On oxidation of aldehydes produces carboxylic acid functional group.
The product of oxidation of aldehydes depends upon whether the reaction occur in acidic medium or alkaline condition.
If oxidation of aldehydes occurs under acidic condition the product is carboxylic acid but if oxidation of aldehydes occurs under alkaline condition then reduce as well as oxidized product obtained which is known as disproportional product.
The oxidation of aldehydes occur through potassium dichromate, potassium permanganate or many more. The oxidation of aldehydes in the presence of base is known as cannizzaro's reaction.
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Answer:
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Explanation:
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From the chemical formula, 1 formula unit of KAl (SO4)2.12H2O
encompasses 1 atom of Al = 4 * 2 atoms of O in KAl (SO4)2 + 12 atoms of O in
12H2O which is equal to 20 atoms of O.
So, if you have 1.3 × 10^21 Al atoms, you have 20 * 1.3 × 10^21 O atoms will
now be equal to 2.6 * 10^22 atoms of O.
The acid and alkali that would react to make sodium sulfate is
alkali= sodium hydroxide
acid = sulphuric acid
<u><em>explanation</em></u>
Acid react react with alkali to form salt and water.
For sodium sulfate salt to be formed a sulfate acid is required to react with a hydroxide of sodium .
therefore Sodium hydroxide (alkali) react with sulphuric acid (an acid) to form sodium sulfate (salt) and water.
<u>Answer:</u> The ionic compound formed is (barium fluoride)
<u>Explanation:</u>
Ionic compound is formed by the complete transfer of electrons from 1 atom to another atom. The cation is formed by the loss of electrons by metals and anions are formed by gain of electrons by non metals.
Taking the metal as barium and non-metal as fluorine.
Barium is the 56th element of the periodic table having electronic configuration of
This element will loose 2 electrons and will form ion
Fluorine is the 9th element of the periodic table having electronic configuration of
This element will gain 1 electron and will form ion
By criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.
Hence, the ionic compound formed is (barium fluoride)