Answer : The concentration of
and
at equilibrium is, 0.0158 M and 0.00302 M respectively.
Explanation :
First we have to calculate the concentration of ![H_2, I_2\text{ and }HI](https://tex.z-dn.net/?f=H_2%2C%20I_2%5Ctext%7B%20and%20%7DHI)
![\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DH_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DH_2%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D%3D%5Cfrac%7B0.00623mol%7D%7B1.50L%7D%3D0.00415M)
![\text{Concentration of }I_2=\frac{\text{Moles of }I_2}{\text{Volume of solution}}=\frac{0.00414mol}{1.50L}=0.00276M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DI_2%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DI_2%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D%3D%5Cfrac%7B0.00414mol%7D%7B1.50L%7D%3D0.00276M)
![\text{Concentration of }HI=\frac{\text{Moles of }HI}{\text{Volume of solution}}=\frac{0.0244mol}{1.50L}=0.0163M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7DHI%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7DHI%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D%3D%5Cfrac%7B0.0244mol%7D%7B1.50L%7D%3D0.0163M)
Now we have to calculate the value of equilibrium constant (K).
The given chemical reaction is:
![2HI(g)\rightleftharpoons H_2(g)+I_2(g)](https://tex.z-dn.net/?f=2HI%28g%29%5Crightleftharpoons%20H_2%28g%29%2BI_2%28g%29)
Initial conc. 0.0163 0.00415 0.00276
At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)
As we are given:
Concentration of
at equilibrium = 0.00467 M
That means,
(0.00415+x) = 0.00467
x = 0.00026 M
Concentration of
at equilibrium = (0.0163-2x) = (0.0163-2(0.00026)) = 0.0158 M
Concentration of
at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M
The answer would be C. hope this helped.
Answer:
Do you mean Tennessine?
Explanation:
It's in the 7th period, and has the shorthand of Rn. And it's 5 into the row.
Halogens and Alkali react aggressively.
The answer is (2) 2, because a total of 6 electrons is transferred. We can know the electron transferred number form Mg to Mg2+ ion. There is 3 Mg involved and two electrons per atom. So total is 6,