Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

creekrat48

  • 4 years ago

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

  • This Question is Closed
  1. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    how many half lifes is that?

  3. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I am still reading these half lifes in book so confused

  4. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  5. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just got that as you did

  6. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  7. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    now ^

  8. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  10. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  12. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh my

  13. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Does that look like anything you have ever done?

  14. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  15. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not that complex

  16. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  17. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  18. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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!

  19. creekrat48
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  20. zbay
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  21. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy