Solution to Set 4: Difference between revisions

From PhyWiki
Jump to navigation Jump to search
Line 4: Line 4:


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 1 cm^3} times {1 cm^3\over 5\times 10^6 m^3} times {1 mol\over 63.55 g} = 1.40\times 10^5{mol\over m^3}</math>
<math>{8.885 g\over 1 cm^3} {1 cm^3\over 5\times 10^6 m^3} {1 mol\over 63.55 g} = 1.40\times 10^5{mol\over m^3}</math>

Revision as of 17:43, 20 February 2009

Problem 1

Cu, density , atomic mass , fcc structure

a. In , the number of moles is