B. Heat of neutralization for NaOH (aq) (AH(neut. NaOH(aq)) Mass NaOH = 1.022 g (From part 1) Moles NaOH = Mass of solution in cup Final temp. of solution Initial temp. of solution AT solution. (AH(neut. NaOH (aq)) = = AH(neut. NaOH (aq)) per mole NaOH = Calculation AH (neut. NaOH (s)) AH (neut. NaOH (s)) per mole NaOH = = 0.026 600 mL 30°C 24°C C. Heat of neutralization for NaOH (s) (AH(neut. NaOH (s)) Mass of solution in cup 30mL Final temp. of solution in cup Initial temp. of solution in cup AT solution in cup Calculation = 30°C-24°C 6°℃ moles 420 25.0°C 42°C-25°C = 14°C

B. Heat of neutralization for NaOH (aq) (AH(neut. NaOH(aq)) Mass NaOH = 1.022 g (From part 1) Moles NaOH = Mass of solution in cup Final temp. of solution Initial temp. of solution AT solution. (AH(neut. NaOH (aq)) = = AH(neut. NaOH (aq)) per mole NaOH = Calculation AH (neut. NaOH (s)) AH (neut. NaOH (s)) per mole NaOH = = 0.026 600 mL 30°C 24°C C. Heat of neutralization for NaOH (s) (AH(neut. NaOH (s)) Mass of solution in cup 30mL Final temp. of solution in cup Initial temp. of solution in cup AT solution in cup Calculation = 30°C-24°C 6°℃ moles 420 25.0°C 42°C-25°C = 14°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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