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anonymous

  • 5 years ago

How many atoms of 12C are in a 3.50 g sample of this particular isotope?

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  1. anonymous
    • 5 years ago
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    12C: 12g 1mol actual: 3.5g xmol x=3.5/12=something (mol) number of atom = something*Avogadro number

  2. anonymous
    • 5 years ago
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    Well there are many ways of doing this, x=3.5/12= unknown thats stands for (mol) but really 3.5g is xmol number of it=unknown. Do you get it?

  3. anonymous
    • 5 years ago
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    Kind of. So for 3.5/12= 482.4?

  4. anonymous
    • 5 years ago
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    These are the options I have 2.11 x 1024 atoms 2.07 x 1024 atoms 8.01 x 1023 atoms 1.52 x 1023 atoms 1.76 x 1023 atoms

  5. anonymous
    • 5 years ago
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    Would it be the first one?

  6. anonymous
    • 5 years ago
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    Yes that is close enough

  7. anonymous
    • 5 years ago
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    This is the answer 1.76 x 1023 atoms.

  8. anonymous
    • 5 years ago
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    the last one is right. You calculate x then take the result times avogadro constant

  9. anonymous
    • 5 years ago
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    Can you guys help me out with this one? A sample of (N2H5)2C3H4O4 contains 1.084 ×1024 carbon atoms. How many moles of hydrogen atoms are in the same sample?

  10. anonymous
    • 5 years ago
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    Yeah, I got that.

  11. anonymous
    • 5 years ago
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    It's 10^24*

  12. anonymous
    • 5 years ago
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    I know man lol

  13. anonymous
    • 5 years ago
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    this is dead easy. can you count how many H and how many C in the formula?

  14. anonymous
    • 5 years ago
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    14 H and 3 C

  15. anonymous
    • 5 years ago
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    good, now do some rough work H C 14 3 x atoms 1.084 *10^24 atoms so you can find x=1.084*10^24*14/3=blabla number of mol of H = blabla/Avogadro constant

  16. anonymous
    • 5 years ago
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    ~8.4 mol I'd say

  17. anonymous
    • 5 years ago
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    That's what I got but it's not in the choices I have. These are the choices! 34.7 g 1.56 × 10-21 g 937 g 6.43 × 1020 g 31.2 g

  18. anonymous
    • 5 years ago
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    those choices are in g, not in mol, I doubt that's not the right choices for this qs

  19. anonymous
    • 5 years ago
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    I think they are because it's online through a school program.

  20. anonymous
    • 5 years ago
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    Do I have to convert it to mol to grams now?

  21. anonymous
    • 5 years ago
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    Never mind, sorry. I mixed up a question. This is the right answer.

  22. anonymous
    • 5 years ago
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    lol so it's correct yeah? my brain is not really getting old lol

  23. anonymous
    • 5 years ago
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    Do you mind helping me with more?

  24. anonymous
    • 5 years ago
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    Haha...

  25. anonymous
    • 5 years ago
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    ok, just 1 more.

  26. anonymous
    • 5 years ago
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    The mass of 5.20 mol of glucose, C6H12O6 is? A-34.7 g B-1.56 × 10-21 g C-937 g D-6.43 × 1020 g E-31.2 g

  27. anonymous
    • 5 years ago
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    I get 482.2g

  28. anonymous
    • 5 years ago
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    It's 482.4g. But it doesn't make sense.

  29. anonymous
    • 5 years ago
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    you need to find the mass formula for glucose first. 6mols of C, 12mols H, 6mols O MC=12 MH=1 MO=16 so 6*12+12*1+6*16=180(g/mol) there are 5.2 mols so mass m=180*5.2=936g

  30. anonymous
    • 5 years ago
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    Oh... ok got it. I'm making harder than it is :P

  31. anonymous
    • 5 years ago
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    you know it's good to remember the mass of 1 mol of some common substance. I've done these in secondary school and I still remember them now. I'd say that's big advantage. e.g H=1, O=16, N=14, Cl=35.5, C=12... those commonly used you should remember. Save you lots of time in exam.

  32. anonymous
    • 5 years ago
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    I actually do remember these basic ones. Thanks for the tip. If you have any other tips, shoot'em.

  33. anonymous
    • 5 years ago
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    lol I wasn't that great at chemistry though. remember this too K - Na - Ca - Mg -Al - Zn - Fe- Ni - Sn - Pb - H - Cu - Hg - Ag- Pt - Au That's extremely helpful when you deal with acid and base And always try to make a table of information first, make the problem clearer. And don't get scared of those Chem teachers. When they die, they'll just turn into HCON, no more no less =))

  34. anonymous
    • 5 years ago
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    good luck. I'm off to bed.

  35. anonymous
    • 5 years ago
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    Thanks a lot :)

  36. anonymous
    • 5 years ago
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    no problem man

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