Question #8b2ee

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Question #8b2ee

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Answer 1 · 2016-12-10T11:54:11

The mass of one atom of iron is $(9.2741105 \times 10^{- 23})$ $(9.2741105xx10^-23)$ .

Explanation:

The formula we are using is $m = \frac{M}{n}$ $m=M/n$ . Where...
=> $m$ $m$ is the mass in grams.
=> $M$ $M$ is the molar mass in $\frac{g}{m o l}$ $g/(mol)$ .
=> $n_{A}$ $n_"A"$ is Avogadro's number - $6.02 \times 10^{23}$ $6.02xx10^23$ .

$m = \frac{M}{n_{A}}$ $m=M/n_"A"$

$= \frac{55.85}{6.02 \times 10^{23}}$ $=55.85/(6.02xx10^23)$

$= (9.2741105 \times 10^{- 23})$ $=(9.2741105xx10^-23)$

The mass of one atom of iron is $(9.2741105 \times 10^{- 23})$ $(9.2741105xx10^-23)$ .

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