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How many grams of K2Cr2O7 are required to prepare 500 ml of a 0.13 N solution?
12 years ago

Answers

Devron
0.13N= # of equivalents/500 x10^-3 L, solving for # of equivalents


# of equivalents in 0.13N=(0.13N)(500 x 10^-3L)

# of equivalents in 0.13N= g of solute/49.03g, solve for grams of solute,


(49.03g, solve )*(# of equivalents in 0.13N)= g of solute
12 years ago

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