net ionic equation for AgNO3 + NA2CO3?
363082
2Ag+ + CO_33-2 -> Ag_2CO_3
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
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.
782516
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)
797533
Providing that the H3C6H5O7 is citric acid, that balanced equation is:\n\nH3C6H5O7 (s) + 3 NaOH (aq) --> Na3C6H5O7 (aq) + 3 H2O (l)\n\nThe net ionic equation is:\n\n3 H+ (aq) + 3 (OH-) (aq) --> 3 H2O (l)
564401
Silver Carbonate  - AgCO3 (3 being the small number on teh O)\nSilver Nitrate - AgNO3\nSilver Sulphate - Ag2SO4
787481
NaHCO2 + H2SO4 --> Na2SO4 + CO2(g) + H2O(L)\n\nSodium bicarbonate + sulphuric acid equals sodium sulfate + carbon dioxide + water
474059
Gold metal won't react normally with sodium carbonate at all.  \nIf you're talking about a gold ion, though:\nAu+ + CO3(2-) ---> Au2CO3\nThe sodium is always soluble, so it is ignored in net ionic equations, which makes the first answer to this question incorrect.
862596
K2SO4\nCa(NO3)2\n\npotassium has a +1 charge and sulfate has a -2 so therefore you need 2 potassiums so that you can have a net zero charge.\ncalcium has a +2 charge and nitrate has a -1 charge so you need two nitrates to balance the charges
164971
BaCl2 + Na2SO4 --> BaSO4 + 2NaCl is the chemical reaction. \nNet ionic equation: Ba(2+) + SO4(2-) --> BaSO4 \nTotal ionic equation: Ba2+ 2Cl- + Na+ + 2SO4(2-) --> BaSO4(s) + 2Na+ + Cl-\nBarium sulfate is a classic precipitate.  \n\nAll of the masses of these compounds can be calculated by adding the individual ions' masses together, which is the same as atomic masses on a periodic table.   \nBarium chloride: 208.25 g/mol\nSodium sulfate: 142.04 g/mol (anhydrous)\nBarium sulfate: 233.43 g/mol \nSodium chloride: 58.4 g/mol
924773
(1) NaHCO3    --->  Na(+) + HCO3-\n(2) Ca(NO3)2  ---> Ca2(+) + 2NO3-\n(3) Al2(CO3)3  ---> 2Al(3+) + 3CO3(2-)\n\nThis is all taking place in water (ideally), so you should consider that too.
261798
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
824269
If you complete the reaction:\n\nCa(OH)2 + 2HCl(aq) -----> CaCl2(s) + 2H2O\n\nThe net ionic equation is:\n\nCa(2+) + 2OH(-) + 2H(+) + 2Cl(-) ------> CaCl2(s) + 2H2O\n\nstatements a) b) are true. c)d)e) are false.\n\nHope it helps!\n\nGood luck!
913031
Standard electrode potential for Ag(NH3)2 = 0.37V
921926
mg + ion\no -ion\nk +ion\ns -ion\nwhen they are 'added' together (MgO and K^2S) the charges are balanced, that means that the net charge is 0.
954955
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
1049588
I think that you are trying to find the net ionic equation. \nWhich in this case would be \nAg+ (aq) + Br- arrow AgBr (solid)\nif you are trying to find a single replacement reaction?\nthen you need to put Br- instead of NO3- and cancel out the NO3 - on the product side? \nTried my Best.