Asked by Confused
Three samples of an unknown acid are titrated with 0.4901 M NaOH. The average equivalent weight of the acid using the data below is g/equiv
1 2 3
mass of unknown acid (g) 1.020 0.522 0.742
vol NaOH used (ml) 27.72 14.22 20.02
1 2 3
mass of unknown acid (g) 1.020 0.522 0.742
vol NaOH used (ml) 27.72 14.22 20.02
Answers
Answered by
DrBob222
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