what is the caloric value of propane gas?
152006
Like, calories? You want to consume it? Why? I think it's poisonous. Don't do it. And sorry if that isn't a good answer, I'm failing chemistry.
793717
Glycerol is also known as propane-1,2,3-triol, formula C3H5(OH)3, so the answer is three.
998949
Matt gave the correct answer for propane VAPOR. Oil field trash is correct. Propane in liquid form does not ignite, only the vapor or "fumes" as some would call it.
643566
In the Environmental Lab, we used the Parr Oxygen Bomb Calorimeter to measure the change of heat or capacity of heat for different substances.  What I basically mean is that we put the substance into a reaction vessel and then into the reaction container(bomb).  The bomb is pressurized,  filled with Oxygen, ignited and the data from the reaction is extrapolated to give us the Caloric Value for the substance.  \n\nWe took in Hazardous Waste for incineration in an industrial Kiln.  It was important to know the Caloric Value for these substances as it would determine how much other fuel (Usually natural gas) we would have to use to maintain the optimum temperature for the Kiln to operate.
192284
http://energy21.freeservers.com/bookrep.html  Want to know something else? I've found a way to convert gasoline into Propane! And it can never be marketed! All Engines from 1996 on up must run at an Air to Fuel Ratio of 14.7 Parts of Air to 1 Part of Fuel! My System goes 100 to 1 Air to Fuel Ratio, but it will never happen! You want something interesting to Research ? This is it!
741437
http://en.wikipedia.org/wiki/Calorimeter
509763
Try using the calculator at the webs site listed below. \nEthanol at 57.5 deg C has a heat of vaporization = 40.23 KJ/mole.\n\nThis web site has calculators for a number of other physical properties.\n\nBest wishes !
1059927
liquid propane and natural gas
179193
The alcohol itself - 7 calories a gram -
569971
This problem only has enough information if you assume it is an ideal gas.  In this case, PV=nRT is the equation of state.\n\nn and R are constants in this case, so PV/T must also be constant.  These values must also all be in absolute scales - the celcius must be converted to kelvin or any other scale based on absolute zero.  This means:\n\nP(i)V(i)/T(i)=P(f)V(f)/T(f)\nV(f)=T(f)P(i)V(i)/[P(f)T(i)]\n    =(250+273)K(740mmHg)(410mL)/[(680mmHg)(27+273)K]\n    =[523*740*410/(680*300)]mL\n    =777.83mL\n\nBy the significant digits you've presented, this can only be evaluated to any accuracy as 780mL.  (two significant digits)
491381
Fat=9.1KJ per mole\nProtein=5.4 KJ per mole\nCarbohydrates= 4.2 KJ per mole
521266
This sounds like homework, so I'll give you the ideas but not the data.\n\nThe enthalpy of the mixture at standard conditions is the sum of standard heats of formation of the components of the mix. It will be given in kJ/mol (Ar is 0 kJ/mol being an element, the other two you can look up in tables or on internet). \n\nThen you need to adjust for the sensible energy difference compared to standard conditions : you will also need to look up the specific heat of each component (usually expressed in kJ / mol ºK) and multiply the specific heat of each component by the amount and by the difference in temperature. Make sure you have everything in the same units.\n\nRemember if you are working in moles its easier because to a first approximation, one mole of any gas occupies the same volume (if you undrestand how to allow for deviations from the ideal gas laws, you wouldn't be asking this question) and I assume that the percentages are volume percent, not by weight\n\nEnjoy
855413
Latent heat of fusion of ice is only 80 calories/gm whereas the latent heat of vaporisation of water is as high s 536 cal/gm.Therefore C is correct
618904
Not Glucose, that is a carbohydrate.\n\nHeptane (C7H16) has a molecular weight of 100.23\nAny longer chain hydrocarbons obviously are more.\nC8H18\nC(n)H(n+2)
170755
Specific heat 0.13 J/gK
865194
It's not, that is only a certain method of measuring a calorie.  A calorie is the amount of energy required to raise the temperature of 1 gram of water 1 degree celsius at 1 atmospheric pressure.  \n14.5 and 15.5 (celsius) are merely the temperatures chosen to start and end at.  There is also the 20 degree calorie (19.5-20.5) and 4 degree calorie (3.5-4.5). The differences in energy required from different temperatures vary in the thousandths of a joule.
102536
First use the ideal gas law to find out how much oxygen gas reacts with the butane.\n\nPV=nRT where P is pressure in Pa, V is volume in meters cubed (m^3), n is amount in moles, T is temperature in Kelvin (K), and R is the ideal gas constant 8.314 J/mol K.\n\nn=PV/RT=[(101325 Pa)(0.015 m^3)]/[(8.314 J/mol K)(298 K)]\nn=0.613 moles O2\n\nNow write the balanced equation for butane combustion.\n\n2 C4H10 + 13 O2 --> 8 CO2 + 10 H2O\n\nNow use stoichiometry to find the moles of butane.\n\n(0.613 mol O2)(2 mol C4H10/13 mol O2)= 0.094 moles C4H10.\n\nNow convert moles C4H10 to grams by multiplying by molar mass. The molar mass for butane is 58 g/mol.\n\nMass butane = (0.094 moles)(58 g/mol) = 5.47 grams butane