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anonymous

  • one year ago

Linear substitution application

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  1. anonymous
    • one year ago
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    I need help balancing a chemical equation (now before you send me off to the chemistry section), the way to solve it involves math: I have to balance this equation: Cu + HNO3 = Cu(NO3)2 + NO + H2O I find that the answer is 3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O But I have no idea how to get there I found a wesite talking about using a linear substittuion, basically rewriting the equation like this a*Cu + b*HNO3 = c*Cu(NO3)2 + d*NO + e*H2O Since the same number of atoms exist on both sides, it becomes: Cu: a=c H: b=2e N: b=2c+d O: 3b=6c+d+e

  2. anonymous
    • one year ago
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    @nincompoop @phi @TheSmartOne Any help please?

  3. sparrow2
    • one year ago
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    what is the question?

  4. anonymous
    • one year ago
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    are you short an equation? b/c there are 5 unknowns

  5. anonymous
    • one year ago
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    How to solve for each unknown

  6. anonymous
    • one year ago
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    @peachpi I was thinking the same thing.

  7. anonymous
    • one year ago
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    Here is where I found the technique http://www.chemicalaid.com/articles.php/view/1/how-to-balance-chemical-equations

  8. Elsa213
    • one year ago
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    @TheЅmartOne

  9. Loser66
    • one year ago
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    need the matrix method to solve it???

  10. anonymous
    • one year ago
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    I don't have 5 equations though

  11. anonymous
    • one year ago
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    the example cleverly had enough eqautions to solve the problem.

  12. Loser66
    • one year ago
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    I arrange it as \(x_1(Cu) +x_2(HNO_3)\rightarrow x_3(Cu(NO_3)_2)+x_4(NO)+x_5(H_2O)\) Left hand side Righ hand side Cu \(x_1\) = \(x_3\) Hydrogen \(x_2\) = \(2x_5\) Nitron \(2x_2\) = \(2x_3+x_4\) Oxygen \(3x_2\) = \(6x_3+x_4+x_5\)

  13. anonymous
    • one year ago
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    Cu: a=c H: b=2e N: b=2c+d O: 3b=6c+d+e

  14. anonymous
    • one year ago
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    a*Cu + b*HNO3 = c*Cu(NO3)2 + d*NO + e*H2O

  15. anonymous
    • one year ago
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    Where do we go from there

  16. Loser66
    • one year ago
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    It gives you: \(x_1-x_3=0\\x_2-2x_5=0\\2x_2-2x_3-x_4=0\\3x_2-6x_3-x_4-x_5=0\) hence the matrix is \[\left[\begin{matrix}1&0&-1&0&0&0\\0&1&0&0&-2&0\\0&2&-2&-1&0&0\\0&3&-6&-1&-1&0\end{matrix}\right]\]

  17. anonymous
    • one year ago
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    When I do rref, I get -.25 -2 -.25 and -3.5

  18. Loser66
    • one year ago
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    Take row reduce echelon form, you have \(x_1=(1/4)x_5\\x_2=2x_5\\x_3=(1/4)x_5\\x_4=(7/2)x_5\) hence if \(x_5= 4\) then \(x_1= 1\\x_2=8\\x_3=1\\s_4=14\)

  19. anonymous
    • one year ago
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    Where did you ger that that x_1=1/4x_5

  20. Loser66
    • one year ago
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    Now, replace all \(1(Cu) +8(HNO_3)\rightarrow 1(Cu(NO_3)_2)+14(NO)+4(H_2O)\)

  21. anonymous
    • one year ago
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    That's not correct. It's 3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O

  22. Loser66
    • one year ago
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    from the matrix, take row reduce echelon form, you have the first row 1 0 0 0 -1/4 0 that is x1 -1/4 x5 =0

  23. anonymous
    • one year ago
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    oh ok

  24. Loser66
    • one year ago
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    I may make mistake at somewhere, but the technique is that. Do it carefully and you can get the answer

  25. anonymous
    • one year ago
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    Thank you very much

  26. Loser66
    • one year ago
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    ok

  27. anonymous
    • one year ago
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    It' in the matrix, you put for Nitrogen, 2x instead of just x

  28. Loser66
    • one year ago
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    Yes!!!!!!!!!!! it is. My bad. I am sorry

  29. Loser66
    • one year ago
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    then the rref is \(x_1=(3/4)x_5\\x_2= 2x_5\\x_3=(3/4)x_5\\x_4=(1/2)x_5\) if \(x_5=4\) then \(x_1=3\\x_2=8\\x_3=3\\x_4=2\) and the equation is the correct one as you had.

  30. Loser66
    • one year ago
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    Thanks for the question!! I have an opportunity to recall my knowledge. :)

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