Please use the values in the resources listed below instead of the textbook values. From the following information, determine the missing value of AS298 (in J/K/mol). H(g) + Br(g) → HBr(g) AS298 = ? H2(g) + Br,(g) → 2 HBr(g) AS298 = 21.2 J/K/mol H,(g) → 2 H(g) AS298 98.7 J/K/mol Br,(g) → 2 Br(g) AS 298 104.5 J/K/mol %3D 4.0 J/K/mol

Please use the values in the resources listed below instead of the textbook values. From the following information, determine the missing value of AS298 (in J/K/mol). H(g) + Br(g) → HBr(g) AS298 = ? H2(g) + Br,(g) → 2 HBr(g) AS298 = 21.2 J/K/mol H,(g) → 2 H(g) AS298 98.7 J/K/mol Br,(g) → 2 Br(g) AS 298 104.5 J/K/mol %3D 4.0 J/K/mol

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

Chapter6: Thermochemistry

Section: Chapter Questions

Problem 68E: In a coffee-cup calorimeter, 1.60 g NH4NO3 is mixed with 75.0 g water at an initial temperature of...

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The heat of neutralization, Hneut, can be defined as the amount of heat released (or absorbed), q, per mole of acid (or base) neutralized. Hneut for nitric acid is -52 kJ/mol HNO3. At 27.3C, 50.00 mL of 0.743M HNO3 is neutralized by 1.00 M Sr(OH)2 in a coffee-cup calorimeter. (a) How many mL of Sr(OH)2 were used in the neutralization? (b) What is the final temperature of the resulting solution? (Use the assumptions in Question 11.)
Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer.
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