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amandax014

  • one year ago

how many molecules of glucose are in a 2.0 mL of a 1.02 mM solution of glucose

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  1. aaronq
    • one year ago
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    find the amount of moles using the formula for molarity: Molarity = moles/Liters of solution then multiply the number of moles by avogadros number

  2. aaronq
    • one year ago
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    ps. convert to mM to M. 1 mM = 0.001 M

  3. amandax014
    • one year ago
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    so calculate the mass C6H12O6? then divide that by 1.02?

  4. aaronq
    • one year ago
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    no, you don't need the molar mass since you're not converting to grams

  5. amandax014
    • one year ago
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    i'm just lost on the moles then how do i find that

  6. aaronq
    • one year ago
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    with the molarity and the volume of solution given

  7. aaronq
    • one year ago
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    molarity is a form of expressing concentration (in terms of moles per liter).. so basically by multiplying the molarity by the volume, you find the moles but you have to covert to molarity because mM meaning millimolar, 1 molar = 0.001 millimolar

  8. amandax014
    • one year ago
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    so like \[\frac{ 2mL }{ 1mm} X \frac{ 1mm }{ 0.001 } ?\]

  9. aaronq
    • one year ago
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    nope convert volume to liters .. 2mL = 0.002 L convert millimolarity to molarity 1.02 mM = 0.00102 M 0.00102 M x 0.002 L =

  10. amandax014
    • one year ago
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    2.04X10^-6

  11. aaronq
    • one year ago
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    looks right, i don't have a calculator around to check though

  12. amandax014
    • one year ago
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    ok thank you so much

  13. aaronq
    • one year ago
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    no problem

  14. aaronq
    • one year ago
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    ps you still need to multiply by avogadros number

  15. amandax014
    • one year ago
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    yea i figured that I'm trying to get a better number I would take that equations answer 2.04 ... and multiply it by avogadro's number right

  16. aaronq
    • one year ago
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    yep, it's 2.04x10^-4 though don't forget

  17. amandax014
    • one year ago
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    why 10^-4 i got -6

  18. aaronq
    • one year ago
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    sorry -6 lol my mistake

  19. amandax014
    • one year ago
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    haha its ok thank you again

  20. aaronq
    • one year ago
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    lol no prob, good luck

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