What is the oxidation number of an element?
1 answer:
You might be interested in
Answer:
17.934 kg of water
Explanation:
If balanced equation is not given; this format can come in handy.
For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:
2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O
For butane:
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)
2 moles of butane gives 10 moles of water.
1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)
Mass of 1 mole of any substance is equal to it's molar mass
So, if 2 x N molecules of butane gives 10 x 18 g of water.
Then 1.2 x 10²⁶ molecules will give:
= 17.934 x 10³ g of water
= 17.934 kg of water
Answer : The partial pressure of at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar
Solution : Given,
Initial pressure of = 1.42 bar
Initial pressure of = 2.87 bar
= 0.036
The given equilibrium reaction is,
Initially 1.42 2.87 0
At equilibrium (1.42-x) (2.87-3x) 2x
The expression of will be,
Now put all the values of partial pressure, we get
By solving the term x, we get
From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.
Thus, the partial pressure of at equilibrium = 2x = 2 × 0.287 = 0.574 bar
The partial pressure of at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar
The partial pressure of at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar
The total pressure at equilibrium = Partial pressure of + Partial pressure of
+ Partial pressure of
The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar