Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
Answer:
A) 29.9g
Explanation:
first find the weight of 1 staple.
then multiply with 225
Answer:
%
Explanation:
The ethanol combustion reaction is:
→
If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:
Dividing the previous equation by x:
We would need 3.30 oxygen moles per ethanol mole.
Then we apply the composition relation between O2 and N2 in the feed air:
Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:
Calculate the number of moles of CO2 and water considering the same:
The total number of moles at the reactor output would be:
So, the oxygen mole fraction would be:
%
Oxygen and Sulfur are in the same group.
Sulfur is the next element in the group, and it reacts with hydrogen gas (H2) in a manner similar to oxygen.
Answer:The best phrase which supports the eruption would be high-viscosity magma lava broken into fragments.
