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anonymous

  • one year ago

For the reaction 2A+B=C+2D, the starting [A]=4.0M and [B]=3.0. At equilibrium the [C]=1.0M. Calculate the value of the equilibrium constant, K(c). I tried the formula but go no where because of "2D". Thank-you. a.)0.5 b.)2.0 c.)3.0 d.)3.5

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  1. anonymous
    • one year ago
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    @Lady_

  2. anonymous
    • one year ago
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    @jebonna

  3. anonymous
    • one year ago
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    @Cuanchi

  4. cuanchi
    • one year ago
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    you have to do an ICE table, are you familiar with that?

  5. anonymous
    • one year ago
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    No I am not, i'm sorry.

  6. cuanchi
    • one year ago
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    |dw:1444511844065:dw|

  7. cuanchi
    • one year ago
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    I= initial C= change E= equilibrium

  8. anonymous
    • one year ago
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    Oh! yes, yes! but I am to given one. Only that statement.

  9. cuanchi
    • one year ago
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    |dw:1444511977301:dw|

  10. anonymous
    • one year ago
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    But how did you come up with those numbers?

  11. anonymous
    • one year ago
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    Wait are those the coefficients?

  12. anonymous
    • one year ago
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    So [1][2]/[2]^2[1]=.5?????

  13. cuanchi
    • one year ago
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    the problem said that you have at the equilibrium 1M of C. You have to assume that was 0 M of C at the beginning and 1 mole was produced. Then by the stoichiometry of the reaction you can calculate how much of A and B has to be consume to produce 1 mol of C.

  14. cuanchi
    • one year ago
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    no

  15. cuanchi
    • one year ago
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    |dw:1444512283220:dw|

  16. anonymous
    • one year ago
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    OMG thank-you!!!!! so simple.

  17. anonymous
    • one year ago
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    @Cuanchi can I ask you another question?

  18. cuanchi
    • one year ago
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    sure if it fast

  19. anonymous
    • one year ago
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    1 Attachment
  20. cuanchi
    • one year ago
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    |dw:1444513373209:dw|

  21. cuanchi
    • one year ago
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    |dw:1444513578346:dw|

  22. Jhannybean
    • one year ago
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    I was trying to upload a picture of what I did as well xD

  23. cuanchi
    • one year ago
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    isolate the "x" that is just math

  24. cuanchi
    • one year ago
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    i got to go sorry

  25. anonymous
    • one year ago
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    thank you so much!

  26. cuanchi
    • one year ago
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    |dw:1444514849777:dw|

  27. cuanchi
    • one year ago
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    |dw:1444514968806:dw|

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