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A student needs to dilute a 0.23 M Pb(NO3)2 solution to make 53.0 mL of 0.14 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 0.23 M Pb(NO3)₂ solution that is needed. 53.0 mL 0.23 M X Answer Bank 0.14 M 0.14 mL 53.0 L 0.23 mL = mL

A student needs to dilute a 0.23 M Pb(NO3)2 solution to make 53.0 mL of 0.14 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 0.23 M Pb(NO3)₂ solution that is needed. 53.0 mL 0.23 M X Answer Bank 0.14 M 0.14 mL 53.0 L 0.23 mL = mL

Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter15: Solutions
Section: Chapter Questions
Problem 63QAP
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A student needs to dilute a 0.23 M Pb(NO3)2 solution to make 53.0 mL of 0.14 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 0.23 M Pb(NO3)₂ solution that is needed. 53.0 mL 0.23 M Answer Bank 0.14 M 0.14 mL 53.0 L 0.23 mL mL
Transcribed Image Text:A student needs to dilute a 0.23 M Pb(NO3)2 solution to make 53.0 mL of 0.14 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 0.23 M Pb(NO3)₂ solution that is needed. 53.0 mL 0.23 M Answer Bank 0.14 M 0.14 mL 53.0 L 0.23 mL mL
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