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lyudmila [28]
2 years ago
10

Xenon-135 is produced in a nuclear reactor by two primary methods. one is directly from fission, the other is from the decay of:

_______
Chemistry
1 answer:
LekaFEV [45]2 years ago
6 0

Xenon-135 is produced from the  the decay of iodine-135.

<h3>What is a nuclear reactor?</h3>

The term nuclear reactor refers to an instrument where nuclear reaction takes place. Now we know that nuclear reactions can be used for several purposes and one of those purposes is the generation of electricity.

Now, Xenon-135 is produced in a nuclear reactor by two primary methods. one is directly from fission, the other is from the decay of iodine-135.

Learn more about nuclear reactor:brainly.com/question/13869748

#SPJ1

Missing parts

Xenon-135 is produced in a reactor by two primary methods. One is directly from fission; the other

is from the decay of...

A. cesium-135.

B. iodine-135.

C. xenon-136.

D. iodine-136.

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The half-life of radium-226 is 1590 years. if a sample contains 100 mg, how many mg will remain after 1000 years?
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64.52 mg.

Explanation:

The following data were obtained from the question:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Final amount (N) =.?

Next, we shall determine the rate constant (K).

This is illustrated below:

Half life (t½) = 1590 years

Rate/decay constant (K) =?

K = 0.693 / t½

K = 0.693/1590

K = 4.36×10¯⁴ / year.

Finally, we shall determine the amount that will remain after 1000 years as follow:

Half life (t½) = 1590 years

Initial amount (N₀) = 100 mg

Time (t) = 1000 years.

Rate constant = 4.36×10¯⁴ / year.

Final amount (N) =.?

Log (N₀/N) = kt/2.3

Log (100/N) = 4.36×10¯⁴ × 1000/2.3

Log (100/N) = 0.436/2.3

Log (100/N) = 0.1896

Take the antilog

100/N = antilog (0.1896)

100/N = 1.55

Cross multiply

N x 1.55 = 100

Divide both side by 1.55

N = 100/1.55

N = 64.52 mg

Therefore, the amount that remained after 1000 years is 64.52 mg

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4 years ago
Suppose that 0.25 mole of gas C was added to
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Answer:

Pi = 0.25[P(TTL)/n(TTL)]

Explanation:

Let Total Pressure = P(TTL) and Total moles =n(TTL) => n(i)/n(TTL) = P(i)/P(TTL)

Given n(i) = 0.25 => Pi = 0.25[P(TTL)/n(TTL)]

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