Appendix D
CHECKPOINT ANSWERS
D-1
D-2
APPENDIX D
Chapter 1
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(p. 6) The symbol 8 S represents 8 individual atoms of sulfur. The symbol
S8 represents 8 atoms of sulfur bonded together to form a single molecule.
(p. 8) 529 mg; 259 cm
(p. 10) 4.72 * 10 - 7; 1.03 * 1022; 0.0000000754; 3,668,000
(p. 13) 5; 2; 3; 1; 4
(p. 13) In 10,000 atoms of lithium you would expect to find approximately
742 6Li atoms and 9258 atoms of 7Li.
(p. 14) The ratio of the natural abundance of 1H to 2H is 99.985>0.015 =
6.7 * 103; 2 significant figures
(p. 16) NaOH: Na⫹ and OH⫺; K2SO4: 2 K⫹ and SO42 - ; BaSO4: Ba2⫹ and
SO42 - ; Be3(PO4)2: 3 Be2⫹ and 2 PO43⫺
(p. 17) F: 9; Pb: 82; 24: Cr; 74:W
(p. 22) H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) ¡
Na + (aq) + Cl - (aq) + H2O(l)
Chapter 2
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(p. 33) K atomic weight 39.098 amu; molar mass 39.098 g/mol; U atomic
mass 238.03 amu; 238.03 g/mol.
(p. 35) 2.2 * 1022 Al atoms
(p. 40) There are two atoms of carbon in one molecule of C2H2; two moles
of carbon atoms in one mole of C2H2; and 1.20 * 1024 atoms of carbon in
one mole of C2H2.
(p. 43) CaC2. The molecular weight must be known to find the molecular
formula.
(p. 48) Four moles of H2; 2.4 * 1024 molecules of H2; 4.8 * 1024 atoms
of H.
(p. 49) Two moles of CO require 4.0 moles of H2. This equals 8.06 grams
of H2.
(p. 53) In the balanced chemical equation I2 and Mg react in a one-to-one
mole ratio. Four grams of I2 equals 0.016 moles. Four grams of Mg equals
0.16 moles. There is 10 times more Mg present than I2. I2 is the limiting
reagent.
(p. 58) Moles are generally more useful to chemists than mass when relating how much of one chemical substance will react with another.
Chapter 3
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(p. 73) The radius of a typical atom would have to be 10,000 cm. This is
equivalent to 100 m.
(p. 75) The frequency of microwaves covers a range of approximately
3 * 1010 to 3 * 1012 Hz.
(p. 79) The higher the temperature of an object, the higher the frequency of
electromagnetic radiation emitted by the object.
(p. 85) The first ionization energy of Sr would be greater than that of Rb.
Sr has one more proton in its nucleus than Rb and therefore attracts its electrons more strongly than Rb. The first ionization energy of Cl is greater than
D-3
CHECKPOINT ANSWERS
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Energy (MJ/mol)
Relative Intensity
433
48.5
39.2
5.44
3.24
0.77
0.63
2
2
6
2
6
1
2
Electrons are removed from the subshells of Sc in the following order: 4s,
3d, 3p, 3s, 2p, 2s, 1s.
Relative number of electrons
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that of Br. Bromine has one more shell of electrons than Cl, and therefore
its electrons are further from the nucleus and more easily removed.
(p. 86) Carbon has a core charge of ⫹4, and fluorine has a core charge of ⫹7.
(p. 87) The first ionization energy of sodium is much less than the first ionization energy of neon. If there were 9 electrons in the second level, we
would expect these electrons to be held more tightly by the 11 protons in
the nucleus of sodium than the 8 valence electrons held by the 10 protons
in the nucleus of neon. This would give sodium a larger first ionization than
neon rather than the observed smaller ionization energy.
(p. 88) Sodium has three shells, whereas lithium only has two. The first ionization energy of sodium is much less than the first ionization energy of
lithium. Therefore, the radius of the valence shell of sodium is larger than
that of lithium.
(p. 89) The number of peaks in a photoelectron spectrum is determined by
the number of different energies that the electrons in an atom have.
(p. 90) There is only one peak in the PES of He because both of its electrons have the same energy.
(p. 92) Neon has 2 electrons in its n ⫽ 1 shell and 8 electrons in its n ⫽ 2
shell. Of the 8 electrons in the n ⫽ 2 shell, 2 are in the s subshell and 6
are in the p subshell.
(p. 94) The energies and the relative intensities of the PES of Sc are as follows:
39.2
433
3.24
5.44
48.5
55
50
45
40
35
8
6
0.77 0.63
4
2
0
Ionization Energy (MJ/mol)
The subshells of Sc are filled with electrons in the following order: 1s, 2s, 2p,
3s, 3p, 4s, 3d. There is a difference in the order in which electrons are added
to and taken away from Sc. Electrons are added to the lowest energy subshell
(the most stable subshell closest to the nucleus) first and then to subsequent
subshells in order of increasing energy until the 3p subshell is filled. After the
3d subshell is filled, the 4s is filled followed by the lower energy 3d subshell.
D-4
APPENDIX D
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Electrons are removed from the highest energy subshell (the subshell farthest
from the nucleus) first and then from lower energy subshells.
(p. 96) Three quantum numbers are required to describe three dimensions
in space. The numbers describe n-size, l-shape, and m-orientation.
(p. 99) Li, B, and N all have at least one unpaired electron and would therefore be expected to have magnetic properties. Be has no unpaired electrons
and would therefore have no magnetic properties and would not be deflected
in a Stern-Gerlach experiment.
(p. 102) C: 2; N: 3; Ne: 0; F: 1; C, N, and F all have at least one unpaired
electron and would therefore be expected to have magnetic properties.
(p. 105) A metallic radius is one-half the distance between two adjacent
nuclei of two like atoms measured in a solid crystal lattice. A covalent radius
is one-half the distance between two adjacent nuclei of two like atoms of a
molecule. The molecule may be in the solid, liquid, or gaseous state.
(p. 107) The ionic radius of Si4⫹ would be less than the ionic radius of
Al3⫹. Si4⫹ and Al3⫹ have the same number of electrons, but Si4⫹ has one
more proton that pulls the electrons more closely to the nucleus.
(p. 108) Ionization energies obtained from PES are for the removal of only one
electron. That electron may be located in any shell or subshell. Second, third,
and fourth ionization energies describe the energy to sequentially remove a second, third, and fourth electron from the outermost shells of an atom.
(p. 110) The maximum positive charge on atoms in Group IVA is ⫹4. For
example, Sn has the electron configuration [Kr] 5s2 4d10 5p2. It can lose
four electrons to form a Sn4⫹ ion with the electron configuration: [Kr] 4d10.
(p. 111) AVEEs increase from left to right across a period because each subsequent atom in a period has one more positive charge in the nucleus that
attracts the electrons more tightly. As you move from the bottom to the top
of a group, there are fewer shells containing electrons. The electrons are
closer to the nucleus and therefore held tightly by the nucleus. This is the
same trend followed by first ionization energy. The trend for AVEEs is the
reverse of the trend followed by atomic radii.
Chapter 4
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(p. 125) An atom in Group IVA has 4 valence electrons.
(p. 127) H2⫹ has only one electron to form the bond between the two hydrogen atoms. A covalent bond is formed by the attraction between the nucleus
of one atom and the electron(s) of another atom. Because there is only one
electron in H2⫹, both nuclei share the single electron. In addition, there is less
electron density to shield the repulsion between the two nuclei in H2⫹. Therefore the nuclei in H2⫹ are not pulled as closely together as the nuclei in H2.
(p. 128) In CO2 the nonbonding electrons are located in the nonbonding
electron domains of the oxygen atoms. In CO2 each oxygen has two nonbonding domains. There are two electrons in each nonbonding domain, but
four electrons are located in each of the bonding domains between carbon
and oxygen. Four electrons are required to form a double bond.
(p. 129) Hydrogen needs only two electrons in its outer shell in order to
achieve the electron configuration of He. Therefore, hydrogen is capable of
forming only one bond and cannot serve as the central atom in a Lewis
structure.
D-5
CHECKPOINT ANSWERS
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(p. 129) A skeleton structure shows the relative positions of the atoms. A
Lewis structure shows the positions of the electrons.
(p. 131)
(p. 133) In order for an atom to expand its octet, it must have free, unoccupied orbitals with an energy close to the energy of its valence electrons
(within the same shell). Nitrogen and oxygen have valence electrons in the
second shell. The second shell contains one s orbital and three p orbitals.
These four orbitals can contain a maximum of eight total electrons. Because
there are no more subshells in the second shell, no more orbitals are available to hold any additional electrons.
(p. 136) The oxygen–oxygen bond length would be greatest in hydrogen
peroxide because single bonds are longer than multiple bonds when comparing bond lengths between the same atoms.
(p. 138) Isomers have the same molecular formula but differ in the arrangement of the atoms. All of the atoms in the two structures shown for acetate
are in the same positions. Only the location of a multiple bond differs in the
two Lewis structures. This makes them resonance structures, not isomers.
(p. 138)
H
H
H
H
C
C
C
C
H
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C
C
H
H
H
H
C
C
C
C
C
C
H
H
H
The two resonance Lewis structures of benzene have alternating single and
double bonds in two different arrangements. The actual structure of benzene
is an average of these two structures. Therefore, we would expect each bond
in benzene to behave like a bond and a half (an average of a single bond
and a double bond). The bond length of each C¬C bond will be greater
than the C“C double bond of ethylene but less than the C¬C single bond
length of ethane.
(p. 140) The bond in NO would be polar covalent because N and O have
different electronegativities. The bond in O2 would be pure covalent because
both oxygen atoms have the same electronegativity.
(p. 142) The most electronegative atom will have the negative partial charge.
If the electronegativity of B is greater than that of A, B would have the negative partial charge.
(p. 143) The formal charges are: double-bonded oxygen 0, sulfur ⫹1, singlebonded oxygen -1.
(p. 149) Each carbon in the Lewis structure of benzene has one C¬C single
bond, one C¬C double bond, and one C¬H single bond. Therefore, each
carbon atom has three electron-pair domains.
(p. 154) The shape of a molecule can be found using Table 4.3. The shape
of a molecule with three bonding domains and one nonbonding domain
around the central atom is trigonal pyramidal.
D-6
APPENDIX D
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(p. 155) The C1¬C2¬C3 is 109⬚.
The C1¬C2¬Hg is 109⬚.
The Hg¬C1¬O is slightly less than 109°.
(p. 156) There are three bonding electron domains around each carbon.
Therefore, the bond angles would be approximately 120°.
(p. 156) Chloroform would have a dipole moment because the polarity of
the C¬H bond would be different from that of the three C¬Cl bonds. The
polarities of the bonds would not cancel and would therefore result in a
dipole moment for the molecule.
Chapter 5
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(p. 179) Ca 6 K 6 Rb
(p. 181) Atoms in Group IIA have two electrons in their valence shell that
can be more easily lost than their inner-shell electrons. Group IA elements
have only one valence shell electron. Therefore, an electron must be
removed from an inner shell in order to lose a second electron. These electrons are closer to the nucleus and are more difficult to remove.
(p. 189) The most common ion of Sr is Sr2⫹. The peroxide anion is O22 - .
The formula is SrO2.
(p. 192) The electrons in metallic bonds are delocalized and therefore free
to move from one atom to another and conduct an electric current. The electrons in both covalent and ionic bonds are localized and therefore not free
to move to another atom or ion.
(p. 195) The difference in electronegativity between Zn and S is 0.99. Based
only on the difference in electronegativity, we would predict that ZnS is
polar covalent.
(p. 199) Using the bond-type triangle, we find ZnS located directly on the
dividing line between covalent and ionic bonding. We would predict that
ZnS would have approximately equal covalent and ionic character.
(p. 201) BaSi2 is primarily metallic, BaBr2 is primarily ionic, and GaBr3 is
primarily covalent. BaBr2 would be the best insulator that would also stand
up to high heat.
(p. 207) Formal charge and oxidation numbers are tools that have been
developed by chemists. Formal charge treats all bonds as if they are pure
covalent bonds. Oxidation numbers treat all bonds as if they were pure ionic
bonds. Both of these are useful tools for making predictions about chemical structures and changes that occur during reactions.
(p. 209) The first reaction is not an oxidation–reduction reaction. None of
the atoms undergoes a change in oxidation state. The second reaction is an
oxidation–reduction reaction. The oxidation state of N changes from ⫺3 in
NH3 to ⫹2 in NO. The oxidation state of O changes from zero in O2 to ⫺2
in both NO and water. The third reaction is an oxidation–reduction reaction.
The oxidation number for carbon in CH3OH changes from ⫺2 to ⫺3, while
the oxidation number of carbon in CO changes from ⫹2 to ⫹3.
Chapter 6
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(p. 224) If the liquid water and gaseous water are both at the same temperature, then the average kinetic energy of the water molecules should be the same.
CHECKPOINT ANSWERS
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(p. 228) The pressure exerted on the floor by the spike heel will be greater.
If the same person wears one shoe with a spike heel and the other with a
normal heel, the same total force is exerted on the floor. However, in the case
of the spike heel, that force is exerted over a much smaller area of the floor.
(p. 232) No, the diameter of the tube does not change the relationship
between the pressure and volume of the gas.
(p. 233) PV r T
(p. 234) Boyle’s law described the relationship between pressure and volume, and Amonton’s law described the relationship between pressure and
temperature. By combining these laws, we can obtain a relationship between
temperature and volume. This is Charles’ law.
(p. 244) If we assume that 1 L of wet air and 1 L of dry air are at the same
temperature and pressure, then there must be the same number of moles in
both containers. The gas in the container of dry air has an average molecular
weight of 29.0 g/mol. The container of wet air contains both air and water
vapor. Because it contains both air and water vapor but has the same total number of moles of gas as the dry air, there are fewer moles of dry air (29 g/mol)
in this container and more moles of water vapor (18 g/mol) than in the container of dry air. Therefore, the container of wet air is lighter than the container
of dry air because the container of wet air has fewer heavy molecules (N2 and
O2) and more light molecules (H2O) of gas than the container of dry air.
(p. 247) As the temperature of a gas increases, the molecules of gas begin
to move more rapidly, resulting in their having more kinetic energy.
(p. 250) The heavier gas molecules will be moving at a slower velocity than
the lighter gas molecules. Because kinetic energy is dependent on both mass
and velocity, a heavy molecule moving at a slow velocity can have the same
kinetic energy as a less massive molecule moving at a greater velocity.
Chapter 7
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(p. 268) No bonds are broken in the reactants. One bond is formed in the
products. Therefore, more bonds are made than broken. This reaction will
release energy because it is a bond-making process.
(p. 269) When heat enters a balloon, the temperature of the gas particles
increases. This results in an increase in the motion of the gas particles, causing them to strike the sides of the balloon more frequently and with more
force. If the balloon is elastic, it will expand, increasing the volume of the
gas while the pressure remains constant.
(p. 271) Specific heats are based on the mass of a substance, and molar heat
capacities are based on moles. One mole of iron has a much larger mass
than one mole of aluminum. Given two substances with similar structures
and properties, the one with the higher molecular weight will have a higher
molar heat capacity and a lower specific heat compared to the substance
with the lower molecular weight.
(p. 274) While at rest on the 10th floor, the crate will have only potential
energy. During the fall the crate will have both potential and kinetic energy.
As the crate falls its velocity increases, thus increasing its kinetic energy.
As it moves closer to the ground, its potential energy in relation to the
ground decreases. When the crate strikes the ground, it can release its energy
in the form of sound and heat or by doing work on the ground (making a
hole or indentation).
D-7
D-8
APPENDIX D
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(p. 275) According to the first law of thermodynamics, the total energy
change must equal zero. This requires that the sum of the change in energy
of the system and the change in energy of the surroundings equals zero. If
we assume the brick to be our system and the water to be the surroundings,
the brick can gain energy only if the surroundings supply that energy. Therefore, the first law does not prevent the brick from becoming hotter and the
water from becoming colder. The first law requires only that what is gained
by one part of the universe must be lost by another part of the universe.
(p. 278) The temperature of the gas will increase the most for the gas in a
fixed-volume piston and cylinder. All of the heat transferred to this gas is
used to increase the temperature of the gas. Some of the heat transferred to
the gas in the movable piston and cylinder is used to do work.
(p. 282) Table sugar can provide or give off energy. This is an exothermic
process, and the bond-making process must provide more energy than is
required by the bond-breaking process. The products must therefore be more
stable than the reactants. The bonds formed are stronger than the bonds broken.
(p. 283) All of the reactions listed in Table 7.2 release or give off energy.
Therefore, these are exothermic reactions, and the sign of the change in
enthalpy is negative.
(p. 288) ¢H373ⴰ represents the change in enthalpy for a process that occurs
under standard conditions and at 373 K.
N(g) ¡ N(g)
(p. 290)
H(g) ¡ H(g)
N(g) + 3 H(g) ¡ NH3(g)
ⴰ
¢Hac for N(g) and H(g) are both zero because both are isolated atoms in the
gaseous state. The ¢Hacⴰ for NH3 is ⫺1171.76 kJ/molrxn. ¢Hacⴰ for NH3
represents the formation of NH3 from its gaseous atoms. No bonds are broken in the reactants, but bonds are formed by the products; therefore, ¢Hacⴰ
is negative.
(p. 293)
Gaseous atoms
Energy to
break
bonds
Energy to
form
bonds
Products
Change in
enthalpy
Reactants
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This is an endothermic reaction. The energy required to break bonds is
greater than the energy given off in forming bonds. Therefore, the total
strength of the bonds is greatest in the reactants.
(p. 295) Bond breaking: 2 CO(g) ¢H° = 2(1076.38 kJ/molrxn)
= 2152.76 kJ/molrxn
O2(g) ¢H° = 498.34 kJ/molrxn
Total bond breaking ¢H° = 2651.10 kJ/molrxn
Bond making: 2 CO2(g) ¢H° = 2(-1608.53 kJ/molrxn) = - 3217.06 kJ/molrxn
¢Hrxn° = (2651.10 kJ/molrxn) + (-3217.06 kJ/molrxn) = - 565.96 kJ/molrxn
The reaction is exothermic.
D-9
CHECKPOINT ANSWERS
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(p. 297) The enthalpy of atom combination of liquid ethanol is greater than
the enthalpy of atom combination of gaseous ethanol. More energy is
released in the formation of liquid ethanol than in the formation of gaseous
ethanol from its gaseous atoms. The change in state from liquid ethanol to
gaseous ethanol requires an input of energy. There are forces of attraction
between the molecules of liquid ethanol that are not present in gaseous
ethanol.
(p. 299) The sum of the bond strengths can be estimated using the values
from Exercise 7.9 and the discussion that follows this exercise.
Si¬F = 597 kJ/mol
Si¬O = 466 kJ/mol
O¬H = 463 kJ/mol
3 * 597 kJ/mol + 466 kJ/mol + 463 kJ/mol = 2720 kJ/mol
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(p. 299) When similar atoms are bonded together, the enthalpy of atom combination decreases with increasing bond length. BrF would be expected to
have weaker bonds (a smaller enthalpy of atom combination) than ClF
because BrF is a larger molecule with a larger bond length.
Chapter 8
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(p. 319) As an atom becomes larger, the electrons farther from the nucleus
are more easily induced into a temporary dipole. As the size of an atom
increases, there is an increase in the dispersion force between atoms.
Xe 7 Kr 7 Ar 7 Ne 7 He.
(p. 320) Hydrogen bonds between NH3 molecules are weaker than those
between H2O molecules because oxygen is more electronegative that nitrogen. This gives a larger partial charge to the H in the OH bond.
(p. 322) (a) These three substances have very different molecular weights.
We would predict that the boiling point should increase with an
increase in molecular weight due to increasing dispersion forces.
Therefore the boiling point of tetrabromobutane (373.6 g/mol)
is greater than the boiling point of carbon tetrachloride (154
g/mol), and acetone (58 g/mol) has the lowest boiling point.
(b) These three substances all have very similar molecular weights
(142 to 144 g/mol). When comparing the boiling points of substances with similar molecular weights, it is necessary to examine their intermolecular forces. Octanoic acid is capable of
hydrogen bonding and will therefore have the highest boiling
point. Nonanal is polar and therefore has dipole–dipole intermolecular forces. Decane is nonpolar and will have only
dispersion forces. Therefore, decane is expected to have the
lowest boiling point.
(p. 325) The phase change from liquid to gas requires more enthalpy
(enthalpy of vaporization 44 kJ/molrxn) than the phase change from solid to
liquid (enthalpy of fusion 6 kJ/molrxn). The solid state has the strongest
intermolecular forces. Some of these forces must be overcome to melt the
solid, and the remaining forces must be overcome to cause the liquid to
change to a gas.
(p. 327) If the volume of the container is decreased, the ideal gas law tells us
that the pressure of the vapor in the container should increase. However, if
D-10
APPENDIX D
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the pressure of the vapor in the container is at its vapor pressure (the maximum pressure of vapor for a given temperature), the pressure cannot increase.
Therefore, as the volume is decreased, condensation of the vapor will occur
so that the pressure of the vapor remains constant at the vapor pressure.
(p. 338) In order for a side chain to be hydrophilic, it must be able to interact with water. This means that the side chain should have intermolecular
forces similar to the intermolecular forces of water (hydrogen bonding). The
side chain on alanine contains carbon and hydrogen atoms that have similar
electronegativities. This side chain is nonpolar. The side chain on cysteine
is somewhat polar due to the presence of sulfur with its two nonbonding
pairs of electrons, but it is not capable of hydrogen bonding. These two side
chains are therefore hydrophobic (they do not form hydrogen bonds with
water). Lysine contains an NH3 group, and serine contains an OH group.
Both lysine and serine are capable of forming hydrogen bond intermolecular forces. Therefore, both of them will be hydrophilic.
(p. 342) The bond-breaking process for one mole of BaCl2(s) is ⫹1282
kJ/molrxn. The formation of attractive forces between one mole of Ba2⫹ ions
and water is –718 kJ/molrxn. The formation of attractive forces between two
moles of Cl⫺ ions and water release (2) ⫻ (⫺289 kJ/molrxn). The change
in enthalpy associated with the dissolution of BaCl2(s) is ⫺14 kJ/molrxn.
(p. 343) The bond-breaking process for AgCl(s) is ⫹533.30 kJ/molrxn. The
formation of attractive forces between one mole of Ag⫹ ions and water is
⫺178.97 kJ/molrxn. The formation of attractive forces between one mole of
Cl⫺ ions and water is ⫺288.838 kJ/molrxn. The change in enthalpy associated with the dissolution of AgCl(s) is ⫹65.49 kJ/molrxn.
Chapter 9
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(p. 371) On the atomic scale, molecular solids are composed of individual molecules held together by intermolecular forces. Network covalent
solids are composed of a three-dimensional array of atoms held together
by covalent bonds. On the macroscopic scale network covalent solids
have high melting points, and molecular solids have relatively low melting points.
(p. 374) Both the strength of the attraction of an atom for its valence electrons and the energy difference between valence subshells within an atom
contribute to the AVEE and electronegativity. Atoms that have a relatively
small attraction for their valence electrons and have small differences in
energy between valence subshells have low AVEE and electronegativity.
Atoms with these characteristics can lose their valence electrons to form
positive cations. The lost electrons can move easily between available subshells from cation to cation. This results in metals being good conductors
of electricity.
(p. 381) Diffusion of atoms can produce the formation of metallic bonds
between different metals.
(p. 382) On the atomic scale molecular solids are composed of individual molecules held together by intermolecular forces. Ionic solids are
composed of cations and anions in a three-dimensional array held
together by ionic bonds. Ionic solids have a higher melting point than
molecular solids. Ionic solids conduct an electric current when melted.
Molecular solids do not.
CHECKPOINT ANSWERS
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(p. 384) The lattice energy of MgF2 is greater than that of MgCl2 because
F⫺ is a smaller ion than Cl⫺. The smaller fluoride ion is held more closely
and tightly to the Mg2⫹.
(p. 384) Metallic solids are composed of positively charged metal cations
in a three-dimensional array. The metal cations are surrounded by a sea of
delocalized electrons that form metallic bonds. Ionic solids are composed
of cations and anions in a three-dimensional array held together by localized electrons in ionic bonds.
Metallic solids have a wide range of melting points compared to ionic
solids which have relatively high melting points. Metallic solids can conduct electricity in the solid state but ionic solids cannot.
(p. 384) In order for a substance to be classified as a body-centered cubic
cell, all of the positions in the unit cell must be occupied by the same type
of atom or ion.
(p. 385) B4C has an average EN of 2.3 and a ¢ EN of 0.5. This places it in
the covalent region of a bond-type triangle. MoC has an average EN of 2.0
and a ¢ EN 1.1 and is in the ionic region. Both compounds could serve as
insulators because they would not conduct heat or electricity.
(p. 392) The simple cubic cell has the simplest relationship between the radii
of the ions and the length of the edge of the unit cell. In a simple cubic cell
the length of the edge of the cell is equal to r ⫹ r. In a face-centered cube,
2(r ⫹ r) ⫽ a21>2 where a is the length of the edge of the unit cell. For a
body-centered cubic cell, the relationship is 2(r ⫹ r) ⫽ a31>2.
Chapter 10
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(p. 411) After the reaction has reached equilibrium, both N2O4 and NO2 will
be present in the container.
(p. 412) Using Table 10.1, we see that at equilibrium the ratio of trans-2butene to cis-2-butene is 0.559 to 0.441 which equals 1.27.
(p. 415) An increase in the rate constant of a reaction will result in an
increase in the rate of the reaction. Therefore, the rate of the reaction will
increase with temperature.
(p. 418) If the rate of the reverse reaction is greater than the rate of the forward reaction at a given moment in time, this does not mean that the reverse
rate constant is greater than the forward rate constant. The rate of a reaction is dependent on both the rate constant and the concentrations of the
reactants. If the rate of the reverse reaction is greater than the forward, this
may be due to a higher concentration of products (reactants for the reverse
reaction) than reactants for the forward reaction.
(p. 420) The equilibrium constant can be determined from the division of
the forward rate constant by the reverse rate constant. For this reaction the
equilibrium constant would be 2.
(p. 427) The stoichiometry of the reaction is 1:1:1. For every mole of PCl3
that reacts, one mole of PCl5 is produced and one mole of Cl2 is used up.
Therefore, if the concentration of PCl3 decreased by 0.96 mol/L, the concentration of Cl2 also decreased by 0.96 mol/L, leaving only 0.04 mol/L.
The concentration of PCl5 must increase from zero to 0.96 mol/L.
(p. 430) The equilibrium concentrations found in Exercise 10.8 give the correct Kc (1.27) within ; 1 in the last significant figure. 0.559> 0.441 ⫽ 1.27
D-11
D-12
APPENDIX D
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(p. 431) The change in concentrations that takes place during a chemical
reaction must follow the stoichiometry of the balanced chemical equation.
(p. 434) ¢ C will be small when the equilibrium constant is significantly smaller
than one. ¢ C would be small for K values of 1.0 ⫻ 10⫺5 and 1.0 ⫻ 10⫺10.
(p. 441) If more P2 is added to the system at equilibrium, the system will
shift to use up the added P2. The system will shift to the right to produce
more P4. If the pressure increases while the temperature remains constant,
a reaction will shift in a direction to relieve the increase in pressure by
decreasing the number of moles of gas. For this reaction the equilibrium
will shift to the right to produce P4 and use up P2.
Chapter 11
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(p. 470) An Arrhenius base must ionize to give OH⫺ when it dissolves in
water. Mg(OH)2 ionizes to give Mg2⫹ and 2 OH⫺. It is therefore an Arrhenius base. HNO3 and CH3CO2H both ionize to produce H⫹ and are therefore Arrhenius acids.
(p. 473) H3PO4 (aq) ⫹ H2O(l) uv H2PO4⫺(aq) ⫹ H3O⫹(aq). The conjugate base is H2PO4⫺. Phosphoric acid is a polyprotic acid, meaning that it can
further dissociate to lose an additional proton and produce additional H3O⫹.
C6H5NH2(aq) ⫹ H2O(l) uv C6H5NH3⫹(aq) ⫹ OH⫺
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The conjugate acid of aniline is C6H5NH3⫹.
(p. 478) The Kw will still equal 10⫺14. The addition of acid will result in a
higher overall concentration of H3O⫹, even though some of the hydronium
ion from the dissociation of water may shift back to the left. What is most
significant about the shift in the equilibrium of water back to the left is the
decrease in the concentration of hydroxide ion. This decrease in concentration of the hydroxide ion multiplied by the increase in concentration of the
hydronium ion will therefore equal Kw, 10⫺14.
(p. 481) pH is ⫺log[H3O⫹]. As the concentration of hydronium ion
increases, there will be a decrease in the pH.
Kw = [OH - ][H3O + ]
(p. 484)
[OH - ][H3O + ]
Ka =
[H2O]
The Ka expression includes H2O as a reactant.
(p. 485) Both solutions have a low concentration of acid. The acid with the
larger Ka will be a stronger acid and will dissociate more than the acid with
the small Ka. The solution with the larger Ka will produce more hydronium
ion and will therefore have a smaller pH.
(p. 489) A stronger acid produces a weaker conjugate base. HOCl is the
stronger acid; therefore its conjugate base, OCl⫺ will be the weaker base.
OBr⫺ will be the stronger base.
(p. 491) H2S is a stronger acid than H2O. Sulfur is a larger atom than
oxygen. The sulfur–hydrogen bond is longer and therefore weaker than the
oxygen–hydrogen bond. The sulfur–hydrogen bond is more easily broken
than the oxygen–hydrogen bond.
(p. 493) The only difference between the two acids (HOCl and HOI) is the
oxygen bonded to Cl in one acid and to I in the other. Because Cl is more
electronegative than I, it will pull electron density away from the
CHECKPOINT ANSWERS
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oxygen–hydrogen bond more than I. Thus the oxygen–hydrogen bond will
be weaker for HOCl than for HOI, and the hydrogen will more easily be
lost from HOCl. Therefore, HOCl is a stronger acid than HOI.
(p. 494) When a strong acid is added to water, it is assumed to completely dissociate to produce hydronium ion. For a monoprotic acid the
hydronium concentration will be equal to the initial concentration of
strong acid.
(p. 497) It is not possible to determine which of the two solutions has the
higher pH. Solution 1 contains a stronger acid and therefore dissociation is
greater than for the acid in solution 2, but the concentration of the acid in
solution 2 is greater than the concentration of the acid in solution 1. It is
possible that solution 2 could produce more hydronium ion because it is
more concentrated, even though it is a weaker acid.
(p. 499) The 1.0 M solution is more acidic.
(p. 500) A large Kb indicates a stronger base that would react with water
to accept more hydrogen ions and produce more hydroxide ions. Since
both bases have the same concentration, the base with the larger Kb is
more basic.
(p. 504) NO2 - is the conjugate base of HNO2. Therefore, Kb ⫽ Kw> Ka.
Kb = (1.0 * 10 - 14)>(5.1 * 10 - 4) = 2.0 * 10 - 11
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(p. 510) If acid is added to a buffer, the pH of the buffer will decrease
slightly, but the decrease will not be as great as when the same amount
of acid is added to water. If base is added to a buffer, the pH of the buffer
will increase slightly but not as much as when the same amount of base
is added to water. The resulting pH is dependent on how much acid or
base is added.
(p. 511) Lactic acid produced in the body can react with HCO3⫺ by the following reaction.
HC3H5O3(aq) + HCO3 - (aq) uv C3H5O3 - (aq) + H2CO3(aq)
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(p. 520) As NaOH is slowly added to a solution of HCl, the pH will gradually increase until just enough NaOH has been added to completely react
with the HCl. At this point the pH will increase rapidly for several pH
units. As more NaOH is added, the pH will resume a more gradual
increase. A plot of pH versus volume of NaOH will yield a typical titration curve.
Chapter 12
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(p. 542) Oxidation is the loss of electrons, whereas reduction is the gain of
electrons. One process cannot occur without the other.
(p. 544) Fe = +3, Cl = -1; C = -4, H = +1; Mn = +7, O = - 2;
C = + 2, H = + 1, O = - 2, N = - 3
(p. 550) The anode is negative, and the cathode is positive.
(p. 551) The number of atoms is balanced, but the charge is not. There is a
charge of ⫹2 on the reactant side and a charge of ⫹3 on the product side
of the chemical equation. The equation is not balanced.
D-13
D-14
APPENDIX D
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(p. 553) If the reaction is reversed, the sign of the cell potential is changed.
(p. 554) Sn(s) is oxidized and nitrogen is reduced. Sn is the reducing agent,
and the HNO3 is oxidizing agent.
(p. 567) The standard potential corresponds to 1 M concentrations of both
Cu2⫹ and Ni2⫹. A concentration of 1.5 M Cu2⫹ on the reactant side and a
concentration of 0.010 M Ni2⫹ on the product side will result in a greater
tendency for the reaction to proceed toward the products. This will result in
a greater (more positive) value for the potential (E) under these conditions
than at standard conditions.
(p. 587) The student received no credit because the reaction was balanced
in base rather than in acid.
Chapter 13
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(p. 596) Dissolving salt in water is a spontaneous process. Even though heat
is absorbed during the process, the salt dissolves because the system (water
and salt) becomes more disordered as the salt dissolves.
(p. 600) It is possible for a process to proceed if the entropy of the system
decreases as long as the entropy of the surroundings increases even more,
causing the entropy of the universe to increase.
(p. 601) At absolute zero, motion at the molecular and atomic level
approaches zero. A substance would have minimum entropy when there is
no motion.
(p. 603) Like cyclopentane being converted to pentene, glucose would have
greater freedom of motion (more disorder) in its open-chain form. The sign
of ¢ S would be positive.
(p. 604) Both molecules consist of the same kind and number of atoms.
Their reactions of atom combination would involve the same gaseous atom
reactants. Because the entropy of atom combination of isobutane is more
negative than the entropy of atom combination of butane, the formation of
isobutane results in the loss of more freedom of motion.
(p. 609) If ¢ H for a spontaneous process, such as dissolving table salt in
water, is positive, the entropy of the process must be positive enough to
make the change in Gibbs Free energy negative. The increase in disorder
must be enough to make up for the endothermic process.
(p. 611) No, the hardening of cement has a negative ¢ S. The cement
becomes more ordered. In order for this to be a spontaneous process, it must
therefore be exothermic (releasing heat).
(p. 621) If Qp 7 Kp the reaction must proceed toward the reactants. In order
for Qp to decrease to Kp, the numerator (product partial pressures) of the
expression must decrease and the denominator (reactant partial pressures)
must increase.
(p. 629) If ¢ H is 7 0 (endothermic), the reactants are favored. If ¢ S is
7 0, the products are favored. A change in temperature will change the
equilibrium constant. In the relationship between K, ¢ H, and ¢ S, it is
the enthalpy term that is dependent on temperature. As temperature
decreases, the term containing enthalpy becomes more negative and
therefore K becomes smaller. Thus the equilibrium is shifted toward the
reactants.
If ¢ H is 7 0 (endothermic), the reactants are favored. If ¢ S is 6 0, the
reactants are favored. These circumstances will have the same result for a
D-15
CHECKPOINT ANSWERS
decrease in temperature as in the previous question because ¢ H is still 7 0. Therefore, a decrease in temperature will shift the equilibrium toward the reactants.
Chapter 14
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(p. 641) The kinetics of the reaction is so slow that during a human lifetime there will be no observable change of diamond to graphite.
(p. 645) If a large value for ¢ t is used, there will be a significant change
in rate during the time interval. The resulting rate would then be more of
an average rate during the time period rather than an instantaneous rate at
some moment in time.
(p. 645) As the reaction progresses, the concentration of phenolphthalein
decreases as it reacts. The concentration term in the rate law decreases, and
therefore the rate of the reaction decreases. As the reaction proceeds, phenolphthalein is used up, and there is a decrease in the probability of a collision between phenolphthalein molecules and base molecules. Thus the
reaction rate decreases.
(p. 646) The rate describes how quickly the reactants are used up and products are produced with respect to time. The rate constant is a proportionality constant used in the rate law. The rate of most reactions decreases with
respect to time and concentration of reactants. However, the rate constant
remains constant with respect to time and concentration.
(p. 646) The units of k will be 1/(M # sec) ⫽ L/(sec # mol).
(p. 647) The rate of disappearance of HI is twice as fast as the rate of formation of H2.
rateHI ⫽ 2(ratehydrogen)
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(p. 655) The rate of a zero-order reaction is independent of the concentration of the reactants. The rate of reaction of a first-order reaction is
dependent on concentration. As a reaction proceeds, the rate of reaction
of a zero-order reaction will remain constant, while a first-order reaction
will slow down.
(p. 657) In trials 2 and 3 the concentrations of both NH4+ and NO2- change.
Therefore, there will be two unknowns (m and n) in the equation. One of
the concentrations must remain constant between two trials so that its effect
will cancel out of the equation. Trials 1 and 2 cannot be used to find the
order with respect to NH4+ because the concentration of NH4+ does not
change in these trials. However, the concentration of NO2- does change. Any
change in the rate will therefore be due to NO2- not NH4+ .
(p. 657) For trial 1 the following are known: (NH4+ ) ⫽ 5.00 * 10⫺2, (NO2- )
⫽ 2.00 * 10⫺2, m ⫽ 1, n ⫽ 1, rate ⫽ 2.70 * 10⫺7. Solving the rate law
equation for the rate constant, k, we get 2.70 * 10⫺4 M⫺1sec⫺1.
(p. 662) The rate constant for the decay of 15O is less than the rate constant
for 19O. Therefore, the rate of decay of 15O will be slower, and it will have
a longer half-life.
(p. 662) For a zero-order reaction the half-life for the second reaction will
be twice as large as that for the first trial. For a first-order reaction the halflife will not change. The half-life of a first-order reaction is independent of
concentration. For a second-order reaction, doubling the concentration will
decrease the half-life by a factor of two.
D-16
APPENDIX D
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(p. 669) No, this is not true for an endothermic reaction.
Ea reverse
Energy
Ea forward
Products
ΔH
Reactants
Reaction coordinate
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(p. 671) ¢ H is the difference between the energy of the products and the
energy of reactants in both diagrams. ¢ H will be the same in both diagrams
because the enthalpies of products and reactants have not changed.
(p. 672) The Ea term is a negative in the Arrhenius equation. Therefore, if
Ea is large, k will be small. A small k indicates a slow reaction rate. Because
a catalyst decreases Ea, the value of k will increase, leading to a faster reaction rate.
(p. 673) Because the external pressure is less at the top of a mountain, the
water will boil at a lower temperature than at sea level. Therefore, the egg
will not be as hot in the boiling water. Cooking the egg to make it hardboiled involves a chemical reaction. According to the Arrhenius equation,
the rate constant for this reaction will be smaller at the lower temperature
of the boiling water at the top of the mountain.
Chapter 15
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(p. 693) Both mass number and charge must be conserved.
(p. 695) 35S contains 19 neutrons and 16 protons.
Chapter 16
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(p. 729) In the first structure individual methyl groups (CH3) are attached
to the second and third carbons. In the second structure both methyl groups
are attached to the second carbon. Because the methyl groups are bonded
to different atoms in the two structures, the structures must be isomers.
In the second set of structures the methyl groups are bonded to the same
carbons (carbon 2 and 3). The only difference is the relative position of the
two methyl groups. Because there is free rotation around carbon–carbon single bonds, the two structures represent the same compound.