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Everything posted by Tartaglia

  1. Ecoli - Many synthetic chemists spend a large proportion of their time drying solvents as the compounds they work with are moisture sensitive. Rdaniel - Drying ethanol is reasonably easy with the right equipment, as you just dissolve Na in it and distill it off under N2. Distilling solvents is however where most accidents happen in the lab.
  2. ecoli - nmr machines were used in the 1960's and Nobel prizes were awarded for work on nmr in the 1940's and 1950's
  3. Ammine complexes of ammonia are not unknown but tend to be adducts rather than cationic complexes. Some of the reasons outlined in my first post do not apply to aluminium as it doesn't have a partially filled d shell. Aluminium is incredibly oxophilic as the thermodynamic sink of Al is Al2O3. Al3+ is extremely polarising due to its small size and large charge. It will find ripping OH- or O2- out of water a lot easier than ripping NH2- or even N3- out of ammonia as the Pka of H2O is a lot lower than for NH3. Thus it prefers aqua ligands
  4. GradGrrl- I know about it. I used to use 1H and 13C nmr a lot when I was a synthetic chemist. Its not particularly easy to explain though. What exactly don't you understand?
  5. That's where experience comes into it. You would have to collate data from similar situations in the past or guess
  6. Your estimate of p(A) would be a prior estimate. After applying Bayes theorem you have your posterior estimate P(A|C)
  7. Clearly if the police are just rounding up suspects P(A) is a lot less than 0.5, but if they are making a single arrest after a lot of detective work then P(A) could be a lot more than 0.5
  8. You have forgotten a bracket - assuming P(A) = P(B) = 0.5 (which is not necessarily a good assumption) then your answer should be 1/1.01
  9. P(A|C) = P(C|A)*P(A)/(P(C|A)*P(A) + P(C|B)*P(B)) where P(C|A) = 1, P(C|B) =0.01 and P(A) and P(B) are chosen appropriately
  10. Dak - I made a mistake when I put it up. In my calculation they would be the same which clearly would not be true, but this does lead you into how to demonstrate the fallacy mathematically. When I put the post up I was thinking in terms of testing a long line of suspects
  11. Assuming only one person is arrested and tested then probability of guilt is given by Bayes theorem P(A|C) = P(A and C)/ (P(A and C) + P(B and C)) = 1/1.01 However if a whole bunch of people are tested until one is found positive then other distributional assumptions need to be made and the calculation is more complicated. A geometric distribution could be made use of here Edit - Mistake in this post see post below
  12. I suspect entropy is not the major factor, but the polarising power of metal ions increases with charge and decreases with ionic radius and the more polarising the metal ion the more the solvent molecules will affected beyond the immediate metal coordination sphere so there could be entropy differences
  13. Anything titled "Engineering Maths" or "Advanced Engineering Maths", "Mathematics for Scientists" will help and are commonly available on ebay for a few quid. Authors that spring to mind Stroud, James, Stephenson, Jeffrey, O Neil, Kreyszig For the lower end you might also consider buying some of the A level further mechanics module booklets published by the A level boards Edexcel, Ocr or AQA. They are usually called "Mechanics 3 and 4" or "Mechanics 5 and 6" or variations on this theme
  14. This isn't really my field of expertise, but it is probably different change in Gibbs free energy of adsorbtion on the surfaces. You will have the eqm CO2(aq) <> CO2 (adsorbed on surface) <> CO2(g) and the position of eqm will depend on the strength of intermolecular interaction between the CO2 and the solid surface There may also be a kinetic effect too
  15. This is actually a remarkably difficult question to answer. It is actually a thermodynamic question with a lot of potential contributions from a lot of different, physical, chemical and electronic influences. I will however give a number of pointers. Symmetry of the complex matters The oxidation state matters The ionic radius Whether it is a 3d, 4d or 5d metal Whether the complex is high spin or low spin -ie crystal field stabilisation energy Whether the ligand is strong field or weak field Whether ligands can pi accept or pi donate Solubility considerations, enthalpy, entropy Whether bridging dimetal or polymetal complexes are possible eg with hydroxide ligands and probably a lot of other things which haven't occured to me yet
  16. The tautomer of a 1,2 hydroxy alkene is an alpha hydroxy ketone. Magnesium is a strong enough metal to react with water to produce hydrogen (all be it slowly) and so should be able to act as a reducing agent to give a 1,2 dihydroxy alkane. Again no surprise there
  17. Oxidation of 1,2 dihydroxyalkenes, by transtion metals is not a particularly unusual reaction. You form a chelating 5 membered pi system which is easy to push electrons around
  18. V = High - low = 0.34-(-.13) = 0.47 V Faradays used = 2 as Pb is oxidised to dication E = V*Q = 0.47 * 2 *96500 J
  19. Oxygen is a diatomic molecule and hence has 6 degrees of freedom, 3 translational, 2 rotational and 1 vibrational. Both rotational degrees of freedom will contribute to Cv but the vibrational one won't (at least at normal temperatures). For a diatomic Cv = 5/2R, Cp =7/2R
  20. Ksp can actually contain terms for the concentration of solids, its just they are defined as 1 and then usually left out. As for the eqm of solid and aqueous salt, it does actually exist, its just that you only need a infinitesimal amount of precipitate to set up the eqm as the solid automatically attains its defined concentration of 1. Thus there is no observable precipitate
  21. You need a table of standard electrode potentials, but in this case it will reduce only silver ions
  22. Tartaglia


    Boranes can be quite treacherous, particularly the lower ones. The higher caged type boranes/carboranes I have used in the past and are much more stable.
  23. 1,4 dibromobenzene can be made by direct Friedel Krafts halogenation, but it would have to be separated from the ortho product. para dinitrobenzene would have to be made in a roundabout fashion eg using the sandmeyer reaction as nitro is a meta director C6H6 ---> C6H5NO2 ---> C6H5NH2 ---> p + o C6H4(NO2)(NH2) p-C6H4(NO2)(NH2) ---> p-C6H4(NO2)(N2)+ ---> p-C6H4(NO2)2 For C6H5CH2NH2, direct chloro methylation can be carried out using HCHO +HCl with a ZnCl2 catalyst will give benzylchloride which will easily undergo SN1 displacement by NH3 m-C6H4BrCO2H can got from the Gatterman Koch formylation of benzene to produce benzaldehyde, followed by oxidation followed by Friedel Krafts bromination The last one I'll have to think about
  24. Let small room be n*n and large (n+m)*(n+m) ie (n+m)^2 -n^2 = 101 or m*(2n+m) = 101 ie m is a factor of 101, which is prime therefore m = 1, n =50
  25. If you have a single crystal of graphite then the anisotropic conductivity can be observed, with conduction within layers and insulation between layers
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