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Mass and atoms

  1. Sep 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Lead has a mass of 11.35 g per cubic centimeter of volume, and the mass of one of its atoms is 3.439×10-25 kg. If the atoms are spherical and tightly packed, what is the volume of an individual atom? ( m3)

    2. Relevant equations
    conversions

    3. The attempt at a solution
    For this problem I converted the 11.35 g/cm ^3 into 11350 g/m^3 and then converted the atom from 3.439*10^-25 kg into 3.439*10^-22 g. Then I divided it by the 11350 to get m^3 and the result was 3.03*10^-26 but it appears to be wrong.
     
  2. jcsd
  3. Sep 27, 2016 #2

    Bystander

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  4. Sep 27, 2016 #3

    PeroK

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    How many ##cm## are there in a ##m##?
     
    Last edited: Sep 27, 2016
  5. Sep 27, 2016 #4
    100 so it would be cubed to get m^3 and it would be 3.03*10^-29 which is the right answer
     
  6. Sep 27, 2016 #5
    If 11,35 grams fit into 1 cm3 then 3.439×10-22 g will fitt into
    3.439×10-22g/ 11.35= 3.03 cm3
     
  7. Sep 27, 2016 #6
    * 3.03*10-23 cm3
     
  8. Sep 27, 2016 #7
    But you're looking for m^3
     
  9. Sep 27, 2016 #8
    1m3= 1.000.000cm3
    ? = 3.03*10 - 23cm3
     
  10. Sep 27, 2016 #9

    PeroK

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    Are you helping him now? o_O
     
  11. Sep 27, 2016 #10
    'crosswise multiplication table' i don't know if it is an english term, i translated it literally from dutch.
     
  12. Sep 27, 2016 #11
    Right. so it would be 11.35 g/cm^3 * 1000000cm^3/1 m^3 to give 11350000 g/m^3 which you divide into 3.439*10^-22 g
     
  13. Sep 27, 2016 #12
    I just call it a conversion
     
  14. Sep 27, 2016 #13
    keep in mind that 1 cm3 is not equalle to 1.000.000 m3
     
  15. Sep 27, 2016 #14

    PeroK

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    I don't think you're helping here. The OP has got the correct answer. Leaving aside the question of the basic volume of a sphere as opposed to the volume it occupies in a 3D array.
     
  16. Sep 27, 2016 #15

    PeroK

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    I assume that's the answer on your answer sheet?
     
  17. Sep 27, 2016 #16
    you are right i made a typo in my calculator, sorry for that.
     
  18. Sep 27, 2016 #17
    Yes it was a test question I got wrong so I was trying to find the correct method to the answer
     
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