Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
Answer:
No, there is not because it would form H2 instead of methane if hydrogen bonded with itself.
Explanation:
from the shape of methane which is tetrahedral it's evident there's no hydrogen bond only C-H bond.
1000 mL=1L
25 mL = 0.025 L
125 mL = 0.125 L
M1V1=M2V2
0.15(0.125) = M2(0.025)
0.01875 = M2(0.025)
0.75 = M2
0.75 M
Answer:
H2S(g) + 2OH^-(aq) --------> S^2-(aq) + 2H2O(l)
Explanation:
We know that the net ionic equation shows the major reaction that occurs in the reaction system.
The molecular reaction equation is;
H2S(g) + 2NaOH(aq) ------> Na2S(aq) + 2H2O(l)
The complete ionic equation is;
H2S(g) + 2Na^+(aq) + 2OH^-(aq) --------> 2Na^+(aq) + S^2-(aq) + 2H2O(l)
Net ionic equation;
H2S(g) + 2OH^-(aq) --------> S^2-(aq) + 2H2O(l)