Compute the change in temperature for 4.50 g of CaCl2 dissolved in 90.0mL of H2O. Presume that the calorimeter absorbs no heat. (that is Ccal=0) (remember use dimensional analysis) information given CaCl2(s) = Ca 2+(aq) + 2Cl-(aq) H=-82.8 kJ/mol
Density as well as specific heat of water is 1.0g/mol and 4.18Jg-1oC-1 respectively