 kJ
10๏ญ3 J
๏ด 3 = 4.98 ๏ด 10โ4 kJ
10 J
1 mJ
1.136. a.
7.19 ยตg ๏ด
10๏ญ6 g
1 mg
๏ด ๏ญ3 = 7.19 ๏ด 10โ3 mg
1 ฮผg
10 g
b.
104 pm ๏ด
1ร
10๏ญ12 m
๏ด ๏ญ10
= 1.04 ร
1 pm
10 m
c.
0.010 mm ๏ด
d.
0.0605 kPa ๏ด
1 ฮผs
10-9 s
๏ด -6 = 4.6 ๏ด 10โ3 ยตs
10 s
1 ns
1 kHz
106 Hz
๏ด 3
= 9.01 ๏ด 104 kHz
1 MHz 10 Hz
1 cm
10๏ญ3 m
๏ด ๏ญ2
= 1.0 ๏ด 10โ3 cm
1 mm
10 m
1 cPa
103 Pa
๏ด ๏ญ2
= 6.05 ๏ด 103 cPa
1 kPa
10 Pa
3
3
๏ฆ 103 m ๏ถ
1L
๏ฆ 1 dm ๏ถ
1.137. Volume = 12,230 km ๏ด ๏ง
= 1.2230 ๏ด 1016 L
๏ท ๏ด ๏ง ๏ญ1 ๏ท ๏ด
3
10
m
1
km
1
dm
๏จ
๏ธ
๏จ
๏ธ
3
3
3
๏ฆ 103 m ๏ถ
1L
๏ฆ 1 dm ๏ถ
1.138. Volume = 3.50 km ๏ด ๏ง
= 3.50 ๏ด 1012 L
๏ท ๏ด ๏ง ๏ญ1 ๏ท ๏ด
3
10
m
1
km
1
dm
๏จ
๏ธ
๏จ
๏ธ
3
1.139. First, calculate the volume of the room in cubic feet.
Volume = LWH = 10.0 ft ๏ด 11.0 ft ๏ด 9.0 ft = 990 ft3
Next, convert the volume to liters.
3
3
1L
๏ฆ 12 in ๏ถ
๏ฆ 2.54 cm ๏ถ
V = 990 ft x ๏ง
= 2.80 ๏ด 104 L = 2.8 ๏ด 104 L
๏ท ๏ด๏ง
๏ท ๏ด 3
10 cm3
๏จ 1 ft ๏ธ
๏จ 1 in ๏ธ
3
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
23
1.140. First, calculate the volume of the cylinder in cubic feet.
Volume = ๏ฐr2l = ๏ฐ x (15.0 ft)2 x 5.0 ft = 3534 ft3
Next, convert the volume to liters.
3
3
1L
๏ฆ 12 in ๏ถ
๏ฆ 2.54 cm ๏ถ
= 1.00 ๏ด 105 L
V = 3534 ft3 ๏ด ๏ง
๏ท x๏ง
๏ท ๏ด 3
10 cm3
๏จ 1 ft ๏ธ
๏จ 1 in ๏ธ
= 1.0 ๏ด 105 L
1.141. Mass = 384 carats ๏ด
200 mg 10๏ญ3 g
๏ด
= 76.80 g = 76.8 g
1 mg
1 carat
1.142. Mass = 49.6 ๏ด 106 troy oz ๏ด
31.10 g
1 ton
๏ด 6 = 1.542 ๏ด 103 ton
1 troy oz 10 g
= 1.54 ๏ด 103 ton
1.143. The adhesive is not permanent, is easily removable, and does no harm to the object.
1.144. The scientific question Art Fry was trying to answer was: โIs there an adhesive that will not
permanently stick things together?โ
1.145. Chromatography depends on how fast a substance moves in a stream of gas or liquid, past a
stationary phase to which the substance is slightly attracted.
1.146. The moving stream is a gaseous mixture of vaporized substances plus a gas such as helium, the
carrier. The gas is passed through a column containing a stationary phase. As the gas passes
through the column, the substances are attracted differently to the stationary column packing and
thus are separated. The separated gases pass through a detector, and the results are displayed
graphically.
โ SOLUTIONS TO STRATEGY PROBLEMS
1.147. 5 x 10โ2 mg = 0.05 mg. So 4.7 mg โ 0.05 mg = 4.65 mg = 4.7 mg
1.148. V =
m
33.0 g
=
= 41.405 = 41.4 cm3
d
0.797 g/cm3
mA
124 g
+ 40.8 mL =
+ 40.8 mL = 41.33 mL + 40.8 mL
3.00 g/mL
dA
= 82.13 = 82.1 mL
1.149. V = VA + VB =
1.150. a. r =
b.
d
832 mi
=
= 39.619 mi/h = 40. mi/h
21 h
t
832 mi 1.609 km
๏ด
= 63.75 = 64 km/h
21 h
1 mi
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24
Chapter 1: Chemistry and Measurement
c.
1.151
1 qt
832 mi 1.609 km 1 gal
๏ด
๏ด
๏ด
= 11.41 km/L = 11 km/L
0.9464 L
1 mi
31 gal
4 qt
Here it is necessary to convert the data into comparable units, mL for example. Converting 0.100
qt to mL gives ( 0.100 qt = 0.106 L) 106 mL. Next, using the density relationship, and the fact
that 1 mL = 1 cm3, 50.0 g of Pb corresponds to 4.42 mL Pb. The last quantity is first converted to
gram mass units, and then to mL using the density relationship. In this case 0.0250 lb = 11.3 g Pb
= 1.00 cm3 = 1.00 mL Pb. Thus, ranking from smallest volume to greatest:
0.0250 lb Pb (or 1.00 mL) < 50.0 g Pb (or 4.42 mL) < 50.0 mL Pb < 0.100 qt Pb (or 106 mL)
3
3
103 mL 1 cm3
๏ฆ 1 in ๏ถ
๏ฆ 1 ft ๏ถ
7
7 3
๏ด
๏ด๏ง
1.152. 7.6 x 10 L ๏ด
๏ท ๏ด๏ง
๏ท = 2.684 ๏ด 10 = 2.7 ๏ด 10 ft
1L
1 mL
๏จ 2.54 cm ๏ธ
๏จ 12 in ๏ธ
8
2
1.153. 2 converters ๏ด
2
5.0 x 103 beads 1.0 x 106 cm 2
๏ฆ 1m ๏ถ
๏ฆ 1 km ๏ถ
2
๏ด
๏ด๏ง
๏ท ๏ด ๏ง 3 ๏ท = 1.00 = 1.0 km
100
cm
10
m
bead
converter
๏จ
๏ธ
๏จ
๏ธ
3
1 cm3 ๏ฆ 10๏ญ2 m ๏ถ
โ4
3
1.154. First convert 200.0 mL to m : 200.0 mL ๏ด
๏ด๏ง
๏ท = 2.000 ๏ด 10 m
1 mL ๏จ 1 cm ๏ธ
3
Mass does not depend on temperature like volume does. However, if the mass is determined at 20
o
C using the density at 20 oC, i.e., m = d20 x V20,the volume of the water at 80 oC can be calculated
m
using V80 =
d80
m = d20 ๏ด V20 =
998 kg
x 2.000 ๏ด 10-4 m3 = 0.1996 kg
m3
200.0 mL
m
0.1996 kg
=
๏ด
= 205.3 mL = 205 mL (as expected, an
3
2.000 x 10๏ญ4 m3
d80 972 kg/m
expansion with increasing temperature is noted here).
V80 =
Because water is more dense at 20 oC than at 80 oC, 1.0 L of water will contain more mass of
water at the lower temperature, i.e., 998 g vs. 972 g. Since the number of molecules of water is
directly proportional to mass, the 1.0 L of water at 20 oC contains more water molecules.
3
1025 kg ๏ฆ 10๏ญ2 m ๏ถ
1000 g
1.155. 47.0 cm ๏ด
๏ด๏ง
= 48.18 = 48.2 g sample of ocean water originally
๏ท ๏ด
3
m
1 kg
๏จ 1 cm ๏ธ
3
Assuming a density of 1.0 g/mL for the evaporated water, and recognizing the equivalency
between cm3 and mL units, when 4.1 mL of water evaporates, the sample loses 4.1 g in mass and
4.1 cm3 in volume. The density of the partially evaporated ocean water sample is obtained by
dividing the remaining mass (48.2 g โ 4.1 g = 44.1 g) by the remaining volume (47.0 cm3 โ 4.1
cm3 = 42.9 cm3).
d=
mass
44.1 g
๏ฆ 1030 kg ๏ถ
=
= 1.03 g/cm3 ๏ง ๏ฝ
๏ท
3
m3 ๏ธ
42.9 cm
volume
๏จ
As one would expect, the remaining salt solution is now slightly more dense than the original
ocean water sample.
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
25
1.156. First, recognizing 5.5 mL = 5.5 cm3 , determine the mass of liquid gas present.
Mass = d x V = 0.75 g/cm3 x 5.5 cm3 = 4.13 g
Next, use the density formula to determine the density of the vaporized gas.
d=
mass
4.13 g
=
= 1.375 = 1.4 g/L, or 1.4 ๏ด 10-3 g/mL (or g/cm3)
volume
3.00 L
1.157. a.
Since there is the same number of atoms of the gas in each container, the mass is the same
in each container.
b.
Since d = m/V, for the same mass, when the volume is smaller, the density is greater. Since
the volume is less in container A, the density is greater.
c.
If the volume of container A was doubled, the density would decrease and become equal to
the density in container B
1.158. a.
b.
1.159.
d=
m
39.45 g
=
= 0.92011 = 0.920 g/cm3
V
(3.50 cm)3
400.4 mL 1 cm3
39.45 g
๏ด
๏ด
= 368.41 = 368 g
1
1 mL
(3.50 cm)3
300 million people 1 ๏ด 1010 miles 1.609 km 1000 m
1 lightyear
๏ด
๏ด
๏ด
๏ด
1 person
1
9.46 ๏ด 1015 m
1 mi
1 km
= 5.102 ๏ด 105 = 5 ๏ด 105 lightyears
3
1.160.
1.31 g
1 bucket
1L
1 mL
๏ฆ 2.54 cm ๏ถ
๏ฆ 12 in ๏ถ
๏ด
๏ด
๏ด
๏ด๏ง
๏ท ๏ด๏ง
๏ท
3
1000 mL 1 cm
1 bucket
4.67 L
๏จ 1 in ๏ธ
๏จ 1 ft ๏ธ
2.38 ๏ด 103 ft 3
๏ด
= 1.8904 ๏ด 104 g = 1.89 ๏ด 104 g
1
3
1.161. a.
Since d = m/V, an increased mass for the same volume means a higher density for the
solution.
b.
There would be less water for the same mass of salt. Since the salt is more dense than
water, the density would be higher than in part a.
c.
Since there would be a greater volume of water for the same mass of salt, the density would
be lower than in part a.
1.162. The water will have a lower density than the salt solution. It will not conduct electricity. It will
have different chemical properties, such as reaction with silver nitrate solution; and different
physical properties, such as the boiling point. Also, you could boil away the water from the salt
solution and recover the salt.
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26
Chapter 1: Chemistry and Measurement
โ SOLUTIONS TO CAPSTONE PROBLEMS
1.163. The mass of hydrochloric acid is obtained from the density and the volume.
Mass = density x volume = 1.096 g/mL ๏ด 54.3 mL = 59.51 g
Next, from the law of conservation of mass,
Mass of marble + mass of acid = mass of solution + mass of carbon dioxide gas
Plugging in gives
11.1 g + 59.51 g = 65.7 g + mass of carbon dioxide gas
Mass of carbon dioxide gas = 11.1 g + 59.51 g โ 65.7 g = 4.91 g
Finally, use the density to convert the mass of carbon dioxide gas to volume.
Volume =
4.91 g
mass
=
= 2.731 L = 2.7 L
density 1.798 g/ L
1.164. The mass of sulfuric acid is obtained from the density and the volume.
Mass = density x volume = 1.153 g/mL ๏ด 50.0 mL = 57.65 g
Next, from the law of conservation of mass,
Mass of ore + mass of acid = mass of solution + mass of hydrogen sulfide gas
Plugging in gives
10.8 g + 57.65 g = 65.1 g + mass of hydrogen sulfide gas
Mass of hydrogen sulfide gas = 10.8 g + 57.65 g โ 65.1 g = 3.35 g
Finally, use the density to convert the mass of hydrogen sulfide gas to volume.
Volume =
mass
3.35 g
=
= 2.40 L = 2.4 L
density 1.393 g/ L
1.165. First, calculate the volume of the steel sphere.
3
๏ฆ 2.54 cm ๏ถ
3
V = (4/3)๏ฐr3 = (4/3)๏ฐ ๏ด (1.58 in)3 ๏ด ๏ง
๏ท = 270.7 cm
1
in
๏จ
๏ธ
Next, determine the mass of the sphere using the density.
Mass = density x volume = 7.88 g/cm3 ๏ด 270.7 cm3 = 2133 g = 2.13 ๏ด 103 g
1.166. First, calculate the volume of the balloon. Note that the radius is one-half the diameter, or 1.75 ft.
3
3
1L
๏ฆ 12 in ๏ถ
๏ฆ 2.54 cm ๏ถ
V = (4/3)๏ฐr3 = (4/3)๏ฐ x (1.75 ft)3 ๏ด ๏ง
= 635.7 L
๏ท ๏ด๏ง
๏ท ๏ด 3
10 cm3
๏จ 1 ft ๏ธ
๏จ 1 in ๏ธ
Next, determine the mass of the helium using the density.
Mass = density ๏ด volume = 0.166 g/L ๏ด 635.7 L = 105.5 g = 106 g
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
27
1.167. The area of the ice is 840,000 mi2 โ 132,000 mi2 = 708,000 mi2. Now, determine the volume of
this ice.
Volume = area ๏ด thickness
2
3
๏ฆ 5280 ft ๏ถ
๏ฆ 12 in ๏ถ
๏ฆ 2.54 cm ๏ถ
= 708,000 mi2 ๏ด 5000 ft ๏ด ๏ง
๏ท ๏ด๏ง
๏ท ๏ด๏ง
๏ท
๏จ 1 mi ๏ธ
๏จ 1 ft ๏ธ
๏จ 1 in ๏ธ
3
= 2.794 ๏ด 1021 cm3
Now use the density to determine the mass of the ice.
Mass = density x volume = 0.917 g/cm3 ๏ด 2.794 ๏ด 1021 cm3 = 2.56 ๏ด 1021 g = 3 ๏ด 1021 g
1.168. The height of the ice is 7500 ft โ 1500 ft = 6000 ft. Now, determine the volume of the ice.
Volume = area ๏ด thickness
2
3
๏ฆ 5280 ft ๏ถ
๏ฆ 12 in ๏ถ
๏ฆ 2.54 cm ๏ถ
= 5,500,000 mi2 ๏ด 6000 ft ๏ด ๏ง
๏ท ๏ด๏ง
๏ท ๏ด๏ง
๏ท
๏จ 1 mi ๏ธ
๏จ 1 ft ๏ธ
๏จ 1 in ๏ธ
3
= 2.605 ๏ด 1022 cm3
Now use the density to determine the mass of the ice.
Mass = density ๏ด volume = 0.917 g/cm3 ๏ด 2.605 ๏ด 1022 cm3 = 2.38 ๏ด 1022 g = 2.4 ๏ด 1022 g
1.169. Let x = mass of ethanol and y = mass of water. Then, use the total mass to write
x + y = 49.6 g, or y = 49.6 g โ x. Thus, the mass of water is 49.6 g โ x. Next,
Total volume = volume of ethanol + volume of water
Since the volume is equal to the mass divided by density, you can write
Total volume =
mass of ethanol
mass of water
+
density of ethanol
density of water
Substitute in the known and unknown values to get an equation for x.
54.2 cm3 =
x
49.6 g ๏ญ x
+
3
0.789 g/cm
0.998 g/cm3
Multiply both sides of this equation by (0.789)(0.998). Also, multiply both sides by g/cm3 to
simplify the units. This gives the following equation to solve for x.
(0.789)(0.998)(54.2) g = (0.998) x + (0.789)(49.6 g โ x)
42.678 g = 0.998 x + 39.134 g โ 0.789 x
0.209 x = 3.544 g
x = mass of ethanol = 16.95 g
The percentage of ethanol (by mass) in the solution can now be calculated.
Percent (mass) =
mass of ethanol
1 6.95 g
๏ด 100% =
๏ด 100% = 34.1% = 34%
mass of solution
49.6 g
To determine the proof, you must first find the percentage by volume of ethanol in the solution.
The volume of ethanol is obtained using the mass and the density.
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
28
Chapter 1: Chemistry and Measurement
Volume =
mass of ethanol
1 6.95 g
=
= 21.48 cm3
density of ethanol
0.789 g/ cm3
The percentage of ethanol (by volume) in the solution can now be calculated.
Percent (volume) =
volume of ethanol
21.48 cm3
๏ด 100% =
๏ด 100% = 39.63%
volume of solution
54.2 cm3
The proof can now be calculated.
Proof = 2 ๏ด Percent (volume) = 2 ๏ด 39.63 = 79.27 = 79 proof
1.170. Let x = mass of gold and y = mass of silver. Then, use the total mass to write
x + y = 9.35 g, or y = 9.35 g โ x. Thus, the mass of silver is 9.35 g โ x. Next,
Total volume = volume of gold + volume of silver
Since the volume is equal to the mass divided by density, you can write
Total volume =
mass of gold
mass of silver
+
density of gold
density of silver
Substitute in the known and unknown values to get an equation for x.
0.654 cm3 =
x
9.35 g ๏ญ x
+
3
19.3 g/cm
10.5 g/cm3
Multiply both sides of this equation by (19.3)(10.5). Also multiply both sides by g/cm3 to
simplify the units. This gives the following equation to solve for x.
(19.3)(10.5)(0.654) g = (10.5) x + (19.3)(9.35 g โ x)
132.53 g = 10.5 x + 180.45 g โ 19.3 x
8.8 x = 47.92 g
x = mass of gold = 5.445 g
The percentage of gold (by mass) in the solution can now be calculated.
Percent (mass) =
mass of gold
5.445 g
๏ด 100% =
๏ด 100% = 58.2% = 58%
mass of jewlrey
9.35 g
The relative amount of gold in the alloy can now be calculated. The fraction of gold in the alloy is
58.2% / 100% = 0.582. Thus,
Proportion of gold = 24 karats x 0.582 = 13.9 karats = 14 karats
1.171. The volume of the mineral can be obtained from the mass difference between the water displaced
and the air displaced, and the densities of water and air.
Mass difference = 18.49 g โ 16.21 g = 2.28 g
Volume of mineral =
=
mass difference
density of water ๏ญ density of air
2.28 g
= 2.286 cm3
0.9982 g/cm ๏ญ 1.205 ๏ด 10๏ญ3 g/cm3
3
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
29
The mass of the mineral is equal to its mass in air plus the weight of the displaced air. The weight
of the displaced air is obtained from the volume of the mineral and the density of air.
Mass of displaced air = density x volume
= 1.205 g/L x 2.286 cm3 ๏ด
1L
= 2.755 ๏ด 10โ3 g
3
3
10 cm
Mass of mineral = 18.49 g + 2.755 ๏ด 10โ3 g = 18.4927 g
The density of the mineral can now be calculated.
Density =
18.4927 g
mass
=
= 8.089 g/cm3 = 8.09 g/cm3
2.286 cm3
volume
1.172. The volume of the mineral can be obtained from the mass of the water displaced and the density
of water.
Mass difference = 7.35 g โ 5.40 g = 1.95 g
Volume of mineral =
mass difference
density of water ๏ญ density of air
Volume of mineral =
1.95 g
= 1.955 cm3
0.9982 g/cm ๏ญ 1.205 x 10๏ญ3 g/cm3
3
The mass of the mineral is equal to its mass in air plus the weight of the displaced air. The weight
of the displaced air is obtained from the volume of the mineral and the density of air.
Mass of displaced air = density x volume
= 1.205 g/L x 1.955 cm3 ๏ด
1L
= 2.356 ๏ด 10โ3 g
103 cm3
Mass of mineral = 7.35 g + 2.356 ๏ด 10โ3 g = 7.352 g
The density of the mineral can now be calculated.
Density =
7.352 g
mass
=
= 3.760 g/cm3 = 3.76 g/cm3
volume 1.955 cm3
1.173. The volume of the object can be obtained from the mass of the ethanol displaced and the density
of ethanol.
Mass of ethanol displaced = 15.8 g โ 10.5 g = 5.3 g
Volume of object =
mass of ethanol
5.3 g
=
= 6.717 cm3
density of ethanol
0.789 g/ cm3
The density of the object can now be calculated.
Density =
15.8 g
mass
=
= 2.352 g/cm3 = 2.4 g/cm3
3
6.717 cm
volume
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
30
Chapter 1: Chemistry and Measurement
1.174. The volume of the metal can be obtained from the mass of the mercury displaced and the density
of mercury.
Mass of mercury displaced = 255 g โ 101 g = 154 g
Volume of object =
mass of mercury
154 g
=
= 11.323 cm3
3
density of mercury 13.6 g/cm
The density of the metal can now be calculated.
Density =
255 g
mass
=
= 22.51 g/cm3 = 22.5 g/cm3
volume 11.323 cm3
1.175. The first set of measurements, although quite close to each other, yields an average that is
significantly and consistently higher than the expected outcome. Thus, it is precise but
inaccurate. A systematic error is likely present that could be corrected to yield a more
favorable outcome.
The second set yields an average that is quite close to the expected outcome, thus it is
accurate. However, the precision of the data is relatively poor compared to the data sets 1
and 3.
The third set results in an average that is quite close to the expected outcome with each
measurement very close to one another. It would be described as both accurate and precise,
and the most favorable outcome of all the sets.
The last data set gives an average that is significantly higher than the expected outcome.
Compared to data sets 1 and 3, the precision is also very poor. It would be described as
inaccurate and imprecise, and the least favorable outcome of all the sets.
The following table summarizes these arguments:
Data
Accurate?
Precise?
38.74, 38.75, 38.76
No
Yes
37.15, 37.44, 37.75
Yes
No
37.44, 37.46, 37.48
Yes
Yes
39.43, 37.45, 38.64
No
No
1.176. The first data set gives an average that is quite close to the expected outcome with each
measurement very close to one another. It would be described as both accurate and precise,
and the most favorable outcome of all the sets.
The second data set yields an average that is quite close to the expected outcome, thus it is
accurate. However, the precision of the data is relatively poor compared to the data sets 1
and 3.
The third set of measurements, although quite close to each other, yields an average that is
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1: Chemistry and Measurement
31
significantly and consistently lower than the expected outcome. Thus, the data has precision
but is inaccurate. A systematic error is likely present that could be corrected to yield a more
favorable outcome.
The last data set gives an average that is significantly lower than the expected outcome.
Compared to the first and third sets of data, the precision is also very poor. It would be
described as inaccurate and imprecise, and the least favorable outcome of all the sets.
The following table summarizes these arguments:
Data
Precise?
Accurate?
32.00, 32.01, 31.99
Yes
Yes
29.50, 32.00, 34.50
No
Yes
29.00, 29.01, 29.02
Yes
No
25.00, 27.00, 29.00
No
No
1.177. (a) For the moment, assume the liquid to be water. Multiplying the volume in gallons by the
density in lbs/gal and then converting to kg would give the mass of the water in kg as
follows:
8.35 lb
1 kg
m H 2O ๏ฝ VH 2O x d H 2O ๏ฝ (24,500 gal) x (
)x(
) ๏ฝ 9.278 x 104 kg
gal
2.205 lb
Since the liquid in the barge is denser than pure water by a factor of 1.01, the mass of liquid
being carried is obtained by multiplying the calculated mass of water by the specific gravity
of the liquid, i.e.,
mliquid ๏ฝ 1.01 x mH2O ๏ฝ 1.01 x 9.278 x 104 kg = 9.371 x 104 kg = 9.37 x 10 4 kg
(b) Water bodies at river mouths that are adjacent to the ocean have specific gravities very close
to that of average seawater (specific gravity = 1.03); hence, the bargeโs liquid is less dense than
the surrounding water. Considering the leaked liquid does not mix well with the surrounding
seawater, the less dense liquid would float on the seawater. Thus, assuming weather conditions do
not serious work to disperse the spill, surface booms would likely be effective tools in cleanup
efforts.
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32
Chapter 1: Chemistry and Measurement
1.178. Step 1: Graph using complete data set
Step 2: Graph from Step 1 scaled to a maximum of 100 g/100 g H 2O
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Chapter 1: Chemistry and Measurement
33
Step 3: Graph from Step 1 scaled to a maximum of 25 g/100 g H 2O
Step 4:
(a) Those salts which exhibit an increasing trend in solubility as the temperature increases
include NaCl, KCl, K 2Cr2O7, KNO3, KNO2, and PbI2.
(b) Those salts which exhibit a decreasing trend in solubility as the temperature increases
include Ce2(SO4)3 ยท 2H2O and Nd2(SO4)3.
(c) MnSO4 initially exhibits a slight increase in solubility as the temperature increases but
reverses this trend above 20 oC. It is also noted that NaCl and PbI 2 do not exhibit
substantial increases over the temperature range of the data.
Step 5:
(a) Reasonable estimates (to 2 significant figures) of the requested solubility values would be
as follows:
solubility of Ce2(SO4)3.2H2O at 10oC : 16 g / 100 g H2O
solubility of Ce2(SO4)3.2H2O at 50oC : 4.8 g / 100 g H2O
solubility of KNO2 at 50oC
: 340 g / 100 g H2O
(b) The only salt that would likely be considered insoluble in the current context would be
PbI2.
ยฉ 2017 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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