We have 495 ml, but we need to convert this to 1000, so we divide by 495 and multiply by 1000. We also do this to 21.1g. 21.1/495*100 is 42.626g. KI's molecular weight is 166g/mol. 42.626/166 is 0.2568 molar.
The compound is called Potassium Iodide (-ide.. not -ine).
somehow. the potassium iodide act as a acidifying agent,
A precipitate of Lead iodide and Potassium nitrate are formed
no
It's the same thing. It is a solution of water that has been saturated to the max with potassium, then iodide is added (a form of iodine). So it is a potassium iodide solution.
Find moles potassium iodide first.2.41 grams KI (1 mole KI/166 grams) = 0.01452 moles KIMolarity = moles of solute/Liters of solution ( 100 ml = 0.1 Liters )Molarity = 0.01452 moles KI/0.1 Liters= 0.145 M KI solution================
A precipitate of Lead iodide and Potassium nitrate are formed
somehow. the potassium iodide act as a acidifying agent,
The compound is called Potassium Iodide (-ide.. not -ine).
The solution of potassium iodide (if it is not extremely diluted) is more dense.
Silver nitrate + Potassium iodide ----> Silver iodide + Potassium nitrate AgNO3 + KI ----> AgI + KNO3
Reddish precipitate of Mercuric iodide and clear solution of Potassium chloride is produced
no
A yellow precipitate of silver iodide (AgI).
In aqueous solution they would not react. They would form a solution of ferric ions, chloride ions, potassium ions, and iodide ions.
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.