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anonymous

  • one year ago

A sample of gas with a mass of 8.40 g occupies a volume of 4.50 L at 298 K and 1.10 atm. What is the molar mass of the gas? Show all work used to find your answer. *Best answer gets medal, fan and testimony*

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  1. aaronq
    • one year ago
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    Use the ideal gas law, substitute in the definition of moles, solve for M, molar mass. Ideal gas law: \(\sf PV=nRT\) moles: \(\sf n=\dfrac{m}{M}\) \(\sf \huge PV=\dfrac{m}{M}RT\rightarrow M=\dfrac{mRT}{PV}\)

  2. anonymous
    • one year ago
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    May you help me plug in the variables?

  3. aaronq
    • one year ago
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    I can check it for you if you give it a try

  4. anonymous
    • one year ago
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    Ok. Give me a second.

  5. anonymous
    • one year ago
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    I have: 1.10*4.50 L = n(0.08206atm)(298K) And that's all I can do. Unfortunately.

  6. aaronq
    • one year ago
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    that's right! you can now solve for n, then divide the mass 8.4 g by the moles.

  7. aaronq
    • one year ago
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    the mass is given in the question, it's 8.4 grams

  8. anonymous
    • one year ago
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    So it would be: 1.10 atm * 298 K = 24.45388 8.4 g / 24.45388 = 0.3435

  9. aaronq
    • one year ago
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    not quite, first solve this equation: 1.10*4.50 L = n(0.08206atm)(298K) divide the mass by the answer (moles)

  10. anonymous
    • one year ago
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    4.95 = n(0.08206*298) 4.95 = n(24.45388) 4.95/24.45388 = 0.202421... n = 0.202421

  11. aaronq
    • one year ago
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    okay, looks good! now divide the mass by n

  12. anonymous
    • one year ago
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    244.5398

  13. aaronq
    • one year ago
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    There you go

  14. anonymous
    • one year ago
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    WOOHOO! How would I properly write with sig figs?

  15. aaronq
    • one year ago
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    all the numbers used have 3 sig figs, so make the answer have 3 figures.

  16. anonymous
    • one year ago
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    244. Thank you so much! You're a life saver :D

  17. aaronq
    • one year ago
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    that's not quite right. look at the fourth digit, you have to round up

  18. anonymous
    • one year ago
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    Oh... 245, correct?

  19. aaronq
    • one year ago
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    yeah. you should give this a look https://www.physics.uoguelph.ca/tutorials/sig_fig/SIG_dig.htm

  20. anonymous
    • one year ago
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    Will do.

  21. anonymous
    • one year ago
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    Got it. Thanks :)

  22. aaronq
    • one year ago
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    no problem!

  23. anonymous
    • one year ago
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    @aaronq there is problem in this question

  24. aaronq
    • one year ago
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    whats the problem?

  25. anonymous
    • one year ago
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    should the answer be around 41 to 42 because we are dividing 8.4 by n

  26. anonymous
    • one year ago
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    @aaronq

  27. aaronq
    • one year ago
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    you're right, the answer is 41.4974933333333333 I guess this person plugged the numbers into the calc wrong

  28. anonymous
    • one year ago
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    oh thank so much

  29. aaronq
    • one year ago
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    np!

  30. aaronq
    • one year ago
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    thanks for double checking lol

  31. anonymous
    • one year ago
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    np!

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