net ionic equation for gold + sodium carbonate?
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.
294143
add:\n\nH2S2O8(ag)+ NH3(aq) + ? -->(NH4)2S208(aq)  + ?\n\nAmmonia can be prepared in the laboratory by treating ammonium chloride with sodium hydroxide:\n\nthe net ionic equation to this reaction is:\n\n NH4+(aq) +  OH-(aq) ---> NH3(g) + H2O(l) in system 1\n \nand\n\n NH3(g)   = NH3(aq) in solution in system 2\n\nyou finish it ,if you need more help just e-mail me
708721
Here is the right equation for nitric acid and sodium bicarbonate\n\n\nHNO3 + NaHCO3 ---> NaNO3 + H2O + CO2\n\nIf you do not like the sodium:\n\nHNO3 + HCO3(-) --> NO3(-) + H2O + CO2
176528
http://www.chem1.com/CQ/ionbunk.html\n\nskim this article.
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)
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!
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
47432
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
353745
Net ionic is\n\nAg+ + Cl- => AgCl\n\nThis is double replacement.
1008973
HF+ KOH  ---> KF +H2O can also be written as \nH+F- + K+OH- . ---> K+F- + H2O\nWe can canel out K+ and F-  since they appear on both side of the equation.  Thus the net ionic equation is H + OH ---> H2O.  The answer is D.
425080
NaCO3(s) -->  2Na+(aq)  + CO3 2-(aq)
715971
Gold would be deposited on the surface of the lead.\n\n3Pb + 2Au(NO3)3 ---> 3Pb(NO3)2 + 2Au
750179
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
409421
n CaCl2= 0.0362*0.158=5.7196.10^(-3) (mol)\nn Na2CO3 = 0.0375*0.149=5.5875.10^-3 (mol)\nwe have   CaCl2+Na2CO3 = 2 NaCl + CaCO3 \nmol :      1  --> 1            --> 1  \n--> mol CaCO3 = min (n CaCl2 , n Na2CO3) = 5.5875.10^(-3) (mol)\n( that's means : 0.55875 g)
793702
Firstly,\nNaCl dissociates into Na+(aq) + Cl-(aq)\n1faraday of charge is required to produce 23 g Na from molten NaCl .\ntherefore production of 100000 g of Na will require charge = \n(1/23)*1000000 Faradays = 43478.26 Faradays = 43478.26 * 96500 C      [ as 1F = 96500 c]\n=4.1956 * 10^9 C\nBut, Q=IT \nso, T=Q/I = (4.1956 * 10^9)/300000 = 13.9855 * 10^3 secs = 3.88486 hrs.
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.
887064
The main thing you need to know to figure out the ionic equation is whether the product is soluble or not.  In this case, PbSO4 and Mg(NO3)2 are your products, and lead salts are insoluble.  Thus, PbSO4 and any ion that is used to make this molecule is removed from the ionic equation:\n\nPb(+2) + 2(NO3)(-1) + Mg(+2) + (SO4)(-2) -> PbSO4 + Mg(NO3)2 (aq)\n\nbecomes:\n\n2(NO3)(-1) + Mg(+2) -> Mg(NO3)2 (aq)