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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brainly.com/question/16856280</h3><h3 /><h3>#SPJ4</h3>
For stainless steel different kinds of compositions are used. Based on that different series of stainless steel has been coined.
1. Series 200 - Iron alloyed with <span>chromium, nickel and manganese.
2. Series 300 - It has
a. Stainless Steel 304 - it has composition of 18% chromium and 8% Nickel
b. </span>Stainless Steel 316 - This has 18% chromium and 10% Nickel
Each kind of stainless steel is of different cost and has different applications.
Data:
M (molarity) = 1.75 M (mol/L)
m (mass) = 35 g
MM (molar Mass) of NaCl = 58.44 g/mol
V (volume) = ? (in liters)
Formula:

Solving:





Answer:
% of n-propyl chloride = 43.48 %
Explanation:
There are 2 secondary hydrogens and 6 primary hydrogens
The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen
probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)
Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)
Total probability = 3.9
% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %