Does anyone know the DISSOCIATION EQUATIONS?
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.
823368
Of what?\n\nπ= i*C*R*T\n\nT= absolute temperature (in Kelvin)\nR= the gas constant in whatever units you need to express osmotic pressure (e.g. if you want π in atm then R=0.082 L*atm/(mole*K))\nC= the concentration of your solute in mole/L\n\nand...\n\ni is the van't Hoff coefficient.\nFor non-electrolytes i=1\nFor strong electrolytes i= the number of ions that are produced by the dissociation according to the molecular formula e.g for NaCl you have 2 ions (1 Na+ and 1 Cl-) so i=2, for CaCl2 3 ions (1 Ca+2 and 2 Cl-) so i=3.\nFor weak electrolytes, if n is the number of ions coming from the 100% dissociation according to the molecular formula and a the degree of dissociation then i=(1-a)+na. E.g. if we assume for CH3COOH a=80% i=(1-0.8)+2*0.8= 0.2+1.6=1.8
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
106704
If you go to a search engine, like google, you can just type in "balancing chemical equations" and it will give you tons of info. We were doing this in chemistry a few weeks ago, and thats how I found alot of help online. \n\nGood luck!!
757438
CaC2 + 2 H2O ---> Ca(OH)2 + HCCH
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
970544
Here is what you do: \nDo the same operations as I did on the equations with the H values and you will be OK (Do not forget to use the same multipliers on the H as I did on the reactions).\n\n10PCl3 + 5O2 ---> 10Cl3PO\n10PCl2 + 10Cl2 --- 10PCl5\nSubstract the equations to get (keep coeff positives by moving to the other side):\n\n5O2 + 10PCl5 ---> 10Cl3PO + 10Cl2\nP4 + 5O2 ---> P4O10\nSubstract and get\n10PCl5 + P4O10 ---> P4 + 10Cl3PO + 10Cl2\nP4 + 6Cl2 ---> 4PCl3\n\nSum the two to get (P4 goes):\n\n10PCl5 + P4O10 ---> 10Cl3PO + 4Cl2 + 4PCl3\n\n4 PCl3 + 4Cl2 ---> 4PCl5\n\nSum the two equations\n\n10PCl5 + 6P4O10 ---> 10 Cl3PO.
1123465
Na2S + 2 HCl > H2S + 2 NaCl  \n\nH2SO4 + BaOH > H20 + BaSO4  \n\nNH4Cl + NaOH > NaCl + NH4OH\n\nAgNO3 + NaCl > AgCl + NaNO3\n\nnot sure about complete or reversible..:)
564798
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
557422
First, let's calculate the pOH of the solution:\npOH = -log [OH-] \nBecause KOH is a strong base (very soluble and complete dissociation), the concentration of KOH is the same as the concentration of the OH- because all of the KOH becomes K+ and OH- ions.  \npOH = -log [OH-] = -log 4.8 x 10^-2 M = 1.32\n\nNow, let's calculate the pH.  \npOH + pH always adds up to 14 under normal conditions.  \n1.32 + pH = 14\npH=12.68\n\nNow, we can calculate the H+ concentration. \npH=12.68=-log[H+]\n12.68 = -log[H+]\n-12.68 = log[H+]\nTake the antilog of both sides, which is the same as ten raised to the both sides. \n10^(-12.68)= 10^(log[H+])\n10^(-12.68)= [H+]\n2.1 x 10^-13 M= [H+]
4678
to find this answer....assuming ideal mixtures and complete disassociation.  \n\nM1V1=M2V2\n\nso take .450*300/4 = 33.75
390546
I'll give you a hint.\n\nDelta G = Delta H - T Delta S\n\nUsing that formula, you can get 1\n\n(B) you can look up in any physical chemistry text:\n\nG(2) - G(1) = a formula I don't have available
729557
_NaOH(aq) + _Fe(ClO4)3(aq) => _Fe(OH)3(s) + _NaClO4(aq)\n\n3  NaOH(aq) + 1  Fe(ClO4)3(aq) => 1  Fe(OH)3(s) + 3  NaClO4(aq)\nThe coefficients of one could be excluded but I put one which is also equivalent.\n\n_((NH4)2)SO4(aq) + _Pb(NO3)2(aq) => _PbSO4(s) + _NH4NO3(aq)\n\n1  ((NH4)2)SO4(aq) + 1  Pb(NO3)2(aq) => 1  PbSO4(s) + 2  NH4NO3(aq)\n\nThese are only the balanced equations.\n\nThe following is the ionic equations:\n(Na)+ (aq) + (OH)- (aq) + (Fe)3+ (aq) + (ClO4)- (aq) => Fe(OH)3(s) + (Na)+ (aq) + (ClO4)- (aq)\nand\n(NH4)+ (aq) + (SO4)2- (aq) + (Pb)2+ (aq) + (NO3)- (aq) => PbSO4(s) + (NH4)+ (aq) + (NO3)- (aq)\n\nThe ionic equations are already balanced.  It is simply the breaking down of the equations into ions.  The solid remains intact while the compounds in aqueous solution become ions.\n\nThe + and - signs are the charge of the ions.  I can't find any other way to write them so I separated the compound with parentheses.  Sorry if it's confusing.\n\nThe net ionic equations are "simplified" ionic equations.  Basically, the things that are repeated on both sides get crossed off.  What is remaining is simply the net ionic equation.\n\n(OH)- (aq) + (Fe)3+ (aq) => Fe(OH)3(s)\nand\n(SO4)2- (aq) + (Pb)2+ (aq) => PbSO4(s)
533621
(delta)H(rxn) = sum of heat of formation of products - sum of heat of formation of reactants.\n\na.  Hf Br2(g) = 30.91 kJ/mol, Hf Br2(l) = 0, so (delta)H = -30.91 kJ/mol\n\nb.  Hf CaCO3(s) = -1207 kJ/mol, Hf CaO(s) = -635.5 kJ/mol, Hf CO2(g) = -393.5 kJ/mol, so (delta)H = -635.5 - 393.5 + 1207 = +178 kJ/mol\n\nc.  Hf NO(g) = 90.25 kJ/mol, Hf O2(g) = 0, Hf NO2(g) = 33.2 kJ/mol, so (delta)H = 2 * 33.2 - 2 * 90.25 = -114.1 kJ/mol