Can oxygen be deliberated from all oxygen containing compound?
88765
yea - but then it would be "blank" containing compound
720001
strictly speaking, substances undergo combustion in the presence of oxygen and 'burn' would not be precise. \nhowever answering the question, oxygen is an oxidising substance and hence it is able to oxidise the organic subs. and usually converting them into carbon di oxide and water, and this is accompanied by the release of energy in the form of heat
394104
Yes. Everything is a chemical element.\n\nCSO2 I believe.
570980
If an element combines with another element to form more than one compound, the various weights of one compound which combine with the fixed weight of the other element bear a simple ratio to one another eg\nCarbon Monoxide C 12  + O 16\ncarbon Dioxide  C 12  + O 32\nRatio of 1 to 2
65886
Synthesis - a compound forms from simpler reactants
77405
Boron, Carbon, Nitrogen, Oxygen
800603
Peroxide. The simplest related neutral compound is hydrogen peroxide, H2O2, H-O-O-H. Take off 2H+, and you have O-O with a (-) on each O.
937834
Well, you are getting in to a bit more complicated chemistry. For the question you are asking, the simple answer is No. However there are multi-valence ions, such as Oxygen. Normally Oxygen would gain 2 electrons making it O(2-) but it has another valence of 1 where it only gains one electron and becomes O(1-). By the sounds of it you are not at this point so don't worry about but you will get there eventually. In order to understand multi-valence you need a Periodic Table of Elements that lists all the valences of each element. This makes bonding harder, etc,etc so I wouldn't worry about it if I were you, yet.\n\nSo to simply answer your question, No, it is impossible for Magnesium
359329
In truth, it is not an element. However, oxygen occurs in nature as O2. Therefore, we call it elemental oxygen.
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
860447
There's no separation of molecular oxygen to atomic oxygen as previously written:\nsimply \nOx: Sr -> Sr2+ + 2e \nRed: O2 + 4e -> 2 O2-
979598
No. Combustion occurs b/c something reacts with oxygen. So unless the helium is contaminated with oxygen, no combustion can take place.
22780
Thank you punnet pundit. I was thinking the same thing. \n\nWhen combining Hydrogen and Oxygen one will get water or hydrogen peroxide. Simple enough. But that's not all there is to it. Just mixing the two doesn't cut it. The mixture must first gain enough energy to react. This is called it's Activation Energy. Which is the amount of energy it needs to absorb in order to carry out a chemical reaction. In the case of water the reaction is spontaneous and once started will continue until there is no longer enough reactants to form products. \n\nIn the case of water the reaction is explosive and quite violent. Liquid Oxygen and Liquid Hydrogen are used in space ships, because the reaction is enough to give the rocket a signifigant amount of upward force. The product is steam really because the water is way above 212 degrees F. but if this is condensed onto something below that temperature you get pure water.\n\nHydrogen peroxide slowly over time can break down in light. That's all I really know about it.
506214
Ans. C) lead (IV) oxide acts as a catalyst and speeds up the reaction, so that more O2 is evolved per unit time.\n\nAns. D) --by increasing the temperature
898426
http://en.wikipedia.org/wiki/Chemical_oxygen_generator
727114
Yes, you can split the C and O from CO2, as said above its done by plants all the time.  There are also several other chemical reactions (the lithium based CO2 scrubbers on the space shuttle).  The problem that that poses, is that in an absolutely ideal world (which this isnt) it would require as much energy to break the CO2 as it gave off when you made it.  For example when you burn coal (pure carbon) you combine it with oxygen and it gives off heat (energy) and CO2.  This is because CO2 is at a much lower chemical energy state.  In order to break the CO2 you would have to add in the exact same amount of energy in order to make it Coal and Oxygen again.  And in the real world, you lose on both sides because the reactions are't 100% efficient.
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!