What would the solubility product expression fo Ag2Co3- be and how do you find the answer?
967201
Write the dissociation reaction:\n\nAg2Co3(s) <--> 2Ag+1(aq) + CO3-2(aq)\n\nSince this is a solubility equilibrium, the solubility product expression is just the equilibrium expression.\n\nKsp=[Ag2]^2[CO3]\n\nTo find the actual value, you'll need to look it up in a table.\nBTW, it's 8.1*10^-12
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)
197904
based on the net inoic equation:\n\n\nCu2+(aq)---> Cu(s)\n\nAg+(aq)---> Ag(s) --- zero grams\n\nexcess Cu(s) or Cu(aq)\n\nif, Cu(s) then it is  3.78 grams of Ag
874983
1. Molarity = concentration and it expressed as mol per dm^3\n    Change the mass to mole and the 851 ml to dm^3. Then find   \n    the ratio of the number of moles to volume in dm^3\n\n2. Soluble ionic compounds breaks up the water molecules in H^+ ions and OH^- ions. Sodium carbonate give Na^+ ions and CO3^2- ions......Other ions that are present are HCO3^- ions called hydrogen carbonate ions. I have found that a solution of sodium carbonate is alkaline: implying that there are more OH^- ions to H^+ ions, due to the formation of HCO^- ions.\n3. OH^- (aq) + H^+(aq) ----> H2O (l)\n4. H^+ ions (ie. hydrogen ions), OH^- ions (ie. hydroxyl ions),\n   Ca^2+ calcium ions and Cl^- chloride ions\n5. Hydrogen ions, hydroxyl ions. ammonium ions and sulphate ions.\n6. Actually the calcium metal reacts with water to give hydrogen gas and calcium hydroxide. At the same time, calcium reacts with nitric acid to give calcium nitrate, water and brown nitrogen dioxide gas.\nHence there is no simple ionic equation for your question\nYou may want to consider this one:\nCa^0  +  2H^+   + 2OH^-  -----> H2(g)  + Ca^2+  + 2OH^-\n7. Hydrogen ions, hydroxyl ions, iron (III) ions Fe^3+  and sulphate ion SO4^2-
611161
Silver carbonate, Ag2CO3.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(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
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
36586
Ag+ + SO4(2-) ---> Ag2SO4\nYeah, it's double replacement, and according to my solubility rules, AgSO4 is insoluble.
793636
shawty's NOT good at stoichiometry...has made a fundamental error in the final line so end result is rubbish...\nhow can you end up with MORE Cu than you started with when it's 2 moles : 2moles? \n\n(8/159)x 127 = 6.4 g
350981
This isn't too bad.\n\nFirst, find the enthalpy of formation for each of the reactants and products.\n\nFe2O3(cr) = -272 kJ/mol\nCO(g) = -110.525 kJ/mol\nCO2(g) = -393.509 kJ/mol\nFe = 0 kJ/mol\n\nThen add the enthalpies of formation for the reactants and subtract the enthalpies of formation for the products:\n\nFe2O3 + 3CO - (3CO2 + 2Fe)\n\n-272 kJ/mol + 3(-110.525 kJ/mol) - 3(-393.509 kJ/mol) - 2(0) = -603.575 kJ/mol - -1180.53 = -576.952 kJ/mol \n\nThat should do it.
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
791597
You will get Silver Chromate as the precipitate.  It is red in color.\n2AgNO3(aqueous) + K2CrO4(aqueous) -> Ag2CrO4(solid) + 2KNO3(aqueous)
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.
830588
If I understand your question, you start out with 2.15 g of AgNO3 and you isolate 1.16 g of AgCL.\n\n(It saves time if you include the M.Wt. of the materials.)\n\nAg = 107.87g/mole\nNO3- has 14.0067 + 3*15.9994 = 62.005 g/mole\nCL = 35.453 g/mole\nAgNO3 = 169.875 g/mole\nAgCL = 143.323 g/mole\n\nIn theory, for every mole of AgNO3 you should get 1 mole of AgCl\n\n(2.15 g AgNO3)*(1 mole AgNO3/169.875g) = 0.0126 mole AgNO3, and also the same number of moles of AgCL\n\n0.0126 mole AgCL * (143.323 g AgCl/mole AgCl) = 1.81 g AgCl in theory (rounded)\n\nYou isolated only 1.16 g, therefore\n\n  (1.16/1.81)*100% = 64 % yield.\n\nYou say you want % error, I don' t know if that is the correct term but here is what the error term is:\n\n [abs (1.16 - 1.81)/1.81]*100% = 35.9%\n\nplease check for errors
138848
Here is one example ,\n\nAgNO3(aq) + NaCl(aq)-> AgCl(s)white precipitate + NaNO3(aq)
843441
The answer is c.\nC12H22O11+12O2-->12CO2+11H2O