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>{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> | <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> | ||
b. Atoms in <math>1 m^3</math> | |||
<math>{1.40\times 10^5 mol\over m^3} {6.022\times 10^23\over 1 mol}= 8.43\times 10^28</math> | |||
Revision as of 17:46, 20 February 2009
Problem 1
Cu, density , atomic mass , fcc structure
a. In , the number of moles is
b. Atoms in
