Trial 1 2150 joules 2150 kilojoules trial 2 2144

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Trial 1 215.0 joules = .2150 kilojoules Trial 2 214.4 joules = .2144 kilojoules Trial 3 215.0 joules = .2150 kilojoules Converting to KJ / mol Molar mass of NH NO = 80.0 g / mol Trial 1 .2150 KJ / 80.0 g/ mol = .00269 KJ / mol Trial 2 .2144 KJ / 80.0 g/mol = .00268 KJ / mol Trial 3 .2150 KJ / 80.0 g / mol = .00269 KJ / mol Follow up question If you were to dissolve a mol of hydrated salt , the temperature change would be smaller than if you were to dissolve a mol of a non hydrated salt. Specifically the temperature of the water would go up in with both salts, but it would go up less if you were to use a hydrated salt than if you were to use a non hydrated salt. This is because less energy is released with hydrated salts then if they are not. Conclusion By using the data that I collected during the lab I was able to calculate the heat of the solution
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for the salt that I tested. I used the ammonium nitrate salt and found that in both the first and last trial the heat of the solution using that salt was .00269 KJ / mol. For the second trial using the same salt I found that the heat of solution was .00268 KJ / mol. Because these numbers are so consistent I concluded that the Heat of a solution tested for Ammonium nitrate is very close to . 00268 to .00269 KJ / mol. Lab Title: Heat of Solution Total Earned Total Possibl e Category Comments Lab Work 2.0 2.0 Purpose 1.0 1.0 Procedure 4.0 4.0 Data Tables 3.0 3.0 Analysis See below. 0.5 4.0 Error Analysis 2.0 2.0 Conclusions 1.0 1.0 P & A At least the sig figs are correct . . . 0.5 2.0 F & P 1.0 1.0 15.00 20.00 Lab Total You've used an improper equation to find the heat (q). Where in the world did you get the idea that you should subtract the temperature from the other variables?!?! Look at the units - that makes zero sense!!! Unfortunately this majorly impacts your final answer making it totally invalid. Your conversion into kJ/mol is also seriously flawed. THINK!!!!
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