Please help with as many as you can, this is a test. :(

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.

60.66% is the answer

Heat capacity = mcδtheat capacity = 2.676×10^3 j ..the heat capacity given is for 7.00 g of kcn,potassium cyanide which is equal to 0. mole of mole = mass  ÷ mr of kcn is 7  ÷ 65 =0.therefore: 0. of kcn liberates 2.676×10^3 j of heat1 mole   (2.676×10^3    ÷  0.107 )  × 1 =do the calculation to obtain 25009 j = 25kj