we can precipitate out solid lead (II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. if we have 500. mL of 0.632 M lead(II) nitrate solution how much 1.50 M potassium iodide must we add to exacltly precipitate all the lead (II) ions? 0.211 L 461 mL 0.632 L 0.421 L 0.316 L or 1.19 L
we can precipitate out solid lead (II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. if we have 500. mL of 0.632 M lead(II) nitrate solution how much 1.50 M potassium iodide must we add to exacltly precipitate all the lead (II) ions? 0.211 L 461 mL 0.632 L 0.421 L 0.316 L or 1.19 L
Introductory Chemistry: A Foundation
8th Edition
ISBN:9781285199030
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 126AP: You pour 150.0 mL of a 0.250 M lead(ll) nitrate solution into an empty 500-mL flask. What is the...
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we can precipitate out solid lead (II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. if we have 500. mL of 0.632 M lead(II) nitrate solution how much 1.50 M potassium iodide must we add to exacltly precipitate all the lead (II) ions?
0.211 L
461 mL
0.632 L
0.421 L
0.316 L
or
1.19 L
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