Chapter 16

Chapter 16 - 16.1. Model: Recall the density of water is...

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16.1. Model: Recall the density of water is 1000 kg/m 3 . Solve: The mass of lead ( )( ) 33 Pb Pb Pb 11,300 kg m 2.0 m 22,600 kg mV ρ == = . For water to have the same mass its volume must be 3 water water 3 water 22,600 kg 22.6 m 1000 kg m m V = Assess: Since the lead is 11.3 times as dense we expect the water to take 11.3 times the volume.

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16.2. Model: Assume the nucleus is spherical. Solve: The volume of the uranium nucleus is () 3 31 5 4 2 3 44 33 7.5 10 m 1.767 10 m VR ππ −− == × The density of the uranium nucleus is 25 17 3 nucleus nucleus 42 3 nucleus 4.0 10 kg 2.3 10 kg m 1.767 10 m m V ρ × = × × Assess: This density is extremely large compared to the typical density of materials.
16.3. Solve: The volume of the aluminum cube is 10 3 m 3 and its mass is Al Al Al mV ρ == ( )( ) 33 3 2700 kg m 1.0 10 m 2.7 kg ×= The volume of the copper sphere with this mass is () 3 43 Cu Cu Cu 3 Cu 1 3 Cu 4 2.7 kg 3.027 10 m 3 8920 kg m 3 3.027 10 m 0.042 m 4 m Vr r π = = × ⎡⎤ × ⎢⎥ ⇒= = ⎣⎦ The diameter of the copper sphere is 0.0833 m = 8.33 cm. Assess: The diameter of the sphere is a little less than the length of the cube, and this is reasonable considering the density of copper is greater than the density of aluminum.

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16.4. Model: The volume of a hollow sphere is () 33 out in 4 3 Vr r =− p Solve: We are given 0 690 kg, m. = out 0050±m , r. = and we know that for aluminum 3 2700 kg/m . = r Solve the above equation for in r . 3 3 in out 4 3 3 3 out 4 3 3 3 3 4 3 / 0 690 kg/2700 kg/m 0.050 m 0040±m V rr m r . . = p r p p So the inner diameter is 8.0 cm. Assess: We are happy that the inner diameter is less than the outer diameter, and in a reasonable range.
16.5. Solve: The volume of the aluminum cube 63 8.0 10 m V and its mass is 36 3 (2700 kg/m )(8.0 10 m ) 0.0216 kg 21.6 g MV ρ == × = = One mole of aluminum ( 27 Al) has a mass of 27 g. The number of atoms is () 23 23 6.02 10 atoms 1 mol 21.6 g 4.8 10 atoms 1 mol 27 g N ⎛⎞ × × ⎜⎟ ⎝⎠ Assess: This is almost one mole of atoms, which is a reasonable value.

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16.6. Solve: The volume of the copper cube is 8.0 × 10 6 m 3 and its mass is 36 3 (8920 kg/m )(8.0 10 m ) 0.07136 kg 71.36 g MV ρ == × = = 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 () 1 mol 71.36 g 1.1 mol 64 g n ⎛⎞ ⎜⎟ ⎝⎠ Assess: This answer is in the same ballpark as the previous exercise.
16.7. Solve: (a) The number density is defined as , N V 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 () 5 1 mol 2700 kg 1.00 10 mol 0.027 kg n ⎛⎞ == × ⎜⎟ ⎝⎠ The number of Al atoms in 1.00 × 10 5 mols is ( ) ( ) 52 3 2 8 A 1.00 10 mol 6.02 10 atoms mol 6.02 10 atoms Nn N == × × Thus, the number density is 28 28 3 3 6.02 10 atoms 6.02 10 atoms m 1 m N V × × (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 1 mole 11,300 kg 0.207 kg n = The number of Pb atoms is thus N = nN A , and hence the number density is ( ) 23 28 A 33 1 mol 11,300 kg atoms atoms 6.02 10 3.28 10 0.207 kg mol 1 m m N VV × = × Assess: We expected to get very large numbers like this.

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16.8.
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This note was uploaded on 11/26/2010 for the course MTH 232 taught by Professor Smith during the Spring '10 term at UConn.

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Chapter 16 - 16.1. Model: Recall the density of water is...

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