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How much of a 4.45 M CaBr2 solution can be prepared if one has 79.6 g of CaBr2 available?
Wiki User
∙ 12y ago
mole=moles/liters -> moles=mole*liters
moles=7.65 * 4.65 = 35.5725
Molar mass= [40.007+(2)(79.904)]*1 g/mol = 199.815 g/mol
35.5725*199.815= 7107.9190875 g
7108 grams of CaBr2 are needed
Wiki User
∙ 16y ago
Wiki User
∙ 12y ago.0894 L
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