chemistry

So the question I have is:
Calculate the pressure of dry hydrogen collected. (Use the water temperature).

And the evidence that we obtained was:

Length of magnesium
3.8Cm
Mass of 1 m of magnesium
1.254g
Volume of hydrogen collected
67mL
Air temperature
27C
Water temperature
26C
Air pressure
95.97kPa
Vapour pressure of water
3.36kPa

I really don't understand it, thank you!

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  1. Remember Dalton's Law of Partial Pressures; i.e., the total pressure of a system is the sum of the partial pressures of each gas. So
    Total P = pH2O vapor + pH2(dry) gas.
    You know total pressure--that's air pressure of 95.97 kPa. You are given the pH2O at that temperature of 3.36 kPa. That lets you calculate the pH2. That will be the pressure of dry H2O since you've subtracted the pressure of the water vapor.

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  2. Thank you! (: I'm also confused about the 2nd question:

    Correct your dry hydrogen volume to a volume at STP (Assume the temperature of the hydrogen in your collection tube was the same as the air temperature).

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  3. I thought that would be the next question. :-)
    Use P1V1 = P2V2. That will give you the volume at STP if you plug in values for STP of 273 K and 101.325 kPa.

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  4. Sorry for all the questions. So the equation I ended up with was:

    101.325 x 0.067 / 95.97 = V2

    But I'm assuming that's not right, so instead of x0.067 would it be x 273K?

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  5. No but that's my mistake. I didn't write all of the equation.
    It should be (P1V1/T1) - P2V2/T2)
    Substitute and solve for V2.

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