A calorimeter was used to measure the heat change when an ionic compound dissolves in water. The final mass of the contents of the calorimeter was 50.54 g50.54 g and the change in temperature was 3.05 °C.3.05 °C. Calculate the heat change of the calorimeter contents, given that its specific heat is 4.14 J/(g⋅°C.)4.14 J/(g⋅°C.) q1=q1= JJ Calculate the heat change of the calorimeter, given that its heat capacity is 7.37 J/°C.7.37 J/°C. q2=q2= JJ Calculate the heat change of the solution process.

Is a logical series of steps

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.