Nitric oxide,   NO,   reacts instantly with oxygen gas to give nitrogen dioxide,   NO2,   a dark brown gas.?
957244
2NO + O2 --> 2NO2\nMr of NO = 14 + 16 = 30\nMoles of NO = (29.03/30) = 0.9677 (to 4 s.f.)\nMr of O2 = 32\nMoles of O2 = (19.35/32) = 0.6047 (to 4 s.f.)\n\nComparing the moles of NO and O2 given, you should notice that NO is the limiting reagent.\nHence moles of NO2 produced = moles of NO used = 0.9677\n(since the ratio of NO2 to NO is 2:2)\nMass of NO2 produced = 0.9677 x (Mr of NO2) = 0.9677 x 46 = 44.5 g (to 3 s.f.)\n\nHope this helps :)
712891
In one experiment 32.40 g of NO is mixed with 19.20 g of O2.\n\n\nThe reaction goes: 2NO + O2 -> 2NO2\n\n\nFirst, you have to find out which is the limiting reactant.\n\nno of moles of NO\n= mass / molar mass\n= 32.40 / (14 + 16)\n= 32.40 / 30\n= 1.08\n\nno of moles of O2\n= mass / molar mass\n= 19.20 / (2 x 16)\n= 0.6\n\nGoing by the equation, there must be twice the no of moles of NO than O2. However, the no of moles of O2 supplied is more than half of the no of moles of NO. Thus, O2 is in excess and NO is the limiting reactant.\n\nWorking with the no of moles of NO,\nno of moles of NO2\n= no of moles of NO\n= 1.08\n\nmass of NO2\n= (no of moles) x (molar mass of NO2)\n= 1.08 x (14 + 2 x 16)\n= 1.08 x 46\n= 49.68g\n\nAnd there you have it! :D
1035732
Nitrous oxide, also known as laughing gas
412358
You could skip the Nitrogen Oxide and use Nitrous Oxide.   :o)
668488
Unleaded fuels-so no lead\nnitrogen compounds come as a result of air combustion in your engine (remember air is approx. 80% Nitrogen).  and you gotta have air to make your engine run.\nSulfur gases can also be a result of this combustion if there is high sulfur content in the fuel. (Shell had a problem with this last year)\nPS. Don't buy into the greenhouse hype, at this point there is NO real evidence either way that fuel combustion makes any difference in the grand scheme of climate dynamics
339232
Yes, the reaction would spontaneously occur because Sn is higher on the activity series than Ag.\n\nSn(s) + 2AgNO3(aq) => Sn(NO3)2(aq) + 2 Ag(s)\nThe net ionic is\nSn(s) + 2Ag+(aq) => Sn+2(aq) + 2 Ag(s)\n\nI don't remember the oxidation number for Sn.  2 sounds about right.\n\nIt would turn a silver nitrate solution to a tin nitrate solution, and you'd see metallic silver precipitate out.
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
641390
no..nitric acid is not produced by plants...nitric acid is  produced by the reaction between water and nitrogen di oxide..due to it's acidic nature plants would die very soon if they were to produce it .
214694
The creation of stratospheric ozone is chiefly due to what is called the Chapman cycle which represents the production and subsequent annihilation of ozone by solar UV radiation . Destruction of ozone also results from catalytic reactions with hydrogen atoms, hydroxyl radicals (symbolised by HO-, derived from water H2O), nitric oxide (NO, from the exhausts of high-altitude aircraft), and so-called halogens, notably chlorine and bromine compounds. A catalytic reaction allows a single molecule of the triggering chemical to repeatedly react with the source chemical until its reservoir becomes depleted. However, there are complications, such as the inhibition of the destructive effects of both ClO (a compound of chlorine and oxygen) and NO2 (from NO) by their forming chlorine nitrate (ClONO2) instead. Chlorine nitrate does not destroy ozone, but it has the potential to destroy ozone since it is a reservoir of ClO.
793971
According to the equation:\n\n2NO + O2 --> 2NO2\n\n2 mol of NO reacts with 1 mol O2, so\n0.886 mol NO reacts with 0.886/2 = 0.443 mol O2\n\nWe have 0.503 mol O2 > 0.443 mol. So O2 is in excess and NO is the limiting reactant. It reacts fully and produces 0.886 mol NO2 (because 2 mol NO give 2 mol NO2)
629324
Nitrogen Dioxide is red. Nitrogen gas is colourless, otherwise you won't be able to see anything, since air is mainly made up of nitrogen.
377149
Nitrogen (N2) 78.084 % by volume\nOxygen(O2)20.9476 % by volume\nArgon(Ar)0.934 % by volume\nCarbon Dioxide(CO2)0.0314 % by volume
521349
Nitric acid is a very strong oxidizing agent, so on reaction with metals, it gives salt (nitrate), nitrogen dioxode and oxygen.\n\nE.g: Cu + HNO3 ---> Cu(NO3)2 + NO2 + O2\n\nHNO3 self undergoes reduction to NO2 and metals reacting are oxidized.\n\nMetal with HCl give salt and hydrogen and with dilute sulphric acid give salt and hydrogen too, but if you use concentrated sulphuric acid with some metals, then SO2 can be released, because again with more concentration, sulphuric acid becomes a stronger oxidizing agent and gets reduced to SO2 in a redox reaction.\n\nEDIT: The reaction above for copper occurs with concentrated nitric acid; if dilute nitric acid is used, NO would be released instead of NO2.
1083534
Nitrogen(I)oxide - N2O- is used as a mild anaesthetic for minor surgical operations, such as in dentistry and as a detonator \n\nNitrogen monoxide [2 oxide]- NO- is used in the production of nitrosyls by combining with transition metal derivatives; it combines with nitrous oxide to form oxynitride gates in CMOS devices. It is an intermediate of the Ostwald's process for the manufacture of nitric acid. It is also used in detecting surface radicals on polymers.\n\nNitrogen dioxide [4 oxide]- NO2- is used as a oxidizing agent and as an intermediate in the formation of nitric acid. The liquid acts as a good solvent and anhydrous nitrates can be prepared in the solvent.
875082
a. N2O4\nb.PCl5\nc.NF3\nd.S2Cl2
954211
Your first conclusion is correct, but your second isn't.  Since you already have the mechanism given to you, you can see that the 2nd step is the slow step.  So, the rate law should be:\n\nrate = k (N2O2) (H2)\n\nSince N2O2 is an intermediate, it can't appear in the overall rate law, so you have to get rid of that.  From the first reaction, you see that N2O2 is formed from the collision of 2 NO.  So, the concentration of N2O2 is proportional to [NO]^2.  That gives you the overall rate law that you had, and the overall order of the reaction is 3rd order.\n\nYour second conclusion is incorrect because the slow step requires the collision of N2O2 and H2.  The slow step is just a second order reaction.  \n\nSo, you've got things pretty well in hand...Hope this helps a little...
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)