## anonymous 3 years ago 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,

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. 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.

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

yes

9. anonymous

Okay.

10. anonymous

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

11. anonymous

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

12. anonymous

ill post a print screen of the problem

13. anonymous

14. anonymous

sorry i should have probably posted the problem earlier

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

i have 5 attempts to get it right

19. anonymous

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

20. anonymous

That's to 4 significant numbers

21. anonymous

ok im gonna try that see if its right

22. anonymous

nope

23. anonymous

No it's not.

24. anonymous

25. anonymous

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

26. anonymous

no we dont

27. anonymous

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

28. anonymous

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

29. anonymous

yes

30. anonymous

the question does say that :P

31. anonymous

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

32. anonymous

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

33. 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.

34. anonymous

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

35. anonymous

36. anonymous

first year chemisty college not incredibly basic but 101

37. 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

38. Mertsj

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

39. 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.

40. anonymous

@Mertsj got you covered.

41. anonymous

so would you |dw:1359339344388:dw|

42. anonymous

and square the speed of light

43. anonymous

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

44. anonymous

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

45. anonymous

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

46. anonymous

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

47. anonymous

then i divide 1554000/2.9*10^8

48. Mertsj

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

49. anonymous

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

50. anonymous

i really am not getting this

51. Mertsj

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

52. anonymous

yea ive done that before is that just it?

53. anonymous

then you convert from kg to g?

54. Mertsj

Multiply by 1000

55. Mertsj

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

56. anonymous

nope thats not it

57. anonymous

oops i messed up

58. anonymous

hold on i forgot to square

59. Mertsj

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

60. anonymous

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

61. anonymous

well thanks a lot for your help

62. Mertsj

63. anonymous

1.72*10^-8

64. Mertsj

yep. I thought so.

65. anonymous

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

66. Mertsj

yw