hmm.. let me re-do that.. i think i screwed up somewhere... go linear algebra

quote:

Originally posted by Evan Almae I was just about to say that! Nice try though! Thanks everyone for helping by the way! Didn't want to leave that unsaid, your help is much appreciated!

i'll leave the chemistry to those that took notes...talk to you guys later hehehe...and enjoy slaming this answer in your teachers face tomorrow...give a big BOOOYA for me hehehe...

quote:

Originally posted by kewlness hmm.. let me re-do that.. i think i screwed up somewhere... go linear algebra

Yea, sorry though, after that long explanation of yours too. But your method is still probably right, just did a mistake maybe like you mentioned.

kewlness: I still don't get what you mean by variables assigned as x and a number. Shouldn't it be like x, y, z, etc?

okay, to make this any harder, is it legal to use decimal for the coefficients???

hmm, should have posted this last year and i would have solved it. all i can remember now is using oxidation numbers and adding water or oxygen to balance out the equation.

The Real correct answer


3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O

How I did it:
assign the unknown coefficients as unknowns x1,x2,x3,x4,x5
x1Cu + x2HNO3 -> x3Cu(NO3)2 + x4NO + x5H2O

Now you know that
x1 - x3 = 0 for Copper
x2 - 2x5 = 0 for Hydrogen
3x2 - 6x3 - x4 -x5 = 0 for Oxygen
x2 - 2x3 - x4 = 0 for Nitrogen

Solving for the system of equations (i used linear algebra but you can use any method you want), I got the solutions:

x1= 3/4 t
x2 = 2t
x3= 3/4 t
x4= 1/2 t
x5= t

where t is a parameter
Now you have to choose a value t so that x1,x2,x3,x4,x5 are integers (1,2,3,4,5,6, etc...) while still being in the lowest denominator... The number that fits this category is where t=4

so therefore x1 = 3, x2 = 8 x3 = 3 x4 = 2 x5 = 4

I win for sure this time. You all lose

lol, I just took chem last year, (and did well too), and I've forgotten everything,

lol, kewlness, i was just about there! I had everything but the 2 infront of the NO! Well done. Thanks for the help!

Crap, I'm late!

I'm an ox-redux whore. Give me another one!

___________________
"Trance is dead. Ferry Corsten killed it." -Ishkur

quote:

Originally posted by kewlness The Real correct answer 3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO + 4H2O How I did it: assign the unknown coefficients as unknowns x1,x2,x3,x4,x5 x1Cu + x2HNO3 -> x3Cu(NO3)2 + x4NO + x5H2O Now you know that x1 - x3 = 0 for Copper x2 - 2x5 = 0 for Hydrogen 3x2 - 6x3 - x4 -x5 = 0 for Oxygen x2 - 2x3 - x4 = 0 for Nitrogen Solving for the system of equations (i used linear algebra but you can use any method you want), I got the solutions: x1= 3/4 t x2 = 2t x3= 3/4 t x4= 1/2 t x5= t where t is a parameter Now you have to choose a value t so that x1,x2,x3,x4,x5 are integers (1,2,3,4,5,6, etc...) while still being in the lowest denominator... The number that fits this category is where t=4 so therefore x1 = 3, x2 = 8 x3 = 3 x4 = 2 x5 = 4 I win for sure this time. You all lose

Well I got the right answer too, just did the last step wrong. Oh well...

quote:

Originally posted by Turbonium I get the initial part correct, but restoring it to normal eqn yield incorrect result, weird... 3Cu + 8H(+) + 2NO3(-) ---> 3Cu(2+) + 2NO + 4H2O full eqn (if I did it right, which according to this, I didn't): 3Cu + 2HNO3 + 6H(+) ---> 3Cu(NO3)2 + 2NO + 4H2O

Obviously it was 8HNO3, stupid mistake, meh, I aced it on the exams, that's all that matters

quote:

Originally posted by armandzadza Crap, I'm late! I'm an ox-redux whore. Give me another one!

Gimme a few minutes. I'm doing a paper, the ones I can't figure out I'll post. Or you can just do my homework for me???lol!