Fisica Serway 4 Edicion Solucionario

© 2000 by Harcourt College Publishers. All rights reserved.
Chapter 1 Even Answers
2. 623 kg/m3
4.
4πρ (r3
2 – r3
1)
3
6. 7.69 cm
8. 8.72 × 1011 atoms/s
10. (a) 72.6 kg (b) 7.82 × 1026 atoms
12. equation is dimensionally consistent
16. The units of G are: m3/kg ⋅ s2
18. 9.19 nm/s
20. (a) 3.39 × 105 ft3 (b) 2.54 × 104 lb
22. 8.32 × 10–4 m/s
24. 9.82 cm
26. (a) 6.31 × 104 AU (b)1.33 × 1011 AU
28. (a) 1.609 km/h (b) 88.5 km/h (c) 16.1 km/h
30. (a) 3.16 × 107 s/yr (b) 6.05 × 1010 yr
32. 2.57 × 106 m3
34. 1.32 × 1021 kg
36. (a) 2.07 mm (b) 8.62 × 1013 times as large
38. (a) 13.4 (b) 49.1
40. rAl = rFe 3 (ρFe/ρAl)
42. ~ 106 km
44. ~ 109 drops
46. time required ≅ 50 years or more;
advise against accepting the offer
48. ~ 105 tons
50. (a) 2 (b) 4 (c) 3 (d) 2
52.(a) 797 (b) 1.1 (c) 17.66
54. (a) 3 (b) 4 (c) 3 (d) 2
56. 5.2 m3, 2.7%
58. 1.79 × 10–9 m
60. 24.6°
62. (b) Acylinder = πR2, Arectangular solid = lw
64. 0.141 nm
66. 289 μm
68. (a) 1000 kg (b) 5.2 × 10–16 kg 0.27 kg (d) 1.3 × 10–5 kg
70. Aluminum: 2.75
g
cm3 (table value is 2% smaller)
Copper: 9.36
g
cm3 (table value is 5% smaller)
Brass: 8.91
g
cm3
Tin: 7.68
g
cm3
Iron: 7.88
gcm3 (table value is 0.3% smaller)
2 Chapter 1 Even Answers
© 2000 by Harcourt College Publishers. All rights reserved.
© 2000 by Harcourt College Publishers. All rights reserved.
Chapter 1 Solutions
*1.1 With V = (base area) · (height)
V = π r2 · h
and ρ =
m
V , we have
ρ =
m
π r 2 h
=
1 kg
π (19.5 mm)2 39.0 mm


 
109 mm3
1 m3
ρ = 2.15 × 104 kg/m3
1.2 ρ =
M
V =
M
43 πR3
ρ =
3(5.64 × 1026 kg)
4π (6.00 × 107 m)3 = 623 kg/m3
1.3 VCu = V0 − Vi =
4
3 π (r3
o – r3i
)
VCu =
4
3 π [(5.75 cm)3 – (5.70 cm)3] = 20.6 cm3
ρ = m
V
m = ρV = (8.92 g/cm3)(20.6 cm3) = 184 g
1.4 V = Vo – Vi =
4
3 π (r3
2 – r3
1 )
ρ =
m
V , so m = ρV = ρ 

 
4
3 π (r3
2 – r3
1) =
4πρ (r3
2 – r3
1)
3
*1.5 (a) The number of moles is n = m/M, and the densityis ρ = m/V. Noting that we have 1 mole,
V1 mol =
mFe
ρFe
=
nFe MFe
ρFe
=
(1 mol)(55.8 g/mol)
7.86 g/cm3 = 7.10 cm3
5.7cm
0.05 cm
2 Chapter 1 Solutions
© 2000 by Harcourt College Publishers. All rights reserved.
(b) In 1 mole of iron are NA atoms:
V1 atom =
V1 mol
NA
=
7.10 cm3
6.02 × 1023 atoms/mol = 1.18 × 10–23 cm3
= 1.18 × 10-29 m3
(c) datom =
3
1.18 × 10–29 m3 = 2.28 ×10–10 m = 0.228 nm
(d) V1 mol U =
(1 mol)(238 g/mol)
18.7 g/cm3 = 12.7 cm3
V1 atom U =
V1 mol U
NA
=
12.7 cm3
6.02 × 1023 atoms/mol = 2.11 × 10–23 cm3
= 2.11 × 10-29 m3
datom U = 3
V1 atom U =
3
2.11 × 10–29 m3 = 2.77 × 10–10 m = 0.277 nm
*1.6 r2 = r1
3
5 = (4.50 cm)(1.71) = 7.69 cm
1.7 Use m = molar mass/NA and 1 u = 1.66 × 10-24 g
(a) For He, m =
4.00 g/mol
6.02 × 1023 mol-1= 6.64 × 10-24 g = 4.00 u
(b) For Fe, m =
55.9 g/mol
6.02 × 1023 mol-1 = 9.29 × 10-23 g = 55.9 u
(c) For Pb, m =
207 g/mol
6.02 × 1023 mol-1 = 3.44 × 10-22 g = 207 u
Chapter 1 Solutions 3
© 2000 by Harcourt College Publishers. All rights reserved.
Goal Solution
Calculate the mass of an atom of (a) helium, (b) iron, and (c) lead. Give your answers in atomic
mass units and in grams. Themolar masses are 4.00, 55.9, and 207 g/mol, respectively, for the
atoms given.
Gather information: The mass of an atom of any element is essentially the mass of the protons
and neutrons that make up its nucleus since the mass of the electrons is negligible (less than a
0.05% contribution). Since most atoms have about the same number of neutrons as protons, the
atomic mass is approximatelydouble the atomic number (the number of protons). We should also
expect that the mass of a single atom is a very small fraction of a gram (~10–23 g) since one mole
(6.02 × 1023) of atoms has a mass on the order of several grams.
Organize: An atomic mass unit is defined as 1/12 of the mass of a carbon-12 atom (which has a
molar mass of 12.0 g/mol), so the mass of any atom in atomic mass units is...