how to solve   Cu + Pb(NO3)2?
831216
Cu (s) + Pb(NO3)2 (aq) ---------> No Reaction. \nCopper will not replace lead.\n\nreaction VI in the attached link.
978387
You should have gotten elemental copper via the following equation:\n\n3CuCl2 + 2Al ---> 3Cu + 2AlCl3\n\nThe water is not part of the reaction.
782516
a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
853340
KCl + Cu(Cr2O7)\n\nDichromate has 2 Chromium +6s to counter the 7 oxygen -2s.  +2 + -2 = 0\n\nPotassium Chlorite and Copper Dichromate
903130
Na2SO4(aq)+ 2 AgNO3(aq)-->Ag2SO4(s)+ 2 NaNO3(aq).
367815
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
945429
Equation becomes CuCl2 + 2KI --> CuI2 + 2KCI\n\nProducts are Cuprous Iodide and Potassium Chloride
750179
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
857898
2Al + 3*(CuO) -> Al2O3 + 3*Cu\n\nI2 + 2*(Na2S2O3) -> 2*(NaI) + Na2S4O6\n\nI2 + 5*(Cl2) -> 2*(ICl5)
387399
Pb2+ + Cl- ---> PbCl2\n\nWe know that the lead in Pb(NO3)2 has to have a 2+ charge because a nitrate ion has a 1- charge, so two of them is a 2- charge.  The entire compound must have no charge, so a 2+ charge is assigned to the lead. \n\nBoth compounds Pb(NO3)2 and NaCl dissociate (they are both soluble), and because nitrate and Na+ ions are always soluble, they can be ignored in a net ionic equation.
771965
I can't tell you the name because i can't remember that stuff but the product will be K(NO3) PbI2\nif you're writing it in the long form it will be\n2KI + Pb(NO3)2 yeilds (arrow) 2K(NO3) + PbI2
748872
Cu2+ (aq)  +  Zn (s) -->  Cu (s) + Zn2+ (aq)
842717
Using the atomic mass of all the elements from your periodic table...\nPb(NO3)2 has the mass of (207+14x2+16x6)= 331 g/mol\nK2CrO has the mass of (83.8x2+52+16) = 235.6 g/mol\n\n0.005 mol Pb(NO3)2 x (331 g/mol) = 1.66 g Pb(NO3)2\n0.005 mol K2CrO x (235.6 g/mol) = 1.18 g K2CrO
971239
When the two solutions are mixed, you'll get a bright yellow precipitant.\n\nPb(NO3)2 + 2 NaI ---> PbI2 + 2 NaNO3
700367
1) Balance the reaction equation:\nPb(NO3)2  + 2 LiBr -> PbBr2 + 2LiNO3\n\n2) Calculate the amounts we start with\nPb(NO3)2 :  2.05 L * 0.0550 mol/L = 0.1128 mol\nLiBr : 1.52 L * 0.0618 mol/L = 0.0939 mol\nAs we need 2 moles of the latter for 1 mole of the former, the reaction is limited by the amount of LiBr\n\n3) Calculate the amount of PbBr2 produced\n0.0939 mol of LiBr will react with 0.470 mol of Pb(NO3)2 to produce 0.0470 mol of PbBr2. This is 0.0470 * (207.2 + 2* 79.9) =17.249 gram\n\n4) Calculate the yield\nThe yield is 15.784 / 17.249 = 91.51%
429906
Never trust materials science majors ...\n\n1. NaCl (s) --> Na+ (aq) + Cl- (aq)\n2. Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g)\n3. Ca(OH)2 (s) --> H2O (g) + CaO (s)