If we represent the equilibrium as: cocl42-(aq) co2+(aq) + 4cl-(aq)we can conclude that: 1. this reaction is: a. exothermicb. endothermicc. neutrald. more information is needed to answer this question.2. when the temperature is decreased the equilibrium constant, k: a. increasesb. decreasesc. remains the samed. more information is needed to answer this question.3. when the temperature is decreased the equilibrium concentration of cocl42-: a. increasesb. decreasesc. remains the samed. more information is needed to answer this question.

Ok so co(g) is not pronoused that. the only clue is that youpronouse it the oppisote way. 

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.

Hello! data: p (pressure) = 1.25 atmv (volume) = ? (in l)t (temperature) = 298 kn (number of mols) = 0.900 molesr (gas constant) = 0.082 (atm*l/mol*k)apply the data to the clapeyron equation (ideal gas equation), see: i hope this . =)