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OpenStudy (anonymous):
Determine the number of atoms of nitrogen present in 0.5g of calcium nitrate, Ca(NO3)2. (Ca=40.0 N=14.0 o=16.0)
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OpenStudy (anonymous):
\[.5 g Ca(NO_3)_2 (\frac{1 mole Ca(NO_3)_2}{164 gCa(NO_3)_2 })=3.05^-3 moles \]
OpenStudy (anonymous):
\[3.05^-3 moles Ca(NO_3)_2(\frac{6.022x10^23 molecule Ca(NO_3)_2}{1 moleCa(NO_3)_2 })=1.84x10^23 molecules \] That should be done in 1 2 step factor label but i forgot what i was looking for
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