Answer:
the answer should be henry's law
Answer:
1. 4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. 6 moles of Cl2
Explanation:
1. The balanced equation for the reaction. This is illustrated below:
4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2
2. Determination of the number of mole of Cl2 produce when 4 moles of FeCl3 react with 4 moles. To obtain the number of mole of Cl2 produced, we must determine which reactant is the limiting reactant.
This is illustrated below:
From the balanced equation above,
4 moles of FeCl3 reacted with 3 moles of O2.
Since lesser amount of O2 (i.e 3 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of FeCl3, therefore FeCl3 is the limiting reactant and O2 is the excess reactant.
Finally, we can obtain the number of mole Cl2 produced from the reaction as follow:
Note: the limiting reactant is used as it will produce the maximum yield of the reaction since all of it is used up in the reaction.
From the balanced equation above,
4 moles of FeCl3 will react to produced 6 moles of Cl2.
<h3><u>Answer</u>;</h3>
1.0875 x 10-2 atm
<h3><u>Explanation;</u></h3>
2O3(g) → 3O2(g)
rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t
The average rate of disappearance of ozone ... is found to
be 7.25 × 10–3 atm over a certain interval of time.
This means (ignoring time)
∆[O3]/∆t = -7.25 × 10^–3 atm
(it is disappearing, thus the negative sign)
rate = -(1/2)∆[O3]/∆t
rate = -(1/2)*(-7.25 × 10^–3 atm)
= 3.625 × 10^–3 atm
Now use the other part of the expression:
rate = +(1/3)∆[O2)∆t
3.625 × 10–3 atm = +(1/3)∆[O2)/t
∆[O2)/∆t = (3)*(3.625× 10^–3 atm)
= 1.0875 x 10-2 atm over the same time interval
Answer:
The relative rates of diffusion for methane and oxygen is 1.4142.
Methane gas will be able to travel 1.4142 meter in the same conditions.
Explanation:
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:
We are given:
Molar mass of methane gas, m = 16 g/mol
Molar mass of oxygen gas,m' = 32 g/mol
By taking their ratio, we get:
The relative rates of diffusion for methane and oxygen is 1.4142.
If oxygen gas travels 1 meters in time t.
Rate of diffusion of oxygen =
If methane gas travels travels in y meters in time t.
Rate of diffusion of methane=
y = 1.4142 m
Methane gas will be able to travel 1.4142 meter in the same conditions.
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.
