Consider the following statements: I) Real gases act more like ideal gases as the temperature increases. II) When n and T are constant, a decrease in P results in a decrease in V . III) At 1 atm and 273 K, every molecule in a sample of a gas has the same speed. IV) At constant T, CO2 molecules at 1 atm and H2 molecules at 5 atm have the same average kinetic energy. Which of these statements is true

A1 a 2 a 3 ooh lol i think that's it

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 about 50 % of an original isotope's number of atoms remain