A coffee-cup calorimeter initially contains 125 g water at 24.28C. Potassium bromide (10.5 g), also at 24.28C, is added to the water, and after the KBr dissolves, the final temperature is 21.18C. Calculate the enthalpy change for dissolving the salt in J/g and kJ/mol. Assume that the specific heat capacity of the solution is 4.18 J/8C

The solution is given below

Explanation:

Heat, q= mc∆T

q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C

q=  -1619.75J

NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.

Enthalpy Change, ∆H = 1619.75 7/ 10.5 g

                                     = 154.26 J/g

No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr

                                =10.5g/119gmol-1

                                =0.088 mol

∆H= 1619.75 J/ 0.088 mol

      = 18.41 kJ/mol  

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.