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moongazer Group Title

K2Cr2O7 + S8 ----> K2O + Cr2O3 + SO2 (the numbers are subscripts) find the oxidation number

  • one year ago
  • one year ago

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  1. Callisto Group Title
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    Oxidation number of ..?

    • one year ago
  2. moongazer Group Title
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    all elements there

    • one year ago
  3. Callisto Group Title
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    K -> Group I metal. Either 0, or +1. Here, it's an ion. So, what is the oxidation number of K?

    • one year ago
  4. Callisto Group Title
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    Btw, I think it's K2Cr2O7 on the left?

    • one year ago
  5. moongazer Group Title
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    yes, there is a typo :)

    • one year ago
  6. Callisto Group Title
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    Okay. What have you got for K?

    • one year ago
  7. moongazer Group Title
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    +1

    • one year ago
  8. moongazer Group Title
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    could you check my answers?

    • one year ago
  9. moongazer Group Title
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    I'll just post it :)

    • one year ago
  10. Callisto Group Title
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    Yup. For both sides, there is no oxidation number change for K. So, K on both sides are +1. Sure :)

    • one year ago
  11. moongazer Group Title
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    I'll just answer it :)

    • one year ago
  12. moongazer Group Title
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    Isn't it that if the subscript of O is greater than 2, it will have an oxidation number of -2 ?

    • one year ago
  13. Callisto Group Title
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    O has an oxidation number of -2 - You're right! :)

    • one year ago
  14. moongazer Group Title
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    but if "O" only it will have -1 and O2 it will have 0?

    • one year ago
  15. Callisto Group Title
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    No, not really. O is in group 6, it need 2 more electrons. So, usually, its oxidation number is -2 when it is bonded to other atoms. If it is an element (i.e. O2), its oxidation number is 0 as element in its own form. But remember, it's just 'usually' when its oxidation number is of -2. There are some exceptions. For instance, if I remember correctly, O in F2O is +2 since F is a more electronegative element and it has an oxidation number of -1. Since there are two F, oxidation number of O has to be +2 in order to get a net oxidation number of the compound equal to zero.

    • one year ago
  16. moongazer Group Title
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    ohh, thanks for the info. Here are my answers :) K2(+1)Cr2(+6)O7(-2) + S8(0) ----> K2(+1)O(-2) + Cr2(+3)O3(-2) + S(0)O2(0) The oxidation number is enclosed by parenthesis.

    • one year ago
  17. moongazer Group Title
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    Is it correct ?

    • one year ago
  18. Callisto Group Title
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    Almost. S(0)O2(0) <- not correct (for both)

    • one year ago
  19. moongazer Group Title
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    K2(+1)Cr2(+6)O7(-2) + S8(0) ----> K2(+1)O(-2) + Cr2(+3)O3(-2) + S(-2)O2(+1)

    • one year ago
  20. moongazer Group Title
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    I made the oxidation number of O2 the unknown. Is it correct now?

    • one year ago
  21. Callisto Group Title
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    Not really. For SO2, O is -2 (you can take a look at O first as it is more electronegative than S!) O2 is NOT an element in this case, so O doesn't have a oxidation number of 0!

    • one year ago
  22. moongazer Group Title
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    K2(+1)Cr2(+6)O7(-2) + S8(0) ----> K2(+1)O(-2) + Cr2(+3)O3(-2) + S(+4)O2(-2) I think this is correct already :)

    • one year ago
  23. Callisto Group Title
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    Nice :)

    • one year ago
  24. moongazer Group Title
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    I'll say something. I'll just fix the laptop :)

    • one year ago
  25. moongazer Group Title
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    please wait :)

    • one year ago
  26. Callisto Group Title
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    Take your time!

    • one year ago
  27. moongazer Group Title
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    Whew! The battery almost got dead. It keeps on saying plugged in, not charging.

    • one year ago
  28. moongazer Group Title
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    So it means, If you have a molecule with an oxygen,to find the oxidation number of the elements in it. You will identify first the more electronegative element and made the charge of it equal to its oxidation number. Then make the oxidation number of the other element the unknown.

    • one year ago
  29. moongazer Group Title
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    you will solve for the unknown oxidation number

    • one year ago
  30. Callisto Group Title
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    Hmm.. you can say so. But when you consider electronegativity of the element, that would be talking about the group V - group VII For metal, it usually has positive oxidation number. If I remember correctly, for group I metal, when it does not exist in element state, it should be of an oxidation number of +1, and looks like it works for most of the group II metal too. But for transition metals, that's another case...

    • one year ago
  31. Callisto Group Title
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    One little practice for you (if you want one lol) Find the oxidation number of H, O and Cl in the compound HOCl.

    • one year ago
  32. moongazer Group Title
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    H(+1)O(-2)Cl(+1)

    • one year ago
  33. Callisto Group Title
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    Yup :D

    • one year ago
  34. moongazer Group Title
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    I hope it is correct. :)

    • one year ago
  35. moongazer Group Title
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    Yey! thanks!

    • one year ago
  36. Callisto Group Title
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    Welcome :)

    • one year ago
  37. Callisto Group Title
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    There are three definitions....

    • one year ago
  38. moongazer Group Title
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    is it the loss of electrons and reactions directly with O2 ?

    • one year ago
  39. moongazer Group Title
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    transfer of electrons?

    • one year ago
  40. Callisto Group Title
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    Definition of oxidation: 1. addition of oxygen to a substance 2. a process in which a substance loses electrons 3. a process in which the oxidation number of an element in a substance increases.

    • one year ago
  41. moongazer Group Title
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    and reduction is the opposite of that?

    • one year ago
  42. Callisto Group Title
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    Yup...

    • one year ago
  43. moongazer Group Title
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    In my notes it says below oxidation it says: +O -H What does "H" have to do with oxidation and reduction ?

    • one year ago
  44. Callisto Group Title
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    -H?!

    • one year ago
  45. moongazer Group Title
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    idk I just saw in my notes. I was late during our class so I didn't know what is H there

    • one year ago
  46. moongazer Group Title
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    Anyway, thank you very much in your time and effort helping me. :)

    • one year ago
  47. Callisto Group Title
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    You're welcome.. Sorry that I couldn't help :(

    • one year ago
  48. moongazer Group Title
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    It's alright you helped me a lot. :)

    • one year ago
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