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If 32.0 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.545 g precipitate, what is
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2. asked by Hel P.
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The class will react 50.00 mL samples of .200 mol/L potassium phosphate with an excess of 0.120 mol/L lead II nitrate solution.w
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2. asked by Anon
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50.00 mL samples of 0.200 mol/L potassium phosphate with an excess of 0.120 mol/L lead II nitrate solution.what is the minimum
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2. asked by Michelle
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I would be very grateful to have these answered with showing workIn order to produce a lead (II) chromate precipitate, lead (II)
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2. asked by Stoich failure
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the solubility product of lead(ii)sulphate in water is 1.6x10^-8 mol2 dm^-6. calculate the solubility of lead(ii)sulphate in
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2. asked by frank
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4. A solution contains 100 ppm of lead. Determine the mass of lead present in 0.500 kg of the solution.16. While cleaning a
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2. asked by carter
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One way to remove lead ion from water is to add a source of iodide ion so that lead iodide will precipitate out of solution:Pb2+
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2. asked by Jane
3. 1,740 views
A solution containing excess lead(ii) nitrate is reacted with 380.0ML of 0.250mol/L potassium iodide solution. A bright yellow
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2. asked by Robyn
3. 772 views
Suppose 0.490g of lead(II) acetate is dissolved in 50 mL of a 57.0 m M aqueous solution of ammonium sulfate. Calculate the final
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2. asked by Nadia
3. 124 views
a class reacts 50.00ml samples of 0.200 mol/L potassium phosphate solution with an excess of 0.120 mol/L lead(II) nitrate
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2. asked by tryme00
3. 681 views