text:
What is Y-aromaticity? Is the trinitromethanide anion aromatic?
The wikipedia article on trinitromethane claims that:

There is some evidence that the anion (which obeys the $4n+2$ Hückel rule) is aromatic.

Unfortunately the citation is behind a paywall so I can't read it but my first obvious question is how can the trinitromethane anion obey Hückel's rule? Two of the requirements for the rule are that the molecule must be planar (or nearly planar) and must be cyclic and I fail to see how either of these conditions are satisfied since the anion is roughly trigonal pyramidal around the carbon and definitely not cyclic.
This appears to about Y-aromaticity. All the papers linked are behind paywalls for me so can someone please explain what Y-aromaticity is and how it applies to the example given. I found this previous answer which gives some discussion of Y-aromaticity in nitrate ions. If someone could provide a similar MO analysis of the trinitromethane anion this might be very interesting.

Related: Y-aromaticity: why is the trimethylenemethane dication more stable than the butadienyl dication? J. Org. Chem. 2005, 70 (19), 7605–7616. DOI: 10.1021/jo0508090.

maybe this: https://books.google.pl/books?id=z76sVepirh4C&pg=PA538&lpg=PA538&dq=Y-aromaticity&source=bl&ots=jePDS-evv3&sig=HRgFVo4kiTwV6cIOoEK9nVub9Tk&hl=pl&sa=X&ei=f-ZtVaWOO8O9ygPItYPgDA&ved=0CB8Q6AEwADgU#v=onepage&q=Y-aromaticity&f=false

Introduction
It is fairly obvious that the statement given by Wikipedia cannot be entirely correct.

There is some evidence that the anion (which obeys the 4n+2 Hückel rule) is aromatic.

The trinitromethane anion cannot obey the Hückel (4n+2) rule because it is neither monocyclic nor planar. And assigning a number of π electrons is fuzzy at best, so the (4n+2) part is also questionable. Hückel's rule is a very strict set, but an aromatic compound does not necessarily have to fulfil it. The easiest compound to understand this is naphtalene, which is clearly aromatic, but does not fulfil the criteria of Hückel's rule, i.e. it is not monocyclic. In extension to this, helicene is an aromatic compound that fails two of the given conditions. Another case is thiophene, which due to the included sulfur certainly does not entirely consist of trigonally hybridised (sp2) atoms (see Bent's rule).[1]
But can the anion still be regarded as aromatic? In order to explain that, we need to look at the concept of Y-aromaticity.
What is Y-aromaticity?
The term and the concept is not without controversial discussion. It is an attempt to give the "traditional" concept of resonance energy an new, fancy and more catchy name. It is often used to explain the extraordinary stability of the guanidinium cation, which has tree equivalent resonance structures, i.e. the electronic ground state is heavily delocalised.
Conventional Lewis structures employ two electron two centre bonds only, which therefore cannot represent the highly delocalised electronic structures, e.g. of aromatic compounds. Hence the concept of resonance had to be introduced to account for such stabilising effects. Obviously the same holds true for any compounds, that exhibit excessively delocalised electronic structures.
The term Y-aromaticity has been used for compounds, that are Y-shaped, with a trigonal planar coordinated (carbon) centre. In addition to that the π orbitals hold six electrons.[2] These compounds have a similar orbital picture and could therefore be grouped.
The electronic structure of Y-delocalised π systems
The π orbitals of a trigonal planar coordinated centre can be generalised by the following orbital picture, which was generated for the nitrate anion at the DF-BP86/def2-SVP level of theory.

What we can clearly see is, that there is one totally symmetric bonding orbital. Two degenerate (orthogonal) slightly bonding (or non-bonding) orbitals and another totally symmetric anti-bonding orbital.
The remarkable stabilisation comes from the lowest lying π orbital, which is delocalised over all atoms, just like in the more obvious cases like benzene. Additionally, such compounds often retain their planarity by undergoing substitution reactions, rather than addition reactions.
Pro and contra of Y-aromaticity
The concept itself allows the discussion of certain electronic properties in quite simple terms. It is related enough to the general concept of aromaticity so that it can easily be understood at the same level of teaching. Like aromaticity, it allows predictions - to a certain degree - about the reactivity of such compounds.
Peter Grund makes the following assessment, arguing in favour of the concept of acyclic aromaticity.[3]

Despite recent trends, it would appear useful to designate systems as "aromatic" if they represent a delocalized system exhibiting peculiar chemical stability, in order to focus attention on such systems even when they are buried in complex structures. Since guanidine and its derivatives appear to possess really exceptional stability, and since their physical and chemical properties appear to be dominated by the tendency to retain the closed-shell, Y-delocalized 6 pi-electron configuration, we suggest that it is profitable to consider such substances as possessing a special type of "aromatic character," despite their acyclic nature.

For a general understanding of this feature it certainly does not hurt to keep it in the back of ones head.[4]
However, the applicability of this concept is limited to a quite small subset of compound (the same holds true for Hückel's rule, too). It therefore seems to be quite a superfluous concept, since it is in a more general sense already covered by resonance stabilisation.
After all, electronic structures are often not well enough described with simple models and in some cases these simple models can lead to wrong assumptions. This is one of these cases, where one can only use this concept as long as its restrictions are well known.
Some people might argue, that the naming itself is the problem and it may lead to confusion. I can partly agree with that. That does not change the fact, that the description in resonance, valence bond and molecular orbital theory terms will produce this result. In general I think the term aromaticity is unfortunate right in the beginnings. The concept has changed and extended quite a few times already. I often see problems when the term "Hückel rule" is used synonymously with aromaticity.
There is no true theoretical and at the same time all general explanation for aromaticity. It is often determined on a case by case basis experimentally. This limitation also has to apply to any derivations of it.
Electronic structure of the trinitromethane anion (TNM⊖)
Trinitromethane is a quite acidic substance, which can in part be attributed to the presence of the well delocalised π system. Cioslowski et. al. have investigated the anion and came to this conclusion. In their paper they also use the term Y-aromaticity.[5]

Our calculations, however, show that at the HF/6-31G* level, the carbon atom is coplanar with the nitrogens while the oxygens of the nitro groups are rotated out of plane. The result is a structure with $D_3$ symmetry. The $\ce{C(CN)3^-}$  anion is likewise found to be planar, while the $\ce{CF3^-}$ ion exhibits a
  pyramidal geometry. In addition, we observe in $\ce{C(CN)3^-}$ a shortening of the carbon-carbon bond and a lengthening of the carbon-nitrogen bond relative to $\ce{CH(CN)3}$. We also find an analogous shortening of the carbon-nitrogen bond and lengthening of the nitrogen-oxygen bond in $\ce{C(NO2)3^-}$ relative to $\ce{CH(NO2)3}$. In contrast, the $\ce{C-F}$ bond length increases upon going from $\ce{CHF3}$ to $\ce{CF3^-}$. The planarity of the $\ce{C(NO2)3^-}$ and $\ce{C(CN)3^-}$ ions and the changes in bond lengths are obvious consequences of Y-aromaticity in these systems.

Obviously the bonding situation in the trinitromethane anion is a little more complicated than in guanidine, as the nitro group itself cannot be described by a single Lewis structure. The out of plane rotation of these groups further complicate things, but since the perpendicular $C_2$ axis is retained it is still somewhat possible to compare the resulting orbitals to a planar species. Therefore when you look at the orbitals you still find a remarkable resemblance to the picture above. For this purpose I have included the same schematic drawings next to the orbitals that resemble these best.

It is noteworthy, that because of the propeller structure there is no strict separation between σ and π orbitals, since the latter require a horizontal mirror plane, which is absent here.
But that is not all. Try counting the number of π electrons. This is far from trivial and it is in no case that makes sense six.[6] Don't let your brain explode about that fact, we are clearly dealing with a borderline case at this point.
And we have not even started to talk about the possibility of bonds between the oxygens of the nitro group.[7]
Take home message for TNM⊖
The trinitromethane anion has a strongly delocalised electronic system. It is extraordinarily stable due to this. If you accept the concept of Y-aromaticity as valid, then it is still a borderline case, where additional considerations have to be taken into account.
Notes, References and further reading

Bent's Rule

Bent, H. A. Chem. Rev. 1961, 61 (3), 275–311. DOI: 10.1021/cr60211a005.
What is Bent's rule?
Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?

The six π electrons are probably the reason for the statement in the Wikipedia article.
Gund, P. J. Chem. Educ. 1972, 49 (2), 100. DOI: 10.1021/ed049p100.
I have used this concept already various times to explain certain electronic structures. From a personal point of view I find it quite helpful.

Delocalization of pi electrons in nitrate ion
Why is the bond order in the SO₃ molecule 1.33 and not 2?
Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

Cioslowski, J.; Mixon, S. T.; Fleischmann, E. D. J. Am. Chem. Soc. 1991, 113 (13), 4751–4755. DOI: 10.1021/ja00013a007.
There are several possibilities, that can be considered to count electrons towards the number of π electrons.

The probably easiest way would be to flatten the molecule for this thought experiment, then count electrons perpendicular to the plane. This gives you two electrons for the central carbon (sp2), no electrons from the nitrogens (sp2) and two electrons each from the oxygens (sp). That gives you a total of 14 π electrons.
Maybe as simple as this, but somewhat the easy way out would be to disregard all contributions from the nitro groups, leaving only 2 π electrons.
Regard the nitro groups as π-backdonor. Since only one of the intrinsic π orbitals can effectively overlap with the central carbon, each of them would donate two electrons (maybe less, because backdonation). Plus two from the central carbon, makes up to eight electrons. This certainly is a very fuzzy approach.
Count four electrons per oxygen, which is kind of cheeky, but it could work. This would give 26 π electrons.
Write a Lewis structure and count only the double bonds (× 2), which gives a total of eight electrons.
I am almost certain, there are still more ways to count the electrons.

In [5] they analysed the anion with the quantum theory of atoms in molecules. They found bond critical points between the oxygen atoms. How these can be explained and what they mean is certainly a story for another time.

Actually, we cannot say that thiophene doesn't obey Huckel's rule because the hybridisation state of the sulfur atom could be sp2. Its hybridisation state remains ambiguous as we do not know the electronic geometry of the sulfur atom.

@TanYongBoon I disagree. Hybridisation is an entirely arbitrary mathematical concept and can be applied in any way you might want to choose, that does not make it more or less correct, it simply makes it different. As such, it is far beyond its original intention to justify anything with it. The original derivation of the set of rules by Hückel follows a very narrow set of molecules, none of them include hetero atoms. As such thiophene stretches this theory far beyond its means. Aromaticity is a far too complex concept to be accurately covered by Hückel's rules. Stop using them for everything.

Perhaps you indirectly answered this and I missed it - does Y-aromaticity provide additional stability beyond the resonance stability of the isolated pi systems? If there is stability beyond the isolated pi systems, as in benzene vs. cyclohexatriene, that might be some support in favor of considering Y-aromaticity a valid concept beyond just a curious case of resonance.

The Wikipedia article refers to a paper discussing so-called Y-aromaticity, a proposed extension of the idea of aromaticity to particularly stable non-cyclic structures.
This concept was proposed originally to explain the uncommon stability of the guanidinium cation. You can find some references in the Wikipedia article on aromaticity under "Atypical aromatic compounds". Whether this type of aromaticity really exists, however, is still somewhat controversial: other explanations have been proposed for the extra stability of guanidinium that do not necessarily require creating a new "type of aromaticity".

Can you explain in more detail what Y-aromaticity is because I can't look at any of those references.

The conjugate base of trinitromethane is flat, symmetrical (trigonal), mesomerically stabilised and it has $4n+2$ electrons. Every nitro group has $4 π$ electrons in resonance (an oxygen lone pair and a double bond), and together with the anionic lone pair on carbon, makes for a total of $14 π$ electrons ($14 = 4\times3+2$). Therefore it may be considered Y-aromatic.
Generally, Y-aromaticity is a controversial concept, which says that in Y-shaped molecules additional stabilisation is present, analogous to the planar, cyclic compounds with $4n+2$ delocalised π electrons). Examples of such molecules include square and Y-shaped dications with 4 carbon atoms and one double bond. It seems that it still hasn't been proven as to whether it is really a valid concept.
Additionally, as mentioned by Martin, in this case the molecule is distorted from planar geometry by steric clash of oxygen atoms - it's shaped like a propeller, but it only partially hinders delocalisation.

What is Y-aromaticity though? Why does it happen?

I generally agree with the assessment, but the  molecule is not flat. As for the assignment of the pi electrons, that is a bit fuzzy, see my answer.


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Is passivation of stainless steel feasible in the field?
Does anyone have input or experience on passivation of stainless steel, particularly in outdoor environments where submersion of components is not practical due to location or size? Nitric acid (and solutions with sodium dichromate) typically call for submersion for extended periods of time based on the concentration of the solution. Citric acid has been referenced as a more suitable alternative and is more environmentally friendly but the literature on non-submersion passivation techniques is limited. Any help or direction is appreciated.

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Hi, could you provide an example? what do you mean by field, purpose etc.

@JaroslavKotowski  I am looking at the post-manufacturing passivation for AISI 301LN ¼ Hard and 1/2 Hard Stainless Steel sheets, specifically around areas where spot welding occurred.  The sheets are attached to framework with spot welding and were initially passivated, though not to an acceptable level, as rust is appearing after 1-2 months of normal outdoor exposure (no notable chemical exposures).  The passivation would need to be done outdoors and submersion is not an option due to the size of structure (up to 100' long).

That could be problematic, any solutions I am aware of require concentrated acids. They could be sprayed onto the surface and neutralise afterwards.

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How does a "sea of electrons" hold metal atom together?
How do delocalized electrons hold metal atoms together? What are the forces of attraction between the atoms and the electrons called?

Related http://chemistry.stackexchange.com/questions/25003/clarification-for-metallic-bond?rq=1

This is a homework question. Please share your thoughts towards a solution otherwise it will be closed in accordance with our homework policy. A "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.

I understand how @bon would think this is homework, while the comments in the answer demonstrate genuine curiosity. Might be a good idea if you edit your question and say what you think those forces are/aren't.

Agreed. If you edit your thoughts into your question then this shouldn't be closed (or will be reopened if it is closed).

I think you will find the more appropriate answer to your question by applying the modern band theory.

Metallic crystals are held together by metallic bonds, electrostatic interactions between cations and delocalized electrons.

The metal atoms lose electrons to form positive metal ions and a sea of delocalised electrons. The positive ions are attracted to the oppositely charged electrons and form an electrostatic attraction.

i thought electrostatic attraction are usually between ions that don't move?

delocalised electrons are always moving so how is it possible that it is electrostatic attraction? the word static means non-moving?

Do you believe that non-delocalised electrons are precisely fixed? Because they aren't. The delocalised electrons are still in bonding 'orbitals' - they just aren't confined.

It's another definition of static: concerned with bodies at rest or forces in equilibrium.

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What turns clear liquid soap blue?
Liquid soap that sits in metal pump handles turns blue after a week or so, as in this photo comparing the soap in its initial clear state and then after pumping from the dispenser.

The pumps are presumably brass with either nickel or chrome electroplating on the exterior.  Can't say for sure whether or to what degree the interior where the soap makes contact is plated.  They have been in use for years.
The ingredients of this particular soap are:

water
sodium laureth sulfate
cocamidopropyl betaine
decyl glucoside
sodium chloride
"fragrance"
DMDM hydantoin
polyquaternium-7
tetrasodium EDTA
citric acid
sodium sulfate
PEG-120 methyl glucose dioleate
PEG-7 glyceryl cocoate
poloxamer 124
D&C red #33
FD&C blue #1

The color of the precipitate is strongly reminiscent of copper(II) hydroxide. I hypothesize:

Chloride corrosion of copper from exposed brass (ref 1 || ref 2)
a. Per ref 2 above, sulfate may also participate
Alkaline precipitation of $\ce{Cu(OH)2}$ in the neutrally-buffered soap
Insufficient $\ce{Na_4EDTA}$ to complex the amount of $\ce{Cu}$ corroded from the brass

It's also possible the blue material is $\ce{Ni(II)}$ complexed in such a way that its color is changed from the typical green-blue of $\ce{Ni(OH)2}$.

Is it clear where the free chlorine would come from?  The system is not exposed to any electrical charge and is held at a stable room temperature.  (Don't assume I know anything special about the chemistry of solutions like this.  E.g., if sodium chloride has some steady-state dissociation rate, or something in there is or could be ionic, I don't know about it.)

You don't need free chlorine for brass to corrode. Chloride ion is sufficient.  See the new "ref 2", added above in-edit. Quote: "It was reported
previously that ions as thiozionates, bromides, and iodides did not participate in dezincification
process, while chlorides and sulphates enhanced dezincification process." Etc.

Good finds on those references!

I guess, to cover my bases, it's possible there are chloride-resistant varieties of brass I don't know about.  But, there are definitely chloride-susceptible alloys.


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Anomalous trends in ionization energy
I want to address two exceptions in the trend of ionization energy across the period, that are causing me problems:
Taking the second period as an example :
The two exceptions from the general trend are the ionization energies of B lesser than Be and that of O less than N.
My teacher told me the reason to both was that half filled and fully filled orbitals of N and Be are more stable and hence require more energy to pull off an electron. He also said that this affects only the next element in the period for a change in trend( he never said why though).
This place and my textbook tells me that the reasons are:
1) Be,B :
The electron is to be removed from the 2p orbital in B and the 2s orbital in Be. The 2s orbital penetrates more into the nucleus and hence it is more difficult to remove an electron form Be than  from B.
Question: In that case why doesn't this apply to C? Shouldn't it be easier to remove one of the electrons from C's 2p orbital than Be's 2s orbital just like the B's 2p orbital? And so shouldn't the ionization energy of Be must be more than C too?
2) N,O:
N 1s2 2s2 2px1 2py1 2pz1
  O 1s2 2s2 2px2 2py1 2pz1
The reason I have understood here is that there are two electrons in the 2px orbital of O and because of electron-electron repulsion, it is easier to remove an electron from here as opposed to the 2px orbital of N which had only one electron.
Question: Why doesn't this apply to F?
F 1s2 2s2 2px2 2py2 2pz1
F has paired electrons too like O in the px and py subshells. Wouldn't there be repulsion here too? And hence why isn't it easier to remove electrons from F than is it to from N which doesn't have any paired electrons at all? Why is the ionization energy of F more than N instead?
The same problems can be seen in the next period.I hope they can be sorted out with the same logic that i am missing for this period.
Finally, was what my teacher said just a simplification of everything I found in the link and the textbook? Was he trying to convey the same thing? Or are they two complete different explanations?

As you said, the ionisation energies of all the atoms after B, such as C, N, etc. are lowered from what they would be if we were to take out a 2s electron from them. However you're missing the big picture. If you ignore the small dips going from B to C and from N to O, you get an overall increase going from Li to Ne. Why is that so?

Also, chemguide is correct. The explanation involving half- and fully-filled orbitals involves something called exchange energy, but these effects are not as important as the 2s -> 2p change going from Be to B and the pairing of the electrons (leading to greater repulsion) going from N to O. Exchange energy is more commonly used to rationalise the ground-state electronic configurations of Cr and Cu.
Also my previous comment should of course have said Be to B, not B to C. I am very sleepy today, it seems. Sorry.

@orthocresol  The overall increase is because the atomic radius decreases across the period.  But however small the dips are , they are dips.  Why does it affect only the immediate to the right  elements in the period?

Well, it's a simple case of balancing two factors. Going from Be to B, 1) IE increases due to a decrease in atomic radius 2) IE decreases because you are going from 2s->2p. In this case factor no. 2 wins out. Going from Be to C, you again have the same two factors, but factor 1 wins out. And that is because the decrease in atomic radius is larger going from Be to C than it is going from Be to B, because you are adding two protons instead of just one.

To put it another way, going from B to C there should definitely be an increase because the nuclear charge increases (leading to a smaller atomic radius). This increase is large enough to counteract the decrease going from Be to B, such that going from Be to C the IE increases.

@orthocresol, you said that when going from N to O, the formation of a half-filled orbital is a minor factor. Could you share any data to confirm this? I found literature that mentioned both factors, but they didn't say about their relative contribution.

OK. Lets get a few things cleared up first. The main factor affecting ionizing energy (IE) is the attraction between the nucleus and the electrons. The more the attraction, the more energy is needed to remove an electron from it's orbital and hence higher the IE. The other factors modify this main factor and thus change the expected IE values to the anomalous observed values. Now coming to your questions:

As we go to the right of a row, size of the atoms get smaller. $\ce{C}$ is smaller than $\ce{Be}$ and $\ce{B}$ and hence has higher IE.
The same thing happens in case of $\ce{F}$. The size of $\ce{F}$ is smaller compared to $\ce{N}$ and this factor over powers the repulsion between the paired $p$ orbital electrons.


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Calculating enthalpy change and energy required to convert lead sulphide to lead

Consider the following:
  $$\begin{alignat}{2}
\ce{PbS(s) + 3/2O2(g) &-> PbO(s) + SO2(g)}\qquad&&\Delta H^\circ = -413.7~\mathrm{kJ} \\
\ce{PbO(s) + C(s) &-> Pb(s) + CO(g)}\qquad&&\Delta H^\circ = +106.8~\mathrm{kJ}
\end{alignat}$$
  a. Calculate the enthalpy of the following process:
  $$\ce{PbS(s) + 3/2O2(g) + C(s) -> Pb(s) + SO2(g) + CO(g)}$$
b. What is the energy needed to make $454\ \mathrm g$ of $\ce{PbS}$ into $\ce{Pb}$?

My efforts:
a.
$\ce{PbS(s) + 3/2O2(g) -> PbO(s) + SO2(g)}$
$\ce{PbO(s) + C(s) -> Pb(s) + CO(g)}$
$\Delta H_\mathrm r^\circ = \Delta H_{\mathrm r(1)}^\circ + \Delta H_{\mathrm r(2)}^\circ$
$\Delta H_\mathrm r^\circ = -413.7~\mathrm{kJ~mol^{-1}} + 106.8~\mathrm{kJ~mol^{-1}} = -306.9~\mathrm{kJ~mol^{-1}}$
b. How do I find the energy needed?
Molecular mass of $\ce{PbS}$ is   $239.2650~\mathrm{g~mol^{-1}}$
so we have $\frac{454~\mathrm{g}}{239.2650~\mathrm{g~mol^{-1}}} = 1.897~\mathrm{mol}$ of $\ce{PbS}$
So energy required is $1.897~\mathrm{mol} \times 306.9~\mathrm{kJ~mol^{-1}} = 582.2~\mathrm{kJ}$

Part 1:
It's easier to imagine \begin{equation}\ce{A + B -> C + D} \tag{1}
\end{equation}
\begin{equation}\ce{D + E -> F + G} \tag{2}\end{equation}
\begin{equation}\end{equation}
Now if we wanted to write a 3rd equation where both the reactions 1 and 2 happen, we'd get
\begin{equation}\ce{A + B + E -> C + F + G} \tag{3}\end{equation}
Why?

D is both a reactant and a product. So, it's a reaction intermediate. It's like it's produced in 1 and again consumed in 2. So, if both reactions take place, it's neither on the reactants' side nor the products.
The conventional way of writing reactions doesn't show the mechanism of the reaction. You have to image the reaction happening, to completely understand why D isn't written.

So, as the third reaction is reaction 1 and 2 happening at the same place, $\Delta H_{r_3}$ is the algebraic sum of $\Delta H_{r_2}$ and $\Delta H_{r_1}$. Thus, your final result is correct: $$\mathbf{-413.7 ~kJ \times mol^{−1} +106.8 ~kJ \times mol^{−1}=−306.9 ~kJ \times mol^{−1}}$$
Part 2:
We have to take a more careful look at the unit of $\Delta H$. What does it mean? It means that for each mole of the reactant ($\ce{PbS}$), one mole of product ($\ce{Pb}$) is formed. (With regards to their stoichiometric coefficients in reaction 3) So, we've got to simply 'punch' the numbers in the formula $n= \frac{m}{M}$ (Where $n$ is the number of moles, $m$ is the given mass and $M$ is the molar mass) And then multiply that number by the $\Delta H_{r_3}$. Thus, your final answer is correct too.


text:
Why is grease from cooking hard to remove from wood?
Cooking oil that jumps from a frying pan in use and over time accumulates around the frying pan on a wooden surface is sticky and hard to remove.
My question is why. Why is it hard to remove? What is the chemistry behind it? Does it bond with the wood somehow? If so, how is it different from oil used to give the wood a protective coating? Why is one sticky and the other not sticky?

This site explains the use of oil on wood fairly well:

Oil is made of molecules small enough to seep down into the wood rather than merely sit on top. As a result, oil makes wood look richer and more translucent without adding a film on the surface. There are two different types of oils that woodworkers use: drying and non-drying oils. Drying oils will change from liquid to a solid film when exposed to oxygen in the air. Nut oils (boiled linseed, tung, etc.) are drying oils, but vegetable (peanut, olive) and mineral oils are non-drying. Edible mineral oil is popular on food contact items, like cutting boards. However, non-drying oils stay wet indefinitely, and they will wash off when the board is scrubbed with soap and water. Because they do not dry to a solid film, non-drying oils are considered a wood treatment, but not a finish.

I know this is old, but it seems non-trivial to simply scrub off and remove with "soap and water". This leads me to believe that even non-drying oils undergo some chemical process that make it more sticky, without making it completely solid like a drying oil.


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Initial rate of reaction from the graph?

I tried finding the slope between two points but I keep getting wrong values, how does one find the initial rate by using such graphs?
So I started by trying to find a slope of two points, $(0,0.4)$ and $(10,0.2)$ after calculations I got  $-0.02$, which is the opposite of the right answer in sign. I tried to check myself by trying to find the slope for other points like: $(10,0.2)$ and $(20,0.1)$ which gives me $-0.01$, I'm confused as whether my initial idea is right or not.

This standard exam question is a first order decay with half life (from the graph) of about 13 s. So the rate constant, k, is $(ln2)/t_{1/2}$, which is about $0.053 s^{-1}$. Hence the initial slope is -k times 0.4 M, giving $0.021 M s^{-1}$ as the magnitude of the initial slope. Rounded to one significant figure, this is $0.02 M s^{-1}$, as per answer A.

@EdV Well could you elucidate on the fact how you found out the half-life to be of 13s from the graph, I mean that's exactly I've been searching for! Is this a previously done question? I mean yeah it's an exponential curve but how are we being so sure of it?

@EdV Well I was actually asking how are you so sure about the coordinates? The graph lines don't intersect the graph exactly at any point other than at $(0,0.4)$, there's always a chance that $t_{{1/2}}$ =14s instead of 13 right? And that's exactly why one can't be so sure that the $t_{{1/2}}$ is constant or not. Being an exam question from your previous knowledge, you're probably right, beginners aren't made to fuss, they stick to order 0,1 or at max 2. But we should try to be general whenever possible, for e.g. here nothing's been mentioned!

@EdV No point in arguing, I just personally feel that those dashes don't count as a scale of measure, but that doesn't change the answer or the method itself. Since you've been teaching such standard exam questions, you're surely better than me here. Well you could actually put this up as an answer, people could upvote it!

It seems like you have the right idea to me,
For the "Initial Rate," you're taking a curve and doing a straight approximation of the curve over a small portion, as you appear to be doing.
Your logic also appears correct to me, a decrease in concentration should correspond to a negative rate.
My guess is that the question, (a poorly worded one?) is actually referring to the rate of product creation, and asking you to assume that there is a 1-1 relationship between the decrease in reactant, and the appearance of product. Thus if your reactant is disappearing at $0.02~M/s$ (as indicated by $rate= -0.02 ~M/s$) , then your product should be appearing at $0.02~M/s$, making A the correct answer.

yeah, but since it's a decelerating curve it wont have the same slope all the time right? how am I to find the right one if the slope is always different...

I agree, the question in the textbook seems very vague

@Jx1 That's correct! That's where the approximation and initial keywords come in.
You're approximating the slope at a given point, meaning that you're using a line that has about the same slope as the curve over a specific region.
Initial indicates that this region as at the very beginning. (The initiation of the reaction.)
So you're drawing a straight line that's the same as the line at the very beginning of the reaction.

This is a standard exam question pertaining to the kinetics material that is typically presented in a first year general chemistry course. Usually the coverage is restricted to discussion of zero, first and second order kinetics and this is what gets tested on exams. From the photo of the exam question, the decay is apparently first order, i.e., $$\mathrm C(t) = C(0)e^{-kt}$$ where $\mathrm C(t)$ is the molar concentration of whatever is decaying away, $\mathrm C(0)$ is the initial molar concentration (i.e., at $\mathrm t = 0$) of whatever is decaying away, and $k$ is the decay rate constant in units of $\mathrm s^{-1}$. In the present case, $\mathrm C(0) = 0.4 M.$
But, is the decay really exponential? To determine this, without curve fitting data pairs or linearizing the plot, it suffices to see if the half-life, $\mathrm t_{1/2}$, is constant. From the photo of the graph, the initial concentration declines to $\mathrm 0.2 M$ at about $\mathrm t = 13 s$. Note that there are $\mathrm 10$ dash marks every $\mathrm 10 s$ along the time axis, so this is an aid, albeit imperfect, in estimating times along the time axis. The concentration declines to $\mathrm 0.1 M$ at about the $\mathrm 26 s$ mark and further declines to about $\mathrm 0.05 M$ at about the $\mathrm 39 s$ mark.
So, $\mathrm t_{1/2}$ is constant and approximately equal to $\mathrm 13 s$. Therefore, the decay process is a first order exponential, as shown in the equation at the top. Note that the only other alternatives, in the restricted context of typical course material coverage and exam coverage of said material, are zero and second order decays. But these do not have constant half-lives: for a zero order decay, every subsequent half-life is halved and for a second order decay, every subsequent half-life is doubled. The graph clearly eliminates these two possibilities. It is also obvious that the decay cannot be first order since then the concentration versus time plot would be linear. If the decay was second order, the concentration could not be down to $\mathrm 0.1M$ in less than $\mathrm 30s$, even if the first $\mathrm t_{1/2}$ was clearly underestimated as $\mathrm 10s$.
The relationship between $\mathrm k$ and $\mathrm t_{1/2}$ is $\mathrm k = (ln2)/t_{1/2}$. This is easily verified by substitution into the exponential equation, resulting in $$\mathrm C(t = t_{1/2}) = C(0)/2$$ Hence, with $\mathrm t_{1/2}$ estimated as $\mathrm 13s$, the rate constant estimate is $\mathrm k = 0.053 s^{-1}$.
To find the initial rate of decay, simply differentiate the exponential equation and evaluate it at $\mathrm t = 0$. The derivative is $$\mathrm dC(t)/dt = -kC(0)e^{-kt}$$ so at $\mathrm t = 0$, the slope is simply $\mathrm -kC(0)$ and the magnitude of the slope is $\mathrm kC(0)$. With $\mathrm k = 0.053 s^{-1}$ and $\mathrm C(0) = 0.4 M$, this yields $\mathrm 0.021 M s^{-1}$ as the initial rate magnitude. Rounded to one significant figure, which is all this exam problem deserves, yields $\mathrm = 0.02 M s^{-1}$. This is exam answer A.
One last thing: a small error in estimating the half-life will not matter because the initial decay rate is inversely proportional to it, so it would have to be high by a factor of two in order to get answer B on the exam.

No calculus? No problem! Since $\mathrm e^{-x} \approx 1-x$ for $\mathrm x << 1$, assume $\mathrm 0 < t$ and $\mathrm kt << 1$. Then $\mathrm C(t) \approx C(0)[1-kt] = C(0) - C(0)kt$. Subtract $\mathrm C(0)$ from both sides and divide both sides by $\mathrm t$. Then the approximate initial slope is $\mathrm [C(t) - C(0)]/t$, which equals $\mathrm -C(0)k$, as in the answer. N.B. Taking the limit, as $\mathrm t$ goes to zero, gives the true derivative.

This is the correct answer, not the one that was accepted. Win some, lose some.

The idea is that you want to find the instantaneous rate of change (rise over run)at the initial value (time = 0). but since this gives .4/0 (can't be done) just use a close number to the initial value to get a good approximation of the slope at the value. So for initial so (.4-.3)/(0-5) =.02 you use the slope because the units are Moles(rise)/Seconds(run). since it doesn't specify the rate of appearance or disappearance then the magnitude is what matters not sign.

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Firstly, you're right that the slopes at all points on this curve will be negative as it's a downward sloping curve. The rate of reaction w.r.t. reactants is the negative of the rate of consumption of reactants provided there's 1:1 stoichiometric ratio between the reactants and products. But in this question, nowhere is it mentioned that this curve depicts the concentration vs time curve for reactants. Even if we consider it to be so, there's no mention of a function that describes this graph, from which we could have found out the order and rate constant of the reaction.
Thus we have to proceed in a way @Dan pointed out, you need to use the slope a line that has approximately the same slope as the curve at the initial state of the reaction. But for that we need 2 exact coordinates of very closely spaced points on this graph, just like you tried to do, but as you see, the graph doesn't pass through $(10,0.2)$, actually there's no other point than $(0,0.4)$ whose coordinates we know for sure. But if you look closely, you'd find $(5,0.3)$ is a better match (as far as my eyes tell me). So using that, one can find the slope of line between them to be $\frac{0.3-0.4}{5-0}=-0.02$
Thus the rate of reaction w.r.t. reactants could be-
Rate $=\frac{\mathrm{-d}[\text{reactants}]}{\mathrm{d}t}=-(-0.02)=0.02 \pu{mol L^-1 s^-1}$
Even though my stand is that this is a poor question that lacks a lot of necessary detail. So it's better not to waste time on this one, but yeah there's always a chance of learning something out of everything.

@EdV You're welcome :) Yeah actually I found that "Don't write EDIT" part from that post itself. I prefer that post over all other formatting posts, because it's mostly community agreed. But you could have posted this comment on your post too, that would have been better right?

@EdV You're right, officially we are supposed to avoid such things, but it's mostly confined to Q&A, where we should avoid "Thanks for any help" or "Hope that this helps", but informally we're quite lax on doing that in comments, until it leads to an extended discussion (and eventually moved to chat). Chem SE is a small community and we mostly know each other, so we're kinda family at times, we thank others if they find out a typo in our posts, or give some advice on formatting.

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text:
How to calculate the equilibrium pressure when ammonia and hydrogen gasses react in a closed system?

There is a container that has a volume of $5~\mathrm{L}$ on one side and $3~\mathrm{L}$ on the other and is separated in the middle. There are $2~\mathrm{mol}$ of $\ce{N2}$ on one side at $0~^\circ\mathrm{C}$ and $4~\mathrm{mol}$ of $\ce{H2}$ on the other at $0~^\circ\mathrm{C}$. When the valve is turned and the gases react, what will the total pressure be?

The balanced reaction is $$\ce{N2 + 3H2 <=> 2NH3}$$
My thinking was that the reaction occurs at STP so the pressure would stay at 1  atm?
Or that you could use
\begin{align}
pV &= nRT\\
p(8~\mathrm{L}) &=6(.082)(273.15)\\
p &= 0.06
\end{align}
The $8~\mathrm{L}$ comes from adding the two volumes, and the 6 from adding the number moles together.

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

You are neglecting the fact, that these gasses react with each other. Also the temperature is given as $0~^\circ\mathrm{C}$, so STP it is not. I feel like there is an equilibrium constant missing in the question.

Thank you for adding your attempt! This makes it much easier to help you.

That is all the information that was given for the problem. And I thought that STP was 1 atm and 0 degrees C, thats why I said that.

Note: Not all the reactants will react to form the products, but rather exist in equilibrium. However since the question doesn't mention anything about the equilibrium constant, I will assume that the reaction goes to completion as otherwise it will be impossible to do this question.
If you are aware of the various gas laws, doing this question is actually really simple and doesn't involved many calculations.
Step 1: Finding Initial Pressure
Firstly, the question doesn't specify what the initial pressure of the container is. Therefore you can't just assume that it is $1\ \mathrm{atm}$. So the first step would be finding the initial pressure: $$p = \frac{nRT}{V} = \frac{6\ \mathrm{mol}\times 8.3125\ \mathrm{L\ kPa\ mol^{-1}\ K^{-1}}\times273\ \mathrm{K}}{8\ \mathrm{L}} = 1702.4\ \mathrm{kPa} = 16.8\ \mathrm{atm}$$
Step 2: Find Moles of Products
Looking at the stoichiometry of the chemical equation, $4$ moles of gas react to form $2$ moles of gas. Therefore the ratio is $2:1$. In this case, we originally had $6$ moles of gas. Therefore at the end of the reaction, there should be $3$ moles of gas.
Step 3: Find Final Pressure
Now we know that the total number of moles of gas has halved (from $6$ moles to $3$ moles). One of the gas laws state that the number of moles of gas is directly proportional to the pressure of the system. This is clearly obviously by looking at the ideal gas equation. Therefore, by halving the original amount of moles of gas should result in halving of the original pressure of the system.
Therefore the final pressure will be: $8.4\ \mathrm{atm}$


text:
Alkylation of 1-naphthol in trifluoroethanol versus in DMSO

I am supposed to work out which product is formed in which solvent when naphthalen-1-olate is treated with benzyl iodide.
DMSO is a polar aprotic solvent, as it does not possess an acidic hydrogen. So, the negative charge on oxygen will be concentrated as the anion will be poorly solvated. Hence, it should act as a hard nucleophile. Since alkyl halides such as BnI are soft electrophiles, I think that the substituent should preferentially add to the ring.
In 2,2,2-trifluoroethan-1-ol (TFE), I think the reaction would proceed via an SN1 mechanism. Now, the carbocation is a hard electrophile, and thus O-alkylation might be observed.
But considering the better solvation of the negatively charged oxygen in TFE, it would also be more hindered, which suggests that C-alkylation might also be seen.
Which products are actually formed, and am I on the right track?

Echoing jerepierre's comment (which has since been edited into bon's answer), I don't find any examples of 1-naphthol alkylation with BnX on Reaxys. One can only conclude that the question is made up, which is quite unfortunate, as it seems to teach that there is a black-and-white difference in reactivity when the solvent is switched. Real life is nearly never as clear-cut as some of these questions would have one believe.

The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:

Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.
In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.
EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."

.But there isn't any scope of O-alkylation in the protic medium, because being in a highly polar and protic medium, oxygen will be highly solvated. The upcoming electrophile can not approach the Oxygen atom because of steric effects by the solvent, and as far as breaking aromaticity is concerned, if reaction were just to be relied on that, How do we have electrophilic aromatic substitution on benzene ring. All intermediates involved there requires breaking of aromaticity which is regained in the final product

These solvents should give identical products.
DMSO is considered the quintessential polar aprotic solvent. Given that the negative oxygen is going to fairly nucleophilic and that the second reactant has iodine on it (and that iodine is a leaving group of the Gods), my guess is that this reaction will proceed via SN2 mechanism to form the product they show for A in answer (A).
Similarly, fluorinated ethanol is very polar and protic. This reaction will proceed via SN1, with the iodine leaving and the carbocation stabilized by the phenyl ring it is attached to. I don't believe there should be significant carbocation rearrangement, since the stability of the product with the aromatic ring intact should outweigh the any added stability from the carbocation being on the ring, but don't quote me on that. So this will give a product identical to the one for DMSO. I would say that this mechanism can result in stereochemical differences, but the carbon being attacked by the oxide is not chiral.

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text:
How does the human body process Bromelain in pineapple? (A meat tenderising/dissolving enzyme)
It turns out pineapple juice can dissolve a jello set, and turn a beef roast to paste.
It turns out this is caused by Bromelain, an enzyme sometimes used in meat tenderising.
I'm trying to work out how you can eat pineapple without dissolving on the spot.
My question is: How does the human body process Bromelain in pineapple? (A meat tenderising/dissolving enzyme)

The enzyme is slow acting, so it does not dissolve your mouth, just slightly tenderises it, which is the sandpaper feel that you get on your tongue after eating it. When the pineapple reaches your stomach the enzyme gets dissolved by your stomach acids, so cannot harm you any longer.


text:
When the heat is absorbed by the ice then ice will melt to form liquid water

My new try:
Given data
Energy given by reaction =1690 kJ
Initial temperature = 0 oC
Final temperature = 50 oC
C water = 4.18 Jmol-1 K-1
?Hfus ice = 6.01 kJ*mol-1
Calculate Mass of ice = ?
Ice melts at constant temperature of 0 degree C and changes the phase from solid to liquid .
Then liquid water absorbs the energy and changes from 0 degree C to 50 degree C
Change in temperature of the water is ?T = (50 oC - 0 oC )= 50 oC
Let’s set up the equation as follows to calculate the mass of ice
q= (m*?Hfus) + (m*Cwater * ?T)
where , q = energy , m = mass , ?T = (Tf-Ti)
convert given energy from unit kJ to Joules
(1690 kJ * 1000 J)/1kJ = 1690000 J
Convert ?H fus from kJmol-1 to Jg-1
6.01kJ*mol-1 (1000 J/1kJ) = 6010 Jmol-1
6010 Jmol-1 / 18.0148 gmol- = 333.6 J *g-1
Convert C water from J per mol to J per g
4.18 Jmol-1 oC /18.0148 g = 0.232 J*g-1 oC-1
Now lets put the values in the formula and calculate the mass of ice
q= (m*?Hfus) + (m*Cwater * ?T)
1690000 J = (m X 333.6 Jg-1) + (m X 0.232 Jg-1 *oC-1 * 50 oC )
1690000 J = m X 345.2 J*g-1
1690000 J /345.2 J*g-1 = m
4895.7 g = m
Please help me!!

Most of your manipulations seem correct, but I think the question has inaccurate data. The specific heat of water is $4.18 \ \mathrm{J\ \mathbf{g^{-1}}\ K^{-1}}$, not $4.18 \ \mathrm{J\ mol^{-1}\ K^{-1}}$.

My teacher said that the  information is correct

@NicolauSakerNeto is right, 1 mole of water contains ~20 grams, so using the wrong units will give you an answer that's off from the real answer by a factor of ~20. (However maybe it's wrong in the Homework's answer as well, which seems to happen sometimes)

Energy needed for melting and heating up the water
$$
Q = \Delta H_{\text{fus}} \cdot m + m \cdot c_p \cdot \Delta T
$$
solve for $m$:
$$
m = \frac{Q}{\Delta H_{\text{fus}} + c_p \cdot \Delta T }
$$
The molar heat capacity for water is around $75\,\frac{\text{J}}{\text{mol K}}$. You've been given the specific heat capacity which is around $4.18\,\frac{\text{kJ}}{\text{kg K}}$. All that's left to do is convert the enthalpy of fusion to $\frac{\text{kJ}}{\text{kg}}$ and you can solve the equation.
$$
\Delta H_{\text{fus}} = \frac{6.01\,\frac{\text{kJ}}{\text{mol}}}{18.02\,\frac{\text{g}}{\text{mol}}} = 0.3335\,\frac{\text{kJ}}{\text{g}} = 333.5\,\frac{\text{kJ}}{\text{kg}}
$$
Calculate $m$:
$$
m = \frac{1690\,\text{kJ}}{333.5\,\frac{\text{kJ}}{\text{kg}} + 4.18\,\frac{\text{kJ}}{\text{kg K}} \cdot 50\,\text{K}} = 3.12\,\text{kg}
$$
The combustion of 1 mol of propane supplies enough heat to melt 3.12$\,$kg of ice and and heat the resulting water up to 50$\,$°C


text:
How does the MG811 sensor work?
The MG811 is an electrochemical sensor that measures $\ce{CO2}$. I think is something like a redox reaction but i don't understant how it produces $\ce{e-}$ so i can read them and get the $\ce{CO2}$ concentration.
I copy you the reaction that makes the sensor measures
When the sensor exposed to $\ce{CO2}$，the following electrodes reaction occurs：

Cathodic reaction: $\ce{2Li+ + CO2 + 1/2O2 + 2e- -> Li2CO3}$
Anodic reaction：$\ce{2Na+ + 1/2O2 + 2e- -> Na2O}$
Overall chemical reaction：$\ce{Li2CO3 + 2Na+ -> Na2O + 2Li+ + CO2}$

The datasheet says that it workings adopting the electrolyte cell Principle and that it's composed by the following cells:

Air，$\ce{Au}$|NASICON|| carbonate|$\ce{Au}$, air，$\ce{CO2}$

I need you to explain me how actually the sensor is working and with which material are made the electrodes.

This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

The link you gave in the question, actually shows the equation of how this sensor works. You don't need to produce any electrons, you just need to measure the voltage.
Try to write the Nernst equation for the above overall reaction and that will show you what is going on.

I meant how does the voltage measured is produced in the reaction.

The potential in any electrochemical reaction is produced because the electron is more stable in some molecules than others. This creates an energy difference that we explain by using the concept of electrochemical potential.
The Nernst equation describes the details of this potential.

@Burak_Ulgut I got that last. So between Li2CO2 and Na2O there's a electrical potencial, righy? One more thing why in the cell structure says NASICON and AU but in the reaction Li2CO3 and 2Na?


text:
Why is the less substituted alkene favoured in Hofmann elimination?
In the reaction, $\ce{OH-}$ group attacks a hydrogen ion in an E2 reaction, forming a transition state in which simultaneously $\ce{NCH3}$ is removed. In this reaction, the $\ce{OH-}$ should attack, like in other case of E2 so as to form the Zaitsev product, but due to "steric hindrance" the less substituted alkene is formed. But surely the $\ce{OH-}$'s attack is not affected by $\ce{CH3}$ groups? It can be affected by the $\ce{NCH3}$ group - but whichever way it attacks, it's going at the beta carbon - so is at the same distance from that group in all cases. How can steric hindrance affect its attack? Why is the Hofmann product (less substituted product) favoured?

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Also can you give the actual reaction that is taking place here. I'm not quite sure what the substrate is.

@bon I imagine it is something like this

Yes, it is that reaction.

There are 3 main types of elimination reactions, the E1, E2 and E1cb.  Here is a generalized picture of the transition state for the E2 elimination reaction.

Breaking of the $\ce{C-H}$ bond and the $\ce{C-LeavingGroup}$ bond are roughly equally advanced, as is formation of the double bond. Because the 2 bonds are broken to roughly the same degree, it can be said that the E2 reaction is in the "middle" of the possible elimination mechanisms.
If we change this pattern and make the $\ce{C-H}$ bond breaking more advanced than the $\ce{C-LeavingGroup}$ bond breaking (we can do this by making the leaving group a poor leaving group [i.e. groups like ammonium, $\ce{-NR3^+}$, sulfonium, $\ce{-SR2^+}$, etc.]), then we shift the mechanism towards the E1cb side of the elimination spectrum.

In the E1cb case, because breaking of the $\ce{C-H}$ bond is advanced, partial negative charge develops on the carbon where the proton is leaving from.  Now primary carbanions are more stable than secondary carbanions, which, in turn, are more stable than tertiary carbanions (electron releasing alkyl groups will destabilize a negative charge).  Therefore, when we have a choice of which proton to remove, we remove the proton on the least substituted carbon as this carbon will better stabilize the developing negative charge.  As the LeavingGroup departs and the reaction completes, this will produce the olefin with the least substituted double bond, the Hofmann elimination product.
Sometimes it is argued that base size is responsible for elimination patterns and that when a large base is used the proton on the least hindered carbon is abstracted.  This is true but the effect is relatively small.  Changing the base from methoxide to t-butoxide only increases the amount of Hoffmann product by a modest 10-20% in simple haloalkanes.
So while sterics play a role, it is usually a small role with the electronic factors discussed above being more important.
For completeness, let me briefly mention the E1 elimination which is on the "other side" of the elimination spectrum of mechanisms. In this case, breaking of the $\ce{C-LeavingGroup}$ bond is advanced and partial positive charge develops on the carbon where the LeavingGroup is departing from.

Elimination of a proton from this intermediate can often proceed in two directions.  Since more highly substituted double bonds are more stable than less substituted double bonds and since the transition state for proton elimination involves partial double bond character, the transition state  leading to the more stable double bond (e.g. the more highly substituted double bond) will be lower in energy.  Hence, the product with the more highly substituted double bond (Zaitsev product) will form in E1 reactions.
Edit: Response to OP's comment

I meant the pyrolytic elimination of quaternary ammonium hydroxide,
  which follows E2 mechanism

I wasn't saying that this reaction (the thermal Hofmann elimination) goes through an E1cb mechanism, rather I said "we shift the mechanism towards the E1cb side of the elimination spectrum."  In other words, it is an E2 reaction but $\ce{C-H}$ bond breaking is somewhat more advanced than $\ce{C-LeavingGroup}$ bond breaking. Therefore the primary $\ce{C-H}$ bond is more likely to be involved than the secondary or tertiary $\ce{C-H}$ bond, for the reasons outlined above.
Here's another approach.  Instead of electronic factors, let's look at steric factors - in this reaction they both lead to the same conclusion.
E2 eliminations prefer a trans or antiperiplanar geometry.  Below are Newman projections of the 2 conformations with this geometrical arrangement of the substituents.

Conformation A is set up to eliminate towards the least substituted carbon, the methyl group (Hofmann), while conformation B depicts elimination towards the more highly substituted carbon (Zaitsev).  The trialkyammonium group is extremely large and generates an unfavorcbale steric interaction in conformation B.  This makes conformation B higher in energy, hence elimination from conformation A to produce the less-substituted double bond will be preferred.

What about the stochastical argument? Creating a terminal double bond one has the choice of three protons which one as a base can abstract, while (assuming something like a dimethylcyclohexane derivate) the other direction can have as little as one hydrogen left, meaning a $3:1$ ratio assuming ideal equivalence.

Yes, statistical factors play a role in all reactions where a choice of pathways is available..  However, in mechanistically switching from Zaitsev to Hofmann type elimination they are not the controlling factor.

I did not mean the E1cb mechanism. I meant the pyrolytic elimination of quaternary ammonium hydroxide, which follows E2 mechanism. E2 mechanism forms Zaitsev product normally. In this reaction, the Hofmann product is formed instead.

If one tries to look at the molecule entirely in one go - instead of as two different projections - then both the alpha H in the antiperiplanar configuration seem to be at the same distance from the bulky N group. Then the approaching base should suffer similar repulsion if it attacks either H?

If I understand your question correctly, then the answer is "yes".


text:
Why does magnesium have an exceptionally low melting point?
Why does magnesium have a lower melting point than both calcium and beryllium? It does not seem to fit into the group trend.

I'm not sure I'd consider 922 K (Mg) relative to 1112 K (Ca) exceptionally low. And given Sr (1041 K), Ba (1002 K) and Ra (973 K) perhaps the question might be why is calcium's melting point a tad higher than you might expect. Now if magnesium's melting point were 250 K, that might be exceptionally low compared with the rest of the column.

Following up on my comment, even though I'm not convinced that this is a 'real' answer to the question.  Consulting the SGTE database of elemental free energies [1] (with various updates over the years), one can obtain the melting temperatures for the hcp, fcc, and bcc phases of beryllium, magnesium, calcium, strontium, and barium (rough guesses looking at plots — consider good to approx. $\pm\pu{5 K}):$
$$
\begin{array}{lrrrcc}
\hline
\text{Element} & T_\mathrm{m}(\text{hcp})/\pu{K} & T_\mathrm{m}(\text{bcc})/\pu{K} & T_\mathrm{m}(\text{fcc})/\pu{K} & \text{Phase at RT} & \text{Melt phase} \\
\hline
\ce{Be} & 1544 & 1560 & 1018 & \text{hcp} & \text{bcc} \\
\ce{Mg} & 922 & 755 & 705 & \text{hcp} & \text{hcp} \\
\ce{Ca} & 940 & 1114 & 1063 & \text{fcc} & \text{bcc} \\
\ce{Sr} & 900 & 1050 & 1030 & \text{fcc} & \text{bcc} \\
\ce{Ba} & 635 & 1000 & 709 & \text{bcc} & \text{bcc}\\
\hline
\end{array}
$$
Several things to note, amongst them the variety of which crystal structure is most stable for the different elements at different temperatures. Also, the hcp phase's melting point does almost(!) monotonically decrease going down the table.  The biggest anomaly would actually seem to be the relative instability of the bcc and fcc phases of magnesium (it is the only one that remains hcp up to the melting point).
Overall, it would appear that the group 2 elements display a wide variety of preferences for crystal structure, and hence bonding energy configurations.  Is $\ce{Mg}$ unusual? Not really…
References

Dinsdale, A. T. SGTE Data for Pure Elements. Calphad 1991, 15 (4), 317–425. DOI: 10.1016/0364-5916(91)90030-N.


text:
Are there non-transparent gases?
Are there gases that are not transparent at room temperature (i.e. at temperature below the point where the substance starts to radiate visible light due to heating)?

Also pressure may be important.

I think, thick enough iodine or bromine fumes could be non-transparent albeit at sli~ghtly elevated temperatures.

@Jan what about mercury?

Gaseous mercury is colourless.

@gsurfer04 what about supercritical one?

I don't know, are we talking about above just Tc or above Pc, too? Because I've only heard "supercritical fluid" for the latter, the former just being gas.

Radiating light doesn't make gases opaque - the substance of the sun is transparent (i.e. some of the photons we see originate from below the surface, and pass through the layers above them) http://en.wikipedia.org/wiki/Limb_darkening

related: https://chemistry.stackexchange.com/a/102492/23561

First, a little bit of background. Transparency is not an absolute property of a material. Every substance is opaque, so long as light has to pass through enough of it, and opacity also changes according to ambient conditions.  Some substances, such as most metals, are opaque even in $100\ \mathrm{nm}$ thin films, while many gasses will let a small amount of light through mostly unperturbed even after several kilometres. For example, here is a measure of how far photons of different energies can travel on average in the pure elements before interacting once with their atoms (the graph is actually for high energy x-rays rather than visible wavelengths, but they're all photons nonetheless). It makes intuitive sense that a gas should let more light through, as the atoms in it are much more spaced than in a solid or liquid.
The two general processes that work to make materials non-transparent are photon absorption and photon scattering (reflection being included as an example of scattering). When we say a material is "transparent" colloquially, what we generally mean is that the material, at the thickness observed, does not scatter much light and thus allows a coherent image to pass through the material (basically, you can see through it to the other side without a lot of distortion, that is, the object is "see-through" or clear). Interestingly, this means that a clear material need not be colourless, as correctly pointed out by Geoff, because photons of a given colour may be absorbed while letting most other wavelengths pass through without scattering.
Gasses in general are highly transparent because they are both highly colourless (absorb little light) and very clear (scatter little light) unless you're looking through several kilometres of gas, as I mentioned previously. However, there are examples of highly clear but coloured gasses, as it is possible for them to absorb a significant amount of photons of a certain energy due to low energy electronic transitions, while allowing the rest of the visible spectrum to pass through unimpeded. While some gasses are coloured and others are not, I believe all pure gasses must be very clear (small amounts of gas will scatter almost no light), and they can only scatter a fair amount of light across a short distance in conditions where they are probably not best described as gasses anymore (plasmas, supercritical fluids, etc).
Perhaps the most classic example of a coloured gas is nitrogen dioxide, $\ce{NO2}$, a strongly brown-red noxious gas which is easily formed by decomposition of nitric acid, among several other ways. The unpaired electron in its structure is somewhat uncommon, and it can be easily excited to a higher electronic state upon absorption of photons in the blue region of the visible spectrum. Since blue light is strongly absorbed even in relatively thin samples (a few centimetres of $\ce{NO2}$ gas), the rest of the white light passes through with essentially no scattering, forming a perfect image of the other side but with a strong red tinge.
Even more interestingly, $\ce{NO2}$ reacts with itself at low temperatures or high pressures and dimerizes to form dinitrogen tetroxide, $\ce{N2O4}$, a colourless solid/liquid/gas (depending on conditions). In other words, the two substances are in a reversible equilibrium:
$$\ce{2 NO2 (g) <=> N2O4 (g) + energy}$$
Coincidentally, this equilibrium is characterized by an equilibrium constant which is close to 1 at ambient conditions, and this constant can easily be changed by varying temperature or pressure to higher or lower values. This means that under relatively easily achieved conditions, it is possible to study the reversible change of a coloured gas into a colourless one!

(Source)
Though I focused on $\ce{NO2}$ because it is a remarkable case, there are several other examples of clear but coloured gasses. As Geoff mentioned, the halogens all form coloured gasses, though bromine is a volatile liquid and iodine is a solid at room temperature, so a small amount of heating is necessary. Ozone, $\ce{O3}$, is somewhat blue (source), though its colour is best observed in the liquid phase. The same goes for oxygen gas, $\ce{O2}$, which is slightly blue, though it is not the cause for the blue sky. There are probably a few more examples out there.

I like your answer! After reading your first paragraph it occurred to me that we take it for granted that the concept of transparency always applies to visible light. It seems obvious in retrospect that all substances should block light over some wavelengths, even if it is not obvious that they would block light over all wavelengths.

You mentioned metals being opaque in films 100nm thick. Would a film of, say, iron actually be transparent if the film were only a few atoms thick?

@BrianGordon Yes, a layer of iron only a few tens of atoms thick will be transparent. As part of my work I regularly deposit very thin metal films by evaporation, often on the order of 10-100 nm thick, to form conductive contacts for circuits. I have not personally worked with iron, but I can tell you that I have deposited transparent silver films with thicknesses of 10-20 nm, which is around 100 atoms thick. This is often exploited to make transparent electrical contacts. Similar gold films are also both clear and colourless.

I'd separate transparent and colorless.
Most gases are transparent or very nearly so because the concentration is low and absorptions are often weak.
Chlorine, though is yellow-green, and has a noticeable color (from Wikipedia)

Other halogens such as bromine and iodine do have observable colors as vapor, although as mentioned in the comments, you often need slightly elevated temperature for significant vapor to form.

Are there gases with metallic or non-metallic luster? Will supercritical mercury have it?

It might, but I've never seen an image of supercritical mercury. It's electrically conductive, but you need high pressure and temperature.

Under high enough pressure will chlorine become non-transparent or will become liquid/solid earlier?

I've never even contemplated the existence of supercritical mercury until Geoff's comment.  Cool stuff!

@CurtF. Neither did I until asked. But I figured a little searching might be worthwhile. :-)

The answer to your question is yes, there are non-transparent gases, however, it depends upon the wavelength at which you are observing and how much gas you are looking through. At some wavelengths the gas is opaque at others transparent. The amount of light absorbed depends on its concentration, the path-length through which the light passes, and how strongly a molecule absorbs the light. It is quite easy to have conditions that absorbs 99.9% of all light at a given wavelength.
If you are observing in the infra red then simple gases such as oxygen or nitrogen have no transitions which absorb photons as they have no dipole which can 'capture' (in a vibrational transition or rotational) the energy of a photon. (Technically weak transitions can occur by other perturbations but for the present discussion we ignore these details).
Oxygen has low lying electronic transitions, which is when an electron is promoted from the ground state to an excited state, but these are in the near infra red and cannot be observed by eye. NO$_2$ also has visible electronic transitions as it is brown and so do the halogen gases and vapours. Mercury vapour also absorbs in the visible and this can be seen as shadows of vapour from the liquid when illuminated by light also from a mercury lamp. The transition is so intense that the vapour becomes opaque.
At shorter wavelengths such as the ultraviolet, oxygen, nitrogen and water vapour and most all other vapours and gases become opaque at relatively low pressures and path-lengths. i.e. if our eyes were sensitive to uv light only the air would look black.
At very short wavelengths, x-rays are absorbed by a molecules atoms (not by the molecule itself) but they are also elastically scattered (no energy imparted to atom as in diffraction) or inelastically scattered (some energy imparted).


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Why amino acids (Zwitterion) become either negative or positive at low and high pH solutions?
The amino acids are Zwitterions. In neutral pH, an Amino acid's amino group has a postive charge and Carboxyl group has negative charge. They cancel each others charge thanks to the $Hydrogen$ that's roaming between the two groups.
But, I am not able to understand why Amino acids have positive charges at a low pH and negative at high pH? Given amino acids are Zwitterions, how can both terminals have the same charge (either positive or negative) at low and high pH?
Can someone explain this from general chemistry and pH perspective (acid/base)?
PS: first posted in Biology site but community recommended it to be moved here.

related http://chemistry.stackexchange.com/questions/24713/how-to-rationalise-the-protonation-state-of-amino-acids, http://chemistry.stackexchange.com/questions/10416/amphoteric-amino-acid-titration

http://en.wikipedia.org/wiki/Amino_acid#Zwitterions this should explain well enough

Firstly a note about terminology. The word "terminus" is reserved for the N- or C-termini of a polypeptide chain. For a free amino acid, you should refer to the carboxyl and amino groups as the $\alpha$-$\ce{COOH}$ and $\alpha$-$\ce{NH2}$ groups respectively.
Anyway, the -$\ce{COOH}$ group is acidic; above a certain pH, typically around 2, it can be deprotonated to form a negatively charged -$\ce{COO-}$ group. On the other hand, the -$\ce{NH2}$ group is basic; below a certain pH, typically around 9, it can be protonated to form a positively charged -$\ce{NH3+}$ group. I made a short table describing the protonation states of the two groups.
$$\begin{array}{c|c|c|c}
\textbf{pH} & \textbf{Carboxyl group exists as} & \textbf{Amino group exists as} & \textbf{Net charge} \\
\hline
<2 & \ce{-COOH}\text{ (neutral)} & \ce{-NH3+}\text{ (positively charged)} & +1 \\
>2 & \ce{-COO^-}\text{ (negatively charged)} & \ce{-NH3+}\text{ (positively charged)} & 0 \\
>9 & \ce{-COO^-}\text{ (negatively charged)} & \ce{-NH2}\text{ (neutral)} & -1 \\
\end{array}
$$
For more information you can read Wikipedia or any biochemistry textbook - the first few chapters will often be devoted to discussing chemistry like this.

To be frank, when it comes to academic related things (science), Wikipedia hasn't been an easy diegest feed. I found out the video, that I commented below. Also thanks to your explanation, I also checked Morrison & Boyd, plus a lecture from MIT open courseware.

@bonCodigo - RE: Wikipedia - You're right. When Wikipedia first started out the articles were written at a fairly low level. Then the articles got rewritten by experts so that the articles were absolutely technically correct. The result many times is a lot of jargon and complicated formulas which are difficult for a beginning student to understand.

By increasing the pH you are actually increasing the concentration of hydrogen ions present.
By the simplistic definition,
$pH = -\log\ce{[H+]}$
Thus at a low pH you have lots of $\ce{[H+]}$ present. This can protonate your amino acid resulting in a positive charge overall. The inverse is true for high pH (low concentration of $\ce{[H+]}$)

My question : How can both terminals get protonated to result in one charge? Can you explain the chemical process of how each NH2, COOH terminals getting protonated and deprotonated?

Both terminals will not be protonated (under reasonable conditions). To start off with you have a single positive charge and a single negative charge. This is your zwitterion with no net charge. If you have lots of H+ you will protonate, removing a minus. So overall you have a single positive charge.

At physiological pH (7.4) the amino group (which has a pKa of around 9) is protonated.

I watch this video, things became much clearer. Isn't pH 7.4 considered to be in neutral range?

By increasing the pH you are actually increasing the concentration of hydrogen ions present - you meant higher pH refers to decrease [H+] and increase [OH-] present (-log...)? (e.g. pH 12 vs pH 2)...

Look for the $pK_a/pK_b$ values for each amino acid. When the pH of the solution is lower than both $pK_a$s, the amino acid will be in its protonated, non-zwitterionic form. When the pH is greater than both $pK_a$s, the amino acid will be in its deprotonated, anionic form. When the pH is between the two $pK_a$s (around neutral pH), the amino acid will be in its zwitterionic form.

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How to compute the NMR shift in Gaussian09 for only one specific atom in a bigger unit cell?
I am trying to compute the NMR shift for a specific atom in the middle of the surrounding crystal unit cell (1 surrounding anion sphere and 1 cation sphere) in Gaussian09. I only need the chemical shift for the middle atom.
How can I only compute the shift for this atom, and leave out the rest? And would it even save any computation time if I did?

Short answer, you can't.
It might save a bit of time, but probably not much. The calculation would still need to go through the normal SCF on the entire system to calculate the NMR shifts.
In principal, in evaluating the property calculation, you could save a small amount of time by only computing the one atom, but for most systems, I would think the overall SCF calculation would be vastly greater than computing a few properties.

As far as I understand the very procedure that is used to calculate NMR chemical shifts, that should not be possible. The magnetic field is basically treated as perturbation and a set of coupled-perturbed SCF (HF or DFT) equations is solved to find the response of the density matrix with respect to the field.

Agreed. I would think it's possible to perform the evaluation at one point in space (i.e. one atom) rather than all atoms. It's not implemented in Gaussian or other programs and I doubt it would have much performance savings.


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Looking for a Chemistry Demo Resource Group
I'm looking for some kind of email list, forum, or group that discusses chemistry demonstrations. I'm working on an improvement to a demo that's been used for a few decades at my institution and I'm looking for a other few people (that aren't at my institution) to try it out and give feedback. Any ideas?


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How to temporarily hide hydrogen atoms in Molden?
I am visualizing some molecules with Molden. Is there a way to hide hydrogens, without deleting the hydrogen entries from the structure file?

I don't see any such feature in the manual or command-line flags. Offhand, I'd say no.

@Geoff It has and had this option for a while at least. I checked the history and it came sometime in Molden3.2 (added a button "H" to undisplay hydrogens).

As long as you are not using an ancient version of Molden, this should be no problem. (The oldest one I found in my software archive was Moleden5.1 from January 2014 and it already contains this feature.)
In the Molden control window there is a button with an italic H. This will open a menu with "Excl. All Hydrogens", "Excl. H except H-bond", "(Re)calculate H-bond", "H-bond parameters", and "Opt. H positions". I'll leave it to you to play around with it, below you find a small animation of the first two modes for the base pair GC.

And because it is a little bit too tiny in the gif, here is a crop of the Molden control window.

AFAIK there is no such an option in molden. Take another (free) program (f.e. molekel)

This is false (and likely was false at the time of answering), see my answer for how to do it.


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homogeneous mixture - Are they truly uniform in composition and proportion?
If 5ml alcohol is dissolved in 1 L of water ( the purpose is to highlight the low quantity of one of the liquids) , how is it possible for 5 ml to evenly dissolve in 1 L of water ? There is not enough of alcohol to sit between water molecules , isn't it ? So how is the mixture's proportion/composition is uniform ?
Or is there anything that I miss to understand in the term - composition?

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Your question is one of my favorite question. The concept of microscopic homogeneity and macroscopic homogeneity can explain this. Like if you add sugar and sand together and mix them well, you can see a homogeneous mixture but if you take a random sample you will find more sugar than sand or vice versa and even if your sample size is so small that it contains only one particle than it might be either sand or sugar. So the consistency of your system depends on the sample size. It has statistical significance also. In statistics if you want to make an average of large pile of data (Whatever data you have), you have to make sure that the sample you are analyzing is a true representative of the whole pile. In statistical mechanics microscopic description is used to find out the macroscopic behavior and so this simple question has lot of potential. Now this very concept is also used to find the freezing point depression or colloidal behavior.

Thank you.Quoting you "So the consistency of your system depends on the sample size. ". Thank you for this , this is exactly I was thinking as well. However , there is a difference in the way a homogeneous mixture behaves (macro scope) right ? I am more curious to know what enables asugar or salt to dissolve in such a way that it is not heterogeneous.

Because salt dissolves as separate ion and in case of sugar it dissolves because of weak van-der-waals type forces between polar solvent and apolar solute.

Maybe it is just a question of what you mean by "evenly dissolved". True enough, in your example, the proportion of alcohol to water is not "even" - rather the solution is indeed mostly water containing only a little bit of alcohol. But for a true solution such as this, the alcohol is dispersed evenly (uniformly) throughout the bulk of the solution, so that if you repeatedly withdrew a sample and analyzed it, you would get the same result no matter which part of the liquid you sampled.
That would not be the case for oil and water - they would exist in separate phases and which phase you sampled would give a different analysis result.

Thank you for the response. Quoting  you - "so that if you repeatedly withdrew a sample and analyzed it, you would get the same result no matter which part of the liquid you sampled." - The question is , how is that possible ? Lets increase the volume of alcohol to 250 ml. There would not be sufficient alcohol to spread over 1 L. Could you provide a specific measure that you could analyze - such as concentration ? Wont it depend on the sample as mamun described below?

If not , what property of bonding between alcohol and water molecules( not the chemical bond since its mixture but the weak bonding that exists between molecules)  results in same result in an analysis of concentration ?

(1) I agree that if you combine 1 liter of water and 250 ml of alcohol, the resulting solution will probably have a volume greater than 1L (note that it is not always true that when you dissolve A in B you get an increase in volume).  The alcohol present will distribute itself evenly throughout the mixture.

(2) To answer your question about bonding, the alcohol and water molecules are mutually attracted due primarily to "hydrogen bonding" - where the hydrogens on water or on the OH group of alcohol are attracted to the oxygen on another water or alcohol molecule. Oil and water, on the other hand, do not hydrogen-bond together, so they separate into two phases.

(3) Yes, sample size is important, as Mamun points out in his answer. If you took a series of extremely small samples of a water-alcohol mixture, you could get inconsistent results - some would have more alcohol and some less. Or, if you had an extremely dilute solution of alcohol in water, even if you took normal sized samples, you could see a variation in the amount of alcohol present in the samples.


text:
Why isn't the silicon analogue of benzene flat?
Why isn't the silicon analogue of benzene flat? As far as I understand, silicon and carbon make similar structures and it doesn't make sense that its benzene analogue should be substantially different.

Quoted from the article:

Chemists in the UK have constructed a structural analogue of benzene made from silicon atoms. The molecule is not flat like benzene, but it reveals a new type of aromatic stabilisation.

This 'new' type of aromatic stabilisation is said to occur only around the seat of the chair structure.
I am thinking that silicon's bonds are too weak to be kept in a double bond, and therefore it doesn't have enough energy to change the shape of the bonding and is thus quite unstable.

Well, for starters, I wouldn’t consider the two silicon atoms outside of the plane to take part in aromaticity, because they have four bonds each. So the actual aromatic part would be only four silicons and flat.

@Jan Yep. Hence my comment on the seat of the chair structure.

It doesn't look like new type - 6 electrons in flat ring...  True hexasilabenzene would be unstable as Si-Si pi bonds  are weaker than sigma.

strongly related http://chemistry.stackexchange.com/questions/28702/how-come-there-arent-any-silicon-analogs-of-alkenes-or-alkynes

At some level, an equivalent question is why silicon does not form a graphite-like phase. It really likes those tetrahedral bonds and will accept no substitutes! Note that the 'chair' structure is present in the diamond cubic structure.

The linked article is a short summary of a Science article: "A Tricyclic Aromatic Isomer of Hexasilabenzene" by Kai Abersfelder, Andrew J. P. White, Henry S. Rzepa, and David Scheschkewitz; Science 2010: 327 pp. 564-566.
They propose a new term, "dismutational aromaticity":

The highly dispersed $\ce{^29Si}$ nuclear magnetic resonance shifts in solution ranging from +125 to −90 parts per million indicate an inhomogeneous electron distribution due to the dismutation of formal oxidation numbers as compared with that of benzene. Theoretical analysis reveals nonetheless the cyclic delocalization of six mobile electrons of the $\pi$-, $\sigma$- and non-bonding type across the central four-membered ring.

They point out that while $\ce{Si=Si}$ bonds were ignored for a long time, multiple compounds have been prepared, even including delocalization, like 1,2-disilabenzenes.
They prepared $\ce{Si6R6}$ with "Tip" or 2,4,6-triisopropylphenyl groups.

The surprisingly stable 3a can be exposed to air for hours as a solid or for minutes in solution without detectable changes.

Long story short, there is delocalization around the central 4-center ring:

Note that there is a 1,2-shift occurring, and that 2 $\ce{Si}$ atoms have two substituents, 2 have one substituent, and 2 have no substituents.
The paper argues that a fully planar hexasilabenzene might be possible, but:

the choice of precursor may be important in any synthetic approach.

Presumably the choice of substituent is also relevant to prevent the rearrangement here.


text:
Phenyl group in proton NMR

Suppose you have this spectrum and the molecular formula of $\ce{C10H12O}$
The structure is

My problem is with the peak at ~7 which is an H-Benzene signal, with quartet splitting and a relative area of 4.
Surely this means that:

The benzene would have only one Hydrogen environment
with 4 Hydrogens in that environment
and 3 hydrogens on the adjacent carbon atoms

But the answer has 2 H-Benzene environments.
How come this goes against normal? Please can answers be not too technical as I am a pre-university student

Pleas use the \ce command when formatting chemical formulae. For more information on correct formatting please see here and here.

@bon i have done that now

As a starting note, its not a quartet, its a doublet of doublets. I'll be back in a bit after dinner to write up an answer if someone doesn't beat me to it.

@Jan please explain things as if I am an idiot in your answer!

@Cobbles Good luck. Btw you don't actually need to analyse the splitting pattern of that peak cos they give you the 13C data as well, but it is interesting though :D. (also it's not on the syllabus but you probably already know that).

Guys, guys! Getting a little off-topic! Please come to the chatroom to further continue the discussion, if you wish.

You asked to keep it simple, which is why I’m skipping explaining some details. Check the bottom of the answer to see the parts with superscript numbers explained.

The signal at $7\,\mathrm{ppm}$ is not a quartet. It would only be a (typical; there are exceptions, however not in typical organic chemistry) quartet if both of the following two conditions are met:

The signal ratio is $1\,:\,3\,:\,3\,:\,1$ and
The distance between all four signals is equal (they have the same coupling constant[1])

Both reasons stem from the way quartets are created due to coupling.
In your case, the signal is approximately $3.5\,:\,6.5\,:\,6.5\,:\,3.5$ (violating the first condition) and the distances are (enlarging the image on my screen and measuring with a ruler, so only a very rough approximation) $\mathrm{7\,mm, 12\,mm, 7\,mm}$ (violating the second condition).
What you are in fact observing is a pair of doublets. (It could be a doublet of doublets[2] as suggested by bon, but it is not.) You can think of it as two signals, each being a doublet, that are very close together. And that fits perfectly with the theory: Each signal would have an integral of $2$ and would correspond to one set of aromatic protons:

The two on the aldehyde’s side at lower field (slightly higher shift)
The two on the ethyl group’s side at higher field (slightly lower shift).

They are doublets because each one couples with one hydrogen of the other group. All that is left to explain is why one of the two signals of the doublet is higher than the other, creating a quartet-like effect. This is due to an effect called roofing or tenting. The bottom line about roofing is that it occurs if two coupling protons have close chemical shifts, and that once you have identified its existence, you can basically ignore it (it doesn’t give you any information that you couldn’t have derived otherwise).[3] You can observe the same effect in the triplet at $\mathrm{1\,ppm}$ and the quartet at $\mathrm{2.5\,ppm}$: The right-hand side signals of the quartet are slightly stronger than those on the left-hand side and vice-versa for the triplet.

[1] As you know, coupling stems from two protons being close, and connected via a chain of bonds — usually three for hydrogens. The reason that nuclei couple is because they have a spin. For hydrogen (and carbon), the spin can be ‘up’ or ‘down’. Consider two protons next to each other: Their spins can point in the same or in the opposite direction. Both cases have a slight difference in energy which gives rise to slightly different chemical shifts — the coupling you observe.
For reasons that I won’t go into detail into here, this difference does not depend on the strength of your NMR magnet (while the difference between the signals of different protons does). The ppm-scale has been chosen to remove the differences between different proton’s signals (so $2\,\mathrm{ppm}$ will always be $2\,\mathrm{ppm}$) and therefore the coupling splits will vary with the strength of the NMR magnet, which is why they aren’t given in ppm, but rather in Hz.
If you have multiple protons of the same type, the coupling value (one says: coupling constant) will always be the same, so $7\,\mathrm{Hz}$ three times in the event of a true quartet. The reason why the signals then need to be $1\,:\,3\,:\,3\,:\,1$ lies within Pascal’s triangle: Starting from a central signal split it into two halfes with a certain distance from the centre: You will get two $1\,:\,1$ signals. Do the same thing with each signal and the same distance: You will get a $1\,:\,2\,:\,1$ signal, because the two central ones will overlap. Do the same again: Voilà, $1\,:\,3\,:\,3\,:\,1$. So all true quartets that stem from this will obey the $1\,:\,3\,:\,3\,:\,1$ rule.
[2] If you have two different protons on a neighbouring atom instead of two identical ones, you will receive a doublet of doublets instead of a triplet. Think of it this way: One proton has a coupling constant of (say) $\mathrm{5\,Hz}$ with the proton of interest, and would give a doublet. The second one has a coupling constant of (say) $\mathrm{7\,Hz}$ with the same proton, and by itself would give a doublet again. Together, these two doublets would not meet in the middle as in the above $1\,:\,2\,:\,1$ case, because they have different distances. Instead, you will see a $1\,:\,1\,:\,1\,:\,1$ set of four signals, where the distance between the first and the second corresponds to $\mathrm{5\,Hz}$ while the distance between the first and the third corresponds to $\mathrm{7\,Hz}$.
But this doesn’t work here, because we are only coupling to one proton (and I would estimate a coupling constant of $\mathrm{19\,Hz}$ for the second coupling which is impossible in the structure given, and isn’t seen anywhere else in the spectra. Estimate was based on the aromatic coupling being $\mathrm{7\,Hz}$.)
[3] Two spins pointing in the same direction are different in energy from two pointing in different directions I just pointed out. However, if the chemical shifts of both protons that couple are similar, the difference is also noticeable without an external magnet, so that even without the sample being in the NMR magnet the two energy levels would differ — hence the protons are slightly more aligned per se than even distribution would suggest. This in turn means that one of the sides of the signal (corresponding to an energy difference) shows stronger absorption than the other, and because it is always the inner one, one can assume to see a roof — hence roofing.
The only information that roofing can give you — that two protons are close by in the molecule and their shifts are similar — can be derived from other information in the spectrum as well (information that corresponds to numbers! Real, mathematically physical hard numbers! Hence preferred by scientists): The proximity by the coupling constants and 2D-COSY and the similarity of the shifts … well, science doesn’t care too much.

Is there any way to tell the difference between a doublet of doublets and two doublets? They look basically the same to me. You state that it is not a doublet of doublets (and I believe you) but how would you know this without the structure being given?

@bon: For one if the coupling constant is ‘out of range’ (as would be here). Also, a doublet of doublets needs to couple to two distinct protons in COSY. And finally, most doublets of doublets are 1:1:1:1, not fitting this pattern. A beautiful example of a true doublet of doublets can be seen in styrene systems.

@Jan thank you very much for a great answer keeping the more technical stuff aside but still available

Roofing or tenting are the two common terms used for the asymetric populations of multiplet transtions. Roofing can give you two very important pieces of information. First, which side of the peak the coupled partner lives - the roof peak leans towards its coupled partner. Great when looking for coupled partners in busy spectra. Second, they can provide a better-than-rough approximation of the chemical shift of the coupled partner. The roof slope is a function of chemical shift difference, and this can be extrapolated using some simple geometry to locate coupled partners.

Ah yes, no disagreement, but a good example of information that is readily available from a basic 30second 1D proton without having to run further experiments. Too often we resort to 2D to gather data that is readily accessible in 1D. A +1 answer, by the way.


text:
Are there compounds that melt under their own weight?
Is there a compound that can melt under its own weight? That is, a block of such material remains solid when put horizontally but starts to melt when put vertically or remains solid in microgravity, but melts in normal gravity?
The question is motivated by a documentary which I saw on TV some years ago where they claimed that a metal block (indium or its alloy) melts under its own weight. It is suspicious though because it was quite a crank documentary which featured alleged "red mercury" for instance.

Are you mixing plasticity with phase change?

@Mithoron I am asking exactly about melting. Plastic deformation can lead to rise in temperature though which in turn can lead to melting. I wonder whether there are such compounds.

Then it's turning of potential energy into kinetic with temperature rise -  you only need  big enough amount in right shape and high enough starting temperature.

@Mithoron so are there compounds that exhibit this property, particularly with blocks under 20 cm in length?

By big enough I'm thinking rather about kilometers

Sure. Even discounting non-equilibrium conditions, any substance which expands upon freezing will by symmetry melt under pressure, even if the pressure is caused by supporting its own weight. There's a simple example: water! A sufficiently tall self-supporting block of ice standing on a flat surface will create enough pressure at its bottom that it could melt in temperatures between approximately $-21.9\ ^\circ \mathrm{C}$ and $0\ ^\circ \mathrm{C}$, according to the phase diagram for water. This has very important implications for the study of glaciers, for example. However, this process requires a very large amount of pressure, and thus a very tall block of ice. Using an approximate value for the freezing point depression under pressure of $0.0072\ ^\circ \mathrm{C\ atm^{-1}}$, it would take an ice block about $7.5\ \mathrm{km}$ tall for the base to melt at a temperature of $-5\ ^\circ \mathrm{C}$ or higher.
I thought indium was among the substances which expanded upon freezing, but it actually turns out to contract by 2.5% when frozen, so this mechanism would not work (though there could be something else to consider).  Some other materials which could theoretically melt under their own weight in some regime of temperatures and pressures are gallium, bismuth and silicon. There might be some very specific conditions under which this effect could be achieved in a column of around $20\ \mathrm{cm}$, though pressure usually only slightly alters the freezing point, so it's likely that the material would have to be kept only slightly under its melting point at standard pressure for this to happen.

Brilliant answer! It really brings out the inevitable symmetry inherent in thermodynamic equilibria. (+1)

Don't forget germanium and antimony as elements that densify on melting.  Not that it helps for this question...