Pb2+ + I- --> PbI2(s)potassium and acetate ions are left out of the equation, because they don't react (stay unchanged in solution)
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
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)
Bromine and Potassium iodide react to form Potassium bromide and Iodine.
Pb2+ + I- --> PbI2(s)potassium and acetate ions are left out of the equation, because they don't react (stay unchanged in solution)
In the reaction: Lead (Ⅱ) Nitrate + Potassium Iodide → Potassium Nitrate + Lead (Ⅱ) Iodide.. all nitrates are soluble and lead(ii)iodide is insoluble.
The chemical equation is:2 KI + Pb(NO3)2 = 2 KNO3 + PbI2(s)
It produces Potassium nitrate and Lead iodide
A precipitate of Lead iodide and Potassium nitrate are formed
The balanced equation is 2 KI + Pb(NO3)2 -> 2 KNO3 + PbI2.
potassium iodide
lead nitrate(Pb(NO3)2 + potassium iodide(KI) = lead iodide(PbI) + potassium nitrate (KNO3)
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Equation: Pb(NO3)2 + KI ----> PbI2 + KNO3
Worded Equation; Potassium Iodide + Calcium Chloride ------> Potassium Chloride + Calcium Iodide Chemical Equation; KI (l) + CaCl (l) -----> KCl (aq) + Ca(I)2 (aq) Note Answer is only correct if proper states are applied and used.
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3