Chapter 4, Problem 113b
The arsenic in a 1.22-g sample of a pesticide was converted to AsO43- 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.
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A sample of 8.69 g of Zn1OH22 is added to 155.0 mL of 0.750 M H2SO4. (c) How many moles of ZnSO4 are present after the reaction is complete?
In 2014, a major chemical leak at a facility in West Virginia released 28,390 L of MCHM (4-methylcyclohexylmethanol, C8H16O) into the Elk River. The density of MCHM is 0.9074 g/mL. (a) Calculate the initial molarity of MCHM in the river, assuming that the first part of the river is 2.00 m deep, 90.0 m wide, and 90.0 m long.
The arsenic in a 1.22-g sample of a pesticide was converted to AsO43- 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?
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.)