If the mixture given was 2: 1 caffeine to binder, is the mass of caffeine that was recovered an accuratereflection of the original composition?
a. no, it is too low –there should be 100% caffeine in the original mixture
b. no, it is too low–there should be 67% caffeinein the original mixture
c. yes, it gives a percent composition of37.5% caffeine, 62.5% binder
d. yes, it gives a percent composition of 62.5% caffeine, 37.5% binder

Ibelieve y y even though i've only taken chemistry for one year

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.