## Chemistry 12th Edition

We can find the number of atoms in 1.10g H as $1.10g\space H\times \frac{1mol\space H}{1.008g\space H}\times \frac{6.022\times 10^{23}H\space atoms}{1mol\space H}=6.57\times 10^{23}H\space atoms$ Similarly the number of atoms in 14.7g of Chromium can be determined as $14.7g\space Cr\times \frac{1mol\space Cr}{52.00g\space Cr}\times\frac{6.022\times 10^{23}Cr\space atoms}{1mol\space Cr}=1.70\times 10^{23}Cr\space atoms$