Cu + 2AgNO3 => Cu(NO3)2 + 2Ag\n\nIn a particular reaction between copper metal and silver nitrate, 12.7 g Cu produced 38.1 g Ag. What is the % yield of silver in this reaction?\n\nPlease help i have no idea how to do this question
198870
You must first determine the maximum amount of silver that can be formed from 12.7g of copper.  I don't have a periodic table in front of me, but you would determine how many moles of copper you have and use that to determine how many moles of silver could be formed in a perfect reaction.  example:  you find that 12.7g copper is 0.25mol and from that you know that it would form 0.50mol of silver because every mol of copper yields two moles of silver.  With the 0.50mol silver you multiply by the atomic weight of silver to get the maximum yield.  In order to get the percent yield you simply divide the actual yield by the max yield.\n\nNow I'm going to get the numbers for you :)\n\nAtomic weight of Cu is 63.54g/mol\nAtomic weight of Ag is 107.9g/mol\n\nSo... 12.7g Cu is 0.200mol which means the max yeild is 0.400mol of Ag (from the balanced equation 1mol Cu forms 2mol Ag.)  \n\nNext find the max yeild which is 0.400mol Ag times the atomic weight, 107.9g/mol to get 43.16g Ag.\n\nFinally, percent yeild is actual over max yield or 38.1/43.16 x 100% = 88.3% yield
530377
Okay, it's been a looooooong time since I had to use my chemistry skills, so lets see what I have:\n\nCopper is more magnetic with electrons than silver is because it's electrons (with fewer of them) are close to the nucleus (or whatever you call it) of the element.  Silver is more apt to lose its electrons with them being so far away.  \n\nCopper is more often found in a +2 state.  So I'd say that's a safe bet to go with.  I'd only assume a +1 state if your teacher said so.\n\nSo therefore:  2AgNO3 + Cu -> Cu(NO3)2 + 2 Ag would probably be your best bet.\n\nBut once again, it's been a long time since I used this so I don't know how right I might be.
512925
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
347740
KCl(aq) + AgNO3(aq)  --> KNO3(aq) + AgCl(s)\n\nNet ionic: Cl-(aq) + Ag+(aq)  --> AgCl(s)\n\nNa2SO4(aq) + Ba(NO3)2  --> 2NaNO3(aq) + BaSO4(s)\n\nnet ionic: SO4(-2)(aq) + Ba(+2)(aq)  --> BaSO4(s)
292808
I hope you mean this...\n\n8Cu + 2AgNO3 -> 2Ag + CU(NO3)2 + 7Cu
265409
A) CuSO4 and Na2CO3\n\nB) CuCO3 (solid)\n\nC) Na2SO4 (aqueous)\n\nD) CuSO4 (aq) + Na2CO3 (aq) = CuCO3 (s) + Na2SO4 (aq)\n\nE) Cu +2 + SO4 -2 + 2(Na +1) + CO3 -2 = CuCO3 + Cu +2 + 2Na +1\n\nF) Copper (II) and Na+1\n\nG) Cu +2 + CO3 -2 = CuCO3
537551
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
38967
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
230790
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
665249
sounds like an AP question.  So you are trying to calculate Ksp for silver acetate. \n\nAgC2H3O2 <--> Ag+  + C2H3O2-\n\nSolubility will be darn low so from there set up your ice chart and go to town.
147226
Half reactions are:\nOxidation: Cu ----> Cu2+ + 2e    Eo = -0.34V\nReduction: NO3- + 4H+ + 3e --->  NO + 2H2O   Eo = +0.96V\n\nAdd the two voltages (do not multiply this number by the coefficients you use to cancel the number of electrons out) and you get the +0.62V for the Eo of the cell.\n\nThe Nernst equation will take this form:\nE = Eo - 0.0592/n * log {(PNO)^2 * [Cu2+]^3 / [NO3-]^2 * [H+]^8}\n\nn will equal 6 here, not 8, as that will be the number of electrons that will be cancelled out when you add the oxidation and reduction reactions together (multiplied by the correct number to cancel those electrons out).\n\nThe concentration's portion simplifies, once you plug all of the numbers in, to about 18.94, whose log is 1.278.  Multiply that by 0.0592 and divide by 6 and you get 0.01, so only 0.01 V will be lost from the standard potential for that reaction at those concentrations.
466646
Well, for a start your equation shows copper(II) sulphide, not sulphate, and your equation is wrong even then.  That makes a bit of a difference!\n\nLet us imagine that you've got the formula right, but the name wrong.\n\nThe equation is simply Cu + S ---> CuS\n\nThen 64g will give (64+32)g.\n\nThus 64g will give 96g.\n\nSo 64 x (200/96)g of Cu will give 200g of product.
324994
Ag+ will precipitate in HCL solution as AgCl, but as for Ag(NO3), i am not so sure.
471669
AgNO3 + NaCl => AgCl(white ppt) + NaNO3\nAgNO3 + KI => AgI( yellow ppt) + KNO3\n\nthere is a change of color because AgI is produced
432576
Copper is more reactive than silver, hence displacement of silver occurs.\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nSum of the coefficients? 6...\n\nHope this helps :)
370512
I have forgotten all the chemical formulas & whatnot... but I'm going to do this assuming that I'm right for the following:\n\n(note that if i get any of these wrong you'll have to make appropriate changes, this is just to show you HOW to do this)\nCu = Copper\nS = Sulfur\nCuS2 = Copper Sulfide\n\nSo first step is to balance the equation\n\nCu + 2S -> CuS2\n\nso then you can set up a problem like this:\n\n0.90 mol Cu * 1 mol CuS2  = # mol copper sulfide yielded\n(this is because there was 1 mole of Copper in 1 mole of CuS2.. so if CuS2 is not the correct formula change the "1 mol CuS2" accordinginly)\n\n\nThen add up the # of grams in 1 mol of copper sulfide\n\ndo that by taking the # of grams in 1 mol of copper + the number of grams in 2 mol of sulfur (again, this is just assuming that copper sulfide IS CuS2).  Once you've gotten that figure, multiply it by the number found from the earlier problem (# mol copper sulfide yielded).  This gives you the amount of grams.  To change this number to kilograms, divide it by 1000.
329417
fe 2+ , SO4 2-, Ag+, NO3 -\n\nthe products:\n\nFe(NO3)2 + Ag2SO4