When we balance the given equation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We will get
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
Solution:
Balancing the given equaation
SF₆(g) + SO₃(g) → SO₂F₂(g)
We have to balance the given number of O
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
We get balanced equation
SF₆(g) + 2SO₃(g) → 3SO₂F₂(g)
The reaction quotient will be
Qc = [product] / [reactant]
Qc = [SO₂F₂(g)] / [SF₆(g) + SO₃(g)]
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As,
Water has a pkw=14
so it can be represented as,
[H+] [OH-] = 1*10^-14
If [H+] = 3*10^-5M
[OH-] = (1*10^-14) / ( 3*10^-5)
[OH-] = 3.3*10^-9 M
MnCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of MnCl2 that dissolves.
MnCl2(s) --> Mn+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.92 mol MnCl2/1L × 2 mol Cl⁻ / 1 mol MnCl2 = 1.8 M
The answer to this question is [Cl⁻] = 1.8 M
Answer:
use secondary data. the normal method to use
