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1 2 3 4 5 6 7 8 9 10 045 09 08 07 06 05 04 03 02 01
CHALLENGE PROBLEMS
Contents
Chapter 1
Production of Chlorofluorocarbons, 1950–1992 . . . . . . . . . 1
Chapter 2
Population Trends in the United States . . . . . . . . . . . . . . . . 2
Chapter 3
Physical and Chemical Changes . . . . . . . . . . . . . . . . . . . . . 3
Chapter 4
Isotopes of an Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Chapter 5
Quantum Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Chapter 6
Döbereiner’s Triads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Chapter 7
Abundance of the Elements . . . . . . . . . . . . . . . . . . . . . . . . 7
Chapter 8
Comparing the Structures of Atoms and Ions . . . . . . . . . . . 8
Chapter 9
Exceptions to the Octet Rule . . . . . . . . . . . . . . . . . . . . . . . . 9
Chapter 10 Balancing Chemical Equations . . . . . . . . . . . . . . . . . . . . . 10
Chapter 11 Using Mole-Based Conversions . . . . . . . . . . . . . . . . . . . . 11
Chapter 12 Mole Relationships in Chemical Reactions . . . . . . . . . . . . 12
Chapter 13 Intermolecular Forces and Boiling Points . . . . . . . . . . . . . 13
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Chapter 14 A Simple Mercury Barometer . . . . . . . . . . . . . . . . . . . . . . 14
Chapter 15 Vapor Pressure Lowering . . . . . . . . . . . . . . . . . . . . . . . . . 15
Chapter 16 Standard Heat of Formation . . . . . . . . . . . . . . . . . . . . . . . 16
Chapter 17 Determining Reaction Rates . . . . . . . . . . . . . . . . . . . . . . . 17
Chapter 18 Changing Equilibrium Concentrations in a Reaction . . . . . 18
Chapter 19 Swimming Pool Chemistry . . . . . . . . . . . . . . . . . . . . . . . . 19
Chapter 20 Balancing Oxidation–Reduction Equations . . . . . . . . . . . . 20
Chapter 21 Effect of Concentration on Cell Potential . . . . . . . . . . . . . 21
Chapter 22 Structural Isomers of Hexane . . . . . . . . . . . . . . . . . . . . . . 22
Chapter 23 Boiling Points of Organic Families . . . . . . . . . . . . . . . . . . 23
Chapter 24 The Chemistry of Life . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Chapter 25 The Production of Plutonium-239 . . . . . . . . . . . . . . . . . . . 25
Chapter 26 The Phosphorus Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
Answer Key . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . T27
Challenge Problems
Chemistry: Matter and Change
iii
iv
Chemistry: Matter and Change
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Name
CHAPTER
Date
1
Class
CHALLENGE PROBLEMS
C
hlorofluorocarbons (CFCs) were first produced in
the laboratory in the late 1920s. They did not
become an important commercial product until some
time later. Eventually, CFCs grew in popularity until
their effect on the ozone layer was discovered in the
1970s. The graph shows the combined amounts of two
important CFCs produced between 1950 and 1992.
Answer the following questions about the graph.
Amount of CFCs
(billion kilograms)
Production of
Chlorofluorocarbons, 1950–1992
400
350
300
250
200
150
100
50
0
1950
1960
Use with Chapter 1,
Section 1.1
1970
Year
1980
1990
1. What was the approximate amount of CFCs produced in 1950? In 1960? In 1970?
2. In what year was the largest amount of CFCs produced? About how much was produced
that year?
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
3. During what two-year period did the production of CFCs decrease by the greatest
amount? By about how much did their production decrease?
4. During what two-year period did the production of CFCs increase by the greatest
amount? What was the approximate percent increase during this period?
5. How confident would you feel about predicting the production levels of CFCs during the
odd numbered years 1961, 1971, and 1981? Explain.
6. Could the data in the graph be presented in the form of a circle graph? Explain.
Challenge Problems
Chemistry: Matter and Change • Chapter 1
1
Name
Date
CHAPTER
2
Class
CHALLENGE PROBLEMS
Population Trends in the
United States
Use with Chapter 2,
Section 2.4
T
he population of the United States is becoming more diverse. The circle graphs below show the
distribution of the U.S. population among five ethnic groups in 1990 and 2000. The estimated
total U.S. population for those two years was 2.488 108 in 1990 and 2.754 108 in 2000.
U.S. Population Distribution
African American
11.8%
Hispanic American
9.0%
Asian American
2.8%
Native American
0.70%
1990
2000
African American
12.2%
Hispanic American
11.8%
Caucasian
75.7%
Asian American
3.8%
Native American
0.70%
Caucasian
71.4%
(Percentages may not add up to 100% due to rounding.)
1. By how much did the total U.S. population increase between 1990 and 2000? What was
2. Calculate the total population for each of the five groups for 1990 and 2000.
3. Make a bar graph that compares the population for the five groups in 1990 and 2000. In
what ways is the bar graph better than the circle graphs? In what way is it less useful?
2
Chemistry: Matter and Change • Chapter 2
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
the percent increase during this period?
Name
CHAPTER
Date
3
Class
CHALLENGE PROBLEMS
Physical and Chemical
Changes
Use with Chapter 3,
Section 3.2
P
hysical and chemical changes occur all around us. One of the many places in which
physical and chemical changes occur is the kitchen. For example, cooking spaghetti in a
pot of water on the stove involves such changes. For each of the changes described below, tell
(a) whether the change that occurs is physical or chemical, and (b) how you made your choice
between these two possibilities. If you are unable to decide whether the change is physical or
chemical, tell what additional information you would need in order to make a decision.
1. As the water in the pot is heated, its temperature rises.
2. As more heat is added, the water begins to boil and steam is produced.
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
3. The heat used to cook is produced by burning natural gas in the stove burner.
4. The metal burner on which the pot rests while being heated becomes red as its
temperature rises.
5. After the flame has been turned off, a small area on the burner has changed in color from
black to gray.
6. A strand of spaghetti has fallen onto the burner, where it turns black and begins to
smoke.
7. When the spaghetti is cooked in the boiling water, it becomes soft.
Challenge Problems
Chemistry: Matter and Change • Chapter 3
3
Name
Date
CHAPTER
4
Class
CHALLENGE PROBLEMS
Isotopes of an Element
mass spectrometer is a device for separating
atoms and molecules according to their
mass. A substance is first heated in a vacuum and
then ionized. The ions produced are accelerated
through a magnetic field that separates ions of different masses. The graph below was produced
when a certain element (element X) was analyzed
in a mass spectrometer. Use the graph to answer
the questions below.
30
Percent abundance
A
Use with Chapter 4,
Section 4.3
25
20
15
10
5
0
190 192 194 196 198 200 202 204 206 208 210
Atomic mass (amu)
1. How many isotopes of element X exist?
2. What is the mass of the most abundant isotope?
3. What is the mass of the least abundant isotope?
4. What is the mass of the heaviest isotope?
5. What is the mass of the lightest isotope?
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
6. Estimate the percent abundance of each isotope shown on the graph.
7. Without performing any calculations, predict the approximate atomic mass for element
X. Explain the basis for your prediction.
8. Using the data given by the graph, calculate the weighted average atomic mass of
element X. Identify the unknown element.
4
Chemistry: Matter and Change • Chapter 4
Challenge Problems
Name
Date
5
CHAPTER
Class
CHALLENGE PROBLEMS
Quantum Numbers
Use with Chapter 5,
Section 5.2
T
he state of an electron in an atom can be completely described by four quantum numbers,
designated as n, , m, and ms. The first, or principal, quantum number, n, indicates the
electron’s approximate distance from the nucleus. The second quantum number, , describes
the shape of the electron’s orbit around the nucleus. The third quantum number, m, describes
the orientation of the electron’s orbit compared to the plane of the atom. The fourth quantum
number, ms, tells the direction of the electron’s spin (clockwise or counterclockwise).
The Schrödinger wave equation imposes certain mathematical restrictions on the quantum
numbers. They are as follows:
n can be any integer (whole number),
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
can be any integer from 0 to n 1,
m can be any integer from to , and
1 or 1
ms can be 2
2
As an example, consider electrons in the first energy level of an atom, that is, n 1. In
this case, can have any integral value from 0 to (n 1), or 0 to (1 1). In other words,
must be 0 for these electrons. Also, the only value that m can have is 0. The electrons in
1 or 1 for m . These restrictions agree with the
this energy level can have values of s
2
2
observation that the first energy level can have only two electrons. Their quantum numbers
1 and 1, 0, 0 1 .
are 1, 0, 0, 2
2
Use the rules given above to complete the table listing the quantum numbers for each
electron in a boron atom. The correct quantum numbers for one electron in the atom is
provided as an example.
Boron (B)
Electron
n
m
ms
1
1
0
0
1
2
2
3
4
5
Challenge Problems
Chemistry: Matter and Change • Chapter 5
5
Name
Date
CHAPTER
6
Class
CHALLENGE PROBLEMS
Döbereiner’s Triads
Use with Chapter 6,
Section 6.2
O
ne of the first somewhat successful attempts to arrange the elements in a systematic way
was made by the German chemist Johann Wolfgang Döbereiner (1780–1849). In 1816,
Döbereiner noticed that the then accepted atomic mass of strontium (50) was midway between
the atomic masses of calcium (27.5) and barium (72.5). Note that the accepted atomic masses
for these elements today are very different from their accepted atomic masses at the time
Döbereiner made his observations. Döbereiner also observed that strontium, calcium, and barium showed a gradual gradation in their properties, with the values of some of strontium’s
properties being about midway between the values of calcium and barium. Döbereiner eventually found four other sets of three elements, which he called triads, that followed the same pattern. In each triad, the atomic mass of the middle element was about midway between the
atomic masses of the other two elements. Unfortunately, because Döbereiner’s system did not
turn out to be very useful, it was largely ignored.
Set 1
Element
Melting Point (°C)
219.6
Fluorine
Chlorine
Set 2
Calculated:
Element
Actual:
Boiling Point (°C)
Krypton
153
Calculated:
Element
Tin
Actual:
6
Calcium
Lead
Chemistry: Matter and Change • Chapter 6
Calculated:
Strontium
1384
Set 6
Melting Point (°C)
937
Calculated:
Element
Boiling Point (°C)
Beryllium
Magnesium
Actual:
62
1107
Actual:
39.098
Germanium
Boiling Point (°C)
Magnesium
Set 5
Element
Radon
Calculated:
Potassium
Set 4
Xenon
6.941
Element
Actual:
7.2
Bromine
Atomic Mass
Lithium
Sodium
Set 3
1285
Calculated:
Actual:
327
Calcium
851
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Had Döbereiner actually discovered a way of identifying trends among the elements?
Listed below are six three-element groups in which the elements in each group are consecutive
members of the same group in the periodic table. The elements in each set show a gradation in
their properties. Values for the first and third element in each set are given. Determine the missing value in each set by calculating the average of the two given values. Then, compare the values you obtained with those given in the Handbook of Chemistry and Physics. Record the
actual values below your calculated values. Is the value of the property of the middle element
in each set midway between the values of the other two elements in the set?
Name
CHAPTER
Date
7
Class
CHALLENGE PROBLEMS
Abundance of the Elements
Use with Chapter 7,
Section 7.1
T
he abundance of the elements differs significantly in various parts of the
universe. The table below lists the abundance of some elements in various
parts of the universe. Use the table to answer the following questions.
Abundance (Number of atoms per 1000 atoms)*
Element
Universe
Hydrogen
927
Helium
71.8
Solar System
Earth
863
Earth’s Crust
Human Body
30
606
610
257
135
Oxygen
0.510
0.783
500
Nitrogen
0.153
0.0809
24
Carbon
0.0811
0.459
106
Silicon
0.0231
0.0269
140
210
Iron
0.0139
0.00320
170
19
* An element is not abundant in a region that is left blank.
1. What percent of all atoms in the universe are either hydrogen or helium? What percent of
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
all atoms in the solar system are either hydrogen or helium?
2. Explain the relatively high abundance of hydrogen and helium in the universe compared
to their relatively low abundance on Earth.
3. Only the top four most abundant elements on Earth and in Earth’s crust are shown in the
table. Name two additional elements you would expect to find among the top ten elements both on Earth and in Earth’s crust. Explain your choices.
4. Name at least three elements in addition to those shown in the table that you would
expect to find in the list of the top ten elements in the human body. Explain your choices.
Challenge Problems
Chemistry: Matter and Change • Chapter 7
7
Name
Date
CHAPTER
8
Class
CHALLENGE PROBLEMS
Comparing the Structures of
Atoms and Ions
Use with Chapter 8,
Section 8.1
T
he chemical properties of an element depend primarily on its number of valence electrons in
its atoms. The noble gas elements, for example, all have similar chemical properties
because the outermost energy levels of their atoms are completely filled. The chemical properties
of ions also depend on the number of valence electrons. Any ion with a complete outermost
energy level will have chemical properties similar to those of the noble gas elements. The fluoride ion (F), for example, has a total of ten electrons, eight of which fill its outermost energy
level. F has chemical properties, therefore, similar to those of the noble gas neon.
Shown below are the Lewis electron dot structures for five elements: sulfur (S), chlorine (Cl),
argon (Ar), potassium (K), and calcium (Ca). Answer the questions below about these structures.
S
Cl
Ar
K
Ca
1. Write the atomic number for each of the five elements shown above.
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
2. Write the electron configuration for each of the five elements.
3. Which of the above Lewis electron dot structures is the same as the Lewis electron dot
structure for the ion S2? Explain your answer.
4. Which of the above Lewis electron dot structures is the same as that for the ion Cl?
Explain your answer.
5. Which of the above Lewis electron dot structures is like that for the ion K? Explain
your answer.
6. Name an ion of calcium that has chemical properties similar to those of argon. Explain
your answer.
8
Chemistry: Matter and Change • Chapter 8
Challenge Problems
Name
CHAPTER
Date
9
Class
CHALLENGE PROBLEMS
Exceptions to the Octet Rule
Use with Chapter 9,
Section 9.3
T
he octet rule is an important guide to understanding how most compounds are formed.
However, there are a number of cases in which the octet rule does not apply. Answer the
following questions about exceptions to the octet rule.
1. Draw the Lewis structure for the compound BeF2.
2. Does BeF2 obey the octet rule? Explain.
3. Draw the Lewis structure for the compound NO2.
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
4. Does NO2 obey the octet rule? Explain.
5. Draw the Lewis structure for the compound N2F2.
6. Does N2F2 obey the octet rule? Explain.
7. Draw the Lewis structure for the compound IF5.
8. Does IF5 obey the octet rule? Explain.
Challenge Problems
Chemistry: Matter and Change • Chapter 9
9
Name
Date
CHAPTER
10
Class
CHALLENGE PROBLEMS
Balancing Chemical
Equations
Use with Chapter 10,
Section 10.1
E
ach chemical equation below contains at least one error. Identify the error or errors and
then write the correct chemical equation for the reaction.
1. K(s) 2H2O(l) 0 2KOH(aq) H2(g)
2. MgCl2(aq) H2SO4(aq) 0 Mg(SO4)2(aq) 2HCl(aq)
3. AgNO3(aq) H2S(aq) 0 Ag2S(aq) HNO3(aq)
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
4. Sr(s) F2(g) 0 Sr2F
5. 2NaHCO3(s) 2HCl(aq) 0 2NaCl(s) 2CO2(g)
6. 2LiOH(aq) 2HBr(aq) 0 2LiBr(aq) 2H2O
7. NH4OH(aq) KOH(aq) 0 KOH(aq) NH4OH(aq)
8. 2Ca(s) Cl2(g) 0 2CaCl(aq)
9. H2SO4(aq) 2Al(NO3)3(aq) 0 Al2(SO4)3(aq) 2HNO3(aq)
10
Chemistry: Matter and Change • Chapter 10
Challenge Problems
Name
Date
11
CHAPTER
Class
CHALLENGE PROBLEMS
Using Mole-Based
Conversions
Use with Chapter 11,
Section 11.3
T
he diagram shows three containers, each of which holds a certain mass of the
substance indicated. Complete the table below for each of the three substances.
UF6 (g)
CCl3CF3 (l)
Pb (s)
225.0 g
200.0 g
250.0 g
Substance
Mass (g)
Molar Mass
(g/mol)
Number of
Moles (mol)
Number of Representative
Particles
UF6(g)
CCl3CF3(l)
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Pb(s)
1. Compare and contrast the number of representative particles and the mass of UF6 with
the number of representative particles and mass of CCl3CF3. Explain any differences
you observe.
2. UF6 is a gas used in the production of fuel for nuclear power plants. How many moles of
the gas are in 100.0 g of UF6?
3. CCl3CF3 is a chlorofluorocarbon responsible for the destruction of the ozone layer in
Earth’s atmosphere. How many molecules of the liquid are in 1.0 g of CCl3CF3?
4. Lead (Pb) is used to make a number of different alloys. What is the mass of lead present
in an alloy containing 0.15 mol of lead?
Challenge Problems
Chemistry: Matter and Change • Chapter 11
11
Name
Date
CHAPTER
12
Class
CHALLENGE PROBLEMS
Mole Relationships in
Chemical Reactions
Use with Chapter 12,
Section 12.2
T
he mole provides a convenient way of finding the amounts of the substances in a chemical
reaction. The diagram below shows how this concept can be applied to the reaction
between carbon monoxide (CO) and oxygen (O2), shown in the following balanced equation.
2CO(g) O2(g) 0 2CO2(g)
Use the equation and the diagram to answer the following questions.
Moles of
CO
3
Particles of
CO
1
6
2
4
Grams of
CO
Moles of
CO2
5
7
Particles of
CO2
Grams of
CO2
1. What information is needed to make the types of conversions shown by double-arrow 1
in the diagram?
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
2. What conversion factors would be needed to make the conversions represented by
double-arrow 2 in the diagram for CO? By double-arrow 6 for CO2?
3. What information is needed to make the types of conversions represented by
double-arrows 3 and 7 in the diagram?
4. What conversion factors would be needed to make the conversions represented by
double-arrow 3 in the diagram for CO?
5. Why is it not possible to convert between the mass of a substance and the number of
representative particles, as represented by double-arrow 4 of the diagram?
6. Why is it not possible to use the mass of one substance in a chemical reaction to find the mass
of a second substance in the reaction, as represented by double-arrow 5 in the diagram?
12
Chemistry: Matter and Change • Chapter 12
Challenge Problems
Name
CHAPTER
Date
13
Class
CHALLENGE PROBLEMS
Intermolecular Forces and
Boiling Points
he boiling points of liquids depend partly on the mass of the
particles of which they are made. The greater the mass of
the particles, the more energy is needed to convert a liquid to a
gas, and, thus, the higher the boiling point of the liquid. This pattern may not hold true, however, when there are significant forces
between the particles of a liquid. The graph plots boiling point
versus molecular mass for group 4A and group 6A hydrides. A
hydride is a binary compound containing hydrogen and one other
element. Use the graph to answer the following questions.
100
Boiling point (°C)
T
Use with Chapter 13,
Section 13.3
H2O
H2Te
0
H2Se
H2S
100
0
0
Group 6A
hydrides
SiH4
CH4
SnH4
GeH4
Group 4A
hydrides
50
100
Molecular mass
150
1. How do the boiling points of the group 4A hydrides change as the molecular masses of
the hydrides change?
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
2. What are the molecular structure and polarity of the four group 4A hydrides?
3. Predict the strength of the forces between group 4A hydride molecules. Explain how
those forces affect the boiling points of group 4A hydrides.
4. How do the boiling points of the group 6A hydrides change as the molecular masses of
the hydrides change?
5. What are the molecular structure and polarity of the four group 6A hydrides?
6. Use Table 9-4 in your textbook to determine the difference in electronegativities of the
bonds in the four group 6A hydrides.
Challenge Problems
Chemistry: Matter and Change • Chapter 13
13
Name
Date
CHAPTER
14
Class
CHALLENGE PROBLEMS
A Simple Mercury Barometer
I
n Figure 1, a simple mercury barometer is made by filling a long
glass tube with mercury and then inverting the open end of the
tube into a bowl of mercury. Answer the following questions about
the simple mercury barometer shown here.
Use with Chapter 14,
Section 14.1
Glass tube
Mercury
column
1. What occupies the space above the mercury column in the
Bowl of
mercury
barometer’s glass tube?
At sea level
At 500 meters
above sea level
Figure 1
Figure 2
2. What prevents mercury from flowing out of the glass tube into the bowl of mercury?
3. When the barometer in Figure 1 is moved to a higher elevation, such as an altitude of
4. Suppose the barometer in Figure 1 was carried into an open mine 500 meters below sea
level. How would the height of the mercury column change? Explain why.
5. Suppose the liquid used to make the barometer was water instead of mercury. How would
this substitution affect the barometer? Explain.
6. Suppose a tiny crack formed at the top of the barometer’s glass tube. How would this
event affect the column of mercury? Explain why.
14
Chemistry: Matter and Change • Chapter 14
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
5000 meters, the column of mercury changes as shown in Figure 2. Why is the mercury
column lower in Figure 2 than in Figure 1?
Name
CHAPTER
Date
15
Class
CHALLENGE PROBLEMS
Vapor Pressure Lowering
Use with Chapter 15,
Section 15.3
Y
ou have learned that adding a nonvolatile solute to a solvent
lowers the vapor pressure of that solvent. The amount by
which the vapor pressure is lowered can be calculated by means of
a relationship discovered by the French chemist François Marie
Raoult (1830–1901) in 1886. According to Raoult’s law, the vapor
pressure of a solvent (P) is equal to the product of its vapor pressure
when pure (P0) and its mole fraction (X) in the solution, or
P P0X
The solution shown at the right was made by adding 75.0 g of
sucrose (C12H22O11) to 500.0 g of water at a temperature of 20°C.
Answer the following questions about this solution.
Solution
Water
molecule
Sucrose
molecule
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
1. Why do the sugar molecules in the solution lower the vapor pressure of the water?
2. What is the number of moles of sucrose in the solution?
3. What is the number of moles of water in the solution?
4. What is the mole fraction of water in the solution?
5. What is the vapor pressure of the solution if the vapor pressure of pure water at 20°C is
17.54 mm Hg?
6. How much is the vapor pressure of the solution reduced from that of water by the
addition of the sucrose?
Challenge Problems
Chemistry: Matter and Change • Chapter 15
15
Name
Date
CHAPTER
16
Class
CHALLENGE PROBLEMS
Standard Heat of Formation
C(s) O2(g)
H
H 110 kJ/mol
CO(g) 1
O (g)
2 2
H 393 kJ/mol
Enthalpy
ess’s law allows you to determine the
standard heat of formation of a compound
when you know the heats of reactions that lead
to the production of that compound. The first
diagram on the right shows how Hess’s law can
be used to calculate the heat of formation of
CO2 by knowing the heats of reaction of two
steps leading to the production of CO2. Use this
diagram to help you answer the questions below
about the second diagram.
Use with Chapter 16,
Section 16.4
H 283 kJ/mol
The equations below show how NO2 can be
formed in two ways: directly from the elements
or in two steps.
H 33 kJ/mol
1
1
N2(g) O2(g) 0 NO(g)
2
2
H 91 kJ/mol
1 O (g) 0 NO (g)
NO(g) 2
2 2
H 58 kJ/mol
CO2(g)
C
NO(g) 1/2 O2(g)
1. On the diagram at the right, draw arrowheads
to show the directions in which the three lines
labeled 1, 2, and 3 should point.
2. Write the correct reactants and/or products on
2 H 58 kJ/mol
each of the lines labeled A, B, and C.
1 H 91 kJ/mol
each number on the diagram.
Enthalpy
3. Write the correct enthalpy change next to
B
NO2(g)
3 H 33 kJ/mol
A
16
Chemistry: Matter and Change • Chapter 16
1/2 N2(g) O2(g)
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
1
N2(g) O2(g) 0 NO2(g)
2
or
Name
Date
CHAPTER
17
Class
CHALLENGE PROBLEMS
Determining Reaction Rates
initrogen pentoxide decomposes to produce
nitrogen dioxide and oxygen as represented
by the following equation.
2N2O5(g) 0 4NO2(g) O2(g)
The graph on the right represents the concentration of N2O5 remaining as the reaction proceeds
over time. Answer the following questions about
the reaction.
1.6
Concentration (mol/L)
D
Use with Chapter 17,
Section 17.1
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0
0 1 2 3 4 5 6 7 8 9 10
Time (h)
1. What is the concentration of N2O5 at the beginning of the experiment? After 1 hour?
After 2 hours? After 10 hours?
2. By how much does the concentration of N2O5 change during the first hour of the
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
reaction? Calculate the percentage of change the concentration undergoes during the
first hour of the reaction.
3. The instantaneous rate of reaction is defined as the change in concentration of reactant
during some specified time period, or instantaneous rate of reaction = [N2O5]/t. What is
the instantaneous rate of reaction for the decomposition of N2O5 for the time period
between the first and second hours of the reaction? Between the second and third hours?
Between the sixth and seventh hours?
4. What is the instantaneous rate of reaction for the decomposition of N2O5 between the sec-
ond and fourth hours of the reaction? Between the third and eighth hours of the reaction?
5. How long does it take for 0.10 mol of N2O5 to decompose during the tenth hour of the reaction?
6. What is the average rate of reaction for the decomposition of N2O5 overall?
Challenge Problems
Chemistry: Matter and Change • Chapter 17
17
Name
Date
CHAPTER
18
Class
CHALLENGE PROBLEMS
R
eversible reactions eventually reach an equilibrium
condition in which the concentrations of all reactants
and products are constant. Equilibrium can be disturbed,
however, by the addition or removal of either a reactant or
product. The graph on the right shows how the concentrations of the reactants and product of a reaction change
when equilibrium is disturbed. Use the graph to answer the
following questions.
Concentration (mol/L)
Changing Equilibrium
Concentrations in a Reaction
8
7
6
5
4
3
2
1
0
Use with Chapter 18,
Section 18.1
SO2
SO2
O2
O2
SO3
SO3
0 1 2 3 4 5 6 7 8 9 10
Time (sec)
1. Write the equation for the reaction depicted in the graph.
2. Write the equilibrium constant expression for the reaction.
3. Explain the shapes of the curves for the three gases during the first 2 minutes of the
4. At approximately what time does the reaction reach equilibrium? How do you know
equilibrium has been reached?
5. What are the concentrations of the three gases at equilibrium?
6. Calculate the value of Keq for the reaction.
7. Describe the change made in the system 4 minutes into the reaction. Tell how you know
the change was made.
8. At what time does the system return to equilibrium?
18
Chemistry: Matter and Change • Chapter 18
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
reaction.
Name
CHAPTER
Date
19
Class
CHALLENGE PROBLEMS
Swimming Pool Chemistry
Use with Chapter 19,
Section 19.2
T
he presence of disease-causing bacteria in swimming pools is a major health concern.
Chlorine gas is added to the water in some large commercial swimming pools to kill
bacteria. However, in most home swimming pools, either solid calcium hypochlorite
(Ca(OCl)2) or an aqueous solution of sodium hypochlorite (NaOCl) is used to treat the
water. Both compounds dissociate in water to form the weak acid hypochlorous acid
(HOCl). Hypochlorous acid is a highly effective bactericide. By contrast, the hypochlorite
ion (OCl) is not a very effective bactericide. Use the information above to answer the
following questions about the acid-base reactions that take place in swimming pools.
1. Write an equation that shows the reaction between hypochlorous acid and water. Identify
the acid, base, conjugate acid, and conjugate base in this reaction.
2. Write an equation that shows the reaction that occurs when the hypochlorite ion (OCl),
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
in the form of calcium hypochlorite or sodium hypochlorite, is added to water. Name the
acid, base, conjugate acid, and conjugate base in this reaction.
3. What effect does the addition of hypochlorite ion have on the pH of swimming pool water?
4. The effectiveness of hypochlorite ion as a bactericide depends on pH. How does high pH
affect the equilibrium reaction described in question 2? What effect would high pH have
on the bacteria?
5. In the presence of sunlight, hypochlorite ion decomposes to form chloride ion and oxygen
gas. Write an equation for this reaction and tell how it affects the safety of pool water.
Challenge Problems
Chemistry: Matter and Change • Chapter 19
19
Name
Date
CHAPTER
20
Class
CHALLENGE PROBLEMS
Balancing Oxidation–
Reduction Equations
Use with Chapter 20,
Section 20.3
S
cientists have developed a number of methods for protecting
metals from oxidation. One such method involves the use of a
sacrificial metal. A sacrificial metal is a metal that is more easily
oxidized than the metal it is designed to protect. Galvanized iron, for
example, consists of a piece of iron metal covered with a thin layer
of zinc. When galvanized iron is exposed to oxygen, it is the zinc,
rather than the iron, that is oxidized.
Water heaters often contain a metal rod that is made by coating
a heavy steel wire with magnesium or aluminum. In this case, the
magnesium or aluminum is the sacrificial metal, protecting the iron
casing of the heater from corrosion.
The diagram shows a portion of a water heater containing
a sacrificial rod. Answer the following questions about the diagram.
Steel wire
Sacrificial
metal
Iron
casing
Water
1. In the absence of a sacrificial metal, oxygen dissolved in water may react with the iron
2. Balance the oxidation–reduction equation for this reaction:
Fe(s) O2(aq) H2O 0 Fe(OH)2(aq)
3. Write the two half-reactions for this example of corrosion.
4. Suppose the sacrificial rod in the diagram above is coated with aluminum metal. Write
the balanced equation for the reaction of aluminum with oxygen dissolved in the water.
(Hint: The product formed is aluminum hydroxide (Al(OH)3).
5. Write the two half-reactions for this example of corrosion.
6. Suppose that some iron in the casing of the water heater is oxidized, as shown in the
equation of question 2 above. The sacrificial metal (aluminum, in this case) immediately
restores the Fe2 ions to iron atoms. Write two half-reactions that represent this situation.
20
Chemistry: Matter and Change • Chapter 20
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
casing of the heater. One product formed is iron(II) hydroxide (Fe(OH)2). Which element
is oxidized and which is reduced in this reaction?
Name
Date
21
CHAPTER
Class
CHALLENGE PROBLEMS
Effect of Concentration on
Cell Potential
Use with Chapter 21,
Section 21.1
I
n a voltaic cell where all ions have a concentration of 1M, the cell potential is
equal to the standard potential. For cells in which ion concentrations are greater or
less than 1M, as shown below, an adjustment must be made to calculate cell potential.
That adjustment is expressed by the Nernst equation:
[product ion]x
0.0592 log Ecell E 0cell n
[reactant ion]y
In this equation, n is the number of moles of electrons transferred in the reaction,
and x and y are the coefficients of the product and reactant ions, respectively, in the
balanced half-cell reactions for the cell.
Voltmeter
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Ag
Ag
1.0 102M
Cu
Cu2
1.0 103M
1. Write the two half-reactions and the overall cell reaction for the cell shown above.
2. Use Table 21-1 in your textbook to determine the standard potential of this cell.
3. Write the Nernst equation for the cell.
4. Calculate the cell potential for the ion concentrations shown in the cell.
Challenge Problems
Chemistry: Matter and Change • Chapter 21
21
Name
Date
CHAPTER
22
Class
CHALLENGE PROBLEMS
Structural Isomers of Hexane
Use with Chapter 22,
Sections 22.1 and 22.3
T
he structural formula of an organic compound can sometimes be written in a
variety of ways, but sometimes structural formulas that appear similar can
represent different compounds. The structural formulas below are ten ways of
representing compounds having the molecular formula C6H14.
a. CH3
CH2
e. CH3
CH2
CH2
CH2
CH3
CH2 CH3
CH
i. CH2
CH2
CH3
CH3
CH2
CH2
CH3
CH3 CH3
CH2
CH
CH2
CH3
b. CH3
CH
f. CH3
CH2
CH2
CH3
CH
CH
CH3
CH3 CH3
j.
CH2
CH3
c.
g. CH2
CH3
CH3
CH
CH
CH3
CH3
CH
CH2
CH3
CH3
CH3
h.
CH3
CH3
C
CH2
CH3
CH3
CH3
CH3
CH
CH2
CH2
CH3
1. In the spaces provided, write the correct name for each of the structural formulas, labeled
a–j, above.
a.
e.
i.
b.
f.
j.
c.
g.
d.
h.
2. How many different compounds are represented by the structural formulas above? What
are their names?
22
Chemistry: Matter and Change • Chapter 22
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
d.
Name
CHAPTER
Date
23
Class
CHALLENGE PROBLEMS
Boiling Points of Organic
Families
he most important factor determining the boiling point of
a substance is its atomic or molecular mass. In general,
the larger the atomic or molecular mass of the substance, the
more energy is needed to convert the substance from the liquid
phase to the gaseous phase. As an example, the boiling point
of ethane (molecular mass 30; boiling point 89°C) is
much higher than the boiling point of methane (molecular
mass 16; boiling point 161°C).
Intermolecular forces between the particles of a liquid also
can affect the liquid’s boiling point. The graph shows trends in
the boiling points of four organic families: alkanes, alcohols,
aldehydes, and ethers. Use the graph and your knowledge of
intermolecular forces to answer the following questions.
100
Boiling point (°C)
T
Use with Chapter 23,
Section 23.3
50
0
50
30
40
50
60
70
Molecular mass
alkane
alcohol
80
aldehyde
ether
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
1. For any one family, what is the relationship between molecular mass and boiling point?
2. For compounds of similar molecular mass, which family of the four shown in the graph
has the lowest boiling points? Which family has the highest boiling points?
3. Find and list the boiling points for ethanol (molecular mass 46) and dimethyl ether
(molecular mass 46) on the graph. Why would you expect these two compounds to
have relatively similar boiling points?
4. Find the aldehyde with a molecular mass of about 58. Name that aldehyde and write its
chemical formula.
5. Can this aldehyde form hydrogen bonds? Can other aldehydes form hydrogen bonds?
Explain.
Challenge Problems
Chemistry: Matter and Change • Chapter 23
23
Name
Date
24
CHALLENGE PROBLEMS
The Chemistry of Life
Use with Chapter 24,
Section 24.4
P
roteins are synthesized when RNA molecules
translate the DNA language of nitrogen bases
into the protein language of amino acids using a
genetic code. The genetic code is found in RNA molecules called messenger RNA (mRNA), which are synthesized from DNA molecules. The genetic code
consists of a sequence of three nitrogen bases in the
mRNA, called a codon. Most codons code for specific
amino acids. A few codons code for a stop in the synthesis of proteins. The table shows the mRNA codons
that make up the genetic code. To use the table, read
the three nitrogen bases in sequence. The first base is
shown along the left side of the table. The second base
is shown along the top of the table. The third base is
shown along the right side of the table. For example,
the sequence CAU codes for the amino acid histidine
(His). The table gives abbreviations for the amino
acids. Answer the following questions about the
genetic code.
The Genetic Code
Second base
U
C
A
G
}
}
C
UCU
Phe
UCC
UCA
Leu
UCG
CCU
CCC
Leu
CCA
CCG
ACU
Ile
ACC
ACA
Met ACG
GCU
GCC
Val
GCA
GCG
A
UAU
UAC
Ser
UAA
UAG
CAU
CAC
Pro
CAA
CAG
AAU
AAC
Thr
AAA
AAG
GAU
GAC
Ala
GAA
GAG
G
}Tyr
Stop
Stop
} His
} Gln
} Asn
} Lys
} Asp
} Glu
UGU
UGC
UGA
UGG
CGU
CGC
CGA
CGG
AGU
AGC
AGA
AGG
GGU
GGC
GGA
GGG
} Cys
}
}
U
C
Stop A
Trp G
U
C
Arg
A
G
U
Ser
C
A
Arg
G
U
C
Gly
A
G
Third base
First base
U
UUU
UUC
UUA
UUG
CUU
CUC
CUA
CUG
AUU
AUC
AUA
AUG
GUU
GUC
GUA
GUG
1. What amino acid is represented by each of the following codons?
a. CUG
b. UCA
2. Write the sequence of amino acids for which the following mRNA sequence codes.
-C-A-U-C-A-C-C-G-G-U-C-U-U-U-U-C-U-U-
3. Errors sometimes occur when mRNA molecules are synthesized from DNA molecules.
Nitrogen bases may be omitted, an extra nitrogen base may be added, or a nitrogen base
may be changed during synthesis. The two mRNA sequences shown below are examples
of such errors. In each case, tell how the mRNA sequence shown differs from the correct
mRNA sequence given in question 2.
a. -C-A-U-C-A-C-C-G-G-U-U-C-U-U-U-U-C-U-U-
b. -C-A-U-U-A-C-C-G-G-U-C-U-U-U-U-C-U-U-
4. Write the amino acid sequence for each of the mRNA sequences shown in question 3.
a.
b.
24
Chemistry: Matter and Change • Chapter 24
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
CHAPTER
Class
Name
CHAPTER
Date
25
Class
CHALLENGE PROBLEMS
The Production of
Plutonium-239
Use with Chapter 25,
Section 25.4
45p
75n
W
hen nuclear fission was first discovered, only two
isotopes, uranium-233 and uranium-235, were
known of being capable of undergoing this nuclear change.
Scientists later discovered a third isotope, plutonium-239,
also could undergo nuclear fission. Plutonium-239 does not
occur in nature but can be made synthetically in nuclear
reactors and particle accelerators.
92p
143n
1n
0
92p
143n
1n
0
1n
0
A
92p
146n
Source
of
neutrons
The diagram shows the process by which plutonium-239
is made in nuclear reactors. Answer the questions about the
diagram.
0
0
C
1n
0
0
–1
D
0
–1
1. Identify the isotope whose nucleus is labeled A in the
diagram.
B
F
48p
77n
E
G
2. Name the type of nuclear reaction that occurs when a
neutron strikes nucleus A.
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
3. Identify the isotope whose nucleus is labeled B.
4. Besides fragmented nuclei, what else is produced when a neutron strikes nucleus A?
5. Identify the isotope whose nucleus is labeled C.
6. Write the nuclear equation for the reaction that occurs when a neutron strikes nucleus C.
Identify the product D formed in the reaction.
7. Write the nuclear equation for the decay of nucleus D. Identify isotope E formed in the
reaction.
8. Write a balanced nuclear equation for the decay of nucleus E. Identify isotope F formed
in the reaction.
9. Name the type of nuclear reaction that occurs when a neutron strikes nucleus F.
10. Write the nuclear equation for the reaction that occurs when a neutron strikes nucleus F.
Identify isotope G formed in the reaction.
Challenge Problems
Chemistry: Matter and Change • Chapter 25
25
Name
Date
CHAPTER
26
Class
CHALLENGE PROBLEMS
The Phosphorus Cycle
Use with Chapter 26,
Section 26.4
P
hosphorus is an important element both in organisms and in the lithosphere. In
organisms, phosphorus occurs in DNA and RNA molecules, cell membranes, bones
and teeth, and in the energy–storage compound adenosine triphosphate (ATP). In the lithosphere, phosphorus occurs primarily in the form of phosphates, as a major constituent of
many rocks and minerals. Phosphate rock is mined to produce many commercial products,
such as fertilizers and detergents. When these products are used, phosphates are returned to
the lithosphere and hydrosphere. Thus, phosphorus—like carbon and nitrogen—cycles in the
environment. Use the diagram of the phosphorus cycle to answer the questions below.
Phosphate
rocks
Phosphate rocks
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Geological uplift
1. By what methods does phosphorus get into soil?
2. By what method do plants obtain the phosphorus they need?
3. By what method do animals obtain the phosphorus they need?
4. In what way is the phosphorus cycle different from the carbon and nitrogen cycles you
studied in the textbook?
5. The phosphorus cycle has both short-term and long-term parts. Use different colored
pencils to show each part on the diagram.
26
Chemistry: Matter and Change • Chapter 26
Challenge Problems