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what weight of na2c03 of 95% purity would be required to neutralize 45.6ml of 0.235N acid?

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  • chemistry - ,

    mL x N x mew = grams (at 100%)
    Then grams/0.95 = grams Na2CO3 at 95%.
    Note: mew Na2CO3 = molar mass/2

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