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,

- anonymous

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- anonymous

how would you start the conversion would you do this |dw:1359337417354:dw|

- anonymous

im really confused as to how to do this conversion although i thought i had this stuff down :(

- anonymous

thats how i did the above conversion btw |dw:1359337550325:dw|

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## More answers

- 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

this if from my chemistry book all its asking to do is conversion

- anonymous

the formula is given just have to apply it and i have no clue

- anonymous

1.554 × 103 kJ ?
Is it 1.554 x 10^3 kJ.

- anonymous

yes

- anonymous

Okay.

- anonymous

i converted it to J thats why its 1.554*10^6 in my attempt

- anonymous

SO it's this?
\[1.554\times 10^6 \times \frac{ kgm^x }{ s^x }\]

- anonymous

ill post a print screen of the problem

- anonymous

##### 1 Attachment

- anonymous

sorry i should have probably posted the problem earlier

- anonymous

Okay, Do you have the correct answer? Like an answer sheet or something?

- anonymous

no i cant get the solution it would be 0 points after that :)

- anonymous

i can try random answer to see if it gives me hint

- anonymous

i have 5 attempts to get it right

- anonymous

I got some weird number. I got
\[4.564\times 10^{13}\]

- anonymous

That's to 4 significant numbers

- anonymous

ok im gonna try that see if its right

- anonymous

nope

- anonymous

No it's not.

- anonymous

not the right answer

- anonymous

Don't you have to find the mass of the water?

- anonymous

no we dont

- anonymous

we are just trying to do some conversion this is Chapter 2 should be some basic stuff

- anonymous

Wait mate. You're trying to find the change in mass?

- anonymous

yes

- anonymous

the question does say that :P

- anonymous

Man, heat of combustion keeps popping up in my head for some reason.

- anonymous

so would you do something like mole so 12.096/1.008 and 96/16 and get the value in mole form

- 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

yea its nothing complicated like that i have just started to dip into basic chemisty

- anonymous

What year/grade is this?

- anonymous

first year chemisty college not incredibly basic but 101

- 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

Just plug into the equation E = mc^3. The energy is given, you know the speed of light. Calculate mass

- 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

@Mertsj got you covered.

- anonymous

so would you |dw:1359339344388:dw|

- anonymous

and square the speed of light

- anonymous

then change from g to kg by dividing by 1000 or would you conver to mole

- anonymous

ANd then I think you might subtract what you get from the mass of the oxygen and hydrogen molecules...

- anonymous

so heres what im doing right now since m = e/c^2

- anonymous

first i change the 10.965/1.001 and 87/16 to convert it to mole

- anonymous

then i divide 1554000/2.9*10^8

- Mertsj

E should be in joules, m in kilograms and c in meters per second

- anonymous

sorry to interrupt but can someone help me after you are done helping this person with their question please?

- anonymous

i really am not getting this

- Mertsj

\[\frac{1.554\times 10^6}{(3\times 10^8)^2}=.1727 \times 10^{-10} kg\]

- anonymous

yea ive done that before is that just it?

- anonymous

then you convert from kg to g?

- Mertsj

Multiply by 1000

- Mertsj

That's what I think it is. Because the equation tells you how much mass is needed to produce that much energy.

- anonymous

nope thats not it

- anonymous

oops i messed up

- anonymous

hold on i forgot to square

- Mertsj

\[.1727 \times 10^{-10}kg=.1727 \times 10^{-7}g = 1.727 \times 10^{-8}g\]

- anonymous

yea but i ran out of turns and that was the answe r:(

- anonymous

well thanks a lot for your help

- Mertsj

What was the answer?

- anonymous

1.72*10^-8

- Mertsj

yep. I thought so.

- anonymous

yea im gonna have to hit the books again and learn why i didnt get it thanks a lot

- Mertsj

yw

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