Nitrogen --> nitrate?
356333
Leguminous plants contain root nodules that contain symbiotic Rhizobium bacteria, responsible for nitrogen fixation i.e. it fixes the nitrogen in air into the soil. The soil contains nitrifying bacteria - nitrosomonas and nitrobacter - that bring about the conversion of nitrogen to nitrate.\nThe nitrogen in the soil is converted to ammonia. Nitrosomonas bacteria convert this to ammonium compounds in the form of nitrites; nitrobacter bacteria (chemosynthetic) then oxidize nitrites to nitrates using up energy from this oxidation for their benefit.\nThe nitrates thus formed are fixed into the root nodules of leguminous plants by rhizobium bacteria and taken up.
439393
The nitrogen gas in air gets converted to 1) ammonia 2) nitrates (from nitrites)
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
788534
N2 + 3H2 -> 2NH3\n\nN2 has a gfm of 28g/mol and H2 has a gfm of 2g/mol.  (you do the math to find the moles I will make a rough guess) you've got about .5 mols of N2 and about 1 mole of H2, so H2 is your limiting reagent.  your ratio of H2 to NH3 is 2molNH3/3moleH2 and you multiply that by the one mole of H2.\n\nNH3's gfm is 17g/mol so .667molNH3 x 17g/molNH3 ~ 12g\n\n*edit* make sure you do the math over because I was guessing and estimating where you should not *edit*
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.
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
683249
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
693408
It's not nitrogen... It's hydrogen. They called my car the "Hindenburg" after it burst into flames when I pumped in a few pounds too many way back when in Lakehurst, New Jersey... Oh, the humanity!
600524
Give the mass of one reagent in a reaction, and ask for the mass of another. A word equation or chemical equation must be provided.\n\neg:\n\nWhat mass of nitrogen gas is required to produce 45.0 g of ammonia, according to the reaction below ?\n\n         N2   + 3 H2   ---> 2 NH3\n\n\nTo solve the problem\n\na) convert mass of NH3 to moles (2.65 mol)\nb) use stoichiometry - find moles of N2 needed (1.32 mol) \nc) convert moles of N2 to mass (37.0 g)
706856
Acidic.\nBecause sodium nitrate is a salt solution, and is neutral. Since you are adding an acid to a neutral soution, the end reault will be an acidic mixture.
823698
too many nitrates is not good, because then you get a ton of algae growth and the algae uses up all of the oxygen causing the other organisms to die\ndont know bout the other question
691741
Total Volatile Nitrogen\nhttp://www.omegaproteininc.com/products-natural-nautic.html
814902
The main source for ammonium nitrate  is high-nitrogen fertilizers. Ammonium nitrate is industrially produced through an acid-base reaction of ammonia with nitric acid using anhydrous ammonia gas and concentrated nitric acid.  It is commonly used in agriculture, and it has also been used as an oxidizing agent in explosives, especially improvised explosive devices.♥
840588
First of all, you need a chemical equation,\n\nNO + O2 ----> NO2\n\nThen, balance it,\n\nNO + 1/2 O2 -----> NO2\n\nNow, what this balanced equation tells you is that the mole ratio between NO and O2 is 2:1. i.e., you need 2 parts of NO to react with every 1 part of O2.\n\nNow, you need to calculate the total number of moles of NO and O2 respectively,\n\n# mole of NO = 17.26/(14+16) = 0.575 mole\n\n# mole of O2 = 13.15/(16+16) = 0.411 mole\n\nAs said before, for every 1 mole of O2, it takes 2 moles of NO to react. From the number of moles of NO and O2, it is clear to see that NO is the limiting reagent. Why? Because if you react all 0.575 mole of NO, it takes only 0.575/2 mole (0.2875 mole) of O2, which you have a lot more than that.\n\nThe limiting reagent is the substance that is NOT in excess, which then determines how much product will be produced (because once this substance is all gone, there will be no reactants to react and give product).\n\nNow, from the balanced equation, you know the mole ratio between NO and NO2 is 1:1. Therefore, the number of moles of NO2 that be formed will be 0.573 moles.\n\nMass of NO2 = # mole * molecular weight\n= 0.573 * (14 + 16 + 16) = 26.465 g
716053
in NH3 , N is -3, each H is +1\n\nin NO2- N is +3     each O is -2\n\nin NO3- N is +5, each O is-2
325149
3NH3(aq) + 3H20(l)--> 3NH4+(aq) + 3OH-(aq)\n AlPO4(s) --> Al3+(aq) + PO4 3-(aq)\n\nequal:\n\nAlPO4(s) +3NH3(aq)+3H2O(l)-->Al(OH)3(s) + (NH4)3PO4(aq)\n\nliquid fertilizer is  (NH4)3PO4(aq)
852232
Pv = nRT\nn = m/M (M Molecular weight)\nPV = (m/M)RT\nP = (m/V)RT/M\nPM/RT = m/v = Density.\nPlug the numbers, watch the units and get the density.\n\nOh a numerical answer, yes, ...42.