You can show the presence of Sulphate in a mixture by adding barium chloride solution followed by dil HCl. A white precipitate that does not dissolvev in dil HCl indicates the presence of sulphate ioins.
Ba2+ (aq) + SO4^2- (aq) ---------> BaSO4 (s)
Iodide ions can be shown by adding to the suspected iodide a solution of lead nitrate where upon a brilliant yellow ppt is thrown down
Pb2+ (aq) + 2 I- (aq) --------> PbI2 (aq)
Also addition of silver nitrate will give a yellow ppt which is insoluble in ammonia solution.
Ag+ (aq) + I- (aq) ----------> AgI (s)
When iron sulfate reacts with potassium iodide, a double displacement reaction occurs where potassium sulfate and iron(II) iodide are formed. The balanced chemical equation for this reaction is FeSO4 + 2KI → FeI2 + K2SO4. This reaction is characterized by a color change from yellow (iron sulfate) to brown (iron(II) iodide).
To separate chalk from a mixture of chalk and copper sulfate solution, you can add water to dissolve the copper sulfate, leaving behind the chalk. The water-soluble copper sulfate will form a blue solution, allowing you to filter out the chalk residue, which is insoluble in water.
When you add sodium sulfate to barium chloride, a white precipitate of barium sulfate forms. This is due to the reaction between sodium sulfate and barium chloride, which forms insoluble barium sulfate.
It is easier to separate a mixture of seashells and marbles because they have distinct physical properties, such as size, shape, and color, which allow for manual sorting. On the other hand, separating copper sulfate solution is more difficult because the components are both in liquid form and have similar physical properties, requiring a process like distillation or filtration to separate them effectively.
Tin has two oxidation states (II and IV), and exhibits approximately equal stability in both its II and IV oxidation state. The chemical formula Tin (II) Iodide is SnI2. The chemical formula for Tin (IV) Iodide is SnI4.
When iron sulfate reacts with potassium iodide, a double displacement reaction occurs where potassium sulfate and iron(II) iodide are formed. The balanced chemical equation for this reaction is FeSO4 + 2KI → FeI2 + K2SO4. This reaction is characterized by a color change from yellow (iron sulfate) to brown (iron(II) iodide).
Neither, iodized salt is just a mixture of sodium chloride and potassium iodide (both neutral salts).
To separate chalk from a mixture of chalk and copper sulfate solution, you can add water to dissolve the copper sulfate, leaving behind the chalk. The water-soluble copper sulfate will form a blue solution, allowing you to filter out the chalk residue, which is insoluble in water.
When barium sulfate reacts with copper sulfate, a double displacement reaction occurs. Barium sulfate and copper sulfate exchange ions to form barium sulfate and copper sulfate precipitates. The reaction results in the formation of a white solid of barium sulfate, which is insoluble in water.
No reaction because the anion is common in both compounds.
No, iodide refers to the ion I-, whereas potassium iodide is a chemical compound made up of both potassium (K+) and iodide ions (I-). When iodide is combined with potassium as in potassium iodide, the resulting compound has different properties and uses compared to just iodide alone.
Potassium iodide reacts similarly to sodium iodide, as both compounds are halides. They display similar chemical properties and reactivity when involved in reactions that involve iodine.
Take a few drops of both samples and add some lead nitrate. A yellow precipitate indicates lead iodide and it gives the inference that it contains iodide ions, hence the solution of sodium iodide.
When you add sodium sulfate to barium chloride, a white precipitate of barium sulfate forms. This is due to the reaction between sodium sulfate and barium chloride, which forms insoluble barium sulfate.
The isoelectronic of iodide (I-) is xenon (Xe), because both species have the same number of electrons.
It is easier to separate a mixture of seashells and marbles because they have distinct physical properties, such as size, shape, and color, which allow for manual sorting. On the other hand, separating copper sulfate solution is more difficult because the components are both in liquid form and have similar physical properties, requiring a process like distillation or filtration to separate them effectively.
If the bulb lights up when copper sulfate and vinegar acetic acid are tested for conductivity, it indicates that both solutions are conductive. This means that they contain ions that can carry electrical current. Conductivity in these solutions is likely due to the presence of dissolved ions from the copper sulfate and acetic acid.