The sample of oxygen gas was collected through water displacement. So, the gas collected will be a mixture of oxygen and water vapor.
Given that the total pressure of the mixture of gases containing oxygen and water vapor = 749 Torr
Vapor pressure of pure water at 
=25.81mmHg
= 
According to Dalton's law of partial pressures,
Total pressure = Partial pressure of Oxygen gas + Partial pressure of water
749 Torr = Partial pressure of Oxygen gas + 25.81 Torr
Partial pressure of Oxygen gas = 749 Torr - 25.81 Torr = 723.19 Torr
Therefore the partial pressure of Oxygen gas in the mixture collected will be 723.19 Torr
The answer to your question is false.
Answer
is: activation energy of this reaction is 212,01975 kJ/mol.<span>
Arrhenius equation: ln(k</span>₁/k₂) =
Ea/R (1/T₂ - 1/T₁).<span>
k</span>₁
= 0,000643 1/s.<span>
k</span>₂
= 0,00828 1/s.
T₁ = 622 K.
T₂ = 666 K.
R = 8,3145 J/Kmol.
<span>
1/T</span>₁ =
1/622 K = 0,0016 1/K.<span>
1/T</span>₂ =
1/666 K = 0,0015 1/K.<span>
ln(0,000643/0,00828) = Ea/8,3145 J/Kmol ·
(-0,0001 1/K).
-2,55 = Ea/8,3145 J/Kmol </span>· (-0,0001 1/K).<span>
Ea = 212019,75 J/mol = 212,01975 kJ/mol.</span>
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