Suppose you heat a metal object with a mass of 30.3 g to 96.9 oC and transfer it to a calorimeter containing 100.0 g of water at 18.0 oC. The water and metal reach a final temperature of 22.4 oC. What is the specific heat of the metal in J/g-oC?

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.

Try adding 700% because of the x in y

Descriptive investigations are mainly questions to uncover information. it does not have an hypothesis. it's mainly gathering information, it is not repeatable as you are just asking questions.