Which equation shows the proper setup to find the volume of a 0.250 M CaCl2 solution that contains 39.5 g of CaCl2? 0.250 Molarity upper C a upper C l subscript 2 equals StartFraction 39.5 grams upper C a upper C l subscript 2 over V EndFraction. V equals 39.5 grams upper C a upper C l subscript 2 time StartFraction 1 mole upper C a upper C l subscript 2 over 11098 grams upper C a upper C l subscript 2 EndFraction times StartFraction 1 liter over 0.250 moles upper C a upper C l subscript 2 EndFraction. V equals StartFraction 39.5 grams upper C a upper C l subscript 2 over 0.250 moles upper C a upper C l subscript 2 EndFraction.

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.

Answer ozone (o3) is made of three atoms of oxygen, it is a molecule;

Self-ionizes to form an equilibruium system in which [ h_{3} o^{+} ]=[ oh^{-} ]=10^{-7} m