The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (a) What is the oxidation state of As in AsO₄^3-?

The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (b) Name Ag₃AsO₄ by analogy to the corresponding compound containing phosphorus in place of arsenic.

The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (c) If it took 25.0 mL of 0.102 M Ag⁺ to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Potassium superoxide, KO₂, is often used in oxygen masks (such as those used by firefighters) because KO₂ reacts with CO₂ to release molecular oxygen. Experiments indicate that 2 mol of KO₂(s) react with each mole of CO₂(g). (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to: NH₃₍aq) + HCl(aq) → NH₄Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (a) How many grams of NH₃ were drawn into the acid solution?

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to: NH₃₍aq) + HCl(aq) → NH₄Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (b) How many ppm of NH₃ were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.)