Tartaglia

Woelen - No offence but you strike me as an amateur. I used to have an Office next to Norman Greenwood and I know Alan Earnshaw too. In the eight years in which I made a living as a synthetic chemist I saw people leave labs when others were using Selenium. A horrendously toxic element. I personally would not recommend that anybody does an unnecessary syntheses using selenium

The toxicity of selenium would put me off doing the last prep more than heating conc H2SO4

Assume bag is negligible weight Work given by /6 | m(t)*g*(dh/dt)*dt /0 where t = time (minutes) m(t) = 144-12t (lb) dh/dt = 3ft/min horrendous units though

Oxalic acid is fairly easily oxidised. Just cooking rhubarb destroys it. This may make it difficult to extract. As it is a potential chelating ligand it may be possible to precipitate out a metal complex of it from rhubarb juice. Very cheap to buy though and more easily made by other procedures

p= 24.8638977 q=17.13963918

Sin A + Sin B = 2 Sin((A+B)/2) * Cos((A-B)/2) = sqrt(2)/2 (1) CosA + CosB = 2Cos((A+B)/2)*Cos((A-B)/2) = sqrt(6)/2 (2) (1)/(2) gives tan((A+B)/2) = 1/sqrt(3) therefore (A+B)/2 = pi/6 therefore A+B = pi/3 Sin(A+B) = sqrt(3)/2

The rules are basically rules of thumb and I believe the Pauling scale is not the only electronegativity scale used. Most chemists have an idea of electronegativity difference but are rarely dogmatic in applying the differences quantitatively. Appreciably ionic means more or less fully ionic with the cation polarising the electron density of the anion to some extent

The reaction is not really totally different at high and low PH because the first intermediate formed in the haloform reaction is CH3C(O)CH2I. The mechanism is also broadly similar in that at high PH the species CH3C(O-)=CH2 atacks I2 to give CH3C(O)CH2I and at low pH the vinyl alcohol CH3C(OH)=CH2 attacks I2 to give [CH3C(=OH+)CH2I} which then elininates H+ to give CH3C(O)CH2I. At low pH the rate determining step will be the formation of CH3C(OH)=CH2 which involves H+ and CH3C(O)CH3 and not I2. Therefore the reaction is relatively slow and the order wrt I2 is zero

Evo - The one I put up is an obvious square (x+y-z)^2 = X^2+y^2+z^2+2xy-2xz-2yz. That's why I think there may be a mistake in your original

You are right in assuming that heptane has the larger BP and your reasoning is pretty good. The larger the molecule the larger the surface area the greater the attraction due to induced dipoles. However, it is really the vapour pressure that is important. The vapour pressure is the pressure exerted by gaseous molecules when in contact with their liquid phase. The lower the boiling point the larger the vapour pressure. The larger the vapour pressure the greater the rate at which you can evapourate the liquid for a given temperature (p is proportional to n because PV = nRT). The faster it evapourates the faster the liquid left behind cools. Therefore hexane will cool faster than heptane

I suspect there may be a mistake in the question. If the coefficient of B^4 is 1 instead of 2 it factorises a^{4} + b^{4} + c^{4} - 2a^{2}c^{2} - 2a^{2}b^{2} + 2b^{2}c^{2} = (-a^2+b^2+c^2)^2

Expanding (z-w0)*(z-w1)*(z-w2)* ...*(z-wn-1) = z^n -(w0+w1+w2+w3 ....+wn-1)z^(n-1) + (-1)^n*(w0*w1*w2*... *wn-1) with all other coefficients of z less than n-1 being zero. equating coefficents i = -(w0+w1+....+wn-1) -10 = (-1)^n* (w0*w1*w2*...wn-1) therefore sum of roots = -i, product = -10 if n is even or +10 if n is odd

P(X = 2) = 6^2*e^-6/2! = 0.0446 P(X >=2) = 1 - P(X=0) - P(X=1) = 1 - e^-6 - 6e^-6 = 0.98265 P(x = 3 | X >= 2) = P(x = 3)/ P(X >= 2) P(X =3) = 3^3*e-3/3! = 0.22404 P(X >= 2) = 1 - P(X =0) -P(X = 1) = 1 - e^-3 - 3e^-3 = 1 - 4*e^-3 = 0.8008517 P(x = 3 | X >= 2) = 0.22404/0.8008517 = 0.27975

Let x = tanh-1® ie x = 0.5*ln ((1+r)/(1-r)) x should be approximately normally distributed with mean 0 and variance 1/(n-3). This will allow you to work out p using z tables This is called a Fisher transformation Alternatively r*sqrt(n-2) / sqrt(1-r^2) should be distributed as a t distribution with n-2 degrees of freedom. Use t tables for p value

P (winner < x) = x^2n therefore F(x) = X^2n therefore f(x) = 2n*x^(2n-1) E(X) = integrate between 1 and 0 f(x) * x dx E(X) = 2n/2n+1