Asked by Fai

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



Who Helps
Answered by DrBob222
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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