The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 19.00-mL portion of the diluted solutionwas refluxed with 40.00 mL of 0.04601 M KOH:CH3 СОО С2H; + ОН — СH3 COO + C2H; ОНAfter cooling, the excess OH was back-titrated with 3.02 mL of 0.05056 M H2 SO4. Calculate the amount of ethyl acetate (88.11 g/mol) in the original samplein grams.Amount of ethyl acetate =

The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 19.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04601 M KOH: CH; СОО С> Н, + ОН — СH; CОО + С, H; ОН After cooling, the excess OH was back-titrated with 3.02 mL of 0.05056 M H2 SO4. Calculate the amount of ethyl acetate (88.11 g/mol) in the original sample in grams. Amount of ethyl acetate g

The ethyl acetate concentration in an alcoholic solution was determined by diluting a 10.00-mL sample to 100.00 mL. A 19.00-mL portion of the diluted solution was refluxed with 40.00 mL of 0.04601 M KOH: CH; СОО С> Н, + ОН — СH; CОО + С, H; ОН After cooling, the excess OH was back-titrated with 3.02 mL of 0.05056 M H2 SO4. Calculate the amount of ethyl acetate (88.11 g/mol) in the original sample in grams. Amount of ethyl acetate g

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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