What is the conversion from Nm3 to litres?
816566
Nm3 means nominal meter cube means quantity (volume) equivalent of 1 m3\ncm=centimeter\ncc= cubic centimeter.\n1 Litre= 10x10x10 cm3=1000 cc\n1 m3 = 100x100x100 cm3=1000,000 cc\ntherefore 1m3=1Nm3=1000 Litres
428451
1 cubic foot is 28.32 litres\n\nMultiply 2000 by 28.32\n\nMinutes is in the denominator of both units, so no conversion is needed for that aspect.\n\nKeep in mind that if you're working with gases that gases are compressible.
977807
M = moles/L\nso, 0.220M = 0.220 moles per liter and there are 2 liters so there are...\n0.440 moles of Mg(NO3)2\n\nBUT... notice that there are 2 ions of NO3 for each molecule, so double it again and get 0.880 moles of NO3
585074
Generally its between 1.5 and 2.5 litres
384394
Well, let's see:  you're releasing 1000 kg/hr of liquid N2, which equilibrates to 1 atmosphere and 20C (STP).\n\nAt STP, one mole of a gas occupies 22.4 liters;\nyou're releasing 1000 kg/hr = (1000 kg*1000g/kg)/28g/mole = 35,714 moles of N2 per hour, and thus:\n\n22.4 liters/mole * 35,714 moles = 800,000 liters (at STP) per hour
773826
the answer that i got is 3.10 M. \n\n3.25 g NaOH/40.00 g NaOH = .0813 mol NaOH\n\n26.2 mL / 1000mL = .0262 L\n\nM= .0813 mol / .0262 L = 3.10 M
649619
you mean litre?  32oz?
847153
PV = nRT  .... n/R = P/RT\n\n800 torr x (1.3158×10^−3 atm/1 torr) = 1.053 atm\n\nn/R = (1.053 atm)/(8.206 x 10^-2 L atm/Kmol  x 298 K)\nn/R = 0.043 mol/L\n\nNH3 mass is (14 + 1x3)=17 g/mol\n\ndensity = n/R but we need it in g/mL\nn/R = 0.043 mol/L x (17 g/mol NH3) x (1L/1000mL)\nn/R = 7.31 x 10^-4 g/mL
808549
M=molesof solute/Liter of solution\n\nM= molarity
315401
Use PV=nRT ideal gas equation.\nP=0.400 atm\nV=?\nn=3 mol\nR=0.0821 L atm /K/Mol\nT=300K\n\nDo yourself now...
797268
MM = mass * 22.4 liters/volume\nMM = .125 * 22400/93.3 = 30.01
164131
1 milliliter = 0.001 liters
215877
1 liter is a cubic decimeter. There are 1000 dm3 per m3, 264.17205 gallons per m3 and 3.7854118 liters per gallon.\n\nVolume can be calculated by the equation:\narea*height. So if you have a nice symmetrical tank just try to calculate its bottom (or top) area and then multiply with the height.\n\nArea is calculated with either:\nside*lenght of the tank for squares or for circles:\nradius^2*pi\n\nIf your tank is neither cylindrical nor cubical you may have your work cut out for you. Determine which shape it resembles the most.\nA sphere:\nV=(4*pi*r^3)/3\n\nA prism:\nV=B*h\nwhere h is height and B is the bottomarea of the parallell trapeze:\nB=(h*(a+b))/2\n\nFor a pyramid:\nV=(B*h)/3\n\nFor a straight circular cone:\nV=(pi*r^2*h)/3\n\nIf all else fails just fill that sucker with water (if it´s not a monstersized tank). You need to check it for leaks anyway. Weigh it empty and subtract that from the weight of the tank full. 1 liter (dm3)= 1 kg.
190428
22.4 l=>1 mole N2O\n0.1 l=> 0.0045 mole N20\nsince 1 mole NH4NO3 gives 1 mole N20,\n0.0045 mole of NH4NO3 is required.\nsince 1 mole NH4N03=>80 grams,\nmass of NH4N03 required is 80*0.0045=0.3571 grams\n\nsince 4 moles of Fe reacts with 3 moles of 02\nand 52 g of Fe=1 mole Fe,\n0.75 mole of 02 required.\nsince 1 mole 02=>22.4 l,\nvolume of 02 required is 16.8 litres\n\nhope that helps
919420
3.2 ppm is 3.2 mg per liter.  Multiply by 0.47 liters (470ml) to get 1.5 mg or 0.0015 g.
699263
V=nRT/P\nn(O2)=3/2(0.0022mol of KClO3) =0.0033 gmol\nR=0.0826\nT=294 k\np = p (O2) + p (water vapour)\np(O2) = 0.926 Atm\nV=(0.0033)(0.0826)(294)/0.926=0.087 liter = 87 ml
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!