How many formula units are there in one mole of silver cyanide?
605313
1 mole of anything is always 6.022 E23 of that thing.\n\n1 mole of formula units is 6.022 E23 formula units, it does not matter whether it is formula units of Silver Cyanide or what-have-you.
913880
There are 18.06x10^23 molecules and its weight is 132.0285 or the weight of 3 moles of carbon +6 moles of oxygen
498666
Aaaaaaaaalright... Formulas depend on what you're solving for but since they all involve using the ratio of moles it shouldn't be too different from equation to equation.\n\nLet's go over the mole first to help you better understand what's going on in stoichiometry.  The mole is the amount of a substance that contains as many particles as there are atoms in exactly 12g of carbon-12.\n\nWHOA there... hold on for a second.  The mole is simply a counting unit, just like a dozen is.  We don't usually buy 12 or 24 doughnuts; we order half dozen, one dozen, or two dozen.  Similarly, a chemist may want 1 mol of carbon, or 2 mol of iron, or 2.567 mol of calcium.  \n\nThe number of particles in a mole is known as Avogadro's number, named after an Italian chemist Avogadro. (Go figure)\n\nAvogadro's number is 6.022 x 10^23... which is the number of particles in exactly one mole of a pure substance.  One mol of a substance is going to equal its AMU on the periodic table, or, the mass. \n\nFor example, one mol of Helium is 4g.  If you wanted to find out how many grams are in 3 mols of Helium, you would do a conversion like this:\n\n3 mols He x 4g He/1 mol He\n\nIn words that would be... 3 mols of Helium multiplied by however many grams per mol Helium is, which is 4 grams per mole.  The mols would cancel out leaving us with the units in grams.  Do 3x4 and you get 12 grams of Helium in 3 mols of it.\n\nIf you want to find out how many mols of something you have you would take the amount of grams you have of it and set up another conversion like this:\n\nAmount of g x 1 mol/gram in that mol\n\nFor example, if you have 20 grams of Helium and you want to find out how many mols of it you have, the equation would look like this:\n\n20g Helium x 1 mol/4g He per mol\n\nThe grams would cancel out, leaving the answer in mols.  You would then take 20 divided by 4 and get 5 mols of Helium.\n\n(Hoping this is making sense)\n\nNow stoichiometry is just pretty much comparisons of stuff. (Mols)\n\nAlright, when you're asked to calculate the amount in moles of one substance that will react with or be produced from the given amont in moles of another substance, you must use this simple mole conversion problem.\n\namount of given substance (mols) ----> amount of unknown substance (mols)\n\nIn order to solve this type of equation, use this conversion factor.\n\ngiven quantity x mole ratio = unknown quantity\n\n\nLet's try an example now.\n\nCOsub2 + 2LiOH ---> Lisub2COsub3\nsub=subscript\n\nLet's calculate how many moles of Lithium Hydroxide are required to react with 20 mol of COsub2\n\nGiven information = amount of COsub2 = 20 mols\nUnknown = amount of LiOH in moles\n\nSince the equation is balanced, we need to use the ratio that would give us LiOH in the final answer, therefore, meaning LiOH must be on the top of the ratio.\n\nThe correct ratio would be (mols LiOH/mols COsub2)\n\nIn order to calculate this, set up the equation now. (given q x ratio = unknown)\n\nmols COsub2 x (mols LiOH/mols COsub2) = mols LiOH \n\nsince the mols of COsub2 are on the bottom, it cancels out with the mols of COsub2 on the other side of the multiplication side because that's the rule in math, therefore leaving you with the units of LiOH.\n\nsubstitute the values into the equation and calculate the answer.\n\n20 mols COsub2 x (2mols LiOH/1 mol COsub2) = 40 mol LiOH\n\nsince the mols of COsub2 cancel out, you're left with 20 x 2 mols LIOH which equals 40 mols of LiOH.\n\nThe ratio of LiOH/COsub2, 2:1, was made from the balanced equation because for every mol of COsub2 you have you need 2 mols of LiOH to react with it to produce the right amount of products.\n\nHope this helped, good luck.
903130
Na2SO4(aq)+ 2 AgNO3(aq)-->Ag2SO4(s)+ 2 NaNO3(aq).
183009
Get the CAS number.  Look it up on Chemfinder.
890894
C12H22O11+12O2-------12CO2+11H2O+E
464799
I assume that you have an equation such as A and B react together to give C. Work out the molar ratios of each, i.e. is it 1 equivalent of A reacted with 1 equivalent of B to give 1 equivalent of C?\nAssuming it is a 1:1:1 ratio, then look at the amounts you actually reacted - chances are that you did not use 1 mole of each of the reactants. Work out the molar quantities of each reactant. (mass used/molecular weight of compound)\n\nIf you used 0.5 moles of each, then you would get 0.5moles of product and just convert this to a mass.\n\nIf one of the reactants is in excess, e.g. 0.5 moles of A and 0.8moles of B was used, ignore B because as there is only 0.5 moles of A you would only make a maximum of 0.5 moles of C.\n\nHope this helps
778656
1 mol of Mg + 2 mol of OH = 1 mol of Mg(OH)2 because there is only one Mg used and 2 OH molecules
755325
The first thing we want to account for is that there are 6.02 x 10^23 molecules of H2O in one mole of H2O.  This is Avagadro's number, and it applies for the number of molecules in a mole, the number of atoms in a mole, etc.  So we want to see the problem up as follows:\n\nWe will start with our given quantity of molecules (2.48 x 10^26 molecules of H2O) and multiply this number by the ratio 1 mole of H2O/6.02 x 10^23 molecules H2O.  Essentially we are dividing 2.48 x 10^26 by 6.02 x 10^23.  The answer will be left with moles of H2O because the molecules of H2O will cancel each other, and we will be left with an answer.  \n\nFor a frame of referance, this answer is 4.12 x 10^2, or is it's not put into scientific notation, it is 412. molecules of H2O.
36042
n-hexanoic acid\n\nThere are a LOT of compounds with that chemical formula.  Check chemfinder.com for more.
844628
a) 53.4\n34gCO  x 1mol CO/28gCO x 2molCO2/2molCO x44gCO2/1molCO2 = 53.4gCO2\nThe 2 moles CO cancel the 2 moles CO2 (from the balanced equation).
876756
Depends on what base was used.  Was it NaOH?\n\nOne mole of anything is 6.02 x 1023 (Avogadro's number) of that thing. One mole of an element is 6.02 x 1023 atoms of that element and is a mass of that element equal to its atomic weight in grams. One mole of a compound is 6.02 x 1023 molecules of that compound and equals its formula weight (the sum of the atomic weights of all the atoms in the formula) in grams. For example, 1 mole of C = 12 g C = 6.02 x 1023 atoms; 1 mole CO2 = 44 g CO2 = 6.02 x 1023 molecules of CO2. And (grams of compound) × (formula weight) = number of moles of compound (number of moles of compound) × (formula weight) = grams of compound.\n\n\nNa = 23\nO = 16\nH = 1\n1 mole of NaOH = 40 grams
46188
the formula is CH3CH2OH,the structural formula for ethyl alcohol. The molecular formula is C2H5OH\n\nOh sorry!the answer should be: CnH2n+1OH(general formula)
753121
Ok, now this one I can do:\n\nMg3N2  ---> 3Mg (s) + N2 (g)
470003
First you need to compare the standard reduction potentials of each half-reaction. (See URL below if you don't have values.) The more positive value will be the reduction half-equation and must therefore be the cathode (and the positive terminal for a voltaic cell). The other half equation will be the oxidation half-equation, and must therefore be the anode (and the negative terminal). You need to reverse this equation. Before you can add the two half-equations together, you must ensure that each half-equation has the same number of electrons.\n\na) has a problem; it is not balanced. Either the silver atom on the right hand side should have a single positive charge OR there should be 2 electrons on the left hand side.\n\nb) The MnO2 reduction half-equation has a more positive value than AgBr reduction half-equation, therefore is the cathode etc ie\n\nReduction: MnO2 + 4H^+ + 2e^- = Mn^2+ + 4H2O (Cathode, +ve)\nOxidation: Ag + Br^- =  AgBr + e^-           (Anode, -ve)\n\nNow multiply the oxidation half-equation by 2 so that the number of electrons in each half equation is the same: \n\n2Ag + 2Br^- =  2AgBr + 2e^-\nMnO2 + 4H^+ + 2e^- = Mn^2+ + 4H2O \n---------------------------------\n2Ag + 2Br^- + MnO2 + 4H^+ =  2AgBr + Mn^2+ + 4H2O \n\nNote: the electrons cancel out.\n\nc)  i hope that the above will allow you to complete this one on you own.
606075
Cu(s) + 4HNO3 (aq)--> Cu(NO3)2(aq) + 2H2O(l) +2NO2(g)
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