Answer:
Kp = 0.049
Explanation:
The equilibrium in question is;
2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)
Kp = p SO₃² / ( p SO₂² x p O₂ )
The initial pressures are given, so lets set up the ICE table for the equilibrium:
atm SO₂ O₂ SO₃
I 3.3 0.79 0
C -2x -x 2x
E 3.3 - 2x 0.79 - x 2x
We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:
p SO₂ = 3.3 -0.47 atm = 2.83 atm
p O₂ = 0.79 - (0.47/2) atm = .56 atm
Now we can calculate Kp:
Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )
Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.
Carbon dioxide and oxygen are removed from the air.
Explanation:
When air is passed through aqueous sodium hydroxide solution the carbon dioxide is removed from the air.
First the carbon dioxide will dissolve and react with water to form carbonic acid ( H₂CO₃) :
CO₂ + H₂O → H₂CO₃
The the carbonic acid will react with sodium hydroxide to form sodium carbonate (Na₂CO₃):
H₂CO₃ + 2 NaOH → Na₂CO₃ + 2 H₂O
After this by passing the air over heated cooper the oxygen is removed.
2 Cu + O₂ → 2 CuO
Learn more about:
neutralization reaction
brainly.com/question/2632201
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N(CH₃OH)=3,62·10²⁴/6·10²³ 1/mol = 6,033 mol
m(CH₃OH) = 6,033 mol · 32 g/mol (molar mass) = 193,06 g.
