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Chemistry 11 Online
OpenStudy (anonymous):

how many molecules are there in 57.3g of ethane?

OpenStudy (anonymous):

given mass / molar mass * (6.022*10^23) will give u the no of molecules

sam (.sam.):

\[57.3gC2H6 \times \frac{6.02 \times 10^{23} moleculesC2H6}{30gC2H6}\]

OpenStudy (callisto):

Chemical formula of ethane = C2H6 no. of mole of ethane = 57.3 / (12.0x2+1.0x6) = 57.3/30 = 1.91 mole no. of molecules required = no. of mole x Avogadro's Constant = 1.91 x (6.02x10^23) = ?

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