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Lead Ions Can Be Precipitated from Aqueous Solutions by the Addition

question 226

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Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide:   (aq)  +   (aq)  →   (s)  Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many millilitres of 3.550 mol L<sup>-1</sup> HI(aq)  must be added to a solution containing 0.700 mol of   to completely precipitate the lead? A)  197 mL B)  0.197 mL C)  394 mL D)  0.394 mL E)  2.54 × 10<sup>-3</sup> mL (aq) + Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide:   (aq)  +   (aq)  →   (s)  Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many millilitres of 3.550 mol L<sup>-1</sup> HI(aq)  must be added to a solution containing 0.700 mol of   to completely precipitate the lead? A)  197 mL B)  0.197 mL C)  394 mL D)  0.394 mL E)  2.54 × 10<sup>-3</sup> mL (aq) → Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide:   (aq)  +   (aq)  →   (s)  Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many millilitres of 3.550 mol L<sup>-1</sup> HI(aq)  must be added to a solution containing 0.700 mol of   to completely precipitate the lead? A)  197 mL B)  0.197 mL C)  394 mL D)  0.394 mL E)  2.54 × 10<sup>-3</sup> mL (s) Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many millilitres of 3.550 mol L-1 HI(aq) must be added to a solution containing 0.700 mol of Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide:   (aq)  +   (aq)  →   (s)  Lead iodide is virtually insoluble in water so that the reaction appears to go to completion. How many millilitres of 3.550 mol L<sup>-1</sup> HI(aq)  must be added to a solution containing 0.700 mol of   to completely precipitate the lead? A)  197 mL B)  0.197 mL C)  394 mL D)  0.394 mL E)  2.54 × 10<sup>-3</sup> mL to completely precipitate the lead?


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