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A student needs to dilute a 0.40 M Pb(NO3)₂ solution to make 75.0 mL of 0.16 M Pb(NO3)2. Set up the calculation by placing the values with the correct units into the equation. Then, calculate the volume, in milliliters, of the 0.40 M Pb(NO3)2 solution that is needed.

A student needs to dilute a 0.40 M Pb(NO3)₂ solution to make 75.0 mL of 0.16 M Pb(NO3)2. Set up the calculation by placing the values with the correct units into the equation. Then, calculate the volume, in milliliters, of the 0.40 M Pb(NO3)2 solution that is needed.

Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter14: Chemical Equilibrium
Section: Chapter Questions
Problem 14.98QE
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A student needs to dilute a 0.40 M Pb(NO3)₂ solution to make 75.0 mL of 0.16 M Pb(NO3)2. Set up the calculation by placing the values with the correct units into the equation. Then, calculate the volume, in milliliters, of the 0.40 M Pb(NO3)₂ solution that is needed. 75.0 mL 75.0 L Answer Bank 0.40 M 0.16 mL 0.16 M 0.40 mL = mL
Transcribed Image Text:A student needs to dilute a 0.40 M Pb(NO3)₂ solution to make 75.0 mL of 0.16 M Pb(NO3)2. Set up the calculation by placing the values with the correct units into the equation. Then, calculate the volume, in milliliters, of the 0.40 M Pb(NO3)₂ solution that is needed. 75.0 mL 75.0 L Answer Bank 0.40 M 0.16 mL 0.16 M 0.40 mL = mL
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