Hi Albert! Yes, with MSN Groups closing, I'll either have to give up doing a chemistry help site, or move to another service. The MSN groups can be moved to the Multiply site, but it does not have the message board format like Groups does, so I am considering other options like Yuku or Aimoo that are better suited for this purpose. Well, it was fun while it lasted! There are some other good chemistry groups out there, so I'm also considering just joining one of those.
Meanwhile, let's see what you've got here....
1. First you need to calculate the moles of N2 in 70 L at 1 atm and 20°C. Use PV = nRT for that, to get n = 2.91 mol. Then, look at the reaction to see how many moles of NaN3 are needed. From the coefficients in the balanced reaction, we can see that moles of NaN3 will be 2/3 times the moles of N2. Finally, multiply moles of NaN3 by its molar mass (65.01 g/mol) to get grams of NaN3. I got 126 g.
2. The work w = –PoppΔV, where Popp, the opposing pressure, is 1000000 Pa. Since 1 atm = 101325 Pa, this is 9.86923 atm. Using atm for pressure and L for volume, –PoppΔV = �?9.86923 atm)(0.020 L) = �?.197385 L atm. We can covert to joules using the coversion 1 L atm = 101.325 J:
�?.197385 L atm X 101.325 J / (L atm) = �?0 J. 
Free Forum Hosting
Important Announcement
Prev Message
