Solution to Set 4: Difference between revisions

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Cu, density <math> \rho= 8.885{g\over cm^3}</math>, atomic mass <math> m_a = 63.57 amu</math>, fcc structure
Cu, density <math> \rho= 8.885{g\over cm^3}</math>, atomic mass <math> m_a = 63.57 amu</math>, fcc structure


a. In <math>1 m^3</math>, the number of moles is  
a. In <math>1 m^3</math>, the number of moles is
<math> 8.885{g\over cm^3} {1 cm^3/over (1e-6 m^3)} x {1 mol/over (63.55 g)} = 1.40e5 {mol/over m^3}</math>

Revision as of 17:25, 20 February 2009

Problem 1

Cu, density , atomic mass , fcc structure

a. In , the number of moles is