Calculate the values of DG° and E°cell for FeO(s) + H2(g) ---> Fe(s) + H20(l)?
374219
For the cell reaction: FeO(s) + H2(g) = Fe(s) + H2O(l) \n\nDG0(298 K) = DGf0 Fe(s) + DGf0 H2O(l) – DGf0 FeO(s) – DGf0 H2(g)\n\nFrom any table of Thermodynamic Properties:\nDGf0 Fe(s) = 0 KJ/mol\nDGf0 H2O(l) = -237.192 KJ/mol\nDGf0 FeO(s) = -251.4 KJ/mol\nDGf0 H2(g) = 0 KJ/mol\n\nThen: \nDG0(298 K) = 14.204 KJ/mol\n\nFurthermore: E0 = -DG0/(nF) = -14204/(2 x 96500) = -0.074 V vs NHE (normal hydrogen electrode), where n: number of electrons in each half reaction (FeO/Fe and 2H+/H2, n=2) and F is the Faraday constant (96500 C/mol).
582466
1 goes in front of CO2.\n\nFe2O3 + CO ---> 2FeO + CO2\n\nAluminium chloride, AlCl3 because\nAl3+ --- group 3\nCl- --- group 7
749374
Here is the balanced equation:\n\n2Fe2O3 + 3C = 4Fe + 3CO2\n44g          2.86g\n\nGet the moles of Fe2O3:\nmoles= 44/159.69\n          = 0.2755 moles of Fe2O3\n\n0.2755 moles Fe2O3 (4 moles Fe/2 moles Fe2O3) = 0.551 moles Fe\n\ngrams of Fe = 0.551 moles Fe (55.845 g Fe/1 mole of Fe)\n\n                     =  30.774 g of Fe
496291
You need to take more care in writing equations when you post your questions. \n\nPart a) is difficult to decipher - you will need to re post it.\n\nPart b) is presumably:\n\nCH3COOH(l) + 2O2(g) --> 2CO2(g) + 2H2O(l)\n\nPart c) is presumably:\n\nFe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)\n\nI would next use heats of formation to calculate the enthalpy change in the above reactions using the formula:\n\ndelta Ho = sum(heats of formation of products) - sum (heats of formation of reactants).\n\nIf you have a data book or data sheet, these values can be found there (Note: heats of formation of elements are zero.) \n\nIf you want web sites that might help you, try googling "enthalpy change problems" +tutorial.\n\nI hope this helps you.
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
187220
W=F*d,  d=W/F,  d=600J/400N = 1.5m
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
116913
I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
906651
2KOH=>K(2)O+H(2)O\n\nFeO+C=>CO+Fe\n\nCa+2AlCl=>CaCl(2)+2Al\n\nP(2)O(5)+3H(2)O=>2H(3)PO(4)\n\n2KI+Pb(NO3)2=>2KNO(3)+PbI(2)
890894
C12H22O11+12O2-------12CO2+11H2O+E
412413
I'm assuming you're asking for the reactions.  Here are a couple:\nH2 + F2 → 2HF \n\n4Fe + 3O2 → 2 Fe2O3
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
782406
°oO****Oo°°oO****Oo°°oO****Oo°\n\n\n 4 ∫[g(x)] g'(x) dx = 4 x 1/2g(x)² +c = 2g(x)²+c\n\nits the first suggestion that u made but i think you forgot the number 4...\n\n\n°oO~<<_P~'*'_ I~'*'_M~'*'_O~'*'~L _~'*'_A>>~Oo°
400981
first, you find out the equation of Keq and delta G. I think it should be Keq = e^ [-deltaG/(RT)].\nThis equation and constant R can be found in any general chemistry textbook. And T is usu. 25+273=298K\nthen you can find out deltaG for the equation.\n\nwe also have deltaG= [- delta G for Ag(NH3)2^+1(aq) + (-32.0)] - [-26.2 - 3.8]\n\nand there comes the answer!!! ;)
730844
OK...You can start with the equation of DG (I'm using D instead of writing "delta").\n\nDG = DH -TDS  You've seen that equation, right?  OK..Since the boiling point of water is 100 degreed C at 1 atm pressure, we are really talking about an equilibrium situation.  So, DG= 0.\n\nConsequently, DH = TDS.  \n\nSince you know DH and you know T (in Kelvin), you can calculate DS.  Make sure that you convert DH into Joules (instead of kilojoules) before you finish the problem.
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
52130
E=mc∆T is the equation you want to use for this situation, where E is energy, m is mass, c is the specific heat, and ∆T is the change in temperature.  Now, the heat lost by the alloy is equal to the heat gained by the water, right?  So look what we can do:\nmc∆T=mc∆T\nwhere the left side of the equation is talking about the metal, and the right side of the equation is talking about the water.  Let's start pluggin in values.  We'll leave the left c value as a variable because that's what we're solving for.  \n36.0 g *c*(178 oC -30.6 oC)=185.0 g*(4.184 J/g*K)*(30.6 oC-25.0 oC)\nThe trickiest thing about that step is the temperatures: the metal started at 178 and goes down to 30.6; the water starts at 25 and goes up to 30.6.  \nNow evaluate that equation and solve for c. \nc=.817 J/g*K