I am looking for a ppb level for N2 and Ar in Bulk Gases (H2, O2, Ar, He and N2)analyzer.
209028
It probably requires a mass spec technique.  Infrared might work.  Your detection level will be the problem.
442187
N2O would be a good guess.
157104
Try here:\n\nhttp://www.valueoptions.com/provider/contractspecific/voc/monitoring.pdf
475316
The first answer does give an indication, but it is not a definitive test.  Take a sample of the substance, drop acid on it as before, but hold a wet piece of blue litmus paper over the bubbles.  If the litmus turns red, then the gas is CO2, and the substance is a carbonate.
283565
Chemexper.com is a great place to look this up and you can browse various companies to buy from as well as search by the specs that you need.
370199
yes, PV=nRT\n\nyeah, 5.8g / 44g = # of moles\n\n^the guy above me is talking about milli-moles... but your R value is probably given in moles
487292
use dalton's law to get the pressure of CH4(total pressure - H2 pressure)\nthen use ideal gas law to get the moles of CH4(PV=nRT)\n*watch your units*\nchange moles of CH4 to grams of CH4(moles x molar mass)
319535
This is a stochiometry problem. You first want to write out a balanced equation. Then, determine what is your limiting reagent. From your values, N2 is the limiting reagent. It does not matter how much of the H2 you have, because once the N2 is depleted you will not be able to form any more product (ammonia gas). So, you have 0.07 moles of N2 x 2 moles of NH3 divided by 1 mol of N2 (this is because you have 2 moles of NH3 for every mol of N2. So now you have 0.14 moles of NH3 x 17 g NH3/mol NH3 = 2.38 g of NH3
246053
This is a test of the basic principle of partial pressures.  The pressure of each gas is independent of the other gasses.  If you fill a 2 liter container with Neon to 2 atm, the neon will exert 2 atm no matter what other gas you put in the container.  This means that if the N2 originally exerted 4 atm, it will exert a 4 atm partial pressure when combined with the other gasses and the overall pressure will be 2+3+4 = 9 atm.
654632
The easiest way to find out the answer to that question is to look up the MSDS.  It gives all the information you need about it's toxicity, reactivity, etc.  A lot of chemical supply companies have MSDS on their websites, and I think there might be a government website that has MSDS accessible online, but I'm not sure.
326149
Here is their products page.\n\nhttp://www.cpchem.com/enu/products.asp
58999
This is a limiting reagent problem. You must calculate the moles of NH3 produces from the given amount of grams of N2 and H2.\n\n3.00g N2 x 1mol/28.02g x 2 mol NH3/1 mol N2 = .214 mol NH3\n\n.54g H2 x 1 mol/2.02g H2 x 2 mol NH3/3 mol H2 = .178 mol NH3\n\nFrom this calculation you now know that the H2 is the limiting reagent. The next part to calculate the grams of NH3 goes like this:\n\n.178 mol NH3 x 17.04g/1 mol = 3.03 g NH3
483240
Use the state equation of the ideal gases and solve for volume V:\n\np*V = n*R*T, V = n*R*T/p\n\nNow p = 657 Torr = 657/760 atm = 0.86 atm and R = 0.082 L*atm/mol*K, T = 283.8 oC = (283.8 + 273) K = 556.8 K so:\n\nV = 2.82*0.082*556.8/0.86 = 148.940 L
654670
Normal air is composed of about 78% nitrogen, 21% oxygen, .93% argon, and around .03-.05% carbon dioxide, among other gases.\n\nGood luck!
441298
Ideal gas law states:  PV = nRT\nSolving for pressure, P = (nRT)/(V)\nR = 0.0821 L-atm/mol-K\n\n\nNumber 76\n\nP = (nRT)/(V)\nP = [(2.3mol)(0.0821 L-atm/mol-K)(294K)] / (0.15L)\nP = 370.11 atm = 370 atm (observing sig figs)\n\n\nNumber 78\n\nFirst determine # of moles in 15.0g of CO2\n(15.0g)/(44 g/mol)=0.34 mol\n\nThen use the manipulated ideal gas law equation to determine pressure\nP = (nRT)/(V)\nP = [(0.34mol)(0.0821 L-atm/mol-K)(310K)] / (0.30L)\nP = 28.84 atm\n\nNow convert the pressure from atm to mm Hg (1atm=760 mm Hg)\n(28.84 atm)(760 mm Hg/atm) = 21918.4 mm Hg (22000 mm Hg observing sig figs)
550844
ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
383866
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l