Tartaglia

Its pretty simple really, you divide it in two. With one lot you determine the Cu(II) in the solution. With the other you oxidise it with a suitable weak oxidant so that all is converted into Cu(II). Precipitate and titrate again The Cu(I) can be determined by difference This is a typical A level question

I would probably precipitate out all the copper as CuI which releases I2 and titrate the iodine with thiosulphate using a starch indicator Cu2+(aq) + 2I-(aq) -> CuI(s) + 1/2 I2(aq) I2 + 2S2O32- -> S4O62- + 2I-

Perfluorates are unlikely to exist as F has no available d orbitals required for pi bonding

Use benzene, p-toluenesulphonic acid and azeotrope the water out with a dienstark apparatus

Because the trial contains mice that died of other causes, I would probably analyse this as a markov process and determine rates rather than probabilities. The rates can then be turned into probabilities Probability of survival to time t = exp[-integrate between t and 0 mu(s) ds], where mu(t) is the death rate at time t Maximum likelihood estmates of rates and the variances are probably easily calculated from exposure times and no of deaths

This almost certainly EDTA. In basic aqueous soltion it will knock most ligands of "hard type metal centres" with ease. The 2nd law of thermodynamics always favours chelation particularly with hexadentate ligands like edta

The most important point here is the ability of the fluorenyl and indenyl ligands to go eta-3 rather than eta-5. This occurs because the aromatic stabilisation energy of the fused benzene ring(s) is returned. This gives rise to an associative mechanism which will be orders of magnitudes faster than the dissociative reactions that cp and cp* will give. Cp* makes the metal centre more electron rich and thus will give rise to complexes that differ in rates of substitution but probably not by more than one order of magnitude. It could increase or decrease rates depending on whether the other ligands are pi acceptors or pi donors. Also steric effects may differ cp is much smaller than cp* in terms of occupied cone angle

It is to do with the symmetry of the p orbitals that form the pi system of the ring. Linear combination of these orbitals lead to three bonding and three antibonding orbitals which are more or less spread over the 6 carbons (I simplify here) and that leads to thermodynamic stabilisation and equivalence of bonds

Quite right uncool - I do apologise. I didn't consider that there are two types of numbers after one has been chosen.Those whose positions are taken and those whose positions aren't

The working is Prob(>= 1) = 1 -P(0) = 1 - (n-1!/n!) = 1 - (1/n) for n> 1 In the case of n = 1, P(>= 1) = 1 - 0/1 = 1. The failure comes from definition of 0! = 1

1-(1/n)

1 Cheap available starting materials 2 Yield 3 Purity 4 Non toxic, non explosive starting materials 5 Safe chemical procedures 6 Short procedures ie few steps 7 Scalability 8 Easy workup procedures 7 Little waste that is easy to dispose off etc etc

The discrepency between yours and mine is due to you missing out the second term in the Taylor's expansion, which is really the whole basis of Ito calculus. I've been looking for a second substitution, but I haven't found an appropriate one yet. Substituting exp(0.25x) instead of exp(x) gives you a Martingale (ie no drift) and this may be useful. My advice is to take this to a financial economics or options forum

Thanks Atheist

The effect here is to turn X(t) which has a large negative drift and a large volatility when X(t) becomes negative into a process F(X) which has constant volatility and positive drift which depends inversely on F(X)

Latex is giving me a little agravation here but here are the first few steps Wt is a Wiener process or Brownian motion not Browns motion let F(X) = exp(X) F'(X) = exp(X) = F(X) F''(X) = exp(X) = F(X) dF(X) = F'(X) dX +0.5 F''(X) (dX)^2 Bearing in mind dWt - N(0,dt) (dX)^2 = 4*sigma^2*exp(-2X)dt substituting and simplifying gives dF(x) = {1.5*sigma^2dt}/F(X) + 2*sigma*dWt

[math] \frac{\partial F}{\partial X} = F(X,t) \frac{\partial^2 F}{\partial X^2} = F(X,t) dF(X, t) = \frac{\partial F}{\partial X}*dX(t) +0.5*\frac{\partial^2 F}{\partial X^2}* (dX(t))^2 (dX(t))^2 = 4\sigma^2 e^{-2X(t)} dt [/math]

Just testing [math] \frac{\partial F}{\partial X} [/math]

I've just worked through this and get Variance of theta^2/28 The important points here are the pdf of the median m ia given by f(x) = 1/theta*(1-x/theta)^2*(x/theta)^2*5!/(2!2!1!) the first part 1/theta is the pdf of the rectangular uniform solution, the (1-x/theta)^2 is the probability of two others above x, the (x/theta)^2 is the probabilty of 2 below x and the 5!/(2!2!11) is the arrangements of x, two above and two below evaluating the integral of x*f(x) between theta and 0 gives E(M) = theta/2 as required evaluating the integral of x^2*f(x) between theta and 0 gives E(M^2) = 2*(theta^2)/7 Variance = theta^2/28 as required More generally for an odd sample size 2n-1 a general formula can be derived by substituting y = x/theta to convert to a Beta style pdf and then by manipulating the integral to equal 1.

R(0,theta) is almost certainly a rectangular uniform distribution from 0 to theta. The expectancy of the median is therefore theta/2. The expectancy of the median squared can be determined by considering the probability of picking a number x and having two others above and two below this number. ie a binomial. Then integrating x^2*f(x) between theta and 0 and then fiddling about with a substitution and factorials to convert the integration into a Beta distribution pdf. Then Var(M) = E(M^2)-E(M)^2 I am unable to decide whether theta is known or not and this would obviously have a significant bearing on the result

I can't see one on Amazon. If its just for passing an exam then a 1994 edition will be plenty recent enough. If its for research then I doubt you would be asking on here

Crow is the usual undergraduate text http://www.amazon.co.uk/Principles-Applications-Electrochemistry-D-R-Crow/dp/0748743782/ref=sr_1_1/202-5649812-5211002?ie=UTF8&s=books&qid=1181830062&sr=8-1

Distilled water invariably contains dissolved CO2

How about toluene to p methyl cumene (via FK alkylation), autooxidation to give p-cresol followed by exhaustive FK alkylation

Yes a Michael followed by an intramolecular aldol, followed by dehydration