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Use the data in the following image to answer this question. a.The reaction involved the neutralization of xxxmol of HCl, therefore, calculate the molar heat of neutralization.

Use the data in the following image to answer this question. a.The reaction involved the neutralization of xxxmol of HCl, therefore, calculate the molar heat of neutralization.

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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Use the data in the following image to answer this question.

a.The reaction involved the neutralization of xxxmol of HCl, therefore, calculate the molar heat of neutralization.

DETERMINATION OF AHneu FOR HCI It is essential to keep track of which quantitites are actually measured in this exercise and which ones are given. Suppose we mix samples of measured volumes and measured temperatures and given concentrations of HCI and NaOH VNaOH = VHI = 50.0 mL 55.0 mL TNGOH = 24.2°C Tна 24.0°C MNaOH = 1.30M MHa = 1.20 M We measure temperature change at t = 0 graphically. The result is that AT = 4.75C° .We need to know how much of the essential reactant (HCI) has been consumed. That requires determining the limiting reagent. Using the measured volumes and given concentrations of HCl and NAOH, we can calculate the number of moles of each reactant. NHCI = 0.0600 mol %3D Мна NNOOH = VNOOH * = 0.0715 mol MNeOH This makes HCI the limiting reagent. Only 0.0600 mol HCl can be neutralized. Using the given heat capacity, given densities and measured volumes of the solutions and the measured temperature change, we can analyze the heat exchange, q, and correct for the heat loss to the calorimeter using the Calorimetric Ccal, alrea determined. Heat Capacity = 3.97 J/g Density = 1.02 g/mL Ccal = 42.0 J/C° Tot Vol of Sol = 50.0 + 55.0 = 105.0 mL
Transcribed Image Text:DETERMINATION OF AHneu FOR HCI It is essential to keep track of which quantitites are actually measured in this exercise and which ones are given. Suppose we mix samples of measured volumes and measured temperatures and given concentrations of HCI and NaOH VNaOH = VHI = 50.0 mL 55.0 mL TNGOH = 24.2°C Tна 24.0°C MNaOH = 1.30M MHa = 1.20 M We measure temperature change at t = 0 graphically. The result is that AT = 4.75C° .We need to know how much of the essential reactant (HCI) has been consumed. That requires determining the limiting reagent. Using the measured volumes and given concentrations of HCl and NAOH, we can calculate the number of moles of each reactant. NHCI = 0.0600 mol %3D Мна NNOOH = VNOOH * = 0.0715 mol MNeOH This makes HCI the limiting reagent. Only 0.0600 mol HCl can be neutralized. Using the given heat capacity, given densities and measured volumes of the solutions and the measured temperature change, we can analyze the heat exchange, q, and correct for the heat loss to the calorimeter using the Calorimetric Ccal, alrea determined. Heat Capacity = 3.97 J/g Density = 1.02 g/mL Ccal = 42.0 J/C° Tot Vol of Sol = 50.0 + 55.0 = 105.0 mL
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