How do I utalize the oxidizing propperties of potassium nitrate so I can feed o2 into the flame?
954959
Potassium nitrate, or any other nitrate, is an oxidizer.  Combustion is an oxidation, so you need something oxidizable (combustable).  This would be called a reducing agent.  A typical reducing agent is any carbon molecule with lots of hydrogens.  \n\nYou need to know the reaction to get the right proportions.  Potassium nitrate mixtures can behave unpredictably, and can be dangerous.  A common one is black gunpowder.  It is a mixture of salpeter (potassium nitrate), charcoal, and sulfur.
394069
3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nFirst you must calculate the (average)oxidation number of central atom of main compounds in reactants.\nEx;Cr oxidation no=+6 in K2Cr2O7\n   C  oxidation no=-2 in C2H5OH\nThen in the products.\nEx;Cr oxidation no=+3 in Cr2(SO4)3\n   C  oxidation no=0 in CH3COOH\n\nThen you must evalute how many electrons must be transferred.\nEx;Oxidation no of Cr is changing from +6 to +3 \n   So it is reduced.\n   SO each atom of Cr must receive 3 electrons.\n   So totally 6 electrons must be received by K2Cr2O7\n   \n   Oxidation no of C is changing from -2 to 0\n   SO each atom of C must donate 2 electrons.\n   So totally 4 electrons must be donated by C2H5OH\n\nThen you must write now a explantion reaction.\nEx;you must write oppsite (simple)propotion of electrons in front of each compound\n  3C2H5OH + 2K2Cr2O7\n \nThen you can calculate how many Cr,C(,K) atoms should be there\n So, 3C2H5OH + 2K2Cr2O7=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nThen you can calculate the medium(SO42-) should be there\n i think 8 SO4 should be there\n\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n\nFinally you must equilize H,O atoms(balanced by water)\nSo,3C2H5OH + 2K2Cr2O7 + 8H2SO4=3CH3COOH + 2K2SO4 +2Cr2(SO4)3\n                                        +11H2O\n\nSecond one ................\nC2H5OH+4I2+6NaOH=CHI3+HCOONa\n                  +5NaI+5H2O
206901
Add heat.\nAdd a catalyst.
131270
para-nitroacetophenone, in other words.\n\nMy first inclination is a Friedel-Craft acylation followed by a nitration with nitric acid in sulfuric acid (the nitronium ion). But the addition of the acetyl group would direct the nitro group to the meta position.\n\nMaybe if you nitrate the ring first and reduce the nitro group to an amino group. Then you could Friedel-craft to get the ketone at the para position (with some ortho isomerism). You'd have to find a way to oxidize the amino group to a nitro group ... I don't know how.\n\nAnother method would involve oxidizing toluene with KMn04, to get benzoic acid. You could then esterify the benzoic acid. Using DIBALH (Diisobutyllithium aluminium hydride), you could get an aldehyde (1-phenyl-methanal). You could use a Grignard reagent (methyl magnesium bromide) to get the alcohol with a methyl group added. You could then oxidize the alcohol with PCC to get a ketone. Finally, you could nitrate the ring with nitric acid and sulfuric acid.\n\nI'm not sure if either synthesis is perfect; My second one is long and rather expensive, in terms of reagents. Also, PCC is toxic (chromium) and many of the reducing agents are water sensitive.
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.
396644
Like this?\nhttp://www.webelements.com/webelements/elements/media/pics/rainbow.jpg\n\nTo quote the web site,\n"The picture above shows the result of adding different metal salts to a burning reaction mixture of potassium chlorate and sucrose. The red colour originates from strontium sulphate. The orange/yellow colour originates from sodium chloride. The green colour originates from barium chlorate. The blue colour originates from copper (I) chloride. The lilac colour that should be evident from the potassium chlorate is washed out by the other colours, all of which are more intense."\n\nBasically, certain salts will color the flames various colors due to how the electrons are excited (more complicated than that, look into spectroscopy).
828284
The first thing that happens is that it melts.  If you continue heating the molten salt, you will see bubbles appearing.  These are bubbles of oxygen:\n2 KClO3(l)   --->   3 O2(g)   +   2 KCl(l)\nThe main product left behind is potassium chloride, but I believe that you also get some potassium perchlorate (tetraoxochlorate(VII)), produced by disproportionation:\n4 KClO3(l)   --->   3 KClO4(l)   +   KCl(l)\n\nIt wouldn't surprise me if the reaction also produced very small amounts of oxides of chlorine, such as chlorine dioxide, ClO2.\n\nAllan Deeds.
231663
_KCl3 _KCL + _o2? K2Cl4o2 thats the answer i think.:)
536199
follow the above.. to get gas.. electrolysis is the safest bet\n\nbut if you want to have fun???\n\nput some soap in the water and stir it around before you put the electrodes, this creates hydrogen bubbles that float and when put next to fire make a sweet pop sound.. do it safely, in a lab\n\nits a good demo for a highschool chem class
113583
The common name for potassium nitrate is saltpeter.  A suitable substitute would be sodium nitrate, which is also called nitrate of soda or chili saltpeter.  (This assumes that you are making something like gun powder).\nIf you are feeding plants potassium phosphate might make a good substitute.
1120784
Well to start off any reaction that produces oxygen is potentially combustible, especially one that requires energy (endothermic) so be careful.\n\nTwo suggestions: \nFirst, electrolysis, this is where you break the covalent bond between oxygen and hydrogen with electricity (energy).  The by products are oxygen and hydrogen gas.\n\nMy second idea is simple rust.  Put iron (Fe) into water (H2O) and you will produce Iron-oxide or rust.  The Oxygen combines with the iron.\n\nGood luck.
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!
862596
K2SO4\nCa(NO3)2\n\npotassium has a +1 charge and sulfate has a -2 so therefore you need 2 potassiums so that you can have a net zero charge.\ncalcium has a +2 charge and nitrate has a -1 charge so you need two nitrates to balance the charges
919726
Sodium and Potassium are correct.  They are extremely reactive with water, forming Hydroxides and releasing Hydrogen gas, which is usually ignited from the exothermic heat of the reaction.\n\n2 Na + 2 H2O -> 2 NaOH + H2\n\n2 K + 2H2O -> 2 KOH + H2
12446
Both group 1 and 2 nitrates give oxides on heating.\ngroup 1 carbonates are thermally very stable but group 2 carbonates are thermally unstable, the unstability increasing with cationic size.
652769
Galvinisation with zinc. \n\nChroming\n\nboth work off the basis of reduction and their poteintal to be oxidised\n\nlook in an chemistry A-Level text book. Or if your american look at a college text book, for exact details
549468
Add some baking soda first and then pour on some vinegar. You will observe some bubble forming and the bublbles will rise very quickly. If you want to make fire instead, add some glycerine to potassium permagnate. Be very careful if you want to do this.