Question #a292a

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Question #a292a

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Answer 1 · 2017-05-14T03:44:52

$113 g$ $113 g$

Explanation:

We can solve this using simple dimensional analysis.

Start with the given quantity $1.70 \times 10^{25}$ $1.70 xx 10^(25)$ atoms $H e$ $He$ , and recognize that there are $4.003 g H e$ $4.003 g He$ in $6.02 \times 10^{23}$ $6.02 xx 10^(23)$ atoms $H e$ $He$ :

$g H e = 1.70 \times 10^{25} a t o m s H e \frac{4.003 g H e}{6.02 \times 10^{23} a t o m s H e} = 113 g H e$ $g He = 1.70 xx 10^(25) at oms He(4.003 g He)/(6.02 xx 10^(23) at omsHe) = 113 g He$

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Question #a292a

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