Answer:
Explanation:
<u>Manganese (VII) ion (an anion) has the formula MnO₄⁻</u>. A polyatomic ion is an ion that is made up of more than one atom. For example, MnO₄⁻ and NH₄⁺. Since the ion provided in the question is an anion, the polyatomic ion that would react with it will have to be a cation (positively charged).
<u>The polyatomic cation that will react with MnO₄⁻ to form a neutral compound is NH₄⁺ (ammonium ion) to form NH₄MnO₄ (Ammonium permanganate).</u>
IT forms because they are highly reactive elements.
Answer:
The equivalent weight of M is approximately 31.8 g
The equivalent weight of N is approximately 27.98 g
Explanation:
The given parameters are;
The percentage of the the metal M in in the chloride = 47.25%
Where by the chemical formula for the metal chloride is MClₓ, we have;
47.25% of the mass of MClₓ = Mass of M = W
Therefore, we have;
0.4725 × (W + 35.5·x) = W
0.4725·W + 0.4725×35.5×x = W
W - 0.4725·W = 16.77·x
0.5275·W = 16.77·x
W/x = 16.77/0.5275 = 31.799 = The equivalent weight of M
The equivalent weight of M = 31.799 ≈ 31.8 g
Given that 1 gram of M is displaced by 0.88 gram of N, then the equivalent weight of N that will displace 31.799 = 0.88 × 31.799 ≈ 27.98 g
The equivalent weight of N = 27.98 g.
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
