Chemistry Mole question?
955015
45.6 grams x 1 mol/323.2 g = 0.141 moles \n\n\nFormula weights \n\nPb= 207.2 x 1 = 207.2\nCr= 52 x 1 =         52.0\nO = 16 x4  =         64.0\n\nPbCrO4 =          323.2
895568
a mole is 6.022×10^23 particles.\n(1.45 * (10^24)) / (6.02200 * (10^23)) = 2.41 moles\n\nN = 14 atomic mass\nO = 16\n\n(14 * 2) + (16 * 5) = 108 g/mole\n(2.41 mole) * 108 (g / mole) = 260.28 grams
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
730581
I love these kinds of problems...lol...\n\nFirst you need to find the empirical formula of the compound.  To do this, pretend that you have 100 grams of the compound.  From your elemental analysis, that means that in that 100 grams, you have 65.45 grams of C, 5.49 grams of H and 29.05 grams of O.  Convert each of those amounts into moles by dividing by each elements atomic weight.  That gives you 5.45 moles of C, 5.49 moles of H and 1.80 moles of O.  Now divide all of those by the smallest number.  (What you're trying to find is the ratio of atoms of each element in the formula.).  WHen you do that you find the empirical formula of the compound is C3H3O. From that empirical formula, the molar mass is 55.\n\nNow, go to the gas law part of the problem.  You need to somehow get to the actual molecular weight of the compound.  Using PV=nRT, you can calculate the number of moles of the compound you have.  (Remember to use the right units for each of the numbers you're given.)  I got something like 0.05 moles.\n\nNow, to get the actual molar mass, divide the mass you have in the flask by the number of moles you just calculated.  This gives a molar mass of 110 g/mol.\n\nSince your empirical formula has a mass of 55, the compound has a formula of C6H6O2.\n\nHope this makes sense...
619470
o.k., in short, d < a < c < b. This is basic material, so I'll give a generously detailed explanation.\n\na) 2 moles N2 * (28 g N2/ mole N2) = 56g N2\n\nb) 1 mole KCl * [(39.1gK +35.5gCl)/ mole KCl] = 74.6 g KCl\n\nc) 9 x 10^23 molecules CO2 * (1 mole CO2/ 6.02 x 10^23)*([12gC + 16gO + 16gO]/ mole CO2) = 66g CO2\n\nd) 2 moles LiF * [(6.94g Li + 19gF)/ mole LiF] = 51.88 g LiF
952068
a) 0.01 mol\n\n1 mole of gas, at standard tremperature and pressure occupies 22.4 l.  You have exactly 1/100 of that, therefore 1/100 of a mole.
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
408650
a mole is the amount of a substance equal to its molecualr weight or 6.02 x 10^23 particles.  Molarity is the number of moles dissolved in a litre of solvent....if memory serves
792924
A.\n2*22.4 l.\n\nB.\n1*22.4 l.\n\nOne mole of any gas occupy 22.4 l. at STP.
781614
Na2SO4 (aq) + Ba(NO3)2 (aq) --> 2 NaNO3 (aq) + BaSO4 (s)\n\nThe moles of Barium = 5.00 L * 0.15 mol/L = 0.75 moles\n\nThis is also the moles of sulfate needed:\n\n0.75 mol / 0.36 mol/L = 2.08 L of Na2SO4
919031
to find Moles use the equation:\nMolarity = moles/Litres\nmolarity = 3.0M\nmoles = ?\nLitres = 0.25L\nso rearrange the equation to get\nMolarity X Litres = moles\n3.0M     X  0.25L = moles\n                           = 0.75\nTherefore there are 0.75 moles of CaCl2 in the solution
371296
well seeing that the molarity of the two solution is a direct ratio of how many molecules there are in the solution, i think that the ration would be 0.15/0.20\n\nso assuming that the reation is 1 to 1 (i cant remember so if it isnt 1 to 1, then i'm wrong) then you would need 18.75 mL of the 0.200 M HCl
313383
Molarity= moles/liters
764973
(0.09/6.022)*(22.414/6.022) atm\n\n0.09 /6.022= n= no. of moles (n=no. of molecules/avogadro no.)\navogadro no. =6.022*10^23
384774
CH3CH2OH\n\nC ~ 12\nO ~ 16\nH ~ 1\n\n2*C + 1*O + 6 * H = 2*12 + 1*16 + 6*1 = 46 g/mol
749866
From the final mass of the MgO, calculate the number of moles of Mg that are present.  Then use a little algebra.  3 times the number of moles of Mg3N2 plus the number of moles of MgO equals the total number of moles of Mg.  Convert all of the moles to the appropriate mass, and solve for the number of moles of Mg3N2.  I leave the solution as an exercise for the student.
342540
Take the mass of the substance of interest, divide it by the total mass, mulitply by 100.\n\nIf you need to find the percent of an element within a compound, you convert mass of the compound to moles, and then determine the number of moles. \n\nThe number of moles of the compound will be equal to the number of moles of the element multiplied by however many atoms of that element are in the compound.\n\nFor example, if you have 3 moles of NH4OH, then you have 3 moles of N in that compound, and 15 moles of H in the same compound.\n\nOnce you have the number of moles of your element of interest, you convert it back to grams. Then divide that by the total weight of the compound and multiply by 100.\n\nHope I haven't been too confusing.