Question

Arsenic-74 has a half life of 18 days. If the initial sample was 1g, how much is left after 54 days?

A. 0.125g
B. 0.063g
C. 0.5g
D. 0.25g​

Answer 1

Answer: After 54 days, the amount of Arsenic-74 remaining from an initial sample of 1g can be calculated using the half-life formula as 0.125g.

Explanation: The half-life of Arsenic-74 is 18 days, which means that after every 18 days, the amount of Arsenic-74 is reduced to half. Therefore, after 18 days, the amount remaining is 0.5g, after 36 days, it is 0.25g, and after 54 days, it is 0.125g.

Answer 2

A. 0.125 g

Step-by-step explanation:

Since Arsenic-74 has a half-life of 18 days, this means that after each 18-day period, the amount of Arsenic-74 remaining will be reduced by half.

So, after the first 18 days, the amount of Arsenic-74 remaining will be:

1 gram / 2 = 0.5 grams

After the next 18 days (a total of 36 days), the amount of Arsenic-74 remaining will be:

0.5 grams / 2 = 0.25 grams

Finally, after 54 days (a total of three half-lives), the amount of Arsenic-74 remaining will be:

0.25 grams / 2 = 0.125 grams

Therefore, the answer is option (a) 0.125 grams.