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anonymous

  • one year ago

How many liters of fluorine gas can react with 32.0 grams of sodium metal at standard temperature and pressure? Show all of the work used to find your answer. 2Na + F2 yields 2NaF

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  1. anonymous
    • one year ago
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    I have: 32 g NaF x \[\frac{ 1 mol NAF }{ 41.98 g NaF}\] x\[\frac{ 1 mol F _{2} }{ 2 mol NaF }\] x \[\frac{ 1 L F _{2} }{ 1 mol F _{2} }\] = .381

  2. anonymous
    • one year ago
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    Is that correct?

  3. anonymous
    • one year ago
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    @gabylovesu

  4. anonymous
    • one year ago
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    @karatechopper

  5. anonymous
    • one year ago
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    @mathstudent55

  6. Vocaloid
    • one year ago
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    @FutureVet one small mistake the last step should be 22.4L F2/1 mol F2, because 1 mol of gas at standard temperature and pressure occupies 22.4L

  7. Vocaloid
    • one year ago
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    other than that, everything else looks ok

  8. anonymous
    • one year ago
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    Thanks I realized that before I Submitted it.

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