Answer
A) $ 1.16\times 10^{23}$ atoms Sr
B) $2.81\times 10^{23}$ atoms Fe
C) $2.846\times 10^{22}$ atoms Bi
D) $7.43\times 10^{23}$ atoms P
Work Step by Step
a) $ 16.9 \textrm{g Sr}\times \frac{1\textrm{mol Sr}}{87.62\textrm{g Sr}}\times \frac{6.022^{22}\textrm{atoms Sr}}{1\textrm{mol Sr}} = 1.16\times 10^{23} \textrm{atoms Sr} $
b) $ 26.1 \textrm{g Fe}\times \frac{1\textrm{mol Fe}}{55.85\textrm{g Fe}}\times \frac{6.022^{22}\textrm{atoms Fe}}{1\textrm{mol Fe}} = 2.81\times 10^{23}\textrm{atoms Fe} $
c) $ 8.55 \textrm{g Bi}\times \frac{1\textrm{mol Bi}}{208.98\textrm{g Bi}}\times \frac{6.022^{22}\textrm{atoms Bi}}{1\textrm{mol Bi}} = 2.846\times 10^{22} \textrm{atoms Bi} $
d) $ 38.2 \textrm{g P}\times \frac{1\textrm{mol P}}{30.97\textrm{g P}}\times \frac{6.022^{22}\textrm{atoms P}}{1\textrm{mol P}} = 7.43\times 10^{23} \textrm{atoms P} $
