You have a 20.0-g sample of silver metal. You are given 10.0 g of another metal and told that this sample contains twice the number of atoms as the sample of silver metal. Is this possible?

The moles of silver in the 20.0 g sample are \( \frac{20.0 \, \text{g}}{107.87 \, \text{g/mol}} \) which gives us the number of silver atoms as \( \text{Moles of silver} \times (6.022 \times 10^{23} \, \text{atoms/mol}) \). The other metal sample has twice the number of atoms, so we have \( \frac{\text{Number of other metal atoms}}{6.022 \times 10^{23} \, \text{atoms/mol}} \) moles of the other metal. To find the molar mass of the other metal, we use \( \frac{10.0 \, \text{g}}{\text{Moles of other metal}} \). Compare the calculated molar mass of the other metal to known values of molar masses in the periodic table. If it closely matches the molar mass of an existing element, then it is possible; otherwise, it is not possible.