Answer:
The value of dissociation constant of the monoprotic acid is 
.
Explanation:
The pH of the solution = 2.46
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
![2.46=-\log[H^+]](https://tex.z-dn.net/?f=2.46%3D-%5Clog%5BH%5E%2B%5D)
![[H^+]=0.003467 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.003467%20M)
Initially
0.0144 0 0
At equilibrium
(0.0144-x) x x
The expression if an dissociation constant is given by :
![K_a=\frac{[A^-][H^+]}{[HA]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BA%5E-%5D%5BH%5E%2B%5D%7D%7B%5BHA%5D%7D)
![x=[H^+]=0.003467 M](https://tex.z-dn.net/?f=x%3D%5BH%5E%2B%5D%3D0.003467%20M)
The value of dissociation constant of the monoprotic acid is .
Answer:
2NO(g) + O2(g) --> 2NO2(g)
now 400 ml of NO × 2 mol of NO2/2 mol of NO
= 400 ml of NO2
now 500 ml of O2 × 2 mol of NO2/1 mol of O2
= 1000 ml of NO2
now 400 ml of NO2 × 1 mol of O2/2 mol of NO
= 200 ml
subtract that from 500 ml of total i.e. 500-200 =300 ml
The total volume of the reaction mixture is 1000 ml -300ml = 700 ml
Answer:
k ≈ 9,56x10³ s⁻¹
Explanation:
It is possible to solve this question using Arrhenius formula:
Where:
k1: 1,35x10² s⁻¹
T1: 25,0°C + 273,15 = 298,15K
Ea = 55,5 kJ/mol
R = 8,314472x10⁻³ kJ/molK
k2 : ???
T2: 95,0°C+ 273,15K = 368,15K
Solving:
<em>k ≈ 9,56x10³ s⁻¹</em>
I hope it helps!
A pi bond... hope this helps!!!!!
Valence.
The electrons in the outer shell of an atom are called valence electrons.
Valence electrons determine whether the an element is ready form compounds. These electrons can be gained, lost, or shared in the formation of compounds.
