What is the composition in atom percent of an alloy that consists of 46.5 lb of silver?

The composition in atom percent of an alloy that consists of 46.5 lb of silver can be calculated by converting the mass of silver to moles and then using the molar masses and atomic masses to determine the atom percent. The molar mass of silver is 107.87 g/mol. To convert the mass of silver from pounds to grams, we can use the conversion factor 1 lb = 453.6 g. Therefore, the mass of silver in grams is 46.5 lb x 453.6 g/lb = 21042.6 g. To convert the mass of silver from grams to moles, we can use the molar mass of silver. Therefore, the number of moles of silver is 21042.6 g / 107.87 g/mol = 195.01 mol. Finally, to calculate the composition in atom percent, we can use the formula:

atom percent = (number of moles / total moles) x 100

Since the alloy consists only of silver, the total moles is equal to the number of moles of silver. Therefore, the composition in atom percent is:

atom percent = (195.01 mol / 195.01 mol) x 100 = 100%