Test Bank for Financial Statement Analysis and Security Valuation, 5th Edition
Test Bank for Financial Statement Analysis and Security Valuation, 5th Edition provides in-depth questions and solutions to reinforce key concepts. Start practicing today!
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1
REVIEW
MEASUREMENT AND CALCULATIONS IN CHEMISTRY
Questions
1. A random error has equal probability of being too high or too low. This type of error occurs
when estimating the value of the last digit of a measurement. A systematic error is one that
always occurs in the same direction, either too high or too low. For example, this type of
error would occur if the balance you were using weighed all objects 0.20 g too high, that is, if
the balance wasn’t calibrated correctly. A random error is an indeterminate error, whereas a
systematic error is a determinate error.
2. 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 that are imprecise implies that all the numbers are not close to each
other. If the numbers aren’t reproducible, then all 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 the data are
accurate; a better description is that the data takers are extremely lucky.
3. Accuracy refers to how close a measurement or series of measurements are to an accepted or
true value. Precision refers to how close a series of measurements of the same item are to
each other (reproducible). The results, average = 14.91 ±0.03%, are precise (are close to each
other) but are not accurate (are not close to the true value).
4. Volume readings are estimated to one 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. 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
REVIEW
MEASUREMENT AND CALCULATIONS IN CHEMISTRY
Questions
1. A random error has equal probability of being too high or too low. This type of error occurs
when estimating the value of the last digit of a measurement. A systematic error is one that
always occurs in the same direction, either too high or too low. For example, this type of
error would occur if the balance you were using weighed all objects 0.20 g too high, that is, if
the balance wasn’t calibrated correctly. A random error is an indeterminate error, whereas a
systematic error is a determinate error.
2. 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 that are imprecise implies that all the numbers are not close to each
other. If the numbers aren’t reproducible, then all 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 the data are
accurate; a better description is that the data takers are extremely lucky.
3. Accuracy refers to how close a measurement or series of measurements are to an accepted or
true value. Precision refers to how close a series of measurements of the same item are to
each other (reproducible). The results, average = 14.91 ±0.03%, are precise (are close to each
other) but are not accurate (are not close to the true value).
4. Volume readings are estimated to one 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. 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
1
REVIEW
MEASUREMENT AND CALCULATIONS IN CHEMISTRY
Questions
1. A random error has equal probability of being too high or too low. This type of error occurs
when estimating the value of the last digit of a measurement. A systematic error is one that
always occurs in the same direction, either too high or too low. For example, this type of
error would occur if the balance you were using weighed all objects 0.20 g too high, that is, if
the balance wasn’t calibrated correctly. A random error is an indeterminate error, whereas a
systematic error is a determinate error.
2. 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 that are imprecise implies that all the numbers are not close to each
other. If the numbers aren’t reproducible, then all 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 the data are
accurate; a better description is that the data takers are extremely lucky.
3. Accuracy refers to how close a measurement or series of measurements are to an accepted or
true value. Precision refers to how close a series of measurements of the same item are to
each other (reproducible). The results, average = 14.91 ±0.03%, are precise (are close to each
other) but are not accurate (are not close to the true value).
4. Volume readings are estimated to one 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. 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
REVIEW
MEASUREMENT AND CALCULATIONS IN CHEMISTRY
Questions
1. A random error has equal probability of being too high or too low. This type of error occurs
when estimating the value of the last digit of a measurement. A systematic error is one that
always occurs in the same direction, either too high or too low. For example, this type of
error would occur if the balance you were using weighed all objects 0.20 g too high, that is, if
the balance wasn’t calibrated correctly. A random error is an indeterminate error, whereas a
systematic error is a determinate error.
2. 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 that are imprecise implies that all the numbers are not close to each
other. If the numbers aren’t reproducible, then all 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 the data are
accurate; a better description is that the data takers are extremely lucky.
3. Accuracy refers to how close a measurement or series of measurements are to an accepted or
true value. Precision refers to how close a series of measurements of the same item are to
each other (reproducible). The results, average = 14.91 ±0.03%, are precise (are close to each
other) but are not accurate (are not close to the true value).
4. Volume readings are estimated to one 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. 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
2 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
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 multi-
plication/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).
6. In both sets of rules, the least 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. In a subtraction, the result gets smaller, but the uncertainties add. If the two numbers are very
close together, the uncertainty may be larger than the result. For example, let’s assume we
want to take the difference of the following two measured quantities, 999,999 ±2 and 999,996
±2. The difference is 3 ±4. Because of the uncertainty, subtracting two similar numbers is
poor practice.
8. Consider gold with a density of 19.32 g/cm3. The two possible conversion factors are:3
cm1
g32.19
org32.19
cm1 3
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 in grams to the volume of gold.
When using conversion factors, concentrate on the units crossing off.
9. 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 32F.
For the TC vs. TK plot:
TC = TK 273
Y = m x + b
The slope of the plot is 1, and the y-intercept is 273C.
10. 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 1F
would correspond to a smaller temperature change than 1C or 1 K.
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 multi-
plication/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).
6. In both sets of rules, the least 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. In a subtraction, the result gets smaller, but the uncertainties add. If the two numbers are very
close together, the uncertainty may be larger than the result. For example, let’s assume we
want to take the difference of the following two measured quantities, 999,999 ±2 and 999,996
±2. The difference is 3 ±4. Because of the uncertainty, subtracting two similar numbers is
poor practice.
8. Consider gold with a density of 19.32 g/cm3. The two possible conversion factors are:3
cm1
g32.19
org32.19
cm1 3
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 in grams to the volume of gold.
When using conversion factors, concentrate on the units crossing off.
9. 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 32F.
For the TC vs. TK plot:
TC = TK 273
Y = m x + b
The slope of the plot is 1, and the y-intercept is 273C.
10. 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 1F
would correspond to a smaller temperature change than 1C or 1 K.
REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 3
11. 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. Electrolysis 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
12. a.
b.
13. The law of conservation of energy states that energy can be converted from one form to
another but can neither be destroyed nor created; the energy content of the universe is
constant. The two categories of energy are kinetic energy and potential energy. Kinetic
energy is energy due to motion and potential energy is stored energy due to position. For
example, the energy stored in the bonds between atoms in a molecule is potential energy.
14. Counting by weighing utilizes the average mass of a particular unit of substance. Consider a
large sample size of marbles which will contain many different individual masses for the
various marbles. However, the very large sample size will have an average mass so that thegas element (monoatomic)
atoms/molecules far apart;
random order; takes volume
of container
atoms/molecules close
together; somewhat
ordered arrangement;
takes volume of container
liquid element solid element
atoms/molecules
close together;
ordered arrangement;
has its own volume2 compounds compound and element (diatomic)
11. 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. Electrolysis 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
12. a.
b.
13. The law of conservation of energy states that energy can be converted from one form to
another but can neither be destroyed nor created; the energy content of the universe is
constant. The two categories of energy are kinetic energy and potential energy. Kinetic
energy is energy due to motion and potential energy is stored energy due to position. For
example, the energy stored in the bonds between atoms in a molecule is potential energy.
14. Counting by weighing utilizes the average mass of a particular unit of substance. Consider a
large sample size of marbles which will contain many different individual masses for the
various marbles. However, the very large sample size will have an average mass so that thegas element (monoatomic)
atoms/molecules far apart;
random order; takes volume
of container
atoms/molecules close
together; somewhat
ordered arrangement;
takes volume of container
liquid element solid element
atoms/molecules
close together;
ordered arrangement;
has its own volume2 compounds compound and element (diatomic)
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4 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
marbles behave as if each individual marble has that average mass. This assumption is valid
as long as the sample size is very large. When a large sample of marbles is weighed, one
divides the total mass of marbles by the average mass of a marble, and this will give a very
good estimate of the number of marbles present.
Because we can’t count individual atoms, we “count” the atoms by weighing; converting the
sample mass in grams to the number of atoms in the sample by using the average mass of the
substance. The mole scale of atoms is a huge number (6.022 × 1023 atoms = 1 mole), so the
assumption that a weighable sample size behaves as a bunch of atoms, each with the same
average mass, is valid and allows one to count atoms by weighing. The mole scale also gives
easier numbers to deal with. So instead of saying you have 12 × 1023 atoms of helium, you
can instead say that you have 2.0 moles of helium.
Exercises
Significant Figures and Unit Conversions
15. a. exact b. inexact
c. exact d. inexact (π has an infinite number of decimal places.)
16. 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. one S.F.
17. a. 6.07 ×15
10 ; 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. 300; 1 S.F.
g. 301; 3 S.F. h. 300.; 3 S.F.
18. 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. 0.00480; 3 S.F.
g. 4.80 ×3
10 ; 3 S.F. h. 4.800 ×3
10 ; 4 S.F.
19. 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.
a. 3.42 ×4
10 b. 1.034 × 104 c. 1.7992 × 101 d. 3.37 × 105
20. a. 4 × 105 b. 3.9 × 105 c. 3.86 × 105 d. 3.8550 × 105
21. Volume measurements are estimated to one place past the markings on the glassware. The
first graduated cylinder is labeled to 0.2 mL volume increments, so we estimate volumes to
the hundredths place. Realistically, the uncertainty in this graduated cylinder is ±0.05 mL.
The second cylinder, with 0.02 mL volume increments, will have an uncertainty of ±0.005
marbles behave as if each individual marble has that average mass. This assumption is valid
as long as the sample size is very large. When a large sample of marbles is weighed, one
divides the total mass of marbles by the average mass of a marble, and this will give a very
good estimate of the number of marbles present.
Because we can’t count individual atoms, we “count” the atoms by weighing; converting the
sample mass in grams to the number of atoms in the sample by using the average mass of the
substance. The mole scale of atoms is a huge number (6.022 × 1023 atoms = 1 mole), so the
assumption that a weighable sample size behaves as a bunch of atoms, each with the same
average mass, is valid and allows one to count atoms by weighing. The mole scale also gives
easier numbers to deal with. So instead of saying you have 12 × 1023 atoms of helium, you
can instead say that you have 2.0 moles of helium.
Exercises
Significant Figures and Unit Conversions
15. a. exact b. inexact
c. exact d. inexact (π has an infinite number of decimal places.)
16. 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. one S.F.
17. a. 6.07 ×15
10 ; 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. 300; 1 S.F.
g. 301; 3 S.F. h. 300.; 3 S.F.
18. 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. 0.00480; 3 S.F.
g. 4.80 ×3
10 ; 3 S.F. h. 4.800 ×3
10 ; 4 S.F.
19. 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.
a. 3.42 ×4
10 b. 1.034 × 104 c. 1.7992 × 101 d. 3.37 × 105
20. a. 4 × 105 b. 3.9 × 105 c. 3.86 × 105 d. 3.8550 × 105
21. Volume measurements are estimated to one place past the markings on the glassware. The
first graduated cylinder is labeled to 0.2 mL volume increments, so we estimate volumes to
the hundredths place. Realistically, the uncertainty in this graduated cylinder is ±0.05 mL.
The second cylinder, with 0.02 mL volume increments, will have an uncertainty of ±0.005
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REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 5
mL. The approximate volume in the first graduated cylinder is 2.85 mL, and the volume in
the other graduated cylinder is approximately 0.280 mL. The total volume would be:
2.85 mL
+0.280 mL
3.13 mL
We should report the total volume to the hundredths place because the volume from the first
graduated cylinder is only read to the hundredths (read to two decimal places). The first
graduated cylinder is the least precise volume measurement because the uncertainty of this
instrument is in the hundredths place, while the uncertainty of the second graduated cylinder
is to the thousandths place. It is always the least precise measurement that limits the
precision of a calculation.
22. a. Volumes are always estimated to one position past the marked volume increments. The
estimated volume of the first beaker is 32.7 mL, the estimated volume of the middle
beaker is 33 mL, and the estimated volume in the last beaker is 32.73 mL.
b. Yes, all volumes could be identical to each other because the more precise volume
readings can be rounded to the other volume readings. But because the volumes are in
three different measuring devices, each with its own unique uncertainty, we cannot say
with certainty that all three beakers contain the same amount of water.
c. 32.7 mL
33 mL
32.73 mL
98.43 mL = 98 mL
The volume in the middle beaker can only be estimated to the ones place, which dictates
that the sum of the volume should be reported to the ones place. As is always the case,
the least precise measurement determines the precision of a calculation.
23. 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).
24. 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.
mL. The approximate volume in the first graduated cylinder is 2.85 mL, and the volume in
the other graduated cylinder is approximately 0.280 mL. The total volume would be:
2.85 mL
+0.280 mL
3.13 mL
We should report the total volume to the hundredths place because the volume from the first
graduated cylinder is only read to the hundredths (read to two decimal places). The first
graduated cylinder is the least precise volume measurement because the uncertainty of this
instrument is in the hundredths place, while the uncertainty of the second graduated cylinder
is to the thousandths place. It is always the least precise measurement that limits the
precision of a calculation.
22. a. Volumes are always estimated to one position past the marked volume increments. The
estimated volume of the first beaker is 32.7 mL, the estimated volume of the middle
beaker is 33 mL, and the estimated volume in the last beaker is 32.73 mL.
b. Yes, all volumes could be identical to each other because the more precise volume
readings can be rounded to the other volume readings. But because the volumes are in
three different measuring devices, each with its own unique uncertainty, we cannot say
with certainty that all three beakers contain the same amount of water.
c. 32.7 mL
33 mL
32.73 mL
98.43 mL = 98 mL
The volume in the middle beaker can only be estimated to the ones place, which dictates
that the sum of the volume should be reported to the ones place. As is always the case,
the least precise measurement determines the precision of a calculation.
23. 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).
24. 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.
Loading page 6...
6 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
a.2.26352.26
1.01
2730.08210.102
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 ×3
10 × 7.34993 × 102 = 1.476 × 105 = 1.5 × 105
d.1212
7
6
1067.6106766.6
1000.3
1000.2
25. 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.4326.0
705.80
623.0
470.0
1.3
526.2
= 0.8148 + 0.7544 + 186.558 = 188.1
b. Here, the mathematical operation requires that we apply the addition/subtraction rule
first, then apply the multiplication/division rule.6.1
91.2404.6
1.177.18
91.2404.6
= 12
c. 6.071 ×5
10 8.2 ×6
10 0.521 ×4
10 = 60.71 ×6
10 8.2 ×6
10 52.1 ×6
10
= 0.41 ×6
10 = 4 ×7
10
d.26
12
13
1212
1313
1312
1312
103.6
1076
1024
1063104
100.41038
103.6104
100.4108.3
e.4
755.19
4
175.38.21.45.9
= 4.89 = 4.9
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.925.8
002.0
100
925.8
905.8925.8
× 100 = 0.22
26. a. 6.022 × 1023 × 1.05 × 102 = 6.32 × 1025
b.17
9
834
1082.7
1054.2
10998.2106262.6
a.2.26352.26
1.01
2730.08210.102
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 ×3
10 × 7.34993 × 102 = 1.476 × 105 = 1.5 × 105
d.1212
7
6
1067.6106766.6
1000.3
1000.2
25. 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.4326.0
705.80
623.0
470.0
1.3
526.2
= 0.8148 + 0.7544 + 186.558 = 188.1
b. Here, the mathematical operation requires that we apply the addition/subtraction rule
first, then apply the multiplication/division rule.6.1
91.2404.6
1.177.18
91.2404.6
= 12
c. 6.071 ×5
10 8.2 ×6
10 0.521 ×4
10 = 60.71 ×6
10 8.2 ×6
10 52.1 ×6
10
= 0.41 ×6
10 = 4 ×7
10
d.26
12
13
1212
1313
1312
1312
103.6
1076
1024
1063104
100.41038
103.6104
100.4108.3
e.4
755.19
4
175.38.21.45.9
= 4.89 = 4.9
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.925.8
002.0
100
925.8
905.8925.8
× 100 = 0.22
26. a. 6.022 × 1023 × 1.05 × 102 = 6.32 × 1025
b.17
9
834
1082.7
1054.2
10998.2106262.6
Loading page 7...
REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 7
c. 1.285 ×2
10 + 1.24 ×3
10 + 1.879 ×1
10
= 0.1285 ×1
10 + 0.0124 ×1
10 + 1.879 ×1
10 = 2.020 ×1
10
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.27
2323 1029.2
1002205.6
00138.0
1002205.6
)00728.100866.1(
e.1
2
2
2
22
101.8100
10875.9
10080.0
100
10875.9
10795.910875.9
f.3
10625.110824.010942.0
3
10625.110234.81042.9 333322
= 1.130 × 103
27. a. 8.43 cm ×mm3.84
m
mm1000
cm100
m1 b. 2.41 × 102 cm ×cm100
m1 = 2.41 m
c. 294.5 nmcm10945.2
m
cm100
nm101
m1 5
9
d. 1.445 × 104 m ×m
km
1000
1 = 14.45 km e. 235.3 m ×m
mm1000 = 2.353 × 105 mm
f. 903.3 nmmμ9033.0
m
mμ101
nm101
m1 6
9
28. a. 1 Tgkg101
g1000
kg1
Tg
g101 9
12
b. 6.50 × 102 Tmnm1050.6
m
nm101
Tm
m101 23
912
c. 25 fgkg102.5kg1025
g1000
kg1
fg101
g1 1718
15
d. 8.0 dm3 ×3
1
dm
L = 8.0 L (1 L = 1 dm3 = 1000 cm3 = 1000 mL)
e. 1 mLLμ101
L
Lμ101
mL0001
L1 3
6
f. 1 μgpg101
g
pg101
gμ101
g1 6
12
6
c. 1.285 ×2
10 + 1.24 ×3
10 + 1.879 ×1
10
= 0.1285 ×1
10 + 0.0124 ×1
10 + 1.879 ×1
10 = 2.020 ×1
10
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.27
2323 1029.2
1002205.6
00138.0
1002205.6
)00728.100866.1(
e.1
2
2
2
22
101.8100
10875.9
10080.0
100
10875.9
10795.910875.9
f.3
10625.110824.010942.0
3
10625.110234.81042.9 333322
= 1.130 × 103
27. a. 8.43 cm ×mm3.84
m
mm1000
cm100
m1 b. 2.41 × 102 cm ×cm100
m1 = 2.41 m
c. 294.5 nmcm10945.2
m
cm100
nm101
m1 5
9
d. 1.445 × 104 m ×m
km
1000
1 = 14.45 km e. 235.3 m ×m
mm1000 = 2.353 × 105 mm
f. 903.3 nmmμ9033.0
m
mμ101
nm101
m1 6
9
28. a. 1 Tgkg101
g1000
kg1
Tg
g101 9
12
b. 6.50 × 102 Tmnm1050.6
m
nm101
Tm
m101 23
912
c. 25 fgkg102.5kg1025
g1000
kg1
fg101
g1 1718
15
d. 8.0 dm3 ×3
1
dm
L = 8.0 L (1 L = 1 dm3 = 1000 cm3 = 1000 mL)
e. 1 mLLμ101
L
Lμ101
mL0001
L1 3
6
f. 1 μgpg101
g
pg101
gμ101
g1 6
12
6
Loading page 8...
8 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
29. 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 ×kg4536.0
lb1 = 8.62 lb; 0.62 lb ×lb
oz16 = 9.9 oz
Baby’s weight = 8 lb and 9.9 oz or, to the nearest ounce, 8 lb and 10. oz.
51.4 cm ×cm54.2
in1 = 20.2 in ≈ 20 1/4 in = baby’s height
b. 25,000 mi ×mi
km61.1 = 4.0 × 104 km; 4.0 × 104 km ×km
m1000 = 4.0 × 107 m
c. V = 1 × w × h = 1.0 m ×
dm10
m1
dm1.2
cm100
m1
cm6.5 = 1.2 ×2
10 m3
1.2 ×2
10 m3 ×3
3
dm
L1
m
dm01
= 12 L
12 L ×33
cm54.2
in1
L
cm1000
= 730 in3; 730 in3 ×3
in12
ft1
= 0.42 ft3
30. a. 908 oz ×lb
kg4536.0
oz16
lb1 = 25.7 kg
b. 12.8 L ×qt4
gal1
L9463.0
qt1 = 3.38 gal
c. 125 mL ×L9463.0
qt1
mL1000
L1 = 0.132 qt
d. 2.89 gal ×L1
mL1000
qt057.1
L1
gal1
qt4 = 1.09 × 104 mL
e. 4.48 lb ×lb1
g6.453 = 2.03 × 103 g
f. 550 mL ×L
qt06.1
mL1000
L1 = 0.58 qt
31. a. 1.25 mi ×mi
furlongs8 = 10.0 furlongs; 10.0 furlongs ×furlong
rods40 = 4.00 × 102 rods
4.00 × 102 rods ×cm100
m1
in
cm54.2
yd
in36
rod
yd5.5 = 2.01 × 103 m
29. 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 ×kg4536.0
lb1 = 8.62 lb; 0.62 lb ×lb
oz16 = 9.9 oz
Baby’s weight = 8 lb and 9.9 oz or, to the nearest ounce, 8 lb and 10. oz.
51.4 cm ×cm54.2
in1 = 20.2 in ≈ 20 1/4 in = baby’s height
b. 25,000 mi ×mi
km61.1 = 4.0 × 104 km; 4.0 × 104 km ×km
m1000 = 4.0 × 107 m
c. V = 1 × w × h = 1.0 m ×
dm10
m1
dm1.2
cm100
m1
cm6.5 = 1.2 ×2
10 m3
1.2 ×2
10 m3 ×3
3
dm
L1
m
dm01
= 12 L
12 L ×33
cm54.2
in1
L
cm1000
= 730 in3; 730 in3 ×3
in12
ft1
= 0.42 ft3
30. a. 908 oz ×lb
kg4536.0
oz16
lb1 = 25.7 kg
b. 12.8 L ×qt4
gal1
L9463.0
qt1 = 3.38 gal
c. 125 mL ×L9463.0
qt1
mL1000
L1 = 0.132 qt
d. 2.89 gal ×L1
mL1000
qt057.1
L1
gal1
qt4 = 1.09 × 104 mL
e. 4.48 lb ×lb1
g6.453 = 2.03 × 103 g
f. 550 mL ×L
qt06.1
mL1000
L1 = 0.58 qt
31. a. 1.25 mi ×mi
furlongs8 = 10.0 furlongs; 10.0 furlongs ×furlong
rods40 = 4.00 × 102 rods
4.00 × 102 rods ×cm100
m1
in
cm54.2
yd
in36
rod
yd5.5 = 2.01 × 103 m
Loading page 9...
REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 9
2.01 × 103 m ×m1000
km1 = 2.01 km
b. Let's assume we know this distance to ±1 yard. First, convert 26 miles to yards.
26 mi ×ft3
yd1
mi
ft5280 = 45,760. yd
26 mi + 385 yd = 45,760. yd + 385 yd = 46,145 yards
46,145 yard ×yd5.5
rod1 = 8390.0 rods; 8390.0 rods ×rods40
furlong1 = 209.75 furlongs
46,145 yard ×cm100
m1
in
cm54.2
yd
in36 = 42,195 m; 42,195 m ×m1000
km1 = 42.195 km
32. a. 1 ha ×22
m1000
km1
ha
m000,10
= 122 km10
b. 5.5 acre ×22
cm100
m1
in
cm54.2
yd
in36
rod
yd5.5
acre
rod160
= 2.2 × 104 m2
2.2 × 104 m2 ×24 m101
ha1
= 2.2 ha; 2.2 × 104 m2 ×2
m1000
km1
= 0.022 km2
c. Area of lot = 120 ft × 75 ft = 9.0 × 103 ft2
9.0 × 103 ft2 ×2
2
rod160
acre1
yd5.5
rod1
ft3
yd1
= 0.21 acre;acre
000,31$
acre21.0
500,6$
We can use our result from (b) to get the conversion factor between acres and hectares
(5.5 acre = 2.2 ha.). Thus 1 ha = 2.5 acre.
0.21 acre ×acre5.2
ha1 = 0.084 ha; the price is:ha
000,77$
ha084.0
500,6$
33. a. 1 troy lb ×g1000
kg1
grain
g0648.0
pw
grains24
oztroy
pw20
lbtroy
oztroy12 = 0.373 kg
1 troy lb = 0.373 kg ×kg
lb205.2 = 0.822 lb
b. 1 troy oz ×grain
g0648.0
pw
grains24
oztroy
pw20 = 31.1 g
1 troy oz = 31.1 g ×g200.0
carat1 = 156 carats
2.01 × 103 m ×m1000
km1 = 2.01 km
b. Let's assume we know this distance to ±1 yard. First, convert 26 miles to yards.
26 mi ×ft3
yd1
mi
ft5280 = 45,760. yd
26 mi + 385 yd = 45,760. yd + 385 yd = 46,145 yards
46,145 yard ×yd5.5
rod1 = 8390.0 rods; 8390.0 rods ×rods40
furlong1 = 209.75 furlongs
46,145 yard ×cm100
m1
in
cm54.2
yd
in36 = 42,195 m; 42,195 m ×m1000
km1 = 42.195 km
32. a. 1 ha ×22
m1000
km1
ha
m000,10
= 122 km10
b. 5.5 acre ×22
cm100
m1
in
cm54.2
yd
in36
rod
yd5.5
acre
rod160
= 2.2 × 104 m2
2.2 × 104 m2 ×24 m101
ha1
= 2.2 ha; 2.2 × 104 m2 ×2
m1000
km1
= 0.022 km2
c. Area of lot = 120 ft × 75 ft = 9.0 × 103 ft2
9.0 × 103 ft2 ×2
2
rod160
acre1
yd5.5
rod1
ft3
yd1
= 0.21 acre;acre
000,31$
acre21.0
500,6$
We can use our result from (b) to get the conversion factor between acres and hectares
(5.5 acre = 2.2 ha.). Thus 1 ha = 2.5 acre.
0.21 acre ×acre5.2
ha1 = 0.084 ha; the price is:ha
000,77$
ha084.0
500,6$
33. a. 1 troy lb ×g1000
kg1
grain
g0648.0
pw
grains24
oztroy
pw20
lbtroy
oztroy12 = 0.373 kg
1 troy lb = 0.373 kg ×kg
lb205.2 = 0.822 lb
b. 1 troy oz ×grain
g0648.0
pw
grains24
oztroy
pw20 = 31.1 g
1 troy oz = 31.1 g ×g200.0
carat1 = 156 carats
Loading page 10...
10 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
c. 1 troy lb = 0.373 kg; 0.373 kg ×g3.19
cm1
kg
g1000 3
= 19.3 cm3
34. a. 1 grain ap ×apdram
g888.3
scruples3
apdram1
apgrain20
scruple1 = 0.06480 g
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 ×g1.31
*troyoz1
apdram
g888.3
apoz
apdram8 = 1.00 oz troy; *see Exercise 33b.
c. 5.00 × 102 mg ×apdram
scruples3
g888.3
apdram1
mg1000
g1 = 0.386 scruple
0.386 scruple ×scruple
apgrains20 = 7.72 grains ap
d. 1 scruple ×apdram
g888.3
scruples3
apdram1 = 1.296 g
35. 15.6 g ×g65.0
capsule1 = 24 capsules
36. 1.5 teaspoons ×teaspoon50.0
acetmg.80 = 240 mg acetaminophenkg454.0
lb1
lb24
acetmg240
= 22 mg acetaminophen/kgkg454.0
lb1
lb35
acetmg240
= 15 mg acetaminophen/kg
The range is from 15 to 22 mg acetaminophen per kg of body weight.
37. warp 1.71 =h/yd2000
knot1
h
min60
min
s60
m
yd094.1
s
m1000.3
00.5
8
= 2.95 × 109 knotsh
min60
min
s60
km609.1
mi1
m1000
km1
s
m1000.3
00.5
8
= 3.36 × 109 mi/h
38.s58.9
m.100 = 10.4 m/s;h
min60
min
s60
m1000
km1
s58.9
m.100 = 37.6 km/hyd
ft3
m
yd0936.1
s58.9
m.100
= 34.2 ft/s;h
min60
min
s60
ft5280
mi1
s
ft2.34 = 23.3 mi/h
c. 1 troy lb = 0.373 kg; 0.373 kg ×g3.19
cm1
kg
g1000 3
= 19.3 cm3
34. a. 1 grain ap ×apdram
g888.3
scruples3
apdram1
apgrain20
scruple1 = 0.06480 g
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 ×g1.31
*troyoz1
apdram
g888.3
apoz
apdram8 = 1.00 oz troy; *see Exercise 33b.
c. 5.00 × 102 mg ×apdram
scruples3
g888.3
apdram1
mg1000
g1 = 0.386 scruple
0.386 scruple ×scruple
apgrains20 = 7.72 grains ap
d. 1 scruple ×apdram
g888.3
scruples3
apdram1 = 1.296 g
35. 15.6 g ×g65.0
capsule1 = 24 capsules
36. 1.5 teaspoons ×teaspoon50.0
acetmg.80 = 240 mg acetaminophenkg454.0
lb1
lb24
acetmg240
= 22 mg acetaminophen/kgkg454.0
lb1
lb35
acetmg240
= 15 mg acetaminophen/kg
The range is from 15 to 22 mg acetaminophen per kg of body weight.
37. warp 1.71 =h/yd2000
knot1
h
min60
min
s60
m
yd094.1
s
m1000.3
00.5
8
= 2.95 × 109 knotsh
min60
min
s60
km609.1
mi1
m1000
km1
s
m1000.3
00.5
8
= 3.36 × 109 mi/h
38.s58.9
m.100 = 10.4 m/s;h
min60
min
s60
m1000
km1
s58.9
m.100 = 37.6 km/hyd
ft3
m
yd0936.1
s58.9
m.100
= 34.2 ft/s;h
min60
min
s60
ft5280
mi1
s
ft2.34 = 23.3 mi/h
Loading page 11...
REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 11
1.00 × 102 yd ×m.100
s58.9
yd0936.1
m1 = 8.76 s
39.km
mi6214.0
h
km65 = 40.4 = 40. mi/h
To the correct number of significant figures (2), 65 km/h does not violate a 40 mi/h speed
limit.
40. 112 km ×mi65
h1
km
mi6214.0 = 1.1 h = 1 h and 6 min
112 km ×gal
L785.3
mi28
gal1
km
mi6214.0 = 9.4 L of gasoline
41.euro
32.1$
lb2046.2
kg1
kg
euros45.2 = $1.47/lb
One pound of peaches costs $1.47.
42. For the gasoline car:
500. migal
50.3$
mi0.28
gal1 = $62.5
For the E85 car:
500. migal
85.2$
mi5.22
gal1 = $63.3
The E85 vehicle would cost slightly more to drive 500. miles as compared to the gasoline
vehicle ($63.3 versus $62.5).
43. Volume of lake = 100 mi2 ×2
mi
ft5280
× 20 ft = 6 × 1010 ft3
6 × 1010 ft3 ×mL
μg0.4
cm
mL1
in
cm2.54
ft
in12
3
3
= 7 × 1014 μg mercury
7 × 1014 μgg101
kg1
gμ101
g1
36
= 7 × 105 kg of mercury
44. Volume of room = 18 ft × 12 ft × 8 ft = 1700 ft3 (carrying one extra significant figure)
1700 ft33
333
m48
cm100
m1
in
cm54.2
ft
in12
48 m3COgμ101
COg1
m
COgμ000,400
63
= 19 g = 20 g CO (to 1 sig. fig.)
1.00 × 102 yd ×m.100
s58.9
yd0936.1
m1 = 8.76 s
39.km
mi6214.0
h
km65 = 40.4 = 40. mi/h
To the correct number of significant figures (2), 65 km/h does not violate a 40 mi/h speed
limit.
40. 112 km ×mi65
h1
km
mi6214.0 = 1.1 h = 1 h and 6 min
112 km ×gal
L785.3
mi28
gal1
km
mi6214.0 = 9.4 L of gasoline
41.euro
32.1$
lb2046.2
kg1
kg
euros45.2 = $1.47/lb
One pound of peaches costs $1.47.
42. For the gasoline car:
500. migal
50.3$
mi0.28
gal1 = $62.5
For the E85 car:
500. migal
85.2$
mi5.22
gal1 = $63.3
The E85 vehicle would cost slightly more to drive 500. miles as compared to the gasoline
vehicle ($63.3 versus $62.5).
43. Volume of lake = 100 mi2 ×2
mi
ft5280
× 20 ft = 6 × 1010 ft3
6 × 1010 ft3 ×mL
μg0.4
cm
mL1
in
cm2.54
ft
in12
3
3
= 7 × 1014 μg mercury
7 × 1014 μgg101
kg1
gμ101
g1
36
= 7 × 105 kg of mercury
44. Volume of room = 18 ft × 12 ft × 8 ft = 1700 ft3 (carrying one extra significant figure)
1700 ft33
333
m48
cm100
m1
in
cm54.2
ft
in12
48 m3COgμ101
COg1
m
COgμ000,400
63
= 19 g = 20 g CO (to 1 sig. fig.)
Loading page 12...
12 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
Temperature
45. a. TC =9
5 (TF 32) =9
5 (459F 32) = 273C; TK = TC + 273 = 273C + 273 = 0 K
b. TC =9
5 (40.F 32) = 40.C; TK = 40.C + 273 = 233 K
c. TC =9
5 (68F 32) = 20.C; TK = 20.C + 273 = 293 K
d. TC =9
5 (7 × 107F 32) = 4 × 107C; TK = 4 × 107C + 273 = 4 × 107 K
46. 96.1°F ±0.2°F; first, convert 96.1°F to °C. TC =9
5 (TF 32) =9
5 (96.1 32) = 35.6°C
A change in temperature of 9°F is equal to a change in temperature of 5°C. So the
uncertainty is:
±0.2°F ×F9
C5
= ±0.1°C. Thus 96.1 ±0.2°F = 35.6 ±0.1°C.
47. a. TF =5
9 × TC + 32 =5
9 × 39.2°C + 32 = 102.6°F (Note: 32 is exact.)
TK = TC + 273.2 = 39.2 + 273.2 = 312.4 K
b. TF =5
9 × (25) + 32 = 13°F; TK = 25 + 273 = 248 K
c. TF =5
9 × (273) + 32 = -459°F; TK = 273 + 273 = 0 K
d. TF =5
9 × 801 + 32 = 1470°F; TK = 801 + 273 = 1074 K
48. a. TC = TK 273 = 233 273 = 40.°C
TF =5
9 × TC + 32 =5
9 × (40.) + 32 = 40.°F
b. TC = 4 273 = 269°C; TF =5
9 × (269) + 32 = 452°F
c. TC = 298 273 = 25°C; TF =5
9 × 25 + 32 = 77°F
d. TC = 3680 273 = 3410°C; TF =5
9 × 3410 + 32 = 6170°F
49. TF =5
9 × TC + 32; from the problem, we want the temperature where TF = 2TC.
Substituting:
Temperature
45. a. TC =9
5 (TF 32) =9
5 (459F 32) = 273C; TK = TC + 273 = 273C + 273 = 0 K
b. TC =9
5 (40.F 32) = 40.C; TK = 40.C + 273 = 233 K
c. TC =9
5 (68F 32) = 20.C; TK = 20.C + 273 = 293 K
d. TC =9
5 (7 × 107F 32) = 4 × 107C; TK = 4 × 107C + 273 = 4 × 107 K
46. 96.1°F ±0.2°F; first, convert 96.1°F to °C. TC =9
5 (TF 32) =9
5 (96.1 32) = 35.6°C
A change in temperature of 9°F is equal to a change in temperature of 5°C. So the
uncertainty is:
±0.2°F ×F9
C5
= ±0.1°C. Thus 96.1 ±0.2°F = 35.6 ±0.1°C.
47. a. TF =5
9 × TC + 32 =5
9 × 39.2°C + 32 = 102.6°F (Note: 32 is exact.)
TK = TC + 273.2 = 39.2 + 273.2 = 312.4 K
b. TF =5
9 × (25) + 32 = 13°F; TK = 25 + 273 = 248 K
c. TF =5
9 × (273) + 32 = -459°F; TK = 273 + 273 = 0 K
d. TF =5
9 × 801 + 32 = 1470°F; TK = 801 + 273 = 1074 K
48. a. TC = TK 273 = 233 273 = 40.°C
TF =5
9 × TC + 32 =5
9 × (40.) + 32 = 40.°F
b. TC = 4 273 = 269°C; TF =5
9 × (269) + 32 = 452°F
c. TC = 298 273 = 25°C; TF =5
9 × 25 + 32 = 77°F
d. TC = 3680 273 = 3410°C; TF =5
9 × 3410 + 32 = 6170°F
49. TF =5
9 × TC + 32; from the problem, we want the temperature where TF = 2TC.
Substituting:
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REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 13
2TC =5
9 × TC + 32, (0.2)TC = 32, TC =2.0
32 = 160C
TF = 2TC when the temperature in Fahrenheit is 2(160) = 320F. Because all numbers when
solving the equation are exact numbers, the calculated temperatures are also exact numbers.
50. TC =9
5 (TF – 32) =9
5 (72 – 32) = 22C
TC = TK – 273 = 313 – 273 = 40.C
The difference in temperature between Jupiter at 313 K and Earth at 72F is 40.C – 22C =
18C.
51. a. A change in temperature of 160°C equals a
change in temperature of 100°A.
SoA100
C160
is our unit conversion for a
degree change in temperature.
At the freezing point: 0°A = 45°C
Combining these two pieces of information:
TA = (TC + 45°C) ×C160
A100
= (TC + 45°C) ×C8
A5
or TC = TA ×A5
C8
45°C
b. TC = (TF 32) ×9
5 ; TC = TA ×5
8 45 = (TF 32) ×9
5
TF 32 =5
9
45
5
8
TA =25
72
TA 81, TF = TA ×A25
F72
49F
c. TC = TA ×5
8 45 and TC = TA; so TC = TC ×5
8 45,5
3TC = 45, TC = 75°C = 75°A
d. TC = 86°A ×A5
C8
45°C = 93°C; TF = 86°A ×A25
F72
49°F = 199°F = 2.0 × 102°F
e. TA = (45°C + 45°C) ×C8
A5
= 56°A
52. a. A change in temperature of 140C is equal to 50X. Therefore,X50
C140
o
o is the unit con-
version between a degree on the X scale to a degree on the Celsius scale. To account for
the different zero points, 10 must be subtracted from the temperature on the X scale to
get to the Celsius scale. The conversion between X to C is:
TC = TX X50
C140
o
o 10C, TC = TX X5
C14
o
o 10C100oA 115oC
160oC
-45oC
100oA
0oA
2TC =5
9 × TC + 32, (0.2)TC = 32, TC =2.0
32 = 160C
TF = 2TC when the temperature in Fahrenheit is 2(160) = 320F. Because all numbers when
solving the equation are exact numbers, the calculated temperatures are also exact numbers.
50. TC =9
5 (TF – 32) =9
5 (72 – 32) = 22C
TC = TK – 273 = 313 – 273 = 40.C
The difference in temperature between Jupiter at 313 K and Earth at 72F is 40.C – 22C =
18C.
51. a. A change in temperature of 160°C equals a
change in temperature of 100°A.
SoA100
C160
is our unit conversion for a
degree change in temperature.
At the freezing point: 0°A = 45°C
Combining these two pieces of information:
TA = (TC + 45°C) ×C160
A100
= (TC + 45°C) ×C8
A5
or TC = TA ×A5
C8
45°C
b. TC = (TF 32) ×9
5 ; TC = TA ×5
8 45 = (TF 32) ×9
5
TF 32 =5
9
45
5
8
TA =25
72
TA 81, TF = TA ×A25
F72
49F
c. TC = TA ×5
8 45 and TC = TA; so TC = TC ×5
8 45,5
3TC = 45, TC = 75°C = 75°A
d. TC = 86°A ×A5
C8
45°C = 93°C; TF = 86°A ×A25
F72
49°F = 199°F = 2.0 × 102°F
e. TA = (45°C + 45°C) ×C8
A5
= 56°A
52. a. A change in temperature of 140C is equal to 50X. Therefore,X50
C140
o
o is the unit con-
version between a degree on the X scale to a degree on the Celsius scale. To account for
the different zero points, 10 must be subtracted from the temperature on the X scale to
get to the Celsius scale. The conversion between X to C is:
TC = TX X50
C140
o
o 10C, TC = TX X5
C14
o
o 10C100oA 115oC
160oC
-45oC
100oA
0oA
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14 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
The conversion between C to X would be:
TX = (TC + 10C)C14
X5
o
o
b. Assuming 10C andC14
X5
o
o are exact numbers:
TX = (22.0C + 10C)C14
X5
o
o = 11.4X
c. Assuming exact numbers in the temperature conversion formulas:
TC = 58.0X X5
C14
o
o 10C = 152C
TK = 152C + 273 = 425 K
TF =C5
F9
o
o 152C + 32F = 306F
Density
53. Mass = 350 lblb
g6.453
= 1.6 × 105 g; V = 1.2 × 104 in33
in
cm54.2
= 2.0 × 105 cm3
Density =3
35
5
g/cm80.0
cm100.2
g101
volume
mass
Because the material has a density less than water, it will float in water.
54.3
333
cm52.0
g0.2
d;cm52.0)cm50.0(14.3
3
4
r
3
4
V
= 3.8 g/cm3
The ball will sink.
55.333
3
53 cm104.1
m
cm100
km
m1000
km100.714.3
3
4
r
3
4
V
Density =333
36
cm104.1
kg
g1000
kg102
volume
mass
= 1.4 × 106 g/cm3 = 1 × 106 g/cm3
56. V = l × w × h = 2.9 cm × 3.5 cm × 10.0 cm = 1.0 × 102 cm3
d = density =332 cm
g2.6
cm100.1
g0.615
The conversion between C to X would be:
TX = (TC + 10C)C14
X5
o
o
b. Assuming 10C andC14
X5
o
o are exact numbers:
TX = (22.0C + 10C)C14
X5
o
o = 11.4X
c. Assuming exact numbers in the temperature conversion formulas:
TC = 58.0X X5
C14
o
o 10C = 152C
TK = 152C + 273 = 425 K
TF =C5
F9
o
o 152C + 32F = 306F
Density
53. Mass = 350 lblb
g6.453
= 1.6 × 105 g; V = 1.2 × 104 in33
in
cm54.2
= 2.0 × 105 cm3
Density =3
35
5
g/cm80.0
cm100.2
g101
volume
mass
Because the material has a density less than water, it will float in water.
54.3
333
cm52.0
g0.2
d;cm52.0)cm50.0(14.3
3
4
r
3
4
V
= 3.8 g/cm3
The ball will sink.
55.333
3
53 cm104.1
m
cm100
km
m1000
km100.714.3
3
4
r
3
4
V
Density =333
36
cm104.1
kg
g1000
kg102
volume
mass
= 1.4 × 106 g/cm3 = 1 × 106 g/cm3
56. V = l × w × h = 2.9 cm × 3.5 cm × 10.0 cm = 1.0 × 102 cm3
d = density =332 cm
g2.6
cm100.1
g0.615
Loading page 15...
REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 15
57. a. 5.0 caratg51.3
cm1
carat
g200.0 3
= 0.28 cm3
b. 2.8 mLg200.0
carat1
cm
g51.3
mL
cm1
3
3
= 49 carats
58. For ethanol: 100. mL ×mL
g789.0 = 78.9 g
For benzene: 1.00 LmL
g880.0
L
mL1000 = 880. g
Total mass = 78.9 g + 880. g = 959 g
59. V = 21.6 mL 12.7 mL = 8.9 mL; density =mL9.8
g42.33 = 3.8 g/mL = 3.8 g/cm3
60. 5.25 gg5.10
cm1 3
= 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 is more dense than water. Note: 1 mL = 1 cm3.
1.0 mLmL
g6.13
= 14 g of mercury; 1.0 mLmL
g998.0
= 1.0 g of water
c. Same; both represent 19.3 g of substance.
19.3 mLmL
g9982.0
= 19.3 g of water; 1.00 mLmL
g32.19
= 19.3 g of gold
d. 1.0 L of benzene (880 g versus 670 g)
75 mLmL
g96.8
= 670 g of copper; 1.0 LmL
g880.0
L
mL1000 = 880 g of benzene
62. a. 1.50 qtmL
g789.0
L
mL1000
qt0567.1
L1 = 1120 g ethanol
b. 3.5 in33
3
cm
g6.13
in
cm54.2
= 780 g mercury
63. 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. a. H2(g): V = 25.0 g ×g000084.0
cm1 3 = 3.0 × 105 cm3 [H2(g) = hydrogen gas.]
b. H2O(l): V = 25.0 g ×g9982.0
cm1 3 = 25.0 cm3 [H2O(l) = water.]
57. a. 5.0 caratg51.3
cm1
carat
g200.0 3
= 0.28 cm3
b. 2.8 mLg200.0
carat1
cm
g51.3
mL
cm1
3
3
= 49 carats
58. For ethanol: 100. mL ×mL
g789.0 = 78.9 g
For benzene: 1.00 LmL
g880.0
L
mL1000 = 880. g
Total mass = 78.9 g + 880. g = 959 g
59. V = 21.6 mL 12.7 mL = 8.9 mL; density =mL9.8
g42.33 = 3.8 g/mL = 3.8 g/cm3
60. 5.25 gg5.10
cm1 3
= 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 is more dense than water. Note: 1 mL = 1 cm3.
1.0 mLmL
g6.13
= 14 g of mercury; 1.0 mLmL
g998.0
= 1.0 g of water
c. Same; both represent 19.3 g of substance.
19.3 mLmL
g9982.0
= 19.3 g of water; 1.00 mLmL
g32.19
= 19.3 g of gold
d. 1.0 L of benzene (880 g versus 670 g)
75 mLmL
g96.8
= 670 g of copper; 1.0 LmL
g880.0
L
mL1000 = 880 g of benzene
62. a. 1.50 qtmL
g789.0
L
mL1000
qt0567.1
L1 = 1120 g ethanol
b. 3.5 in33
3
cm
g6.13
in
cm54.2
= 780 g mercury
63. 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. a. H2(g): V = 25.0 g ×g000084.0
cm1 3 = 3.0 × 105 cm3 [H2(g) = hydrogen gas.]
b. H2O(l): V = 25.0 g ×g9982.0
cm1 3 = 25.0 cm3 [H2O(l) = water.]
Loading page 16...
16 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
c. Fe(s): V = 25.0 g ×g87.7
cm1 3 = 3.18 cm3 [Fe(s) = iron.]
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 ×g57.22
cm1 3 = 44.3 cm3
44.3 cm3 = 1 × w × h = 4.00 cm × 4.00 cm × h, h = 2.77 cm
66. V = 22 g ×g96.8
cm1 3 = 2.5 cm3; V = πr2 × l, where l = length of the wire
2.5 cm3 = π ×2
2
mm25.0
×2
mm10
cm1
× l, l = 5.1 × 103 cm = 170 ft
Classification and Separation of Matter
67. A gas has molecules that are very far apart from each other, whereas 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 ele-
ments
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.
c. Fe(s): V = 25.0 g ×g87.7
cm1 3 = 3.18 cm3 [Fe(s) = iron.]
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 ×g57.22
cm1 3 = 44.3 cm3
44.3 cm3 = 1 × w × h = 4.00 cm × 4.00 cm × h, h = 2.77 cm
66. V = 22 g ×g96.8
cm1 3 = 2.5 cm3; V = πr2 × l, where l = length of the wire
2.5 cm3 = π ×2
2
mm25.0
×2
mm10
cm1
× l, l = 5.1 × 103 cm = 170 ft
Classification and Separation of Matter
67. A gas has molecules that are very far apart from each other, whereas 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 ele-
ments
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.
Loading page 17...
REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 17
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.)
b. homogeneous (hopefully; if you live in a heavily polluted area, air may be
heterogeneous.)
c. homogeneous d. homogeneous (hopefully, if not polluted)
e. heterogeneous f. heterogeneous
70. a. heterogeneous b. homogeneous
c. heterogeneous d. homogeneous (assuming no imperfections in the glass)
e. heterogeneous (has visibly distinguishable parts)
71. a. pure b. mixture c. mixture d. pure e. mixture (copper and zinc)
f. pure g. mixture h. mixture i. mixture
Iron and uranium are elements. Water (H2O) is a compound because it is made up of two or
more different elements. Table salt is usually a homogeneous mixture composed mostly of
sodium chloride (NaCl), but will usually contain other substances that help absorb water
vapor (an anticaking agent).
72. Initially, a mixture is present. The magnesium and sulfur have only been placed together in
the same container at this point, but no reaction has occurred. When heated, a reaction occurs.
Assuming the magnesium and sulfur had been measured out in exactly the correct ratio for
complete reaction, the remains after heating would be a pure compound composed of
magnesium and sulfur. However, if there were an excess of either magnesium or sulfur, the
remains after reaction would be a mixture of the compound produced and the excess reactant.
73. Chalk is a compound because it loses mass when heated and appears to change into another
substance with different physical properties (the hard chalk turns into a crumbly substance).
74. Because vaporized water is still the same substance as solid water (H2O), no chemical
reaction has occurred. Sublimation is a physical change.
75. 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).
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.)
b. homogeneous (hopefully; if you live in a heavily polluted area, air may be
heterogeneous.)
c. homogeneous d. homogeneous (hopefully, if not polluted)
e. heterogeneous f. heterogeneous
70. a. heterogeneous b. homogeneous
c. heterogeneous d. homogeneous (assuming no imperfections in the glass)
e. heterogeneous (has visibly distinguishable parts)
71. a. pure b. mixture c. mixture d. pure e. mixture (copper and zinc)
f. pure g. mixture h. mixture i. mixture
Iron and uranium are elements. Water (H2O) is a compound because it is made up of two or
more different elements. Table salt is usually a homogeneous mixture composed mostly of
sodium chloride (NaCl), but will usually contain other substances that help absorb water
vapor (an anticaking agent).
72. Initially, a mixture is present. The magnesium and sulfur have only been placed together in
the same container at this point, but no reaction has occurred. When heated, a reaction occurs.
Assuming the magnesium and sulfur had been measured out in exactly the correct ratio for
complete reaction, the remains after heating would be a pure compound composed of
magnesium and sulfur. However, if there were an excess of either magnesium or sulfur, the
remains after reaction would be a mixture of the compound produced and the excess reactant.
73. Chalk is a compound because it loses mass when heated and appears to change into another
substance with different physical properties (the hard chalk turns into a crumbly substance).
74. Because vaporized water is still the same substance as solid water (H2O), no chemical
reaction has occurred. Sublimation is a physical change.
75. 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).
Loading page 18...
18 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
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 that wash away.
c. This is a physical change since all that is happening during the boiling process 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.
76. 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
77. Avogadro’s number of dollars = 6.022 × 1023 dollars/mol dollarspeople107
dollarsmol
dollars106.022
dollarsmol1
9
23
= 8.6 × 1013 = 9 × 1013 dollars/person
1 trillion = 1,000,000,000,000 = 1 × 1012; each person would have 90 trillion dollars.
78. Density 3
cm32.0
g384.0
volume
mass 1.2 g/cm3; from the table, the other ingredient is caffeine.
79. Because each pill is 4.0% Lipitor by mass, for every 100.0 g of pills, there are 4.0 g of Lipitor
present. Note that 100 pills is assumed to be an exact number.
100 pillsg1000
kg1
pillsg0.100
Lipitorg0.4
pill
g5.2 = 0.010 kg Lipitor
80. A chemical change involves the change of one or more substances into other substances
through a reorganization of the atoms. A physical change involves the change in the form of
a substance, but not its chemical composition.
a. physical change (Just smaller pieces of the same substance.)
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 that wash away.
c. This is a physical change since all that is happening during the boiling process 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.
76. 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
77. Avogadro’s number of dollars = 6.022 × 1023 dollars/mol dollarspeople107
dollarsmol
dollars106.022
dollarsmol1
9
23
= 8.6 × 1013 = 9 × 1013 dollars/person
1 trillion = 1,000,000,000,000 = 1 × 1012; each person would have 90 trillion dollars.
78. Density 3
cm32.0
g384.0
volume
mass 1.2 g/cm3; from the table, the other ingredient is caffeine.
79. Because each pill is 4.0% Lipitor by mass, for every 100.0 g of pills, there are 4.0 g of Lipitor
present. Note that 100 pills is assumed to be an exact number.
100 pillsg1000
kg1
pillsg0.100
Lipitorg0.4
pill
g5.2 = 0.010 kg Lipitor
80. A chemical change involves the change of one or more substances into other substances
through a reorganization of the atoms. A physical change involves the change in the form of
a substance, but not its chemical composition.
a. physical change (Just smaller pieces of the same substance.)
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REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 19
b. chemical change (Chemical reactions occur.)
c. chemical change (Bonds are broken.)
d. chemical change (Bonds are broken.)
e. physical change (Water is changed from a liquid to a gas.)
f. physical change (Chemical composition does not change.)
81. Total volume =
m
cm100
m.300
m
cm100
m.200 × 4.0 cm = 2.4 × 109 cm3
Volume of topsoil covered by 1 bag =
in
cm54.2
in0.1
in
cm54.2
ft
in12
ft.10
22
2
= 2.4 × 104 cm3
2.4 × 109 cm334 cm104.2
bag1
= 1.0 × 105 bags topsoil
82. 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 Lg1000
kg1
cm
g32.91
L
cm0001
3
3
= 19.32 kg (= 42.59 lb)
It wouldn't be easy to play catch with the idol because it would have a mass of over 40
pounds.
83. 1 light year = 1 yrs
mi000,186
min
s60
h
min60
day
h24
yr
day365 = 5.87 × 1012 miles
9.6 parsecskm
m1000
mi
km609.1
yrlight
mi1087.5
parsec
yrlight26.3 12
= 3.0 × 1017 m
84.mi
ft5280
h
mi65
min60
h1
s60
min1
s1 = 95.3 ft = 100 ft
If you take your eyes off the road for one second traveling at 65 mph, your car travels
approximately 100 feet.
85. 18.5 cm ×cm25.5
F0.10 o = 35.2F increase; Tfinal = 98.6 + 35.2 = 133.8F
TC = 5/9 (133.8 – 32) = 56.56C
b. chemical change (Chemical reactions occur.)
c. chemical change (Bonds are broken.)
d. chemical change (Bonds are broken.)
e. physical change (Water is changed from a liquid to a gas.)
f. physical change (Chemical composition does not change.)
81. Total volume =
m
cm100
m.300
m
cm100
m.200 × 4.0 cm = 2.4 × 109 cm3
Volume of topsoil covered by 1 bag =
in
cm54.2
in0.1
in
cm54.2
ft
in12
ft.10
22
2
= 2.4 × 104 cm3
2.4 × 109 cm334 cm104.2
bag1
= 1.0 × 105 bags topsoil
82. 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 Lg1000
kg1
cm
g32.91
L
cm0001
3
3
= 19.32 kg (= 42.59 lb)
It wouldn't be easy to play catch with the idol because it would have a mass of over 40
pounds.
83. 1 light year = 1 yrs
mi000,186
min
s60
h
min60
day
h24
yr
day365 = 5.87 × 1012 miles
9.6 parsecskm
m1000
mi
km609.1
yrlight
mi1087.5
parsec
yrlight26.3 12
= 3.0 × 1017 m
84.mi
ft5280
h
mi65
min60
h1
s60
min1
s1 = 95.3 ft = 100 ft
If you take your eyes off the road for one second traveling at 65 mph, your car travels
approximately 100 feet.
85. 18.5 cm ×cm25.5
F0.10 o = 35.2F increase; Tfinal = 98.6 + 35.2 = 133.8F
TC = 5/9 (133.8 – 32) = 56.56C
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20 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
86. Massbenzene = 58.80 g 25.00 g = 33.80 g; Vbenzene = 33.80 gg880.0
cm1 3
= 38.4 cm3
Vsolid = 50.0 cm3 38.4 cm3 = 11.6 cm3; density =3
cm6.11
g00.25 = 2.16 g/cm3
87. 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.
88. Circumference = c = 2πr; V =2
333
6
c
2
c
3
4
3
r4
Largest density =2
3
π6
)in00.9(
oz25.5 =3
in3.12
oz25.5 =3
in
oz427.0
Smallest density =2
3
π6
)in25.9(
oz00.5 =3
in4.13
oz00.5 =3
in
oz73.0
Maximum range is:3
in
oz)427.0373.0( or 0.40 ±0.03 oz/in3
Uncertainty is in 2nd decimal place.
89. V = Vfinal Vinitial; d =333 cm4.3
g90.28
cm4.6cm8.9
g90.28
= 8.5 g/cm3
dmax =min
max
V
mass ; we get Vmin from 9.7 cm3 6.5 cm3 = 3.2 cm3.
dmax =33 cm
g0.9
cm2.3
g93.28 ; dmin =max
min
V
mass =333 cm
g0.8
cm3.6cm9.9
g87.28
The density is 8.5 ±0.5 g/cm3.
86. Massbenzene = 58.80 g 25.00 g = 33.80 g; Vbenzene = 33.80 gg880.0
cm1 3
= 38.4 cm3
Vsolid = 50.0 cm3 38.4 cm3 = 11.6 cm3; density =3
cm6.11
g00.25 = 2.16 g/cm3
87. 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.
88. Circumference = c = 2πr; V =2
333
6
c
2
c
3
4
3
r4
Largest density =2
3
π6
)in00.9(
oz25.5 =3
in3.12
oz25.5 =3
in
oz427.0
Smallest density =2
3
π6
)in25.9(
oz00.5 =3
in4.13
oz00.5 =3
in
oz73.0
Maximum range is:3
in
oz)427.0373.0( or 0.40 ±0.03 oz/in3
Uncertainty is in 2nd decimal place.
89. V = Vfinal Vinitial; d =333 cm4.3
g90.28
cm4.6cm8.9
g90.28
= 8.5 g/cm3
dmax =min
max
V
mass ; we get Vmin from 9.7 cm3 6.5 cm3 = 3.2 cm3.
dmax =33 cm
g0.9
cm2.3
g93.28 ; dmin =max
min
V
mass =333 cm
g0.8
cm3.6cm9.9
g87.28
The density is 8.5 ±0.5 g/cm3.
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REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 21
90. We need to calculate the maximum and minimum values of the density, given the uncertainty
in each measurement. The maximum value is:
dmax =33 cm03.0cm00.25
g002.0g625.19
=3
cm97.24
g627.19 = 0.7860 g/cm3
The minimum value of the density is:
dmin =33 cm03.0cm00.25
g002.0g625.19
=3
cm03.25
g623.19 = 0.7840 g/cm3
The density of the liquid is between 0.7840 and 0.7860 g/cm3. These measurements are
sufficiently precise to distinguish between ethanol (d = 0.789 g/cm3) and isopropyl alcohol
(d = 0.785 g/cm3).
91. Let x = mass of copper and y = mass of silver.
105.0 g = x + y and 10.12 mL =5.1096.8
yx ; solving:
5.10
0.105
96.8
12.10 xx
× 8.96 × 10.5, 952.1 = (10.5)x + 940.8 (8.96)x
(carrying 1 extra sig. fig.)
11.3 = (1.54)x, x = 7.3 g; mass % Cu =g0.105
g3.7 × 100 = 7.0% Cu
92. In general, glassware is estimated to one place past the markings.
a. 128.7 mL glassware b. 18 mL glassware c. 23.45 mL glassware
read to tenth’s place read to one’s place read to two decimal places
Total volume = 128.7 + 18 + 23.45 = 170.15 = 170. (Due to 18, the sum would be
known only to the ones place.)130
129
128
12730
20
1023
24
90. We need to calculate the maximum and minimum values of the density, given the uncertainty
in each measurement. The maximum value is:
dmax =33 cm03.0cm00.25
g002.0g625.19
=3
cm97.24
g627.19 = 0.7860 g/cm3
The minimum value of the density is:
dmin =33 cm03.0cm00.25
g002.0g625.19
=3
cm03.25
g623.19 = 0.7840 g/cm3
The density of the liquid is between 0.7840 and 0.7860 g/cm3. These measurements are
sufficiently precise to distinguish between ethanol (d = 0.789 g/cm3) and isopropyl alcohol
(d = 0.785 g/cm3).
91. Let x = mass of copper and y = mass of silver.
105.0 g = x + y and 10.12 mL =5.1096.8
yx ; solving:
5.10
0.105
96.8
12.10 xx
× 8.96 × 10.5, 952.1 = (10.5)x + 940.8 (8.96)x
(carrying 1 extra sig. fig.)
11.3 = (1.54)x, x = 7.3 g; mass % Cu =g0.105
g3.7 × 100 = 7.0% Cu
92. In general, glassware is estimated to one place past the markings.
a. 128.7 mL glassware b. 18 mL glassware c. 23.45 mL glassware
read to tenth’s place read to one’s place read to two decimal places
Total volume = 128.7 + 18 + 23.45 = 170.15 = 170. (Due to 18, the sum would be
known only to the ones place.)130
129
128
12730
20
1023
24
Loading page 22...
22 REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY
93. Ball A: PE = mgz = 2.00 kg ×2
s
m81.9 × 10.0 m =2
2
s
mkg196 = 196 J
At point I: All this energy is transferred to ball B. All of B's energy is kinetic energy at this
point. Etotal = KE = 196 J. At point II, the sum of the total energy will equal
196 J.
At point II: PE = mgz = 4.00 kg ×2
s
m81.9 × 3.00 m = 118 J
KE = Etotal PE = 196 J − 118 J = 78 J
94. 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) =mL0.10
g5171.14 = 1.45 g/mL
Density of methanol =mL00.10
g9125.7 = 0.7913 g/mL
Density of sand particles (air absent) =mL6.7
g5171.14 = 1.9 g/mL
ChemWork Problems
95. 4,145 mi ×fathom100
lengthcable1
ft6
fathom1
mi
ft5280 = 3.648 × 104 cable lengths
4,145 mi ×km
m1000
mi0.62137
km1 = 6.671 × 106 m
3.648 × 104 cable lengths ×lengthscable10
milenautical1 = 3,648 nautical miles
96.km1
m1000
mi0.6214
km1
s119.2
mi1.25 = 16.9 m/s
97. TC =9
5 (TF 32) =9
5 (134°F 32) = 56.7°C ; phosphorus would be a liquid.
93. Ball A: PE = mgz = 2.00 kg ×2
s
m81.9 × 10.0 m =2
2
s
mkg196 = 196 J
At point I: All this energy is transferred to ball B. All of B's energy is kinetic energy at this
point. Etotal = KE = 196 J. At point II, the sum of the total energy will equal
196 J.
At point II: PE = mgz = 4.00 kg ×2
s
m81.9 × 3.00 m = 118 J
KE = Etotal PE = 196 J − 118 J = 78 J
94. 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) =mL0.10
g5171.14 = 1.45 g/mL
Density of methanol =mL00.10
g9125.7 = 0.7913 g/mL
Density of sand particles (air absent) =mL6.7
g5171.14 = 1.9 g/mL
ChemWork Problems
95. 4,145 mi ×fathom100
lengthcable1
ft6
fathom1
mi
ft5280 = 3.648 × 104 cable lengths
4,145 mi ×km
m1000
mi0.62137
km1 = 6.671 × 106 m
3.648 × 104 cable lengths ×lengthscable10
milenautical1 = 3,648 nautical miles
96.km1
m1000
mi0.6214
km1
s119.2
mi1.25 = 16.9 m/s
97. TC =9
5 (TF 32) =9
5 (134°F 32) = 56.7°C ; phosphorus would be a liquid.
Loading page 23...
REVIEW MEASUREMENT AND CALCULATIONS IN CHEMISTRY 23
98.324
3
12
3 cm101.4
m
cm100
pm
m101
pm6914.3
3
4
rπ
3
4
V
Density =324
23
cm101.4
g103.35
volume
mass
= 24 g/cm3
99. a. False; sugar is generally considered to be the pure compound sucrose, C12H22O11.
b. False; elements and compounds are pure substances.
c. True; air is a mixture of mostly nitrogen and oxygen gases.
d. False; gasoline has many additives, so it is a mixture.
e. True; compounds are broken down to elements by chemical change.
100. The rusting of iron is the only change listed where chemical formulas change, so it is the only
chemical change. The others are physical properties. Note that the red glow of a platinum
wire assumes no reaction between platinum and oxygen; the red glow is just hot Pt.
98.324
3
12
3 cm101.4
m
cm100
pm
m101
pm6914.3
3
4
rπ
3
4
V
Density =324
23
cm101.4
g103.35
volume
mass
= 24 g/cm3
99. a. False; sugar is generally considered to be the pure compound sucrose, C12H22O11.
b. False; elements and compounds are pure substances.
c. True; air is a mixture of mostly nitrogen and oxygen gases.
d. False; gasoline has many additives, so it is a mixture.
e. True; compounds are broken down to elements by chemical change.
100. The rusting of iron is the only change listed where chemical formulas change, so it is the only
chemical change. The others are physical properties. Note that the red glow of a platinum
wire assumes no reaction between platinum and oxygen; the red glow is just hot Pt.
Loading page 24...
24
CHAPTER 1
CHEMICAL FOUNDATIONS
Questions
11. 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.
12. a. At 8 a.m., approximately 57 cars pass through the intersection per hour.
b. At 12 a.m. (midnight), only 1 or 2 cars pass through the intersection per hour.
c. Traffic at the intersection is limited to less than 10 cars per hour from 8 p.m. to 5 a.m.
Starting at 6 a.m., there is a steady increase in traffic through the intersection, peaking at
8 a.m. when approximately 57 cars pass per hour. Past 8 a.m. traffic moderates to about
40 cars through the intersection per hour until noon, and then decreases to 21 cars per
hour by 3 p.m. Past 3 p.m. traffic steadily increases to a peak of 52 cars per hour at 5
p.m., and then steadily decreases to the overnight level of less than 10 cars through the
intersection per hour.
d. The traffic pattern through the intersection is directly related to the work schedules of the
general population as well as to the store hours of the businesses in downtown.
e. Run the same experiment on a Sunday, when most of the general population doesn’t
work and when a significant number of downtown stores are closed in the morning.
13. 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 turns into a theory (model). However, scientists continue to
perform experiments to refine or replace existing theories. Hence, science is a dynamic or
active process, not a static one.
14. A compound will always contain the same numbers (and types) of atoms. A given amount of
hydrogen will react only with a specific amount of oxygen. Any excess oxygen will remain
unreacted.
15. Law of conservation of mass: Mass is neither created nor destroyed. The total mass before a
chemical reaction always equals the total mass after a chemical reaction.
CHAPTER 1
CHEMICAL FOUNDATIONS
Questions
11. 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.
12. a. At 8 a.m., approximately 57 cars pass through the intersection per hour.
b. At 12 a.m. (midnight), only 1 or 2 cars pass through the intersection per hour.
c. Traffic at the intersection is limited to less than 10 cars per hour from 8 p.m. to 5 a.m.
Starting at 6 a.m., there is a steady increase in traffic through the intersection, peaking at
8 a.m. when approximately 57 cars pass per hour. Past 8 a.m. traffic moderates to about
40 cars through the intersection per hour until noon, and then decreases to 21 cars per
hour by 3 p.m. Past 3 p.m. traffic steadily increases to a peak of 52 cars per hour at 5
p.m., and then steadily decreases to the overnight level of less than 10 cars through the
intersection per hour.
d. The traffic pattern through the intersection is directly related to the work schedules of the
general population as well as to the store hours of the businesses in downtown.
e. Run the same experiment on a Sunday, when most of the general population doesn’t
work and when a significant number of downtown stores are closed in the morning.
13. 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 turns into a theory (model). However, scientists continue to
perform experiments to refine or replace existing theories. Hence, science is a dynamic or
active process, not a static one.
14. A compound will always contain the same numbers (and types) of atoms. A given amount of
hydrogen will react only with a specific amount of oxygen. Any excess oxygen will remain
unreacted.
15. Law of conservation of mass: Mass is neither created nor destroyed. The total mass before a
chemical reaction always equals the total mass after a chemical reaction.
Loading page 25...
CHAPTER 1 CHEMICAL FOUNDATIONS 25
Law of definite proportion: A given compound always contains exactly the same proportion
of elements by mass. For example, water is always 1 g H for every 8 g oxygen.
Law of multiple proportions: When two elements form a series of compounds, the ratios of
the mass of the second element that combine with 1 g of the first element always can be
reduced to small whole numbers. For CO2 and CO discussed in Section 1.4, the mass ratios
of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio.
16. Yes, 1.0 g H would react with 37.0 g 37Cl, and 1.0 g H would react with 35.0 g 35Cl.
No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37Cl and 1 g H/35 g Cl for 35Cl.
As long as we had pure 37Cl or pure 35Cl, the ratios will always hold. If we have a mixture
(such as the natural abundance of chlorine), the ratio will also be constant as long as the
composition of the mixture of the two isotopes does not change.
17. Natural niacin and commercially produced niacin have the exact same formula of C6H5NO2.
Therefore, both sources produce niacin having an identical nutritional value. There may be
other compounds present in natural niacin that would increase the nutritional value, but the
nutritional value due to just niacin is identical to the commercially produced niacin.
18. a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons
and neutrons, which can be broken down into quarks. For our purpose, electrons,
neutrons, and protons are the key smaller parts of an atom.
b. All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have
0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a
neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions
were included, then different ions/atoms of H could have different numbers of electrons.
c. Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2
protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and
a helium atom. Tritium (3H) and 4He both have 2 neutrons. Assuming neutral atoms, then
the number of electrons will be 1 for hydrogen and 2 for helium.
d. Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2 is always 1
g hydrogen for every 16 g of O present. These are distinctly different compounds, each
with its own unique relative number and types of atoms present.
e. A chemical equation involves a reorganization of the atoms. Bonds are broken between
atoms in the reactants, and new bonds are formed in the products. The number and types
of atoms between reactants and products do not change. Because atoms are conserved in
a chemical reaction, mass is also conserved.
19. J. J. Thomson’s study of cathode-ray tubes led him to postulate the existence of negatively
charged particles that we now call electrons. Ernest Rutherford and his alpha bombardment of
metal foil experiments led him to postulate the nuclear atoman atom with a tiny dense center
of positive charge (the nucleus) with electrons moving about the nucleus at relatively large
distances away; the distance is so large that an atom is mostly empty space.
Law of definite proportion: A given compound always contains exactly the same proportion
of elements by mass. For example, water is always 1 g H for every 8 g oxygen.
Law of multiple proportions: When two elements form a series of compounds, the ratios of
the mass of the second element that combine with 1 g of the first element always can be
reduced to small whole numbers. For CO2 and CO discussed in Section 1.4, the mass ratios
of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio.
16. Yes, 1.0 g H would react with 37.0 g 37Cl, and 1.0 g H would react with 35.0 g 35Cl.
No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37Cl and 1 g H/35 g Cl for 35Cl.
As long as we had pure 37Cl or pure 35Cl, the ratios will always hold. If we have a mixture
(such as the natural abundance of chlorine), the ratio will also be constant as long as the
composition of the mixture of the two isotopes does not change.
17. Natural niacin and commercially produced niacin have the exact same formula of C6H5NO2.
Therefore, both sources produce niacin having an identical nutritional value. There may be
other compounds present in natural niacin that would increase the nutritional value, but the
nutritional value due to just niacin is identical to the commercially produced niacin.
18. a. The smaller parts are electrons and the nucleus. The nucleus is broken down into protons
and neutrons, which can be broken down into quarks. For our purpose, electrons,
neutrons, and protons are the key smaller parts of an atom.
b. All atoms of hydrogen have 1 proton in the nucleus. Different isotopes of hydrogen have
0, 1, or 2 neutrons in the nucleus. Because we are talking about atoms, this implies a
neutral charge, which dictates 1 electron present for all hydrogen atoms. If charged ions
were included, then different ions/atoms of H could have different numbers of electrons.
c. Hydrogen atoms always have 1 proton in the nucleus, and helium atoms always have 2
protons in the nucleus. The number of neutrons can be the same for a hydrogen atom and
a helium atom. Tritium (3H) and 4He both have 2 neutrons. Assuming neutral atoms, then
the number of electrons will be 1 for hydrogen and 2 for helium.
d. Water (H2O) is always 1 g hydrogen for every 8 g of O present, whereas H2O2 is always 1
g hydrogen for every 16 g of O present. These are distinctly different compounds, each
with its own unique relative number and types of atoms present.
e. A chemical equation involves a reorganization of the atoms. Bonds are broken between
atoms in the reactants, and new bonds are formed in the products. The number and types
of atoms between reactants and products do not change. Because atoms are conserved in
a chemical reaction, mass is also conserved.
19. J. J. Thomson’s study of cathode-ray tubes led him to postulate the existence of negatively
charged particles that we now call electrons. Ernest Rutherford and his alpha bombardment of
metal foil experiments led him to postulate the nuclear atoman atom with a tiny dense center
of positive charge (the nucleus) with electrons moving about the nucleus at relatively large
distances away; the distance is so large that an atom is mostly empty space.
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CHAPTER 1 CHEMICAL FOUNDATIONS26
20. The atom is composed of a tiny dense nucleus containing most of the mass of the atom. The
nucleus itself is composed of neutrons and protons. Neutrons have a mass slightly larger than
that of a proton and have no charge. Protons, on the other hand, have a 1+ relative charge as
compared to the 1 charged electrons; the electrons move about the nucleus at relatively large
distances. The volume of space that the electrons move about is so large, as compared to the
nucleus, that we say an atom is mostly empty space.
21. The number and arrangement of electrons in an atom determine how the atom will react with
other atoms. The electrons determine the chemical properties of an atom. The number of
neutrons present determines the isotope identity.
22. Density = mass/volume; if the volumes are assumed equal, then the much more massive
proton would have a much larger density than the relatively light electron.
23. For lighter, stable isotopes, the number of protons in the nucleus is about equal to the number
of neutrons. When the number of protons and neutrons is equal to each other, the mass
number (protons + neutrons) will be twice the atomic number (protons). Therefore, for
lighter isotopes, the ratio of the mass number to the atomic number is close to 2. For
example, consider 28Si, which has 14 protons and (28 – 14 =) 14 neutrons. Here, the mass
number to atomic number ratio is 28/14 = 2.0. For heavier isotopes, there are more neutrons
than protons in the nucleus. Therefore, the ratio of the mass number to the atomic number
increases steadily upward from 2 as the isotopes get heavier and heavier. For example, 238U
has 92 protons and (238 – 92 =) 146 neutrons. The ratio of the mass number to the atomic
number for 238U is 238/92 = 2.6.
24. Some elements exist as molecular substances. That is, hydrogen normally exists as H2
molecules, not single hydrogen atoms. The same is true for N2, O2, F2, Cl2, Br2, and I2.
Exercises
Development of the Atomic Theory
25. a. The composition of a substance depends on the numbers of atoms of each element
making up the compound (depends on the formula of the compound) and not on the
composition of the mixture from which it was formed.
b. Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at
constant temperature and pressure. H2(g) + Cl2(g) → 2 HCl(g). From the balanced
equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted.
26. Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at
constant temperature and pressure. Here, 1 volume of N2 reacts with 3 volumes of H2 to
produce 2 volumes of the gaseous product or in terms of molecule ratios:
1 N2 + 3 H2 2 product
In order for the equation to be balanced, the product must be NH3.
20. The atom is composed of a tiny dense nucleus containing most of the mass of the atom. The
nucleus itself is composed of neutrons and protons. Neutrons have a mass slightly larger than
that of a proton and have no charge. Protons, on the other hand, have a 1+ relative charge as
compared to the 1 charged electrons; the electrons move about the nucleus at relatively large
distances. The volume of space that the electrons move about is so large, as compared to the
nucleus, that we say an atom is mostly empty space.
21. The number and arrangement of electrons in an atom determine how the atom will react with
other atoms. The electrons determine the chemical properties of an atom. The number of
neutrons present determines the isotope identity.
22. Density = mass/volume; if the volumes are assumed equal, then the much more massive
proton would have a much larger density than the relatively light electron.
23. For lighter, stable isotopes, the number of protons in the nucleus is about equal to the number
of neutrons. When the number of protons and neutrons is equal to each other, the mass
number (protons + neutrons) will be twice the atomic number (protons). Therefore, for
lighter isotopes, the ratio of the mass number to the atomic number is close to 2. For
example, consider 28Si, which has 14 protons and (28 – 14 =) 14 neutrons. Here, the mass
number to atomic number ratio is 28/14 = 2.0. For heavier isotopes, there are more neutrons
than protons in the nucleus. Therefore, the ratio of the mass number to the atomic number
increases steadily upward from 2 as the isotopes get heavier and heavier. For example, 238U
has 92 protons and (238 – 92 =) 146 neutrons. The ratio of the mass number to the atomic
number for 238U is 238/92 = 2.6.
24. Some elements exist as molecular substances. That is, hydrogen normally exists as H2
molecules, not single hydrogen atoms. The same is true for N2, O2, F2, Cl2, Br2, and I2.
Exercises
Development of the Atomic Theory
25. a. The composition of a substance depends on the numbers of atoms of each element
making up the compound (depends on the formula of the compound) and not on the
composition of the mixture from which it was formed.
b. Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at
constant temperature and pressure. H2(g) + Cl2(g) → 2 HCl(g). From the balanced
equation, the volume of HCl produced will be twice the volume of H2 (or Cl2) reacted.
26. Avogadro’s hypothesis (law) implies that volume ratios are equal to molecule ratios at
constant temperature and pressure. Here, 1 volume of N2 reacts with 3 volumes of H2 to
produce 2 volumes of the gaseous product or in terms of molecule ratios:
1 N2 + 3 H2 2 product
In order for the equation to be balanced, the product must be NH3.
Loading page 27...
CHAPTER 1 CHEMICAL FOUNDATIONS 27
27. From the law of definite proportions, a given compound always contains exactly the same
proportion of elements by mass. The first sample of chloroform has a total mass of 12.0 g C
+ 106.4 g Cl + 1.01 g H = 119.41 g (carrying extra significant figures). The mass percent of
carbon in this sample of chloroform is:totalg41.119
Cg0.12
× 100 = 10.05% C by mass
From the law of definite proportions, the second sample of chloroform must also contain
10.05% C by mass. Let x = mass of chloroform in the second sample:x
Cg0.30
× 100 = 10.05, x = 299 g chloroform
28. A compound will always have a constant composition by mass. From the initial data given,
the mass ratio of H : S : O in sulfuric acid (H2SO4) is:02.2
00.64
:
02.2
07.32
:
02.2
02.2
= 1 : 15.9 : 31.7
If we have 7.27 g H, then we will have 7.27 × 15.9 = 116 g S and 7.27 × 31.7 = 230. g O in
the second sample of H2SO4.
29. Hydrazine: 1.44 ×1
10 g H/g N; ammonia: 2.16 ×1
10 g H/g N; hydrogen azide:
2.40 ×2
10 g H/g N. Let's try all of the ratios:0240.0
144.0
= 6.00;0240.0
216.0 = 9.00;0240.0
0240.0 = 1.00;144.0
216.0 = 1.50 =2
3
All the masses of hydrogen in these three compounds can be expressed as simple whole-
number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios
6 : 9 : 1.
30. The law of multiple proportions does not involve looking at the ratio of the mass of one
element with the total mass of the compounds. To illustrate the law of multiple proportions,
we compare the mass of carbon that combines with 1.0 g of oxygen in each compound:
compound 1: 27.2 g C and 72.8 g O (100.0 27.2 = mass O)
compound 2: 42.9 g C and 57.1 g O (100.0 42.9 = mass O)
The mass of carbon that combines with 1.0 g of oxygen is:
compound 1:Og8.72
Cg2.27 = 0.374 g C/g O
compound 2:Og1.57
Cg9.42 = 0.751 g C/g O1
2
374.0
751.0
; this supports the law of multiple proportions because this carbon ratio is a whole
number.
27. From the law of definite proportions, a given compound always contains exactly the same
proportion of elements by mass. The first sample of chloroform has a total mass of 12.0 g C
+ 106.4 g Cl + 1.01 g H = 119.41 g (carrying extra significant figures). The mass percent of
carbon in this sample of chloroform is:totalg41.119
Cg0.12
× 100 = 10.05% C by mass
From the law of definite proportions, the second sample of chloroform must also contain
10.05% C by mass. Let x = mass of chloroform in the second sample:x
Cg0.30
× 100 = 10.05, x = 299 g chloroform
28. A compound will always have a constant composition by mass. From the initial data given,
the mass ratio of H : S : O in sulfuric acid (H2SO4) is:02.2
00.64
:
02.2
07.32
:
02.2
02.2
= 1 : 15.9 : 31.7
If we have 7.27 g H, then we will have 7.27 × 15.9 = 116 g S and 7.27 × 31.7 = 230. g O in
the second sample of H2SO4.
29. Hydrazine: 1.44 ×1
10 g H/g N; ammonia: 2.16 ×1
10 g H/g N; hydrogen azide:
2.40 ×2
10 g H/g N. Let's try all of the ratios:0240.0
144.0
= 6.00;0240.0
216.0 = 9.00;0240.0
0240.0 = 1.00;144.0
216.0 = 1.50 =2
3
All the masses of hydrogen in these three compounds can be expressed as simple whole-
number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios
6 : 9 : 1.
30. The law of multiple proportions does not involve looking at the ratio of the mass of one
element with the total mass of the compounds. To illustrate the law of multiple proportions,
we compare the mass of carbon that combines with 1.0 g of oxygen in each compound:
compound 1: 27.2 g C and 72.8 g O (100.0 27.2 = mass O)
compound 2: 42.9 g C and 57.1 g O (100.0 42.9 = mass O)
The mass of carbon that combines with 1.0 g of oxygen is:
compound 1:Og8.72
Cg2.27 = 0.374 g C/g O
compound 2:Og1.57
Cg9.42 = 0.751 g C/g O1
2
374.0
751.0
; this supports the law of multiple proportions because this carbon ratio is a whole
number.
Loading page 28...
CHAPTER 1 CHEMICAL FOUNDATIONS28
31. For CO and CO2, it is easiest to concentrate on the mass of oxygen that combines with 1 g of
carbon. From the formulas (two oxygen atoms per carbon atom in CO2 versus one oxygen
atom per carbon atom in CO), CO2 will have twice the mass of oxygen that combines per
gram of carbon as compared to CO. For CO2 and C3O2, it is easiest to concentrate on the
mass of carbon that combines with 1 g of oxygen. From the formulas (three carbon atoms per
two oxygen atoms in C3O2 versus one carbon atom per two oxygen atoms in CO2), C3O2 will
have three times the mass of carbon that combines per gram of oxygen as compared to CO2.
As expected, the mass ratios are whole numbers as predicted by the law of multiple
proportions.
32. Compound I:Qg00.1
Rg67.4
Qg00.3
Rg0.14 ; compound II:Qg00.1
Rg56.1
Qg50.4
Rg00.7
The ratio of the masses of R that combine with 1.00 g Q is:56.1
67.4 = 2.99 ≈ 3
As expected from the law of multiple proportions, this ratio is a small whole number.
Because compound I contains three times the mass of R per gram of Q as compared with
compound II (RQ), the formula of compound I should be R3Q.
33. Mass is conserved in a chemical reaction because atoms are conserved. Chemical reactions
involve the reorganization of atoms, so formulas change in a chemical reaction, but the
number and types of atoms do not change. Because the atoms do not change in a chemical
reaction, mass must not change. In this equation we have two oxygen atoms and four
hydrogen atoms both before and after the reaction occurs.
34. Mass is conserved in a chemical reaction.
ethanol + oxygen water + carbon dioxide
Mass: 46.0 g 96.0 g 54.0 g ?
Mass of reactants = 46.0 + 96.0 = 142.0 g = mass of products
142.0 g = 54.0 g + mass of CO2, mass of CO2 = 142.0 – 54.0 = 88.0 g
35. To get the atomic mass of H to be 1.00, we divide the mass of hydrogen that reacts with 1.00
g of oxygen by 0.126; that is,126.0
126.0 = 1.00. To get Na, Mg, and O on the same scale, we do
the same division.
Na:126.0
875.2 = 22.8; Mg:126.0
500.1 = 11.9; O:126.0
00.1 = 7.94
H O Na Mg
Relative value 1.00 7.94 22.8 11.9
Accepted value 1.008 16.00 22.99 24.31
For your information, the atomic masses of O and Mg are incorrect. The atomic masses of H
and Na are close to the values given in the periodic table. Something must be wrong about
the assumed formulas of the compounds. It turns out the correct formulas are H2O, Na2O,
and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H.
31. For CO and CO2, it is easiest to concentrate on the mass of oxygen that combines with 1 g of
carbon. From the formulas (two oxygen atoms per carbon atom in CO2 versus one oxygen
atom per carbon atom in CO), CO2 will have twice the mass of oxygen that combines per
gram of carbon as compared to CO. For CO2 and C3O2, it is easiest to concentrate on the
mass of carbon that combines with 1 g of oxygen. From the formulas (three carbon atoms per
two oxygen atoms in C3O2 versus one carbon atom per two oxygen atoms in CO2), C3O2 will
have three times the mass of carbon that combines per gram of oxygen as compared to CO2.
As expected, the mass ratios are whole numbers as predicted by the law of multiple
proportions.
32. Compound I:Qg00.1
Rg67.4
Qg00.3
Rg0.14 ; compound II:Qg00.1
Rg56.1
Qg50.4
Rg00.7
The ratio of the masses of R that combine with 1.00 g Q is:56.1
67.4 = 2.99 ≈ 3
As expected from the law of multiple proportions, this ratio is a small whole number.
Because compound I contains three times the mass of R per gram of Q as compared with
compound II (RQ), the formula of compound I should be R3Q.
33. Mass is conserved in a chemical reaction because atoms are conserved. Chemical reactions
involve the reorganization of atoms, so formulas change in a chemical reaction, but the
number and types of atoms do not change. Because the atoms do not change in a chemical
reaction, mass must not change. In this equation we have two oxygen atoms and four
hydrogen atoms both before and after the reaction occurs.
34. Mass is conserved in a chemical reaction.
ethanol + oxygen water + carbon dioxide
Mass: 46.0 g 96.0 g 54.0 g ?
Mass of reactants = 46.0 + 96.0 = 142.0 g = mass of products
142.0 g = 54.0 g + mass of CO2, mass of CO2 = 142.0 – 54.0 = 88.0 g
35. To get the atomic mass of H to be 1.00, we divide the mass of hydrogen that reacts with 1.00
g of oxygen by 0.126; that is,126.0
126.0 = 1.00. To get Na, Mg, and O on the same scale, we do
the same division.
Na:126.0
875.2 = 22.8; Mg:126.0
500.1 = 11.9; O:126.0
00.1 = 7.94
H O Na Mg
Relative value 1.00 7.94 22.8 11.9
Accepted value 1.008 16.00 22.99 24.31
For your information, the atomic masses of O and Mg are incorrect. The atomic masses of H
and Na are close to the values given in the periodic table. Something must be wrong about
the assumed formulas of the compounds. It turns out the correct formulas are H2O, Na2O,
and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H.
Loading page 29...
CHAPTER 1 CHEMICAL FOUNDATIONS 29
36. If the formula is InO, then one atomic mass of In would combine with one atomic mass of O,
or:Og000.1
Ing784.4
00.16
A
, A = atomic mass of In = 76.54
If the formula is In2O3, then two times the atomic mass of In will combine with three times
the atomic mass of O, or:Og000.1
Ing784.4
00.16)3(
A2
, A = atomic mass of In = 114.8
The latter number is the atomic mass of In used in the modern periodic table.
The Nature of the Atom
37. From section 1-7, the nucleus has “a diameter of about 1013 cm” and the electrons “move
about the nucleus at an average distance of about 108 cm from it.” We will use these
statements to help determine the densities. Density of hydrogen nucleus (contains one proton
only):
Vnucleus =3403143 cm105)cm105()14.3(
3
4
r
3
4
d = density =315
340
24
g/cm103
cm105
g1067.1
Density of H atom (contains one proton and one electron):
Vatom =32438 cm104)cm101()14.3(
3
4
d =3
324
2824
g/cm0.4
cm104
g109g1067.1
38. Because electrons move about the nucleus at an average distance of about 1 ×8
10 cm, the
diameter of an atom will be about 2 ×8
10 cm. Let's set up a ratio:cm102
cm101
modelofdiameter
mm1
atomofdiameter
nucleusofdiameter
8
13
; solving:
diameter of model = 2 × 105 mm = 200 m
39.C101.602
chargeelectron1
C1093.5 19
18
= 37 negative (electron) charges on the oil drop
40. First, divide all charges by the smallest quantity, 6.40 ×13
10 :13
12
1040.6
1056.2
= 4.00;640.0
68.7 = 12.0;640.0
84.3 = 6.00
36. If the formula is InO, then one atomic mass of In would combine with one atomic mass of O,
or:Og000.1
Ing784.4
00.16
A
, A = atomic mass of In = 76.54
If the formula is In2O3, then two times the atomic mass of In will combine with three times
the atomic mass of O, or:Og000.1
Ing784.4
00.16)3(
A2
, A = atomic mass of In = 114.8
The latter number is the atomic mass of In used in the modern periodic table.
The Nature of the Atom
37. From section 1-7, the nucleus has “a diameter of about 1013 cm” and the electrons “move
about the nucleus at an average distance of about 108 cm from it.” We will use these
statements to help determine the densities. Density of hydrogen nucleus (contains one proton
only):
Vnucleus =3403143 cm105)cm105()14.3(
3
4
r
3
4
d = density =315
340
24
g/cm103
cm105
g1067.1
Density of H atom (contains one proton and one electron):
Vatom =32438 cm104)cm101()14.3(
3
4
d =3
324
2824
g/cm0.4
cm104
g109g1067.1
38. Because electrons move about the nucleus at an average distance of about 1 ×8
10 cm, the
diameter of an atom will be about 2 ×8
10 cm. Let's set up a ratio:cm102
cm101
modelofdiameter
mm1
atomofdiameter
nucleusofdiameter
8
13
; solving:
diameter of model = 2 × 105 mm = 200 m
39.C101.602
chargeelectron1
C1093.5 19
18
= 37 negative (electron) charges on the oil drop
40. First, divide all charges by the smallest quantity, 6.40 ×13
10 :13
12
1040.6
1056.2
= 4.00;640.0
68.7 = 12.0;640.0
84.3 = 6.00
Loading page 30...
CHAPTER 1 CHEMICAL FOUNDATIONS30
Because all charges are whole-number multiples of 6.40 ×13
10 zirkombs, the charge on one
electron could be 6.40 ×13
10 zirkombs. However, 6.40 ×13
10 zirkombs could be the
charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of
an electron is 6.40 ×13
10 zirkombs or an integer fraction of 6.40 ×13
10 zirkombs.
41. Z is the atomic number and is equal to the number of protons in the nucleus. A is the mass
number and is equal to the number of protons plus neutrons in the nucleus. X is the symbol
of the element. See the front cover of the text which has a listing of the symbols for the
various elements and corresponding atomic number or see the periodic table on the cover to
determine the identity of the various atoms. Because all of the atoms have equal numbers of
protons and electrons, each atom is neutral in charge.
a.Na23
11 b.F19
9 c.O16
8
42. The atomic number for carbon is 6. 14C has 6 protons, 14 6 = 8 neutrons, and 6 electrons in
the neutral atom. 12C has 6 protons, 12 – 6 = 6 neutrons, and 6 electrons in the neutral atom.
The only difference between an atom of 14C and an atom of 12C is that 14C has two additional
neutrons.
43. a.79
35 Br: 35 protons, 79 35 = 44 neutrons. Because the charge of the atom is neutral,
the number of protons = the number of electrons = 35.
b.81
35 Br: 35 protons, 46 neutrons, 35 electrons
c.239
94 Pu: 94 protons, 145 neutrons, 94 electrons
d.133
55 Cs: 55 protons, 78 neutrons, 55 electrons
e.3
1 H: 1 proton, 2 neutrons, 1 electron
f.56
26 Fe: 26 protons, 30 neutrons, 26 electrons
44. a.235
92 U: 92 p, 143 n, 92 e b.27
13 Al: 13 p, 14 n, 13 e c.57
26 Fe: 26 p, 31 n, 26 e
d.208
82 Pb: 82 p, 126 n, 82 e e.86
37 Rb: 37 p, 49 n, 37 e f.41
20 Ca: 20 p, 21 n, 20 e
45. a. Element 8 is oxygen. A = mass number = 9 + 8 = 17;17
8 O
b. Chlorine is element 17.37
17 Cl c. Cobalt is element 27.60
27 Co
d. Z = 26; A = 26 + 31 = 57;57
26 Fe e. Iodine is element 53.131
53 I
f. Lithium is element 3.7
3 Li
Because all charges are whole-number multiples of 6.40 ×13
10 zirkombs, the charge on one
electron could be 6.40 ×13
10 zirkombs. However, 6.40 ×13
10 zirkombs could be the
charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of
an electron is 6.40 ×13
10 zirkombs or an integer fraction of 6.40 ×13
10 zirkombs.
41. Z is the atomic number and is equal to the number of protons in the nucleus. A is the mass
number and is equal to the number of protons plus neutrons in the nucleus. X is the symbol
of the element. See the front cover of the text which has a listing of the symbols for the
various elements and corresponding atomic number or see the periodic table on the cover to
determine the identity of the various atoms. Because all of the atoms have equal numbers of
protons and electrons, each atom is neutral in charge.
a.Na23
11 b.F19
9 c.O16
8
42. The atomic number for carbon is 6. 14C has 6 protons, 14 6 = 8 neutrons, and 6 electrons in
the neutral atom. 12C has 6 protons, 12 – 6 = 6 neutrons, and 6 electrons in the neutral atom.
The only difference between an atom of 14C and an atom of 12C is that 14C has two additional
neutrons.
43. a.79
35 Br: 35 protons, 79 35 = 44 neutrons. Because the charge of the atom is neutral,
the number of protons = the number of electrons = 35.
b.81
35 Br: 35 protons, 46 neutrons, 35 electrons
c.239
94 Pu: 94 protons, 145 neutrons, 94 electrons
d.133
55 Cs: 55 protons, 78 neutrons, 55 electrons
e.3
1 H: 1 proton, 2 neutrons, 1 electron
f.56
26 Fe: 26 protons, 30 neutrons, 26 electrons
44. a.235
92 U: 92 p, 143 n, 92 e b.27
13 Al: 13 p, 14 n, 13 e c.57
26 Fe: 26 p, 31 n, 26 e
d.208
82 Pb: 82 p, 126 n, 82 e e.86
37 Rb: 37 p, 49 n, 37 e f.41
20 Ca: 20 p, 21 n, 20 e
45. a. Element 8 is oxygen. A = mass number = 9 + 8 = 17;17
8 O
b. Chlorine is element 17.37
17 Cl c. Cobalt is element 27.60
27 Co
d. Z = 26; A = 26 + 31 = 57;57
26 Fe e. Iodine is element 53.131
53 I
f. Lithium is element 3.7
3 Li
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