text:
Galvanic Cells, theoritical voltage
This reaction proceeds in the galvanic cell:
$$\ce{Zn + Pb^{2+} -> Zn^{2+} + Pb}$$
I need the equation for the half-reactions of this cell and the theoretical voltage produced by this cell if concentration of $\ce{Zn^{2+}}$ is $\mathrm{0.1\ mol/L}$ and concentration of $\ce{Zn^{2+}}$ is $\mathrm{10^{-4}\ mol/L}$.

$$Pb^{2+} + 2e^{-}->Pb  E_{0}(Pb^{2+}) = -0.1263$$
$$Zn->Zn^{2+} +2e^{-}   E_{0}(Zn^{2+}) = -0.76 $$
$$[Zn^{2+}] = 0.1 mol/L [Pb^{2+}] = 10^{-4} mol/L$$
$$Qc= \frac{[Zn^{2+}]}{[Pb^{2+}]}$$
$$E_{0}(cell) = E_{0}(Pb^{2+})- E_{0}(Zn^{2+})= -0.1263 - (-0.76) $$
$$ E(cell)= E_{0}(cell) - \frac {RT ln Q_{c}}{nF}$$
Small piece of advice : Tag homework and exercises on these type of questions , read through "Help Center" specifically "Asking" . Try to attempt the question by yourself and show your own attempt of an answer. This site isnt "Solve my homework site " good luck cheers

Thank you, I tried and submit, teacher said it is wrong ,thats why i wrote it here.

Well if you did something post that here , only then people will try to answer i think . anyway this is still in beta so i guess moderators arent here yet .or your post would have been downvoted/removed


text:
Why is Cu2HgI4 filtered while it is hot?
Why is it required that the solution of copper (I) tetraiodomercurate (II), which has a thermochromic property, be filtered while it is hot?

Copper tetraiodomercurate, $\ce{Cu2HgI4}$, is typically prepared by heating combined suspensions of copper(I) iodide, $\ce{CuI}$ and mercury(II) iodide, $\ce{HgI2}$.
The iodides are prepared by (a) reacting copper(II) sulfate, $\ce{CuSO4*5 H2O}$ with potassium iodide, $\ce{KI}$, and (b) mercury(II) nitrate, $\ce{Hg(NO3)2}$ with with potassium iodide.
In order to avoid contamination of the final reaction product with one (or both) of these iodides, it is essential to filter off while the reaction mixture is still hot.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
IUPAC name for a branched alkane
What is the systematic name of this compound:

My try: 3-propylpentane.

You need to take the longest chain first, i.e. the basis of the hydrocarbon skeleton is hexane. It has a ethyl group at position 3, so the IUPAC name will be 3-ethylhexane.

First of all, make sure to find the longest chain. In this case, it is 6 starting from top right carbon to bottom right carbon. Then, number the chain. You will find that the C atom bonded to the branch has the number 3.It is not 4 because the branch should have the lowest locant and so the numbering should start from top right C atom and not from the bottom one. So, the name of the compound is 3-ethyl hex-ane.

The most important thing to do is to be able convert the representation like those in your question to line-angle drawing and then number those carbons from 1 to the end in all directions. Try numbering then completely: For instance, you are talking about this compound. Then it helps to see where every thing is and where there are branches.
In other words:

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How does a catalytic poison reduce a catalyst's activity?
How does a catalytic poison reduce a catalyst's activity, such as palladium used in the hydrogenation of alkenes or any other such catalyst?

Basically the catalytic poison reduces the activity by slowing or inhibiting the reaction by occupying the surface of the catalyst and therefore the surface area available for the reactants reduces. You can get a general idea from the following text.

Substances which themselves are not catalysts, but when mixed in small quantities with the catalysts increase their efficiency are called as promoters or activators.
Catalyst poisoning refers to the partial or total deactivation of a catalyst caused by exposure to a range of chemical compounds. Poisoning may be desirable when it results in improved selectivity (e.g.Lindlar's catalyst) but may be undesirable when the catalyst is rendered ineffective (e.g. in catalytic converters). Poisoning refers specifically to chemical deactivation, rather than other mechanism of catalyst degradation such as thermal decomposition or physical damage.
Poisoning involves compounds which bonds chemically to the active surface sites of a catalyst. This may have two effects: the total number of catalytic sites or the fraction of the total surface area that has the capability of promoting reaction always decreases, and the average distance that a reactant molecule must diffuse through the pore structure before undergoing reaction may increase. Poisoned sites can no longer accelerate the reaction with which the catalyst was supposed to catalyze. Large scale production of substances such as ammonia in the Haber–Bosch process include steps to remove potential poisons from the product stream.

For more info, read the Wikipedia article on Catalytic poisoning.


text:
Why is a reaction more favorable if one of the products is not soluble/sparingly soluble?
I have heard people go so far as to say that the "driving force" of a reaction is the formation of the precipitate. Why is this so? Whilst the process is almost certainly exothermic (I think - please clarify) the entropic effects are surely unfavorable.

If the reaction product is removed from the phase (liquid solution for example) in which the equilibrium reaction is occurring, significant amount of product can be obtained even if the equilibrium constant for the forward reaction is relatively small ($K<1$).
Aside from precipitation, product could also be removed by vaporization, with subsequent condensation (distillation) to collect the product.  Anything that reduces the product concentration in the phase where the equillibrium occurs favors the forward reaction.

Thank you very much. Could you possibly take a look at this question for me: http://chemistry.stackexchange.com/questions/21805/how-is-it-possible-to-quantify-the-strength-of-a-lewis-acid-base.


text:
What distinguishes a Lewis Complex from an Adduct or are they synonymous?
I cannot think of a distinction - the terms seem to be used interchangeably when talking about products of Lewis acid-base reactions. However, I am concerned that I may have overlooked a subtlety. Please can you clarify?


text:
How is it possible to quantify the strength of a Lewis acid/base?
When the $pK_a$ scale is not applicable - when there is no proton/charged species transfer - what should be used to quantify the strength of an acid?

see Gutmann–Beckett method http://en.wikipedia.org/wiki/Gutmann%E2%80%93Beckett_method

$$\ce{A~  + :\!\!B -> \ \ A\!\leftarrow:\!\!B}$$
$$K_f = \dfrac{\ce{[A\!\leftarrow:\!\!B]}}{\ce{[A][B]}}$$
See sections 4.1 and 4.2 in the following reference for more information:
http://goicoechea.chem.ox.ac.uk/teaching/abse/abse_lecture2.pdf
But $K_f$ is specific to an acid-base pair.
A more universal scale is formed by the Gutmann–Beckett method by observing the phosphorous NMR spectrum of triethylphosphine oxide as it interacts with a sample lewis acid.
An Acceptor Number (AN) scale is formed with AN=0 corresponding to triethylphosphine oxide in hexane (P chemical shift of 41), and AN = 2.21 (d-41), where d is the chemical shift of triethylphosphine oxide interacting with the sample. The greater AN is, the more acidic the compound.


text:
Which is the parent chain in Neopentyl chloride? and why?
Which is the parent chain in this structure?
and since all have 3 carbons so should any of them be the parent chain, if not ..
then which should be and why? is it a rule in IUPAC naming?

(source: chemspider.com)

A chemist would say that all 3 methyl groups are equivalent (identical by symmetry).  Therefore they all produce the same chain, and that chain is the longest chain in the molecule.  Consequently, since the methyl groups are all the same, it doesn't matter which one you pick.

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


text:
Why is gold unreactive when only one electron is in the outer shell?
I've been trying to answer my (high school) daughter's questions about the periodic table, and the reactivity series, but we keep hitting gaps in my knowledge.
So I showed that the noble gases have a full outer shell, which is why they don't react with anything. And then over the other side of the periodic table we have potassium and sodium, which have only one electron in their outer shell, which is what makes them so reactive, and at the top of our reactivity list. (And the bigger they get, the more reactive, which is why we were not allowed to play with caesium in class...)
But then we looked up gold, which is at the bottom of the reactivity series, and found it also has only one electron in its outermost shell (2-8-18-32-18-1).
Is there an easy explanation for why gold doesn't fizz like potassium when you drop it in water?
(This question could be rephrased as "What properties of each element decide their ranking in the metal reactivity series?" if you prefer; that was the original question we were trying to answer.)

First off, gold does react. You can form stable gold alloys and gold compounds. It's just hard, mostly for reasons explained by the other answer
The reason bulk gold solid is largely unreactive is because the electrons in gold fall at energies which few molecules or chemicals match (i.e., due to relativistic effects).
A nice summary of some work by Jens K. Norskov can be found here:
http://www.thefreelibrary.com/What+makes+gold+such+a+noble+metal%3F-a017352490

In their experiments, they distinguished between gold atoms' ability to break and form bonds and the ease with which they form new compounds, such as gold oxides. The two qualities are related: To make a compound, gold atoms must bond with other atoms, yet they cannot do so until they have sundered their bonds with neighboring gold atoms.

I think this is a nice succinct explanation. You always have this trade-off in reactions, but in gold, you don't get much energy in the new compound formation, and you're losing the gold-gold interactions.
You can, of course, react gold with aggressive reagents like aqua regia, a 3:1 mix of $\ce{HCl}$ and $\ce{HNO3}$.
If properly done, the product is $\ce{HAuCl4}$ or chloroauric acid.

Relativistic effects account for gold's lack of reactivity.  Gold has a heavy enough nucleus that its electrons must travel at speeds nearing the speed of light to prevent them from falling into the nucleus.  This relativistic effect applies to those orbitals that have appreciable density at the nucleus, such as s and p orbitals.  These relativistic electrons gain mass and as a consequence, their orbits contract.  As these s and (to some degree) p orbits are contracted, the other electrons in d and f orbitals are better screened from the nucleus and their orbitals actually expand.
Since the 6s orbital with one electron is contracted, this electron is more tightly bound to the nucleus and less available for bonding with other atoms.  The 4f and 5d orbitals expand, but can't be involved in bond formation since they are completely filled. This is why gold is relatively unreactive.
If you'd like to see the formulas and math behind this (it's not all that complicated) see here.  Also note that similar arguments explain mercury's anomalous properties.

It's not good to talk about "relativistic mass" as it implies a change in the intrinsic properties that doesn't exist.

Why should the 5p orbitals contract? They are empty anyway, plus they have a node at the core.

@Karl  The $\ce{5p}$ orbital is occupied in gold ($\ce{5p^6}$). Unlike d and f orbitals, s orbitals, and to a lesser degree p orbitals, have an appreciable electron density near the nucleus. As a consequence, in heavier elements where they are strongly attracted by the nucleus they can attain relativistic speeds and experience other relativistic effects such as orbital contraction.  This earlier answer may be useful.

(I meant 6p, sorry.) Ah well, 5p6 belongs to the [Xe] core of the gold atom. The 6s shrinks a lot, and actually lets the 5p expand, so they become of similar energy, which means it can take part in the chemistry. Which is what you say in that earlier answer. I see no reason to say 5p shrinks, it is well below the 4f14 and 5d10 anyway.

@Karl   p orbitals shrink due to relativistic effects, albeit not as much as s orbitals - see here or google something like "relativistic contraction of p orbitals" and see some of the links provided


text:
How is it possible to account for the fact that NH3 is a weak base in water but a strong base in HCl?
I understand that the solvents differ in their acidity but why does this affect the basicity of the solute?

The difference between a weak base and a strong base is that a strong base is completely converted to ions in a given solvent, whereas a weak base is not. Although this depends on the actual basicity of the solute, it is also very dependent on the solvent involved.
In water, $\ce{NH3}$ can react to form $\ce{NH4+}$ and $\ce{OH-}$. However, most of the ammonia remains as $\ce{NH3}$. This is due to the difference in pKas:1
$\ce{NH4+}$ = 9.2
$\ce{H2O}$ = 15.7
Because the ammonium ion is more acidic than neutral water, very little $\ce{NH4+}$ will be present at equilibrium.
In contrast, the relevant pKas for the $\ce{HCl}$ solution are:
$\ce{NH4+}$ = 9.2
$\ce{HCl}$ = -8
This tells us that $\ce{HCl}$ is much more acidic than the ammonium ion, and therefore essentially all of the ammonia will be protonated when dissolved in $\ce{HCl}$. As a result, it is considered a strong base in this case.
1Evans' pKa tables

I'm guessing because ammonium chloride $\ce{(NH4Cl)}$ is ionic and ammonium hydroxide $\ce{(NH4OH)}$ is not. You simply get more dissolution with the former.

Ammonium hydroxide is ionic: it consists of NH4+ and OH-.


text:
Is there a simple and inexpensive way to generate nitrogen gas?
It's fairly easy to generate a significant yield of hydrogen gas simply by reacting zinc metal with $\ce{HCl}$ or aluminum with $\ce{NaOH}$ solution. But is there a simple and inexpensive chemical reaction that generates a significant yield of pure nitrogen gas?

Liquid nitrogen is rather available, I think from welding supply stores.  Some tire stores inflate tires with nitrogen and will have compressed gas available.

@trb456 really had the best answer regarding this question as I don't consider heating or even working with azides as a safe and practical means. The potential $NO$ and $HNO_3$ byproducts are poisonous and so its probably a good idea to provide the means suggested to scrub these gases from the gas produced.

Per Wikipedia, we have this laboratory method:
$$\ce{NH4Cl(aq) + NaNO2(aq) → N2(g) + NaCl(aq) + 2 H2O (l)}$$
Also per Wikipedia, this method generates small quantities of impurities of $\ce{NO}$ and $\ce{HNO3}$, which can be eliminated by passing the gas through potassium dichromate $\ce{(K2Cr2O7)}$ dissolved in sulfuric acid $\ce{(H2SO4)}$
Both ammonium chloride and sodium nitrite are reasonably available chemicals, and the other products (table salt and water) are undeniably safe. Sulfuric acid might a bit tougher to get depending on where you live, and potassium dichromate will be the most difficult given that the use of hexavalent chromium is discouraged (and you should research procedures to reduce the $\ce{Cr2O7^{-2}}$ to $\ce{Cr2O3}$). But otherwise, this looks like the best approach.

You could try a simple two step process. Take some air, burn some carbon-rich material in it until no further burning occurs. Then pass the remaining gas through an alkaline solution to remove the carbon dioxide. The remaining gas will be more than 95% nitrogen.
And you get this with no expensive chemicals.

clever approach. I suppose you could increase the purity by selecting the right material to burn, and controlling the temperature. For $CO$ that might result, are there other inexpensive filtering methods that could remove it?

@docscience There are lots of industrial methods (e.g. passing the gas over copper oxide) and many physical methods (if the concentration is small, activated carbon will work).

How about using an O2 scavenger, depending on the tank volume you could also get mostly nitrogen; also add a co2 scavenger for more effect. I do this at home and it works for me.

One more classic method for $\ce{N2}$ preparation in the lab that hasn't been mentioned yet is oxidative dehydrogenation of ammonia to nitrogen, typically with the help of transition metal oxides, e.g.:
$$\ce{2 NH3 + 3 CuO ->[\pu{500 .. 550^\circ C}] 3 Cu + N2 ^ + 3H2O ^} \tag{1}$$
Reaction with copper(II) oxide is considered canonical and is often listed in textbooks and lab manuals, for example A Text-book of Practical Chemistry [1, p. 6] from 1921 describes this process as follows:

Nitrogen is obtained by passing air first through ·880 ammonia and then over red-hot copper turnings. The oxygen of the air is absorbed by the copper and the resulting copper oxide is at once reduced back to copper by the ammonia. The frequent renewals of the contents of the copper tube are thus entirely avoided.
\begin{align}
\ce{2 Cu + O2 & = 2 CuO},\\
\ce{2 NH3 + 3 CuO & = N2 + 3 H2O}
\end{align}
The gas should be washed with sulphuric acid (1 : 4) to free from excess of ammonia, and dried by strong sulphuric acid. The use of dilute acid to free it from ammonia is to avoid the danger of a great rise in temperature should a large excess of ammonia come over.

Method of obtaining nitrogen via thermal decomposition of $\ce{NH4NO2}$ is criticized as a rather inconvenient one that may potentially result in an explosion.

Small quantities of nitrogen may be prepared by carefully heating a solution of ammonium nitrite (sodium nitrite and ammonium chloride in molecular proportions). [...] The method is not recommended as it is difficult to control, and the nitrogen is generally contaminated with oxides of nitrogen, especially when the action becomes violent.

Instead, one can utilize "volcano demo" (thermal decomposition of ammonium dichromate) as somewhat safer alternative:
$$\ce{(NH4)2Cr2O7 ->[\pu{170 .. 190^\circ C}] Cr2O3 + N2 ^ + 4 H2O ^} \tag{2}$$
These days various sources provide slightly deviating techniques for reaction type (1). Other metal oxides such as $\ce{TiO2}$, $\ce{V2O5}$, $\ce{MnO2}$, their binary and ternary mixtures on silica- and alumina-supported substrates doped with rare-earths (Y, mostly) are capable of selective catalytic oxidation at lower temperatures [2, 3].
References

Hood, G. F.; Carpenter, J. A. A Text-Book of Practical Chemistry; J. & A. Churchill, 1921. (Google Books).
Li, Y.; Armor, J. N. Applied Catalysis B: Environmental 1997, 13 (2), 131–139 DOI: 10.1016/S0926-3373(96)00098-7.
Gang, L.; van Grondelle, J.; Anderson, B. G.; van Santen, R. A. Journal of Catalysis 1999, 186 (1), 100–109 DOI: 10.1006/jcat.1999.2524.

There is one incredible simple, cheap and easy way to make nitrogen gas with near 99 percent purity if you don't need it compressed.  I've used this method in my chemistry experiments when I needed just modest amounts of nitrogen or when I need to do reactions in the absence of reactive oxygen.
Simply buy handwarmers from Walmart or many other sources. Use a heavy duty, large resealable bag and fill it as best you can with regular, uncontaminated air. Drop the hand warmer in there and it will absorb almost all oxygen, leaving you with about 1 percent "all other gases" that are naturally in the atmosphere. Personally, I glue an airtight small rubber tube (a hose from aquarium bubblers, also found at Walmart or any pet store) then clamp it shut with a small but strong spring clamp (although other clamps will due). I've found the hand warmer can dry out before it loses its capacity to absorb O2, so I cut one end open, seal it with laundry clamps (just to keep the contents from spilling out, making a mess) and spray it LIGHTLY now and then with mist from a stray bottle. However, if your not worried about the carbon, iron powder, and table salt contained in these packets (all very harmless) you can cut the packet open and just dump it in the bottom of the bag, and spray as described above once in a while. You would be surprise how long they last and how many bags full of nitrogen you can get from just one hand warmer packet. If you're not worried about adding excess heat to your O2 purged experiment, just put the handwarmer in the bag containing the enclosed beaker. Otherwise, you can purge air out of a different bag, vessel by inserting the hose, and squeezing the nitrogen filled bag.

How do they work? As I can see, they burn charcoal. So the result won't be nitrogen, but an $\ce{N2}$ + $\ce{CO2}$ + $\ce{CO}$ mix.

(+1) This is a nice example of thinking outside the box! I am going to try it and up vote right after this is posted. Thanks!

According to Wikipedia you can generate pure nitrogen by heating metal azides, which gives out
nitrogen and the metal. Sodium azide is not really expensive.
$$\ce{2NaN_3 ->2Na + 3N_2}$$
However, as you may expect, this may not be a very safe reaction as the first reaction that came to my mind when you asked for a reaction with large yield of nitrogen gas is that of TNT, though that would not give a pure yield as you have required. Azides are also explosive. Sodium azide is used in car airbag and airplane escape chutes where this exact reaction is used to quickly inflate these safety features. The metallic sodium formed is then further reacted to less dangerous sodium compounds.

This can be an extremely violent decomposition reaction.  It's the reaction that generates enough gas in an instant to fill an airbag...

Please nobody use this. Not only is sodium azide explosive, it's very toxic.

Even though the reaction and the reactant are rather unpleasant, this is one of the methods that produce purest nitrogen at a lab scale.

It's also interesting that Brauer's Handbuch der präparativen anorganischen Chemie recommends this method exclusively for the nitrogen generation in the lab.

To generate nitrogen you could mix potassium nitrate, hydrochloric acid, and zinc.
It would generate nitrogen and the byproducts will be zinc chloride, ammonium chloride, and potassium chloride.
The nitrogen would be mixed with some HCl vapors, which you could remove through bubbling it in water.
The combustion of dinitrogen monoxide can also produce nitrogen.

Thermal decomposition of ammonium nitrite:
$$\ce{NH4NO2 -> N2 + H2O}$$

Please elaborate more. At Chem.SE, we like answers to include reasoning and explanation, otherwise they are likely to be downvoted and/or deleted.

I guess it's worth mentioning that solid ammonium nitrite decomposes with an explosion.

Ammonium Dichromate reacted with heat produces only one gaseous product, Nitrogen. FIlter through a bubbler to get particles out of it. See Wikipedia under Ammonium Dichromate for the reaction equation and stoichiometry.

It's considered a good practice to include an actual reaction, conditions and original source (use ACS bibliography style) in your question. Avoid citing encyclopedias and link-only answers. Also, chemical names are generally starting with low-case letters unless it's a first word in a sentence.

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


text:
How do atoms of noble gases bond with each other?
Noble gases have full electron shells, which virtually blocks any other element from bonding with it. However, I've heard about cases where they bond to each other  - for example, helium can apparently form a dimer $\ce{He2}$.
How is this possible?

related https://chemistry.stackexchange.com/questions/518/do-noble-gasses-besides-helium-form-diatomic-molecules-at-low-temperatures

Noble gases usually do not form strong bonds between their atoms - it takes a fair amount of energy to dimerise them into excimers, but those are short-lived excited molecules. Thanks to excitation, shells of the atoms aren't closed and they react, but very quickly they lose energy and become separate atoms.
With exception of weakly bound $\ce{Xe2^+}$ cation, stable compounds created from noble gases (mainly xenon) only feature bonds with other elements, typically fluorine or oxygen, but there are also, for example,  organoxenon compounds.
As mentioned in the comments, you may have heard about detection of so called van der Waals molecules of helium, which aren't "true" molecules, but very weakly bound pairs of atoms. In fact helium vdW dimer may have weakest bond even among them, and it was an achievement to observe it.

Noble gases do form dimers under low energy conditions, but they are extremely weakly bound and thus difficult to observe with traditional spectroscopic techniques. The bond strength in dihelium is about 7 orders of magnitude lower than that in dihydrogen. As antibonding orbitals are more antibonding than bonding orbitals are bonding we would not expect these to form. The attractions which overcome this partial antibond are dispersion forces.
Experimental verification of dihelium can be found in this paper.

There's also helium dimer article on wikipedia now

@J.LS Strictly speaking the helium dimer is no molecule in the sense of the definition as it lacks the ro-vibrational mode.

Stable [Xe2]+ ions are known in solution - https://www.researchgate.net/publication/250816500_Formation_of_the_dixenon_cation_Xe2_in_fluoroantimonateV_media_by_oxidation-reduction_methods_Spectroscopic_properties_of_the_ion

The reason why noble gases form molecules is because of London forces, which are very weak. The electrons in noble gas atoms are on the move all the time, so there will be a moment when they will spend more time in an area, which generates a temporary dipole and allows for interaction with another atom.
An addendum regarding noble gas compounds in general: the first noble gas compound synthesised was $\ce{XePtF6}$. When $\ce{[O2+][PtF6-]}$ was discovered by the oxidation of dioxygen with platinum hexafluoride, it was hypothesised that $\ce{XePtF6}$ could be made via the oxidation of xenon with platinum hexafluoride, because of the similar ionisation energies of dioxygen and xenon.


text:
What effects would an electron in the nucleus have?
Under the understanding of quantum chemistry:

There is a very miniscule chance that an electron could appear inside the nucleus of an atom.

What could possibly happen if this "phenomenon" were to occur?  Would it wreak havoc upon the atom and tear it apart?  Or will it bind with the protons and create a temporary vacuum?
Under my current research, the electron would jump out of the nucleus as soon as possible because this phenomenon should not occur for more than a fraction of a second. Though it shouldn't leave without some adverse effects.

In electron capture electron does get to nucleus.

I'm not referring to electron capture, but rather quantum mechanics. The possibility that an electron can appear inside the nucleus.

In quantum chemistry, having two things simultaneously at the same place is not such a big deal.

An electron being at/in the nucleus results in the Fermi Contact Interaction.
This interaction can be observed experimentally through NMR, EPR, and electron capture.
In NMR of paramagnetic compounds, chemical shifts can be far outside the usual diamagentic range. The contribution to chemical shift of the electron density at the nucleus is referred to as the "contact shift".
For more information on Fermi contact, see "Origin and meaning of the Fermi contact interaction" Theor Chim Acta (1988) 73:173-200.

If you do a (simple) calculation of a quantummechanical rotator in a spherical potential — i.e. calculate where an electron would be in the vicinity of an nucleus — you will find that the eigenfunction $\left | \Psi_1 \right \rangle$ corresponding to the lowest energy $E_1$ is dependent only on the distance of the electron from the nucleus ($\left | \Psi_1 \right \rangle = f(r)$ — use spherical coordinates because it makes your job a million times easier!) — which gives us a sphere like the s-orbitals are often drawn.
If you carry on, you realise that $\left | \Psi_1 \right \rangle$ is more or less a decaying exponential function, so its value is highest for $r = 0$. The probability of an electron being at a specific point is typically given as $\left \langle \Psi_1 \middle | \Psi_1 \right \rangle$ and evaluates as $\Psi^2_1$ in this case. Since the function has only one phase, the highest $\Psi_1^2$ value is — in the nucleus.
So the nucleus already carries the highest single probability of having the electron right inside it. It’s not even ‘special’ for the electron to be in the nucleus.
However, check Dave’s answer for the physical implications, which I didn’t even know about!


text:
Halogen bond definition
I was looking for an accurate definition of halogen bonding.
I was able to find quite a few good ones, but none of them would explain if a X---H intermolecular interaction would count as a form of halogen bonding. Where X is any halogen and H is a hydrogen from another molecule. Does this count as halogen or hydrogen bonding? Or maybe as both?
Thank you,

I would say no, it is only a hydrogen bond, not a halogen bond.
To be a halogen bond, the halogen atom must accept electron density from the other member of the bond.  If the other member of the bond is a hydrogen atom bonded to a more electronegative element, I don't see have the halogen atom could be an acceptor of electron density.
If you can get access see "Halogen Versus Hydrogen" Science Vol. 321 no. 5891 pp. 918-919.  Footnote 7 there explains: "A halogen bonding donor is a species that contains an electrophilic halogen that can become a member of a halogen bond. In the literature on halogen bonding, the reader should be alert to how the words 'donor' and 'acceptor' are used. In a complex RX---B, RX is the halogen bond donor but the electron acceptor (Lewis acid); B is the electron donor and halogen bond acceptor (Lewis base)."

Thanks for your answer DavePhD. I completely agreed with what you said at the first glance. But later, I started thinking of the interaction in details. The nucleus of the halogen (chlorine for example) is much stronger at withdrawing electrons than the hydrogen nucleus. That means the chlorine atom is actually accepting the hydrogen's electron (and that's why causes their interaction at small distances). Thinking in this way suggests it is a halogen bond. Unless I have missed something during those deep thoughts! -oh and thanks for the reference btw- :)

If you look at the definition of hydrogen bond quoted in my answer here: http://chemistry.stackexchange.com/questions/21773/why-is-hydrogen-bonding-generally-defined-to-include-only-three-period-two-eleme it requires that the acceptor of the hydrogen bond is "an electron rich region", which directly contradicts the requirement of a halogen bond, that the halogen act electrophilicly.

Yes I think I got what you are saying. All clear now!
Appreciated.


text:
Why is aluminium(III) considered a hard acid if it forms covalently bonded compounds like aluminium oxide?
If something is a hard acid, it tends to form ionic bonds as it is polarizing. So why is the aluminium ion a hard acid when it forms bonds with covalent character. A similar argument follows for why carbon dioxide is considered a hard acid?

The hydrogen proton is the hardest acid, yet I can't think of any ionic compounds with the hydrogen proton as a component. Perhaps you may need to reexamine your rule of thumb.

what is the rule of thumb then?

Short summary - HSAB theory has it's limits of applicability, and this is a good example.
More recent discussion of acid-base theory center on the frontier orbital concept, e.g. W. B. Jensen, The Lewis Acid-Base Concepts (1980) pp. 112-155.
In this concept, an "acid-base" reaction creates new HOMO-LUMO orbitals of the product. So it's about energy matching between the acid (empty LUMO) and base (filled HOMO) (and also symmetry and orbital overlap). It's still very much a Lewis picture, since you have the electron pair on the base and an electron acceptor on the acid.
For example, if we react $\ce{NH3 + H+ -> NH4+}$ we combine the HOMO of $\ce{NH3}$ (i.e., a lone pair) with an empty LUMO on $\ce{H+}$ to create the adduct.
IMHO, this is a nice concept, since the orbital energy match between Al and O is good, you get covalent character to the resulting product. (Now you also get ionic character, but that's a different question.)

The Orbital Molecular Theory explains it very well; the symmetry and orbital overlap defined by the wave functions gives a complete approach. I add some thoughts to the question qualitatively:
The concept of covalent and ionic bonds must be taken carefully when talking of acidity, a property normally refereed to aqueous solution where other interactions (hydration enthalpy) take place.
The $\ce{Al^3+}$ ion is very polarizing, and has a high charge and low ionic radius leading to high effective nuclear charge. In solution the cation must have many shells of hydration, coordinating or polarizing water molecules and attaching $\ce{OH-}$ groups in order to compensate the positive charge, thus increasing the net amount of available $\ce{H+}$.
Similarly, carbon dioxide is a gas and it can be seen as a separated phase on a soda drink (bubbles); so first it must be dissolved in water. The dissolution of a $\ce{CO2 (g)}$ molecule in water increases the acidity, $\ce{[H+]}$, as can be tasted when leaving a glass of soda open in equilibrium with air for a while. I copy here the equations found at this link which explains why carbon dioxide is acidic. If you follow the link, it has a brief explanation:
$\ce{CO2 (g) <=> CO2 (aq)}$
$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)}$
$\ce{H2CO3 (aq) <=> HCO3- (aq) + H+ (aq)}$
$\ce{HCO3- (aq) <=> CO3^2- (aq) + H+ (aq)}$


text:
Why is the electronegativity of hydrogen 2.20?
Well, I was doing research about the ways we can measure electronegativity and I just realized there must be an arbitrary electronegativity set for an specific element to give us the relative amounts of the other ones. Pauling used hydrogen and suggested 2.1 for its electronegativity which later was altered to 2.20. Now, the question is, why was the electronegativity of hydrogen suggested as 2.1 and why is it 2.20 right now? What are the secrets in those numbers?

I might be wrong, but I remember my teacher saying that the Pauling scale was based on the arbitrary value of 4.0 for Florine, not 2.2 for Hydrogen.

Sorry to turn you down, but according to IUPAC, Wiki and many other reliable sources, H is the chosen element for the arbitrary electronegativity. However, in IUPAC gold book the value is documented as 2.1 as a sign of respect for Pauling's decision.

No problem. That's good to know.

Interestingly, the first paper from Pauling on electronegativity "The Nature of the Chemical Bond. IV. The Energy of Single Bonds and The Relative Electronegativity Of Atoms" published in 1932 gives a different scale with H = 0.0 and F = 2.0!
Later, he changed to 2.1 for H and 4.0 for F, e.g. http://scarc.library.oregonstate.edu/coll/pauling/bond/notes/sci5.001.14-notes-01.html
If we look through his book, on p. 90 we find the following text:

Following these there are given the values obtained by adding 2.05 to them...

The resulting numbers are then rounded to one decimal, from H = 2.05 to H = 2.1 and F = 4.0.
His notes are online (e.g., here) but there's no obvious explanation I can find yet.
My hunch is that he started with H = 0.0, and realized that the alkali metals would be negative numbers. The shifted scale gave Li ~ 1.0 and F ~ 4.0, which seem like "nice" numbers.
I tried to hunt down the reason for the move to 2.2 as the standard for hydrogen. The first reference seems to be by Huggins (J. Am. Chem. Soc., 1953, 75 (17), pp 4123–4126) but the paper itself doesn't really explain the reasoning.
Wikipedia suggests a paper by Allred, but that cites the Huggins paper for the 2.2 number.
Presumably more could be found with a very careful search through the papers of Pauling and Huggins, but I've never heard any clear explanations and the documents online don't give them either.


text:
About the meaning of the names of f orbitals
Our chemistry teacher encouraged us to study the history and naming of the orbitals on the web. (actually, our textbook did, but that's irrelevant to the problem) I easily could find the reason behind their naming, and what their names stood for. However, some sources stated that f as in f orbitals stands for fine, while others claimed it to be fundamental. Which of these two is correct? And why did scientists gave f orbitals that name? (as you're aware, they did it due to the f block elements' emission spectrum; my question is, what was the specification of their emission spectrum?) I'm sorry if the question looks stupid, but I can't figure it out myself.

I would have to say "f" overwhelmingly stands for "fundamental" (not "fine").
An early (1915) reference is "Some Recent Discoveries in Spectrum Series" Astrophysical Journal, vol. 41, p.323

Within each of these three systems there are, in general, four different types of series, known as "principle", "sharp" (second subordinate), "diffuse" (first subordinate), and "fundamental" (Bergmann).  These names are not always very appropriate; for instance the "sharp" series is often diffuse, the "fundamental" may be composed of the faintest lines in the spectrum, but this is not vital.

The fundamental series was discovered by Arno Bergmann in 1907, and is sometimes called the Bergmann series. It was named "fundamental" by William M. Hicks, I beleive in A Critical Study of Spectral Series. Part I: The Alkalies H and He (1911): "F is the symbol for the new series (fundamental) as P.S.D. stand for those already known" (page 57) and "It was my first impression that this would be the same for all the alkalies and the .987 of the earlier discussion. It would be the basis of a fundamental series for each
element, and the letter F which has been attached to it had its origin in this idea." (page 94).
The names existed before orbital theory.  There is no need to investigate f-block elements to observe excitations to f-orbitals.  Electrons of any elment can be excited to an f-orbital. In fact, Bergmann discovered the series in the sodium spectrum.  What was orginially known as the sharp series, were transition to s-orbitals, the primary series transitions to p-orbitals, etc.
See The Origin of the s, p, d, f Orbital Labels for additional information.


text:
Why is Graphene So Strong?
There has been a lot of news about Graphene since its discovery in 2004. And as we are all told it is a revolutionary material which is very strong, conductive and transparent; even in some cases it has revolutionized especial laws of physics (see this page).
But what is it about the structure of Graphene which makes it so strong?

If I understood what strong means, then graphene has high strength because of the 2-D layer structure of carbon atoms. The strength can also be attributed to its $sp^{2}$ hybridisation.
Any crystalline material will have a lattice structure. When we mean strength of a material its usually the ability to break apart the bond between two lattice atoms. But the observed strength is usually 2 or 3 orders lower than theoretically predicted strength due to flaws (voids, dislocations etc). But when graphene has a near flawless 2-D structure we can reach its theoretical strength. Though its is a very expensive process to actually realize "flawless". Even with flaws its strength is really high compared to steel.
So its essentially the 2-D one atom layer structure that can be responsible for its high strength.
One important thing is though graphene has high strength and elasticity (ability to get stretched)  it is relatively brittle (fracture toughness ~4 MPa$\sqrt{m}$).

Imagine carbon as some sort of wool.
Graphite would be just some wool without any pressure on it.
Diamond would be a wool ball with a structure and a shape.
Graphene would be a textile knitted with the strings of wool.
Wool is soft, a wool ball is more durable to pressure and a textile could carry some weight.
Now imagine graphene as a canvas fabric knitted with the finest strings the wool has and all strings are equally long. The pressure would be splitted equally on each of this strings and you wouldn't have a weak point. You would carry an elephant with it.


text:
How do sunscreens protect the skin from UV rays?
I was unsure to the largest extent about whether I should post this question in chemistry Q & A or bio Q & A until I just read that sunscreens "absorb" UV rays, not allowing the most of those rays from penetrating into the skin. So when this is about the combination, it must be chemistry. Anyhow, is there any special name for these "absorbents"? And how exactly, do they "absorb" UVA & UVB? (I'm not really definite if this absorbing is really a scientific expression or not, and that's why I use quotations when using the word) I, myself, reckon that the sunscreen products form a special compound on the skin or they make the body produce something as a shield against UV rays, but I have got no clues.

Apart from titanium dioxide, the active species in sunscreens are organic compounds from different substance classes, such as cinnamates, benzophenones, benzimidazoles, and triazines.
The do absorb in the 280-315 (UVB) and/or 315-380 nm (UVA) range. This "uptake" of energy results in an electronic excitation, the compounds undergo a transition from the $S_0$ (ground) state to an excited $S_1$ state.
The excited singlet state ($S_1)$ may

deactivate radiationless via internal conversion (IC), the energy of the photon is released as heat or
undergo intersystem crossing (ISC) to a triplet state (usually $T_1)$ with different spin multiplicity

The latter state then might again loose its energy via IC or undergo a chemical transformation, through which the suncreen might be turned to a species that not longer is able to absorb UV radiation.

Just to add, Zinc (like Titanium) is another UV filter that reflects UV light rather than absorbing it.  I believe ever other UV filter, at least those approved in the U.S., absorb the UV rays.


text:
How to find the possible valence states of elements in the lower end of the d block and the f block?

For silver, why don't the electrons in the 4d orbital move to the 5p orbital?
Is it possible for any elements to move their electrons between $n$d and $(n+1)$p orbitals?
When the f orbital is completely filled with 14 electrons, should we take it in consideration to find the valence states?
When Z=58, can you show me the electronic configuration and how do we form the valence state(s)?

[For silver], why don't the electrons in the 4d orbital move to the 5p orbital?

In the case of silver, this is likely because the energy separation between the 4d and 5s orbitals are quite large (absorbing in ultraviolet region, which is why silver appears white/silver), so the energy separation between 4d and 5p would be even larger.
In general, when you write down the electron configuration you are writing down the ground state, i.e. the most stable configuration. While it is possible in principle to move electrons from 4d to 5p, it will not give you the most stable configuration as moving that electron requires a large amount of energy.
For inferring the likely valence states, you can only use orbitals with unpaired electrons, and/or move the electrons that are not separated by large energy differences between different orbitals to make hybrid orbitals with unpaired electrons. In this case, as you cannot move 4d electrons to 5p (which would make some kind of spd hybrid orbital), you can only have a valence state of 1 from the unpaired electron in 5s.

Is it possible for any elements [to move their electrons between $n$d and $(n+1)$p orbitals]?

Yes, but only for elements nearer the top of the periodic table (and not at the very top, as you will then not have any accessible d orbitals). Period 3 and period 4 (I think) would be most likely to do this. For example, phosphorus can have a valence state of 5, and sulfur can have a valence state of 6, both by involving their accessible 3d orbital, to make hybrid $\ce{sp^3d}$ and $\ce{sp^3 d^2}$ orbitals respectively.

When the f orbital is [completely] filled with 14 electrons, should we take it in consideration to find the valence states?

In general, no. When the f block is completely filled, the electrons in the f orbital are generally assumed to be core-like, i.e. stable, or low in energy, therefore will have a large energy gap to the other electrons in 6s, 5d and 6p (or 7s, 6d and 7p), and do not tend to participate in bonding/valency. There are possibly one or two exceptions, but I cannot recall them right now.

When Z=58, can you show me the electronic [configuration] and how do we form the valence state(s)?

Z = 58 is Cerium, which has the following electronic configuration: $\ce{[Xe] 4f^1 5d^1 6s^2}$.
On paper, it will appear that due to the diffuseness of the orbitals 4f, 5d and 6s, it is probably unusual to form covalent compounds to cerium, and you would expect a valence state of 3 (if you only use the 5d and 6s electrons), or a valence state of 4 (if you also use in addition the 4f electron).
It was found the 4f is actually near the energy of 5d and 6s, giving rise to a variable electron structure and a valence state change of 3 to 4 when cerium is cooled or compressed. Moreover, it is possible to form organocerium compounds with a valency of 3. By the way, this is not something that you can deduce simply by looking at the electronic configuration. You have to find this out empirically, using experiments!


text:
Stability variation in heavy nuclei
The heavier the atom, the more unstable it gets, right?
That is not true about Uranium and we know it.
I wondered why. A brief explanation stated that since more neutrons are there in the nucleus there is more nuclear force to 238U than 235U. Seems legitimate, but then that should have been true about the other radioactive species. Again, we know that's not true.
So, either the explanation about the Uranium exception is not totally true or the principle of the heavier-atom instability (of course, exceptions occur at elements like U) is not a principle. Or maybe there is a relation between these two (nuclear force and atomic weight) which is beyond me.
My hunch is that the latter is true; searched some sites which had an answer to the question, but no acceptable results gained.

Yes, it is true that nuclear physics can not be reduced to a brief explanation.  This might be better over in Physics...

Protons and neutrons are fermions like electrons, i.e. they obey the Pauli exclusion principle. If there's a sufficiently large excess of one type, the highest energy proton/neutron has to stay far above the highest energy other particle, which makes it favourable for some type of decay to happen and even the counts out more.

If the ratio of neutrons to protons in the nucleus is too high, the nucleus is unstable to beta decay.  A neutron emits an electon and becomes a proton.
If the ratio of neutrons to proton is too low, the nucleus is unstable to positron decay and/or electron capture.
But how high/low is too high/low, and why does this occur?
We need to look at the Semi-empirical mass formula.
$$E = vA -sA^{2/3} - c\frac{Z^2}{A^{1/3}} - a\frac{N-Z}{A} - \delta (A,Z) $$
Where $E$ is the binding energy of the nucleus, $N$ is number of neutrons, $Z$ is number of protons, $A$ is $N + Z$, and $v$, $c$, $s$ and $a$ are empirical constants.
Only the $c$ (coulomb) and $a$ (asymmetry) terms involve the relative number of protons and neutrons.
The asymmetry term is zero when the number of protons and neutrons are equal, and relates to the Pauli exclusion princple (neutrons and protons being fermions).
Except for the coulomb term (that protons electrostatically repel each other), the ideal ratio of protons to neutrons would be 1 to 1.  Electrostatic repulsion of protons explains why somewhat more neutrons than protons is favorable.

The principle that "heavier nuclei are more stable" is a very crude approximation. Theoretical explanations of nuclear stability are complex. While the transuranium elements become progressively more unstable with increasing atomic mass, this trend is not predicted to continue indefinitely; there is an island of stability around copernicium and flerovium, and the as-of-yet unsynthesized copernicium-291 has been predicted to have a half-life of 1200 years or so. You may be interested this paper for details; figure one is a nice illustration of how stability is affected by neutron/proton number towards the end of the periodic table.

Well, it's not exactly incorrect; there are no non-radioactive isotopes of any element after lead. It's just a very broad trend, and you can't use it to judge the relative stability of the metastable isotopes of the later elements accurately.

"The heavier the atom, the more unstable it gets, right?" Wrong. Nickel-62 is the "most stable" isotope, measured by binding energy per nucleon. Since in the core of massive stars, it is subject to photodissociation, Fe-56 is often considered to be the "most stable" isotope. This discrepancy shows that context matters. Take a look at the graph of binding energy/nucleon vs nucleon number https://en.wikipedia.org/wiki/Nuclear_binding_energy#/media/File:Binding_energy_curve_-_common_isotopes.svg  You can see that the graph is nearly flat around ~±10 nucleons of Fe-56, so there's no value in making broad statements, imho. Even if unlike the OP, the statement was accurate, the devil is in the details (context, environment, what is meant by "stability" , etc.). Why did anyone bother to resurrect this old question, I wonder? Heavier nuclei are neither or both more and less stable than lighter nuclei. It isn't (imho) a "very crude approximation", it is simply wrong.  A more accurate "crude" approximation is to say that low mass (not weight, for God's Sake!) elements tend to release less and less fusion energy up to approximately iron, and also past iron, the tendency is a slow increase in the fission energy released. (but details matter, especially the exact balance between the various forces). You simply can not predict the stability of an isotope by using only its mass (at least, not below ~ 250 or 300 amu).


text:
Ratio of MbO2/Mb in a satured solution of dissolved oxygen from dry air

What is the ratio of $\ce{MbO2/Mb}$ if the solution is saturated with air?
Details of the question:
  $$
\begin{align}
\ce{Mb + O2 (aq) &-> MbO2}\\
\Delta G' &= -30.3~\mathrm{kJ/mol}\\
T &= 298~\mathrm{K}\\
\ce{pH} &= 7\\
\text{Henry's Constant for }\ce{O2}  &= 42 \cdot 10^3~\mathrm{bar}\\
\text{Content of }\ce{O2}\text{ in dry air} &= 21\%\\
\end{align}
$$

Attempt at Solution:
So, basically this is a two part question, as I see it. First I need to figure out the concentration of dissolved $\ce{O2}$ using Henry's Law. Then I can simply plug this value into the $$\Delta G' = -RT\ln K$$ equation where $$K = \frac{\ce{[MbO2]}}{\ce{[O2][Mb]}}.$$
Therefore:
$$
\begin{align}
P &= K\cdot m\\
0.21~\mathrm{bar} &= (42 \cdot 10^3)\cdot m\\
m &= 5 \cdot 10^{-6} \pu{mol/kg}\\
-30.300~\mathrm{J/mol} &= -8.314 \cdot 298 \cdot \ln\left(\frac{\ce{[MbO2]}}{[5\cdot10^{-6}]\ce{[Mb]}}\right)
\end{align}
$$
After dividing and raising both sides to $e^x$ (to get rid of $\ln$) I get $$\frac{\ce{MbO2}}{\ce{Mb}} = 1.02.$$
However the answer sheet says it is $21.6$. I cannot figure out why this is so.

I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details. || For the ones unfamiliar (like me) with this kind of chemistry, it would be nice, if you could introduce your variables. What is $m$ or $P$? As a general remark: You should include units in every step.

Thanks for the edit, I didn't know how to use the markup. Also units have been included where necessary (except for m which I have now included). I will, however, try to be more clear in the future

You are very welcome. As for the units, you should include them always and everywhere, they are usually the first hint that something went wrong, when they don't cancel correctly.

I don't have time to answer properly, but I am inclined to think that the mistake is in the Henry's Law calculation.  Usually when the constant is given as a pressure, the formula calculates a mole fraction rather than a molality.  The answer you were given works back to a molarity of 1.1x10^-4M for oxygen, and water is 55 mol/L, so that works out nicely.  @Martin It's the units that point me in that direction.

While I admit that there is an error in using molality instead of mole fractions, even after using mole fractions I am getting an answer of 2.78 x 10^-4 Mol/L for the O2 concentration


text:
What kind of carbon ions exist?
I think there are $\ce{C^{4-}}$ and $\ce{C^{2-}}$ existing. How can we find the charge of ions in the elements? Should we look at the group or valence?

Chemistry of group XIV , especially the all notorious Carbon {:)))} is not something with definite clear-cut rules (At least, that's what I have reached with studying this element's properties.) Carbon usually forms Carbocations and not Carbanions. I think it's better for you to do some research yourself rather than directly asking it. That way this broadly answerable question of yours won't be answered with short answers, and you will understand how to deal with C ions better. Hope I've helped.

Could you please be more specific with your question. Can you give some examples of where you think these ions exist?

Many different carbon ions exist in the gas phase.
In Electronic spectra of linear carbon anions, spectra of $\ce{C_2^-}$, $\ce{C_3^-}$... $\ce{C_10^-}$ are discussed.
Dianions such as $\ce{C_7^{2-}}$ have also been observed.
In the solid phase, aluminum carbide and berylium carbide have more than 50% ionic character according to the Handbook of Refractory carbides and nitrides, and correspond to $\ce{C^{4-}}$.  Calcium carbide, which could be considered a salt of acetylene, corresponds to  $\ce{C2^{2-}}$  rather than $\ce{C^{2-}}$


text:
Why do some solid chemicals dissolve better in colder water?
I just discovered that some solid chemicals, such as $\ce{Na2SO4}$, dissolve better in cold water than hot water from here and would like to know if there is any particular reason as to why. Is it their structure, their charge, etc.?

possible duplicate of Why does sodium sulfate have an unusual solubility-temperature curve?

As with every process, dissolving can be described by the thermodynamical equation:
$$\Delta G = \Delta H - T \Delta S$$
Which the process being spontaneous if $\Delta G$ is negative.
We have two components here, enthalpic and entropic. The enthalpic component $\Delta H$ is a constant of the salt in question. The entropic component $\Delta S$ can usually be considered positive because a highly ordered crystal lattice is broken to give dissolved, much less ordered fragments — hence why for most salts they dissolve easier in hot water.
If the solubility decreases with higher temperatures, then $\Delta S$ must be negative. The salt will then only dissolve if $\Delta H$ is negative enough to give an overall negative result. This seems to be the case for $\ce{Na2SO4}$
It is impossible to predict $\Delta S$ and $\Delta H$ a priori without any computer quantum calculations. One has to resort to experiments to determine whether a salt will dissolve better or worse in the heat.

While I have used the term ‘salt’ most of the time, the same is true for molecules. Take sucrose: I was asked in a physical chemistry lab to guess with my colleague whether sugar would dissolve endothermicly or exothermicly. We decided, that the degree of order be a lot higher in dissolved sugar due to a high amount of water surrounding it rather tightly, that the entropy must decrease, so that the enthalpy has to be negative and sugar has to dissolve exothermically. Hence sugar should dissolve more easily in cold water, too.
According to a comment by Nicolau Saker Neto, our reasoning was false and sucrose displays increasing solubility at higher temperatures. Apparantly the testing faculty member approved of it, though.

Regarding your last paragraph, glucose and sucrose both show monotonically increasing solubility in water with increasing temperature.

Well the theory behind it was right, as can be seen from the theoretical development in your answer, it's just the quantification that did not go so well. But then you couldn't have reasonably been expected to find the right answer! Solvation is difficult.


text:
Delocalization of π-electrons in nitrate ion
Benzene and nitrate ion are given in my textbook as examples for the delocalization of π-electrons. Benzene, due to symmetry of its resonating structures, is simple enough. We assume that σ-electrons are localized and π-electrons are delocalized in the ring.
Each carbon atom promotes one electron from its $\mathrm{s}$ orbital to the empty $\mathrm{2p}$ orbital. It hybridizes two $\mathrm{p}$ orbitals with the $\mathrm{s}$ to form three $\mathrm{sp^2}$ orbitals which it then uses to form three σ-bonds, two with other carbons and one with a hydrogen.
The remaining unhybridized $\mathrm{p}$ orbital then sticks out above and below the plane of the ring. The electrons from all the six unhybridized $\mathrm{p}$ orbitals of the six carbons are then delocalized above and below the plane of the ring.
However, when I try to apply a similar reasoning to the nitrate anion, problems arise. The resonance Lewis structures of nitrate ion are:

Now, assuming again that only the π-electrons are delocalized, we would expect that only two electrons are delocalized (since there is only one double bond). But my textbook claims that each atom is $\mathrm{sp^2}$-hybridized, and as in benzene, one unhybridized $\mathrm{p}$ orbital per atom sticks out above and below the plane of the molecule.
Assuming this to be true, I presumed that two electrons would be localized and tried to arrange the remaining 22 electrons into a Lewis structure with $\mathrm{sp^2}$ hybridized atoms. I couldn't find any viable structures with only three lone pairs around each atom.
What is the exact mechanism for delocalization of electrons in nitrate? Is there a general scheme to this mechanism that applies to all such similar molecules such as $\ce{CO3^2-}?$

Unfortunately the key to understanding the delocalisation of electrons lies in understanding rudimentary molecular orbital theory. It is also necessary to understand that hybridisation is a mathematical concept which can be used to describe bonding. It is no necessity to have bonding.
Your explanation of benzene is the most common description, but it is somewhat incomplete. First, you are right, one can consider the carbon atoms having three (roughly) sp2 hybrid orbitals, giving rise to two σ bonds to the neighbouring carbons and one σ bond to hydrogen. The remaining 6 perpendicular p orbitals form the delocalised π system. These orbitals form 6 new orbitals and due to symmetry constraints, i.e. D6h, three of them are bonding and three of them are antibonding. In the π system, there are six electrons, so only the bonding orbitals will be occupied. This makes the extraordinary stability of these compounds. The following picture schematises this on the left side, while on the right side these are the orbitals obtained by a quantum chemical calculation. (At the DF-BP86/def2-SVP level of theory.)

The nitrate ion $\ce{NO3-}$ is what I like to call a Y-aromat since it has some similarity in the electronic structure. The bonding situation is very comparable to benzene.
The resonance structures, however, look much more complicated.

In quite good approximation, you can regard the nitrogen as having three sp2 hybrid orbitals (and one p orbital) and the oxygen as (roughly) having two sp (and 2 p) orbitals. You follow the same scheme, therefore you obtain four equivalent nitrogen oxygen σ bonds. Each oxygen has one lone pair in the in-plane p orbital and one lone pair in the sp hybrid. The remaining out-of-plane p orbitals of nitrogen and oxygen build up the delocalised π system. This time there are only four orbitals. Because of symmetry constraints, i.e. D3h, one is bonding, two are (close enough to) non-bonding, and one is antibonding.* With the remaining six electrons you only fill the bonding and non-bonding orbitals. The concept is similar. Below you find the schemes like the ones above for benzene.

Since the carbonate anion $\ce{CO3^{2-}}$ is isoelectronic to nitrate, approximately the same orbital picture will result.

* Strictly speaking, there are no non-bonding orbitals. The classification can only be with respect to a bonding axis, and there are only two options: 1. there is no nodal plane (or surface) perpendicular to that bonding axis; it is then called bonding. 2. There is a nodal plane (or surface) perpendicular to that bonding axis; it is the called antibonding. A molecular orbital can be bonding respectively to one bonding axis, and antibonding with respect to another.
For example, the E" orbitals of nitrate have this feature, since the nodal surface curves through the nitrogen atom. The above depicted orbital on the left is bonding with respect to one of the $\ce{N-O}$ bonds and antibonding towards the other. (It is also bonding with respect to a $\ce{O\bond{~}O}$ interaction.) The orbital on the right is exactly opposite (orthogonal) to that.


text:
A misunderstanding about the energy profile of reactions with a catalyst involved
All of us are aware of the importance of the catalysts in bio-chemistry. For a high school learner like me, catalysts ,and therefore, enzymes play a bridge-like role that connect high school bio to high school chemistry.
Yet, I got baffled about the two common energy profiles drawn for an anonymous reaction with a presence of a catalyst. The question may seem rudimentary, I agree, but the possible answers to my question were too technical for me to understand.
One of the energy profiles had several curves in the pathway for $E$, and the other one has only one curve in the pathway where the catalyst should have affected.

My guess is that the first energy profile is about a net change in $E_\mathrm{a}$ but the second one demonstrates the reality when an exothermic reaction occurs.
 Now, here are the questions:

Why are the two energy profiles different? Please be as if you're teaching thermo-chemistry to a little kid!
Why do those curves occur in the latter energy profile and is there a way to know how many of those curves will occur in the presence of a catalyst?

Image credits: chemwiki.ucdavis.edu (page here) and wiki (page here).

The upper graph is just the most simple way to visualize the effect of a catalyst on a reaction $\ce{S -> P}$:
The activation energy is lowered.
The activation energy for the reaction in that direction is the difference of the energies of the starting material $S$ and a transition state $TS^\#$.
Since it is the same starting marterial in the presence or absence of the catalyst, the energy of the transition state is different.
Can the same transition state have two different energies - just through the remote distance magical action of a catalyst located somewhere? Probably not!
It is much more plausible that - in the absence and presence of a catalyst - two completely different (different in structure and different in energy) transition states are involved.
Exactly this situation is described in the second graph!
The catalyst "reacts" with the starting material, either by forming a covalent bond, by hydrogen bonding, etc. and thus opens a different reaction channel with different intermediates and a transition state not found in in the non-catalyzed reaction.
In the end, the same product $P$ is formed and the catalyst is regenerated, but this doesn't mean that the catalyst wasn't heavily involved in the reaction going on.

1# Curve 1 is just for a simple case where it's showing the comparison between a theoretical pathway  for a catalyzed and non-catalyzed reaction but it has an another significant which I will discuss in part two.
2# How many of these curves can occur: Infinite (Theoretically) but in practical case you have to consider all types of possible elementary reaction in your model. An elementary reaction is a reaction which can occur in one step or the reactant, product and TS has well defined position on potential energy surface. For example if you have a simple ketone hydrogenation reaction, you can get an overall energy profile for gas phase reaction but on a catalytic surface there are lots of possibilities. At first the ketone and hydrogen will adsorb on the surface and then 1st hydrogen addition can occur either on C of the ketone group or on the hydrogen of the ketone group. They are intuitively main reaction pathway but there can be much more.

In the above image I was working on LA hydrogenation and you can see there are two different pathway for the hydrogenation and there might be some other more pathway. And if you want to capture the real chemistry you have to make sure you are including all important intermediates. So your energy profile may look like this

Here the picture becomes quite complicated with lots of elementary reaction. To determine the overall activation energy of a catalyzed reaction scientists follow two methods
method-1: Energetic span theory: Given by Shaik et al. I am adding the link to that paper here,
http://pubs.acs.org/doi/ipdf/10.1021/ar1000956
2) Apparent activation barrier: This is an age old method. Here the sensitivity of turn over frequency (TOF) with the change of temperature is measured and defined as apparent activation barrier.
$$E_a=-R\frac{\delta~ln(TOF)}{\delta(\frac{1}{T})}$$
Now you can plot the first E profile with your gas phase activation energy and apparent activation energy of the catalyzed reaction.

A reaction involving more than one elementary step has one or more intermediates being formed which, in turn, means there is more than one energy barrier to overcome. In other words, there is more than one transition state lying on the reaction pathway. As it is intuitive that pushing over an energy barrier or passing through a transition state peak would entail the highest energy, it becomes clear that it would be the slowest step in a reaction pathway. However, when more than one such barrier is to be crossed, it becomes important to recognize the highest barrier which will determine the rate of the reaction. This step of the reaction whose rate determines the overall rate of reaction is known as rate determining step or rate limiting step. The height of energy barrier is always measured relative to the energy of the reactant or starting material. Different possibilities have been shown in figure 6

From the link you gave on energy profile i found this. Basically there is no "net change" in $E_{a}$ . In the first graph in question the curve which is at the top is the representing how the reaction proceeds without a catalyst and the one below is of the reaction with catalyst. In the second graph there are 4 intermediate reactions/transition states before the product is formed. Hence you have 4 peaks of different activation energy as shown.
A catalyst basically provides alternate routes to a reaction by forming the "transition/metastable/intermediate state at a lower energy than the original reaction. You will study later that this has to do with basic Collision theory and molecular dynamics.
In the given graphs a catalyst can make the R->P through just one route where only one transition happens (as in the graph 1) or for 4(or even more) transition states (as in the graph 2).
 For one set of conditions(P,T,V etc) I am guessing the number of these peaks will be unique. It is also possible that for a certain catalyst and reactant the number of peaks / intermediate states is constant.

authors: شعبان علي كامل محمد


text:
Why do Grignard reagents react with epoxides but not THF?
THF is used as a solvent for Grignard reagents, but epoxides react with Grignard reagents. There is only a small difference between epoxides and THF, what causes the difference in reactivity towards Grignard reagents?

The difference between THF and epoxides is perhapse bigger than you might think. Due to the three-membered-ring of an epoxide, a lot of energy is 'stored' in ring strain. This destabilizes the stucture and is a driving force for a Grignard reaction (or any substitution on an epoxide).
The THF is a five-membered-ring, which is far more stable. This bond is simply an ether, that happens to be in a cyclic structure. Therefore it also reacts like an ether: not easily attacked by nucleophiles.


text:
What makes R-MgX a more reactive Grignard reagent than R-Cu?
What makes $\ce{R-MgX}$ a more reactive Grignard reagent than $\ce{R-Cu}$? Has it to do with solvation, bond enthalpy, electronegativity?

I'm really not sure, but it could be because of the size of the Copper versus Magnesium? The alkyl mostly does nucleophilic attack, so it might need the space.

The difference in electronegativity between copper (1.9) and magnesium (1.3) is the key difference.
Since copper is more electronegative than magnesium its electronegativity is much closer to that of carbon (2.55).  This results in carbon-copper bonds being less polarized and more covalent than carbon-magnesium bonds.  The electrons in a carbon-copper bond are better shared between the two atoms, whereas the electron distribution in a carbon-magnesium bond places more electron density on carbon.
As a result of this difference in electron density on the carbon atoms the carbon portion of a cuprate is a softer nucleophile, while the carbon portion of a Grignard is a harder nucleophile.  These difference in nucleophilicity produce different reactivity patterns for the two reagents.  For example, the softer cuprate will undergo 1,4 addition to an $\alpha,\beta$-unsaturated carbonyl compound (a softer electrophile), while the Grignard adds directly to the carbonyl (which is a harder electrophile).

Would you mind explaining what you mean by a soft/hard nucleophile please? Thanks for the great answer by the way.

In a nutshell, hard nucleophiles are generally small, highly charged atoms.  Because of this they tend to be more reactive.  They prefer to react with similar (hard) electrophiles.  Their reactions are usually kinetically controlled with early transition states and are governed by electrostatic interactions. Soft nucleophiles are generally larger systems with a more diffuse charge.   Because of this they tend to be less reactive (more discriminating).  They prefer to react with similar (soft) electrophiles.

Because they are more discriminating their reactions are usually thermodynamically controlled with later transition states and are governed more by orbital interactions.

The question is if it is actually less reactive or more reactive.
Addition of $\ce{RMgX}$ catalized by copper (I) salts is an 1,4 addition (characteristic for organocopper) meaning that active particle of $\ce{R2CuM}$ is MORE reactive than $\ce{RMgX}$ Ron, care to refine your answer?


text:
DFT Calculations, Atomic Ionization Potentials -- Which Exchange-Correlation Functional to Use, to Preserve Koopmans Theorem?
I have a program which can perform density-functional calculations for atoms, given a density functional.
Of course the simplest form of exchange potential to use is one relevant for a uniform electron gas (i.e. the original Kohn-Sham exchange, proportional to $n_e^{1/3}$). A good correlation functional is also available. However when I performed calculations for some atoms the energy of the highest occupied orbital differs greatly from the ionization potential (off by a few eVs!).
That the energy of the highest occupied orbital equals the ionisation potential of the neutral atom is known as Koopmans' Theorem. I get a feeling from the literature that DFT can be quite accurate. So what's wrong?
Interestingly when I use the Slater exchange functional, coupled with a correction suggested by Skillman (i.e. replacing the potential by $1/r$ when the overall potential drops below that value), the results improve significantly. (See the book "Atomic Structure Calculations" by Herman Skillman). Well, maybe I should follow this recipe, but it seems the procedure is quite ad-hoc.
My question is, are there any functionals which will give reasonable values for the atomic ionisation potentials? I do not want to implement methods like the optimised effective potential method since the latter is not readily generalised to a finite-temperature scenario. Thanks.

The only thing which is wrong is that you somehow misinterpret the fact that DFT, as you said, can be quite accurate. Yes, it can, but it is not guaranteed to. In general the accuracy of DFT depends on many factors, but for prediction of -IP by HOMO eigenvalues it is well known that DFT performs rather poorly: absolute errors of few eVs are in fact quite usual.

See, e.g. this paper and references there in.

Thanks for the paper. Very informative indeed. It seems that KMLYP is best from the point of view of conformance to Koopmans' theorem. You are right; absolute errors of a few eVs are in fact quite usual.

http://journals.aps.org/prb/abstract/10.1103/PhysRevB.90.075135

I think the most widely-used approach at the moment comes from Roi Baer and Leeor Kronik and others, e.g.

Phys. Rev. Lett. (2010) 105, art. 266802
Phys. Rev. B (2011) 84, art. 075144

Basically the idea is to tune the range-separation parameter $\gamma$ between short-range and long-range electrostatic effects in a range-separated hybrid functional. You find a match such that Koopmans' theorem holds.
The result is an "optimally-tuned" range-separated hybrid DFT functional (OT-RSH).
It's not perfect, for example, it violates size consistency: J. Chem. Phys. (2013) 138, 204115

Have yet to read the paper. But from what you said, "You find a match such that Koopmans' theorem holds", it seems that the method is even more computationally intensive than the original, naive DFT. (For each potential obtain the self-consistent solutions (for both the atom and the ion?); for each pair of solutions, compute the ionization potential, and the energy of the highest occupied orbital (of the atom); if the two do not agree, tune the parameter; otherwise stop.)

This may be true in general, but it seems that for comparisons across related molecules, one can find a reasonable $\gamma$. The authors probably cringe at that suggestion, but errors ~ 0.1 eV for an intermediary parameter in the second paper, if I remember correctly.

My other comment is that Koopmans' theorem may not always hold in absolute terms, but there's usually a linear correlation between eigenvalues and ionization potentials with low residual errors.


text:
In E1 reactions, does carbocation stability have impact on the final product?

In this question, I believe there is a mistake for iii). I think that there the tertiary carbocation would be formed (more stable), which would lead to the tetra substituted alkene (even more stable). Am I wrong?

The answer above only shows one of the two possible isomers that could be formed by elimination from the tertiary carbocation intermediate.

I suspect that both isomeric cyclohexenes with tri-substituted double bonds would probably be produced in roughly equal amounts, along with a smaller amount of the exo-methylene compound.
Edit:  Thanks to Bon and Yolo123 for their comments
Elimination towards the isopropyl group is also a possibility, but two factors argue against this being the major elimination pathway.  First there is a statistical factor;  there is only proton on the central isopropyl carbon available for elimination, whereas the are 4 (2+2) protons available alpha to the carbocation center in the 6-membered ring.  Second, if you look at a model and consider that the proton to be eliminated must be lined up with the carbocation p-orbital you see that each methylene group adjacent to the carbocation has one hydrogen well lined up for elimination, whereas to properly line up the isopropyl proton results in some adverse steric interactions between the isopropyl methyl hydrogens and the ring methylene hydrogens.
My guess is that the gem-dimethyl group in the cyclohexane ring will not favor one in-ring elimination pathway over the other - my model doesn't show them exerting any adverse steric effect one way or the other.

Hi ron, imagine that the base pulls off an H from the isopropyl group. That forms tetrasubstituted

@yolo123 There is a fair amount of steric hindrance on the central carbon of the isopropyl group so removal of a proton there is much less likely than at the other two positions

@bon, I doubt so. The "ortho-like" positions would also be very much sterically hindered, if there were so much steric hindrance at the central C of the i-Pr group.

@ron For the second isomer you have given in your answer, is it possible that there would be some steric hindrance due to the adjacent methyl groups and so the first isomer would predominate.

@bon, let's see what ron has to say about the tetra-substitution.

@yolo123 I think they would be less hindered than the isopropyl position because there are only two hindering groups as opposed to three


text:
How to calculate the heat of dissolution from a calorimeter experiment?

A $\pu{1.50 g}$ sample of $\ce{KCl}$ is added to $\pu{35.0 g}$ $\ce{H2O}$ in a styrofoam cup and stirred until dissolved. The temperature of the solution drops from $24.8$ to $\pu{22.4 ^\circ C}$. Assume that the specific heat and density of the resulting solution are equal to those of water, $\pu{4.18 J g-1 ^\circ C-1}$ and $\pu{1.00 g mL-1}$, respectively, and assume that no heat is lost to the calorimeter itself, nor to the surroundings.
$$\ce{KCl (s) + H2O (l) -> KCl (aq)} \qquad \Delta H = ?$$
a) (2 points) Is the reaction endothermic or exothermic (circle the correct answer)?

Endothermic

b) (4 points) What is the heat of solution of $\ce{KCl}$ expressed in kilojoules per mole of $\ce{KCl}$?

$$q_\mathrm{rxn} = -q_\mathrm{cal}$$
I multiplied the sample $\pu{1.50 g}$ by $\pu{4.18 J} \cdot (-2.4) = \pu{-15.048 J}$
Divided that by $1000 = -0.015048$; thus, $0.015048$
However, my answer seems to be wrong. I know that reaction is endothermic since the temp drops, but I am wondering which values I should be using to correctly determine the "Heat of Solution".

You have multiplied the mass of the sample, 1.50g, by temperature change and heat capacity.
However, the water provides most of the heat for the reaction.
The total mass of the solution is 1.50g + 35.0g = 36.5g.
You should be multiplying  36.5g by the temperature change and heat capacity.
Then, you need to consider how many moles 1.50g KCl is.
Divide the change in enthalpy of the solution by the number of moles of KCl to determine the molar heat of solution of KCl.

I have done that. The answer is still wrong. If I were use 36.5g, my answer would be 0.366kj; however, my professors answer is 18.3kj

is his/her answer 18.3 kJ or 18.3 kJ/mol?

@user137452 if you want the answer as "per mole of KCl" you need to divide by the moles of KCl in the sample.

Here's the calculation, step by step:
$$q_\mathrm{cal} = 36.5\cdot 4.18\cdot (-2.4) = \pu{-366 J}$$
$$q_\mathrm{rxn} = -q_\mathrm{cal} = \pu{366 J}$$
$$n(\ce{KCl}) = \frac{\pu{1.50 g}}{\pu{74.55 g mol-1}} = \pu{0.0201 mol}$$
$$\frac{\pu{366 J}}{\pu{0.0201 mol}} = \pu{18.209 J mol-1} = \pu{18.2 kJ mol-1}$$

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


text:
Molality of solution having more than one solute?
So here is my question,

Question: A water sample contains 9.5% $\mathrm{MgCl_2}$ and 11.7% $\mathrm{NaCl}$ (by weight). Assuming 80% ionization of each salt. Boiling point of water will be ________. ($\mathrm{K_b}$ for water $\mathrm{=0.52}$)

You know, this question isn't that tough to bother anyone (definitely not!). What's all confusing here is the molality of two solutes.
I know that,
$$\Delta \mathrm{T_b=K_b\cdot m}$$
Degree of dissosiation $(\mathrm{\alpha}) = 0.8$
So I know very well that molality of a solution having only one solute is,
$$\mathrm{m=\frac{n\ (no.\ of\ moles\ of\ solute)}{y\ (mass\ of\ solvent\ in\ kg)}}$$
But what about the solution having more than one solute? Do they have any molality (I am sure that there will surely be, because it is there in the question). If yes then what's the method of calculating?
Also if you are solving this, here is the answer to the question (as per my answer key),

377 K

The boiling point elevation fomula assumes that all particles are equal, only the number of particles in the solution matters.
So I would use:
m = (0.2 + (3)(0.8))(molality of $\ce{MgCl_2}$) + (0.2 + (2)(0.8))(molality of $\ce{NaCl}$)
To find molality of $\ce{MgCl_2}$ and molality of $\ce{NaCl}$, consider a 1kg sample of the solution.  There are 95g  $\ce{MgCl_2}$, 117g $\ce{NaCl}$ and 788g water.
Calculate the number of moles $\ce{MgCl_2}$ in 95g $\ce{MgCl_2}$, and divide this number by 0.788kg.  This is the molality of $\ce{MgCl_2}$ in the solution.
Repeat for $\ce{NaCl}$


text:
Titration of sodium carbonate and nitric acid
I am trying to find the pH of the following problem (Answer key is $\mathrm{pH}=10.77$)

Calculate the $\mathrm{pH}$ of a $\pu{100 mL}$ solution containing $\pu{40.0 g}$ of $\ce{Na2CO3}$ after   $\pu{400 mL}$ of $\pu{1.00 M}$ $\ce{HNO3}$ has been added. Ignore  the volume  of water    produced    by  the reaction

My approach:
$$\ce{Na_2CO_3~(aq)~+~2HNO_3~(aq)\rightarrow CO_2~(g)~+~H_2O~+~2NaNO_3~(aq)}$$
I start with $\pu{0.377 mol}$ $\ce{Na2CO3}$ and $\pu{0.4 mol}$ $\ce{HNO3}$.  Because of the 2:1 ratio, I use up all of the $\ce{HNO3}$, and create $\pu{0.4 mol}$ $\ce{NaNO3}$ and have $\pu{0.177 mol}$ of $\ce{Na2CO3}$ left over.
I am unsure how to go about setting up the ice table.  Should I use that
$$\ce{NaNO3 + H2O -> HNO3 + Na+ + OH-}$$
where I have the initial amount of $\ce{NaNO3}$?

To solve this problem, you need to know the pKas of carbonic acid.
The pKas are 6.35 and 10.33.
If you add 1 equivalent of acid to $\ce{Na_2CO_{3}}$, the product will be $\ce{NaHCO_{3}}$ (sodium bicarbonate).
In this case, you are going 0.4-.377 = 0.023 moles past the point of 1 equivalent.
I would model the final solution as 0.354 moles of $\ce{HCO_{3}^-}$ and 0.023 moles $\ce{H_2CO_{3}}$ , and use the pKa of 6.35 and the Henderson-Hasselbalch equation to calculate pH.
The proposed answer 10.77 is incorrect.


text:
Friedel-Crafts Alkylation Using Alkenes: Mechanism
What is the mechanism by which benzene, cyclohexene and $\ce{AlCl3}$ form cyclohexylbenzene?
I believe that cyclohexene will form a bond with Al with its double bond. However, how do we get rid of the $\ce{AlCl3}$ now attached to the previous cyclohexene after the benzene attaches to the module?

The Friedel-Crafts alkylation is another example of an electrophilic aromatic substitution. I've drawn the mechanism below showing the formation of cyclohexylbenzene.  The reaction is typically run in the presence of an acid and the aluminum chloride serves as a catalyst to enhance the electrophilic properties of the proton, at the end of the reaction the aluminum chloride has been regenerated.

In the insert box I've shown that when the reaction is carried out with n-propylchloride isopropyl benzene is produced, not n-propylbenzene.  A carbocation intermediate is generated in thr Friedel-Crafts alkylation and will rearrange to a more stable carbocation if possible.  In the cyclohexene case, rearrangement of the intermediate carbocation just regenerates the same cyclohexyl carbocation, so the rearrangement is "invisible" in this case (unless we label the starting cyclohexene).

Hi ron, I was particularly asking for the case where there is no acid. My textbook says the reaction can be done with only AlCl3, which is what was confusing me. Does this make sense?

If you mix aluminum chloride and an alkene a carbocation will still be generated and the product will be a mixture of the various possible alkene isomers weighted according to their thermodynamic stability.  In this Friedel-Crafts example (without a proton source) the cyclohexyl carbocation is weakly bound with the aluminum chloride (ion pair).  After this carbocation attacks benzene, the resultant cyclohexylbenzenonium carbocation will just transfer what was a benzene proton up to the cyclohexyl ring releasing (and regenerating ) the aluminum chloride.

Why would the cyclohexyl carbocation be weakly bound? Would it not covalently share its double bond with the AlCl3?

Could you show the transfer of the proton too? Thanks in advance.

weakly bound - I just mean that the carbon-aluminum bond is not particularly strong.

I bet that this reaction has never been done with rigorous exclusion of water. Water will hydrolyze AlCl3 to give the catalytic HCl necessary for Ron's top mechanism.

@jerepierre  Good point.

MECHANISM OF ALKYLATION: At first, electrophile methyl carbonium ion ($\ce{CH3}$) is formed from $\ce{CH3Cl3}$ and $\ce{AlCl3}$, which, at the second step, attracted by $\pi$-electron of the benzene ring to form sigma complex. At last step, the approach of nucleophile like, $\ce{AlCl4-}$, one proton is given out from sigma complex to form stable product toluene


text:
What is the origin of the naming of deoxycorticosterone acetate?
Deoxy- means the compound misses an oxygen comparing to the name without the prefix.
Acetate means ethanoate in IUPAC language. Many biologists still abide old rules and call it acetate. And finally, corticosterone. I just threw away thoughts and checked for the formula  online.
What is the law behind naming "corticosterone"? Is it relative to corticosteroids? Please answer in a detailed way!
Maybe it is corti+coster+one? Since I know that there exists cortisone and cortisol.

Naming complex chemical compounds in a straightforward systematic fashion is usually not very easy. And often there are trivial names, that existed way before this systematic naming scheme was introduced. Especially when it comes to naturally occurring compounds like steroids. The IUPAC defines steroids as the following:

Naturally occurring compounds and synthetic analogues, based on the cyclopenta[a]phenanthrene carbon skeleton, partially or completely hydrogenated; there are usually methyl groups at C-10 and C-13, and often an alkyl group at C-17. By extension, one or more bond scissions, ring expansions and/or ring contractions of the skeleton may have occurred. Natural steroids are derived biogenetically from triterpenoids.

(source: iupac.org)

There is even a quite lengthy article about the nomenclature of steroids: [1] G. P. Moss, Pure & App. Chem. 1989, 61 (10), 1783-1822.
The name Deoxycorticosterone acetate is itself derived from a trivial name. The underlying principle compound is corticosterone and its systematic steroid name is 11β,21-dihydroxypregn-4-ene-3,20-dione, see [1] Table 2. It can be derived from cortisol, which has an additional hydroxy group in position 17. The systematic name therefore is (11β),17,21-trihydroxypregn-4-ene-3,20-dione.
You assigned correctly, that the prefix de(s)oxy- refers to losing an oxygen containing functional group. In this case a hydroxy group at position 11. The systematic name for 11-deoxycorticosterone would therefore be 21-hydroxypregn-4-ene-3,20-dione.
Finally the acetate refers to an acetyl ester group at position 21 (in the sidechain). Interestingly the systematic name for cortisol acetate is 21-O-acetylcortisol. Since it has already a trivial name, see [1], 3S-4.3:

When the steroid acid has a trivial name that already specifies the hydroxyl group that is esterified, the prefix must be O-acyl-.

Applying the same logic, the systematic name for Deoxycorticosterone acetate should be 21-O-acetyldeoxycorticosterone. Which makes little to no sense. A more systematic approach is described again in  [1], 3S-4.3

Esters of steroid alcohols
  Esters of steroid alcohols are named by the appropriate steroid substituent group followed by that of the acyloxy group in its anionic form. The steroid name is formed by replacing the terminal -e of the hydrocarbon name by -yl, -diyl, etc. and inserting this before the locant with a Greek letter to designate position and configuration. When necessary locants should be used with the anionic part of the name. Examples: [...]
  The prefix form acyloxy-, with the appropriate locants, is used if there is a preferred functional group as the suffix (e.g. an acid or lactone). Examples: [...]

So in this case since you have a ketone, which is probably more important, I would choose the second part of this rule. Hence the systematic name becomes 21-acetoxypregn-4-ene-3,20-dione.
As you can see, naming is quite a big, often confusing subject and sometimes the origins are not clear at all. And your analysis was completely correct in every way. Trivial names are often like the vocabulary of a specific language - if you don't know it, you have to look it up.

Additional fun fact:
The name acetate is a retained name which can be used in a semisystematic way.

What is the law behind naming "corticosterone"?

There is no law, since the name indicate where the compounds are produced. Corticosterone and corticosteroid are synonyms for a groups of compounds (steroids) that are produced in the cortex of the adrenal glands. Cortisone and cortisol refer to 2 of the corticosteroids.