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For the following problems, be sure to use an appropriated formula (i.g half lif

ID: 903653 • Letter: F

Question

For the following problems, be sure to use an appropriated formula (i.g half life or intergrade law), not any short cuts

Polonium-214 has a relatively short half-life of 164 s. How many seconds would it take for 8.0 g of this isotope to decay to 0.25 g?

The half life of argon is 6.32 days, how much of argon- 35 would be left after 50.56 days when there was initially 126.35 grams.

The radioisotope radon-222 has a half-life of 3.8 days. How much of an initial 20.0-g sample of radon-222 would remain after 15.2 days?

7. The mass of cobalt-60 in a sample is found to have decreased to 0.200 g in a period of 1015.5 years. The half life of cobalt is 425.26 years, what was the initial amount of the sample of cobalt-60?

Explanation / Answer

1)

half -life = t1/2 = 164 s

k = 0.693/ t1/2

    = 0.693 / 164

    = 4.22 x 10^-3 s-1

k = (2.303/t )( log a/a-x )

4.22 x 10^-3 = (2.303 / t)( log 8/0.25 )

     t = 821.4 sec

time taken = 821.4 sec

2) The half life of argon is 6.32 days, how much of argon- 35 would be left after 50.56 days when there was initially 126.35 grams.

half -life = t1/2 = 6.32 days

k = 0.693/ t1/2

    = 0.693 / 6.32

    = 0.1096 day-1

k = (2.303/t )( log a/a-x )

0.1096 = (2.303 / 50.56)( log (126.35/ a-x ))

(126.35/ a-x ) = 254.68

a - x = 0.496

remained = 0.496 g

3)

half -life = t1/2 = 3.8 days

k = 0.693/ t1/2

    = 0.693 / 3.8

    = 0.1823 day-1

k = (2.303/t )( log a/a-x )

0.1823 = (2.303 / 15.2)( log (20.0/ a-x ))

(20.0/ a-x ) = 15.96

a - x = 1.25 g

remained = 1.25 g

4)

The mass of cobalt-60 in a sample is found to have decreased to 0.200 g in a period of 1015.5 years. The half life of cobalt is 425.26 years, what was the initial amount of the sample of cobalt-60?

half -life = t1/2 = 425.26 days

k = 0.693/ t1/2

    = 0.693 / 425.26

    = 1.63 x 10^-3 year -1

k = (2.303/t )( log a/a-x )

1.63 x 10^-3 = (2.303 / 1015.5)( log (a/ 0.2))

(a/ 0.2) = 5.22

a = 1.045 g

initial is 1.045 g

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