If you've got 8.5 grams of NH3 (Amoniac) and you eliminate 1.5x10^23 molecules...?
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!
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
897929
convert 995.0 g CH4 into moles and then multiply that answer with 882kJ/mole -- make sure your units cancel
104563
My answer is:\n\n.217N
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!
860532
nitrogen & hydrogen react to give ammonia according to the reaction \nN2 + 3H2 -> 2NH3\naccording to the above reaction 1 & 3 moles of N2 & H2 respectively react to give 2 moles of NH3.\nso, 6 moles ammonia can be produced by reacting 3 moles N2 & 9 moles H2.
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.
879654
multiply the moles by the number 6.02 x 10^23.....altho it would be a very big and unmanageable number but thats how u find it
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
213015
NH3  +  CH3COOH = CH3COO–  +  NH4+\n\nThe gas would be NH4?
243948
Yes, using the David Hasselhoff equation is much easier. The HH equation is:\n\npH = pKa + log(base/acid)\n\nFill in the information you know (either from the problem or a table).\n\n9.00 = 9.25 + log(0.10/x)\n\nSolve for the concentration of NH4.\n\n0.178 = [NH4+]\n\nSince there is 2.0L of solution, use that to solve for the number of moles.\n\n0.178M * 2.0L = 0.356 moles NH4Cl.\n\n\n\nIn case you were wondering, you have to look up the Kb of ammonia (NH3) and convert it to the Ka of ammonium.\n\nKw/Kb = Ka\n\n1e-14/1.78e-5 = Ka = 5.62e-10\n\nTake the -log of that to find the pKa of ammonium, which is needed in the equation.\n\n\nHope this helped.
756159
n   = pV/RT = 1.03*1.02/284*0.0821 = 0.045moles\n\nmoles of Nitrogen = 0.045*0.78=0.0351\n\nmoles of oxygen = 0.022*0.045 = 0.0098mole\n\nweight of the gases = 0.0098*32 + 0.0351*28.02 =0.316+ 0.986=1.3g\n\ndensity 1.3 g/l
773122
n(H2O) = m/M = 8.02 / 18.02 = 0.4451 mols\n\nno. of molecules = n(H2O) x Avogadro's Number = 0.4451 x (6.02 x 10^23) = 2.68 x 10^23\n\n1 molecule of H20 = 2 atoms of H and 1 atom of 0\n\nno. of atoms of H = 2 x no. of molecules of H2O = 2 x (2.68 x 10^23) = 5.36 x 10^23
663581
NH4NO2 --------------> N2 + 2H20 \n\nYOU NEED TO HEAT THE SOLUTION IN ORDER TO DECOMPOSE THE COMPOUND \n \nHOPE THIS HELPS
412358
You could skip the Nitrogen Oxide and use Nitrous Oxide.   :o)
670103
Assuming that NO2 is an ideal gas is these conditions (which is a reasonable assumption at such a low P), the number of moles into that flask is: n = PV/(RT) where P = 432 mmHg = 0.568 atm, V = 0.5 L, T = 60 C = 333 K, R = 0.082 L atm/(mol K).\nThen: n = 0.01 mol and, taking into account the Avogadro Number (in number of particles/mol) the number of molecules is NM = 0.01*6.022x10^23 = 6.022x10^21.
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).