What amount of Sodium Nitrate is present in 31.6 grams of the compound?
回答 (1)
2016-06-13 8:20 am
Molar mass of NaNO₃ = (22.99 + 14.01 + 16.00×3) g/mol = 85.00 g/mol
No. of moles of NaNO₃ = (31.6 g) / (85.00 g/mol) = 0.372 mol
No. of moles of NaNO₃ = (31.6 g) / (85.00 g/mol) = 0.372 mol
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