Answer:
0.42%
Explanation:
<em>∵ pH = - log[H⁺].</em>
2.72 = - log[H⁺]
∴ [H⁺] = 1.905 x 10⁻³.
<em>∵ [H⁺] = √Ka.C</em>
∴ [H⁺]² = Ka.C
∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.
<em>∵ Ka = α²C.</em>
Where, α is the degree of dissociation.
<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>
<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>
Answer:
Explanation:
The formation of ammonia that occurred by the reaction of nitrogen and ammonia is expressed as:
⇄ ![2NH_{3(g)}](https://tex.z-dn.net/?f=2NH_%7B3%28g%29%7D)
where;
The reactants are:
Hydrogen and nitrogen
The product is ammonia.
For the reaction, the equilibrium constant can be expressed as:
![K = \dfrac{[product]}{[reactants]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cdfrac%7B%5Bproduct%5D%7D%7B%5Breactants%5D%7D)
![K = \dfrac{[NH_3]^2}{[N_2]^3[H_2]^3}](https://tex.z-dn.net/?f=K%20%3D%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5E3%5BH_2%5D%5E3%7D)
From the equilibrium constant conditions, the formation of ammonia and its decomposition due to its reversible reaction back to hydrogen and nitrogen are equal. It implies that the rate of the forward reaction is also equal to that of the backward reaction.
Thus, during when equilibrium is obtained;
Hydrogen, Nitrogen, and Ammonia are present.
Answer: Electronic transition moments are defined as the probability for a given excitation energy transition to take place. It should be evident that the transition moment depends upon the spin-orbit coupling of the electrons in both the ground and excited states.
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Explanation:
K. from Latin name kalium