66.8g to mg is 66800 hope this helps
The volume of O₂ : 21 L
<h3>Further explanation</h3>
Given
8.7 grams of C₂H₄
Required
Volume O₂
Solution
Reaction
C₂H₄ + 3 O₂ ⇒ 2 CO₂ + 2 H₂O
mol C₂H₄(MW= 28 g/mol) :
= mass : MW
= 8.7 g : 28 g/mol
= 0.311
From the equation, mol O₂ :
= 3/1 x mol C₂H₄
= 3/1 x 0.311
= 0.933
At STP, 1 mol gas=22.4 L, so for 0.933 mol :
= 0.933 x 22.4 L
= 20.899 L ≈ 21 L
Answer:- Solubility of the gas at 759 torr is 0.00753 g/L.
Solution:- From given data, 0.00327 g of gas is soluble in 0.376 L of water at 876 torr.
Solubility of gas at 876 torr pressure is = 0.00327g/0.376L = 0.00869 g/L
Solubility of gases is directly proportional to the pressure. It means, grater is the pressure, more is the solubility of gases.
So, 0.00869/876 = X/759
Where, X is the solubility of the gas at 759 torr.
X = 0.00869(759)/876
X = 0.00753 g/L

- From the first mechanism, it can be seen that the Aluminum is reduced at cathode and Florine is oxidized at anode.
- The ratio of aluminium and fluorine is the ratio of their change in oxidation number, i.e. 2:3
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Answer:
Axis
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:) gimme a crown plz