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Analytical chemisty

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

  • Analytical chemisty -

    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

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