A student measured the molar heat of neutralization of a monoprotic acid with NaOH and the following data was obtained. Calculate the AHrx per mole of acid. Ti = 24.3°C V(acid) = 50.0 mL V(base) = 50.0 mL T(final) = 32.2°C [acid] = 1.02 M [NaOH] = 1.05 M C(cal) = 8.5 J/°C
A student measured the molar heat of neutralization of a monoprotic acid with NaOH and the following data was obtained. Calculate the AHrx per mole of acid. Ti = 24.3°C V(acid) = 50.0 mL V(base) = 50.0 mL T(final) = 32.2°C [acid] = 1.02 M [NaOH] = 1.05 M C(cal) = 8.5 J/°C
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter8: Thermochemistry
Section: Chapter Questions
Problem 11QAP: When one mol of KOH is neutralized by sulfuric acid, q=56 kJ. (This is called the heat of...
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A student measured the molar heat of neutralization of a monoprotic acid with NaOH and the following data was obtained. Calculate the AHrx per mole of acid. Ti = 24.3°C V(acid) = 50.0 mL V(base) = 50.0 mL T(final) = 32.2°C [acid] = 1.02 M [NaOH] = 1.05 M C(cal) = 8.5 J/°C
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