LiBr (s) -> Li^+ (aq) + Br^- (aq) delta H=? In a coffee cup calorimeter you diss

Question

LiBr (s) -> Li^+ (aq) + Br^- (aq) delta H=? In a coffee cup calorimeter you dissolve 19.5 g of this salt in 120. g of water at 25.1°C. The final temperature of the solution formed is 43.9°C. What is the molar enthalpy of this reaction, in kJ/mol of salt? Assume the specific heat of the solution is 4.184 J/g°C, and that there is no heat loss to the surroundings. Please enter your answer to the correct number of significant figures. Consider the following reaction: LiBr (s) Li+ (aq) Br (aq) AH In a coffee-cup calorimeter you dissolve 19.5 g of this salt in 120. g of water at 25.1 oC. The final temperature of the solution formed is 43.9 oC. What is the molar enthalpy of this reaction, in kJ/mol of salt? Assume the specific heat of the solution is 4.184 J/g°C, and that there is no heat loss to the surroundings. Please enter your answer to the correct number of significant figures. kJ/mol 128 submit Answer Incorrect. Tries 1 Previous Tries Send Feedbac

Explanation / Answer

The mass of solution = 120 g + 19.5g = 139.5 g

the rise in temperature = 43.9-25.1= 18.8C(K)

specific heat of solution = 4.184 J/g.C

The heat given by substance = heat absorbed by the calorimeter

= mass x specific heat x rise in temperature

= 139.5g x 4.184J/g.C x 18.8C

= 10972.95 J

Molar mass of LiBr = 86.845 g /mol

If 19.5 g of salt gives -10972.95 J

86.845 g gives = 86.845 g/molx 10972.95J /19.5g

= 48869 .05 j

= 48.869kJ /mol

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