Testo

I suspect you could only get this kind of information (catalytic amounts, turnover number etc) from Journals on this topic. Hydrogenations that I've done don't generally need much (ca. 10% loading) and is good to use 3/4 times again (but with decreasing performance).

I'm sure DMSO will dissolve it pretty well, however it is somewhat toxic.

It would entirely depend on what the molecule is exactly but I wouldn't bet on it

Well borodin, the double distillation is slightly confusing but I don't think the last distillation is to remove water as there shouldn't be much in a fraction collected at 57-75C, thats why it's dreid over Sodium carbonate. I think it's just to purify the acetone again because the fraction collected wasn't very pure, that's the only reason i would do two distillations - if the first one failed to give me my product! Unusual to see that in a procedure however...

Well an IR absorption spectrum would be capable of detecting the transistions from the ground state to the first excited state and assuming the rigid-rotor (for roational spectra) model you can calculate the bond length from the transistions. And yeah it will have to be in the gas phase and free of other gases of course =)

Okay if you have carried out the reaction and distilled it you should have found that acetone (product) boils at around 60C and propan-2-ol (any unreacted starting material) at around 85C. So the first collection should contain the product. You wont have any salt in there arter the distillation, but u may have some water, thats why you need to dry it. Acetone has a distinct smell from propan-2-ol, thats the easiest way of knowing which is which.

You seem to be under the impression that I have invented this system, i have not! And i'm very capable of counting electrons thank you, i think you need to get over the fact that this is nomenclature not 'how many outer electrons does the elements in the group have'. i agree that it is a silly numbering system but there it is. In fact here is somebody whining about it; http://homepages.paradise.net.nz/totomark/PeriodicTable/MarksBrosPT1994.html

I think you mean sodium acetate, and no theres no faster way to remove solvents than with heating them in a vaccum (just depends how good the vaccum is). You do need higher atmospheres to get liquid CO2 and no theres no cheap way to do it. Also what the hell do you want it for?

Zn has a full 4s orbital, Cu has an unpaired electron in the 4s - the reason for why Zn has a higher ionisation energy is due to the pairing energy difference i.e. you need to put in more energy to unpair an electron (also compare Ag with Cd and Au with Hg). But as to why the radius is larger I don't really know...

Multiply by two

1. Well its a bit fuzzy actually, your right in saying 4s is lower in energy but there are several exceptions to when the 3d is filled preferentially, most notably a half filled d-subshell has an element of stability (see hunds rules) so u get 4s13d5 instead of 4s23d4 etc. 2. It's just an old system of naming the groups, groups 1-18 are used now for clarity 3. Assuming u want the S- ion then yeah thats your answer.

Well you can either buy it from sigma-aldrich or you can make it by adding sodium bicarb (backng soda) with acetic acid (in vinegar) and then boiling off the water and crystallising the sodium acetate.

Well I can't really answer your question without more information but heres the IUPAC rules in full... http://www.acdlabs.com/iupac/nomenclature/

I'm not sure where he quotes them as being F block (although the links to wikipedia do), he writes the lanthanides and actinides underneath the rest of the table which is quite common so i guess it's a little unfair to call him a moron. Perhaps Peter Atkins is also a moron for following the same convention?

Yeah i know i learn't the halogens as group 7 but now they are group 17 (new IUPAC) and told to teach them as such... http://www.dayah.com/periodic/

You are correct in saying that the tertiary bromide will form in the presence of light and heat via a radical mechanism. However NaOEt will likely not do a substitution reaction on a tertiary haloalkane, but would instead do an elimination (E2) of HBr by removing a proton beta to the bromide. This would furnish a tri-substituted alkene (1-methyl-cyclohexene). If this was a primary haloalkane, you would likely get substitution unless you had a sterically hindered base such as tBuONa.

I think it's; ΔN = sqrt [ ( (-a/y)*exp(-x/y)*Δx )² + ( (ax/y²)*exp(-x/y)*Δy )² ]

Maybe your observation is correct depends what you have read. But no the amounts depend totally on which acid and base are being used for example alot of acetic acid would be needed to neutralise a concentrated sodium hydoroxide solution, a much smaller amount of sulphuric acid would be needed to reach neutralisation.

http://www.webelements.com/webelements/compounds/text/As/As1I3-7784454.html

Okay my bad, i should have said you can percipitate out the calcium sulphate (along with strontium sulphate), filter it and the use an EDTA titration on the filtrate which contains the Mg2+ ions to determine magnesium content.

For this to work you would need to use hot, concentrated (98%) H2SO4 to form MgSO4 (liberation of HF) so you couldnt get the MgSO4 to dissolve without the rest of the metal sulpahtes also doing so. Please note that HF is an extremly corrosive acid and I wouldnt consider trying this unless you are suitably skilled. So maybe you could co-percipitate the Ca and Mg ions using strontium sulphate collect the filtrate and determine magnesium content using an EDTA titration to determine the Ca content first (EDTA will chelate with Ca and Mg) using the fact that the Ca2+ is more tightly bound to EDTA then Mg2+. Consequently, the colour change using Calmagite as an indicator (has a red colour when complexed with Mg2+ and a blue color when Mg2+ is chelated by the EDTA) for Mg2+ occurs after all Ca2+ is chelated. Then you can subtract this from the total amout of Ca and Mg to get the Mg concentration

Essentially you need to plug the 1s wavefunction (radial and angular parts) into the 3D time independant schrodinger equation and show that both sides are equal. Theres some tricky partial differenciation involved here but this should also allow you to find the eigenvalue for this eigenfunction. For the vibrational levels you will need the anharmonic equation G = w(v+1/2)-xw(v+1/2)^2 where w = (1/2pi)x(k/reduced mass)^1/2 and x = w/4D(e) then let v = 0,1,2,3 (four lowest energy levels) The last part is simply subtraction of these levels

Well, I think what your referring to is the uncertainty principle which states that you can't know the exact position and its speed (actually momentum) at the same time. Say you know an objects exact location, you have no idea of what its speed is and vice versa. The reasons behind this principle is quite complicated however. As to whether everything is a vibration of energy, i'm not sure what you mean because you need energy for something to vibrate. But i think your question comes from another theory called 'string theory' which basically states that everything is made up of vibrating strings.

You may have wanted to mention that this is an enzyme kinetics question. In a lineweaver/burk plot u plot 1/[substrate concentration] along the x-axis and 1/velocity on the y-axis, giving u a linear plot. The gradient of this is Km/Vmax and the y-intercept is 1/Vmax you now have everything you need to get both quantities.

For determining quantum numbers; n = 1,2,3,4... l = 0,1,2,3...(n-1) ml = 0, +/-1, +/-2, +/-3...+/-l ms = +/-0.5 So for 5g, n=5 and l=4 (0=s, 1=p, 2=d etc.) so ml values can be -4 up to +4 (nine orbitals) each with 2 electrons with +0.5 and -0.5, max occupancy is 18 electrons Term symbols are more tricky; the letter represents the orbital quantum number l, the left superscript is the spin multiplicity = 2s+1 where s is the total spin angular momentum quantum number. A single electron has an s of 0.5 so 2(0.5)+1 = 2. Two electrons has two combinations S = 0, 1 (0.5+0.5=1, 0.5-0.5=0) which gives Either 1 (singlet state) or 3 (triplet state) spin multiplicities and so on.... The right subscript is the total angular quantum number J = L+S, L+S-1, ... |L-S| The best way to show this is an example e.g. H 1s^1; s=0.5 => 2s+1 = 2 l=0 => S J= 0+0.5 = 0.5 For He 1s^2 s=0, 1 => 2s+1 = 1 AND 3 l=0 => S J= 0+0 = 0 AND 0+1 = 1 So helium has a triplet S and a singlet S state