In this experiment you will place a sample of your salt in water in a constant pressure calorimeter with a heat capacity of 29.4 J/℃. You will determine the enthalpy change for the dissociation of your salt, Delta Hdiss. A sample of 4.368 grams of the salt SrCl2 was placed in 35.5 g water, the initial temperature was 20.00℃ and the final temperature was 27.086365℃. Now that we have the enthalpy of reaction, q (reaction) = – 1390.399457 J, and the number of moles 0.035491131 mol SrCl2 , we can get the enthalpy of reaction. Give Delta H (rxn) in correct sig figs.
In this experiment you will place a sample of your salt in water in a constant pressure calorimeter with a heat capacity of 29.4 J/℃. You will determine the enthalpy change for the dissociation of your salt, Delta Hdiss. A sample of 4.368 grams of the salt SrCl2 was placed in 35.5 g water, the initial temperature was 20.00℃ and the final temperature was 27.086365℃. Now that we have the enthalpy of reaction, q (reaction) = – 1390.399457 J, and the number of moles 0.035491131 mol SrCl2 , we can get the enthalpy of reaction. Give Delta H (rxn) in correct sig figs.
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter5: Principles Of Chemical Reactivity: Energy And Chemical Reactions
Section5.2: Specific Heat Capacity: Heating And Cooling
Problem 5.1CYU: You did an experiment in which you found that 59.8 J was required to raise the temperature of 25.0 g...
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In this experiment you will place a sample of your salt in water in a constant pressure calorimeter with a heat capacity of 29.4 J/℃. You will determine the enthalpy change for the dissociation of your salt, Delta Hdiss. A sample of 4.368 grams of the salt SrCl2 was placed in 35.5 g water, the initial temperature was 20.00℃ and the final temperature was 27.086365℃. Now that we have the enthalpy of reaction, q (reaction) = – 1390.399457 J, and the number of moles 0.035491131 mol SrCl2 , we can get the enthalpy of reaction. Give Delta H (rxn) in correct sig figs.
Delta H (rxn) = q / n = – 1390.399457 J / 0.035491131 mol SrCl2
Delta H (rxn) = – 39175.96895 J / mol SrCl2
Delta H (rxn) = – 39.17596895 kJ / mol SrCl2
a. – 39.17 kJ / mol
b. – 39.17596895 kJ / mol
c. – 39 kJ / mol
d. – 39.2 kJ / mol
e. none of these
f. – 4 x 10^1 kJ / mol
g. – 39.1 kJ / mol
h. – 39.18 kJ / mol
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