Calculate the energy change for the reaction\nK(g) + Br(g) --> K+(g) + Br-(g)given the following ionization enthalpy (IE) and electron affinity (EA) values.\nK : IE = 419 KJoule/mol ; EA = - 48 KJoule/mol\nBr : IE = 1140 KJoule/mol ; EA = - 324 KJoule/mol
138308
Just add the IE for K and the EA for Br and subtract the EA for K and the IE for Br
160015
Try finding the standard reduction potential for Cu and NO first then use those poteintial rxn equ to form the above full reaction afterwhich simply add or substract the poteinital u find to get E for the above reaction. \n\nSince E is proportional to the rate constate, u can use the above concentration as well as partial pressure to calculate the rate constant and hence E. Taking care that rate constant is the concentration/partial pressure of the reactant over the product!
208982
E = Fd = (2E-9 N) (200E-12 m) = x J\nx J (1 kJ / 1000 J) = xE-3 kJ\n\nI think this is already per mol...
28470
-278.0 + 2*(-321.0) + 2*(-784.0) = -2488 kJ\n\nI am assuming that the enthalpy was actually given in kJ/mol.
219453
AgBr-------->Ag+ + Br-\n(a) Ksp=[Ag][Br] solubility=s=[Ag]=[Br]=Ksp^0.5\ns=(6*10^-7)^0.5\n(b) [Br]=0.01 (because all of kBr was dissuciated)\ns=Ksp/[Br]=(6*10^-7)/0.01 mol/lit
479859
Do you have a chart indicating the electronegativities of atoms?\n\nThe more electronegative atom will be at the point of the arrow.\n\nH has an EN value of 2.1 and Br has an EN value of 2.8.\n\nSo, in this case, the point of the dipole arrow points to Br.
513735
They measured it.  The first electron has an ionization energy is 786 kJ/mol.  The fourth one has an ionization energy of 4360 kJ/mol.  It takes more energy to pull off the fourth electron because the net charge on the atom is positive (+3, in fact).
177286
i already answered parts a and b in the last question like this.\n\nc) enthalpy = mass of water x specific heat of water x change in temp\n\nE=0.1*4.18*3.5\nE=1.463KJ\n\nmultiply by two to get answer per mole = 2.926KJ/mol\n\nd) main source of error is the glass beaker since it is leak out the heat produced by the reaction, the reaction would be better carried out in a calorimeter or in a polystyrene beaker with a lid.\n\ne)  the temperature change would be the same as 1mol of KOH reacts with 1mol of HCl therefore with the extra HCl the KOH will act as a limiting reagent and u will get the same reaction as before.
359737
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
510063
Ans 1- Au+3 (Auric ion) +3e-(electron) ---> Au(s) [reduction]\nAns 2- V - 2e-(electron) ---> V+2(Vandium ion) [Oxidation]\nAns 3- Mg - 2e-(electron) ----> Mg+2 (magnesium ion) [Oxidation]
122259
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
167487
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
123069
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O + 726 kJ/mol
366895
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
364634
You missed the electron changes.\n\nThe bismuth electon change is +3 to 0 for a change of 3\n\nThe Sn electron change is +2 to +4 for a change of 2\n\nSo all the Bi species need a 2  and the Sn species need a 3.\nThen fill in the O and H.
463629
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
343336
1) Balance the reaction equation:\nPb(NO3)2  + 2 LiBr -> PbBr2 + 2LiNO3\n\n2) Calculate the amounts we start with\nPb(NO3)2 :  2.05 L * 0.0550 mol/L = 0.1128 mol\nLiBr : 1.52 L * 0.0618 mol/L = 0.0939 mol\nAs we need 2 moles of the latter for 1 mole of the former, the reaction is limited by the amount of LiBr\n\n3) Calculate the amount of PbBr2 produced\n0.0939 mol of LiBr will react with 0.470 mol of Pb(NO3)2 to produce 0.0470 mol of PbBr2. This is 0.0470 * (207.2 + 2* 79.9) =17.249 gram\n\n4) Calculate the yield\nThe yield is 15.784 / 17.249 = 91.51%