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arlik [135]
3 years ago
9

Item5 Time Remaining 56 minutes 30 seconds00:56:30 Item 5Item 5 Time Remaining 56 minutes 30 seconds00:56:30 A pure sample of tr

itium, 3H, was prepared and sealed in a container for a number of years. Tritium undergoes β decay with a half-life of 12.32 years. How long has the container been sealed if analysis of the contents shows there are 5.25 mol of 3H and 6.35 mol of 3He present? Multiple Choice A. 14.1 y B. 25.6 y C. 2.34 y D. 3.38 y E. 9.77 y
Mathematics
1 answer:
trapecia [35]3 years ago
7 0

Answer:

option A

Step-by-step explanation:

given,

sample of tritium, ³H

t_{1/2} = 12.32 years

contents present of ³H = 5.25 mol

content of ³He = 6.35 mol

reaction

 ^3H\rightarrow \ ^3He

A₀ is the initial concentration

At is the concentration after time t

A₀ = 5.25 + 6.35 = 11.6 mol

At = 5.25

now,

  k = \dfrac{0.693}{t_{1/2}}

  k = \dfrac{0.693}{12.32}

           k = 0.0563 /year

ln(\dfrac{A_o}{A_t}) = k t

t = \dfrac{1}{k} ln(\dfrac{A_o}{A_t})

t = \dfrac{1}{0.0563} ln(\dfrac{11.6}{5.25})

t = 14.1 yr

hence, the correct answer is option A

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- 1.3 is the number we get when we subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5. This can be obtained by finding sum separately and then subtracting them.

<h3>What is the required number:</h3>

Here in the question it is given that,

subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5

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⇒ sum of -5/6 and -1 3/5  

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⇒ sum of 2 2/3 and -6 2/5

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subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5

= (sum of 2 2/3 and -6 2/5) - (sum of -5/6 and -1 3/5)

= (- 56/15) - (- 73/30 )  

= - 56/15 + 73/30

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= - 39/30

= - 13/10

= - 1.3

Hence - 1.3 is the number we get when we subtract the sum of -5/6 and -1 3/5 from the sum of 2 2/3 and -6 2/5.

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