When `CO_(2)` is bubbled through a cold pasty solution of barium peroxide in water, `H_(2)O_(2)` is obtained. <br> `BaO+CO_(2)+H_(2)OtoBaCO_(3)+H_(2)O_(2)` Barium carbonate being insoluble is filtered off. This is known as Merck's process.
<h3>What is meant by Perhydrol?</h3>
perhydrol (countable and uncountable, plural perhydrols) A stabilised solution of hydrogen peroxide.
<h3>What is Merck's Perhydrol?</h3>
Uses: Perhydrol is used as an antiseptic for wounds, and also acts as a germicide to kill bacteria and germs.
Being a strong oxidizing agent it has bleaching properties and acts as a ripening agent.
Learn more about merck's process here:
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28%
Explanation:
mass of solute(KBr) = 3.73g
mass of solvent(H2O) = 131g
mass of solution = mass of solute + mass of solvent
= 3.73 + 131
= 134.73g

<h2>Answer : Option B) Hydrogen</h2><h3>Explanation :</h3>
All fossil fuels contains hydrocarbons in it. Amongst the given options hydrogen is the correct answer. Except that it contains carbon in it. Hydrocarbons are those class of compounds which contains hydrogen and carbon as element in it.
They are considered to be good fuels because they naturally bring out complete combustion as they contain hydrogen and carbon in its compound form.
Answer:
The value of Q must be less than that of K.
Explanation:
The difference of K and Q can be understood with the help of an example as follows
A ⇄ B
In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]
In this case ratio of [B] and [A] that is
K = [B] / [A] which is called equilibrium constant.
But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.
Q = [B] / [A] . It is clear that Q < K before equilibrium.
If Q < K , reaction will proceed towards equilibrium or forward reaction will
proceed .