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Ni(NO3)2 + 2NH4F = NiF2 + 2NH4NO3 a. How many mL of a 0.245 M Ni(NO3)2...

Ni(NO3)2 + 2NH4F = NiF2 + 2NH4NO3

a. How many mL of a 0.245 M Ni(NO3)2 solution are required to consume 25.0 mL of 0.623 M NH4F?

b. If 17 mL of a 0.198 M NH4F solution reacts with 12.0 mL of a 0.215 M Ni(NO3)2 solution which reactant is in excess? How many moles?

c. What is the ionic and net ionic equations?

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