solucionario de zemansky 12 problemas

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1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D to find the mass of one H 2 O molecule: 18.015 u ¿ 1.661 ¿ 10 –27 kg/u = 2.992 ¿ 10 –26 kg/molecule. (70 kg/2.992 ¿ 10 –26 kg/molecule) = 2.34 ¿ 10 27 molecules. (Assuming carbon to be the most common atom gives 3 ¿ 10 27 molecules.

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Page 1: Solucionario de Zemansky 12 Problemas

1.60: Assume a 70-kg person, and the human body is mostly water. Use Appendix D to find the mass of one H2O molecule: 18.015 u ¿ 1.661 ¿ 10–27 kg/u = 2.992 ¿ 10–26 kg/molecule. (70 kg/2.992 ¿ 10–26 kg/molecule) = 2.34 ¿ 1027 molecules. (Assuming carbon to be the most common atom gives 3 ¿ 1027 molecules.