The chemical equation is:
(K+ + I-)(aq) + (Ag+ + [NO3]-)(aq) --> AgI (s) + (K+ + [NO3]-)(aq)
or
The chemical equation is:
K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
Ag(NO3)(aq) + KI(aq) ---> K(NO3)(aq) + AgI(s)
Lead iodide (Pb2I) precipitates as a yellow solid, leaving a solution of potassium and nitrate ions.
A yellow precipitate of silver iodide (AgI).
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
A precipitate of Lead iodide and Potassium nitrate are formed
Potassium iodide + silver nitrate --> Silver iodide and potassium nitrate The chemical equation is: K+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + K+[NO3]- (aq)
Ag(NO3)(aq) + KI(aq) ---> K(NO3)(aq) + AgI(s)
Yes. The equation is Pb(NO3)2(aq) + 2KI(aq) ==> PbI2(s) + 2 KNO3(aq)
no
Lead iodide (Pb2I) precipitates as a yellow solid, leaving a solution of potassium and nitrate ions.
potassium nitrate would be left was an aqueous solution and lead iodide would be the precipitate
Pour a solution of Sodium(or Potassium) Iodide over Lead nitrate solution. The Lead iodide will be precipitated out as a yellow solid
A yellow precipitate of silver iodide (AgI).
This is a double displacement reaction. 2KI + Pb(NO3)2 --> 2KNO3 + PbI2 Potassium iodide + Lead(II) nitrate --> Potassium nitrate + Lead(II) iodide A bright yellow precipitate will form when these two react.