Question

the 1/2 life of U-235 is 7 x 10^8 years. How much of an original 25.0g sample of U-235 would after 2.1 x 10^9 years ?

Answer 1

devide 2.1*10^9 by 1/2life to get how many half lifes the sample has undergone
then use this equation W=w(1/2)^n
where n is the number of halflives
W=the weight at the final
w= weight at the begining
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or u could just try it steo by step

Answer 2

huh ? can u solved for me please

Answer 3

i dont get it

Answer 4

n= 2.1*10^9 / 7*10^8
n=3
W = 25g 8(1/2)^3
W = 25/8 g

Answer 5

alternatively, if you know the halflife of Uran, you can use the information to solve the equation \[1/2 w_0 = w_0 \exp(-k T_{1/2}) \] for k.

once you've got k you can use the equation again to calculate the weight after \(t = 2.1 \times 10^9\)

Answer 6

thanks guy

Answer 7

no prob

Answer 8

if only 0.050mg of PU-239 remains after 6 half lives have past ,how much did the original sample weigh ?

Answer 9

can u guys help me this one too any idea?

Answer 10

double the amount 6 times ;)

the idea is if there is only half of a substance left after halflife, then there was twice as much there at the beginning.

Answer 11

2 (0.050)^6 ?

Answer 12

after 6 halflifes you've got 0.05, so you've got 0.1 after 5 hl. 0.2 after 4, 0.4 after 3, 0.8 after 2, 1.6 after 1 and 3.2 at the beginning.
so it's basically 0.05*2^6
if you use for formular the 0.05^6 gets very small.