In the laboratory, copper(I) Iodide is prepared by simply mixing an aqueous solutions of potassium iodide and a soluble copper(II) salt such copper sulphate. : :: Cu2+ + 2I− → CuI2 The CuI2 immediately decomposes to iodine and insoluble copper(I) iodide, releasing I2. : :: 2 CuI2 → 2 CuI + I2
Potassium sulfate would be colorless; copper iodide is probably blue or green (I don't know which off the top of my head, but copper salts are generally either blue or green).
2K(I) + CuCl2 => 2KCl + Cu(I)2 The (I) is iodide, because the capital I and lowercase l look the same. This is a double replacement reaction, where copper (II) iodide comes out of solution, because it's not soluble, assuming that both of those are in solution.
When copper sulfate solution is mixed with potassium iodide, a solid precipitate of copper iodide is formed, while potassium sulfate remains in solution. This reaction is a double displacement reaction. The balanced chemical equation is CuSO4 + 2KI → CuI2 + K2SO4.
React a solution of Copper(II) sulfate with Sodium Iodide. It will produce Copper(II) Iodide and Sodium sulfate. This Copper(II) iodide, due to instability, spontaneously decomposes, producing Iodine and Copper(I) Iodide.The Copper(I) iodide comes out as a yellowish brown precipitate. So the overall reaction seems to be:2 CuSO4 + 4 NaI ----> 2 CuI + 2 Na2SO4 + I2
Copper iodide is sparingly soluble in water, meaning it dissolves only to a small extent. It is more soluble in concentrated aqueous ammonia and potassium iodide solutions.
The amount of excess potassium iodide depends on the stoichiometry of the reaction between potassium iodide and copper sulfate. One equivalent of potassium iodide is needed to react with one equivalent of copper sulfate. Excess potassium iodide would be any amount added beyond this stoichiometric ratio.
Potassium sulfate would be colorless; copper iodide is probably blue or green (I don't know which off the top of my head, but copper salts are generally either blue or green).
2K(I) + CuCl2 => 2KCl + Cu(I)2 The (I) is iodide, because the capital I and lowercase l look the same. This is a double replacement reaction, where copper (II) iodide comes out of solution, because it's not soluble, assuming that both of those are in solution.
The balanced chemical equation for the reaction between potassium iodide and copper sulfate is: 2KI(aq) + CuSO4(aq) → CuI2(s) + K2SO4(aq). In this reaction, a double displacement occurs leading to the formation of insoluble copper iodide and soluble potassium sulfate.
When copper sulfate solution is mixed with potassium iodide, a solid precipitate of copper iodide is formed, while potassium sulfate remains in solution. This reaction is a double displacement reaction. The balanced chemical equation is CuSO4 + 2KI → CuI2 + K2SO4.
When iron reacts with potassium iodide, it forms potassium iodide. The reaction can be written as 2Fe + 2KI → 2K + 2FeI. Potassium iodide is a salt and is commonly used as a nutritional supplement.
React a solution of Copper(II) sulfate with Sodium Iodide. It will produce Copper(II) Iodide and Sodium sulfate. This Copper(II) iodide, due to instability, spontaneously decomposes, producing Iodine and Copper(I) Iodide.The Copper(I) iodide comes out as a yellowish brown precipitate. So the overall reaction seems to be:2 CuSO4 + 4 NaI ----> 2 CuI + 2 Na2SO4 + I2
Copper iodide is sparingly soluble in water, meaning it dissolves only to a small extent. It is more soluble in concentrated aqueous ammonia and potassium iodide solutions.
When potassium iodide reacts with lead nitrate, a double displacement reaction occurs. The potassium ion and the lead ion switch places to form potassium nitrate and lead iodide. This reaction results in the formation of a yellow precipitate of lead iodide.
Silver iodide (AgI), a precipitate insoluble in water, don't react with potassium nitrate.
Nothing happens
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