what precipitate is formed by mixing K2CO3 and AgNO3? what is the formula?
611161
Silver carbonate, Ag2CO3.
564401
Silver Carbonate  - AgCO3 (3 being the small number on teh O)\nSilver Nitrate - AgNO3\nSilver Sulphate - Ag2SO4
857082
1.  Split up the elements and give them their original charges\n\nAg +\nNO3 -\nK +\nBr -\n\n2.  Combine an element with an element that they were not with, putting them side by side will help\n\nAg + with Br -\nNO3 - with K +\n\n3. So now you have 2 new compounds\n\nAgBr & KNO3-\n\n4.  Their should be a chart in your book that tells you which compound would be the precipitate but i don't think their is a precipitate for this reaction'\n\nso both will dissolve in a solution
98695
My answer is,\n \nNH4NO3(ag) + KOH(aq) -> NH4OH(aq) + KNO3(aq)
744573
You added 0.04 liter * 0.07891 mol/liter = 3.156*10^-3 moles of Ag as AgNO3 to the dissolved unknown sample.  \n\nHowever, there was excess Ag added.  The "extra" Ag amounted to:\n\n0.01127 liter * 0.1 mol/liter = 1.127*10^-3 moles Ag\n\n\nSo the net amount of Ag that reacted with the AsO4(3-) ion was \n\n3.156*10^-3 - 1.127*10^-3 = 2.029*10^-3 moles Ag.\n\nIt takes three moles of Ag for every one mole of As to make Ag3AsO4, so the precipitate consisted of \n\n2.029*10^-3 moles/3 = 6.765*10^-4 moles Ag3AsO4, which, of course, contains 6.765*10^-4 moles of As.\n\nThe molecular mass of As2O3 is 197.8414 gm/mol, but each formula unit of this compound contains 2 As atoms, while the Ag3AsO4 precipitate only contains 1 As atom per formula unit.  If we rewrite the formula of As2O3 as AsO1.5, it would have a molecular mass of 197.8414/2 gm/mol = 98.921 gm/mol.\n\nThe precipitate contains 6.765*10^-4 moles of As, and this amount of As as an oxide would have a mass of:\n\n6.765*10^-4 moles * 98.921 gm/mol =  gm As2O3\n\nIf the initial sample had a mass of 1.223 grams, then this mass of As2O3 constituted a mass fraction of:\n\n6.692*10^-2/1.223 = 5.472*10^-2 = 5.472 wt %
769198
the above is correct EXCEPT the AgSO3 is silver sulfite, not sulfide, just AgS would be silver sulfide
904363
(NH4)2CO3+MgSO4->(NH4)2SO4+MgCO3
339176
The result is Ag2S: Ag+ is monovalent, S2- is the sulphide ion.\n\nA solid silver surface will blacken (slowly) when exposed to material containing bound (negatively charged) sulphur atoms (such as the amino acid cystein and proteins containing it). Dissolved silver will precipitate.
1185928
put your object in a concentrated solution of HCl (muriatic acid) , the silver will precipitate as silver chloride, you can now filter the solution and keep the silver chloride. Then add ammonia to the silver chloride to solubilise the AgCl. Then put a copper wire in the solution, the silver will precipitate on the copper, put copper until no silver is appering. you can now filter the liquid to obtain pure silver
150925
Reaction One (CO is Co I'm assuming ... they are NOT equivalent) is a double replacement. The products will be aqeuous KNO3 and insoluble Co(OH)2\n\nReaction 2 is a redox. The product will probably be Mg2N3 (Magnesium Nitride) but I'm not entirely sure if that chemical formula is correct. The magnesium is oxidized and the nitrogen is reduced.\n\nAg + H2SO4 ---> Maybe Ag2SO4 as a redox.\n\nF2 + KBr ---> KF + BR2 Single Replacement
910345
The formula would be Al2(CO3)3; the numbers are subscripts. \n\nLearn your charges for the polyatomic ions and for the other elements, it will really help.
970329
Its a double replacement reaction. It is really simple to do. All you do is add the aqueous forms of both reactants and the reaction happens instantaneously.\n\nInsoluble PbI is formed and KNO3 is also formed and precipitates out.. Any compound with K+ or NO3- in it is soluble so it remains aqueous.
327609
If I've understood what you're asking then:\nA\nCd(s) + Ag2+ -> Cd2+ + Ag(s) \n\nCadmium occurs at Anode and the Silver would deposit on the Cathode\n\nB (I hate Manganese equations and the partial above is incorrect)\n\nMnO2 + 4H+ + 2AgBr -> Mn2+ 2Br- + 2Ag + 2H2O\n\nMnO2 @ Anode\nAgBr @ Cathode\n\nC\n\n(PtCl4)2- + 4Ag -> Pt + 4AgCl \nAg @ anode\n(PtCl4)2- @ Cathode
167726
potassium can form many compounds. such as KCN KCl KOH et cetera.
760435
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. :)
339232
Yes, the reaction would spontaneously occur because Sn is higher on the activity series than Ag.\n\nSn(s) + 2AgNO3(aq) => Sn(NO3)2(aq) + 2 Ag(s)\nThe net ionic is\nSn(s) + 2Ag+(aq) => Sn+2(aq) + 2 Ag(s)\n\nI don't remember the oxidation number for Sn.  2 sounds about right.\n\nIt would turn a silver nitrate solution to a tin nitrate solution, and you'd see metallic silver precipitate out.
760563
N=aM where a is the valency (depending on the type of chemical and reaction, e.g. number of H+/OH- for acid/bases per molecule, number of electrons gained/lost in redox reactions).\n\nFor AgNO3 a=1 (it is Ag+ so for redox or precipitation reactions it gives one equivalent group per molecule)\n\nSo M= N/a = N/1 =N\n\n1 M is 169.87 g/L =16.987g/100mL =16.987%\n\nUse M1*V1=M2*V2\n16.987*V1=1*V2\nV1 is the volume in mL of the stock solution that you will use to prepare V2 mL of 1%\nYou didn't specify how much you want to prepare...\n\nAlternatively you could have converted 1% to mole/L and do the math with molarities.