A calorimeter contains 28.0mL of water at 15.0?C . When 1.90g of X (a substance

Question

A calorimeter contains 28.0mL of water at 15.0?C . When 1.90g of X (a substance with a molar mass of 58.0g/mol ) is added, it dissolves via the reaction X(s)+H2O(l) - X(aq) and the temperature of the solution increases to 25.5?C Calculate the enthalpy change, ?H, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g??C)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Explanation / Answer

heat given to the water:
dH = m C dT
dH = (29.5 total mass) (4.184 J/g-C water) (15.0 C rise in temp)
dH = 1789.7 Joules released to the water

find joules per gram
1789.7 Joules released to the water / by 1.50 g of X = 1193 Joules released / gram X

find joules / mole X

(58.0 grams X / mole ) times ( 1193 Joules released / gram X) = 69,194.5 joules released / mole X

aka
69.194kJ/mol X

in your 3 sig fig problem, this gets rounded to
70.0 kJ released per mole X

your answer, for this exothermic reaction is:
dH reaction = - 70.0 kJ / mol X

process is same just check the calulation and rate me frnd

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