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vera_ewing

  • one year ago

chem

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  1. vera_ewing
    • one year ago
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  2. vera_ewing
    • one year ago
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    @JoannaBlackwelder A or B?

  3. JoannaBlackwelder
    • one year ago
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    Yep, I get one of those.

  4. anonymous
    • one year ago
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    Sorry been a while, I am more of organic, but I figured it out. I got 1.42m

  5. JoannaBlackwelder
    • one year ago
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    Hm, that's not what I got.

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  6. JoannaBlackwelder
    • one year ago
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    Did you do it like this?

  7. anonymous
    • one year ago
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    2HI--> H2 +I2 I 1.6m 0 0 C -x X X E 1.6x X X k=x^2/ (1.6-x)^2 k=1.42m

  8. anonymous
    • one year ago
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    Make snese

  9. JoannaBlackwelder
    • one year ago
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    No, I get the equilibrium expression for HI to be 1.6-2x

  10. JoannaBlackwelder
    • one year ago
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    Since 2 moles of it decompose for each 1 mole of H2 and I2

  11. anonymous
    • one year ago
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    Remember 2 moles HI so ()^2

  12. anonymous
    • one year ago
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    Look at my work

  13. anonymous
    • one year ago
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    2HI intial, when you have any number of moles you take it to the the power, example 3HI then (HI)^3

  14. anonymous
    • one year ago
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    @n648c788 what chemistry classes have you taken?

  15. anonymous
    • one year ago
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    The decompsition just means you will make the HI -x, and H x, Ix

  16. anonymous
    • one year ago
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    All but Organic is what I know best, but If I see examples I can solve old problems

  17. anonymous
    • one year ago
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    HI would be -2x not -x

  18. anonymous
    • one year ago
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    Yeah you are right little error, but it is worked the same way

  19. anonymous
    • one year ago
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    Sorry internet is bad, but you did it, you used quadratic equation at end right?

  20. anonymous
    • one year ago
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    yes

  21. anonymous
    • one year ago
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    Awseome job, sorry its been a while

  22. anonymous
    • one year ago
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    Are you ok? seems the task is done

  23. anonymous
    • one year ago
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    @vera_ewing is offline

  24. anonymous
    • one year ago
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    Oh ok thanks I appreciate it!

  25. anonymous
    • one year ago
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    Yes do you need any help

  26. anonymous
    • one year ago
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    ya I have a chem question posted above

  27. anonymous
    • one year ago
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    http://openstudy.com/study#/updates/5569209be4b01de5673b1c3d

  28. anonymous
    • one year ago
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    Ok

  29. vera_ewing
    • one year ago
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    Joanna is correct.

  30. JoannaBlackwelder
    • one year ago
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    :-) Thanks for the feedback, Vera.

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