tin hydroxide would be formed if water was present during the experiment which would make it impossile to make the desired product.
The reaction between concentrated sulfuric acid and solid potassium iodide does not produce hydrogen iodide. Instead, it generates sulfur dioxide gas, potassium sulfate, and iodine. Therefore, this reaction is unsuitable for obtaining hydrogen iodide.
Hydrogen iodide is made by combining hydrogen gas and iodine vapor in a reaction vessel under controlled conditions. The reaction forms hydrogen iodide gas, which can be further purified and isolated through various methods such as fractional distillation.
When solutions of lead nitrate and potassium iodide are mixed, a yellow precipitate of lead iodide is formed. This reaction is a double displacement reaction where the lead from lead nitrate reacts with the iodide from potassium iodide to form the insoluble lead iodide.
Ethyl iodide will undergo an SN2 reaction with potassium acetate to form ethyl acetate and potassium iodide. This reaction involves the substitution of the iodine atom in ethyl iodide with the acetate ion from potassium acetate.
The reaction of 1-bromo-cyclopentane with sodium iodide in acetone is a nucleophilic substitution reaction known as the Finkelstein reaction. In this reaction, the bromine atom is replaced by an iodine atom, resulting in the formation of sodium bromide, sodium iodide, and cyclopentyl iodide.
The reaction between concentrated sulfuric acid and solid potassium iodide does not produce hydrogen iodide. Instead, it generates sulfur dioxide gas, potassium sulfate, and iodine. Therefore, this reaction is unsuitable for obtaining hydrogen iodide.
Hydrogen iodide is made by combining hydrogen gas and iodine vapor in a reaction vessel under controlled conditions. The reaction forms hydrogen iodide gas, which can be further purified and isolated through various methods such as fractional distillation.
When solutions of lead nitrate and potassium iodide are mixed, a yellow precipitate of lead iodide is formed. This reaction is a double displacement reaction where the lead from lead nitrate reacts with the iodide from potassium iodide to form the insoluble lead iodide.
No reaction because the anion is common in both compounds.
Ethyl iodide will undergo an SN2 reaction with potassium acetate to form ethyl acetate and potassium iodide. This reaction involves the substitution of the iodine atom in ethyl iodide with the acetate ion from potassium acetate.
The reaction of 1-bromo-cyclopentane with sodium iodide in acetone is a nucleophilic substitution reaction known as the Finkelstein reaction. In this reaction, the bromine atom is replaced by an iodine atom, resulting in the formation of sodium bromide, sodium iodide, and cyclopentyl iodide.
The reaction between vanadium (III) sulfate (V2(SO4)3) and barium iodide (BaI2) would form barium sulfate (BaSO4) and vanadium (III) iodide (V2I3). This reaction is a double displacement reaction.
The reaction between lead(II) nitrate and sodium iodide is a double replacement reaction. The products of this reaction are lead(II) iodide and sodium nitrate, formed through the exchange of ions.
The reaction between bromine and aqueous strontium iodide is a double displacement reaction. Bromine replaces iodide in strontium iodide, forming strontium bromide and iodine. The balanced chemical equation is: Br2 + SrI2 --> SrBr2 + I2.
The reaction between silver nitrate and potassium iodide forms silver iodide precipitate and potassium nitrate. This reaction is a double displacement reaction where the silver ions from silver nitrate switch places with the potassium ions in potassium iodide.
There is no reaction, because silver iodide is very insoluble.
When bleach is mixed with potassium iodide, it undergoes a redox reaction. The bleach oxidizes the iodide ions to produce iodine, which can be observed as a brown color. This reaction can be used to test for the presence of bleach in a solution.