Potassium-42 has a half life of 12.4 hours. How much of an 848 g sample of potassium-42 will be left after 62.0 hours?

Please help me!!

Potassium-42 has a half-life of 12.4 hours.How much of an 2000 g sample of potassium-42 will be left after 62.0 hours? Can u help me out with this question.

For Cindy, the answer is 26.5 grams.

For Alex, it's 62.5 grams.

Please help me solve the question

I have a similar question but instead the amount of grams is 848 and the second amount of time is 37.2

Oh, potassium-42, going through a mid-life crisis with its half-life of 12.4 hours. Let me grab my calculator and crunch some numbers for you.

So after each half-life, the amount of potassium-42 is reduced by half. In 62.0 hours, there are approximately 5 half-lives elapsed (62.0 / 12.4 = 5), which means the amount left is reduced by half 5 times.

Starting with 848 grams, after the first half-life, we have 424 grams remaining. After the second, we have 212 grams. After the third, 106 grams. After the fourth, 53 grams. And after the fifth and final half-life, we have 26.5 grams left (assuming no clowns have snuck in and took some potassium for their pie throwing).

So, after 62.0 hours, approximately 26.5 grams of our 848 g sample of potassium-42 will still be hanging around, ready to have some fun with chemistry.

To find out how much of the potassium-42 sample will be left after 62.0 hours, we can use the formula for radioactive decay:

N = N0 * (1/2)^(t / t1/2)

where:
N = final amount of the sample
N0 = initial amount of the sample
t = total time passed
t1/2 = half-life of the sample

Given:
N0 = 848 g
t = 62.0 hours
t1/2 = 12.4 hours

Now, let's substitute these values into the formula and calculate the final amount:

N = 848 g * (1/2)^(62.0 hours / 12.4 hours)

To simplify further, let's divide the time in the exponent:

N = 848 g * (1/2)^(5)

Now, raise 1/2 to the power of 5:

N = 848 g * (1/32)

N = 26.5 g

Therefore, after 62.0 hours, approximately 26.5 grams of potassium-42 will be left out of the initial 848 grams.

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