# Fisica cap 1

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Chapter 1 Solutions

*1.1

With V = (base area) · (height) V = π r2 · h and ρ = m , we have V
9 3 m 1 kg  10 mm  = π r2 h π (19.5 mm)2 39.0 mm  1 m3 

ρ=

ρ = 2.15 × 104 kg/m3
1.2

ρ=

M M = V 4 πR3 3 3(5.64 × 1026 kg) = 623 kg/m3 4 π (6.00 × 107 m) 3 4 3 3 π (ro – ri ) 3
5.7cm 0.05 cm

ρ=

1.3

VCu = V0 − Vi =

VCu =

4 π [(5.75 cm)3 – (5.70 cm)3] = 20.6 cm3 3

ρ=mV
m = ρV = (8.92 g/cm3)(20.6 cm3) = 184 g 1.4 V = Vo – Vi = 4 3 3 π (r2 – r1 ) 3
3 3

ρ=
*1.5 (a)

4 πρ (r2 – r1 ) m 4 3 3 , so m = ρV = ρ  π  (r2 – r1) = V 3 3  The number of moles is n = m/M, and the density is ρ = m/V. Noting that we have 1 mole, V1 mol = mFe nFe MFe (1 mol)(55.8 g/mol) = = = 7.10 cm3 ρFe ρFe 7.86 g/cm3

Chapter 1 Solutions

(b)

In 1 mole of iron are NA atoms: V1 atom = V1 mol 7.10 cm3 = = 1.18 × 10–23 cm3 NA 6.02 × 1023 atoms/mol

= 1.18 × 10-29 m3
3

(c) (d)

datom =

1.18 × 10–29 m3 = 2.28 × 10–10 m = 0.228 nm (1 mol)(238 g/mol) = 12.7 cm3 18.7 g/cm3 V1 mol U 12.7 cm3 = = 2.11 × 10–23 cm3 NA 6.02 × 1023 atoms/mol

V1 mol U =

V1 atom U =

= 2.11 × 10-29 m3
3 3datom U =
3

V1 atom U =

2.11 × 10–29 m3 = 2.77 × 10–10 m = 0.277 nm

*1.6 1.7

r2 = r1

5 = (4.50 cm)(1.71) = 7.69 cm

Use m = molar mass/NA and 1 u = 1.66 × 10-24 g (a) For He, m = 4.00 g/mol = 6.64 × 10-24 g = 4.00 u 6.02 × 1023 mol-1

(b)

For Fe, m =

55.9 g/mol = 9.29 × 10-23 g = 55.9 u 6.02 × 1023 mol-1 207 g/mol -22 g = 207 u 23 -1 = 3.44 × 10 6.02 × 10 mol

(c)

ForPb, m =

Chapter 1 Solutions

3

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. The molar 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 themass 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 approximately double 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 simply the numerical value of the molar mass. The mass in grams can be found by multiplying the molar mass by the mass of one atomic mass unit (u): 1 u = 1.66 × 10–24g. Analyze: For He, m = 4.00 u = (4.00 u)(1.66 × 10–24 g/u) = 6.64 × 10–24 g For Fe, m = 55.9 u = (55.9 u)(1.66 × 10–24g/u) = 9.28 × 10–23 g For Pb, m = 207 u = (207 u)(1.66 × 10–24 g/u) = 3.44 × 10–22 g Learn: As expected, the mass of the atoms is larger for bigger atomic numbers. If we did not know the conversion factor for atomic mass units, we could use the mass of a proton as a closeapproximation: 1u ≈ mp = 1.67 × 10–24 g. ∆m 3.80 g – 3.35 g = = 0.00228 mol M 197 g/mol

*1.8

∆n =

∆N = (∆n)NA = (0.00228 mol)(6.02 × 1023 atoms/mol) = 1.38 × 1021 atoms ∆t = (50.0 yr)(365 d/yr)(24.0 hr/d)(3600 s/hr) = 1.58 × 109 s ∆N 1.38 × 1021 atoms = = 8.72 × 1011 atoms/s ∆t 1.58 × 109 s 1.9 (a) m = ρL3 = (7.86 g/cm3)(5.00 × 10-6 cm)3 = 9.83 × 10-16 g NA   = (9.83 × 10-16 g)(6.02 × 1023atoms/mol) Molar mass 55.9 g/mol

(b)

N=m

= 1.06 × 107 atoms

4

Chapter 1 Solutions

1.10

(a)

The cross-sectional area is A = 2(0.150 m)(0.010 m) + (0.340 m)(0.010 m) = 6.40 × 10-3 m2

15.0 cm

1.00 cm

36.0 cm

The volume of the beam is V = AL = (6.40 × 10-3 m2)(1.50 m) = 9.60 × 10-3 m3 Thus, its mass is...