Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:
Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:
The expression for equilibrium constant is:
Now put all the given values in this expression, we get:
The concentration of 
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
Therefore, the pH of the solution is, 2.67
Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:
2.08L
Explanation:
V1 = 2.00L
P1 = 101kPa
P2 = 97.0kPa
V2 = ?
from Boyle's law,
P1V1 = P2V2
101 × 2 = 97 × V2
V2 = 202/97 = 2.08L
Therefore, the new volume is 2.08L
Percent yield is expressed as the ratio of the actual yield and the theoretical yield of the reaction multiplied by 100 to get the percent value. The actual yield is usually given in a problem. The theoretical yield is calculated from the reaction. For this problem, it cannot be solved since we cannot obtain the theoretical yield.
