Answer:
0.3023 M
Explanation:
Let Picric acid = 
So,
+
⇄
+ 
The ICE table can be given as:
+
⇄
+ 
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
=
OR 
=
OR 
=
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [
] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
Answer is: concentration of ammonium ions are 7,14·10⁻¹⁴ M.
Chemical reaction: 2NH₃(l) → NH₄⁺(am) + NH₂⁻(am).
Kam = 5,1·10⁻²⁷.
[NH₄⁺] · [NH₂⁻] = x; equilibrium concentration of cations and anions.
Kam = [NH₄⁺] · [NH₂⁻].
Kam = x².
x = [NH₄⁺] = √5,1·10⁻²⁷.
[NH₄⁺] = 7,14·10⁻¹⁴ M.
Answer:
See explanation
Explanation:
Now , we have the equation of the reaction as;
2H2S(g) + 302(g)------->2SO2(g) + 2H2O(g)
This equation shows that SO2 gas is produced in the process. Let us recall that this same SO2 gas is the anhydride of H2SO4. This means that it can dissolve in water to form H2SO4
So, when SO2 dissolve in rain droplets, then H2SO4 is formed thereby lowering the pH of rain water. This is acid rain.
Answer:
B)
Explanation:
they are not the same animal