What is the Kdissociation values for Ag[NH_3_]2+???
209943
FORMATION CONSTANT IS :\n\nkf = 1.5*10^7
484918
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
279993
(NH4)2 SO4  + (FeSO4)2 + 2H2O____(\nNH4)2SO4Fe2(SO4)3.24H2O\n\nAmmonium ferric sulphate
352521
1.   Ag + KNO3 --> AgNO3 + K\n      single replacement means that a single element breaks a compound apart and joins with part of it--but which part--the positive or negative part?  It depends on what the single element is.  If it's a metal (#1 above), it will join with the negative part of the compound.  If it's a nonmetal (# 4 below), it will join with the positive part of the compound.\n4.  Cl2 + 2 KI --> 2 KCl + I2\nI'll leave you to have fun with the rest!
534322
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
348168
a. Silver nitrate\nb. isopropanol? methanol?\nc. Potassium chloride\nd. Benzene\ne. Carbon monoxide\nf. Sodium hydroxide
177137
first, you find out the equation of Keq and delta G. I think it should be Keq = e^ [-deltaG/(RT)].\nThis equation and constant R can be found in any general chemistry textbook. And T is usu. 25+273=298K\nthen you can find out deltaG for the equation.\n\nwe also have deltaG= [- delta G for Ag(NH3)2^+1(aq) + (-32.0)] - [-26.2 - 3.8]\n\nand there comes the answer!!! ;)
466512
Here is what you need.\n\n2HF(aq) + Mg(s) ---> MgF2(s) + H2(g)
387093
You are looking at a displacement reaction, in which the more reactive metal (Cu) is displacing the less reactive metal (Ag) from its salt solution (AgNO3).\nCu (s) + 2AgNO3 (aq) --> 2Ag (s) + Cu(NO3)2 (aq)\n\nHope this helps. :)
445789
This compound is an amine.  You can identify it by the NH2 (amine group) at the end.  You did a fair job with the structure considering the limits of the text here.  I am sure you realize that two of your carbons do not have the correct number of bonds.
239375
67.  AgCl <---> Ag+ + Cl-  Ksp = [Ag+][Cl-] = 1.8 x 10^-10\n[Ag+] = 0.001/2 = 0.0005 M; [Cl-] = 0.0025 M; [Ag+][Cl-] = 1.25 x 10^-6, as this is greater than Ksp, a precipitate will form.\n\n68.  SrCrO4; Ksp = [Sr2+][CrO42-] = 3.5 x 10^-5; molar solubility = [Sr2+] = [CrO42-] = sqrt (3.5 x 10^-5) = 0.0059 M\n\n69.  Total volume = 3L  [Fe2+] = 0.150 * 1/3 = 0.050M; [OH-] = 0.0333 * 2 / 3 = 0.0222 M.  Ksp Fe(OH)2 = 4 x 10^-38.  (0.050) * (0.0222)^ 2 >>> 4 x 10^-38, so a precipitate will form.\n\n73.  Pb3(AsO4)2 <---> 3 Pb2+ + 2 AsO43-.  If solubility = x, then [Pb2+] = 3x, [AsO43-] = 2x.\n4.0 x 10^-36 = [3x]^3[2x]^2 = 108 x^5 ==> x = 3.26 x 10^-8
403482
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)
426249
Just an initial note: NH3 is ammonia\n\nA. \nInitial math:\nSince you have 8.5 grams of NH3, you have to convert grams mols, then into molecules. \nSo you do: 8.5 grams of NH3 / 17 grams in 1 mole of NH3 and X 6.022 x 10^23 molecules in one mole\nYou get 3.011 x 10 23 molecules, and from that you subtract 1.5 x 10^23 molecules, and you get: \n<b>1.511 x 10^23</b>\n\nB. Then, you convert 1.5 x 10^23 to moles by dividing by 6.022 x 10^23 moles, which equals .249 moles of NH3. \nThen you convert 8.5 grams of NH3 to moles too by dividing 17 grams of NH3 per mole again, and get: .5 moles! \nso: .5 mol - .249 mol = <b>.241 mol</b>\n\nC. \nFrom before, you know 1.5 x 10^23 molecules of NH3 is .249 moles. You convert this to grams by multiplying .249 by 17 grams per mol and get: 4.233 grams. \nSo: 8.5 grams - 4.233 grams = <b>4.267 grams</b>\n\nD. From before, you know you have .241 moles of ammonia left. So, you multiply this by avocado's number (6.022 x 10^23) and get: 1.45 x 10^23 molecules of ammonia. Then, you can see that there are 3 atoms of hydrogen of per molecule of ammonia, and multiply the number of molecules by 3 and get: \n<b> 4.354 x 10^23</b> \n\nI think those should be right. Sorry about the sig figs though!
440498
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)
434591
The temperature difference is 500 K.  If we add 26/deg for NH3, and subtract 0.5 x 27/deg for N2 and 1.5 x 29/deg for H2, we will net out at -0.5 -4.5 = -5.0 J/K per mole of product, so (if i've got the signs right) we will be at -46,000 -2,500 joules/mole, or -48,500 joules per mole.  It is more straightforward to compute the enthalpy changes separately and add them up, but by saving the multiplication till later I can do the numbers in my head.
368903
The answer to the first part:\n1) You need to find the pKa ---> use pKa+pKb = 14\nso 14-pKb = pKa\n14-4.75 = 9.25\n\n2) work out the final Molarity of each\n[NH3] = 0.400M/0.480L = 0.833M\n[NH4Cl] = 0.100M/0.520L = 0.192M\n\n3) now use the henderson-hasselbach equation\npH = pKa + log([base]/[acid])\n= 9.25 + log([0.833]/[0.192])\n= 9.89
468688
NH4NO3 --->  N20 + H2O\n\n NH4NO3 --->  N20 + 2H2O\n2N 4H  3O        2N      4H   3O