text:
Finding the percentage of pure CaCO3
I found this in a textbook; it seems that the question is poorly worded. It goes as follows:

$1.5~\mathrm{g}$ of chalk were treated with $10~\mathrm{ml}$ of $4~\mathrm{N}$ $\ce{HCl}$. The chalk was dissolved and the solution made to $100~\mathrm{ml}$. $25~\mathrm{ml}$ of this solution required $18.75~\mathrm{ml}$ of $0.2~\mathrm{N}$ $\ce{NaOH}$ solution for complete neutralization. Calculate the percentage of pure $\ce{CaCO3}$ in the sample of chalk.

I don't get it when the question says the chalk was dissolved and the solution made to $100~\mathrm{ml}$. Does this refer to the chalk which already reacted with $\ce{HCl}$, or was another $1.5~\mathrm{g}$ sample of $\ce{CaCO3}$ taken?

It is a little poorly worded but still comprehensible. It means that the chalk dissolved upon treatment with HCl and the resultant solution was diluted.

By the way, see my edit for an example of how to format mathematical or chemical expressions properly.

The question is a bit poorly written. But it also gives a sense to the procedure of dissolution of chalk upon treatment with $\ce{HCl}$ and dilution of the solution.
The reaction that takes place after the treatment with $\ce{HCl}$ is;
$\ce{CaCO3}$+ 2$\ce{HCl}$ → $\ce{CaCl2}$ + $\ce{H2CO3}$
$\ce{H2CO3}$ decomposes into $\ce{H2O}$ and $\ce{CO2}$.
$\ce{CO2}$  is a gas. It bubbles up out of the solution.
$\ce{CaCl2}$ is soluble in water. So it should be dissolved by adding water.Further, it should then be diluted to the required volume.
"The chalk was dissolved" might signify the addition of water to dissolve the water soluble $\ce{CaCl2}$ produced during chemical reation with $\ce{HCl}$.

Water is already present inside the HCl solution.

Obviously it will be present in HCl solution. But it might be needed in case of conc ones.


text:
Could altered silica gel which is releasing its water content when in contact with fire be used as a fire retardant?
Silica gels absorb moisture. I intend to use silica gels as fire retardants. My question is - Is it possible to chemically alter silica gels to release water content when in contact with fire, thus suppressing the fire?
Here's my idea. Since silica gels are essentially silicon dioxides and since they are tetrahedral in nature, replacing some of the silicon atoms with carbon atom will produce carbon dioxide which is useful in putting out fires.

"replacing some of the silicon atoms with carbon" How do you plan to do that? If everything was that easy I would've made diamonds out of the dirt in kitchen.

You don't have to alter silica gel to make it release its water content upon heating. It already does that. Carbon dioxide is another story, though.

Gypsum wallboard soled this problem a long time ago.

Silica gel would be a bad fire retardant. You do not need to modify it to make it release water in fire: it happens on itself.
The reason water works well to put out fires is because of its endothermic vaporisation. When you put liquid water in fire it vaporises and takes a lot of heat from the fire. Silica gel can only absorb a certain amount of water which is not that much. You will need a lot of silica gel, so much it will become impractical.
Yes, silica is tetrahedral but you can't just replace silicon with carbon. It does not work like this. Carbon oxides are gaseous, not solid like silica. The reason why carbon dioxide works against fires is that it's not oxygen, so that suppresses a critical component for fires. Even if you could store somehow carbon in solid dioxide, there will not be enough to actually do anything.

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text:
Finding out the composition of seawater/brine?
As part of a project, I am working on precipitating salts from seawater and brine. I have a sample of seawater and I need to measure the concentrations of $\ce{Mg^2+, Ca^2+ and Li+}$ present in the sample. As well as being able to measure $\ce{MgCl2, CaCl2 and LiCl}$.
What analytical instrument would be best for this experiment? We have a decent budget but dont want to invest in equipment too large (i.e ICP-AES or ICP-MS) as the laboratory isnt very big.
I have looked into ion-chromatography and can get one for about 20k USD, anyone else got any better ideas or reasons not to go for ion chromatograph?

What about gravametric methods?

20k US will buy a lot of analysis by people with expertise in whatever analysis you choose.  They can even help you choose the best analytical technique.  And no reputable lab would try to "oversell" you a more expensive technique if a cheaper one were available.  Also, no lab does it all, but a good lab that can do some of your analysis should be able to direct you to another reputable lab for whatever analysis they do not perform.  As you may know by now if you've gone this route, operating and maintaining an IC is far from free...20k just gets you in the door.

Your budget should allow you to measure  $\ce{Mg^{2+}, Ca^{2+}, Li+ and Cl-}$ in-house using ion specific electrode analysis. There will certainly be some training and maintenance involved, but anyone with a bit of a chemistry background will be able to learn this technique. Adding to the simplicity, and thus the ease of training, is the fact that your samples are all of a fairly similar composition.
Ion chromatography, which would basically require one set up for anions and one for cations, would break your bank in a hurry unless you have an experienced IC analyst with some time to spare and access to good used equipment.  Frankly, this just doesn't seem like a realistic approach.
One thing to keep in mind is that your budget will buy a lot of analysis by people with expertise in whatever analysis you choose to perform. They will even help you choose the best analytical technique for your needs.
My advice is to look at outsourcing first, particularly if you are going to be analyzing fewer than 10-20 samples per day.  If you will be analyzing a large number of samples, or if you need very quick turnaround times, then I suggest looking into ion specific electrode analysis.

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Why are Kc and Kp interchangeable in the free energy-equilibrium equation?
In this equation that relates Standard free energy and the equilibrium constant:
$\Delta Gº=-RT \ln K_{eq}$
My textbook (and my teacher) say both $K_p$ (constant related to pressure) and $K_c$ (constant related to concentration) can be used in the place of $K_{eq}$.
When using the equation that relates $K_p$ with $K_c$
$K_p=K_c(RT)^{\Delta n}$
We would get two different $\Delta Gº$ values, which shouldn't be possible.
My question is, why can we use both constants if their values are different?

Also the same question http://chemistry.stackexchange.com/questions/40567/which-equilibrium-constant-is-appropriate-to-use/40586#40586

The equilibrium constant is defined by activities:

The activity of a species in a liquid solution can be approximated to: $a=c/c^0$ where c is the concentration of the species and $c^o$ is the standard concentration ($1M$). This approximation leads to $K_{eq}=K_c$. The activity of gasses is called fugacity and it can be approximated to: $f=p/p^0$ where p is the pressure of the gas and $p^0$ is the standard pressure ($1$ $bar$). This leads to $K_{eq}=K_p$. Therefore, if a reaction takes place in e.g. water, one should use $K_c$, but if the reaction takes place in the gaseous state, $K_p$ should be used when calculating $dG$.

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Mechanism for reaction between hydrochloric acid and magnesium
Balanced equation:
$$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}$$
However, Cl- is a spectator ion so the actual reaction equation is:
$$\ce{Mg + 2H+ -> Mg^2+ + H2}$$
Here is my proposed reaction mechanism:

A magnesium atom reacts with a $\ce{H+}$ ion: $\ce{Mg + H+ -> Mg+ + H}$
The $\ce{Mg+}$ intermediate produced reacts with another $\ce{H+}$ ion: $\ce{Mg+ + H+ -> Mg^2+ + H}$
The two hydrogen atoms (from step 1 & 2) react together: $\ce{H + H -> H2}$

However, my problem arises when I compare this to my rate law that I found experimentally ($\ce{Mg}$ is not in the rate law as it is a solid and has no concentration):
$$\mathrm{rate} = k[\ce{H+}]^2$$
None of these mechanism steps above conform to my observed rate law (or do they?), could any of you enlighten me to any other mechanisms?
There are only two mechanisms I can think of: a one-step mechanism where it is literally $\ce{Mg + 2H+ -> Mg^2+ + H2}$ (though I've read that a one-step reaction is unlikely) and the other way is via an intermediate (as described above).

Please visit this page, this page and this one on how to format your posts better. Alternatively, visit this chatroom for further formatting guidance.

I suspect the reaction mechanism is much more complicated than what you have proposed. Have tried searching any publications? If so it would be helpful if you could include this into your post. For example, protons in aqueous environment are hydrated and even $\ce{H3+O}$ is already an approximation. I also doubt the existence of hydrogen atoms in liquid phase. From there on in it probably gets even more complicated.

In a similar question on this site it is proposed that the HCL dissociates in water to form H3O+ ions which then oxidize the magnesium.

What rate law do you expect from your proposed mechanism(s)? (Remember to replace the concentration of any intermediates with terms based on their generation, so the final expression only contains substrate concentration terms.)

Martin: I can't find any official publications on the mechanism unfortunately :( I've tried to think of a mechanism involving H3O+, however, this just ends up with the same general mechanism as in the opening post. It also leaves 2H2O on the right hand side of the equation when cancelled down (is having H2O when cancelled down allowed?).
Joel: HCl is already dissociated in solution so is not in the mechanism. I believe the Mg does not get oxidised straight away and instead is done via multiple/compicated steps e.g. intermediates. (Although it would conform to my mechanism, I think(!))

R.M: First step = [H+] (no Mg as it is a solid)
Second step = k1[H+]
Third step = 2k1[H+]

oops, didnt mention: @Martin-マーチン

oops, didnt mention: @Joel

oops, didnt mention: @R.M.

Based on the three step mechanism you have provided, what if the slow step is your step 3 which is H + H --> H2  so rate law is now k3[H]^2. Since H is an intermediate, you can substitute for this intermediate from earlier steps and you will end up with
Rate = (k3.k1.k2.[Mg][H+]^2)/[Mg2+]   and if [Mg]/[Mg2+] is constant as the reaction of a solid always happens at the surface, you can get a rate law = k [HCl]^2

The chemistry of magnesium is somewhat more complicated than you might think. First, Mg metal is always covered with OH groups - Mg(OH)2 is quite insoluble, and so a piece of magnesium does not react with water (pH = 7). Well, not very rapidly, anyway - we say that the magnesium is passivated and corrosion is stifled by its insoluble hydroxide coating.
Magnesium anodes are used in glass-coated steel hot water heaters to delay corrosion of the steel if the glass cracks, exposing the outer casing. Sometimes the magnesium is alloyed with aluminum to change the corrosion chemistry of the anode. So, while magnesium doesn't react with water rapidly, it can be anodic enough to protect steel, especially if it is warmer than room temperature (Mg(OH)2 is more soluble in hot water than in cold).
So, the first step in the reaction of magnesium with water must be the dissolution of some Mg(OH)2. This is also the case with making Grignard reagents: magnesium ribbon hardly ever reacts immediately when you try to make a Grignard. You crack the ribbon under the solvent, you grind away at it, then finally, it just takes off and reacts rapidly. Sometimes you have to add a more reactive organic halide, or iodine. But the idea is to break thru the passive coating to expose bare metal, which is quite reactive.
Magnesium powder can be pyrophoric in air. Once you get it burning, it is difficult to put it out - water won't do it. CO2 won't do it. Dry sand might.
Experimentally, you may determine a rate and develop some formula to describe the rate, but establishing the mechanism will have to take into account the passive layer and how it dissolves. In fact, that may be the only thing you can examine, because when you break thru the passive layer to bare magnesium metal, the reaction rate should zoom up and then be limited by diffusion, stirring, removal of H2 gas bubbles, and who knows what else.


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Reasoning questions on Solubility
Why is solute dissolved faster in amorphous state and at high temperature?
I don't have any idea about the dissolving of a solute in amorphous state but for higher temperature it might be because the solution is more unaaturated; but I am not sure.

Things at higher temperature dissolve faster because the molecules move faster with heat, increasing the speed of the reaction. Things in an amorphous state have their molecules arranged randomly, their structure is less stable giving them greater solubility than crystalline structures.


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Relationship between inhibition constant Ki and interaction energy?
Inhibition of an enzyme $E$ by an inhibitor $I$:
$$
E + I \rightleftharpoons EI
$$
The inhibition constant $K_i$ is defined as follows:
$$
K_i = \frac{[E][I]}{EI}
$$
where $[E]$ is the enzyme concentration, $[I]$ is the inhibitor concentration, and $[EI]$ is the enzyme-inhibitor-complex concentration. Inhibition is strong when $K_i$ is low; that is, when the equilibrium above is shifted toward the product.
Can the strength of inhibition be approximated by the relative interaction energy: $E_{Interaction} = E_{EI complex} - E_{Enzyme} - E_{Inhibitor}$? This seems like a logical and valid hypothesis to me, but I cannot find any sources confirming that this is justified. If the enzyme-inhibitor complex is very stable, I would think that the equilibrium above naturally is shifted toward the product, given a relatively fast kinetics. In the case of a commercial drug inhibiting the HIV-1 protease, for example, where it is safe to assume fast kinetics, I would model the strength of inhibition as the relative interaction energy: stronger interaction $\Rightarrow$ lower $K_i$ value.
The context is a simple theoretical modelling of a protein-ligand interaction using density functional theory.

What do you mean by interaction energy? and what are $E_{Enzyme}$ and others? Free energy?

One complication of enzymatic inhibition constants is that there a multiple mechanisms to inhibition, and the type of inhibition can change how the inhibition constant relates to binding energy.
On one side of the spectrum you have competitive inhibition, which means that the $EI$ complex is functionally inactive, and cannot bind substrate. (The easiest way of getting competitive inhibition of an enzyme is to have the inhibitor bind in the active site, competing with the substrate binding. But that's not the only way. You can also have allosteric inhibitors which exhibit "competitive inhibition", if they bind the apo enzyme and by doing so prevent substrate binding.) This is the sort of inhibition you're assuming in your question, but isn't the only type of inhibition.
On the other side of the spectrum, you have uncompetitive inhibition. (N.B. uncompetitive - noncompetitive inhibition is something different.) This means that the inhibitor requires the substrate to be bound to the enzyme prior to binding itself, and it's the $ESI$ complex which is functionally inactive.
And then you have a whole range of variation where you have a mix of effects (binding to both the apo and substrate-bound forms, with various affinities), of which noncompetitive inhibition is a particular type.
You're correct in that for pure competitive inhibition, the $K_i$ is equal to the $K_D$ of the inhibitor. The problem arises when you have one of those mixed effect or uncompetitive inhibitors, where catalytic steps and substrate concentrations get folded (partially) into the $K_i$.
Personally, I would recommend Enzyme Kinetics and Mechanism by Paul F. Cook and W.W. Cleland, which goes into great details about reaction mechanism, kinetics and inhibition constants. It should include derivations and relations for all the relevant scenarios.


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Carbon numbering in carbohydrates

Carbons have been numbered differently in two different pictures. What are the rules of carbon numbering?

Actually, the numbering is the same, at least for the fructose portion of the molecule (which is the right-hand monosaccharide in your sucrose image - the one on the left is glucose).
The key point in carbohydrate numbering is to look for the carbon which is an aldehyde or a ketone (the carbonyl carbon). This will be the end of the monosaccharide which is numbered lowest. For aldoses, this carbon would be carbon 1. For ketoses, it's the lowest number that can be reached in the linear, ring-open form (e.g. for fructose, it's 2). The rest of the carbons are numbered accordingly.
When you have a linear Fischer projection (like the "D-fructose (linear)" images on the second image) the carbonyl carbon should be obvious. In the ring form, it's less obvious.
Instead of looking for an aldehyde or a ketone in the cyclic form, you're looking for an acetal or hemiacetal. That is, you're looking for a carbon with multiple bonds to oxygens. In the first image, this is the carbon which is making the oxygen bridge between the two monomers (the one on the left of the furanose) - the oxygen of the bridge is one oxygen, and the oxygen of the furanose ring is the other. In the second picture, it's the carbon on the far right. You have the ring oxygen of the furanose, and then a free hydroxyl oxygen. Both of these carbons are numbered "2", and if you imagine rotating the furanose ring 180 degrees to flip over that acetal position, you'll see the other carbons match up as well.
(One thing that might be tripping you up is the directionality of the 1 carbon with respect to the 2 carbon. Keep in mind that the hemiacetal form can interconvert with the ring-open carbonyl form, and the "anomeric position" of the monosaccharide is free to invert in chirality - that is, it can readily flip between "up" and "down". It's only when it's in the bonded acetal form that it's locked into either the alpha or beta configuration.)


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Identify what alloy of metal tool
The title is more general than my particular question, in particular I bought a "allen key" and it broke after doing a lot of force with it trying to turn a hard screw.
In order to get a harder tool I need to know the material of this one. Tell me how to test if this allen key is chrome vanadium. If i should weight it, measure volume thus density, and this picture:
broken allen
The other broke like this, it deformed:
deformed allen key, because forcing
I want to be also a little educated on this topics: probably the tool is chrome vanadium but of low quality, or maybe it is extremely expensive or hard to test for me if the new tools I bought are of another material and if it's harder than the broken ones, the new ones are chrome vanadium and "stanley" (a corporation).

I think if they were of chrome vanadium steel, they wouldnt have broke or deformed. I made like 50 kg force of force. I'm not a chemist.

You need a key that is both tough (deforms rather than shatters) and strong (does not bend easily). To some extent, these qualities conflict with each other. BTW, often a tight bolt can be removed by an impact wrench, or a quick snap of the key... the same is true of the lug-bolts of a car wheel.

You can tell by the dark gray appearance that it is not vanadium chrome, but rather an alloy of hardened steel.  If it were made of vanadium chrome, it would not have broken, and it would also have a silver appearance to it.  Density may help tell if you are close, but with alloys the density does not necessarily conform to the rule of mixtures.

How do you know if have not broken? What would have happened, bended?

Would have likely stripped the hex-hole out


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Are heterolytic dissociations barrier-less or do they have an activation energy?
When a bond is broken into ionic fragments, and I'm specifically thinking about organometallic or coordination chemistry, the two products carry opposite charges and the potential energy curve should, in my humble opinion, be attractive everywhere (at every point where the distance is larger than its equilibrium value).
Does this mean that the dissociation of a compound like ferrocene into $\ce{FeCp+}$ and $\ce{Cp-}$ (in the gas phase) occurs without a barrier?

What is IMHO? If the potential energy is a curve, does it not already imply an energy barrier?

Yeah, what you say implies exactly the opposite.

IMHO = "in my humble opinion". The fact that the potential energy curve is attractive everywhere just means that the energy rises the further you pull the two ions apart. A barrier implies that the energy rises up to the transition state and then falls again as you approach the product valley. You could re-phrase my question as: Is there a transition state?

In my theoretical research I have often seen that, indeed, there is no barrier for such reactions. In modeling the kinetics you take the bond dissociation energy as the forward barrier and assume a zero barrier going backward. This will also be true for radical recombination reactions.

The discussion stands and falls with your definition of a barrier. Let me note that your’s is a slightly more obscure one in my opinion. I would have understood a barrier as any movement that requires an energy input to proceed.
In your example of a heterolytic dissociation, there is — as you noted — always an attractive force between the compounds you are separating, and thus there is always an incentive for the system to return to the undissociated state. With increased separation, the energy difference per separation length grows smaller, so at the beginning you need more force to separate a smaller distance. To me, that is a barrier.
If you really define a barrier as a local maximum on the potential energy surface, then please note that all dissociations would be barrierless. The typical homolytic dissociation curve does not have any local minimum; it only converges to a constant value at infinite separation.


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What does gases at constant pressure mean?
My teacher told me that if we keep some weight on a piston containing 2 different gases, both gases will be at constant pressure, and hence the volume occupied by each gas will be proportional to number of moles of each gas according to the ideal gas law ($pV=nRT$)
My doubt is that won't the 2 gases exert different pressures and the sum of the two different pressures will be equal and opposite to the pressure exerted by the weight? Also, if the two gases exert different pressures, then how can we say that both gases are at constant pressure? Do gases at constant pressure exert equal pressure?

Without a clearer setup it's difficult to say whether someone misspoke, whether you misunderstood precisely what your teacher was saying, or whether your teacher was incorrect.
If the setup is that there is a single piston that contains two gases, and a weight is placed upon it, then:

The volume of each of the gases in that piston is equal to the entire volume of the piston.  The volume of any gas is the entire volume of its container.  If the gases are ideal, then the volume of the piston will be the sum of the volumes of the piston if each of the gases were kept in it individually.
The pressures of the two gases sum to the pressure exerted on the piston by the weight (and the outside atmosphere.)  The pressures will be constant as long as the amount of gas present, the applied pressure, and the temperature are kept constant.

If the setup is that there are two pistons that each contain a gas, and an identical weight is placed upon each (or a single piston filled with two gases, one at a time) then:

The volume of the two pistons will be proportional to the number of moles of ideal gas that it contains, as long as both are also at the same temperature (we're already assuming the same pressure due to the identical weight.)
The pressures of the gases in the two pistons will be equal, because they are opposing the same weight + outside atmospheric pressure.

As to your other questions:
Gases at different pressures do indeed exert constant pressure, assuming things like constant temperature, constant volume, and constant number of moles of gas.  The pressure of a gas is the result of huge numbers of gas particles striking the walls of their container every second.  The rate at which those particles strike and rebound and the force with which they do so determines the pressure.  Because there are sooooo many gas particles in any reasonable quantity of gas, the fluctuation in the rate and intensity of those collisions is nearly constant (to many, many decimal places.)
Gases at constant pressure do not necessarily exert equal pressures.  If I pump up the front tire of a bicycle to twice the pressure as the back tire, each has a constant pressure, but they are different.  If I put two moles of gas A and one mole of gas B into a sealed container, the pressure due to A will be twice that of the pressure due to B, but both will be constant, as will their sum, the total pressure of the container.

Thankyou for ypur reply.I didn't understand the line:"gases at different pressure do indeed exert constant pressure."What is the difference between the pressure exerted by gas and gas at any pressure?

@Raksh23 I meant to say that although the partial pressures (the individual pressures of two or more gases in a mixture) are probably different, they are still constant.  If I have 1 atm of mixed nitrogen and oxygen, with a 2:1 ratio of N2 to O2, the pressure of N2 will be 2/3 atm and the pressure of O2 will be 1/3 atm.  Though they are different, those values will remain constant (unless other factors like temperature, volume, etc change.)  I won't find that both the N2 and the O2 are suddenly producing 1/2 atm, for example.

Saying that the gases are at constant pressure means that the pressures do not vary with spatial position within the cylinder.  Both gases occupy the entire volume, and, as you said, they exert unequal pressures.  The pressure that a gas exerts within a mixture is called its partial pressure.  The sum of the partial pressures of the gases add up to the total pressure, as you also said.  The partial pressures are unequal, and the partial pressure of each gas is equal to the total pressure in the cylinder times the mole fraction of that gas.  The total pressure supports the weight of the piston, so it is equal to the weight of the piston divided by its cross sectional area.  I don't know what your teacher said, but your own understanding of what is happening is right on target.


text:
Calcium hydroxide and why there are parenthesis
Basics but in calcium hydroxide Ca(OH)2 ,
*Where the 2 is a subscript what do the parenthesis symbolize? Why isnt it just CaOH2??

$\ce{CaOH2}$ has one calcium, one oxygen and two hydrogen atoms. $\ce{(OH)2}$ means that there are two OH groups.

Parenthesis in a chemical formula indicate that the elements inside them form a group. If followed by a sub-scripted number then there are that number of the group attached.
So Ca(OH)2 is different from CaOH2. It has an extra oxygen.


text:
How was the composition of DNA studied before the knowledge of its structure?
In a large polymer molecule (such as a DNA chain) which is made from multiple component monomers containing many atoms, how was it determined that the chain was made up from the four groups of specific combinations of monomers of elements (in this case, adenine, thymine, guanine and cytosine)?
In other words how were they able to determine that the DNA contained the four specific bases before they knew the structure of the DNA molecule?
In general, doesn't breaking up a large molecule into its constituent atoms also break up the monomers?

In the early experiments performed by Erwin Chargaff and co-workers to determine the composition of DNA, leading to the deduction of the Chargaff's rules, the total DNA was extracted, chemically digested and the products were analysed using paper chromatography. The DNA was not broken down to the level of atoms which is certainly impossible using the standard chemical treatment procedures.
This is an excerpt from one of the earliest papers by Chargaff [1]:

Two procedures were compared. In one, which will be referred to here
  as Procedure 1, the nitrogenous components were liberated and
  determined by the methods described previously from this laboratory
  (5, 8, 12, 13); i.e., the purines were set free by N H2S04 (1 hour,
  100o) and chromatographed in aqueous butanol-diethylene glycol
  (NH3 atmosphere); the pyrimidines were liberated with
  concentrated formic acid (2 hours, 175o) after removal of the purines
  as the hydrochlorides by treatment with methanolic HCl, and
  chromatographed in aqueous butanol.
The second method, Procedure 2, is considerably simpler and permits the
  use of even smaller quantities of DNA (2 to 6 mg. for a complete
  analysis) than was possible heretofore; its use was briefly mentioned
  in a recent paper (14). It is based on the simultaneous liberation of
  all purines and pyrimidines by the hydrolysis of the DNA with
  concentrated formic acid and on their separation, on the same
  chromatogram, by means of ammoniacal butanol. When it is desired to
  establish complete balances, i.e. to express the molar distribution of
  individual nitrogenous constituents not only in terms of P present in
  the hydrolysate, but also in relation to the total N and P of the
  nucleic acid, about 6 mg. of DNA are used in the analysis. The sample
  (dried in a high vacuum at 60o for 3 hours) was heated in a sealed
  small bomb tube (about 220 × 6.5 mm.) with 0.8 cc. of concentrated
  formic acid (98 to 100 per cent) at 175o for 2 hours. The hydrolysate
  was transferred quantitatively to a 1 cc. volumetric flask and
  adjusted to volume by means of small portions of distilled water which
  served to rinse the hydrolysis vessel. Aliquots, measured with an
  accurate microburette in the usual manner (12), were employed for the
  calorimetric P determination (0.04 cc. of hydrolysate) and for the
  chromatographic separation (0.01 cc.). When very small quantities
  of DNA were to be analyzed, 2 mg. of the substance were heated with
  0.2 cc. of concentrated formic acid and the hydrolysate was, without volume adjustment, analyzed for P and for individual purines and
  pyrimidines. In this case, all computations were based on the ratio of
  moles of nitrogenous constituent to moles of phosphorus present in the
  hydrolysate (13, 14). The chromatographic and spectroscopic techniques
  were, in general, similar to those described previously (8, 12). The
  paper sheets (16 × 46 cm.), carrying five lanes 3 cm. wide, were left
  for 20 minutes in an ammonia atmosphere in order to neutralize the
  acid. The solvent used for the chromatographic separation consisted of
  a mixture of 6 volumes of n-butanol and 1 volume of 0.6 N aqueous
  ammonia. Two beakers, one containing the solvent mixture, the other
  0.6 N ammonia, were placed along- side each paper sheet at the bottom of the jar used for chromatography; the paper sheets were suspended,
  the solvent trough being left empty, and allowed to equilibrate with
  the solvent atmosphere for 1 hour in the closed vessel. Then the
  trough was filled with the solvent mixture and the separation
  permitted to proceed for 18 to 20 hours, at which time the solvent
  front had passed over the lower edge of the paper. After the papers
  had dried in air, the position of the separated components was marked
  under a Mineralight lamp or a similar ultraviolet lamp.4 The
  components, in the order of increasing distance from the starting
  point, were aligned as follows (the distance of thymine arbitrarily
  taken as 100) : guanine (26), cytosine (47), adenine (71), thymine.
  The quantitative determinations were carried out as described before
  (12).

 Chargaff, Erwin, et al. "The composition of the desoxyribonucleic acid of salmon sperm." Journal of Biological Chemistry 192.1 (1951): 223-230.


text:
Does the salt in the ocean act as a binding agent on sand?
When I walk on the beach, there appears to be an upper layer of the sand that is 'crusty'.

This appears to be the opposite of what you see in sand in a dessert setting, where the individual particles are not bound together, and blow to form 'dunes'.
I'm trying to figure out what causes the sand to bind together like this.
To me the crystalline properties of salt could cause this. (Ie the way that salt binds together to form crystals could cause this salt-binding that leads to 'crustiness'.
My question is: Does the salt in the ocean act as a binding agent on sand?

The saltwater recrystallizing could definetly aid in forming the crust on top of the sand. There would be no true bonding between the SiO2 and the salts present in the water though they are just crystallizing together.

Sounds plausible to me. Try it at home, both with salted and pure water; then you'll be able to tell for sure. Chemistry is an experimental science, after all.

Good observations!
Seawater contains salts in small amounts such as such as $\ce{LiCl}$ and $\ce{MgCl2}$ that are quite hygroscopic, which could keep the sand wet longer.
There is also dissolved organic matter (http://www.eoearth.org/view/article/154471/) in seawater that might act as a hydrating agent or as a wetting agent, helping to hold onto the water and to bond it to the sand.
You might try an experiment comparing the appearance of (washed) sand after being wet with plain water, with $\ce{NaCl}$ salt water of the same concentration as seawater, and with seawater itself to see if there is a difference. Though this seems like a casual test, this could be of significance in cases where it's necessary to drive on a sandy shore.


text:
Why is the molecular peak not observed in the mass spectrum of 2,2-dimethylbutane?
Why isn't the molecular weight of 2,2-dimethylbutane on the mass spectrum? I can tell that a $\ce{-CH3}$ group is removed just by the math, but I do not know the reason why. The molecular weight is 86 and I know it has to do with fragmentation.

The complete spectrum shown on http://webbook.nist.gov/cgi/cbook.cgi?ID=C75832&Mask=200#Mass-Spec actually shows a very small peak at m/z = 86 (relative intensity about 0.1 %).

The answer probably has to do with how the mass spectrum was obtained.
Most mass spectra that are provided online, including the ones on the NIST WebBook, are electron ionisation mass spectra. That means that you put the sample in a chamber, accelerate electrons to a really high speed (giving them approximately $70~\mathrm{eV}$ of energy), and then bombard the sample with those high-energy electrons in order to ionise the sample.
The ionisation of these organic molecules doesn't take that much energy - $15~\mathrm{eV}$ is a good approximation. So you have an extra $55~\mathrm{eV}$ of energy in each molecule. How can the molecule get rid of this excess energy? The simplest way is to break chemical bonds; a typical bond strength lies between $3$ and $10~\mathrm{eV}$.
Branched compounds in general do tend to fragment more easily than straight-chain compounds. The reason is simple. If you think about n-hexane, no matter which bond you cleave, you're going to end up with a primary cation (or radical). However, if you cleave at one of the branching points of 2,2-dimethylbutane, you'll get a tertiary cation (or radical). A more stable product means that fragmentation occurs more readily.
However, the fact that the molecular ion is not observed is not really unique to 2,2-dimethylbutane. Lots of EI mass spectra don't have the $\ce{M+}$ peak. If you want to observe the $\ce{M+}$ peak, it's a better idea to use a gentler ionisation technique, such as chemical ionisation or electrospray ionisation.

I'll add that once ionized, molecules can also relocate the charge to form a more stable ion.  It just depends on the molecule and where it was ionized.


text:
What is the structure of iodide of Millon's base?
In qualitative analysis, $\ce{NH4+}$ ions are confirmed using Nessler's         reagent:-
$$\ce{NH4+ + 2[HgI4]^{2−} + 4OH− → HgO·Hg(NH2)I ↓ + 7I^{−} + 3H2O}$$
The brown precipitate formed is called iodide of Millon's base.  It is also written as  $\ce{3HgO.Hg(NH3)2I2}$ and as $\ce{NH2.Hg2I3}$. (wikipedia).
But its structure is present differently in different sites:-

Source 1 - ionic form

Source 2  - Here the reaction is given differently:-

$$\ce{([K2]HgI4 <=> 2KI + HgI2) * 2}$$
$$\ce{HgI2 + NH3 -> Hg[NH2]I + NH4I}$$
$$\ce{Hg[NH2]I + HgI2 -> NH2Hg2I3 }$$
$$\ce{[K2]HgI4 + 2NH3 -> NH2Hg2I3 + 4KI + KH4I }$$
structure -

Source 3- non ionic form

Source 4 - Hydrate form; Here also the reaction is different:-
$$\ce{[K2]HgI4 -> 2KI + HgI2}$$
$$\ce{HgI2 + NH3 -> Hg[NH2]I + HI}$$

This site also says that its structure is still uncertain. Various other structure has been proposed:-

So, what is the correct reaction, correct formula and correct structure of iodide of Millon's base?

http://scripts.iucr.org/cgi-bin/paper?S0365110X51000453 https://en.wikipedia.org/wiki/Mercuric_amidochloride

@permeakra, the paper showed the structure of millon base. Not the iodide of millon's base.

the second link provides structure for similar compounds, chloride of milton's base . It should give an idea, I believe. Anyway, it is absolutely clear, that non of structures provided in the question has anything to do with real life: the iodide is definitely polymeric.

First, ignore any monomeric formula. The compound is definitely polymeric.
The literature contains hints for existence of two separate lines of compounds. The first one is salts of polyamidomercury: $\ce{[Hg(NH2)]_{n}A_{n}}${1} with mercury-amide chains in the structure, and $\ce{(Hg2N)_{n}A_{n}}$ with structure based on crystoballite {2}{3}
It seems thatn the lines are not distinguished in earlier literature, but recent researchers use the words "Millon's base salts" for $\ce{(Hg2N)_{n}A_{n}}$ only.


text:
Coordinate Bond...(Dative Bond)
Is NH4Cl Dative or Ionic or Covalent ?
I made the Lewis structure but I was not able to make it...
After many tries I posted this question
Some where I saw that they are made up of Ions but contain Covalent bond..

It is all of the above.

The bonding between $\ce{NH4+}$ and $\ce{Cl-}$ is ionic.
The bonding between $\ce{NH3}$ and $\ce{H+}$ in $\ce{NH4+}$ is a dative bond. (Since $\ce{H+}$ ion has no electrons, 2 electrons from $\ce{N}$ atom in $\ce{NH3}$ form the coordinate bond with $\ce{H+}$, thus forming $\ce{NH4+}$)
The bonding between $\ce{N}$ atom and $\ce{H}$ atom in $\ce{NH3}$ is covalent.


text:
Chemical kinetics of a reaction, rate constant, and activation energy
From the Arrhenius equation in kinetics of reactions,
$$k = A\exp{\left(-\frac{E_\mathrm{a}}{RT}\right)}$$
Which tells us about the temperature dependence on rate constant of a reaction.
Activation energy is dependent on the temperature (as it is the energy difference between the average energy of the reactants and threshold energy, and since the average energy of reactants depends on the temperature, activation energy also should depend on temperature) Is my assumption that activation energy depends on temperature correct?
Moreover, is there any case where activation energy is zero or negative? (like in the case of spontaneous reaction or $E_\mathrm{a} = 0$) I feel that whenever activation energy is zero, all the molecules undergoing collisions should form products successfully, but this case won't occur.
But when we substitute $E_\mathrm{a} = 0$, in the Arrhenius equation, we get the result as $k = A$, i.e. rate constant will be equal to pre-exponential factor.

The rate constant and the activation energy tell us how fast the reaction proceeds (moles/sec), not whether it is spontaneous. The higher the temperature, the higher the rate constant and, for specified concentrations of reactants, the faster the reaction. The rate of a reaction always increases with increasing temperature (higher rate of collisions), so the activation energy is always positive.  As the temperature increases, the rate constant approaches the pre-exponential factor (E/T approaches zero). For some reactions, the activation energy is very low, so even at lower temperatures, the rate constant approaches the pre-exponential factor. The activation energy has a weak dependence on temperature, and, in practice, this temperature dependence is usually  neglected.

But when we consider the reaction between two oppositely charged gaseous ions, this reaction almost doesn't require any kind of activation energy, as they readily attract each other making the reaction spontaneous.
So, for these kind of spontaneous reactions, can we assume that activation energy is zero.

Moreover other than the kinetic and potential energies, what are the other energies that sum up in the Activation energy?
Lastly what exactly is meant by reaction coordinate that is plotted on the X-axis in the energy diagram.
Please clarify these doubts.

With regard to your first comment, the answer is yes.  With regard to your second comment, my background does not permit me to respond about all the different contributions to the activation energy, but certainly, potential and kinetic energies are part of it.  Regarding the energy diagram, it  is only schematic, and the x axis only qualitatively represents the progress in moving from the reactants to the products.

So, when there is an increase in temperature of a reaction ,what can be concluded about the activation energy of the reaction ?
Decreases or it remains constant....
Because according to the concept the average energy of the reactant molecules increase as the temperature increases, the transition state or the activated complex state is not changed. so, theoretically  the activation energy decreases but practically it should remain constant as the transition state is unaltered.
Please explain this point.

E doesn't increase (substantially), but E/T decreases, so k increases.

Does that means activation energy remains constant, even when temperature is increased?

Very nearly.  There is a slight temperature dependence that is usually neglected in practice.

Although Chester Miller has very well answered the question, I would like to add a (rather long) comment to the case of zero barrier. His answer says that

(...) [t]he rate of a reaction always increases with increasing temperature (higher rate of collisions), so the activation energy is always positive. (...) [emphasis added]

Although I believe this to be perfectly correct in the context of Arrhenius (and Eyring) theory, Marcus theory allows (slightly) different behaviour. Here, the rate constant for an electron-transfer reaction follows the equation:
$$k_{et} = \frac{2 \pi}{\hbar} |H_{AB}|^2 \frac{1}{\sqrt{4 \pi \lambda k_B T}} \exp \left(-\frac{(\lambda + \Delta G^\circ)^2}{4 \lambda k_B T}\right)$$
$\lambda$ is the so called total reorganization energy accompanying a charge redistribution. There might be cases where $\Delta G^\circ$ (total Gibbs free energy change for the electron-transfer) equals $-\lambda$, and we get something like the following for a series of similar reactions:

Figure 1. Intramolecular electron-transfer rate constants as a function of free energy change. Electrons are transferred from biphenyl ions to different acceptor groups (image source, J. Am. Chem. Soc. 1984, 106, 3049-3050).
The maximum above is associated to an essentially barrierless electron-transfer. Thus, to answer (part of) the original question,

(...) Moreover, is there any case where activation energy is zero or negative? (...)

Marcus theory allows zero, but not negative, barriers.


text:
Why aren't people pouring ethanol my unskinned chicken to remove excessive fat?
I see a lot of people putting cooked chicken thighs on paper towels to absorb the excessive fat. This gave me an idea: if you don't care about taste, why can't we rinse cooked chicken with ethanol and pour away the mixture?
The ethanol will quickly evaporate and the excessive grease from the chicken will be rinsed away.

Sounds like this might work, but it is way more expensive than the conventional method.

I can't help but consider the possibility of fried chicken which can get you drunk. Now there's something with the potential to take off...

And, perhaps some of us do care about the taste? Heck, we could soak them in carbon tetrachloride - that would degrease them nice and quick...

Poulet Flambé, anyone?

@JonCuster Too toxic

@Sparrowcide - yes, I'm quite familiar with the health issues with carbon tet - that was tongue-in-cheek. But I'm not about to start pouring grain alcohol over my chicken either - the remaining fat comes off in a paper towel just fine, which then gets thrown away. Your solution causes additional disposal problems with no net gain.


text:
Utility of Delta G
What is the main utility of delta G? As far as I could understand, if we do not know whether the reaction we are dealing with occurs in nature or not, we would not be able to tell in which direction it will proceed. Delta G will only tell us whether the reaction with compounds on one side being A + B  and compound on other side be C + D, whether A+B will react to give C+D or whether C+D react to give A+B. Am I right?
(If possible please explain in simple language. Not a chemistry student)

$\Delta G$ is a convenient way of determining the spontaneity of a reaction that takes place in constant temperature and pressure. As you might know, a positive $\Delta G$ would mean that the reaction would not be spontaneous under normal conditions, and the opposite would go for a negative $\Delta G$.
Your assumption about what we determine the outcome of the reactions to be is only half-correct; there are always direct or indirect methods to compute enthalpies and entropies.
Thermodynamics isn't everything though. There is kinetics.

Thermodynamics should, as Professor Zare noted, really be called "thermostatics." Thermodynamics is not about things moving and changing but instead about how stable they are in one state versus another, while kinetics is about how quickly or slowly species react. It is dangerously easy to confuse thermodynamic quantities like free energy with kinetic ones like activation energy. Probably for this reason, thermodynamics and kinetics are often taught separately. – Stanford University Site

Imagine a mixture of hydrogen and oxygen gas. Thermodynamics favors that hydrogen and oxygen react:
$$\ce{2H2 +O2->2H2O}$$
However, this reaction won't happen unless you provide activation energy in the form of some heat, spark etc. IOW, this is a famous reaction for having a high kinetic barrier despite thermodynamic favorability.
Take into account the reverse reaction: $$\ce{2H2O->2H2 +O2}$$
This reaction is non-spontaneous. Does that mean it won't happen? It means it won't happen by the pathway deduced. But you can use electrolysis to make it happen. So being $\mbox{non-spontaneous}$ only means the reaction won't occur as you now have written it.
To sum up, the only thing $\Delta G$ can tell you is that whether by your proposed mechanism the reaction will proceed or not. How fast that might be is pure kinetics. Note that the usual famous $\Delta G = \Delta H - T\Delta S$ only holds for constant temperature and pressure.


text:
How many bonds can Hydrogen make?
Recently I saw a video which told me that Hydrogen has a valency of 1, i.e, Hydrogen can only bond with 1 other atom. But since hydrogen wants a complete shell, it can have 2 covalent bonds with 2 electrons. Please explain.

Your sentences are a bit unclear. Did the video mention that hydrogen makes two covalent bonds? Well, that is the case in some three-centered bonds, but I don't think that's what they thought.

Hydrogen has already one own electron in the shell. So it admits only one extra electron from some other atom.

1+1=2. Hydrogen has 1 electron. When it make a bond, it "gets" the second bonding electron, ergo has 2 and thus a complete shell.

Apart from some Group 13 weirdness, hydrogen can only make one bond. Covalent bonds require pairs of electrons and hydrogen can only have two electrons bound in one covalent bond.

Actually it's more than just group 13. Some carbocations, the trihydrogen cation etc. all have 3c-2e bonds involving hydrogen.

It's even more than that.  The hydrogen bonding in water may be interpreted as delocalization of the O-H vonds onto neighboring molecules.  Ditto for ammonia, hydrogen fluoride, DMSO to which a strong acid has been added to make solvated protons, etc.  With no inner shells to encumber it, hydrogen is making delocalized bonds in many condensed media.

Hydrogen forms 1 single bond when there is a s-s overlap which is the greatest overlap but it can also form 2 single bond where there are 2 s-p overlaps such as is
Even in compounds like $\ce{B2H6}$, H seems to make 2 bonds but it doesn't.

As you can see the two Hydrogen atoms in the centre make 2 bonds each with Boron.
The number of a bonds that hydrogen can make is dependent on the type of bonding.
It can form only a single ionic bond.

It can form two bonds when there is overlap involved (covalent bond)

It can form a single covalent bond.

If stability can be achieved it can form a lot of coordinate bonds, but due to it's small size and only a single proton I would think a max of 1 coordinated bond can be formed if we being with an ionised H-atom.

(-1) Hydrogen does not form two "single bonds" in B2H6

It does, two s-p overlaps

It is a 3c2e bond and has absolutely nothing to do with the atomic orbitals that are overlapping

3c-2e is a classification just like 4c-3e bond on I2Cl6 but the number of bonds hydrogen make is 2. I will confirm this by tomorrow and give you a more accurate reply.

@AgyeyArya Hydrogen makes two bonds (sort of) but they are not single bonds.

After further reading on the matter. I would like to revoke what I said. Hydrogen make half a bond and hence not 2 bonds.

Since ionic bonds have no direction you cannot count them. Please do not use strike through commands. If it is wrong or not important, delete it. Strike through makes it really hard to follow and understand what you are trying to convey.


text:
Naming a 4 branched cyclic compound
If this compound exists, what will be its name?

Can it be called bicyclo[9.2.2.2] hepta-decane ?

Related: Cubane systematic name and How to name tricyclo carbon compounds?

Your selection of the main ring and the bridges that results in the bridge lengths ‘[9.2.2.2]’ and your counting of the total number of skeletal atoms that results in the name ‘heptadecane’ are correct.
For your proposed name ‘bicyclo[9.2.2.2]heptadecane’, you obviously followed the nomenclature of similar bicyclic systems, such as bicyclo[9.2.2]pentadecane.

This is not a bad idea since the nomenclature of polycyclic hydrocarbon ring systems is actually based on the nomenclature of bicyclic systems.
However, the structure given in the question is not a bicyclic ring system because it includes one additional secondary bridge; therefore, the structure actually is a tricyclic system.

The additional steps for the naming of tricyclic hydrocarbons are described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

P-23.2.5 Naming and numbering tricyclic alicyclic hydrocarbons
P-23.2.5.1 Tricyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system (…). Rings not described by the bicyclic system are defined by citing the number of atoms in the independent secondary bridge as an arabic number. The locants of the two attachment points of the independent secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower number is cited first) separated by a comma.
The name of the tricyclic system is then constructed by citing:
(a) the prefix ‘tricyclo’, in place of ‘bicyclo’, indicating the presence of three rings in the polyalicyclic system;
(b) numbers indicating the bridge lengths, starting with the two branches of the main ring (…), followed by the main bridge (…), and the secondary bridge (with superscript locants separated by commas indicating its points of attachment to the main ring), all separated by full stops and placed in brackets,
(c) the name of the acyclic hydrocarbon having the same total number of skeletal atoms.

Therefore, the correct name for the structure given in the question is tricyclo[9.2.2.21,11]heptadecane.


text:
What is the meaning of the "-osic" suffix on a metal (e.g., cobaltosic oxide)?
I just came across the term cobaltosic for the first time in the abstract of Ind Eng Chem Res 52(18): 6076, 2013 (emphasis added):

A nanocomposite of cobaltosic oxide and nitrogen-doped graphene ($\ce{Co3O4}$/N-G) was prepared by the facile hydrothermal method.

I assume this is a historical name; Wikipedia's article on the cobalt spinel mentions cobaltosic oxide as an "other name", but provides no other historical information.  Google searches on manganosic oxide and ferrosic oxide turn up the sort of results I would expect, though not all consistent:
Manganese

Analogous to cobalt, manganosic oxide appears to be the $\ce{Mn3O4}$ spinel; e.g., see the abstract here (PDF at link).

Iron

The term ferrosic appears to refer (at least) to a mixed-oxidation state hydroxide, $\ce{Fe3(OH)8}$; e.g., here.
However, it is also used to refer to the iron $\ce{Fe3O4}$ spinel (magnetite); e.g., here.

What is the meaning of this "-osic" suffix?

In the course of searching for examples of usage of the "-osic" suffix, I think I found the answer to my own question.  From the last link in the question, a web store selling pigments (emphasis added):

The source of natural black oxide usually is from an iron ore called magnetite. Magnetite, also known as lodestone, is a ferrous ferric oxide (ferrosic oxide) that is a heavy, black color and opaque.

So, it would appear that the suffix indicates a compound of mixed oxidation state: a combination of -ous and -ic forms of the metal.


text:
Chemical potential and it's implications in an open system
Take the fundamental equation for the Gibbs free energy of a system:
$$dG=VdP-SdT+\sum_i\mu_idn_i$$
Thus, the chemical potential can be defined in terms of the Gibbs free energy as follows:
$$\mu_i=(\frac{\partial G}{\partial n_i})_{P,T}$$
Now, assuming that the chemical potential is positive (i'm not sure if that is always true) if I add $dn$ of component $i$, $dG>0$ which means (given that the Gibbs free energy is the capacity of a system to do non PV work) I have enabled the system to do more work. Consequently, I have done work on the system by adding more substance to it because i think I have increased it's potential - like raising a mass to a greater height.
It makes sense to me that adding a substance to a system would increase it's capacity to do work however, it doesn't make sense to me that work must be done to add moles of a substance. It's more clear when I think about the converse...
If I remove a small amount of the substance I reduce it's capacity for work and the system must therefore give out energy. This is confusing to me because surely i'd have to do work to remove a portion of a substance from a system - breaking the intramolecular forces and so on. However, for an ideal gas, there aren't any intramolecular interactions so unless the chemical potential for an ideal gas is zero (which I assume it's not) I must be missing something?

Too long for a comment, so here goes.
It's quite subtle and I'm really not sure how to explain in words, but the gist is that the chemical potential does not reflect a real, physical process of adding a substance that can be carried out. If that were the case, then the chemical potential would be the difference in the Gibbs free energies of the two systems:

System 1: $n$ moles of substance and $\mathrm{d}n$ moles of substance, separated by some imaginary barrier
System 2: $n + \mathrm{d}n$ moles of substance

and measuring the chemical potential would simply mean removing the imaginary barrier, allowing intermolecular attractions to form, etc. But that's not the case.
The chemical potential is the difference between these two systems:

System 1: $n$ moles of substance
System 2: $n + \mathrm{d}n$ moles of substance

That's all there is to it. Where do the extra $\mathrm{d}n$ moles come from? It doesn't come from anywhere. There's no real physical process going from System 1 to 2 here, so don't think of it as one.
Since Gibbs free energy is an extensive property, the $\mathrm{d}n$ moles of substance carry their own Gibbs free energy, just like they carry their own internal energy. That is why the chemical potential is positive - yes, the formation of intermolecular forces, etc. will affect the exact value of the chemical potential, and in a sense that is why the chemical potential is a partial derivative - but remember just simply having more of the substance means that there is more Gibbs free energy.

Surely there must be a physical process? I'm not sure what you mean by there isn't one

No, there isn't. Physical processes must obey conservation laws - in going from system 1 to system 2 you are "adding extra substance that comes from nowhere". Don't think of it as having to represent some process, because by virtue of its mathematical definition, it just doesn't. It's weird!


text:
Why is the eclipsed conformer more stable than the bisected conformer in 1-butene?
I have just started learning conformational analysis, and a major doubt came into my mind.
In simple alkanes such as ethane, the staggered conformer is much more stable than the eclipsed conformer. This fact may be explained as follows:

In the staggered conformer, the repulsion between the hydrogens is minimized as they are "anti" with respect to each other.
We have a case of favorable overlap between sigma bonding MO of the C-H bond, and sigma anti bonding orbital of the C-H bond of the second Carbon.

I have no issues with this part. Then, I started reading about alkenes. In the simple example of propene, two cases may arise, as shown in the figure.

Note that (2) can be obtained from (1) by rotating the C-C bond by 60 degrees. Unlike in the case of alkanes, the eclipsed form is more stable, not the bisected form. This can be easily accounted for:
In the eclipsed form, the Hydrogen is oriented in the right way to allow for effective Hyperconjugation. This involves delocalization of electrons from C-H sigma bond to the C=C pi antibonding orbital.
My main question is about when it becomes more complicated. Let me take the example of 1-butene. Here, we have twice the number of eclipsed and bisected conformers. I have labelled them as E1, E2, B1 and B2 (E= eclipsed;B=bisected)

The correct order of stability is E2 > E1 > B1 > B2.
I do not understand this order. When I attempted to find the order of stability, I got: E2 > B1 > B2 > E1. I based this mainly on steric effects.

E2 is clearly the most stable as we have Hyperconjugation due to the C-H bond which is oriented in just the right way.
I couldn't get the rest of the order. Why is E1 the second most stable? I understand that it is eclipsed, but shouldn't that lead to greater torsional strain? Is Hyperconjugation the reason for stability, again?
Can Hyperconjugation take place when we have a C-C bond(when there is no alpha Hydrogen with respect to the double bond), or is it restricted to C-H bonds(where we have an alpha Hydrogen)?

Related: Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered

Satisfactory answer, but doesn't explain the "new bond interactions" that come about in 1-butene. Only talks about Hyperconjugation facilitated by presence of an alpha Hydrogen with respect to the double bond. What if this alpha Hydrogen is absent?

There are still $\alpha $-hydrogens in 1-butene

Hyperconjugation makes the Es more stable than the Bs and steric effects make E2 more stable than E1 and B1 more stable than B2. Both C-C and C-H bonds can participate in hyperconjugation.

In alkanes staggered conformers (all anti) are local minima, while eclipsed conformers represent transition states. The first is "stable", while the latter is not. The rotational barrier has been measured and calculated often, it is around 12 kJ/mol.[1]
In alkenes the eclipsed conformers are local minima, while the staggered, or more accurately bisected, conformers are transition states. The rotational barrier for propene is only about 8 kJ/mol.[2]
I have performed a full, relaxed rotational scan of 1-butene around the central carbon-carbon bond axis (blue). The minima and maxima of this scan have been reoptimised and thermal corrections at 298.15 K and 1 atm have been added (orange).

The chosen level of theory, DF-BP86+D3(BJ)/def2-SVP, is not the most accurate, but it will give you a general idea. You can see the bisected conformations are maxima along the rotational coordinate and the eclipsed ones are minima.
There is dispersion correction included, which makes the E1 conformer slightly more stable than the E2 conformer.
Including the thermal corrections, the energy difference between these two structures is negligible.
$$\begin{array}{crrr}
\text{Conformer} &
\angle_\mathrm{D} &
\Delta E~/\mathrm{kJ\,mol^{-1}} &
\Delta G~/\mathrm{kJ\,mol^{-1}}\\\hline
\mathrm{E_1}     &   0.0             &     0.0  &   0.0   \\
\mathrm{B_2}     &  60.0             &     9.9  &  10.1   \\
\mathrm{B_1}     & 120.0             &     1.2  &  -0.2   \\
\mathrm{E_1}     & 180.0             &     9.7  &   9.4   \\\hline
\end{array}$$
The eclipsed forms are stabilised by two π-σ/π-σ hyperconjugation interactions. Here is the visualisation for E1 in the natural bond orbital picture. Occupied orbitals are depicted blue/orange, virtual orbitals are depicted red/yellow. The phases are assigned randomly.

In the B2 conformer the overlap between those orbitals is not ideal. The nodal planes in the anti-bonding orbitals are in the way.

The representations for E2 are analogue to E1, and B1 is similar to B2. Because of time reasons I have not been able to create the pictures yet.
Taking the maximum difference in the energy we can approximate the rate of the reaction with transition state theory (assuming $\kappa=1$):
$$k = \kappa\frac{k_\mathrm{B}T}{h}e^{\frac{- \Delta G^{\ddagger }}{RT}}
\approx 1\times10^{11}~\mathrm{\frac{1}{s}}$$
This is ultrafast and at room temperature we consider it a free rotation. With spectroscopic methods we will not see a difference. We would probably have to cool the system down to below 1 K to actually see two (E1, E2) conformations.
Steric effects are often electronic effects in disguise, they offer tempting explanations, but are often rooted in a more complex electronic explanation. In this example the possibility of hyperconjugation explains the slightly higher energy the bisected forms compared to the staggered forms.
Hyperconjugation can come in different forms, the most common explanation is the overlap of an empty anti-bonding orbital with a filled (non-)bonding orbital. In this scenario it does not matter which elements are involved in these interactions.
There is even negative hyperconjugation, but this is a little beyond the scope of this post - if there are more questions about the topic, you are always welcome to ask them here.

Footnotes

For a quite recent molecular orbital study see: Ramiro F. Quijano-Quiñones,  Mariana Quesadas-Rojas,  Gabriel Cuevas and  Gonzalo J. Mena-Rejón, Molecules 2012, 17(4), 4661-4671. (open access)
For an educational approach see: Henry Eyring, D. M. Grant and H. Hecht
J. Chem. Educ. 1962, 39 (9), 466.
I did not have the time to search thoroughly, but the following publication contains the value. Leland Cullen Allen, Eugene Scarzafava, J. Am. Chem. Soc. 1971, 93 (2), 311–314.

I'm trying to make sense of this picture, but I can't. I need to check that again.

Both E1 and E2 are made more stable by hyperconjugation.
Hyperconjugation can involve electrons in a sigma bond (either C-H or C-C, no restriction) with an adjacent empty (or partially filled) non-bonding p-orbital, antibonding sigma or pi orbital, or filled pi orbital, to give an extended molecular orbital that increases the stability of the system.
Between the two, E1 and E2, as you can probably guess E2 is better because of steric effects.


text:
How to differentiate the notation of a compounds charge and an anions oxidations state?
I have had the case a few times. Being given something of this form:
$$\ce{Cr2O7^2-}$$
How do I know whether the $2-$ is referring to the oxidation of the oxygen anion or the charge of the whole compound?

Because charge is written like that?

So  you never write oxidation state in the upper right corner?

https://en.wikipedia.org/wiki/Oxidation_state

Oxidation state: first the sign then the number: -2. Charge: first the number then the sign: 2-.

@Marko Roman numerals are used in formulas and names, so there's no mistake.

Sorry, I have mixed ox. state with ox. number.

make sure to read up about that topic. The concept of charge and oxidation state are closely related but aren't the same.
Usually oxidation state is written with roman numbers (that's how I learned it) and are not explicitely written like in the example you showed.
Charges on the other hand are always written (if mentionned) on the upper right side of the anion.
I am far from being an expert, but thinks this might be an answer to your question.

Superscripted Roman numerals are (I think?) a somewhat newer notation, but are in my experience becoming quite common.


text:
Boiling point and Intermolecular forces
After looking for proper reasons for boiling point orders, nobody could even explain why $\ce{CCl4}$ has a higher boiling point than $\ce{SiCl4}$, but after looking for patterns, in lots and cases of $\ce{CH2X2}$, $\ce{CH3X}$, $\ce{CHX3}$ has more boiling point than similar silicon derivatives.
In some places some explanation related to electronegativity difference is given, but how is that relevant and since Si is less electronegative hence there's more partial charge on Cl which results in more polarisability and stronger London forces; which are quite powerful as evident from various examples like $\ce{HCl < HBr < HI}$.
Order of boiling point (leaving HF due to Hydrogen bonding) London forces even dominate dipole-dipole , which I observed in various cases like the above one and they increase rapidly with polarisability and number of polarisable electrons.
Now, in comparing $\ce{CH2F2,CH3F,CHF3,CHF4}$ we can see that it follows dipole's magnitude order as polarization is negligible.
So my questions are

Is there a good way with no real calculation to compare polarisability/effectiveness of polarisable electrons of molecule to compare magnitude of vanderwall forces or any other way to do it?
And does polarising power of central atom play a role?
Why do halomethanes have more boiling point than their respective silicon derivatives?
While comparing the boiling points of $\ce{CH4}$ and $\ce{CF4}$, as fluorine is less polarisable, methane should have more boiling point as both are non-polar (my reasoning works for higher homologous for $n\geq 2$ in $\ce{C_{n}H_{2n +2}}$ and $\ce{C_{n}F_{2n + 2}}$)

If somebody could include nitrogen family too, that would be great.

There was already such question: http://chemistry.stackexchange.com/questions/43334/why-do-chlorinated-silanes-have-lower-boiling-points-than-their-methane-analogs but is unanswered

could you tell me how would I compare london forces in two molecules in general with an example? like same surrounding atoms then same central atom would be sufficient (exceptions like this excluded)

I think shape of molecules can be important here.


text:
How to dissolve the build-up from a paraffin based oil?
I have a CNC lathes that use "oil" for cutting fluid during the machining of mostly metal materials.  Typically a heavy paraffinic based cutting oil (petroleum distillate) is used during machining.  My company switched suppliers of this cutting fluid and went to another cutting oil, supposedly with compatible chemistry.  This happened months ago.  Within weeks of making the switch the oil (the two oils mixed together) became slightly "sticky" or tacky.  This negatively affected the machine tools and processes and the decision was quickly made to go back to the original product from the original supplier.  Over a period of several months the reaction between the two products has left a byproduct of build-up of some type inside the machine tools.  I'm thinking it's the paraffin separating from the oil (somewhat) and along with other components of the chemical make-up that's left this brown, sort of sticky, almost "glue-like" material all over the internal surfaces of the machines.  I'm looking for some solvent(s) that will quickly dissolve this material.  Acetone seems to work but by itself it's not a great option.  The fumes are a problem, it evaporates too quickly.  Any other ideas?

Hexanes (or other long chain hydrocarbon solvents) work excellently for cleaning paraffin grease  (like dissolves like). They're also a bit less volatile than acetone.

Speak to the suppliers of both products, they could possibly have encountered your situation all too often over the years and have a simple remedy for cleaning when moving over to their product.

Naphtha, paint thinner or mineral spirits are common names for the type of solvent suggested by @NotNicolau, and should be effective. Caution: all are flammable and could be ignited by a spark from machining, from a motor or from a switch. These solvents are considered only mildly toxic, though.
Chlorinated solvents such as trichloroethylene are less flammable and even more effective at removing this "varnish" (probably not paraffin, which would be greasy but not tacky) but they are more toxic and the fumes are more annoying.
Ventilate the area well both during cleaning and afterwards until all solvent has evaporated and blown away.


text:
Why is the cyclic voltammetry of my ferrocene compound not showing a standard shape?

I am trying to set-up a new cyclic voltammetry measurement station in my lab but the results are very poor. I am using the following parameters:
working electrode: platinum wire, 0.25mm diameter
counter electrode: platinum wire, 0.25mm diameter
reference electrode: none
solvent: acetonitrile
electrolyte: ferrocenecarboxaldehyde
support electrolyte: tetrabutylammonium hexafluorophosphate
scan rate: ~600mV/s

Platinum wires each have a submerged length of about 2mm to 5mm and are separated by about 2cm. I was previously using tetrabutylammonium iodide, but that is probably not a good support because the iodide ions will become iodine and forming a blocking layer on my electrode?
I also tried using graphite rods as the electrodes but obtained similar poor-looking CV shapes.

What should I change to obtain the standard CV shape of an outer-sphere redox species?

For starters, you might want to try a much lower scan rate—maybe 50 mV/s, and it would be helpful to know the concentrations of your analyte and supporting electrolyte. The area of the counter electrode should also be much larger than that of the working electrode in most cases, so try submerging it more or coiling it.

Without a proper reference electrode, knowing what potential your ferrocene/ferrocenium couple will show up is quite impossible. You should sweep a larger potential window, until the solvent breakdown potential on either end.
As Michael Dryden suggested, you should sweep slower as well. That will help with the signal getting resolved easier.


text:
Is the caffeine in coffee an oxidation agent?
I heard a friend say:

You have to watch out for coffee, the caffeine is an oxidation agent.

Now I remember carbon Chemistry from year 12.

You can oxidise ethanol to ethanal, and then to ethanoic acid.
But I don't understand how this would apply to caffeine.

My question is: Is the caffeine in coffee an oxidation agent?

No, caffeine is not known as an oxidizing agent.

Guess it could oxidize $\ce{LiAlH4}$ or something of the sort, but in any case this has absolutely nothing to do whatsoever with the health hazard it presents.

Yeah it could be made to lose its oxygen or electrons but couldn't any compound containing oxygen? I think because it doesn't do this readily in normal conditions like H2O2 or another common oxidizing agent it is not classed as an oxidizing agent. I was going to add that in the answer but thought it would go off topic.


text:
How can the formation of an alkene be explained during the hydrolysis of an ester when the alkoxy moiety is tertiary?
$$ \ce{RCOOR' + H2O -> RCOOH + R'OH} $$
But when $\ce{R'}=3^\circ$ it may result in the formation of alkene. But by the normal mechanism given in books and in various sites I am not able to explain its formation. I think that somewhere $\ce{R'^{+}}$ must be formed for alkene to be the product.

By the normal mechanism given in books and in various sites I am not able to explain its formation

That's right, and it means that there must be a different mechanism predominating.
This is the "standard" mechanism for ester hydrolysis in acid:

If you look closely at the colour-coding, you will notice that it is the blue bond in the ester which connects the $\ce{C=O}$ to the $\ce{OR'}$ that is broken. This is called the acyl bond. The rate-determining step involves two molecules - it is a bimolecular reaction. Therefore, this ester hydrolysis mechanism is called $\mathrm{A_{AC}2}$: $\mathrm{A}$ for acid, $_\mathrm{AC}$ for acyl bond, and $2$ for bimolecular.
Let's use $\ce{R'} = \ce{C(CH3)3}$, the prototypical example of a tertiary alkyl group. The alternative mechanism goes as follows:

Notice here it is the green bond, between the $\ce{O}$ and the $\ce{R'}$, that is broken. This is called the alkyl bond. The slow step is unimolecular. So, following the same logic as before, this hydrolysis is called an $\mathrm{A_{AL}1}$ mechanism. (It doesn't matter which of the red and blue bonds ends up as the carbonyl group and which one ends up as the $\ce{OH}$ group - they are interchangeable via a simple proton transfer.)
As you correctly suspected, the formation of the $\ce{C(CH3)3}$ carbocation occurs. Depending on the reaction conditions, it could be attacked by a nucleophile (e.g. if you use ethanol as the solvent, you might get some ether byproduct), or it could just simply lose a proton to give isobutene.
What factors may work to favour the $\mathrm{A_{AL}1}$ pathway? Well, the most obvious thing is the formation of the relatively stable tertiary carbocation. Another possibility is steric hindrance arising from the bulky $\ce{R'}$ group, which disfavours nucleophilic attack of water in the "normal" pathway. In fact, t-butyl esters like the one shown above are so resistant to nucleophilic attack (under basic conditions) that they are sometimes used as protecting groups for carboxylic acids!


text:
What would happen if Carbon-14 was in a molecule, and decayed into Nitrogen?
This interested me because of Melvin Calvin's experiment with photosynthesis where he used radioactive Carbon-14. If a plant used it as a reactant to make glucose, and that Carbon-14 decayed into nitrogen, what would happen to the molecule?

I'm closing this question as it was cross-posted to biology.se. Please consider removing the duplicate and flag for reopening.

@Martin-マーチン I think this is more of chemistry than biology. Vote on close by this one: http://biology.stackexchange.com/questions/42239/what-would-happen-if-carbon-14-was-in-a-glucose-molecule-and-decayed-into-nitrog

@inf3rno It does not matter if it is about biology or chemistry, as long as there are two identical posts in the network. I closed it as a courtesy to our partner site. This community can still decide if it is on or off-topic. The question here, should/will only be reopened if it is deleted at biology.se.


text:
What proportion of a combustion engine's output energy comes from the air?
When trying to compare the efficiency and range of electric engines vs gasoline engines, it seemed at first that gasoline had a higher energy density than batteries. But then I realiced that a significant proportion of the energy of combustion must come from the oxygen in the air, whereas in an electric engine all the energy comes from the batteries, so it's not a fair comparison.
So, assuming a typical air-fuel ratio of 14:1, what proportion of the energy produced by combustion comes from the air and what proportion from the fuel itself?
PS: Please assume usual values of any other variables if you need to.

@Martin-マーチン, thanks, no homework question, I'm too old for that. I'm just curious about this because after comparing efficiency and range of electric engines vs gasoline engines I realiced that a significant proportion of the energy of combustion must come from the air, whereas in an electric engine all the energy comes from the batteries. So gasoline by itself is not so miraculous as one would think.

Please [edit] the question with this kind of context and address your concern more specifically. In its current form it reads like a homework question - see the link I posted. It might get closed in the process.

@GetFree Perhaps, when comparing energy densities, it would be better to look at the masses of fuel and air instead of their energy contribution. Thus, consider what if the vehicle had to carry the weight of the fuel and the air (e.g. in separate tanks).

@Loong, at that air-fuel ratio, it would be 14 times more mass of air than fuel. But that doesn't resolve my doubt because I'm interested in how much energy comes from the fuel alone.

Energy density is a practical measure, which you implicitly recognize when you list efficiency and range as assessment criteria.  Given this, the reason gasoline has such a high energy density (i.e., is such a mass-efficient and volume-efficient fuel for cars)  is that 14.7/15.7 = 95% ( by mass) of the reactants needed for combustion come from air (the oxygen), and thus don't have to be carried by the vehicle.   So I don't think it's unfair to assess (if one is being practical) gasoline's energy density using the gasoline alone, as it's only the gas that needs to be carried and replenished.

[contined]  Now if, by contrast, you were using gasoline as rocket fuel, then you would need to account for the mass of the oxygen, since rockets don't breath air, so you'd need to carry the oxygen with you* (and would, attendantly, find gasoline is no longer as performant). *For this very reason, there has been work on air-breathing rockets, like SABRE (https://en.wikipedia.org/wiki/SABRE_(rocket_engine)), since the initial flight of most rockets is through the atmosphere, but it's still in the development stages.

None of the energy come from the air: all of it comes from the reaction of air with fuel. It doesn't make sense to partition the contributions (unless you are in a vacuum and can't access air). Besides, there are batteries that also use air (admittedly not used in cars).

@matt_black, I could conclude from your words that the energy from combustion is created at the moment the reaction takes place. But we all know that's not possible. The energy in a chemical reaction comes from the reactants. There's no creation of energy.

In combustion the "energy" is really the heat that is released. This heat causes the expansion that moves the piston upwards.
However, the fuel and the oxygen did not enter the engine with energy, and then leave without it. The fuel and the oxygen did not leave at all, different molecules did. The heat that was released was due to the reaction that changed the molecules into different molecules, it was not "brought" and handed off by the molecules. I don't think you can divvy this energy up, since it would not have happened without either the fuel or the oxygen. 0% of the reaction would have happened without the oxygen, but that doesn't mean the oxygen was responsible for 100% of the heat generated.
If the oxygen had entered at one temperature, and left at another, you could calculate the heat the oxygen released or absorbed by its change in enthalpy. However, this does not happen in combustion, the oxygen that is responsible for the reaction does not leave as oxygen.
In fact, if anything, the uncombusted oxygen absorbs heat that could otherwise have been used in the expansion of the piston and this quantity would be measurable. So the only thing I think you can quantify is the amount of work not done because air is used (same goes for all the nitrogen that goes in cold and leaves hot). If pure oxygen was used rather than air, engines would get more expansion work since less heat would be absorbed by  nitrogen.

"the fuel and the oxygen did not enter the engine with energy". Are you saying that energy is created during combustion?

When fuel and Oxygen react, energy is released as heat. This energy cannot be divvied up into contributions from the oxygen and contributions from the fuel. It is the energy due to the reaction. That is all. There can be no attributing some amount of it to the oxygen. Energy is neither created nor destroyed during combusion, it is redistributed to $CO_2$, $CO$, $H_2O$ and heat. Energy in equals energy out, but different molecules are coming out. We can calculate their energy due to some reference point, but in thermo we only care about changes and that is all we need too.

They enter with energy of an unknown amount. We only know how much energy they have relative to a reference point, we don't know absolutes... it is like the temperature scale... Celcius and Fahrenheit both have a 0, but it is entirely arbitrary. Whereas the Kelvin scale has an absolute and actual 0 degree. In thermo Energy scales are like Celcius and Fahrenheit, we don't have an absolute scale. As as result we can only measure differences from our "made up" reference points.

TL;DR. The current state of science doesn't allow to calculate how much energy comes from the oxygen and how much from the fuel. I realize now that this question should've been asked in a Quantum Mechanics website.

This is the proper place to ask it, chemist's do qm as well.


text:
Reaction between 2 hydrogen halides
I know that hydrogen halides can react with each other to produce halogen halides.
As an example here is a reaction between $\ce{HCl}$ and $\ce{HF}$:
$$\ce{HCl + HF -> H2 + ClF}$$
As for the mechanism it is like this (taking into account that chlorine is hypervalent but fluorine isn't):
$$\ce{HCl + FH -> HCl\bond{...}F + H. -> H. + ClF + H. -> H2 + ClF}$$
I know that both of these are acids and that $\ce{HF}$ is weaker than $\ce{HCl}$. However with the halogen bond the negative charge on the fluorine (partial negative after the formation of the halogen bond) is more delocalized. This stabilizes the $\ce{H-Cl\bond{...}F}$ intermediate and makes the dissociation easier for the hydrogen and harder for the fluorine. And yes the hydrogen radical is very reactive but that is exactly why $\ce{H2}$ forms in this process.
But I could see some problems with this.

It would have to be in a vacuum for it to not be explosive(most reactions involving fluorine or fluorine compounds are explosive because of fluorine's high electronegativity (same reason reactions with $\ce{O2}$ burn things))
If it is in air $\ce{O2}$ could react very violently with both the chlorine and the fluorine giving you these products:
$\ce{ClO}$, $\ce{ClO2}$, $\ce{ClO3}$, $\ce{ClO4}$, $\ce{Cl2O}$, $\ce{Cl2O2}$,
$\ce{Cl2O3}$, $\ce{Cl2O4}$, $\ce{ClO6}$, $\ce{ClO7}$, $\ce{OF2}$, $\ce{O2F2}$, $\ce{ClO2F}$, $\ce{ClO3F}$
Chlorofluorocarbons could also form from reactions with carbon based compounds in the air like methane and $\ce{CO2}$.

And this is all because the halogens are very reactive.
Similar things would be true for other halogen fluorides.
So is there a more practical way to form halogen halides than reacting $\ce{HX}$ with $\ce{HY}$ where both $\ce{X}$ and $\ce{Y}$ are halogens?

The reaction you are proposing is very unlikely to happen under standard conditions. Taken average bond energies from UC Davis ChemWiki,[1] the approximated heat of reaction would be $\Delta_\mathrm{R}H = 307~\mathrm{kJ\,mol^{-1}}$.
$$\ce{HCl + HF <=> H2 + FCl}$$
The reaction mechanism you are proposing is also not quite feasible. If we would ignore the energetics of this reaction, and we speak hypothetical about this reaction, we would still have to consider an activation step. What we also need to consider is the high affinity of fluorine towards electrons and therefore have to exclude anything that would enhance the heterolytic bond cleavage of hydrogen fluoride. Hence this reaction can only take place in gas phase in an inert atmosphere.
A feasible starting point would be the thermolytical or photolytical cleavage of one of the bonds.
$$\begin{align}
\ce{HCl &->[\Delta T/ h\nu] H + Cl}\\
\ce{HF  &->[\Delta T/ h\nu] H + F} \\
\end{align}$$
Then we would need to consider propagation of the reaction.
$$\begin{align}
\ce{HCl + H/Cl/F &->{} [ HCl.(H/Cl/F)]^{\ddagger} &&
-> H(H/Cl/F)/Cl(H/Cl/F) + H/Cl}\\
\ce{HF + H/Cl/F &->{} [ HF.(H/Cl/F)]^{\ddagger} &&
-> H(H/Cl/F)/F(H/Cl/F) + H/F}\\
\end{align}$$
And then finally recombination of the remaining free radicals.
We are basically looking at the reverse mechanism of the reaction of hydrogen and chlorine.[2]
The practical approach is given in orthocresol's answer. Interhalgen compounds, i.e. halogen halides, are formed by reacting the pure halogen gasses. For the reaction of chlorine and fluorine we find the approximate enthalpy of reaction to be $\Delta_\mathrm{R}H = -113~\mathrm{kJ\,mol^{-1}}$.
$$\ce{F2 + Cl2 <=> 2FCl}$$
Side products also include $\ce{ClF3}$ and $\ce{ClF5}$. The reaction itself is exothermic and ignoring entropy should proceed spontaneously.
$$\begin{align}
\ce{Cl2 &->[h\nu] 2Cl}    &&\text{(start)}\\
\ce{Cl + F2 &-> ClF + F}  &&\text{(propagation)}\\
\ce{F + Cl2 &-> ClF + Cl}\\
\ce{F + Cl &-> ClF}       &&\text{(stop)}\\
\end{align}$$
All reactants are highly reactive and may decompose explosively when other reactants are involved. Even in diluted systems these compounds react with almot everything, see also Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?
Conclusion
The formation of interhalogen compounds via their hydrogen halides is energetically not feasible. They can be synthesised from their pure compounds under moderate conditions.
The mechanisms for these reactions can only develop in a closed system as radical chains, which need an initiation step.
Appendix

Average bond energies (Bond dissociation energies; BDE) from UC Davis ChemWiki:
$$\displaystyle\begin{array}{lr}
&\text{BDE}/\mathrm{[kJ\,mol^{-1}]}\\\hline
\ce{HCl} & 427\\
\ce{HF}  & 565\\
\ce{H2}  & 432\\
\ce{FCl} & 253\\\hline
\ce{F2}  & 154\\
\ce{Cl2} & 239\\\hline
\end{array}$$
A. L. Marshall, J. Phys. Chem. 1925, 29 (7), 842–852.
Carl M. Furgason and John W. Moore, J. Chem. Educ. 1943, 20 (1), 41.

But halogens are more reactive than oxygen group elements which are more reactive than organic compounds.  This has to do with valence electrons and increased stability in the anion state(which is why oxygen hates being positive and thus hydronium ions don't last nearly as long as hydroxide ions, however extending this to the oxide anion decreases stability and increases reactivity which is how come the oxide anion is a super strong nucleophile)

@Caters In this scenario, anions are very unlikely, if not even impossible, since it is in the gas phase.

I think the term you want is "interhalogen compounds", not halogen halides.
Personally, I have never heard of them being prepared via the reaction of the hydrogen halides. Most common synthetic routes involve direct reaction of the elements $\ce{X2 + Y2}$, or reaction of one element with a different interhalogen compound, such as $\ce{X2 + XY$_n$}$.
Chlorine monofluoride is not a great example because no matter what synthetic route you take, it is going to be kind of dangerous. Greenwood & Earnshaw's Chemistry of the Elements 2nd ed. (p 825) gives two routes:
$$\ce{Cl2 + F2 -> 2ClF}$$
(the product must be purified from the by-product $\ce{ClF3}$, as well as unused reactants)
$$\ce{Cl2 + ClF3 -> 3ClF}$$
(must be purified from excess $\ce{ClF3}$).
As for the exclusion of oxygen or organic compounds, that's not difficult at all, as long as you set up your experiment properly.

Halogen halide is perfectly appropriate for this because of how the compounds are named(That is X (mono, di, tri, tetra, penta, hexa, hepta) Y-ide where X and Y are halogens). This is the same reason why ionic compounds with halogens can be considered halides and organic compounds with halogens are also considered halides(Such as methyl chloride(more commonly known as chloromethane to the organic chemist)).

I'm not saying it's not appropriate or technically wrong - the fact that I can understand what you are saying shows that the name is descriptive enough. It's just not used, to the best of my knowledge. All major inorg texts I have seen speak of interhalogen compounds and not halogen halides.


text:
Does autoionization occur in steam?
It is well known that H2O spontaneously autoionizes to H+ and OH- when it is in liquid phase. I was wondering whether it does the same in gas phase, either to the same extent as in liquid phase or at all.

Yes, water is ionized to some extent in all phases. One study even extends to 1,000 K!
A concise discussion of the change in ionization with temperature at Chemguide also explains the variation of pH of neutral water with temperature. [The concept of pH 7 being neutral is based on Kw (in mol2 dm-6) of ~10-14 at ~300 K.]
In general, the higher the temperature, the more a compound is ionized.

Thing is, in actual vapour and low temp. auto-ionisation is as good as non-existent.

@Mithoron, read the answer: "water is ionized to some extent". "as good as non-existent," is not none.

OP asked to what extent. Later you talk about liquid water. While it's true that raising temp. raised deg. of autoionization, vaporisation lowers it extremely.


text:
half-reaction in a battery
A half-reaction on one of the electrodes in a battery produces free electrons (for example) and consumes anions (or produces cations). I understand why, if there's an external wire for electrons to travel on, there needs also to be a salt bridge for ions to travel on inside the battery, and if there isn't, the electrons will soon stop flowing due to separation of charge building up.
However, since each half-reaction on its own is charge-balanced, what stops the half-reaction from just continuing to happen even when the circuit is open? Why for example in an alkaline battery Zn won't continue to react with OH- abundantly coming from the electrolyte?

The half-reaction stops primarily because it's still producing/consuming electrons, and there's only so many free electrons an electrode (the local environment) can accept/supply.
Just like chemical species, electrons have an equivalent to concentration. (Whereas the generalization of the concentration is the "chemical potential" of the species, the "concentration" of electrons is related to the electrical potential or voltage). If the "concentration" of electrons gets too high or drops too low, the reaction rate of the half reaction will drop off - just like it does if the concentration of a chemical product/reactant gets too high/drops too low. In normal operation, the external conductor allows electrons to move, renormalizing the "concentration" of electrons and allowing the reaction to proceed further.
Note that if you have a poorly made half-cell, you may indeed have a pathway for the electrons to travel to the other half cell, or to react in some other (half-)reaction. If this occurs, then your battery will indeed continue to run even when disconnected, draining its stored chemical energy. But decently made batteries control for this, and the only way for the half cells to receive/dispose of electrons is through the external circuit.


text:
Are coordination numbers of elements fixed?
Can we say that the coordination number of a particular element is fixed? If not please give an example where an element exhibits variable coordination numbers.

$\ce{Fe3O4}$ would do: some $\ce{Fe}$ have the coordination number of 4 and some have 6.

@Ivan that's the beginning of a very good answer.

Heh, better to ask if there's any element which doesn't.

No. Coordination numbers of no element are fixed. For example, if you add chloride ions to a pink solution of octahedral $\ce{[Co(H2O)6]^3+}$, a blue, tetrahedral complex is generated according to the following equation:
$$\ce{[Co(H2O)6]^3+ + 4 Cl- <=>> [CoCl4]- + 6 H2O}$$
In fact, if you study the mechanisms of different metal-catalysed organic reactions, you will notice that different coordination numbers of the central metal and specific switching between them is the key to understand these mechanisms.
Finally, for some complexes even establishing their coordination is hard. Zinc(II) forms aquacomplexes like any other cation. But while those complexes of transition metals with any number of d-orbitals unequal to zero or ten are often well-defined and established, that is not the case for zinc. Zinc(II) aquacomplexes rapidly equilibrate from tetracoordinated and pentacoordinated to hexacoordinated species.

Can we say that for a particular ligand, the C.N. of a certain central metal ion is fixed? e.g., $\ce{[CoCl4]^2-}$ exist but does $\ce{[CoCl6]^4-}$ exist (for the same $\ce{Co^2+}$ metal ion)?

@Apurvium No. In organic chemistry, a number of $\ce{[Pd_x(dba)_y]}$ complexes are known and characterised including $\ce{[Pd2(dba)3]}$ (three ligands per palladium centre) and $\ce{[Pd(dba)2]}$ (two chelating ligands per palladium centre).


text:
How to know when subscripts refer to charges or not
My question is very easy. Sometimes to determine to charge of a compound you multiply the charge by the number of ions you have of that element. But sometimes it seems as if you don't do it this way. Example:

$\ce{Ba(BrO2)NO2} = \ce{Ba^2+ + BrO2^0 + NO2^-}$: Here you multiply the subscript by the charge of 1- in order to balance it
$\ce{KNaCO3} = \ce{K+ + Na+ CO3^2-} $: here you don't multiply the subscript 3 by the charge 2- for it to be balanced.

Could someone please explain the difference? When do you mdo a multiplication and when not?

The answer to when do you multiply is this: you don't. $\ce{BrO2}$ in this compound (if we assume it exists) is charged -1, not 0. You do multiply, however, when the whole thing is in brackets, but that's another story.

@IvanNeretin, How did you check whether $\ce{BrO2}$ had a charge of -1 ? I thought compounds with an oxygen in their anion always had a charge -2? Or is this only for acids? (I am a beginner in chemistry)
You do a multiplication only when the whole thing is between brackets or there is a coefficient specified?

You do the multiplication only when the whole thing is in brackets and there is a coefficient specified after the closing bracket. Come to think of it, you also do the multiplication when there are no brackets but the "thing" in question is just one atom, like in $\ce{Ca^{2+}Cl^{-}2}$. $\ce{BrO2}$ must be -1 to balance the whole thing, because $\ce{Ba}$ is definitely +2 and $\ce{NO2}$ is certainly not -2. Anions with oxygen in them are pretty diverse; some have the charge of -1, some -2, some -3 and so on.

@IvanNeretin. Well there you have it "$\ce{NO2}$ is certainly not $-2$". How do you know that? I was quite sure it was the case.
Concerning the example from calcium and Chlore. You do the multiplication because the anion is composed just of 1 ion $\ce{Cl2^-}$. If it was $\ce{ClO2^-}$ you wouldn't do such a multiplication? (just made up a random example to clarify what you meant, don't even know if it makes sense)

No, you do the multiplication because the anion is composed of one atom $\ce{Cl^{-}}$, and there are two of them (two anions). (Also, it is not good to write things like $\ce{Ca^{2+}Cl^{-}2}$ the way I did in the previous comment, for it may be confusing.) As for $\ce{ClO2^{-}}$, it is one anion, so there is no reason to multiply by 2, unless there are two of them, which would look like $\ce{\dots(ClO2)2}$. As for $\ce{NO2}$, well, I just know that there is $\ce{NO2^{-}}$ and there is no $\ce{NO2^{2-}}$. There are also $\ce{NO2+}$ and neutral $\ce{NO2}$, but that's another story.

@IvanNeretin, a response to these 2 questions will clarify everything: How would you write it if the compound $\ce{ClO2}$ had a charge of -1? How would you write the formula from $\ce{ClO2}$ if only oxygen had a charge of -1? I think this is the issue I am having: the subscript refers only to that one single element (unless it is written between brackets), while the charges usually refers to the whole compound.

How would I write what? If you want a compound with anion $\ce{ClO2^{-}}$, look at $\ce{KClO2}$; if I'd like to specify the charges, I would write $\ce{K+ClO2^{-}}$. An example of a compound where you'd better multiply the charge by the subscript is $\ce{CaCl2}$. Not sure how to write $\ce{CaCl2}$ with the charges in a non-ambiguous way; probably $\ce{Ca^{2+}(Cl^{-})2}$ would do.