Tartaglia

Le Chatalier's principle can of course in certain circumstances be given in a mathematical form. eg the Van't Hoff equation is a statement of how the equilibrium constant varies with change in temperature and is basically a mathematical version of Le Chatalier's Principle d(lnK)/dT = deltaH/RT^2 This allows quantitative as well as qualitative evaluation of the change of temperature on the eqm position

Evergreen - I think you may be getting at the anomeric effect which occurs when there is an oxygen in the ring adjacent to the OMe

You get two electrochemical reactions Fe -> Fe2+ + 2e (Anodic reaction) 1/2 O2 + H20 +2e -> 2 OH- (Cathodic) This will give rise to current due to electrons moving through the metal from the anodic region to the cathodic region and presumably the OH- move the other way through the water completing the circuit. (Or Fe2+ move through the water the same way) Fe2+ +2OH- -> Fe(OH)2 There is then a subsequent further oxidation to give Fe (III) The cathodic equation tells you why both O2 and H2O are required for rusting. The fact that salt speeds up rusting so quickly tells you there must be movement of ions through the water involved. Salt increases the conductivity of water considerably Although anodic and cathodic regions are distinct I doubt there would be miles between them

The second and third moment about the mean are the variance and skewness respectively, but in order to answer the question you need to specify which moment you want about 10. This should help E(X) = 5 E(X^2) - 25 =20 therefore E(X^2) = 45 E(X^3) - 3E(X^2)E(X) + 2 *(E(X))^3 = 140 therefore E(X^3) = 140 +3*45*5 -2*5^3 = 565

Chuihen - I've just noticed this. The point here is not to use a strong oxidising agent as that will probably cause oxidation beyond the 1,2 diol required. Organic chemistry is a subtle and sophisticated science, limiting yourself to the classic schoolboy reagents will not get you very far

These questions can be tricky. In general you want most groups equatorial rather than axial. This is due to a small 1, 3 axial clash. However it is more important to have bigger groups equatorial as the clash can then be quite serious. In the first one, the NH2 should be the only group axial. In the second, the OMe should go equatorial as it is the most likely to clash, but if it does both the Me and Cl must go axial. This is probably a close call but OMe axial is probably favoured allowing Me and Cl to go equatorial In the third it is best to have the OMe and Cl that are next to each other axial and the other 2 OMe and Me equatorial. The other conformer would have three groups axial with a horrendous 1,3 OMe, OMe clash

Woelen - I agree a few mls is not a danger Collection of chlorinated and non chlorinated solvents separately has been going on for decades. This predates any modern disposal and environmental concerns. It was introduced as good labratory practice to prevent explosions. We do not seem to have gotten off to friendly start. I do in actual fact admire your chemistry knowledge and I don't doubt you are a competent electronic engineer, but I have seen far too many silly and serious accidents not to point out what most practising chemists know

Woelen standard laboratory procedure, known by most chemists However here is a reference http://www.southalabama.edu/environmental/chemicalhygieneplanappendic2003.DOC

It is not in general a good idea to mix chloroform and acetone. The reason laboratories have a chlorinated solvent waste and a non chlorinated solvent waste is precisely to stop people mixing these two solvents. In the presence of a base it is an explosive mixture

Although the iron atom is a good nucleophile for some electrophilic aromatic substituions it is unlikely to be the case in nitration. For instance Friedel Krafts acylation will almost certainly occur by formation of a Fe-C(O)R moiety and the acyl group will attack the ring in an endo fashion. Any likely nitrating conditions, even as mild as using for example [NO2][bF4] will almost certainly oxidise the ferrocene to the ferrocinium ion [Fe(eta-C5H5)2]+. This electron poor nature of this cationic species will then make electrophilic attack extremely unlikely

An acquaintance of mine was killed when a leak in a phosphine delivery system sprayed him with flames. He took about two weeks to die. I have witnessed and come across some other horrendous accidents in my time, including a lad who loss half his intestines when a peroxide explosion impaled him on a retort stand, somebody who blew half his hand off in a perchlorate explosion and 4 solvent explosions mainly ether but one was thf. In one of these the person concerned suffered 40% body burns. When I look at this forum I often wince. There are obviously a lot of incompetentents who are heading for serious accidents, but this is exacerbated by poor advice handed out by people whose descriptions rather overstate their expertise. Often the dangers are not obvious, such as repeated exposure to toxins. My PhD supervisor's, supervisor's supervisor was the great boron chemist Leonard Stock who managed to poison about 30 of his students when they were developing the mercury diffusion pump The reality of academic science is unfortnately not self sacrificing individuals trying to extend the extent of human knowledge but rather a lot of egotistical, self serving glory hunters, whose rewards are often not worth the risks their unsung students take

Iridium has been extensively studied particularly low oxidation state iridium complexes such as Vaskas complex. This is largely because many of its compound undergo oxidative addition reactions which were of tremendous interest due to their similarity to Wilkinson's catalyst. Higher oxidation "Werner type" complexes were almost certainly studied in the early part of the 20th century and probably revisited later when spectroscopic methods became much more sophisticated in the 1960's and 1970's. One name springs to mind - "Kurt Dehnicke". He is the sort of chemist who looked at all sorts of high oxidation type metal complexes and has just published his 1250th paper One point to note is that in colours of transition metal complexes the strict d-d transition is formally quantum mechanically forbidden. This however is relaxed often due to slight lowering of symmetry. Thus really intense colours are not generally due to d-d transitions, but to charge transfer bands which have much higher extinction coefficients

This is not something I have really looked at for 15 years but the formation of these things is essentially multiple ring closures and I believe they are affected by rotational entropy considerations and may become more and more difficult as the ball becomes larger. I might add this is only my feeling and I can't really back it up with suitable references

It is due to d-d splitting of an incomplete d shell. When an electron is promoted from an occupied to an unoccupied orbital it will absorb light with photon energy equal to the diiference in energy of the two states. You will see the complementary colour. The energy splitting will depend on the metal, its oxidation state the ligand and whether it is a pi donor or pi acceptor etc etc, but the symmetry has a major effect too. [Cu(H2O)6]2+ is roughly octahedral (Oh) but since Cu2+ is d9 it will suffer a Jahn Teller distortion which lowers the symmetry to D4h. This will lower the degeneracy of some of the d orbitals

Tom - My mind doesn't actually work that way. I just thought Jordan's original idea was smart and unusual The point about the tan substitution is that regardless as to how you do it the final arctan term comes from a direct or implied tan substitution. I didn't really mean to suggest the tan^6(A) substitution was the only one possible

Ethanol may well oxidise under those conditions to give choral hydrate. Elimination will be discouraged by more water and less ethanol, but will depend on whether there are hydrogens alpha to the chloro group. Ring closure to form epoxide may occur but they should then hydrolyse fairly easily to give 1,2 diol

If you do a long division first it is obvious [math] \frac{(x^3-x)+(x +2)}{x^3-x} = 1 + \frac{x+2}{x^3-x} [/math]

No it will be mainly N2 Try making CO2 from baking soda and vinegar

Your expression is for dx not dt/dx and so it is tha same as mine only upside down. You are also obviously unfamiliar with the laws of logs as ln(2 + 2tan(x/2)) = ln2 + ln(1+tan(x/2)) and the ln2 disappears into the integration constant

Substitute sinx = 2t/(1+t^2) cosx = (1-t^2)/(1+t^2) where t = tan(x/2), dt/dx = 1/2*(1+t^2) rearranging gives / | dt/(1+t) = ln(1 + tan(x/2)) + c /

The first metal metal quadruple bond was made by F. A. Cotton in the mid 1960's. He also use to claim credit for the first metal-metal covalent bond too, but that was published in an obscure Russian journal. This didn't however stop him from doing this for years until my PhD supervisor, who still has a "God like" status in the chemistry world told him not to.

Since the answer is 6x^(1/6) + 6*arctan(x^(1/6)) + C I would say a tan substitution is necessary

Good spot by Jordan With x = tan^6(A) the integral simplifies to / |6tan^5(A)sec^2(A) dA/((tan^3(A)*(1+tan^2(A)))= / / |6tan^2(A)dA / 1/(1+cos(x)+sin(x)) can be done by substituing sinx = 2t/1+t^2 and Cosx = (1-t^2)/(1+t^2) where t = tan(0.5x)

Platinum's stability is probably due to the lanthanide contraction. This makes it much smaller than you would expect and so the outer electrons are much more tightly bound

http://www.pnas.org/cgi/content/abstract/81/8/2592 http://www.tntech.edu/chemistry/Inorganic/Chem4110/Student/08%20MMQBShow.ppt http://en.wikipedia.org/wiki/Covalent_bond Look well down the page of the last