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Answer to Question #68640 in General Chemistry for tyler presley
Question #68640
what mass of calcium bromide CaBr2 is needed to prepare 150 mL OF A 2.50M solution?
Expert's answer
CM = n/V n = CM·V
n = m/M m = n·M
n (CaBr2) = CM·V = 2.50 · 0.150 = 3.75 mol
M (CaBr2) = 199.9 g/mol
m (CaBr2) = 3.75 · 199.9 = 749.6 g
n = m/M m = n·M
n (CaBr2) = CM·V = 2.50 · 0.150 = 3.75 mol
M (CaBr2) = 199.9 g/mol
m (CaBr2) = 3.75 · 199.9 = 749.6 g
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