How much oxygen does a regular open-flame candle burn in a minute/hour?
554441
The "burn rate" of a candle depends on the type of wax (or other material being burned, e.g., "gel candles"), the type of wick, the geometry of the candle and wick, etc.  In general, though, burn rates for typical candles are on the order of 2-8 grams of paraffin/hour.\n\nParaffin wax is composed of straight-chain hydrocarbons with the general formula CnH2n+2, where n is greater than 20.  As the size of paraffin molecules increases, the carbon content approaches the limiting value of 85.7 percent asymptotically. For the sake of argument, let's say the wax in a candle is 85% carbon by weight (the rest is hydrogen).  This corresponds to a hydrocarbon chain containing about 20 carbons.\n\nIf the wax is burned completely, the overall chemical reaction involved is:\n\n  30.5*O2 + C20H42 -> 21*H2O + 20*CO2\n\nor, for each molecule of C20H42 burned, one needs 30.5 molecules of O2 gas.  For each atom of carbon in the wax that is burned, one needs 30.5/20 = about 1.5 atoms of oxygen gas.\n\nOxygen is the only gas in air that is consumed by the reaction, and if the candle burns prefectly cleanly, the only things produced are carbon dioxide (CO2) and water.  In reality, candles don't burn perfectly, and they produce some unburned hydrocarbons, soot, and carbon monoxide.  In addition, candles commonly have additives, such as colorants and fragrances, that are released.\n\nA candle that burns at a rate of 5 grams of paraffin per hour releases about 5*0.85 = 4.25 grams of garbon/hour (85% of the total mass).  The atomic weight of carbon is 12 grams/mole, so 4.25 grams/hour = 0.354 moles C/hour.  \n\nEach mole of carbon needs 1.5 times as many moles of O2 gas, so oxygen is consumed at a rate of 0.531 moles/hour.\n\nAt room temperature and pressure, a mole of gas occupies about 22.4 liters, so the candle uses up 11.9 liters of O2/hour.  As you point out, air is only 20% oxygen, so this is the oxygen contained in 11.9/.2 = 59.5 liters of air.  \n\n59.5 liters air/hour = (approximately) 1 liter air/minute.
290344
first figure out the number of moles of each substance, in this case: ethane = 0.038 and oxygen = 0.39, so oxygen is in excess and will not limit the reaction.\n\nThe standard enthalpy of combustion of ethane = -1423KJ/mol\n\nas you have 0.038mols of ethane the energy change is 0.038x-1423 = -54.07JK/mol
574060
The biggest problem likely will be that the filament must glow hot enough to give off light and it will quickly burn up if there is any oxygen present in the bulb.  Since your bulb must last only a minute at one candle power, perhaps you can burn up the oxygen in the "bulb" with a small lit candle before "sealing" it.  Many radio vacuuum tubes used "getters" to remove traces of oxygen from the vacuum - just like your candle.  You might try a glass jar with a small twist lid which you would twist tight after the candle goes out.  The filament might be a strand of steel wool held between two copper wires that enter the jar (from the top) through the cap (glued to exclude oxygen).  Start with one cell then add more cells in series (bumper-to-bumper with none reversed) to get the right voltage without melting the steel wool.  Results will depend on how fine or course the steel wool is and how long the strand is (to obtain enough resistance to current flow).  The longer it is (a loop?) the more light.  You can twist off the cap and cover the jar to try other filaments or replaced melted ones.  Good luck.
51504
if you breath you only get about 20% of the total amount of oxygen out of the air. The remaining oxygen you blow onto the coal which helps it to re-ignite for a sec.
121351
Are you tryn to have us do your hw!?!?\n\nThe eqn should be in your book.  I dont feel like digging mine out, but its something like;\n\nCp*g*(T2-T1)=energy\n\nSo what u want to find is (t2-t1) Ima call it x\n\n400*50*(x)=800  do the math.  I think thats it.\n\nfor the 2nd question, this one is easy.  The combustion is energy is for 1 mol (per mol).  If you only have .2 moles, the combustion amount is .2* 1370 .
8516
The heat of the flame melts the wax. The wick will transport liquid wax into the flame zone (by capillary action), where it can evaporate. The flame consists of two different zones, an oxygen-rich outer layer where carbon monoxide burns with a transparent blue flame and an oxygen-poor inner core where soot particles give off the warm orange glow of a candle flame.\n\nChemically, you have a reaction of the wax (esters of fatty acid) with the air oxygen, forming carbon dioxide and water, and depending on the quality of the wick, soot from incomplete combustion.
455855
The temperature of the candle is measured in kelvin scale . it varies between 1800 to 1850 degrees kelvin.Zero Kelvin (not zero degree Kelvin) corresponds to -273.15 degrees Celsius, so ice melts at +273.15 K and water boils at +373.15 K.( 100 degrees celsius) .so in degrees celsius it would range from 1576.85 to 1526.85°celsius..hope this answers your question.
931550
adjust the ratio of gas that mixes with air for combustion
191989
ever 1 mole of CO2 and one mole of H20 produces 1 mole of O2 for plants. A fire needs only one molecule of H20 to create a fire
694526
read this http://www.physlink.com/Education/AskExperts/ae673.cfm\n\nOn earth, as we all have seen and experienced, when you light a flame, such as a candle or a match, the flame points "up". But why? If we answer that question, we will be quite a bit closer to answering the actual question here.\n\nWhen you have combustion, such as a match, what happens is that oxygen gets consumed, and other gases (such as carbon dioxide) get produced. In addition, of course, energy is liberated in the form of heat and light. The produced gases are hotter than the environment due to the generated energy, and they expand, as heated gases do. Then, of course, since their density is lower than the surrounding atmosphere, the gas tends to rise. This has two effects. First, while it moves up, it carries energy with it, thus the flame ends up pointing "up". Second, since the products of combustion move up, they make more oxygen available to make the flame continue to burn.
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.
333447
Well, a standard "candle" is a defined unit of measurement:\nCandle: An obsolete unit of luminous intensity, originally defined in terms of a wax candle with standard composition and equal to 1.02 candelas\n\nCandela:\nA unit of luminous intensity equal to 1/60 of the luminous intensity per square centimeter of a blackbody radiating at the temperature of solidification of platinum (2,046°K)\n\nSo your torch is about 1 million times brighter than a bit of really hot plantinum!\n\nPresumably, someone has worked out exactly how bright this is, and can also measure how bright your torch is and can work out how much brighter the torch is than the standard "candle"
278240
C6H12O2 + 6O2 --> 6CO2 + 6H2O\n\nis the combustion reaction for glucose.
3360
The inner most part of t he burning candle is 800 degrees centigrades. The outer parts are somewhat less hot by a 100-200 degrees.
936398
To understand this question, you need to know what kind of product will be produced during a combustion. Generally, for a complete combustion of a hydrocarbon (contains only H and C), the product will be H2O and CO2. Depending on the C and H content in the hydrocarbon, you will produce different amount of H2O and CO2. \n\nOn the other hand, if it is an incomplete combustion (with limited supply of O2), apart from CO2 and H2O, you may also produce C and CO. The relative amount between them varies depending on the amount of O2 that is made available. With very limited amount of O2, more C (soot) will be produced. This leads to the "sooty" flame you are seeing.\n\nIn order to explain why generally alkenes require less O2 to burn compared to alkanes, let's compare ethane (2 C alkane) and ethene (2 C alkene). \n\nEthane has a formula C2H6 while ethene has a formula C2H4. Their corresponding balanced equation for combustion are:\n\nC2H6 + 7/2 O2 → 2 CO2 + 3 H2O\nC2H4 + 3 O2 → 2CO2 + 2 H2O\n\nAs you see, since both compounds have same number of C, they both produce 2 mole of CO2 for every mole of compound. However, since they have different number of H, the number of moles of H2O produced will be different. Since H2O has a fixed formula, in order to produce it, you need to supply O2. That's why the two compounds require different amount of O2.\n\nThe reason that you see a sooty flame is due to the fact that you don't have enough oxygen supply during the burning process. You can easily confirm this by using just a Bunsen burner (but be very carefully!!!). First, turn on your Bunsen burner with the air hole closed. The flame you see at the moment should be a luminous flame. It is golden yellow color. Now, if you carefully put a glass plate (hold with clamp or other device) on top of the flame and slowly move towards the tip of the burner (but don't cover it completely), you should see the plate will get "darker" by the flame. This is the soot produced during the combustion. Now, take away the plate, open the air hole in the burner to the max. You should see a the flame turns to blue. If you put the glass plate (with the soot marks) back to the flame, you can see the flame can slowly "clean up" the plate (if enough O2 is supplied with the flame) and the soot will disappear. The soot is turn back to CO2 due to the excess of O2.
796598
Wow, we're going to need A LOT more info here. What shape is the candle, how tall is it, what is it's density?
518927
(46.2 grams)/(32 grams per mole) = 1.44 moles of Oxygen (O2)