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-BARSIC- [3]
3 years ago
5

Name two methods that can be used to break down compounds into simpler substance

Chemistry
2 answers:
Nezavi [6.7K]3 years ago
8 0
<span>The two methods that can be used do break down compounds into simpler substances or smaller units, are known as Hydrolysis and Photolysis. 

Hydrolysis uses water to split a molecule. Water is used to break the bonds of a molecule. There are three main types of hydrolysis:</span>1. acid hydrolysis2. base hydrolysis 3. salt hydrolysis

Photolysis uses energy from light to split molecules.  This energy is called photons. As in hydrolysis, photons are used to break the bonds of a molecule. Photolysis is a chemical reaction. 
kotykmax [81]3 years ago
7 0
<span>Photolysis and hydrolysis. These are two methods that can be used to break down a compound into simpler substances and smaller units.

Water which is used to break the bonds of molecules and split molecules is used from hydrolysis. Hydrolysis is made of three types which include;
1. salt hydrolysis.
2. acid hydrolysis.
3. Base hydrolysis.
 Photolysis is well known to use energy from light to split the molecule and the same energy is referred to as photons which are used to break builds of molecules.</span>
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In the reaction 2H2O (1)+ 2Cl^- (aq)= H2(g)+Cl2 (g)+ 2OH^-(aq), which substance is reduced?
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Answer:- C. H

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As per the rules, oxidation number of hydrogen in its compounds is +1(except metal hydrides) and the oxidation number of oxygen in its compounds is -2.

The oxidation number in elemental form is zero.

In H_2O , the oxidation number of H is +1 and oxidation number of O is -2. Oxidation number of Cl in Cl^- is -1. On product side, the oxidation number of hydrogen in H_2 is zero and in OH^- the oxidation number of H is +1 and that of O is -2. Oxidation number of Cl in Cl_2 is 0.

From above data, Oxidation number of O is -2 on both sides so it is not reduced.

Oxidation number of Cl is changing from -1 to 0 which is oxidation.

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<u>a) Answer: </u>

<em>Number of molecules in 1 mole</em>

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<u>b)  Answer: </u>

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