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wondering_

How many milligrams of C atoms are in 2.04 x 10^20 molecules of ethanol?

  • one year ago
  • one year ago

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  1. zepp
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    Ethanol: \(\large \text{C}_2\text{H}_6\text{O}\)

    • one year ago
  2. zepp
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    So first question, how many moles does 2.04 x 10^20 molecules represent?

    • one year ago
  3. wondering_
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    its not CH(Sub 6)O?

    • one year ago
  4. wondering_
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    why did u give Carbon2?

    • one year ago
  5. zepp
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    Because that's the formula for Ethanol

    • one year ago
  6. zepp
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    It's a 2-carbon alcohol.

    • one year ago
  7. wondering_
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    \[(6.02 \times 10^{23})(2.04 \times 10^{20})?\]

    • one year ago
  8. wondering_
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    oh ok

    • one year ago
  9. zepp
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    That's incorrect; \[\large 1 \text{ mole}= 6.02×10^{23}\]

    • one year ago
  10. zepp
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    \[\large ? \text{ mole}=2.04×10^{20}\]

    • one year ago
  11. zepp
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    That's a proportion and you should have this: \(\LARGE \frac{2.04×10^{20}}{6.02×10^{23}}\)

    • one year ago
  12. wondering_
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    oh so its divide instead of multiply?

    • one year ago
  13. zepp
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    Yes, that's algebraic manipulation.

    • one year ago
  14. zepp
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    Once you get that, you'll get the number of moles this quantity of molecules represent, are afterward, you may convert that easily to its mass through the molar mass.

    • one year ago
  15. wondering_
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    |dw:1349138080842:dw|

    • one year ago
  16. zepp
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    Why 1000?

    • one year ago
  17. wondering_
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    for milligrams?

    • one year ago
  18. zepp
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    Our units are in moles...

    • one year ago
  19. zepp
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    \[\large \frac{1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=\frac{x \text{ mole}}{2.04 * 10^{20}\text{ atoms}}\]Multiply both sides by 2.04*10^20\[\large \frac{2.04*10^{20}\text{ atoms}*1 \text{ mole}}{6.02*10^{23} \text{ atoms}}=x \text{ mole}\]

    • one year ago
  20. zepp
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    atoms and atoms simplify out, leaving the mole on both sides.\[\large \frac{2.04*10^{20}}{6.02*10^{23}}\text{mole}=x \text{ mole}\]

    • one year ago
  21. wondering_
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    ok i see

    • one year ago
  22. wondering_
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    i got .0003 correct?

    • one year ago
  23. zepp
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    You'll get as answer: \(\large x\text{ moles}\) Now you may proceed to transform this using \(\LARGE n=\frac{m}{M}\), where \(\large \text{n}\) is the number of moles, \(\large \text{m}\) is the mass in \(\large \textbf{grams}\) (so you'll need to retransform it into milligrams as you're asked to do) and \(\large \text{M}\) is the molar mass.

    • one year ago
  24. wondering_
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    Molar mass is 46.07?

    • one year ago
  25. zepp
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    Yes

    • one year ago
  26. wondering_
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    well for ethanol it self right?

    • one year ago
  27. wondering_
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    would i multiply that by .0003

    • one year ago
  28. zepp
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    Exact.

    • one year ago
  29. wondering_
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    46.68->46.7 sorry

    • one year ago
  30. wondering_
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    so i got .0140 because i used 46.68*.0003

    • one year ago
  31. wondering_
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    thats just grams right?

    • one year ago
  32. zepp
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    Yep

    • one year ago
  33. wondering_
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    .0140 x 10^-3 ?

    • one year ago
  34. zepp
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    Good, now you'll need to set up another proportion:\[ \frac{\text{Mass of Carbon in 1 mole of Ethanol}}{\text{Molar mass of Ethanol}}=\frac{\text{Mass of Carbon (Unknown) }}{\text{Mass of the Ethanol given to you}}\]

    • one year ago
  35. wondering_
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    (?/46.68)=(?/.0140x10^-3?)

    • one year ago
  36. zepp
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    Mass of carbon in 1 mole of ethanol is a known number, 2*Molar mass of carbon, 12, 2*12 = 24.

    • one year ago
  37. wondering_
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    oh ok i see where you got that from

    • one year ago
  38. wondering_
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    (24/46.68) = (?/.0140x10^-3?)

    • one year ago
  39. wondering_
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    was i correct on the second half too?

    • one year ago
  40. zepp
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    I don't know, I didn't push the values in my calculator, I've shown you how to do these kind of problem, all the other mathematical manipulations/calculations should be done by you; the right method is shown so you can take down any question that looks like this.

    • one year ago
  41. wondering_
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    I was asking about for the mass of the ethanol that the answer i found before right?

    • one year ago
  42. zepp
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    Like I said, I don't know :)

    • one year ago
  43. wondering_
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    Ok so now i simple mult across top and bottom and cancel out to get my answer

    • one year ago
  44. wondering_
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    simply*

    • one year ago
  45. wondering_
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    i think i have it

    • one year ago
  46. wondering_
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    thank you

    • one year ago
  47. zepp
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    np

    • one year ago
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