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Challenge
Problems
A Glencoe Program
Hands-On Learning:
Laboratory Manual, SE/TE
Forensics Laboratory Manual, SE/TE
CBL Laboratory Manual, SE/TE
Small-Scale Laboratory Manual, SE/TE
ChemLab and MiniLab Worksheets
Review/Reinforcement:
Study Guide for Content Mastery, SE/TE
Solving Problems: A Chemistry Handbook
Reviewing Chemistry
Guided Reading Audio Program
Applications and Enrichment:
Challenge Problems
Supplemental Problems
Teacher Resources:
Lesson Plans
Block Scheduling Lesson Plans
Spanish Resources
Section Focus Transparencies and Masters
Math Skills Transparencies and Masters
Teaching Transparencies and Masters
Solutions Manual
Technology:
Chemistry Interactive CD-ROM
Vocabulary PuzzleMaker Software,
Windows/MacIntosh
Glencoe Science Web site:
science.glencoe.com
Assessment:
Chapter Assessment
MindJogger Videoquizzes (VHS/DVD)
Computer Test Bank, Windows/MacIntosh
Copyright © by The McGraw-Hill Companies, Inc.
All rights reserved. Permission is granted to reproduce the material contained herein
on the condition that such material be reproduced only for classroom use; be provided
to students, teachers, and families without charge; and be used solely in conjunction
with the Chemistry: Matter and Change program. Any other reproduction, for use or
sale, is prohibited without prior written permission of the publisher.
Send all inquiries to:
Glencoe/McGraw-Hill
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ISBN 0-07-824533-8
Printed in the United States of America.
1 2 3 4 5 6 7 8 9 10 045 09 08 07 06 05 04 03 02 01
CHALLENGE PROBLEMS
Contents
To the Teacher . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv
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
To the Teacher
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Students can take their learning one step beyond their textbooks with
Challenge Problems. The worksheets in this supplement to Chemistry:
Matter and Change challenge students to apply their knowledge of
chemistry to new situations, analyze and interpret those situations, and
synthesize responses. Whether analyzing experimental results or investigating
a hypothetical situation, these worksheets encourage students to use chemical
concepts along with their critical thinking skills to solve problems.
iv
Chemistry: Matter and Change
Challenge Problems
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.
C
0
0
1n
0
0
–1
0
–1
1. Identify the isotope whose nucleus is labeled A in the
diagram.
B
D
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
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
CHALLENGE PROBLEMS
Answer Key
1
CHALLENGE PROBLEMS
Date
Chemistry: Matter and Change
400
350
300
250
200
150
100
50
0
1950
1960
1990
Challenge Problems
Challenge Problems Answer Key
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Chemistry: Matter and Change • Chapter 1
into categories, not how a variable changes over time.
A circle graph could not be used because it shows how a total amount is divided
6. Could the data in the graph be presented in the form of a circle graph? Explain.
regarding the production level during 1981 might be less accurate.
1
production levels were less regular between 1974 and 1988. Therefore, predictions
the production levels during 1961 and 1971 from data in the graph. However, the
and decreased regularly after 1988. Thus, one might be able to predict accurately
The production of CFCs increased regularly from 1950 to 1974 (except in 1958)
odd numbered years 1961, 1971, and 1981? Explain.
5. How confident would you feel about predicting the production levels of CFCs during the
(305 billion kg 240 billion kg)/240 billion kg 27%.
Between 1970 and 1972; the percent increase was approximately
amount? What was the approximate percent increase during this period?
4. During what two-year period did the production of CFCs increase by the greatest
Between 1988 and 1990; the decrease amounted to about 140 billion kg.
amount? By about how much did their production decrease?
3. During what two-year period did the production of CFCs decrease by the greatest
1988; about 375 billion kg
that year?
2. In what year was the largest amount of CFCs produced? About how much was produced
1950: less than 10 billion kg; 1960: about 50 billion kg; 1970: about 240 billion kg
1980
Use with Chapter 1,
Section 1.1
Class
1970
Year
1. What was the approximate amount of CFCs produced in 1950? In 1960? In 1970?
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.
C
Production of
Chlorofluorocarbons, 1950–1992
CHAPTER
Name
Amount of CFCs
(billion kilograms)
T28
2
CHALLENGE PROBLEMS
Date
Use with Chapter 2,
Section 2.4
Class
2
Native American
0.70%
Asian American
3.8%
African American
12.2%
Hispanic American
11.8%
U.S. Population Distribution
Caucasian
71.4%
2000
(Percentages may not add up to 100% due to rounding.)
Caucasian
75.7%
1990
Chemistry: Matter and Change • Chapter 2
difficult to show data for the latter groups clearly.
Challenge Problems
large differences in populations for Caucasian and non-Caucasian groups make it
may be more difficult to estimate than the values given on the circle graph. The
Unless the numerical values for each bar are actually written on the graph, they
The bar graph may make the changes from 1990 to 2000 show up more clearly.
what ways is the bar graph better than the circle graphs? In what way is it less useful?
3. Make a bar graph that compares the population for the five groups in 1990 and 2000. In
American: 3.25 107; Native American: 1.9 106; Asian American: 1.0 107.
2.754 108, or Caucasian: 1.97 108; African American: 3.4 107; Hispanic
7.0 106; Native American: 1.7 106; for 2000, total percent for group African American: 2.94 107; Hispanic American: 2.2 107; Asian American:
For 1990, total percent for group 2.488 108, or Caucasian: 1.88 108;
2. Calculate the total population for each of the five groups for 1990 and 2000.
percentage increase 2.66 107 2.488 108 10.7%.
Total increase 2.754 108 2.488 108 2.66 107;
the percent increase during this period?
1. By how much did the total U.S. population increase between 1990 and 2000? What was
Hispanic American
9.0%
Asian American
2.8%
Native American
0.70%
African American
11.8%
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.
T
Population Trends in the
United States
CHAPTER
Name
Challenge Problems Answer Key
Chemistry: Matter and Change
T29
3
4
4
4. What is the mass of the heaviest isotope? 204 amu
Challenge Problems
Chemistry: Matter and Change • Chapter 3
Physical change; the composition of the spaghetti has not changed.
7. When the spaghetti is cooked in the boiling water, it becomes soft.
evidence of a chemical change.
Chemical change; the formation of smoke and a permanent color change are
smoke.
6. A strand of spaghetti has fallen onto the burner, where it turns black and begins to
material.
without comparing the composition of the gray material with the original black
It is not possible to know whether this is a physical change or a chemical change
black to gray.
5. After the flame has been turned off, a small area on the burner has changed in color from
composition does not. When the metal cools, it will no longer be red.
Physical change; the appearance of the metal changes as it is heated, but its
temperature rises.
4. The metal burner on which the pot rests while being heated becomes red as its
Chemical change; the composition of the natural gas changes as it burns.
Atomic mass (amu)
Chemistry: Matter and Change • Chapter 4
atomic mass of 200.59 amu.
Challenge Problems
value of 200.3 amu is mercury. Element X is probably mercury, which has an
The element in the periodic table with an atomic mass closest to the calculated
Mass of X 200.3 amu
60.2 amu 13.9 amu
Mass of X 0.27 amu 19.8 amu 33.4 amu 46.2 amu 26.5 amu
(0.231)(200 amu) (0.132)(201) (0.298)(202 amu) (0.068)(204 amu)
Mass of X (0.0014)(196 amu) (0.10)(198 amu) (0.168)(199 amu)
element X. Identify the unknown element.
8. Using the data given by the graph, calculate the weighted average atomic mass of
the two most abundant isotopes.
The atomic mass of element X is between 200 and 202 amu, the masses of
X. Explain the basis for your prediction.
7. Without performing any calculations, predict the approximate atomic mass for element
201: about 13%; 202: about 30%; 204: about 7%
196: less than 1%; 198: about 10%; 199: about 17%; 200: about 23%;
6. Estimate the percent abundance of each isotope shown on the graph.
5. What is the mass of the lightest isotope? 196 amu
3. What is the mass of the least abundant isotope? 196 amu
its composition.
3. The heat used to cook is produced by burning natural gas in the stove burner.
Use with Chapter 4,
Section 4.3
Class
190 192 194 196 198 200 202 204 206 208 210
0
5
10
15
20
25
30
2. What is the mass of the most abundant isotope? 202 amu
1. How many isotopes of element X exist? 7
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.
A
Date
CHALLENGE PROBLEMS
Isotopes of an Element
CHAPTER
Name
Physical change; the water changes from liquid to gas, but there is no change in
2. As more heat is added, the water begins to boil and steam is produced.
Physical change; the composition of the water does not change.
1. As the water in the pot is heated, its temperature rises.
3
Use with Chapter 3,
Section 3.2
Class
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.
P
Date
CHALLENGE PROBLEMS
Physical and Chemical
Changes
CHAPTER
Name
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Percent abundance
T30
Chemistry: Matter and Change
Challenge Problems Answer Key
5
Use with Chapter 5,
Section 5.2
Class
1
1
Challenge Problems
5
4
3
2
n
Electron
0
m
Chemistry: Matter and Change • Chapter 5
1
2
ms
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
0
Boron (B)
5
For boron, the first two
electrons have quantum
numbers 1,0,0,1/2 and
1,0,0,1/2. The second two
electrons have quantum
numbers 2,0,0,1/2 and
2,0,0,1/2. The fifth electron
may have any one of six
possible quantum numbers:
2,1,1,1/2; 2,1,1,1/2;
2,1,0,1/2; 2,1,0,1/2;
2,1,1,1/2; 2,1,1,1/2.
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.
can be any integer from 0 to n 1,
n can be any integer (whole number),
The Schrödinger wave equation imposes certain mathematical restrictions on the quantum
numbers. They are as follows:
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).
T
Date
CHALLENGE PROBLEMS
Quantum Numbers
CHAPTER
Name
6
Use with Chapter 6,
Section 6.2
Class
Set 1
108
Actual:
6
Lead
Tin
Germanium
Element
Potassium
Sodium
Lithium
Element
Chemistry: Matter and Change • Chapter 6
62
108
Calculated:
Xenon
Radon
153
Krypton
7.2
100.98
113.4
Boiling Point (°C)
Set 4
Actual:
Calculated:
219.6
Melting Point (°C)
Element
Bromine
Chlorine
Fluorine
Element
midway in value.
39.098
22.990
23.019
Actual:
Calculated:
327
232
632
937
Melting Point (°C)
Set 5
Actual:
Calculated:
6.941
Atomic Mass
Set 2
Calcium
Magnesium
Beryllium
Element
Strontium
Calcium
Magnesium
Element
1107
1384
484
1246
1285
851
650
1068
Challenge Problems
Actual:
Calculated:
Boiling Point (°C)
Set 6
Actual:
Calculated:
Boiling Point (°C)
Set 3
No, only some of the properties of some of the middle elements of triads are
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?
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.
O
Date
CHALLENGE PROBLEMS
Döbereiner’s Triads
CHAPTER
Name
Challenge Problems Answer Key
Chemistry: Matter and Change
T31
7
CHALLENGE PROBLEMS
Date
0.0811
0.0231
0.0139
Nitrogen
Carbon
Silicon
Iron
0.00320
0.0269
0.459
0.0809
0.783
135
863
Solar System
170
140
500
Earth
19
210
610
30
Earth’s Crust
CHALLENGE PROBLEMS
Ar
K
Challenge Problems
Chemistry: Matter and Change • Chapter 7
potassium (#9), chlorine (#10), and phosphorus (#7).
important roles in the human body include sodium (#8 in list of top ten),
Students’ answers may vary. Elements discussed in Section 7.1 as having
expect to find in the list of the top ten elements in the human body. Explain your choices.
4. Name at least three elements in addition to those shown in the table that you would
many kinds of rocks with oxygen, silicon, and carbon.
choices. Aluminum is the most abundant metal in Earth’s crust. Calcium forms
Chemistry: Matter and Change • Chapter 8
Challenge Problems
Ca2; like argon, Ca2 has 18 electrons, eight of which are valence electrons.
your answer.
6. Name an ion of calcium that has chemical properties similar to those of argon. Explain
eight of which are valence electrons.
The Lewis electron dot structure for argon; both argon and K have 18 electrons,
your answer.
5. Which of the above Lewis electron dot structures is like that for the ion K? Explain
eight of which are valence electrons.
Explain your answer.
4. Which of the above Lewis electron dot structures is the same as that for the ion Cl?
eight of which are valence electrons.
The Lewis electron dot structure for argon; both argon and S2 have 18 electrons,
structure for the ion S2? Explain your answer.
3. Which of the above Lewis electron dot structures is the same as the Lewis electron dot
K: 1s22s22p63s23p64s1; Ca: 1s22s22p63s23p64s2
S: 1s22s22p63s23p4; Cl: 1s22s22p63s23p5; Ar: 1s22s22p63s23p6;
2. Write the electron configuration for each of the five elements.
S: 16; Cl: 17; Ar: 18; K: 19; Ca: 20
Ca
Based on the information in Section 7.1, aluminum and calcium are possible
8
Cl
1. Write the atomic number for each of the five elements shown above.
S
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.
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.
T
Use with Chapter 8,
Section 8.1
Class
The Lewis electron dot structure for argon; both argon and Cl have 18 electrons,
7
8
Date
Comparing the Structures of
Atoms and Ions
CHAPTER
Name
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.
3. Only the top four most abundant elements on Earth and in Earth’s crust are shown in the
gravitational attraction.
Hydrogen and helium are gases of low density that have escaped from Earth’s
to their relatively low abundance on Earth.
2. Explain the relatively high abundance of hydrogen and helium in the universe compared
solar system: 863 H 135 He 1000 100% 99.8%
universe: 927 H 71.8 He 1000 100% 99.9%;
all atoms in the solar system are either hydrogen or helium?
106
24
257
606
Human Body
Use with Chapter 7,
Section 7.1
Class
1. What percent of all atoms in the universe are either hydrogen or helium? What percent of
* An element is not abundant in a region that is left blank.
0.510
0.153
Oxygen
71.8
927
Hydrogen
Helium
Universe
Element
Abundance (Number of atoms per 1000 atoms)*
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.
T
Abundance of the Elements
CHAPTER
Name
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
T32
Chemistry: Matter and Change
Challenge Problems Answer Key
9
CHALLENGE PROBLEMS
Date
F
F
F
F
F
Challenge Problems
Chemistry: Matter and Change • Chapter 9
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
the central atom (I).
IF5 does not obey the octet rule because there are 12 electrons attached to
8. Does IF5 obey the octet rule? Explain.
7. Draw the Lewis structure for the compound IF5.
by four electrons pairs.
N2F2 obeys the octet rule because every atom of the molecule is surrounded
6. Does N2F2 obey the octet rule? Explain.
F N N F
5. Draw the Lewis structure for the compound N2F2.
not surrounded by four electron pairs.
NO2 does not obey the octet rule because the central atom (N) of the molecule is
4. Does NO2 obey the octet rule? Explain.
O N O
3. Draw the Lewis structure for the compound NO2.
is not surrounded by four electron pairs.
BeF2 does not obey the octet rule because the central atom (Be) of the molecule
2. Does BeF2 obey the octet rule? Explain.
F Be F
1. Draw the Lewis structure for the compound BeF2.
9
Use with Chapter 9,
Section 9.3
Class
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.
T
Exceptions to the Octet Rule
CHAPTER
Name
10
The equation is not balanced. 2K(s) 2H2O(l) 0 2KOH(aq) H2(g)
Chemistry: Matter and Change • Chapter 10
6HNO3(aq)
Challenge Problems
The equation is not balanced. 3H2SO4(aq) 2Al(NO3)3(aq) 0 Al2(SO4)3(aq) 9. H2SO4(aq) 2Al(NO3)3(aq) 0 Al2(SO4)3(aq) 2HNO3(aq)
An incorrect formula is given for calcium chloride. Ca(s) Cl2(g) 0 CaCl2(aq)
8. 2Ca(s) Cl2(g) 0 2CaCl(aq)
equation can be written.
The products are the same as the reactants, so no reaction occurred and no
7. NH4OH(aq) KOH(aq) 0 KOH(aq) NH4OH(aq)
water is not given. LiOH(aq) HBr(aq) 0 LiBr(aq) H2O(l)
The coefficients are not expressed in lowest terms. Also, the physical state of
6. 2LiOH(aq) 2HBr(aq) 0 2LiBr(aq) 2H2O
the equation is not balanced. 2NaHCO3(s) 2HCl(aq) 0 2NaCl(s) 2CO2(g) 2H2O(l)
A term must be missing because there is no product containing hydrogen. Therefore,
5. 2NaHCO3(s) 2HCl(aq) 0 2NaCl(s) 2CO2(g)
Sr(s) F2(g) 0 SrF2(s)
not balanced. Also, the physical state of strontium fluoride is not given.
An incorrect formula is given for strontium fluoride. Therefore, the equation is
4. Sr(s) F2(g) 0 Sr2F
The equation is not balanced. 2AgNO3(aq) H2S(aq) 0 Ag2S(aq) 2HNO3(aq)
3. AgNO3(aq) H2S(aq) 0 Ag2S(aq) HNO3(aq)
not balanced. MgCl2(aq) H2SO4(aq) 0 MgSO4(aq) 2HCl(aq)
An incorrect formula is given for magnesium sulfate. Therefore, the equation is
2. MgCl2(aq) H2SO4(aq) 0 Mg(SO4)2(aq) 2HCl(aq)
1. K(s) 2H2O(l) 0 2KOH(aq) H2(g)
10
Use with Chapter 10,
Section 10.1
Class
ach chemical equation below contains at least one error. Identify the error or errors and
then write the correct chemical equation for the reaction.
E
Date
CHALLENGE PROBLEMS
Balancing Chemical
Equations
CHAPTER
Name
Challenge Problems Answer Key
Chemistry: Matter and Change
T33
11
352.03
187.37
207.2
225.0
200.0
250.0
UF6(g)
CCl3CF3(l)
Pb(s)
1.207
1.067
0.6392
Number of
Moles (mol)
7.266 1023 atoms
6.423 1023 molecules
3.848 1023 molecules
Number of Representative
Particles
250.0 g
Pb (s)
Use with Chapter 11,
Section 11.3
Class
Challenge Problems
Chemistry: Matter and Change • Chapter 11
0.15 mol Pb 207.2 g Pb/1.0 mol Pb 31 g Pb
in an alloy containing 0.15 mol of lead?
4. Lead (Pb) is used to make a number of different alloys. What is the mass of lead present
11
1.0 g CCl3CF3 1.0 mol CCl3CF3/187.37 g CCl3CF3 6.02 1023 molecules CCl3CF3/
1.0 mol CCl3CF3 3.2 1021 molecules CCl3CF3
Earth’s atmosphere. How many molecules of the liquid are in 1.0 g of CCl3CF3?
3. CCl3CF3 is a chlorofluorocarbon responsible for the destruction of the ozone layer in
100.0 g UF6 1 mol UF6/352.03 g UF6 0.2841 mol UF6
the gas are in 100.0 g of UF6?
2. UF6 is a gas used in the production of fuel for nuclear power plants. How many moles of
molecule has a significantly greater mass than a single CCl3CF3 molecule has.
the molar mass of UF6 is significantly greater than that of CCl3CF3. A single UF6
tive containers, there are fewer molecules of UF6 than of CCl3CF3. That is because
Even though the mass of UF6 is greater than the mass of CCl3CF3 in their respec-
the number of representative particles and mass of CCl3CF3. Explain any differences
you observe.
1. Compare and contrast the number of representative particles and the mass of UF6 with
Molar Mass
(g/mol)
Mass (g)
200.0 g
Substance
CCl3CF3 (l)
UF6 (g)
225.0 g
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.
T
Date
CHALLENGE PROBLEMS
Using Mole-Based
Conversions
CHAPTER
Name
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
12
Use with Chapter 12,
Section 12.2
Class
4
Grams of
CO
2
Moles of
CO
5
1
Grams of
CO2
6
Moles of
CO2
7
Particles of
CO2
12
Chemistry: Matter and Change • Chapter 12
Challenge Problems
The masses of the substances do not react with each other in definite ratios.
of a second substance in the reaction, as represented by double-arrow 5 in the diagram?
6. Why is it not possible to use the mass of one substance in a chemical reaction to find the mass
Masses of different substances contain different numbers of representative particles.
representative particles, as represented by double-arrow 4 of the diagram?
5. Why is it not possible to convert between the mass of a substance and the number of
6.02 1023 particles CO/1 mol CO and 1 mol CO/6.02 1023 particles CO
double-arrow 3 in the diagram for CO?
4. What conversion factors would be needed to make the conversions represented by
6.02 1023
the number of representative particles in a mole, or Avogadro’s number:
double-arrows 3 and 7 in the diagram?
3. What information is needed to make the types of conversions represented by
1 mol CO2/44.01 g CO2
28.01 g CO/1 mol CO and 1 mol CO/28.01 g CO; 44.01 g CO2/1 mol CO2 and
double-arrow 2 in the diagram for CO? By double-arrow 6 for CO2?
2. What conversion factors would be needed to make the conversions represented by
the coefficients of the substances in the balanced chemical equation
in the diagram?
1. What information is needed to make the types of conversions shown by double-arrow 1
Particles of
CO
3
Use the equation and the diagram to answer the following questions.
2CO(g) O2(g) 0 2CO2(g)
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.
T
Date
CHALLENGE PROBLEMS
Mole Relationships in
Chemical Reactions
CHAPTER
Name
T34
Chemistry: Matter and Change
Challenge Problems Answer Key
13
CHALLENGE PROBLEMS
Date
0
0
100
0
100
SiH4
150
CHALLENGE PROBLEMS
Figure 1
At sea level
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Chemistry: Matter and Change • Chapter 13
14
Chemistry: Matter and Change • Chapter 14
Challenge Problems
pressure exerted on the mercury in the bowl.
Challenge Problems
tube. Air pressure exerted on the mercury in the tube would equal the air
Air would enter the glass tube, causing the mercury column to flow out of the
event affect the column of mercury? Explain why.
6. Suppose a tiny crack formed at the top of the barometer’s glass tube. How would this
water than of mercury.
dense than mercury, air pressure is able to support a much longer column of
The column of liquid in the tube would be longer. Because water is much less
this substitution affect the barometer? Explain.
5. Suppose the liquid used to make the barometer was water instead of mercury. How would
pressure would be pushing on the surface of the mercury in the bowl.
The height of the mercury column would increase in the mine because more air
level. How would the height of the mercury column change? Explain why.
4. Suppose the barometer in Figure 1 was carried into an open mine 500 meters below sea
of the mercury in the bowl in Figure 2.
pressure at the lower elevation in Figure 1, there is less pushing on the surface
Because air pressure at the higher elevation in Figure 2 is less than the air
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?
3. When the barometer in Figure 1 is moved to a higher elevation, such as an altitude of
in the glass tube.
Air pressure pushing on the surface of the mercury in the bowl keeps the mercury
Figure 2
At 500 meters
above sea level
Bowl of
mercury
Mercury
column
Glass tube
2. What prevents mercury from flowing out of the glass tube into the bowl of mercury?
the mercury column in the glass tube.
Hg vapor due to the presence of a vacuum above
The space is occupied only by small amounts of
barometer’s glass tube?
1. What occupies the space above the mercury column in the
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.
I
Use with Chapter 14,
Section 14.1
Class
H—Te bond: en 0.1
13
14
Date
A Simple Mercury Barometer
CHAPTER
Name
bonds in the four group 6A hydrides.
H—O bond: en 1.2; H—S bond: en 0.4; H—Se bond: en 0.4;
6. Use Table 9-4 in your textbook to determine the difference in electronegativities of the
All group 6A hydride molecules are bent, polar molecules.
5. What are the molecular structure and polarity of the four group 6A hydrides?
point than any other group 6A hydride.
do the boiling points of the hydrides. Water, however, has a much higher boiling
With the exception of water, as the molecular masses of the hydrides increase, so
the hydrides change?
4. How do the boiling points of the group 6A hydrides change as the molecular masses of
forces between their molecules have little effect on their boiling points.
Because group 4A hydrides are nonpolar, the relatively weak intermolecular
those forces affect the boiling points of group 4A hydrides.
3. Predict the strength of the forces between group 4A hydride molecules. Explain how
All group 4A hydride molecules are regular tetrahedrons and are nonpolar.
2. What are the molecular structure and polarity of the four group 4A hydrides?
hydrides.
As the molecular masses of the hydrides increase, so do the boiling points of the
the hydrides change?
H2Te
SnH4
GeH4
Group 4A
hydrides
H2Se
Group 6A
hydrides
50
100
Molecular mass
CH4
H2S
H2O
Use with Chapter 13,
Section 13.3
Class
1. How do the boiling points of the group 4A hydrides change as the molecular masses of
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.
T
Intermolecular Forces and
Boiling Points
CHAPTER
Name
Boiling point (°C)
15
CHALLENGE PROBLEMS
Date
Water
molecule
Class
Chemistry: Matter and Change
Challenge Problems
Chemistry: Matter and Change • Chapter 15
reduction 17.54 mm Hg 17.4 mm Hg 0.1 mm Hg
addition of the sucrose?
6. How much is the vapor pressure of the solution reduced from that of water by the
P P 0X; P (17.54 mm Hg)(0.992) 17.4 mm Hg
17.54 mm Hg?
5. What is the vapor pressure of the solution if the vapor pressure of pure water at 20°C is
27.75 mol/(27.75 mol 0.219 mol) 0.992
Mole fraction of water moles solvent/(moles solvent moles solute)
4. What is the mole fraction of water in the solution?
500.0 g water 1 mol water/18.02 g water 27.75 mol water
3. What is the number of moles of water in the solution?
75.0 g sucrose 1 mol sucrose/342.3 g sucrose 0.219 mol sucrose
2. What is the number of moles of sucrose in the solution?
molecules able to escape into the vapor phase.
number of water molecules at the surface, thus reducing the number of water
The presence of sugar molecules at the surface of the solution reduces the
Sucrose
molecule
15
Solution
Use with Chapter 15,
Section 15.3
1. Why do the sugar molecules in the solution lower the vapor pressure of the water?
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.
P P0X
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
Y
Vapor Pressure Lowering
CHAPTER
Name
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
16
CHALLENGE PROBLEMS
Date
H 58 kJ/mol
H 91 kJ/mol
H 33 kJ/mol
16
Chemistry: Matter and Change • Chapter 16
each number on the diagram.
3. Write the correct enthalpy change next to
each of the lines labeled A, B, and C.
2. Write the correct reactants and/or products on
to show the directions in which the three lines
labeled 1, 2, and 3 should point.
1. On the diagram at the right, draw arrowheads
1 O (g) 0 NO (g)
NO(g) 2
2 2
1
1
N2(g) O2(g) 0 NO(g)
2
2
1
N2(g) O2(g) 0 NO2(g)
2
or
The equations below show how NO2 can be
formed in two ways: directly from the elements
or in two steps.
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.
H
1
O (g)
2 2
1 H 91 kJ/mol
Challenge Problems
3 H 33 kJ/mol
B NO2(g)
2 H 58 kJ/mol
NO(g) 1/2 O2(g)
A 1/2 N2(g) O2(g)
C
CO2(g)
CO(g) H 110 kJ/mol
Use with Chapter 16,
Section 16.4
Class
H 283 kJ/mol
H 393 kJ/mol
C(s) O2(g)
Standard Heat of Formation
CHAPTER
Name
Enthalpy
Enthalpy
Challenge Problems Answer Key
T35
17
CHALLENGE PROBLEMS
Date
Chemistry: Matter and Change
0
0 1 2 3 4 5 6 7 8 9 10
Time (h)
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
After 2 hours? After 10 hours?
Challenge Problems
Challenge Problems Answer Key
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Chemistry: Matter and Change • Chapter 17
(1.50 mol/L 0.18 mol/L)10 h 0.13 mol/L/h
6. What is the average rate of reaction for the decomposition of N2O5 overall?
x (0.10 mol/L)/(0.20 mol/L/h) 5 hours
0.20 mol/L/h 0.10 mol/L/x
5. How long does it take for 0.10 mol of N2O5 to decompose during the tenth hour of the reaction?
(0.70 mol/L 0.25 mol/L)/5 h 0.09 mol/L/h
(0.90 mol/L 0.55 mol/L)/2 h 0.18 mol/L/h;
ond and fourth hours of the reaction? Between the third and eighth hours of the reaction?
4. What is the instantaneous rate of reaction for the decomposition of N2O5 between the sec-
0.20 mol/L/h; (0.35 mol/L 0.30 mol/L)/1 h 0.050 mol/L/h
(1.20 mol/L 0.09 mol/L)/1 h 0.30 mol/L/h; (0.90 mol/L 0.70 mol/L)/1 h 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?
3. The instantaneous rate of reaction is defined as the change in concentration of reactant
(1.50 mol/L 1.20 mol/L)/1.50 mol/L .20 100% 20%
1.50 mol/L 1.20 mol/L 0.30 mol/L;
reaction? Calculate the percentage of change the concentration undergoes during the
first hour of the reaction.
2. By how much does the concentration of N2O5 change during the first hour of the
1.50 mol/L; 1.20 mol/L; 0.90 mol/L; 0.18 mol/L
17
Use with Chapter 17,
Section 17.1
Class
1. What is the concentration of N2O5 at the beginning of the experiment? After 1 hour?
The graph on the right represents the concentration of N2O5 remaining as the reaction proceeds
over time. Answer the following questions about
the reaction.
2N2O5(g) 0 4NO2(g) O2(g)
initrogen pentoxide decomposes to produce
nitrogen dioxide and oxygen as represented
by the following equation.
D
Concentration (mol/L)
Determining Reaction Rates
CHAPTER
Name
18
CHALLENGE PROBLEMS
Date
0
8
7
6
5
4
3
2
1
18
Chemistry: Matter and Change • Chapter 18
at 6 minutes
8. At what time does the system return to equilibrium?
suddenly at that time.
Challenge Problems
SO2 was added to the system because the concentration of SO2 increased
the change was made.
7. Describe the change made in the system 4 minutes into the reaction. Tell how you know
Keq [1.0]2/[6.0]2[4.5] 1.0/(36)(4.5) 0.0062
6. Calculate the value of Keq for the reaction.
[SO2]eq 6.0 mol/L; [O2]eq 4.5 mol/L; [SO3]eq 1.0 mol/L
5. What are the concentrations of the three gases at equilibrium?
gases become constant at that time.
Equilibrium is reached at about 3 minutes; the concentrations of all three
equilibrium has been reached?
4. At approximately what time does the reaction reach equilibrium? How do you know
up to produce SO3; the concentration of SO3 increases.
The concentrations of the two reactants, SO2 and O2, decrease as they are used
reaction.
SO3
O2
SO2
0 1 2 3 4 5 6 7 8 9 10
Time (sec)
SO3
O2
SO2
Class
Use with Chapter 18,
Section 18.1
3. Explain the shapes of the curves for the three gases during the first 2 minutes of the
Keq [SO3]2/[SO2]2[O2]
2. Write the equilibrium constant expression for the reaction.
2SO2(g) O2(g) 3 2SO3(g)
1. Write the equation for the reaction depicted in the graph.
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.
Changing Equilibrium
Concentrations in a Reaction
CHAPTER
Name
Concentration (mol/L)
T36
Challenge Problems Answer Key
Chemistry: Matter and Change
T37
19
CHALLENGE PROBLEMS
Date
20
Sacrificial
metal
Steel wire
Chemistry: Matter and Change • Chapter 19
20
Chemistry: Matter and Change • Chapter 20
Challenge Problems
Al(s) 0 Al3(aq) 3e; Fe2(aq) 2e 0 Fe(s)
more HOCl to break down, leaving less HOCl to kill bacteria.
Challenge Problems
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.
6. Suppose that some iron in the casing of the water heater is oxidized, as shown in the
Al(s) 0 Al3(aq) 3e; O2(aq) 2H2O(l) 4e 0 4OH(aq)
5. Write the two half-reactions for this example of corrosion.
the balanced equation for the reaction of aluminum with oxygen dissolved in the water.
(Hint: The product formed is aluminum hydroxide (Al(OH)3).
4Al(s) 3O2(aq) 6H2O(l) 0 4Al(OH)3(aq)
4. Suppose the sacrificial rod in the diagram above is coated with aluminum metal. Write
O2(aq) 2H2O(l) 4e 0 4OH(aq)
Fe(s) 0 Fe2(aq) 2e;
3. Write the two half-reactions for this example of corrosion.
2Fe(s) O2(aq) + 2H2O 0 2Fe(OH)2(aq)
Fe(s) O2(aq) H2O 0 Fe(OH)2(aq)
2. Balance the oxidation–reduction equation for this reaction:
Iron is oxidized, and oxygen is reduced.
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?
Water
Iron
casing
Use with Chapter 20,
Section 20.3
Class
1. In the absence of a sacrificial metal, oxygen dissolved in water may react with the iron
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.
S
Date
CHALLENGE PROBLEMS
Balancing Oxidation–
Reduction Equations
CHAPTER
Name
Removal of OCl drives the reaction described in question 2 to the left, causing
gas. Write an equation for this reaction and tell how it affects the safety of pool water.
2OCl(aq) 0 2Cl(aq) O2(g)
5. In the presence of sunlight, hypochlorite ion decomposes to form chloride ion and oxygen
more OCl in solution, making it easier for bacteria to survive.
reaction to the left. As a result, there will be relatively less HOCl and relatively
A high pH value indicates a high concentration of OH, which tends to drive the
affect the equilibrium reaction described in question 2? What effect would high pH have
on the bacteria?
4. The effectiveness of hypochlorite ion as a bactericide depends on pH. How does high pH
therefore increases pH.
The addition of hypochlorite ion increases the concentration of OH and
3. What effect does the addition of hypochlorite ion have on the pH of swimming pool water?
conjugate base is OH.
The acid is H2O, and the base is OCl. The conjugate acid is HOCl, and the
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.
OCl(aq) H2O(l) 3 HOCl(aq) OH
2. Write an equation that shows the reaction that occurs when the hypochlorite ion (OCl),
conjugate base is OCl.
The acid is HOCl, and the base is H2O. The conjugate acid is H3O, and the
the acid, base, conjugate acid, and conjugate base in this reaction.
HOCl(aq) H2O(l) 3 H3O(aq) OCl(aq)
1. Write an equation that shows the reaction between hypochlorous acid and water. Identify
19
Use with Chapter 19,
Section 19.2
Class
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.
T
Swimming Pool Chemistry
CHAPTER
Name
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
T38
Chemistry: Matter and Change
Challenge Problems Answer Key
21
CHALLENGE PROBLEMS
Date
Cu2
1.0 103M
Ag
1.0 102M
CHALLENGE PROBLEMS
CH3
CH3
CH3
C
CH3
CH
CH3
CH2
CH2
CH2
CH3
CH
CH2
CH2
CH3
CH3
CH3
CH2
CH3
h.
CH3
CH3
g. CH2
f. CH3
e. CH3
CH
CH3
CH2
CH
CH3
CH
CH3
CH
CH2
CH2
CH3
CH3
CH3
CH
CH2
CH2 CH3
CH3
CH3
j.
CH2
CH3
CH3
i. CH2
CH2
CH3
CH2
CH
CH2
CH3 CH3
CH2
hexane
2,3-dimethylbutane
2-methylpentane
h.
g.
f.
e.
3-methylpentane
2-methylpentane
3-methylpentane
2,3-dimethylbutane
j.
i.
hexane
3-methylpentane
Challenge Problems
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Chemistry: Matter and Change • Chapter 21
Chemistry: Matter and Change • Chapter 22
Challenge Problems
and 3-methylpentane
22
five; hexane, 2-methylpentane, 2,3-dimethylbutane, 2,2-dimethylbutane,
question 1.)
4. Calculate the cell potential for the ion concentrations shown in the cell. 0.16 V
are their names?
2. How many different compounds are represented by the structural formulas above? What
d. 2,2-dimethylbutane
c.
b.
a.
a–j, above.
1. In the spaces provided, write the correct name for each of the structural formulas, labeled
d.
c.
CH3
CH
b. CH3
CH2
a. CH3
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.
T
Use with Chapter 22,
Sections 22.1 and 22.3
Class
two moles of electrons are transferred in the overall cell reaction (See answer to
21
22
Date
Structural Isomers of Hexane
CHAPTER
Name
Ecell 0.46 V 0.0592/2log 1.0 103/(1.0 102)2; n 2 in this case because
3. Write the Nernst equation for the cell.
2. Use Table 21-1 in your textbook to determine the standard potential of this cell. 0.46 V
2Ag(aq) Cu(s) 0 2Ag(s) Cu2(aq)
Cu(s) 0 Cu2(aq) 2e
Ag(aq) 1e 0 Ag(s)
Class
Use with Chapter 21,
Section 21.1
1. Write the two half-reactions and the overall cell reaction for the cell shown above.
Cu
Ag
Voltmeter
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.
[product ion]x
0.0592 log Ecell E 0cell n [reactant ion]y
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:
I
Effect of Concentration on
Cell Potential
CHAPTER
Name
23
T
Challenge Problems Answer Key
30
50
0
50
80
Chemistry: Matter and Change
Challenge Problems
aldehydes lack OH groups.
Chemistry: Matter and Change • Chapter 23
Neither propanal nor any other aldehyde can form hydrogen bonds because
Explain.
5. Can this aldehyde form hydrogen bonds? Can other aldehydes form hydrogen bonds?
The aldehyde is propanal, CH3CH2CHO.
chemical formula.
4. Find the aldehyde with a molecular mass of about 58. Name that aldehyde and write its
because they have the same molecular mass.
25°C; the two compounds would be expected to have similar boiling points
Boiling point of ethanol: about 78°C; boiling point of dimethyl ether: about
(molecular mass 46) on the graph. Why would you expect these two compounds to
have relatively similar boiling points?
3. Find and list the boiling points for ethanol (molecular mass 46) and dimethyl ether
points.
23
The alkanes have the lowest boiling points; the alcohols have the highest boiling
has the lowest boiling points? Which family has the highest boiling points?
2. For compounds of similar molecular mass, which family of the four shown in the graph
As the molecular mass increases, so does the boiling point.
aldehyde
ether
50
60
70
Molecular mass
alkane
alcohol
40
Use with Chapter 23,
Section 23.3
Class
1. For any one family, what is the relationship between molecular mass and boiling point?
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.
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).
100
CHALLENGE PROBLEMS
Date
Boiling point (°C)
Boiling Points of Organic
Families
CHAPTER
Name
24
G
A
C
U
UUU
UUC
UUA
UUG
CUU
CUC
CUA
CUG
AUU
AUC
AUA
AUG
GUU
GUC
GUA
GUG
Leucine
b. UCA
Serine
} His
} Gln
} Asn
} Lys
} Asp
} Glu
Stop
Stop
}Tyr
24
b.
a.
Chemistry: Matter and Change • Chapter 24
-His-Tyr-Arg-Ser-Phe-Leu-
-His-His-Arg-Phe-Phe-Ser-
4. Write the amino acid sequence for each of the mRNA sequences shown in question 3.
C (cytosine) at position #4 has been changed to U (uracil).
b. -C-A-U-U-A-C-C-G-G-U-C-U-U-U-U-C-U-U-
An extra U (uracil) has been added at position #10.
a. -C-A-U-C-A-C-C-G-G-U-U-C-U-U-U-U-C-U-U-
UGU
UGC
UGA
UGG
CGU
CGC
CGA
CGG
AGU
AGC
AGA
AGG
GGU
GGC
GGA
GGG
U
C
Stop A
Trp G
U
C
Arg
A
G
U
Ser
C
A
Arg
G
U
C
Gly
A
G
}
}
} Cys
Challenge Problems
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.
3. Errors sometimes occur when mRNA molecules are synthesized from DNA molecules.
-His-His-Arg-Ser-Phe-Leu-
-C-A-U-C-A-C-C-G-G-U-C-U-U-U-U-C-U-U-
2. Write the sequence of amino acids for which the following mRNA sequence codes.
a. CUG
UAU
UAC
Ser
UAA
UAG
CAU
CAC
Pro
CAA
CAG
AAU
AAC
Thr
AAA
AAG
GAU
GAC
Ala
GAA
GAG
A
Second base
C
G
Use with Chapter 24,
Section 24.4
Class
The Genetic Code
UCU
UCC
UCA
UCG
CCU
CCC
Leu
CCA
CCG
ACU
Ile
ACC
ACA
Met ACG
GCU
GCC
Val
GCA
GCG
} Phe
} Leu
U
1. What amino acid is represented by each of the following codons?
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.
P
Date
CHALLENGE PROBLEMS
The Chemistry of Life
CHAPTER
Name
First base
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Third base
T39
T40
Chemistry: Matter and Change
Challenge Problems Answer Key
25
uranium-235 (235
92U)
silver-114 (114
47Ag)
48p
77n
Source
of
neutrons
1n
0
1n
0
45p
75n
F
G
A
92p
143n
0
–1
1n
0
1n
0
E
C
0
–1
0
0
92p
143n
92p
146n
B
uranium-238 (238
92U)
0
0
1
239
93Np; neptunium-239
Challenge Problems
Chemistry: Matter and Change • Chapter 25
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Identify isotope G formed in the reaction.
1n 239Pu 0 125Cd 115Pd; palladium-115
0
94
48
46
10. Write the nuclear equation for the reaction that occurs when a neutron strikes nucleus F.
nuclear fission
9. Name the type of nuclear reaction that occurs when a neutron strikes nucleus F.
in the reaction.
239Np 0 0
239Pu; plutonium-239
1
93
94
8. Write a balanced nuclear equation for the decay of nucleus E. Identify isotope F formed
239U
92
reaction.
7. Write the nuclear equation for the decay of nucleus D. Identify isotope E formed in the
Identify the product D formed in the reaction.
1n 238U 0 0 239U; The product nucleus is uranium-239.
0
92
0
92
6. Write the nuclear equation for the reaction that occurs when a neutron strikes nucleus C.
5. Identify the isotope whose nucleus is labeled C.
D
25
Use with Chapter 25,
Section 25.4
Class
4. Besides fragmented nuclei, what else is produced when a neutron strikes nucleus A? neutrons
3. Identify the isotope whose nucleus is labeled B.
neutron strikes nucleus A. nuclear fission
2. Name the type of nuclear reaction that occurs when a
diagram.
1. Identify the isotope whose nucleus is labeled A in the
The diagram shows the process by which plutonium-239
is made in nuclear reactors. Answer the questions about the
diagram.
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.
W
Date
CHALLENGE PROBLEMS
The Production of
Plutonium-239
CHAPTER
Name
26
26
Chemistry: Matter and Change • Chapter 26
Challenge Problems
Short-term cycle: animals excreting phosphates; decomposing organisms; animals
eating plants; plants extracting phosphates from the soil; mining; use of fertilizers, detergents, and other synthetic products containing phosphorus. Long-term
cycle: geological uplift, phosphates leaching from rocks, phosphate rock forming
under the oceans
pencils to show each part on the diagram.
5. The phosphorus cycle has both short-term and long-term parts. Use different colored
The phosphorus cycle has no atmospheric component.
studied in the textbook?
4. In what way is the phosphorus cycle different from the carbon and nitrogen cycles you
by eating plants or other animals
3. By what method do animals obtain the phosphorus they need?
by extracting phosphates from the soil
2. By what method do plants obtain the phosphorus they need?
Rainfall leaches phosphates from rocks.
Animals excrete phosphates. Dead organisms release phosphates as they decay.
1. By what methods does phosphorus get into soil?
Geological uplift
Phosphate
rocks
Phosphate rocks
Use with Chapter 26,
Section 26.4
Class
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.
P
Date
CHALLENGE PROBLEMS
The Phosphorus Cycle
CHAPTER
Name
CREDITS
Art Credits
Copyright © Glencoe/McGraw-Hill, a division of the McGraw-Hill Companies, Inc.
Navta Associates: 1, 2, 4, 11, 12, 13, 15, 16, 17, 18, 20, 21, 23, 24; MacArt Design: 14, 25, 26
Challenge Problems
Chemistry: Matter and Change
T41
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