Trial 1: 5.000g NaCl was placed in a calorimeter with 40g of water at 23.8ºC. Af

Question

Trial 1:
5.000g NaCl was placed in a calorimeter with 40g of water at 23.8ºC. After the water reacted with the NaCl, the final temperature was 22.7ºC.

Trial 2:
4.975g NaCl was placed in a calorimeter with 40g of water at 23.0ºC. After the water reacted with the NaCl, the final temperature was 22.0ºC.

Did I correctly calculate the heat flows of water and salt?

Also, I can't find how to calcualte enthalpy of formations in my lab manual, so how will I solve them? Just working it out for 1 of the 2 trials is fine as long as I have adequate information to solve for the other one.

Determination of AH of a cation nula of salt sample: Nac Formula of salt sample: Trial 1 Trial 2 Mass of salt: atc Moles of salt: Mass of water: 23.8 22.7 10Yol 23.0°C 2200 Initial temperature of salt and water: Final temperature of salt and water Heat flow of water (include correct sign): //--ir.w/ Heat flow of salt (include correct sign): Enthalpy of solution for the salt: Enthalpy of formation of the salt: Enthalpy of formation of the anion: Enthalpy of formation of the cation: Average Enthalpy of formation of the cation c comation of the cation

Explanation / Answer

NaCl(s) + H2O(l) -------> Na+(aq) + Cl-(aq)

Specific heat of water = 4.184 J/g/0C

Now, heat lost by water = mass*specific heat*drop in temperature = 40*4.184*1.1 = 184.096 J

Heat lost by salt = mass*specific heat of salt*drop in temperature = 5*specific heat*1.1 = 5.5*s J

Total heat of reaction = heat lost by water + heat lost by salt

Now, heat lost in the reaction = enthalpy of formation of Na+ + enthalpy of formation of Cl- - enthalpy of formation of H2O - enthalpy of formation of NaCl

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