What is the best way to separate Oxygen from chlorine while in a gas form?
476397
bubbling the O2/Cl2 mixture through aqueous NaOH will remove the Cl2.  it reacts to give bleach ... NaOCl ... (and salt and water).  assuming you didn't want the Cl2 ;-)
157864
The following two links may be of some use regarding your question, you may just have to find a substitute for the chloroform extraction buffer.
652740
Girl Chemists are soO SexY!  We need more of them. =)\n\nThe level of active chlorine is called the free chlorine residual and should measure 1.5 - 3.0 ppm, regardless of size or volume of the pool.  How to achieve the correct ppm depends on the chlorine product you choose to use.  Once you pick your pool chlorinating product it should have a detailed instruction to make the correct concentration of pool water.  There are kits with paper indicators that you can dip into your pool water after you add Chlorine solution or tablets to see if you have the correct amount of chlorine dissolved in the water.  If you still need help, you know where to come back and ask.
338399
The chlorine gas can be damaging to your lungs if inhaled. Be careful and always make sure the area is well ventilated.
1119175
First step\nKnow the properties of gases\nCarbon dioxide will turn into a solid while cooling air, so it should be removed. \nAll remaining gases will become liquid when they are cooled. \n\nStep 2\nThe only step to separate gases in the air is to cool them. There are  2 principal ways: By reducing the pressure from standard pressure to lower presure values. OR by decreasing temperature. \n\nMETHOD: Fractional distillation of air should be used. It involves pumping air into a container, after having removed carbon dioxide from it. \n\nCarbon dioxide, being an acidic gas, is removed by passing air through an alkali by bubbling the air through the alkali. \n\nThe air rid of carbon dioxide is then passed through a special and sealed container, where the pressure will be reduced for the gas components to appear as a liquid. On the basis of recorded boiling points of gases, each gas is collected in turn. \n\nCONCLUSION:Thus you have separated the gas mixtures from air. You can use them to propel rockets.
113508
If "salt" is Sodium Chloride, then you will be making Hydrogen gas (H2) and (Oxygen gas (O2) and/or Chlorine gas (Cl2)).\n\nUnless you plan on very carefully controlling the potential difference across your electrolytic cell to use just the bare minimum voltage, you are likely to make Chlorine gas at the anode instead of Oxygen.\nOxygen gas has a reduction potential of 1.23 volts, where are Chlorine gas has a reduction potential of 1.36 volts.\nThis means that if you supply just a tiny bit of extra voltage across the cell, instead of just producing Oxygen gas from the water, you will be oxidizing the Chloride ions in solution to make Chlorine gas.
405797
heat the pure gas in a sealed tube in absence of air. it breaks up into its components. you've got solid yellow sulphur.
585164
Well, you can combine it with oxygen to make hydrochloric acid, but it gives off pure hydrogen as a by-product, so I wouldn't try this at home....
198671
Electrolysis. By applying electricity, you can break H2O into H2 and O2 gas.\n\n2H2O --> 2H2 + O2
960502
Since the oxygen is collected over water, you must subtract out the water vapor pressure to determine the pressure due to just the oxygen.\n\n722.2 mmHg - 21.1 mmHg = 701.1 mmHg = 0.922 atm due to O2\n\n0.276g KClO3 = 0.00175 moles KClO3 (molar mass = 158 g/mol)\nSince KClO3 decomposes to form O2 in a 2:3 ratio, the number of moles of O2 produced = 0.00262 moles O2\n\nPV=nRT\n(0.922 atm) V  = 0.00262 mol (0.08206)(296.2K)\n\nSolving for the volume due to the O2,  V = 691 mL
310893
The best thing would be a bit of bleach.  Do the math - you should shoot for about 2 to 4 parts per million hypochlorite (since commercial preparations are usually 5% - which is 50,000 parts per million - YOU NEED TO DILUTE IT 25,000 FOLD BEFORE YOU DRINK IT - SEE MY NOTE AT THE BOTTOM).  You should also try to store the water in the dark, in an opaque container, or at least in a brown or green glas container since algae growth is photosynthetically driven (i.e. needs light)\n\nBE CAREFUL HANDLING BLEACH - AT HIGH CONCENTRATIONS IT IS TOXIC AND CORROSIVE.
185385
Basically you need to oxidise ammonia and then  adsorb the gases into water.\n\nSee the site below.
254723
zinc sulfide is insoluble in water, zinc chloride is.. simply add water, mix it up, and what ever settles to the bottom is zinc sulfide, the zinc chloride will solublize in the water... decant off the water, or filter the water somehow and collect the solids that remain, theres your zinc sulfide. if you want the zinc chloride too, simply boil off the water from your filtrate (the water you passed through a filter)... provided there are no other impurities in your original solid mixture, and you used distilled or deionized water for your separation..... the solid that remains after you boil off the water should be mostly, or completely zinc chloride.
1168680
To purify (separate) the product you want from impurities. For example, if you wanted to separate a mix of alcohol and water, you would use the distillation flask along with a heating mantle (like a Bunsen burner, but designed for flasks) to heat the mixture; because alcohol has a lower boiling point than water, it would evaporate faster and re-condense in the collection flask, leaving water in the distillation flask.
798162
Timothy gives good advice.  If you are mixing helium with O2, then it must come through a SCUBA shop, else you risk disaster!\n\nFor other uses, none of those gases should be restricted for sale in any way.  A welding supply would hook you up.
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!
744587
Hydrogen powered autos is the way to go. The sun's energy can be used to separate Hydrogen from Oxygen, so you definitely have something environmentally friendly.