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Answer to Question #18113 in Inorganic Chemistry for Whitney
Question #18113
How many moles are in 733 g of calcium carbonate (CaCO3)?
Expert's answer
n =m/M
n(CaCO3) = 733 g / 100g/mol = 7.33mol.
n(CaCO3) = 733 g / 100g/mol = 7.33mol.
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