Small quantities of oxygen gas are sometimes generated by heating   KClO3   in the presence of   MnO2   as a c
830712
The MW of KClO3 is 122.55 (ref.), so 0.274 g KClO3 is .0022358 M. This will evolve 3/2 * .0022358 = .0033537 M O2 * 22.4 L/M = .075124 L O2 at STP. Correcting for temperature, pressure and partial pressure H2O, multiply by 295/273 * 760 / (726.2 -19.8) and volume at given temp. and press. = 0.087337 L.
179426
well i think \nCH3COOH -kmno4--> 2COOH\n         \nAS KMNO4 IS AN OXIDISING AGENT\ni hope so ask ur teach and plz mail me with a correct answer
413674
Some of the elements while in gaseous state.
772976
Combustion generally gives off CO2 and H2O vapor.
363430
Carbon with 3 bonds to oxygen\n\nC-O\n\nwith 3 of the - signs\n\nIt makes carbon monoxide
864677
carbon monoxide = CO\nIron (II) Chlorate= Fe(ClO3)2\nLead (IV) oxide = PbO4\nNitric acid = HNO3\nSulfuric Acid = H2SO4\n\nall numbers are subscripts.
775000
2 MnO4-(aq ) + 5 H2C2O4(aq ) + 6 H3O+(aq ) --> 2Mn2+(aq ) + 10 CO2(aq ) + 14 H2O
211591
Ca3PO4 and CuOH\n\n^^those are the answers
504488
it is a chemical reaction and the gas is c02
1022360
- known oxidation levels of the participants\n- conditions under which the reaction will take place (i.e. possible subsequent reactions which may tip the equilibrium elsewhere, like precipitation of MnO2 in alkaline medium rather than reduction to Mn(2+) from KMnO4).\n-  electron balance. Write the partial reactions for a single component giving up or gaining electrons, then multiply so that you get the smallest possible whole number for all factors.
875082
a. N2O4\nb.PCl5\nc.NF3\nd.S2Cl2
169686
manganese oxide and peroxide
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2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
532416
It is CS2+3(O2)=CO2+2(SO2)
728598
2 KClO3--->  KCl + 3O2 \n\nthe molecular weight of KClO3 is 122.5\n\nso you can see that 2*122.5 = 245g of KClO3 corresponds to 3 moles of O2\n\n0.261 g corresponds to 0.261*3/122.5 = 0.064 moles of O2\n\nto calculate the volume you use the  formula  pV = nRT ; \nT = 293 K  and 1mmHg = 133 Pa\n\nV = 0.064*8.31 * 293/ 722.7*133 =0.0016m^3 = 1.6 liter
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
895580
Mercury(II) oxide decomposes into its elements when heated.\nThere is no way that you would know this unless you had been taught about thermal decomposition. So all you need to worry about is the correct formula of oxygen gas.\n\nTherefore 2HgO --> 2Hg + O2.