What do chemists use percent yield calculations for in the real world?
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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
Data:
Arsenic Molar Mass = 74,9216 ≈ 75 u (<span>atomic mass unit)</span>
Solving:
1 mole of arsenic → 75g ------------ 6,02*10²³ molecules
..................................X -------------- 1 molecule
6,02*10²³X = 75
Arsenic Molar Mass = 74,9216 ≈ 75 u (<span>atomic mass unit)</span>
Solving:
1 mole of arsenic → 75g ------------ 6,02*10²³ molecules
..................................X -------------- 1 molecule
6,02*10²³X = 75