anonymous
  • anonymous
Linear substitution application
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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
anonymous
  • anonymous
@nincompoop @phi @TheSmartOne Any help please?
sparrow2
  • sparrow2
what is the question?

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anonymous
  • anonymous
are you short an equation? b/c there are 5 unknowns
anonymous
  • anonymous
How to solve for each unknown
anonymous
  • anonymous
@peachpi I was thinking the same thing.
anonymous
  • anonymous
Here is where I found the technique http://www.chemicalaid.com/articles.php/view/1/how-to-balance-chemical-equations
Elsa213
  • Elsa213
@TheĐ…martOne
Loser66
  • Loser66
need the matrix method to solve it???
anonymous
  • anonymous
I don't have 5 equations though
anonymous
  • anonymous
the example cleverly had enough eqautions to solve the problem.
Loser66
  • Loser66
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 \(3x_2\) = \(6x_3+x_4+x_5\)
anonymous
  • anonymous
Cu: a=c H: b=2e N: b=2c+d O: 3b=6c+d+e
anonymous
  • anonymous
a*Cu + b*HNO3 = c*Cu(NO3)2 + d*NO + e*H2O
anonymous
  • anonymous
Where do we go from there
Loser66
  • Loser66
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 \[\left[\begin{matrix}1&0&-1&0&0&0\\0&1&0&0&-2&0\\0&2&-2&-1&0&0\\0&3&-6&-1&-1&0\end{matrix}\right]\]
anonymous
  • anonymous
When I do rref, I get -.25 -2 -.25 and -3.5
Loser66
  • Loser66
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 \(x_1= 1\\x_2=8\\x_3=1\\s_4=14\)
anonymous
  • anonymous
Where did you ger that that x_1=1/4x_5
Loser66
  • Loser66
Now, replace all \(1(Cu) +8(HNO_3)\rightarrow 1(Cu(NO_3)_2)+14(NO)+4(H_2O)\)
anonymous
  • anonymous
That's not correct. It's 3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O
Loser66
  • Loser66
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
anonymous
  • anonymous
oh ok
Loser66
  • Loser66
I may make mistake at somewhere, but the technique is that. Do it carefully and you can get the answer
anonymous
  • anonymous
Thank you very much
Loser66
  • Loser66
ok
anonymous
  • anonymous
It' in the matrix, you put for Nitrogen, 2x instead of just x
Loser66
  • Loser66
Yes!!!!!!!!!!! it is. My bad. I am sorry
Loser66
  • Loser66
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.
Loser66
  • Loser66
Thanks for the question!! I have an opportunity to recall my knowledge. :)

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