## sat_chen 2 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. sat_chen

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

2. sat_chen

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

3. sat_chen

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

4. Azteck

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. sat_chen

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

6. sat_chen

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

7. Azteck

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

8. sat_chen

yes

9. Azteck

Okay.

10. sat_chen

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

11. Azteck

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

12. sat_chen

ill post a print screen of the problem

13. sat_chen

14. sat_chen

sorry i should have probably posted the problem earlier

15. Azteck

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

16. sat_chen

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

17. sat_chen

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

18. sat_chen

i have 5 attempts to get it right

19. Azteck

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

20. Azteck

That's to 4 significant numbers

21. sat_chen

ok im gonna try that see if its right

22. sat_chen

nope

23. Azteck

No it's not.

24. sat_chen

25. Azteck

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

26. sat_chen

no we dont

27. sat_chen

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

28. Azteck

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

29. sat_chen

yes

30. sat_chen

the question does say that :P

31. Azteck

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

32. sat_chen

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

33. Azteck

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. sat_chen

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

35. Azteck

36. sat_chen

first year chemisty college not incredibly basic but 101

37. sat_chen

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. Azteck

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. Azteck

@Mertsj got you covered.

41. sat_chen

so would you |dw:1359339344388:dw|

42. sat_chen

and square the speed of light

43. sat_chen

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

44. Azteck

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

45. sat_chen

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

46. sat_chen

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

47. sat_chen

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. snapcracklepopz

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

50. sat_chen

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. sat_chen

yea ive done that before is that just it?

53. sat_chen

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. sat_chen

nope thats not it

57. sat_chen

oops i messed up

58. sat_chen

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. sat_chen

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

61. sat_chen

well thanks a lot for your help

62. Mertsj

63. sat_chen

1.72*10^-8

64. Mertsj

yep. I thought so.

65. sat_chen

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

66. Mertsj

yw