Corrections to Ch. 1–6 of the Student Solutions Manual
for Introduction to General, Organic, and Biochemistry, Eighth Edition,
by Bettelheim, Brown, Campbell, and Farrell
Chapter 1 Matter, Energy, and Measurement
1.1
Multiplication: (a) 4.69 x 105
Division:
(a) 1.94 x 1018
(b) 2.8 x 10-15
(b) 1.37 x 105
1.19 (a) 6.65 x 107
(b) 1.2 x 101
(c) 3.89 x 10-16
1.21 (a) 1.31 x 105
(b) 9.40 x 104
(c) 5.139 x 10-3
1.65 d brain =
(d) 3.5 x 10-23
1.0 lb ⎛ 453.6 g ⎞
⎜
⎟ = 0.73 g/mL
620 mL ⎝ 1 lb ⎠
Specific gravity brain =
0.73 g/mL
d brain
= 0.73
=
d H2O
1.0 g/mL
1.69 Convert European car’s fuel efficiency of 22 km/L into mi/gal, then compare:
22 km ⎛ 1 mi ⎞ ⎛ 3.785 L ⎞
Fuel efficiency European =
⎟ = 52 mi/gal
⎜
⎟⎜
L ⎝ 1.609 km ⎠ ⎝ 1 gal ⎠
The European car is more fuel efficient by 22 miles per gallon
1.71 Shivering is a form of kinetic energy which generates heat by way of frictional effects
within body tissues.
⎛ 1.609 km ⎞ ⎛ 1 hr ⎞
1.75 Travel time = 1490 mi ⎜
⎟⎜
⎟ = 11 hr
⎝ 1 mi ⎠ ⎝ 220 km ⎠
Heat
+ T1
1.83 T2 =
SH × m
60.0 J
o
1o
+
20
C
=
2
×
10
C
T2 =
10.0 g 1.339 J/g ⋅o C
(
11/21/06
)(
)
Page 1 of 10
Chapter 2 Atoms
2.31 (a) An ion is an atom or group of bonded atoms having an unequal number of
protons and electrons.
(b) Isotopes are atoms with the same number of protons in their nuclei but a
different number of neutrons.
2.33 Rounded to three significant figures, the calculated value is 12.0 amu. The value
given in the Periodic Table is 12.011 amu.
⎛ 98.90
⎞ ⎛ 1.10
⎞
× 12.000 amu ⎟ + ⎜
× 13.003 amu ⎟ = 12.01 amu
⎜
⎝ 100
⎠ ⎝ 100
⎠
2.37 Americium-241 (Am) has atomic number 95. This isotope has 95 protons, 95
electrons and 241 - 95 = 146 neutrons.
2.61 Following are the ground-state electron configurations of Mg atom, Mg+, Mg2+, and
Mg3+.
Mg
Mg+ + eIE = 738 kJ/mol
Electron
configuration
1s22s22p63s2
1s22s22p63s1
Electron
configuration
Mg+
Mg2+ +
Electron
configuration
1s22s22p63s1
Mg2+
1s22s22p6
e-
IE = 1450 kJ/mol
Mg3+ + e-
IE = 7734 kJ/mol
1s22s22p6
1s22s22p5
The first electron is removed from the 3s orbital. The removal of each subsequent
electron requires more energy because, after the first electron is removed, each
subsequent electron is removed from a positive ion, which strongly attracts the
remaining extra nuclear electrons. The third ionization energy is especially large
because the electron is removed from the filled second principal energy level,
meaning that it is removed from an ion that has the same electron configuration as
neon.
2.75 Rounded to four significant figures, the atomic weight of naturally occurring boron
is 10.81. The value given in the Periodic Table is 10.811.
⎛ 19.9
⎞ ⎛ 80.1
⎞
× 10.013 amu ⎟ + ⎜
× 11.009 amu ⎟ = 10.81 amu
⎜
⎝ 100
⎠ ⎝ 100
⎠
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Page 2 of 10
Chapter 3 Nuclear Chemistry
3.5 Barium-122 (10.0 g) has decayed through 5 half-lives, leaving 0.31 g:
10.0 g → 5.0 g → 2.5 g → 1.25 g → 0.625 g → 0.31 g
(Careful consideration of significant figures ignored)
3.7 The intensity of any radiation decreases with the square of the distance:
I1
d2
= 22
I2
d1
300 mCi
(3.0 m) 2
=
I2
(0.010 m) 2
(300 mCi)(0.010 m ) 2
I2 =
= 3 × 10-3 mCi
2
(3.0 m )
3.39 (a)
(b)
(c)
(d)
(e)
(f)
(g)
Amount of radiation absorbed by tissues
Effective dose absorbed by humans
Amount of radiation delivered
Intensity of radiation
Amount of radiation absorbed by tissues
Intensity of radiation
Effective dose absorbed by humans
294 mrem/yr
× 100 = 82%
359 mrem/yr
39 mrem/yr
× 100 = 15%
(b)
359 mrem/yr
0.5 mrem/yr
× 100 = 0.1%
(c)
359 mrem/yr
3.65 (a)
3.67 X-rays will cause more ionization than radar because X-rays are more energetic.
3.69 1000/432 ~ 2.3 half-lives: 1/2 x 1/2 = 1/4 so a little less than 25% of the original
americium will be around after 1000 years.
241
4
237
95 Am → 2 He + 93 Np Decay Product
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Page 3 of 10
Chapter 4 Chemical Bonds
4.17 (a) By losing one electron, Li becomes Li+, a helium electron configuration.
Li: 1s22s1 → Li+: 1s2 (filled s shell)
(b) By gaining one electron, Cl becomes Cl-, an argon electron configuration.
Cl 1s22s22p63s23p5 → Cl-: 1s22s22p63s23p6 (octet)
(c) By gaining three electrons, P becomes P3-, an argon electron configuration.
P 1s22s22p63s23p3 → P3-: 1s22s22p63s23p6 (octet)
(d) By losing three electrons, Al becomes Al3+, a neon electron configuration.
Al: 1s22s22p63s23p1 → Al3+: 1s22s22p6 (octet)
(e) By losing two electrons, Sr becomes Sr2+, a krypton electron configuration.
Sr: 1s22s22p63s23p64s23d104p65s2 → Sr2+: 1s22s22p63s23p64s23d104p6 (octet)
(f) By gaining two electrons, S becomes S2-, an argon electron configuration.
S: 1s22s22p63s23p4 → S2-: 1s22s22p63s23p6 (octet)
(g) By gaining four electrons, Si becomes Si4-, an argon electron configuration.
Si: 1s22s22p63s23p2 → Si4-: 1s22s22p63s23p6 (octet)
or by losing four electrons, Si becomes Si4+, a neon electron configuration.
Si: 1s22s22p63s23p2 → Si4+: 1s22s22p6 (octet)
(h) By gaining two electrons, O becomes O2-, a neon electron configuration
O: 1s22s22p4 → O2-: 1s22s22p6 (octet)
4.19 (a) H: 1s1 + 1 electron → H+ 1s2 (filled 1s shell)
(b) Al: 1s22s22p63s23p1 → Al3+: 1s22s22p6 (octet) + 3 electrons
4.23 Only (f) Cs+ will be stable because it has a nobel gas electron configuration. The
other ions have a partially filled shell or a charge that is too high.
4.37 Sodium chloride in the solid state has each Na+ cation surrounded by six Cl- anions
and each Cl- anion surrounded by six Na+ cations.
4.71 (a) There are 16 valence electrons present in N2O.
(b) The two contributing structures can be represented as follows:
N
N
O
N
N
O
(c) The central nitrogen atom has 10 electrons around it.
4.81 Use differences in electronegativity to predict the polarity of the bond.
(a) Nonpolar covalent
(b) Nonpolar covalent
(c) Polar covalent
(d) Ionic
(e) Nonolar covalent
(f) Polar covalent
(g) Nonpolar covalent
(h) Ionic
4.83 Calcium dihydrogen phosphate monohydrate, calcium phosphate, and calcium
carbonate.
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Page 4 of 10
4.93 (a) ClO2 has 19 valence electrons. In the Lewis structure, the chlorine atom must
have an odd number of valence electrons. Being a third period electron,
chlorine can also have an expanded octet.
(b) Lewis structure of ClO2:
O
Cl
O
F
F
C
F
11/21/06
C
F
Page 5 of 10
Chapter 5 Chemical Reactions
5.1 (a) Ibuprofen, C13H18O2
C 13 x 12.0 amu = 156 amu
H 18 x 1.01 amu = 18.9 amu
O 2 x 16.0 amu = 32.0 amu
C13H18O2 = 206 amu
5.3
(b) Barium Phosphate, Ba3(PO4)2
Ba 3 x 137.3 amu = 411.9 amu
P 2 x 31.0 amu = 62.0 amu
O 8 x 16.0 amu = 128 amu
Ba3(PO4)2 = 602 amu
⎛ 78.05 g Na 2S ⎞
⎟ = 222. g Na 2S
2.84 mol Na 2S ⎜
⎜1 mol Na 2S ⎟
⎝
⎠
5.17 (a) KCl = 74.6 amu
⎛ 68,000 amu
5.29 ⎜
⎝ 1 molecule
(b) Na3PO4 = 163.9amu
⎞ ⎛ 1.6605 × 10-24 g ⎞
-19
⎟⎜
⎟ = 1.1 × 10 g/molecule
1 amu
⎠
⎠⎝
5.53 Using the balanced equation: CH 3CH 3 (g) + Cl2 (g)
Theoretical yield of ethyl chloride:
⎛ 1 mol Ethane ⎞ ⎛ 1 mol CH 3CH 2 Cl
5.6 g Ethane ⎜
⎟⎜
⎜ 30.1 g Ethane ⎟ ⎜ 1 mol Ethane
⎝
⎠⎝
Actual yield of CH3CH2Cl = 8.2 g
% Yield of CH3CH 2 Cl =
5.63 (a)
(b)
(c)
(d)
(e)
(c) Fe(OH)2 = 89.9 amu
% yield =
→ CH 3CH 2Cl(l) + HCl(g)
⎞ ⎛ 64.5 g CH CH Cl
3
2
⎟⎜
⎟ ⎜ 1 mol CH 3CH 2 Cl
⎠⎝
⎞
⎟ = 12 g
⎟
⎠
Actual yield
× 100%
Theoretical yield
8.2 g CH3CH 2 Cl
× 100% = 68%
12 g CH 3CH 2 Cl
MgCl2 (soluble): most compounds containing Cl- are soluble
CaCO3 (insoluble): most compounds containing CO32- are insoluble
Na2SO3 (soluble): all compounds containing Na+ are soluble
NH4NO3 (soluble): all compounds containing NO3- and NH4+ are soluble
Pb(OH)2 (insoluble): most compounds containing OH- are insoluble
5.77 Fluoride reacts with the Ca10(PO4)6(OH)2 in enamel, by exchanging with the OHions, forming a less soluble Ca10(PO4)6F2 under the acidic conditions found in the
mouth.
5.81 N 2 O5 (g) + H 2O(l) → 2HNO3 (aq)
5.87 MWchlorophyll =
11/21/06
24.305 g Mg /mol
0.0272 g Mg /1 g chlorophyll
= 894 amu
Page 6 of 10
5.89 8.00 g Pb(NO3)2 added to 2.67 g AlCl3 yielded 5.55 g PbCl2
Mass of aluminum chloride required based on 8.00 g Pb(NO3)2:
⎛ 1 mol Pb(NO3 ) 2 ⎞ ⎛ 2 mol AlCl
⎞
3
8.00 g Pb(NO3 ) 2 ⎜
⎟⎜
⎟ = 1.61 × 10-2 mol AlCl3
⎜ 331.2 g Pb(NO3 ) 2 ⎟ ⎜ 3 mol Pb(NO3 ) 2 ⎟
⎝
⎠⎝
⎠
⎛ 133.3 g AlCl ⎞
3
1.61 × 10-2 mol AlCl 3 ⎜
⎟ = 2.15 g AlCl3 needed
⎜ 1 mol AlCl3 ⎟
⎝
⎠
Mass of lead(II) nitrate required based on 2.67 g AlCl3:
⎛ 1 mol AlCl3 ⎞ ⎛ 3 mol Pb(NO ) ⎞
3 2
2.67 g AlCl3 ⎜
⎟⎜
⎟ = 3.00 × 10-2 mol of Pb(NO3 ) 2
⎜ 133.3 g AlCl3 ⎟ ⎜ 2 mol AlCl3 ⎟
⎝
⎠⎝
⎠
⎛ 331.2 g Pb(NO ) ⎞
3 2
3.00 × 10-2 mol Pb(NO3 ) 2 ⎜
⎟ = 9.94 g Pb(NO3 ) 2 needed
⎜ 1 mol Pb(NO3 ) 2 ⎟
⎝
⎠
(a) Pb(NO3)2 is the limiting reagent.
(b) Theoretical yield of PbCl2:
⎛ 1 mol Pb(NO3 ) 2
8.00 g Pb(NO3 ) 2 ⎜
⎜ 331.2 g Pb(NO3 ) 2
⎝
⎞ ⎛ 3 mol PbCl
2
⎟⎜
⎟ ⎜ 3 mol Pb(NO3 ) 2
⎠⎝
⎞
⎟ = 2.42 × 10-2 mol PbCl2
⎟
⎠
⎛ 278.1 g PbCl ⎞
2
2.42 × 10-2 mol PbCl2 ⎜
⎟ = 6.73 g PbCl2
⎜ 1 mol PbCl2 ⎟
⎝
⎠
% Yield =
11/21/06
5.55g PbCl2
Actual yield
× 100 = 82.5% PbCl2
=
Theoretical yield
6.73 g PbCl2
Page 7 of 10
Chapter 6 Gases, Liquids, and Solids
6.9 Heat of vaporization of water = 540 cal/g
⎛ 1000 cal ⎞ ⎛ 1 g H 2 O ⎞
45.0 kcal ⎜
⎟⎜
⎟ = 83 g H 2 O vaporized
⎝ 1 kcal ⎠ ⎝ 540 cal ⎠
6.11 According to the phase diagram of water (figure 6.18), the vapor will first condense
to a liquid and then crystallize to a solid, finally chilling further to -30 C.
6.23 Complete this table:
V1
546 L
43 mL
4.2 L
1.3 L
Use the
T1
43oC
-56oC
234 K
25oC
P1V1
PV
= 2 2 equation.
T1
T2
P1
6.5 atm
865 torr
0.87 atm
740 mm Hg
V2
2.0 x 103L
48 mL
3.2 L
1.2 L
T2
65oC
43oC
29oC
0oC
P2
1.9 atm
1.5 atm
1.5 atm
1.0 atm
6.33 Using the PV=nRT Ideal Gas Law equation, the following equation is derived:
MW =
(mass)RT
(8.00 g)(0.0821 L i atm i mol-1 i K -1 )(273 K )
= 4.00 g/mol
=
PV
(2.00 atm )(22.4 L )
⎛ 0.21 L O 2 ⎞
6.39 5.5 L air ⎜
⎟ = 1.16 L O 2
⎝ 1 L air ⎠
Moles of O 2 = n =
(
)(
)
)(
1.1 atm 1.16 L
PV
=
RT
0.0821 L i atm imol-1 i K -1 310 K
(
)
= 0.050 mol O 2
⎛ 6.02 × 1023 molecules O ⎞
2
0.050 mol O 2 ⎜
⎟ = 3.0 × 1022 molecules O 2
⎜
⎟
1 mol O 2
⎝
⎠
⎛ 64.1 g ⎞ ⎛ 1 mol SO 2 ⎞
6.43 (a) dSO2 = ⎜
⎟⎜
⎟ = 2.86 g/L
⎜ 1 mol SO 2 ⎟ ⎜ 22.4 L ⎟
⎝
⎠⎝
⎠
⎛ 16.0 g ⎞ ⎛ 1 mol CH 4 ⎞
(b) d CH4 = ⎜
⎟⎜
⎟ = 0.714 g/L
⎜ 1 mol CH 4 ⎟ ⎜ 22.4 L ⎟
⎝
⎠⎝
⎠
⎛ 2.02 g ⎞ ⎛ 1 mol H 2 ⎞
(c) d H2 = ⎜
⎟⎜
⎟ = 0.0902 g/L
⎜ 1 mol H 2 ⎟ ⎜ 22.4 L ⎟
⎝
⎠⎝
⎠
11/21/06
Page 8 of 10
⎛ 4.00 g ⎞ ⎛ 1 mol He ⎞
(d) d He = ⎜
⎟ = 0.179 g/L
⎟⎜
⎝ 1 mol He ⎠ ⎝ 22.4 L ⎠
⎛ 44.0 g ⎞ ⎛ 1 mol CO 2 ⎞
(e) d CO2 = ⎜
⎟⎜
⎟ = 1.96 g/L
⎜ 1 mol CO 2 ⎟ ⎜ 22.4 L ⎟
⎝
⎠⎝
⎠
Gas comparison: SO2 and CO2 are denser than air (See problem 6.41); He, H2 and
CH4 are less dense than air.
6.47 The densities would be the same. The density of a substance depends on both mass
and volume. Partitioning the gas to two smaller compartments would equally
partition the mass as well.
6.49 (a) PT = PN + PO + PAr
2
2
PN2 = (0.7808)(760 mm Hg) = 593 mm Hg
PO2 = (0.2095)(760 mm Hg) = 159 mm Hg
PAr = (0.0093)(760 mm Hg) =
7 mm Hg
PT = 759 mm Hg
(b) The total pressure exerted by the components is the sum of their partial
pressures: 759 mm Hg (≈ 1.00 atm)
6.53 Intramolecular covalent bonds are stronger than intermolecular hydrogen bonds.
Covalent bonds involve the sharing of electrons, whereas hydrogen bonds involve
weaker electrostatic interactions.
6.65 (a) ~80 mm Hg
(b) ~130 mm Hg
(c) ~330 mm Hg
6.73
⎛ 1.49 g Freon-11 ⎞ ⎛ 1 mol Freon-11 ⎞
-2
1.00 mL Freon-11 ⎜
⎟⎜
⎟ = 1.08 × 10 mol Freon-11
⎜ 1 mL Freon-11 ⎟ ⎜ 137.4 g Freon-11 ⎟
⎠
⎝
⎠⎝
⎛ 6.42 kcal ⎞
-2
1.08 × 10-2 mol Freon-11 ⎜
⎟ = 6.96 × 10 kcal
1
mol
Freon-11
⎝
⎠
11/21/06
Page 9 of 10
6.97 Use the PV = nRT equation after converting some of the units:
⎛ 1 mol NH ⎞
3
Mol of NH 3 = 60.0 g NH 3 ⎜
⎟ = 3.52 mol NH 3
⎜ 17.03 g NH 3 ⎟
⎝
⎠
⎛ 25.4 mm Hg ⎞ ⎛
⎞
1 atm
P (in atm) = 77.2 inch Hg ⎜
⎟⎜
⎟ = 2.58 atm
⎜ 1 inch Hg ⎟ ⎜ 760 mm Hg ⎟
⎝
⎠⎝
⎠
T=
(
)(
)
2.58 atm 35.1 L
PV
=
nR
3.52 mol 0.0821 Liatm i mol-1 iK -1
(
)(
)
= 313 K (40.o C)
⎛ 1 atm ⎞
6.101 (a) Pressure on body = 100 ft ⎜
⎟ = 3.0 atm due to H2O
⎝ 33 ft ⎠
(b) At 1.00 atm, PN = 593 mm Hg (0.780 atm) and thus makes up 78.0% of the
2
gas mixture, which does not change at the depth of 100 feet. At a depth of
100 feet, the total pressure on the lungs, which is equalized by pressure of air
delivered by the SCUBA tank, is 4.0 atm.
⎛
0.78 atm PN 2
PN2 (at 100 ft) = 4.0 atm total pressure ⎜
⎜ 1 atm total pressure
⎝
⎞
⎟ = 3.12 atm PN 2
⎟
⎠
(c) At 1.00 atm, PO2 = 158 mm Hg (0.208 atm) and thus makes up 20.8% of the
gas mixture, which does not change at the depth of 100 feet. At a depth of
100 feet, the total pressure on the lungs, which is equalized by pressure of air
delivered by the SCUBA tank, is 4.0 atm.
⎛
0.21 atm PO 2
PO2 (at 100 ft) = 4.0 atm total pressure ⎜
⎜ 1 atm total pressure
⎝
⎞
⎟ = 0.84 atm PO 2
⎟
⎠
(d) At 100 feet, the partial pressure of N2 in the blood and fatty tissue is
significantly increased. It takes time to release this N2 by transfer to the blood
and then to the lungs for exhalation. If a diver rises too quickly, too much N2
remains in the blood and fatty tissue, affecting the nervous system. In
addition, if bubbles of N2 form in blood vessels, they can impede the flow of
blood, which reduces the flow of O2 to tissues.
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Page 10 of 10