anonymous
  • anonymous
It was pointed out that mass and energy are alternate aspects of a single entity called mass-energy. The relationship between these two physical quantities is Einstein's equation, E = mc2, where E is energy, m is mass, and c is the speed of light. In a combustion experiment, it was found that 10.965 g of hydrogen molecules combined with 87.000 g of oxygen molecules to form water and released 1.554 × 103 kJ of heat. Use Einstein's equation to calculate the corresponding mass change in this process,
Chemistry
katieb
  • katieb
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anonymous
  • anonymous
how would you start the conversion would you do this |dw:1359337417354:dw|
anonymous
  • anonymous
im really confused as to how to do this conversion although i thought i had this stuff down :(
anonymous
  • anonymous
thats how i did the above conversion btw |dw:1359337550325:dw|

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anonymous
  • anonymous
I haven't thoroughly studied physics yet. I think this is better off in the physics section. But I think you're meant to convert the mass into kg's first before doing anything else. And the kilojoules to joules.
anonymous
  • anonymous
this if from my chemistry book all its asking to do is conversion
anonymous
  • anonymous
the formula is given just have to apply it and i have no clue
anonymous
  • anonymous
1.554 × 103 kJ ? Is it 1.554 x 10^3 kJ.
anonymous
  • anonymous
yes
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
i converted it to J thats why its 1.554*10^6 in my attempt
anonymous
  • anonymous
SO it's this? \[1.554\times 10^6 \times \frac{ kgm^x }{ s^x }\]
anonymous
  • anonymous
ill post a print screen of the problem
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
sorry i should have probably posted the problem earlier
anonymous
  • anonymous
Okay, Do you have the correct answer? Like an answer sheet or something?
anonymous
  • anonymous
no i cant get the solution it would be 0 points after that :)
anonymous
  • anonymous
i can try random answer to see if it gives me hint
anonymous
  • anonymous
i have 5 attempts to get it right
anonymous
  • anonymous
I got some weird number. I got \[4.564\times 10^{13}\]
anonymous
  • anonymous
That's to 4 significant numbers
anonymous
  • anonymous
ok im gonna try that see if its right
anonymous
  • anonymous
nope
anonymous
  • anonymous
No it's not.
anonymous
  • anonymous
not the right answer
anonymous
  • anonymous
Don't you have to find the mass of the water?
anonymous
  • anonymous
no we dont
anonymous
  • anonymous
we are just trying to do some conversion this is Chapter 2 should be some basic stuff
anonymous
  • anonymous
Wait mate. You're trying to find the change in mass?
anonymous
  • anonymous
yes
anonymous
  • anonymous
the question does say that :P
anonymous
  • anonymous
Man, heat of combustion keeps popping up in my head for some reason.
anonymous
  • anonymous
so would you do something like mole so 12.096/1.008 and 96/16 and get the value in mole form
anonymous
  • anonymous
I'm not sure but it's worth a try. because heat of combustion is sort of what this experiment sort of is. Because you got your energy/heat released and then you got your mass changes and stuff.
anonymous
  • anonymous
yea its nothing complicated like that i have just started to dip into basic chemisty
anonymous
  • anonymous
What year/grade is this?
anonymous
  • anonymous
first year chemisty college not incredibly basic but 101
anonymous
  • anonymous
so far we have just learned how to name basic stuff, learning to name polyatomic ions etc emperical formula stuff like that really easy stuff i aced all the other problem but this one is tripping me up
Mertsj
  • Mertsj
Just plug into the equation E = mc^3. The energy is given, you know the speed of light. Calculate mass
anonymous
  • anonymous
Okay. I don't think I can help. I still have one year until universtiy/college. I'm afraid I can't help you.
anonymous
  • anonymous
@Mertsj got you covered.
anonymous
  • anonymous
so would you |dw:1359339344388:dw|
anonymous
  • anonymous
and square the speed of light
anonymous
  • anonymous
then change from g to kg by dividing by 1000 or would you conver to mole
anonymous
  • anonymous
ANd then I think you might subtract what you get from the mass of the oxygen and hydrogen molecules...
anonymous
  • anonymous
so heres what im doing right now since m = e/c^2
anonymous
  • anonymous
first i change the 10.965/1.001 and 87/16 to convert it to mole
anonymous
  • anonymous
then i divide 1554000/2.9*10^8
Mertsj
  • Mertsj
E should be in joules, m in kilograms and c in meters per second
anonymous
  • anonymous
sorry to interrupt but can someone help me after you are done helping this person with their question please?
anonymous
  • anonymous
i really am not getting this
Mertsj
  • Mertsj
\[\frac{1.554\times 10^6}{(3\times 10^8)^2}=.1727 \times 10^{-10} kg\]
anonymous
  • anonymous
yea ive done that before is that just it?
anonymous
  • anonymous
then you convert from kg to g?
Mertsj
  • Mertsj
Multiply by 1000
Mertsj
  • Mertsj
That's what I think it is. Because the equation tells you how much mass is needed to produce that much energy.
anonymous
  • anonymous
nope thats not it
anonymous
  • anonymous
oops i messed up
anonymous
  • anonymous
hold on i forgot to square
Mertsj
  • Mertsj
\[.1727 \times 10^{-10}kg=.1727 \times 10^{-7}g = 1.727 \times 10^{-8}g\]
anonymous
  • anonymous
yea but i ran out of turns and that was the answe r:(
anonymous
  • anonymous
well thanks a lot for your help
Mertsj
  • Mertsj
What was the answer?
anonymous
  • anonymous
1.72*10^-8
Mertsj
  • Mertsj
yep. I thought so.
anonymous
  • anonymous
yea im gonna have to hit the books again and learn why i didnt get it thanks a lot
Mertsj
  • Mertsj
yw

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