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amandax014

  • 2 years 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
    • 2 years 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
    • 2 years ago
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    ps. convert to mM to M. 1 mM = 0.001 M

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

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

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

  6. aaronq
    • 2 years ago
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    with the molarity and the volume of solution given

  7. aaronq
    • 2 years 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
    • 2 years ago
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    so like \[\frac{ 2mL }{ 1mm} X \frac{ 1mm }{ 0.001 } ?\]

  9. aaronq
    • 2 years 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
    • 2 years ago
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    2.04X10^-6

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

  12. amandax014
    • 2 years ago
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    ok thank you so much

  13. aaronq
    • 2 years ago
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    no problem

  14. aaronq
    • 2 years ago
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    ps you still need to multiply by avogadros number

  15. amandax014
    • 2 years 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
    • 2 years ago
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    yep, it's 2.04x10^-4 though don't forget

  17. amandax014
    • 2 years ago
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    why 10^-4 i got -6

  18. aaronq
    • 2 years ago
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    sorry -6 lol my mistake

  19. amandax014
    • 2 years ago
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    haha its ok thank you again

  20. aaronq
    • 2 years ago
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    lol no prob, good luck

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