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anonymous

  • one year ago

HAlf Life problem Radon-222 has a half-life of 4.0 d (abbreviation for “diem”—Latin for “day”). If the initial mass of the sample of this isotope is 6.8 g, calculate the mass of radon-222 remaining in the sample after: a) 8 d b) 16 d c) 32d the awnsers are : a) 1.7g b)0.62g c) 0.026g i can't figure out how to do it

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  1. anonymous
    • one year ago
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    You mean 0.42 for b?

  2. anonymous
    • one year ago
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    no it is 0.62 for 6 this question is from a text book with given awnsers

  3. anonymous
    • one year ago
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    Check it again.

  4. anonymous
    • one year ago
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    i will show you a picture one second

  5. anonymous
    • one year ago
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    Kk.

  6. anonymous
    • one year ago
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    1 Attachment
  7. anonymous
    • one year ago
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    its a word file

  8. anonymous
    • one year ago
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    Is that the only you cant figure or is it all of them?

  9. anonymous
    • one year ago
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    It is probably a typo as I know I got a and c right. If b is between the ones I got right, then I am sure it is a mistake

  10. anonymous
    • one year ago
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    Is that really a textbook? If so, what publisher is it?

  11. anonymous
    • one year ago
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    i am not sure about the publisher i had a pdf version

  12. anonymous
    • one year ago
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    and i can't figure out how to do all 3

  13. anonymous
    • one year ago
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    i think it is called nature of matter 11 thre could have possible been a mistake, they are common in textbooks

  14. anonymous
    • one year ago
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    \[Half life period=\frac{ \ln2 }{ k }\] Mass after some days is given as y(t) where t is the time in days y0 would be initial mass. Formula for mass after some days is \[y(t)=y _{0}e ^{-kt}\]

  15. anonymous
    • one year ago
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    Try it and would get a and c right but not b. Strange.

  16. anonymous
    • one year ago
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    okay ty i will try it and see if i get the awnser right

  17. anonymous
    • one year ago
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    KK, I can wait.

  18. anonymous
    • one year ago
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    in the formula you put -kt what does that represent?

  19. anonymous
    • one year ago
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    as the expoenet

  20. anonymous
    • one year ago
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    There would be a negative sign because it is decaying. k is the constant from knowing the half life period. t is the time in days

  21. anonymous
    • one year ago
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    with the formula i am doing y(8)=6.8 and im am confused for what i do after

  22. anonymous
    • one year ago
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    because i am still getting the wrong answer

  23. anonymous
    • one year ago
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    can you do a question so i know how its done

  24. anonymous
    • one year ago
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    you forgot yo use the e^-kt part.

  25. anonymous
    • one year ago
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    to*

  26. anonymous
    • one year ago
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    You done e before in Algebra right?

  27. anonymous
    • one year ago
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    yeah it's the exponent right? i didnt know what values would go for the k

  28. anonymous
    • one year ago
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    Read my explanation above for the formula.

  29. anonymous
    • one year ago
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    lol i did but i still dont understand

  30. anonymous
    • one year ago
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    we briefly learned this in class so i am very confused

  31. anonymous
    • one year ago
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    Half life period is the period it takes for an element to lose half its mass

  32. anonymous
    • one year ago
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    which was four days

  33. anonymous
    • one year ago
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    so is the k value supposed to be 4

  34. anonymous
    • one year ago
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    Yes! Now solve for k

  35. anonymous
    • one year ago
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    No. Read my 2 formulas above.

  36. anonymous
    • one year ago
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    what does the IN mean?

  37. anonymous
    • one year ago
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    Oh man, I thought algebra 2 was needed to take chemistry. ln is natural logarithm. But dont worry about its meaning since you probably havent took algebra 2 yet. It is in a scientific calculator. Do you see it. Its actually lowercase of L then n, not in btw.

  38. anonymous
    • one year ago
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    yeah i found it now i can solve for k using this

  39. anonymous
    • one year ago
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    btw this is highschool chemistry fir grade 11 and you only need grade 10 scince to be able to take it :P

  40. anonymous
    • one year ago
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    is k value 1.45?

  41. anonymous
    • one year ago
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    How?

  42. anonymous
    • one year ago
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    beacuse in*2 *1.45=4

  43. anonymous
    • one year ago
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    \[4=\frac{ \ln2 }{ k} \] Cross multiply \[4k=\ln2\] Divide both sides by 4 \[4=\frac{ \ln2 }{ k }\]

  44. anonymous
    • one year ago
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    Dont forget to close the parentheses inside the ln

  45. anonymous
    • one year ago
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    i am getting 0.17

  46. anonymous
    • one year ago
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    if this isnt correct i will ask my chemistry teacher at school tom

  47. anonymous
    • one year ago
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    im very confused right now

  48. anonymous
    • one year ago
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    Yea you are correct. But makes but includes the other decimals. I got 0.1732

  49. anonymous
    • one year ago
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    yeah i rounded

  50. anonymous
    • one year ago
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    Now do a).

  51. anonymous
    • one year ago
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    yeah i fiannly got the right awnser

  52. anonymous
    • one year ago
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    tysm for helping me

  53. anonymous
    • one year ago
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    No problem! Just remember the formula and its meanings and you're all good.

  54. anonymous
    • one year ago
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    Okay thank you

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