Solution to Set 4: Difference between revisions

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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>\left 8.885{g\over cm^3}\right {1 cm^3/over (1 x 10^-6 m^3)} x {1 mol/over (63.55 g)} = 1.40 x 10^5 {mol/over m^3}</math>
<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