The half-life of arsenic-74 is about 18 days. If a sample initially contains 5.13 x 10 4 mg arsenic-74, what mass (in mg) would be left after 80 days?

Here it says the half life of Arsenic 70 4 is about 18 days. If a sample initially contains 5.13 times 10 to the 4 milligrams of arsenic 74, what mass in milligrams will be left after 80 days? Alright, so they're asking how much is left, so they're looking for final concentration. Here we'd say LN of my final concentration equals negative kt plus ln of my initial concentration. What do we know from the question? Alright. We know half life, and we'll see how that fits in later. We know initially how much, so we know the initial concentration, And we know days, which would have to be our time t. So let's plug in what we know. We're looking for final, so we don't know that. Our time is 80 days, plus ln of my initial which is 5.13 times 10 to the 4. But look, we can't determine what our final amount is because we also don't know what our k value is, our decay constant. Well, remember we're also given half life. We can use the half life equation to isolate k. Rearranging this equation, k equals ln2 over half life. Plug in these values. So ln2 over 18 days, that's going to give me 0.03851 days inverse. We can plug that in over here. Days and days inverse will cancel out. Here we'd have ln of my final concentration equals 7.76479. Take the inverse of both sides, so my final concentration would equal e to 7.76479, which comes out to be 2.36 times 10 to the 3 milligrams. So this is how much of our arsenic 74 will be left after 80 days.

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21. Nuclear Chemistry

Radioactive Half-Life

General Chemistry

21. Nuclear Chemistry

Radioactive Half-Life

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Multiple Choice

The half-life of arsenic-74 is about 18 days. If a sample initially contains 5.13 x 10⁴ mg arsenic-74, what mass (in mg) would be left after 80 days?

2.36 x 10³ mg

7.02 x 10² mg

1.43 x 10³ mg

1.14 x 10⁸ mg

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Verified step by step guidance

Identify the initial mass of arsenic-74, which is 5.13 x 10^4 mg, and the half-life, which is 18 days.

Determine the number of half-lives that have passed in 80 days by dividing the total time by the half-life: \( \frac{80 \text{ days}}{18 \text{ days/half-life}} \).

Calculate the remaining mass after each half-life. Use the formula: \( \text{Remaining mass} = \text{Initial mass} \times \left(\frac{1}{2}\right)^{\text{number of half-lives}} \).

Substitute the known values into the formula: \( \text{Remaining mass} = 5.13 \times 10^4 \text{ mg} \times \left(\frac{1}{2}\right)^{\frac{80}{18}} \).

Simplify the expression to find the remaining mass of arsenic-74 after 80 days.