3.15 mol of an unknown solid is placed into enough water to make 150.0 mL of solution. The solution's temperature increases by 19.2°C. Calculate ∆H for the dissolution of the unknown solid. (The specific heat of the solution is 4.18 J/g･°C and the density of the solution is 1.20 g/mL).

Enthalpy change for the dissolution of the unknown solid = 4.6 Kj/mole

Explanation:

Using Q = m x Cs x ΔT ................................(1)

where Q = Amount of heat absorbed

            m = Mass of solution

           Cs = Specific heat capacity of solution

          ΔT = Change in temperature

Given Density of solution = 1.20 g/ml

And volume of solution = 150 ml

Mass of solution = density x volume

                           = 1.2 x 150

                           = 180 g

From equation (1)

              Q = 180 x 4.18 x 19.2 = 14446.08 = 14.4 Kj

So, ΔH  of the dissolution of the unknown solid = = = 4.6 kj/mole

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: