what is the chemical formula of liquified natural gas(LNG) and compressed natural gas(CNG)?
552992
ch4 for both.\nOne carbon atom and 4 hydrogen atoms. Simple!
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.
795424
First, use the ideal gas law PV=nRT to solve for n, the number of moles.  You are given P,V, and T.  R is the gas constant 62.364 mmHg*L/(mol*K).  Now you have the number of moles of the unknown and you were given its mass.  Thus, you can find the molar mass of your unknown.  This will help to determine your molecular formula when you compare this value to that of the empirical formula.\n\nTo determine the empirical formula: Assume a 100g sample.  Then you'll have 65.45 g C, 5.49 g H, and 29.06 g O.  Determine the molar ratios of each compound and divide by the least common multiple.  In this case, you get a ratio of C:H:O of roughly 3:3:1.  So your empirical formula is C3H3O.  Compare the molar mass of the empirical formula (55 g/mol) with the molar mass calculated above.  Multiply the components of the empirical formula by a scalar as necessary to match the molar mass of the unknown.
159750
Commercially:\n=============\nFractional distillation of liquified air. This method involves:\n\n1. Liquifying air which is mostly Oxygen and Nitrogen\n2. Heating liquid air.\n3. Cooling at different temperatures to collect different gases.\n\nOne of the major bi-products of this method is Nitrogen. \n\nLaboratory:\n===========\nHeating some oxide in presence of a catalyst or electrolysis.
286140
96% Carbon Dioxide\n4% Nitrogen\n\n(closer to 3.5% nitrogen with assorted other gasses making up the last 0.5%)
445028
Carbondioxide gas, Oxygen gas and Nitrogen gas
873085
88.02 g of CO2 are equivalent to 88.02/44 moles of CO2 = 2 moles\nAccording to the equation 1 mol LiOH reacts with 1 mol CO2,\n\n2 moles CO2 will require 2 mol LiOH.\n\nTo find out the gram amount multiply 2 moles of LiOH by its molecular weight\n\n2 x 23.95 = 47.9 grams LiOH.
912510
That is the molecular formula for nitrous acid, written HNO2.
46596
Use mole concept. Heres how it's done:\n1)Neon's atomic number's 10 and it has acomplete orbit. therefore it's molecular formula is Ne.\n2)1 mole of Ne weighs 20.18g&i mole is 22.4 l\n3) use the following equation and then cross multiply:\n\n    22.4   litres    weigh          20.18 a.m.u\n     2     litres    weigh          x a.m.u.\n\nNow ur equation is 'x'= (2 X 20.18)22.4\nwhich is equal to 1.801 gms or a.m.u
1486
Usually the composition f natural gas is as follows:\nComponent   Typical Analysis(mole %)  Range(mole %)\nMethane          94.9                  87.0 - 96.0\nEthane           2.5                   1.8 - 5.1\nPropane          0.2                   0.1 - 1.5 \niso - Butane     0.03                  0.01 - 0.3\nnormal - Butane  0.03                  0.01 - 0.3\niso - Pentane    0.01                  trace - 0.14\nnormal - Pentane 0.01                  trace - 0.04\nHexanes plus     0.01                  trace - 0.06\nNitrogen         1.6                   1.3 - 5.6\nCarbon Dioxide   0.7                   0.1 - 1.0\nOxygen           0.02                  0.01 - 0.1\nHydrogen         trace                 trace - 0.02\n  \nSpecific Gravity 0.585                 0.57 - 0.62\nGHV(MJ/m3), dry* 37.8                  36.0 - 40.2
981642
First you have to find the empirical formula.\nThe compound has only C and H, so the percentage of H is 100-85.62= 14.38%\nDivide each percentage by the respective atomic mass\nC: 85.62/12 =7.135\nH: 14.38/1= 14.38\nDivide with the smallest number\nC: 7.135/7.135=1\nH: 14.38/7.135=2.01=2\n\nTherefore the empirical formula is CH2. This means that the molecular formula is (CH2)x or CxH2x and the moleular weight\nMW=12x+2x=14x. We need to find x in order to find the molecular formula.\n\nPV=nRT and since n=mass/MW\nPV=massRT/MW= massRT/14x =>\nx=massRT/(14PV) =>\nx =2.58*0.082*324 / (14*0.98*1)= 4.99=5\nTherefore the gas is C5H10
777427
C:H = 85.63:14.37 (mass ratio)\n= 85.63/12.01:14.37/1.008 (molar ratio)\n= 7.13:14.26\n= 1:2\nSo the empirical formula is CH2 and its empirical molecular weight ≈ 14.03\n\nP1V1/T1 = P2V2/T2\n0.980 * 1/(273.2 + 51.7) = 1.V2/273.2 (at STP of 1 atm and 0°C)\nSo V2 = 0.980 * 273.2/324.9\n= 0.824 L\nThus 3.09 g occupies 0.824 L\nSo 3.09 * 22.4/0.824 g occupies 22.4 L at STP\nie 1 g mole = 84\nThus MW of this gas is 84\nSo number of empirical units = 84/14.03 ≈ 6\nSo the molecular formula is C6H12
942319
This is what you should be looking at:\n\nHCl(gas) + NH3(gas)  -->  NH4Cl (solid)
786324
i would guess D ethanol ch3ch2oh \nc is a gas propane\nand b diethyl ether is more volitile the d
444489
Yes the two are different in their chemical compositions. Liquid petrolium gas, commonly known as LPG is a derivative of petroleum and is generally used as cooking gas.
940427
Convert Liters of CH4 into moles of CH4 with PV=nRT. n= (2 L x 1 atm)/(.082058 x 273.15 K) = .08923 moles CH4. Then, using the mole ratio of 2 moles of O2 per 1 mole of CH4, you multiply .08923 by two to get .17846 moles O2. Plug that into the PV=nRT. V= (.17846 x .082058 x 273.15 K)/1 atm, which equals 4.
787380
PV=nRT\nP=pressure\nV=volume\nn=moles\nR=constant(.08026atm*L/mol*K)\nT=temperature\n\nThis is the ideal gas law.