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4 Avogadro's Number and the Mole

4 Avogadro's Number and the Mole

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ANSWERS TO GO FIGURE

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 more

than one type of atom, and molecules of an element are composed of

only one type of atom. Figure 1.6 Earth is rich in silicon and

aluminum; the human body is rich in carbon and hydrogen

Figure 1.7 The relative volumes are in direct proportion t to the

number of molecules Figure 1.14 The separations are due to

physical processes of adsorption of the materials onto the column

Figure 1.17 True Figure 1.18 1000 Figure 1.22 The darts would

be scattered widely (poor precision) but their average position would

be at the center (good accuracy).



Figure 5.1 In the act of throwing, the pitcher transfers energy to

the ball, which then becomes kinetic energy of the ball. For a given

amount of energy E transferred to the ball, Equation 5.1 tells us that

the speed of the ball is v = 22E>m where m is the mass of the ball.

Because a baseball has less mass than a bowling bowl, it will have a higher

speed for a given amount of energy transferred. Figure 5.2 When she

starts going uphill, kinetic energy is converted to potential energy and

her speed decreases. Figure 5.3 The electrostatic potential

energy of two oppositely charged particles is negative (Equation 5.2). As

the particles become closer, the electrostatic potential energy becomes

even more negative—that is, it decreases. Figure 5.4 Yes, the system

is still closed—matter can’t escape the system to the surroundings unless the piston is pulled completely out of the cylinder. Figure 5.5 If

Efinal = Einitial, then ΔE = 0. Figure 5.6



Figure 2.3 We know the rays travel from the cathode because

the rays are deflected by the magnetic field as though coming from

the left. Figure 2.4 The electron beam would be deflected downward because of repulsion 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 The beta rays,

whose path is diverted away from the negative plate and toward the

positive plate, consist of electrons. Because the electrons are much

less massive than the alpha particles, their motion is affected more

strongly by the electric field. Figure 2.9 The beam consists of alpha

particles, which carry a + 2 charge. 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 fitting this pattern are H and Cl. Figure 2.15 The

metals are in the form of solid blocks, or relatively large pieces, as opposed to powders. They have a metallic sheen that is missing from

the nonmetals. The nonmetals more readily form powders, and are

more varied in color and consistency than metals. Figure 2.17 The

ball-and-stick model. Figure 2.18 The elements are in the following

groups: Ag + is 1B, Zn2+ is 2B, and Sc3+ is 3B. Sc3+ has the same number of electrons as Ar (element 18). Figure 2.21 It is a difference in

a physical property, color. Figure 2.22 Removing one O atom from

the perbromate ion gives the bromate ion, BrO3 - .



Chapter 3

Figure 3.4 There are two CH4 and four O2 molecules on the

reactant side, which contain 2 C atoms, 8 H atoms and 8 O atoms.

The number of each type of atom remains the same on the product

side as it must. Figure 3.8 The flame gives off heat and therefore

the reaction must release heat. Figure 3.9 As shown, 18.0 g

H2O = 1 mol H2O = 6.02 * 1023 molecules H2O. Thus, 9.00 g

H2O = 0.500 mol H2O = 3.01 * 1023 molecules H2O. Figure 3.12

(a) The molar mass of CH4, 16.0 g CH4/1 mol CH4. (b) Avogadro’s

number, 1 mol CH4 >6.02 * 1023 formula units CH4, where a formula

unit in this case is a molecule. Figure 3.17 If the amount of H2 is

doubled then O2 becomes the limiting reactant. In that case

17 mol O22 * 12 mol H2O>1 mol O22 = 14 mol H2O would be

produced.



Chapter 4

Figure 4.3 NaCl(aq) Figure 4.4 K+ and NO3- Figure 4.9 Two

moles of hydrochloric acid are needed to react with each mole of

Mg1OH22. Figure 4.12 Two. Each O atom becomes an O2- ion.

Figure 4.13 One, based on the reaction stoichiometry. Figure 4.14

Because the Cu(II) ion produces a blue color in aqueous solution.

Figure 4.18 The volume needed to reach the end point if Ba1OH221aq2

were used would be one-half the volume needed for titration with

NaOH(aq), because there are two hydroxide ions for every barium ion.



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Final

state

Internal energy, E



Chapter 2



Mg(s) + Cl2(g)



ΔE > 0



Initial

state



MgCl2(s)



Figure 5.7 ΔE = 50 J + (-85 J) = -35 J

Figure 5.10 The battery is doing work on the surroundings, so

w 6 0. Figure 5.11 We need to know whether Zn(s) or HCl(aq)

is the limiting reagent of the reaction. If it is Zn(s), then the addition of more Zn will lead to the generation of more H21g2 and

more work will be done. Figure 5.16 Endothermic—heat is being

added to the system to raise the temperature of the water. Figure

5.17 Two cups provide more thermal insulation so less heat will

escape the system. Figure 5.18 The stirrer ensures that all of the

water in the bomb is at the same temperature. Figure 5.20 The

condensation of 2 H2O1g2 to 2 H2O1l2 Figure 5.21 Yes, ΔH3

would remain the same as it is the enthalpy change for the process

CO1g2 + 12 O21g) ¡ CO21g2. Figure 5.23 Grams of fat



Chapter 6

Figure 6.3 Wavelength = 1.0 m, frequency = 3.0 * 108 cycles>s.

Figure 6.4 Longer by 3 to 5 orders of magnitude (depending on what

part of the microwave spectrum is considered). Figure 6.5 The

hottest area is the white or yellowish white area in the center.

Figure 6.7 The energy comes from the light shining on the surface. Figure 6.12 The n = 2 to n = 1 transition involves a larger

energy change than the n = 3 to n = 2 transition. (Compare the

space differences between the states in the figure.) If the n = 3 to

n = 2 transition produces visible light, the n = 2 to n = 1 transition

must produce radiation of greater energy. The infrared radiation has

lower frequency and, hence, lower energy than visible light, whereas

ultraviolet radiation has higher frequency and greater energy. Thus,

the n = 2 to n = 1 transition is more likely to produce ultraviolet

radiation. Figure 6.13 The n = 4 to n = 3 transition involves a

smaller energy difference and will therefore emit light of longer

wavelength. Figure 6.17 The region of highest electron density is

where the density of dots is highest, which is near the nucleus.



Answers to Go Figure

Figure 6.18 The fourth shell 1n = 42 would contain four subshells,

labeled 4s, 4p, 4d, and 4f. Figure 6.19 There would be four maxima

and three nodes. Figure 6.23 (a) The intensity of the color indicates

that the probability of finding the electron is greater at the interior of

the lobes than on the edges. (b) 2px. Figure 6.24 The dz2 orbital has

two large lobes that look like those of a p orbital, but also has a “doughnut” around the middle. Figure 6.25 The 4d and 4f subshells are not

shown. Figure 6.26 In this pictorial view of spin, there are only two

directions in which the electron can spin 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 Yes, because of the peak near the nucleus in the 2s curve there is a probability,

albeit small, that a 2s electron will be closer to the nucleus than a 1s

electron. Figure 7.7 Bottom and left Figure 7.8 They get larger,

just like the atoms do. Figure 7.10 900 kJ>mol 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 5A 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. 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 of the line. Figure 7.16 No. The Na+ and NO3ions will simply be spectator ions. The H + 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 in the periodic table, 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 H21g2.

This could be confirmed by carefully testing the bubbles with a

flame—there should be popping as the hydrogen gas ignites.

Figure 7.26 Water does not decompose on sitting the way that

hydrogen peroxide does. Figure 7.27 A regular octagon.

Figure 7.28 I2 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 molecules

(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 8.2 Yes, the

same sort of reaction should occur between any of the alkali metals

and any of the elemental halogens. Figure 8.3 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 Cl- , so we would infer that

the larger green spheres represent the chloride ions and the smaller

purple spheres represent the sodium ions. Figure 8.4 The distance

between ions in KF should be larger than that in NaF and smaller

than that in KCl. We would thus expect the lattice energy of KF to be

between 701 and 910 kJ/mol. Figure 8.6 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 8.7 The electronegativity decreases

with increasing atomic number. Figure 8.9 m will decrease

Figure 8.10 The bonds are not polar enough to cause enough excess

electron density on the halogen atom to lead to a red shading.

Figure 8.12 The lengths of the bonds of the outer O atoms to the

inner O atom are the same. Figure 8.13 Yes. The electron densities

on the left and right parts of the molecule are the same, indicating

that resonance has made the two O ¬ O bonds equivalent to one

another. Figure 8.14 The dashed bonds represent the “half bonds”



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that result when the two resonance structures are averaged.

Figure 8.15 Exothermic Figure 8.17 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 of the atoms involved (see Section 7.3).

Figure 9.3 Octahedral. Figure 9.4 No. We will get the same bentshaped geometry regardless of which two atoms we remove.

Figure 9.7 The electron pair in the bonding domain is attracted to

two nuclei, whereas the nonbonding pair is attracted to only one nucleus. Figure 9.8 90°. Figure 9.9 The C ¬ O ¬ H bond involving

the right O because of the greater repulsions due to the nonbonding

electron domains. The angle should be less 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 The

Cl 3p orbital extends farther in space than does the H 1s orbital. Thus,

the Cl 3p orbital can achieve an effective overlap with a H 1s orbital at

a longer distance. Figure 9.14 At very short internuclear distances

the repulsion between the nuclei causes the potential energy to rise

rapidly. Figure 9.16 The hybrid orbitals have to overlap with

the 2p orbitals of the F atoms. The large lobes of the hybrid orbitals

will lead to a much more effective overlap. Figure 9.17 120°.

Figure 9.18 The pz orbital. Figure 9.19 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 9.20

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.

Figure 9.23 They have to lie in the same plane in order to allow the

overlap of the p orbitals to be effective in forming the p bond.

Figure 9.24 Acetylene should have the higher carbon-carbon

bond energy because it has a triple bond as compared to the double

bond in ethylene. Figure 9.26 It has six C ¬ C s and six

C ¬ H s bonds. Figure 9.32 Zero. A node, by definition,

is the place where the value of the wave function is zero.

Figure 9.33 The energy of s1s would rise (but would still be below the

energy of the H 1s atomic orbitals). Figure 9.34 The two electrons

in the s1s MO. Figure 9.35 s*1s and s*2s. Figure 9.36 The end-on

overlap in the s2p MO is greater than the sideways overlap in the p2p.

Figure 9.42 The s2p and p2p MOs have switched order.

Figure 9.43 F2 contains four more electrons than N2. These electrons go into the antibonding p*2p orbitals, thus lowering the bond

order. 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 10.5 1520 torr or 2 atm Figure 10.6 Linear

Figure 10.9 one Figure 10.10 Chlorine, Cl2. Figure 10.13 About

one-sixth Figure 10.14 O2 has the largest molar mass, 32 g>mol,

and H2 has the smallest, 2.0 g>mol. Figure 10.16 n, moles of gas

Figure 10.20 True Figure 10.22 It would 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 11.5 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 same molecular weights. Thus, the change in

boiling point moving left to right is due mainly to the increasing

dipole-dipole attractions.



A-40



Answers to Go Figure



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 SnH4.

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 H2O 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 ¬ O g H bond angle

is approximately 109°. Figure 11.13 The O atom is the negative end

of the polar H2O molecule; the negative end of the dipole is attracted

to the positive ion. Figure 11.14 There are no ions present, but

when we ask, “Are polar molecules present?” we make a distinction

between the two molecules because SiH4 is nonpolar and SiH2Br2

is polar. 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 energy of freezing.

Figure 11.21 Because we are dealing with a state function, the energy

of going straight from a solid to a gas must be the same as going from

a solid to a gas through an intermediate liquid state. Therefore, the

heat of sublimation must be equal to the sum of the heat of fusion and

the heat of vaporization: ΔHsub = ΔHfus + ΔHvap.

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 All liquids 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 12.5 There is not a centered square lattice, because if you 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 corners. 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 corners of the unit cell so there is only

8 * 11>82 = 1 Sm atom per unit 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 * 11>22 + 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 14s12. 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 112 * 1>42 and

four F- ions 18 * 1>8 + 6 * 1>22 per unit cell. In MgF2 there are

two Mg2+ ions 18 * 1>8 + 12 and four F- ions 14 * 1>2 + 22 per

unit cell. In ScF3 there is one Sc3 + ion 18 * 1>82 and three F- ions

112 * 1>42 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 If you doubled

the amount of doping in panel (b), the amount of blue shading the

conduction band would also double. Figure 12.44 Decrease. As

the quantum dots get smaller, the band gap increases and the emitted

light shifts to shorter wavelength. Figure 12.47 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 electron-rich O atom of the

H2O molecule, which is the negative end of the dipole, is attracted to

the positive Na+ ion. Figure 13.3 The negative end of the water dipole (the O) is attracted to the positive Na+ ion, whereas the positive

end of the dipole (the H) is attracted to the negative Cl- ion.

Figure 13.4 For exothermic solution processes the magnitude of

ΔHmix will be larger than the magnitude of ΔHsolute + ΔHsolvent

Figure 13.6 237.6 g>mol, which includes the waters of hydration. Figure 13.7 If the solution wasn’t supersaturated, solute

would not crystallize from it. 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 slopes increase as the molecular weight increases. The larger the molecular weight, the greater the

polarizability of the gas molecules, leading to greater intermolecular

attractive forces between gas molecules and water molecules.

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 H2O, whereas NaCl has a solubility of about 39 g>100 g

H2O. 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 14.2 No. The surface area of a steel nail is much smaller than

that for the same mass of steel wool, so the reaction with O2 would

not be as vigorous. Depending on how hot it is, it might not burn at

all. Figure 14.3 Our first guess might be half way between the values at 20 s and 40 s, namely 0.41 mol A. However, we also see that the

change in the number of moles of A between 0 s and 20 s is 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 of A is

between 0.41 and 0.30 mol. Figure 14.4 The instantaneous rate



Answers to Go Figure

decreases as the reaction proceeds. Figure 14.8 The reaction is first

order. Figure 14.10 At the beginning of the reaction when both

plots are linear or nearly so. Figure 14.13 The reaction that causes

the light occurs more slowly at lower temperatures than at higher

ones. Figure 14.14 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 14.16 He would not need to hit the ball as hard; that is,

less kinetic energy would be required if the barrier were lower.

Figure 14.17 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 14.18 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 14.20 As shown, it is easier to convert to

products because the activation energy is lower in that direction.

Figure 14.21 It can’t be determined. If the amount of backup

at the two toll plazas is roughly the same in the two scenarios,

the time to get from point 1 to point 3 will be about the same.

Figure 14.22 The color is characteristic of molecular bromine, Br2,

which is present in appreciable quantities only in the middle photo.

Figure 14.23 The intermediate is at the valley in the middle of the blue

curve. The transition states are the peaks, one for the red curve and two

for the blue curve. Figure 14.26 Grinding increases the surface area,

exposing more of the catalase to react with the hydrogen peroxide.

Figure 14.27 Substrates must be held more tightly so that they can

undergo the desired reaction. Products are released from the active site.



Chapter 15

Figure 15.1 The same because once the system reaches equilibrium

the concentrations of NO2 and N2O4 stop changing. Figure 15.2

Greater than 1 Figure 15.6 The boxes would be approximately the

same size. Figure 15.7 It would decrease. To reestablish equilibrium

the concentration of CO21g2 would need to return to its previous

value. The only way to do that would be for more CaCO3 to

decompose to produce enough CO21g2 to replace what was lost.

Figure 15.9 High pressure and low temperature, 500 atm and 400 °C

in this figure. Figure 15.10 Nitrogen must react with some of the

added hydrogen to create ammonia and restore equilibrium.

Figure 15.14 (a) the energy difference between the initial state and

the transition state. Figure 15.15 About 5 * 10-4



Chapter 16

Figure 16.2 Hydrogen bonds. Figure 16.4 O2-1aq2 + H2O1l) ¡

2 OH -1aq2. Figure 16.5 Basic. The mixture of the two solutions

will still have 3H+4 6 3OH-4. 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 7 10. Figure 16.9

Bromothymol 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 7 6, while phenolphthalein is not

sensitive to pH changes when pH 6 8, so neither changes color

at pH = 7. Figure 16.12 Yes. The equilibrium of interest is

H3CCOOH Δ 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 The acidic hydrogens belong to carboxylate (—COOH) groups, whereas the fourth proton bound to oxygen is

part of a hydroxyl (—OH) group. In organic acids, like citric acid, the

acidic protons are almost always part of a carboxylate group.



A-41



Figure 16.14 The nitrogen atom in hydroxylamine accepts a proton

to form NH3OH+. As a general rule, nonbonding electron pairs on

nitrogen atoms are more basic than nonbonding electron pairs on

oxygen atoms. Figure 16.16 The range of pH values is so large that

we can’t show the effects using a single indicator (see Figure 16.8).

Figure 16.18 Yes. HI is a strong acid, which is consistent with the

trends shown in the figure. Figure 16.19 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 Yes. Any indicator that changes

color between pH = 3 and pH = 11 could be used for a strong

acid–strong base titration. Methyl red changes color between pH

values of approximately 4 and 6. Figure 17.22 ZnS and CuS would

both precipitate on addition of H2S, preventing separation of the two

ions. 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 The peak value is about 5 * 1012 molecules per cm3.

If 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 * 10-9 mole>L.

Figure 18.7 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 18.9 CaSO31s2

Figure 18.11 1168 W>m22>1342 W>m22 = 0.49; that is, about 50%

of the total. (Keep in mind that the IR radiation absorbed by Earth’s

surface is not directly solar radiation.) Figure 18.13 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 18.15 Evaporation from sea water, evaporation from freshwater; evaporation and transpiration from land.

Figure 18.16 The variable that most affects the density of water in

this case is the change in temperature. As the water grows colder, its

density increases. Figure 18.19 Water is the chemical species that

is crossing the membrane, not the ions. The water flows oppositely

to the normal flow in osmosis because of the application of high

pressure. Figure 18.20 To reduce concentrations of dissolved

iron and manganese, remove H2S and NH3, and reduce bacterial

levels.



Chapter 19

Figure 19.1 Yes, the potential energy of the eggs decreases as they

fall. Figure 19.2 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 19.3 The freezing of liquid water to ice is exothermic. Figure 19.4 To be truly reversible, the temperature change

dT must be infinitesimally small. Figure 19.8 There are two other

independent rotational motions of the H2O molecule:



A-42



Answers to Go Figure



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 S° by 40–45 J/mol@K. Based on

this observation, we would predict that S°1C4H102 would 310–315

J/mol@K. Appendix C confirms that this is a good prediction:

S°1C4H102 = 310.0 J>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

Figure 20.1 (a) The bubbling is caused by the hydrogen gas formed in

the reaction. (b) The reaction is exothermic, and the heat causes the

formation of steam. Figure 20.2 The permanganate, MnO4- , is reduced, as the half-reactions in the text show. The oxalate ion, C2O42- ,

acts as the reducing agent. Figure 20.3 The blue color is due to

Cu2+1aq2. 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 20.5 The electrical balance is maintained in two ways: Anions migrate into the halfcell, and cations migrate out. Figure 20.9 As the cell operates, H+ is

reduced to H2 in the cathode half-cell. As H+ is depleted, the positive

Na+ 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 Ni2+1aq2 ions and the cations in the salt bridge

migrate toward the cathode. The NO3-1aq2 ions and the anions

in the salt bridge migrate toward the anode. Figure 20.19 The

cathode consists of PbO21s2. Because each oxygen has an oxidation state of - 2, lead must have an oxidation state of +4 in

this compound. Figure 20.20 Zn Figure 20.21 Co3+ . The

oxidation number increases as the battery charges Figure

20.24 O21g) + 4 H+ + 4 e- ¡ 2 H2O1g2 Figure 20.25 The

oxidizing agent of O21g2 from the air Figure 20.29 0 V



Chapter 21

Figure 21.2 From Figure 21.2 we see that the belt of stability for a

nucleus containing 70 protons lies at approximately 102 neutrons.

Figure 21.4 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 21.4 that have an odd number of protons have more

than two stable isotopes. Figure 21.6 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 14CO2 to

14

C@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 of the

gas atoms. Figure 21.13 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. Figure 21.14

16. Each of the eight neutrons would split another uranium-235 nucleus, releasing two more neutrons. Figure 21.15 Critical without

being supercritical so that the release of energy is controlled.

Figure 21.19 Because large quantities of water are needed to condense the secondary coolant once it passes through the turbine.

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 22.5 Beaker on the right is warmer. Figure 22.6 HF is the most

stable, SbH3 the least stable. Figure 22.8 More soluble in CCl4 — the

colors are deeper. Figure 22.9 CF2 Figure 22.10 Redox

reactions: The halides are being oxidized. Figure 22.13 No

Figure 22.15 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! Figure 22.20 They have been converted into

water. Figure 22.21 Formally they could both be +2. If we consider

that the central sulfur is like SO42- , however, then the central sulfur

would be +6, like SO42- , and then the terminal sulfur would be - 2.

Figure 22.22 Nitrite Figure 22.23 Longer. (There is a triple bond

in N2.) Figure 22.25 The NO double bond Figure 22.27 In both

compounds the electron domains around the P atoms are tetrahedral.

In P4O10 all the electron domains around the P atoms are bonding.

In P4O6 one of the electron domains about each P atom is nonbonding. Figure 22.32 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 23.5 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+ , Ti 4+ , V 5+ , Cr 6+

and Mn7+ . Figure 23.6 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 3Fe1H2O2643+1aq) + SCN - 1aq2 ¡ 3Fe1H2O25

1SCN242+1aq2 + H2O1l2 Figure 23.10 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 Four for both (assuming no other ligands coordinate to the

metal). Figure 23.15 The coordination number is 6. Five nitrogen

atoms (four from the heme and one from the protein) and one oxygen

atom (from the O2 molecule) coordinate iron. Figure 23.16 The

peak at 660 nm. Figure 23.20 3Fe1NH3251NO2242+ and pentaamminenitroiron(III) ion for the complex on the left. 3Fe1NH3251ONO242+

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. Figure 23.28 dx2 - y2 and dz2. Figure 23.29 Convert the

wavelength of light, 495 nm, into energy in joules using the relationship E = hc>l. Figure 23.30 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 dx2 - y2 orbital points

directly at the ligands.



Answers to Go Figure



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 24.5 CnH2n, because there are

no CH3 groups, each carbon has two hydrogens. Figure 24.7 Just

one Figure 24.9 Intermediates are minima and transition states are

maxima on energy profiles. Figure 24.13 Both lactic acid and citric

acid Figure 24.14 No, because there are not four different groups



A-43



around any carbon Figure 24.17 Those labeled “basic amino acids,”

which have basic side groups that are protonated at pH 7 Figure

24.18 Two. Figure 24.24 The long hydrocarbon chains, which are

nonpolar Figure 24.26 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 24.28 Negatively charged, because of the charge on the phosphate groups.

Figure 24.30 GC because each base has three hydrogen bonding

sites, whereas there are only two in AT.



ANSWERS TO SELECTED PRACTICE EXERCISES

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,

(d) 0.00422 g.

Sample Exercise 1.3

Practice Exercise 2: (a) 261.7 K,



(b) 6.0 km, (c) 4.22 * 10-3 g,



(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: Five, as in the measurement 24.995 g, the uncertainty being in the third decimal place.

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 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 of Ar.

Sample Exercise 2.10

Practice Exercise 2: (a) Na3PO4,



(b) ZnSO4,



Sample Exercise 2.11

Practice Exercise 2: BrO- and BrO2Sample Exercise 2.12

Practice Exercise 2: (a) ammonium bromide,

oxide, (c) cobalt(II) nitrate



Sample Exercise 1.8

Practice Exercise 2: 9.52 m>s (three significant figures).



Sample Exercise 2.13

Practice Exercise 2: (a) HBr,



(b) H2CO3



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 2.14

Practice Exercise 2: (a) SiBr4,



(b) S2Cl2,



Sample Exercise 1.10

Practice Exercise 2: 804.7 km



(c) Fe21CO323



(b) chromium(III)



(c) P2O6.



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.



Chapter 3



Sample Exercise 1.11

Practice Exercise 2: 12 km>L.

Sample Exercise 1.12

Practice Exercise 2: 1.2 * 104 ft.



Sample Exercise 3.1

Practice Exercise 2: (a) C2H4 + 3 O2 ¡ 2 CO2 + 2 H2O. (b) Nine

O2 molecules



Sample Exercise 1.13

Practice Exercise 2: 832 g



Sample Exercise 3.2

Practice Exercise 2: (a) 4, 3, 2; (b) 2, 6, 2, 3; (c) 1, 2, 1, 1, 1



Chapter 2



Sample Exercise 3.3

Practice Exercise 2: (a) HgS1s2 ¡ Hg1l2 + S1s2,

(b) 4 Al1s2 + 3 O21g2 ¡ 2 Al2O31s2



Sample Exercise 2.1

Practice Exercise 2: (a) 154 pm,



(b) 1.3 * 106 C atoms



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 3.4

Practice Exercise 2: C2H5OH1l2 + 3 O21g2 ¡ 2 CO21g2 + 3 H2O1g2

Sample Exercise 3.5

Practice Exercise 2: (a) 78.0 amu, (b) 32.0 amu, (c) 211.0 amu



Sample Exercise 2.3

Practice Exercise 2: 208

82Pb



Sample Exercise 3.6

Practice Exercise 2: 16.1%



Sample Exercise 2.4

Practice Exercise 2: 28.09 amu



Sample Exercise 3.7

Practice Exercise 2: 1 mol H2O 16 * 1023 O atoms2 6 3 * 1023

molecules O3 19 * 1023 O atoms2 6 1 mol CO2112 * 1023 O atoms2



Sample Exercise 2.5

Practice Exercise 2: Na, atomic number 11, is a metal; Br, atomic number 35, is a nonmetal.



Sample Exercise 3.8

Practice Exercise 2: (a) 9.0 * 1023,



Sample Exercise 2.6

Practice Exercise 2: B5H7



Sample Exercise 3.9

Practice Exercise 2: 164.1 g>mol



A-44



(b) 2.71 * 1024



Answers to Selected Practice Exercises

Sample Exercise 3.10

Practice Exercise 2: 55.5 mol H2O.



Sample Exercise 4.10

Practice Exercise 2: Zn and Fe



Sample Exercise 3.11

Practice Exercise 2: (a) 6.0 g,



Sample Exercise 4.11

Practice Exercise 2: 0.278 M



(b) 8.29 g.



Sample Exercise 3.12

Practice Exercise 2: (a) 4.01 * 1022 molecules HNO3,

(b) 1.20 * 1023 atoms O



Sample Exercise 4.12

Practice Exercise 2: 0.030 M

Sample Exercise 4.13

Practice Exercise 2: (a) 1.1 g,



Sample Exercise 3.13

Practice Exercise 2: C4H4O

Sample Exercise 3.14

Practice Exercise 2: (a) CH3O,



(b) 76 mL



Sample Exercise 4.14

Practice Exercise 2: (a) 0.0200 L = 20.0 mL, (b) 5.0 mL, (c) 0.40 M

(b) C2H6O2



Sample Exercise 3.15

Practice Exercise 2: (a) C3H6O,



Sample Exercise 4.15

Practice Exercise 2: (a) 0.240 g,



(b) C6H12O2



(b) 0.400 L



Sample Exercise 4.16

Practice Exercise 2: 0.210 M



Sample Exercise 3.16

Practice Exercise 2: 1.77 g



Sample Exercise 4.17

Practice Exercise 2: (a) 1.057 * 10-3 mol MnO4- ,

(b) 5.286 * 10-3 mol Fe2+, (c) 0.2952g, (d) 33.21%



Sample Exercise 3.17

Practice Exercise 2: 26.5 g

Sample Exercise 3.18

Practice Exercise 2: (a) Al,



(b) 1.50 mol,



Chapter 5



(c) 0.75 mol Cl2



Sample Exercise 3.19

Practice Exercise 2: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn



Sample Exercise 5.1

Practice Exercise 2: (a) 1.4 * 10-20 J, (b) 8.4 * 103 J



Sample Exercise 3.20

Practice Exercise 2: (a) 105 g Fe,



Sample Exercise 5.2

Practice Exercise 2: + 55 J



(b) 83.7%



Sample Exercise 5.3

Practice Exercise 2: 0.69 L@atm = 70 J



Chapter 4

Sample Exercise 4.1

Practice Exercise 2: (a) 6,



A-45



Sample Exercise 4.3

Practice Exercise 2: (a) Fe1OH23, (b) Fe21SO4231aq2 + 6 LiOH1aq2

¡ 2 Fe1OH231s2 + 3 Li2SO41aq2



Sample Exercise 5.4

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 4.4

Practice Exercise 2: 3 Ag +1aq2 + PO43-1aq2 ¡ Ag3PO41s2



Sample Exercise 5.5

Practice Exercise 2: - 14.4 kJ



Sample Exercise 4.5

Practice Exercise 2: The diagram would show 10 Na+ ions, 2 OH- ions,

8 Y - ions, and 8 H2O molecules.



Sample Exercise 5.6

Practice Exercise 2: (a) 4.9 * 105 J,

(b) 11 K decrease = 11 °C decrease



Sample Exercise 4.6

Practice Exercise 2: C6H6O6 (nonelectrolyte) 6 CH3COOH (weak

electrolyte, existing mainly in the form of molecules with few

ions) 6 NaCH3COO (strong electrolyte that provides two ions, Na+

and CH3COO-2 6 Ca1NO322 (strong electrolyte that provides three

ions, Ca2 + and 2 NO3- 2



Sample Exercise 5.7

Practice Exercise 2: -68,000 J>mol = -68 kJ>mol



(b) 12,



Sample Exercise 4.2

Practice Exercise 2: (a) insoluble,



(c) 2,



(d) 9



(b) soluble,



(c) soluble



Sample Exercise 4.7

Practice Exercise 2:

(a) H3PO31aq2 + 3 KOH1aq2 ¡ 3 H2O1l2 + K3PO31aq2,

(b) H3PO31aq2 + 3 OH -1aq2 ¡ 3 H2O1l2 + PO33-1aq2.

1H3PO3 is a weak acid and therefore a weak electrolyte, whereas KOH, a

strong base, and K3PO3, an ionic compound, are strong electrolytes.)

Sample Exercise 4.8

Practice Exercise 2: (a) +5,



(b) - 1,



(c) + 6,



(d) +4,



Sample Exercise 4.9

Practice Exercise 2:

(a) Mg1s2 + CoSO41aq2 ¡ MgSO41aq2 + Co1s2;

Mg1s2 + Co2+1aq2 ¡ Mg2+1aq2 + Co1s2,

(b) Mg is oxidized and Co2+ is reduced.



(e) - 1



Sample Exercise 5.8

Practice Exercise 2: (a) -15.2 kJ>g,



(b) - 1370 kJ>mol



Sample Exercise 5.9

Practice Exercise 2: +1.9 kJ

Sample Exercise 5.10

Practice Exercise 2: -304.1 kJ

Sample Exercise 5.11

Practice Exercise 2: C1graphite2 + 2 Cl21g2 ¡ CCl41l2;

ΔH°f = - 106.7 kJ>mol.

Sample Exercise 5.12

Practice Exercise 2: -1367 kJ

Sample Exercise 5.13

Practice Exercise 2: -156.1 kJ>mol

Sample Exercise 5.14

Practice Exercise 2: (a) 15 kJ>g,



(b) 100 min



A-46



Answers to Selected Practice Exercises



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. The

lower wave has the longer wavelength (lower frequency) and would

be the red light.

Sample Exercise 6.2

Practice Exercise 2: (a) 1.43 * 1014 s-1,

Sample Exercise 6.3

Practice Exercise 2: (a) 3.11 * 10-19 J,

(c) 4.2 * 1016 photons



(b) 2.899 m

(b) 0.16 J,



Sample Exercise 6.4

Practice Exercise 2: (a) ΔE 6 0, photon emitted, (b) ΔE 7 0,

photon absorbed



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,



Cl



(b) 3;



C



Cl



H

Sample Exercise 8.7

Practice Exercise 2: (a) [ N



Sample Exercise 6.5

Practice Exercise 2: 7.86 * 102 m>s

Sample Exercise 6.6

Practice Exercise 2: (a) 5p;



H



(b)



O ]+,



(b) H



C



H

Sample Exercise 8.8

Practice Exercise 2: (a) [ O



(c) 1, 0, - 1



Sample Exercise 6.7

Practice Exercise 2: (a) 1s22s22p63s23p2,



O ]–



Cl



(b)



3–



P



O



O



Sample Exercise 6.8

Practice Exercise 2: group 4A

Sample Exercise 6.9

Practice Exercise 2: (a) 3Ar44s23d7 or 3Ar43d74s2,

or 3Kr44d105s25p1



H

O



O



(b) two



H



C



Sample Exercise 8.9

Practice Exercise 2:

(b) 3Kr45s24d105p1



–2



(a)



Chapter 7

Sample Exercise 7.1

Practice Exercise 2: P- Br

Sample Exercise 7.2

Practice Exercise 2: C 6 Be 6 Ca 6 K



[N



0



+1



C



O



–1



]–



(i)



Sample Exercise 7.5

Practice Exercise 2: Ca

Sample Exercise 7.6

Practice Exercise 2: Al lowest, C highest

Sample Exercise 7.7

Practice Exercise 2: (a) 7

Sample Exercise 7.8

Practice Exercise 2:

CuO1s2 + H2SO41aq2 ¡ CuSO41aq2 + H2O1l2



0



C



O] –



0



0



–1



[N



C



O ]–



(ii)



(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: H



Sample Exercise 7.3

Practice Exercise 2: S2Sample Exercise 7.4

Practice Exercise 2: Cs+



[N



0



C



O







O

Sample Exercise 8.11

Practice Exercise 2:

(a) C, (b) F Xe



H



C



O







O



F



Sample Exercise 8.12

Practice Exercise 2: -86 kJ



Chapter 9

Sample Exercise 9.1

Practice Exercise 2: (a) tetrahedral, bent; (b) trigonal planar, trigonal planar



Sample Exercise 7.9

Practice Exercise 2: P4O61s2 + 6 H2O1l2 ¡ 4 H3PO31aq2



Sample Exercise 9.2

Practice Exercise 2: (a) trigonal bipyramidal, T-shaped; (b) trigonal

bipyramidal, trigonal bipyramidal.



Sample Exercise 7.10

Practice Exercise 2: 2K1s2 + S1s2 ¡ K2S1s2



Sample Exercise 9.3

Practice Exercise 2: 109.5°, 180°



Chapter 8

Sample Exercise 8.1

Practice Exercise 2: ZrO2



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 8.2

Practice Exercise 2: Mg2 + and N3 -



Sample Exercise 9.5

Practice Exercise 2: tetrahedral, sp3



Answers to Selected Practice Exercises

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 s bonds and two p bonds

Sample Exercise 9.7

Practice Exercise 2: SO2 and SO3, as indicated by the presence of two

or more resonance structures involving p bonding for each of these

molecules.

Sample Exercise 9.8

Practice Exercise 2: s1s2s*1s1; 12

Sample Exercise 9.9

Practice Exercise 2: (a) diamagnetic, 1;



A-47



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 rodlike.



(b) diamagnetic, 3



Chapter 12



Chapter 10



Sample Exercise 12.1

Practice Exercise 2: 0.68 or 68%



Sample Exercise 10.1

Practice Exercise 2: (D) 1.6 M



Sample Exercise 12.2

Practice Exercise 2: a = 4.02 Å and density = 4.31 g>cm3



Sample Exercise 10.2

Practice Exercise 2: 807.3 torr



Sample Exercise 12.3

Practice Exercise 2: smaller.



Sample Exercise 10.3

Practice Exercise 2: 5.30 : 103 L



Sample Exercise 12.4

Practice Exercise 2: A group 5A element could be used to replace Se.



Sample Exercise 10.4

Practice Exercise 2: 2.0 atm



Chapter 13



Sample Exercise 10.5

Practice Exercise 2: 3.83 * 103 m3

Sample Exercise 10.6

Practice Exercise 2: 27 °C



Sample Exercise 13.1

Practice Exercise 2: C5H12 6 C5H11Cl 6 C5H11OH 6 C5H101OH22

(in order of increasing polarity and hydrogen-bonding ability)



Sample Exercise 10.7

Practice Exercise 2: 5.9 g>L



Sample Exercise 13.2

Practice Exercise 2: 1.0 * 10-5 M



Sample Exercise 10.8

Practice Exercise 2: 29.0 g>mol



Sample Exercise 13.3

Practice Exercise 2: 90.5 g of NaOCl



Sample Exercise 10.9

Practice Exercise 2: 14.8 L



Sample Exercise 13.4

Practice Exercise 2: 0.670 m



Sample Exercise 10.10

Practice Exercise 2: 2.86 atm



Sample Exercise 13.5

Practice Exercise 2: (a) 9.00 * 10-3,



Sample Exercise 10.11

Practice Exercise 2: 1.0 * 103 torr N2, 1.5 * 102 torr Ar, and 73 torr

CH4



Sample Exercise 13.6

Practice Exercise 2: (a) 10.9 m, (b) XC3H8O3 = 0.163, (c) 5.97 M



Sample Exercise 10.12

Practice Exercise 2: (a) increases,



(b) no effect,



(c) no effect



Sample Exercise 10.13

Practice Exercise 2: 1.36 * 103 m>s



Sample Exercise 13.7

Practice Exercise 2: 0.290

Sample Exercise 13.8

Practice Exercise 2: -65.6 °C

Sample Exercise 13.9

Practice Exercise 2: 0.048 atm



Sample Exercise 10.14

Practice Exercise 2: rN2 >rO2 = 1.07

Sample Exercise 10.15

Practice Exercise 2: (a) 7.472 atm,



(b) 0.505 m



Sample Exercise 13.10

Practice Exercise 2: 110 g>mol

(b) 7.181 atm



Chapter 11

Sample Exercise 11.1

Practice Exercise 2: chloramine, NHCl



Sample Exercise 13.11

Practice Exercise 2: 4.20 * 104 g>mol



Chapter 14

Sample Exercise 14.1

Practice Exercise 2: 1.8 * 10-2 M>s



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) CH3CH2OH



Sample Exercise 14.2

Practice Exercise 2: 1.1 * 10-4 M>s



Sample Exercise 11.3

Practice Exercise 2: - 20.9 kJ - 33.4 kJ - 6.09 kJ = -60.4 kJ



Sample Exercise 14.3

Practice Exercise 2: (a) 8.4 * 10-7 M>s, (b) 2.1 * 10-7 M>s



Sample Exercise 11.4

Practice Exercise 2: about 340 torr (0.45 atm)



Sample Exercise 14.4

Practice Exercise 2: 2 = 3 6 1



A-48



Answers to Selected Practice Exercises



Sample Exercise 14.5

Practice Exercise 2: (a) 1,



Sample Exercise 15.9

Practice Exercise 2: Qp = 16; Qp 7 Kp, and so the reaction will proceed from right to left, forming more SO3.



(b) M -1 s-1.



Sample Exercise 14.6

Practice Exercise 2: (a) rate = k3NO423H24,

(c) rate = 4.5 * 10-4 M>s



(b) k = 1.2 M -2 s-1



Sample Exercise 15.10

Practice Exercise 2: 1.22 atm



Sample Exercise 14.7

Practice Exercise 2: 51 torr



Sample Exercise 15.11

Practice Exercise 2: PPCl5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm



Sample Exercise 14.8

Practice Exercise 2: 3NO24 = 1.00 * 10-3 M



Sample Exercise 15.12

Practice Exercise 2: (a) right, (b) left, (c) right, (d) left



Sample Exercise 14.9

Practice Exercise 2: (a) 0.478 yr = 1.51 * 107 s,

half-lives, 210.478 yr2 = 0.956 yr



(b) it takes two



Sample Exercise 14.10

Practice Exercise 2: 2 6 1 6 3 because, if you approach the barrier

from the right, the Ea values are 40 kJ>mol for reverse reaction 2,

25 kJ>molfor reverse reaction 1, and 15 kJ>mol for reverse reaction 3.

Sample Exercise 14.11

Practice Exercise 2: 2 * 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) Mo1CO25.

Sample Exercise 14.13

Practice Exercise 2: (a) Rate = k3NO423Br24,

molecular reactions are very rare.



(b) No, because ter-



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.

Sample Exercise 14.15

Practice Exercise 2: 3Br4 = a



Sample Exercise 15.1

2



3H243I24



,



Sample Exercise 16.1

Practice Exercise 2: H2SO3, HF, HPO42-, HCO+

Sample Exercise 16.2

Practice Exercise 2: O2 - 1aq2 + H2O1l2 ¡ OH-1aq2 + OH-1aq2.

The OH- is both the conjugate acid of O2- and the conjugate base

of H2O.

Sample Exercise 16.3

Practice Exercise 2: (a) left,



Sample Exercise 16.4

Practice Exercise 2: (a) basic, (b) neutral,

Sample Exercise 16.5

Practice Exercise 2: (a) 5 * 10-9 M,

(c) 7.1 * 10-9 M



(b) Kc =



3CdBr42-4

2+

- 4



3Cd 43Br 4



Sample Exercise 16.12

Practice Exercise 2: 3.41



Sample Exercise 15.4



Sample Exercise 16.13

Practice Exercise 2: (a) 3.9%,



3



= 1.51 * 10



Sample Exercise 15.5

Practice Exercise 2:

1PH224

3Cr3 + 4

(a) Kc =

,

(b)

K

=

p

1PH2O24

3Ag +43

Sample Exercise 15.6

Practice Exercise 2: H21g2

Sample Exercise 15.7

Practice Exercise 2: 1.79 * 10-5

Sample Exercise 15.8

Practice Exercise 2: 33



9



(b) 2.4 * 10-3 M



Sample Exercise 16.10

Practice Exercise 2: 1.5 * 10-5



Sample Exercise 15.3

Practice Exercise 2: At the lower temperature because Kp is larger at

the lower temperature



1.04 * 10-4



(b) 1.0 * 10-7 M,



Sample Exercise 16.9

Practice Exercise 2: (a) 7.8 * 10-3 M,



Sample Exercise 16.11

Practice Exercise 2: 2.7%



Practice Exercise 2:



(c) acidic



Sample Exercise 16.6

Practice Exercise 2: (a) 3.42, (b) 3H+4 = 5.3 * 10-9 M, so pH = 8.28



Sample Exercise 15.2

Practice Exercise 2: 0.335



154.02



(b) right



Sample Exercise 16.8

Practice Exercise 2: 0.0046 M



Chapter 15

Practice Exercise 2: (a) Kc =



Chapter 16



Sample Exercise 16.7

Practice Exercise 2: 3H+4 = 6.6 * 10-10



1/2

k1

3Br24b

k-1



3HI4



Sample Exercise 15.13

Practice Exercise 2: ΔH° = 508.3 kJ; the equilibrium constant will

increase with increasing temperature.



(b) 12%



Sample Exercise 16.14

Practice Exercise 2: (a) pH = 1.80,



(b) 3C2O42-4 = 6.4 * 10-5 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 16.16

Practice Exercise 2: 0.12 M

Sample Exercise 16.17

Practice Exercise 2: (a) PO43-1Kb = 2.4 * 10-22,

(b) Kb = 7.9 * 10-10

Sample Exercise 16.18

Practice Exercise 2: (a) Fe1NO323 ,

(d) NH4NO3



(b) KBr,



(c) CH3NH3Cl,



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