ZOOM IN ON WEB BROWSER TO READ QUESTION. THANKS! The following data was collecte
ID: 954821 • Letter: Z
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ZOOM IN ON WEB BROWSER TO READ QUESTION. THANKS!
The following data was collected after running a calorimetric experiment using three salts. The data collected for each experiment is as follows. Experiment 1: 2.6S g of unknov/n salt added to 150. mLof water resulted in a AT value of-0.2 degree C. Experiment 2: 2.69 g of unknown salt added to 150. mL of water resulted in a AT value of -2.0 degree C. Experiment 3: 2.69 g of unknov/n salt added to 150. mLof water resulted in a AT value of 0.8 degree C. Calculate the value of q for each of the experiments. (Assume that the heat capacity of water is 4.1801 J/g degree C.) Based on the magnitude and sign of the q values calculated, determine the identity of the salt used in each experiment and the resulting Delta H based on that deduction. (You can also calculate to see if your deduction makes sense when compared to the known values given above.)Explanation / Answer
The heat liberated in calorimetry q is calculated using the realtion , q= m.s.delta T where m is mass of water, s is the heat capacity(specific heat) for gram of water and delat T is the rise in temperature.
Experiment 1
q= 150 g x 4.1801J/g x (-0.2K) = -125.4 JK
Experiment 2
q =150 g x 4.1801J/g x (-2K) = - 1254.0 JK
Experiment 3
q =150 g x 4.1801J/g x 0.8K = 501.6 JK
Identity of salts
1. LiNO3 since 2.69g of salt produces -125.4 J of heat one mole of LiNO3 (69g/mol) produces -3.21KJ /mol .
This is closer to the given value of 2.51kJ.
2 . LiI since 2.69g of salt produces -1254 J ofheat one mole of LiI (134g/mol) produces- 62.46kJ/mol.
This is very near to the given value of -62.6KJ
3. The salt is LiF . As 2.69g of salt takes 501.6 J , one mole of LiF (26g /mol) takes 4.848 kJ /mol
This is nearer to 4.73 given value.
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