can you draw the spacious form of the NO2 molecular?
401004
Yes, see for example the page below:
712728
choose the noble gas that comes before the element.   in Mn(2+)'s case, that's Ar\n\nwrite the noble gas down in brackets, and continue on with the electron config. as normal\n\n[Ar}4s23d3\n\nFor O(2-), it turns into Neon so it would just be [Ne]
504488
it is a chemical reaction and the gas is c02
937526
If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)\nFill in the atomic masses of H = 1.00797 and N = 14.0067:\n[3 x 2 x (1.00797)] +[ 2 x (14.0067)]  = 2 x (14.0067) x 3 x (1.00797) = 34.06122\nThis is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:\n2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.
555626
The Lewis dot structure would be that N in the middle with branches off to oxygen and 2 Fs and then one single electron will be on nitrogen and it is a radical I think. It is very reactive because of this one electron. The electron geometry will be something with sp3 orbital, I am not sure since it only has one electron. I would check up on that. Hope that helps
856026
Use the gas law\nPV=nRT
328129
just draw a triangle has right angle which dimensions 1,sqrt(3),and hypothenus 2. if you want it, i can send to you
412358
You could skip the Nitrogen Oxide and use Nitrous Oxide.   :o)
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
436740
CH2Cl2\nThe molecule shape is tetraedric with the carbon in the middle
806618
The only difference I can see would be the molecular geometry, due to the number of H's bonded onto the anion. Also, when drawing the Lewis dot structures, NH3 (N) would have one lone pair of electrons, HF (F) and H2O (O) would have two. This would dictate their structures according to the VSEPR rules.
644051
See if this article has a diagram\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1999/Apr/abs542.html
956863
Gotta love wiki\n\nhttp://en.wikipedia.org/wiki/Octane\n\nclick on each one of the 18 isomers, then you can draw. Basically, you shift a methyl group from carbon 1 on down the line so you don't have a straight hydrocarbon chain.
788575
Ozone is trigonal planar.  The lone pairs on the terminal oxygens give the trigonal structure, and the molecule is planar (2 dimensional) mainly due to the double bonds in the most stable resonance structure of ozone.\n\nH2Se should be a bent molecule.  The lone pairs take up maximum available space forcing the tiny hydrogen atoms closer together than they's like to be at an angle of about 91 deg from the Se atom.
1015209
BECOZ of gases and light
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
18683
Look at your periodic table. \n\nHydrogen - 1s1\nHelium - 1s2\nLithium - 1s2 2s1\nGo to Neon - 1s2 2s2 2p6\nArgon - 1s2 2s2 2p6 3s2 3p6\nKrypton - 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6\n\nSo, let's say you need Silver (Ag)\n[Kr] 5s2 4d9\n\nIt's perfectly acceptable to use a noble gas in brackets to represent the configuration.\n\nI hope this give you an idea how to write it all.