Which gas diffuses faster, NH3 or NO2? How many times faster?
595896
This is similar to your other question.  Apply the same equation that the relatively speeds are the square root of the ratio of the molar masses\n\nNH3 has a mass of 17.  NO2 has a mass of 46.\n\nrate1/rate2 = sqrt(46/17) = 1.64
672620
2. An oxygen molecule moving at 560 meters/second.
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...
496847
N2- has a bond order of 3.\n\nhttp://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html
1023754
Diffusion of gas occurs through the one cell layer thick aveoli in the lungs. The speed of diffusion is dependent on different factors, one being the surface area and second the size of the diffusing particles. In the lungs there is ample surface and Oxygen and Co2 are small particles so the rate of diffusion is great. Hemoglobin has a high affinity for O2 in the lungs and a low affinity for CO2. In the tissues the role reverses and hemoglobin has a low affinity for O2 and a greater affinity for CO2. Now in the tissues since they are small molecules O2 and CO2 readily diffuse across the cell membranes and require no special transport mechanisms. This is simple diffusion where gasses are diffusing down their concentration gradient.
839202
Last week I wrote a detailed answer to a very similar question.  Try looking at that first, and follow the same procedure for your ammonia oxidation question.
813106
Just an initial note: NH3 is ammonia\n\nA. \nInitial math:\nSince you have 8.5 grams of NH3, you have to convert grams mols, then into molecules. \nSo you do: 8.5 grams of NH3 / 17 grams in 1 mole of NH3 and X 6.022 x 10^23 molecules in one mole\nYou get 3.011 x 10 23 molecules, and from that you subtract 1.5 x 10^23 molecules, and you get: \n<b>1.511 x 10^23</b>\n\nB. Then, you convert 1.5 x 10^23 to moles by dividing by 6.022 x 10^23 moles, which equals .249 moles of NH3. \nThen you convert 8.5 grams of NH3 to moles too by dividing 17 grams of NH3 per mole again, and get: .5 moles! \nso: .5 mol - .249 mol = <b>.241 mol</b>\n\nC. \nFrom before, you know 1.5 x 10^23 molecules of NH3 is .249 moles. You convert this to grams by multiplying .249 by 17 grams per mol and get: 4.233 grams. \nSo: 8.5 grams - 4.233 grams = <b>4.267 grams</b>\n\nD. From before, you know you have .241 moles of ammonia left. So, you multiply this by avocado's number (6.022 x 10^23) and get: 1.45 x 10^23 molecules of ammonia. Then, you can see that there are 3 atoms of hydrogen of per molecule of ammonia, and multiply the number of molecules by 3 and get: \n<b> 4.354 x 10^23</b> \n\nI think those should be right. Sorry about the sig figs though!
876773
The first reaction is simply wrong.  Nitrous acid does not ionize to nitrate.\n\nThe second reaction is a precipitation reaction, showing the net ionic equation.  The compound is ionic.\n\nThe third reaction is an example of SN2 displacement, not elimination.  Only single bonds are formed.
207024
It depends on the permeation time of the smell (ie. dense gases spread faster in a closed environment than less condesed ones.)  Wind helps as well if it is an open environment.\n\nTo put it simply smell has too many variables to create a unified theory of odor permeation.\n\nBut I do know my farts seems to move fasters than other odors.
592234
3g of hidrogen (plus 14 g of nitrogen) make up 17g of ammonia (H+H+H+N=1+1+1+14), so:\n\n3g  H____17g NH3\n50g H____=283.33g NH3\n\nI assumed a 100% efficient reaction (that means every bit of reactant will actually react).
714001
1. NO2-\n\n2. tellurite\n\n3. pernitrate\n\n4.NH4NO3\n\n5. OH-
873492
oxygen 21%\nnitrogen 78%\nco2 0.03%\nrest - inert gas and water vapours
239315
im pretty sure that would be a very volatile compound but im not sure what the combination or final chemical compound would create but I think the best place to find ammonia nitrate is in ceratin types of fertilizers
439393
The nitrogen gas in air gets converted to 1) ammonia 2) nitrates (from nitrites)
731423
The most probable speed is the highest point on each curve, so b. is true.\n\nc. is also true, because SOME of the molecules of gas III have little or no velocity.\n\nd.  This is true, if the gases are ideal gases.\n\nYou are correct about a and e being false.
829049
Ammonia (NH3) is the catalyst. Are u in 10th class?
877279
First you need to determine your limiting reagent in the first step.\n\nMolecular mass of NH3 = 17 g/mol, O2 = 32 g/mol\nMol of NH3 = 5,700,000 g/17 g/mol\nMol of NH3 = 3.35E5 mol\n\nMol of O2 = 57,000,000 g/32 g/mol\nMol of O2 = 1.78E6 mol\n\nSo based on the first equation (1), NH3 is the limiting reagent.\nFor every 4 mol of NH3, one gets 4 mol of NO.  So the first reaction produced 3.35E5 mol of NO.\n\nThe second reaction produces 2 mol of NO2 for every 2 mol of NO used, so the second reaction produced 3.35E5 mol of NO2.\n\nThe third reaction produces 2 mol of HNO3 for every 3 mol of NO2 used, so this final reaction will produce 2/3 x 3.35E5 mol of HNO3 = 2.23E5 mol.\n\nThen use molecular mass of HNO3 to calculate how much of it (mass) that you made.\nHNO3 molecular mass = 63 g/mol\nmass = mol x molar mass\nmass = 2.23E5 x 63 g/mol\nmass = 1.41E7 g of HNO3\n\nOf course this assumes 100% yields in all the above noted reactions.