Question #60c07

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Question #60c07

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Answer 1 · 2018-01-17T23:42:37

$3.107 \cdot 10^{21} atoms Mn$ $3.107*10^21 "atoms Mn"$

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:

$8.100 g \cdot 3.5 % = 0.2835 g Mn$ $8.100"g" * 3.5% = 0.2835"g Mn"$

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

$0.2835 g Mn \cdot \frac{1 mol Mn}{54.94 g Mn} = 0.005160 mol Mn$ $0.2835"g Mn" * ( 1 "mol Mn")/(54.94"g Mn" ) = 0.005160"mol Mn"$

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

$0.005160 mol Mn \cdot \frac{6.022 \cdot 10^{23} atoms Mn}{1 mol Mn} = 3.107 \cdot 10^{21} atoms Mn$ $0.005160"mol Mn" * (6.022*10^23 "atoms Mn")/(1"mol Mn") = 3.107*10^21 "atoms Mn"$

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Question #60c07

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