Write the net ionic equation for the following reaction? make sure to include the states(s,l,aq,g) and oxidati
610699
Pb(NO3)2(aq) + H2SO4(aq/l)[depending on concentration] --> PbSO4(s) + 2HNO3(aq)\nPb2+(aq) + SO42-(aq) --> PbSO4(s)\n\nI don't see how an redox reaction could occur.
546125
The general equation will often include all of the ions and elements.  The net equation takes away the ions and elements that didn't change with the reaction.
863456
I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622
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.
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
866210
2NO+O2--->2NO2\n\nCaO+H2O--->Ca(OH)2\n\nCaCO3--->CaO+CO2
752218
2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O
519991
1, 1, 2, HNO3, 1, PbSO4.\n\nTechnically, the HNO3 is aqueous and stays in ion form, but it doesn't seem like you're looking for a net ionic.
976540
a) +1\nb) 0   (in molecule state)\nc) 0\nd) -3\ne) 0\nf) +4 of Carbon\ng) +2 of Mg\nh) +1 of Chlorine (interesting!)\ni) +6 of Sulphur
651011
H2O2 + OC(NH2)2 -> HCOOH + NH4OH -> NH4(HCOO)
425080
NaCO3(s) -->  2Na+(aq)  + CO3 2-(aq)
129663
Please do your own home work.\n\nI suspect you are missing a symbol, it probably should be,\nN2O4  <==>  2NO2\n\nThink about how each of the changes will effect the equilibrium of the reaction.
502184
5H2O2(aq)+2KMnO4(aq)+\n3H2SO4(aq)-->\n2MnSO4(aq)+K2SO4(aq)+\n8H2O(l)+5O2(g)
539827
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
868500
Oxidation :\n\nCu > Cu+2 + 2e-\n\nReduction :\n\n2H+ + e- + NO3- > NO2 + H2O\n\nMultiply the 2nd equation by 2, you get the complete balanced redox equation :\n\nCu + 4H+ + 2NO3- > 2NO2 + 2H2O + Cu+2
83000
The net ionic equation is:\n\nFe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s)
702491
Mn2+ +2NH3 +2H2O  <-->  Mn(OH)2 + 2NH4+ (+:sup, 4:sub)\n4Mn(OH)2 + O2 + 2H2O  <-->  4Mn(OH)3\n2Mn(OH)2 + O2  <-->  2(MnO2 . H2O)\n\nMn(OH)2 white sediment\nMn(OH)3 brown sediment\nThe oxidization is faster if we add H2O2