Question #60c07

1 Answer
Jan 17, 2018

3.107*10^21 "atoms Mn"3.1071021atoms Mn

Explanation:

First you need to find the mass of Manganese (Mn) in the coin. Since you know what percent of the coin is Mn and you have the mass of the coin, you can multiply those together to find the mass of Mn:

8.100"g" * 3.5% = 0.2835"g Mn"8.100g3.5%=0.2835g Mn

Now that we have the mass of Mn, we can find how many mols of Mn there are by using its molar mass, 54.94"g Mn/mol Mn"54.94g Mn/mol Mn:

0.2835"g Mn" * ( 1 "mol Mn")/(54.94"g Mn" ) = 0.005160"mol Mn"0.2835g Mn1mol Mn54.94g Mn=0.005160mol Mn

Finally we can use Avogadro's Number, 6.022*10^23"particles/mol"6.0221023particles/mol to determine how many particles of Mn there are in the coin:

0.005160"mol Mn" * (6.022*10^23 "atoms Mn")/(1"mol Mn") = 3.107*10^21 "atoms Mn"0.005160mol Mn6.0221023atoms Mn1mol Mn=3.1071021atoms Mn