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Atomic properties
This may be a very broad question.
I always asked myself how scientists manage to find out the exact molecular structures of for instance water or carbohydrates. How do they know? How do they know the angle between the bonds in water? Is there a classical example demonstrating and substantiating current knowledge on molecular level of structure?
The if we go to atoms, how do people know what charge an ion has? How can be know we have atoms with 7 protons for instance?
Probably there are zillion ways required to find out, but could you give me some general examples?

An answer discussing a bit of chemical history might be interesting. It took us a surprisingly long time to figure out even the atomic composition of simple molecules such as water (John Dalton, for example, proposed in 1808 that water was $\ce{OH}$). Once we got a firm grasp on the concept of the mole, we started identifying the simplest compounds via simple mass balances. Then came electrochemistry, the quantum mechanical model of the atom, x-ray crystallography, and computational methods, and many many others in between, building upon and cross-validating the information we had.

Nuclei charge can be measured through Moseley's law. The concept of bond length is slightly subtle. It can be defined in various ways $r_e$, $r_o$, $\langle r \rangle$, etc, which are measured by rovibrational spectroscopy, electron differaction, X-ray spectroscopy, etc. Bond angle can be defined/measured similarly. http://www.chem.helsinki.fi/~sundholm/winterschool/lecture_notes_2010/Stanton_1b.pdf p27 "What do we mean by “molecular structure”?" discussed some subtlety of the bond length.

The title seems a bit different from the content of the question.

First it is very important to know that molecules don't freeze at one geometry and stay there forever.  Even at absolute zero temperature, a molecule can go through any geometry imaginable, and what you call molecular structure is one of the most likely structures, which would be more accurately termed equilibrium structure.  Even more sophisticated definitions are also present, depending on wether you measure the vibrational average or thermal average or the actually most statically likely geometry, but I believe this is beyond the scope of this question.
With finite temperature this is more of a problem, in many cases the notion of a single structure may be completely inappropriate.
There are a lot of ways to determine molecular structure.  However, they all have limitations.  You may be surprised to learn that for many molecules, the structures are not yet determined.
Various experiments can provide information regarding equilibrium structure.   For small molecules, microwave spectroscopy can be used to get geometry information.   Other spectroscopy methods that can achieve high enough precision to resolve rotational states can also be used to obtain a direct measurement of molecular structure.  These are especially useful if the molecule in question is difficult to prepare or unstable, such as reaction intermediates, ions and free radicals. Examples include Slow Photoelectron Velocity-Map Imaging, cryogenic IR spectroscopy and cryogenic ion spectroscopy.  Note that these method only work for isolated molecules in gas phase, and all these are limited to very small molecules.
For larger molecules, if you can obtain a crystal of it, you can use X-ray crystallography to determine the coordinates of atoms inside.  However, it is important to realize that crystal structure of a molecule can be qualitatively different from the structure in gas phase or solution; also, it is not always possible to obtain a crystal.
If you absorb an molecule on a surface, you can use atomic force microscopy to directly measure the molecule. However, similar to crystallization process, absorption and AFM itself may disturb the structure, and you cannot just put anything you want on a surface.  AFM is also limited to the atoms on the surface and cannot see atoms wrapped inside, so it is also limited to rather small molecules.
In solution, the situation is usually much more complicated and there is not even a clear definition of equilibrium geometry anymore.  For example, even a relatively small number of water molecules can form thousands of different water clusters, each having different equilibrium structure; in bulk water that number is astronomical.  There will be no measurement of a quantity if you cannot even clearly define it.
On the other hand, you can use quantum chemistry methods to calculate the geometry of a molecule.  This can be done fairly accurately for a moderately large molecule, provided that the electronic structure of that molecule is not too complicated.
Currently one of the most reliable method of predicting molecule structure is coupled cluster method. For small molecules (let us say up to 100 atoms) CCSDT/CBS is usually describe as "gold standard" and in many cases can provide similar precision with experimental measurements.
For large molecules (for example, up to a few hundred atoms) one usually use density functional theory, which less reliable but in many cases can still be surprisingly accurate.  However, if not used correctly it can easily provide results that are not even remotely correct.
For even larger molecules such as protein, molecular mechanics methods or a mixture of quantum and classical mechanics can be used, which by the way won the 2013 Nobel Prize in Chemistry.   This is, however, so dependent on the details of implementation and expertise of the users that any results must be viewed with utmost discretion.

Is there any analytic gradient for local correlated (can be used for 100 atom system) CCSD(T) in common (can be obtained outside developer) program package?

@user26143 I don't know of any. I guess if such package exist it will have to be MOLPRO.

Seems the current version only has up to LQCISD...

@user26143 It might be coming out in the near future.


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Plastics for food packaging
There are many types of plastics that can be used for food packaging, for example:

Polyethylene terephthalate (PET)
Polyethylene (PE)
Polyvinyl chloride (PVC)
Polypropylene (PP)

What are the differences between them?
Which one is most suitable for packaging pastries, for example, biscuits?

Not for your question but only for the title "plastics-for-food-packaging": some fruit ripen (rotten) faster when in contact with ethylene, which ripening fruits produce. Putting those in airtight plastic bags is not a good idea. That's why some plastic packing for fruits have deliberately made holes.

This question is a little bit broad, but thankfully, there are many resources available.
In answer to part of your question

Which one is most suitable for packaging pastries, for example, biscuits?

The Natural Resources Defense Council webpage Food Storage Containers, they state that the safest plastics are PET, HDPE, LDPE, and PP.
For comparing the various types of plastic, the Plasticology 101 website compares the types of plastic according to:

Temperature tolerances
Clarity
Chemical, scratch and impact resistance
Rigidity
Food contact acceptability
Recyclable material

They suggest that PET, HDPE, PVC, LDPE, PP and PS are suitable for food, but temperature tolerances would be a major factor in what food they store.  Another chemical comparison that needs to be considered for the storage of food is information about the $\ce{O2}$, $\ce{N2}$, $\ce{CO2}$ gas and moisture transmission rates are included in Evergreen's document The barrier performance of common plastic film.
Specifically for biscuits, and any other baked goods, according to sevral websites, it depends on what the ingredients are - things to consider are the dairy content, whether the plastic will react with some of the ingredients etc.

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How can it be certain that some bonds are 180 degrees?
Take acetylene, for example.

It is defined that the two outer bonds are at 180 degrees.
How is it possible to theoretically define all $\ce{C-H}$ bonds in acetylene as being 180 degrees?

With a protractor? — Kidding aside, we have no idea what's your background, what sort of explanation you want (theoretical? experimental?), and at what level. Thus this question can hardly be answered in its present form.

Theoretical, I mean, why can't those outer bonds be, say, 60 degrees?

Again: what's your level of understanding of molecules and their shapes?

Not a whole lot.  I have a working knowledge of the octet rule and the purpose of bonds, but the measure of those bonds is what I do not understand

You may want to read on VSEPR theory… that's the model most appropriate at your level, and it predicts the geometry of a great number of molecules (including acetylene).

I think this question is clear, and shouldn't be on hold. There is a simple and definite answer: acetylene is perfectly linear when you 'freeze' its vibrations, because there is nothing pushing H's to bend. Look at its charge density at http://www.uam.es/departamentos/ciencias/qfa/DAM/acetylene.html

If we look at the molecule we see it is symmetrical. The forces acting upon the 2 hydrogen atoms are identical. As you know, when hydrogen forms a covalent bond with carbon, it's electron is drawn more strongly by the carbon than by the hydrogen, which leaves the hydrogen with a net positive charge (carbon is more electronegative than hydrogen).
Also, the 3 covalent bonds between the carbons are kind of bowed outward in the middle. The 3 bonds between the carbons constitute the most negative part of the molecule.
Naturally, the positive and negative charged moieties are drawn toward each other as much as possible. But, given that the force of attraction is equal on all sides of the hydrogen, it will be aligned equidistant on all sides to the negative triple bond.
I don't know if I am painting a good word picture here, but the forces exerted on the hydrogens place them exactly on a straight line, i.e. their bonds are 180 degrees.
Of course there are other forces in play, namely the forces of neighboring molecules. However, they will also align themselves as much as possible so as to minimize the repulsion of having hydrogen atoms close to one another and to maximize proximity of hydrogen atoms to the triple bonds of neighboring molecules. This arrangement will be preserved as much as possible as the molecules move about. So, the net forces upon the hydrogen-carbon bonds will always force those bonds to remain 180 degrees.

How would this explanation work for a water molecule that has a 120 degree bond angle?

In addition to the two hydrogens, the oxygen atom has 2 pairs of unbonded electrons (lone pairs) that point towards the remaining two vertices of a tetrahedron with the oxygen in the center. Because the lone pairs are less localized than the electrons in the oxygen-hydrogen bond, they take up a little more room, causing the H-O-H angle to deviate away from the ideal 109.5 degrees.

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A chemistry approach to environmentally friendly yoga mats… NBR, NR, TPE, PER
I'm currently seeking your expertise while creating a line of environmentally friendly yoga mats, and I apologize in advance if any chemistry knowledge of mine has been absent since college and I lack the technical capability of this forum. I typically have a scientific approach to problem solving, and I'm hoping I can apply the same methodology to finalizing the material composition of my product.
Instead of blindly accepting the industry suggestions of natural rubber, NBR, TPE, and PER, I wanted to start with the physical characteristics required and work backwards to determine the ideal (perhaps innovative) material that would work best. My requirements are: 1) Eco-Friendly, 2) high fidelity printing compatibility via heat transfer
1) Environmentally friendly: PVC is not an option for the reasons I'm sure you would know. One would assume natural rubber would be the obvious choice, however if it degrades and has to be replaced every 2 years or so, I'd think the long term production effects might actually be more harmful then something more sustainable. From reading an article on PER it seems like it is essentially phthalate free PVC, and although processed different to be phthalate free the vinyl chloride monomer still exists and could be viewed as toxic, although to a lesser degree then PVC. Nitrile Rubber (NBR) is another option but it sounds like although it might be more heat resistant then natural rubber, it degrades similar to natural rubber. Various types of TPEs are also an option, but I've found minimum answers towards their actual eco-friendliness.
2) Heat resistant printing - Sublimation printing has typical been used on polyester, but from what I've researched should be compatible with any polymer. Unfortunately it requires a heat transfer of about 400 deg F, which I believe would scorch/melt natural rubber.
I know its a stretch, but I would love to get your input on the above materials or any other polymer that might be an option. I don't want to blindly commit to an "eco-friendly" material without fully investigating their chemical properties.
Thank you in advance for your help, and I'm definitely open to all of your respected critiques and suggestions.
Regards,
Shane

What's the question?  As it stands this is an open ended opinion question, which is off topic for the site.

I suggest you provide clear criteria. What do you mean by "eco-friendly"?

You will need a natural woven fiber. Any film will not help the skin in contact with it.


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Explaining redox reactions to a 6 year old?
In a similar vein and for pretty much the same reason as I post this question on the Physics.SE site, what would be a an effective means to promote understanding about redox reactions to a 6 year old?  In particular, what (if any) simulations are available?

Why specifically redox reactions? Seems quite a challenge for a six year old to understand even what a chemical reaction is in general.

Long story behind this, but after seeing a 'redox front' in a road cross section, this child got interested in what caused it.

I might suggest you take a look at What's Chemistry All About by Frith, Gillespie and Larkum.  (ISBN: 1409547078)  It is a very approachable book that in my mind pays close attention to the science jargon introduced to its audience.  Redox reactions are described in the section "Let's make things happen" under the types of reactions.  A bit of history about where the name oxidation comes from and the safety symbol that you will see on an oxidizing agent are both included in the brief discussion.
The example reactions given are combustion and firecrackers, which I think are appropriate and accessible to a child.
As for experiments to do, I'll suggest another book, the Illustrated Guide to Home Chemistry Experiments by Thompson. (ISBN: 0596514921) There is a chapter on electrochemistry with two experiments that might be of interest to a young learner: the electrolysis of water to make oxygen and hydrogen gas, and the oxidation of iron.  Of course, there's also the "make your battery" type experiments which are always fun.
As for simulations, I would suggest heading on over to the PHet website and have a stroll through all the activities there.  I don't recall off hand what they have for electrochemistry, but there's plenty to keep a budding scientist interested.

I was taught using the OIL RIG acronym, where oxidation is loss, and reduction is gain. I would also utilise a picture of some sort showing how a oxidant is reduced, and a reducer is oxidised. I would also mention that most oxidising agents have oxygen in them, whereas reducing agents can have hydrogen. Good luck.

Let's speak the Redox with an exempla : Corrosion off a piece of Iron.And Oxygene around you and me. As child of 6 years old had probably seen this, it will directly rely to everyday 's live.
Say this to him : You cannot see the "messy movements" in this Iron because your eyes are too big.......and because you can see the moon and stars.......but not all stars in the universe because we are too small.
The idea off beeing somewhere in the middle off "The Scale" is an important step to start imaging the all around and inside off you. Back the Redox.
The iron, as all the rest is made of 3 sweeties, called Neutron...not an easy name but this is exiting for a child the knonw some specials words. As 3 new words is fine at C yers old,continue with Electron, an other one to love and Proton. Those 3 sweeties are bricks to create all you can see, fill and  smell.
Redox is when sweeties are given and received.I can not receive if nothing is given. Both are related and if one does not happens, the other one too.
A child should ask how it is possible to have only 3 sweeties too have all around with différences. More or less you have always those
3 sweeties together exept when speaking e.g. nuclear explosion. Just catch his meaning with powerfull stuffs. Back on redox,now to have Iron you need those 3 sweeties with 25 other teams of those 3 members: E-, P+ and N°. This make 26 teams to make Iron, the very kind Iron can give some Electrons too Oxygene we need to be alive. And the volume off Iron increase as the oxygene who received some electrons must stay around the iron or leave. As they are now togheter, they will stay togheter as long as you do not found and other team off sweeties able to give back electrons too Iron or to stoll the electron owned by oxygene. Use off this ugly stuff called chlorine can make it true.
Next : Explain the exeption off nuclear phenomena ?
Regards.


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When temperature is decreased, why do reactions occur at all?
I admit that my knowledge of collision theory may be lacking, but, as I understand it, when particles collide, a reaction will not occur without overcoming the activation energy.
That being said, as the temperature of the environment in which the collision takes place is decreased, I believe it is logical that the kinetic energy of these particles will also decrease. Hence, to me, it would make sense if none of these particles had the energy required to react.
So, my question is: Why do reactions still occur when the environment's temperature around a collision decreases? That is, shouldn't there be a point (such as the temperature in a freezer, perhaps?) in which the activation energy cannot be overcome at all?

The activation energy varies a lot for various reactions. As long as T > 0 K, there is still disordered kinetic energy in the molecules. But the reason gasoline doesn't spontaneously combust, is that the temperature isn't high enough to overcome the activation energy - a spark is needed. So you are onto something, just remember that the activation energy varies with each reaction.

Collision theory is not perfect and cannot describe well some types of reaction, making them appear to have a negative activation energy. Tunnelling is also a factor in the kinetics of some reactions, so reaction rates might be higher than expected at very low temperatures. If you wait a long, long time, a reaction can proceed to equilibrium even arbitrarily close to 0 K simply by tunnelling effects.

Besides tunnel effect, the kinetic energy of molecules at equilibrium is a distribution of various values (e.g. Boltzman). At low but not zero temperature, there are fewer but still some molecules with high velocity, which could pass the barrier.

Very interesting comments from all of the above - thank you. @Brian perhaps you could post something like that as your answer

On the contrary of what you are saying exothermic reactions on decreasing temperature are more forward favorable.

@AyushPateria While your comment is true, it deals with thermodynamical equilibrium, not kinetics. Even if decreasing the temperature shifts the equilibrium in the direction of the products, the reaction rate may slow down tremendously.

@NicolauSakerNeto Okay :)

Negative activation energies are not physically possible in a unimolecular or bimolecular reaction. They can be observed in complex reactions due to adding/subtracting individual energies in a complex rate equation. Tunnelling will be come the dominant mechanism at low temperatures but only for H/D .

The activation energy varies a lot for various reactions. As long as T > 0 K, there is still disordered kinetic energy in the molecules. But the reason gasoline doesn't spontaneously combust, is that the temperature isn't high enough to overcome the activation energy - a spark is needed. So you are onto something, just remember that the activation energy varies with each reaction.


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Spontaneity of an exothermic reaction when temperature is decreased
How is the spontaneity of an exothermic reaction changed when temperature is decreased?
Assuming the entropy is negative.
Using Gibbs free energy equation, $\Delta G=\Delta H-T\Delta S$, $\Delta G$ should become less negative as temperature is decreased meaning the reaction is less spontaneous. However, using Le Chatelier's principle, as temperature decreases the reaction should shift towards the right towards the products, becoming more spontaneous.
What's wrong here?

They are not directly correlated. The Le Chatelier's principle counts from one equilibrium state, considers the changing due to different condition. The Gibbs free energy counts from reactant (haven't start reaction), considers the amount of free energy to reach equilibrium. They could be related if one starts from Gibbs free energy-> set up van't Hoff equation at spefic condition-> derive Le Chatelier's principle. (I don't remember some details, though)

Here, $H$ and $S$ are constants for particular reactions, so $T$ is the only variable, which is always positive (measured in K). As $S$ is said to be negative, $-TS$ will be positive. Now, $H$ is negative (exothermic reaction). Therefore, we can write $G=H+(-TS)$ as $G=(-)+(+)$, as $T$ increases, negative value decreases, so reaction will tend to be less spontaneous, if $T$ value decreases, negative value increases, so reaction will tend to be spontaneous. So, Le Chatelier's principle is in agreement with gibbs energy. Anyway, you have good thinking in linking concepts.

The key is $\Delta G = -RT \ln K$.
The $T$ specifically is causing your reasoning to be flawed, as it complicates the link between temperature dependence of the equilibrium constant and $\Delta G$.
$-RT \ln K = \Delta H-T\Delta S$
$-R \ln K = \frac{\Delta H}{T}-\Delta S$
In this form you can see that (at least to the extent that $\Delta H$ and $\Delta S$ are temperature independent), the sign of $\Delta H$ (meaning whether the reaction is endothermic or exothermic) determines whether or not temperature change favors the forward or the reverse reaction. You can also see that the sign of $\Delta S$ is irrelavent.
For an exothermic reaction, $\Delta H$ is negative.  If $T$ decreases, $\frac{\Delta H}{T}$ and becomes a larger magnitude negative number, $-R\ln K$ decreases, $R\ln K$ increases therefore $K$ increases and the reaction shifts forward.


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Light sensitive AgF decomposition reaction
Silver compounds such as silver fluoride are light sensitive (as are all other silver halides) so they will decompose when exposed to light.
I'm stuck with a problem : I want to expose silver(I)fluoride to light to then produce elemental silver and fluorine gas - I'm worried that because of the high reactivity of fluorine silver(I)fluoride will be formed again...
So how could I successfully split silver(I)fluoride by light into elemental silver and fluorine gas without the fluorine reacting again with the silver?

If you want to actually keep the fluorine, you could pull the fluorine gas away with a vacuum and have it pass through a liquid nitrogen trap or two (fluorine boils only 8 K above LN2, however, so you'd need a system with excellent heat exchange). Though liquid fluorine is probably not something I would ever like to see without some serious PPE. It could also damage the pump irreversibly. If for some reason you want to split AgF and not keep the fluorine, it's pretty easy, pretty much any substance will react with fluorine gas preferentially to silver, so you can put some Al foil nearby.

and could teflon be used (or any other fluorine-containing 'polymer') to transport the fluorine gas away? (I suspect that it won't react with fluorine?)

Teflon or other saturated perfluorinated polymers such as FEP are indeed resistant to corrosion by fluorine. I don't know how they behave at cryogenic temperatures, though. Do you have the proper equipment and training to handle fluorine? If not, don't try to isolate it. Many people have died or been severely poisoned in the process.

Actually scratch that, what I suggested won't work. Under low pressure, fluorine's boiling point will definitely be lowered below nitrogen's, so it won't condense at all. I should've realized that sooner. It's probably best if you find some references for this sort of stuff, though I don't know where you might find them.

I think that you understood that I wanted to store the fluorine for a long time - it is just to reuse it in another reaction (so just transporting it to another reaction area to then let it react with something else)

Then if the fluorine is going to be consumed, that will shift the equilibrium in the formation of fluorine gas. You shouldn't even need a pump, just connect a flask with AgF to another flask with the reactants, then begin decomposing the AgF. Are you sure the fluoride decomposes with light, though? It should be the least willing to do so among the halides.

You are right about that- maybe the decomposition could be speeded up with catalysts?

I have no idea about possible catalysts. If you have plenty of AgF you could first do a test by putting Al foil in the other flask, exposing the AgF to light, then checking if the aluminium corrodes. If nothing happens then the photolysis may not be working, or the fluorine may be produced too slowly, or it may be escaping, or it may be reacting with something else present in the system (impurities, glass, etc)

Do you know of a way to predict this theoretically (the decomposition speed of AgF on photolysis)?

No idea. You should be able to get a feel for the spontaneity of the reaction by drawing a thermodynamic cycle and using the Gibbs' free energy state function properties, but I don't know how to obtain kinetic information. The reaction speed may simply be limited by the incident photon flux with sufficient energy to drive the reaction, though, and in that approximation all you have to do to increase speed is find a more powerful source of light of the right wavelengths.

As discussed in the comments, the asker only wanted to create some fluorine in situ for use in another flask, with no desire to store it. Chemistry with fluorine is difficult due to its dangerously high reactivity. Do not attempt anything without the proper safety procedures.
I actually still have doubts whether it is possible to obtain fluorine gas reliably from the exposure of silver fluoride to sufficiently high energy photons, and if there is any simple way to produce photons with the required energy. It is well known that the other silver halides easily suffer photolysis, especially for the heavier halides, but reactions which produce fluorine gas tend to be extremely endergonic, and therefore don't happen except in rather severe situations.
In case it is somehow possible to produce fluorine gas directly by photolysis of silver fluoride, then it is possible to transport the gas to another container where it can react, simply by connecting the flasks with a pipe. Few materials are capable of withstanding exposure to fluorine, however, so the containers must all be made of very well dried gas. Unless the asker has access to a very high power light source, I suspect that the rate at which silver halides decompose is quite slow, which may preclude their efficient usage as halogen generators. This is especially true for silver fluoride, which likely decomposes the slowest, and the fluorine formed in very low concentrations could easily be lost by reaction with impurities or residual water before even reaching the target. I do not know whether there are any catalysts which can be added to increase the photolysis speed of silver fluoride. Presumably some semicondutor with a high bandgap, such as titanium dioxide, could absorb the light and cause photoreduction of the silver/photooxidation of fluoride, but the catalyst would be quickly consumed by the fluorine formed.
If fluorine gas can be formed by silver halide photolysis fast enough, then by simply linking it to another flask with the desired reactants should cause the fluorine to diffuse and react, followed by a continuous equilibrium shift in the production of more fluorine in the flask containing the silver fluoride, and its consumption in the flask with the other reagents.


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Which functional group is -CSSH?
If we replace two oxygen in $\ce{-COOH}$ what is the name of the compound we get?

Does that functional group even exist? If so, do you have an example of a molecule containing it?

Yes the compound exists, it is formed when grignard reagent reacts with thioxide.

Is a dithioic acid.
For example, a benzene with two -CSSH hubstituents in 1, 2 positions is a benzene-1,2-dicarbodithioic acid.

Here's a neat reference containing all sorts of information about carboxylic acid analogues contaning heavier chalcogens.

isn't it just a thioic acid? that thio there means sulfur so why is it a dithioic acid instead of just thioic acid?

@Caters, likely because '-thioic acid' would imply replacement of only one of the oxygens with a sulfur, instead of both.


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Difference between sodium ion and a transition metal ion dissolving in water?
In class, I learned that when $\ce{NaCl}$ is dissolved in water, the partially negative oxygen part of the water molecules surround the $\ce{Na+}$ ion and energy is given off (hydration enthalpy)
Likewise, when a transition metal ion like $\ce{Fe^2+}$ is dissolved in water, water molecules surround the ion. However, in this case, we say that a complex ion is formed.
Can someone please tell if there is really a difference in the two cases and if yes, could someone specify them.

I think that in transition metal complexes the donor atom (the center transition metal ion in the complex) has a large amount of ligands and that sodium has much less ligands (however  (Na(NH3)6)+ is a complex)

There is a slight difference between solvated ions and aqua complexes: The nature of the bond.
When solvating a sodium ion, the electron pairs of the oxygen in water never really interact with the sodium in a way that creates a bond of some sort: It's all just electrostatic interactions.
However, when a transition metal ion gets surrounded with these water molecules, the lone pair electrons of the oxygen form a bond with the ion. This is because they interact with the $d$ orbitals of the transition metal (which usually also contains some electrons), thus changing the energetic state of the ion itself. First and second group elements don't have that energy-stabilizing effect because they don't have electrons in the orbitals that get split up by the formation of the complex.
It is not unimaginable that there then exist further electrostatic interactions between the transition metal aqua complex and the surrounding water molecules, thus leading to the solvation of said complex.


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How can normal potentials be explained?
For these two metals, the standard reduction potentials are:
$$
\ce{2e-} + \ce{Cd^{2+}} \rightarrow \ce{Cd}:\mbox{ } E^0 = -0.403 V\\
\ce{2e-} + \ce{Ni^{2+}} \rightarrow \ce{Ni}:\mbox{ } E^0 = -0.25 V
$$
But the simple ionization energies are:
$$ \ce{Ni} \rightarrow  \ce{Ni^{2+}} \mbox{   } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2490 \ce{kJ. mol^{-1}}$$ $$ \ce{Cd} \rightarrow  \ce{Cd^{2+}} \mbox{   } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2500\ce{kJ. mol^{-1}}$$
The difference in $E^\theta$ is greater than the difference in ionization energy of both elements.
How can this be explained?

Can you be more specific regarding your query? Why shouldn't they be different?

If I get asked, for which of these two equations, the normal potential will be higher, how should I go about answering that?

Without looking at result, it's not obviously for me which one will be higher. I have a guess based on this post. There is a choice of filling electron from d$^8$ to d$^{10}$ vs s$^0$ to s$^2$. The prior one will cause much more electrostatic repulsion (like why 3d$^3$ is not the electronic configuration of Sc). However, this argument will predict Ni$^{2+}$ higher than Zn$^{2+}$.

If I go down one period, the smaller $Z/r$ of Cd$^{2+}$ will also make the $E^0$ more positive, which is an opposite trend. It's hard to say how large this effect. From the experimental $E^0$, the downing period expected to be a smaller contribution. Anyway, all these are hand-waving argument, quantitative calculations are needed to justify this "explanation".

I was trying to find reports of calculations but searching for "Calculation of normal potentials" doesn't provide any success.

There are many results for googling "standard standard electrode potential quantum chemistry", though not directly relevent to your question. You may ask your question (somehow make it more technical, like computational chemistry) at CCL mailing list as well ccl.net/chemistry

I now found a theoretical workaround. The standard potential can in principle be calculated from a thermodynamic cycle in which an electron is transfered from a metal to $\frac{1}{2}H_2(g)$. In principle, I believe, all properties along the cycle should be calculable.

You are acting under two false assumptions

Electrode potentials are measured relative to real zero. They are not, there is no practical procedure to measure absolute electron potential. All standard potentials are measured relative to standard hydrogen electrode (well, actually some more pragmatic standard electrode, but recalculated to be against standard hydrogen electrode)
That values got for gas has anything to do with values for condensed matter. Hydration energy can and does shift electrode potentials greatly, as well as energy of crystal lattice.

+1. I'd make one slight amendment to your answer here. You want "Solvation energy," since often electrochemistry is not in water.


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What's the difference between alpha-glucose and beta-glucose?
What's the difference between $\alpha$-D-glucose and $\beta$-D-glucose? Are they isomers? Or do they differ in their elemental composition?

beta D-glucose units makes up the structure of cellulose polysaccharides while alpha D-glucose units makes up the structure of polysaccharides starch.

$\alpha$-D-glucose and $\beta$-D-glucose are stereoisomers - they differ in the 3-dimensional configuration of atoms/groups at one or more positions.

$\alpha$-D-glucose

$\beta$-D-glucose

Note that the structures are almost identical, except that in the $\alpha$ form, the $\ce{OH}$ group on the far right is down, and, in the $\beta$ form, the $\ce{OH}$ group on the far right is up.
More specifically, they are a class of stereoisomer called an anomer. Anomers are capable of interconverting in solution. All cyclic structures of monosaccharides exhibit anomeric $\alpha$ (down) and $\beta$ (up) versions. These differences occur at the anomeric acetal carbon (the only carbon with two $\ce{C-O}$ bonds.
These two forms exist because all monosaccharides also have an open-chain form with one fewer stereocenter. When the chain closes to the cyclic structure, the aldehyde or ketone carbon becomes a stereocenter, and it can do so in either configuration. One configuration is preferred ($\beta$), but both exist.

Open chain form of glucose:

In the presence of acid or base (although water can fulfill this role if need be), the two anomers interconvert through the open form until dynamic equilibrium is established. The mechanism below starts with $\alpha$ in the upper left and finishes with $\beta$ in the lower right. The open-chain form is in the middle.

Just to add, in the L-configuration, the situation is reversed, since you draw the CH2OH below the ring in the Haworth projection. In both cases, the structure is ß when the relevant groups are on the same side of the ring, and α when they're on opposite sides. Had a question on a test that hinged on that understanding. Understanding the concept of the "anomeric carbon" and the "reference carbon" is also helpful.

It's worth noting that while $\beta$-glucose is more stable, the stability of $\alpha$-glucose is higher than one would expect it to be (anomeric effect) and also that the stabilities are reversed for mannose, i.e. $\alpha$-mannose is more stable than $\beta$-mannose.

which anomer is more stable? and why?


text:
Does unsubstituted benzoic acid not show resonance due to steric effects?
My book says that benzoic acid does not show resonance as the carboxylate anion and the benzene ring are not in the same plane due to steric effects.
But there aren't any large groups in the ortho positions to cause steric inhibition of resonance. Then, is the info given in the book wrong? If not, where does the steric effect come from?

I have never heard of this effect occurring with an ordinary, unsubstituted benzoic acid. There is, however, a phenomenon known as the ortho effect that occurs when bulky groups occupy the ortho positions on the ring. In that case, the steric hindrance between the bulky groups and the carboxylic acid moiety forces the carboxylic acid group to rotate out of the plane of the aromatic ring, thereby breaking the conjugation.
One of the consequences of this, which may not be readily apparent, is that this often produces a substantial increase in the acidity of the carboxylic acid, even when the substituents are not electron-withdrawing (viz. 2,6-dimethylbenzoic acid, which is disubstituted at the two ortho positions, having a $\mathrm{p}K_\mathrm{a}$ roughly one unit lower than ordinary benzoic acid, while similar para- and meta-substituted benzoic acids have higher $\mathrm{p}K_\mathrm{a}$ values — that is, are less acidic).
While I've seen references to this effect in multiple sources, I've never actually found a thorough explanation for it. I think it could be argued that the benzene ring is electron-donating within the $\pi$-system (i.e., by resonance), which would only serve to destabilize the carboxylate conjugate base already bearing a negative charge. By rotating out of the plane, the electron-donating effect is removed. Rather than consider the stability of the resulting carboxylate, I suppose one could also consider the stability of the carboxylic acid directly. Here, I'd expect that the reactivity of the carboxylic acid flanked by bulky substituents should be greater, as the loss of $\pi$-conjugation with the carboxylic acid moiety is inherently destabilizing. My reasoning may be flawed, though, so if anybody has some corrections or a better explanation to offer, that would be entirely welcome.
Whether this effect occurs with ordinary benzoic acid or not, I can't be certain. At least one good reason I can think of to doubt that it does is the observed regioselectivity of electrophilic aromatic substitution reactions on benzene rings conjugated to carbonyl groups. They are known to be deactivating (i.e., they withdraw electron density) via their conjugation with the $\pi$ system of the aromatic ring. If they were not coplanar with the ring, the conjugation would be impossible, and the deactivating effect would not be observed.

So, you are saying that the $COOH$ group experiences steric hindrance (and hence, rotates out of plane) when there are bulky groups in the ortho positions, right?

@Ris97, to the best of my knowledge, yes.

@GregE. You are correct in saying that only if there are bulky groups on o-position, than the carboxylic group goes out of plane . "Electrophilic aromatic substitution reaction will take place mainly in 3-position due to the electron-withdrawing carboxylic group; i.e. benzoic acid is meta directing." this from wikipedia suggests meta direction so resonance is present . Also if only inductive effect had been responsible for this , para would have been major . Anyway your answer covers it all . +1

The answer to this is by something known as the first ortho effect, from Wikipedia:

When a group is present ortho to carboxylic acid group in substituted benzoic acid, the steric hindrance forces the carboxyl group to twist out of the plane of the benzene ring. This inhibits the resonance of carboxyl group with phenyl ring which increases the acidity of carboxyl group which was otherwise reduced due to destabilizing cross conjugation. In-fact this destabilizing cross conjugation is also accounted as the reason for decreased acidity of benzoic acid as compared to formic acid.

Source

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


text:
How do I determine the temperature at which a dissolved substance evaporates?
I would like to understand the temperature at which a substance will vaporize when dissolved in a liquid. I have researched this online for hours, but haven't found a conclusive answer. Is it the boiling point of the dissolved substance? I'm attempting to find the temperature at which caffeine vaporizes when dissolved in water or other vegetable glycerin. Many thanks for your help.

Look for phase equilibria and boiling point of binary mixtures. Once you make a mixture, you cannot speak of boiling point of one component, but only of the mixture. And then of vapor/liquid equilibrium.

First, it is a good idea to clarify the meaning of the word vaporize from here:

(Physics / General Physics) to change or cause to change into vapour or into the gaseous state
to evaporate or disappear or cause to evaporate or disappear, esp suddenly
(Physics / General Physics) to destroy or be destroyed by being turned into a gas as a result of extreme heat (for example, generated
  by a nuclear explosion)

Chemists should be careful when using this word because it can be used to describe evaporation, sublimation, and boiling.  Note the third definition, which suggests that the vaporization of matter typically involves a large amount of energy.
Since the mixture you refer to in your question is in a liquid state, I am going to focus on liquid boiling.  (Note, we can say that boiling is the rapid vaporization of a liquid.)  The boiling point of a liquid is the point at which the vapor pressure of the liquid is equal to the surrounding pressure.  The boiling point of a liquid is affected by intermolecular forces such as hydrogen bonding, dispersion and dipole-dipole interactions.
When a solute is dissolved in a pure liquid (your example of caffeine dissolved in water), these intermolecular forces are altered.  The end result is an elevation of the boiling point.  This change can be predicted quite accurately using concepts typically covered in the first year of a college-level general chemistry class.
So, we know now that a mixture of caffeine and water will have a boiling point higher than pure water; however what happens as the solution continues to boil?  I ignored an important point in the previous paragraph; boiling point elevation due to addition of a solute will only occur if the solute is nonvolatile. Caffeine is such a solute.  Therefore, as you boi your mixture, the water will be turned into a gas, but the caffeine will be left behind.  The volume of solvent is decreasing; therefore the solution concentration is increasing.  Once the caffeine solubility limit is reached, it will start to crystallize.
When you are left with a mound of caffeine you can continue to heat it.  As you heat it above about $230\,^{\circ}\mathrm{C}$ it will start to sublime, or in more general terms, vaporize.
For those interested, the experimental procedure for the extraction and purification of caffeine from tea leaves is described at an undergraduate level here.

@bobthechemist, thank you. What would happen if the mixture were directly exposed to a 230 degree heat source as opposed to a gradual heating process?

@user3904 Rapid heating would still result in the water boiling first.


text:
How can I avoid crystallisation of sugars in vinegar at low temperatures?
I produce high density balsamic vinegar, when it's winter and the temperatures are very low, high density products tends to crystallize. think this is a problem of over saturation of the glucose. The main sugars are glucose and fructose the ratio is 1:0.95.
I try to make products that are lower than 70 brix, or 1,35 density and 6% acidity, composed by $\ce{CH3COOH}$ and tartaric acid.
How can I exactly calculate the solubility threshold of this mixture? How can I measure it?
How can I prevent the crystallization to happen?
Thank you very much
I can't sleep at night because of this problem !!!

It is difficult to correctly measure the solubility, especially if it goes to multicomponent solutions, as your vinegar. It is possible to find tables with some components of your mixture, but you won't get the full picture, and it looks as the moment, when the hands-on experience is most important.
Nonetheless, I found some facts for you, namely:
the solubility of a-D-glucose in all the solvents increases with
increasing temperature and decreases with increasing volume fraction
of acetic acid, from Solubilities of {α-d-glucose in water + (acetic acid or propionic acid)} mixtures at atmospheric pressure and different temperatures
Another point is that the supersaturated solution can be stabilized by various means, and I think it already happens, as your vinegar contains lot of "impurities" from the grapes.
As for the second point, how to prevent it, the simplest solution will be to keep the temperature high enough, so that the crystallization does not occur. If this is not possible, the supersaturated solution will crystallize, sooner or later. If you than ask, how to slow down the process, until spring comes, filtration could help to remove the crystallization nuclei.


text:
How does one separate ethanol from acetic acid using aqueous extraction?
Would you add in $\ce{NaOH}$, deprotonating the acid, and then wash the mixture in water, thereby removing the $\ce{Na^+}$ and $\ce{CH3O^-}$ ions in an aqueous layer? But then wouldn't the ethanol be dissolved in the water as well, as it is polar? In general, how do you separate two polar substances from each other using extraction?
If you don't use extraction, how else would you separate the two, aside from distillation?
All answers are appreciated.

Well, distillation is almost certainly the best method. However, since you've explicitly ruled that out, there are only a few alternatives I can think of:

Ethanol can be separated from water with reasonable efficiency (at least on the microscale) by a process of "salting out." The ethanol/water mixture is saturated with $\ce{NaCl}$. The solvation of the salt ions is a highly thermodynamically favorable process, and as that occurs the molecules of water become increasingly less able and available to form hydrogen bonds with ethanol. I've never tried this with acetic acid specifically, but I can't think of any reason why it wouldn't work.
I would imagine with the right choice of eluent, separation by column chromatography using silica or alumina should be feasible.
Fractional freezing may be possible in theory, although I have no idea as to its efficacy for this specific case.

I'm also assuming here that you're looking strictly for methods of physical separation, not chemical (so, strictly as a general example: adding something like calcium oxide to consume water in the generation of insoluble calcium hydroxide would be out of the question).

Take the mixture of AcOH and ethanol and make it super saturated with  bicarbonate,  extract with ethyl acetate to remove ethanol  and then acidify aqueous layer acidic and extract with 5-10% DCM in MeOH to remove acetic acid.


text:
what is a difference between outer electrons and unpaired electrons
I know outer electrons include (n-1)d...+ all n'ss.. But i don't understand, never been told what unpaired electron is supposed to be? is it the same thing.
so in the following electron configuration:
$4s^2\: 3d^{10}\: 4p^3$:
OuterElectrons: $15$
ValenceElectrons: $5$
from what I remember someone told me unpaired electrons are $15$ too for this...which is the same as  outer electrons so I guess it's the same thing. But I am not sure so please help.

Outer electrons and unpaired electrons are not the same. Outer electrons are, as you said, $(n-1)d$ and all $n$'s. But unpaired electrons are different. To understand what they are, you must know what is electron spin (which, I hope you do). If you don't, read on.
Every electron in an atom has a quantum spin state (either clockwise or anticlockwise, nothing else) denoted by $+\frac{1}{2}$ and $-\frac{1}{2}$. These spin quantum numbers have no classical analogue, i.e., you can't relate them to any phenomenon in real life situations that you can visually perceive.
So, you see, every electron is paired with another electron with opposite spin. Together, the two of them fill one orbital, as for example, $2p_x$, $3d_{xy}$ etc. These orbitals are found out by solving the Schrodinger Equation.
Now, whenever an electron in an orbital does not have another electron with opposite spin, it is called an unpaired electron.
The electronic configuration that is given is $4s^2\: 3d^{10}\: 4p^3$. As, you can see, the $4p$ is filled with $3$ electrons whereas its maximum capacity is $6$. $4p$ has $3$ orbitals and so, each orbital is filled with one electron each, as given (according to Hund's Rule). So, these three are unpaired electrons. All other shells are fully filled and so, there are no more unpaired electrons.
So, number of unpaired electrons is 3 not 15.

how do you know 4p has three orbitals

well, I just know. If you aren't convinced, you should know that s has 1 orbital, p has 3, d has 5, f has 7 etc. Learn atomic structure and you will know.

1,3,5,7...hmm..nature sometimes make me wonder things...:D

@MuhammadUmer To be more precise, a subshell with secondary quantum number $l$ is composed of $2l+1$ orbitals. These conditions arise while solving the angular part of the Schrödinger equation for a hydrogen-like atom.


text:
Where do the electrons not in the S1 orbital get this "extra energy" from?
In this video, Sal Khan says that an electron reaches a higher energy state if the electron possesses more energy. So I assume that, say in the case of carbon, the electrons not in the S1 orbital have more energy. Where does this energy come from?

In simple terms: The further apart the negatively charged electrons are from the positively charged nucleus, the higher is the potential energy of said electrons. When increasing the quantum number $n$, you automatically get further away from the nucleus. That is why the $2s$ electrons have a higher energy than $1s$ electrons. (That, and adding that the $1s$ electrons shield some of the nuclear charge.)


text:
Covalent bond deformation
When do covalent bonds of a molecule $M$ extend or compress while no bonds in $M$ are broken? I can conceive of some possibilities but I don't know how common they are:

Temperature changes
Inter-molecular forces around $M$ which may stretch or compress $M$ in various directions
The transition from $M$ into a different conformation isomer (perhaps some stretching and twisting is required to overcome the energy barrier for a large change in conformation?)
Perhaps bonds can store energy like springs?

How likely are the above to happen, and are there other situations which induce bond deformation? Could I also have some references or topics to search on related topics? As far as I know, covalent bonds are relatively rigid and don't deform much.
Edit: one answer. Upon some literature review, it seems that rotation about $\sigma$-bonds and molecular inversion involve temporary bond deformations because the intermediate states (eclipsed conformations for the former, and the planar/linear form for the latter) involve a change in potential energy. After the transformation, the bond lengths revert to the original.

The most simple example I can think of is the interaction between a specific radiation (mostly IR/Near IR); in this particular case bonds are shown to behave like a spring or a Quantum Mechanical object (the second one is way truer than the first). Strictly speaking, when the energy contained inside the radiation $E_i=h\nu$ meets a equal energy gap inside a bond $e^-\to e^{-*}\iff\Delta E=E_i$, the electrons are excited and fly to the next accessible quantum state, stretching the bond.

I'm sure pressure have a lot of influence, IR spectra are the most suitable instrument to see the shift causes from temperature and pressure variation as you stated changing this parameters change the resonance frequence of the oscillator.

I know from coordination chemistry that the bond length of CO changes according to electronic effects, depending on other ligands in the complex.

There's a slight bending in cyclopentane. So, tension arising from cyclical molecule structure must be accounted for. This also includes molecules forced into a plane by a pi bond (ideally, by an aromatic ring)

@JanDvorak am I right to say that "bending" is the alteration of the usual tetrahedral angle between bonds? But does it involve stretching of the bonds?

There are two distinct phenomena, which are somewhat related.

Dynamical effect of molecular vibration

Bonds do vibrate like springs, and as you said they can store energy by changing its vibrational state. Like a string, the more energy stored in the vibrational motion, the larger the magnitude of such vibration.  Different from a spring, such vibrations are associated with discrete quantum numbers, and even at absolute zero temperature, such vibration still possess a zero-point energy and can never stop.

Non-dynamical effect that change the statistical average of bond length

Molecular vibration often (but not always) occurs at an extremely short time frame, and the vibration itself is often (again not always) of a very small magnitude, so when we refer to bond length we often mean the average bond length over time.
    Normally a molecule stays near the bond length where the energy is the lowest.  There can be a few reasons for a the average bond length to change while keeping the connectivity of the molecule intact, with some examples:

Each conformational isomers will have different bond lengths.  Usually this is a small number.  Examples of large bond length change between conformational isomers with the same bond connectivity are those with Jahn-Teller effect.
By changing the vibrational state of the molecule, the average bond length can be different.  Usually, because of anharmonicity, the higher the vibrational state, the higher the average bond length.  Temperature change, pressure change, collision or photon excitation (especially infrared light) are the most common causes of a change in vibrational state.
Through inter-molecular interactions or external forces, the potential energy surface of the molecule can be changed so that the energy minimum happens at a different distance.  In this case, the bond length will change, but the energy stored in the bond will not change.  This can be achieved through external electronic field, or intermolecular forces such as electrostatic force from an ion, hydrogen bonds, charge transfer, electron donator/acceptor interaction, etc.


text:
Why is this chair conformation the most stable and which leaving group is best?

I'm confused about how this chair conformation was constructed and why that particular Cl leaves. Why is one of the Cl's drawn axial while the others are drawn equatorial (it seems like the others could have been drawn axial as well)? And, to reiterate, why is it that Cl that's favored to leave over the other two?

In cyclohexane, the equatorial position is energetically favored over the axial position. So the lowest energy conformer is the one where the most substituents are in equatorial position. (Or rather: Where you minimize the energy according to the A Value).
The axial Cl is favored as leaving group because of the elimination reaction mechanism. Normally, we see $\beta_{1,2}$ elimination in an anti-fashion. Which is what is proposed by your image.

Can you explain more clearly why that Cl leaves as opposed to the others?

As said, bimolecular elimination "likes" anti-periplanar arrangement of leaving groups. And the axial chlorine is the only one, for which is this possible. The other two would need synclinal, or syn-periplanar, which is not that favored.

What does anti-perpiplanar mean?

@BillyJoel, it means there's roughly a 150-180 degree angle between the beta-hydrogen being abstracted by the base and the chlorine leaving group. See the IUPAC gold book entry on torsion angle.

The chair conformation is the most stable due to the following reasons:

The carbons in cyclohexane have sp3 hybridization and hence they tend to have an angle of 109.5 degrees. The chair conformation allows it to have the closest possible angle which would be 110.9 degrees.Due to this the strain reduces and results in a lowered energy. The two chair conformations have the lowest total energy, and are therefore the most stable.

Since cyclohexane's planar form results into bond angle being 120 degrees
(which means a higher deviation from 109.5). Hence the energy of the planar form is higher and it always has the tendency to convert to chair form.

Due to steric hindrance in the axial location, substituent groups prefer to be equatorial and that chair conformer predominates in the equilibrium.


text:
Is it possible to make liquid nitrogen with a bottle of CO2 gas and a freezer?
I really want to make liquid nitrogen(or oxygen) without using dry ice, a specially made freezer, or other chemicals.
I have a mixture of some sponges, rotten grass, sugar and water, which makes pressure in the bottle(produces $\ce{CO2}$). Which temperatures can I expect if I freeze it and then quickly opens the bottle after 24 hours of freezing. Is it possible to make liquid nitrogen out of it?

Short answer, no.
The method you are suggesting here is to use Joule-Thompson expansion to cool the air as it escapes and hope that it liquefies. This is a valid way to make liquid nitrogen and is used in both lab-scale and industrial air liquefaction. Unfortunately, you will be many orders of magnitude off from where you need to be. Plans I've seen for liquefying air on a lab scale require starting pressures of around 3000 psi. That air is sent through a long copper tube where it is rapidly expanded through a nozzle into an insulated vessel to atmospheric pressure (14.7 psi). Even then it does not liquefy. The cooled air is vented through another tube concentric with the first tube, where it chills the pressurized gas entering the vessel. My recollection is that if you start with a full-sized cylinder of nitrogen, after about 1/2 the tank is gone, you will start to condense liquid nitrogen and when the tank is empty, you'll be left with about a half-liter of liquid nitrogen. Those numbers are probably off, but the order of magnitude is correct. (Note: if you try this based on these "instructions", you will probably get yourself killed given the pressures involved).
Nitrogen liquefies at 77 K, or -196 C. Dry ice is only −78.5 C, and my freezer at home sits at about -5 C. I think a PET Coke bottle gets pressurized to 80 psi and a soda keg is rated to 120 psi. As you can see, the temperatures, pressures, and volumes you're likely to be able to achieve at home are off by one or more orders of magnitude.
Sorry.

Complete agreement with craigim for your unhappy lack of liquid Nitrogen and  your containment vessel will lose pressure in your freezer. Your "pressurized bottle" (I am assuming you wanted to use something like a 2-Liter bottle of coke) will start at around ~20 psi (without rupture) after cooling about 20 deg C in a freezer you could expect a ~8-10% drop in pressure. Your final temperature will be perhaps .1 C lower than the temperature in your freezer. You will heat it up more by holding it than you did by releasing your pressure.


text:
Where do things connect in EAS reactions?
If you have an aromatic ring that requires substitution of some molecule, how do you know where this molecule is supposed to substitute on the ring?

In (i) and (ii) how do we select the ring and position that will receive the molecule (HSO3 in (i) and Cl in (ii))? In (iii), how do we know where to attach the I?

http://encyclopedia2.thefreedictionary.com/Orientation+Rules

The position of substitution in EAS is determined by the electron-donating/-withdrawing effects of existing substituents. The effects can be inductive (the electronegativity of heteroatoms polarize the bonds), or they can be resonating (free electron pairs are donated to a double bond in the transition state).
1. rule; resonance effects overrule inductive effects.
If a ring has both a resonating and an inductive substituent, or one substituent with both effects, the resonating effect will control the reaction.
2. rule; donating effects activate the ring (more electrons for the electrophile), withdrawing effects deactivate the ring (less electrons for the electrophile).
Strong donating groups: -OH, -OR, -NH2, -NHR, -NR2, -SH, -SR, -F, -Cl, -Br, -I, -CH=CH-X (X is any donating group,
Neutral/weak donating groups: -CH3, -(CH2)R, -CHR2, -CR3 (R is any hydrocarbon), Ph.
Strong withdrawing groups: -(C=O)R, -NO2, -SO3,
Weak withdrawing groups:  -(CH2)X (X is any atom more electronegative than carbon),
3. rule; activated rings favor para and ortho positions, deactivated rings favor meta position (para and ortho are deactivated).
Deactivated rings can still react, but much higher temperatures are required, and the reaction is much slower.
4. rule; para is favored over ortho.
Because the existing substituent is "in the way" of the attacking electrophile in ortho position.
An Example
In electrophilic aromatic substitution the aromatic ring has a positive charge in the transition state (because an electrophile (E+) has "borrowed" to two electrons). The electrophile can, and will, attack all three possible positions, but this first step is reversible, and any charge on a carbon is highly unfavorable. Therefore, if the charge cannot be placed on in the bottom path in the example below, the electrophile will be ejected quickly, while the other two paths are stabilized long enough for a base (B-) to extract the proton, leaving the electrons used to bond with the proton to the ring.

rule; Oxygen is inductivly withdrawing, but at the same time donating by resonanse. => donating.
rule; Donating => activated ring.
rule; Activated => ortho/para.
rule; both ortho and para are available. => para favored

Your example
Question (i), the -(C=O)R is a withdrawing group on the left ring, while the RO- group has two electron pairs that it can share with the right ring. ==> para position on the right ring.
Question (ii), the -NO2 on the left ring is deactivating for that ring, while -CH2R is weakly activating for both ring. ==> para position on the right ring.
Question (iii), the ring has a strong and a weak withdrawing group, and is thus highly deactivated. Also, as ssavec pointed out, NaI is no electrophile.
Still let us assume that you had some strong electrophile; Both substituents deactivate their ortho positions, while having little effect on their meta positions. Since -NO2 is a stronger withdrawing group, it rules, and the reaction would happen meta to the -NO2 group (ortho to the -CH2Br).

In (i), it is marked correctly, in (ii) Cl will go only to the "right" benzene, not to the nitro- substituted.
In (iii), I do not see any reasonable reaction.

electrophilic aromatic substitution - there is no Lewis acid.
nucleophilic aromatic substitution - not enough activation by electron-withdrawing substituents.
nucleophilic substitution of Br of $\ce{-CH_2Br}$ - not theromdynamically feasible. Maybe under Finklestein conditions. http://en.wikipedia.org/wiki/Finkelstein_reaction


text:
How can I generally determine which bonds are longest/shortest in a molecule?

My task is to arrange the bonds from longest to shortest. Although I have the answer (C,D,B,A), I'm having a very hard time understanding how to derive this answer. Any clues would be helpful, especially those that help me in other related problems.

From periodic table, the atomic radius $r_C>r_N>r_F$. Therefore C>B,D>A. From conjugation effect, we expect nitrogen in B and D are sp2 and sp3. The more s, the shorter. Therefore D>B.

I have my final exam tomorrow and this was on a previous exam. So it's not homework, but I really need to understand the "why" behind the problem.

Is my explanation reasonable?

I'm not sure about the "conjugation effect" part, do you mind clearing that up?

If the orbitals in nitrogen (the near benzene one) and benzene are coplanar, they will conjugate each other. If the nitrogen adopts sp2 hybridazation for s and two p, one p left. That p will be coplanar with the benzene ring. Is this explanation reasonable?

Coplanar atoms are the atoms that have the double bond, and all of the atoms directly attached to them.

Because atomic radii of elements tend to decrease across a period (source), and bond length increases with atomic radius, the bond lengths are expected to decrease in the following order: C (C-C bond) > B, D (C-N) > A (C-F).
The difference between B and D is that in D, the nitrogen is $sp^3$-hybridized, while in B, it is $sp^2$-hybridized due to conjugation of its lone pair with the aromatic ring 2. This results in partial double bond character of the C(ring 2)-N bond, similar to aniline:

More $s$ character leads to shorter bonds, and therefore the C-N bond B is shorter than D.
In summary, the bond lengths decrease in the following order: C > D > B > A.


text:
Nomenclature for stereoisomer example possible wrong?
I understand the principal of the Cis-Trans naming system (at least I think so), until I saw this example in my learning material. My teacher doesn't know either so I thought I would post it here.
This is the molecule:

There are two ethyl groups both on the bottom side of the double bond, so it could be cis right?

Apparently not, the book says this is the answer:

Could someone please explain the reasoning behind this? This particular example is unusual because there are three ethyl groups attached to the double bond so I cannot determine which one should be the parent chain. The book doesn't explain.

Going from right to left...at the three carbon there are two potential ways to go...both lead only to two more carbons in either direction. either one is correct. Don't think that because only one is numbered that is "the" correct parent chain. Just because one is written in condensed form (CH2CH<3) do not think that there is any difference between them. They are identical. Either number is is exactly the same constitutive name. (That is a terrible book)

Also, I notice you mention that there are three ethyl groups on the double bond. You must number parent chains so that the olefin pi bond is in the parent chain...is that is so there is no way to incorporate all the carbons in a molecule into the parent. Furthermore, if the ethyl group on 3 were missing...it still would not be "trans"...this book is archaic and mistaken...organic chemists do not use cis and trans to describe olefins..those are RELATIVE terms. Organic chemists correctly use E and Z to designate alkene configuration.

In my previous (deleted) answer, I overlooked the fact that this molecule has indeed three identical substituents and not two. Therefore, I would consider Greg E's answer as correct.
Also, C3 has 2 identical substituents on the same sp2 carbon, so cis/trans is not possible.
Example from organic chemistry 6th edition Paula Yurkanis Bruice:

So, your book is wrong!

Thank you very much for your answer, I will mark it correct because you took the time to include a source as such.

Unless I'm very much mistaken, this molecule should not be labelled with any stereochemical descriptors, whether they be relative (cis/trans) or absolute (E/Z). The double bond has three identical ethyl substituents, therefore there is no stereoisomerism possible. By way of explanation, imagine if the ethyl group and the hydrogen on carbon #4 were interchanged: the resulting molecule would be completely identical to the one pictured. This is always the case when an alkene bears two identical substituents on the same carbon atom.
To correctly use cis/trans or E/Z nomenclature, it is necessary that there not be two identical groups on the same carbon of the alkene; if there are, the labels are meaningless, since interchanging the two groups results in an identical molecule.

This is such an incredible relief to hear. I spent so much time trying to understand this since it didn't seem to make sense. Thank you very much for confirming this.

Cis/trans nomenclature for aklenes is usable only for the simplest cases, so you can use it safely when there are two hydrogens on the double bond. If you want to be fully correct, use the E/Z, it's the only bulletproof way.
Cis/trans is also widely used, if you have e.g. chlorine on the double bond, which makes the E/Z nomenclature counter-intuitive (but still correct).

So you're telling me that my text book is wrong and that there is no way to classify this specific example using the Cis/Trans system?

In your very example, it does not even matter, because the two substituents on "left" carbon are the same. To put it clear: http://goldbook.iupac.org/C01092.html

Any time there are two of the same substituent on a carbon, that carbon makes the alkene non-stereogenic. Consider any terminal alkene (they have two hydrogens atoms on the same terminal carbon). Then no unambiguous relationship can be made between those hydrogens and the groups on the other carbon.
What was circled in the (Khan Academy lol) video is a cis relationship, but the molecule as a whole is not stereogenic and no diastereomer of it exists.


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Why isn’t azete aromatic?

I’m confused, because it seems like Hückel’s rule is satisfied (two electrons from the nitrogen lone pair and four from the double bonds) and that the molecule should be aromatic. Where am I going wrong here?

The compound is anti-aromatic. While counting the number of π-electrons, you count the electrons which are delocalized over the ring. In this case the nitrogen lone pair is localised and does not participate in resonance. The nitrogen lone pair is in an $\mathrm{sp^2}$ orbital (red) which is orthogonal to the π system (blue):

So, the total number of π-electrons is only four: two from each double bond.
Please read this for a better understanding : http://www.chem.ucla.edu/harding/ec_tutorials/tutorial04.pdf

The rule for lone pairs in this situation goes like this:  you put one lone pair into the conjugated circuit if the atom otherwise has no pi bond into the ring.  Any other lone pairs after you've done that are localized in outward pointing orbitals.  Here the nitrogen already has a pi bond into the ring, so the lone pair is outward pointing and you can't get to the six conjugated ring electrons you wanted.

Wouldn't a better explanation be that the nitrogen atom here with a pi bond can be either sp or sp2 hybridized, and in most cases, such as this one, an sp hybridization would cause too much ring strain, leading to an sp2 hybridization, which makes the nitrogen lone pair localized and thus unable to participate in conjugation? I don't see why theoretically every nitrogen with a pi bond has to have it's lone pair in a localized orbital if something like ring strain is absent, though I'm guessing that's rarely (or maybe even never) the case practically.

Even if sp hybridization were allowed it still would not work.  Pi orbitals can be conjugated into the ring only if they are properly oriented, and when you have only s and p valence orbitals to work with there is only one properly orbital per atom.  So if you have a pi orbital into the ring you're done.

Ahh, I get it now.


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Should the 'a' in the acid dissociation constant (Ka or pKa) be capitalised?
In science, the negative logarithm of the acid dissociation constant is denoted $\mathrm{p}K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{A}$ depending on the source (lowercase "a" or uppercase "A"). Since it is related to the acid dissociation constant, defined as
$$K_\mathrm{A} = \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]}$$
I would write $\mathrm{p}K_\mathrm{A}$ (the LaTeX package chemmacros renders it so, but doesn't give any source it in its documentation). It seems to me that $\mathrm{p}K_\mathrm{a}$ is the "modern" version of pKa, which was written like this in the last century because of the lack of subscript character on typesetting machines.
I cannot find any source or styling guide saying explicitly which one should be used today (like the IUPAC Green Book), and why.

The subscript is used to indicate that the pK value of an acid (hence the a/A) is meant. In German I use an uppercase S (for Säure, German for acid) and in English I use a lowercase a. I do this because of the way the nouns are written in their respective language. But of course I also understand $\text{p}K_\text{A}$, I just never write it that way.
I don't think there's a "correct" way to write it.

Good go-to references for this kind of problem are the IUPAC books (Gold, Green, etc.). I haven't found $\mathrm{p}K_\mathrm{a}$ directly but here you can find their convention for the acidity constant. They use a lower-case letter, i.e. $K_\mathrm{a}$. But I don't think using an upper-case letter would be wrong; I'd say it's a matter of taste.

TL;DR: for the documents written in English use $\mathrm{p}K_\mathrm{a}$ (upright lowercase "a").

Current (2017Q3) chemmacros' documentation (p. 9) includes a reference for how acid-base module handles \pKa[...]:

\pKa: $\mathrm{p}K_\mathrm{a}$, \pKa[1]: $\mathrm{p}K_\mathrm{a1}$, depends on language settings, see section 6.5 starting on page 30. The translations can be adapted. [...]
As you can see the default subscripts of \Kw, \Ka and \Kb are lowercase letters. The literature is inconclusive about if this is the right way or if uppercase letters should be preferred. In textbooks the uppercase variant usually seems to be used while journals seem to prefer the lowercase variant. chemmacros’ default follows the usage in The IUPAC Green Book [Coh+08].

From IUPAC Green Book [1, p. 59]:

The equilibrium constant of dissolution of an electrolyte (describing the equilibrium between excess solid phase and solvated ions) is often called a solubility product, denoted $K_\mathrm{sol}$ or $K_\mathrm{s}$ (or $K_\mathrm{sol}^⦵$ or $K_\mathrm{s}^⦵$ as appropriate). In a similar way the equilibrium constant for an acid dissociation is often written $K_\mathrm{a}$, for base hydrolysis $K_\mathrm{b}$, and for water dissociation $K_\mathrm{w}$.

Additionally, "a" should be typed as an upright letter [1, p. 5]:

Subscripts and superscripts that are themselves symbols for physical quantities or for numbers should be printed in italic type; other subscripts and
  superscripts should be printed in Roman (upright) type.

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen R. E., Ed.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.

I don't know why but \pKa generates pK_{A} in chemmacros..(at least in my version)

@HernanMiraola Which document language(s) have you set (via babel/polyglossia)?

spanish..\usepackage[spanish,es-noquoting]{babel}

@HernanMiraola Weird, I tried \pKa with your settings and it still renders $\mathrm{p}K_\mathrm{a}$. Probably you might want to ask this  as a separate question on TeX.SE, just remember to include MWE (minimal working example).

It might be my old version (4.6), but that's the current version in linux. Thanks..!

@HernanMiraola I have chemmacros 5.8b, so yes, this might be an issue. Still, I'd ask on TeX.SE :)

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


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Reducing temperature on dissolved Pb(NO3)2
I have pretty much zero experience in chemistry whatsoever but I wonder about how much crystal will I get from fully dissolved element when I reduce water temperature.
Variables

at $\mathrm{40^\circ C,\ 171}\pu{g}$ of $\ce{Pb(NO3)2}$ is fully dissolved in water
at $\mathrm{18^\circ C,\ }\ce{Pb(NO3)2}$ - How much is dissolved now? Everything that was not dissolved turned to crystals?

I know that it probably has something to do with this graph right here but I can't turn my mind to right direction. Can someone explain this problem and how can I find solution?

You need to know how much water is being used, not just the amount of the salt.

@NicolauSakerNeto Well since it's fully dissolved I calculated that it is dissolved in 244,28ml of water.

Being fully dissolved and being fully dissolved in the least amount of water are two different things. If you know the solution is saturated (i.e. it cannot dissolve any more of the salt), then you are correct.

@NicolauSakerNeto So what are next moves if we know that it is dissolved with the last amount of water possible?

Assuming that the solution is saturated, meaning that if you add more salt to the solution it will not dissolve, and neglecting eventual evaporation of water, we have a mass of water of about 228g for 171g of salt (assuming solubility of 75g in 100g at 40°C).
Now, when cooling down to 18°C the solubility drops to ~55g / 100g water. This means that in our solution we still have
$$228~\text{g}~\ce{H2O} \times \frac{55~\text{g}~\ce{Pb(NO3)2}}{100~\text{g}~\ce{H2O}}=125.4~\text{g}~\ce{Pb(NO3)2}\; . $$
The remaining 45.6g of the compound will have precipitated. Whether they form nice crystals or not depends on various other things, see this.
Now, if the solution is not saturated (meaning your salt can still be fully dissolved, but there may be room for more in the solution) the calculation is impossible if you don't know the mass of the water at the beginning.


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Order of positive mesomeric effect of halides
I learned that the order of $+M$ effect (mesomeric effect) for halogens is as follows: $\ce{F} > \ce{Cl} > \ce{Br} > \ce I$.
I wonder why this is so?

Can you briefly describe what should than be observable result of such effect?

I am confused why would Florine being more EN and smallest will have the highest tendency to donate lone pair than Iodine.

Found the answer, it is because of size difference that smaller halogens are able to make more effective p orbital overlap with carbon.

But shouldn't fluorine have a lower urge to use its electron pair for resonance due to smaller radius and higher electronegativity and won't the bond me more polarized in case of fluorine as compared to that of iodine, leading to a weaker covalent bond...?

In general, mesomeric effects are greater than inductive effects, as mesomeric effects change the nature of a bond, while mesomeric effects only polarize them.

$\ce{2p-2p}$ orbitals overlaps better than $\ce{2p-3p}$, $\ce{2p-4p}$, $\ce{2p-5p}$. Although $\ce{F}$ is more electronegative, but it is also smaller in size. Both $\ce{F}$ and $\ce{C}$ (next to it) have $\ce{2p}$ orbitals to overlap.

Basically electronegativity is related to ability of atom to attract BOND PAIR electron, so donation of Lone Pair is not concerned with electronegativity. This leaves us to look for overlapping, generally we talk about donating lone pair to carbon atom which has 2p orbital and fluorine too have lone pair in 2p orbital as 2p - 2p overlapping in better than 2p -3p and so on this +M of F>Cl>Br>I.


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Calculating photolysis time of AgF
How can I theoretically predict the time needed to photolyze n moles of $\ce{AgF}$? (with UV) (this is not a homework question)
The photolysis of $\ce{AgF}$ requires UV light - but I don't know which wavelength the UV must be at for the optimal photolysis. Then if the wavelength is known I can calculate the energy of one UV photon of a specific wavelength. After that I can calculate the required amounts of photons required for the reaction
I was thinking of this method:

First I need to find out which wavelength the UV must be at for the optimal photolysis of $\ce{AgF}$ (How?)
Then I need to calculate the amount of energy required for the photolysis of AgF : on wolframalpha I entered this equation: $\ce{2AgF -> 2Ag + F2}$ and it gave me a $\Delta{}H=409.2\:\mathrm{kJ/mol}$

And somehow I must be able to then figure out which lamp I could use and how long I should use it.

Have you read this? I think it will take a very long time until everything is photolyzed.

Unless its initially in very thin film, it will take infinitely long as the silver blocks the ultraviolet and you need separate the fluorine and silver otherwise it just will just react again.


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Nomenclature of Dihydroxyacetone Phosphate
Why the dihydroxyacetone phosphate is called that way when it has only one OH ?

Because it is the phosphoester of dihydroxyacetone. One of the -OH groups was used in the esterification reaction.
See Here

Mar 16 2020 ... Link is broken, any chance to update to a working one? Thanks!


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Why Cl⁻ can't act as bidentate ligand?
$\ce{Cl-}$ is a monodentate ligand. It has four lone pairs then why it can't donate two lone pairs to central atom and act as a bidentate ligand?

A bidentate ligand is one that forms bonds to the metal center through two different atoms, not through two lone pairs. While chloride does have more than one pair of valence electrons, it only has one atom. Thus it cannot be bidentate.
If the chloride shares a second pair of electrons with the metal center, there would still only be one bond site between the ligand and the metal:
$$\ce{[M+]-Cl <-> [M]=Cl+}$$
A bidentate ligand needs two lone pairs on two separate atoms. Consider ethylenediamine (en), one of the classic bidentate ligands: $\ce{H2NCH2CH2NH2}$. Each nitrogen atom has a lone pair that can be used to form a bond to the metal center, creating two bonds from two different sites on the ligand. See the structure of the tris(ethylenediamine)cobalt(III) ion from tris(ethylenediamine)cobalt(III) chloride $(\ce{Co(en)3Cl3})$.

Update:
As Nicolau points out in a comment, chloride can bridge two metal centers. This is not the denticity phenomenon, which is one ligand with two bonding sites to the same metal. This is one ligand with bonds to two metal centers. See the structure of niobium pentachloride (really $\ce{Nb2Cl10}$ or better $\ce{(NbCl5)2}$.

Also of some relevance is that chlorine can make two bonds with different metal centres, forming a bridged compound. Again, this is not an example of bidenticity.

Thanks! Does that means in some compounds chlorine can form multiple bonds with a metal?

@AyushPateria That may be relatively tricky to find (I know of no such compound in ordinary conditions, at least). Chlorine is one of the most electronegative elements, whilst most metals are on the other end of the scale, so there probably isn't a strong tendency for a chlorine atom to donate electrons to a metal to the point where it acquires a significant partial positive charge. Also, atoms outside the second period tend to form somewhat weaker $\pi$ bonds, and will generally find a more stable structure without multiple bonding, often by dimerization.

It is probably best to think that halogens can, in certain situations, display some multiple bond character. A very simple and interesting example is boron trifluoride, $\ce{BF3}$. Boron is quite a strong Lewis acid, and is able to partially stabilize resonance structures where the fluorine atoms donate an extra bit of electron density to the boron, thus creating bonds which are shorter than expected for a boron-halogen single bond. This may also happen in some structures with metals and halogens, as implied by Ben's first equation displaying canonical structures with double bonding.

Shouldn't chlorine with one bond have a negative formal charge?

Chlorine ion is monodentate.For bidentate ligand it should have atleast two bonding sites.In chlorine it has only one donatable lone pair.So it can't act as bidentate ligand.


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Do eta-4 benzene ligands participate in Diels Alder or other cycloadditions?
While refreshing myself on denticity to answer another question, I started to read about hapticity, which is the coordination of a ligand through multiple continuous atoms. A famous case is ferrocene (bis[$\eta^5$-cyclopentadienyl]iron, $\ce{Cp2Fe}$), where all five carbon atoms in each cyclopentadienyl ligand (Cp) are coordinated to the iron center.
Benzene can also be a polyhaptic ligand, forming $\eta^4$ or $\eta^6$ complexes depending on the oxidation state of the metal. In particular, bis(benzene)ruthenium complexes can undergo redox shuttling, where one benzene ligand switches between $\eta^4$ and $\eta^6$.

In the $\eta^4$ complex on the right, the top benzene ligand is now bent out of planarity, implying a break of aromaticity. This compound looks like a good dienophile for a Diels-Alder reaction. The article linked in the Wikipedia page on this switching phenomenon is published in Angewandte Chemie, International Edition, which is behind a paywall for me.

Does coordination of benzene to a metal in the $\eta^4$ mode activate it toward cycloadditions? Because of paywalls, I would appreciate summaries of articles in addition to links to them. My attempts to find articles using Google Scholar led me to articles about the construction of benzenes by metal mediated cycloadditions, and not to reactions of benzene itself.

Coordinated benzene as a dienophile
It would be logical to suggest that an η4-benzene ligand could behave as a dienophile. The reason why benzene is characteristically unreactive, both as a dienophile and as a diene, is because of its aromaticity; if aromaticity is lost, one would expect it to become more reactive.
However, to the best of my knowledge (and SciFinder-fu), such reactivity does not seem to have been achieved. :(
In the rest of this answer I'll try to summarise some of the research that is (somewhat) related to what you were asking about.

Benzene as a dienophile - in general
Benzene itself (and some substituted benzenes) can act as a dienophile under some conditions:

with 1,2,4,5-tetrazines which act as electron-poor dienes (i.e. inverse demand Diels–Alder).1 Presumably, the reaction is favoured due to the loss of $\ce{N2}$ gas.

Under high pressure.2

Intramolecularly, when constrained to be in close proximity with a reactive diene.3

With incredibly reactive dienophiles (cyclobutadienes).4

(This list is probably not exhaustive.)

Coordinated benzene, but as a diene
In ref 4, Inagaki et al. also summarised the ways in which benzene has been activated towards Diels–Alder reactions:

Currently, there are only a limited number of options to activate benzene sufficiently, including coordination of a transition metal complex to benzene, naphthalene with small resonance energy and the use of benzene derivatives with a sterically induced strained geometry [...] (emphasis mine)

The only reference given for the bolded sentence is the same article that Klaus has linked in his answer.5 It describes the reaction of η2-benzene as a diene, with a hypothesis very similar to what you have written in the question:

In the proper coordination environment, both osmium and rhenium can form stable complexes with arenes in which only two of the six carbons are coordinated. Once complexed in a dihapto fashion, the uncoordinated portion of the arene more closely resembles a conjugated diene. We therefore hypothesized that the complexation should increase the propensity of arenes to undergo cycloaddition reactions.

In this reaction scheme taken from the paper, the carbons numbered 1 through 4 are acting as the diene component in a Diels–Alder reaction with N-methylmaleimide:

References

Seitz, G.; Hoferichter, R.; Mohr, R. Benzol und linear anellierte Arene als Dienophile in der Diels-Alder-Reaktion mit inversem Elektronenbedarf. Angew. Chem. 1987, 99 (4), 345–346. DOI: 10.1002/ange.19870990413.
Jarre, W.; Bieniek, D.; Korte, F. Benzene as Dienophile in the Diels-Alder Reaction. Angew. Chem. Int. Ed. 1975, 14 (3), 181–182. DOI: 10.1002/anie.197501812.
Saito, K.; Omura, Y.; Maekawa, E.; Gassman, P. G. The Addition of 1,3-Diphenylisobenzofuran to Cyclooctatetraene and to Dimethyl Cyclooctatetraene-1,2-dicarboxylate. The Benzene Ring as a Dienophile in an Intramolecular Diels–Alder Reaction. Bull. Chem. Soc. Jpn. 1990, 63 (2), 395–400. DOI: 10.1246/bcsj.63.395.
Inagaki, Y.; Nakamoto, M.; Sekiguchi, A. A Diels–Alder super diene breaking benzene into C2H2 and C4H4 units. Nat. Commun. 2014, 5. DOI: 10.1038/ncomms4018.
Chordia, M. D.; Smith, P. L.; Meiere, S. H.; Sabat, M.; Harman, W. D. A Facile Diels−Alder Reaction with Benzene: Synthesis of the Bicyclo[2.2.2]octene Skeleton Promoted by Rhenium. J. Am. Chem. Soc. 2001, 123 (43), 10756–10757. DOI: 10.1021/ja011689q.

I'm definitely not an expert in organometallic chemistry, and I don't know which aspect ($\eta^{\mathbf{4}}$, any other hapticity, or benzene as a dienophile or diene) is more important for you.
Anyway, here's a JACS paper, which describes the the Diels-Alder reaction of N-methyl maleimide with an $\eta^2$-rhenium complex of benzene as the diene.

I was not aware that $\eta^2$ was an allowable binding mode for benzene. This is definitely within the spirit of the question.


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Acid base titration curve
Why we don't get sharp point at equivalence points in case of acid base titration other than the case of strong acid and strong base titration?

In the case of strong acid-strong base titration at equivalence point with the addition of very small amount of titrant (like a drop) the pH change is very large. That means the range over which indicator changes its color is likely to be passed in a drop of titrant, resulting in a sharp end point.
Whereas in the other cases the change of pH at equivalence point is less dramatic and there will be a smaller change in pH at equivalence point. So, the indicator will change its color slowly resulting in a wider end point.


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Bond Lengths in FCC, BCC and HCP Structures
I was wondering what the crystal structures of FCC, BCC and HCP actually look like, and where atoms are placed in their configuration?
Specifically I'm looking at elemental nickel, iron and cobalt in their lowest temperature lattice type. (Below their curie temperature to be exact.)

If you do an image google search for something like FCC you get a bunch of pictures.  Is this what you are looking for?

No, You get a lot of bad diagrams that don't explain much about where the atoms actually are in the structure, and how to figure out their placement from the shape and lattice constants.


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How does the ambient temperature affect the rate of redox reactions in a salt lake?
Many years ago, I used to work as a gold exploration geologist.  One of the main locations was a salt lake in Western Australia detailed in the document WMC Resources Exploration Successes inN Lake Terrains - Applications of Element Dispersion, Kambalda W.A (Carey and Dusci, 1993).
In the article, and in my research at the time, it is stated that

gold disperses laterally along iron redox fronts that tend to occur higher in the regolith profile

Now the area is arid (up to 50C or 122F in summer), but was not always like this, fossil records suggest far milder conditions occurred in the past.
My question is, how does the ambient temperature affect the rate of redox reaction in a salt lake, such as in the example provided?

The salt lakes that occur in Western Australia are touted by the article Geochemical characteristics of naturally acid and alkaline saline lakes
in southern Western Australia (Bowen and Benison, 2009) are unique environments, in that they possess

an extreme and rare
  range of natural hydrogeochemical conditions. Many WA lake
  waters and groundwaters are hypersaline and acidic.

The article discusses observations of effects due to temperature, especially cycles of dry and wet.  The high ambient temperatures would cause increased evaporation, and hence increase local evapoconcentrations of ions, resulting in a lowering of the local pH, storm/inundation events have the opposite effect.  This leads to further observations by the authors that found that

Field observations of lake pH during storm events
  illustrated that fluids that have been freshened due to meteoric input
  are able to regain acidity rapidly. Part of the acidity may be locally
  sustained by ongoing $\ce{Fe}$ cycling and redox reactions that
  generate $\ce{H+}$ ions. The amount of $\ce{Fe}$ increases significantly in the
  more acid waters, and authigenic $\ce{Fe} mineralization (oxides
  and sulfates, e.g., hematite and jarosite) is abundant throughout
  the acid lake sediments.