
An inorganic compound (A) in its aqueous solution produces a white ppt, with $ NaOH $ , which gets dissolved in excess of $ NaOH $ . The aqueous solution of (A) also produced white ppt, with $ N{H_4}OH $ . Also, its aqueous solution produced light yellow ppt, with $ AgN{O_3} $ solution, soluble in dil. Identity (A).
Answer
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Hint :An inorganic compound is a substance that does not contain both carbon and hydrogen. Examples are water and hydrochloric acid. In general, there are four groups of inorganic compound types. They are divided into bases, acids, salts, and water. We need to find the inorganic white crystalline salt (for white ppt) and it is highly soluble in water and its solution is mildly acidic (to produce ppt).
Complete Step By Step Answer:
Aluminium bromide (A) reacts with sodium hydroxide to give a white precipitate of aluminium hydroxide. Then the chemical reaction can be written as
$ AlB{r_3} + 3NaOH \to \mathop {Al{{\left( {OH} \right)}_3}}\limits_{white\;ppt} + 3NaBr $
The precipitate dissolves in excess of NaOH to form sodium meta aluminate. Then the chemical reaction can be written as
$ Al{\left( {OH} \right)_3} + NaOH \to Na\left( {Al{O_2}} \right) + 2{H_2}O $
Where $ Na\left( {Al{O_2}} \right) $ is sodium meta aluminate.
Similarly, aluminium bromide reacts with ammonium hydroxide to produce aluminium hydroxide and ammonium bromide.
$ AlB{r_3} + 3N{H_4}OH \to Al{\left( {OH} \right)_3} + 3N{H_4}Br $
Aluminium bromide reacts with silver nitrate to form a yellow precipitate of silver bromide. Then the chemical reaction can be written as
$ AlB{r_3} + 3AgN{O_3} \to Al{\left( {N{O_3}} \right)_3} + 3AgBr $
Hence the inorganic compound (A) is Aluminium bromide ( $ AlB{r_3} $ ).
Note :
All living organisms contain carbon-based compounds, making them organic. Organic compounds contain carbon. There are at least four important organic compounds, lipids, proteins, carbohydrates, and nucleic acids. Inorganic compounds do not contain carbon and usually are more simple compounds.
Complete Step By Step Answer:
Aluminium bromide (A) reacts with sodium hydroxide to give a white precipitate of aluminium hydroxide. Then the chemical reaction can be written as
$ AlB{r_3} + 3NaOH \to \mathop {Al{{\left( {OH} \right)}_3}}\limits_{white\;ppt} + 3NaBr $
The precipitate dissolves in excess of NaOH to form sodium meta aluminate. Then the chemical reaction can be written as
$ Al{\left( {OH} \right)_3} + NaOH \to Na\left( {Al{O_2}} \right) + 2{H_2}O $
Where $ Na\left( {Al{O_2}} \right) $ is sodium meta aluminate.
Similarly, aluminium bromide reacts with ammonium hydroxide to produce aluminium hydroxide and ammonium bromide.
$ AlB{r_3} + 3N{H_4}OH \to Al{\left( {OH} \right)_3} + 3N{H_4}Br $
Aluminium bromide reacts with silver nitrate to form a yellow precipitate of silver bromide. Then the chemical reaction can be written as
$ AlB{r_3} + 3AgN{O_3} \to Al{\left( {N{O_3}} \right)_3} + 3AgBr $
Hence the inorganic compound (A) is Aluminium bromide ( $ AlB{r_3} $ ).
Note :
All living organisms contain carbon-based compounds, making them organic. Organic compounds contain carbon. There are at least four important organic compounds, lipids, proteins, carbohydrates, and nucleic acids. Inorganic compounds do not contain carbon and usually are more simple compounds.
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