List the Van't Hoff factor for each of the following solutes and rank their conductivities in solution. Whoever helps me first gets brainliest!!! Ca(OH)2, SO2, LiCL

hopefully this helps is not i have more just let me know

Explanation: #1 Identify the colligative property.

A) vapor pressure lowering

B) freezing point depression

C) boiling point elevation

D) osmotic pressure

E) all of the above

#2 is applied to roads in wintertime so that the ice will melt at a lower temperature.

A) Antifreeze

B) Water

C) Salt

D) Sugar

E) Oil

#3 Give the reason that antifreeze is added to a car radiator.

A) The freezing point is lowered and the boiling point is elevated.

B) The freezing point is elevated and the boiling point is lowered.

C) The freezing point and the boiling point are elevated.

D) The freezing point and the boiling point are lowered.

E) None of the above.

4# Place the following aqueous solutions of nonvolatile, nonionic compounds in order of decreasing osmotic pressure.

I. 0.011 M sucrose

II. 0.00095 M galactose

III. 0.0060 M glycerin

A) I > III > II

B) I > II > III

C) II > III > I

D) III > I > II

E) II > I > III

5# Place the following solutions in order of increasing osmotic pressure. I. 0.15 M C2H6O2 II. 0.15 M BaCl2 III. 0.15 M NaI A) III < I < II B) II < III < I C) I < II < III D) II < I < III E) I < III < II

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.