# To Drbob222

I don't get how to get N, please show me cause I don't understand

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k = 0.693/t1/2
Then
ln(No/N) = kt
No = 120 ug
N = solve for this
k = from above
t = 120 sec.

First, you substitute the half life of this material which is listed as 20 seconds and that allows you to solve for k. You need k to solve the equation below it in the solution above.
Then you substitute 120 ug for No (No is what you have at the beginning). N is what you have at the end and that is what you're solving for. N is the only unknown in the equation. Then substitute k from the first calculation above. t in the problem is 2 minutes but since the half life was given in seconds this 2 min need to be convert to seconds which is 120 s. So the final equation looks like this
ln(120/N) = k(120). I don't remember k but that is what you obtained from above. Solve for N.

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2. Um so is it like this :
0.693 x 120 = 83.16
120/n=83.16
=9979.2?

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3. No, you aren't following the math.
The half life is 20 seconds.
k = 0.693/t1/2 = 0.693/20 = 0.03465.
Then ln(120 ug/N) = 0.03465*120
Solve for N.
By the way, that ln (which you don't have anywhere in your work) in front of the (No/N) is not for looks. That means to take the natural log (that's the log base e) of No/N.

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4. Alright, my final answer is 166.32
is this correct?

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5. No. If the sample starts out at 125 ug and it loses half of the sample every 20 seconds, how can it have more than it started with at the end of 2 minutes. It MUST be less than 125 ug.

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6. Yah that's wut I thought so I tried again and I know I sound but is it .0115 ?

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7. K this was my equation 120in/n=1.386
How do I solve for n?

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8. Much closer but not quite.
I think part of the problem is the 1.386. I think that should be 4.158.
k = 0.693/20 = 0.03465
ln(125/N) = 0.3465*120
ln(125/N) = 4.158
Take the antilog of both sides.
antilog ln(125/N) = 125/N
antilog 4.158 = 63.94
So 125/N = 63.94
and N = 125/63.94 = 1.95 ug

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9. Oh now I get it! Thank u so much for ur help:)

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