The half-life of Th-227 is 18.2 day. How many days are required for 0.70g of a given sample to decay to 0.04375 g?

73 days

Explanation:

From the question given above, the following data were obtained:

Half life (t½) = 18.2 day.

Original amount (N₀) = 0.70 g

Amount remaining (N) = 0.04375 g

Time (t) =.?

Next, we shall determine the rate of decay of the isotope. This can be obtained as follow:

Half life (t½) = 18.2 day.

Decay constant (K) =.?

K = 0.693 / t½

K = 0.693 / 18.2

K = 0.038 /day

Finally, we shall determine the time taken for the isotope to decay to 0.04375 g. This can be obtained as follow:

Original amount (N₀) = 0.70 g

Amount remaining (N) = 0.04375 g

Decay constant (K) = 0.038 /day

Time (t) =.?

Log(N₀/N) = kt /2.303

Log (0.70/0.04375) = (0.038 × t) /2.303

Log 16 = (0.038 × t) /2.303

1.2041 = (0.038 × t) /2.303

Cross multiply

1.2041 × 2.303 = 0.038 × t

Divide both side 0.038

t = (1.2041 × 2.303) / 0.038

t = 72.97 ≈ 73 days

Therefore, it will take approximately 73 days for the isotope to decay to 0.04375 g

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.