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1. Calculations: a. Using the average concentration of vinegar, calculate the mass percentage of acetic acid in vinegar b. Using the average concentration of the Windex solution, calculate the mass percentage of ammonium hydroxide in Windex.

1. Calculations: a. Using the average concentration of vinegar, calculate the mass percentage of acetic acid in vinegar b. Using the average concentration of the Windex solution, calculate the mass percentage of ammonium hydroxide in Windex.

Chemistry
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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1. Calculations:

a. Using the average concentration of vinegar, calculate the mass percentage of acetic acid in vinegar

b. Using the average concentration of the Windex solution, calculate the mass percentage of ammonium hydroxide in Windex.

Indicator #1 Trial 1 Trial 3 Trial 2 2 mL Vol HCl 2 mL 2 mL = 0.002 L = 0.002 L = 0.002 L Estimated = 3.42 mL 3.49 mL 3.34 mL volume of = 0.00342 L = 0.00349 L 0.00334 L Windex solution* Mol of HCl 0.10 х 0.002 0.10 x 0.002 0.10 х 0.002 = 0.0002 0.0002 0.0002 = 2 x 10-4 Mol NaOH (1:1) | 2 x 104 -2х 104 2 х 104 = 2 x 10-4 %3D 2 x 104 Concentration of (0.0002/0.00342) | (0.0002/0.00349) (0.0002/0.00334) = 0.0573 Windex solution = 0.0585 0.0599 Average (0.0585 + 0.0573 + 0.0599)/3 = 0.0586 Indicator #2 Trial 1 Trial 2 Trial 3 Vol HCl 2 mL 2 mL 2 mL = 0.002 L = 4.09 mL 0.002 L 0.002 L Estimated = 3.91 mL 4.23 mL volume of = 0.00391 L = 0.00423 L = 0.00409 L Windex* Mol of HCl 0.10 x 0.002 0.10 x 0.002 0.10 х 0.002 = 0.0002 = 0.0002 = 0.0002 = 2 x 10-4 Mol NaOH (1:1) 2 x 104 = 2 x 10-4 2 x 104 = 2 x 10-4 2 x 104 Concentration of (0.0002/0.00391) | (0.0002/0.00423) (0.0002/0.00409) = 0.0479 M Windex solution 0.0512 M = 0.0489 M (0.0512 + 0.0479 + 0.0489)/3 = 0.0493 M Average
Transcribed Image Text:Indicator #1 Trial 1 Trial 3 Trial 2 2 mL Vol HCl 2 mL 2 mL = 0.002 L = 0.002 L = 0.002 L Estimated = 3.42 mL 3.49 mL 3.34 mL volume of = 0.00342 L = 0.00349 L 0.00334 L Windex solution* Mol of HCl 0.10 х 0.002 0.10 x 0.002 0.10 х 0.002 = 0.0002 0.0002 0.0002 = 2 x 10-4 Mol NaOH (1:1) | 2 x 104 -2х 104 2 х 104 = 2 x 10-4 %3D 2 x 104 Concentration of (0.0002/0.00342) | (0.0002/0.00349) (0.0002/0.00334) = 0.0573 Windex solution = 0.0585 0.0599 Average (0.0585 + 0.0573 + 0.0599)/3 = 0.0586 Indicator #2 Trial 1 Trial 2 Trial 3 Vol HCl 2 mL 2 mL 2 mL = 0.002 L = 4.09 mL 0.002 L 0.002 L Estimated = 3.91 mL 4.23 mL volume of = 0.00391 L = 0.00423 L = 0.00409 L Windex* Mol of HCl 0.10 x 0.002 0.10 x 0.002 0.10 х 0.002 = 0.0002 = 0.0002 = 0.0002 = 2 x 10-4 Mol NaOH (1:1) 2 x 104 = 2 x 10-4 2 x 104 = 2 x 10-4 2 x 104 Concentration of (0.0002/0.00391) | (0.0002/0.00423) (0.0002/0.00409) = 0.0479 M Windex solution 0.0512 M = 0.0489 M (0.0512 + 0.0479 + 0.0489)/3 = 0.0493 M Average
Indicator #1 Trial 1 Trial 2 Trial 3 Vol NaOH 2 mL 2 mL 2 mL = 0.002 L = 0.002 L = 0.002 L Estimated = 2.29 mL = 2.43 mL = 2.11 mL volume of = 0.00229 L = 0.00243 L = 0.00211 L vinegar* Mol of NaOH 0.10 x 0.002 0.10 х 0.002 0.10 х 0.002 = 0.0002 = 0.0002 = 0.0002 = 2 x 10-4 2х 104 = 2 x 10-4 2х 104 = 2 x 10-4 2х 104 Mol vinegar (1:1) Concentration of (0.0002/0.00229) vinegar (0.0002/0.00243) (0.0002/0.00211) = 0.0823 M = 0.0873 M = 0.0948 M Average (0.0873 + 0.0823 + 0.0948)/3 = 0.0881 M Indicator #2 Trial 1 Trial 2 2 mL Trial 3 Vol NaOH 2 mL 2 mL = 0.002 L = 0.002 L = 0.002 L Estimated volume of = 0.32 mL = 0.40 mL = 0.34 mL = 0.00032 L = 0.00040 L = 0.00034 L vinegar ** Mol of NaOH 0.10 х 0.002 0.10 х 0.002 0.10 х 0.002 = 0.0002 = 0.0002 = 0.0002 = 2 x 10-4 2 x 104 = 2 x 10-4 2 x 10-4 = 2 x 10-4 2 x 10-4 Mol vinegar (1:1) Concentration of (0.0002/0.00032) (0.0002/0.00040) (0.0002/0.00034) vinegar = 0.625 M = 0.500 M = 0.588 M Average (0.625 + 0.500 + 0.588)/3 = 0.571 M
Transcribed Image Text:Indicator #1 Trial 1 Trial 2 Trial 3 Vol NaOH 2 mL 2 mL 2 mL = 0.002 L = 0.002 L = 0.002 L Estimated = 2.29 mL = 2.43 mL = 2.11 mL volume of = 0.00229 L = 0.00243 L = 0.00211 L vinegar* Mol of NaOH 0.10 x 0.002 0.10 х 0.002 0.10 х 0.002 = 0.0002 = 0.0002 = 0.0002 = 2 x 10-4 2х 104 = 2 x 10-4 2х 104 = 2 x 10-4 2х 104 Mol vinegar (1:1) Concentration of (0.0002/0.00229) vinegar (0.0002/0.00243) (0.0002/0.00211) = 0.0823 M = 0.0873 M = 0.0948 M Average (0.0873 + 0.0823 + 0.0948)/3 = 0.0881 M Indicator #2 Trial 1 Trial 2 2 mL Trial 3 Vol NaOH 2 mL 2 mL = 0.002 L = 0.002 L = 0.002 L Estimated volume of = 0.32 mL = 0.40 mL = 0.34 mL = 0.00032 L = 0.00040 L = 0.00034 L vinegar ** Mol of NaOH 0.10 х 0.002 0.10 х 0.002 0.10 х 0.002 = 0.0002 = 0.0002 = 0.0002 = 2 x 10-4 2 x 104 = 2 x 10-4 2 x 10-4 = 2 x 10-4 2 x 10-4 Mol vinegar (1:1) Concentration of (0.0002/0.00032) (0.0002/0.00040) (0.0002/0.00034) vinegar = 0.625 M = 0.500 M = 0.588 M Average (0.625 + 0.500 + 0.588)/3 = 0.571 M
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