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It was pointed out that mass and energy are alternate aspects of a single entity called massenergy. 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,
 one year ago
 one year ago
It was pointed out that mass and energy are alternate aspects of a single entity called massenergy. 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,
 one year ago
 one year ago

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sat_chenBest ResponseYou've already chosen the best response.0
how would you start the conversion would you do this dw:1359337417354:dw
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
im really confused as to how to do this conversion although i thought i had this stuff down :(
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
thats how i did the above conversion btw dw:1359337550325:dw
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
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.
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
this if from my chemistry book all its asking to do is conversion
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
the formula is given just have to apply it and i have no clue
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
1.554 × 103 kJ ? Is it 1.554 x 10^3 kJ.
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
i converted it to J thats why its 1.554*10^6 in my attempt
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
SO it's this? \[1.554\times 10^6 \times \frac{ kgm^x }{ s^x }\]
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
ill post a print screen of the problem
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
sorry i should have probably posted the problem earlier
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Okay, Do you have the correct answer? Like an answer sheet or something?
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
no i cant get the solution it would be 0 points after that :)
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
i can try random answer to see if it gives me hint
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
i have 5 attempts to get it right
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
I got some weird number. I got \[4.564\times 10^{13}\]
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
That's to 4 significant numbers
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
ok im gonna try that see if its right
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Don't you have to find the mass of the water?
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
we are just trying to do some conversion this is Chapter 2 should be some basic stuff
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Wait mate. You're trying to find the change in mass?
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
the question does say that :P
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Man, heat of combustion keeps popping up in my head for some reason.
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
so would you do something like mole so 12.096/1.008 and 96/16 and get the value in mole form
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
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.
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
yea its nothing complicated like that i have just started to dip into basic chemisty
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
What year/grade is this?
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
first year chemisty college not incredibly basic but 101
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
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
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
Just plug into the equation E = mc^3. The energy is given, you know the speed of light. Calculate mass
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
Okay. I don't think I can help. I still have one year until universtiy/college. I'm afraid I can't help you.
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
@Mertsj got you covered.
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
so would you dw:1359339344388:dw
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
and square the speed of light
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
then change from g to kg by dividing by 1000 or would you conver to mole
 one year ago

AzteckBest ResponseYou've already chosen the best response.0
ANd then I think you might subtract what you get from the mass of the oxygen and hydrogen molecules...
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
so heres what im doing right now since m = e/c^2
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
first i change the 10.965/1.001 and 87/16 to convert it to mole
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
then i divide 1554000/2.9*10^8
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
E should be in joules, m in kilograms and c in meters per second
 one year ago

snapcracklepopzBest ResponseYou've already chosen the best response.0
sorry to interrupt but can someone help me after you are done helping this person with their question please?
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
i really am not getting this
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
\[\frac{1.554\times 10^6}{(3\times 10^8)^2}=.1727 \times 10^{10} kg\]
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
yea ive done that before is that just it?
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
then you convert from kg to g?
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
That's what I think it is. Because the equation tells you how much mass is needed to produce that much energy.
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
hold on i forgot to square
 one year ago

MertsjBest ResponseYou've already chosen the best response.1
\[.1727 \times 10^{10}kg=.1727 \times 10^{7}g = 1.727 \times 10^{8}g\]
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
yea but i ran out of turns and that was the answe r:(
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
well thanks a lot for your help
 one year ago

sat_chenBest ResponseYou've already chosen the best response.0
yea im gonna have to hit the books again and learn why i didnt get it thanks a lot
 one year ago
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