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anonymous
 3 years 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,
anonymous
 3 years 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,

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how would you start the conversion would you do this dw:1359337417354:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thats how i did the above conversion btw dw:1359337550325:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0this if from my chemistry book all its asking to do is conversion

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the formula is given just have to apply it and i have no clue

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01.554 × 103 kJ ? Is it 1.554 x 10^3 kJ.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i converted it to J thats why its 1.554*10^6 in my attempt

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ill post a print screen of the problem

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry i should have probably posted the problem earlier

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, Do you have the correct answer? Like an answer sheet or something?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no i cant get the solution it would be 0 points after that :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i can try random answer to see if it gives me hint

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i have 5 attempts to get it right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got some weird number. I got \[4.564\times 10^{13}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's to 4 significant numbers

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok im gonna try that see if its right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Don't you have to find the mass of the water?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we are just trying to do some conversion this is Chapter 2 should be some basic stuff

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait mate. You're trying to find the change in mass?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the question does say that :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Man, heat of combustion keeps popping up in my head for some reason.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'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
 3 years ago
Best ResponseYou've already chosen the best response.0yea its nothing complicated like that i have just started to dip into basic chemisty

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What year/grade is this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first year chemisty college not incredibly basic but 101

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.1Just plug into the equation E = mc^3. The energy is given, you know the speed of light. Calculate mass

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Mertsj got you covered.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so would you dw:1359339344388:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and square the speed of light

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then change from g to kg by dividing by 1000 or would you conver to mole

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ANd then I think you might subtract what you get from the mass of the oxygen and hydrogen molecules...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so heres what im doing right now since m = e/c^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0first i change the 10.965/1.001 and 87/16 to convert it to mole

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then i divide 1554000/2.9*10^8

Mertsj
 3 years ago
Best ResponseYou've already chosen the best response.1E should be in joules, m in kilograms and c in meters per second

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i really am not getting this

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea ive done that before is that just it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then you convert from kg to g?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hold on i forgot to square

Mertsj
 3 years ago
Best ResponseYou've already chosen the best response.1\[.1727 \times 10^{10}kg=.1727 \times 10^{7}g = 1.727 \times 10^{8}g\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea but i ran out of turns and that was the answe r:(

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well thanks a lot for your help

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