10 mL of water was added to each of three test tubes. 8 grams of solid #1 dissolved completely in test tube #1. Only 1 gram of solid #2 dissolved completely in test tube #2. None of solid #3 dissolved in test tube number three. The relative solubilities of the three solid in water can be described as:

The answer is 1.82 j/g°c                  

about what?
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Use a magnet to separate iron, evaporate water, i'm not sure how to separate the other two i'm sorry but good luck

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.