What are the conditions needed to do a GC of hydrocarbons (C20-C30)?
372647
No sample prep is really needed, just make sure the samples are pure.  Your injector needed will be a very small syringe, you will only need about 2 microlitres of sample.  The column will be small with a flow rate of about 1cm/sec.  Carrier gas would be helium flowing at a rate of about 100psi.  Detector would be a general detector that would detect organics, nothing special.  And your temp would need to be high enough to vaporize your highest boiling point hydrocarbon.  Probably somewhere around 80 to 100 degrees centigrade.  Good luck!
851681
Write the balanced equation - you need the mole ratios.\n\nChange 1g hexane to moles of hexane.\n\nSee how many moles of oxygen are USED and how many moles of  CO2 and H2O are formed.\n\nYour total number of moles in the reaction mixture would then be:\n\nmoles of Oxygen NOT used + moles of CO2 (we will assume the water formed consenses to a liquid at 20 degrees).\n\nUse the Ideal Gas Law to figure out the volume.\n\nPV = nRT\n\nwhere P = given pressure - water vapor pressure\nwhere V = your unknown\nwhere n = your total moles of gas\nwhere R = the appropriate Universal Gas Law constant\nwhere T = the temperature in KELVIN
823117
That is a pretty tall order.  You will find very little, outside of governmental regulations.\n\nIf this is for the US area, OSHA and the Coast Guard have regulations.  Also American Bureau of Ships\n\nOthers are Safety to life at Sea (SOLAS)  and Intergovernmental Maritime Consultive Organization.\n\nYou can find good information on hydrocarbon processing in documents from Americal Petroleum Institute, Oil Insurance Association, and regulations of freign governments.\n\nYou may find a document called "Mobile Offshore Drilling Units."  It was written to be considered by various governments as a standard regulation\n\nA good book on hydrocarbon processes is "Fire Protection Manual", by Vervalin\n\nIf you have access to a good library, a talk with a librarian would be a good place to start.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
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
1127301
I leave the H's:\nC-C-C-C-O butanol-2\nC-C-CO-C  butanol-2\nC-O-C-C-C propyloxy methane\nnot shown: isopropyloxy methane\nC-C-O-C-C ethoxy ethane
723847
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
854658
I'm assuming that you mean refined petroleum because you would have had more detail if it was a specific oil.\nThe answer is negative 150 degrees Fahrenheit
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
444491
http://www.faqs.org/faqs/autos/gasoline-faq/part2/preamble.html\n\nhttp://www.atsdr.cdc.gov/toxprofiles/tp72-c3.pdf\n\nhttp://www.elmhurst.edu/~chm/vchembook/514gasoline.html
8185
In refining, the component chemicals of petroleum are separated by distillation. Products based on refined crude oil include kerosene, benzene, gasoline, paraffin wax, asphalt, etc. \n\nHere are some of the main products obtained from crude oil and the temperatures at which the different products boil off of the crude oil. Boiling ranges of petroleum atmospheric pressure distillation fractions in degrees Celsius:\n\npetrol ether: 40 - 70 °C (used as solvent)\nlight petrol: 60 - 100 °C (automobile fuel)\nheavy petrol: 100 - 150 °C (automobile fuel)\nlight kerosene: 120 - 150 °C (household solvent and fuel)\nkerosene: 150 - 300 °C (jet engine fuel)\ngas oil: 250 - 350 °C (Diesel fuel/ heating)\nlubrication oil: > 300 °C (engine oil)\nremaining fractions: tar, asphalt, residual fuel
813571
325N/9.8m/s^2=33.16 Kg=33163g\n265N/9.8m/s^2=27.041Kg=27041g\n\n33163g-27041g=6122 g of water displaced.  Since water's density is 1 g/cc it is 6122 ccs displaced by the object, that is it's volume. 33163g/6122cc=4.417 g/cc, that is your density.\n\nIn oil the object weighs 295 N\n295n/9.8m/s^2=30.102Kg=30102g\n\n33163g-30102g=3061 g of oil displaced.  The volume of the object stays the same (unless you add some additional information on pourous material) so if 3061 g are displaced by 6122cc, the oils density is 3061g/6122cc=0.5 g/cc
954811
C3H8 + 5O2 --> 3CO2 + 4H2O
166454
I can supply you said chemicals.\n(C2H5NH)(CH3)C6H3OH \nPlease contact me by email: halsliu@gmail.com
1019349
This site may lead you to the information you seek.\n\nhttp://www.petrochemistry.net/index.asp
165982
Propane - C3H8 = 44 grams/mole\nButane - C4H10 = 58 grams/mole\n1 cc = .001 liter\ngas constant R = .08206\n\nA)\n\nPV=nRT\n(1 atm)(50x.001L) = n(.08206)(273K)\nn=.002 mol\n\n.002 mol propane x (44 g/mol) = .088 g\n.002 mol butane x (58 g/mol) = .116 g\n\n50% propane by mole\n43% propane by mass\n\nB)\n\n(150 kPA)/(101.325 kPA) = 1.48 atm\nPt = P1 + P2\n1.48 atm total = .74 atm propane + .74 atm butane\n\nC)\n\nThe propane should effuse faster because its molecular mass is lower.
194284
To solve this problem we need to think about limiting reactants.  This one is fairly simple because I already have moles and the ratio is 1:1:1.  If I only have 3 kmols of C, that is the most I can make, because going beyond that, I would be out of C.  Thus, I can only make 3kmols of CO2 with 2 kmols of O2 left unreacted.  These amounts are theoretically find for a combustion reaction.