Can someone help me with net ionic equations?
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
897947
For electrolytes \nπ = iCRT\n\nwhere i is the Van't Hoff coefficient and (for simplicity) it is equal to the number of ions that a strong electrolyte dissociates into according to its molecular formula.\n\nFor NaCl i=2, so\nπ =2*0.15RT= 0.3RT\n\nFor Na2SO4 i=3, so\nπ =3*0.05RT= 0.15RT\nπ =3*0.10RT= 0.3RT\nπ =3*0.15RT= 0.45RT\n\nSo 0.10 M Na2SO4 is isotonic to 0.15 M NaCl
836198
OK....You need to understand how this is all working, and why.  So, the formula equation for this reaction would be something like:\n\nKOH(aq) + HNO3(aq) --> H2O(l) + KNO3(aq)\n\nSince KOH, HNO3 and KNO3 are all strong electrolytes, they are totally ionized when they're dissolved in water.  So, written in complete ionic form, that equation would be:\n\nK+(aq) + OH-(aq) + H+(aq) + NO3-(aq) --> H2O(l) + K+(aq) + NO3-(aq)\n\nTo get to the net ionic equation, you just cross out the things that are the same on both sides of the arrow.  K+(aq) and NO3-(aq) are on both sides, so the net ionic equation is:\n\nH+(aq) + OH-(aq) --> H2O(l)
437730
See http://members.aol.com/logan20/elecspon.html for the equation to use.
255966
Use the hendeson-hasselbach equation\n\npH = pKa + log([A-]/[HA])\n\nand that you know the toatal concentration of ions\n.2 = [HA] + [A-]\n\nyou know the pKa (found in textbook or given) 4.18\nthe pH is 4.00\n\nso -.19 = log(A-/HA)  therefore .6457 = A-/HA\nand .2 = [HA] = [A-]\n\n... solve simultaneous equation\n\n[HA] = .122 and [A-] = .078\n\ndouble check\n\nK = ([H+][A-])/[HA]\n\nK = 10^-pKa = 6.45E-5\n[H+] = 10^-pH = 1E-4\n...and it does check out
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.
785530
Answer D.\n\nThis is the Nernst equation at standard conditions.
1103124
See if these help\n\nhttp://www.rit.edu/~pac8612/webionex/website/\n\nhttp://www.dartmouth.edu/~chemlab/chem3-5/ionx1/overview/procedure.html\n\nhttp://www.culliganmiami.com/pf7.html\n\nhttp://www.wqa.org/sitelogic.cfm?id=1085
746933
You could use the Van't Hoff equations to calculate this.\nOP = i * c * RT\nwhere i is the number of ions per particle and the membrane will be retaining the ions.\n\nBecause RT will be constant for all of them, you only need to compare the product i *c for each solution.  So OP/RT for the solutions above will be.\n\nNa2S ===> 0.30 * 3 = 0.9\nMgCl2 ===> 0.2 * 3 = 0.6\nEtOH ===> 0.50 * 1 = 0.5\nKCl ===> 0.50 * 2 = 1.0\nso the answer is KCl, Na2S, MgCl2, EtOH.
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
855087
Ca2+=[Ar]\nMg2+=[Ne]\nI-=[Xe]\nF-=[Ne]
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
893830
ok anna.\n\nthe ionic product of water = [H+] [OH-]\nthe value of this is constant as is 1*10(-14) mol2dm-3\nthus \n\n1*10^-14=[4.2*10^-9]*[OH-] (which u are tryina find)\n1*10(-14)/4.2*10(-9)= [OH-]\n[OH-]=whatever answer u get when u put this in the calc\n\npH = -log [H+] REMEBER THE NEGATIVE!\n      = - log 4.2*10(-9)\n      = whatever answer u get when u put thisin ur calc (lol)\n\npOH= -log [OH-] REMEMBER THE NEGATIVE\n        = -log [the answer from the first part]\n\nits basic\npH is MORE THAN 7!!
429906
Never trust materials science majors ...\n\n1. NaCl (s) --> Na+ (aq) + Cl- (aq)\n2. Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g)\n3. Ca(OH)2 (s) --> H2O (g) + CaO (s)
23536
is your balace wrong?\nI only figure below balance:\nMnO2+4HCl = MnCl4+2H2O
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)
947778
You need to make sure that there is the same number and kind of atoms on each side of the arrow.\n\nCa(OH)2 + MgCl2 ---> Mg(OH)2 + CaCl2, both Mg and Ca stay+2\n\nBaO2(s) + H2SO4 ---> BaSO4(s) +H2O2 \n\nAg2S --->2 Ag+  +  S-2\n\nHC2H3O2 + MgCO3 (s) ---> H2O + CO2 + Mg+C2H302-\n\nI may not have written net ionic form, but I did get all the atoms on each side I am unsure of the reaction producing hydrogen peroxide, but you do have to account for all atoms.