Solution to Set 4: Difference between revisions

Revision as of 17:21, 20 February 2009

Cu, density $\rho =8.885{g \over cm^{3}}$ , atomic mass $m_{a}=63.57amu$ , fcc structure

a. In $1m^{3}$ , the number of moles is $8.885{g \over (cm^{3})}x{1cm^{3}/over(1x10^{-}6m^{3})}x{1mol/over(63.55g)}=1.40x10^{5}{mol/overm^{3}}$

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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 cm^3} x {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)} x {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>