When a 6.00 g sample of KBr is dissolved in water in a calorimeter that has a total heat capacity of 2.71 kJ ⋅ K − 1 , the temperature decreases by 0.370 K . Calculate the molar heat of solution of KBr .

Molar heat enthalpy of KBr = -19.89 kJ/mol

Explanation:

Change in temperature (Δt) = 0.370 K

Heat capacity = 2.71 kJ ⋅ K^-1

Heat absorbed by calorimeter = heat capacity × change in temperature

                                               = 2.71× 0.370

                                               = 1.0027 kJ

Molar mass of KBr = 119 g/mol

No. of moles of KBr = 6.00/119

                               = 0.0504 mol

Heat absorbed by the calorimeter is given by KBr.

Now calculate the heat released by per mol of KBr as follows:

Heat released by per mol of KBr = 1.0027 kJ / 0.0504 mol

                                                   =19.89 kJ/mol

Heat is released therefore, sign will be negative.

Molar heat enthalpy of KBr = -19.89 kJ/mol

the correct name for the n3 ion is azide. azide is a polyatomic anion with a -1 charge and is written as n3 -1. it consists of three nitrogen atoms bonded together. hope i . let me know if you need anything else.