what is the burn ratio for diffrent gasses?
412413
I'm assuming you're asking for the reactions.  Here are a couple:\nH2 + F2 → 2HF \n\n4Fe + 3O2 → 2 Fe2O3
1184525
Depends on what is being split and where the combustion takes place.  The combustion alone will produce changes of oxygen, hydrogen and any other chemical in the air or around the combustion.  The remaining products will more than like be water, carbon min-oxide, etc.
187568
You can come up with a molecular weight of air - use the percentages of the different molecules comprising air and the molecular weight of the components.  The answer is around 29 g/mol (I think).
694912
I do not know where you got your formula for cellulose.  Cellulose is usually considered as having variable length of anywhere from 2000 to 14000 units of -(C6H10O5)-\n\nUsing C52H84O22, combustion would be:\n\n62 O2 (g) + C52H84O22  (s) -->  52 CO2 (g) + 42 H2O (g)
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
209057
It depends on the size and type.\n\nhttp://www.hanford.gov/fire/safety/extingrs.htm
880815
1 therm is equal to 100,000 BTUs. Therms are a measurement of heat. The cu. ft. is a measure of the gas volume. 1 cu. ft. natural gas is equal to approximately 1025 BTUs @ 1/4 lb/sq. ft. of pressure.
699057
The burning of fossil fuels containing sulphur and nitrogen:\n\nS + O2 ---> SO2\n\nN2 + 2O2 ---> 2NO2
320739
Based on what you have given us, there are no coefficients in front of any of the compounds/elements, so the ratio would be 1:1:1.\n\nI'm guessing that you didn't provide all the information need, though.\n\nGenerally these questions give you a reaction that you must first balance to determine the mole ratios.
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.
324991
Mole ratio is the percentage of different molecules in a substance.\nexample H2O molecular wight is 18,and hydrogen-1,and oxygen-16. so mole ratio of hydrogen=2/18(100)=11.11%\nsimilarly for oxygen M/R=16/18(100)=88.89 This is mole ratios of Hydrogen and oxygen in water
892149
This depends on how exact your answer needs to be.  You can use the 68-95-99.7 rule to determine that in the 95% of the time you will fall within two standard deviations (i.e., bet. 6 and 8).  Half of the remaining 5% is usage below 6 liters, so only 2.5% of the time do you run out of gas.\n\nTo be more exact, you can look up the z-score given by:\n(X - mu) / (sigma) = (8 - 7) / .5 = 2 and find that the result is that you run out of gas 2.2275% of the time.\n\nIf you need more explanation, just add that to your additional details.  Good luck!
880114
flare up bigger than normal..\noxygen is not flammable..but the presence of it is needed for combustion\n\nCH4 + 2O2 --> CO2 + 2H2O\n\nwith burning in atmosphere, the flame is smaller because of incomplete burning,*there is only about 21% of oxygen in atmosphere..
756159
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l
262778
73.93 grams\nFirst you use the Ideal Gas Law: PV=nRT >P is pressure in atm., V is volume, n is moles, R is the constant: .0821, T is the temperature in Kelvins> (1atm.)(3.16L)=(n)(.0821)(305K) > .126195563 moles. Now you use a proportion. 9.33grams/.126195563moles = x grams/ 1 mole > 73.93 grams
422875
U have not mentioned the gas gravity to determone deviation factor to have an exact answer...generally..taking z=1\npsia ( absolute value) will also be used \nPsc*Vsc/Tsc=PV/T\n\nso: Vsc/V=PTsc/PscT=53.1/T \n\nThis is the simplest correct answer.... make is easier but worse that the process is isothermal so\n\nVsc/V=102 --> 1cu ft at TVD= 102 SCF\n\nThe most exact solution can be provided if u want..with taking temperatire and deviation,composition in account
28516
Breath them in one at a time for 5 min each.  If you die it was CO2 if you're alive it's O2.\n\nor fill 2 jars with the different gasses. stick a lit candle in them. if the candle goes right out its CO2 if it stays lit its O2.