Answer:
47.36mL
Explanation:
Using Boyles law equation, which states that:
P1V1 = P2V2
Where;
V1 = initial volume (mL)
V2 = final volume (mL)
P1 = initial pressure (atm)
P2 = final pressure (atm)
Based on the provided information, V1 = 25.3mL, P1 = 152 kPa, V2 = ?, P2 = 0.804atm
First, we need to convert 152kPa to atm by dividing by 101
1kPa = 0.0099atm
152kPa = 1.505atm
P1V1 = P2V2
1.505 × 25.3 = 0.804 × V2
38.08 = 0.804V2
V2 = 38.08/0.804
V2 = 47.36mL
Water is the component that is produced here.
I’m assuming in this scenario that 2.5 mols of Fe2O3 is reacting with excess reactant. Based on the balanced equation, you can see that there’s a 1:2 ratio between Fe2O3 and Fe. This means that to find the mols of Fe formed, you need to multiply the mols of Fe2O3 by 2.
2.50mol Fe2O3 * 2mol Fe/1 mol of Fe2O3 = 5.00mols Fe.
A. Electrons are in a cloud around a cluster of proteins and electrons
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
