How many moles are present in $7.25 \times 10^{25}$ $7.25xx10^25$ $carbon dioxide molecules$ $"carbon dioxide molecules"$ ?

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How many moles are present in $7.25 \times 10^{25}$ $7.25xx10^25$ $carbon dioxide molecules$ $"carbon dioxide molecules"$ ?

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Answer 1 · 2017-02-02T14:31:47

Well one mole of stuff, any stuff, including carbon dioxide, specifies $6.022 \times 10^{23}$ $6.022xx10^23$ individual items of that stuff.

Explanation:

And thus we work out the quotient:

$\frac{7.2 \times 10^{25} \cdot carbon dioxide molecules}{6.022 \times 10^{23} \cdot carbon dioxide molecules \cdot m o l^{- 1}}$ $(7.2xx10^25*"carbon dioxide molecules")/(6.022xx10^23*"carbon dioxide molecules"*mol^-1)$

$≅ 120 \cdot m o l$ $~=120*mol$ $carbon dioxide$ $"carbon dioxide"$ .

This is dimensionally consistent, because we get an answer with units $\frac{1}{m o l^{- 1}} = \frac{1}{\frac{1}{mol}} = m o l$ $1/(mol^-1)=1/(1/"mol")=mol$ as required.

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How many moles are present in $7.25 \times 10^{25}$ $7.25xx10^25$ $carbon dioxide molecules$ $"carbon dioxide molecules"$ ?

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How many moles are present in $7.25 \times 10^{25}$ $7.25xx10^25$ $carbon dioxide molecules$ $"carbon dioxide molecules"$ ?