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.
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
Step by step
Solved in 3 steps