what happens when you add FeCl3, KSCN, and AgNO3 together?
375272
My guess is this:\n\nFeCl3 is a Lewis Acid. KSCN is a lewis base. AgNO3 is neither and it's a spectator.\n\nFeCl3 + KSCN ---> Fe(SCN)3 + KCl\n\nFe(SCN)3 may well be an orange color, depending on the field strength of SCN. The D orbitals in the iron are split in a 2X3 arrangement. The bottom 3 will each have one electron, according to Hund's rule. The top orbital can be reached, if SCN is a weak field ligand (I don't feel like looking up the spectrochemical series to find out). What's nice is that the total number of unpaired electrons will not change, so you don't have to worry about any kind of Spin-forbidden transition, that would render the compound colorless.
670034
Ag+ will precipitate in HCL solution as AgCl, but as for Ag(NO3), i am not so sure.
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
791597
You will get Silver Chromate as the precipitate.  It is red in color.\n2AgNO3(aqueous) + K2CrO4(aqueous) -> Ag2CrO4(solid) + 2KNO3(aqueous)
828284
The first thing that happens is that it melts.  If you continue heating the molten salt, you will see bubbles appearing.  These are bubbles of oxygen:\n2 KClO3(l)   --->   3 O2(g)   +   2 KCl(l)\nThe main product left behind is potassium chloride, but I believe that you also get some potassium perchlorate (tetraoxochlorate(VII)), produced by disproportionation:\n4 KClO3(l)   --->   3 KClO4(l)   +   KCl(l)\n\nIt wouldn't surprise me if the reaction also produced very small amounts of oxides of chlorine, such as chlorine dioxide, ClO2.\n\nAllan Deeds.
727659
1.when sodium and chlorine combine they lose their harmful properties and become altogether new substance\n\n2.carbon di oxide\n\n3.bauxite(alumina)
851188
when a piece of Ag is dropped in FeSO4 sol. no reaction will take place. it is because Ag is less reactive than Fe so it cannot displace Fe in FeSO4
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
51552
AgNO3 + NaCl -----> AgCl + NaNO3\n\ntotal ionic equation\n\nAg+  +  NO3-  +  Na+  +  Cl-\n\n                      ----->\n\n                            AgCl  +  Na+  +  NO3-\n\nnet ionic equation\n\n    Ag+  +  Cl-  ----->  AgCl
904155
It certainly should.  Zinc is much higher in the "electromotive series" than is silver.  This means it would displace silver.\n\nZn (s) + 2 AgNO3 (aq)---> Zn(NO3)2 (aq) + 2 Ag (s)\n\nSolid zinc dropped into silver nitrate solution will precipitate silver.
882018
1) 2Al + Fe2O3=Al2O3 + 2Fe\n2) 8.53g BaCl2*(1mole BaCl2/ 208.3G BaCl2)=.0410mole BaCl2\n.0410mole BaCl2*(2moles KCl/1moleBaCl2)=.0820 mole KCl2\n.0820mole KCl2*(110.1gKCl2/1mole KCl2)=9.03gKCl2\n3)i don't know this one\n4) 2NaOH + H2 is reduced\n5) i don't know this one\n6) 5gNa*(1moleNa/23gNa)=.217mole Na\n.217mole Na*(6.02x10^23/1mole Na)=1.31x10^23 atoms Na
729557
_NaOH(aq) + _Fe(ClO4)3(aq) => _Fe(OH)3(s) + _NaClO4(aq)\n\n3  NaOH(aq) + 1  Fe(ClO4)3(aq) => 1  Fe(OH)3(s) + 3  NaClO4(aq)\nThe coefficients of one could be excluded but I put one which is also equivalent.\n\n_((NH4)2)SO4(aq) + _Pb(NO3)2(aq) => _PbSO4(s) + _NH4NO3(aq)\n\n1  ((NH4)2)SO4(aq) + 1  Pb(NO3)2(aq) => 1  PbSO4(s) + 2  NH4NO3(aq)\n\nThese are only the balanced equations.\n\nThe following is the ionic equations:\n(Na)+ (aq) + (OH)- (aq) + (Fe)3+ (aq) + (ClO4)- (aq) => Fe(OH)3(s) + (Na)+ (aq) + (ClO4)- (aq)\nand\n(NH4)+ (aq) + (SO4)2- (aq) + (Pb)2+ (aq) + (NO3)- (aq) => PbSO4(s) + (NH4)+ (aq) + (NO3)- (aq)\n\nThe ionic equations are already balanced.  It is simply the breaking down of the equations into ions.  The solid remains intact while the compounds in aqueous solution become ions.\n\nThe + and - signs are the charge of the ions.  I can't find any other way to write them so I separated the compound with parentheses.  Sorry if it's confusing.\n\nThe net ionic equations are "simplified" ionic equations.  Basically, the things that are repeated on both sides get crossed off.  What is remaining is simply the net ionic equation.\n\n(OH)- (aq) + (Fe)3+ (aq) => Fe(OH)3(s)\nand\n(SO4)2- (aq) + (Pb)2+ (aq) => PbSO4(s)
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
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
971239
When the two solutions are mixed, you'll get a bright yellow precipitant.\n\nPb(NO3)2 + 2 NaI ---> PbI2 + 2 NaNO3
607593
K2PtCl6 only reacts with Ferric ions of Iron i.e. Fe3+ this is rate determining reaction
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6