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 | <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> | ||
Revision as of 17:24, 20 February 2009
Problem 1
Cu, density , atomic mass , fcc structure
a. In , the number of moles is Failed to parse (syntax error): {\displaystyle \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}}
