Analysis of ClA 0.4892 g sample of the chromium compound was dissolved in water and excess silver nitrate was added to precipitate AgCl. 1.0042 g of AgCl was obtained. Calculate the mass of Cland then % Cl- . Show work below. Answers: g Cl- % ClAnalysis of NH3 A 0.2502 g sample of the unknown chromium compound was dissolved in water, and 42.56 mL of 0.1089 M HCl was added, converting all the NH3 to NH4 with HCl left over. The excess HCl was titrated with 0.1022 M NaOH, requiring 10.28 mL Moles HCl added moles Moles NaOH added moles Moles HCl USED moles Moles NH3 moles Grams NH3 grams % NH3 % Determination of Empirical Formula Knowing % Cr3 , % NH3, and % Cl- , find the empirical formula. Show work below. Write the balanced equation for the preparation of the compound. Determine which of the original reagents was limiting and calculate the % yield. Show work below

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.

The answer is 3.

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