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Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb* ions required to form 0.87 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 0: 15.999 g/mol Na: 22.989769 g/mol 1:126.90447 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) Round your answer to 4 decimal places.

Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb* ions required to form 0.87 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 0: 15.999 g/mol Na: 22.989769 g/mol 1:126.90447 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) Round your answer to 4 decimal places.

General Chemistry - Standalone book (MindTap Course List)
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 4.146QP
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Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb2+ ions required to form 0.87 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 0: 15.999 g/mol Na: 22.989769 g/mol 1: 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) Round your answer to 4 decimal places.
Transcribed Image Text:Lead (II) nitrate is reacted with sodium iodide. Calculate the number of Pb2+ ions required to form 0.87 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 0: 15.999 g/mol Na: 22.989769 g/mol 1: 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) Round your answer to 4 decimal places.
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