Answer:
the electrolysis reaction is a non- spontaneous reaction
Explanation:
Since electrons flow from it, the anode in an electrolytic cell is positive, while the cathode is negative when electrons flow into it. The device functions like a galvanic cell in that direction. In an electrolytic cell, an external voltage is applied and that is what causes a non spontaneous reaction
A weak Bronsted-Lowry base is a weak proton acceptor, where the proton is in the form of H+, so the conjugate acid formed contains one more H atom and an extra positive charge.
Hope this helps!
Answer:
Ka = ( [H₃O⁺] . [F⁻] ) / [HF]
Explanation:
HF is a weak acid which in water, keeps this equilibrium
HF (aq) + H₂O (l) ⇄ H₃O⁺ (aq) + F⁻ (aq) Ka
2H₂O (l) ⇄ H₃O⁺ (l) + OH⁻ (aq) Kw
HF is the weak acid
F⁻ is the conjugate stron base
Let's make the expression for K
K = ( [H₃O⁺] . [F⁻] ) / [HF] . [H₂O]
K . [H₂O] = ( [H₃O⁺] . [F⁻] ) / [HF]
K . [H₂O] = Ka
Ka, the acid dissociation constant, includes Kwater.
Answer:
9.4
Explanation:
The equation for the reaction can be represented as:
+ 
⇄ 
The ICE table can be represented as:
+ 
⇄
Initial 0.27 0.49 0.0
Change -x -2x x
Equilibrium 0.27 - x 0.49 -2x x
We can now say that the concentration of at equilibrium is x;
Let's not forget that at equilibrium = 0.11 M
So:
x = [] = 0.11 M
[] = 0.27 - x
[] = 0.27 - 0.11
[] = 0.16 M
[] = (0.49 - 2x)
[] = (0.49 - 2(0.11))
[] = 0.49 - 0.22
[] = 0.27 M
![K_C = \frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_C%20%3D%20%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
= 9.4
∴ The equilibrium constant at that temperature = 9.4
Answer:
The oxidizing agent is the MnO₄⁻
Explanation:
This is the redox reaction:
10 I⁻ (aq) + 2 MnO₄⁻ (aq) + 16 H⁺ (aq) → 5 I₂ (s) + 2 Mn²⁺ (aq) + 8 H2O (l)
Let's determine the oxidation and the reduction.
I⁻ acts with -1 in oxidation state and changes to 0, at I₂.
All elements in ground state has 0 as oxidation state.
As the oxidation state has increased, this is the oxidation, so the iodide is the reducing agent.
In the permanganate (MnO₄⁻), Mn acts with +7 in oxidation state and decreased to Mn²⁺. As the oxidation state is lower, we talk about the reduction. Therefore, the permanganate is the oxidizing agent because it oxidizes iodide to iodine
