CH₂COOH (aq) + KOH (aq) → H₂O (1) + CH₂COOK (aq) If 3.45 mL of vinegar (CH3COOH) needs 24.5 mL of 0.155 M KOH to reach t

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CH₂COOH (aq) + KOH (aq) → H₂O (1) + CH₂COOK (aq) If 3.45 mL of vinegar (CH3COOH) needs 24.5 mL of 0.155 M KOH to reach t

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Ch Cooh Aq Koh Aq H O 1 Ch Cook Aq If 3 45 Ml Of Vinegar Ch3cooh Needs 24 5 Ml Of 0 155 M Koh To Reach T 1
Ch Cooh Aq Koh Aq H O 1 Ch Cook Aq If 3 45 Ml Of Vinegar Ch3cooh Needs 24 5 Ml Of 0 155 M Koh To Reach T 1 (33.98 KiB) Viewed 65 times
CH₂COOH (aq) + KOH (aq) → H₂O (1) + CH₂COOK (aq) If 3.45 mL of vinegar (CH3COOH) needs 24.5 mL of 0.155 M KOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt sample of this vinegar? (20 points) 2. The molar mass of a volatile substance was determined by the Dumas-bulb method. The unknown vapor had a mass of 0.846 g, the volume of the bulb was 534 cm³, pressure 527 torr, and temperature 100° C. Calculate the molar mass of the unknown vapor. (15 points)
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