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Section 2.1 – Atomic Structure
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Problem 2.1 Solution
Use density of gold ρ gold = 19.28 g / cm3 and atomic mass of gold = 196.967 amu
N AV
atomic weight
π (2.0 mm) 2 0.6023 × 10 24 atoms
1 cm 3
19.28 g
100
mm
×
×
×
×
=
4
196.967 g
1000 mm 3
cm 3
= 1.85×1022 atoms of gold
N atoms = ρV
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Problem 2.3 Solution
Use density of SiO2 = 2.20 g / cm3, atomic mass of silicon = 28.086 amu, and atomic mass of
oxygen = 15.999 amu.
molecular weight of SiO2 = (28.086) + 2(15.999) = 60.084 g / mol
mol SiO2 =
ρV
molecular weight
2.20 g
mol
1 cm 3
2
80
nm
1
mm
×
×
×
×
60.084 g (10 7 nm )(100 mm 2 )
cm 3
= 2.9292 moles of SiO2
=
Silicon:
2.9292 × 10
−13
1 gram - atom Si 0.6023 × 10 24 atoms
×
= 1.76×1015 atoms
mol SiO 2 ×
1 mol SiO 2
gram - atom
Oxygen:
2.9292 × 10 −13 mol SiO 2 ×
2 gram - atom O 0.6023 × 10 24 atoms
×
= 3.53×1015 atoms
1 mol SiO 2
gram - atom
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Problem 2.5 Solution
Use density of Al2O3 = 3.97 g / cm3, atomic mass of aluminum = 26.982 amu, and atomic mass
of oxygen = 15.999 amu.
molecular weight of Al2O3 = 2(26.982) + 3(15.999) = 101.961 g / mol
mol Al2O3 =
ρV
molecular weight
§ 1 cm ·
π (1.0 µ m) 2
mol
3.97 g
¸
×
×
×
× ¨¨
25
m
µ
=
3
4
101.961 g © 10,000 µ m ¸¹
cm
= 7.645×10í13 moles of Al2O3
3
Aluminum:
7.645 × 10 −13 mol Al 2 O 3 ×
2 gram - atom Al 0.6023 × 10 24 atoms
= 9.21×1011 atoms
×
1 mol Al 2 O 3
gram - atom
Oxygen:
7.645 × 10 −13 mol Al 2 O 3 ×
3 gram - atom O 0.6023 × 10 24 atoms
= 1.38×1012 atoms
×
1 mol Al 2 O 3
gram - atom
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Problem 2.8 Solution
Let
x = atomic fraction of Cl35
1 − x = atomic fraction of Cl37
Then
35 x + 37(1 − x) = 35.453
− 2 x + 37 = 35.453
35.453 − 37
= 0.7735
x =
−2
Therefore, naturally occurring chlorine is 77.35% Cl35 and 22.65% Cl37.
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Section 2.2 – The Ionic Bond
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Problem 2.13 Solution
Fc = − k o
Z 1 qZ 2 q
2
1.3824
1
V⋅m·
−18
= − §¨ 9 × 10 9
nN - nm 2
¸ 0.16 × 10 C (3)( −2) 2 =
2
2
C
a
a
a
©
¹
a [nm]
0.200
0.250
0.300
0.350
0.400
0.450
0.500
0.550
0.600
0.650
0.700
(
)
Force [nN]
34.6
22.1
15.4
11.3
8.64
6.83
5.53
4.57
3.84
3.27
2.82
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Problem 2.16 Solution
Use rFe 2+ = 0.087 nm and rO 2− = 0.132 nm
a = 0.087 nm + 0.132 nm = 0.219 nm
Z qZ q
Fc = − k o 1 2 2
a
2
1
§
−18
9 V⋅m·
¸(0.16 × 10 C ) ( 2)( −2)
= − ¨ 9 × 10
C ¹
(0.219 nm ) 2
©
= 19.2 nN
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Problem 2.17 Solution
Use rFe 2+ = 0.087 nm and rO 2− = 0.132 nm
a o = 0.087 nm + 0.132 nm = 0.219 nm
a = 2a = 0.3097 nm
Z qZ q
Fc = − k o 1 2 2
a
2
1
§
9 V⋅m·
−18
¸(0.16 × 10 C ) ( 2)( 2)
= − ¨ 9 × 10
C ¹
(0.219 nm ) 2
©
= í 9.61 nN
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Problem 2.18 Solution
Use rFe 2+ = 0.087 nm and rO 2− = 0.132 nm
a o = 0.087 nm + 0.132 nm = 0.219 nm
a = 2a = 0.3097 nm
Z qZ q
Fc = − k o 1 2 2
a
2
1
§
−18
9 V⋅m·
¸(0.16 × 10 C ) ( −2)( −2)
= − ¨ 9 × 10
C ¹
(0.219 nm ) 2
©
= í 9.61 nN
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Section 2.3 – The Covalent Bond
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Problem 2.34 Solution
Use
atomic weight of C = 12.011 amu
atomic weight of H = 1.0079 amu
Ethylene is C2H4.
Molecular weight = 600[2(12.011 amu) + 4(1.0079 amu)]
= 16,800 amu
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Problem 2.35 Solution
Use
atomic weight of C = 12.011 amu
atomic weight of H = 1.0079 amu
atomic weight of O = 15.999 amu
Methyl methacrylate is C5H8O2.
Molecular weight = 450[5(12.011 amu) + 8(1.0079 amu) + 2(15.999 amu)]
= 45,050 amu
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Problem 2.36 Solution
Use
atomic weight of C = 12.011 amu
atomic weight of H = 1.0079 amu
atomic weight of O = 15.999 amu
Methyl methacrylate is C5H8O2.
Low molecular weight = 300[5(12.011 amu) + 8(1.0079 amu) + 2(15.999 amu)]
= 30,030 amu
High molecular weight = 800[5(12.011 amu) + 8(1.0079 amu) + 2(15.999 amu)]
= 80,090 amu
The molecular weights range from 30,030 amu to 80,090 amu.
Problem 2.37 Solution
Use
atomic weight of C = 12.011 amu
atomic weight of H = 1.0079 amu
atomic weight of O = 15.999 amu
C = C bond energy = 680 kJ / mol
C – C bond energy = 370 kJ / mol
mol. wt. of C5H8O2 = 5(12.011 amu) + 8(1.0079 amu) + 2(15.999 amu)
= 100.1162 amu
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When 1 mole of C5H8O2 is polymerized, 1 mole of C = C bonds are changed to 2 moles of C – C
bonds.
Energy = 13 g ×
mol
kJ
= 7.79 kJ
× ( 2 × 370 − 680 )
100.1162 g
mol
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Problem 2.39 Solution
The mer unit for polystyrene is
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(a) The polymer structure is
(b) When 1 mole of the mer unit is polymerized, 1 mole of C=C bonds are changed to 2 moles of
C – C bonds.
C = C bond energy = 680 kJ / mol
C – C bond energy = 370 kJ / mol
Reaction energy = 2 × 370 − 680
kJ
= 60 kJ / mol
mol
(c) One mer unit has 8 carbon atoms and 8 hydrogen atoms.
atomic weight of C = 12.011 amu
atomic weight of H = 1.0079 amu
molecular weight = 500[8(12.011 amu) + 8(1.0079 amu)]
= 52,075 amu
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Section 2.4 – The Metallic Bond
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Section 2.5 – The Secondary, or van der Waals, Bond
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Problem 2.50 Solution
Calculate the number of moles of N2 gas per kg of silica gel
PV = nRT
9.42 × 10 6 mm 3 mol - K
1
PV
×
×
= 4.20297 mol / kg
= (1 atm) ×
n =
RT
100 g
8.314 J 273.15 K
Assume 0.162 nm2 per N2 molecule.
Surface area =
4.20297 mol 0.6023 × 10 24 molecules 0.162 nm 2
×
×
= 410,000 m2 / kg
kg
mol
molecule
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Problem 2.51 Solution
Use R = 8.314 J/mol - K, S1 = 5.51×1023 atoms/(m3 – atm), T1 = 298.15 K, and T2 = 523.15 K.
Find the ratio of the solubility at the two temperatures to eliminate the unknown constant.
S 0 exp[− ∆H /( RT1 )]
S1
=
S2
S 0 exp[− ∆H /( RT2 )]
ª − ∆H § 1 1 ·º
¨¨ − ¸¸»
S 2 = S1 exp «
R
© T2 T1 ¹¼
¬
ª 3.96 kJ mol - K §
1
1
1
·º
§
·
23
×
−
¸
¸ exp «
¨
= ¨ 5.51 × 10
3
8.314 J © 523.15 K 298.15 K ¹»¼
m − atm ¹
©
¬ mol
= 2.77×1023 atoms/(m3 – atm)
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Problem 2.52 Solution
Use R = 8.314 J/mol - K, S1 = 9.07×1023 atoms/(m3 – atm), T1 = 298.15 K, and T2 = 523.15 K.
Find the ratio of the solubility at the two temperatures to eliminate the unknown constant.
S 0 exp[− ∆H /( RT1 )]
S1
=
S2
S 0 exp[− ∆H /( RT2 )]
ª − ∆H
S 2 = S1 exp «
¬ R
§ 1 1 ·º
¨¨ − ¸¸»
© T2 T1 ¹¼
ª 6.70 kJ mol - K §
1
1
1
§
·
·º
23
×
−
¸ exp «
¸
¨
= ¨ 9.07 × 10
3
8.314 J © 523.15 K 298.15 K ¹»¼
m − atm ¹
©
¬ mol
= 2.84×1023 atoms/(m3 – atm)
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