Chapter 1
page s (a) 100 (b) atoms page JO Water is composed of two
types of atoms: hydrogen and oxygen . Hydrogen is co mposed
only of hydrogen atoms, and oxygen is composed only of oxygen
ato ms. Therefore, hydrogen and oxyge n are elements and water is
a compound . page 12 Density is an intensive property. Because
it is m easured per unit of volume, it is independent of how much
of the material is present. page 13 (a) Chemical change: Carbon
dioxide and water are different compound s than sugar. (b ) Physical
change: Water in the gas phase becomes water in the solid phase
(frost). (c) Physical change: Gold in the solid state becomes liquid
and then resolidifies . page 16 Doubling its speed will increase th e
kinetic energy by a factor of four, while doubling its mass will only
double the kinetic energy. page 17 The chemical potential energy
of the battery decreases as it is discharged . page 18 The candela,
cd. page 19103 page 22 2.5 X 102 m3 is, because it has units of
length to the third power. page 26 (b) Mass of a penny page 27
The mass should be reported to five significant figures, two figures
to the left of the decimal point and three to the right. page 29 The
mass of the water can be reported as 5.5 g with a total of two signilicant figures. The number of significant figures to the right of the
decimal point is limited by the precision to which the mass of the
empty beaker is known.
page78
H
H
H
H
I I I I
H—
C—
C—
C—
c—
H
I I I I
H
H
H
H
H
H
I
H
I
H
—c —c —CI—H
I
H —C —H
Chapter 3
page 92 Each Mg( OH )z has 1 Mg, 2 0, and 2 H; thu s, 3 Mg( OH )2
represents 3 Mg, 6 0, and 6 H. page 97 The product is an ionic
co mpound involving Na+ and
and its chemical formula is
therefore Na2S. page 103 (a) A mole of glucose. By inspecting their
chemical formulas we find that glucose has more atoms of H and
0 than water and in addition it also has C atoms . Thus, a molecule
of glucose has a greater mass than a molecule of water. (b) They
both contain the same number of molecules because a mole of each
substance contains 6.02 X 1023 molecules. page 108 No, chemical analysis cannot distinguish co mpounds that have different
molecular formulas but the sa me empirical formula. page 112
3.14 mol because 2 mol I-12 a; 1 mol
based on the coefficients
in the balanced equation page 113 To obey the law of con servation of mass the mass of the products consumed must equal the
mass of reactants formed, therefore the mass of I-120 produced is
1.00 g + 3.59 g - 3.03 g = 1.56 g I-120
Chapter 2
Chapter 4
page 49 (a) The law of multiple proportions. (b ) The second compound must contain two oxygen atoms fo r each carbon atom (that
is, twice as many carbon atoms as the first compound). page 51
(b) The particles that make up cathode rays must be present in all
metals. page 53 As the thickness of the gold foil increases there is a
higher probability that the alpha particle will collide with a nucleus
as it passes through the foil, increasing the number of alpha particles
sca ttered at large angles. page 54 (a) The atom has 15 electrons
because atoms have equal numbers of electrons and protons . (b)
The protons reside in the nucleus of the atom. page 57The mass
of an electron is 5.5 x
amu, so you wou ld need to express
the mass of an oxygen atom to four places past the decim al, givin g
a total of six significant figures, to detect a change in mass upon
losing an electron. page 58 The atomic weight of boron, 10.811
amu, is closer to 11B which mean s 11B is the more abundant isotope. page62 (a) Cl, (b) third period and group 7A, (c) 17,
(d) nonmetal page 63 N204 and C12H22011 can be only molecular
formulas; their empirical formu las would be N02 and Cl-IzO.
NH3 and CO2 could be empirical formulas or molecular formulas.
No fo rmula can be only an empirical formu la; there would always be
a moleculeof thatcomposition. page 65 (a) C2H6, (b) CH3, (c) No,
you cannot determine bond an gles or relative bond distances from
a stru ctural formula. page 69 W03 page 71 MnO page 72
The endings convey the number of oxygen atoms bound to the
nonmetallic element page 75 (a) Ca (HC03)z, (b) KI-1S04, LiH2P04
page 75 Bromic acid, by analogy to the re lationship between the
chlorate ion and chloric acid page 77 No, it contains three
different elements.
page 135 (a) K+(aq) and CW(aq) (b) Na+(aq) and
(aq)
page 137 NaOH because it is the on ly solute that is a strong electrolyte page 141 Na+( aq) and N03-( aq) page 143Three. Each
COOH group will partially ionize in water to form H+( aq).
page 144 Only soluble metal hydroxides are classified as strong
bases and Al (OH )3 is insoluble. page 149 S02 (g) page 151
(a) -3, (b) +5 page 155 (a) Yes, nickel is below zinc in the activity
series so Ni2+(aq) will oxidize Zns to form Nis and Zn2+(aq).
(b) No reaction will occur beca use the Zn2+(aq) ions cannot be further oxidized . page 160 The concentration is halved to 0.25 M.
Chapter 5
page 183 Open system. Humans exchange matter and energy with
their surroundings. page 186 Endothermic page 188 Your
weight. page 190 No. If t:,. V is zero, then the expression w = -Pt:,. V
is also zero. page 191 t:,.H is positive; the fact that the flask (part
of the surroundings) gets cold means that the system is absorbing
heat, meaning that qp is positive. (See Figure 5.8.) Because the process occurs at constant pressure, qp = t:,.H. page 193 No. Because
only half as much matter is involved, the value of t:,.H would be
½(- 483.6 kJ) = - 241.8 kJ. page 197 Hg(T). RearrangingEquation
q
. When q and m are constant for a series of
5.22 gives t:,.T =
C, X m
constant
substances, then t:,. T = ~ - Therefore, the element with the
cS
smallest C, inTable 5.2 has the largest t:,.T, Hg/. page201 (a) The
sign of t:,.H changes. (b) The magnitude of t:,.H doubles . page205
No. Because 0 3(g) is not the most stable form of oxygen at 25 °C, 1
atm [ 02 (g) is ], t:,.Hr for 0 3(g) is not necessarily zero . In Appendix C
1203
1204
we see that it is 142.3 kJ / mo!. page208To trans form a Cl2 molecule
into two separate Cl atoms the Cl - Cl bond mustbe broken which
requires an input of energy. Therefore, the Cl2 molecule will have a
lower energy and be more stable. The difference in energy between
the two is the Cl Cl bond dissociation energy, 242 kJ/mol.
page 212 Reacting 1 mol of glucose releases 2803 kj of energy. The
molar mass ofglucose is 180.6 g, so the amountof energy released
when 1 gis reacted is 2803 kJ/18 .6 g = 15.56 kJ/g. This value is close
to the average value for carbohydrates (17 kJ/g) as expected.
page 213 Fats, because they have the largest fuel value of the three
—
Chapter 6
page 235 No. Both visible light and X-rays are forms of electromagnetic radiation . They therefore both travel at the speed of light, c.
Their differing ability to penetrate skin is due to their different
energies, which we will discu ss in the next section. page 237 The
notes on a piano go in "jumps"; for example, one ca n't play a note
between B and C on a piano. In this analogy, a violin is co ntinuous in principle, one can play any note (such as a note halfway
between B and C). page238 (a) Electrons will be ejected. Ifphotons of red light have sufficient energy to eject electrons, then photons of green light, which have a higher frequ ency and hence energy
than red light, will also be able to eject electrons. page 239 Wavelike behavior. page241 The energies in the Bohr model have only
certain specific allowed values, much like the positions on the steps
in Figure 6.6. page242As the potential energy of the electron
decreases (becomes more negative) it's attraction to the nucleus
increases and the radius ofits orbit decreases . page243 LlE is
negative for an emission . The negative sign yields a positive number
that corresponds to the energy of the emitted p hoton . page 243
True. The energy of the emitted photon increases as the initial value
of n increases and the fin al value of n decreases. In the limit n r = 1
and n; goes to infinity, which gives a photon energy, Ephoton = hcRH'
page 246 Yes, all moving objects produce matter waves, but the
wavelengths associated with macroscopic objects, such as the
baseball, are too small to allow for any way of observing them.
page 247 Less important. page 249 In the first statement, we
know exactly where the electron is. In the second statement, we
are saying that we know the probability of the electron being at a
point, but we don't know exactly where it is. The second statement
is consistent with the Uncertainty Principle. page 250 In the Bohr
model electrons move in orbits, each of which has a specific fixed
distance from the nucleus. In the modern picture of the atom electrans are found in orbitals and cannot be said to orbit the nucleus
at a specific distance. page 255 The two orbitals have the same
principal quantum number, n = 3, and th e same orbital angular
momentum quantum number, I = 1, but different values of the
magnetic quantum number. page 257 In this case we cannot
unambiguously determine the relative energies of the two orbitals
because the 4s orbital has a higher value of n, while the 3d orbital
has a higher value of I, and the energies of electron s in multielectron
atoms depend upon both quantum numbers. page262The 6s
orbital, which starts to hold electrons at element 55, Cs. page 267
We ca n't co nclude anything! Eac h of the three elements has a different valence electron configuration for its (n - l )d and ns subshells:
For Ni,
for Pd, 4d10; and for Pt, 5d96s1.
—
Chapter 7
page 283 Co and Ni as well as Te and I are other pairs of elements whose atomic weights are out of order compared to their
atomic numbers. page 286 The Zp electron in a Ne atom would
experience a larger Zerr than the 3s electron in Na, due to the greater
shielding of the 3s electron of the Na atom by all the 2s and 2p
electrons. page 288 These trends work against eac h other: Zeff
increasing would imply that the valence electrons are pulled tighter
in to m ake the atom smaller, while orbital size "increasing" wou ld
imply that atomic size wou ld also increase. The orbital size effect
is larger: As you go down a column in the periodic table, atomic
size generally increases. page 293 It is harder to remove another
electron from Na+, so the process in Equation 7.3 would require
more energy and, hence, shorter-wavelength light (see Sections 6.1
and 6.2). page293/z for a carbon atom corresponds to ionizing
an electron from c +, which has the same number of electrons as a
neutral B atom. Zerr will be greater for c + than for B, so 12 for a carbon atom will be greater than 11 fo r a boron atom. page 296 The
same, [Ar)3d 3. page 298 The magnitudes are the same, but they
have opposite signs. page299 No. The oxidation state ofAs will
be positive when combined with Cl, and negative when combined
with Mg. page 302 Because the melting point is so low, we would
expect a molecular rather than ionic compound. Thus, A is more
likely to be P than Sc because PC13 is a compound of two nonmetals and is th erefore more likely to be molecular. page 305 Its low
first ionization energy. page 308 In the acidic environment of the
stomach, metal carbonates can react to give carbonic acid, which
decomposes to water and carbon dioxide gas. Thus, calcium carbonate is much more soluble in acidic solution than it is in neutral
water. page 310 (b) The fact that bromine is a liquid and chlorine is a gas. page 312 Based on the trends in the table, we might
expect the radius to be about 1.5 A, and the first ionization energy to
be about 900 kJ / mo!. In fact, its bonding radius is indeed 1.5 A, and
the experim ental ionization energy is 920 kJ/mo!.
Chapter 8
page 328 No. Cl has seven va lence electrons. The first and second
Lewis symbols are both correct they both show seven valence
electrons, and it doesn't matter which of the four sides has the
single electron. The third symbol shows only five electrons and is
incorrect . page 330 No . This reaction corresponds to the lattice
energy for KCl, which is a large positive number. Therefore, the
reaction will cost energy, not release energy. page 334 Rhodium,
Rh page 335 Weaker. In both H2 and Ht the two H atoms are
principally h eld together by the electrostatic attractions between
the nuclei and the electrons concentrated between them . Ht has
only one electron between the nuclei whereas H2 has two and this
results in the H H bond in l-I2 being stron ger. pagt:,.&:J.7 Triple
bond . CO2 has two C 0 double bonds. Because the C 0 bond •·, :
in carbon monoxide is shorter, it is likely to be a triple bond.
page 337 Electron affinity measures the energy released when
an isolated atom gains an electron to form a 1- ion and has units
of energy. Electronegativity has no units, and is the abili ty of the
atom in a molecule to attract electrons to itself within that molecule. page 340 IF; beca use th e difference in electronega tivity
between I and F is greater than that between Cl and F, the magnitude of Q should be greater for IF. In addition, because I has a larger
atomic radius than Cl, the bond length in IF is longer than th at
in ClF. Thu s, both Q and rare larger for IF and , therefore, µ, = Qr
will be larger for IF. page 342 Smaller dipo le moment for C H.
The magnitude of Q should be similar for C H and H
bonds
because the difference in electronegativity for each bond is 0.4.
The
bond length is 1.1 A and the H I bond length is 1.6 A.
Therefo re µ, = Qr will be greater for H ] because it has a lon ger
bond (larger r) . page 343 Os04. The data suggest that the yellow
substance is a molecular species with its low melting and boiling
—
—
—
—
—
—
—
` ``).
1205
points. Os in Os04 has an oxidation number of +8 and Cr in Cr203
has an oxidation number of +3. In Section 8.4, we learn that a compound with a metal in a high oxidation state should show a high
degree of covalence and Os04 fits thi s situation. page 346 There
is probably a better choice of Lewis structure than the one chosen.
Because the formal charges must add up to O and the formal charge
on the F atom is + 1, there must be another atom that has a formal
charge of -1. Because F is the most electronegative element, we
don't expect it to carry a positive forma l charge. page 349 The
ozo n e 00 bond length sh ould be longer than the 00 bond in oxygen. page 349 Weaker. page 350 No, it will not have multiple
resonance structures . We can't "move" the double bonds, as we did
in benzene, because the positions of the hydrogen atoms dictate
specific positions for the double bonds. We can't write any other
reasonable Lewis structures for the molecule. page 351 The forma! charge of each atom is shown here:
F.C.
N=O
N=O
0
-]
0
+1
The first structure shows each atom with a zero formal charge and
therefore it is the dominant Lewis structure. The second one shows a
positive formal charge for an oxygen atom, which is a highly electronegative atom, and this is not a favorab le situation.
Chapter 9
page 372 Removal of two opposi te atoms from an octahedral
arrangement would lead to squ are-planar shape. page 373 No, the
molecule does not satisfy the octet rule because there are ten electrons around the atom A. Each of th e atoms B does satisfy the octet
rule. There are four electron domains around A: Two single bonds,
one double bond, and one nonbonding electron domain around
A. page 377Yes. Because there are three equivalent dominant resonance structures, each ofwhich puts the double bond between the
N and a different 0, the average structure has the same bond order
for all three N 0 bonds. Thus, each electron domain is the same
and the angles are predicted to be 120°. page 379ln a square planar arrangement of electron domains, each domain is 90° from two
other domains (and 180° from the third domain). In a tetrahedral
arrangement, each domain is 109.5° from the other three domains.
Domains always try to minimize the number of 90° interactions, so
the tetrahedral arrangement is favored. page 383 No. Although the
bond dipoles are in opposi te directions, they will not have the same
magnitude because the C S bond is not as polar as the C O bond.
Thus, the sum of the two vectors will not equal zero and the mo!ecule is polar. page 387They are both oriented perpendicular to
the F Be F axis. page 388 None. All of the 2p orbitals are used
in constructing the sp3 hybrid orbitals. page 394 There are three
electron domains about each N atom, so we expect sp2 hybridization
at each ofthe N atoms. The 1--I N-N angles should therefore be
roughly 120°, and the molecule is not expected to be linear. In order
for the .,,. bond to form, all four atoms would have to be in the same
plane. page 398 sp hybridization. page 401 The excited molecu le would fa ll apart. It would have an electron configuration crl,crf,1
and therefore a bond order of 0. page 403 Yes, it will have a bond
order of½- page409No. Ifthe crzv MO were lower in energy than
the .,,.Zp MOs we would expect the last two electrons to go into the .,,.2p
MOs with the same spin, which would cause C2 to be paramagnetic.
—
—
—
——
—
Chapter 10
page 431 No . The heaviest gas in Table 10.1 is S02, whose molar
mass, 64 g/mol, is less than half that of Xe, whose molar mass is
131 g/mol. page432 (a) 6.2 X 10'' N, (b) 1.5 X 103 N page432
(a) 745 mm Hg, (b) 0.980 atm, (c) 99.3 kPa, (d) 0.993 bar page 435
page 436 No because the absolute temperature
is not halved, it only decreases from 373 K to 323 K. page 438
28 .2 cm page 443 Less dense because water has a smaller molar
mass, 18 g/mol, than N2, 28 g/mol. page 445 The partial pressure
of N2 is not affected by the introduction of another gas, but the
total pressure will increase. page 449 Slowest HCl < 02 < 1-12
fastest. page 451 u,111,/ U111p =
This ratio will not change
as the temperature changes and it will be the same for all gases.
page 454 (a) Decrease, (b) have no effect. page 455 (b) 100 K and
5 atm page 456 The negative deviation is due to attractive intermolecular forces .
It wou ld be halved.
Chapter 11
page 474 I-I20 (g) ; during boiling, energy is provided to overcome
intermolecular forces between I-120 molecules allowing the vapor
to form. page 476 CH4 < CCl4 < CBr4. Because all three molecules are nonpolar, the strength of dispersion forces determines th e
relative boiling points. Polarizability increases in order of increasing molecular size and molecular weight, CJ-14 < CCl4 < CBr4;
hence, the dispersion forces and boiling points increase in the same
order. page 480 Mainly hydrogen bonds, which hold the individual I-120 molecules together in the liquid . page 481 Ca (N03)z
in water, because calcium nitrate is a strong electrolyte that forms
ions and water is a polar molecule with a dipole moment. Ion-dipole
forces cannot be present in a CI-1301-I/I-120 mixture because CH30I-I
does not form ions. page 484 Decrease. page 487Melting (or
fusion), endothermic page 489The intermolecular attractive
forces in I-120 are much stronger than those in I-I2S because 1-120 can
form hydrogen bonds. The stronger intermolecular forces results in
a higher critical temperature a nd pressure. page 491 CCl4. Both
compounds are nonpolar; therefo re, only dispersion forces exist
between the molecules. Because dispersion forces are stronger for
the larger, heavier CBr4, it has a lower vapor pressure than CC14. The
substance with the larger vapor pressure at a given temperature is
more volatile.
Chapter 12
page514 Tetragonal. There are two three-dimensional lattices
th at have a square base with a third vector perpendicular to the
base, tetragonal and cubic, but in a cubic lattice the a, b, and c lattice vectors are all of the same length. page 517 Ionic solids are
composed of ions. Oppositely-charged ions slipping past each
other could create electrostatic repu lsions, so ionic solids are britt ie. page 522 The packing efficiency decreases as the number of
nearest neighbors decreases. The structures with the highest packing
efficiency, hexagonal and cubic close packing, both have atoms with
a coordination number of 12. Body-centered cubic packing, where
the coordination number is 8, has a lower packing efficiency, and
primitive cubic packing, where the coordination number is 6, has a
lower packing efficiency sti ll. page 523 Interstitial, because boron
is a small nonmetal atom that can fit in the voids between the larger
palladium atoms. page 529 Gold, Au should have more electrons
in antibonding orbitals. Tu ngsten, W, lies near the middle of the
transition metal series where the bands arising from the d orbitals
and the s orbital are approximately half-filled. This electron count
shou ld [ill the bonding orbitals and leave the antibonding orbitals mostly empty. Because both elements have similar numbers of
electrons in the bonding orbitals but tungsten has fewer electrons in
antibonding orbitals, it will have a higher melting point. page 530
No. In a crystal the lattice points must be identical. Therefore, if
an atom lies on top of a lattice point, then the same type of atom
1206
must lie on alllattice points. In an ionic compound there are at
least two different types of atoms, and only one can lie on the lattice points. page 532 Four. The empirical formula of potassium
oxide is K20. Rearranging Equation 12.1 we can determine the
potassium coordination number to be anion coordination number x (number of anions per formula unit/number of cations per
formula unit) = 8(1/2) = 4. page543A condensation polymer.
The presence of both
COOH and
NH2 groups allow molecules
to react with one another forming C N bonds and eliminating
H20. page 545 As the vinyl acetate content increases more side
chain branching occurs which inhibits the formation of crystalline
regions thereby lowering the melting point. page 548 No. The
emitted photons have energies that are similar in energy to the band
gap of the semiconductor. If the size of the crystals is reduced into
the nanometer range, the band gap will increase. However, because
340-nm light falls in the UV region of the electromagnetic spectrum,
increasing the energy of the band gap will only shift the light deeper
into the UV.
—
—
—
Chapter 13
page S69 Entropy increases as the ink molecules disperse into
water. page S70 The lattice energy of NaCl(s) must be overcome
to separate Na+ and c1- ions and disperse them into a solvent.
C6H14 is nonpolar. Interactions between ions and nonpolar molecules tend to be very weak. Thus, the energy required to separate
the ions in NaCl is not recovered in the form of ion-C6H14 interactions . page S71 (a) Separating solvent molecules from each other
requires energy and is therefore endothermic. (b) Forming the solute-solvent interactions is exothermic. page S73 The added solute
provides a template for the solid to begin to crystallize from solution, and a precipitate will form. page S76 The solubility in
water would be considerably lower because there would no longer
be hydrogen bonding with water, which promotes solubility.
page S81 230 ppm (1 ppm is 1 part in106); 2.30 X 105 ppb (1 ppb
is 1 part in 109) . page S82 For dilute aqueous solutions the molality will be nearly equal to the molarity. Molality is the number of
moles of solute per kilogram of solvent, whereas molarity is the
number moles of solute per liter of solution. Because the solution
is dilute, the mass of solvent is essentially equal to the mass of the
solution. Furthermore, a dilute aqueous solution will have a density
of 1.0 kg/L. Thus, the number of liters of solution and the number of
kilograms of solvent will be essentially equal. page S8S A greater
extent, lowering of the vapor pressure depends on the total solute
concentration (Equation 13.11) . One mole of NaCl (a strong electrolyte) provides 2 mo! of particles (1 mol of Na+ and 1 mo! Cl) , whereas
one mole of (a nonelectrolyte) provides only 1 mol particles .
page S89 Not necessarily; if the solute is a strong or weak electrolyte, it could have a lower molality and still cause an increase of
0.51 °C. The total molality of all the particles in the solution is 1
m. page S91 The 0.20-m solution is hypotonic with respect to the
0.5-m solution. (A hypotonic solution will have a lower concentration and hence a lower osmotic pressure.) page S92 They would
have the same osmotic pressure because they have the same concentration of particles. (Both are strong electrolytes that are 0.20 M in
total ions.) page S96 No, hydrophobic groups would face outward
to make contact with hydrophobic lipids.
Chapter 14
page 614 Increasing the partial pressure increases the number of
collisions between molecules. For any reaction that depends on
collisions (which is nearly all of them), we would expect the rate
to increase with increasing partial pressure . page 618 You can
see visually that the slope of the line connecting O s and 600 s is
smaller than the slope at O s and larger than the slope at 600 s.
The order from fastest to slowest is therefore (ii) > (i) > (iii) .
page 621 No, the reaction rate usually depends on concentration but the rate constant does not. page 621 Generally no . Rate
always has units of M/s. The units of the rate constant depend on
the specific rate law, as we shall see throughout this chapter.
page 622 The reaction rate will double because the rate law is
first order in [I-12] . page 630 After 3 half-lives, the concentration will be 1 /8 of its original value, so 1.25 g of the substance
remains. page 631 Yes, using t = t1 ,2 and [A], =
the halflife
of a zero-order reaction is t112 =
page 634 According to
the figure, the energy barrier is lower in the forward direction than
in the reverse. Thus, more molecules will have energy sufficient to
cross the barrier in the forward direction. The forward rate will be
greater page 635 No. If B can be isolated, it can't correspond to
the top of the energy barrier. There would be transition states
for each of the individual reactions above page 638 Bimolecular. page 643 The likelihood of three molecules colliding at exactly
the same time is vanishingly small. page 648 A heterogeneous catalyst is in a different phase than the reactants and is therefore fairly
easy to remove from the mixture. The removal of a homogeneous
catalyst can be much more difficult as it exists in the same phase as
the reactants. page 650 True. Enzymes act as catalysts and therefore increase the rate of the reaction .
Chapter 15
page 674 (b) the forward and reverse reaction rates page 675
True page 678 It does not depend on starting concentrations.
page 679 Yes, Kc = Kp when the number of moles of gaseous
products and the number of moles of gaseous reactants are equal.
page 680 0.00140 page 681 Kc = 91 page 682 It is cubed.
page684Kp =
page686Kc =
page 694 (a) to the right, (b) to the left page 694 (bottom) It
will shift to the left, the side with a larger number of moles of gas.
page 697 Endothermic. Raising the temperature of an endothermic
reaction shifts the equilibrium to the right, increasing the vapor
pressure of the gaseous product. page 700 It will stay the same.
The presence of a catalyst can accelerate the reaction but does not
alter the value of K, which is what limits the amount of product
produced.
Chapter 16
page 716 The H+ ion for acids and the OH- ion for bases.
page 718 CH3NH2 is the base because it accepts a H+ from H2S as
the reaction moves from the left-hand to the right-hand side of the
equation. page 721 As the conjugate base ofa strong acid, we
would classify c104- as having negligible basicity. page 725 pH
is defined as -log[H+]. This quantity will become negative ifthe
H+ concentration exceeds 1 M, which is possible. Such a solution
would be highly acidic. page 726 pH = 14.00 - 3.00 = 11.00.
This solution is basic because pH > 7.0. page 729 Both NaOH
and Ba (OH)z are soluble hydroxides. Therefore, the hydroxide concentrations will be 0.001 MforNaOH and 0.002Mfor
Ba (OH)z. Because the Ba (OH)z solution has a higher [OH- ], it is
more basic and has a higher pH. page 730 Because CH3- is the
conjugate base of a substance that has negligible acidity, CH3must be a strong base. Bases stronger than OH- abstract H+ from
water molecules: CH3- + H20
CH4 + OH- . page 731
Oxygen. page 735pH < 7. We must consider the [H+] that is
1207
due to the autoionization ofwater. The additional [H+) from the
very dilute acid solution will make the solution acidic. page 738
I--IPo/- (aq)
I--I +( aq) + Poi- (aq) page745 4. page746
Nitrate is the conjugate base of nitric acid , HN03. The conjugate base
o f a strong acid does not act as a base, so N03- ions will not affect the
pH. Carbonate is the conjugate base of hydrogen carbonate, I--IC03- ,
which is a weak acid. The conjugate base ofa weak acid acts as a weak
base, so co}- ions will increase the pH. page 752 I--IBr03.
For an oxyacid, acidity increases as the electronegativity of the
central ion increases, which would make
more acidic than
I--1102. Acidity also increases as the number of oxygens bound to
the central atom increases, which wou ld make I--IBr03 more acidic
than HBr02. Combining these two relationships we can order
these acids in terms of increasing acid-dissociation constant,
I--1102 < I--IBr02 < I--IBr03. page 754 Basic. page 755 It must have
an empty orbital that can interact with the lone pair on a Lewis base
Chapter 17
page 774 (top) The c 1- ion is the on ly spectator ion. The pH is determined by the equilibrium
NH3(aq) + H20(/)
Ol-I- (aq) + NH/ (aq).
page 775 (bottom) HN03 and N03- . To form a buffer we need
comparable concentrations ofa weak acid and its conjugate base.
HN03 and N03- will not form a buffer because HN03 is a strong acid
and the N03- ion is merely a spectator ion. page 776 (a) The Ofr
of NaOH (a strong base) reacts with the acid member of the buffer
(CH3COOH), abstracting a proton. Thus, [CH3COOH] decreases
and [CH3Coo-] increases. (b) The H+ of I-IC! (a strong acid) reacts
with the base member of the buffer [ CI-13COO- ]. Thus, [
decreases and [CI-13COOI-I] increases. page 780I-ICIO would be
more suitable for a pH = 7.0 buffer solution. To make a buffer we
would also need a salt containing Clo- , such as NaClO. page 786
The pH = 7 . The neutralization o f a strong base with a strong acid
gives a salt solution at the equivalence point. The salt contains ions
that do not change the pH of water. page 790 The conjugate base
of the weak acid is the majority species in solution at the equivalence
point, and this conjugate base reacts with water in a Kb reaction
to produce OH- . Therefore, the pH at the equivalence point for a
weak acid/strong base titration is greater than 7.00. page 794 The
14
points, A and B . The first point, A, is reached at a pH of about 9:
Na2C03 (aq) + HCl(aq)
NaCl(aq) + NaHC03( aq)
HC03- is weakly basic in water and is a weaker base than the carbonate ion. The second point, B, is reached at a pH of about 4:
NaHC03(aq) + HCl(aq)
—•
NaCl(aq) + C02(g) + H20(1)
H2C03, a weak acid, forms and decomposes to carbon dioxide and water. page 79S AgCI. Because all three compounds produce the same
number of ions, their relative stabilities correspond directly to the K,p
values, with the compound with the largest K,p value being the most
soluble. page 810 Silver, because AgCl is re latively insoluble.
Chapter 18
page 825 We would expect that helium is relatively more abundant
at the higher elevation, because Earth's gravitational field would
exert a greater downward force on the heavier argon atoms. As a
result the helium to argon ratio would be higher in the mesosphere
than in the troposphere. page 829 Photoionization causes an
electron to be ejected from an atom or molecule leading to the formation ofa cation. It does not genera lly result in the breaking ofa
bond. page 830 From their Lewis structures we predict bond orders
of 2 and 1.5 for 0 2and 0 3, respectively. Bond strength increases with
increasing bond order so we would expect stronger 0
O bonds in
0 2. page 832 Yes; Cl is neither a product nor a reactant in the overall reaction, and its presence does speed the reaction up. page 834
S03 page 836 N02 photodissociates to NO and O; the O atoms
react with 02 in the atmosphere to form ozone, which is a key ingredient in photochemical smog. page 839 Higher humidity means
there is more water in the air. Water absorbs infrared light, which we
feel as heat. After sundown, the ground that has been warmed ea rlier
in the day reradiates heat out. ln locations with higher humidity, this
energy is absorbed somewhat by the water and in turn is reradiated
to some extent back to the Earth, resu lting in warmer temperatures
compared to a low-humidity location. page 840 You can see from
the phase diagram that to pass from the solid to the gaseous state we
need to be below water's critical point. Therefore, to sublime water
we need to be below 0.006 atm. A wide range of temperatures will
work for sublimation at this pressure-the most environmentally
relevant ones are -50 °C to O0C. page 844 Nitrogen and phosphorous . page 850 With a catalyst, the reaction is always faster,
therefore costing less energy to run. In addition, with a catalyst the
reaction may occur readily at a lower temperature, also costing less
energy. page 852 Because the by-products of the reaction can be
used to synthesize additiona l reactants in principle no atoms are
wasted. page 852 sp before reaction; sp2 after reaction.
—
Chapter 19
25
50
Volume HCl (mL)
above titration curve shows the titration of 25 mL of Na2C03 with
I-IC!, both with 0.1 Mconcentrations. The overall reaction between
thetwo is
Na2C03( aq) + I-ICl (aq)
2NaCl(aq) + C02(g) + I-120(/)
The initial (pI-I sodium carbonate in water only) is near 11 because
co} - is a weak base in water. The graph shows two equivalence
page 868 No. Just because the system is restored to its original
condition doesn't mean that the surroundings have likewise been
restored to their original condition, so it is not necessarily reversible. page 870 n.S depends not merely on q but on q,ev· Although
there are many possible paths that could take a system from its
initial to final state, there is always only one reversible isothermal
path between two states . Thus, n.S has only one particular value
regardless of the path taken between states. page 873 Because
rusting is a spontaneous process, n.Suniv must be positive. Therefore,
th e entropy of the surroundings must increase, and that increase
must be larger than the entropy decrease of the system. page 875
S = 0, based on Equation 19.5 and the fact that In 1 = 0.
page 876 No. Argon atoms are not attached to other atoms, so
they cannot undergo vibrational motion. page 883 n.S,urr always
increases. For simplicity, assume that the process is isothermal. The
1208
change in entropy of the surroundings in an isothermal process is
-qsys
=
. Because the reaction is exothermic, -qsys is a posiT
tive number. Thus, 1',.Ssurr is a positive number and the entropy of the
surroundings increases. page 885 (a) In any spontaneous process
the entropy of the universe increases. (b) In any spontaneous process operating at constant temperature, the free energy of the system
decreases. page 886 It indicates that the process to which the thermodynamic quantity refers has taken place under standard conditions, as
summarized in Table 19 .2. page 890 Above the boiling point, vaporization is spontaneous, and /',.G < 0. Therefore, /',.H - T/',.S < 0, and
/',.H < T/',.S. Therefore T/',.S is greater in magnitude. page 893 No. K
is the ratio of product concentration to reactant concentration at equilibrium; K could be very small but is not zero.
Chapter 20
page 913 Oxygen is first assigned an oxidation number of -2.
Nitrogen must then have a +3 oxidation number for the sum of oxidation numbers to equal -1, the charge of the ion. page 916
No. Electrons should appear in the two half-reactions but cancel
when the half-reactions are added properly. page 923 Yes. A
redox reaction with a positive standard cell potential is spontaneous
under standard conditions. page 924 l atm pressure of Cl2(g)
and 1 M concentration of Ci-(aq) page 927 Ni page 930 Sn2+
page 947Al, Zn. Both are easier to oxidize than Fe.
Chapter 21
page 968 The mass number decreases by 4 . page 971 Beta
decay page 975 From Figure 21.3 we can see that each of these four
elements has only one stable isotope, and from their atomic numbers
we see that they each have an odd number of protons. Given the rarity of stable isotopes with odd numbers of neutrons and protons, we
expect that each isotope will possess an even number of neutrons.
From their atomic weights we see that this is the case: F (10 neutrons),
Na (12 neutrons), Al (14 neutrons), and P (16 neutrons). page 981
It doubles as well. The number of disintegrations per second is proportional to the number of atoms or the radioactive isotope. page
986 Any process that depends on the mass of the molecule, such as
the rate of gaseous effusion (Section 10.8) page 988 The values in
Table 21.7 only reflect the mass of the nucleus, while the atomic mass
is the sum of the mass of the nucleus and the electrons. So the atomic
mass of iron-56 is 26 X me larger than the nuclear mass. page 989
No. Stable nuclei having mass numbers around 100 are the most
stable nuclei. They could not form a still more stable nucleus with an
accompanying release of energy. page 999 The absorbed dose is
equal to O.lOJ X (l rad/1 X
= 10 rads. The effective dosage
is calculated by multiplying the absorbed dose by the relative biological effectiveness (RBE) factor, which is 10 for alpha radiation. Thus,
the effective dosage is 100 rems.
Chapter 22
page 1013 No . There is a triple bond in N2. P does form triple bonds,
as it would have to in order to form P2. page1016 H- , hydride. page
1017 + l for everything except I-[2, for which the oxidation state of H
is 0. page1023 0 for Cl2; -1 for c1- ; +1 for Clo- page1024 They
should both be strong, since the central halogen is in the +5 oxidation state for both of them. We need to look up the redox potentials to
see which ion, Bro3- or c103- , has the larger reduction potential. The
ion with the larger reduction potential is the stronger oxidizing agent.
Br03- is the stronger oxidizing agent on this basis (+ 1.52 V standard
reduction potential in acid compared to + 1.47 V for
pagel026 HI03 page1029S03(g) + H20 (I)
H2S04(/)
page1034 (a) +5 (b) + 3 page1036 In PF3, phosphorus in the +3
oxidation state. The reaction with water produces the oxyacid with
phosphorus in that oxidation state, H3P03. Thus, the reaction is analogous to Equation 22.48, which also shows a trihalide reacting with
water. page 1041 C02(g) page 1046 Silicon is the element, Si. Silica
is Si02. Silicones are polymers that have an 0 -Si 0 backbone and
hydrocarbon groups on the Si. page 1047 +3
—
Chapter 23
page 1062 Sc has the largest radius. page 1063 Because titanium only
has 4 valence electrons you would have to remove a core electron to
create a Ti5+ ion . page 1064 The larger the distance, the weaker the
spin-spin interactions . page 1065 Yes, it is a Lewis acid-base interaction; the metal ion is the Lewis acid (electron pair acceptor) .
page1071 The nonbonding electron pairs in H20 are both located
on the same atom, which makes it impossible for both pairs to be
donated to the same metal atom. To act as a chelating agent the
nonbonding electron pairs need to be on different atoms that are not
connected to each other. page 1072 Bidentate. page 1079 No,
ammonia cannot engage in linkage isomerism the only atom that
can coordinate to a metal is the nitrogen. page 1079 No, a trigonal
planar molecule cannot exhibit geometrical isomerism. page 1082
(b) The way polarized light is rotated on passing through the substance. page 1084 (a) Co is [Ar] 4s23d7. (b) Co3+ is [Ar] 3d6. The Co
atom has 3 unpaired electrons and the Co3+ ion has 4 unpaired electrons, assuming all five d orbitals have the same energy. page 1087
Because Ti(IV) ions have an
3d0 electron configuration there are
no d-d transitions that can absorb photons of visible light. page
1089 d4, d5, d6, d7. page 1091 Because the ligands are located in
the xy plane, electrons in a dxy orbital, which has its lobes in the xy
plane, feel more repulsion than electrons in the dx, and dy, orbitals.
—
Chapter 24
—
page 1109 C=N, because it is a polar double bond. C H and C
C bonds are relatively unreactive. (The CN double bond does not
have to fully break to be reactive.) page 1111 Two C H bonds
and two C C bonds page 1112 The isomers have different properties, as seen in Table 24 .3 (e.g., different melting points and different boiling points) . page 1115 Yes, since cyclopropane is more
strained . page 1118 Only two of the four possible C = C bond
sites are distinctly different in the linear chain of five carbon atoms
with one double bond. page 1124
—
—
—
OON02OO
N02
page 1129
page 1129 The generic formula ofan
ether is ROR', while an ester is RCOOR'. If you carbonylate or oxidize
an ether, you can get an ester. page 1133 All four groups must be
different from one another. page 1137 No . Breaking the hydrogen
bonds between N H and O = C groups in a protein by heating
causes the a-helix structure to unwind and the {3-sheet structure to
separate. page 1142 The alpha form of the C o
C linkage.
Glycogen serves as a source of energy in the body, which means
that the body's enzymes must be able to hydrolyze it to sugars. The
enzymes work only on polysaccharides having the a linkage .
—
——
Chapter 1
Chapter 5
Figure 1.1 Aspirin. It contains 9 carbon atoms. Figure 1.4 Vapor
(gas) Figure 1.5 Molecules of a compound are composed of
m ore than one type of atom, and molecules of an element are composed of only one type of atom. Figure 1.6 It would get larger
because hydrogen atoms are lighter than the atoms of any other
element. Figure 1.7 (c) Eac h water molecule contains one oxygen
atom and two hydrogen atom s. Figure 1.14 The separations are
due to physical processes of adsorption of the materials onto th e
column Figure 1.19 True Figure 1.20 1000 Figure 1.24 The
darts would be scattered widely (poor precision) but their average
position would be at the center (good accuracy) .
Figure 5.2 The electro static potential energy of two oppositely
charged particles is negative (Equation 5.2). As the particles become
closer, the electrostatic potential energy becomes even m ore
negative-that is, it decreases. Figure 5.3 Attractions between
oppositely charged ions would cause the ions to move closer
together reducing the potential energy of the system. The lost
potential energy is converted to kinetic energy needed to move the
ions. Figure 5.4 The number o f molecules will change because
three reactant molecules (2 f-12 and 1
are needed to make two
product (H20) molecules. However, in a closed system like this one
the total mass will not change. Figure 5.5 If Er;11a1 = E;11;,ia1> then
6.E = 0. Figure 5.6
Chapter 2
Figure 2.3 Cathode rays would still be generated but without
th e fluorescent screen you would not be able to see them.
Figure 2.4 The electron beam would be deflected downward
because of repu lsion by the upper negative plate and attraction
toward the positive plate. Figure 2.5 No, the electrons are of
negligible mass compared with an oil drop. Figure 2.7 Beta rays
are equivalent to electrons. (a) They are lighter. Figure 2.9 The
beam con sists of alpha particles, which carry a +2 charge. They are
repelled from the positively charged go ld nuclei. Figure 2.10 10- 2
pm Figure 2.13 Based on the periodic trend, we expect that elements that precede a nonreactive gas, as F does, will also be reactive
nonmetals. The elements fittin g this pattern are H and Cl.
Figure2.16 All of the metals shown here are solids, that exhibit
varying degrees of luster to their appearance. In contrast, bro mine is
a liquid and none of the nonmetals shown here are lustrou s.
Figure2.18 The ball-and-stick model. Figure2.19 The elements
are in the fo llowing groups: Ag+ is 1B, Zn2+ is 2B, and Sc3+ is 3B. Only
Sc3+ has the same number of electrons as a noble gas, Ar (element
18). Figure 2.20 No it is not, ionic solids like NaCl do not contain
discrete molecules. Figure 2.23 Removing one O atom fro m the
perbromate ion gives the bromate ion, Br03- .
Chapter 3
Ftgure 3. 4 There are two CH4 and four
molecules on the reactant side, which contain 2 C atoms, 8 1-1 atoms and 8 0 atoms. The
number of each type of atom remains the same on the product side
as it must. Ftgure 3.8 The flame gives off heat and therefore the
reaction must release heat . Figure 3.9 Avogadro 's number
Figure 3.12 g/mol and mo1- 1 Figure 3.17 If the amount of
H2 is doubled then
becomes the limiting reactant. In that case
(7 mol02) X (Z moll-120 / 1
= 14 mol I-120 would be produced.
Chapter 4
Figure 4.3 NaCl(aq) Figure 4.4 K+ and N03- Figure 4.9 Mg
and water Figure 4.12 Two. Each O atom becomes an 0 2ion . Figure 4.13 One, based on the reaction stoichiometry.
Figure 4.14 Because the Cu(Il) ion produces a blue color in aqueous
solution. Figure 4.18 The volume needed to reach the end point
if Ba (OH )z(aq) were used would be one-halfthe volume needed for
titration with NaOH (aq) , because there are two hydroxide ions for
every barium ion .
Mg(s) + Cl2(g)
Final
s tate
/1E > 0
Initial
s tate
MgCl2(s)
FigureS.7 !:iE = 50] + (-851) = - 35] FigureS.10 Thebattery
is doing work on the surroundin gs, so w < 0. Figure 5.11 The
system does work on the surroundings to move the piston upward,
so w < 0. Figure 5.17 Endothermic heat is being added to the
system to raise the temperature of the water. Figure 5.18 Two
cups provide more thermal insulation so less heat will escape the
system. Fig,tre 5.19 In a calorimeter the system is defined as the
reactants and products, so the water inside the calorimeter is part of
the surroundings. Figure 5.21 The condensation of 2 H20(g) to
2 I-120 (/) Figure 5.22 Yes, !:iH3 would remain the same as it is the
enthalpy change fo r the process CO (g)
COz(g).
Figure 5.24 Exothermic, because the enthalpy of the products is
lower than that of the reactants. Figure 5.25 Grams of fat
—
—.
Chapter 6
Figure 6.3 Wavelength = 1.0 m , frequency = 3.0 X 108 cycles/s .
Figure 6. 4 Lon ger by 3 to 5 orders of magnitude (depending on
what part of the microwave spectrum is considered). Figure 6.5
The part of the nail glowing yellow is hotter than the part glowing red. Figure 6.7 As the frequency of the incoming light is
increased the photon ene rgy will increase and the kinetic energy
of the ejected electron s will increase. Figure 6.12 The n = Z to
n = l transition invo lves a larger energy change than the n = 3 to
n = Z transition. (Compare the lengths of the arrows connecting the
states in th e figure.) If the n = 3 to n = Z transition produces visible
light, the n = Z to n = l transition must produce radiation of greater
energy. Of the two choices only ultraviolet radiation has higher
frequency and greater energy than visible light. Figure 6.13
The n = 4 to n = 3 transition invo lves a smaller energy difference
and will therefore emit light of lon ger wavelength. Figure 6.17
The region of highest electron density is where the density of dots
1209
1210
is highest, which is near the nucleus. Figure 6.18 For a single
electron atom like hydrogen the energies of the orbitals are the same
as the energies of the orbits in the Bohr model. Figure 6.19 There
would be four maxima and three nodes . Figure 6.23 The Px
orbital. Figure 6.24 There are two nodal planes for each of the
d-orbitals . For the d,Ay.• orbital the nodal planes are the xz and yz
planes. Figure 6.25 The 4d and 4(subshells are not shown.
Figure 6.31 Osmium
Chapter 7
Figure 7.1 These three metals do not readily react with other elements, especially oxygen, so they are often found in nature in the elemental form as metals (such as gold nuggets) Figure 7.4 Because
of the peak near the nucleus in the 2s curve there is a higher probability of finding a 2s within 0.5 Aof the nucleus. In a multielectron
atom an electron in a 2s orbital will have a lower energy than one
in a 2p orbital. Figure 7.7 Bottom and left Figure 7.8 They get
larger, just like the atoms do. Figure 7.10 900 kJ / mo!
Figure 7.11 There is more electron-electron repulsion in the case of
oxygen because two electrons have to occupy the same orbital.
Figure 7.12 The added electron for the Group 4A elements leads to
a half-filled np3 configuration. For the Group SA elements, the added
electron leads to an np4 configuration, so the electron must be added
to an orbital that already has one electron in it, and thus experiences
more electron-electron repulsion. Figure 7.13 They are opposite:
As ionization energy increases, metallic character decreases, and
vice versa. Figure 7.15 Anions are above and to the right of the
line; cations are below and to the left ofthe line. Figure 7.16 No.
The Na+ and N03- ions will simply be spectator ions. The I-r+ ions
of an acid are needed in order to dissolve NiO. Figure 7.17 No .
As seen in the photo, sulfur crumbles as it is hit with a hammer,
typical of a solid nonmetal. Figure 7.21 Because Rb is below K m
the periodic table, and has a lower first ionization energy we expect
Rb to be more reactive with water than K. Figure 7.23 Lilac (see
Figure 7.22). Figure 7.25 The bubbles are due to H2(g). This
could be confirmed by carefully testing the bubbles with a flame
there should be popping as the hydrogen gas ignites. Figure 7.26
Yes, because water does not decompose on sitting the way that
hydrogen peroxide does. Figure 7.27 A regular octagon.
Figure 7.28 12 is a solid whereas Cl2 is a gas. Molecules are more
closely packed together in a solid than they are in a gas, as will be discussed in detail in Chapter 11.
—
Chapter 8
Figure 8.1 We would draw the chemical structure of sugar mo!ecules (which have no charges and have covalent bonding between
atoms in each molecule) and indicate weak intermolecular forces
(especially hydrogen bonding) between sugar molecules.
Figure 8.3 Yes. Figure 8.4 Cations have a smaller radius than
their neutral atoms and anions have a larger radius. Because Na and
Cl are in the same row of the periodic table, we would expect Na+
to have a smaller radius than c1-, so we would infer that the larger
green spheres represent the chloride ions and the smaller purple
spheres represent the sodium ions. Figure 8.5 The distance
between ions in KF should be larger than that in NaF and smaller
than that in KC!. We would thus expect the lattice energy of KF
to be between 701 and 910 kJ/mol. Figure 8.7 The repulsions
between the nuclei would decrease, the attractions between the
nuclei and the electrons would decrease, and the repulsions between
the electrons would be unaffected. Figure 8.8 The electronegativity decreases with increasing atomic number. Figure 8.10 µ, will
decrease. Figure 8.11 The bonds are not polar enough to cause
enough excess electron density on the halogen atom to lead to a
red shading. Figure 8.13 The lengths of the bonds of the outer
0 atoms to the inner O atom are the same. Figure 8.14 Yes. The
electron densities on the left and right parts of the molecule are the
same, indicating that resonance has made the two 0
O bonds
equivalent to one another. Figure 8.15 The dashed bonds represent the delocalized electrons that result when the two resonance
structures are averaged. Figure 8.16 It should be halfway between
the values for single and double bonds, which we can estimate to be
about 280 kJ/mol from the graph.
—
Chapter 9
Figure 9.1 The radii ofthe atoms involved (see Section 7.3).
Figure 9.2 NF3 Figure 9.3 Octahedral. Figure 9.4 No. We
will get the same bent-shaped geometry regardless of which two
atomsweremove. Figure9.8 90°. Figure9.9 TheC 0 H
bond involving the right O because of the greater repulsions due
to the nonbonding electron domains. The angle should be Jess
than the ideal value of 109.5°. Figure 9.10 Zero . Because they
are equal in magnitude but opposite in sign, the vectors cancel
upon addition. Figure 9.13 0.74 A is the bond length, 436 kJ/
mol is the bond strength. Figure 9.16 120· Figure 9.17 The
Pz orbital. Figure 9.18 No. All four hybrids are equivalent and
the angles between them are all the same, so we can use any of the
two to hold the nonbonding pairs. Figure 9.19 Because the P 3p
orbitals are larger than the N 2p orbitals, we would expect somewhat
larger lobes on the hybrid orbitals in the right-most drawing. Other
than that, the molecules are entirely analogous. Figrire 9.22
They have to lie in the same plane in order to allow the overlap of
the 1r orbitals to be effective in forming the 1r bond . Figure 9.23
Acetylene should have the higher carbon-carbon bond energy
because it has a triple bond as compared to the double bond
in ethylene . Figure 9.25 It has six C C a- and six C H u
bonds. Figure 9.31 Zero. A node, by definition, is the place where
the value of the wave function is zero . Figure 9.32 The energy
ofa-1, would rise (but would still be below the energy ofthe H ls
atomic orbitals) . Figure 9.34 a-1,2, a-*1,1. Figure 9.35 a-fs and
Figure 9.36 The end-on overlap in the <rzp MO is greater than
the sideways overlap in the 1rzp• Figure 9.42 The <rzp and 1rzp MOs
have switched order. Figure 9.43 02 and F2 Figure 9.45 N2 is
diamagnetic so it would not be attracted to the magnetic field. The
liquid nitrogen would simply pour through the poles of the magnet
without "sticking." Figure 9.46 11. All the electrons in the n = 2
level are valenceshell electrons.
——
—
—
Chapter 10
Figure 10.2 It will increase. Figure 10.4 Decreases Figure 10.S
1520 torr or 2 atm Fig,tre 10.8 One Figure 10.9 Chlorine, Cl2.
Figure 10.12 About one-sixth Figure 10.13 02 has the largest molar mass, 32 g/mol, and
has the smallest, 2.0 g/mol.
Figure 10.15 n, moles ofgas Figure 10.19 True Figure 10.21
Itwould increase .
Chapter 11
Figure 11.2 The density in a liquid is much closer to a solid than
it is to a gas. Figure 11.3 The distance within the molecule (the
covalent bond distance) represented by the solid black line is smaller
than the intermolecular distance represented by the red dotted
line. Figure 11.S The halogens are diatomic molecules and have
much greater size and mass, and therefore greater polarizability,
than the noble gases, which are monatomic. Figure 11.8 They
stay roughly the same because the molecules have roughly the
1211
same molecular weights. Thus, the change in boiling point moving
left to right is due mainly to the increasing dipole-dipole attractions. Figure 11.9 Both compounds are nonpolar and incapable
of forming hydrogen bonds. Therefore, the boiling point is determined by the dispersion forces, which are stronger for the larger,
heavier AsH3. Figure 11.10 The non-hydrogen atom must possess
a nonbonding electron pair. Figure 11.11 There are four electron
pairs surrounding oxygen in a water molecule. Two of the electron
pairs are used to make covalent bonds to hydrogen within the H20
molecule, while the other two are available to make hydrogen bonds
to neighboring molecules. Because the electron-pair geometry is
tetrahedral (four electron domains around the central atom), the
H 0 · · · H bond angle is approximately 109°. Figure 11.13 The
0 atom is the negative end of the polar H20 molecule; the negative
end ofthe dipole is attracted to the positive ion. Figure 11.14
Yest although not likely. Figure 11.19 Wax is a hydrocarbon that
cannot form hydrogen bonds. Therefore, coating the inside of tube
with wax will dramatically decrease the adhesive forces between
water and the tube and change the shape of the water meniscus to
an inverted U-shape. Neither wax nor glass can form metallic bonds
with mercury so the shape of the mercury meniscus will be qualitatively the same, an inverted U-shape. Figure 11.20 Because energy
is a state function, the energy to convert a gas to a solid is the same
regardless of whether the process occurs in one or two steps. Thus,
the energy of deposition equals the energy of condensation plus
the
of freezing. Figure 11.21 Because water has stronger
intermolecular interactions. Figure 11.22 The temperature of the
liquid water is increasing. Figure 11.24 Increases, because the
molecules have more kinetic energy as the temperature increases and
can escape more easily Figure 11.25 Allliquids including ethylene
glycol reach their normal boiling point when their vapor pressure
is equal to atmospheric pressure, 760 torr. Figure 11.27 It must be
lower than the temperature at the triple point.
—
Chapter 12
Figure 12.S There is not a centered square lattice, because ifyou
tile squares and put lattice points on the corners and the center of
each square it would be possible to draw a smaller square (rotated by
45°) that only has lattice points on the comers. Hence a "centered
square lattice" would be indistinguishable from a primitive square
lattice with a smaller unit cell. Figure 12.12 Face-centered cubic,
assuming similar size spheres and cell edge lengths, since there are
more atoms per volume for this unit cell compared to the other two.
Figure 12.13 A hexagonal lattice Figure 12.15 The solvent is
the majority component and the solute the minority component.
Therefore, there will be more solvent atoms than solute atoms.
Figure 12.17 The samarium atoms sit on the comers of the unit
cell so there is only 8 x (1/8) = 1 Sm atom perunit cell. Eight of
the nine cobalt atoms sit on faces of the unit cell, and the other sits
in the middle of the unit cell so there are 8 X ( 1 /2) + 1 = 5 Co
atoms per unit cell. Figure 12.19 P4, S8, and Cl2 are all molecules,
because they have strong chemical bonds between atoms and have
well-defined numbers of atoms per molecule. Figure 12.21 In the
fourth period, vanadium and chromium have very similar melting
points. Molybdenum and tungsten have the highest melting points
in the fifth and sixth periods, respectively. All of these elements are
located near the middle of the period where the bonding orbitals
are mostly filled and the antibonding orbitals mostly empty.
Figure 12.22 The molecular orbitals become more closely spaced
in energy. Figure 12.23 Potassium has only one valence electron
per atom (4s1 ). Therefore we expect the 4s band to be approximately
half full. If we fill the 4s band halfway a small amount of electron
density might leak over and start to fill the 3d orbitals as well. The 4p
orbitals should be empty. Figure 12.24 Ionic substances cleave
because the nearest neighbor interactions switch from attractive to
repulsive if the atoms slide so that ions of like charge (cation- cation
and anion-anion) touch each other. Metals don't cleave because the
atoms are attracted to all other atoms in the crystal through metallic bonding. Figure 12.25 No, ions of like charge do not touch
in an ionic compound because they are repelled from one another.
In an ionic compound the cations touch the anions. Figure
12.27 In NaF there are four Na+ ions (12 X 1 /4 ) and four F- ions
( 8 x 1 /8 + 6 x 1 /2) per unit cell. ln MgF2 there are two Mg2+ ions
( 8 x 1 /8 + 1 ) and four F- ions (4 X 1 /2 + 2) perunit cell. In ScF3
there is one Sc3+ ion (8 x 1/8) and three F- ions (12 X 1/4) per
unit cell. Figure 12.28 The intermolecular forces are stronger in
toluene, as shown by its higher boiling point. The molecules pack
more efficiently in benzene, which explains its higher melting
point, even though the intermolecular forces are weaker.
Figure 12.30 The band gap for an insulator would be larger than
the one for a semiconductor. Figure 12.31 Ifyou doubled the
amount of doping in panel (b) , the amount of blue shading the
conduction band would also double. Figure 12.45 Decrease. As
the quantum dots get smaller, the band gap increases and the emitted light shifts to shorter wavelength. Figure 12.49 Each carbon
atom in C60 is bonded to three neighboring carbon atoms through
covalent bonds. Thus, the bonding is more like graphite, where
carbon atoms also bond to three neighbors, than diamond, where
carbon atoms bond to four neighbors.
Chapter 13
Figure 13.1 Gas molecules move in constant random
motion. Figure 13.2 Opposite charges attract. The electronrich O atom of the H20 molecule, which is the negative end of
the dipole, is attracted to the positive Na+ ion. Figure 13.4 For
exothermic solution processes the magnitude of M-lmix will be larger
than the magnitude of !:,_I-f,01,,te + !:,_I-f,0tvent Figure 13.7 46 g.
Figure 13.12 If the partial pressure of a gas over a solution is
doubled, the concentration of gas in the solution would double.
Figure 13.13 The highest. Figure 13.15 Looking at where the
solubility curves for KCl and NaCl intersect the 80 °C line, we see
that the solubility of KCl is about 51 g/100 g H20, whereas NaCl
has a solubility of about 39 g/100 g H20 . Thus, KCl is more soluble
than NaCl at this temperature. Figure 13.16 N2 has the same
molecular weight as CO but is nonpolar, so we can predict that its
curve will be just below that of CO. Figure 13.22 The water will
move through the semipermeable membrane toward the more
concentrated solution. Thus, the liquid level in the left arm will
increase. Figure 13.23 Water will move toward the more concentrated solute solution, which is inside the red blood cells, causing
them to undergo hemolysis. Figure 13.26 The negatively charged
groups both have the composition - CO2- . Figure 13.28 Recall
the rule that likes dissolve likes. The oil drop is composed of nonpolar molecules, which interact with the nonpolar part of the stearate
ion via dispersion forces.
Chapter 14
Figure 14.1 No. The surface area of a steel nail is much smaller
than that for the same mass of steel wool, so the reaction with Oz
would not be as vigorous. Depending on how hot it is, it might not
burn at all. Figure 14.2 Our first guess might be half way between
the values at 20 s and 40 s, namely 0.42 mol A. However, we also see
that the change in the number of moles of A between O s and 20 s is
1212
greater than that between 20 s and 40 s-in other words, the rate of
conversion gets smaller as the amount of A decreases. So we would
guess that the change from 20 s to 30 s is greater than the change
from 30 s to 40 s, and we would estimate that the number of moles
ofA is between 0.42 and 0.30 mo!. Figure 14.3 The instantaneous
rate decreases as the reaction proceeds. Figure 14.7 The reaction
is first order. Figure 14.9 At the beginning of the reaction when
both plots are linear or nearly so. Figure 14.13 No, it will not turn
down. The rate constant increases monotonically with increasing
temperature because the kinetic energy of the colliding molecules
continues to increase. Figure 14.15 The difference in height
between the ball and the top of the barrier. Figure 14.16 As
shown, the magnitude of energy needed to overcome the energy barrier is greater than the magnitude of the energy change in the reaction. Figure 14.17 It would be more spread out, the maximum of
the curve would be lower and to the right of the maximum of the red
curve, and a greater fraction of molecules would have kinetic energy
greater than Ea than for the red curve. Figure 14.19 The rate of
converting intermediates to products will be faster, because this
reaction has a lower barrier than the reaction that converts intermediates back to reactants. Figure 14.20 No, because the bottleneck,
or rate determining step, is passing through toll plaza A. Figure
14.21 The color is characteristic of molecular bromine, Br2, which is
an intermediate in this reaction. Figure 14.22 There is one valley
corresponding to the formation of intermediates, so the reaction has
a total of two steps. Figure 14.25 Grinding increases the surface
area, exposing more of the catalase to react with the hydrogen peroxide. Figure 14.26 Substrates must be held more tightly so that
they can undergo the desired reaction. Products are released from
the active site .
Chapter 15
Figure 15.2 The same because once the system reaches
equilibrium the concentrations of N02 and N204 stop changing.
Figure 15.3 Because the concentration of reactants is decreasing,
which reduces the frequency of collisions and hence the rate of the
forward reaction. Figure 15.4 Yes. The reaction stoichiometry
dictates that
disappears at three times the rate that N2 disappears. Figure 15.6 Experiment 4 Figure 15.7 The boxes would
be approximately the same size. Figure 15.8 It would decrease.
To reestablish equilibrium the concentration of C02(g) would need
to return to its previous value . The only way to do that would be for
more CaC03 to decompose to produce enough C02(g) to replace
what was lost. Figure 15.10 High pressure and low temperature, 500 atm and 400 °C in this figure. Figure 15.11 Nitrogen
must react with some of the added hydrogen to create ammonia
and restore equilibrium. Figure 15.16 (a) the energy difference
between the initial state and the transition state.
Figure 15.17 About 5 x 10- 4
Chapter 16
Figure 16.2 Hydrogen bonds. Figure 16.4
0 2-(aq) + H20 (1)
2 OH -( aq). Figure 16.5 Basic. The
mixture of the two solutions will still have [ H+] < [OH-].
Figure 16.6 Lemon juice. It has a pH of about 2 whereas black coffee has a pH of about 5 . The lower the pH, the more acidic the solution. Figure 16.8 Phenolphthalein changes from colorless, for pH
values less than 8, to pink for pH values greater than 10. A pink color
indicates pH > 10. Figure 16.9 Bromthymol blue would be most
suitable because it changes pH over a range that brackets pH = 7.
Methyl red is not sensitive to pH changes when pH > 6, while phe-
nolphthalein is not sensitive to pH changes when pH < 8, so neither changes color at pH = 7. Figure 16.12 Yes. The equilibrium
of interest is H3CCOOH ;;==::= H+ + H3Ccoo- . If the percent dissociation remained constant as the acid concentration increased, the
concentration of all three species would increase at the same rate.
However, because there are two products and only one reactant,
the product of the concentrations of products would increase faster
than the concentration of the reactant. Because the equilibrium
constant is constant, the percent dissociation decreases as the acid
concentration increases. Figure 16.13 3. Figure 16.15 The
nitrogen atom in hydroxylamine accepts a proton to form NH30H+.
As a general rule, nonbonding electron pairs on nitrogen atoms
are more basic than nonbonding electron pairs on oxygen atoms.
Figure 16.17 The range ofpH values is so large thatwe can't
show the effects using a single indicator (see Figure 16.8) . Figure
16.20 The HOY molecule on the left because it is a weak acid. Most
of the HOY molecules remain undissociated.
Chapter 17
Figure 17.6 The pH will increase upon addition of the base
Figure 17.7 25.00 mL. The number of moles of added base needed
to reach the equivalence point remains the same. Therefore, by doubling the concentration of added base the volume needed to reach
the equivalence point is halved . Figure 17.9 The volume of base
needed to reach the equivalence point would not change because
this quantity does not depend on the strength of the acid. However,
the pH at the equivalence point, which is greater than 7 for a weak
acid-strong base titration, would decrease to 7 because hydrochloric acid is a strong acid. Figure 17.11 The pH at the equivalence
point increases (becomes more basic) as the acid becomes weaker.
The volume of added base needed to reach the equivalence point
remains unchanged . Figure 17.12 No; it would change color by
pH 6 and miss pH 7. Figure 17.15 H2P03- at first equivalence
point, HPo/- at the second. Figure 17.22 Because the solution is
basic. Figure 17.23 Yes. CuS would precipitate in step 2 on addition of H2S to an acidic solution, while the Zn2 + ions remained in
solution.
Chapter 18
Figure 18.1 About 85 km, at the mesopause. Figure 18.3
The atmosphere absorbs a significant fraction of solar radiation.
Figure 18.4 (c) UV Figure 18.S The peak value is about
5 X 1012 molecules per cm3. Jf we use Avogradro's number
to convert molecules to moles, and the conversion factor of
1000 cm3 = 1000 mL = 1 L, we find that the concentration of
ozone at the peak is 8 X 10- 9 mole/L. Figure 18.8 The major
sources of sulfur dioxide emission are located in the eastern half
of the United States, and prevailing winds carry emissions in an
eastward direction. Figure 18.10 CaS03(s) Figure 18.12 The
Earth's surface absorbs 492 W/m2. Of that amount 390 W/m2or
79% is radiated back toward the atmosphere. Figure 18.14 The
increasing slope corresponds to an increasing rate of addition of
CO2 to the atmosphere, probably as a result of ever-increasing burning of fossil fuels worldwide. Figure 18.16 Evaporation from sea
water, evaporation from freshwater; evaporation and transpiration
from land. Figure 18.17 As the temperature of water goes down
its density increases. Therefore, density increases as depth increases
(note the pressure also increases which also results in an increase
in density). Figure 18.21 To reduce concentrations of dissolved
iron and manganese, remove I-12S and NI-13, and reduce bacterial
levels.
1213
Chapter 19
Figure 19.1 Yes, the potential energy of the eggs decreases as
they fall. Figure 19.2 The freezing of liquid water to form ice is
exothermic. Figure 19.3 To be truly reversible, the temperature
change 8T must be infinitesimally small. Figure 19.S Because
the final volume would be less than twice the volume of Flask A, the
final pressure would be greater than 0.5 atm. Figure 19.8 There
are two other independent rotational motions of the H20 molecule:
’ >
Figure 19.9 Ice, because it is the phase in which the molecules are
held most rigidly Figure 19.11 The decrease in the number of
molecules due to the formation of new bonds. Figure 19.12 During a phase change, the temperature remains constant but the
entropy change can be large as molecules increase their degrees of
freedom and motion. Figure 19.13 Based on the three molecules
shown, the addition of each C increases S0 by 40-45 J/mol-K. Based
on this observation, we would predict that S0(C4HJO) would 310-315
J/mol-K. Appendix C confirms that this is a good prediction:
S0(C4H10) = 310.0J/mol-K. Figure 19.14 Spontaneous Figure
19.15 If we plot progress of the reaction versus free energy, equilibrium is at a minimum point in free energy, as shown in the figure.
In that sense, the reaction runs "downhill" until it reaches that
minimum point
Chapter 20
Figure20.1 Exothermic. Figure20.2 The permanganate ion,
Mn04- , is reduced, as the half-reactions in the text show. The
oxalate ion, C2ol- , acts as the reducing agent. Figure 20.3 The
blue color is due to Cu2+(aq). As this ion is reduced, forming Cu(s),
its concentration decreases and the blue color fades. Figure
20.4 The Zn is oxidized and, therefore, serves as the anode of the
cell. Figure 20.S The electrical balance is maintained in two ways:
Anions migrate into the half-cell, and cations migrate out via the salt
bridge. Figure 20.9 As the cell operates, H+ is reduced to H2 in the
cathode half-cell. As H+ is depleted, the positive
ions are drawn
into the half-cell to maintain electrical balance in the solution
Figure 20.11 Yes. Figure 20.13 The variable n is the number
of moles of electrons transferred in the process. Figure 20.14
The cathode. Figure20.19 The cathode consists of Pb02(s) .
Because each oxygen has an oxidation state of-2, lead must have
an oxidation state of +4 in this compound . Figure20.20 Zn
Figure 20.21 Co3+. The oxidation number increases as the battery
charges. Figure20.24 0 2(g) + 4H+ + 4e_
2H20(g)
Figure 20.25 The oxidizing agent of 02(g) from the air.
Figure 20.29 0 V
Chapter 21
Figure 21.1 From Figure 21 .1 we see that the belt of stability for
a nucleus containing 70 protons lies at approximately 102 neutrons. Figure 21.2
%1Pa + -~e Figure 21.3 Only
three of the elements with an even number of protons have fewer
than three isotopes: He, Be, and C. Note that these three elements
are the lightest elements that have an even atomic number. Because
they are so light, any change in the number of neutrons will change
the neutron/proton ratio significantly. This helps to explain why
they do not have more stable isotopes. None of the elements in
Figure 21.3 that have an odd number ofprotons have more than
two stable isotopes. Figure21.6 6.25 g. After one half-life, the
amount of the radioactive material will have dropped to 25.0 g. After
two half-lives, it will have dropped to 12.5 g. After three half-lives,
it will have dropped to 6.25 g. Figure 21.7 Plants convert 14C02
to 14C-containing sugars via photosynthesis. When mammals eat
the plants, they metabolize the sugars, thereby incorporating 14C in
their bodies. Figure 21.8 Gamma rays. Both X rays and gamma
rays consist of high-energy electromagnetic radiation, whereas alpha
and beta rays are streams of particles. Figure 21.9 Ionization
energy. Detection depends on the ability of the radiation to cause
ionization ofthe gas atoms. Figure21.14 The mass numbers are
equal on both sides. Remember that this does not mean that mass is
conserved mass is lost during the reaction, which appears as energy
released. Figure21.15 16 . Each of the eight neutrons would split
another uranium-23 5 nucleus, releasing two more neutrons .
Figure 21.16 Critical without being supercritical so that the release
of energy is controlled . Figure 21.23 Alpha rays are less dangerous when outside the body because they cannot penetrate the skin.
However, once inside the body they can do great harm to any cells
that are nearby.
—
Chapter 22
Figure 22.S Beaker on the right is warmer. Figure 22.6 HF is the
most stable, Sb!-I3 the least stable. Figure22.8 More soluble in
CC14 - the colors are deeper. Figure 22.9 CF2 Figure 22.12
No Figure 22.14 Based on this structure yes, it would have a
dipole moment. In fact, if you look it up, hydrogen peroxide's dipole
moment is larger than water's! Figure22.19 Formally they could
both be +2. If we consider that the central sulfur is like sol- , however, then the central sulfur would be +6, like sol- , and the terminal sulfur would be -2. Figure 22.21 Nitrite Figure 22.22
Longer. (There is a triple bond in N2.) Figure22.24 The NO
double bond Figure 22.26 In both compounds the electron
domains around the P atoms are tetrahedral. In
all the electron
domains around the P atoms are bonding. In P406 one of the electron domains about each P atom is nonbonding. Figure 22.31
The minimum temperature should be the melting point of silicon;
the temperature of the heating coil should not be so high that the
silicon rod starts to melt outside the zone of the heating coil.
—
Chapter 23
Figure 23.3 No. The radius decreases first and then flattens out
before increasing on moving past group 8B, while the effective
nuclear charge increases steadily on moving left to right across the
transition metal series. Figure 23.4 Zn2+. Figure 23.S The 4s
orbitals are always empty in transition metal ions, so all of the ions
shown in this table have empty 4s orbitals. The 3d orbitals are only
empty for those ions that have lost all of their valence electrons:
Sc3+, Ti4+, ys+, Cr6+, and Mn 7+ . Figure23.6 The electron spins
would tend to align with the direction of the magnetic field.
Figure 23.8 No. If you start with a chloride ion on one vertex of
the octahedron and then generate structures by placing the second
chloride ion on any of the other five vertices you will get one complex that is the trans isomer and four complexes that are equivalent
to the cis isomer shown in this figure . Figure 23.9 Neither the
coordination number nor the oxidation state of iron change in this
reaction. Figure23.10 The solid wedge represents a bond coming out of the plane of the page. The dashed wedge represents a
bond that is going into the plane of the page. Figure 23.13 There
1214
are 20 carbon atoms in porphine, all of which have sp2 hybridization . Figure 23.15 The coordination number is 6 . The blue
atoms in the heme are nitrogen atoms. Figure23.16 Thepeak
at 660 nm. Figure23.20
and pentaamminenitroiron(II]) ion for the complex on the left. [Fe(NH3)5(ONO)]2+
and pentaamminenitritoiron(III) ion for the complex on the
right. Figure 23.21 The cis isomer. Figure 23.24 Larger, since
ammonia can displace water. Figure 23.26 The peak would stay
in the same position in terms of wavelength, but its absorbance
would decrease. Figure23.28
and
Figure23.29 Convert the wavelength of light, 495 nm, into energy in joules using the
relationship E = he/ A. Figure23.30 The absorption peak would
shift to shorter wavelengths absorbing green light and the color
of the complex ion would become red. An even larger shift would
move the absorption peak into the blue region of the spectrum and
the color would become orange. Figure 23.34 Only the dx'-y'
orbital points directly at the ligands .
Chapter 24
Figure 24.1 Tetrahedral Figure 24.2 The OH group is polar
whereas the CH3 group is nonpolar. Hence, adding CH3 will (a)
reduce the substance's solubility in polar solvents and (b ) increase
its solubility in nonpolar solvents . Figure 24.S C,,H2n,
because there are no CH3 groups, each carbon has two hydrogens. Figure24.7 Just one Figure24.9 Yes, spontaneous . Figure24.13 Both lactic acid and citric acid Figure24.14
No, because there are not four different groups around any carbon Figure24.16 Those labeled "basic amino acids," which have
basic side groups that are protonated at pH 7 Figure 24.17
Two. Figure24.23 The long hydrocarbon chains, which are nonpolar Figure 24.24 The polar parts of the phospholipids seek to
interact with water whereas the nonpolar parts seek to interact with
other nonpolar substances and to avoid water. Figure 24.26
1-2-3-4 Figure 24.27 GC because each base has three hydrogen
bonding sites, whereas there are only two in AT.
Chapter 1
Sample Exercise 1.1
Practice Exercise 2: It is a compound because it has constant composition and can be separated into several elements.
Sample Exercise 1.2
Practice Exercise 2: (a) 1012 pm, (b) 6.0 km,
(d) 0.00422 g.
(c) 4.22 X 10-3 g,
Sample Exercise 1.3
Practice Exercise 2: (a) 261. 7 K, (b) 11.3 °F
Sample Exercise 1.4
Practice Exercise 2: (a) 8.96 g/ cm3,
(b ) 19.0 mL,
(c) 340 g.
Sample Exercise 1.5
Practice Exercise 2: 2.29 X
Sample Exercise 1.6
Practice Exercise 2: No. The number of feet in a mile is a defined
quantity and is therefore exact, but the distance represented by one
foot is not exact, although it is known to high accuracy.
Sample Exercise 1. 7
Practice Exercise 2: (a) four,
(b) two,
(c) three.
Sample Exercise 1.8
Practice Exercise 2: 9.52 m / s (three significant figures).
Sample Exercise 1.9
Practice Exercise 2: No. Even though the mass of the gas would then
be known to four significant figures , the volume of the container
would still be known to only three.
Sample Exercise 2. 6
Practice Exercise 2: B5H7
Sample Exercise 2. 7
Practice Exercise 2: 34 protons, 45 neutrons, and 36 electrons
Sample Exercise 2.8
Practice Exercise 2: (a) 3+,
(b ) 1-
Sample Exercise 2.9
Practice Exercise 2: (a) Rb is from group 1, and readily loses one
electron to attain the electron configuration of the nearest noble
gas element, Kr. (b) Nitrogen and the halogens are all nonmetallic elements, which form molecular compounds with one another. (c) Krypton, Kr, is a noble gas element and is chemically inactive
except under special conditions. (d) Na and K are both from group
1 and adjacent to one another in the periodic table. They would be
expected to behave very similarly. (e) Calcium is an active metal
and readily loses two electrons to attain the noble gas configuration
ofAr.
Sample Exercise 2.10
Practice Exercise 2: (a) Na3P04,
(b) ZnS041
(c) Fez (C03)3
Sample Exercise 2.11
Practice Exercise 2: Bro- and Bro2Sample Exercise 2.12
Practice Exercise 2: (a) ammonium bromide,
oxide, (c) cobalt(II) nitrate
(b) chromium(III)
Sample Exercise 2.13
Practice Exercise 2: (a) HBr,
(b ) H2C03
Sample Exercise 1.10
Practice Exercise 2: 804.7 km
Sample Exercise 2.14
Practice Exercise 2: (a) SiBr4,
(b) S2Cl2,
Sample Exercise 1.11
Practice Exercise 2: 12 km/L.
Sample Exercise 2.15
Practice Exercise 2: No, they are not isomers because they have different molecular formulas. Butane is C4H10, whereas cyclobutane is
C4H8·
Sample Exercise 1.12
Practice Exercise 2: 1.2 X 104 ft.
(c) P206.
Sample Exercise 1.13
Practice Exercise 2: 832 g
Chapter 3
Chapter 2
Sample Exercise 3.1
Practice Exercise 2: (a) C2H4 + 3 0 2 -------'> 2 CO2 + 2 H20.
(b ) Nine 02 molecules
Sample Exercise 2.1
Practice Exercise 2: (a) 154 pm,
(b ) 1.3 X 106 C atoms
Sample Exercise 3.2
Practice Exercise 2: (a) 4, 3, 2;
(b) 2, 6, 2, 3;
Sample Exercise 2.2
Practice Exercise 2: (a) 56 protons, 56 electrons, and 82 neutrons,
(b ) 15 protons, 15 electrons, and 16 neutrons.
Sample Exercise 3.3
Practice Exercise 2: (a) HgS (s)
(b) 4 Al (s) + 3 02 (g)
2
Sample Exercise 2.3
Practice Exercise 2:
Sample Exercise 3.4
Practice Exercise 2: C2H50H (I) + 3 0 2(g)
Sample Exercise 2.4
Practice Exercise 2: 28 .09 amu
Sample Exercise 3.5
Practice Exercise 2: (a) 78 .0 amu,
Sample Exercise 2.5
Practice Exercise 2: Na, atomic number 11, is a metal; Br, atomic number 35, is a nonmetal.
Sample Exercise 3.6
Practice Exercise 2: 16.1%
(c) 1, 2, 1, 1, 1
Hg (/) + S(s), or k
2 C02 (g) + 3 H20 (g)
(b) 32.0 amu,
(c) 211 .0 amu
1215
1216
Sample Exercise 3. 7
Practice Exercise 2: 1 mol H20 (6 X 1023 0 atoms) < 3 X 1023 molecules 03 (9 x 1023 0 atoms) < 1 mol C02(12 X 1023 0 atoms)
Sample Exercise 3.8
Practice Exercise 2: (a) 9.0
X
1023,
(b) 2.71
1024
X
Sample Exercise 3.9
Practice Exercise 2: 164.1 g/ mol
Sample Exercise 4.8
Practice Exercise 2: (a) +5,
Sample Exercise 3.10
Practice Exercise 2: 55.5 mo! H20.
Sample Exercise 3.11
Practice Exercise 2: (a) 6.0 g,
Sample Exercise 4. 7
Practice Exercise 2:
(a) H3P03 (aq) + 2KOH (aq) 2H20(1) + K2HP03 (aq),
(b) H3P03 (aq) + 2 OW (aq)
2 H20(1) + HPO/-. (H3P03 is
a weak acid and therefore a weak electrolyte, whereas KOH, a strong
base, and K3P03, an ionic compound, are strong electrolytes. In spite
of its formula, H3P03 is a diprotic acid, better written as
(b) 8.29 g.
Sample Exercise 4.10
Practice Exercise 2: Zn and Fe
Sample Exercise 3.13
Practice Exercise 2: C4H40
Sample Exercise 4.11
Practice Exercise 2: 0.278 M
Sample Exercise 3.15
Practice Exercise 2: (a) C3HGO,
Sample Exercise 4.13
Practice Exercise 2: (a) 1.1 g,
(b) C6H1202
Sample Exercise 3.17
Practice Exercise 2: 26 .S g
Sample Exercise 3.19
Practice Exercise 2: (a) AgN03,
(d) 1.52gZn
(c) 0.7S mo1 Cl2
(b ) 1.59 g,
Sample Exercise 3.20
Practice Exercise 2: (a) 105 g Fe,
(c) 1.39 g,
(b ) 5.0 mL,
Sample Exercise 4.15
Practice Exercise 2: (a) 0.240 g, assuming you only need to neutralize
one proton per sulfuric acid molecule, (b) 0.400 L
Sample Exercise 4.16
Practice Exercise 2: 0.210 M
Sample Exercise 4.17
Practice Exercise 2: (a) 1.057 X 10-3 mol Mn04- ,
(b) 5.286 x 10- 3 mol Fe2+, (c) 0.2952g, (d) 33.21%
(b ) 83.7%
Chapter 5
Chapter 4
Sample Exercise 4.1
Practice Exercise 2: (a) 6,
(e) -1
(b ) 76 mL
Sample Exercise 4.14
Practice Exercise 2: (a) 0.0200 L = 20.0 mL,
(c) 0.40M
(b ) 1.SO mol,
(d) +4,
Sample Exercise 4.12
Practice Exercise 2: 0.030 M
(b) C2HG02
Sample Exercise 3.16
Practice Exercise 2: 1. 77 g
Sample Exercise 3.18
Practice Exercise 2: (a) Al,
(c) +6,
Sample Exercise 4.9
Practice Exercise 2:
(a) Mg(s) + CoS04 (aq) MgS04 (aq) + Co (s);
Mg(s) + Co2+( aq) Mg2+ (aq) + Co (s),
(b ) Mg is oxidized and Co2+ is reduced.
Sample Exercise 3.12
Practice Exercise 2: (a) 4.01 X 1022 molecules HN03,
(b ) 1.20 X 1023 atoms 0
Sample Exercise 3.14
Practice Exercise 2: (a) CH30,
(b ) - 1,
(b) 12,
(c) 2,
Sample Exercise 5.1
Practice Exercise 2: +55 J
(d) 9
Sample Exercise 4.2
Practice Exercise 2: (a) insoluble,
(b ) soluble,
Sample Exercise 4.3
Practice Exercise 2: (a) Fe (OHh,
(b)
+ 6 LiOH (aq)
2 Fe(OH)3 (s) + 3 Li2S04(aq)
Sample Exercise 4.4
Practice Exercise 2: 3 Ag+( aq) + POi- (aq)
(c) soluble
Ag3P04(s)
Sample Exercise 4. 5
Practice Exercise 2: The diagram would show 10 Na+ ions, 2 OH- ions,
8 y- ions, and 8 H20 molecules.
Sample Exercise 4.6
Practice Exercise 2: C6H606 (nonelectrolyte) < CH3COOH (weak
electrolyte, existing mainly in the form of molecules with few ions)
< NaCH3COO (strong electrolyte that provides two ions, Na+ and
CH3COO-) <
(strong electrolyte that provides three ions,
Ca2+ and 2Non
Sample Exercise 5.2
Practice Exercise 2: 0.69 L-atm = 70J
Sample Exercise 5.3
Practice Exercise 2: In order to solidify, the gold must cool to below its
melting temperature. It cools by transferring heat to its surroundings.
The air around the sample would feel hot because heat is transferred
to it from the molten gold, meaning the process is exothermic . (YOU
may notice that solidification of a liquid is the reverse of the melting
we analyzed in the exercise. As we will see, reversing the direction of a
process changes the sign of the heat transferred.)
Sample Exercise 5.4
Practice Exercise 2: -14 .4 kJ
Sample Exercise 5.5
Practice Exercise 2: (a) 4.9 x 105J,
(b) 11 K decrease = 11 °C decrease
Sample Exercise 5.6
Practice Exercise 2: - 68,000J/mo! = -68 kJ / mol
1217
Sample Exercise 5. 7
Practice Exercise 2: (a) -15.2 kJ / g,
- 1370 kJ/mol
(b)
Sample Exercise 7. 5
Practice Exercise 2: Ca
Sample Exercise 5.8
Practice Exercise 2: +1.9 kJ
Sample Exercise 7.6
Practice Exercise 2: Al lowest, C highest
Sample Exercise 5.9
Practice Exercise 2: -304.1 kJ
Sample Exercise 7. 7
Practice Exercise 2: (a) [Ar]3d 10,
Sample Exercise 5.10
Practice Exercise 2: C(graphite) + 2 Cl2(g)
!:iHJ = -l06.7 k] / mol
---+
CCl4(/) 1
Sample Exercise 7.8
Practice Exercise 2: CuO (s) + H2S04(aq)
Sample Exercise 7. 9
Practi ce Exercise 2:
Sample Exercise 5.11
Practice Exercise 2: -1367 kJ
+ 6 H20(1)
Sample Exercise 7.10
Practice Exercise 2: 2K(s) + S(s)
Sample Exercise 5.12
Practice Exercise 2: -156.1 kJ / mol
Sample Exercise 5.13
Practice Exercise 2: - 1255 kJ
(b ) [Ar]3d3,
CuS04(aq) + H20(1)
4 H3P03 (aq)
—---->
K2S (s)
------->
Sample Exercise 8.1
Practice Exercise 2: Zr02
(b ) lOO min
Sample Exercise 8.2
Practice Exercise 2: Mg2 + and N3-
Chapter 6
Sample Exercise 6.1
Practice Exercise 2: The expanded visible-light portion of Figure 6.4
tells you that red light has a longer wavelength than blue light.
Sample Exercise 6.2
Practice Exercise 2: (a) 1.43 x 1014 s-1,
Sample Exercise 6.3
Practice Exercise 2: (a ) 3.11 x 10- 19J,
(c) 4.2 X 1016photons
(b) 2.899 m
(b ) 0.16J,
Sample Exercise 6.4
Practice Exercise 2: (a) 1::,.£ < 0, photon emitted,
ton absorbed
Sample Exercise 6.5
Practice Exercise 2: 7.83 X
m/ s
Sample Exercise 6.6
Practice Exercise 2: (a) Sp;
(b ) 3;
Sample Exercise 8.3
Practice Exercise 2: They both show 8 valence electrons; methane has
4 bonding pairs and neon has 4 nonbonding pairs
Sample Exercise 8.4
Practice Exercise 2: Se-Cl
Sample Exercise 8.5
Practice Exercise 2: (a) F,
(b ) 0.11-
Sample Exercise 8.6
Practice Exercise 2: (a) 20,
(b)
(b) t:,.E > 0, pho-
Sample Exercise 8. 7
Practice Exercise 2: (a ) [:N=::: O :]+,
(b) H
H
(c) 1, 0, - 1
Sample Exercise 6. 7
Practice Exercise 2: (a )
(b) two
Sample Exercise 8.8
Practice Exercise 2: (a)
Sample Exercise 6.9
Practice Exercise 2: (a ) [Ar)4s23d7 or [Ar )3d74s2,
(b ) [ Kr )5s24d10Sp1
[Kr )4d1 0Ss2Sp1
Chapter 7
Sample Exercise 7.1
Practice Exercise 2: P-Br
Sample Exercise 7.2
Practice Exercise 2: C < Be < Ca < K
\
/
C= C
/
\
—
[:o-d
.. .. 5:
.. ] (b)
Sample Exercise 6.8
Practice Exercise 2: group 4A
Sample Exercise 7.4
Practice Exercise 2: cs+
------->
Chapter 8
Sample Exercise 5.14
Practice Exercise 2: (a) 15 kJ / g,
Sample Exercise 7.3
Practice Exercise 2: s2-
(c) [Ar]4s2 3d10 4p6
H
H
[\ ]
3-
Sample Exercise 8.9
Practice Exercise 2:
(a)
-2
O
+1
-1
0
0
0
(ii)
(i)
0
-1
(iii)
(b ) Structure (iii), which places a negative charge on oxygen, the most
electronegative element in the ion, is the dominant Lewis structure.
Sample Exercise 8.10
PractIce Exercise 2: IH
—,=\
:Q:
I
-
:o:
1218
Sample Exercise 8.11
Practice Exercise 2: (a) C,
(b) :
Sample Exercise 10.11
Practice Exercise 2: 1.0 X 103 torr N2, 1. 5
:
Sample Exercise 10.12
Practice Exercise 2: (a) increases,
Chapter 9
Sample Exercise 9.1
Practice Exercise 2: (a) tetrahedral, bent;
nalplanar
(b) trigonal planar, trigo-
Sample Exercise 9.2
Practice Exercise 2: (a) trigonal bipyramidal, T-shaped;
nal bipyramidal, trigonal bipyramidal.
Sample Exercise 9.3
Practice Exercise 2: 109.5°, 180°
Sample Exercise 9.4
Practice Exercise 2: (a) polar because polar bonds are arranged in a
seesaw geometry, (b ) nonpolar because polar bonds are arranged in
a tetrahedral geometry
Sample Exercise 9.5
Practice Exercise 2: tetrahed ral, sp3
Sample Exercise 9.6
Practice Exercise 2: (a) approximately 109° around the left C and
180° around the right C; (b) sp3, sp; (c) five u bonds and two 'TT
bonds
Sample Exercise 9. 7
Practice Exercise 2: S02 and S03, as indicated by the presence of two
or more resonance structures invo lving 'TT bonding for each of these
molecules.
Sample Exercise 9.8
l
Practi
ractice Exercise 2:, 61s2UI*1.
S, 2
Sample Exercise 9.9
Practice Exercise 2: (a) diamagnetic, 1;
Chapter 10
Sample Exercise 10.1
Practice Exercise 2: (d) 1.6 M
Sample Exercise 10.2
Practice Exercise 2: 807.3 torr
Sample Exercise 10.3
Practice Exercise 2: 5.30 x 103L
Sample Exercise 10.4
Practice Exercise 2: 2.0 atrn
Sample Exercise 10.5
Practice Exercise 2: 3.83 X 103 rn3
Sample Exercise 10.6
Practice Exercise 2: 27 °C
(b) diamagnetic, 3
102 torr Ar, and 73 torr CH4
(b ) no effect,
(c) no effect
Sample Exercise 10.13
Practice Exercise 2: 1.36 x 103 m /s
Sample Exercise 10.14
Practice Exercise 2: rN,/ro,
(b ) trigo-
X
= 1.07
Sample Exercise 10.15
Practice Exercise 2: (a) 7.472 atm, (b) 7.181 atm
Chapter 11
Sample Exercise 11.1
Practice Exercise 2: chloramine, NHCI
Sample Exercise 11.2
Practice Exercise 2: (a ) CH3CH3 has only dispersion forces, whereas
the other two substances have both dispersion forces and hydrogen
bonds, (b ) CH3CH20H
Sample Exercise 11.3
Practice Exercise 2: -20.9 kJ - 33.4 kJ - 6.09 kJ
=
-60.4 kJ
Sample Exercise 11.4
Practice Exercise 2: about 340 torr (0.45 atm)
Sample Exercise 11.5
Practice Exercise 2: (a) -162 °C; (b) It sublimes whenever the pressure is less than 0.1 atm; (c) The highest temperature at which a
liquid can exist is defined by the critical temperature. So we do not
expect to find liquid methane when the temperature is higher than
-80°C.
Sample Exercise 11.6
Practice Exercise 2: Because rotation can occur about carbon-carbon
single bonds, molecules whose backbone consists predominantly of
C C single bonds are too flexible; the molecules tend to coil in random ways and, thus, are not rod like.
—
Chapter 12
Sample Exercise 12.1
Practice Exercise 2: 0.68 or 68%
Sample Exercise 12.Z
Practice Exercise 2: a = 4.02 Aand density
=
4.31 g/cm3
Sample Exercise 12.3
Practice Exercise Z: smaller.
Sample Exercise 12.4
Practice Exercise 2: A group SA element could be used to replace Se.
Sample Exercise 10. 7
Practice Exercise 2: 5 .9 g/L
Chapter 13
Sample Exercise 10.8
Practice Exercise 2: 29 .0 g/ mol
Sample Exercise 13.1
Practice Exercise 2: C5H12 < C5H11Cl < CsHuOH < C5H10(0H)z
(in order of increasing polarity and hydrogen-bonding ability)
Sample Exercise 10. 9
Practice Exercise 2: 14.8 L
Sample Exercise 10.10
Practice Exercise 2: 2.86 atm
Sample Exercise 13.2
Practice Exercise 2: 1.0 X 10-5 M
1219
Sample Exercise 13.3
Practice Exercise 2: 90 .5 g of NaOCI
Sample Exercise 14.13
Practice Exercise 2: (a) Rate =
termolecular reactions are very rare.
Sample Exercise 13.4
Practice Exercise 2: 0.670 m
Sample Exercise 13.5
Practice Exercise 2: (a) 9.00
X
10-3,
Sample Exercise 13.6
Practice Exercise 2: (a) 10.9 m,
(c) 5.97M
(b) No,
because
Sample Exercise 14.14
Practice Exercise 2: Because the rate law conforms to the molecularity
of the first step, the first step must be the rate-determining step. The
second step must be much faster than the first one.
(b) O.SOS m
Sample Exercise 14.15
(b) Xc, H,o, = 0.163,
Practice Exercise 2: [Br] = (t[Br2] )112
Sample Exercise 13. 7
Practice Exercise 2: 0.290
Chapter 15
Sample Exercise 13.8
Practice Exercise 2: -65.6 °C
Sample Exercise 15.1
Sample Exercise 13.9
Practice Exercise 2: 0 .048 atm
Practice Exercise 2: (a) Kc =
Sample Exercise 13.10
Practice Exercise 2: 110 g/mol
Sample Exercise 15.2
Practice Exercise 2: 0.335
Sample Exercise 13.11
Practice Exercise 2: 4.20 X 104 g/mol
Sample Exercise 15.4
Sample Exercise 14.1
Practice Exercise 2: 1.8
Practi
ractice Exercise 2:
10- 2 M/s
Sample Exercise 14.2
Practice Exercise 2: 1.1 x 10-4 M/s
Sample Exercise 14.3
Practice Exercise 2: (a) 8.4
Sample Exercise 15.S
C
10- 7 M/s,
(b) 2.1 X 10- 7 M/s
Sample Exercise 14.4
Practice Exercise 2: 2 = 3 < 1
Sample Exercise 14.5
Practice Exercise 2: (a) 1,
Sample Exercise 14.6
Practice Exercise 2: (a) rate
(c) rate = 4.5 X 10- 4 M/s
[CdBr42- ]
[cct2+J[Bc]4
(54.0)3
= 1.51 X 109
1.04 X 10- 4
Practice Exercise
ise 2: (a)
( Kc =
X
(b) Kc =
Sample Exercise 15.3
Practice Exercise 2: At the lower temperature because Kp is larger at
the lower temperature
Chapter 14
X
[HI]2
C
[Ag+]3 '
(b) Kp =
(PH2o)4
Sample Exercise 15.6
Practice Exercise 2: H2 (g )
Sample Exercise 15. 7
Practice Exercise 2: 1.79 x 10- 5
(b ) M- 1s- 1.
=
[Cr3+]
Sample Exercise 15.8
Practice Exercise 2: 33
(b) k
=
l.2 M—2
Sample Exercise 14. 7
Practice Exercise 2: 51 torr
Sample Exercise 15.10
Practice Exercise 2: 1.22 atm
Sample Exercise 14.8
Practice Exercise 2: [N02] = 1.00 X 10- 3 M
Sample Exercise 14. 9
Practice Exercise 2: (a) 0.478 yr = 1.51 X 107 s,
half-lives, 2(0.478 yr) = 0.956 yr
Sample Exercise 15.9
Practice Exercise 2: Qp = 16; Qp > Kp, and so the reaction will proceed from right to left, forming more S03.
Sample Exercise 15.11
Practice Exercise 2: Pre,,
(b) it takes two
Sample Exercise 14.10
Practice Exercise 2: 2 < 1 < 3 because, if you approach the barrier
from the right, the Ea values are 40 kJ/mo! for reverse reaction 2,
25 kJ/mo! for reverse reaction 1, and 15 kJ/mo! for reverse reaction 3.
Sample Exercise 14.11
Practice Exercise 2: 3 X 1013 s- 1
Sample Exercise 14.12
Practice Exercise 2: (a) Yes, the two equations add to yield the equation for the reaction. (b) The first elementary reaction is unimolecular, and the second one is bimolecular. (c) Mo (C0)5.
=
0.967 atm, Pre,,
Sample Exercise 15.12
Practice Exercise 2: (a) right,
(b) left,
=
Pc,,
=
(c) right,
0.693 atm
(d) left
Chapter 16
Sample Exercise 16.1
Practice Exercise 2: H2S03, HF, HPO/- , HCO+
Sample Exercise 16.2
PracticeExercise2:02 -( aq) + H20(1)
OH-( aq) + OH-(aq).The
OH- is both the conjugate acid of 02- and the conjugate base of H20.
Sample Exercise 16.3
Practice Exercise 2: (a) left,
(b) right
1220
Sample Exercise 16.4
Practice Exercise 2: (a) basic,
(b) neutral,
Sample Exercise 16.5
Practice Exercise 2: (a) 5 X 10-9 M,
(c) 7.1 X 10- 9 M
Sample Exercise 16.6
Practice Exercise 2: (a) 3.42,
Sample Exercise 16. 7
Practice Exercise 2:
X
=
5.3 x 10- 9 M, so pH
= 8.28
10-lO
(b) 2.4 x 10-3 M
Sample Exercise 16.10
Practice Exercise 2: 1. 5 x 10-s
Sample Exercise 17.9
Practice Exercise 2: (a) 8.21,
(b) 5.28
= 5.0 X
= 1.s x
10-9,
Sample Exercise 17.11
Practice Exercise 2: 1.6 x 1o- 12
mol/ L
Sample Exercise 17.13
Practice Exercise 2: 4.0 X 10- 10 M
Sample Exercise 16.12
Practice Exercise 2: 3 .41
Sample Exercise 16.13
Practice Exercise 2: (a) 3.9%,
Sample Exercise 16.14
Practice Exercise 2: (a) pH
(b) 9.26
Sample Exercise 17.12
Practice Exercise 2: 9 x
Sample Exercise 16.11
Practice Exercise 2: 2.7%
(b) 3.70
Sample Exercise 17.8
Practice Exercise 2: (a) 4.20,
Sample Exercise 17.10
Practice Exercise 2:
(a) K,p =
(b) K,p =
Sample Exercise 16.8
Practice Exercise 2: 0 .0046 M
Sample Exercise 16.9
Practice Exercise 2: (a) 7.8 X 10-3 M,
(b) 1.70
Sample Exercise 17. 7
Practice Exercise 2: (a) 10.30,
(b) 1.0 X 10-7 M,
(b) [ W ]
= 6.6
Sample Exercise 17.6
Practice Exercise 2: (a) 4.68,
(c) acidic
=
Sample Exercise 17.14
Practice Exercise 2:
(a) CuS (s) + H+(aq)
Cu2+ (aq) + HS-(aq) ,
(b) Cu(N3)z(s) + 2W (aq)
Cu2 + (aq) + 2HN3 (aq)
== ==
(b) 12%
1.80,
(b) [ C20/-]
=
6.4 X 10-5 M
Sample Exercise 17.15
Practice Exercise 2: 1 x 10- 16 M
Sample Exercise 16.15
Practice Exercise 2: Methylamine (because it has the larger Kb value of
the two amine bases in the list)
Sample Exercise 17.16
Practice Exercise 2: Yes, CaF2 precipitates because Q
larger than K,0
sp = 3 .9 X
Sample Exercise 16.16
Practice Exercise 2: 0.12 M
Sample Exercise 17.17
Practice Exercise 2: Cu (OH)z precipitates first, beginning when
> 1.5 x 10-9 M. Mg(OH)2 begins to precipitate when
[OW] > 1.9 X 10-5 M.
Sample Exercise 16.17
Practice Exercise 2: (a) Po/- (Kb = 2.4 x 10-2 ),
(b) & = 7.9 X 10-10
Sample Exercise 16.18
Practice Exercise 2: (a)
(d) NH4N03
(b) KBr,
(c) CH3NH3Cl,
Sample Exercise 16.19
Practice Exercise 2: acidic
Sample Exercise 16.20
Practice Exercise 2: (a) HBr,
(b) H2S,
(c) HN03,
(d) H2S03
10-7 is
Chapter 18
Sample Exercise 18.1
Practice Exercise 2: 3.0 X
torr
Sample Exercise 18.3
2.0 X
g CO2 or 20 kg CO2
Chapter 19
Sample Exercise 17.1
Practice Exercise 2: 3 .42
=
9.0 x 10-s, pH
Sample Exercise 17.3
Practice Exercise 2: 4.50
Sample Exercise 17.4
Practice Exercise 2: 4.62 (using quadratic) .
Sample Exercise 17.S
Practice Exercise 2: 0.13 M
X
Sample Exercise 18.2
Practice Exercise 2: 127 nm
Chapter 17
Sample Exercise 17.2
Practice Exercise 2: [HCoo-J
= 4.6
=
1.00
Sample Exercise 19.1
Practice Exercise 2: No, the reverse process is spontaneous at this
temperature.
Sample Exercise 19.2
Practice Exercise 2: - 163 J / K
Sample Exercise 19.3
Practice Exercise 2: no
Sample Exercise 19.4
Practice Exercise 2: (a) 1 mo! of S02(g) at SIP,
atSTP
(b) 2 mol of N02(g)
1221
Sample Exercise 19.5
Practice Exercise 2: 180.39J / K
Sample Exercise 19.6
Practice Exercise 2: t:,.G0 = -14 .7 kJ; the reaction is spontaneous.
Sample Exercise 19. 7
Practice Exercise 2: -800.7 kJ
Sample Exercise 20.14
Practice Exercise 2: (a) 30.2 g of Mg,
=
(b) 3.97 x 103 s
Chapter 21
Sample Exercise 19.8
Practice Exercise 2: More negative
Sample Exercise 19.9
Practice Exercise 2:
(a) M-f° = - 196.6 kJ, t:,.S0 = -189.6 J / K;
Sample Exercise 20.13
Practice Exercise 2: (a) The anode is the half-cell in which [Zn2+]
3.75 x 10- 4 M. (b) The emf is 0.105 V.
Sample Exercise 21.1
Practice Exercise 2:
(b) t:,.G0
= -120.8 kJ
Sample Exercise 19.10
Practice Exercise 2: 330 K
1~N + +?e
Sample Exercise 21.3
Practice Exercise 2: (a) a emission,
Sample Exercise 19.11
Practice Exercise 2: -26.0 kJ/mo!
Sample Exercise 19.12
Practice Exercise 2: t:,.G0 = -106.4 kJ / mo!, K
Sample Exercise 21.2
Practice Exercise 2: 1i O
(b)
/3- emission
Sample Exercise 21.4
Practice Exercise 2: 1go (p, a)
=
4 x 1018
Sample Exercise 21.5
Practice Exercise 2: 3 .12%
Chapter 20
Sample Exercise 21.6
Practice Exercise 2: 2200 yr
Sample Exercise 20.1
Practice Exercise 2: Al(s) is the reducing agent; Mn04- (aq) is the oxidizing agent.
Sample Exercise 21. 7
Practice Exercise 2: 15 .1 %
Sample Exercise 20.2
Practice Exercise 2: Cu(s) + 4 H +(aq) + 2 N03- (aq) ____,,
Cu2 + (aq) + 2 N02(g) + 2 H20(/)
Sample Exercise 20.3
Practice Exercise 2: 2 Cr(OH)J (s) + 6 c 10- (aq) ——>
2 Cro/- (aq) + 3 Cl2(g) + 2 01-r (aq) + 2 H20(1)
Sample Exercise 20.4
Practice Exercise 2: (a) The first reaction occurs at the anode and
the second reaction at the cath ode. (b) Zinc is oxidized to form
Zn2+ as the reaction proceeds and therefore the zinc electrode loses
mass. (c) The platinum electrode is not involved in the reaction
and none of the products of the reaction at the cathode are solids, so
the mass of platinum electrode does not change. (d) The platinum
cathode is positive.
Sample Exercise 21.8
Practice Exercise 2: -3.19 X 1o- 3 g
Chapter 22
Sample Exercise 22.1
Practice Exercise 2: (a) Cs,
Sample Exercise 22.2
Practice Exercise 2: Nal--l (s ) + H20(/)
Sample Exercise 20.6
Practice Exercise 2: 2.20 V
Sample Exercise 22.6
Practice Exercise 2: 6-
Sample Exercise 20.8
Practice Exercise 2: Al(s) > Fe(s ) >
Sample Exercise 20.9
Practice Exercise 2: Reactions (b) and (c) are spontaneous.
Sample Exercise 20.10
Practice Exercise 2: + 77 kJ / mol
Sample Exercise 20.11
Practice Exercise 2: It would increase.
Sample Exercise 20.12
Practice Exercise 2: pH = 4.23 (u sing data from Appendix E to o btain
E0 to three significant figures)
-->
Sample Exercise 22.4
Practice Exercise 2: 2 Br- (aq) + Cl2(aq)
Sample Exercise 22.5
Practice Exercise 2:
4 C02(g) + 12 I--120 (g)
(b) +0.50 V
(c) C,
(d) Sb
Na OI-I(aq) + I-I2 (S)
Sample Exercise 22.3
Practice Exercise 2: trigonal bipyramidal, linear
Sample Exercise 20.5
Practice Exercise 2: -0 .40 V
Sample Exercise 20. 7
Practice Exercise 2: (a) Co ____,, Co2 + + 2 e-;
(b) Cl,
——•
Br2(aq) + 2 Ci-(aq)
5 N204(/) + 4
------->
9 N2 (g) +
Chapter 23
Sample Exercise 23.1
Practice Exercise 2: th ree : [ Co (l-120)6)2 + and two c 1Sample Exercise 23.2
Practice Exercise 2: zero
Sample Exercise 23.3
Practice Exercise 2: (a) triamminetribromomolybdenum(IV) nitrate,
(b) ammonium tetrabromocuprate(ll), (c)
Sample Exercise 23 .4
Prac;tice Exercise 2: two
1222
Sample Exercise 23.5
Practice Exercise 2: No, because the complex is flat. This co mplex ion
does, however, have geometric isomers (for example, the Cl and Br
ligands could be cis or tran s).
Sample Exercise 23.6
Practice Exercise 2: Green, beca use it absorbs the green's complimentary color, red.
Sample Exercise 23. 7
Practice Exercise 2: (Co(NH3)5Cl]2 + is purple which means it must
absorb light near the boundary between the yellow and green regions
of the spectrum . (Co (NH3)6]3+ is orange which means it absorbs blue
light. Because yellow-green photons have a lower energy (longer
wavelength) than blue photons, !:.,. is smaller fo r [Co (NH3)5Cl]2+.
This prediction is consistent with the spectrochemical series because
c i- ligands are at the low end of the spectrochemical series. Thus replacing NH3 with c 1- should make t.,. smaller.
Sample Exercise 23 .8
Practice Exercise 2: No it is not possible to have a diamagnetic co mplex with partially filled d orbitals if the complex is high spin, and all
tetrahedral complexes are high spin.
Chapter 24
Sample Exercise 24.1
Practice Exercise 2: 2,4-dirnethylpentane
Sample Exercise 24.2
Practice Exercise 2:
CH3 CH3
I - I
CH3CH
—
CHCH2CH2CH3
or
CH3CH(CH3)CH(CH3)CH2CH2CH3
Sample Exercise 24.3
Practice Exercise 2: five (1-hexene, cis-2-hexene, trans-2-hexene,
cis-3- hexene, trans-3-hexene)
Sample Exercise 24.4
Practice Exercise 2:
Sample Exercise 24.S
Practi ce Exercise 2: propene
Sample Exercise 24.6
Practice Exercise 2:
?
Sample Exercise 24. 7
Practice Exercise 2: serylaspartic acid; Ser-Asp, SD
Sample Exercise 24.8
Practice Exercise 2: alcohol, ether