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ChallengeProblems
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ISBN 0-07-824533-8Printed 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
A Glencoe Program
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Challenge Problems Chemistry: Matter and Change iii
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
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
Contents
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iv Chemistry: Matter and Change Challenge Problems
Students can take their learning one step beyond their textbooks withChallenge Problems. The worksheets in this supplement to Chemistry:Matter and Change challenge students to apply their knowledge ofchemistry to new situations, analyze and interpret those situations, andsynthesize responses. Whether analyzing experimental results or investigatinga hypothetical situation, these worksheets encourage students to use chemicalconcepts along with their critical thinking skills to solve problems.
To the Teacher
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Challenge Problems Chemistry: Matter and Change • Chapter 1 1
Production ofChlorofluorocarbons, 1950–1992Production ofChlorofluorocarbons, 1950–1992
Chlorofluorocarbons (CFCs) were first produced in the laboratory in the late 1920s. They did not
become an important commercial product until sometime later. Eventually, CFCs grew in popularity untiltheir effect on the ozone layer was discovered in the1970s. The graph shows the combined amounts of twoimportant CFCs produced between 1950 and 1992.Answer the following questions about the graph.
CHALLENGE PROBLEMSCHAPTER 1
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 producedthat year?
3. During what two-year period did the production of CFCs decrease by the greatestamount? By about how much did their production decrease?
4. During what two-year period did the production of CFCs increase by the greatestamount? What was the approximate percent increase during this period?
5. How confident would you feel about predicting the production levels of CFCs during theodd numbered years 1961, 1971, and 1981? Explain.
6. Could the data in the graph be presented in the form of a circle graph? Explain.
Year
Am
ou
nt
of
CFC
s(b
illio
n k
ilog
ram
s)
050
100150200250300350400
1950 1960 1970 1980 1990
Use with Chapter 1,Section 1.1
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2 Chemistry: Matter and Change • Chapter 2 Challenge Problems
Population Trends in theUnited StatesPopulation Trends in theUnited States
CHALLENGE PROBLEMSCHAPTER 2
Use with Chapter 2,Section 2.4
1. By how much did the total U.S. population increase between 1990 and 2000? What wasthe percent increase during this period?
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. Inwhat ways is the bar graph better than the circle graphs? In what way is it less useful?
U.S. Population Distribution
(Percentages may not add up to 100% due to rounding.)
Caucasian71.4%Native American
0.70%
Asian American3.8%
Hispanic American11.8%
African American12.2%
Caucasian75.7%
Native American0.70%
Asian American2.8%
Hispanic American9.0%
African American11.8%
20001990
The population of the United States is becoming more diverse. The circle graphs below show thedistribution 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.
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Challenge Problems Chemistry: Matter and Change • Chapter 3 3
Physical and ChemicalChangesPhysical and ChemicalChanges
Physical 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 choicebetween these two possibilities. If you are unable to decide whether the change is physical orchemical, tell what additional information you would need in order to make a decision.
CHALLENGE PROBLEMSCHAPTER 3
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.
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 fromblack to gray.
6. A strand of spaghetti has fallen onto the burner, where it turns black and begins tosmoke.
7. When the spaghetti is cooked in the boiling water, it becomes soft.
Use with Chapter 3,Section 3.2
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4 Chemistry: Matter and Change • Chapter 4 Challenge Problems
Isotopes of an ElementIsotopes of an Element
A mass spectrometer is a device for separatingatoms and molecules according to their
mass. A substance is first heated in a vacuum andthen ionized. The ions produced are acceleratedthrough a magnetic field that separates ions of dif-ferent masses. The graph below was producedwhen a certain element (element X) was analyzedin a mass spectrometer. Use the graph to answerthe questions below.
CHALLENGE PROBLEMSCHAPTER 4
Use with Chapter 4,Section 4.3
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?
6. Estimate the percent abundance of each isotope shown on the graph.
7. Without performing any calculations, predict the approximate atomic mass for elementX. 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.
0
5
10
15
20
25
30
196194192190 198 200 202 204 206 208 210Atomic mass (amu)
Perc
ent
abu
nd
ance
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Challenge Problems Chemistry: Matter and Change • Chapter 5 5
Quantum NumbersQuantum Numbers
CHALLENGE PROBLEMSCHAPTER 5
The 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, �, describesthe shape of the electron’s orbit around the nucleus. The third quantum number, m�, describesthe orientation of the electron’s orbit compared to the plane of the atom. The fourth quantumnumber, ms, tells the direction of the electron’s spin (clockwise or counterclockwise).
The Schrödinger wave equation imposes certain mathematical restrictions on the quantumnumbers. They are as follows:
n can be any integer (whole number),
� can be any integer from 0 to n � 1,
m� can be any integer from �� to ��, and
ms can be � or �
As an example, consider electrons in the first energy level of an atom, that is, n � 1. Inthis 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
this energy level can have values of � or � for ms. These restrictions agree with the
observation that the first energy level can have only two electrons. Their quantum numbers
are 1, 0, 0, � and 1, 0, 0 � .
Use the rules given above to complete the table listing the quantum numbers for eachelectron in a boron atom. The correct quantum numbers for one electron in the atom is provided as an example.
1�2
1�2
1�2
1�2
1�2
1�2
Use with Chapter 5,Section 5.2
Electron n � m� ms
1 1 0 0 �
2
3
4
5
1�2
Boron (B)
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6 Chemistry: Matter and Change • Chapter 6 Challenge Problems
Döbereiner’s TriadsDöbereiner’s Triads
One 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 betweenthe atomic masses of calcium (27.5) and barium (72.5). Note that the accepted atomic massesfor these elements today are very different from their accepted atomic masses at the timeDöbereiner made his observations. Döbereiner also observed that strontium, calcium, and bar-ium showed a gradual gradation in their properties, with the values of some of strontium’sproperties being about midway between the values of calcium and barium. Döbereiner eventu-ally found four other sets of three elements, which he called triads, that followed the same pat-tern. In each triad, the atomic mass of the middle element was about midway between theatomic masses of the other two elements. Unfortunately, because Döbereiner’s system did notturn out to be very useful, it was largely ignored.
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 consecutivemembers of the same group in the periodic table. The elements in each set show a gradation intheir properties. Values for the first and third element in each set are given. Determine the miss-ing value in each set by calculating the average of the two given values. Then, compare the val-ues you obtained with those given in the Handbook of Chemistry and Physics. Record theactual values below your calculated values. Is the value of the property of the middle elementin each set midway between the values of the other two elements in the set?
CHALLENGE PROBLEMSCHAPTER 6
Use with Chapter 6,Section 6.2
Element Melting Point (°C)
Fluorine �219.6
Chlorine Calculated:
Actual:
Bromine �7.2
Set 1
Element Atomic Mass
Lithium 6.941
Sodium Calculated:
Actual:
Potassium 39.098
Set 2
Element Boiling Point (°C)
Magnesium 1107
Calcium Calculated:
Actual:
Strontium 1384
Set 3
Element Boiling Point (°C)
Krypton �153
Xenon Calculated:
Actual:
Radon �62
Set 4
Element Melting Point (°C)
Germanium 937
Tin Calculated:
Actual:
Lead 327
Set 5
Element Boiling Point (°C)
Beryllium 1285
Magnesium Calculated:
Actual:
Calcium 851
Set 6
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Challenge Problems Chemistry: Matter and Change • Chapter 7 7
Abundance of the ElementsAbundance of the Elements
The 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.
CHALLENGE PROBLEMSCHAPTER 7
1. What percent of all atoms in the universe are either hydrogen or helium? What percent ofall atoms in the solar system are either hydrogen or helium?
2. Explain the relatively high abundance of hydrogen and helium in the universe comparedto their relatively low abundance on Earth.
3. Only the top four most abundant elements on Earth and in Earth’s crust are shown in thetable. Name two additional elements you would expect to find among the top ten ele-ments 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 wouldexpect to find in the list of the top ten elements in the human body. Explain your choices.
Use with Chapter 7,Section 7.1
Abundance (Number of atoms per 1000 atoms)*
Element Universe Solar System Earth Earth’s Crust Human Body
Hydrogen 927 863 30 606
Helium 71.8 135
Oxygen 0.510 0.783 500 610 257
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.
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8 Chemistry: Matter and Change • Chapter 8 Challenge Problems
Comparing the Structures ofAtoms and IonsComparing the Structures ofAtoms and Ions
The 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 propertiesof ions also depend on the number of valence electrons. Any ion with a complete outermostenergy level will have chemical properties similar to those of the noble gas elements. The fluo-ride ion (F�), for example, has a total of ten electrons, eight of which fill its outermost energylevel. 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.
CHALLENGE PROBLEMSCHAPTER 8
Use with Chapter 8,Section 8.1
1. Write the atomic number for each of the five elements shown above.
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 dotstructure 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�? Explainyour answer.
6. Name an ion of calcium that has chemical properties similar to those of argon. Explainyour answer.
S Cl Ar K Ca
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Challenge Problems Chemistry: Matter and Change • Chapter 9 9
Exceptions to the Octet RuleExceptions to the Octet Rule
The 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.
CHALLENGE PROBLEMSCHAPTER 9
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.
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.
Use with Chapter 9,Section 9.3
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10 Chemistry: Matter and Change • Chapter 10 Challenge Problems
Balancing ChemicalEquationsBalancing ChemicalEquations
CHALLENGE PROBLEMSCHAPTER 10
Use with Chapter 10,Section 10.1
Each 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)
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)
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Challenge Problems Chemistry: Matter and Change • Chapter 11 11
Using Mole-BasedConversionsUsing Mole-BasedConversions
The 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.
CHALLENGE PROBLEMSCHAPTER 11
1. Compare and contrast the number of representative particles and the mass of UF6 withthe 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 ofthe gas are in 100.0 g of UF6?
3. CCl3CF3 is a chlorofluorocarbon responsible for the destruction of the ozone layer inEarth’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 presentin an alloy containing 0.15 mol of lead?
UF6 (g)
225.0 g
CCl3CF3 (l)
200.0 g
Pb (s)
250.0 g
Use with Chapter 11,Section 11.3
Molar Mass Number of Number of Representative Substance Mass (g) (g/mol) Moles (mol) Particles
UF6(g)
CCl3CF3(l)
Pb(s)
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12 Chemistry: Matter and Change • Chapter 12 Challenge Problems
Mole Relationships inChemical ReactionsMole Relationships inChemical Reactions
The 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.
CHALLENGE PROBLEMSCHAPTER 12
Use with Chapter 12,Section 12.2
1. What information is needed to make the types of conversions shown by double-arrow 1in the diagram?
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 massof a second substance in the reaction, as represented by double-arrow 5 in the diagram?
Moles ofCO
Grams ofCO
Moles ofCO2
Grams ofCO2
Particles ofCO
Particles ofCO2
1
5
2
3
46
7
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Challenge Problems Chemistry: Matter and Change • Chapter 13 13
Intermolecular Forces andBoiling PointsIntermolecular Forces andBoiling Points
CHALLENGE PROBLEMSCHAPTER 13
1. How do the boiling points of the group 4A hydrides change as the molecular masses ofthe hydrides change?
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 howthose 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 ofthe 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 thebonds in the four group 6A hydrides.
100
0
�100
H2O
H2S
CH4
SiH4
GeH4
SnH4
H2Se
H2Te
Group 6Ahydrides
Group 4Ahydrides
Molecular mass
Bo
ilin
g p
oin
t (°
C)
00 50 100 150
Use with Chapter 13,Section 13.3
The 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 agas, and, thus, the higher the boiling point of the liquid. This pat-tern may not hold true, however, when there are significant forcesbetween the particles of a liquid. The graph plots boiling pointversus molecular mass for group 4A and group 6A hydrides. Ahydride is a binary compound containing hydrogen and one otherelement. Use the graph to answer the following questions.
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14 Chemistry: Matter and Change • Chapter 14 Challenge Problems
A Simple Mercury BarometerA Simple Mercury Barometer
In 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 aboutthe simple mercury barometer shown here.
CHALLENGE PROBLEMSCHAPTER 14
Use with Chapter 14,Section 14.1
1. What occupies the space above the mercury column in thebarometer’s glass tube?
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 of5000 meters, the column of mercury changes as shown in Figure 2. Why is the mercurycolumn lower in Figure 2 than in Figure 1?
4. Suppose the barometer in Figure 1 was carried into an open mine 500 meters below sealevel. 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 wouldthis substitution affect the barometer? Explain.
6. Suppose a tiny crack formed at the top of the barometer’s glass tube. How would thisevent affect the column of mercury? Explain why.
Glass tube
Mercury column
Bowl of mercury
At sea level At 500 metersabove sea level
Figure 1 Figure 2
Cop
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Challenge Problems Chemistry: Matter and Change • Chapter 15 15
Vapor Pressure LoweringVapor Pressure Lowering
You 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 MarieRaoult (1830–1901) in 1886. According to Raoult’s law, the vaporpressure of a solvent (P) is equal to the product of its vapor pressurewhen 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 ofsucrose (C12H22O11) to 500.0 g of water at a temperature of 20°C.Answer the following questions about this solution.
CHALLENGE PROBLEMSCHAPTER 15
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 is17.54 mm Hg?
6. How much is the vapor pressure of the solution reduced from that of water by the addition of the sucrose?
Watermolecule
Solution
Sucrosemolecule
Use with Chapter 15,Section 15.3
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16 Chemistry: Matter and Change • Chapter 16 Challenge Problems
Standard Heat of FormationStandard Heat of Formation
Hess’s law allows you to determine the standard heat of formation of a compound
when you know the heats of reactions that leadto the production of that compound. The firstdiagram on the right shows how Hess’s law canbe used to calculate the heat of formation ofCO2 by knowing the heats of reaction of twosteps leading to the production of CO2. Use thisdiagram to help you answer the questions belowabout the second diagram.
CHALLENGE PROBLEMSCHAPTER 16
Use with Chapter 16,Section 16.4
The equations below show how NO2 can be formed in two ways: directly from the elements or in two steps.
N2(g) � O2(g) 0 NO2(g) �H � 33 kJ/mol
or
N2(g) � O2(g) 0 NO(g) �H � 91 kJ/mol
NO(g) � O2(g) 0 NO2(g) �H � �58 kJ/mol
1. On the diagram at the right, draw arrowheadsto show the directions in which the three lineslabeled 1, 2, and 3 should point.
2. Write the correct reactants and/or products oneach of the lines labeled A, B, and C.
3. Write the correct enthalpy change next toeach number on the diagram.
1�2
1�2
1�2
1�2
CO(g) � O2(g)
�H � �110 kJ/mol
C(s) � O2(g)
�H � �393 kJ/mol
�H � �283 kJ/molEnth
alp
y
CO2(g)
12
NO2(g)
�H � �58 kJ/mol
NO(g) � 1/2 O2(g)
�H � 91 kJ/mol
Enth
alp
y
�H � 33 kJ/mol
1
2
3
A
B
C
1/2 N2(g) � O2(g)
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Challenge Problems Chemistry: Matter and Change • Chapter 17 17
Determining Reaction RatesDetermining Reaction Rates
Dinitrogen 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 concen-tration of N2O5 remaining as the reaction proceedsover time. Answer the following questions aboutthe reaction.
CHALLENGE PROBLEMSCHAPTER 17
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 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 reactantduring some specified time period, or instantaneous rate of reaction = [N2O5]/t. What isthe instantaneous rate of reaction for the decomposition of N2O5 for the time periodbetween 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?
Time (h)
Co
nce
ntr
atio
n (
mo
l/L)
0.2
0
0.4
0.6
0.8
1.0
1.2
1.4
1.6
10 2 3 4 5 6 7 8 9 10
Use with Chapter 17,Section 17.1
87654321
10
0 2 3 4 5Time (sec)
Co
nce
ntr
atio
n (
mo
l/L)
6 7 8 9 10
SO2
SO3
SO3
O2O2
SO2
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18 Chemistry: Matter and Change • Chapter 18 Challenge Problems
Changing EquilibriumConcentrations in a Reaction
Reversible 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 orproduct. The graph on the right shows how the concentra-tions of the reactants and product of a reaction changewhen equilibrium is disturbed. Use the graph to answer thefollowing questions.
CHALLENGE PROBLEMSCHAPTER 18
Use with Chapter 18,Section 18.1
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 reaction.
4. At approximately what time does the reaction reach equilibrium? How do you knowequilibrium 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 knowthe change was made.
8. At what time does the system return to equilibrium?
Changing EquilibriumConcentrations in a Reaction
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Challenge Problems Chemistry: Matter and Change • Chapter 19 19
Swimming Pool ChemistrySwimming Pool Chemistry
The 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 thewater. Both compounds dissociate in water to form the weak acid hypochlorous acid(HOCl). Hypochlorous acid is a highly effective bactericide. By contrast, the hypochloriteion (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.
CHALLENGE PROBLEMSCHAPTER 19
1. Write an equation that shows the reaction between hypochlorous acid and water. Identifythe 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�),in the form of calcium hypochlorite or sodium hypochlorite, is added to water. Name theacid, 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 pHaffect the equilibrium reaction described in question 2? What effect would high pH haveon 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.
Use with Chapter 19,Section 19.2
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20 Chemistry: Matter and Change • Chapter 20 Challenge Problems
Balancing Oxidation–Reduction EquationsBalancing Oxidation–Reduction Equations
Scientists 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, forexample, 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, themagnesium or aluminum is the sacrificial metal, protecting the ironcasing 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.
CHALLENGE PROBLEMSCHAPTER 20
Use with Chapter 20,Section 20.3
1. In the absence of a sacrificial metal, oxygen dissolved in water may react with the ironcasing of the heater. One product formed is iron(II) hydroxide (Fe(OH)2). Which elementis oxidized and which is reduced in this reaction?
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. Writethe 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 theequation of question 2 above. The sacrificial metal (aluminum, in this case) immediatelyrestores the Fe2� ions to iron atoms. Write two half-reactions that represent this situation.
Ironcasing
Steel wire
Sacrificialmetal
Water
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Challenge Problems Chemistry: Matter and Change • Chapter 21 21
Effect of Concentration onCell PotentialEffect of Concentration onCell Potential
In 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:
Ecell � E0cell � �log
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 thebalanced half-cell reactions for the cell.
[product ion]x��[reactant ion]y
0.0592�n
CHALLENGE PROBLEMSCHAPTER 21
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.
Voltmeter
Ag Cu
Cu2�
1.0 � 10�3MAg�
1.0 � 10�2M
Use with Chapter 21,Section 21.1
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22 Chemistry: Matter and Change • Chapter 22 Challenge Problems
Structural Isomers of HexaneStructural Isomers of Hexane
The 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.
CHALLENGE PROBLEMSCHAPTER 22
Use with Chapter 22,Sections 22.1 and 22.3
1. In the spaces provided, write the correct name for each of the structural formulas, labeleda–j, above.
a. e. i.
b. f. j.
c. g.
d. h.
2. How many different compounds are represented by the structural formulas above? Whatare their names?
CH3
CH2 CH2 CH2 CH2 CH3
a.
CH3
CH3
CH CH2 CH2 CH3
b.
CH3
CH3
CH3 CH CH CH3
c.
CH3
CH3
CH3 C CH2 CH3
d.
CH3
CH CH2
CH2
CH3
CH3e.
CH3 CH CH CH3
CH3 CH3
f.
CH2
CH2 CH3
CH CH3
CH3
g.
CH3
CH2
CH3 CH CH2
CH3
h.
CH2
CH3
CH2
CH2 CH2
CH3i.
CH3
CH2 CH CH2
CH3 CH3j.
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Challenge Problems Chemistry: Matter and Change • Chapter 23 23
Boiling Points of OrganicFamiliesBoiling Points of OrganicFamilies
The 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, themore energy is needed to convert the substance from the liquidphase to the gaseous phase. As an example, the boiling pointof ethane (molecular mass � 30; boiling point � �89°C) ismuch higher than the boiling point of methane (molecularmass � 16; boiling point � �161°C).
Intermolecular forces between the particles of a liquid alsocan affect the liquid’s boiling point. The graph shows trends inthe boiling points of four organic families: alkanes, alcohols,aldehydes, and ethers. Use the graph and your knowledge ofintermolecular forces to answer the following questions.
CHALLENGE PROBLEMSCHAPTER 23
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 graphhas 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 tohave relatively similar boiling points?
4. Find the aldehyde with a molecular mass of about 58. Name that aldehyde and write itschemical formula.
5. Can this aldehyde form hydrogen bonds? Can other aldehydes form hydrogen bonds?Explain.
30 40 50 60Molecular mass
� alkane� alcohol
Bo
ilin
g p
oin
t (°
C)
70 80
�50
0
50
100
� aldehyde� ether
Use with Chapter 23,Section 23.3
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24 Chemistry: Matter and Change • Chapter 24 Challenge Problems
The Chemistry of LifeThe Chemistry of Life
Proteins are synthesized when RNA molecules translate the DNA language of nitrogen bases
into the protein language of amino acids using agenetic code. The genetic code is found in RNA mole-cules called messenger RNA (mRNA), which are syn-thesized from DNA molecules. The genetic codeconsists of a sequence of three nitrogen bases in themRNA, called a codon. Most codons code for specificamino acids. A few codons code for a stop in the syn-thesis of proteins. The table shows the mRNA codonsthat make up the genetic code. To use the table, readthe three nitrogen bases in sequence. The first base isshown along the left side of the table. The second baseis shown along the top of the table. The third base isshown 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 aminoacids. Answer the following questions about thegenetic code.
CHALLENGE PROBLEMSCHAPTER 24
Use with Chapter 24,Section 24.4
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 basemay be changed during synthesis. The two mRNA sequences shown below are examplesof such errors. In each case, tell how the mRNA sequence shown differs from the correctmRNA 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.
UUU UCU UAU UGU UUUC UCC UAC UGC CUUA UCA UAA UGA AUUG UCG UAG UGG GCUU CCU CAU CGU UCUC CCC CAC CGC CCUA CCA CAA CGA ACUG CCG CAG CGG GAUU ACU AAU AGU UAUC ACC AAC AGC CAUA ACA AAA AGA AAUG ACG AAG AGG GGUU GCU GAU GGU UGUC GCC GAC GGC CGUA GCA GAA GGA AGUG GCG GAG GGG G
Second base
Firs
t b
ase Th
ird b
ase
}} Phe
Leu
Ile
Met
Leu
Val
Pro
Ser
Ala
Thr
}
}}
}}
}}
His
Gln
Tyr
StopStop
Asn
Lys
Asp
Glu
}
}}
Arg
Gly
Cys
StopTrp
Ser
Arg
U
C
A
G
U C A G
The Genetic Code
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Challenge Problems Chemistry: Matter and Change • Chapter 25 25
The Production ofPlutonium-239The Production ofPlutonium-239
When 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 notoccur in nature but can be made synthetically in nuclearreactors and particle accelerators.
The diagram shows the process by which plutonium-239is made in nuclear reactors. Answer the questions about thediagram.
CHALLENGE PROBLEMSCHAPTER 25
1. Identify the isotope whose nucleus is labeled A in the
diagram.
2. Name the type of nuclear reaction that occurs when a
neutron strikes nucleus A.
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 thereaction.
8. Write a balanced nuclear equation for the decay of nucleus E. Identify isotope F formedin 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.
0–1
0–1
00
10n
10n
10n
10n
45p75n
48p77n
92p143n92p
143n
92p146nSource
ofneutrons
B
A
C
D
E
G
F
Use with Chapter 25,Section 25.4
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26 Chemistry: Matter and Change • Chapter 26 Challenge Problems
The Phosphorus CycleThe Phosphorus Cycle
Phosphorus 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 litho-sphere, phosphorus occurs primarily in the form of phosphates, as a major constituent ofmany 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 tothe lithosphere and hydrosphere. Thus, phosphorus—like carbon and nitrogen—cycles in theenvironment. Use the diagram of the phosphorus cycle to answer the questions below.
CHALLENGE PROBLEMSCHAPTER 26
Use with Chapter 26,Section 26.4
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 youstudied 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.
Phosphate rocks
Phosphaterocks
Geological uplift
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CHALLENGE PROBLEMSAnswer Key
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T28 Chemistry: Matter and Change Challenge Problems Answer Key
Nam
eD
ate
Cla
ss
2C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
2C
hal
len
ge
Pro
ble
ms
Popu
lati
on T
rend
s in
the
Uni
ted
Stat
esPo
pula
tion
Tre
nds
in t
heU
nite
d St
atesCH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
2
Use
wit
h Ch
apte
r 2,
Sect
ion
2.4
1.
By
how
muc
h di
d th
e to
tal U
.S. p
opul
atio
n in
crea
se b
etw
een
1990
and
200
0? W
hat w
asth
e pe
rcen
t inc
reas
e du
ring
this
per
iod?
Tota
l in
crea
se �
2.75
4 �
108
�2.
488
�10
8�
2.66
�10
7 ;
per
cen
tag
e in
crea
se �
2.66
�10
7�
2.48
8 �
108
�10
.7%
.
2.
Cal
cula
te th
e to
tal p
opul
atio
n fo
r ea
ch o
f th
e fi
ve g
roup
s fo
r 19
90 a
nd 2
000.
For
1990
, to
tal �
per
cen
t fo
r g
rou
p �
2.48
8 �
108 ,
or
Cau
casi
an: 1
.88
�10
8 ;
Afr
ican
Am
eric
an: 2
.94
�10
7 ; H
isp
anic
Am
eric
an: 2
.2 �
107 ;
Asi
an A
mer
ican
:
7.0
�10
6 ; N
ativ
e A
mer
ican
: 1.7
�10
6 ; f
or
2000
, to
tal �
per
cen
t fo
r g
rou
p �
2.75
4 �
108 ,
or
Cau
casi
an: 1
.97
�10
8 ; A
fric
an A
mer
ican
: 3.4
�10
7 ; H
isp
anic
Am
eric
an: 3
.25
�10
7 ; N
ativ
e A
mer
ican
: 1.9
�10
6 ; A
sian
Am
eric
an: 1
.0 �
107 .
3.
Mak
e a
bar
grap
h th
at c
ompa
res
the
popu
latio
n fo
r th
e fi
ve g
roup
s in
199
0 an
d 20
00. I
nw
hat w
ays
is th
e ba
r gr
aph
bette
r th
an th
e ci
rcle
gra
phs?
In
wha
t way
is it
less
use
ful?
The
bar
gra
ph
may
mak
e th
e ch
ang
es f
rom
199
0 to
200
0 sh
ow
up
mo
re c
lear
ly.
Un
less
th
e n
um
eric
al v
alu
es f
or
each
bar
are
act
ual
ly w
ritt
en o
n t
he
gra
ph
, th
ey
may
be
mo
re d
iffi
cult
to
est
imat
e th
an t
he
valu
es g
iven
on
th
e ci
rcle
gra
ph
. Th
e
larg
e d
iffe
ren
ces
in p
op
ula
tio
ns
for
Cau
casi
an a
nd
no
n-C
auca
sian
gro
up
s m
ake
it
dif
ficu
lt t
o s
ho
w d
ata
for
the
latt
er g
rou
ps
clea
rly.
U.S
. Pop
ulat
ion
Dis
trib
utio
n
(Per
cen
tag
es m
ay n
ot
add
up
to
100
% d
ue
to r
ou
nd
ing
.)Cau
casi
an71
.4%
Nat
ive
Am
eric
an0.
70%
Asi
an A
mer
ican
3.8%
His
pan
ic A
mer
ican
11.8
%
Afr
ican
Am
eric
an12
.2%
Cau
casi
an75
.7%
Nat
ive
Am
eric
an0.
70%
Asi
an A
mer
ican
2.8%
His
pan
ic A
mer
ican
9.0%A
fric
an A
mer
ican
11.8
%
2000
1990
The
pop
ulat
ion
of th
e U
nite
d St
ates
is b
ecom
ing
mor
e di
vers
e. T
he c
ircl
e gr
aphs
bel
ow s
how
the
dist
ribu
tion
of t
he U
.S. p
opul
atio
n am
ong
five
eth
nic
grou
ps in
199
0 an
d 20
00. T
he e
stim
ated
tota
l U.S
. pop
ulat
ion
for
thos
e tw
o ye
ars
was
2.4
88 �
108
in 1
990
and
2.75
4 �
108
in 2
000.
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
11
Prod
ucti
on o
fCh
loro
fluo
roca
rbon
s,19
50–1
992
Prod
ucti
on o
fCh
loro
fluo
roca
rbon
s,19
50–1
992
Chl
orof
luor
ocar
bons
(C
FCs)
wer
e fi
rst p
rodu
ced
in
the
labo
rato
ry in
the
late
192
0s. T
hey
did
not
beco
me
an im
port
ant c
omm
erci
al p
rodu
ct u
ntil
som
etim
e la
ter.
Eve
ntua
lly,C
FCs
grew
in p
opul
arity
unt
ilth
eir
effe
ct o
n th
e oz
one
laye
r w
as d
isco
vere
d in
the
1970
s. T
he g
raph
sho
ws
the
com
bine
d am
ount
s of
two
impo
rtan
t CFC
s pr
oduc
ed b
etw
een
1950
and
199
2.A
nsw
er th
e fo
llow
ing
ques
tions
abo
ut th
e gr
aph.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
1
1.
Wha
t was
the
appr
oxim
ate
amou
nt o
f C
FCs
prod
uced
in 1
950?
In
1960
? In
197
0?
1950
: les
s th
an 1
0 b
illio
n k
g; 1
960:
ab
ou
t 50
bill
ion
kg
; 197
0: a
bo
ut
240
bill
ion
kg
2.
In w
hat y
ear
was
the
larg
est a
mou
nt o
f C
FCs
prod
uced
? A
bout
how
muc
h w
as p
rodu
ced
that
yea
r?
1988
; ab
ou
t 37
5 b
illio
n k
g
3.
Dur
ing
wha
t tw
o-ye
ar p
erio
d di
d th
e pr
oduc
tion
of C
FCs
decr
ease
by
the
grea
test
amou
nt?
By
abou
t how
muc
h di
d th
eir
prod
uctio
n de
crea
se?
Bet
wee
n 1
988
and
199
0; t
he
dec
reas
e am
ou
nte
d t
o a
bo
ut
140
bill
ion
kg
.
4.
Dur
ing
wha
t tw
o-ye
ar p
erio
d di
d th
e pr
oduc
tion
of C
FCs
incr
ease
by
the
grea
test
amou
nt?
Wha
t was
the
appr
oxim
ate
perc
ent i
ncre
ase
duri
ng th
is p
erio
d?
Bet
wee
n 1
970
and
197
2; t
he
per
cen
t in
crea
se w
as a
pp
roxi
mat
ely
(305
bill
ion
kg
�24
0 b
illio
n k
g) /2
40 b
illio
n k
g �
27%
.
5.
How
con
fide
nt w
ould
you
fee
l abo
ut p
redi
ctin
g th
e pr
oduc
tion
leve
ls o
f C
FCs
duri
ng th
eod
d nu
mbe
red
year
s 19
61,1
971,
and
1981
? E
xpla
in.
The
pro
du
ctio
n o
f C
FCs
incr
ease
d r
egu
larl
y fr
om
195
0 to
197
4 (e
xcep
t in
195
8)
and
dec
reas
ed r
egu
larl
y af
ter
1988
. Th
us,
on
e m
igh
t b
e ab
le t
o p
red
ict
accu
rate
ly
the
pro
du
ctio
n le
vels
du
rin
g 1
961
and
197
1 fr
om
dat
a in
th
e g
rap
h. H
ow
ever
, th
e
pro
du
ctio
n le
vels
wer
e le
ss r
egu
lar
bet
wee
n 1
974
and
198
8. T
her
efo
re, p
red
icti
on
s
reg
ard
ing
th
e p
rod
uct
ion
leve
l du
rin
g 1
981
mig
ht
be
less
acc
ura
te.
6.
Cou
ld th
e da
ta in
the
grap
h be
pre
sent
ed in
the
form
of
a ci
rcle
gra
ph?
Exp
lain
.
A c
ircl
e g
rap
h c
ou
ld n
ot
be
use
d b
ecau
se it
sh
ow
s h
ow
a t
ota
l am
ou
nt
is d
ivid
ed
into
cat
ego
ries
, no
t h
ow
a v
aria
ble
ch
ang
es o
ver
tim
e.
Yea
r
Amount of CFCs(billion kilograms)
050100
150
200
250
300
350
400 19
5019
6019
7019
8019
90
Use
wit
h Ch
apte
r 1,
Sect
ion
1.1
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
Challenge Problems Answer Key Chemistry: Matter and Change T29
Nam
eD
ate
Cla
ss
4C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
4C
hal
len
ge
Pro
ble
ms
Isot
opes
of
an E
lem
ent
Isot
opes
of
an E
lem
ent
Am
ass
spec
trom
eter
is a
dev
ice
for
sepa
ratin
gat
oms
and
mol
ecul
es a
ccor
ding
to th
eir
mas
s. A
sub
stan
ce is
fir
st h
eate
d in
a v
acuu
m a
ndth
en io
nize
d. T
he io
ns p
rodu
ced
are
acce
lera
ted
thro
ugh
a m
agne
tic f
ield
that
sep
arat
es io
ns o
f di
f-fe
rent
mas
ses.
The
gra
ph b
elow
was
pro
duce
dw
hen
a ce
rtai
n el
emen
t (el
emen
t X)
was
ana
lyze
din
a m
ass
spec
trom
eter
. Use
the
grap
h to
ans
wer
the
ques
tions
bel
ow.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
4
Use
wit
h Ch
apte
r 4,
Sect
ion
4.3
1.
How
man
y is
otop
es o
f el
emen
t X e
xist
?
2.
Wha
t is
the
mas
s of
the
mos
t abu
ndan
t iso
tope
?
3.
Wha
t is
the
mas
s of
the
leas
t abu
ndan
t iso
tope
?
4.
Wha
t is
the
mas
s of
the
heav
iest
isot
ope?
5.
Wha
t is
the
mas
s of
the
light
est i
soto
pe?
6.
Est
imat
e th
e pe
rcen
t abu
ndan
ce o
f ea
ch is
otop
e sh
own
on th
e gr
aph.
196:
less
th
an 1
%; 1
98: a
bo
ut
10%
; 199
: ab
ou
t 17
%; 2
00: a
bo
ut
23%
;
201:
ab
ou
t 13
%; 2
02: a
bo
ut
30%
; 204
: ab
ou
t 7%
7.
With
out p
erfo
rmin
g an
y ca
lcul
atio
ns,p
redi
ct th
e ap
prox
imat
e at
omic
mas
s fo
r el
emen
tX
. Exp
lain
the
basi
s fo
r yo
ur p
redi
ctio
n.
The
ato
mic
mas
s o
f el
emen
t X
is b
etw
een
200
an
d 2
02 a
mu
, th
e m
asse
s o
f
the
two
mo
st a
bu
nd
ant
iso
top
es.
8.
Usi
ng th
e da
ta g
iven
by
the
grap
h,ca
lcul
ate
the
wei
ghte
d av
erag
e at
omic
mas
s of
el
emen
t X. I
dent
ify
the
unkn
own
elem
ent.
Mas
s o
f X
�(0
.001
4)(1
96 a
mu
) �
(0.1
0)(1
98 a
mu
) �
(0.1
68)(
199
amu
)
�(0
.231
)(20
0 am
u)
�(0
.132
)(20
1) �
(0.2
98)(
202
amu
) �
(0.0
68)(
204
amu
)
Mas
s o
f X
�0.
27 a
mu
�19
.8 a
mu
�33
.4 a
mu
�46
.2 a
mu
�26
.5 a
mu
�60
.2 a
mu
�13
.9 a
mu
Mas
s o
f X
�20
0.3
amu
The
elem
ent
in t
he
per
iod
ic t
able
wit
h a
n a
tom
ic m
ass
clo
sest
to
th
e ca
lcu
late
d
valu
e o
f 20
0.3
amu
is m
ercu
ry. E
lem
ent
X is
pro
bab
ly m
ercu
ry, w
hic
h h
as a
n
ato
mic
mas
s o
f 20
0.59
am
u.
196
amu
204
amu
196
amu
202
amu
7
051015202530
196
194
192
190
198
200
202
204
206
208
210
Ato
mic
mas
s (a
mu
)
Percent abundance
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
33
Phys
ical
and
Che
mic
alCh
ange
sPh
ysic
al a
nd C
hem
ical
Chan
ges
Phy
sica
l and
che
mic
al c
hang
es o
ccur
all
arou
nd u
s. O
ne o
f th
e m
any
plac
es in
whi
ch
phys
ical
and
che
mic
al c
hang
es o
ccur
is th
e ki
tche
n. F
or e
xam
ple,
cook
ing
spag
hetti
in a
pot o
f w
ater
on
the
stov
e in
volv
es s
uch
chan
ges.
For
eac
h of
the
chan
ges
desc
ribe
d be
low
,tel
l(a
) w
heth
er th
e ch
ange
that
occ
urs
is p
hysi
cal o
r ch
emic
al,a
nd (
b) h
ow y
ou m
ade
your
cho
ice
betw
een
thes
e tw
o po
ssib
ilitie
s. I
f yo
u ar
e un
able
to d
ecid
e w
heth
er th
e ch
ange
is p
hysi
cal o
rch
emic
al,t
ell w
hat a
dditi
onal
info
rmat
ion
you
wou
ld n
eed
in o
rder
to m
ake
a de
cisi
on.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
3
1.
As
the
wat
er in
the
pot i
s he
ated
,its
tem
pera
ture
ris
es.
Phys
ical
ch
ang
e; t
he
com
po
siti
on
of
the
wat
er d
oes
no
t ch
ang
e.
2.
As
mor
e he
at is
add
ed,t
he w
ater
beg
ins
to b
oil a
nd s
team
is p
rodu
ced.
Phys
ical
ch
ang
e; t
he
wat
er c
han
ges
fro
m li
qu
id t
o g
as, b
ut
ther
e is
no
ch
ang
e in
its
com
po
siti
on
.
3.
The
hea
t use
d to
coo
k is
pro
duce
d by
bur
ning
nat
ural
gas
in th
e st
ove
burn
er.
Ch
emic
al c
han
ge;
th
e co
mp
osi
tio
n o
f th
e n
atu
ral g
as c
han
ges
as
it b
urn
s.
4.
The
met
al b
urne
r on
whi
ch th
e po
t res
ts w
hile
bei
ng h
eate
d be
com
es r
ed a
s its
te
mpe
ratu
re r
ises
.
Phys
ical
ch
ang
e; t
he
app
eara
nce
of
the
met
al c
han
ges
as
it is
hea
ted
, bu
t it
s
com
po
siti
on
do
es n
ot.
Wh
en t
he
met
al c
oo
ls, i
t w
ill n
o lo
ng
er b
e re
d.
5.
Aft
er th
e fl
ame
has
been
turn
ed o
ff,a
sm
all a
rea
on th
e bu
rner
has
cha
nged
in c
olor
fro
mbl
ack
to g
ray.
It is
no
t p
oss
ible
to
kn
ow
wh
eth
er t
his
is a
ph
ysic
al c
han
ge
or
a ch
emic
al c
han
ge
wit
ho
ut
com
par
ing
th
e co
mp
osi
tio
n o
f th
e g
ray
mat
eria
l wit
h t
he
ori
gin
al b
lack
mat
eria
l.
6.
A s
tran
d of
spa
ghet
ti ha
s fa
llen
onto
the
burn
er,w
here
it tu
rns
blac
k an
d be
gins
tosm
oke.
Ch
emic
al c
han
ge;
th
e fo
rmat
ion
of
smo
ke a
nd
a p
erm
anen
t co
lor
chan
ge
are
evid
ence
of
a ch
emic
al c
han
ge.
7.
Whe
n th
e sp
aghe
tti is
coo
ked
in th
e bo
iling
wat
er,i
t bec
omes
sof
t.
Phys
ical
ch
ang
e; t
he
com
po
siti
on
of
the
spag
het
ti h
as n
ot
chan
ged
.
Use
wit
h Ch
apte
r 3,
Sect
ion
3.2
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
T30 Chemistry: Matter and Change Challenge Problems Answer Key
Nam
eD
ate
Cla
ss
6C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
6C
hal
len
ge
Pro
ble
ms
Döb
erei
ner’
s Tr
iads
Döb
erei
ner’
s Tr
iads
One
of
the
firs
t som
ewha
t suc
cess
ful a
ttem
pts
to a
rran
ge th
e el
emen
ts in
a s
yste
mat
ic w
ay
was
mad
e by
the
Ger
man
che
mis
t Joh
ann
Wol
fgan
g D
öber
eine
r (1
780–
1849
). I
n 18
16,
Döb
erei
ner
notic
ed th
at th
e th
en a
ccep
ted
atom
ic m
ass
of s
tron
tium
(50
) w
as m
idw
ay b
etw
een
the
atom
ic m
asse
s of
cal
cium
(27
.5)
and
bari
um (
72.5
). N
ote
that
the
acce
pted
ato
mic
mas
ses
for
thes
e el
emen
ts to
day
are
very
dif
fere
nt f
rom
thei
r ac
cept
ed a
tom
ic m
asse
s at
the
time
Döb
erei
ner
mad
e hi
s ob
serv
atio
ns. D
öber
eine
r al
so o
bser
ved
that
str
ontiu
m,c
alci
um,a
nd b
ar-
ium
sho
wed
a g
radu
al g
rada
tion
in th
eir
prop
ertie
s,w
ith th
e va
lues
of
som
e of
str
ontiu
m’s
prop
ertie
s be
ing
abou
t mid
way
bet
wee
n th
e va
lues
of
calc
ium
and
bar
ium
. Döb
erei
ner
even
tu-
ally
fou
nd f
our
othe
r se
ts o
f th
ree
elem
ents
,whi
ch h
e ca
lled
tria
ds,t
hat f
ollo
wed
the
sam
e pa
t-te
rn. I
n ea
ch tr
iad,
the
atom
ic m
ass
of th
e m
iddl
e el
emen
t was
abo
ut m
idw
ay b
etw
een
the
atom
ic m
asse
s of
the
othe
r tw
o el
emen
ts. U
nfor
tuna
tely
,bec
ause
Döb
erei
ner’
s sy
stem
did
not
turn
out
to b
e ve
ry u
sefu
l,it
was
larg
ely
igno
red.
Had
Döb
erei
ner
actu
ally
dis
cove
red
a w
ay o
f id
entif
ying
tren
ds a
mon
g th
e el
emen
ts?
Lis
ted
belo
w a
re s
ix th
ree-
elem
ent g
roup
s in
whi
ch th
e el
emen
ts in
eac
h gr
oup
are
cons
ecut
ive
mem
bers
of
the
sam
e gr
oup
in th
e pe
riod
ic ta
ble.
The
ele
men
ts in
eac
h se
t sho
w a
gra
datio
n in
thei
r pr
oper
ties.
Val
ues
for
the
firs
t and
thir
d el
emen
t in
each
set
are
giv
en. D
eter
min
e th
e m
iss-
ing
valu
e in
eac
h se
t by
calc
ulat
ing
the
aver
age
of th
e tw
o gi
ven
valu
es. T
hen,
com
pare
the
val-
ues
you
obta
ined
with
thos
e gi
ven
in th
e H
andb
ook
of C
hem
istr
y an
d P
hysi
cs. R
ecor
d th
eac
tual
val
ues
belo
w y
our
calc
ulat
ed v
alue
s. I
s th
e va
lue
of th
e pr
oper
ty o
f th
e m
iddl
e el
emen
tin
eac
h se
t mid
way
bet
wee
n th
e va
lues
of
the
othe
r tw
o el
emen
ts in
the
set?
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
6
Use
wit
h Ch
apte
r 6,
Sect
ion
6.2
No
, on
ly s
om
e o
f th
e p
rop
erti
es o
f so
me
of
the
mid
dle
ele
men
ts o
f tr
iad
s ar
e
mid
way
in v
alu
e.
Elem
ent
Mel
ting
Poi
nt (
°C)
Flu
ori
ne
�21
9.6
Ch
lori
ne
Cal
cula
ted
:�
113.
4
Act
ual
: �
100.
98
Bro
min
e�
7.2
Set
1
Elem
ent
Ato
mic
Mas
s
Lith
ium
6.94
1
Sod
ium
Cal
cula
ted
: 23
.019
Act
ual
:22
.990
Pota
ssiu
m39
.098
Set
2
Elem
ent
Boili
ng P
oint
(°C
)
Mag
nes
ium
1107
Cal
ciu
mC
alcu
late
d:
1246
Act
ual
: 48
4
Stro
nti
um
1384
Set
3
Elem
ent
Boili
ng P
oint
(°C
)
Kry
pto
n�
153
Xen
on
Cal
cula
ted
:�
108
Act
ual
:�
108
Rad
on
�62
Set
4
Elem
ent
Mel
ting
Poi
nt (
°C)
Ger
man
ium
937
Tin
Cal
cula
ted
:63
2
Act
ual
: 23
2
Lead
327
Set
5
Elem
ent
Boili
ng P
oint
(°C
)
Ber
ylliu
m12
85
Mag
nes
ium
Cal
cula
ted
:10
68
Act
ual
:65
0
Cal
ciu
m85
1
Set
6
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
55
Qua
ntum
Num
bers
Qua
ntum
Num
bers
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
5
The
sta
te o
f an
ele
ctro
n in
an
atom
can
be
com
plet
ely
desc
ribe
d by
fou
r qu
antu
m n
umbe
rs,
desi
gnat
ed a
s n,
�,m
�,an
d m
s. T
he f
irst
,or
prin
cipa
l,qu
antu
m n
umbe
r,n,
indi
cate
s th
eel
ectr
on’s
app
roxi
mat
e di
stan
ce f
rom
the
nucl
eus.
The
sec
ond
quan
tum
num
ber,
�,de
scri
bes
the
shap
e of
the
elec
tron
’s o
rbit
arou
nd th
e nu
cleu
s. T
he th
ird
quan
tum
num
ber,
m�,
desc
ribe
sth
e or
ient
atio
n of
the
elec
tron
’s o
rbit
com
pare
d to
the
plan
e of
the
atom
. The
fou
rth
quan
tum
num
ber,
ms,
tells
the
dire
ctio
n of
the
elec
tron
’s s
pin
(clo
ckw
ise
or c
ount
ercl
ockw
ise)
.
The
Sch
rödi
nger
wav
e eq
uatio
n im
pose
s ce
rtai
n m
athe
mat
ical
res
tric
tions
on
the
quan
tum
num
bers
. The
y ar
e as
fol
low
s:
nca
n be
any
inte
ger
(who
le n
umbe
r),
�ca
n be
any
inte
ger
from
0 to
n�
1,
m�
can
be a
ny in
tege
r fr
om �
�to
��,
and
ms
can
be �
or �
As
an e
xam
ple,
cons
ider
ele
ctro
ns in
the
firs
t ene
rgy
leve
l of
an a
tom
,tha
t is,
n�
1. I
nth
is c
ase,
�ca
n ha
ve a
ny in
tegr
al v
alue
fro
m 0
to (
n�
1),o
r 0
to (
1 �
1). I
n ot
her
wor
ds,
�m
ust b
e 0
for
thes
e el
ectr
ons.
Als
o,th
e on
ly v
alue
that
m�
can
have
is 0
. The
ele
ctro
ns in
this
ene
rgy
leve
l can
hav
e va
lues
of
�or
�fo
r m
s. T
hese
res
tric
tions
agr
ee w
ith th
e
obse
rvat
ion
that
the
firs
t ene
rgy
leve
l can
hav
e on
ly tw
o el
ectr
ons.
The
ir q
uant
um n
umbe
rs
are
1,0,
0,�
and
1,0,
0 �
.
Use
the
rule
s gi
ven
abov
e to
com
plet
e th
e ta
ble
listin
g th
e qu
antu
m n
umbe
rs f
or e
ach
elec
tron
in a
bor
on a
tom
. The
cor
rect
qua
ntum
num
bers
for
one
ele
ctro
n in
the
atom
is
prov
ided
as
an e
xam
ple.
1 � 21 � 2
1 � 21 � 2
1 � 21 � 2
Use
wit
h Ch
apte
r 5,
Sect
ion
5.2
Elec
tron
n�
m�
ms
11
00
�
2 3 4 5
1 � 2
Bo
ron
(B
)Fo
r b
oro
n, t
he
firs
t tw
oel
ectr
on
s h
ave
qu
antu
mn
um
ber
s 1,
0,0,
�1 /
2 an
d1,
0,0,
�1 /
2. T
he
seco
nd
tw
oel
ectr
on
s h
ave
qu
antu
mn
um
ber
s 2,
0,0,
�1 /
2 an
d2,
0,0,
�1 /
2. T
he
fift
h e
lect
ron
may
hav
e an
y o
ne
of
six
po
ssib
le q
uan
tum
nu
mb
ers:
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.
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
Challenge Problems Answer Key Chemistry: Matter and Change T31
Nam
eD
ate
Cla
ss
8C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
8C
hal
len
ge
Pro
ble
ms
Com
pari
ng t
he S
truc
ture
s of
Ato
ms
and
Ions
Com
pari
ng t
he S
truc
ture
s of
Ato
ms
and
Ions
The
che
mic
al p
rope
rtie
s of
an
elem
ent d
epen
d pr
imar
ily o
n its
num
ber
of v
alen
ce e
lect
rons
in
its a
tom
s. T
he n
oble
gas
ele
men
ts,f
or e
xam
ple,
all h
ave
sim
ilar
chem
ical
pro
pert
ies
beca
use
the
oute
rmos
t ene
rgy
leve
ls o
f th
eir
atom
s ar
e co
mpl
etel
y fi
lled.
The
che
mic
al p
rope
rtie
sof
ions
als
o de
pend
on
the
num
ber
of v
alen
ce e
lect
rons
. Any
ion
with
a c
ompl
ete
oute
rmos
ten
ergy
leve
l will
hav
e ch
emic
al p
rope
rtie
s si
mila
r to
thos
e of
the
nobl
e ga
s el
emen
ts. T
he f
luo-
ride
ion
(F�
),fo
r ex
ampl
e,ha
s a
tota
l of
ten
elec
tron
s,ei
ght o
f w
hich
fill
its
oute
rmos
t ene
rgy
leve
l. F�
has
chem
ical
pro
pert
ies,
ther
efor
e,si
mila
r to
thos
e of
the
nobl
e ga
s ne
on.
Show
n be
low
are
the
Lew
is e
lect
ron
dot s
truc
ture
s fo
r fi
ve e
lem
ents
:sul
fur
(S),
chlo
rine
(C
l),
argo
n (A
r),p
otas
sium
(K
),an
d ca
lciu
m (
Ca)
. Ans
wer
the
ques
tions
bel
ow a
bout
thes
e st
ruct
ures
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
8
Use
wit
h Ch
apte
r 8,
Sect
ion
8.1
1.
Wri
te th
e at
omic
num
ber
for
each
of
the
five
ele
men
ts s
how
n ab
ove.
S: 1
6; C
l: 17
; Ar:
18;
K: 1
9; C
a: 2
0
2.
Wri
te th
e el
ectr
on c
onfi
gura
tion
for
each
of
the
five
ele
men
ts.
S: 1
s22s
2 2p
6 3s2
3p4 ;
Cl:
1s2 2
s22p
6 3s2
3p5 ;
Ar:
1s2
2s2 2
p6 3
s23p
6 ;
K: 1
s22s
2 2p
6 3s2
3p6 4
s1; C
a: 1
s22s
2 2p
6 3s2
3p6 4
s2
3.
Whi
ch o
f th
e ab
ove
Lew
is e
lect
ron
dot s
truc
ture
s is
the
sam
e as
the
Lew
is e
lect
ron
dot
stru
ctur
e fo
r th
e io
n S2�
? E
xpla
in y
our
answ
er.
The
Lew
is e
lect
ron
do
t st
ruct
ure
fo
r ar
go
n; b
oth
arg
on
an
d S
2 �h
ave
18 e
lect
ron
s,
eig
ht
of
wh
ich
are
val
ence
ele
ctro
ns.
4.
Whi
ch o
f th
e ab
ove
Lew
is e
lect
ron
dot s
truc
ture
s is
the
sam
e as
that
for
the
ion
Cl�
?E
xpla
in y
our
answ
er.
The
Lew
is e
lect
ron
do
t st
ruct
ure
fo
r ar
go
n; b
oth
arg
on
an
d C
l�h
ave
18 e
lect
ron
s,
eig
ht
of
wh
ich
are
val
ence
ele
ctro
ns.
5.
Whi
ch o
f th
e ab
ove
Lew
is e
lect
ron
dot s
truc
ture
s is
like
that
for
the
ion
K�
? E
xpla
inyo
ur a
nsw
er.
The
Lew
is e
lect
ron
do
t st
ruct
ure
fo
r ar
go
n; b
oth
arg
on
an
d K
�h
ave
18 e
lect
ron
s,
eig
ht
of
wh
ich
are
val
ence
ele
ctro
ns.
6.
Nam
e an
ion
of c
alci
um th
at h
as c
hem
ical
pro
pert
ies
sim
ilar
to th
ose
of a
rgon
. Exp
lain
your
ans
wer
.
Ca2
�; l
ike
arg
on
, Ca2
�h
as 1
8 el
ectr
on
s, e
igh
t o
f w
hic
h a
re v
alen
ce e
lect
ron
s.
SC
lA
rK
Ca
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
77
Abu
ndan
ce o
f th
e El
emen
tsA
bund
ance
of
the
Elem
ents
The
abu
ndan
ce o
f th
e el
emen
ts d
iffe
rs s
igni
fica
ntly
in v
ario
us p
arts
of
the
univ
erse
. The
tabl
e be
low
list
s th
e ab
unda
nce
of s
ome
elem
ents
in v
ario
uspa
rts
of th
e un
iver
se. U
se th
e ta
ble
to a
nsw
er th
e fo
llow
ing
ques
tions
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
7
1.
Wha
t per
cent
of
all a
tom
s in
the
univ
erse
are
eith
er h
ydro
gen
or h
eliu
m?
Wha
t per
cent
of
all a
tom
s in
the
sola
r sy
stem
are
eith
er h
ydro
gen
or h
eliu
m?
un
iver
se: 9
27 H
�71
.8 H
e �
1000
�10
0% �
99.9
%;
sola
r sy
stem
: 863
H �
135
He
�10
00 �
100%
�99
.8%
2.
Exp
lain
the
rela
tivel
y hi
gh a
bund
ance
of
hydr
ogen
and
hel
ium
in th
e un
iver
se c
ompa
red
to th
eir
rela
tivel
y lo
w a
bund
ance
on
Ear
th.
Hyd
rog
en a
nd
hel
ium
are
gas
es o
f lo
w d
ensi
ty t
hat
hav
e es
cap
ed f
rom
Ear
th’s
gra
vita
tio
nal
att
ract
ion
.
3.
Onl
y th
e to
p fo
ur m
ost a
bund
ant e
lem
ents
on
Ear
th a
nd in
Ear
th’s
cru
st a
re s
how
n in
the
tabl
e. N
ame
two
addi
tiona
l ele
men
ts y
ou w
ould
exp
ect t
o fi
nd a
mon
g th
e to
p te
n el
e-m
ents
bot
h on
Ear
th a
nd in
Ear
th’s
cru
st. E
xpla
in y
our
choi
ces.
Bas
ed o
n t
he
info
rmat
ion
in S
ecti
on
7.1
, alu
min
um
an
d c
alci
um
are
po
ssib
le
cho
ices
. Alu
min
um
is t
he
mo
st a
bu
nd
ant
met
al in
Ear
th’s
cru
st. C
alci
um
fo
rms
man
y ki
nd
s o
f ro
cks
wit
h o
xyg
en, s
ilico
n, a
nd
car
bo
n.
4.
Nam
e at
leas
t thr
ee e
lem
ents
in a
dditi
on to
thos
e sh
own
in th
e ta
ble
that
you
wou
ldex
pect
to f
ind
in th
e lis
t of
the
top
ten
elem
ents
in th
e hu
man
bod
y. E
xpla
in y
our
choi
ces.
Stu
den
ts’ a
nsw
ers
may
var
y. E
lem
ents
dis
cuss
ed in
Sec
tio
n 7
.1 a
s h
avin
g
imp
ort
ant
role
s in
th
e h
um
an b
od
y in
clu
de
sod
ium
(#8
in li
st o
f to
p t
en),
po
tass
ium
(#9
), c
hlo
rin
e (#
10),
an
d p
ho
sph
oru
s (#
7).
Use
wit
h Ch
apte
r 7,
Sect
ion
7.1
Ab
un
dan
ce (
Nu
mb
er
of
ato
ms
per
1000 a
tom
s)*
Elem
ent
Uni
vers
eSo
lar
Syst
emEa
rth
Eart
h’s
Crus
tH
uman
Bod
y
Hyd
rog
en92
786
330
606
Hel
ium
71.8
135
Oxy
gen
0.51
00.
783
500
610
257
Nit
rog
en0.
153
0.08
0924
Car
bo
n0.
0811
0.45
910
6
Silic
on
0.02
310.
0269
140
210
Iro
n0.
0139
0.00
320
170
19
* A
n e
lem
ent
is n
ot
abu
nd
ant
in a
reg
ion
th
at is
left
bla
nk.
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
T32 Chemistry: Matter and Change Challenge Problems Answer Key
Nam
eD
ate
Cla
ss
10C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
10C
hal
len
ge
Pro
ble
ms
Bala
ncin
g Ch
emic
alEq
uati
ons
Bala
ncin
g Ch
emic
alEq
uati
ons
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
10
Use
wit
h Ch
apte
r 10
,Se
ctio
n 10
.1
Eac
h ch
emic
al e
quat
ion
belo
w c
onta
ins
at le
ast o
ne e
rror
. Ide
ntif
y th
e er
ror
or e
rror
s an
d th
en w
rite
the
corr
ect c
hem
ical
equ
atio
n fo
r th
e re
actio
n.
1.
K(s
) �
2H2O
(l) 0
2KO
H(a
q) �
H2(
g)Th
e eq
uat
ion
is n
ot
bal
ance
d. 2
K(s
) �
2H2O
(l) 0
2KO
H(a
q)
�H
2(g
)
2.
MgC
l 2(aq
) �
H2S
O4(
aq) 0
Mg(
SO4)
2(aq
) �
2HC
l(aq
)A
n in
corr
ect
form
ula
is g
iven
fo
r m
agn
esiu
m s
ulf
ate.
Th
eref
ore
, th
e eq
uat
ion
is
no
t b
alan
ced
. Mg
Cl 2
(aq
) �
H2S
O4(
aq) 0
Mg
SO4(
aq)
�2H
Cl(
aq)
3.
AgN
O3(
aq)
�H
2S(a
q) 0
Ag 2S
(aq)
�H
NO
3(aq
)Th
e eq
uat
ion
is n
ot
bal
ance
d. 2
Ag
NO
3(aq
) �
H2S
(aq
) 0
Ag
2S(a
q)
�2H
NO
3(aq
)
4.
Sr(s
) �
F 2(g)
0Sr
2FA
n in
corr
ect
form
ula
is g
iven
fo
r st
ron
tiu
m f
luo
rid
e. T
her
efo
re, t
he
equ
atio
n is
no
t b
alan
ced
. Als
o, t
he
ph
ysic
al s
tate
of
stro
nti
um
flu
ori
de
is n
ot
giv
en.
Sr(s
) �
F 2(g
) 0
SrF 2
(s)
5.
2NaH
CO
3(s)
�2H
Cl(
aq) 0
2NaC
l(s)
�2C
O2(
g)A
ter
m m
ust
be
mis
sin
g b
ecau
se t
her
e is
no
pro
du
ct c
on
tain
ing
hyd
rog
en. T
her
efo
re,
the
equ
atio
n is
no
t b
alan
ced
. 2N
aHC
O3(
s) �
2HC
l(aq
) 0
2NaC
l(s)
�2C
O2(
g)
�2H
2O(l)
6.
2LiO
H(a
q) �
2HB
r(aq
) 0
2LiB
r(aq
) �
2H2O
The
coef
fici
ents
are
no
t ex
pre
ssed
in lo
wes
t te
rms.
Als
o, t
he
ph
ysic
al s
tate
of
wat
er is
no
t g
iven
. LiO
H(a
q)
�H
Br(
aq) 0
LiB
r(aq
) �
H2O
(l)
7.
NH
4OH
(aq)
�K
OH
(aq)
0K
OH
(aq)
�N
H4O
H(a
q)
The
pro
du
cts
are
the
sam
e as
th
e re
acta
nts
, so
no
rea
ctio
n o
ccu
rred
an
d n
o
equ
atio
n c
an b
e w
ritt
en.
8.
2Ca(
s) �
Cl 2(
g) 0
2CaC
l(aq
)A
n in
corr
ect
form
ula
is g
iven
fo
r ca
lciu
m c
hlo
rid
e. C
a(s)
�C
l 2(g
) 0
CaC
l 2(a
q)
9.
H2S
O4(
aq)
�2A
l(N
O3)
3(aq
) 0
Al 2(
SO4)
3(aq
) �
2HN
O3(
aq)
The
equ
atio
n is
no
t b
alan
ced
. 3H
2SO
4(aq
) �
2Al(
NO
3)3(
aq) 0
Al 2
(SO
4)3(
aq)
�
6HN
O3(
aq)
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
99
Exce
ptio
ns t
o th
e O
ctet
Rul
eEx
cept
ions
to
the
Oct
et R
ule
The
oct
et r
ule
is a
n im
port
ant g
uide
to u
nder
stan
ding
how
mos
t com
poun
ds a
re f
orm
ed.
How
ever
,the
re a
re a
num
ber
of c
ases
in w
hich
the
octe
t rul
e do
es n
ot a
pply
. Ans
wer
the
follo
win
g qu
estio
ns a
bout
exc
eptio
ns to
the
octe
t rul
e.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
9
1.
Dra
w th
e L
ewis
str
uctu
re f
or th
e co
mpo
und
BeF
2.
2.
Doe
s B
eF2
obey
the
octe
t rul
e? E
xpla
in.
BeF
2d
oes
no
t o
bey
th
e o
ctet
ru
le b
ecau
se t
he
cen
tral
ato
m (
Be)
of
the
mo
lecu
le
is n
ot
surr
ou
nd
ed b
y fo
ur
elec
tro
n p
airs
.
3.
Dra
w th
e L
ewis
str
uctu
re f
or th
e co
mpo
und
NO
2.
4.
Doe
s N
O2
obey
the
octe
t rul
e? E
xpla
in.
NO
2d
oes
no
t o
bey
th
e o
ctet
ru
le b
ecau
se t
he
cen
tral
ato
m (
N)
of
the
mo
lecu
le is
no
t su
rro
un
ded
by
fou
r el
ectr
on
pai
rs.
5.
Dra
w th
e L
ewis
str
uctu
re f
or th
e co
mpo
und
N2F
2.
6.
Doe
s N
2F2
obey
the
octe
t rul
e? E
xpla
in.
N2F
2o
bey
s th
e o
ctet
ru
le b
ecau
se e
very
ato
m o
f th
e m
ole
cule
is s
urr
ou
nd
ed
by
fou
r el
ectr
on
s p
airs
.
7.
Dra
w th
e L
ewis
str
uctu
re f
or th
e co
mpo
und
IF5.
8.
Doe
s IF
5ob
ey th
e oc
tet r
ule?
Exp
lain
.IF
5d
oes
no
t o
bey
th
e o
ctet
ru
le b
ecau
se t
her
e ar
e 12
ele
ctro
ns
atta
ched
to
the
cen
tral
ato
m (
I).
Use
wit
h Ch
apte
r 9,
Sect
ion
9.3
F
Be
F
O
N
O
F N
N
F
F
FF
FF
�
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
Challenge Problems Answer Key Chemistry: Matter and Change T33
Nam
eD
ate
Cla
ss
12C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
12C
hal
len
ge
Pro
ble
ms
Mol
e R
elat
ions
hips
inCh
emic
al R
eact
ions
Mol
e R
elat
ions
hips
inCh
emic
al R
eact
ions
The
mol
e pr
ovid
es a
con
veni
ent w
ay o
f fi
ndin
g th
e am
ount
s of
the
subs
tanc
es in
a c
hem
ical
re
actio
n. T
he d
iagr
am b
elow
sho
ws
how
this
con
cept
can
be
appl
ied
to th
e re
actio
nbe
twee
n ca
rbon
mon
oxid
e (C
O)
and
oxyg
en (
O2)
,sho
wn
in th
e fo
llow
ing
bala
nced
equ
atio
n.
2CO
(g)
�O
2(g)
02C
O2(
g)
Use
the
equa
tion
and
the
diag
ram
to a
nsw
er th
e fo
llow
ing
ques
tions
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
12
Use
wit
h Ch
apte
r 12
,Se
ctio
n 12
.2
1.
Wha
t inf
orm
atio
n is
nee
ded
to m
ake
the
type
s of
con
vers
ions
sho
wn
by d
oubl
e-ar
row
1in
the
diag
ram
?
the
coef
fici
ents
of
the
sub
stan
ces
in t
he
bal
ance
d c
hem
ical
eq
uat
ion
2.
Wha
t con
vers
ion
fact
ors
wou
ld b
e ne
eded
to m
ake
the
conv
ersi
ons
repr
esen
ted
by
doub
le-a
rrow
2 in
the
diag
ram
for
CO
? B
y do
uble
-arr
ow 6
for
CO
2?28
.01
g C
O/1
mo
l CO
an
d 1
mo
l CO
/28.
01 g
CO
; 44.
01 g
CO
2/1
mo
l CO
2an
d
1 m
ol C
O2/
44.0
1 g
CO
2
3.
Wha
t inf
orm
atio
n is
nee
ded
to m
ake
the
type
s of
con
vers
ions
rep
rese
nted
by
doub
le-a
rrow
s 3
and
7 in
the
diag
ram
?
the
nu
mb
er o
f re
pre
sen
tati
ve p
arti
cles
in a
mo
le, o
r A
vog
adro
’s n
um
ber
:
6.02
�10
23
4.
Wha
t con
vers
ion
fact
ors
wou
ld b
e ne
eded
to m
ake
the
conv
ersi
ons
repr
esen
ted
by
doub
le-a
rrow
3 in
the
diag
ram
for
CO
?
6.02
�10
23p
arti
cles
CO
/1 m
ol C
O a
nd
1 m
ol C
O/6
.02
�10
23p
arti
cles
CO
5.
Why
is it
not
pos
sibl
e to
con
vert
bet
wee
n th
e m
ass
of a
sub
stan
ce a
nd th
e nu
mbe
r of
re
pres
enta
tive
part
icle
s,as
rep
rese
nted
by
doub
le-a
rrow
4 o
f th
e di
agra
m?
6.
Why
is it
not
pos
sibl
e to
use
the
mas
s of
one
sub
stan
ce in
a c
hem
ical
rea
ctio
n to
fin
d th
e m
ass
of a
sec
ond
subs
tanc
e in
the
reac
tion,
as r
epre
sent
ed b
y do
uble
-arr
ow 5
in th
e di
agra
m?
The
mas
ses
of
the
sub
stan
ces
do
no
t re
act
wit
h e
ach
oth
er in
def
init
e ra
tio
s.
Mo
les
of
CO
Gra
ms
of
CO
Mo
les
of
CO
2
Gra
ms
of
CO
2
Part
icle
s o
fC
OPa
rtic
les
of
CO
2
1 5
2
3
46
7
Mas
ses
of
dif
fere
nt
sub
stan
ces
con
tain
dif
fere
nt
nu
mb
ers
of
rep
rese
nta
tive
par
ticl
es.
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
1111
Usi
ng M
ole-
Base
dCo
nver
sion
sU
sing
Mol
e-Ba
sed
Conv
ersi
ons
The
dia
gram
sho
ws
thre
e co
ntai
ners
,eac
h of
whi
ch h
olds
a c
erta
in m
ass
of th
e su
bsta
nce
indi
cate
d. C
ompl
ete
the
tabl
e be
low
for
eac
h of
the
thre
e su
bsta
nces
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
11
1.
Com
pare
and
con
tras
t the
num
ber
of r
epre
sent
ativ
e pa
rtic
les
and
the
mas
s of
UF 6
with
the
num
ber
of r
epre
sent
ativ
e pa
rtic
les
and
mas
s of
CC
l 3CF 3.
Exp
lain
any
dif
fere
nces
yo
u ob
serv
e.Ev
en t
ho
ug
h t
he
mas
s o
f U
F 6is
gre
ater
th
an t
he
mas
s o
f C
Cl 3
CF 3
in t
hei
r re
spec
-
tive
co
nta
iner
s, t
her
e ar
e fe
wer
mo
lecu
les
of
UF 6
than
of
CC
l 3C
F 3. T
hat
is b
ecau
se
the
mo
lar
mas
s o
f U
F 6is
sig
nif
ican
tly
gre
ater
th
an t
hat
of
CC
l 3C
F 3. A
sin
gle
UF 6
mo
lecu
le h
as a
sig
nif
ican
tly
gre
ater
mas
s th
an a
sin
gle
CC
l 3C
F 3m
ole
cule
has
.
2.
UF 6
is a
gas
use
d in
the
prod
uctio
n of
fue
l for
nuc
lear
pow
er p
lant
s. H
ow m
any
mol
es o
fth
e ga
s ar
e in
100
.0 g
of
UF 6?
100.
0 g
UF 6
�1
mo
l UF 6
/352
.03
g U
F 6�
0.28
41 m
ol U
F 6
3.
CC
l 3CF 3
is a
chl
orof
luor
ocar
bon
resp
onsi
ble
for
the
dest
ruct
ion
of th
e oz
one
laye
r in
Ear
th’s
atm
osph
ere.
How
man
y m
olec
ules
of
the
liqui
d ar
e in
1.0
g o
f C
Cl 3C
F 3?
1.0
g C
Cl 3
CF 3
�1.
0 m
ol C
Cl 3
CF 3
/187
.37
g C
Cl 3
CF 3
�6.
02 �
1023
mo
lecu
les
CC
l 3C
F 3/
1.0
mo
l CC
l 3C
F 3 �
3.2
�10
21m
ole
cule
s C
Cl 3
CF 3
4.
Lea
d (P
b) is
use
d to
mak
e a
num
ber
of d
iffe
rent
allo
ys. W
hat i
s th
e m
ass
of le
ad p
rese
ntin
an
allo
y co
ntai
ning
0.1
5 m
ol o
f le
ad?
0.15
mo
l Pb
�20
7.2
g P
b/1
.0 m
ol P
b �
31 g
Pb
UF 6
(g
)
225.
0 g
CC
l 3C
F 3 (
l)
200.
0 g
Pb (
s)
250.
0 gUse
wit
h Ch
apte
r 11
,Se
ctio
n 11
.3
Mol
ar M
ass
Num
ber
of
Num
ber
of R
epre
sent
ativ
e Su
bsta
nce
Mas
s (g
)(g
/mol
)M
oles
(m
ol)
Part
icle
s
UF 6
(g)
225.
035
2.03
0.63
923.
848
�10
23m
ole
cule
s
CC
l 3C
F 3(l
)20
0.0
187.
371.
067
6.42
3 �
1023
mo
lecu
les
Pb(s
)25
0.0
207.
21.
207
7.26
6 �
1023
ato
ms
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
T34 Chemistry: Matter and Change Challenge Problems Answer Key
Nam
eD
ate
Cla
ss
14C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
14C
hal
len
ge
Pro
ble
ms
A S
impl
e M
ercu
ry B
arom
eter
A S
impl
e M
ercu
ry B
arom
eter
In Fi
gure
1,a
sim
ple
mer
cury
bar
omet
er is
mad
e by
fill
ing
a lo
ng
glas
s tu
be w
ith m
ercu
ry a
nd th
en in
vert
ing
the
open
end
of
the
tube
into
a b
owl o
f m
ercu
ry. A
nsw
er th
e fo
llow
ing
ques
tions
abo
utth
e si
mpl
e m
ercu
ry b
arom
eter
sho
wn
here
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
14
Use
wit
h Ch
apte
r 14
,Se
ctio
n 14
.1
1.
Wha
t occ
upie
s th
e sp
ace
abov
e th
e m
ercu
ry c
olum
n in
the
baro
met
er’s
gla
ss tu
be?
The
spac
e is
occ
up
ied
on
ly b
y sm
all a
mo
un
ts o
f
Hg
vap
or
du
e to
th
e p
rese
nce
of
a va
cuu
m a
bo
ve
the
mer
cury
co
lum
n in
th
e g
lass
tu
be.
2.
Wha
t pre
vent
s m
ercu
ry f
rom
flo
win
g ou
t of
the
glas
s tu
be in
to th
e bo
wl o
f m
ercu
ry?
Air
pre
ssu
re p
ush
ing
on
th
e su
rfac
e o
f th
e m
ercu
ry in
th
e b
ow
l kee
ps
the
mer
cury
in t
he
gla
ss t
ub
e.
3.
Whe
n th
e ba
rom
eter
in F
igur
e 1
is m
oved
to a
hig
her
elev
atio
n,su
ch a
s an
alti
tude
of
5000
met
ers,
the
colu
mn
of m
ercu
ry c
hang
es a
s sh
own
in F
igur
e 2.
Why
is th
e m
ercu
ryco
lum
n lo
wer
in F
igur
e 2
than
in F
igur
e 1?
Bec
ause
air
pre
ssu
re a
t th
e h
igh
er e
leva
tio
n in
Fig
ure
2 is
less
th
an t
he
air
pre
ssu
re a
t th
e lo
wer
ele
vati
on
in F
igu
re 1
, th
ere
is le
ss p
ush
ing
on
th
e su
rfac
e
of
the
mer
cury
in t
he
bo
wl i
n F
igu
re 2
.
4.
Supp
ose
the
baro
met
er in
Fig
ure
1 w
as c
arri
ed in
to a
n op
en m
ine
500
met
ers
belo
w s
eale
vel.
How
wou
ld th
e he
ight
of
the
mer
cury
col
umn
chan
ge?
Exp
lain
why
.
The
hei
gh
t o
f th
e m
ercu
ry c
olu
mn
wo
uld
incr
ease
in t
he
min
e b
ecau
se m
ore
air
pre
ssu
re w
ou
ld b
e p
ush
ing
on
th
e su
rfac
e o
f th
e m
ercu
ry in
th
e b
ow
l.
5.
Supp
ose
the
liqui
d us
ed to
mak
e th
e ba
rom
eter
was
wat
er in
stea
d of
mer
cury
. How
wou
ldth
is s
ubst
itutio
n af
fect
the
baro
met
er?
Exp
lain
.
The
colu
mn
of
liqu
id in
th
e tu
be
wo
uld
be
lon
ger
. Bec
ause
wat
er is
mu
ch le
ss
den
se t
han
mer
cury
, air
pre
ssu
re is
ab
le t
o s
up
po
rt a
mu
ch lo
ng
er c
olu
mn
of
wat
er t
han
of
mer
cury
.
6.
Supp
ose
a tin
y cr
ack
form
ed a
t the
top
of th
e ba
rom
eter
’s g
lass
tube
. How
wou
ld th
isev
ent a
ffec
t the
col
umn
of m
ercu
ry?
Exp
lain
why
.
Air
wo
uld
en
ter
the
gla
ss t
ub
e, c
ausi
ng
th
e m
ercu
ry c
olu
mn
to
flo
w o
ut
of
the
tub
e. A
ir p
ress
ure
exe
rted
on
th
e m
ercu
ry in
th
e tu
be
wo
uld
eq
ual
th
e ai
r
pre
ssu
re e
xert
ed o
n t
he
mer
cury
in t
he
bo
wl.
Gla
ss t
ub
e
Mer
cury
co
lum
n
Bo
wl o
f m
ercu
ry
At
sea
leve
lA
t 50
0 m
eter
sab
ove
sea
leve
l
Figu
re 1
Figu
re 2
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
1313
Inte
rmol
ecul
ar F
orce
s an
dBo
iling
Poi
nts
Inte
rmol
ecul
ar F
orce
s an
dBo
iling
Poi
nts
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
13
1.
How
do
the
boili
ng p
oint
s of
the
grou
p 4A
hyd
ride
s ch
ange
as
the
mol
ecul
ar m
asse
s of
the
hydr
ides
cha
nge?
As
the
mo
lecu
lar
mas
ses
of
the
hyd
rid
es in
crea
se, s
o d
o t
he
bo
ilin
g p
oin
ts o
f th
e
hyd
rid
es.
2.
Wha
t are
the
mol
ecul
ar s
truc
ture
and
pol
arity
of
the
four
gro
up 4
A h
ydri
des?
All
gro
up
4A
hyd
rid
e m
ole
cule
s ar
e re
gu
lar
tetr
ahed
ron
s an
d a
re n
on
po
lar.
3.
Pred
ict t
he s
tren
gth
of th
e fo
rces
bet
wee
n gr
oup
4A h
ydri
de m
olec
ules
. Exp
lain
how
thos
e fo
rces
aff
ect t
he b
oilin
g po
ints
of
grou
p 4A
hyd
ride
s.
Bec
ause
gro
up
4A
hyd
rid
es a
re n
on
po
lar,
the
rela
tive
ly w
eak
inte
rmo
lecu
lar
forc
es b
etw
een
th
eir
mo
lecu
les
hav
e lit
tle
effe
ct o
n t
hei
r b
oili
ng
po
ints
.
4.
How
do
the
boili
ng p
oint
s of
the
grou
p 6A
hyd
ride
s ch
ange
as
the
mol
ecul
ar m
asse
s of
the
hydr
ides
cha
nge?
Wit
h t
he
exce
pti
on
of
wat
er, a
s th
e m
ole
cula
r m
asse
s o
f th
e h
ydri
des
incr
ease
, so
do
th
e b
oili
ng
po
ints
of
the
hyd
rid
es. W
ater
, ho
wev
er, h
as a
mu
ch h
igh
er b
oili
ng
po
int
than
an
y o
ther
gro
up
6A
hyd
rid
e.
5.
Wha
t are
the
mol
ecul
ar s
truc
ture
and
pol
arity
of
the
four
gro
up 6
A h
ydri
des?
All
gro
up
6A
hyd
rid
e m
ole
cule
s ar
e b
ent,
po
lar
mo
lecu
les.
6.
Use
Tab
le 9
-4 in
you
r te
xtbo
ok to
det
erm
ine
the
diff
eren
ce in
ele
ctro
nega
tiviti
es o
f th
ebo
nds
in th
e fo
ur g
roup
6A
hyd
ride
s.H
—O
bo
nd
: �en
�1.
2; H
—S
bo
nd
: �en
�0.
4; H
—Se
bo
nd
: �en
�0.
4;
H—
Te b
on
d: �
en�
0.1
100 0
�10
0
H2O
H2S CH
4SiH
4
GeH
4
SnH
4
H2S
e
H2T
e
Gro
up
6A
hyd
rid
es
Gro
up
4A
hyd
rid
es
Mo
lecu
lar
mas
s
Boiling point (°C)
0 050
100
150
Use
wit
h Ch
apte
r 13
,Se
ctio
n 13
.3
The
boi
ling
poin
ts o
f liq
uids
dep
end
part
ly o
n th
e m
ass
of th
e pa
rtic
les
of w
hich
they
are
mad
e. T
he g
reat
er th
e m
ass
ofth
e pa
rtic
les,
the
mor
e en
ergy
is n
eede
d to
con
vert
a li
quid
to a
gas,
and,
thus
,the
hig
her
the
boili
ng p
oint
of
the
liqui
d. T
his
pat-
tern
may
not
hol
d tr
ue,h
owev
er,w
hen
ther
e ar
e si
gnif
ican
t for
ces
betw
een
the
part
icle
s of
a li
quid
. The
gra
ph p
lots
boi
ling
poin
tve
rsus
mol
ecul
ar m
ass
for
grou
p 4A
and
gro
up 6
A h
ydri
des.
Ahy
drid
e is
a b
inar
y co
mpo
und
cont
aini
ng h
ydro
gen
and
one
othe
rel
emen
t. U
se th
e gr
aph
to a
nsw
er th
e fo
llow
ing
ques
tions
.
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
Challenge Problems Answer Key Chemistry: Matter and Change T35
Nam
eD
ate
Cla
ss
16C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
16C
hal
len
ge
Pro
ble
ms
Stan
dard
Hea
t of
For
mat
ion
Stan
dard
Hea
t of
For
mat
ion
Hes
s’s
law
allo
ws
you
to d
eter
min
e th
e st
anda
rd h
eat o
f fo
rmat
ion
of a
com
poun
dw
hen
you
know
the
heat
s of
rea
ctio
ns th
at le
adto
the
prod
uctio
n of
that
com
poun
d. T
he f
irst
diag
ram
on
the
righ
t sho
ws
how
Hes
s’s
law
can
be u
sed
to c
alcu
late
the
heat
of
form
atio
n of
CO
2by
kno
win
g th
e he
ats
of r
eact
ion
of tw
ost
eps
lead
ing
to th
e pr
oduc
tion
of C
O2.
Use
this
diag
ram
to h
elp
you
answ
er th
e qu
estio
ns b
elow
abou
t the
sec
ond
diag
ram
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
16
Use
wit
h Ch
apte
r 16
,Se
ctio
n 16
.4
The
equ
atio
ns b
elow
sho
w h
ow N
O2
can
be
form
ed in
two
way
s:di
rect
ly f
rom
the
elem
ents
or
in tw
o st
eps.
N2(
g) �
O2(
g) 0
NO
2(g)
�H
�33
kJ/
mol
or
N2(
g) �
O2(
g) 0
NO
(g)
�H
�91
kJ/
mol
NO
(g)
�O
2(g)
0N
O2(
g)�
H �
�58
kJ/
mol
1.
On
the
diag
ram
at t
he r
ight
,dra
w a
rrow
head
sto
sho
w th
e di
rect
ions
in w
hich
the
thre
e lin
esla
bele
d 1,
2,an
d 3
shou
ld p
oint
.
2.
Wri
te th
e co
rrec
t rea
ctan
ts a
nd/o
r pr
oduc
ts o
nea
ch o
f th
e lin
es la
bele
d A
,B,a
nd C
.
3.
Wri
te th
e co
rrec
t ent
halp
y ch
ange
nex
t to
each
num
ber
on th
e di
agra
m.
1 � 2
1 � 21 � 21 � 2
CO
(g)
�
O2(
g)
�H
� �
110
kJ/m
ol
C(s
) �
O2(
g)
�H
� �
393
kJ/m
ol
�H
� �
283
kJ/m
ol
Enthalpy
CO
2(g
)
1 2
NO
2(g
)
�H
� �
58 k
J/m
ol
NO
(g)
� 1
/2 O
2(g
)
�H
� 9
1 kJ
/mo
l
Enthalpy
�H
� 3
3 kJ
/mo
l
1
2
3
A
B
C
1 /2
N2(
g)
� O
2(g
)
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
1515
Vapo
r Pr
essu
re L
ower
ing
Vapo
r Pr
essu
re L
ower
ing
You
hav
e le
arne
d th
at a
ddin
g a
nonv
olat
ile s
olut
e to
a s
olve
nt
low
ers
the
vapo
r pr
essu
re o
f th
at s
olve
nt. T
he a
mou
nt b
y w
hich
the
vapo
r pr
essu
re is
low
ered
can
be
calc
ulat
ed b
y m
eans
of
a re
latio
nshi
p di
scov
ered
by
the
Fren
ch c
hem
ist F
ranç
ois
Mar
ieR
aoul
t (18
30–1
901)
in 1
886.
Acc
ordi
ng to
Rao
ult’s
law
,the
vap
orpr
essu
re o
f a
solv
ent (
P)
is e
qual
to th
e pr
oduc
t of
its v
apor
pre
ssur
ew
hen
pure
(P
0 ) a
nd it
s m
ole
frac
tion
(X)
in th
e so
lutio
n,or
P�
P0 X
The
sol
utio
n sh
own
at th
e ri
ght w
as m
ade
by a
ddin
g 75
.0 g
of
sucr
ose
(C12
H22
O11
) to
500
.0 g
of
wat
er a
t a te
mpe
ratu
re o
f 20
°C.
Ans
wer
the
follo
win
g qu
estio
ns a
bout
this
sol
utio
n.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
15
1.
Why
do
the
suga
r m
olec
ules
in th
e so
lutio
n lo
wer
the
vapo
r pr
essu
re o
f th
e w
ater
?
The
pre
sen
ce o
f su
gar
mo
lecu
les
at t
he
surf
ace
of
the
solu
tio
n r
edu
ces
the
nu
mb
er o
f w
ater
mo
lecu
les
at t
he
surf
ace,
th
us
red
uci
ng
th
e n
um
ber
of
wat
er
mo
lecu
les
able
to
esc
ape
into
th
e va
po
r p
has
e.
2.
Wha
t is
the
num
ber
of m
oles
of
sucr
ose
in th
e so
lutio
n?
75.0
g s
ucr
ose
�1
mo
l su
cro
se/3
42.3
g s
ucr
ose
�0.
219
mo
l su
cro
se
3.
Wha
t is
the
num
ber
of m
oles
of
wat
er in
the
solu
tion?
500.
0 g
wat
er �
1 m
ol w
ater
/18.
02 g
wat
er �
27.7
5 m
ol w
ater
4.
Wha
t is
the
mol
e fr
actio
n of
wat
er in
the
solu
tion?
Mo
le f
ract
ion
of
wat
er �
mo
les
solv
ent /(
mo
les
solv
ent
�m
ole
s so
lute
)
�27
.75
mo
l /(27
.75
mo
l �0.
219
mo
l) �
0.99
2
5.
Wha
t is
the
vapo
r pr
essu
re o
f th
e so
lutio
n if
the
vapo
r pr
essu
re o
f pu
re w
ater
at 2
0°C
is17
.54
mm
Hg?
P�
P0 X
; P�
(17.
54 m
m H
g)(
0.99
2) �
17.4
mm
Hg
6.
How
muc
h is
the
vapo
r pr
essu
re o
f th
e so
lutio
n re
duce
d fr
om th
at o
f w
ater
by
the
addi
tion
of th
e su
cros
e?
red
uct
ion
�17
.54
mm
Hg
�17
.4 m
m H
g �
0.1
mm
Hg
Wat
erm
ole
cule
Solu
tio
n
Sucr
ose
mo
lecu
le
Use
wit
h Ch
apte
r 15
,Se
ctio
n 15
.3
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
T36 Chemistry: Matter and Change Challenge Problems Answer Key
8 7 6 5 4 3 2 1
10
02
34
5Ti
me
(sec
)
Concentration (mol/L)
67
89
10
SO2
SO3
SO3
O2
O2
SO2
Nam
eD
ate
Cla
ss
18C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
18C
hal
len
ge
Pro
ble
ms
Chan
ging
Equ
ilibr
ium
Conc
entr
atio
ns in
a R
eact
ion
Rev
ersi
ble
reac
tions
eve
ntua
lly r
each
an
equi
libri
um
cond
ition
in w
hich
the
conc
entr
atio
ns o
f al
l rea
ctan
tsan
d pr
oduc
ts a
re c
onst
ant.
Equ
ilibr
ium
can
be
dist
urbe
d,ho
wev
er,b
y th
e ad
ditio
n or
rem
oval
of
eith
er a
rea
ctan
t or
prod
uct.
The
gra
ph o
n th
e ri
ght s
how
s ho
w th
e co
ncen
tra-
tions
of
the
reac
tant
s an
d pr
oduc
t of
a re
actio
n ch
ange
whe
n eq
uilib
rium
is d
istu
rbed
. Use
the
grap
h to
ans
wer
the
follo
win
g qu
estio
ns.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
18
Use
wit
h Ch
apte
r 18
,Se
ctio
n 18
.1
1.
Wri
te th
e eq
uatio
n fo
r th
e re
actio
n de
pict
ed in
the
grap
h.2S
O2(
g)
�O
2(g
) 3
2SO
3(g
)
2.
Wri
te th
e eq
uilib
rium
con
stan
t exp
ress
ion
for
the
reac
tion.
Keq
�[S
O3]
2 /[SO
2]2 [
O2]
3.
Exp
lain
the
shap
es o
f th
e cu
rves
for
the
thre
e ga
ses
duri
ng th
e fi
rst 2
min
utes
of
the
reac
tion.
The
con
cen
trat
ion
s o
f th
e tw
o r
eact
ants
, SO
2an
d O
2, d
ecre
ase
as t
hey
are
use
d
up
to
pro
du
ce S
O3;
th
e co
nce
ntr
atio
n o
f SO
3in
crea
ses.
4.
At a
ppro
xim
atel
y w
hat t
ime
does
the
reac
tion
reac
h eq
uilib
rium
? H
ow d
o yo
u kn
oweq
uilib
rium
has
bee
n re
ache
d?
Equ
ilib
riu
m is
rea
ched
at
abo
ut
3 m
inu
tes;
th
e co
nce
ntr
atio
ns
of
all t
hre
e
gas
es b
eco
me
con
stan
t at
th
at t
ime.
5.
Wha
t are
the
conc
entr
atio
ns o
f th
e th
ree
gase
s at
equ
ilibr
ium
?[S
O2]
eq�
6.0
mo
l/L;
[O
2]eq
�4.
5 m
ol/
L; [
SO3]
eq�
1.0
mo
l/L
6.
Cal
cula
te th
e va
lue
of K
eqfo
r th
e re
actio
n.
Keq
�[1
.0]2
/[6.0
]2[4
.5]
�1.
0 /(3
6)(4
.5)
�0.
0062
7.
Des
crib
e th
e ch
ange
mad
e in
the
syst
em 4
min
utes
into
the
reac
tion.
Tel
l how
you
kno
wth
e ch
ange
was
mad
e.SO
2w
as a
dd
ed t
o t
he
syst
em b
ecau
se t
he
con
cen
trat
ion
of
SO2
incr
ease
d
sud
den
ly a
t th
at t
ime.
8.
At w
hat t
ime
does
the
syst
em r
etur
n to
equ
ilibr
ium
?
at 6
min
ute
s
Chan
ging
Equ
ilibr
ium
Conc
entr
atio
ns in
a R
eact
ion
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
1717
Det
erm
inin
g R
eact
ion
Rat
esD
eter
min
ing
Rea
ctio
n R
ates
Din
itrog
en p
ento
xide
dec
ompo
ses
to p
rodu
ce
nitr
ogen
dio
xide
and
oxy
gen
as r
epre
sent
edby
the
follo
win
g eq
uatio
n.
2N2O
5(g)
04N
O2(
g) �
O2(
g)
The
gra
ph o
n th
e ri
ght r
epre
sent
s th
e co
ncen
-tr
atio
n of
N2O
5re
mai
ning
as
the
reac
tion
proc
eeds
over
tim
e. A
nsw
er th
e fo
llow
ing
ques
tions
abo
utth
e re
actio
n.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
17
1.
Wha
t is
the
conc
entr
atio
n of
N2O
5at
the
begi
nnin
g of
the
expe
rim
ent?
Aft
er 1
hou
r?A
fter
2 h
ours
? A
fter
10
hour
s?
1.50
mo
l/L;
1.2
0 m
ol/
L; 0
.90
mo
l/L;
0.1
8 m
ol/
L
2.
By
how
muc
h do
es th
e co
ncen
trat
ion
of N
2O5
chan
ge d
urin
g th
e fi
rst h
our
of th
e re
actio
n? C
alcu
late
the
perc
enta
ge o
f ch
ange
the
conc
entr
atio
n un
derg
oes
duri
ng th
e fi
rst h
our
of th
e re
actio
n.
1.50
mo
l/L
�1.
20 m
ol/
L �
0.30
mo
l/L;
(1.5
0 m
ol/
L �
1.20
mo
l/L)
/1.5
0 m
ol/
L �
.20
�10
0% �
20%
3.
The
inst
anta
neou
s ra
te o
f re
actio
n is
def
ined
as
the
chan
ge in
con
cent
ratio
n of
rea
ctan
tdu
ring
som
e sp
ecif
ied
time
peri
od,o
r in
stan
tane
ous
rate
of
reac
tion
= [
N2O
5]/t
. Wha
t is
the
inst
anta
neou
s ra
te o
f re
actio
n fo
r th
e de
com
posi
tion
of N
2O5
for
the
time
peri
odbe
twee
n th
e fi
rst a
nd s
econ
d ho
urs
of th
e re
actio
n? B
etw
een
the
seco
nd a
nd th
ird
hour
s?B
etw
een
the
sixt
h an
d se
vent
h ho
urs?
(1.2
0 m
ol/
L �
0.09
mo
l/L)
/1 h
�0.
30 m
ol/
L/h
; (0.
90 m
ol/
L �
0.70
mo
l/L)
/1 h
�
0.20
mo
l/L/
h; (
0.35
mo
l/L
�0.
30 m
ol/
L)/1
h �
0.05
0 m
ol/
L/h
4.
Wha
t is
the
inst
anta
neou
s ra
te o
f re
actio
n fo
r th
e de
com
posi
tion
of N
2O5
betw
een
the
sec-
ond
and
four
th h
ours
of
the
reac
tion?
Bet
wee
n th
e th
ird
and
eigh
th h
ours
of
the
reac
tion?
(0.9
0 m
ol/
L �
0.55
mo
l/L)
/2 h
�0.
18 m
ol/
L/h
;
(0.7
0 m
ol/
L �
0.25
mo
l/L)
/5 h
�0.
09 m
ol/
L/h
5.
How
long
doe
s it
take
for
0.1
0 m
ol o
f N
2O5
to d
ecom
pose
dur
ing
the
tent
h ho
ur o
f th
e re
actio
n?
0.20
mo
l/L/
h �
0.10
mo
l/L/
x
x�
(0.1
0 m
ol/
L)/(0
.20
mo
l/L/
h)
�5
ho
urs
6.
Wha
t is
the
aver
age
rate
of
reac
tion
for
the
deco
mpo
sitio
n of
N2O
5ov
eral
l?
(1.5
0 m
ol/
L �
0.18
mo
l/L)
10 h
�0.
13 m
ol/
L/h
Tim
e (h
)
Concentration (mol/L)
0.2 0
0.4
0.6
0.8
1.0
1.2
1.4
1.6
10
23
45
67
89
10
Use
wit
h Ch
apte
r 17
,Se
ctio
n 17
.1
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
Challenge Problems Answer Key Chemistry: Matter and Change T37
Nam
eD
ate
Cla
ss
20C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
20C
hal
len
ge
Pro
ble
ms
Bala
ncin
g O
xida
tion
–R
educ
tion
Equ
atio
nsBa
lanc
ing
Oxi
dati
on–
Red
ucti
on E
quat
ions
Sci
entis
ts h
ave
deve
lope
d a
num
ber
of m
etho
ds f
or p
rote
ctin
g m
etal
s fr
om o
xida
tion.
One
suc
h m
etho
d in
volv
es th
e us
e of
asa
crif
icia
l met
al. A
sac
rifi
cial
met
al is
a m
etal
that
is m
ore
easi
ly
oxid
ized
than
the
met
al it
is d
esig
ned
to p
rote
ct. G
alva
nize
d ir
on,f
orex
ampl
e,co
nsis
ts o
f a
piec
e of
iron
met
al c
over
ed w
ith a
thin
laye
r of
zin
c. W
hen
galv
aniz
ed ir
on is
exp
osed
to o
xyge
n,it
is th
e zi
nc,
rath
er th
an th
e ir
on,t
hat i
s ox
idiz
ed.
Wat
er h
eate
rs o
ften
con
tain
a m
etal
rod
that
is m
ade
by c
oatin
g a
heav
y st
eel w
ire
with
mag
nesi
um o
r al
umin
um. I
n th
is c
ase,
the
mag
nesi
um o
r al
umin
um is
the
sacr
ific
ial m
etal
,pro
tect
ing
the
iron
casi
ng o
f th
e he
ater
fro
m c
orro
sion
.
The
dia
gram
sho
ws
a po
rtio
n of
a w
ater
hea
ter
cont
aini
ng
a sa
crif
icia
l rod
. Ans
wer
the
follo
win
g qu
estio
ns a
bout
the
diag
ram
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
20
Use
wit
h Ch
apte
r 20
,Se
ctio
n 20
.3
1.
In th
e ab
senc
e of
a s
acri
fici
al m
etal
,oxy
gen
diss
olve
d in
wat
er m
ay r
eact
with
the
iron
casi
ng o
f th
e he
ater
. One
pro
duct
for
med
is ir
on(I
I) h
ydro
xide
(Fe
(OH
) 2). W
hich
ele
men
tis
oxi
dize
d an
d w
hich
is r
educ
ed in
this
rea
ctio
n?
Iro
n is
oxi
diz
ed, a
nd
oxy
gen
is r
edu
ced
.
2.
Bal
ance
the
oxid
atio
n–re
duct
ion
equa
tion
for
this
rea
ctio
n:Fe
(s)
�O
2(aq
) �
H2O
0Fe
(OH
) 2(aq
)
2Fe(
s) �
O2(
aq)
+ 2
H2O
02F
e(O
H) 2
(aq
)
3.
Wri
te th
e tw
o ha
lf-r
eact
ions
for
this
exa
mpl
e of
cor
rosi
on.
Fe(s
) 0
Fe2 �
(aq
) �
2e�
;
O2(
aq)
�2H
2O(l
) �
4e�0
4OH
�(a
q)
4.
Supp
ose
the
sacr
ific
ial r
od in
the
diag
ram
abo
ve is
coa
ted
with
alu
min
um m
etal
. Wri
teth
e ba
lanc
ed e
quat
ion
for
the
reac
tion
of a
lum
inum
with
oxy
gen
diss
olve
d in
the
wat
er.
(Hin
t:T
he p
rodu
ct f
orm
ed is
alu
min
um h
ydro
xide
(A
l(O
H) 3)
.4A
l(s)
�3O
2(aq
) �
6H2O
(l) 0
4Al(
OH
) 3(a
q)
5.
Wri
te th
e tw
o ha
lf-r
eact
ions
for
this
exa
mpl
e of
cor
rosi
on.
Al(
s) 0
Al3
�(a
q)
�3e
�; O
2(aq
) �
2H2O
(l)
�4e
�0
4OH
�(a
q)
6.
Supp
ose
that
som
e ir
on in
the
casi
ng o
f th
e w
ater
hea
ter
is o
xidi
zed,
as s
how
n in
the
equa
tion
of q
uest
ion
2 ab
ove.
The
sac
rifi
cial
met
al (
alum
inum
,in
this
cas
e) im
med
iate
lyre
stor
es th
e Fe
2�io
ns to
iron
ato
ms.
Wri
te tw
o ha
lf-r
eact
ions
that
rep
rese
nt th
is s
ituat
ion.
Al(
s) 0
Al3
�(a
q)
�3e
�; F
e2�
(aq
) �
2e�0
Fe(s
)
Iro
nca
sin
g
Stee
l wir
e
Sacr
ific
ial
met
al
Wat
er
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
1919
Swim
min
g Po
ol C
hem
istr
ySw
imm
ing
Pool
Che
mis
try
The
pre
senc
e of
dis
ease
-cau
sing
bac
teri
a in
sw
imm
ing
pool
s is
a m
ajor
hea
lth c
once
rn.
Chl
orin
e ga
s is
add
ed to
the
wat
er in
som
e la
rge
com
mer
cial
sw
imm
ing
pool
s to
kill
bact
eria
. How
ever
,in
mos
t hom
e sw
imm
ing
pool
s,ei
ther
sol
id c
alci
um h
ypoc
hlor
ite(C
a(O
Cl)
2) o
r an
aqu
eous
sol
utio
n of
sod
ium
hyp
ochl
orite
(N
aOC
l) is
use
d to
trea
t the
wat
er. B
oth
com
poun
ds d
isso
ciat
e in
wat
er to
for
m th
e w
eak
acid
hyp
ochl
orou
s ac
id(H
OC
l). H
ypoc
hlor
ous
acid
is a
hig
hly
effe
ctiv
e ba
cter
icid
e. B
y co
ntra
st,t
he h
ypoc
hlor
iteio
n (O
Cl�
) is
not
a v
ery
effe
ctiv
e ba
cter
icid
e. U
se th
e in
form
atio
n ab
ove
to a
nsw
er th
e fo
llow
ing
ques
tions
abo
ut th
e ac
id-b
ase
reac
tions
that
take
pla
ce in
sw
imm
ing
pool
s.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
19
1.
Wri
te a
n eq
uatio
n th
at s
how
s th
e re
actio
n be
twee
n hy
poch
loro
us a
cid
and
wat
er. I
dent
ify
the
acid
,bas
e,co
njug
ate
acid
,and
con
juga
te b
ase
in th
is r
eact
ion.
HO
Cl(
aq)
�H
2O(l
) 3
H3O
�(a
q)
�O
Cl�
(aq
)
The
acid
is H
OC
l, an
d t
he
bas
e is
H2O
. Th
e co
nju
gat
e ac
id is
H3O
�, a
nd
th
e
con
jug
ate
bas
e is
OC
l�.
2.
Wri
te a
n eq
uatio
n th
at s
how
s th
e re
actio
n th
at o
ccur
s w
hen
the
hypo
chlo
rite
ion
(OC
l�),
in th
e fo
rm o
f ca
lciu
m h
ypoc
hlor
ite o
r so
dium
hyp
ochl
orite
,is
adde
d to
wat
er. N
ame
the
acid
,bas
e,co
njug
ate
acid
,and
con
juga
te b
ase
in th
is r
eact
ion.
OC
l�(a
q)
�H
2O(l
) 3
HO
Cl(
aq)
�O
H�
The
acid
is H
2O, a
nd
th
e b
ase
is O
Cl�
. Th
e co
nju
gat
e ac
id is
HO
Cl,
and
th
e
con
jug
ate
bas
e is
OH
�.
3.
Wha
t eff
ect d
oes
the
addi
tion
of h
ypoc
hlor
ite io
n ha
ve o
n th
e pH
of
swim
min
g po
ol w
ater
?
The
add
itio
n o
f h
ypo
chlo
rite
ion
incr
ease
s th
e co
nce
ntr
atio
n o
f O
H�
and
ther
efo
re in
crea
ses
pH
.
4.
The
eff
ectiv
enes
s of
hyp
ochl
orite
ion
as a
bac
teri
cide
dep
ends
on
pH. H
ow d
oes
high
pH
affe
ct th
e eq
uilib
rium
rea
ctio
n de
scri
bed
in q
uest
ion
2? W
hat e
ffec
t wou
ld h
igh
pH h
ave
on th
e ba
cter
ia?
A h
igh
pH
val
ue
ind
icat
es a
hig
h c
on
cen
trat
ion
of
OH
�, w
hic
h t
end
s to
dri
ve t
he
reac
tio
n t
o t
he
left
. As
a re
sult
, th
ere
will
be
rela
tive
ly le
ss H
OC
l an
d r
elat
ivel
y
mo
re O
Cl�
in s
olu
tio
n, m
akin
g it
eas
ier
for
bac
teri
a to
su
rviv
e.
5.
In th
e pr
esen
ce o
f su
nlig
ht,h
ypoc
hlor
ite io
n de
com
pose
s to
for
m c
hlor
ide
ion
and
oxyg
en
gas.
Wri
te a
n eq
uatio
n fo
r th
is r
eact
ion
and
tell
how
it a
ffec
ts th
e sa
fety
of
pool
wat
er.
2OC
l�(a
q) 0
2Cl�
(aq
) �
O2(
g)
Rem
ova
l of
OC
l�d
rive
s th
e re
acti
on
des
crib
ed in
qu
esti
on
2 t
o t
he
left
, cau
sin
g
mo
re H
OC
l to
bre
ak d
ow
n, l
eavi
ng
less
HO
Cl t
o k
ill b
acte
ria.
Use
wit
h Ch
apte
r 19
,Se
ctio
n 19
.2
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
T38 Chemistry: Matter and Change Challenge Problems Answer Key
Nam
eD
ate
Cla
ss
22C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
22C
hal
len
ge
Pro
ble
ms
Stru
ctur
al Is
omer
s of
Hex
ane
Stru
ctur
al Is
omer
s of
Hex
ane
The
str
uctu
ral f
orm
ula
of a
n or
gani
c co
mpo
und
can
som
etim
es b
e w
ritte
n in
a
vari
ety
of w
ays,
but s
omet
imes
str
uctu
ral f
orm
ulas
that
app
ear
sim
ilar
can
repr
esen
t dif
fere
nt c
ompo
unds
. The
str
uctu
ral f
orm
ulas
bel
ow a
re te
n w
ays
of
repr
esen
ting
com
poun
ds h
avin
g th
e m
olec
ular
for
mul
a C
6H14
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
22
Use
wit
h Ch
apte
r 22
,Se
ctio
ns 2
2.1
and
22.3
1.
In th
e sp
aces
pro
vide
d,w
rite
the
corr
ect n
ame
for
each
of
the
stru
ctur
al f
orm
ulas
,lab
eled
a–j,
abov
e.
a.
e.
i.
b.
f.j.
c.g
.
d.
h.
2.
How
man
y di
ffer
ent c
ompo
unds
are
rep
rese
nted
by
the
stru
ctur
al f
orm
ulas
abo
ve?
Wha
tar
e th
eir
nam
es?
five
; hex
ane,
2-m
eth
ylp
enta
ne,
2,3
-dim
eth
ylb
uta
ne,
2,2
-dim
eth
ylb
uta
ne,
and
3-m
eth
ylp
enta
ne
2-m
eth
ylp
enta
ne
2,2-
dim
eth
ylb
uta
ne
3-m
eth
ylp
enta
ne
2,3-
dim
eth
ylb
uta
ne
3-m
eth
ylp
enta
ne
2,3-
dim
eth
ylb
uta
ne
2-m
eth
ylp
enta
ne
hex
ane
3-m
eth
ylp
enta
ne
hex
ane
CH
3
CH
2C
H2
CH
2C
H2
CH
3
a.
CH
3
CH
3
CH
CH
2C
H2
CH
3
b.
CH
3
CH
3
CH
3C
HC
HC
H3
c.
CH
3
CH
3
CH
3C
CH
2C
H3
d.
CH
3
CH
CH
2
CH
2
CH
3
CH
3e.
CH
3C
HC
HC
H3
CH
3C
H3
f.
CH
2
CH
2C
H3
CH
CH
3
CH
3
g.
CH
3
CH
2
CH
3C
HC
H2
CH
3
h.
CH
2
CH
3
CH
2
CH
2C
H2
CH
3i.
CH
3
CH
2C
HC
H2
CH
3C
H3
j.
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
2121
Effe
ct o
f Co
ncen
trat
ion
onCe
ll Po
tent
ial
Effe
ct o
f Co
ncen
trat
ion
onCe
ll Po
tent
ial
In a
volta
ic c
ell w
here
all
ions
hav
e a
conc
entr
atio
n of
1M
,the
cel
l pot
entia
l is
equa
l to
the
stan
dard
pot
entia
l. Fo
r ce
lls in
whi
ch io
n co
ncen
trat
ions
are
gre
ater
or
less
than
1M
,as
show
n be
low
,an
adju
stm
ent m
ust b
e m
ade
to c
alcu
late
cel
l pot
entia
l.T
hat a
djus
tmen
t is
expr
esse
d by
the
Ner
nst e
quat
ion:
Ece
ll�
E0 ce
ll�
log
In th
is e
quat
ion,
nis
the
num
ber
of m
oles
of
elec
tron
s tr
ansf
erre
d in
the
reac
tion,
and
xan
d y
are
the
coef
fici
ents
of
the
prod
uct a
nd r
eact
ant i
ons,
resp
ectiv
ely,
in th
eba
lanc
ed h
alf-
cell
reac
tions
for
the
cell.
[pro
duct
ion]
x�
�[r
eact
ant i
on]y
0.05
92�
n
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
21
1.
Wri
te th
e tw
o ha
lf-r
eact
ions
and
the
over
all c
ell r
eact
ion
for
the
cell
show
n ab
ove.
Ag
�(a
q)
�1e
�0
Ag
(s)
Cu
(s) 0
Cu
2 �(a
q)
�2e
�
2Ag
�(a
q)
�C
u(s
) 0
2Ag
(s)
�C
u2 �
(aq
)
2.
Use
Tab
le 2
1-1
in y
our
text
book
to d
eter
min
e th
e st
anda
rd p
oten
tial o
f th
is c
ell.
3.
Wri
te th
e N
erns
t equ
atio
n fo
r th
e ce
ll.E c
ell
�0.
46 V
�0.
0592
/2�l
og
1.0
�10
�3 /(
1.0
�10
�2 )
2 ; n
�2
in t
his
cas
e b
ecau
se
two
mo
les
of
elec
tro
ns
are
tran
sfer
red
in t
he
ove
rall
cell
reac
tio
n (
See
answ
er t
o
qu
esti
on
1.)
4.
Cal
cula
te th
e ce
ll po
tent
ial f
or th
e io
n co
ncen
trat
ions
sho
wn
in th
e ce
ll.
0.16
V
0.46
V
Vo
ltm
eter
Ag
Cu
Cu
2�
1.0
� 1
0�3 M
Ag
�
1.0
� 1
0�2 M
Use
wit
h Ch
apte
r 21
,Se
ctio
n 21
.1
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
Challenge Problems Answer Key Chemistry: Matter and Change T39
Nam
eD
ate
Cla
ss
24C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
24C
hal
len
ge
Pro
ble
ms
The
Chem
istr
y of
Lif
eTh
e Ch
emis
try
of L
ife
Pro
tein
s ar
e sy
nthe
size
d w
hen
RN
A m
olec
ules
tr
ansl
ate
the
DN
A la
ngua
ge o
f ni
trog
en b
ases
into
the
prot
ein
lang
uage
of
amin
o ac
ids
usin
g a
gene
tic c
ode.
The
gen
etic
cod
e is
fou
nd in
RN
A m
ole-
cule
s ca
lled
mes
seng
er R
NA
(m
RN
A),
whi
ch a
re s
yn-
thes
ized
fro
m D
NA
mol
ecul
es. T
he g
enet
ic c
ode
cons
ists
of
a se
quen
ce o
f th
ree
nitr
ogen
bas
es in
the
mR
NA
,cal
led
a co
don.
Mos
t cod
ons
code
for
spe
cifi
cam
ino
acid
s. A
few
cod
ons
code
for
a s
top
in th
e sy
n-th
esis
of
prot
eins
. The
tabl
e sh
ows
the
mR
NA
cod
ons
that
mak
e up
the
gene
tic c
ode.
To
use
the
tabl
e,re
adth
e th
ree
nitr
ogen
bas
es in
seq
uenc
e. T
he f
irst
bas
e is
show
n al
ong
the
left
sid
e of
the
tabl
e. T
he s
econ
d ba
seis
sho
wn
alon
g th
e to
p of
the
tabl
e. T
he th
ird
base
issh
own
alon
g th
e ri
ght s
ide
of th
e ta
ble.
For
exa
mpl
e,th
e se
quen
ce C
AU
cod
es f
or th
e am
ino
acid
his
tidin
e(H
is).
The
tabl
e gi
ves
abbr
evia
tions
for
the
amin
oac
ids.
Ans
wer
the
follo
win
g qu
estio
ns a
bout
the
gene
tic c
ode.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
24
Use
wit
h Ch
apte
r 24
,Se
ctio
n 24
.4
1.
Wha
t am
ino
acid
is r
epre
sent
ed b
y ea
ch o
f th
e fo
llow
ing
codo
ns?
a.
CU
G
b.
UC
A
2.
Wri
te th
e se
quen
ce o
f am
ino
acid
s fo
r w
hich
the
follo
win
g m
RN
A s
eque
nce
code
s.
-C-A
-U-C
-A-C
-C-G
-G-U
-C-U
-U-U
-U-C
-U-U
-
-His
-His
-Arg
-Ser
-Ph
e-Le
u-
3.
Err
ors
som
etim
es o
ccur
whe
n m
RN
A m
olec
ules
are
syn
thes
ized
fro
m D
NA
mol
ecul
es.
Nitr
ogen
bas
es m
ay b
e om
itted
,an
extr
a ni
trog
en b
ase
may
be
adde
d,or
a n
itrog
en b
ase
may
be
chan
ged
duri
ng s
ynth
esis
. The
two
mR
NA
seq
uenc
es s
how
n be
low
are
exa
mpl
esof
suc
h er
rors
. In
eac
h ca
se,t
ell h
ow th
e m
RN
A s
eque
nce
show
n di
ffer
s fr
om th
e co
rrec
tm
RN
A s
eque
nce
give
n in
que
stio
n 2.
a.
-C-A
-U-C
-A-C
-C-G
-G-U
-U-C
-U-U
-U-U
-C-U
-U-
An
ext
ra U
(u
raci
l) h
as b
een
ad
ded
at
po
siti
on
#10
.
b.
-C-A
-U-U
-A-C
-C-G
-G-U
-C-U
-U-U
-U-C
-U-U
-
C (
cyto
sin
e) a
t p
osi
tio
n #
4 h
as b
een
ch
ang
ed t
o U
(u
raci
l).
4.
Wri
te th
e am
ino
acid
seq
uenc
e fo
r ea
ch o
f th
e m
RN
A s
eque
nces
sho
wn
in q
uest
ion
3.
a.
b.
-His
-Tyr
-Arg
-Ser
-Ph
e-Le
u-
-His
-His
-Arg
-Ph
e-Ph
e-Se
r-
Seri
ne
Leu
cin
e
UU
UU
CU
UA
UU
GU
UU
UC
UC
CU
AC
UG
CC
UU
AU
CA
UA
AU
GA
AU
UG
UC
GU
AG
UG
GG
CU
UC
CU
CA
UC
GU
UC
UC
CC
CC
AC
CG
CC
CU
AC
CA
CA
AC
GA
AC
UG
CC
GC
AG
CG
GG
AU
UA
CU
AA
UA
GU
UA
UC
AC
CA
AC
AG
CC
AU
AA
CA
AA
AA
GA
AA
UG
AC
GA
AG
AG
GG
GU
UG
CU
GA
UG
GU
UG
UC
GC
CG
AC
GG
CC
GU
AG
CA
GA
AG
GA
AG
UG
GC
GG
AG
GG
GG
Seco
nd
bas
e
First base
Third base
}}Phe
Leu
Ile Met
Leu
Val
Pro
Ser
Ala
Thr
} }} }} }}His
Gln
Tyr
Sto
pSt
op
Asn
Lys
Asp
Glu
} }}Arg
Gly
Cys
Sto
pTr
p
Ser
Arg
U C A G
UC
AG
The
Gen
etic
Cod
e
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
2323
Boili
ng P
oint
s of
Org
anic
Fam
ilies
Boili
ng P
oint
s of
Org
anic
Fam
ilies
The
mos
t im
port
ant f
acto
r de
term
inin
g th
e bo
iling
poi
nt o
f a
subs
tanc
e is
its
atom
ic o
r m
olec
ular
mas
s. I
n ge
nera
l,th
e la
rger
the
atom
ic o
r m
olec
ular
mas
s of
the
subs
tanc
e,th
em
ore
ener
gy is
nee
ded
to c
onve
rt th
e su
bsta
nce
from
the
liqui
dph
ase
to th
e ga
seou
s ph
ase.
As
an e
xam
ple,
the
boili
ng p
oint
of e
than
e (m
olec
ular
mas
s �
30; b
oilin
g po
int �
�89
°C)
ism
uch
high
er th
an th
e bo
iling
poi
nt o
f m
etha
ne (
mol
ecul
arm
ass
�16
; boi
ling
poin
t ��
161°
C).
Inte
rmol
ecul
ar f
orce
s be
twee
n th
e pa
rtic
les
of a
liqu
id a
lso
can
affe
ct th
e liq
uid’
s bo
iling
poi
nt. T
he g
raph
sho
ws
tren
ds in
the
boili
ng p
oint
s of
fou
r or
gani
c fa
mili
es:a
lkan
es,a
lcoh
ols,
alde
hyde
s,an
d et
hers
. Use
the
grap
h an
d yo
ur k
now
ledg
e of
inte
rmol
ecul
ar f
orce
s to
ans
wer
the
follo
win
g qu
estio
ns.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
23
1.
For
any
one
fam
ily,w
hat i
s th
e re
latio
nshi
p be
twee
n m
olec
ular
mas
s an
d bo
iling
poi
nt?
As
the
mo
lecu
lar
mas
s in
crea
ses,
so
do
es t
he
bo
ilin
g p
oin
t.
2.
For
com
poun
ds o
f si
mila
r m
olec
ular
mas
s,w
hich
fam
ily o
f th
e fo
ur s
how
n in
the
grap
hha
s th
e lo
wes
t boi
ling
poin
ts?
Whi
ch f
amily
has
the
high
est b
oilin
g po
ints
?
The
alka
nes
hav
e th
e lo
wes
t b
oili
ng
po
ints
; th
e al
coh
ols
hav
e th
e h
igh
est
bo
ilin
g
po
ints
.
3.
Find
and
list
the
boili
ng p
oint
s fo
r et
hano
l (m
olec
ular
mas
s �
46)
and
dim
ethy
l eth
er(m
olec
ular
mas
s �
46)
on th
e gr
aph.
Why
wou
ld y
ou e
xpec
t the
se tw
o co
mpo
unds
toha
ve r
elat
ivel
y si
mila
r bo
iling
poi
nts?
Bo
ilin
g p
oin
t o
f et
han
ol:
abo
ut
78°C
; bo
ilin
g p
oin
t o
f d
imet
hyl
eth
er: a
bo
ut
�25
°C; t
he
two
co
mp
ou
nd
s w
ou
ld b
e ex
pec
ted
to
hav
e si
mila
r b
oili
ng
po
ints
bec
ause
th
ey h
ave
the
sam
e m
ole
cula
r m
ass.
4.
Find
the
alde
hyde
with
a m
olec
ular
mas
s of
abo
ut 5
8. N
ame
that
ald
ehyd
e an
d w
rite
its
chem
ical
for
mul
a.Th
e al
deh
yde
is p
rop
anal
, CH
3CH
2CH
O.
5.
Can
this
ald
ehyd
e fo
rm h
ydro
gen
bond
s? C
an o
ther
ald
ehyd
es f
orm
hyd
roge
n bo
nds?
Exp
lain
.
Nei
ther
pro
pan
al n
or
any
oth
er a
ldeh
yde
can
fo
rm h
ydro
gen
bo
nd
s b
ecau
se
ald
ehyd
es la
ck �
OH
gro
up
s.
3040
5060
Mo
lecu
lar
mas
s
� a
lkan
e�
alc
oh
ol
Boiling point (°C)70
80
�50050100
� a
ldeh
yde
� e
ther
Use
wit
h Ch
apte
r 23
,Se
ctio
n 23
.3
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
T40 Chemistry: Matter and Change Challenge Problems Answer Key
Nam
eD
ate
Cla
ss
26C
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
26C
hal
len
ge
Pro
ble
ms
The
Phos
phor
us C
ycle
The
Phos
phor
us C
ycle
Pho
spho
rus
is a
n im
port
ant e
lem
ent b
oth
in o
rgan
ism
s an
d in
the
litho
sphe
re. I
n or
gani
sms,
phos
phor
us o
ccur
s in
DN
A a
nd R
NA
mol
ecul
es,c
ell m
embr
anes
,bon
esan
d te
eth,
and
in th
e en
ergy
–sto
rage
com
poun
d ad
enos
ine
trip
hosp
hate
(A
TP)
. In
the
litho
-sp
here
,pho
spho
rus
occu
rs p
rim
arily
in th
e fo
rm o
f ph
osph
ates
,as
a m
ajor
con
stitu
ent o
fm
any
rock
s an
d m
iner
als.
Pho
spha
te r
ock
is m
ined
to p
rodu
ce m
any
com
mer
cial
pro
duct
s,su
ch a
s fe
rtili
zers
and
det
erge
nts.
Whe
n th
ese
prod
ucts
are
use
d,ph
osph
ates
are
ret
urne
d to
the
litho
sphe
re a
nd h
ydro
sphe
re. T
hus,
phos
phor
us—
like
carb
on a
nd n
itrog
en—
cycl
es in
the
envi
ronm
ent.
Use
the
diag
ram
of
the
phos
phor
us c
ycle
to a
nsw
er th
e qu
estio
ns b
elow
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
26
Use
wit
h Ch
apte
r 26
,Se
ctio
n 26
.4
1.
By
wha
t met
hods
doe
s ph
osph
orus
get
into
soi
l?
An
imal
s ex
cret
e p
ho
sph
ates
. Dea
d o
rgan
ism
s re
leas
e p
ho
sph
ates
as
they
dec
ay.
Rai
nfa
ll le
ach
es p
ho
sph
ates
fro
m r
ock
s.
2.
By
wha
t met
hod
do p
lant
s ob
tain
the
phos
phor
us th
ey n
eed?
by
extr
acti
ng
ph
osp
hat
es f
rom
th
e so
il
3.
By
wha
t met
hod
do a
nim
als
obta
in th
e ph
osph
orus
they
nee
d?
by
eati
ng
pla
nts
or
oth
er a
nim
als
4.
In w
hat w
ay is
the
phos
phor
us c
ycle
dif
fere
nt f
rom
the
carb
on a
nd n
itrog
en c
ycle
s yo
ust
udie
d in
the
text
book
?
The
ph
osp
ho
rus
cycl
e h
as n
o a
tmo
sph
eric
co
mp
on
ent.
5.
The
pho
spho
rus
cycl
e ha
s bo
th s
hort
-ter
m a
nd lo
ng-t
erm
par
ts. U
se d
iffe
rent
col
ored
pe
ncils
to s
how
eac
h pa
rt o
n th
e di
agra
m.
Sho
rt-t
erm
cyc
le: a
nim
als
excr
etin
g p
ho
sph
ates
; dec
om
po
sin
g o
rgan
ism
s; a
nim
als
eati
ng
pla
nts
; pla
nts
ext
ract
ing
ph
osp
hat
es f
rom
th
e so
il; m
inin
g; u
se o
f fe
rtili
z-er
s, d
eter
gen
ts, a
nd
oth
er s
ynth
etic
pro
du
cts
con
tain
ing
ph
osp
ho
rus.
Lo
ng
-ter
mcy
cle:
geo
log
ical
up
lift,
ph
osp
hat
es le
ach
ing
fro
m r
ock
s, p
ho
sph
ate
rock
fo
rmin
gu
nd
er t
he
oce
ans
Pho
sph
ate
rock
s
Pho
sph
ate
rock
s
Geo
log
ical
up
lift
Nam
eD
ate
Cla
ss
Ch
alle
ng
e Pr
ob
lem
sC
hem
istr
y: M
atte
r an
d C
han
ge
• C
hap
ter
2525
The
Prod
ucti
on o
fPl
uton
ium
-239
The
Prod
ucti
on o
fPl
uton
ium
-239
Whe
n nu
clea
r fi
ssio
n w
as f
irst
dis
cove
red,
only
two
isot
opes
,ura
nium
-233
and
ura
nium
-235
,wer
ekn
own
of b
eing
cap
able
of
unde
rgoi
ng th
is n
ucle
ar c
hang
e.Sc
ient
ists
late
r di
scov
ered
a th
ird
isot
ope,
plut
oniu
m-2
39,
also
cou
ld u
nder
go n
ucle
ar f
issi
on. P
luto
nium
-239
doe
s no
toc
cur
in n
atur
e bu
t can
be
mad
e sy
nthe
tical
ly in
nuc
lear
reac
tors
and
par
ticle
acc
eler
ator
s.
The
dia
gram
sho
ws
the
proc
ess
by w
hich
plu
toni
um-2
39is
mad
e in
nuc
lear
rea
ctor
s. A
nsw
er th
e qu
estio
ns a
bout
the
diag
ram
.
CH
ALLEN
GE P
RO
BLEM
SCH
AP
TER
25
1.
Iden
tify
the
isot
ope
who
se n
ucle
us is
labe
led
A in
the
diag
ram
.
2.
Nam
e th
e ty
pe o
f nu
clea
r re
actio
n th
at o
ccur
s w
hen
a
neut
ron
stri
kes
nucl
eus
A.
3.
Iden
tify
the
isot
ope
who
se n
ucle
us is
labe
led
B.
4.
Bes
ides
fra
gmen
ted
nucl
ei,w
hat e
lse
is p
rodu
ced
whe
n a
neut
ron
stri
kes
nucl
eus
A?
5.
Iden
tify
the
isot
ope
who
se n
ucle
us is
labe
led
C.
6.
Wri
te th
e nu
clea
r eq
uatio
n fo
r th
e re
actio
n th
at o
ccur
s w
hen
a ne
utro
n st
rike
s nu
cleu
s C
.Id
entif
y th
e pr
oduc
t D f
orm
ed in
the
reac
tion.
1 0n �
238 92U
00 0
�23
9 92U
; Th
e p
rod
uct
nu
cleu
s is
ura
niu
m-2
39.
7.
Wri
te th
e nu
clea
r eq
uatio
n fo
r th
e de
cay
of n
ucle
us D
. Ide
ntif
y is
otop
e E
for
med
in th
ere
actio
n.23
9 92U
0�
10 �
239 93N
p; n
eptu
niu
m-2
39
8.
Wri
te a
bal
ance
d nu
clea
r eq
uatio
n fo
r th
e de
cay
of n
ucle
us E
. Ide
ntif
y is
otop
e F
form
edin
the
reac
tion.
239 93N
p 0
�10
�23
9 94Pu
; plu
ton
ium
-239
9.
Nam
e th
e ty
pe o
f nu
clea
r re
actio
n th
at o
ccur
s w
hen
a ne
utro
n st
rike
s nu
cleu
s F.
nu
clea
r fi
ssio
n
10.
Wri
te th
e nu
clea
r eq
uatio
n fo
r th
e re
actio
n th
at o
ccur
s w
hen
a ne
utro
n st
rike
s nu
cleu
s F.
Iden
tify
isot
ope
G f
orm
ed in
the
reac
tion.
1 0n �
239 94Pu
012
5 48C
d �
115 46Pd
; pal
lad
ium
-115
ura
niu
m-2
38 (
238 92U
)
neu
tro
ns
silv
er-1
14 (
114 47A
g)
nu
clea
r fi
ssio
n
ura
niu
m-2
35 (
235 92U
)
0 –1
0 –10 0�
1 0n
1 0n
1 0n
1 0n
45p
75n
48p
77n
92p
143n
92p
143n
92p
146n
Sou
rce
of
neu
tro
ns
B
A
C
D
E
G
F
Use
wit
h Ch
apte
r 25
,Se
ctio
n 25
.4
Cop
yrig
ht ©
Gle
ncoe
/McG
raw
-Hill
,a d
ivis
ion
of th
e M
cGra
w-H
ill C
ompa
nies
,Inc
.
Challenge Problems Chemistry: Matter and Change T41
CREDITS
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