text:
Why does depurination happen at a higher pH (or under less harsh conditions) than depyrimidation?
I've always wondered why the N-glycosidic bond, which connects the adenine or guanine purine bases to their sugars, was the easiest to cleave via acid hydrolysis, or why purines have the lowest ionization potentials?  What is it about the rings / resonance / etc. of purine vs. pyrimidine bases in nucleic acid polymers that makes this so?  For example, according to (Lindahl, Nature 362, pp. 709–715 (1993), http://www.nature.com/nature/journal/v362/n6422/abs/362709a0.html), at least in the context of deoxyribonucleic acid (DNA) at pH ~ 7 (probably the rate is much slower in RNA under the same conditions due to the 2'-OH) the rate of depyrimidation is $\approx 20$ fold slower than the rate of depurination.  Note that the actual base rate of depurination is $\approx 4 \times 10^{-9} \space \mathrm s^{-1}$ at 70 °C and pH ~ 7.4 (Lindahl & Nyberg, Biochemistry 11(19), pp. 3610–3618 (1972), http://pubs.acs.org/doi/abs/10.1021/bi00769a018), implying a base rate for depyrimidation of $\approx 2^{-10} \space \mathrm s^{-1}$ under the same conditions.
If I had to guess, I'd say that the reason has to do with purine bases (or the local context of the N9-glycosidic bond on purine bases) being less electrophilic, which is why I mention generally lower ionization energies for purines vs. pyrimidines.
Quoting from: http://www.nrcresearchpress.com/doi/abs/10.1139/o74-139#.U32-3vmSyDZ
"Contrary to previous suggestions, pyrimidine as well as purine deoxynucleosides are subject to base cleavage, the relative ease of thermolysis being 2′-deoxyguanosine > 2′-deoxyadenosine > 2′-deoxycytidine > 2′-deoxyuridine > thymidine."
Note that this actually seems to match the relative (experimentally measured) ionization energies of the bases --- Guanine (7.77 eV) < Adenine (8.26 eV) < Cytosine (8.68 eV) < Thymine (8.87 eV): Orlov, V. M., Smirnov, A. N., Varshavsky, Y. M. Ionization potentials and electron-donor ability of nucleic acid babes and their analogues. Tetra Lett. 1976;17(48):4377–4378.
If my guess regarding the lower electrophilicity of purine bases is true, it's not clear to me, however, why this is the case?
Update / Thought:
If we look at the four bases (where uracil is just missing the C5 methyl group on thymine) - (http://img.sparknotes.com/figures/7/749a4182b7527e44d289a612e420f40c/dna_bases.gif) - we can notice that the N9 nitrogen connecting purines to their sugars is only ever adjacent to carbon atoms with single- or double-bonds to nitrogen.  However, the N1 nitrogen connecting pyrimidines to their sugars is adjacent to a C2 carbon double-bonded with an oxygen atom.  Given that oxygen is more electronegative than nitrogen (oxygen has an electronegativity of ~3.44 vs. nitrogen's electronegativity of ~3.04) could this be a hint as to why depyrimidation is much less common than depurination?  Are there pyrimidine base analogues with a less electronegative group attached to the C2 carbon, and do depyrimidation rates here match or exceed depurination rates?
This could, of course, be completely off.  For example, maybe N7 protonation on purines presents a route to depurination, since guanine having a greater depurination rate than adenine (or lower ionization energy than adenine) is consistent with a decrease in the N7 nitrogen's basicity.  Maybe pyrimidines are much less prone to destabilizing protonation.  Etc.
*
My objective here: I'd like to know the "most significant" factor for (and hopefully most general reason for) the greater observed rates of depurination vs. depyrimidation.  For example, perhaps the aforementioned oxygen on pyrimidines decreases depyrimidation rates by a factor of two or so (this isn't meant to be an actual guess), but what explains the $\approx 20$ fold difference noted by (Lindahl, 1993)?

Really interesting question!
Mechanism of acid catalyzed hydrolysis of purine nucleosides is known to proceed through unimolecular departure of base moiety(DOI: 10.1021/ja00709a055).

It seems that someone else has also asked the same question: "Why reaction rate with pyrimidines is so much slower?" -and published paper of the reaction mechanism (DOI:10.1039/P29940000309).
In the paper it is suggested that pyrimidines follow different reaction mechanims. This would explain huge differences in reaction rates.

In other words,
  the structure-reactivity correlations lend additional support to
  the previous suggestion, according to which the acid-catalysed
  hydrolysis of 4-oxopyrimidine nucleosides follows an openchain
  mechanism

Open chain mechanims:


text:
HNMR - Peak Splitting
If you have $\ce{CH3-\color{\red}{CH2}-CH2Br}$
Then the $\ce{CH2}$ (in red) would have a peak splitting of $12$ because $(n+1)(m+1)$
where $n$ for the $\ce{CH3}$ is $(3+1=4)$ and m for $\ce{CH2}$ is $(2+1=3)$ so its $4\times3=12$
source:

In the general case, a signal will be split into $(n + 1) \times (m + 1)$
  peaks for an $\ce{H}$ atom that is coupled to a set of $n~\ce{H}$ atoms with one
  coupling constant, and to a set of $m~\ce{H}$ atoms with another coupling
  constant.

How can we determine differences in the coupling constant by just looking at the structure? Otherwise how do I know when to apply $n+1$ OR $(n+1)(m+1)$?
But in the other textbook it says:
$\ce{CH3-\color{\red}{CH2}-CH2-COOH}$
The $\ce{CH2}$ (in red) has a peak splitting of $6$ because $(n+1)$ where $n=5$, $(5+1=6)$ and the $n$ is the $3~\ce{H}$ on $\ce{CH3}$ plus the $2~\ce{H}$ on $\ce{CH2}$.
Source:

Which is the correct method? Or are both right? and why?
Alternatively symmetric adjacent hydrogen groups:
$\ce{CH3–\color{\red}{CH2}–CH3}$
Is the $\ce{CH2}$ split to 4 peaks due to symmetric adjacent hydrogen environments? Due to $n+1$, where $n=3$ so $(3+1=4)$.  I read that it should be split into 7 peaks - but where are they getting 7 from - is it $n+1$, where $n=6$ so $(6+1=7)$. For symmetric adjacent groups do you allow $n$ to be the sum of all hydrogens adjacent/or not?

I am not an expert, but I remember the first rule to be exact, while the second rule is an approximation (meaning the neighbour hydrogens are almost indistinguishable). Given the right nmr, should should be able to resolve the 12 signals. In the (last) symmetric case you have six indistinguishable neighbour hydrogens, hence $n=6$ and a septett splitting. But I am far away from being an expert... (Very nice question, btw.)

Both are correct to some degree.
The first book predicting 12 lines is being very theoretical.  In principle with narrow enough line widths 12 lines would be observed.  However, due to coupling constants (J) being similar from the beta protons to the alpha and gamma protons, in practice with reasonable resolution, only 6 lines are observed (the first book is actually explaining this in fine print).

(source: ucalgary.ca)
Credit: Univ. Calgary

The second text is merely reporting the observed number of peaks, not predicting the splitting pattern you would obtain from this molecule.
So, this is not a case of one being an exact rule, and one being an approximate rule. It is a case of one text describing how to determine the expected splitting pattern (which would be a quartet of triplets, or 12 peaks) and the other simply reporting how many peaks are observed (six peaks). If you understand the method explained in the first text, you can very easily rationalise why the expected 12 peak qt appears as a 6 line signal; if you merely follow the observations described in the second text, you will fail to understand complicated splitting patterns that provide valuable information about molecular structure in 'real' molecules. (No disrespect to C3H7X text book examples intended).
Both of these systems would commonly be referred to as a A3M2X2 system, representing $\ce{CH3CH2CH2X}$. To predict the splitting of the M2 group, simply consider the effect of each coupled group in turn. The A3 group will cause the M2 protons group to split into a quartet. Each of those quartet signals is then split due to coupling to the X2 protons, so each quartet signal becomes a triplet. this gives you the 12 lines, as so elegantly described in the first text.
However, if you then consider that the coupling constants between M and A, (JMA) and  M and X (JMX) are roughly equivalent (let's say roughly 7.5Hz), then you get overlap of some of the 12 lines, and the resulting spectrum appears, roughly, as 6 lines. On a lower field instrument, or with poor resolution, this may be indeed what you see (first spectrum below, simulated at 200MHz). However at high field with good resolution (without having to resolution enhance the data), you can clearly see that there are in fact 12 lines, and the coupling constants are not quite equal (second spectrum, simulated at 800MHz). In fact, coupling constants are closer to JMA=7.8, and JMX=7.0.

Simulated Low Field Spectrum of A3M2X2 Spin System (Simulated at 200MHz, 0.5Hz linewidth)

Simulated High Field Spectrum of A3M2X2 Spin System (Simulated at 800MHz, 0.3Hz linewidth)
As an added note, in both molecules, $\ce{CH3-CH2-CH2OH}$ and $\ce{CH3CH2CH2Br}$, the central $\ce{CH2}$ is prochiral. A prochiral centre is defined as one that will become chiral if a single substituent is changed. Prochirality is a very important property to consider, as it is very common for protons at a prochiral centre to be chemically and magnetically non-equivalent if there is a chiral centre elsewhere in the molecule. This often gives rise to methylene protons of ethyl sidechains appearing as non-equivalent groups in NMR.

Here is a real spectrum at 400 MHz: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/nmr/HNL12.html

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text:
How come the same reaction can be balanced in two distinct ways?
I faced this reaction which can be balanced in $2$ distinct ways which are not multiples of each other.
$$\begin{align}\ce{6XeF4 + 12H2O &-> 2XeO3 + 24HF + 4Xe + 3O2}\tag{1}\\[0.4em]
\ce{4XeF4 + 8H2O &-> 2XeO3 + 16HF + 2Xe + O2}\tag{2}\end{align}$$
The first reaction is as it is given in the books and the second reaction was balanced by me.
I sense that only a mechanism can resolve this. I found a similar question on googling, but the answer is unsatisfactory. (It's Yahoo Answers. What can you expect?)
I also found two independent reactions here which on adding produce equation $(2)$. I am confused as to what exactly happens.

Interesting. Try this out for fun: treat those chemical equations exactly like you would mathematical equations. Subtract the second equation from the first (multiply the equations by constants if necessary) and try to cancel out as many substances as possible (i.e. give them stoichiometric coefficients of 0). If you find negative coefficients, don't be alarmed; you can turn a negative coefficient into a positive one by swapping it from the left-hand side to the right or vice-versa (just like in a mathematical equation, where the reaction arrow represents an equal sign). What do you find?

@NicolauSakerNeto I tried. I got some reactions. But that was a long time ago. Do you want me to post that in the question?

You don't necessarily have to add it to the question. All I mean to say is that if you play with the equations a bit, the answer may become clear. I'd love to give you a deep answer myself, but I lack the time to do so presently. If no one else answers though, I'll be happy to give a short but direct response in about 12 hours. If you do figure it out, though, feel free to answer your own question!

@NicolauSakerNeto Is my reasoning in my answer correct?

As per Nicolau's suggestion, :
$$\{(1-2)\times\color\red{4}\}-2$$
gives
$$2\ce{XeO3}\to2\ce{Xe}+\ce{3O2}$$
Thus, I conjecture that the second and third reaction differ by the fact of decomposition of $\ce{XeO3}$
$2$ can be converted into $1$ by multiplying by $\color \red{\frac 3 2}$ and decomposing one of the $\ce{XeO3} $ molecules.
Wikipedia also says that $\ce{XeO3}$ is unstable.

3 times equation 2 can be converted to 2 times equation 1, by decomposing 2 XeO3 molecules in 3 times equation 2.

Precisely the reaction you had to find! Good job, though I think you missed multiplying reaction $2$ by 2 outside the brackets, and it is reaction $2$ that can be tuned into $1$ by decomposing one $\ce{XeO3}$ for every six $\ce{XeF4}$ . Stoichiometry alone cannot pick the "right" equation of the two. The reaction mechanism and conditions would have to determine which is a better description. With no additional knowledge, perhaps equation $2$ is slightly better, as one might be asked to justify why only a third of the $\ce{XeO3}$ molecules decompose in equation $1$.

Oh yeah... could you please provide a better description?

I'm not sure how I could improve on your answer, really. Do you have a specific question?

@NicolauSakerNeto You said : " The reaction mechanism and conditions would have to determine which is a better description"

Sure. For example, carbon burns in the atmosphere below ~800°C according to $\ce{C + O2 -> CO2}$, while above that temperature $\ce{2C + O2 -> 2CO}$ is predominant. Both equations are stoichiometrically valid, but best describe different situations. With regards to the reaction mechanism, if it were known in detail, then you'd know exactly which intermediates/products are formed and at what step of the reaction. Thus you'd have full knowledge of what compounds would be left at the end. However, I don't suspect this particular reaction mechanism is known.

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text:
How can we rationalise the amount of crystal water in coordination complexes?
I was wondering why specific number of water molecules are attached to salts like $\ce{CuSO4.5H2O}$? Searching gave me a satisfying picture :

I think $\ce{FeSO4.7H2O}$
 should have a similar structure with $6$ water molecules in coordination sphere. But what about $\ce{MgSO4.7H2O}$? Or $\ce{Na2CO3.10H2O}$? How do we explain that?

I have given two answers: one regarding the hydration of metal cations in solution and one regarding salts of metal cations and their hydration extents.
When in solution, the answer lies with size. Think of a BB gun pellet and a basketball. How many basketballs can you stick to the sides of a BB gun pellet? 2, maximum. That's why some metal cations are hydrated with fewer waters than others; these metal cations tend to be the smaller ones.
Also, it is not $\ce{FeSO4}$ that is hydrated but rather just $\ce{Fe^{2+}}$. Iron sulfate is soluble in water so there shouldn't be unionized iron sulfate in solution; in addition, if the iron sulfate were unionized, it would be hard to coordinate water molecules to the metal cation - it's charge density is diluted, so to speak, by the presence of the doubly negative sulfate anion. So the correct formula should be $\ce{[Fe(OH_{2})_6]^{2+}}$.
When not in solution - i.e about the salt examples you posted: outside of water these salts can coordinate that many water molecules likely because there is no water to hydrolyze the metal-sulfate bond. By having the sulfate part of the molecule stick around, this likely facilitates more hydrogen bonding with water molecules - note that the sulfate anion is polarized by the water molecules; the negative charge density resides overwhelmingly on the oxygen molecules.

Consider the above picture of ferric sulfate. Note that there are four oxygens with partial negative charges on the sulfate ion. At least some of these partial negative charges facilitate hydrogen bonding in the solid-state. Also note the charge on the iron atom; this positive charge density also facilitates hydrogen bonding to water molecules.
In addition, we can consider calcium sulfate dihydrate - gyspum, drywall, sheet rock - the stuff your walls are likely made of. From here we clearly see that two of the oxygens on the sulfate anion are available for association with water molecules. Hence the existence of $\ce{CaSO_4*2H_2O}$

For Magnesium Sulphate heptahydrate, again 6 waters coordinate the $\ce{Mg^{2+}}$. The 7th water is hydrogen bonded, but not coordinated to the $\ce{Mg^{2+}}$.
The crystal structure is published as Refinement of the Crystal Structure of Magnesium Sulphate Heptahydrate (Epsomite) by Neutron Diffraction.
Magnesium Sulphate hexahydrate is also known.
In the case of (Di)Sodium Carbonate Decahydrate, again 6 waters coordinate each sodium, but 2 of the waters are shared between the to sodium ions, as two octahedra with a shared edge.
The crystal structure is published as Crystal structure of $\ce{Na2CO3.10H2O}$.

I have never seen complexes of sodium and magnesium. What is the hybridization of Na,Mg in these?


text:
Hydrolysis of cyanohydrin
The question is(See image):

I know that initially a cyanohydrin will be formed. On reaction with $\ce{H2SO4}$, the cyanohydrin will give a carboxylic acid/an amide.
Here's my doubt:
Since the medium is highly concentrated acid medium(95% $\ce {H2SO4}$), the quantity of water in the medium is less, which means lesser of the nucleophile(water in this case). Hence the cyanohydrin should undergo partial hydrolysis to give an amide(because of lesser water!).  Therefore, according to me the answer should be (D).(Had it been dilute acid medium I would have gone with the carboxylic acid product.) However, in this case my answer is not correct...:(
What is incorrect about my thinking?

So G is the expected cyanohydrim, but the second step looks very vigorous (concentrated sulfuric acid and heat).  I would expect two things to happen under these conditions, 1) hydrolysis of the nitrile group to the corresponding carboxylic acid (the reaction conditions are too strong to stop at the intermediate amide, rather the amide is further converted to the corresponding carboxylic acid; keep in mind that an equilibrium will initially exist between the amide and the carboxylic acid, but everytime the acid forms from the amide, ammonia is released and escapes from the system making the reaction irreversible, pushing this reaction further and further to the carboxylic acid side, ultimately only the carboxylic acid will be present.), and 2) acid catalyzed elimination of water from the alcohol portion of the molecule.  Taken together I would expect the final product to be the various cis-trans isomers of "A".

what about my doubt regarding the amount of water?

$\ce{H2SO4}$ is typically used in excess since it is the solvent.  There should be more than enough water to convert the amide to the acid, especially since the ammonia escapes from the system pushing the equilibrium totally to the acid side.  But I agree, the question would have been clearer if the words "excess $\ce{H2SO4}$" had been used.

First, if I carry out the reaction in a closed vessel, $\ce{NH3}$ won't escape. Also,this link says that the hydrolysis requires a dilute acid medium, not a concentrated one to get the carboxylic acid.

Would the ammonia stay in the heated solution, or would it be in the vapor phase above the solution?  Also an additional equivalent of water is generated during the reaction from the elimination step.

BTW, as I read your link, to me it says that hydrolysis can occur with dilute acid; it doesn't say that dilute acid is required nor that concentrated acid fails.

The ammonia will be protonated by the strongly acidic solution giving ammonium ion.

good point!  But either way, protonation or escape, the absence of ammonia in the solution drives the equilibrium to the acid side.

What is the answer given


text:
What's the difference between PBE and B3LYP methods?
I can't find an answer to that question. I was told that in B3LYP, more variables implemented in the method are empirical, but I can't find anywhere if it's true, and I'm sure it's not the only difference.
I'm new to calculations, and it seems so hard and confusing, so please help me.

The Gaussian manual page on how to make it use different functionals is a good source of references on this.

@Basia I updated my answer to say something about locality of PBE and to add references to the seminal papers.

PBE
The PBE functional${}^{[1]}$ belongs to the class of generalized gradient approximation (GGA) functionals for the exchange-correlation energy $E_{\mathrm{xc}}$. Considering that the dependence $E_{\mathrm{xc}}[\rho]$ may be non-local, i.e. $E_{\mathrm{xc}}$ may depend on the density $\rho$ at a given point (locality), but also on $\rho$ nearby (non-locality) the assumption made by the local spin density approximation${}^{[2,3]}$ (LSDA) that $E_{\mathrm{xc}}[\rho]$ is strictly local leaves a lot room for improvement. But still LSDA is a good starting point since it is a rather simple efficient model that gives good results. The GGA functionals improve on LSDA in a way that's pretty much in the spirit of a Taylor series with LSDA as a starting point: When we are at a point, what happens further off depends not only on $\rho$ at that point, but also the gradient of $\rho$ at the point, etc. (much as in a Taylor series: if you want to go further you may need not only the gradient, but also the Laplacian, etc.). So, these functionals are typically constructed by adding gradient corrections to the LSDA functionals. But that is not a trivial thing to do and doing it "naively" leads to wrong results. Therefore, some GGA functionals are parametrized by fitting experimental data. But there are also some GGA functionals that achieve the inclusion of gradient correction without introducing experimentally fitted parameters which makes them valid for a wide range of systems. One example of such a parameter-free GGA functional is the one developped by Perdew, Burke and Ernzerhof (PBE). It is known for its general applicability and gives rather accurate results for a wide range of systems.
Side note: GGA functionals are frequently termed non-local
functionals in the literature. This is a somewhat misleading and actually sloppy terminology that should be avoided and since my above statements might be misread in that direction I want to make one thing perfectly clear: All GGA functionals are perfectly
local in the mathematical sense: the value of the functional at a point $\vec{r}$ depends only on information about the density $\rho(\vec{r})$, its gradient $\nabla\rho(\vec{r})$, and possibly other information at this very point and is absolutely independent of properties of $\rho(\vec{r}^{\, \prime})$  at points $\vec{r}^{\, \prime} \neq \vec{r}$. Calling these functionals 'non-local' is only motivated by the fact that these functionals go beyond the 'local' density approximation and of course the observation that knowledge of
the gradients is the first step towards accounting for the inhomogeneity of the real density.
B3LYP
The B3LYP approach${}^{[4,5,6]}$ belongs to the hybrid approximations for the exchange–hybrid correlation functional. The approximation is famous, because it gives very good results and, therefore, is extremely popular. The distinguishing feature of such hybrid approximations is that they mix in a certain amount of the exact Hartree–Fock exchange energy into the exchange and correlation obtained from other functionals. There is actually some justification for that: At the lower limit ($\lambda = 0$, where $\lambda$ is the coupling constant) of the so called coupling constant integration the exchange-correlation hole is equal to the exact exchange hole. This observation led Becke${}^{[7, 8]}$ to conclude that a fraction of exact exchange should be mixed with GGA exchange and correlation. The simplest such hybrid functional is
\begin{align}
E^{\mathrm{hyb}}_{\mathrm{xc}} = a E_{\mathrm{x}}^{\mathrm{exact}} + (1 − a) E^{\mathrm{GGA}}_{\mathrm{x}} + E^{\mathrm{GGA}}_{\mathrm{c}}
\end{align}
where the constant $a$ can be fitted empirically or estimated theoretically ${}^{[9, 10, 11]}$
as $a \approx \frac{1}{4}$ for molecules.
So far so good, but it smells like a witch's brew for the B3LYP exchange–correlation potential $E_{\mathrm{xc}}$: take the exchange–correlation energy from the LSDA
method, add a pinch (20%) of the difference between the Hartree–Fock exchange energy
$E^{\mathrm{KS}}_{\mathrm{x}}$ (well, in fact, this is Kohn–Sham exchange energy, because the Slater determinant wave function, used to calculate it, is the Kohn–Sham determinant, not the Hartree–Fock one) and the LSDA $E^{\mathrm{LSDA}}_{\mathrm{x}}$. Then, mix well 72% of Becke exchange potential${}^{[12]}$ $E^{\mathrm{B88}}_{\mathrm{x}}$ which includes the 1988 correction, then strew in 81% of the Lee–Yang–Parr correlation potential${}^{[13]}$ $E^{\mathrm{LYP}}_{\mathrm{c}}$. You will like this homeopathic magic potion most if you conclude by putting in 19% of the Vosko–Wilk–Nusair potential$^{[14]}$ $E^{\mathrm{VWN}}_{\mathrm{c}}$:
\begin{align}
  E^{\mathrm{B3LYP}}_{\mathrm{xc}} = E^{\mathrm{LSDA}}_{\mathrm{xc}} + 0.2 (E^{\mathrm{KS}}_{\mathrm{x}} - E^{\mathrm{LSDA}}_{\mathrm{x}}) + 0.72 E^{\mathrm{B88}}_{\mathrm{x}} + 0.81 E^{\mathrm{LYP}}_{\mathrm{c}} + 0.19 E^{\mathrm{VWN}}_{\mathrm{c}}
\end{align}
So, you can see B3LYP contains lots of empirical parameters but that's fine as long as you only want the results and don't want to know why this is working :)
References
${}^{1}$ J. P. Perdew, K. Burke, M. Ernzerhof, "Generalized Gradient Approximation
Made Simple", Phys. Rev. Lett. 1996, 77, 3865; J. P. Perdew, K. Burke,
M. Ernzerhof, "Erratum to Generalized Gradient Approximation Made Simple", Phys. Rev. Lett. 1997, 78, 1396 (E). (Link)
${}^{2}$ W. Kohn, L. J. Sham, Phys. Rev. 1965, 140, A1133. (Link)
${}^{3}$ U. von Barth, L. Hedin, "A local exchange-correlation potential for the spin polarized case", J. Phys. C: Solid State Phys. 1972, 5, 1629–1642. (Link)
${}^{4}$ A. D. Becke, "Density-functional thermochemistry. III. The role of exact exchange", J. Chem. Phys. 1993, 98, 5648-5652. (Link)
${}^{5}$ P. J. Stephens, F. J. Devlin, C. F. Chabalowski, M. J. Frisch, "Ab Initio Calculation of Vibrational Absorption and Circular Dichroism Spectra Using Density Functional Force Fields", J. Phys. Chem. 1994, 98, 11623–11627. (Link)
${}^{6}$ K. Kim, K. D. Jordan, "Comparison of Density Functional and MP2 Calculations on the Water Monomer and Dimer", J. Phys. Chem. 1994, 98, 10089–10094. (Link)
${}^{7}$ A. D. Becke, "A new mixing of Hartree–Fock and local density‐functional theories ", J. Chem. Phys., 1993, 98, 1372. (Link)
${}^{8}$ A. D. Becke, "Density‐functional thermochemistry. III. The role of exact exchange", J. Chem. Phys., 1993, 98, 5648. (Link)
${}^{9}$ J. P. Perdew, M. Ernzerhof, and K. Burke, "Rationale for mixing exact exchange with density functional approximations", J. Chem. Phys., 1996 105, 9982. (Link)
${}^{10}$ C. Adamo and V. Barone, "Toward reliable density functional methods without adjustable parameters: The PBE0 model", J. Chem. Phys., 1999 110, 6158. (Link)
${}^{11}$ M. Ernzerhof and G. E. Scuseria, "Assessment of the Perdew–Burke–Ernzerhof exchange-correlation functional", J. Chem. Phys., 1999, 110, 5029. (Link)
${}^{12}$ A. D. Becke. "Density-functional exchange-energy approximation with correct asymptotic behaviour", Phys. Rev. A 1988, 38, 3098-3100. (Link)
${}^{13}$ C. Lee, W. Yang, R. G. Parr, "Development of the Colle-Salvetti correlation-energy formula into a functional of the electron density", Phys. Rev. B, 1988, 37, 785-789. (Link)
$^{14}$ S. Vosko, L. Wilk, M. Nusair, "Accurate spin-dependent electron-liquid correlation energies for local spin density calculations: a critical analysis", Can. J. Phys. 1980, 58, 1200-1211. (Link)

Many people say that pbe is a "non-local" functional, but that is not true. It depends also on the gradient, which is a local property of the PES on any given point. I also wished people would not rely on B3LYP so much - it was an improvement back in the day, but there are better and more efficient functionals nowadays.// I also want to state that PBE consists of (at least) two functionals: One for the exchange and one for the correlation, the combination of those two is often denoted as PBE0.

Would you mind if I edit in the complete references?

I use pbe1pbe functional with 6-311+G(2d,p) basis set for all my calculations. I'm generally interested in reactions of perhalogenated compounds. Many people say I should switch to b3lyp because it's better and always gives good results, but after I read your answer I think I'll stick with pbe

@Martin I would be thankful if you took the trouble of editing in the references. I refrained from most of it because I thought it should be rather easy to find. Sorry about the "non-local" part: Of course, you are right: PBE is local. But it was late and I wrote the answer before going to sleep so I was a bit sloppy in that point and didn't state explicitly that PBE is not a non-local functional. Also, I didn't want to dwell on too many details.

@Martin I added a paragraph making clear that PBE is local. I hope it's clear now. Are you sure about the PBE0 thing? Isn't the name PBE0 usually used as a synonym for the PBE1PBE hybrid functional?

@Basia I didn't want to preach against B3LYP. Actually there is some physical motivation for mixing in around 20 or 25% of Hartree-Fock exchange into the exchange-correlation functional. And if B3LYP gives you better results than PBE I see little reason not to use it. But as Martin stated there are better functionals than B3LYP out there although he might be better suited than I am to recommend one to you. I will only mention that if hybrid functionals are better suited for the systems you calculate you could try the PBE1PBE hybrid functional (also called PBE0) which has fewer parameters.

@Martin I added the references.

@Philipp: Agreed, whenever I've seen the name PBE0 it's been for the modified hybrid version of PBE that includes weighting factors and a fraction of HF exchange, as opposed to pure (GGA) PBE which doesn't use any HF exchange.

@Aesin I was confused myself - I am very sorry, of course PBE0 is the modified Hybrid functional. @ Philipp I like your answer so very much and while reading it the first time I thought it was not sloppy - I just wanted to point out a very common (very unfortunate) misconception. (And extra kudos for the witch's brew comment.)

@Basia I am not a big fan of b3lyp and I tend towards much more robust (BP86) or much more sophisticated functionals (PBE, Minnesota, ...). As with all calculations they have to account for a certain standard and using one df is usually not enough. The problem is that a functional that is good for describing one bonding situation might be terrible for describing another. There is unfortunately no best functional. Stick with pbe1pbe when you started with it, and confirm results with other functionals or ab initio. I highly recommend switching to a more sophisticated basis set.

@Martin Nice to hear that you like my answer, thanks. And thanks for pointing out the misconception of PBE being called non-local. I think such a question as this is a good place to make other people reading my answer (who are potentially rather new to the subject) aware of that misconception so that they don't fall into that trap.


text:
What is there in this 3D representation of organic molecules I fail to grasp?
I think there's a basic idea I'm missing about 3 Dimensional representation of molecules. I think it may be about converting to Fischer representation. Consider the following two products of bromination of an alkene, the first one being for cis-form and the 2nd one for trans-form.

I know the question may sound stupid, but it's crucial to get the basics. Why is the product II) the meso form and the products I) are the optically active forms. By what I know, the solid wedge goes towards the observer, so in product II), shouldn't the Bromine substituent(or maybe the $CH_3$ substituent) and $H$ atom be on the 'same side', making it the one optically active?

You have drawn the compounds correctly, and yes

a solid wedge means the bond is coming out of the plane of the screen towards you
a dashed wedge means the bond is going behind the plane of the screen away from you
a solid line means the bond lies in the plane of the screen

Here is a Newman projection of your molecules, sometimes Newman projections can make it easier to "see" things.

Build a model (I did) or flip the drawings around, there is no way you can interconvert the isomers labeled "S,S' and "R,R"; they are chiral enantiomers.  The meso compound and its mirror image are easily converted into one another (try rotating the molecule about a $\ce{C_2}$ axis that lies in the plane of the screen, 2) is perpendicular to the mirror plane [the dashed line in my drawing] and 3) bisects the $\ce{C_2-C_3}$ bond in the molecule); they are indeed representations of the same molecule.

The R,R and S,S stereo isomers have $C_2$ symmetry, with the axis lying as you described it for the meso compound. The meso isomer is either of $S_2=C_i$ symmetry (staggered) or $C_s$ symmetry (eclipsed) - it does not have a $C_2$ axis.

@Martin  I did not say that the meso isomer has a $\ce{C_2}$ axis.  In fact I didn't mention the symmetry of the molecules at all.  What I did say was that the two meso drawings could be converted one into the other by rotation about a specific axis external to the molecule (not a molecular symmetry element); and that the R,R and S,S stereoisomers cannot be converted one into the other by rotations or translations.

That is of course absolutely true and that statement just slipped by me. I am sorry if I caused any confusion.


text:
Fractional Bond Order - How to Visualize?
Consider nitric oxide, which has a real bond order of 2.5, contrary to what this UC Davis rendering might imply:

What's the best way to visualize a molecule with a fractional bond order? What exactly is "half a bond"? The electron spends half its time participating in bonding and the other half of the time as a lone electron on nitrogen?

What exactly is a "bond"?

Pairing of two electrons, one from either atom?

Electrons are waves, not necessarily just particles.  Electron density is shared between two atoms which gives rise to attractive forces, not 'one electron from either atom to form a bond'.  (I'm not trying to act superior by saying this).  The Lewis structure is an approximation and can be grossly incorrect.  It is commonly misinterpreted and can lead to problems such as defining what a 'half-bond' is among other things.  In order to understand what is going on here, we need to re-think about what electrons and bonds are.

Are you going to explain?

A "half a bond" does not exist because the Lewis concept of dots being electrons is unphysical.  Therefore, it is difficult to answer your question.  Bonds are attractive forces that arise from electron density surrounding atoms in molecules (and between molecules).

Resonance ${}{}{}$

Dotted line will work just fine.

Maybe some examples will help.  In the picture below, I've drawn some resonance structures for butadiene and acrolein.  What do they tell us?  They suggest that the central "single" bond should have some double bond character, it's more than a single bond, but less than a double bond.  By the same token, the "double"

bonds at the ends of the molecules should both have some single bond character.  These are examples of fractional bond order.  The real world consequences of fractional bond order is a higher barrier to rotation about the central "single" bond in these two molecules than we might have expected, and a lower barrier to rotation in the pi bonds than we might have expected.

Interesting comment about the rotation!

It might be helpful to know that your molecule is a hybrid of all these structures. It is not that these structures exist separately in equilibrium. You have to superimpose all of the structures to understand the molecule.


text:
Global warming kinetics
A rule of thumb in chemistry is that a reaction doubles in rate for every 10 °C increase in temperature. Considering the impact of a 10 °C increase in global temperature, would the same rule apply to kinetic weather processes – for example, would the average wind speed around the world double?

Weather is complex and, as ron guessed, I am confident it is not straight and simple as you said. Counter example is NO oxidation by O2. This reaction can be assumed as trimolecular (very rare) and its kinetic constant decrease with temperature.

Weather is a physical not a chemical process so simple rules about chemical reactions are not going to work. Besides the temperature difference at different points on the earth's surface is >100 degrees and we don't see slower wind speeds, for example, in antarctica than in Singapore.

That is a good rule of thumb.  It is based on the Arrhenius equation $$k = A {\operatorname{e}^{\frac{-{E_{\text{act}}}}{RT}}}$$ You'll find that the "temperature doubles every 10 degrees" rule  only applies if the following assumptions are valid

The reaction must obey the Arrhenius equation

it must have an entropic pre-term based on collision probability
the reaction must be an activated process (e.g. it must have an activation energy)

The Arrhenius equation works well for simple unimolecular reactions like the ring opening of cyclobutene to buta-1,3-diene.
Now you'll need to ask a weather physicist or a meteorologist if weather can be treated as an activated process.  My guess is that the answer is "no", but that's just a guess.  It seems to me that there is no minimum energy required (activation energy) for the rain to fall or for the wind to blow.

Thanks - I appreciate that different processes may or may not have an "activation energy". Formation of a hurricane may be such a process.

This comment is correct but very incomplete, because the "rule of thumb" only applies to a small subset of reactions that obey the Arrhenius equation.  In particular for $k=Ae^{-E/RT}$ and $2k=Ae^{-E/R(T+10)}$, you can solve the equations to get $R \ln{2} = -E_a ((T+10)^{-1}-T^{-1})$.  This equation is a constraint on the energy of activation.  At 298 K, that means (if I did my math right), the energy of activation has to be 52 kJ/mol, or else the rule doesn't work.

It is just an approximation with reasonable activation energies. About weather: Weather is caused by density differences in the atmosphere and gravity. This is directly related to the energy content of the atmosphere. A major component of energy in the atmosphere is the latent heat in its water vapor content. The vapor pressure increases exponentially with the temperature. While not an activation energy an increase in T will have an effect on weather as if it were. The wind speeds will not double; they will increase on the average and last longer.


text:
MacGyvered chlorine gas protection
I was watching Periodic Video's chlorine video, which discusses chlorine's ferocity in stripping electrons whenever possible, and the professors mentioned its consequent use in WWI as a chemical agent. This got me thinking, if I were in that situation (say Syria if they decided to go old-school or a chemical leak), what could I do to protect myself against gaseous chlorine using the chemistry of everyday materials around me?
I imagine a wet towel won't help much to prevent the chlorine from getting in my lungs. Are there any easily obtainable substances I could safely breath through that would neutralize the chlorine? Or are there any (realistic) MacGyver solutions to quickly building a semi-functional gas mask? Just kinda curious what could be done.

Interestingly, the British had such a solution in WWI: the Hypo Helmet. This was simply a bag soaked in sodium thiosulfate solution and worn over the head.  Thiosulfate reduces chlorine (and halogens generally) to chloride (halide) ions:
$\ce{2 S2O3^{2−} + Cl2 → S4O6^{2−} + 2 Cl^{−}}$
As the article notes, this was not the best solution, but I think it fits your criteria for rough and ready protection.

Very interesting! Any ideas how to MacGyver some sodium thiosulfate from everyday chemicals?

It might surprise you that hobbyists still develop black & white film. Hypo is used to "fix" the final print after developing. So it's easy to get. It's also sold to de-chlorinate water. It might not be everyday, but it's readily available.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Conformations of butane
Would these be valid conformations of butane? I tried matching these up with the group but my drawings all seem rotated versus the book.
I'm fairly sure my anti conformer is correct, along with my gauche conformer. I'm not too sure about the two eclipsed conformers; from what I understand, there are two eclipsed conformations - one with strong torsional strain resulting from the methyl group interacting with the other methyl group, and the other resulting from the methyl groups interacting with the hydrogens, right?
Also, can we consider the dihedral angle to be the angle between the methyl groups?

Another problem is that what I've drawn doesn't seem to match with the book's energy diagram too. I'd expect the anti-conformer to be the lowest in energy, but I'd also expect a difference in E for the two eclipsed conformations. One should have more torsional strain from placing the two methyl groups next to each other and the other, less, because the methyl groups are eclipsing the comparatively small hydrogen.

Just to be clear, your drawn figures are for butane and the book's figure is for propane.

You have got them correct. Just your diagrams in eclipsed should have very close bonds. In your third diagram, one $\ce H$ is a bit far for another.
Wikipedia provides a nice description for this.
A diagram too  :

You are correct.

I've edited my text to reflect my beef with the solution manual.

Thank you though for catching the problem with the hydrogens! You're right, they are just a tad bit far apart!

Thank you. Do you know why the diagram in the book provides the wrong graph?

@Dissenter Because it is of propane, not butane,

Ah, good catch again!

That diagram seems very odd to me... (as in qualitatively right, but hand-drawn or something.)

@Martin So what is your point?

Which diagram, Martin?

Some years ago I did a (relaxed) rotational scan of n-butane for the purpose of showing that fatty acids are all-trans. That is also why I found the graph in Awesome's answer some kind of disturbing - it should form a sinus like wave. I include this graph here:

Edited out the error noticed by ron. Thanks again.


text:
How come water would not affect the results of a titration?
We were told to rinse the buret with the base so that the concentration won't be messed up, but when it comes to the flask that is going to hold the acid, there is no need to rinse, because it would not affect the result if it has water in it.
Why is that? Wouldn't the acid be diluted a little bit and that throws off the accuracy?

The volume and concentration of base equals the number of moles of base. This is use to determine the number of moles of acid in the flask. Since you are interested in the number of moles of acid in the flask, why would adding water (which doesn't have any acid) change the number of moles of acid?

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

For a first year student, referring specifically to number of moles probably carries more pedagogical value. "Amount of substance" may not expediently lead them to infer moles are the precise variable in question. Moles are not as much a practiced intuition as the more tangible kilograms that we often use growing up. The consistent English is lost on new students (even more experienced chemists like LDC3 don't consider it), and the explanation will be slightly more opaque for it.

Considering a direct titration:

You will probably use a volumetric pipette to pass the acid solution to the erlenmeyer used on the titration, so the volume of acid is known (the exactly volume of the pipette).

Any distilled water you add to this erlenmeyer will change its volume, but won't change the amount of substance of acid inside it, neither the initial volume you added of the acid solution. Sometimes it's even recommended that you add some water to make the color change more visible.
The objective of the titration is to find the volume of the base (of known concentration) necessary to neutralize the acid, and therefore, since you know the volume of acid added in the erlenmeyer (the pipette volume), calculate the concentration of the acid solution.
Water in the glass of the buret can cause variations in the concentration of the base being used, reason why we rinse it with the base, so we have a good precision titration. The erlenmeyer can be rinsed only with distilled water, since the volume of acid solution used for the calculation is constant.
You will have a equation that looks like this:
$$V(\text{base}) \cdot c(\text{base}) = V(\text{acid}) \cdot c(\text{acid}) \cdot f$$
Where $V(\text{base})$ is the volume of base consumed in the titration, $c(\text{base})$ is the concentration of the base (which is known), $V(\text{acid})$ is the volume of the solution of acid added to the erlenmeyer (the volume of the pipette), $c(\text{acid})$ is the estimated concentration of the acid and $f$ is the correction coefficient (so the real acid concentration is $c(\text{acid}) \cdot f$).

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

Where concentration matters ( burette )
there is rinsing to achieve stable and constant concentration.
Where amounts matters ( flask ) there is no rinsing to achieve stable and constant amount.

If you were to calculate the concentration of acid from the titration, and you add distilled water to the erlenmeyer flask, then yes, you would affect the [H3O+(aq)] since increasing volume with same number of moles decreases the concentration of the solute (or analyte) and thus increases the pH of the solution (because pH is -log(hydronium conc.)), ultimately affecting the color of the indicator since the indicator is affected by the pH of the solution. Since you're using the indicator to determine of the titration has reached its endpoint and therefore the predicted equivalence point, if the indicator is affected by the dilution, the entire result of the titration will be afflicted with error if you were to add distilled water into the erlenmeyer.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". I believe the error you make when taking a reading is worse than a little bit of dilution.


text:
Resonance - Legal Moves
My teacher insists that we can push electrons either from a bond to an atom's valence, or vice versa. Bond to lone pair; lone pair to bond, he insists.
However, what about bond to bond?
Is there a pedagogical reason he might be avoiding bond-to-bond movement? This is a legal move.
Or does he simply not see bond-to-bond movement as a separate entity; could bond-to-bond electron pushing be construed as simply a bond to lone pair and lone pair to new bond move?

Pedagogically speaking, many classes first introduce the concept of Lewis structures before resonance theory.  The lone pairs are very "visible" in Lewis structures and so moving lone pairs before moving bonding electrons around is often a normal progression.  But this is a good question to ask your teacher, it could lead to an interesting conversation.
As to your second question, I've never thought of bond-to-bond electron pushing as bond-to-lone-pair-and-lone-pair-to-new-bond electron pushing.
I think there really is just one rule to drawing resonance structures: You can't change the position of the atoms in resonance structures.  For example in the figure below,

The top line shows 3 resonance structures for nitromethane.  The structure on the right probably doesn't contribute much to the overall description of nitromethane because of the unusually long $\ce{O-O}$ bond length.  In the second row, the $\ce{O-O}$ bond length has been shortened to a reasonable length - but nuclei have been moved - so this is not a resonance structure of the nitromethane drawn to its left, but rather this is an equilibrium.
Aside from the don't move atoms rule, everything else is fair game.  Look at the resonance structures drawn for methane on the bottom line.  They are both valid, but using our chemical "intuition" we would just say that the structure on the right is of such high energy that it doesn't contribute much towards the overall description of methane.

Those triangles I'm seeing aren't actually triangles right? But rather lone pairs on nitrogen?

Yes, they are (triangles) three-membered rings.

What about moving a bonding pair from the double bonded oxygen to the nitrogen's valence?

That would make for a neutral nitrogen, a positive oxygen and a negative oxygen.  Yep, that's a valid resonance structure.  It just wouldn't contribute as much to the real electronic picture of nitromethane as the structures on the left and middle of the first row where we have a positive charge on nitrogen instead of on oxygen.

Resonance is a concept that originates from the incapability of the original Lewis structure to describe delocalised systems. The formulation of these resonance structures are a representation of extreme states that are possible. The true bonding pattern lies between all possible resonance structures.
Consider benzene, where we, among others, consider two major resonance structures. In fact the electron structure is well known to be a delocalised $\pi$ system, therefore better described by the right figure.

In this particular case, electrons appear to be "pushed" from one bond to another. This is not the case, as electrons are occupying one (or more degenerate) orbitals the whole time.
So when it comes to resonance structures, there are no clear rules, what is allowed and what is not allowed, however it has to be a possible representation of the actual bonding pattern. Resonance is a mere concept of describing a very complicated bonding pattern in simpler terms.


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Identification of zinc/aluminium/magnesium with copper sulphate solution
We are doing an experiment and would like to be able to identify zinc/aluminium/ magnesium . Would a copper sulphate solution work ? I hear that it turns black on contact with zinc, stays clear on contact with aluminium and fizzes on contact with magnesium.

http://chemistry.stackexchange.com/questions/11021/identification-of-zinc-aluminium-magnesium-with-copper-sulphate-solution/11026?s=8|0.0466#11026

If you mean identification of the metals in their elemental form, then yes, it might be possible to distinguish between them with an aqueous $\ce{CuSO4}$ solution due to the different reactions.
Zinc in $\ce{CuSO4}$ will be oxidized to colorless $\ce{Zn^2+}$, while $\ce{Cu^2+}$ is reduced to elemental copper, because copper is the more noble metal. The $\ce{Cu}$ precipitate is normally not black, although it can appear dark in the background of the blue solution. It can also form a layer on the piece of zinc, and in the course of the reaction, the initially blue solution will become more and more colorless.
$$\ce{Zn + CuSO4 \rightarrow Cu\downarrow + ZnSO4}$$
Aluminium does not react at all because it is protected by its native oxide layer which is stable in this solution. Magnesium, however, is oxidized under evolution of $\ce{H2}$ gas, which fizzes off from the solution because it is poorly soluble in water. The reaction will gradually become slower when insoluble $\ce{Mg(OH)2}$ passivates the metal surface.
$$\ce{Mg + 2 H2O \rightarrow Mg(OH)2 + H2\uparrow}$$


text:
Energy Diagram for Propane
I've been unable to find an energy diagram for propane either in my book or online via Google. So I resorted to making my own. I suspect that propane isn't covered because it's essentially the same deal as ethane - we can have one staggered conformation and one eclipsed conformation (only in propane it's a methyl eclipsing a hydrogen, not hydrogens eclipsing each other).
So I drew the Newman projections and an energy diagram.

Your diagram is correct.  Here is a link to an on-line energy diagram for propane and here is the diagram itself


text:
Where is the acid in DNA/RNA?
It is well known that the A in both DNA and RNA stands for acid, but where is the acid in chemical formula for the compound, and it is classified so based on what acid-base theory? Like Arrhenius, Brønsted-Lowry or Lewis.
Just to better explain my question, when I say DNA I am essencially thinking about nucleotides linked together, am I missing something in the structure that contains my answer or this is the way to go?

The phosphate ester groups which connect the nucleotides contain one acidic proton at their OH group, and two at the end of each strand. As you can see here, here and in the picture below (with the groups marked turquoise), most of these groups are deprotonated depending on pH. They act as Brønsted-Lowry acids.

Just to better explain my question, when I say DNA I am essencially thinking about nucleotides linked together, am I missing something in the structure that contains my answer or this is the way to go?

Yes, you do. *NA consist of nucleotides, linked by phosphodiester fragments. This leaves one hydroxyl of the phosphate fragment intact, so it can act as an acid.
Look closely:

There is a $\ce{P-O-}$ fragment, which means that there is a cation somewhere nearby. So, *NA are Brønsted acids, usually occuring as salts.

It refers to phosphoric acid. There are three aspects that might make this confusing.

Phophoric acid is a triprotic acid, i.e. it can lose 3 protons.
The phosphoric acid comes in the form of a diester (two ester bonds linking the phosphoric acid with a a sugar-hydroxyl), so now it only has one proton that can dissociate.
At pH = 7, it already lost the proton, so what we see in DNA and RNA at neutral pH is the conjugate base of the phosphate diester.

The resulting negative charge on the phosphate oxygen makes DNA and RNA a polyanion and explains some of its properties (high solubility in water, metal binding, direction it travels in gel electrophoresis, to name just three).


text:
Accuracy of pH meters in MgSO4 solution
I bought a 2 pH meters several months back from Amazon for roughly $10 a piece.  I couldn't tell you how they work, but here's the link on Amazon:  http://www.amazon.com/gp/product/B0087LOWIA/ref=wms_ohs_product?ie=UTF8&psc=1
I also got a neutral buffer solution for calibration.  It worked fine initially but I'm thinking it's off.  It's now being used for pH testing in the salt water of an floatation tank (147 gallons of water + 750 lbs pharmaceutical grade epsom salts and 50 pounds of dead sea salt).  One meter shows a pH of 6 and another of the same kind at about 8.
My question is, does the salt effect the accuracy of the meter?  Are the meters inherently bad, or do they need periodic recalibration?  I didn't want to keep buying litmus paper.

You should recalibrate your pH meter often. Also clean it with [much]$_n$ distilled water in between uses. Check it against solutions you know the pH of. Do not measure directly in the tank (take a sample first). Always make sure, that your solution is stirred very well. Measure multiple times.

For people who have not the slightest idea of these ridiculous units. (147 gallons of water  $\approx 556.5~\mathrm{L}$ + 750 lbs pharmaceutical grade epsom salts  $\approx 340.2~\mathrm{kg}$ and 50 pounds of dead sea salt  $\approx 22.7~\mathrm{kg}$)

The answer is yes, ionic force does influence real pH signal. Because pH meter is sensitive to activities (instead of concentration of H+ only) of ions that modifying it junction potential you must take in account ionic force when it is strength enough to significantly impact activities. There is a lot of limitation with pH-meter: above pH>12 or bellow pH<2 they are not precise at all (see for acidic and basic errors). Or if they are used in different solvents than water. Organic chemists should tell you what pH could mean in hexane. Assessing pH is a complex task that must be addressed with the care it deserves.
Anyway, you should find the user-guide of your pH meter on internet. I am sure the manufacturer will provide limitation and use tips.
There is a detailed chapter on pHmetry in Skoog and West Analytical Chemistry. It may help you to deeper understand those phenomena.


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Sulfuric Acid - Bond Lengths and Charge Separation
So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent.
However, why are the $\ce{S-O}$ and $\ce{S-OH}$ bonds not equivalent then? Why do their lengths differ?
I suspect this has something to do with the addition of the hydrogen on the oxygen atom.
What else could be a cause for the discrepancy?
I ask because my teacher challenged the fact that sulfur might not be hypervalent on the basis that the bond lengths differ.
In retrospect, I think that he:
1) Is placing too much weight on Lewis Structures - which are just rough renditions of what chemicals really are.
2) The $\ce{S-O}$ bonds and $\ce{S-OH}$ bonds may differ in length not because the $\ce{S-O}$ bonds are necessarily double bonded but also because the $\ce{S-OH}$ bonds are different in that S is bound to a hydroxyl group rather than a single oxygen. Also can't there be very minor resonance contributors involving 5-6 bonds on sulfur? It's not that sulfur absolutely can't access its d-orbitals but rather it doesn't to any great extent.

Good questions, let's step through them one at a time.
1) "So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent. "
Look at the resonance structures below.  The one on the left with sulfur hypervalent is the only uncharged resonance structure.  I suspect that that counts for something.  Here is a link that says it is the most important of the structures (don't get confused when you view the link, they use equilibrium arrows (wrong) in their diagram instead of  resonance arrows (correct).

2) "why are the S-O and S-OH bonds not equivalent then? Why do their lengths differ?"
This is a very fundamental question.  My answer is, they are not equivalent because they are different.  Just as you wouldn't expect the same $\ce{C-O}$ bond length in $\ce{CH_3-O-CH_3}$ and $\ce{CH3-OH}$, you wouldn't expect the same bond lengths in an $\ce{S-O}$ bond and an $\ce{S-OH}$ bond.  In fact, since these bonds are not superimposable one on the other by a symmetry operation, we can say that they MUST be different (i.e. have different lengths).  As one of my favorite professors used to say, now if were to actually measure the $\ce{S-O}$ and $\ce{S-OH}$ bond lengths and our measurements gave rise to the same numeric bond length out to 10 decimal places, we could still step back and say that both we and God know that the lengths are not the same, we just need to measure out to more decimal places!  Let me say the converse, the only bonds that can be identical are those that can be interconverted one with another through a symmetry operation.  For example in chloroform, $\ce{CHCl3}$, there is a $\ce{C_3}$ axis containing the $\ce{C-H}$ bond.  Rotation about this axis interconverts the 3 $\ce{C-Cl}$ bonds, therefor they are equivalent bonds.  Their bond lengths MUST be the same.


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Why doesn't sulfate get electrolyzed in sulfuric acid electrolysis?
If I have aqueous sulfuric acid solution as electrolyte and the anode electrode is made out of copper and the cathode electrode is made out of graphite, what are the two reactions at the respective electrodes and why?
\begin{align}
\text{Anode:}  && \ce{Cu &-> Cu^2+ + 2e-}\\
\text{Cathode:}&& \ce{2e- + 2H+ &-> H2 (g)}\\\hline
\text{Total:}  && \ce{2H+ + Cu &-> H2 + Cu^2+}
\end{align}
That should be the reaction, but why? Why isn't sulfuric acid participating? Why does that specific reaction happen?

Related: Reaction of sulfate ion in copper sulfate electrolysis

To analyse this kind of problem, the first step should be identifying all the species you have in solution that could take part in any electrolytic reaction. These are:
$$\begin{array}{}\ce{SO4^2-} & \ce{H+} & \ce{H2O} & \ce{Cu}\end{array}$$
I did not add graphite to this list since it generally never participates in electrolytic processes. The next step would be to identify what can be reduced and what can be oxidised. Thankfully, all species are either in their lowest or highest common oxidation state. See the list below with oxidation states added:
$$\begin{array}{cccc}\ce{\overset{{+VI}}{S}\overset{-II}{O4}^2-} & \overset{\mathrm{+I}}{\ce{H+}} & \ce{\overset{{+I}}{H2}\overset{-II}{O}} & \overset{\pm 0}{\ce{Cu}}\end{array}$$
Obviously, the only candidates for oxidation are oxygen or copper, and the only candidates for reduction are sulfur or hydrogen. I’ll concentrate on the oxidation first, as the equations are simpler. We need to find a redox pair for each component and then decide which is more likely. For copper, the only real possibilty is generating copper(II), while for oxygen the only real possibility is generating oxygen gas. See equations $(1)$ and $(2)$:
$$\begin{align}\ce{Cu &-> Cu^2+ + 2 e-}&&E^0 = +0.337~\mathrm{V}\tag{1}\\
\ce{2H2O &-> O2 + 4 e- + 4 H+}&&E^0 = +1.229~\mathrm{V}\tag{2}\end{align}$$
I added the standard electrode potentials $E^0$, too. We can easily compare which oxidation is more likely: the smaller potential will act first, thus, copper is oxidised. Now on to the reductions. For hydrogen, it is naturally the known equation. What about sulfur in sulfate? Well, the most obvious choice is assuming $\ce{SO3^2-}$ as a reduction target. See equations $(3)$ and $(4)$:
$$\begin{align}\ce{2H+ + 2e- &-> H2 ^}&&E^0 = \pm 0~\mathrm{V}\tag{3}\\
\ce{SO4^2- + 2 e- + 2 H+ &-> SO3^2- + H2O}&&E^0 = +0.17~\mathrm{V}\tag{4}\end{align}$$
The reduction potential is actually given for the target $\ce{SO2 (aq)}$; however, that has the same oxidation state and is in equilibrium with sulfite, so they are comparable. Here, the more positive potential should actually react first. However, using the graphite electrode typically favours hydrogen evolution via overvoltage mechanisms. This is, unfortunately, something you will have to memorise.

Summing up:

There are only four reactions that can happen
of these, two can be chosen as more favourable by their respective electrode potentials
one choice needs to be adjusted due to overvoltage.


text:
Relationship between magnitudes of forward and reverse kinetic rate constants
Consider the reversible unimolecular reaction:
$$\ce{A <=>[k_1][k_2] B}$$
We know that the forward reaction is often considerably more thermodynamically favourable than the reverse reaction, and therefore the relationship $k_1 \gg k_2$ holds between the rate constants. The rate constants are in the same units, and so it is possible to write this relationship. A similar example would be a reaction that is bimolecular on both sides, $\ce{A + B <=> C + D}$, etc.
However, consider this reversible reaction:
$$\ce{A + B <=>[k_1][k_2] C}$$
Assume the reaction proceeds at rate $k_1 [\ce{A}][\ce{B}]$ in the forward direction and $k_2 [\ce{C}]$ in reverse. If the overall reaction rate has units $M s^{-1}$, then $k_1$ necessarily has units $M^{-1} s^{-1}$ and $k_2$ has units $s^{-1}$.
My question is: For this second case, we can no longer impose that $k_1 \gg k_2$, due to difference in units, but can we say anything about their relationship? Does setting $k_1$ to some value constrain the choice of $k_2$ in any way?

A chemical reaction may be a very complicated thing, involving many different paths and elementary reactions. Basically it always leads to the formation of an equilibrium, where the lowest possible energy will be the most favoured. Under certain conditions more than one state will be populated and hence there will be enough potential energy to interconvert molecules in certain states. We usually refer to that as an dynamic equilibrium.
Changes under equilibrium conditions will most likely be elementary reactions (or an infinitely complicated chain of these.)
The population of the various states seems to be static at equilibrium, since the overall change of enthalpy is zero. In a dynamic picture that only means, that forward and backward reaction cancel each other. In that sense they are the same reaction in different directions. Therefore they can be described as one trajectory on the potential energy surface. (If there is a chemical reaction involving many different elementary reactions and the equilibrium still occurs, then the mathematical description will be much more complicated as all elementary reactions will be correlated. However, the main conclusions should stay the same.)
\begin{aligned}
\Delta G_r &= \Delta G_r^\circ + \mathcal{R}T\cdot \ln K = 0\\
\Delta G_r^\circ &= − \mathcal{R}T \cdot \ln K
\end{aligned}
The equilibrium constant may be derived from the mass-action law and should be defined as a product of activities (for the standard state). For $\ce{A + B <=> C}$ this will lead to
$$K^\circ= \frac{a(\ce{C})}{a(\ce{A})\cdot a(\ce{B})}$$
The activities are proportional to concentrations (in first approximation) and are unit less ($c^\circ$ being the standard concentration $1 \:\mathrm{mol/L}$)
$$a=\gamma\frac{c}{c^{\circ}}$$
For reasonable dilutions ($c\to0\:\mathrm{mol/L}$) one can assume that the activity coefficient becomes one $\gamma\approx1$ and therefore rewrite the equilibrium constant with concentrations.
$$K^\circ= \frac{c(\ce{C})/c^\circ}{c(\ce{A})\cdot c(\ce{B})/(c^\circ)^2}$$
At equilibrium the forward and backward reaction are coupled and therefore have to reflect the equilibrium constant.
$$K^\circ= \frac{k_f}{k_b} = \frac{c_{\text{eq}}(\ce{C})/c^\circ}{c_{\text{eq}}(\ce{A})\cdot c_{\text{eq}}(\ce{B})/(c^\circ)^2}$$
As one can see, the constants also need to have the same unit. As they are at equilibrium, they are the same reaction with different populated states.
If you are moving away from the equilibrium, thing will start to get a little bit more messy. On reaction will be faster than the other and rates will not be comparable any more.

I might be missing something here - are the units of $c_{eq}$, $(C)$, and $c^{\circ}$ all $\frac{mol}{L}$?

I see your point now - $c_{eq}(C)$ is the equilibrium concentration of C, and has units of mol/L. The thermodynamic equilibrium coefficient is dimensionless. However, I am not sure that we can relate that to the rate constants, since they are defined from empirical rate laws based on concentration, and therefore do not necessarily have the same units. We are each making different assumptions here about the conditions, I will change my answer to clarify.

@thomij If you are using empiric determined rate laws, then you are far away from equilibrium, and very far away from elementary reactions. Everything I have written would not apply any more, since a reaction that is not in equilibrium can not be described by an equilibrium constant.

if the rate constants and the equilibrium constants are really constant, then we are close enough to equilibrium to relate them. I understand your point that if you use activities, the equilibrium constant is dimensionless (although you might consider editing your answer to talk about using activities for the rate laws, otherwise the units situation gets worse). My point is that even if you don't use activities, it still works.

@thomij I really do not understand what you are trying to say. Especially the units part. And I really do not want to spent any more thought on this, as it is not my field of expertise and I did not want to write an answer in the first place. There was a demand I tried to satisfy that. Now I am more confused than before. That should not happen. And obviously edison does not even care about an answer any more - so it might be appropriate to remove it.

I don't think deleting it is necessary - there is nothing wrong with having different approaches to a problem. What I am trying to say is that the thermodynamic equilibrium constant is, as you said, dimensionless, because it is defined using activities. However, unless the rate constants are also defined using activites, they will have different units, and their ratio cannot be equal to the rate constant. Since edison mentioned being concerned about the units,  I was thinking you would want to explicitly show that in your answer.

The reaction rates can be related to each other through the equilibrium constant.
However, it is important to note that we are making two key assumptions in order to do this:

The solution is nearly ideal - in layman's terms, that means the concentration of one substance does not depend on any of the others. When that is true, we can use an equilibrium constant defined in terms of concentrations, $K_{eq} = \frac{[C]}{[A][B]}$, instead of activities (effective concentrations).
The rate laws are in fact given by $k_1[A][B]$ and $k_2[C]$ - this may be true for elementary reactions, but you cannot in the general case determine the reaction rates solely from a reaction equation. You must find them empirically, or have a good idea of what the underlying mechanism is (and even then you might not get the correct rate law.)

Assuming that these two conditions hold, we can then relate the two rate constants to each other through the equilibrium constant.
At equilibrium, we know that the rate of change in one direction is equal to the rate of change in the other direction. In other words,
$k_1[A][B] = k2[C]$.
This  means that
$\frac{k_1}{k_2} = \frac{[C]}{[A][B]}$,
and since
$K_{eq} = \frac{[C]}{[A][B]}$,
$K_{eq} = \frac{k_1}{k_2}$.
When we say that a reaction is more thermodynamically favorable in the forward direction, we are also saying that $K_{eq}$ is large (relatively speaking) and that $k_1 >> k_2$.
You are right to be suspicious, since it doesn't seem like we should be able to compare two numbers with different units. The key here is that whether we explicitly look at the units or not, they are still there in the equilibrium constant. We are comparing the magnitudes alone, usually in an effort to reduce the complexity of an equation by assuming that some terms are approximately equal to zero.

The reaction constants do have the same unit $\mathrm{s}^{-1}$ as kinetics is defined by activities, which is unitless. With the relation $a=\gamma\frac{c}{c^{\ominus}}$ it is possible to state, that for reasonable concentrations $\gamma\approx1$ and hence $a\approx c$. We would introduce the error and the misconception, because most people forget that the used concentration is unitless.

@Martin, perhaps you could put together your comments to Andrea and thomij, and post a more detailed answer to this question?


text:
What are some incapabilities that the early spectroscopy have when compared to the more modern method?
What are some incapabilities that the earlier spectroscopy (Visible, Ultraviolet and Infra-red) have when compared to the more modern method?

Some of the things that come to mind are:

Resolution  Modern equipment has much higher resolution than older equipment.  Optical component surface smoothness and uniformity, optical component alignment and electrical component stability all permit the user to see more detail and fine structure in the spectrum, which assists in analysis
More stable light sources  Again resulting in more detailed and more reproducible spectra
Improved monochromaticity of the light sources  Further improving spectral resolution
Improved sample cells  Polished glass surfaces provide for better signal to noise making it easier to discern the signal
Signal processing software  Such as Fourier Transform methods which increase the signal to noise ratio making it easier to discern the true peaks in a spectrum
Other computer and software enhancements  That can be applied to sample handling and spectrum processing.  They dramatically reduce the time it takes to record and process a spectrum


text:
Reaction mechanism of a terminal alkene with an anhydride
In a reaction (as given in my textbook),
$$\ce{R-CH=CH2 + (CH3-CO)2O  -> R-CH2=CH-COCH3}$$
an alkene reacts with acetic anhydride to give methyl alkenyl ketone.
I am not sure of how this proceeds, a search on google displays no links to any relevant matter, and I have not successfully found this reaction in any of the other books I possess.
According to me, it should have been an addition reaction (characteristic of alkenes) across the double bond. But this isn't the case here. So, what exactly is the reaction mechanism?

This was the most relevant link I came across but did not have an answer- https://uk.answers.yahoo.com/question/index?qid=20080903023831AACLKxC

This is the  Friedel-Crafts acylation reaction of alkene. Catalyzed by Lewis acid. Addition-elimination mechanism.

"The Friedel–Crafts reactions are a set of reactions developed by Charles Friedel and James Crafts in 1877 to attach substituents to an aromatic ring" - Wikipedia (read the whole Friedel Craft Reactions page), I am not as yet aware of any non-aromatic Friedel-Craft, any pointers or links would be much appreciated, thanks a lot....

This was the only relevant article I found after a long search but that too gives just a 'page review' perhaps you could link something better thanks..   http://pubs.rsc.org/en/content/articlelanding/1972/cs/cs9720100073#!divAbstract

Here is the reference. Only first page is free. But you can see the introduction about that. http://pubs.rsc.org/en/content/articlelanding/1972/cs/cs9720100073#!divAbstract

Here is another link. It is full text. http://www.metzger.chemie.uni-oldenburg.de/download/83_engl.pdf


text:
How does activated carbon bind the adsorbate?
I read some articles and journals about activated carbon and found a problem about how the adsorbate could bind to the activated carbon. Some said it is because of intramolecular forces (van der Waals) and some said it is because of chemisorption. If chemisorption plays a role, what would the bonding be?
And if the bonding is just caused by intramolecular forces, what is the active site of the activated carbon? An answer or illustration for better understanding is greatly appreciated.

Activated carbon works by enlarging the surface area of elemental carbon. Activated carbon can reach surface areas of several hundred square meters to several thousand square meters per gram. Also elemental carbon is fairly non reactive, non toxic and cheap.

As seen in the picture, activated carbon forms a micro structure with incredibly small pores, so that many molecules can be adsorbed (the forces involved decrease with a factor $1/r^{6}$ with distance). The adsorption itself is an application of the London dispersion force (one of the Van der Waals forces).
I don't believe chemisorption has anything to do with this.

Chemisorption means chemical reaction occurs during absorption process, and any bond type other than intermolecular forces (dipole-dipole, H-bond, Van Der Waals) is possible. Activated carbon absorb using intermolecular forces, and I do not think it is chemisorption. The absorption is strong because the surface area is huge, not because of any real chemical bonds.

So what inter-molecular force is responsible? And why carbon and not sulfur for example? I think if you discuss these points too, this would greatly improve your answer.

Not all solid can have such a big surface area like activated carbon. Sulfur for example as find powder do absorb but it is reactive, and not as cheap as activate carbon. If you do a search, there are a lot of physical absorbent that utilize intermolecular forces to absorb. Activated carbon is just the most commonly used one.


text:
Demountable simple fume hood
I found this build-plan recycling on the internet (I believe it was originally published in a book written by the infamous clandestine chemist Hobart Huson) which seemed interesting to me. I do not have the money to pay for a professional fume hood, nor do I have the space to fit such thing. I have to switch the function of my work workplace sometimes because of lack of space (it is also used for woodworking, another hobby of me), so a simple demountable fume hood would be ideal. As said, I found this one with a simple build description, but I was wondering about its effectiveness and safety.

As shown on the picture, the fume hood consists of a demountable frame, a thick plastic sheet covering the frame that is attached to a tube, and a simple exhaust system connected to an other tube that is placed outside the building, so that the fumes can diffuse in the atmosphere. The following four points concern me:

Is the thick plastic sheet enough the prevent most fumes from diffusing in my workplace?
Is this plastic sheet a potential danger in case of fire? Is there some fire resistant (at least until I'm able to pull my fire blanket) plastic sheet that should (must) be used?
Should I be worried about damaging the environment with these exhaust fumes as there is no filtration in this system? Or is reading the MSDS (i.e. the environment section) for the chemicals being used enough?
There is said nothing about the necessity of an explosion proof fan in the build instructions. To what extend should I worry about this when sucking off flammables?

Take note of the fact that we are talking about amateur chemistry here. There are some nasty fumes that need to be removed, like chlorine for example, but no dense fogs of hydrogen cyanide or phosgene.

Since the chemical will not be in contact with the plastic for very long, very little will diffuse through the plastic. If you are concerned about fire, you could wrap the outside (not the inside) with aluminum foil; this should keep the fire contained. Have a container of sand handy in case the fire is extremely hot and the aluminum starts to burn. As long as you have a high air flow and the volatiles are coming from a small area, there should be no problem. I can't comment about the environmental effect unless you provide more detail.

You might want to consider building in a simple filtering system to prevent damage to the pump - as that is most likely the most expensive part of the whole construction. (Unfortunately I cannot provide any idea on how to achieve something like that.)

If you do a little reading about fume cupboards, and in particular, benchtop extraction hoods, you will discover that they are constructed from materials not too dissimilar to your design; consider the difference being amateur use versus professional use. To address your four main concerns:

Is the thick plastic sheet enough the prevent most fumes from diffusing in my workplace?

Provided the plastic is thick enough. So, certainly thicker than cling wrap, and I'd be happy using the types of material common to decent portable greenhouses. It needs to be robust enough to withstand the extraction forces from your exhaust fan. If your exhaust is working correctly, any chemicals will not be in contact with the plastic long enough to be a concern. Be sure you run a post-purge extraction once you have finished your day's chemistry to remove all traces of chemical fumes.

Is this plastic sheet a potential danger in case of fire? Is there some fire resistant (at least until I'm able to pull my fire blanket) plastic sheet that should (must) be used?

Most fume hoods are made from combustible material. I have, in my 20+ year career as a research chemist, seen many fume hoods destroyed by fire. This includes hoods made from wood, various forms of plastic, even those with metal frames and plastic components. If you have a fire, yes the plastic sheet will burn, and depending on the material, it may burn quickly or slowly - you will have to judge that. This, then, will increase the volume of your fire, and the amount of combustible material. So you will need to manage that in your workshop layout. So, yes, this is a hazard, but one that you should be able to plan for, and adequately implement a risk mitigation strategy. But the real concerns are not whether the plastic sheet would be a danger if there is fire, but (a)will the plastic sheet increase the risk of fire, and (b)what effect will your exhaust fan have in the event of fire.  If you are doing chemistry that involves refluxing, as your picture suggests, you need to make sure that the size of the hood is big enough that you are not heating your plastic, and that the air flow is sufficient to extract any hot air from the roof of your hood. You also need to consider that fires in fume hoods are generally more active than fires on open bench areas because they are actively fanned and have a fresh air source driven into the fire zone because of your extraction fan. A fire blanket in your workshop is not a sufficient measure against the type of fire you might have doing chemistry, Buy yourself at least one dry chemical extinguisher, which are sold in all major retailers that sell fire blankets. If you are using chlorine, then you should NOT use the ABC type dry extinguishers that use ammonium based material, as they will react violently with chlorine. You should use a BC type, which is based on sodium bicarbonate.

Should I be worried about damaging the environment with these exhaust fumes as there is no filtration in this system? Or is reading the MSDS (i.e. the environment section) for the chemicals being used enough?

Yes, you should be worried. Reading the MSDS is not enough. You need to read it, understand it, and then take appropriate measures to mitigate any hazardous risk to the local environment. If you are extracting chlorine gas, where is it going? Apart from being pulled through your pump system, which will be corroded pretty badly after only a few uses, what is directly outside at your pump exhaust? If you position your exhaust just outside a window of your building, what local wind channelling exists to push that exhaust straight into the next window of your, or your neighbour's house? You could consider installing a carbon filter like the types used on water systems, as these are not overly expensive, but for many commercial systems, no scrubbing or filtration occurs, but the key issue is where is your exhaust going? So, understand your design, consider your risks, and plan your exhaust appropriately.

There is said nothing about the necessity of an explosion proof fan in the build instructions. To what extend should I worry about this when sucking off flammables?

I'm pretty sure that all building codes (certainly all the ones I have ever encountered) will specify that any motor used in an area containing flammable vapours (and dust, if you need to consider that for your woodworking) should be of the type approved for such an environment. What dust extraction do you use for your woodworking? If you get an arc from the pump motor, you will very likely get ignition. Again, this is something I have seen in research laboratories for all sorts of non-approved electrical equipment, including pumps, air heaters, fridges. So, if you are exhausting flammable vapours, this is a real risk.
The final consideration you haven't really mentioned is the exhaust pump itself. You need to make sure that it is of suitable size to provide correct laminar flow into the front of the hood, around all your equipment, into the exhaust outlet, and pushed back through the exhaust port after the pump. You should test this using a smoke pellet, or something similar. the larger the opening of your hood, the larger the exhaust fan needs to be, which is why you should always lower the sash hood on fume cupboards when doing chemistry.
Good luck.

"If you get an arc from the pump motor, you will very likely get ignition." What exactly do you mean by that? For woodworking I use an adjustable fan that can suck up to 360 cph (cubic meter per hour), which should be more than enough I guess. Here is a picture: https://i.imgur.com/A92uVf8.png (it's a PrimaKlima PK-125-EUP, used for growing manily I believe). Thanks for your thorough answer!

Arcing is something that occurs in electrical equipment where there is a faulty ground connection. It is not all that uncommon, and you may have seen it in hair dryers or battery drills etc. It causes a small spark. Equipment, such as fridges and pumps being used where there is a high level of flammable vapours are specifically designed for such an environment, and are very expensive; you will have to be the judge as to whether your conditions are producing significant flammable vapours. For instance, I would not use a kitchen exhaust hood to extract flammable vapours from refluxing acetone.

With refluxing? Isn't the point of refluxing that as less as possible vapor escapes the system? Also 360 cph seems quite much to me. I don't think I can afford an explosive proof fan.

Yes, absolutely. But with all things, you need to plan for worst case scenarios. I'm not suggesting you should buy a spark proof fan, just be aware of the potential hazard, assess your risk, and plan your procedures accordingly. Nobody can give you the 'you'll have no problems' thumbs up, based on the info provided, we can only make you aware of potential risks. You still need to make your own assessment based on what procedures and chemicals you will be using.

of course and I completely agree with you, but it is not possible for me to buy such thing anytime soon and I do not believe that many other hobby chemist could afford it either. I was just curious why you took the aceton refluxing as example and not something like alcohol solvent evaporation. I could hardly imagine that with a proper reflux condenser any significant vapor that could set off an explosion can escape.

Acetone used as an example because it is far more volatile and has higher flammability than methanol.... an example of a worse case scenario. I don't know what chemistry you plan to do, so I'd hate for you to not at least have considered potential dangers.

Of course, to play devil's advocate, one could ask, if as you quite rightly suggest that only insignificant vapour would be generated from a correctly set up chemical procedure, why would anyone need an exhaust hood anyway! But it is always good to be safe.

Since the chemicals will not be in contact with the plastic for very long, very little will diffuse through the plastic.
If you are concerned about fire, you could wrap the outside (not the inside) with aluminum foil; this should keep the fire contained. Have a container of sand handy in case the fire is extremely hot and the aluminum starts to burn. In fact, sand would be handy for most fires since it will decrease the amount of oxygen getting to the flames.
As long as you have a high air flow and the volatiles are coming from a small area, there should be no problem. Added: The concentration of the volatiles will not be high enough for a spark to ignite.
I can't really comment about the environmental effect unless you provide more detail. If the MSDS doesn't mention a hazardous exposure from the fumes, then you probably don't need a fume hood. You can get some activated charcoal (from a pet store) and pass the fumes through the charcoal, but I don't know how effective that would be.

What do you consider to be high air flow?

It looks like they are using a leaf blower; that would be more than needed.

Is the thick plastic sheet enough the prevent most fumes from diffusing in my workplace?

Yes. Actually, you should care more about the integrity of the tube

Is this plastic sheet a potential danger in case of fire?

Depends. But care more about inflammable and unmeltable working surface (metal or ceramic). Some old institutions work just fine with fume hoods using wood frames.

Is there some fire resistant (at least until I'm able to pull my fire blanket) plastic sheet that should (must) be used?

PVC is self-extinguishing plastic that is extremely hard to lit up. However, when burned, it produce toxic vapours, and is a spark risk.

Should I be worried about damaging the environment with these exhaust fumes as there is no filtration in this system?

Care about your car. It affects environment more than most other fumes you can produce.

There is said nothing about the necessity of an explosion proof fan in the build instructions. To what extend should I worry about this when sucking off flammables?

The engine must not touch the air sucked of, as there is a spark in common electric engines.The fan itself must not touch anything to not produce sparks. Everything should be oiled.