# Quick Question

1. Oct 4, 2006

### helpm3pl3ase

I have a question Iam stuck on. I was wondering if i did the right calculations. Please let me know:

Through decay, Chemical Q loses one-third of its mass every 7 years.
Twenty-five years ago, a container of Chemical Q was buried. The EPA has
just dug it up and found 1,000 grams remaining. How much Chemical Q was
originally buried?

1000=(Q)(1/3)^(x/7) when x = 25

1000=(Q)(1/3)^(25/7)

Q = 1000/(3)^(3/7)/81

Q = 50583???

I am not sure if I did this correct??

2. Oct 4, 2006

### Staff: Mentor

If it loses 1/3 of its mass every 7 years, then 2/3 of its mass is left at the end of 7 years, and (2/3)^2 at the end of 14 years, etc. I think you used 1/3 where you should have used 2/3....?

3. Oct 4, 2006

### helpm3pl3ase

hmm I dont really understand that?? The question called for one third. Iam not understanding wher eyou get 2/3

4. Oct 4, 2006

### daniel_i_l

If it loses 1/3 then you have 2/3 left. so every 7 years the new amount would equal 2/3 the old amount. thats why the equation is
(Q)(2/3)^(x/7) - you're doing (Q)(2/3 * 2/3 * 2/3 * ... x/7 times)
in other words, to take away 1/3 each time you multiply by 2/3

5. Oct 4, 2006

### helpm3pl3ase

O alright.. I get it now. Thank you

So everything thing else is correct, but instead of 1/3 use 2/3?

6. Oct 4, 2006

### HallsofIvy

Staff Emeritus
Yes, use 2/3 instead of 1/3.