What are the values of [SO4^2-], [Pb^2+], and [Sr^2+] in a solution at equilibrium with both substances?
595177
Actually, this is pretty simple.  You have a saturated  solution of lead sulfate (PbSO4) and strontium sulfate (SrSO4).  Probably the biggest hurdle here is to realize that the concentration of sulfate ion will be the same for both salts!  Therefore, the concentration of sulfate ion will be the result of the most soluble of the two salts:\n\nKsp for PbSO4 = [Pb+2]*[SO4-2] = 1.6x10^-8\nKsp for SrSO4 = [Sr+2]*[SO4-2] = 2.8x10^-7\n\nStrontium sulfate is the more soluble of the two salts, so it will control the concentration of sulfate:\n\nSrSO4 ===> Sr+2  + SO4-2, thus [Sr+2] = [SO4-2]\n\nKsp for SrSO4 = 2.8x10^-7 = [SO4-2]^2 , so \n\n[SO4-2] = sqrt(2.8x10^-7) = 5.29x10^-4 = [Sr+2]\n\nNow, to find out the concentration of Pb+2, we use the value we have just calculated for [SO4-2] and the Ksp:\n\nKsp for PbSO4 = [Pb+2] * [SO4-2] = 1.6x10^-8\n\n[Pb+2] = 1.6x10^-8/[SO4-2] = (1.6x10^-8)/(5.29x10^-4)\n[Pb+2] = 3.02x10^-5\n\nand there we have your answers:\n[Sr+2] = [SO4-2] = 5.29x10^-4 moles/liter\n[Pb+2] = 3.02x10^-5 moles/liter
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)
741136
(a): Si+, P2+\n\n(b) Ar+, S-\n\n(c) Ca+, Cl2-\n\nBut the other answers that you gave are also correct.
170129
CuSO4 (s)  --H2O-->  Cu+2 (aq) + SO4-2 (aq)
812798
3Ca(OH)2+2H3PO4>Ca3(PO4)2+6H2O
434926
2Sb+3 + 3S-2 => Sb_2S_3\n\nquepie's probably right.  I'm not going to look up SB(OH)_3.\n\nThen, it'd be\n\n2Sb(OH)_3 + 3S-2 => Sb_2S_3 + 6OH-
696739
The chemicals are not correct\n\n2 (NH4)3PO4+3 Pb(NO3)2-> Pb3(PO4)2+6 NH4NO3?
766811
d. \n\nIt's a double replacement (acid/base) reaction.  The Sr and the H are both positive and they switch.  Since Sr has a 2+ charge and the PO4 has a 3- charge, you use a common multiple of 6.  The other product would be H(OH) or water.
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
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
854997
CuS + H2SO4 ===> H2S + CuSO4
853922
These are the right ones:\n\n2K(s) + 2H2O(l) ---> 2KOH(aq) + H2(g)\nH2SO4(aq) + 2KOH(aq) ---> K2SO4(aq) + 2H2O(l)\nCa(OH)2(aq) + 2HCl(aq) ---> CaCl2(aq) + 2H2O(l)
706351
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)
748872
Cu2+ (aq)  +  Zn (s) -->  Cu (s) + Zn2+ (aq)
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
1123465
Na2S + 2 HCl > H2S + 2 NaCl  \n\nH2SO4 + BaOH > H20 + BaSO4  \n\nNH4Cl + NaOH > NaCl + NH4OH\n\nAgNO3 + NaCl > AgCl + NaNO3\n\nnot sure about complete or reversible..:)