Chapter16 - 16.1. Solve: The mass of lead mPb = Pb VPb =...

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16.1. Solve: The mass of lead mV Pb Pb Pb 33 11,300 kg m 2.0 m 22,600 kg == () ( ) = ρ . For water to have the same mass its volume must be V m water water water 3 3 22,600 kg kg m 22.6 m = 1000
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16.2. Solve: The volume of the uranium nucleus is VR == × () −− 4 3 3 4 3 15 3 42 7 5 10 1 767 10 ππ .. m m 3 The density of the uranium nucleus is ρ nucleus nucleus nucleus 3 3 kg m kg m × × m V 40 10 1 767 10 226 10 25 42 17 . . . Assess: This density is extremely large compared to the typical density of materials.
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16.3. Solve: The volume of the aluminum cube is 10 3 m 3 and its mass is mV Al Al Al = ρ = 2700 kg m m kg 33 () × = 10 10 27 3 .. The volume of the copper sphere with this mass is Vr m r Cu Cu Cu Cu 3 3 Cu 3 2.7 kg 8920 kg m m m 4 m = == = × ⇒= × = 4 3 3 027 10 3 3 027 10 0 04165 3 4 4 1 3 π . . . The diameter of the copper sphere is 0.0833 m = 8.33 cm.
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16.4. Solve: The volume of the hollow sphere is Vr r =− () = [] 4 3 4 3 2 555 10 3 3 4 ππ out 3 in 3 3 0.050 m 0.040 m m . Thus, the density of the shell is ρ == × = M V 0.690 kg m 2700 kg m 3 3 2 555 10 4 . Table 16.1 identifies this density with aluminum.
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16.5. Solve: The volume of the aluminum cube V = 8.0 × 10 6 m 3 and its mass is M = ρ V = (2700 kg/m 3 )(8.0 × 10 6 m 3 ) = 0.0216 kg = 21.6 g One mole of aluminum ( 27 Al) has a mass of 27 g. The number of atoms is N = × () 602 10 482 10 23 23 . . atoms 1 mol 1 mol 27 g 21.6 g atoms
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16.6. Solve: The volume of the copper cube is 8.0 × 10 6 m 3 and its mass is M = ρ V = (8920 kg/m 3 )(8.0 × 10 6 m 3 ) = 0.07136 kg = 71.36 g Because the atomic mass number of Cu is 64, one mole of Cu has a mass of 64 g. The number of moles in the cube is n = () = 1 mol 64 g 71.36 g mol 112 .
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16.7. Solve: (a) The number density is defined as NV , where N is the number of particles occupying a volume V . Because Al has a mass density of 2700 kg/m 3 , a volume of 1 m 3 has a mass of 2700 kg. We also know that the molar mass of Al is 27 g/mol or 0.027 kg/mol. So, the number of moles in a mass of 2700 kg is n = () 2700 kg 1 mol 0.027 kg mol 100 10 5 . The number of Al atoms in 1.00 × 10 5 mols is Nn N == × × A mol atoms mol atoms 1 00 10 6 02 10 6 02 10 52 3 2 8 .. . Thus, the number density is N V = × 602 10 28 28 . . atoms 1 m atoms m 3 3 (b) Pb has a mass of 11,300 kg in a volume of 1 m 3 . Since the atomic mass number of Pb is 207, the number of moles in 11,300 kg is n = 11,300 kg 1 mole 0.207 kg The number of Pb atoms is thus N = nN A , and hence the number density is N V nN V × A 33 11300 kg 0.207 kg atoms mol 1 mol 1 m atoms m 6 02 10 3 28 10 23 28
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16.8. Solve: The mass density is ρ = M / V . The mass M of the sample is related to the number of atoms N and the mass m of each atom by M = Nm . Combining these, the atomic mass is m M N V NN V == = = × ρρ /. . 1750 439 10 3 986 10 28 26 kg/m atoms/m kg / atom 3 3 The atomic mass in m = A u, where A is the atomic mass number. Thus A m × × = 1 3 986 10 1 661 10 24 26 27 u kg kg .
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This homework help was uploaded on 04/02/2008 for the course PHYS 131-133 taught by Professor All during the Spring '08 term at Cal Poly.

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Chapter16 - 16.1. Solve: The mass of lead mPb = Pb VPb =...

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