If you've got 8.5 grams of NH3 (Amoniac) and you eliminate 1.5x10^23 molecules...?
959238
mole of NH3 = 8.5 / 17 = 0.5 mole\nmolecules of NH3 = 0.5 * 6.02 * 10^23 = 3.01 * 10^23 molecules\n\nA.\nmolecules of NH3 left = 3.01 * 10^23 - 1.50 * 10^23 = 1.51 * 10^23 molecules\n\nB.\nmole of NH3 left = 1.51 * 10^23 / 6.02 * 10^23 = 0.25 mole\n\nC.\nmass of NH3 left = 0.25 * 17 = 4.26 grams\n\nD.\nmole of NH3 left = 0.25 mole\nmole of H atoms = 3 * 0.25 = 0.75 mole\natoms of H left = 0.75 * 6.02 * 10^23 = 4.51 * 10^23 atoms\n\npeace\nvixklen
1062523
2NaI + Hg(NO3)2 --> HgI2 + NaNO3\nwe have;\n52 ml and a 0.78M Hg(NO3)\ntherefore we can find # of Moles\nn = CV\nn = 0.78 * 0.052                       Concentration should be in Litres\nn = 0.04056 mol\n\naccording to the equation;\n2 moles of NaI react with 1 mole of Hg(NO3)2\ntherefore 2(0.04056) moles of NaI needed in the reaction\ni.e. 0.08112\nwe know V i.e. Volume equals the ratio of #of moles and Concentration \ntherefore ; \nV= n/C\nV= 0.08112 / 1.60\nV= 0.0507 L\ni.e. 50.7 ml of NaI required in the reaction !!
1013801
Molar mass of NH4NO3 = 14+4+14+48 = 80\n\nSo, 1.4 g is equal to 1.4/80 = 0.0175 mol\n\nNow, mass of soln = 74+1.4 g = 75.4 g\n\nSo, heat absorbed = 75.4x4.18x(25-23.71) = 404.415 J\n\nSo, 0.0175 mol absorbs 404.415 J\n\nSo, 1 mol absorbs 23109.429 J\n\nSo, heat of dissolution of NH4NO3 is 23.109 kJ/mol (+ve as heat is absorbed)\n\nb) Lattice energy = 630+23.109 = 653.109 kJ/mol
412358
You could skip the Nitrogen Oxide and use Nitrous Oxide.   :o)
829049
Ammonia (NH3) is the catalyst. Are u in 10th class?
917667
Part A:\nThe answer is 12 L H2.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(3 mol H2 / 1 mol N2)(22.4 L H2 / 1 mol H2)\nso (4*1*3*22.4) / (22.4*1*1) = 12 L H2\n**note that in in ideal gas there are 22.4 liters/mol**\n\nPart B:\nThe answer is 8 L NH3.  Here is the solution:\n(4 L N2)(1 mol N2 / 22.4 L N2)(2 mol NH2 / 1 mol N2)(22.4 L NH3 / 1 mol NH3)\nso (4*1*2*22.4) / (22.4*1*1) = 8 L NH3\n\nHope this helps!
842976
20 grams\ndissolving compounds and then driving of the solvent is a common way to purify a compound. You will not lose any of the compound unless it is reactive with the solvent you dissolve it in. In the case of water and salt they are not reactive so when you drive off the water you will be left with the same amount of salt that you started with.
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.
149153
This is a limiting reagent problem. You must calculate the moles of NH3 produces from the given amount of grams of N2 and H2.\n\n3.00g N2 x 1mol/28.02g x 2 mol NH3/1 mol N2 = .214 mol NH3\n\n.54g H2 x 1 mol/2.02g H2 x 2 mol NH3/3 mol H2 = .178 mol NH3\n\nFrom this calculation you now know that the H2 is the limiting reagent. The next part to calculate the grams of NH3 goes like this:\n\n.178 mol NH3 x 17.04g/1 mol = 3.03 g NH3
750432
Amount of N2 which reacted = 21/ relative molecular mass of N2\n                                               = 21/ 28\n                                               = 0.75mol.\nMass of H = (0.75 x 6/2) x 2 = 4.5g of H2\nMass of NH3 = (0.75 x 2) x 17 =25.5g of NH3.\n\nThe value in the parentheses are your mol values of H and NH3 respectively.
781038
I agree with birchardv here...........we need the rest of the problem, and you would learn a lot more if you did the homework yourself.  But do whatever floats your boat.  Add the rest of the problem and I'll get back to you.\n\nOkay Chemist, this problem is really long; if you want a good answer with plenty of steps, I'll need more room.  Email me and I'll send you the solution.
457812
Ammonia, baking soda.
518629
not too much, they are both basic, and don't really react. even if it is an aqueous solution of ammonia (thus avoiding the gas problem above)
247481
If I'm not mistaken, they will have the same pressure, as they are in equal number of molecules. 17grams of Amonia or NH3 sum up 1mol of NH3 molecules while 2 grams of Hydrogen, H2 consists of 1mol of Hydrogen molecules. \nAn equation : Pv = nrT, where P= pressure, v=volume, n is the number of molecule/atom, r is a constant, and T is the absolute temperature (in Kelvin). Since they are equal in their number of molecules, it share the same pressure. Hope it helps.
1083881
Reduction - Mn from +7 state to +4 state, N from +5 state to +4 state. Addition of H+ will lead to reduction reaction most of the times.
413959
Well, this is a simple factor labelling problem... I hope you aren't using this to have people do your homework... ;)\n\n1pound x (454 grams/pound) x (1 mol H atoms/1.0079g H atoms) x (6.022x10^23 H atoms/1 mol H atoms) = 2.71 x 10^26 H atoms\n\nnote to PaulB_23: periodic tables are in amu, which is also a ratio of grams per mole, not moles per kg, the 1.008 would just be 1.0079 rounded, I just went w/ a slightly more accurate #
942373
From the reaction\n\nS8(g) ===> 4S2(g)\n\nKp = (pS2(g))4/pS8(g) where p is the partial pressure\n\nGiven the partial pressure of S8(g) at equilbrium 0.25 atm, clearly 0.75 atm of S8(g) has been consumed to produce (from the equation) 4 volumes of S2(g), Given the volume was fixed this mean dS2(g) = 3 atm\n\nSubstituting into Kp yields\n\nKp =3^4/0.25 = 324\n\nThus Kp at 1325 K is 324