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Temperature (°C) 70 65 60 55 50 45 40 35 30 Freezing Point Determination: Pure Solvent 10 20 40 " D 60 80 100 Time (seconds) 120 140 160 Liquid cooling y = -0.3333x + 63.2 R² = 0.9889 Liquid Freezing y = -0.0255x + 44.929 R² = 0.989 ◆ Liquid Cooling Freezing

Temperature (°C) 70 65 60 55 50 45 40 35 30 Freezing Point Determination: Pure Solvent 10 20 40 " D 60 80 100 Time (seconds) 120 140 160 Liquid cooling y = -0.3333x + 63.2 R² = 0.9889 Liquid Freezing y = -0.0255x + 44.929 R² = 0.989 ◆ Liquid Cooling Freezing

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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A) When 1.0105 g of an unknown was added, the freezing point of the solution was determined to be 41.8 °C. Calculate the change in the freezing point, and then calculate the molality of the solution. The Kf of lauric acid is 3.90 °C⁄?. B) Using the determined molality, calculate the number of moles of unknown present, and then calculate the molar mass of the unknown. 9.467 grams of Lauric acid was used
Temperature (°C) 70 65 60 55 50 45 40 35 30 Freezing Point Determination: Pure Solvent 10 20 40 4 D 60 100 80 Time (seconds) 120 140 160 Liquid cooling y = -0.3333x + 63.2 R² = 0.9889 Liquid Freezing y = -0.0255x+44.929 R² = 0.989 ◆ Liquid Cooling I Freezing
Transcribed Image Text:Temperature (°C) 70 65 60 55 50 45 40 35 30 Freezing Point Determination: Pure Solvent 10 20 40 4 D 60 100 80 Time (seconds) 120 140 160 Liquid cooling y = -0.3333x + 63.2 R² = 0.9889 Liquid Freezing y = -0.0255x+44.929 R² = 0.989 ◆ Liquid Cooling I Freezing
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