Answer:
3 g.
Step-by-step explanation:
The following data were obtained from the question:
Original amount (N₀) = 24 g
Half life (t½) = 5000 years
Time (t) = 15000 years
Amount remaining (N) =?
Next, we shall determine the rate constant K. This can be obtained as follow:
Half life (t½) = 5000 years
Rate constant (K) =.?
K = 0.693 / t½
K = 0.693 / 5000
K = 1.386×10¯⁴ / years
Finally, we shall determine the amount remaining (N) after 15000 years as follow:
Original amount (N₀) = 24 g
Time (t) = 15000 years
Rate constant (K) = 1.386×10¯⁴ / years
Amount remaining (N) =?
Log (N₀/N) = kt/2.3
Log (24/N) = 1.386×10¯⁴ × 15000/2.3
Log (24/N) = 2.079/2.3
Log (24/N) = 0.9039
24/N = antilog (0.9039)
24/N = 8.01
Cross multiply
24 = N × 8.01
Divide both side by 8.01
N = 24/8.01
N = 3 g
Therefore, the amount remaining after 15000 years is 3 g.