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Everything posted by Tartaglia

  1. Yes quadruple metal-metal bonds are known. They usually involve one of sigma symmetry (ie no node), 2 of pi symmetry (ie one node) and one of delta symmetry (two nodes). The last is face on face d orbital overlap
  2. The Co3+ will catalyse decomposition H2O2 +2Co3+ --> O2 + 2Co2+ +2H+ The Co2+ will then be reoxidised H2O2 + 2Co2+ + 2H+ --> 2H2O + 2Co3+
  3. H2O2 +2H+ +2e --> 2H2O (H2O2 acting as an oxidant) H2O2 --> 2H+ +2e +O2 (H2O2 acting as a reductant) O2 should be released when H2O2 acts as a reductant, but loads of transition metals catalyse the decomposition of H2O2 and that is where the O2 comes from H2O2 --> H2O +1/2O2
  4. jmarjorie - which alkene are you oxidising? The problem with KMnO4 is that the yield will be low if the alkene has any H substituents (and most do). You should read section 5.6 "Advanced organic chemistry" 3rd Edition Jerry March. There are plenty of references in there. Remember when you use a catalyst you are only using a very small amount. OsO4 may sound exotic to you but it is a standard procedure. Just bear in mind it is volatile and horribly toxic. Lead (IV) acetate will add to acetate groups syn across a double bond which are then easily hydrolysed. There are many methods involving anti addition rather than syn addition. You want to look at the symmetry of the alkene to see whether it matters to you. If anti addition is OK there are a whole host of epoxidation reactions that could be used, which yield 1,2 diols on ring opening hydrolysis
  5. A very common and very useful method. I spent most of my PhD recrystalising compounds this way, but using light petroeum, diethylether, thf, ethanol, methanol, diglyme, acetone, water, chloroform, dichloromethane and just about every combination of these
  6. Usually H2O2 is used with an OsO4 catalyst to give syn addition of two hydroxyl groups. It really depends whether you want a 1,2 diol or to cleave the C=C bond. There are many types of oxidation reactions of alkenes using peroxycarboxylic acids, ozone,lead(IV) compound etc,etc. Some are more agressive than others. Good reference "Advanced Organic Chemistry" Jerry March
  7. Yes but I would consider using a less aggressive oxidant. Practically anything will work including atmospheric oxygen
  8. I can't give a definitive answer but the usual "half filled subshell" explanation is a simplification. The promotion of an electron usually arises when the spin exchange energy gained excedes the energy required to promote. This is most usual when filling or half filling a subshell but doesn't necessarily have to be the final electron making up a half or full shell
  9. Futureless - This isn't a dig at you specifically. Its more a reflection of the English educational system. Logs were taught to 12 year olds in the 1970's and so "ln" isn't advanced at all. I'm sure you can see it on your scientific calculator
  10. Absorbance = ln(100/% transmittance) = E*c*l Where E = extinction coefficient, c = conc, l = path length of cell. You should really be fitting a straight line to ln(transmittance) (yaxis) and concentration (x axis)
  11. Phenol will also react with sodium nitrite in the presence of H2SO4 (7-8C) to give an 80% yield of p-nitrosophenol which can be reduced in the same way as p-nitrophenol. I would however caution against slap dash use of nitrites as they can produce nitrosamines under the right circumstances and they are very carcinogenic
  12. I agree with darkblade. The direct nitration of phenol is a good prep for the ortho nitrophenol as it has an intramolecular hydrogen bond and so is easily separated by steam distillation. It is possible to get reasonably pure para nitrophenol from the residue but there is usually a lot of tarry rubbish as well, making the yield low.
  13. Henry's law should still apply as you are still dealing with the eqm of CO2(g)/CO2(aq) and there is more than one gas anyway as water vapour is present Some considerations - the behaviour of CO2 is not going to be ideal and in water it will react with water to form traces of carbonic acid in an eqm reaction and that will equilibrate giving H+ HCO3-
  14. Looks like a good candidate for an azeotrope to me. If both pyridine and ethanol are wet (and ethanol will almost certainly be wet unless distilled from sodium) then you will probably have quite a complicated phase system.
  15. For more accuracy by far the easiest way is to look at the historical betting odds at the time the selections were made. Betting markets are in general very efficient, although just occasionally they aren't.
  16. I think I should point out that my calculation assumes all are equally likely to win. If you choose favourites the odds will be a lot shorter and if you choose outsiders the odds will be a lot longer. The number of people entered will not affect the probability of picking 8 or 9, but it will affect your chance of getting a prize, but with only 700 entries you stand a pretty good chance of winning or being top equal
  17. Probability all nine correct (1/5)^9 ie Decimal odds 1953125 or fractional odds 1953124/1 Probability 8 correct = (1/5)^8*(4/5)*9 ie decimal odds 54253.476 or fractional odds 54252.47/1
  18. Kirchoff's 2nd law gives E(t) = RI(t) + Q(t)/C or E(t)/R = I(t)+ Q(t)/CR differentiating wrt t and setting dQ(t)/dt = I(t) gives (1/R)*dE(t)/dt = dI(t)/dt + I(t)/RC as required (a) dE(t)/dt = 0 ie dI(t)/dt + I(t)/RC = 0 solving I(t) = I(0)exp(-t/CR) = (Eo/R)*exp(-t/CR) (b) dE(t)/dt = Eo ie dI(t)/dt + I(t)/RC = Eo/R Integrating factor exp(t/CR) I(t)exp(t/CR) = integral (Eo/R)exp(t/CR) I(t)exp(t/CR) = EoC*exp(t/CR) + k put I(0) = Eo/R, t = 0 k = Eo/R-EoC I(t) = EoC{1-exp(-t/CR)}+ (Eo/R)exp(-t/CR) Please note there looks to be an inconsistent use of units in question which makes the answer look odd. In (a) the unit of Eo is volts whereas in (b) unit of Eo is volts per second
  19. Alkoxide anions can act as 4e donors. The second pair is donated through a pi interaction rather than a sigma interaction. There are plenty of others such as the NR group which can also act as a 4e donor. Neutral ligands such as NO and CR act as 3e donors, but if you were to postulate that they were negatively charged then you could look upon them as 4e donors. It is a matter of interpretation of the metals oxidation state 4e electron donating alkyne groups are pretty common too - I've even published some myself. These however donate sideways through the pi CC orbitals and so both carbons are involved ( perhaps not quite what you want)
  20. Try x = 1 +2cost+Cos2t y = 2sint+Sin2t
  21. It really should go something like this Let m(t) be a function of time and let h be a small increment of time m(t+h) - m(t) = hm(t) divide through by h (m(t+h)-m(t))/h = m(t) Take limit of LHS as h tends to zero ie h becomes a infinitesimal change in time dt and m(t+dt) - m(t) = dm(t) ie dm(t)/dt = m(t) I suspect this is some sort of Markov process
  22. O2 possesses a triplet state - ie two unpaired electrons and is therefore a diradical. This makes it substantially more reactive than N2 which is a singlet state ie all electrons are paired.
  23. I find it incredible that I appear to be on te backfoot here when I consider safety to be the most important aspect of synthetic chemistry. I will continue to put safety first. This thread was put up without drawing any attention to selenium's toxicity. The thought of someone, who doesn't know what they are doing, reading this thread and going off a boiling up selenium in concentrated sulphuric acid in their garden shed doesn't bear thinking about. Jdurg - I have a friend with substantial burn scarring from an ether explosion caused by LiAlH4 coming in contact with water. There are many reasons to be careful in a lab. Toxicity is not the only one
  24. The inconsistencies of poison regulations has always been a surprise to me and is nothing new.
  25. Unfortunately price is not correlated with toxicity. Please be my guest and carry on using it. I won't be joining you. The basic problem with selenium is the low ratio of required amount to lethal amount. From my days as a chemist I believe the ratio is the lowest of all required trace elements
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