The equilibrium constant, Kc, for the following reaction is 10.5 at 350 K. 2CH2Cl2(g) CH4(g) + CCl4(g)

Calculate the equilibrium concentrations of reactant and products when 0.277 moles of CH2Cl2 are introduced into a 1.00 L vessel at 350 K

257 grams h2o i think

A. burning woodcombustion is the process of burning something. make brainliest if correct

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.

Ithink carbon dioxide, i am not completely sure.tell me if i am wrong.