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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Answer:
The answer to your question is below
Explanation:
a) HCl 0.01 M
pH = -log [0.01]
pH = - (-2)
pH = 2
b) HCl = 0.001 M
pH = -log[0.001]
pH = -(-3)
pH = 3
c) HCl = 0.00001 M
pH = -log[0.00001]
pH = - (-5)
pH = 5
d) Distilled water
pH = 7.0
e) NaOH = 0.00001 M
pOH = -log [0.00001]
pOH = -(-5)
pH = 14 - 5
pH = 9
f) NaOH = 0.001 M
pOH =- log [0.001]
pOH = 3
pH = 14 - 3
pH = 11
g) NaOH = 0.1 M
pOH = -log[0.1]
pOH = 1
pH = 14 - 1
pH = 13
Answer:
Lens and focal point is the correct answer.
The total number of ions in 38.1 g of SrF₂ is 5.479 x 10²³.
<h3>What are ions?</h3>
Ions are the elements with a charge on them. It happens when they share electrons with other atoms to form a compound.
We have to calculate the total number of ions in 38.1 g of .
The molar mass of SrF₂ = 125.62 g/mol
The number of moles = 38.1 g of 1.0 mol / 125.62 = 0.30329 moles
Given that, total moles of SrF₂ ions in = 1.0 mol of + 2.0 moles of = 3.0 moles
Total moles of ions in 0.30329 moles of
= (0.30329 moles of SrF₂) x 3.0 / 1.0 = 0.90988 mol ions
We know that,
1.0 mole of ions = 6.023 x 10²³ ions
Thus, the number of total ions = ( 0.90988 mol ions) x 6.023 x 10²³ / 1.0 mol = 5.479 x 10²³ ions
Thus, the number of ions is in 38.1 g of 5.479 x 10²³ ions
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Answer:
1. The reaction will proceed backward, shifting the equilibrium position to the left.
2. The reaction will proceed forward, shifting the equilibrium position to the right.
3. Either add more of the products ( H2O or Cl2) or remove the reactant (HCl or O2)
Explanation:
