This is just my homework not exam Question 3Not yetansweredMarked out of6.00FlagquestionIt was found that 24.7 mL of 0.184 N HI solution was required to neutralize 20.0 mL of a Ba(OH) 2 solution.2 HI (aq) + Ba(OH)2 (aq) --> Bal₂(aq) + 2 H₂O (1)Mol wt of Ba(OH) 2: 171,34 g/molMol wt of HI: 127.904 g/molWhat is the molar concentration (molarity) of Ba(OH)2 solution ?O a. 0.227 MO b. 0.0568MC. 0.1136MO d. 0.454 MTime left 1:16:02

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Question 3 Not yet answered. Marked out of 6.00 Flag question It was found that 24.7 mL of 0.184 N HI solution was required to neutralize 20.0 mL of a Ba(OH)2 solution. 2 HI (aq) + Ba(OH)2 (aq) --> Bal2(aq) + 2 H₂O (1) Mol wt of Ba(OH) 2: 171,34 g/mol Mol wt of HI: 127.904 g/mol What is the molar concentration (molarity) of Ba(OH) 2 solution ? a. 0.227 M b. 0.0568M C. 0.1136M O d. 0.454 M Time left 1:16:02

Question 3 Not yet answered. Marked out of 6.00 Flag question It was found that 24.7 mL of 0.184 N HI solution was required to neutralize 20.0 mL of a Ba(OH)2 solution. 2 HI (aq) + Ba(OH)2 (aq) --> Bal2(aq) + 2 H₂O (1) Mol wt of Ba(OH) 2: 171,34 g/mol Mol wt of HI: 127.904 g/mol What is the molar concentration (molarity) of Ba(OH) 2 solution ? a. 0.227 M b. 0.0568M C. 0.1136M O d. 0.454 M Time left 1:16:02

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 14.74QE: Lead poisoning has been a hazard for centuries. Some scholars believe that the decline of the Roman...

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