i need help with my chemistry homework?

Question

1. a radioisotope has a half life of 4 days. How much of a 20 gram sample of this radioisotope remains at the end of each time period.
a. 4 days
b. 8 days
2. the mass of cobalt-60 in a sample is found to have decreased from 0.800 g to 0.200 g in a period of 10.5 years. From this info calculate the half life of cobalt-60

please answer these, i really need help

Answer 1

1. Since the half-life is 4 days, half of the original amount or 10 grams will remain after 4 days. After 8 days, 25% of the original amount will remain = 5 grams.

2. Since the amount of Co-60 remaining is 25% of the original mass, this means that 2 half-lives has past (see #1). Since the time past was 10.5 years, the half-life must be 5.25 years.

Answer 2

for the first one, well, if u have 20 g, and it is halved every 4 days, for a, you have 10 g.
for 8 days, you half the 10 again and get 5 g.
for the last part, construct a table, and try working backwards.
hope this helps!

Answer 3

1.a. 10g
b. 5g
2.let x be the half life of cobalt...
then in x years, the mass of Co-60 is 0.400g and for another x years, its mass will be 0.200g...so you can get 0.200g ofCo-60 in 2xyears and you have gotten 0.200g Co-60 in 10.5 years so,
2x=10.5
the half life of Co-60 is 5.25 years...

Answer 4

The equation for the decay is

N=N0*e^-kt

where N is the quantity remaining from an initial N0 quantity after t time has passed. k is the decay constant.

If the half-life is t1/2 then for N=N0/2 you have

N=N0*e^-kt1/2 =>
N0/2 = N0*e^-kt1/2 =>
1/2= e^-kt1/2 =>
ln2=kt1/2=>
k=(ln2)/t1/2

So we substitute k in our original equation and get

N=N0*e^-[(ln2)(t/t1/2)]

With this equation you don't need tables, etc
For 1a) N= 20*e^-[(ln2) *(4/4)]= 20 *e^-ln2 =20/2 =10
For 1b) N= 20*e^-[(ln2) *(8/4)]= 20 *e^-2ln2= 20* e^-ln4= 20/4= 5
For 2
0.2 = 0.8* e^-[(ln2)*10.5/t1/2]=>
4= e^[(ln2)*10.5/t1/2] =>
ln4 = (ln2)*10.5/t1/2 =>
t1/2 =10.5*ln2/ln4 = 5.25 years