Answer:
Explanation:
<u>1. Equilibrium equation</u>
<u>2. Equilibrium constant</u>
The liquid substances do not appear in the expression of the equilibrium constant.
<u>3. ICE table.</u>
Write the initial, change, equilibrium table:
Molar concentrations:
H₂(g) + Br₂(l) ⇄ 2HBr(g)
I 0.400 0
C - x +2x
E 0.400 - x 2x
<u>4. Substitute into the expression of the equilibrium constant</u>
<u>5. Solve the quadratic equation</u>
- 192,000,000 - 480,000,000x = 4x²
- x² + 120,000,000x - 48,000,000 = 0
Use the quadratic formula:
The only valid solution is x = 0.39999999851M
Thus, the final concentration of H₂(g) is 0.400 - 0.39999999851 ≈ 0.00000000149 ≈ 1.5 × 10⁻⁹M
In chemistry, pH<span> is a numeric scale used to specify the acidity or basicity of an aqueous solution. It is approximately the negative of the base 10 logarithm of the concentration of hydronium ions. We calculate as follows:
pH = -log [H3O+]
pH = -log[</span><span>5.45 × 10–5 M]
pH = 4.3</span>
Answer:
½O 2 + 2e - + H 2O → 2OH.
Explanation:
Redox reactions - Higher
In terms of electrons:
oxidation is loss of electrons
reduction is gain of electrons
Rusting is a complex process. The example below show why both water and oxygen are needed for rusting to occur. They are interesting examples of oxidation, reduction and the use of half equations:
iron loses electrons and is oxidised to iron(II) ions: Fe → Fe2+ + 2e-
oxygen gains electrons in the presence of water and is reduced: ½O2 + 2e- + H2O → 2OH-
iron(II) ions lose electrons and are oxidised to iron(III) ions by oxygen: 2Fe2+ + ½O2 → 2Fe3+ + O2-
Answer: Concentration of in the equilibrium mixture is 0.31 M
Explanation:
Equilibrium concentration of = 0.729 M
The given balanced equilibrium reaction is,
Initial conc. x 0 0
At eqm. conc. (x-2y) M (y) M (3y) M
The expression for equilibrium constant for this reaction will be:
3y = 0.729 M
y = 0.243 M
Now put all the given values in this expression, we get :
concentration of in the equilibrium mixture =
Thus concentration of in the equilibrium mixture is 0.31 M