K(NO3)2 + AgCl + H2
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
AgNO3(aq) + KI(aq) = KNO3(aq) + AgI(s) This is a classic test for halogens, and AgI precipitates down as a yellow solid.
Yes, AgI is insoluble, therefore will form a precipitate.
The net ionic equation for the reaction between Pb(NO3)2 and KI is: Pb2+ (aq) + 2I- (aq) -> PbI2 (s) This represents the formation of solid lead(II) iodide.
The balanced chemical equation for the reaction between KCl (potassium chloride) and AgNO3 (silver nitrate) is: 2AgNO3 + KCl -> 2AgCl + KNO3 This equation shows that two moles of silver nitrate react with one mole of potassium chloride to produce two moles of silver chloride and one mole of potassium nitrate.
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
AgNO3(aq) + KI(aq) = KNO3(aq) + AgI(s) This is a classic test for halogens, and AgI precipitates down as a yellow solid.
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)
Do you want that for Monopotassium phosphate, dipotassium phosphate, or tripotassium phosphate? --------------------------------- To clarify for the previous answerer, ionic compounds inherently don't use the mono-, di-, or tri- system used for molecular compounds. Instead, when a cation and an anion is supplied, the ionic compound assumes the number of cations and anions that will generate a neutral ionic compound. In this case, since K is 1+ and PO4 is 3-, the compound potassium phosphate always refers to K3PO4. Therefore: 3AgNO3 + K3PO4 -> 3KNO3 + Ag3PO4
The balanced equation for the formation of silver iodide is: 2 AgNO3 + 2 KI → 2 AgI + 2 KNO3
Yes, AgI is insoluble, therefore will form a precipitate.
The net ionic equation for the reaction between Pb(NO3)2 and KI is: Pb2+ (aq) + 2I- (aq) -> PbI2 (s) This represents the formation of solid lead(II) iodide.
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)
The balanced chemical equation for the reaction between KCl (potassium chloride) and AgNO3 (silver nitrate) is: 2AgNO3 + KCl -> 2AgCl + KNO3 This equation shows that two moles of silver nitrate react with one mole of potassium chloride to produce two moles of silver chloride and one mole of potassium nitrate.
When silver nitrate and potassium iodide are combined, they undergo a double displacement reaction. Silver iodide is formed as a yellow precipitate, while potassium nitrate remains in solution. The balanced chemical equation for this reaction is: AgNO3 + KI -> AgI + KNO3.
No- KI, Potassium iodide is an ionic compound.