The moles of acid are constant in a dilution process, therefore,
M1V1 = M2V2 = n
V1 = M2V2/M1 plug in values. I and 2 can be either before or after. dilution as long as the molarity and volume are for the same case.
Ellis B.
asked 03/21/22Calculate the volume of 6.82 M acetic acid needed to prepare 325 mL of 0.885 M acetic acid.
The moles of acid are constant in a dilution process, therefore,
M1V1 = M2V2 = n
V1 = M2V2/M1 plug in values. I and 2 can be either before or after. dilution as long as the molarity and volume are for the same case.
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