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anonymous

  • one year ago

acetic acid 0.921% (w/w) 100grams. The acetic acid will be measured by weight. how much is used?

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  1. Jhannybean
    • one year ago
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    What does (w/w) mean?

  2. anonymous
    • one year ago
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    weight in weight

  3. anonymous
    • one year ago
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  4. Jhannybean
    • one year ago
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    @aaronq come in here. So im guessing it means : \[\sf \frac{0.921 ~g~C_2H_3O_2}{100~g~comp}\]

  5. anonymous
    • one year ago
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    i sent the actual question above. with 4 answers..

  6. aaronq
    • one year ago
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    acetic acid 0.921% (w/w) 100 grams. means that there is 0.00921*100 g =0.921 g of acetic acid and 100 g - 0.921 g = 99.079 g of solvent

  7. Jhannybean
    • one year ago
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    oh i see how it works now

  8. aaronq
    • one year ago
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    from that picture posted, it's A and C, because of the units

  9. aaronq
    • one year ago
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    okay, coooool

  10. anonymous
    • one year ago
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    so its a and c? why?

  11. aaronq
    • one year ago
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    because of the units, 0.921 g = 921 mg

  12. anonymous
    • one year ago
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    ok. thank you so much.. both of you ..

  13. anonymous
    • one year ago
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    Are you willing to help me understand other questions?

  14. aaronq
    • one year ago
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    no problem guyses, idk if you're asking me, but if so, you should post in the chem section

  15. anonymous
    • one year ago
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    im asking either of of you

  16. anonymous
    • one year ago
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    if not, would you really consider this chem?

  17. aaronq
    • one year ago
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    yeah

  18. anonymous
    • one year ago
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    its actually pharmacy compounding math

  19. aaronq
    • one year ago
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    pharmacy is based from chem and bio

  20. anonymous
    • one year ago
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    are you willing to help me understand a few other questions ?

  21. aaronq
    • one year ago
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    yep

  22. anonymous
    • one year ago
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    ok thank you i will post the next question

  23. aaronq
    • one year ago
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    kk

  24. anonymous
    • one year ago
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  25. anonymous
    • one year ago
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    if possible explanations please.

  26. anonymous
    • one year ago
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    I am trying to get some of these done before work tomorrow

  27. anonymous
    • one year ago
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    along with understanding them

  28. aaronq
    • one year ago
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    so, v/v means volume by volume, it's pretty straight forward that in 100 mL of 50 % v/v glycerin solution there is 50 mL of glycerin.

  29. anonymous
    • one year ago
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    what about the second question

  30. anonymous
    • one year ago
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    i should have just read that one. that was common sense but im old..

  31. anonymous
    • one year ago
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    is the answer to 17. 6.25ml?

  32. aaronq
    • one year ago
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    sorry, i had to step away from the keyboard. yes, that's the answer. you basically just have to watch the units

  33. anonymous
    • one year ago
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    its ok . i am trying to do this, which most has nothing to do with what i do at work. and thank you so so much

  34. anonymous
    • one year ago
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  35. anonymous
    • one year ago
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    here is another one if you can help thank you

  36. aaronq
    • one year ago
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    which one?

  37. anonymous
    • one year ago
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    i need to do all 3 questions? so what ever you can help with

  38. anonymous
    • one year ago
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    well one is d

  39. anonymous
    • one year ago
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    so 2 and 3

  40. aaronq
    • one year ago
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    b) is 400 mg 0.02*20 g =0.4 g = 400 mg

  41. anonymous
    • one year ago
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    wow thank you.. at least i can see it now

  42. aaronq
    • one year ago
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    for c) you add the whole masses and make a dilution \(2\%*20g=x\%*(20~g+30~g+30~g)\rightarrow x\%=0.5\%\)

  43. aaronq
    • one year ago
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    lol sorry that sounds to weird, i meant "add all the masses" not "the whole.." lol

  44. anonymous
    • one year ago
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    add all the grams

  45. aaronq
    • one year ago
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    yep

  46. aaronq
    • one year ago
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    grams is a unit of mass

  47. anonymous
    • one year ago
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    you are a great help.. can you help with a few more? i dont want to over welcome my help

  48. aaronq
    • one year ago
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    I would but i have to go do laundry and stuff. Hopefully someone else can come help ya, if you can't find help here, try the chem section.

  49. anonymous
    • one year ago
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  50. anonymous
    • one year ago
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    ok thank you so much

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