In a cup run and CuI crystals PbCl2. Did four works of solubility: pKCuI = 11.96, pKCuCl = 6.73, pKPbCl2 = 4.29, pKPbI2 = 8.19. You have 4 unknown concentrations in the solution, but 5 equations (which?). Can you explain what will happen? Documented with calculations.

Law of Hess and heats of dissolution in acid.^б-в Х, КХ

In 298.15 º K are set following the dissolution enthalpy of these compounds in HCl:

B₂O₃.3CaO           -344.86 kJ/mol,

B₂O₃.2CaO           -210.40 kJ/mol,

B₂O₃.CaO  ...