Tartaglia

The bonds in diamond are the classic sp3 sigma bonds and so the bond order is precisely 1 The bonds in graphite are sp2 type bonds with a delocalised pi system across layers. Since the bonds between layers are merely van der waals the average bond order in graphite is 1.333333. Thus within layers they are stronger than diamond (but weaker than benzene which has bond order 1.5).

An allosteric enzyme is an enzyme that does not obey the Michaelis Menten model but obeys a similar model with a different power due to cooperativity between two or more active sites V=Vmax*X^h/(K^h + X^h) K = half max rate conc of substrate V = Rate X = substrate conc h = 1 for Michaelis Menten model , but is not equal to 1 for allosteric enzyme

The attrition rate during the exams in UK is incredibly high

4100 active in UK, so 16000 sounds about right for USA - much easier to qualify in the USA though

YT That wasn't a dig at you - merely a retort to matt's comment on calculators

Matt - I am the last of the log generation and seeing people struggling to do simple arithmetic without calculators astonishes me. My last maths lectures were by one of your colleagues - Judy Holyer in 1983 when I was 18/19

394915 - 391758 = 3157 = 7*11*41 = 7*451

YT - I factorised the difference 41*11*7 and looked for three figure factors

divisor 451, remainder 290

(b) 976 I think 1,2,4,8,16,61,122,244,488,976

Hephaestus, The same thing happened to my flatmate's PhD supervisor when I was a student. Very sad

497+498+499+500 since 1994 has factors 1,2,997 and 1994 and the sum of this arithmetic series is (n/2)*(2a +n-1) then either n/2 or n is a factor. ie n = 1,2, 997, 1994 or 2,4, 1994 or 3988 since the large values of n give negative a and n = 2 must give an odd number and n = 1 is the trivial case n = 4 is the only possibility

(a)97^2

I would be interested and would contribute, but seeing as the "Applied Mathematics" forum has very little activity as it is I doubt it would be a success

The proof or otherwise of this problem revolves around the factors of 1991 Since for an arithmetic progression S(n) = (n/2)*(2a+(n-1)d) either n/2 or n must be a factor of 1991. This means the string of digits can only be 1,11,181 long or 2,22,362.

What about the 11 consecutive integers 176,177, 178,...186?

Don't the 22 consecutive numbers 80,81,82...101 add up to 1991?

In CO2 all the valence electrons for carbon are used for bonding. In CO there is a triple bond due to a dative pi interaction of the oxygen and carbon orbitals. This leaves a lone pair on the carbon. Normally lone pairs on elements which are not especially electronegative like carbon are not good ligands due to the poor sigma donating ability. However the multiple bond of the CO has appropriate pi* antibonding orbitals of the same symmetry as the d orbitals as the metal, this allows pi back donation into the CO pi* orbitals, similtaneously weakening the CO bond and strengthening the Metal -C dative bond. This is what makes it such an effective haemoglobin blocker

Since sec^2(X) = tan^2(x) + 1, the only difference will be in the integration constant

0.0592 = 0.02567 * ln 10 Thats why

the 0.0592 factor should be RT/F which I make 8.314*298/96500C = 0.02567 This would make the emf about 1V less - still substantial

Woelen - Good point the pH will change from standard conditions where [H+] = 1 therefore K will change by a factor of 0.1^2*0.1^3/{0.1*{10^-7}^14} = 10^94 lnQ = ln(10^94) making emf about two volts less ie a lot less favourable

Cr2O72- + 14H+ + 3Cu --> 2Cr3+ +7H2O + 3Cu2+ K = [Cr3+]^2*[H2O]^7*[Cu^2+]^3/{[Cr2O72-]*[H+]^14} Since the solutions are reasonably dilute we can take the activities to be the concentrations Cu is a solid and [Cu] is defined as 1, [H2O] is constant and [H+] is constant at 10^-7 Therefore K will change by a factor of 0.1^2*0.1^3/0.1 = 0.1^4 therefore ln Q = ln {0.1^4}

! means factorial ie 4! = 4*3*2*1 This problem has been unnecessarily complicated. The number of ways of arranging n different objects in n boxes is just n!

nCr is the number of ways of choosing r objects from n objects when the order doesn't matter nPr is the number of ways of choosing r objects from n when the order matters If you choose 4 from 4 and the order matters then 4P4 will work but it is not an efficient way of thinking about this problem. The answer is of course just 4! = 24