text:
Why is the addition of a singlet carbene to an alkene stereospecific?
Why is the addition of a singlet carbene to an alkene stereospecific, but not when a triplet carbene is added?

Here is a diagram that might help explain things.

Carbenes have two electrons that, for the moment, are not being used for bonding.  These two electrons can exist in the same orbital with their spins paired (singlet carbene) or in separate orbitals with their spins unpaired (triplet carbene). Normal hydrocarbons exist as singlets, spins are paired in all bonds, as they must be according to the Pauli exclusion principle.
If we look first at the pi electrons in ethylene, their spins are paired since ethylene is a singlet in the ground state. When we add a singlet carbene to ethylene, a bond is formed by using one electron from the carbene and one from ethylene.  Those two electrons are selected such that their spins will be paired in the newly-formed bond (Pauli exclusion principle).  Either 1) a second bond is formed at the same time using the two remaining spin-paired electrons (one on ethylene and one on the carbene) to form cyclopropane in a spin-conserving, concerted addition or 2) a transient (short-lived), intermediate 1,3-biradical is formed that still contains two spin-paired electrons is formed, since the spins are paired on these two remaining electrons in the biradical and since they are held in relatively close proximity they can close quickly to form a three-membered ring without violating the Pauli exclusion principle.
Let's now contrast this to the triplet carbene case. The spins are unpaired in the triplet carbene. When it forms a bond with one of the ethylene pi electrons, in order to obey the Pauli exclusion principle we must select our electrons so that they are paired in the newly formed bond. This requires that the two remaining electrons in our 1,3-biradical have their spins unpaired.  Before the biradical can form the final bond and close to cyclopropane, we must wait for one of the electron spins to flip so that we have a singlet biradical.  The triplet cannot close directly since this would violate the Pauli exclusion principle.
This "waiting" for the spins to pair in the triplet biradical means that the triplet biradical has an appreciable lifetime. Had we run our reaction using cis-2-butene instead of ethylene. We could see the effects of biradical lifetime by observing the stereochemistry in the resultant products. The singlet biradical, which either reacts in one concerted step or passes through a short-lived, singlet biradical would produce only cis-dimethylcyclopropane. In the case of the triplet carbene, the triplet biradical exists long enough (waiting for one electron spin to flip) that rotation about the various carbon-carbon bonds is now possible. As a result, we obtain both cis- and trans-dimethylcyclopropane.

Can you backup the intermediate biradical with anything? I would expect this mechanism to be concerted, not only for symmetry reasons.

@Martin  I'm going from memory here, but 1) I recall energy surface calculations showing a shallow minimum where an intermediate would exist for some small number of vibrations, 2) in the gas phase, even if we accept a concerted reaction to produce cyclopropane, at least in some cases (depending upon pressure) it would be so hot that it would ring open and vibrate - again the biradical would not live long enough to undergo bond rotation.

@ron then wouldn't there be same products for both if the ring actually breaks up due to heat  as it can happen in either cases

@Anenthusiast  It depends how fast bond rotation is in comparison to ring closure.  If a singlet biradical forms initially or later due to ring reopening, it's closure is symmetry allowed and will happen quickly, likely faster than bond rotation.  On the other hand, it takes longer for the triplet biradical to close (spin inversion required), so now bond rotation can compete with ring closure.

@ron: In your image, the last structure (singlet-spins paired) of triplet addition has erroneously had $\ce{CH3}$ terminal instead of  $\ce{CH2}$. I corrected it. Yet, I'd like you to know the correction since you may not see the erroneous image anymore.

@MathewMahindaratne Thank you for catching that, I've updated my original drawing.


text:
What is the mole fraction in mixture before combustion?
What is the mole fraction of $\ce{C2H6} / \ce{C3H8}$ in mixture, if both compounds reacted with  $\ce{O2}$ in stoichiometric ratio? Mass of produced $\ce{CO2}$ is $1.819$ times larger than the mass of $\ce{H2O}$.
$$\ce{C2H6_{(g)} + O2_{(g)} -> CO2_{(g)} + H2O_{(g)}}$$
$$\ce{C3H8_{(g)} + O2_{(g)} -> CO2_{(g)} + H2O_{(g)}}$$
I balanced the equations and tried the following:
$\frac{X4M(CO_2)+(X-1)4M(CO_2)}{X8M(H_2O)+(X-1)6M(H_2O)}=1.819$
Which led me to the wrong result.
Could anyone kick me in the right direction?

It's always helpful to show as much work as you can so that we can see where you went off track,  In any case it appears you made an effort, so let's start at the beginning.
First we need to write out the balanced equations
\begin{aligned}
\ce{&C2H6(g) + $3.5$~ O2(g) -> 2 CO2(g) + 3 H2O(g)} \\
\ce{&C3H8(g) + 5 O2(g) -> 3 CO2(g) + 4 H2O(g)}
\end{aligned}
Second we're given that
mass ($\ce{CO2}$) = 1.819 mass ($\ce{H2O}$)
we can converting this to a molar ratio
$$\ce{\frac{[mass ~(CO2)]}{[$44.009$]}~=~ $1.819$ * \frac{[mass~ (H2O)]}{[$18.015$]}}$$
and we find that the molar ratio of
$$\ce{\frac{[moles~(CO2)]}{[moles~(H2O)]}~=~ $0.744604$}$$
Now, just to get a feel for things, let's look at the ethane case and we see that we have a mole ratio of $\ce{CO2~/~ H2O}$ of 2/3, whereas in the propane case the ratio is 3/4.  The observed molar ratio of 0.74 indicates that the propane reaction is the major contributor. We can write a general equation
$$\ce{\frac{[2/3 + (r * 3/4)]}{[1+r]}~=~$0.744604$}$$
where "r" is the molar ratio of propane to ethane being burned. Solving for "r" yields
$$\ce{r~=~$13.59$}$$
In other words $$\ce{moles~\frac{[C2H6]}{[C3H8]}~=~ \frac{[1]}{[$13.59$]}}$$

Thank you for your effort. I completely forgot to look at the bigger picture, to see which reaction is the major contributor.
The correct result is $1/9,19$. Could it be the fault of over simplification of $\ce{\frac[2/3 + (r * 3/4)][1+r]~=~$0.74$}$ ?

I get a different value: $\ce{\frac{[C2H6]}{[C3H8]}} \approx \frac{1}{11.26}$. I let $A$ represent the moles of ethane, $B$ represent the moles of propane, and solved $(2A + 3B) \cdot 44.01 = (3A + 4B) \cdot 18.02 \cdot 1.819$ for either variable. Plugging in any valid pair of values for $(A, B)$ derived from the above relation into the $\ce{CO2}$/$\ce{H2O}$ mass ratio equation gives $1.819$, while using the ratio of $\frac{1}{7}$  yields $1.812$.

I did mine with greater precision and now I get 13.59. The answer is obtained by dividing 2 very small numbers, so the answer is very extremely sensitive to the atomic weights you use for the elements and the number of decimal places you carry along.

It really comes down on how you round up the results.I appreciate both your help.

Yes, the instructor or book made a poor choice in setting up a problem that is so sensitive to numeric input choice and rounding.

source: Stack Exchange

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text:
If dark surfaces absorb/store more light/energy than light surfaces, then does dark skin, on average, store energy from light most efficiently?
If darker surfaces absorb/store more light/energy per square inch than light surfaces, then does dark skin, on average, store energy more efficiently than light skin? (edited).

yes dark skin adsorbs more solar light then pale skin. Do you have a problem with that?

What do you mean by store energy from light? I don't think skin has any chlorophyll. If you mean absorb energy, yes.

Melanin is a chemical compound(s) found in the skin and is responsible for skin pigmentation.  The darker the skin tone, the more melanin is present.  In light-toned-skin, melanin is produced as part of the tanning process when the skin is exposed to UV radiation.  Melanin is a very effective filter for UV radiation.  The more melanin present in the skin, the less light is available for light-catalyzed chemical processes that occur in the skin.
As Dissenter noted in his comment, light energy is not stored per se in the human body.  Rather light catalyzes certain reactions in the skin and the products of these reactions can be stored (at least for a while) for use in the body.
Vitamin D production is an example of an extremely important light-mediated process that begins in the skin.  Vitamin D is thought to play a role in a large number of regulatory processes throughout the body ranging from bone health to cancer.  Because melanin is such an effective filter of sunlight, non-white populations are at an increased risk of vitamin D deficiency.

So do darker skin surfaces utilize/absorb energy more efficiently than lighter skin or not? (a yes or no answer).

Yes, darker skin surfaces absorb light more efficiently.

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text:
What are possible side reactions for the generation of hydrogen via electrolysis?
I recently completed making a simple water electrolysis setup to generate hydrogen which uses two graphite rods from a standard AA battery. The power source is a $24~\mathrm{V}$ $6~\mathrm{A}$ adapter. I am using salt water as electrolyte (The salt is iodized). Strangely, a minute or so after running, a black precipitate starts to form on the negative terminal.
Also, besides the hydrogen, another gas is generated which smells sour, not very sour, but somewhat like lemonade. I have no idea what is causing this.
I thoroughly cleaned the graphite rods before using them. Has anybody an idea about what reaction might be happening here?

24 V is way more than you need and is the source of some of the additional reactions. You only need 2 V for the electrolysis. Since you are using $NaCl$, one of the additional reactions produces chlorine gas (on the anode). The production of chlorine gas needs about 3 V. It would be better to use baking soda and vinegar as your electrolytes (unless you want to use sodium hydroxide). http://en.wikipedia.org/wiki/Electrolysis_of_water

First of all, check again your electrodes to see if this 'black precipitate' is formed at the negative electrode as you state; I predict it's actually happening at the positive electrode for the following reason:
Since you're using two graphite electrodes, it isn't a black precipitate that's forming, but is indeed a side reaction as you state. I suggest what you're actually making is oxygen gas at the positive terminal (as one would predict) which then reacts with the graphite electrode surface to generate $\ce{CO2}$. To verify this,take a look at your electrode—it should have started to break up, now having a non-uniform surface. Perhaps even a bit crumbly.
As LDC3 points out in his comment, you're forming $\ce{Cl2}$ as well as $\ce{O2}$ which is the characteristic smell you're noticing. To understand this we need to look at the standard electrode potential for each of these reactions. Forming $\ce{O2}$ from water at constant $\mathrm{pH}$ (compared to the standard hydrogen electrode) requires a potential difference of $-1.23~\mathrm{V}$ (this value is negative since this is the reverse of the tabulated version). Chlorine evolution on the other hand requires a p.d. of $-1.36~\mathrm{V}$. Remember that the larger (the less negative in this case) the reduction potential, the higher the tendency for reduction. So since $\ce{Cl-}$ reduction potential is more negative, it's less likely to undergo reduction and hence more likely to be oxidised (lose electrons) than water, which agrees with our chemical intuition.
So what we say here is that $\ce{Cl-}$ is preferentially oxidised over water, and as a consequence $\ce{Cl2}$ is formed preferentially over $\ce{O2}$. The reason why you have any $\ce{O2}$ forming at all is due to the electrolyte being the limiting reagent. Your supporting solvent (water) is in such abundance, essentially completely surrounding the electrode, that the system simply has to oxidise some water molecules some of the time.
Of course, experiment should rule the day, so revise your experiment to provide evidence for/refute my theoretical framework. Replace your positive terminal with a high purity metal electrode (copper should work and be readily available at high purity). As for the choice of electrolyte, test a couple of salts whose reduction potentials are less negative than water. Hint: perhaps look at sulphate salts and try and find a further reason why these are very useful in electrochemistry.

Actually, using metal anode is a bad idea, unless it is a special metal. Metal anodes are readily dissolved. Carbon anodes may also be destroyed in some circumstances. Common choices for anodes are lead oxide electrodes (lead electrodes passivated by layer of lead oxide), platinum and in some cases titan (passivated by layer of dioxide).

The other thing is that overpotentials are different for different materials, so you wouldn't necessarily be able to transfer your results to carbon electrodes. e.g. Hydrogen is produced very easily on platinum electrodes, but may need more than a volt more negative on a bismuth electrode.

@permeakra, I understand your point, but I think that this is just a home experiment. I want AvZ to understand the general principles first, as we all did when we learnt this, but your additional info is invaluable so thanks.I forgot to mention about overpotentials so thanks to Michael also!

Hello. Do you know a modern reference to Theoretical Electrochemistry? I only know the book of Vetter.


text:
What does a radical mean here?
I'm really confused as none of the definitions of radical on the internet compare to this.
This is from my college textbook of Applied Chemistry-

Radicals
Every inorganic compound is made up of two electrically charged parts
  called radicals. One part has positive charge and is called basic
  radical, while the other part has negative charge and is called acid
  radical. Hence,
An atom or a group of atoms which forms a part of an inorganic compound is called a radical.
For example :

Sodium Chloride is made up of two parts:

(i) Sodium radical (Na+)  and   (ii) Chloride radical (Cl-)
The two radicals combine to form sodium chloride molecule.

Copper sulphate is made up of two parts:

(i) Cupric radical (Cu2+)  and   (ii) Sulphate radical (SO42-)
The two radicals combine to form copper sulphate molecule.
It should be noted that a radical cannot exist independently.

This definitely doesn't mean the same as free radicals. I don't think it means "functional group" because how can sodium be a functional group?!. Functional groups are in organic chemistry, right?
Can someone explain what the term "radical" means here?

A radical, according to the IUPAC, has an unpaired electron. Radicals may or may not be ions.

A molecular entity such as $\ce{.CH3}$, $\ce{.SnH3}$, $\ce{Cl.}$ possessing an unpaired electron.

http://goldbook.iupac.org/R05066.html
Contrary to your book's claim, the sodium ion is not considered a radical. It actually has all its electrons paired. It is isoelectronic with helium.
Regarding your book's usage of "radical:" your book's terminology is outdated. Here is the IUPAC statement on your book's usage of the term:

In the past, the term 'radical' was used to designate a substituent
  group bound to a molecular entity, as opposed to 'free radical', which
  nowadays is simply called radical. The bound entities may be called
  groups or substituents, but should no longer be called radicals.


text:
Auto ionization of water and evaporation of water
I know that the heating of water drives the auto ionization of water. It also drives the evaporation of water. My question is where do the ions go as the water evaporates as it is heated up. Ions can't be isolated in the solid phase. Do the ions enter the gas phase?

I suspect that the hydronium and hydroxide react to regenerate water before it enters the gas phase. I don't have any way to prove that. It just seems to me that if you have pure water, and some of it ionizes, the ions will be too attracted to each other to want to vaporize. I assume if you heated the water up so fast that it instantly vaporized you could get ions into the gas phase, but I don't have any proof for that either. That's why this is a comment.

That's what I suspect too; we'll just have to wait for a better answer!

If an ion has sufficient kinetic energy to reach the gas phase what is stopping it from evaporating?  It should be far enough away from other ions in solution to make the force holding it in solution comparable to hydrogen bonding holding water in the liquid phase.  Cars near the ocean rust faster because of the salt in the air.  Have you ever accidentally inhaled the vapor coming off boiling aqueous NaOH?  It burns.  I'd say the ions enter the gas pahse the same as water molecules.

There is nothing against ions vaporising. It is far more likely for them do so at least as neutral pairs, though, because of the strong attraction between opposing charges. In the case of autoionised water, this would necessarily mean recombination into molecular $\ce{H2O}$. Uncompensated ions in the gas phase are also possible in certain circumstances, but not common.

@NicolauSakerNeto I would suspect, that via the Grotthuss mechanism it is almost impossible to vaporise hydronium and/or hydroxyl ions.

Could you please clarify where your question points to?
If it is possible to evaporate ions (yes, it is, e.g. MALDI, or other MS techniques), or if it will happen if you cook water in your pot?

The auto-ionization is an equilibrium process, means that it works in both direction. As water evaporates, the auto-ionization converts back the ions into water molecules. You can take it as an example to le chatelier's principle. You gradually remove water from the system, therefore the concentration of hydroxide and oxonioum would go up if they do not turn to water.
Ions in gas phase are rather high energy things, because the dielectric properties of air are much less favorable in stabilizing charges than water can.
It is much more easy to evaporate water molecules.


text:
How were the first molecular formulas determined?
How were molecular formulas determined before spectroscopy and spectronomy? I know that John Dalton thought that the formula of water was "HO" but we now know it to be "H2O." How did Dalton even know that water had hydrogen and oxygen in the first place?

Lavoisier burned hydrogen and oxygen together and found that water was produced.  Dalton incorporated this information into his thinking that the most stable arrangement of any two atoms involved a 1:1 combination, therefore water would be "OH".
In slightly more recent times, but prior to modern-day spectroscopy, chemists would chemically chop molecules into identifiable fragments.  They would also combust the molecule and get a C,H,N,O ratio.  Putting all of this together they would (often) come up with remarkably accurate molecular formulas.

How did he know he was burning hydrogen and oxygen?

They had about a dozen relatively pure, known elements at the time; hydrogen and oxygen were among them.

Well how did they know hydrogen was hydrogen?

Heating hydrides and oxides produced gases.

Sorry if I appear flippant, but how did one know, a priori, that a hydride is a hydride and an oxide is an oxide?

They didn't. All they knew was that when a certain mineral was heated, a gas was released. If the gas couldn't be further separated, they called it an element. Sometimes they were right, sometimes not.

@Dissenter Chemists did a great deal of detailed work on the specific properties of the substances they produced. For example, the density of gases and the ratios in which they combined. We now know what the substances are but they worked out the detailed properties and how they combined and this was the foundation of modern knowledge.


text:
Hydrogenation stereochemistry-Pd,Pt, Ni catalysts
When my textbook talks about hydrogenation using Pt, Pd or Ni heterogeneous catalysts, it never mentions if it is anti or syn addition.
It simply jumps on to say that for alkynes, NiB2 (P-2) catalysts and Lindlar's catalysts are syn hydrogenation.
Can someone tell me if using Pt gives syn or anti for alkynes?

Using any transition metal catalyst (Pt, Pd, Ni) for hydrogenation of alkynes results in syn hydrogenation to not alkenes but alkanes. Using Lindlar's catalyst results in syn hydrogenation of an alkyne to an alkene.
In the case of sodium metal (not a transition metal) in ammonia, hydrogenation of alkynes results in a trans alkene.

Pt gives you syn addition of hydrogens to alkynes, but Pt without anything else (Lindlar's catalyst) will hydrogenate your alkyne to an alkane.

Really!? So, if I write a reaction of
CH3CCCH3 --> CH3CH=CHCH3 (   H2,Pt)
I am actually wrong because that will directly undergo another hydrogenation?
My whole life has been a lie :O

Yes if you don't use Lindlar's catalyst with the transition metal catalyst, then the alkyne will be hydrogenated to an alkane. Sure, the alkyne passes through an alkene, but it's hard to isolate this alkene. http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch09/ch9-6.html

@Dissenter Hydrogenation is not a concerted process.

Ok perfect. Now I get the whole point of using Lindlar's or the NiB2 (P-2). GREAT HELP AS USUAL.

@ron i got that from here: http://www.masterorganicchemistry.com/2013/04/02/alkene-addition-pattern-3-the-concerted-pathway/

@Dissenter  I agree with your diagram, 2 hydrogen atoms are not transferred in a single step, the reaction is not concerted.

Is the linked article incorrect or am I interpreting it incorrectly then?

@ron, I retract, that link is correct!

The link says that hydroboration and epoxidation are concerted.  It does not say hydrogenation on a metal catalyst is concerted - such hydrogenations are two-step, not concerted.

@ron I'm referring to the part where he says " That is, products of syn addition as a result of a concerted mechanism?" and goes on to list "Hydrogenation" as his first bullet point in that particular list. So the author's incorrect? He does say that hydrogenation is concerted. Also the title of the article is "The “Concerted” Pathway."

"In the transition state for this reaction, each of the two hydrogen atoms are delivered to the same face of the alkene. " the author of the link also says that; that seems to weakly suggest that hydrogenation is not concerted (note his usage of "each"). I'm on your side ron, but can you provide some paper or research showing that hydrogenation is not concerted?

yes, both hydrogens are delivered in a step-wise fashion, this is definitely not concerted, look at diimide reduction for an example of a  concerted hydrogenation.

One final comment, some metals (Na\NH3 - it's not a transition metal, but still...) produce pure trans hydrogenation of an alkyne.

Ron you are awesome. I forgot about Na/NH3. Thank you!

You should edit in any information that is in addition to your original answer and then remove the comments. It is now really hard to follow. I even do not see any real relationship to the original question. It seems like the answer is only hidden in the comments. In general it should be avoided to discuss anything in the comments.

All catalytic hydrogenations involving solid metal catalysts transfer hydrogen in a syn fashion to the alkyne to produce the corresponding cis-olefin.  The following figure provides a mechanism that illustrates why the

addition must be syn.
But, with alkynes there are problems.  First, reduction of the resultant alkene is faster than reduction of the starting alkyne, so product mixtures of alkenes and alkanes will result unless poisoned catalysts (Lindlar catalyst) are used or the reaction is stopped early.  A second problem often observed is that the resultant alkene usually remains adsorbed on the catalyst surface.  Consequently cis-trans isomerization of the alkene can take place leading to further contamination of the initial cis-product.

"First, reduction of the resultant alkene is faster than reduction of the starting alkyne". Why?

@yolo123  the first pi bond to be reduced in an alkyne is stronger than the second pi (alkene) bond, so once the strong (alkyne) bond is reduced, the second pi bond goes quick.

By reduced, you mean hydrogenated?
By the way, oxidizing something is equivalent to dehydrogenating?
And oxylation= adding oxygen atom?
Sorry, I'm just orgo newbie!

"By reduced, you mean hydrogenated?" Yes; 2) Yes, dehydrogenation is an example of oxidation.

I would expect that the coordination of the alkyne is the rate determining step. After that the alkyne would most likely already be coordinated and therefore be further reduced.

authors: فرسان الدعوة


text:
Chlorination of Water
Does passing chlorine gas through water form hydrochloric acid or bleach? Also, will a mixture of hydrogen and chlorine gas from electrolysis of brine do the same?

Common house hold bleach is actually basic, as it consists of a solution of sodium hypochlorite, among other things.

@Dissenter I think he's asking if it will form one or the other, not implying that hydrochloric acid is also called bleach. That's how I read it anyways.

Cl2 + H2O = HCl + HOCl

Not sure if it will form bleach or not but Chlorine itself can be used as a bleach. See this.

Passing chlorine gas through water yields a solution of $\ce{Cl2}$, hydrochloric acid ($\ce{HCl}$) and hypochlorous acid ($\ce{HOCl}$). The latter two are formed by reaction of chlorine with water under disproportionation.
$$\ce{Cl2 + H2O \rightleftharpoons HCl + HOCl}$$
Bleaches are chemicals which remove color by oxidation of dyes. $\ce{Cl2}$ and $\ce{HOCl}$ themselves are already strong oxidizers, but adding alkali metal hydroxides to an aqueous chlorine solution increases the solubility of $\ce{Cl2}$, and chlorine bleach is obtained, which contains alkali metal hypochlorite as active ingredient.
$$\ce{Cl2 + OH- \rightleftharpoons ClO- + Cl- + H2O}$$
Sodium hypochlorite is produced by this reaction and common household bleach is an aqueous 3-8 mass-% solution of the salt. The necessary chlorine gas is actually produced by electrolysis of $\ce{NaCl}$.
While gaseous $\ce{Cl2}$ itself can act as a bleaching agent by oxidation, the use of hypochlorite solutions has the benefit of avoiding handling of the poisonous gas. $\ce{H2}$ is a much better reducing agent, and therefore unlikely to be useful for bleaching.
A mixture of $\ce{H2}$ and $\ce{Cl2}$ will react to form gaseous hydrogen chloride when exposed to UV light. Hydrochloric acid is obtained when gaseous $\ce{HCl}$ is dissolved in water.
$$\ce{HCl + H2O \rightleftharpoons H3O+ + Cl-}$$


text:
Siliconization for Organic Compounds
I was wondering if it was possible to create Silane from Methane using the process of siliconization to replace the Carbon with Silicon. For example Silicon Carbide is produced by replacing C-C bonds with a Si-C bond.

What do you mean by siliconization and what is replacing bonds? In drawing, or in actual chemical reaction?

A key difference between carbon and silicon (and why you usually can't make an all-silicon version of an organic compound where all its carbons are replaced with silicon) is that long chains of silicon atoms are not stable. Another key difference is that a hydrogen attached directly to silicon, as in silane, is typically more reactive--say toward hydrolysis or oxidation--that one attached to a carbon atom. I do not know of any process to directly "silanize" a carbon compound, as stated in your question.


text:
Reaction of silver nitrate and hydrochloric acid
Consider the reaction of $\ce{AgNO3}$ and $\ce{HCl}$. I read that silver chloride would be formed. But $\ce{H} > \ce{Ag}$ in reactivity, then how could $\ce{Ag}$ displace $\ce{H}$ from $\ce{HCl}$ ?
If there was a reaction, $\ce{AgCl}$ and $\ce{HNO3}$ were to form. But then these would react again to give us the original compounds back. The actual question was to write down the reaction between $\ce{AgNO3}$ and $\ce{HCl}$. I (wrongly) realized that the compounds wouldn't react because of the reasons stated.

Why should they react back again to the original compounds? There must be some driving force for the reaction in the first place.

One of the most difficult parts of chemistry is learning to recognize a type of reaction based solely on its reactants. This has to be done before you can apply a reaction pattern to the problem, and so it is critical to get this step right in the beginning.
In this case, you are looking at the reaction:
$$
\ce{AgNO3 + HCl ->  AgCl + HNO3}
$$
You already know the products, but are questioning how these products can be formed since the activity of H is greater than Ag, implying that Ag is easier to oxidize than H. This is true - but take a look at the oxidation numbers of each species in this reaction:
$\ce{Ag: +1 -> +1}$
$\ce{NO3: -1 -> -1}$
$\ce{H: +1 -> +1}$
$\ce{Cl: -1 -> -1}$
Since there is no overall transfer of electrons, this can't be a redox reaction, and that means activities won't matter here.
That means there must be another driving force for this reaction - another reaction pattern that fits better.
I don't want to give you the answer directly, but I will give you some advice that should help:
Write the full equation - including the phases.
$$
\ce{AgNO3(aq) + HCl(aq) ->  AgCl(s) + HNO3(aq)}
$$
See if you can find another reaction pattern that fits this equation better. Later, I'll update this answer to show how you can identify the correct pattern for aqueous reactions using only the reactants.
Update
As I mentioned, one of the hardest parts of chemistry is learning to identify reaction patterns based only on the reactants. This is a double-displacement (or metathesis, or precipitation) reaction. It is easy to see now, given that the reactants are aqueous and at least one of the products is solid (the precipitate). Formation of the solid is the driving force for this reaction - the quick explanation is that the forces attracting silver and chloride ions together are stronger than the solvation forces between those ions and water, as well as the forces holding them to sodium and nitrate ions.
This is good, but how do we predict that this is a precipitation reaction ahead of time?
The key is to have a good understanding of the three common types of salt/acid/base reactions that occur in aqueous solution, and to learn to recognize clues in the reactants.
Briefly, the three types that are most commonly seen are:

Precipitation
Acid/base (Arrhenius definition)
Redox (single displacement)

You can recognize each of these by the reactants if you know what to look for.
Precipitation - Two soluble salts (ionic compounds), or a salt with an acid or base.
Acid/base - An Arrhenius acid and base (compound containing hydroxide)
Redox - a salt or acid and an elemental metal
There are many other types of reactions that occur in aqueous solution, and many variations of the acid/base and redox category, but these three cover the cases most commonly seen in a classroom.
Once you have identified the likely pattern that the reaction will follow, the next step is to predict the products using that pattern to see if they make sense. If they do, then you have likely chosen correctly.
You can use this algorithm for more advanced chemistry as well - in organic chemistry, for example, one of the major goals is to learn to predict reactions based on functional groups. Once you can identify functional groups and have memorized reaction patterns for them, it becomes possible to predict a huge range of reactions.

I don't know, I'm not good at chemistry. But is that a double-displacement reaction ? Just guessing from the phases.

@Mriganka, yes - this is a double displacement (a.k.a. precipitation, a.k.a. metathesis) reaction. The giveaway in this case is the insoluble product $\ce{AgCl}$. However, the real question is - how do we figure this out from just the reactants? See my edit for more info.

One thing of paramount importance in Chemistry is to visualize concepts.
According to the solubility table, nitrates are always soluble, so the strong ionic bond between silver ions and nitrate ions are broken by water molecules because of ion-dipole attraction. By definition, acids ionize in water to give mobile ions, so hydrogen chloride in aqueous solution gives out hydrogen ions (and form hydronium ions) and chloride ions.
Hence, you can see that there are five types of particles in the solution now: $\ce{H2O}$ molecules, $\ce{Ag+}$ ions, $\ce{NO3-}$ ions, $\ce{H+}$ ions, and $\ce{Cl-}$ ions. (Slight ionization of water is neglected in this case.) They freely bump into each other as they are mobile. However, this freely moving condition is inhibited by the interaction between $\ce{Ag+}$ ions and $\ce{Cl-}$ ions. According to the solubility table, $\ce{AgCl}$ is insoluble in water. When $\ce{Ag+}$ ions and $\ce{Cl-}$ ions bump into each other, they strongly attract each other, in which the strong ionic force cannot be separated by the ion-dipole force between them and $\ce{H2O}$ molecules. As a result, $\ce{AgCl}$ evolves as a white solid.
Although some may speak this of an example of double displacement reaction, this has no direct relationship with reactivity of hydrogen and silver because there is no transfer of electrons (which occurs in redox reactions).

But H>Ag in reactivity, then how could Ag displace H from HCl ?

What do you mean that hydrogen is more reactive than silver? In what context?

If there was a reaction, AgCl and HNO3 were to form. But then these
  would react again to give us the original compounds back.

How do you know? Why would these form? Why not silver nitrate? Why not hydrochloric acid?

How about $\ce{AgNO3 + HCl <=> AgCl + HNO3}$?

What about that equation?

'How do you know? Why would these form? Why not silver nitrate? Why not hydrochloric acid?' is creating too much confusion in your answer. If by this, you meant that they (AgNO3 + HCl & AgCl + HNO3) will be in equilibrium (did I get it right?), consider writing a simple amswer with an equation.

The OP didn't really put effort into the problem the second time around, and another poster has already provided the answers, so these are just thought questions.

I am going to leave this for the time being, but if you put a little bit of explanation after your rhetorical questions that adds to what thomij has already contributed, it should be fine.  Otherwise, it should probably be a comment.

The silver displaces the hydrogen because this is a double replacement equation, they just switch their anion. When the silver would not be able to displace the hydrogen is when It would have been a single replacement equation like Ag + HCl


text:
Making Zn in school lab
I want to make $\ce{N2O}$ in my school chem lab.
I'm thinking of making it from $\ce{Zn_{(s)}}$ in $\ce{HNO3_{(dil)}}$ :
$$\ce{
Zn_{(s)} + HNO3_{(dil)} -> N2O + Zn(NO3)2 + H2O
}$$
I'm not able to get $\ce{Zn_{(s)}}$, so I would like to make it on my own in lab.

This might help (without zinc): http://chemistry.about.com/b/2013/07/27/how-to-make-nitrous-oxide-or-laughing-gas.htm

Zinc is cheap and readily available. Making it is most likely much harder than the reaction you want to use it for (even if that is the best way to achieve your goal.) Most labs and lab suppliers will have it. Unless you want to set up a metal foundry which seems overambitious to obtain something you can buy off the shelf.

Ahem, forget it.
Theoretically speaking, you can get $\ce{Zn}$ from alkaline batteries (see details here), but you have no chance to get pure $\ce{N2O}$ this way and reaction may easily go out of control.
Attention! Detonation hazard !!!
The usual and easy way to produce $\ce{N2O}$ is to heat very carefully widely available $\ce{NH4NO3}$, to $180-200~^\circ\mathrm{C}$. It is an absolute imperative to keep temperature of the reaction mixture below $240~^\circ\mathrm{C}$. Decomposition of ammonia nitrate ($\ce{NH4NO3}$) is exothermic and may proceed explosively. To be on the safe side, use sand bath to heat the reaction flask and put thermometer into the sand, so its temperature never risen above safe limits. Liquid bath with appropriate liquid may be better, but I'm not sure if a school lab would have proper liquid (silicone oil with high flash point). The produced gases should be cooled in Liebig condenser to separate water, and $\ce{N2O}$ may be further dried with $\ce{H2SO4}$. Of course, all joints must be sealed using proper oil. And keep a bath of cold water to cool the reaction mixture in case if temperature raises above safe limits.
It is advised to use lab grade, not fertilized grade ammonia nitrate.
Possible safer, but costlier route
It is reported that it is possible to produce $\ce{N2O}$ by heating mixture of   $5~\mathrm{mL}$ $6~\mathrm{M}~\ce{HNO3}$, $1~\mathrm{g}~\ce{NH4NO3}$ and $0.05~\mathrm{g}~\ce{NaCl}$ on water bath at $\approx 80~^\circ\mathrm{C}$. If this works, it is a safer route, as hot ammonia nitrate is diluted and autospeeding reaction cannot occur. Link

C'mon folks, can we call it ammonium nitrate instead of ammonia nitrate?

If I edit the post, nobody learns anything. If you are going to succeed in chemistry, you have to learn correct nomenclature.


text:
Why are non-polar hydrogens neglected in computational protein-drug binding studies?
Why are (non-polar) hydrogen atoms generally neglected in a protein-ligand (or drug) binding energy calculation? They do have charge and mass. Doesn't that affect the free energy calculation?
The relevant equations are given here:
$$ \Delta G_\text{bind} = -RT\ln K_\text{d} $$
$$ K_\text{d} = \mathrm{\frac{[Receptor][Acceptor]}{[Complex]}} $$
$$ \Delta G_\text{bind} = \Delta G_\text{desolvation} + \Delta G_\text{motion} + \Delta G_\text{configuration} + \Delta G_\text{interaction} $$

A system where this is the case would be the following, note that only the polar hydrogen atoms are explicitly represented.

are you referring to computer simulations of protein-ligand interactions?

@user137: Yes, I want to know while calculating the above mentioned binding energy(free energy difference), why we neglect non-polar hydrogen.

@DevashishDas because their influence is very, very small? But yes, generally it should be considered, especially if zero energy correction is used.

What kind of calculation are you referring to? Quantum chemical or molecular mechanics, molecular dynamics, empirical values, semiempirical, monte carlo simulations, or a mixture of the previous? Are solvent effects considered? How rigid and how large is the system? Could you elaborate a little more about these parameters, please.


text:
Why is the equilibrium constant given in litre per mole?

At $800\:\mathrm{K}$ a reaction mixture contained $0.5\:\mathrm{mol}$ of $\ce{SO2}$, $0.12\:\mathrm{mol}$ of $\ce{O2}$ and $5\:\mathrm{mol}$ of $\ce{SO3}$ at equilibrium. $K_\mathrm c$ for the equilibrium $$\ce{2SO2 + O2 -> 2SO3}$$ is $833\:\mathrm{L/mol}$. If the volume of the container is $1\:\mathrm{L}$, calculate how much $\ce{O2}$ is to be added at this equilibrium in order to get $5.2\:\mathrm{mol}$ of $\ce{SO3}$ at the same temperature.

The answer comes out to be $0.34\:\mathrm{mol}$.
I have understood the whole question and I am getting the answer. I just have a confusion why we are taking $K_\mathrm c$ as $833$ and not $1/833$ which makes it $\mathrm{mol/L}$. I think $K_\mathrm c$ should be in $\mathrm{mol/L}$.

$K_\mathrm c$ is defined as (for more information see here)
\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} = \frac{\prod \limits_{i \in \text{products}} [c_{i}]^{|\nu_{i}|}}{\prod \limits_{j \in \text{reactants}} [c_{j}]^{|\nu_{j}|}}
\end{equation}
where $\nu_{i}$ and $c_{i}$ are the stochiometric coefficient and the concentration of the $i^{\text{th}}$ component in the reaction, respectively.
If you use this definition for the reaction
\begin{equation}
  \ce{2SO2 + O2 -> 2SO3}
\end{equation}
you get
\begin{equation}
 K_\mathrm c = \frac{[\ce{SO3}]^2}{[\ce{SO2}]^{2} [\ce{O2}]}
\end{equation}
where the notation $[i] = c_{i}$ was used.
This equation tells you that $K_\mathrm c$ has the unit
\begin{equation}
 \text{unit of }K_\mathrm c = \frac{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2}}{\left(\frac{\mathrm{mol}}{\mathrm{L}} \right)^{2} \left(\frac{\mathrm{mol}}{\mathrm{L}} \right)} = \frac{\mathrm{L}}{\mathrm{mol}} \ .
\end{equation}


text:
Why do Group 10 transition metals (and Cu) lose an S electron to fill the D shell?
I don't understand why it varies between Group 10 metals (as with Palladium, Platinum and Darmstadtium) and Copper (Group 11). If each of Palladium, Platinum and Darmstadtium entirely filled the 3d/4d/5d shells then it would make more conceptual sense, but both Platinum and Darmstadtium still have +1/2 spin in the final 4d/5d shell.

a similar question is here http://chemistry.stackexchange.com/questions/11069/why-can-cu-have-an-oxidation-number-of-2

That kind of question comes up regularly but unfortunately there is no easy way to answer it. Have a look at this and this question.

@Philipp -- that's fine, I'm okay with weird explanations. I'm glad I'm not just missing a fundamental exception to the Aufbau principle. Or is there a rule?

Well, there is no rule I know of.


text:
How does mixing lemon juice in hooka water affect the composition of the mixture?
The other day I saw one of my friends, who was mixing lemon juice directly into the flavour chimney (or wind guard) and water pot of hooka. When I asked why are you doing as such, he said that it makes inhaling hooka smoke passive.  By passive, I mean easily inhalable without any hiccups or harshness.  I have tried it, it really makes the smoke passive. But how did this happen?
Here is a diagrammatic view of working principle of hooka:

Here is a list of ingredients of hooka flavour which he used, it was of about 200 grams:

glycerol 70 %
18 % mollases
5 % synthetic edible flavours
5 % sodium xanthate
0.05 % nicotine and 1.95 % water

the temperature at which glycerol boils (and hence the smoke appears) is 290 °C,
By adding lemon juice, we are actually adding citric acid to it, so what would be the products obtained heating that flavour at the given temperature which reduces harshness of hooka.
My friend says the water present at the bottom in the hooka pot, serves for the purification of nicotine present in the flavour (which I can't believe), and he further adds that adding lemon juice to it will act as a filter for nicotine, hence reducing its harshness. [I think there shouldn't be any harshness due to nicotine as it is in a very low concentration i.e. measurable in the order of milligrams per puff, see the comments beside the answer below]
So the harshness must be due to other substances present in the mixture, such as glycerol ( which wouldn't cause harshness as it is in the vapour state), mollases and other flavoured substances along with sodium xanthate, here is molecular structure of mollasses

is mollasses causing harshness, then what is the left over debris after all the flavour is exhausted, many individuals say that the left over & exhausted part of the flavour is nothing but the molasses itself.
and coming to the other substances present in the flavour.
the knowledge which i have gathered from online sources is, the reaction between glycerol and citric acid yields a polymer in a controlled and catalysed reaction, called PGC, which is generally obtained at the temperature of 150°C,  but here in this case the temperature is above 290°C i.e. boiling point of glycerol.
so the fact is that, no polymer is obtained while inhaling hooka, as the glycerol is in its vapour state and citic acid being, very acidic {1M citric acid has a pH value of 2}, so it is nearly in its dissociated state {the lemon juice was mixed in water present in the water pot},

Citric acid

Glycerol

I changed the title to be more in line with what you are asking, since the site is not really in a position to evaluate any health claims.

The safest way to inhale a hooka is to not use tobacco (flavour). Harshness is also no very well defined concept, it is very subjective. In your setup you are burning things/stuffs/chemicals, not only boiling/vaporizing them. As there is an unknown number of chemicals present in the flavour, you cannot determine which may or may not cause harshness. And then again, lemon juice is also a mixture of many chemicals. The answer that you are looking for is far beyond easy.

@Martin wow, i didn't know uptill now that hooka can be inhaled without flavour, can you please elaborate it, i am unable to understand what you said

@agharehanabbas The intention behind my post was to tell you not to smoke. Inhaling smoke of any kind is not healthy and even adding lemon juice to anything won't change that.

@Martin ok got it... thanks anyway, but is it possible to calculate molarity of lemon juice domestically, i mean by using kitchen utensils and some other measuring stuffs ??

The acidic water is probably protonating nicotine, causing it to carry a positive charge. Once charged, the resulting nicotinium cation will not be volatile, but rather, highly soluble. Based on this, I would expect that acidifying the water causes a much lower amount of nicotine  to be pulled through the exit hose (because it is trapped in the acidic water).

as you said, the nicotinium ion would contain +ve charge, on which atom would the +ve charge reside then ?? see here1 for its structure                                                             [1]: http://t0.gstatic.com/images?q=tbn:ANd9GcQG8wDV640_Ps-mar_pzn1axP8LU_RbQkdYxz4ACfgDHclp0vbWIw

but would that minimal concentration of nicotine cause harshness ??

as per calculations, there is a total of 0.1 grams of nicotine in total sachet, 0.025 grams in one 50 grams { which was the quantuty taken in the flavour chimney} and that 50 grams would last for nearly 150- 300 puffs, if we consider even distribution of nicotine through the mixture ( which we can't actually) it would result in a value of milligrams, so i wont think it would cause harshness


text:
Is an SN1 reaction possible in halobenzene?

Is an SN1 reaction possible in halobenzene?

Will the phenyl carbocation be stabilized via resonance? Suppose the leaving group is very electronegative and the environment is protic. What then would be a suitable nucleophile?

Yes, that reaction occurs, but generation of the phenyl carbocation was much more difficult than anyone might have guessed. Here is a drawing of the phenyl carbocation.
First note that the 6 p orbitals making up the aromatic pi system are all still intact and overlapping - the aromatic nature of the benzene ring has not been tampered with.  Then notice that the cationic center exists as an empty $\ce{sp^2}$ orbital where the

$\ce{C-H}$ bond used to be. The empty (cationic) $\ce{sp^2}$ orbital is orthogonal to the aromatic pi system; as a consequence there is no resonance interaction or stabilization of any sort between the two systems.  Further, being constrained by the 6-membered ring, the cationic $\ce{sp^2}$ orbital cannot rehybridize.  This results in considerable (low energy) s-character being used to stabilize an empty orbital, and is not available to stabilize other orbitals that do contain electrons! The vinylic carbocation, the ethylenic analogue of the phenyl cation, is much more stable and common (it can even be observed by nmr) because it is not constrained by a ring and can rehybridrize, thereby lowering its energy. It is both this total lack of stabilization and relatively high (higher than need be if rehybridization could occur) energy of the molecule that has made the phenyl carbocation such an elusive target.
To stack the deck in their favor, investigators have run experiments using extremely good leaving groups such as nitrogen and the triflate anion. Magic acid and its variations, a very powerful acid that stabilizes carbocations, was often employed as well.  The nitrogen leaving group route, using thermal decomposition of aryl diazonium cations has been one of the few methods that has allowed access to this transient intermediate.
A few references if anyone would like to read more:

On Attempts at Solvolytic Generation of Aryl Cations - 1976
Attempts at Solvolytic Generation of Phenyl Cations - 2004
A one paragraph review - 2000

Just to add to the last part of the question: Any nucleophile will be suitable, since this mechanism is solely depending on the dissociation of the phenyl moiety and the leaving group. Once that is done, the strength of the nucleophile is of minor importance.

It's so reactive, sometimes it even recaptures the "diazonium" nitrogen - the leaving group!

I have heard that the carbocationic carbon becomes sp hybridised which causes an angle of 180, which is not possible in a ring. Is that correct?

@AdilMohammed Yes, the carbocation carbon would like to be 180 but is constrained by the ring.


text:
Why can the lone pair not align with the phenyl moiety in 2,6-xylidine?
I am told that for 2,6-xylidine (2,6-dimethylaniline), the amino group cannot line up in such a way that its p-orbital is parallel with respect to the p-orbitals of the carbons in the ring.

I've looked for papers and no one has actually proved to me why I should believe that two methyl groups are bulky enough to interfere with the amino group. I've come across various statements regarding this molecule, but no proof. For example, here's a Wiki article on steric hinderance and resonance.

Steric inhibition of resonance is present only in benzene rings. Presence of any group at the ortho position of benzoic acid,it throws the carboxylic acid group out of the plane so it's mesomeric connection w tin benzene ring vanishes thus ortho substituted benzoic acids are stronger than meta and para substituted benzoic acids.

But there are no citations to back any of that up.
The closest I've found is:

Steric effects of ortho methyl groups are base strengthening. This is not due to steric inhibition of resonance since the conformation remains planar in most derivatives. Two ortho methyl groups are necessary to distort the planarity; their steric effect is more than doubled compared with one methyl group.

http://onlinelibrary.wiley.com/doi/10.1002/poc.642/abstract
But note again – there isn't really any demonstration – just a statement.
Am I supposed to do some Euclidean geometry and use van der Waals radii and trigonometry or something?

possible duplicate of Ortho-effect in substituted aromatic acids and bases

@Martin, I'm not sure this is really a duplicate, as the emphasis on proof in the question (assuming I'm interpreting it correctly) seems to be asking for empirical substantiation of the ortho-effect, while the other question is looking for a theoretical justification.

@GregE. I might have misunderstood the question then. Then again I do not really understand, what the OP considers proof. My statement in the linked answer is quite clear: There is an interaction between the nitrogen and the hydrogen of the phenyl moiety, which is shorter than the sum of the van der Waals radii, hence can be seen as an attractive interaction.

@Martin, I think your answer in the linked article is superb, but I understood the question as asking for empirical/experimental proof in some form. I could be mistaken. Certainly, I think your answer in the linked article is very clear and persuasive, and I'm not sure what proof would be sufficient. It would be good if the OP clarified what type or standard of proof would be adequate.

I mean how does one realize this ona piece of paper, without the use of computer models. Do we just use van der waals radii in that case.

"Without the use of computer models"? That, in conjunction with missing experimental data, just boils down to "I draw it like this on paper, thus it is like this in reality". Models get better and better (read: they gain complexity), and at the beginning end of the spectrum we had pen-and-paper models. Now we need powerful calculators to evaluate these models. I suggest you reframe your question with this in mind.

There is no prove for this statement, as it is most likely incorrect for 2,6-xylidene. This is based on an electronic structure theory approach. However, there are non-negligible steric interactions, that affect the basicity as demonstrated in Ortho-effect in substituted aromatic acids and bases.
Whenever a distance between two atoms becomes (significantly) shorter than the sum of their respective van der Waals radii, this has to be considered. Because of the strain in the xylidene molecules, this is the case.
As a very rough estimate you can use the pen and pencil approach with tabulated values for van der Waals radii (german - much better overview) and covalent radii. Another approach is building the molecule with a molecular modeling set (Molekülbaukasten -  german wikipedia), rotate and wiggle bonds, to see if there are any close contacts.
There is also some free software that lets you build molecules and check them, they usually come equipped with a set of standard bond lengths and angles. It is always a good approach to visualise a molecule.
If you have access to quantum chemical tools you can perform calculations, but this is usually too much of an effort. For this educational purpose I ran some quick calculations, though.
The following table is composed from values from the german wikipedia pages (values in Å):
\begin{array}{rrr}\hline
             & \text{van der Waals} & \text{covalent}\\\hline
\ce{H}       & 1.10  & 0.32 \\
\ce{C}       & 1.70  & 0.77 \\
\ce{N}       & 1.55  & 0.71 \\\hline
\sum\ce{H-H} & 2.20  & 0.64 \\
\sum\ce{C-H} & 2.80  & 1.09 \\
\sum\ce{N-H} & 2.65  & 1.03 \\
\sum\ce{C-N} & 3.20  & 1.48 \\
\sum\ce{C-C} & 3.40  & 1.54 \\\hline
\end{array}
Now we can have a look at the optimised (DF-BP86/def2-TZVPP) geometry of the molecule. (Of course the pen and paper approach will result in a different geometry.)

Here you can see, that the $\ce{H\cdots{}H}$ distance is very close to the sum of the van der Waals radii. The $\ce{C\cdots{}H}$ distance with about $2.5~\mathrm{Å}$ and the $\ce{C\cdots{}N}$ distance with about $2.9~\mathrm{Å}$ are already shorter than the sum f their van der Waals radii. While these interactions are not strong, the will effect the behaviour of this molecule in solution (and gas phase). It is very important to understand, that these are only snapshot values. In reality the molecule is very flexible and bonds will rotate and vibrate. Steric interactions are basically electronic and dispersive interactions in disguise - they are most likely to influence the kinetic behaviour of a molecule.
The best evidence for this is the transition state of the rotation of the amine moiety. Here we can clearly see a bonding interaction between the nitrogen and the neighbouring hydrogen (see linked post), which is evidently influencing the basicity of this molecule.

This state is only about $7~\mathrm{kcal/mol}$ higher in energy, which means it is readily accessible at room temperature. This will mean that there are many different conformations present, there is no true one. This affects the reactivity of a molecule (in this case lowered basicity).
In conclusion, your gut feeling was correct. However, there are more effects at play and the whole picture is often not easy to see. It is usually a case by case decision/ interpretation how strong which effects affect reactions and molecules (properties and barriers). When substituents get larger the more dominant steric effects will obviously be.
I hope this answers your question sufficiently.

Amazing answer; thank you @Martin. So the conformer that can exhibit effective resonance (lone pair on N is parallel with p-orbitals in ring) is slightly higher in energy, not lower in energy as you previously suggested?

@Dissenter The conformation where the lone pair of nitrogen delocalises into the aromatic ring is lower in energy. The conformation which has the shortest nitrogen hydrogen interaction is actually a transition state with respect to the rotation of the amine group. (It is exactly the other way around, than I originally presumed.)

Very nice answer. I have not been able to find experimental confirmation anywhere. However, in principle, I think there are probably ways of determining this. $^1\ce{H}$-NMR may be able to distinguish the two conformers at sufficiently low temperatures where rotation of the amine group is inhibited. Experimentally, one may also compare relative relates and product distribution of, for example, electrophilic substitution reactions. If the amine's lone pair is primarily conjugated to the ring's $\pi$-system, it should activate the ring and direct electrophiles to the- para position.

(Continuing from previous comment.) On the other hand, if the predominant conformer has the amine rotated out of the plane and not conjugated, it effectively becomes an electron-withdrawing group by induction, hence deactivating and meta-directing.

That said, I doubt it would be possible to achieve adequate selectivity in the latter experiment, and being a kinetically controlled reaction, distinguishing TS and ground-state energies is tricky, and drawing conclusions about the latter based on the former is questionable.

Ultimately this means that 2,6-xylidene is less basic than aniline (and 2-methylaniline)? I am confused a little, does this all mean that if nitrogen's lone pair is in a position to conjugate with benzene ring, its basicity decreases, and if it is perpendicular to this position, then it is involved in hydrogen bonding? Is this a two way damage to its basicity? I hope I understand this correctly.

@FreezingFire I would conclude like that. If you'd compare it to ammonia, then a phenyl moiety of any kind reduces basicity. Adding more stress and strive cannot enhance it significantly. And again I have to point out that the whole picture might be much more complicated.

There are a number of different experimental parameters that you could check to attempt to find out more about 2,6-xylidine 1. I performed a few literature searches on different ones. For the remainder of this answer, xylidine is always taken to mean 2,6-xylidine.
Crystal structures
Naturally, the first idea of mine was to look up crytal structures. Unfortunately, xylidine is a liquid at room temperature, which may explain why I was unable to find a structure of the delinquent itself. Instead, I found a number of metal-coordinated xylidines. Hu, Mains and Holt investigated copper(I)-iodide clusters, and amoung them solved the structure of $\ce{[Cu4I4(MeCN)2(L)2]}$ 2 ($\ce{L} = \text{xylidine}$).[1] The structure is shown in figure 1.

Figure 1: Structure of $\ce{[Cu4I4(MeCN)2(L)2]}$ 2 as solved by Hu, Mains and Holt.[1]
From their molecular analysis, $\ce{C20H28Cu4I4N4}$, it is evident that xylidine is a neutral (not anionic) ligand in this complex. The structure shows how the $\ce{N\bond{->}Cu}$ is almost perpendicular to the phenyl ring. This suggests a (possibly distorted since the positions of the hydrogen atoms are not determined) planar amino group as in the first of Martin’s structures.
Humphries et al. studied niobium complexes and solved another few relevant structures.[2] In this answer, I will restrict myself to $\ce{[Nb({\eta^5-}C5H5)(CH3)(N-C6H3{-2,6-}Me2)(NH-C6H3{-2,6-}Me2)]}$ 3. This comlex features the parent xylidine structure both as an imido ligand ($\ce{R-N=M}$) and as an amido ligand ($\ce{R-HN-M}$). In this case, we are dealing with the anion of xylidine (one hydrogen abstracted). The structure is shown in figure 2.

Figure 2: Structure of $\ce{[Nb({\eta^5-}C5H5)(CH3)(N-C6H3{-2,6-}Me2)(NH-C6H3{-2,6-}Me2)]}$ 3 as solved by Humphries et al.[2]
This structure, while a position of the xylidide hydrogen has been determined, has less informative value with respect to the configuration of the amino group: The oxidation state of niobium ($\mathrm{+V}$) is rather high making it rather small and the abstraction of a hydrogen means that an $\mathrm{sp^2}$-type hybrid orbital is used for coordination. Altogether, this means that this structure — while displaying an out-of-phenyl-plane lone pair — is less representative of xylidine and should be considered a representative of N-tert-butylxylidine instead. It should not be surprising that that molecule would adopt a perpendicular amino group rather than a planar one.
NMR shifts
After analysing crystal structures proved unsatisfactory, I turned my attention to NMR shifts. Both xylidine[3] 1 and aniline[4] 4 are featured in the spectral database of organic compounds (SDBS). The corresponding chemical shifts are given in table 1.
Table 1: NMR shifts of selected protons in xylidine 1 and aniline 4. Both compounds $0.04~\mathrm{ml}$ of amine in $0.5~\mathrm{ml}\ \ce{CDCl3}$. Spectra recorded at $90~\mathrm{MHz}$.[3,4]
\begin{array}{lccc}\hline
\text{compound} & \delta(\ce{NH2}) & \delta(\ce{{$m$-}CH}) & \delta(\ce{{$p$-}CH}) \\ \hline
\text{xylidine, }\textbf{1} & 3.46 & 6.93 & 6.62 \\
\text{aniline, }\textbf{4} & 3.55 & 7.12 & 6.73 \\ \hline \end{array}
In my opinion, these differences in chemical shift are minor. The slight upfield shift of the xylidine signals may, in my opinion, be attributed to the $+I$-effects of the two additional methyl groups which, according to theory, affect the meta protons most. Otherwise, hardly any difference can be observed. This seems to point to a structure of xylidine with a mostly planar amino group with respect to the phenyl ring.
$\mathrm{p}K_\mathrm{a}$ values
Finally, I decided to analyse the $\mathrm{p}K_\mathrm{a}$ values. The assumption here is that a lower $\mathrm{p}K_\mathrm{a}$ value indicates a greater stability of the unprotonated species which is most likely if the corresponding lone pair participates in resonance with the phenyl ring. The values are summarised in table 2.
$$\textbf{Table 2: }\mathrm{p}K_\mathrm{a}\text{ values of xylidine and selected related compounds.}\\
\begin{array}{lccc}\hline
\text{acid} & \text{base} & \mathrm{p}K_\mathrm{a} & \text{source}\\ \hline
\text{xylidine}\ce{. H+} & \textbf{1} & \phantom{0}3.90 & [5] \\
\text{aniline}\ce{. H+} & \textbf{4} & \phantom{0}4.87 & [6] \\
\ce{Me2HN^+-C6H5} & \textbf{5} & \phantom{0}5.07 & [7] \\
\ce{CH3NH3+} & \textbf{6} & 10.64 & [8] \\
\ce{(CH3)3NH+} & \textbf{7} & \phantom{0}9.81 & [9] \\
\ce{H2C=C(Me)-NHMe2+} & \textbf{8} & \phantom{0}8.28 & [10] \\ \hline \end{array}$$
Somewhat surprisingly, xylidine 1 has the lowest $\mathrm{p}K_\mathrm{a}$ value — surprising because a lower value should indicate a higher acidity which should indicate a better stabilised free base which does not fit with conjugation and the $+I$ effect of the methyl groups. However, we may also notice that the difference betweeeen xylidine 1 and aniline 4 is small, while methylamine 6 (which I chose in place of ammonia due to the presence of a carbon atom) differs by about $5{-}6$ logarithmic units making it substantially different. The significantly increased acidity of aniline 4 with respect to methylamine 6 is often explained with the delocalisation effect of the phenyl ring.
This opinion has, however, been contested. What we often fail to acknowledge when determining $\mathrm{p}K_\mathrm{a}$ values of phenyl-substituted heteroatoms is the significant difference in electronegativity between $\mathrm{sp^2}$ and $\mathrm{sp^3}$ carbons. Thus, one could say I am comparing apples and oranges when lumping methylammonium’s and anilinium’s $\mathrm{p}K_\mathrm{a}$ values into a single analysis. In fact, the difference between these two is lower than the difference between phenol and cyclohexanol with respect to the acidity constant.
For that reason, I included the seemingly unrelated N,N-dimethylaniline 5, trimethylamine 7 and 2-(dimethylamino)prop-1-ene 8 in the list of $\mathrm{p}K_\mathrm{a}$ values. Compound 8 features a nitrogen atom bound to one $\mathrm{sp^2}$-hybridised carbon and two $\mathrm{sp^3}$ hybridised ones. This is the same carbon types as 5 features with the exception that 5 also features a phenyl ring which is predicted to support delocalisation of the lone pair. Indeed, the difference in $\mathrm{p}K_\mathrm{a}$ values between the conjugate acids of 5 and 8 is significant — $3.2$ logarithmic units — which shows that resonance must feature an additional stabilising effect.
Conclusion
Crystal structures were badly unsatisfactory for determining the structure of xylidine 1. The NMR analysis suggests a strong structural homology to aniline 4. Most notably, however, the $\mathrm{p}K_\mathrm{a}$ values of the conjugate acids 1, 4, 5, 6 and 8 as summarised in table 2 show a clear trend which indicates that 1 should be structurally very similar to 4 featuring a delocalised p lone pair and not similar to 6 with a localised lone pair. The high acidity of 1’s conjugate base cannot be explained with an $\mathrm{sp^2}$ carbon alone. I thus am inclined to believe that xylidine’s amino group is complanar with respect to the phenyl ring and that the lone pair participates in the π system.

References:
[1]: G. Hu, G. J. Mains, E. M. Holt, Inorg. Chim. Acta 1995, 240, 559. DOI: 10.1016/0020-1693(95)04583-X.
[2]: M. J. Humphries, M. L. H. Green, R. E. Douthwaite, L. H. Rees, J. Chem. Soc., Dalton Trans. 2000, 4555. DOI: 10.1039/b007178l.
[3]: Spectral Database for Organic Compounds SDBS, 2,6-xylidine. SDBS-no. 1825. Spectrum no. 1825HSP-03-481.
[4]: Spectral Database for Organic Compounds SDBS, 2,6-xylidine. SDBS-no. 905. Spectrum no. 905HSP-03-391.
[5]: B. H. Asghar, Monatsh. Chemie 2008, 139, 1191. DOI: 10.1007/s00706-008-0913-5.
[6]: Wikipedia entry Aniline
[7]: I. Kaljurand, A. Kütt, L. Sooväli, T. Rodima, V. Mäemets, I. Leito, I. A. Koppel, J. Org. Chem. 2005, 70, 1019. DOI: 10.1021/jo048252w.
[8]: Wikipedia entry Methylamine.
[9]: Wikipedia entry Trimethylamine.
[10]: SciFinder entry for ‘1-Propen-2-amine, N,**N-dimethyl-’, CAS-number 22499-75-8. Acid-base constant predicted by Advanced Chemistry Development (ACD/Labs) Software V11.02 (© 1994–2016 ACD/Labs).

A simple space-filling model shows the plausibility of the effect being steric interaction between the methyl groups and the amino-hydrogens.
This view is a non-optimised space-filling picture of the molecule generated by Chemdoodle:

Even this crude picture shows the plausibility of significant spatial interaction between the left hand methyl group and the amino hydrogens. This is likely enough to constrain the orientation of the amino group giving the effect you query.

Surprisingly, a calculation shows, that the state where the nitrogen lone pair overlaps with the ring is much more stable, than where it is out of plane.

What do you mean by overlap with the ring?

@Dissenter You are the most lucky guy, that I came back by accident to this question. Use @ to notify the person you are asking a question (I believe this is not the first time I am telling you this). With overlap I mean of course that the pi orbital of nitrogen is in line with the pi system of the phenyl ring. Hence the $\ce{NH2}$ is in plane with the ring, just as you suggested.

Wait if this calculation suggests that the aforementioned conformed is more stable, does that destroy the premise of my OP? Or does the molecule not usually adopt this conformation for whatever reason? Thank you @Martin

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text:
Why do distilled alcoholic beverages taste different from each other?
I was wondering about vodka vs. whiskey mostly. The way I understand it, a distilled beverage is created by heating a non-distilled beverage, say, beer, and then collecting and cooling the vapors.
Why would these vapors contain anything but alcohol and some water? Certainly the impurities that color, say, whiskey do not vaporize at the low temperatures involved? Why don't all distilled beverages taste and look like vodka (i.e. water and alcohol)?

Two different aspects come into play here. For once, most of the whiskey aroma stems from aging the fresh destillate in oak barrels previously used to store wines, e.g. sherry. Colour an aroma thus, at least in part, stem from a  slow extraction process using the fresh destillate as a solvent.
But thinking in other beverages, particular the clear fruit spirits, there's more in it. The key to a beverage that doesn't smell and taste just like ethanol but has a cherry aroma is the sloppyness of the destillation :) In the lab, a destillation or rectification is performed to isolate a particular compound from a mixture at a sharp boiling point.
In the production of alcoholic beverages, the destillation typically aims to obtain three fractions over different boiling point ranges.

At lower boiling points, methanol, acetone and ethylacetate are removed.
The upper range mostly furnishes long chain alcohols.

These two fractions aren't meant to be digested ;)
The central fraction, however, will contains all the fragrances, ethanol and water.

And the middle fraction, which contains mostly water and ethanol will also contain a lot of volatile organic compounds from the original fermented source, thanks to steam distillation

Ben Norris: AFAIK, the steam distillation is "good" mainly for immiscible liquids. And according to my experimental knowledge (fruit spirits), the nice aroma comes very late in the alcohol fraction, at about ~25-30% alc. in condensate.

You compare incomparable. Vodka is distilled and immediately bottled and sold. Whiskey is stored for >3 years in wooden casks, often used for wine previously, thereby getting lot of fragrances from wood and also from environment (Scotch whiskey from islands, e.g.). The fresh distilled whiskey is clear colorless liquid. As for the color, 99% of the whiskeys contain added colorants, as the customers think that darker color means more matured and therefore more tasty spirit. AFAIK, the Ardbeg distillery does not add them that much, so this is the genuine whiskey color (more like white wine): http://thewhiskyphiles.files.wordpress.com/2014/07/ardbeg-blasda.jpg?w=339
The aromatic compounds are nevertheless distilled together with the spirit, so you can distinguish fruit distillates, Calvados, Williams, Kirsch, etc...
These compounds (mostly esters) have boiling point somewhere between alcohol and water, thereby during batch distillation, you have to carefully watch the process and collect the smelly fraction, as it is immediately followed by acids.


text:
Common etchant for Stainless Steel 304L and Titanium Grade 2
I have Stainless Steel welded to Titanium. I need a common etchant to etch both S.S. and $\ce{Ti}$ so that I can do further analysis (like SEM etc). Right now I'm etching Ti Side separately and S.S. side separately and doing analysis twice, once for each metal. The etchants I'm using are mentioned below:

$\ce{HNO3 + 3HCl}$ - for S.S. 304L
$5\%~\ce{HCl} + 6\%~\ce{HNO3} + 16\%~\ce{HF} + 74\%~\ce{H2O}$ - for $\ce{Ti}$

Is it possible to combine these two etchants to form a common etchant, or are there any standard etchants available to etch both S.S. and $\ce{Ti}$?

What do you want to accomplish?

After several trials, we were able to come up with an etchant, which can etch both Stainless Steel (316L) and Titanium (Ti-6Al-4V) at the same time. We've used the same etchant for Aluminium (1xxx and 6xxx series) and got very good microstructures. The etchant comprises of equal parts of the following :

Ethanol + HCl + HNO$_{3}$ + HF

Note :

First mix the acids together (HCl + HNO$_{3}$ + HF) and then mix it
with ethanol.
The reaction between alcohol and acid is very vigorous. So take precautions as necessary.
It takes around 5 minutes for ethanol to completely react with these acids. If you want a fast reaction, then use propanol or
methanol instead of ethanol. The final solution should be clear and
colorless.

Common etchant for stainless steel is $\ce{FeCl3}$ with $\ce{HCl}$ (up to 3%) heated at 45°C.
The etching solution for $\ce{Ti}$ looks fine.


text:
What is the correct definition of the Gibbs free energy of activation?
Is the following correct?

Gibbs free energy of activation is used in energy profiles where the stabilities of the species are expressed as changes in Gibbs energy, while the activation energy Ea is used in energy profiles where the stabilities of the species are expressed as changes in enthalpies.

There are a lot of possible definitions and it possibly depends very much on the case and the context you are looking at. However, there are the definitions provided by the IUPAC goldbook, which can be considered the official ones.
The first definition you suggested is correct so far and known as (standard) Gibbs energy of activation (standard free energy of activation) $\Delta^\ddagger{}G^\circ{}$:

The standard Gibbs energy difference between the transition state of a reaction (either an elementary reaction or a stepwise reaction) and the ground state of the reactants. It is calculated from the experimental rate constant  $k$ via the conventional form of the absolute rate equation:
  $$\Delta^\ddagger{}G^\circ{} = \mathcal{R} T \left[\ln\left(\frac{\mathcal{k}_\mathrm{B}}{h}\right) − \ln \left(\frac{k}{T} \right)\right]$$
  where  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant and  $h$ the Planck constant ($\frac{\mathcal{k}_\mathrm{B}}{h} = 2.08358\cdot 10^{10}~\mathrm{K^{−1}s^{−1}}$). The values of the rate constants, and hence Gibbs energies of activation, depend upon the choice of concentration units (or of the thermodynamic standard state).

Your second definition refers to the definition of enthalpy of activation, $\Delta^\ddagger{}H^\circ{}$:

The standard enthalpy of activation $\Delta^\ddagger{}H^\circ{}$ is the enthalpy change that appears in the thermodynamic form of the rate equation obtained from conventional transition state theory. This equation is only correct for a first order reaction, for which the rate constant has the dimension reciprocal time. For a second order reaction, for which the rate constant has the dimension (reciprocal time) × (reciprocal concentration), the left hand side should be read as  $k c^\circ$, where $c^\circ$ denotes the standard concentration (usually  $\mathrm{1 mol\cdot dm^{−3}}$).
  $$k = \frac{\mathcal{k}_\mathrm{B} T}{h} \exp\left\{\frac{\Delta^\ddagger{}S^\circ}{\mathcal{R}}\right\} \cdot\exp\left\{\frac{-\Delta^\ddagger{}H^\circ}{\mathcal{R}T}\right\}$$
  The quantity  $\Delta^\ddagger{}S^\circ$ is the standard entropy of activation, and care must be taken with standard states. In this equation  $\mathcal{k}_\mathrm{B}$ is the Boltzmann constant,  $T$ the absolute temperature,  $h$ the Planck constant, and $\mathcal{R}$ the gas constant. The enthalpy of activation is approximately equal to the activation energy; the conversion of one into the other depends on the molecularity. The enthalpy of activation is always the standard quantity, although the word standard and the superscript $^\circ$ on the symbol are often omitted. The symbol is frequently (but incorrectly) written  $\Delta{}H^\ddagger{}$, where the standard symbol is omitted and the  $\ddagger$ is placed after the  $H$.

The activation energy (Arrhenius activation energy) $E_a$ itself is defined as:

An empirical parameter characterizing the exponential temperature dependence of the rate coefficient, $k$,  $$E_a = \mathcal{R} T^2 \frac{\mathrm{d}(\ln k)}{\mathrm{d} T}$$, where $\mathcal{R}$ is the gas constant and $T$ the thermodynamic temperature. The term is also used for threshold energies in electronic potential surfaces, in which case the term requires careful definition.

And while we are at it, entropy of activation $\Delta^\ddagger{}S^\circ$:

The entropy change that appears in the thermodynamic form of the rate equation obtained from conventional transition-state theory.

Additional note and general warning
Although there are the official definitions, the term activation energy is often used as a synonym for many kinds of different barriers. One has to be very careful when reading articles, which definition is used.
In computational chemistry articles you may also find the terms electronic energy of activation, zero-point corrected energy of activation, solvent corrected energy of activation, ..., etc., pp., and many more.

I thought that the enthalpy of activation and Ea are the same.

@Martin how can we derive the first equation in your answer from first principles?

I think that your definitions are correct, if not perhaps slightly trivial.
You might like to look at the Eyring equation and the terms deriving from that to provide some more thought. It might help to provoke an understanding of the physical intuition of these terms in order to know whether your definitions at least make some sense. Learning how to learn, so to speak!
Some further questions to ask, beyond your definition question:
What does an activation energy mean? What does it mean if we have a negative entropy term of activation? Not in terms of numbers but in terms of what the molecules are actually doing.


text:
Why is magic acid a superacid?

The above is a representation of magic acid. The left part is the acid; the right part is the solvent. Why would this combination make for a particularly good acid?
The left part looks like a sulfuric acid derivative, except one of the oxygens has been replaced by an electronegative fluorine. Why can't we replace all the oxygens with fluorines? Wouldn't that make for an even stronger acid through inductive stabilization of negative charge after proton loss?
Also, does the solvent play a role in magic acid's acidity? Does the highly partially positive Sb central atom play a role in stabilizing negative charge on the solute after the solute has been deprotonated?

Fluorosulfuric acid is already stronger than H2SO4 - that means pure HSO3F is already a superacid. Addition of even small amount of SbF5 makes it much stronger, still.

The explanation is that sufficiently strong Lewis acids like $\ce{SbF5}$ will form adducts even with very weak, traditionally non-coordinating Lewis bases like $\ce{SO3F-}$. The resulting anionic complexes are extremely stable, with the negative charge distributed over a large, highly polarizable molecule and inductively stabilized by the fluorine atoms. The formation of this complex, in turn, promotes the autoprotolysis equilibrium:
$$\ce{HSO3F + HSO3F <=> SO3F- + H2SO3F+}$$
The protonated species above is the most active molecule in sufficiently concentrated solutions. (In reality, the composition of these mixtures may be much more complex, with dimers of the form $\ce{Sb2F10(SO3F)-}$ and possibly more complex oligomeric structures. See, for example, this book.)
While there are some more recent and exotic examples of such Brønsted/Lewis acid mixtures, the idea is actually a fairly old one. Perhaps the most common example: sulfur trioxide in sulfuric acid gives a solution known as oleum, which contains a proportion of disulfuric acid molecules. The resulting mixture is considerably more acidic than pure sulfuric acid alone and has various uses in industrial applications as well as laboratory-scale synthetic organic chemistry.
As for why the oxygens cannot be subsituted with fluorines to yield an even stronger acid system, I can only give a speculative answer. It seems to me that a hypothetical molecule such as, e.g., $\ce{SF3OH}$ (if this is what you're imagining) would probably be very liable to decompose:
$$\ce{SF3OH -> SOF2 + HF}$$
In any case, I can find absolutely no relevant literature, which in and of itself says something about the stability of such a molecule (or perhaps my ineptitude, as the case may be).

Isn't sulfuric acid such a strong acid because of the delocalization effect? Replacing oxygen with fluorine would then only decrease the acidity.

@Jori, I was addressing Dissenter's secondary question as to why sulfuric acid derivatives with all but one $\ce{-OH}$ replaced aren't seen (and probably don't exist). Yes, the structure with oxygens has an advantage in polarizability over fluorines, but (to the best of my knowledge) inductive through-bond and through-space effects are chiefly influenced by electronegativity, similarly for any negative hyperconjugation that may exist, and fluorine is dominant there. The molecules in question are purely hypothetical, so I really have no definitive answer.

Why is magic acid a superacid?

Because it meets the definition of a superacid: A medium having a high acidity, generally greater than that of 100 wt.% sulfuric acid

Why would this combination make for a particularly good acid?

I'm simplifying a bit, but let's examine the following equilibrium (it is one of the early equilibria involved in the formation of Magic acid)
$$\ce{HSO_{3}F + SbF5 <=> H^+ + [SbF_{5}-O-SO_{2}F]^-}$$
The extreme stability of the hexacoordinate antimony anion is what drives the equilibrium to the right.

Why can't we replace all the oxygens with fluorines?

First, the oxygen(s) is used to coordinate to the $\ce{SbF_5}$ structures.  In some structures there is coordination to multiple $\ce{SbF5}$ units through multiple oxygens.  In other structures the oxygens have been removed and replaced with fluorine. Also, you need at least one oxygen so you can have a proton. The equilibria that occur in magic acid are very complex and not entirely understood (for example, see p 52)

does the solvent play a role in magic acid's acidity?

There is no solvent.

I thought the antimony complex was the solvent

Also why would the complex be stable ? Induction?

"I thought the antimony complex was the solvent"  I misunderstood, I thought you meant some traditional solvent.  "why would the complex be stable"  With all of the oxygens and fluorines inductive effects must play a role.  I also suspect that the Sb is playing a role.

@ron " I misunderstood, I thought you meant some traditional solvent" Don't you wish to  update your answer according to that misunderstanding?

@GaurangTandon  I think it is a useful answer as is.

@ron Your answer says in the end "There is no solvent." and then you've acknowledged a misunderstanding in the comments...

@GaurangTandon  I understand, but it is still an answer to a related question and can broaden the readers understanding.

@ron What is the solvent if the equilibrium constant of the above reaction is very big?

In addition to the answers of ron and Greg E. I just want to add another concept, that was not mentioned so far: non-coordinating anions. You can stop reading here or continue and see a short summary.
As a second remark, it is important to know, that fluorosulfuric acid is already a superacid.

Acidity is strongly dependent on the strength of the proton-conjugated-base bond. Basically stating, the weaker the bond, the higher is the acidity.
All currently known superacids operate on this principle and "magic acid" is one of the strongest. The reason for the extraordinary acidity is the high potency of antimony pentafluoride as a lewis acid. It will coordinate to almost anything with electrons, especially itself in liquid phase, forming chain oligo-/polymers $\ce{[SbF5]_{n}}$ with a length between $\ce{n=5-10}$.
If there is something more potent than this, it will go for it. Antimony pentafluoride is a very dangerous substance. This is exactly what is happening in magic acid, forming weakly or non-coordinating anions as conjugated base.
$$\ce{SbF5 + 2 HSO3F <=> H2SO3F+ + [SbF5SO3F]^{-}}.$$
Now this is a fairly large anion with a highly delocalised (due to symmetry constraints) charge. With excess of antimony pentafluoride, even larger anions can be formed, which is the case when used as a solvent.
$$\ce{[SbF5SO3F]^{-} + SbF5 <=> [(SbF5)2SO3F]^{-}}$$
In essence, $\ce{SbF5}$ is shifting the equilibrium of autodissociation of $\ce{HSO3F}$ to the right, creating much greater concentration of $\ce{H2SO3F+}$ - that's protonated fluorosulfuric acid - much stronger than the neutral molecule. It's strong enough to protonate aliphatic hydrocarbons to form carbocations.
There is another superacid, using the same concept, which is even stronger: fluoroantimonic acid. It is in principle that, what you already suggested, substituting the oxygens by fluorines. Instead of $\ce{HSO3F}$, hydrofluoric acid $\ce{HF}$ is used, giving rise to the (to date) most powerful acid. It also makes use of the high affinity of $\ce{SbF5}$ towards fluoride anions.
$$\ce{2SbF5 + 2 HF <=> H2F+ + [Sb2F11]- }$$
$\ce{H2F+}$ is even stronger acid than $\ce{H2SO3F+}$ - it can protonate water twice, even if transiently.

This "naked proton" idea is lame, but I also thought it's right years ago :( In fluoroantimonic it's "just" H2F+ (or even bigger cations). And in magic it's still OH, even if perhaps more polarised.

@Mithoron yes, probably. I don't think it completely invalidates the answer though. So, maybe rework it, but that needs time; which I don't have. :(

OK, I edited it. Check if you think it's OK.

@Mithoron thanks! That was not on my radar anymore, and holy cow is that stuff like...

Oh, I stumbled upon it again, because of the new superacidic question.

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text:
Is the valency of carbon satisfied in graphite?
Graphite has hexagonal parallel planes. In a hexagonal structure the carbon atom has three bonds. Since the valancy of carbon is 4 is the valency satisfied in graphite?

Just as much as in benzene or any other $\mathrm{sp^2}$ carbon case.
The concept of valency doesn't mean that the given atom has to form bonds with the given number of neighbors, but refer to a simple way of counting electrons and formed bonds.
In organic chemistry, we often find carbon forming 4 (single) bonds to neighbors. For example you can find four $\ce{C-H}$ bonds from the central carbon in $\ce{CH4}$ or one $\ce{C-C}$ and three $\ce{C-H}$ bonds from each carbon in $\ce{C2H6}$.
However, in a benzene or ethene $(\ce{H2C=CH2})$ molecule there is only three neighbors of each carbons. In ethene the $\ce{C=C}$ is a double bond (unlike $\ce{C-C}$ in ethylene) which is formed by using two-two valence electrons of the carbons. When you count valence, you count these valence electrons, therefore a double bond counts twice.
The situation of benzene is a little trickier, as it does not have a double or a single bond between the carbons. If the carbons would form a single bond, there would be a extra dangling electron remaining on each of them. In reality these electrons are delocalized (shared) over the whole ring. If you count the valence electrons of a carbon, you should count this delocalized electron, too. You can understand graphite along this line: three bonds to the three neighbors plus the remaining one valence electron delocalized over the ring system (the whole graphite sheet) are together gives back the four valence electrons.
You can find more on hybridization in the other answer.

In the hexagonal graphite structure the carbon atoms are $\ce{sp^2}$ hybridized, just like in benzene.  In graphite, this p-orbital is used for bonding just as it is in benzene, resulting in an extended pi system in the graphite structure.  Graphite is really just a large number of benzene rings annelated together to form a continuous, ring structure.  Just as carbon satisfies its valence requirement of 4 in benzene (3 sigma bonds and 1 pi bond), carbon's valency requirements are satisfied the same way in graphite.

With regards to the physical appearance of both elements : if you notice, pure, mined Graphite is infact a bunch of flat layers (slice) combined together. These slices are fragile. And if you look at diamond, very well known hard build structure. Naturally formed diamonds.


text:
Which has higher rate of hydrogenation - methyl propene or trans-2-butene?

Which has higher rate of hydrogenation- methyl propene or trans-2-butene?

I know that cis-2-butene has a higher rate of hydrogenation than both of these due to steric reasons, but here, both of these compounds have the same steric hindrance.
So, when they come in contact with the hydrogen adsorbed catalyst, they both should behave in a similar manner. Is this correct?

We should be able to determine which molecule will hydrogenate faster by comparing the activation energies for hydrogenation of these two molecules.  A catalyst does not alter the reaction mechanism, it merely reduces the activation energy for the process.
The heat of formation of 2-methyl-propene is ca. 1.3 kcal/mol higher than that of trans-2-butene, presumably because of the adverse steric interaction between the geminal methyl groups in 2-methyl-propene.  The transition state for this reaction will start to resemble the corresponding alkane in its eclipsed conformation (due to the cis-hydrogen delivery).  If we only consider methyl-methyl steric interactions, then for 2-methyl-propene as we approach the transition state the two geminal methyl groups, initially separated by a 120 degree ($\ce{sp^2}$) angle, will displace towards a 109 degree ($\ce{sp^3}$) angular separation.  Therefore, the methyl-methyl steric interaction should increase in 2-methyl-propene as we approach the transition state.  2-methyl-propene's  transition state will be more destabilized than the ground state, consequently increasing the activation energy required to reach it.  With trans-2-butene, there are no methyl-methyl steric interactions in the ground state or transition state; its activation energy, the energy separating the ground state and transition state, should not be increased or decreased.
Based on this analysis, the ground state - transition state energy separation would be expected to increase (slightly) for 2-methyl-propene as compared to trans-2-butene.  Consequently, faster hydrogenation of trans-2-butene would be predicted.  However the effect is expected to be very small.

A catalyst does alter the reaction mechanism - if it would not do that, than it would not participate in the reaction. It does not change the thermodynamics of the system, i.e. the relative stability of the products and educts.

The Cis-2-butene has the higher hydrogenation rate because both hydrogens on the double bond are on the same side, so you can get these close to the catalyst and transfer two more hydrogens on relatively easily. You can mimic this by building a molecular model of the cis-2-butene and use a table as your catalyst. Clearly, the hydrogen side of the double bond fits pretty well. Now build the 2-methyl-propene and trans-2-butene, and try to fit those against the surface as well.
It would also be good to keep in mind that the 2-methyl-propene has a terminal alkene. I'm trying not to answer the question outright and make you do the work, but that terminal alkene is probably very important for determining which reacts more readily.

You need to consider an actual hydrogenation system rather than generics. Hydrogenation is often much more difficult than adding a substrate, finely divided metal and then throwing in hydrogen gas for good fun.
Take a look at Wilkinson's catalyst and its corresponding catalytic cycle (an old catalyst, but one which has plenty of literature). I expect tentatively for the transition state of the migratory insertion of the alkene to play a huge part in the relative reaction rates.
If you consider a four-membered transition state at this stage, then we can possibly imagine a build up of charge on the carbon where the hydride inserts (the more hindered end, the metal on the less) which would be more stabilised if there are additional alkyl chains due to hyperconjugation.


text:
How many empty orbitals do elements have?
A student was contemplating a question regarding elements with "no empty valence orbitals." An argument was made that elements all have infinite number of valence orbitals.

Every atom basically has an infinite number of shells. The thing is that almost all of those shells are empty (they don’t have electrons in them).

https://van.physics.illinois.edu/qa/listing.php?id=1224
So it is facile to say that, say, the fluoride anion has no empty valence orbitals on the basis of it having 8 valence electrons and the $2n^2$ rule (where n = 2)?

While it's true that electron orbitals could go on forever, each element has to have a practical limit. For example, a hydrogen atom can easily hold 1 electron because it has 1 proton. A hydride can have 2 electrons and a -1 charge. However I don't think you can jam 10 electrons on a hydrogen even if the orbitals technically exist to hold them. I don't know what the best way to describe the number of orbitals on an atom is, but calling it infinite may confuse some students.

@user137 its important to note that once you have more than 1 electron, the hydrogenic orbitals are an approximation anyway. Though from a simple perspective you could just say that after a certain point those orbitals have a positive orbital energy and that can explain why you wouldn't ever have $\ce{H^{9-}}$. You could access these orbitals by excitations if you only have 1 electron.

In my view there is an important difference between empty orbitals and empty valence orbital.  There are an infinite number of orbitals, a finite number of filled orbitals, but only one valence orbital.
The definition of a valence electron is usually something like, "The electrons in the outermost shell are the valence electrons--the electrons on an atom that can be gained or lost in a chemical reaction."  Therefore, a valence orbital would be the orbital holding those electrons.  If the valence orbital is full, it is full; there are no other valence orbitals.
Based on this understanding, it would be correct to say "the fluoride anion has no empty valence orbitals."

+1, I agree that the word valence is critical there, and it can be made rigorous by reference to the principal quantum number.

In general I agree.  However, there is further confusion when you get to the transition metals and above where there can be several incomplete orbitals.

I generally agree, but I am a little confused, why you used one valence orbital. Did you mean valence shell? I am just uncertain of how to fit degenerate orbitals, like in oxygen, into this. || In computational chemistry we like to avoid the term valence orbital, being the previously stated one reason. It is more practical to think of occupied and unoccupied (or virtual) orbitals only. It is also impossible to measure an unoccupied orbital, so speaking of it is in a fuzzy grey area...

@Martin Yes, "shell" would have been a better choice of words.

@Jon Custer Sometimes it gets a bit tricky just where to draw the line on what information to include and where to focus the answer.

The outermost orbital is locally referencing the valence orbital(=shell) although valence isn't necessary the outermost. @ron isn't that the summary of your argument/statement?

@bonCodigo I'd say, the outermost orbital containing electrons is locally referencing the valence orbital(=shell); "valence isn't necessary the outermost." - are making reference to Jon's point?


text:
Delocalization of positive charge
How is positive charge delocalized? I understand how negative charge can be delocalized. Electrons are mobile and they can move around. But what about positive charge? For example, consider this resonance structure and its composite:

How exactly is that positive charge delocalized? Protons aren't mobile. I can't see any protons moving through the remaining pi-system.

There are no protons moving - protons cannot move in resonance structures.  In resonance structures all nuclei must remain fixed, only electrons can move.  As you progress from the top-left resonance structure to the top-middle resonance structure you have taken the two pi electrons in what I'll call the 3-4 bond (position 1 being where the bromine is attached) and moved them between carbons 2 and 3, creating a pi bond between carbons 2 and 3 and simultaneously moving the electron vacancy (the positive charge, what a physicist would call a "hole") to carbon 4.

I'm no expert but i think of it like this- there is a shortage of electrons, when the electrons try to satisfy two carbons in a double bond(in the empty p orbital), there is always a shortage somewhere else no matter what the electrons do, we can see this in the resonating structures, so finally they try to partially satisfy everyone with a partial double bond. Now think about it, isn't this easier to understand than delocalisation of a negative charge?
Happy learning:)

This is to supplement ron's answer. The positive charge is due to a deficit of electrons in that region, and that "electron hole" may be treated in many ways as if it is a particle (moving around due to certain rules, etc). Non-particle phenomena that functionally behave as particles are dubbed "quasiparticles", and there are a wide range of examples in physics (eg. phonons). In this and many other applications (such a semiconductor behavior), this positive charge caused by lower local electron density may be considered a quasiparticle.