URGENT!! What is the reducible representation of trans (E)-1,2-dichloroethene?
372589
How about making yourself a model, finding its point group then working out the reducible representation?
17774
It doesn't give me room to list them all and explain them on here so let me direct you to a website I found that does a pretty good job:\nhttp://vergil.chemistry.gatech.edu/notes/quantrev/node20.html
1177708
Please expand on which type of "equivalents" you mean as in substitutes or normality equivalents.  For suppliers, go here:\n\nhttp://www.chemcyc.com/ME2/Default.asp
649874
Nope, it's answer A.  The first compound in D has two chlorines on it, not a chiral compound.
843792
Be very careful:\nFirst, look at the electronegativity differences, the presence of groups like OH- or cations (like Na+ of soap)\nThen, you must know molecular geometries, it is highly important:\nFor BF3 this is a  trigonal planar compound, all differences of electronegativity B<F are canceled due to the position of the F atoms (every 120º on the same plane). It is not polar.\n[SnCl3]- this is a trigonal pyramid compound.\nCl\n I\n:Sn-Cl\n I\nCl\nAS you can see, Sn has a lone pair of electrons, making Cl atoms accommodate in the base corners of an imaginary pyramid. Even if Cl cancel each others dipolar momentum on the sides, they all point ↓. This is a polar compound.
806654
F2 + HCl ---> HF + Cl2\n\nThe oxidation numbers are:\nF2 :  0\nHCl: H: +1, Cl: -1\n\nHF: H: +1, F: -1\nCl2 : 0\n\nOxidation is loss of electrons.\nReduction is gain of electrons.\nF2 ---> F(-) gains electrons so it is reduced.  If it is reduced then it is the oxidation agent.\nCl(-1) ---> Cl2 loses electrons so it is oxidized.  If it is oxidized then it is the reducing agent.\n\nTo balance write the two half reactions and balance the atoms undergoing redox:\n\nF2 ---> 2F(-)\n2Cl(-1) ---> Cl2\n\nAdd the electrons to balance the charge:\n\nF2 + 2e ---> 2F(-)\n2Cl(-1) ---> Cl2 + 2e-\n\nSum the two equations:\n\nF2 + 2Cl(-) ---> 2F(-) + Cl2\n\nAdd 2 hydrogens to both sides\n\nF2 + 2HCl ---> 2HF + Cl2\n\nand that's it.
956142
2Rb(s)+Cl2(g)--2RbCl(s)
1138290
Look up stuff like Arrhenius equations. I don't have my phys chem text book to hand.\n\nFor part two, I think it's D, as I'm fairly certain it's not any of the others.
215160
http://www-ec.njit.edu/~skawinsk/chem243/chap11/chap11.htm\nthis may help–near the bottom
837624
Cassandra C has given you a very good answer.\nYou are interested in a single type of reaction: single displacement. It is a redox reaction so you have to think which compound is more happy having extra electrons (more electronegative).\n\nIn the case of A+BX-> AX + B, B(+) in BX is a cation and is missing an electron. If B is a more electronegative negative element than A, then it will steal away the electron from A and the reaction will occur. If A is more electronegative then B is not "strong" enough to steal the electron and the reaction will mot occur.\n\nIn the case of Y+BX -> BY+X you have the opposite. X(-) in BX has already an extra electron. So if X is more electronegative than Y it will not give away the electron and the reaction will not occur.\n\nThis is an oversimplified reasoning. The proper one is that you have to look at the standard reduction potentials (E0) which give the reactivity series. The half-reaction with the highest reduction potential will be the one favoured or more correctly the reaction will occur if the redox potential of the whole reaction is positive.\n\nFor instance you can find a detailed table at\nhttp://www.northland.cc.mn.us/Chemistry/standard_reduction_potentials.htm\n\nLet's see your reaction\nCu+FeSO4 -> CuSO4+ Fe \nIn this case the half reactions would be\nCu - 2e -> Cu(+2)\nFe(+2) +2e-> Fe\n\nFrom the website we get that for\nCu(+2) + 2e->Cu  E01=+0.342 Volt\nFe(+2) +2e-> Fe E02=-0.447 Volt\n\nE01 is more positive than E02 thus Cu(+2) tends to get electrons from Fe and not the other way around so the reaction will not occur \nOR\nYou find the E of the reaction like this.\nThe second half-reaction is in the form we want it. The first is in the reverse. So we just reverse the reaction and the sign for the potential\n\nCu -> Cu(+2) +2e E01'=-0.342\nFe(+2) +2e-> Fe E02=-0.447 and add\n\nCu + Fe(+2)+2e -> Cu(+2)+ Fe+ 2e with E=E01'+E02= -0.342+(-0.447) \n\nTherefore Cu+Fe(+2) -> Cu(+2) + Fe E= -0.789<0 and so the reaction will not occur.
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
502725
That second reaction should be Pb(s) + Cl2(g) ---> PbCl2(s).  Reverse the second reaction and then add the two together and you will get the resultant reaction.\n\nPbCl2(s) ---> Pb(s) + Cl2(g)    H = +359.4 kJ\nPb(s) + 2Cl2(g) ---> PbCl4(s)   H = -329.2 kJ\n\nPbCl2(s) + Cl2(g) ---> PbCl4(s)  H = +30.2 kJ
172042
C4H9N3O2 For the structure goto http://www.cwu.edu/~chem/courses/chem388488f00/coulombe/CREATINE1.GIF
284634
3K4Fe(CN)6 + 4FeCl3 -> 12KCl + Fe4[Fe(CN)6]3
747348
Hi Kitty!  Dr. J to the rescue with a full answer.\n\n1) 4Li + O2 ---> 2Li2O\n\n2) 4HF + SiO2 ---> SiF4 + 2H2O\n\n3) It is simpler if we eliminate the K and the sodium and write the equation ionic because it is a redox reaction and we need to balance.\n\nThe overall redox equation will be:\n\nCr2O7(=) + SO3(=) + H(+) ---> Cr(+3) + SO4(=).\n\nNow you balance by the half reaction method.  The two reactions are:\n\n6e(-) + Cr2O7(=) + 14H(+) ---> 2Cr(+3) + 7H2O\n3SO3(=)  + 3H2O --->  3SO4(=) + 6H(+) + 6e(-)\n\nEliminating the electrons:\n\nCr2O7(=) + 3SO3(=) + 8H(+) ---> 2Cr(+3) + 3SO4(=) + 4H2O\n\nThis one was tough!!!  Thumbs up!\n\nIf you want you could put the K and the Na back.\n\nBest answer?
803448
This is an SN2 reaction because you have a secondary haloalkane and a polar aprotic solvent so inversion of conformation occurs. The answer is B
788070
Well its easy dear.\n\n2Li (s) + Cl2 (g) ----> 2 LiCl (s)