Answer:
pH = 7.46
Explanation:
2H₂O ⇄ H₃O⁺ . OH⁻ Kw = [H₃O⁺] . [OH⁻]
[H₃O⁺] = [OH⁻]
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
- log 3.46×10⁻⁸ = pH → 7.46
I cannot answer your question about a image without being able to see it myself.
Answer:
105 mg
Explanation:
The following data were obtained from the question:
Original amount (N₀) = 420 mg
Half life (t½) =20 days
Time (t) = 40 days
Amount remaining (N) =..?
Next, we shall determine the rate constant K, for the disintegration.
This can be obtained as follow:
Half life (t½) =20 days
Rate constant (K) =?
K = 0.693/ t½
K = 0.693/20
K = 0.03465 /day
Finally, we shall determine the amount remaining after 40 days as follow:
Original amount (N₀) = 420 mg
Rate constant (K) = 0.03465 /day
Time (t) = 40 days
Amount remaining (N) =..?
Log(N₀/N) = kt/2.3
Log (420/N) = (0.03465 x 40)/2.3
Log (420/N) = 1.386/2.3
Log (420/N) = 0.6026
420/N = antilog (0.6026)
420/N = 4
Cross multiply
420 = N x 4
Divide both side by 4
N = 420/4
N = 105 mg
Therefore, the amount remaining after 40 days is 105 mg.
