Middle School

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------------------------------------------------ Sodium-25 was to be used in an experiment, but it took 3.0 minutes to get the sodium from the reactor to the laboratory. If 5.0 mg of sodium-25 was removed from the reactor, how many mg of sodium-25 were placed in the reaction vessel 3.0 minutes later, given that the half-life of sodium-25 is 60 seconds?

Answer :

3 half lives later... 5/2 = 2.5mg, then 2.5/2 = 1.25 mg, then 1.25/2 = .625 mg left

Final answer:

The amount of Sodium-25 that was placed in the reaction vessel three minutes later is roughly 0.625 mg. This is due to the sodium undergoing three half-life periods in which half of its amount decays.

Explanation:

The subject of the question pertains to nuclear physics and radioactivity, which entails understanding the concept of half-lifes. Sodium-25 is a radioactive isotope and its half-life is given as 60 seconds. This means that in 60 seconds, half of any sample of Sodium-25 will decay.

In this scenario, the time delay in transporting the sodium from the reactor to the laboratory is given as 3 minutes, which equals 180 seconds. This is three half-lives of Sodium-25 (180 seconds / 60 seconds per half-life = 3 half-lives).

Since it takes three half-lives for the sodium to reach the laboratory, we need to calculate how much would remain after each step. After the first half-life, half of the 5.0 mg of Sodium-25 would remain, which is 2.5 mg. After the second half-life, half of 2.5 mg would remain, giving us 1.25 mg. And finally, after the third half-life, half of 1.25 mg would remain, which is about 0.625 mg.

Therefore, the amount of Sodium-25 that was placed in the reaction vessel 3 minutes later would be approximately 0.625 mg.

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