Loren has the samples of elements that are listed below at room temperature. He exposes the samples to the same heat source until each sample reaches a temperature of 90.0°C. 10 g of Al(s) ( = 0.897 J/(gi°C)) 10 g of Ag(s) (= 0.234 J/gi°C ) 10 g of Fe(s) ( = 0.450 J/gi°C) 10 g of Zn(s) ( = 0.387 J/gi°C ) From first to last, which lists the order in which these samples will reach 90.0°C? Al, Fe, Zn, Ag Ag, Zn, Fe, Al Al, Fe, Ag, Zn Ag, Al, Zn, Fe

Electrons are emitted

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.