Answer:
M = 0.441 M
Explanation:
In this case, we have two solutions that involves the Manganese II cation;
We have Mn(CH₃COOH)₂ and MnSO₄
In both cases, the moles of Mn are the same in reaction as we can see here:
Mn(CH₃COO)₂ <-------> Mn²⁺ + 2CH₃COO⁻
MnSO₄ <------> Mn²⁺ + SO₄²⁻
Therefore, all we have to do is calculate the moles of Mn in both solutions, do the sum and then, calculate the concentration with the new volume:
moles of MnAce = 0.489 * 0.0283 = 0.0138 moles
moles MnSulf = 0.339 * 0.0125 = 0.0042 moles
the total moles are:
moles of Mn²⁺ = 0.0138 + 0.0042 = 0.018 moles
Finally the concentration: 12.5 + 28.3 = 40.8 mL or 0.0408 L
M = 0.018 / 0.0408
M = 0.441 M
This would be the final concentration of the manganese after the mixing of the two solutions
Explanation:
The given reaction is as follows.
Initial : 0.160 0.160 0
Change : -x -x 2x
Equilibrium: 0.160 - x 0.160 - x x
It is given that ![[H_{2}]](https://tex.z-dn.net/?f=%5BH_%7B2%7D%5D)
= [0.160 - x] = 0.036 M
and, ![[I_{2}]](https://tex.z-dn.net/?f=%5BI_%7B2%7D%5D)
= [0.160 - x] = 0.036 M
so, x = (0.160 - 0.036) M
= 0.124 M
As, [HI] = 2x.
So, [HI] = 
= 0.248 M
As it is known that expression for equilibrium constant is as follows.
![K_{eq} = \frac{[HI]^{2}}{[H_{2}][I_{2}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BHI%5D%5E%7B2%7D%7D%7B%5BH_%7B2%7D%5D%5BI_%7B2%7D%5D%7D)
= 
= 47.46
Thus, we can conclude that the equilibrium constant, Kc, for the given reaction is 47.46.
If the heat is absorbed, then the temperature will increase as well.
<span> Charge and it is surrounded by 5 electrons.</span>
