How much liquid nitrogen do I need to fill up a pressure vessel with 104 cu.meters of nitrogen gas at 5.2 barg
453509
i'm not familiar with the unit barg, but just use the ideal gas equation PV=nRT solved for n so n=(PV)/(RT) with an apropriate value for the universal gas constant (R).  that will give you the number of moles of N2 which you can then determine the volume under standard conditions if you wish using the same equation.
558084
Start with the ideal gas law: PV=nRT.  Since no gas enters or leaves the system, n is constant, and R is the universal gas constant, the product nR has the same value at the beginning and as at the end.  So, collect it to one side, and move the other factors to the other side.\n\nP1 * V1 / T1 = nR = P2 * V2 / T2\n\nSolve for T2\n\nT2 = T1 * (P2 * V2) / (P1 * V1)\n= 300e4 K (.0003e15 atm * .046e-15 cc)/(50e3 atm * .0004e-7 cc)\n=2.07 * 10^7 K
12890
Generally speaking it's not hard to buy liquid nitrogen from a welding supply shop, gas supplier or cryogenics company, but you have to bring your own Dewar and they won't fill anything other than the proper kind (at least they shouldn't). You might find a welding shop that rents Dewars, or you can buy one for about $300. You can contact your nearest gas supplier or cryogenics company. Price depends on quantity but for smaller quantities it usually costs around three dollars a litre.
144053
Pay more attention in class! And assuming it is an ideal gas, use PV=nRT. Since there is not change in volume you can basically drop it from the equation and make P=nRT. If you like algebra you can leave it in but it will cancel out eventually. You could solve for 'n' first using initial P and T, and then plug and chug.
940335
In your crucible, you have a mixture of mostly MgO and some Mg3N2.  That combined mass is 0.470 g.\n\nWhen you add water, all you are really adding is oxygen, since the hydrogen leaves as part of the ammonia gas.\n\nBasically, you are adding oxygen (atomic weight 16) and removing nitrogen (atomic weight 14).\n\nBecause of the formula Mg3N2(s) + H2O(l) -> MgO(s) + NH3(g),\nyou are 'taking out' 2 nitrogen atoms for every 3 oxygens you add (because there are 2 nitrogens to every 3 magnesium atoms, and each magnesium atom uses 1 oxygen to make MgO)\n\nSo you are gaining (0.486 - 0.470) = 0.016 g from the reaction of Mg3N2 to MgO\n\nThe mass from adding the oxygen is 16 g/mol * (3) * X\nThe mass from taking out the nitrogen is -14 g/mol * (2) * X\n\nWhere X is the number of moles of Mg3N2\n\nSo, 16 g * 3 * X - 14 g * 2 * X = 0.016 g\n\nSolving for X, X = 0.016 / (48 - 28) = 0.0008 mol\n\nThe molecular weight of the Mg3N2 is 100.9 g/mol, so the mass of Mg3N2 is 100.9 g/mol * 0.0008 mol = 0.0807 g\n\nSo, the mass percent is .0807 / 0.470 = 17.2 % Mg3N2\n\n.............\n\nAn another note, one reason you might be confused is your second reaction:\n\nThere are actually 2 reactions going on:\nMg + 1/2 O2 -> MgO\n3Mg + N2 ->Mg3N2\n\nThe first reaction reacts much faster than the second, meaning that more MgO will be produced than Mg3N2.\n\nHope this helps.
452634
N2 +3H2 ---> 2NH3\n\nso, H2 Mwt is 2g/mol and NH3 Mwt is 34g/mol\nas per above eqn, 6g of H2 will give u 34g of NH3\nSo, 9.988g of H2 will give 56.6g of NH3\n\nYou got it?
600524
Give the mass of one reagent in a reaction, and ask for the mass of another. A word equation or chemical equation must be provided.\n\neg:\n\nWhat mass of nitrogen gas is required to produce 45.0 g of ammonia, according to the reaction below ?\n\n         N2   + 3 H2   ---> 2 NH3\n\n\nTo solve the problem\n\na) convert mass of NH3 to moles (2.65 mol)\nb) use stoichiometry - find moles of N2 needed (1.32 mol) \nc) convert moles of N2 to mass (37.0 g)
354060
0.2 gal. or 25.6 fluid oz.
750464
My chemistry is pretty rusty, but let's see.\n\nI'll assume that you have an infinite amt of O2, and that you have 68.12 grams of NH3.\n\nThe equation isn't balanced, so you need to do that first (you have 3 hydrogens on one side, but two on the other).  Balance to get:\n\n2(NH3) + (2.5)(O2) --> 2NO + 3(H20)\n\n4(NH3) + 5(O2) --> 4NO + 6(H20)\n\nNow, given that you have 68.12 grams of NH3, you need to convert this into moles of NH3, as Stoichiometry deals with molar ratios.  To get the conversion ratio, look at the periodic table.  Under each element, there should be an atomic mass.  For oxygen is 16.something, Hydrogen 1.008 ( i think ).  Find it for N and H.  (this is where I get hazy on the math).  I believe: take 3 * atomic mass of hydrogen + atomic mass of nitrogen, let's call it M  (it should be in grams/mole)  So to get moles, multiply 68.12 * (1/M).  \n\nNow, our equation above shows that for every 4 moles of NH3, we get 6 of H20.  # moles NH3 * (6/4) = # moles of H20.  Now, do the reverse operation of H20 to get how many grams is produced.  H20 has an atomic mass of something like 18.016.  So, # moles H20 * 18.016 = # grams H20\n\n\nI hope that's somewhat correct, and if not, leads you in the right direction at least
39906
Liquid nitrgen absorbs heat as it boils away. Its boiling temperature at normal atmospheric pressure is about -270 deg F. Almost all mechanical refrigeration uses a boiling substance (refrigerant) to absorb heat. \n\nREK Mechanical engineer
889941
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
755048
The ideal gas equation is pV = nRT.\n\nSo if you double n on one side, you must double something on the other. In your case it is p.
766934
Assume it is an ideal gas and use PV = nRT\n\nP is in atm 1 atm = 760 torr\n    225 torr / 760 torr = 0.296 atm\nV is in L so 5.7 L\nR is 0.0821 atm*L/mol*K\nT is in K so C + 273 = K\n     27 + 273 = 300 K\n\n(0.296)(5.7) = (0.0821)(300)n\nn = 0.0685 mols
942319
This is what you should be looking at:\n\nHCl(gas) + NH3(gas)  -->  NH4Cl (solid)
249344
If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.\n\nWe solve this problem by first writing the equation\nN2 + 3H2 = 2NH3\nN2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g\nso 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent.
856026
Use the gas law\nPV=nRT
917667
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!