50 mL (50 g) of 6.0 M HCl is mixed with 50 mL (50 g) of 6.0 M NaOH in a calorimeter that absorbs no heat. The temperature change is 40°C. Which equation calculates the heat of reaction? Question 18 options: (100 g × 4.184 J g–1 °C–1 × 40°C) / (6.0 mol L–1 × 0.05 L) –(100 g × 4.184 J g–1 °C–1 × 40°C) / (6.0 mol L–1 × 0.05 L) (6.0 mol L–1 × 0.05 L) / (100 g × 4.184 J g–1 °C–1 × 40°C)
50 mL (50 g) of 6.0 M HCl is mixed with 50 mL (50 g) of 6.0 M NaOH in a calorimeter that absorbs no heat. The temperature change is 40°C. Which equation calculates the heat of reaction? Question 18 options: (100 g × 4.184 J g–1 °C–1 × 40°C) / (6.0 mol L–1 × 0.05 L) –(100 g × 4.184 J g–1 °C–1 × 40°C) / (6.0 mol L–1 × 0.05 L) (6.0 mol L–1 × 0.05 L) / (100 g × 4.184 J g–1 °C–1 × 40°C)
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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Question
50 mL (50 g) of 6.0 M HCl is mixed with 50 mL (50 g) of 6.0 M NaOH in a calorimeter that absorbs no heat. The temperature change is 40°C. Which equation calculates the heat of reaction?
Question 18 options:
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|
(100 g × 4.184 J g–1 °C–1 × 40°C) / (6.0 mol L–1 × 0.05 L) |
|
|
–(100 g × 4.184 J g–1 °C–1 × 40°C) / (6.0 mol L–1 × 0.05 L) |
|
|
(6.0 mol L–1 × 0.05 L) / (100 g × 4.184 J g–1 °C–1 × 40°C) |
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