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Chemistry 6 Online
OpenStudy (anonymous):

How many atoms of carbon are there in 100. grams of C30H62

OpenStudy (anonymous):

\[100g(\frac{1 mole C30H62}{422 g C30H62})(\frac{6.02x10^{23} molecules}{1 mole C30H62})(\frac{30 C atoms}{1 molecule})=?\]

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