Lead(II) nitrate reacts with potassium iodide in solution to produce the spectator species potassium nitrate, and a bright-yellow lead(II) iodide precipitate. This is a common reaction in high school chemistry. The balanced equation is: Pb(NO3)2(aq) + 2KI(aq) --> 2KNO3(aq) + PbI2(s)
When lead acetate reacts with potassium iodide then lead iodide PbI2 which is an yellow precipitate is formed. This is a double displacement reaction.The balanced chemical equation for this:-
Pb(CH3COO)2 + 2 KI = PbI2 + 2 K(CH3COO)
Pb(CH3COO)2 + KI ---> (?) aq + aq ---> s + aq
The Solution is a yellow powder in water (you can see it clearly if you filter it).
yes
when added together the solution will turn yellow and after a while leave powdery sediment at the bottom
if the potassium iodide mix with lead II nitrate it will results a brigth yellow color. it means that the two substance are reactive tats why is it results that color.
Yes it does. Because the Fluorine is more reactive than iodine it will form potassium fluoride and iodine.
2KI + F2 --> 2KF + I2
If solutions are mixed, then lead iodide will precipitate.
Pb2+ + 2I- --> PbI2(s)
yes. the combined chemicals create a yellowish slime, a little like oobleck, if you've ever done that experiment. its toxic though, so don't play with it!
Pb(CH3COO)2 + 2 KI = PbI2 + 2 K(CH3COO)
PbI2 is an yellow precipitate.
The products are Silver chloride (a white precipitate) and potassium ethanoate (acetate). NB THis is a classic test for halides.
Pb2+ + I- --> PbI2(s)potassium and acetate ions are left out of the equation, because they don't react (stay unchanged in solution)
A strong electrolyte dissociates completely into ions in aqueous solution. When potassium acetate, a strong electrolyte, is put into water the cations and anions are surrounded by water molecules and the solid dissolves by the following equation:KCH3COO(s) ----> K+(aq) + CH3CO2-(aq)
KC2H3O2 is Potassium acetate. K- Potassium C2H3O2 - acetate
Silver (I) Chloride and Potassium Acetate. All one has to do to solve this problem is to switch the anions and the cations.
The products are Silver chloride (a white precipitate) and potassium ethanoate (acetate). NB THis is a classic test for halides.
Water, potassium and acetate ions. If you're adding equal amounts of both, the final solution will have a pH greater than 7.
Pb2+ + I- --> PbI2(s)potassium and acetate ions are left out of the equation, because they don't react (stay unchanged in solution)
A strong electrolyte dissociates completely into ions in aqueous solution. When potassium acetate, a strong electrolyte, is put into water the cations and anions are surrounded by water molecules and the solid dissolves by the following equation:KCH3COO(s) ----> K+(aq) + CH3CO2-(aq)
A reaction doesn't occur.
Yes. Vinegar will neutralize potassium hydroxide to form a mildly alkaline solution of potassium acetate.
potassium acetate
Pb(CH3COO)2 + 2 KI = PbI2 + 2 K(CH3COO) PbI2 is an yellow precipitate.
There is no reaction, because silver iodide is very insoluble.
The reaction you describe in words is: KCl + AgCH3CO2 → AgCl + KCH3CO2 This type of reaction is known as a "double replacement reaction" aka "double displacement reaction" aka "metathesis reaction." A double replacement reaction is a chemical reaction where two reactant ionic compounds exchange ions to form two new products compounds with the same ions. In this case the ions are K+ , Cl-, Ag+ and C2H3O2−. Note that potassium chloride (KCl), silver acetate (AgCH3CO2) and potassium acetate (KCH3CO2) are all quite soluble in water. Silver chloride (AgCl) is not particularly soluble in water (520 μg/100 g at 50 °C) and will precipitate out as the reaction occurs.
KC2H3O2 is Potassium acetate. K- Potassium C2H3O2 - acetate
These compounds doesn't react.