Mimi_x3
  • Mimi_x3
At room temperate and pressure, 1.0 L of helium gas has a mass of 0.16g. Under the same conditions, what is the mass of 1.0L of hydrogen gas? Can someone pls explain step by step (:
Chemistry
schrodinger
  • schrodinger
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anonymous
  • anonymous
I'm not claiming to be right here but follow me on this. The formula for Pressure is: \[PV=nRT\] The items on the right can be brought to the left leaving 1 on the right. The same ratios could be done for hydrogen, watch: \[\frac{P_1V_1}{n_1R_1T_1}=1=\frac{P_2V_2}{n_2R_2T_2}\] We have stated that the conditions are the same so T_1 will equal T_2. P_1 will equal P_2. R_1 will equal R_2. From the problem V_1 will equal V_2. Rewriting the formula with this in mind: \[\frac{PV}{n_1RT}=\frac{PV}{n_2RT}\] If we cross multiply and cancel like values we get: \[n_1=n_2\] So we know that the moles of atoms are the same. Remember that both helium and hydrogen are diatomic so any calculations of mass will be based on 2 atom molecules. \[0.16gHe_2\frac{1 mol He_2}{8.0052gHe_2}=0.020 mol He_2\] We have the formula above so we know we have 0.020 mol of H_2 as well. \[0.020molH_2\frac{2.0158gH_2}{1molH_2}=0.040gH_2\] Note we are following rules of SF so don't drop the trailing 0.
Mimi_x3
  • Mimi_x3
wow~ tyvm again (: It's a multiple choice question and there's 0.04 that is B , so I assume that it's right
anonymous
  • anonymous
Np :) One thing I recall about pressure based problems is that they are solvable if you remember the pressure formula. You can rearrange it as you like and cancel like values.

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Mimi_x3
  • Mimi_x3
Lols, I haven't learn the pressure formula yet =/
Mimi_x3
  • Mimi_x3
learnt about*
anonymous
  • anonymous
They must be using something simpler then. You will learn about it later then. R is a constant that varies depending on the units.
Mimi_x3
  • Mimi_x3
Yeah idk, I think that there's an easier way since its a multiple choice
anonymous
  • anonymous
The other thing to pay attention to with gases is the diatomic molecules. You can end up with incorrect values if you use mono-atomic calculations.
Mimi_x3
  • Mimi_x3
lols, I haven't learn about gases, diatomic molecules, mono-atomic calculations =/
Mimi_x3
  • Mimi_x3
learnt*
anonymous
  • anonymous
If you look closely at the values in the formula you'll note that the pressure in a system when all else is equal is based on the number of molecules, not the size.
Mimi_x3
  • Mimi_x3
Um okay, ty (:
anonymous
  • anonymous
Which may be where the question is coming from. They state that all else is equal so you're left with the number of molecules being equal which is the moles calculation at the end.
anonymous
  • anonymous
Think of it this way. Imagine the molecules of gas bouncing around inside a balloon. The prssure on the balloon is based on how many molecules are pushing on it, not how hard or fast they are bouncing.
anonymous
  • anonymous
Clear as mud I am sure. Just stash it away for later. It will make more sense when you hit those chapters ;)
Mimi_x3
  • Mimi_x3
Yeah xD That will be next term then, i don't really understand chem hopefully i will pass my chem exam xD
anonymous
  • anonymous
Best of luck. I still have to take Chem 2 in the Spring myself :)
Mimi_x3
  • Mimi_x3
Ty, what's Chem 2?
anonymous
  • anonymous
Second class in the college Chemistry lineup. Not sure where you're at but talking US.
Mimi_x3
  • Mimi_x3
Cool, your in College. Wonder why I don't know what's chem 2 because I live in Australia xD
anonymous
  • anonymous
Like any other class you've taken. These all build on top of each other. Yopur knowledge of moles are the the building blocks for things like the gas questions. I started to get lost when it came to molecule shapes. The next class is going to be interesting ;)
Mimi_x3
  • Mimi_x3
Oh cool (: Hopefully I would understand the topic next term

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