"q=cm\\Delta T = 4.186 \\frac{J}{g\\times ^\\circ C} \\times 250.0 g \\times (36.0 ^\\circ C-20.5 ^\\circ C)= 16220.75 J"
2.Moles of solute:
"2.35 g Mg(OH)_2 \\times (\\frac{1 mole Mg(OH)_2}{58.32 g Mg(OH)_2})= 0.0403 mol"
3.As the temperature is increased then the reaction is exothermic, then:
"\\Delta H_{soln}= \\frac{-q}{n}"
"\\Delta H _{soln} = \\frac{-16220.75 J}{0.0403 mol}=-402500 \\frac{J}{mol}=- 403 \\frac{kJ}{mol}"
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