Having 2SO3(g) > 2SO2(g) + O2(g) the Kc = 3.57 x 10^-3 at 726.85 celcius degrees what is the equilibrium K for the following reaction at 726.85 celcius degrees and 0.25 mol SO3(g)> 0.25 mol SO2(g) + 0.125 mol )O2(g) answers 0.00357 2.64x10^-20 4.46 x10^-4 0.494 0.244

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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.

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Hcn is the correct answer leave a good rate and .