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The half-life of the radioactive element krypton-91 is 10 seconds. If 16 grams of krypton-91 are initially present, when will there be 7 grams of the element remaining? Round to the nearest thousandths.

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Final answer:

To find the time when there will be 7 grams of krypton-91 remaining, we calculate the decay using the half-life concept and an exponential decay formula. The exact time is roughly 13.516 seconds, which is between the first and second half-lives of the substance.

Step-by-step explanation:

The question involves calculating the time when there will be 7 grams of the radioactive element krypton-91 remaining, given that it has a half-life of 10 seconds and the initial amount is 16 grams. To solve this problem, we would use the concept of half-life, which is the time it takes for half of the radioactive nuclei to decay.

Since the initial quantity is 16 grams, after one half-life (10 seconds), there would be 8 grams left. After each subsequent half-life, the quantity remaining is halved again. Thus, we can set up a sequence like this:

  • After 1 half-life (10 seconds) = 16 g / 2 = 8 g
  • After 2 half-lives (20 seconds) = 8 g / 2 = 4 g

We can see that after two half-lives, or 20 seconds, only 4 grams are left, which is less than 7 grams. Therefore, the time at which 7 grams remains must be between 10 and 20 seconds. To find the exact time, we can use the formula for exponential decay:

N = N0(1/2)^(t/T)

where:

  • N is the final amount of the substance (7 grams)
  • N0 is the initial amount of the substance (16 grams)
  • t is the time that has elapsed
  • T is the half-life of the substance (10 seconds)

Plugging in the values we get:

7 = 16(1/2)^(t/10)

To solve for t, we take the natural logarithm (ln) of both sides:

ln(7) = ln(16(1/2)^(t/10))

ln(7) = ln(16) + ln((1/2)^(t/10))

ln(7) - ln(16) = (t/10) * ln(1/2)

Finally, we solve for t:

t = [ln(7) - ln(16)] / [ln(1/2)] * 10

Using a calculator, we find that t is approximately equal to 13.516 seconds, rounded to the nearest thousandths.

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