thermodynamic property of ethylene?
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Standard enthalpy of formation 227 KJ/mol\nStandard enthalpy of combustion -1300 KJ/mol\nStandard entropy 201 J/K/mol
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D.  Based on the ideal gas law, PV=nRT\n\nSolve for R to get: R = (P*V)/(n*T)\n\nfor two different conditions, then, you would have:\n\n(P1*V1)/(n1*T1) = R = (P2*V2)/(n2*T2)  or,\n\n(P1*V1)/(n1*T1) = (P2*V2)/(n2*T2)\n\nor P1*V1*n2*T2 = P2*V2*n1*T1\n\nNote how V1 and T1 can never be in the numerator together?  \n\nIn order for the ratio to work, one must be the numerator, the other the denominator, because volume is directly proportional to temperature -- the higher the temperature (keeping number of moles and pressure equal), the larger the volume.
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Of course you must specify whether you want the thermodynamic properties of water (or chewing gum?).  Good luck.
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The specific heat of liquefied methane is 3.50 J/g/K. The specific heat capacity of LNG is likely to be close to this as LNG is about 90% methane. If you need a more precise value, you will need to find out how much methane, ethane etc is in the LNG and look up the appropriate specific heat values and calculate a weighted average. See page 2 of the URL below.
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Acetone       2.13kJ/kg K or  0.51  btu/lb F\nWater, fresh 4.19kJ/kg K or  1.0   btu/lb F\n\nsee also http://www.engineeringtoolbox.com/specific-heat-fluids-d_151.html
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thermodynamic factors as well as the orientation and relative energies of the pi molecular orbitals involved.
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If your compound has the formula C6H10, then the molecular weight would be 82 grams/mole\n\nYou could have one of 13 different compounds -- \n2,3-dimethyl-1,3-butadiene\n3-3-dimethylbutyne\nCyclohexene\nany one of the five hexadienes\nany one of three hexynes\n4-methyl-1-pentyne\n4-methyl-2-pentyne\n\nWhat is the rest of the question regarding the 5.3*10^-23 odr?
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You want "energy density:"\n\nEthanol (Ethyl Alcohol):\n26.8 MJ / kg\n= 2905 kCal / lb.\n\nCrude Oil:\n45 MJ / kg\n= 4879 kCal / lb\n\nAviation Kerosene:\n43 MJ / kg\n= 4662 kCal / lb\n\nI don't understand the "If measured with a calorimeter, what degree celsius would it reach." Are you talking about the combustion temperature?
199900
The easiest way is to simply find literature listing delta H for the reaction.  Otherwise you have to do one of 2 things.  \n\n1. Find the enthalpy of formation for the product of the reaction which should be what you're looking for.\n\n2. Use the equation G=H-TS, plugging in the Gibbs free energy, temperature, and entropy values for the reaction,then solving for H, the enthalpy
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Yes, knowing these values for the combusion of the fuel used on the internal combustion engine (ICE) will help you determine the maximum amount of energy which may be extracted from the engine and then from that, the engine's efficiency.\n\nThe maximum amount of energy which may be extracted out of a chemical reaction (the max work the reaction can do) can be calculated by finding the reactions change in the Gibbs free energy under the conditions which the reaction is under.\n\nThe Change in the Gibbs Free Energy (delta G) can be calculated as,\n(delta G) = (delta H) - T*(delta S)\nWhere (delta H) is the change in the enthalpy of the reaction, T is the temperature of the reaction in Kelvin, and (delta S) is the change in entropy of the reaction.\n\nIf the value of (delta G) comes out to be <0 (i.e negative), then the reaction is spontaneous under those conditions and work can be extracted up to the magnitude of the change in Gibbs free enery.\nIf the value of (delta G) is greater than zero (positive) then the reaction is not spontaneous and requires that much energy to be put INTO the system in order for the reaction to proceed under those conditions.  In this case, no work can be extracted, work actually needs to be spent in order to get the reaction to occur.
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The easiest way to work this out is from the Ideal Gas Law, PV=nRT. Assuming that they start at the same temperature, this is even easier because both R and T will be constants.\n\nLets make the first gas 1, the second gas 2 and the final mix f so that :\n\nPressure Gas 1 =P1, Volume Gas 1 = V1, Temperature Gas 1 = T1\nPressure Gas 2 =P2, Volume Gas 2 = V2, Temperature Gas 2 = T2\nPressure final mix =Pf, Volume final mix = Vf, Temperature final mix = Tf\n\nthen :  P1V1/RT1+P2V2/RT2=PfVf/RTf\n\ntake out the constant R\n\nP1V1/T1+P2V2/T2=PfVf/Tf\n\nIf T is different for each mix\n\nPf=(Tf*(P1V1/T1+P2V2/T2))/Vf\n\nif T is constant (which I assume from your question)\n\nP1V1+P2V2=PfVf\n\nPf=(P1V1+P2V2)/Vf\n\nOkay?
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heat of combustion = -(enthalpy change for reaction )\n          \n = -(heats of formation of products - reactants)\n \n1 mole of ethanol, when burnt completely, produces 2 moles of CO2 and 3H2O i.e delta H = 2(heat of formation of CO2) + 3(heat of formation of H2O ) \n\n= 2(-394 kJ/mol ) + 3(-286 kJ/mol)\n\n= -1646 kJ/mol\n\nheat of formation of ethanol = -277 kJ/mol\n\nso\n\nheat of combustion = - (-1646 - (-277) )\n                   \n                   = - (-1646 + 277 )\n\n                   = 1369 kJ mol
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http://www.ausetute.com.au/ethene.html\n\nhttp://www.infoplease.com/ce6/sci/A0817790.html\n\nhttp://en.wikipedia.org/wiki/Ethene
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I'm only helping you because you seem to have done the rest of the questions!\n\n1. Butene is C4H8\n\n2. N and O are very close in the Periodic Table\n\n3. NaBr contains a metal and a non-metal\n\n4. Your diagrams have come out as unintelligible. Look for a molecule with a C=C double bond and no oxygen.\n\n5. CO2 contains non-metals only.\n\n6. Ethane is C2H6, and ethene is C2H4.  So they both contain 2 C atoms.\n\n17. See 3.
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You will have to first find the standard enthalpy values of the reactants and products. They're given in the appendices of any standard GenChem or PhysChem books.\n\nThen, you subtract the sum of enthalpies of reactants from the sum of enthalpies of the product to get the enthalpy of the reaction.
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Check out this site: http://www.questconsult.com/~jrm/enthpres.html\n\nUse the component id for oxygen (28) and mole fraction (1.00), press submit form. Makes a nice Enthalpy-Pressure diagram.
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You have Pressure (being asked) You have the volume (5.43L). You have the number of moles (1.82)You have temperature (69.5C or 342.5K). I assume you know R.\nSo yes it is PV=nRT