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Chemistry 16 Online
OpenStudy (anonymous):

from this graph How much nitrogen-14 will be produced from a 200-g sample of carbon-14 after 17,190 years?

OpenStudy (anonymous):

OpenStudy (anonymous):

how many half lifes is that?

OpenStudy (anonymous):

I am still reading these half lifes in book so confused

OpenStudy (anonymous):

3 half lifes meaning there will be an 8th of the atoms left, can you go from mass to atoms?

OpenStudy (anonymous):

I just got that as you did

OpenStudy (anonymous):

No i would calculate the atoms total and then divide them by 3/4 and those would be the ones that are N

OpenStudy (anonymous):

now ^

OpenStudy (anonymous):

I'm sorry I just don't understand with the 200 gram sample

OpenStudy (anonymous):

do you know how to go from mass-->moles-->atoms?

OpenStudy (anonymous):

i have done it before but would have to go back into book to reread again

OpenStudy (anonymous):

\[200g(\frac{1 mole c}{12.01g c})\frac{6.022x10^23 c atom}{1 mole c})\] that will get you atoms,

OpenStudy (anonymous):

oh my

OpenStudy (anonymous):

Does that look like anything you have ever done?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

not that complex

OpenStudy (anonymous):

1.00x10^25 is the total (I think i'm using a jankie calculator) now just multiply that number by .875 and that will be the number of nitrogen

OpenStudy (anonymous):

zbay thank you I guess some people just don't understand that it just doesn't click for me. Another guy was very mean like I am some dumbass.

OpenStudy (anonymous):

No chemistry is a ball buster, it takes a lot of practice and everything builds on previous knowledge. If you keep pluging away at it i'm sure you will get a handle on this subject. Good luck and keep asking questions!

OpenStudy (anonymous):

in 3 half-lifes, 200g carbon-14 will decay to 100-50-25g, so taking that carbon-14 decays to nitrogen-14, 175 g is produced or using the half-life formula, amount of nitrogen-14 produced = 200 - 200*0.5^(17190/5730) = 175 g

OpenStudy (anonymous):

Thats a much easier way to do it, It can be calculated the way i explained it sorry if i added to the confusion

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