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Neutralization Titration In the standardization of HCI solution using 10.00 ml of 0.02   Na,co, according to the equation   M of   Na,CO, 2 HCI 2NaCl + H₂O+CO, 15.0 ml of HCI is required to reach the end point using Bromocresol green indicator the molarity of HCI (mol/L) would be   al 0.032 b) 0.027   20.013   d) 0.008   e) 0.064

Neutralization Titration In the standardization of HCI solution using 10.00 ml of 0.02   Na,co, according to the equation   M of   Na,CO, 2 HCI 2NaCl + H₂O+CO, 15.0 ml of HCI is required to reach the end point using Bromocresol green indicator the molarity of HCI (mol/L) would be   al 0.032 b) 0.027   20.013   d) 0.008   e) 0.064

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Neutralization Titration In the standardization of HCI solution using 10.00 ml of 0.02

 

Na,co, according to the equation

 

M of

 

Na,CO, 2 HCI 2NaCl + H₂O+CO, 15.0 ml of HCI is required to reach the end point using Bromocresol green indicator the molarity of HCI (mol/L) would be

 

al 0.032 b) 0.027

 

20.013

 

d) 0.008

 

e) 0.064

Experiment 3: Neutralization Titration In the standardization of HCI solution using 10.00 ml of 0.02 M of Na,, according to the equation Na₂CO₂+2 HCI---- 2NaCl + H₂O + CO₂ if 15.0 ml of HCI is required to reach the end point using Bromocresol green indicator the molarity of HCI (mol/L) would be a) 0.032 b) 0.027 c) 0.013 d) 0.008 e) 0.064
Transcribed Image Text:Experiment 3: Neutralization Titration In the standardization of HCI solution using 10.00 ml of 0.02 M of Na,, according to the equation Na₂CO₂+2 HCI---- 2NaCl + H₂O + CO₂ if 15.0 ml of HCI is required to reach the end point using Bromocresol green indicator the molarity of HCI (mol/L) would be a) 0.032 b) 0.027 c) 0.013 d) 0.008 e) 0.064
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