suppose an epa chemist tests a 200 ml sample of groundwater known to be contaminated with iron(iii) chloride, which would react with silver nitrate solution like this:
fecl3(aq) +3agn)3 (aq) =3agcl (s) fe(no3)3 (aq)
the chemist adds a 59.0mm silver nitrate solution to the sample until silver chloride stops forming. she then washes, dries, and weighs the precipitate. she finds she has collected 4.3 mg of silver chloride.
calculate the concentration of iron chloride containment in the original groundwater

Sorry i dont understand the question can you rephase 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.

We first assume that this gas is an ideal gas where it follows the ideal gas equation. the said equation is expressed as: pv = nrt. from this equation, we can predict the changes in the pressure, volume and temperature. if the volume and the temperature of this gas is doubled, then the pressure still stays the same.