Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb2+ ions required to form 0.78 g of precipitate. Write the balanced molecular, total ionic, and net ionic equations of the reaction. Atomic Mass: Pb: 207.2 g/mol N: 14.0067 g/mol O: 15.999 g/mol Na: 22.989769 g/mol I: 126.90447 g/mol Note: Use scientific notation with the format (n)e(x) where n = number and x = exponent (Example: 6.022e23 for 6.022 x 1023)
Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb2+ ions required to form 0.78 g of precipitate. Write the balanced molecular, total ionic, and net ionic equations of the reaction. Atomic Mass: Pb: 207.2 g/mol N: 14.0067 g/mol O: 15.999 g/mol Na: 22.989769 g/mol I: 126.90447 g/mol Note: Use scientific notation with the format (n)e(x) where n = number and x = exponent (Example: 6.022e23 for 6.022 x 1023)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter4: Chemical Reactions
Section: Chapter Questions
Problem 4.141QP: Lead(II) nitrate reacts with cesium sulfate in an aqueous precipitation reaction. What are the...
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Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb2+ ions required to form 0.78 g of precipitate. Write the balanced molecular, total ionic, and net ionic equations of the reaction.
Pb: 207.2 g/mol
N: 14.0067 g/mol
O: 15.999 g/mol
Na: 22.989769 g/mol
I: 126.90447 g/mol
Note: Use scientific notation with the format (n)e(x) where n = number and x = exponent (Example: 6.022e23 for 6.022 x 1023)
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