Alcohol, alcohol is the only liquid since copper and aluminum are a solid ammonia is a bacteria
Answer:
12.02 g
Explanation:
From the question given above, the following data were obtained:
Half life (t½) = 2 days
Original amount (N₀) = 96 g
Time (t) = 6 days
Amount remaining (N) =..?
Next, we shall determine the rate of disintegration of the isotope. This can be obtained as follow:
Half life (t½) = 2 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 2
K = 0.3465 /day
Therefore, the rate of disintegration of the isotope is 0.3465 /day.
Finally, we shall determine the amount of the isotope remaining after 6 days as follow:
Original amount (N₀) = 96 g
Time (t) = 6 days
Decay constant (K) = 0.3465 /day.
Amount remaining (N) =.?
Log (N₀/N) = kt / 2.303
Log (96/N) = (0.3465 × 6) / 2.303
Log (96/N) = 2.079/2.303
Log (96/N) = 0.9027
Take the anti log of 0.9027
96/N = anti log (0.9027)
96/N = 7.99
Cross multiply
96 = N × 7.99
Divide both side by 7.99
N = 96 /7.99
N = 12.02 g
Therefore, the amount of the isotope remaining after 6 days is 12.02 g
Answer is: Ksp for silver sulfide is 8.00·10⁻⁴⁸.
Reaction
of dissociation: Ag₂S(s) → 2Ag⁺(aq) + S²⁻(aq)<span>.
</span>s(Ag₂S) = s(S²⁻) = 1.26·10⁻¹⁶ M.
s(Ag⁺) = 2s(Ag₂S) = 2.52·10⁻¹⁶ M; equilibrium concentration of silver cations.
Ksp = s(Ag⁺)² · s(S²⁻).
Ksp = (2.52·10⁻¹⁶ M)² · 1.26·10⁻¹⁶ M.
Ksp = 6.35·10⁻³² M² · 1.26·10⁻¹⁶ M.
Ksp = 8.00·10⁻⁴⁸ M³.
It should be <em>Ultraviolet rays </em>because they are moderate in energy.