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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