The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
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Answer:
![[H^+]=0.00332M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.00332M)
Explanation:
Hello,
In this case, considering the dissociation of valeric acid as:
Its corresponding law of mass action is:
![Ka=\frac{[H^+][C_5H_9O_2^-]}{[HC_5H_9O_2]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH%5E%2B%5D%5BC_5H_9O_2%5E-%5D%7D%7B%5BHC_5H_9O_2%5D%7D)
Now, by means of the change 
due to dissociation, it becomes:
Solving for we obtain:
Thus, since the concentration of hydronium equals , the answer is:
![[H^+]=x=0.00332M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.00332M)
Best regards.
Answer:
%age Yield = 51.45 %
Solution:
Step 1: Convert Kg into g
68.5 Kg CO = 68500 g CO
8.60 Kg H₂ = 8600 g
Step 2: Find out Limiting reactant;
The Balance Chemical Equation is as follow;
CO + 2 H₂ → CH₃OH
According to Equation,
28 g (1 mol) CO reacts with = 4 g (2 mol) of H₂
So,
68500 g CO will react with = X g of H₂
Solving for X,
X = (68500 g × 4 g) ÷ 28 g
X = 9785 g of H₂
It shows 9785 g H₂ is required to react with 68500 g of CO but we are provided with 8600 g of H₂ which is less than required. Therefore, H₂ is provided in less amount hence, it is a Limiting reagent and will control the yield of products.
Step 3: Calculate Theoretical Yield
According to equation,
4 g (2 mol) H₂ reacts to produce = 32 g (1 mol) Methanol
So,
8600 g H₂ will produce = X g of CH₃OH
Solving for X,
X = (8600 g × 32 g) ÷ 4 g
X = 68800 g of CH₃OH
Step 4: Calculate %age Yield
%age Yield = Actual Yield ÷ Theoretical Yield × 100
Putting Values,
%age Yield = 3.54 × 10⁴ g ÷ 68800 g × 100
%age Yield = 51.45 %
