DATA AND CALCULATIONS Show all calculations neatly on an attached sheet. Trial 1 Trial 2 Trial 3 70.6849. 68.0749 2.61g 23.25ML O.65ML 22.6mL Mass of flask + vinegar Mass of empty flask Mass of vinegar used Final buret reading Initial buret reading Volume of NaOH used Moles of NaOH used 0.00226 mol Moles of açetic acid titrated Mass of acetic acid titrated Mass % acetic acid in vinegar 5.0%
DATA AND CALCULATIONS Show all calculations neatly on an attached sheet. Trial 1 Trial 2 Trial 3 70.6849. 68.0749 2.61g 23.25ML O.65ML 22.6mL Mass of flask + vinegar Mass of empty flask Mass of vinegar used Final buret reading Initial buret reading Volume of NaOH used Moles of NaOH used 0.00226 mol Moles of açetic acid titrated Mass of acetic acid titrated Mass % acetic acid in vinegar 5.0%
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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Transcribed Image Text:DATA AND CALCULATIONS
Show all calculations neatly on an attached sheet.
Trial 1
Trial 2
Trial 3
70.6849.
68.0749
2.61g
23.25ML
O.65ML
22.6mL
Mass of flask + vinegar
Mass of empty flask
Mass of vinegar used
Final buret reading
Initial buret reading
Volume of NaOH used
Moles of NaOH used
0.00226 mol
Moles of açetic acid titrated
Mass of acetic acid titrated
Mass % acetic acid in vinegar
5.0%
Expert Solution
Step 1
Solution:- Mass of vinegar used = 2.61 g
volume of NaOH used = 22.6 mL = 0.0226 L
Moles of NaOH used = 0.00226 mol
Now, In titration at equivalence point :- no. of moles of acid = no. of moles of base
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