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CHAPTER 43 - 7

# CHAPTER 43 - 7 - N =(7.50 10...

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4 H 1 1 He 2 4 + 2e + + 2 ν   C 6 12 + H 1 1 N 7 13   N 7 13 C 6 13 + e +   C 6 13 + H 1 1 N 7 14   N 7 14 + H 1 1 O 8 15   O 8 15 N 7 15 + e + N   = (7.50 × 10 25  MeV/h)/[(7.30 MeV)/(4 nuclei)](2 nuclei/molecule)(0.015 × 10 –2 = 1.37 × 10 29  molecules/h. The required mass of water is m   = [(1.37 × 10 29  molecules/h)/(6.02 × 10 23  molecules/mol)](18 g/mol)  = 4.1 × 10 6  g/h =      4.1 × 10 3  kg/h . 37. The number of deuterium nuclei in 1.00 kg of water is N   = [(1.00 × 10 3  g)/(18 g/mol)](6.02 × 10 23  molecules/mol)(2 nuclei/molecule)(0.015 × 10 –2 = 1.00 × 10 22  nuclei. Two deuterium nuclei release 4.03 MeV, so the total energy is E  = (1.00 × 10 22  nuclei)(4.03 MeV)(1.60 × 10 –13  J/MeV)/(2 nuclei) =      3.23 × 10 9  J . This is       65 ×       the energy from burning 1.0 kg of gasoline. 38. ( a ) If we look at the reactions for the carbon cycle, we see that one carbon atom is used in the first  reaction and one carbon atom is produced in the last reaction.  If we add all of the reactions, we get

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