Answer:
8.36 g
Explanation:
From;
0.693/t1/2 = 2.303/t log (No/N)
t1/2 = half life of potassium-40
t = age of the sample
No= initial amount of the sample
N= amount of the sample at time t
Substituting values;
0.693/1.3 × 10^9 = 2.303/2.6 × 10^9 log (No/2.10)
5.33 × 10^-10 = 8.86 × 10^-10 log (No/2.10)
5.33 × 10^-10/8.86 × 10^-10 = log (No/2.10)
0.6 = log (No/2.10)
Antilog (0.6) = (No/2.10)
No = 2.10 × Antilog (0.6)
No= 8.36 g
