chemist mixes 1.00 g of CuCl2 with an excess of (NH4)2HPO4 in dilute aqueous solution. He measures the evolution of 670 J of heat as the two substances react to give Cu3(PO4)2(s). Compute H for the reaction of 1.00 mol of CuCl2 with an excess of (NH4)2HPO4.
−90.1 103 J mol−1
90.1 104 J...