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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
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how would you start the conversion would you do this |dw:1359337417354:dw|
im really confused as to how to do this conversion although i thought i had this stuff down :(
thats how i did the above conversion btw |dw:1359337550325:dw|

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Other answers:

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

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