Chemistry: The half-life of cesium-137 is 30 years. suppose we have a 70-mg sample. (a) find the mass that remains after t years. y(t) = mg (b) how much of the sample remains after 110 years? (round your answer to two decimal places.) mg (c) after how long will only 1 mg remain? (round your answer to one decimal place.) t = yr

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The half-life of cesium-137 is 30 years. suppose

RayshawnBoulware

16.03.2022

Answer :

(a) The mass that remains after 't' years will be, $\frac{70}{2^{(\frac{t}{30})}}mg$

(b) The sample remains after 110 years will be, 5.52 mg

Explanation :

Formula used :

$a=\frac{a_o}{2^n}$ ............(1)

where,

a = amount of reactant left after n-half lives

a_o = Initial amount of the reactant

n = number of half lives

And as we know that,

$n=\frac{t}{t_{1/2}}$ ..........(2)

where,

t = time

$t_{1/2}$ = half-life

Now equating the value of 'n' from (2) to (1), we get:

$a=\frac{a_o}{2^{(\frac{t}{t_{1/2}})}}$ ...........(3)

(a) Now we have to calculate the mass that remains after 't' years.

As we are given that,

a_o = 70 mg

$t_{1/2}$ = 30 years

Now put all the given values in formula (3), we get:

$a=\frac{70}{2^{(\frac{t}{30})}}$

(b) Now we have top calculate the sample remains after 110 years.

As we are given that,

t = 110 years

Now put all the given values in formula (3), we get:

$a=\frac{70}{2^{(\frac{110}{30})}}$

a=5.52mg

As we are given that,

a = 1 mg

Now put all the given values in formula (3), we get:

$1=\frac{70}{2^{(\frac{t}{30})}}$

t=55.0years

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