How do you find the temperature of  the flame of a Bunsen Burner?
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You can get a infrared thermometer. just point it at the flame and it will tell you the temperature. it will also show you the differences when you move to diff rent colors in the flame.. blue being hotter than the yellow.. i bought one on line for 39 dollars from eBay i think
616344
Salt compounds can produce different flame colors when ionized under extreme heat (such as blue bunsen burner flames). At that temperature, it would be a fire hazard to the house, but just for your information, \n\nworks for Chloride AND Nitrate salts.\n\nCalcium-Orangy Red\nStrontium-Pure Dark Red\nLithium-Radioactive Red\nPotassium-Pinkish Purple/Fuschia\nBarium-Green with yellow tint\nSodium-Yellow/Orange\nMagnesium-Light Orange\nCopper-Radioactive Green
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The Cp of Hydrogen gas is 28.82 J/mol/k or 14.30 J/g/k\nTherefore, the Cv = Cp - R\n\nCv = 28.82- 8.3\n     => 20.52 J/mol/k\n\nTo covert this to B.T.U\nhttp://www.gordonengland.co.uk/conversion/specific_heat.htm\n\nRate mine as the best answer if you found this helpful.
454506
100 degrees celcius\n212 degrees Fahrenheit
956325
false - temperature is a measure of heat
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click here to see:http://en.wikipedia.org/wiki/Bunsen_burner#History
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Here you are:\n\nFirst write the combustion equation and balance it:\n\nH3NNH2(l) + O2(g) ---> N2(g) + 2H2O(g).\nThe heat will be the diffrence of enthalpies of formation between the products and the reactants. Do not forget to multiply by the stoichiometric coefficient.\n\nRemember that the enthalpies of formation of O2 and N2 are 0.\n\nYou will also need to find the enthalpy of formation of H2O(g) in the table of your book.  It is -241.8 kJ/mol\n\nHf(H2NNH2) = [2*(-241.8) + 1*0] - [1*50.6 + 1*0]\n\nHf(H2NNH2) = -483.6 - 50.6 = - 534.2 kJ/mol
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temperature of fire was 900°F to 1200°F\n\nMost volatiles ignite at temperatures ranging from 1000º - 1400º F.\n\ncombusted in the fourth and final state at temperatures in excess of 935°F.
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Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
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http://ushotstuff.com/Heat.Scale.htm
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Try Thermo.com/neslab contact them with any questions you have.
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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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a candle flame consists of various regions where different temperatures are assigned to them(temperatures have been measured). \n\nThe outer part has a lower temperature than that at the centre. \n\nThe temperature range can be from 500 to about 1500 degrees celcius.
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2 reasons really, courtesy for the next person, and if you let the alcohol level drop below the wick, the next burning will not be a burning of alcohol, but of the wick itself (and as it burns down into the alcohol burner, it can ignite the fumes, creating an explosion)\nI need to ask.. where are you that your lab is still using alchohol burners instead of bunsen burners?
873405
Flame color depends on several factors. Typically the most important being: blackbody radiation and spectral band emission with both spectral line emission and spectral line absorption playing smaller roles. In the most common type of flame, hydrocarbon flames, the most important factor which determines color is oxygen supply and the extent of fuel oxygen "pre-mixture", which determines the rate of combustion and thus the temperature and reaction paths, different color hues are thereby produced.\n\nIn a laboratory under normal gravity conditions and with a closed oxygen valve, a Bunsen burner burns with yellow flame (also called a safety flame) at around 1,000°C. This is due to incandescence of very fine soot particles which are produced in the flame. With increasing oxygen supply less blackbody-radiating soot is produced due to a more complete combustion and the reaction creates enough energy to excite and ionize gas molecules in the flame, leading to a blue appearance. The spectrum of a premixed (complete combustion) butane flame on the right shows that the blue color arises specifically due to emission of excited molecular radicals in the flame which emit the vast majority of their light well below ~565 nanometers in the blue and green regions of the visible spectrum.
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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?
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B. 93.3\n\nThis will help you:  http://www.onlineconversion.com/temperature.htm