70 Решение задачки по химии №06 774

3. The ester saponification reaction CH3COOC2H5+KOH=CH3SOOK+C2H5OH is a second-order reaction. 10 minutes after the start of the reaction, C(KOH) was equal to 0.04 mol/L. How long does it take for the starting substances to react by 75% if their initial concentrations are the same and equal to 0.1 mol/l?

If you do not open the file - install the RAR archiver. Inside the archive you will find a solution in Word format.