Post a New Question
posted by Fai on Sunday, January 27, 2013 at 2:14pm.
what weight of na2c03 of 95% purity would be required to neutralize 45.6ml of 0.235N acid? Who Helps
mL x N x mew = grams (at 100%) Then grams/0.95 = grams Na2CO3 at 95%. Note: mew Na2CO3 = molar mass/2
More Related Questions