how do you separate oxygen from carbon dioxide?.?
481006
the boiling point of oxygen is -183oC.\n\nthe freezing point of carbon dioxide at atmospheric pressure is -78oC.\n\ntherefore having a mixture of the two in a flask, and cooling with liquid nitrogen (~-170oC) will freeze the CO2 to a solid, and leave the oxygen as a gas.  You can then separate the two.  If you feel -170oC is too close to the condensing point of oxygen, then add some diethyl ether to your liquid nitrogen and this will raise the temperature of your ice-bath to ~ -100oC.
698578
Don't listen to the above post. CO2 is not flammable, and you are not necessarily asking for it to be pressurized. The cheapest way to get bottled CO2 is to catch the products of a combustion. I would try to suck the smoke of burning coal into a bottle. Another way is to make your own carbonic acid by blowing air from your lungs into watter. Of course using these cheap methods you'll never get a good sample (eg, humans breathe out about 20% CO2). Perhaps there is a way to get rid of all the gasses?\n\nActually, there are "air sponges" that absorb certain gasses. If you could pump air though enough of these filters to take out all the gasses you don't want, you could be left with pure CO2.
336167
oxygen and carbon dioxide
198671
Electrolysis. By applying electricity, you can break H2O into H2 and O2 gas.\n\n2H2O --> 2H2 + O2
876161
You're right - as written it's impossible to balance.  But if you write\n\na(Ag2CO3) --> b(Ag2O) + c(CO2) + d(O2) you can balance it with the values \na=1, b=1, c=1, d=0.    Meaning that the reaction doesn't generate any oxygen.  The carbonate falls apart into the oxide and gaseous CO2.
701278
There is a test called an Arterial Blood Gas, using a CoHgb module on the co-oximeter. There is a direct measurement for carbon monoxide in the blood. Using high-flow/ high oxygen concentrations is the fastest most potent way to remove the carbon monoxide from the blood.
232626
Several companies make packs that generate CO2 and N once opened.  They also bind atmospheric oxygen (O2).  If you plate your specimen on an anaerobic media, then place the media in an airtight container with one of these packs, you should acheive the appropriate environment. (You can use an anaerobic indicator to make sure that you've acheived the right air mix.)\n\nWhile plating the organisms, be sure to be quick.  Some organisms are killed by O2, so the longer it takes you to plate them, the more chance there is for O2 to get in and damage them.
991601
You actually form 11g CO2.\n\nPer your description, the O2 is a limiting reactant. I.e., there is a fixed amount of it and more C2H2. So, the balanced equation becomes:\n\n2C2H2 + 5O2 = 4CO2 + 2H2O  (balanced equation)\n\nx = 10 g O2 x 1 mol/32g x 4 mol CO2/5 mol O2 x 44 g CO2/mol\n\nx = 11 g CO2 formed\n\nIf you burn C2H2 to completion the equation changes to:\n\nx = 10 g C2H2 x mol/26g x 4 mol CO2/2 mol C2H2 x 44 g/mol CO2\n\nx = 33.8 g CO2 formed
116688
from the equation: 2CO + O2 -----> 2CO2,\n\n0.69 g CO2 x (1 mol CO2/ 44 g CO2) x (2 mol CO/ 2 mol CO2) x (28 g CO/ 1 mol CO)\n\n\nthe answer is 0.4391 g of CO needs to react with O2 to produce 0.69 g of CO2
446321
Hmmm, a multi-step problem!  Love those!\n\nFirst, you will need to know how many moles of oxygen are generated.  For that, you will need to use the Ideal Gas Law:\n\nPV = nRT\n\nYou need to solve for n = PV/RT, and since you know the volume, pressure, and temperature, and the value of R is a constant, this is an easy calculation.\n\nThen, from the equation:  \n\nKClO3 ===> KCl + 1.5 O2\n\nyou can work backwards to determine the number of moles of KClO3 you started with (one mole of KClO3 generates one and a half moles of oxygen, thus the number of moles of KClO3 you started with will be 2/3 of the number of moles of oxygen generated).  Knowing how many moles of KClO3 you had, you can then determine the starting mass of the KClO3, then find out what percentage that was of your original starting mixture.\n\nGood Luck!
551286
oxygen is made by distilling liquid air. liquid air is regular air cooled to about -200 degrees.  it becomes a liquid.\n\nthe oxygen doesnt need cleaning,  the distillation process removes any impurities.\n\noxygen is put into the cylinders by a machine similar to a super heavyduty air compressor.  the oxygen in a full cylinder is at about 2000 psi.
217530
Good question. . . . Just use W/O convertor ((W/O = WOOD/Oxygen )) it costs u about 2.000 $$ \nafter u use that piece of wood try to grt another piece or u will need to recycle CO2 and convert it to wood using (C/W convertor), don't ask me what's C/W ?? but u should know that's too expensive. you will need a fortune to live not just a piece of WoOd
861843
Start with the Ca then balance the number of PO4 ions, then balance the Oxygen and the hydrogens should balance themselves out.  Its just trial and error. Monkey with it until it works.
826844
H\n |    \nC=O:  (You need 2 lone pairs on Oxygen)\n |   \nH\n\nIt doesnt get much better than that :)
968045
A common lab method to produce small amounts of oxygen is to decompose potassium chlorate.  A little manganese oxide will catalyze the reaction.  Just put the KClO3 in a test tube, add a little MnO2 and heat with a burner.  You can collect the O2 by water displacement.\n\n2KClO3  --> 2 KCl + 3O2
729087
1) this one is already balanced for you!!\n\n2)Count the number of Mg and O on the right side, and count the number of Mg and O on the right side.  On the left, you have 2 O and 1 Mg, and on the right you have 1 Mg and 1 O.  To make the oxygen on both sides equal, you need to double the MgO.  Now on the left, you have 2 O and 1 Mg, and on the right, you have 2Mg and 2 O.  Now you need to multiply the Mg by two so the Mg on both sides is equal.  and you're done!!!! \n2Mg + 02 --> 2MgO\n\n3) On the left, you have 1 Zn, 1H, and 1 Cl.  on the right you have 1 Zn, 2Cl and 2H.  the Zn are balanced and you just need to multply the HCl on the left to balance the whole equ.\nZn + 2HCl --->ZnCl2 + H2\n\n4) On the left, you have 1 P and 2O, and on the left you have 4 P and 10 O. If you multiply the P on the right by 4, and the O on the right by 10, you have 4P and 10 O on both sides\n4P + 5O2 ---->P4O10\n\n5)  On the left, you have 1 Fe 2 Cl, and pn the left, you have 1 Fe and 3 Cl.  The easiest way to balance this is multiply the Cl on the right by the amount of the left one and vice versa. when you do that, you have 1 Fe and 6Cl on the right, and you have 2Fe and 6Cl on the left. To finish the equation, multiply the Fe by two.\n2Fe + 3Cl2---->2FeCl3\n\nVoila!  You have balanced 5 equations!\n\ni hope this was helpful!
402174
You can Use Dean and Stark Apparatus To Separate them \n\nAlso one Source i m providing you which involve the separation using PTC\n\nRefer it \n\nhttp://www.ptcorganics.com/Top10PTC.pdf