A solution is prepared by adding 0.080 mole of Ni(NH3). Cl₂ to 0.50 L of 3.1 M NH,. Calculate 5.5 x 10%. That is, 5.5 x 10 = for the overall reaction NH(NH,),7] [N] [NH] Ni (aq) + 6NH,(aq) NH(NH,),”(aq) [Ni(NH3),*]-[ [N]-C M te [Ni (NH₂)] and [Ni+] a in this solution. Kurall for Ni (NH₂), is

A solution is prepared by adding 0.080 mole of Ni(NH3). Cl₂ to 0.50 L of 3.1 M NH,. Calculate 5.5 x 10%. That is, 5.5 x 10 = for the overall reaction NH(NH,),7] [N] [NH] Ni (aq) + 6NH,(aq) NH(NH,),”(aq) [Ni(NH3),*]-[ [N]-C M te [Ni (NH₂)] and [Ni+] a in this solution. Kurall for Ni (NH₂), is

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter10: Effect Of Electrolytes On Chemical Equilibria

Section: Chapter Questions

Problem 10.16QAP

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