Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>
Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48
K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075
NH3 = 0.44 M
Explanation:
oxidation of Nitrogen in NO2 is +4
4?
Explanation:
i think its 4, i might be wrong.
Answer:
By decreasing the volume, the equilibrium will shift to the side with the smallest amount of particles
Explanation:
Step 1: Data given
The principle of Le Chatelier says:
When the volume is decreased, the equilibrium will shift to the side of the smallest number of particles.
2NOBr(g) ⇌ 2NO(g)+Br2(g)
In the following example, we have 2 moles of NOBr (on the left side) and 3 moles of gas on the right side. This means the left side, the reactant side, has the smallest number of particles.
The equilibrium will here shift to the left side, the side of NOBr.
In the following example
2NO2(g) ⇌ N2O4(g)
We have 2moles of NO2 on the left side and 1 mol of N2O4 on the right side.
By decreasing the volume, the equilibrium will shift to the side of the smallest number of moles. Here this is the right side, the side of N2O4.
