Sodium iodide is highly soluble in water, with a solubility of approximately 184 grams per 100 mL of water at room temperature. This high solubility is due to the strong ionic interactions between the sodium cation and the iodide anion with water molecules.
The reaction between silver nitrate and sodium iodide will result in the formation of silver iodide precipitate and sodium nitrate solution. This is a double displacement reaction where the silver ions from silver nitrate will combine with iodide ions from sodium iodide to form the insoluble silver iodide.
The precipitate formed when sodium hydroxide reacts with potassium iodide is brown.
The chemical name of Nal is Naltrexone.
No, lithium iodide is not soluble in acetone. Lithium iodide is generally soluble in water, but it has limited solubility in organic solvents like acetone.
Sodium chloride is not soluble in benzene.
It is false; sodium iodide is more soluble than sodium chloride in water.
Yes, sodium iodide (NaI) is highly soluble in water. It dissolves readily to form a clear solution.
The filtrate of lead nitrate and sodium iodide would contain soluble sodium nitrate and insoluble lead iodide. Lead iodide is a yellow solid that precipitates out of the solution, while sodium nitrate remains in the filtrate as it is soluble in water.
Benzene is nonpolar, so its molecules do not have any strong attraction to sodium chloride, which is ionic.
Sodium iodide is soluble in water.
Yes. Simple alkali metal salts tend to be soluble with extremely few exceptions; most halides are likewise soluble. An alkali metal halide, such as sodium iodide, should be expected to be extremely soluble in water.
Sodium iodide is highly soluble in water, with a solubility of approximately 184 grams per 100 mL of water at room temperature. This high solubility is due to the strong ionic interactions between the sodium cation and the iodide anion with water molecules.
No, NaCL is polar, benzen is non-polar.
No, sodium sulfate (Na2SO4) is not soluble in benzene because it is a polar compound and benzene is a nonpolar solvent. Therefore, they will not form a homogeneous solution.
The reaction between silver nitrate and sodium iodide will result in the formation of silver iodide precipitate and sodium nitrate solution. This is a double displacement reaction where the silver ions from silver nitrate will combine with iodide ions from sodium iodide to form the insoluble silver iodide.
The ionic compound of sodium iodide is NaI. It is composed of sodium (Na+) ions and iodide (I-) ions held together by ionic bonds.