text:
Why is methyl group more electron-donating than tert-butyl group?
As title says, why is methyl group more electron-donating than tert-butyl group? The context behind this is stabilization of conjugate base. (http://www.khanacademy.org/science/organic-chemistry/organic-structures/acid-base-review/v/stabilization-of-conjugate-base-iii look at 10:51)

Methyl group is not more electron donating than tert-butyl group, as the video provided in the link suggests (11:20 of the video), which is true.
Inductive effect is a permanent effect which results in displacement of electrons along the chain of atoms in the molecule (here, carbon atoms).  The reason for this displacement can be worked out from the electronegativity difference between the atoms.
Consider the tert-butyl group.  In a C-H bond, the electron density is directed towards carbon.  Methyl group, which has three hydrogen atoms attached to a carbon, pushes the electron density towards $\ce{C2}$ of tert-butyl acetic acid.  So do the two other methyl groups.  The amount of partial negative charge accumulated on $\ce{C2}$ of tert-butyl group (has 3 methyl groups) is more than that on methyl group (only has 3 hydrogens) of acetic acid.
The more negative charge on $\ce{C2}$ in tert-butyl acetic acid destabilizes the conjugate base, compared to a simple acetate ion.  Hence acetic acid has less pKa than tert-butyl acetic acid.

On what evidence do you say that the electron density is directed toward the carbon in a C-H bond? (Certainly possible, but I am wondering.)

Electronegativity; EN of carbon is 2.4 (minor changes depending on hybridisation of carbon in the compound); EN of hydrogen is 2.1

I think Inductive effect does not result in permanent displacement of electrons.

Inductive effects do not displace electrons, in as much as a standing probability wave can be displace by anything. However, inductive effects change the electron density distribution of a molecule in a real and permanent (i.e. neither transient nor fluctuating) way.


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How many valence electrons does the azide ion have?
We had a test and and we had a question asking the number of valence electrons in an azide ion, $\ce{N3^{-}}$.  I find out that the answer is 16 but I could not understand how.
 I saw the structure. It says that the end nitrogen have a -1 charge and the middle nitrogen has a +1 charge.
From where does the one of the end nitrogen get the electron assuming the other end nitrogen gets it from the middle nitrogen?

Nitrogen (the element) has 7 electrons. But two of those electrons are in the $1s$ orbital, and so are not considered valence.  Therefore, a single elemental nitrogen has five valence electrons.
There are three nitrogens, and so $5 \times 3 = 15$ electrons for $\ce{N_3}$.
Finally, azide is $\ce{N_3^{-}}$, there is an additional electron beyond the neutral form.  So, the total number of valence electrons is $15 + 1 = 16$.

Nitrogen is a 5 group element hence has valency 5 and AZIDE ion has an negative charge it means it gained an electron hence as AZIDE HAS 3 NITROGENS and a negative charge it will be 5 multiplied by 3 = 15 + one electron gained = 16

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text:
Oxidation State in Tetraaquadichlorochromium(III) chloride
$\ce{[Cr(H2O)4Cl2]Cl}$, Tetraaquadichlorochromium(III) chloride
According to a book, the (III) indicates the oxidation state of chromium, but it doesn't explain why. Cr has 6 valence electrons. How does it get to +3?

I have cleaned up your formula using the chemistry plugin for mathjax (see the [faq] for more info).

I think the easiest way to see this is to count the formal charges on the ligands.
The four waters do not have net charge.  The remaining ligands--the three Cl's--are likely $\ce{Cl-}$, and so their total charge contribution is $-3$.
The only way the complex can be neutral is if the chromium has a $+3$ charge:  hence Chromium (III).

By the way, since chromium is a transition metal, it is able to form multiple ions, such as 2+, +6 or even -1. There are no rules about ionisation on transition metals, and the only way to discover their possible ionisations is going through every specific case of ions.

Eric is correct about the easiest method to determine the most likely oxidation state of Cr. As for the rest of your question, the oxidation state of Cr in this compound means that it has three electrons less than in its neutral state, i.e. three valence electrons instead of six. This +3 oxidation state is the most stable one for chromium, although many other states are also possible. Note that a wide range of oxidation states is characteristic of d-block elements such as chromium.

Most elements have several oxidation states in the everyday life. The simplest one, hydrogen, is usually +1 but it is 0 in H2, or even -1 when bonded to an element less electronegative than itself such as a metal.
Remember that valence is different from oxidation state (see https://en.wikipedia.org/wiki/Valence_(chemistry) ).
Off-topic but always good to know: chromium VI (6+) is extremely toxic and cancerogenous, whereas chromium III (3+) is much less dangerous.

This answer is all useful and true, but it does not address the OP's question of how to determine the oxidation state of chromium in the compound shown.

Well his question was more about "how can chromium be 3+" than about "formulas must be wrong". But I understand your concern. In fact, the "oxidation number" is completely artificial and part of our (poor) understanding of chemistry.

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text:
Are there substances that cannot undergo combustion?
Are there substances that cannot undergo combustion? If so, what are they?

I think this is a bit to broad in its current form.  Could you narrow down your scope a bit?  Are you trying to choose a material for a particular application?

Also, please consider using the "Contact Us" link below to merge your accounts together, as you keep creating new ones each time, that way you can consolidate all of your questions.  Thanks!

On burning "unburnable" substances: Sand won't save you this time

It depends a little on your definition of combustion.
Most combustion we see involves things reacting with the oxygen in air. Most combustion reactions are a self-sustaining exothermic reaction of some substance with oxygen gas. But atmospheric oxygen is relatively dilute (only about 21% of air is oxygen) and this limits what will burn sustainably in air (also, at slightly lower concentrations many commonly flammable things won't stay lit). The requirement for things to be combustible in air is, crudely, a reaction path with oxygen that releases energy.
Examples include many types of compound. Hydrogen reacts with oxygen to give water; hydrocarbons react to give carbon dioxide and water. Many metals react to give metal oxides (eg magnesium, spectacularly when lit; iron, only when hot and finely divided).
Many things won't burn in air for two reasons: there are activation barriers to a potential oxidation meaning a potential reaction can't be sustained (aluminium has a strong oxide layer on its surface protecting the reactive metal from further oxidation, for example); other compounds are already highly oxidised an no further oxidation can happen (eg silicon dioxide, a major ingredient of sand and glass).
But there are many more things that will combust in pure oxygen as Apollo 1 discovered, tragically. And as many demonstrations show spectacularly (see demo here). But highly oxidised things like silica can't be oxidised further and won't burn not matter how concentrated the oxygen is.
But oxygen isn't the only gas that can support exothermic reactions. Rather more things will burn in fluorine, for example. Even relatively unreactive substances can support combustion if there is an available reaction path that releases enough energy and a reaction can get started. One of the best known examples is that magnesium will burn in carbon dioxide (the reaction producing magnesium oxide releases energy) and even nitrogen (producing magnesium nitride). See the explanation here and the video here.
So, to return to the original question, are there any substances that won't combust? Whatever the definition of combustion is the answer is still yes. Noble gases, for example, though some can be coerced into reacting under extreme circumstances, never combust or support combustion.
If we restrict our definition to combust in air, many substances won't combust though many of them might do so under different circumstances.

If one takes combustion to mean redox reaction then it is hard to see how lithium fluoride could take part in any further redox reactions since it is composed of strongest reducing agent and the strongest oxidising agent.

Actually KrF2 is even stronger, at least at normal temp.

If you define combustion as a chemical process that consumes $\ce{O2}$ and produces $\ce{H2O}$, then the answer is: any $\ce{X}$ that does not contain hydrogen is not combustible.
$\ce{X + O2 -> Y + H2O}$

Even if we accept your definition of combustion, it does not follow that the hydrogens in water must come from the combustible material.  They may be picked up from solution, for example (and often are).

If the solvent is providing H's to the reaction, then the solvent is actually part of X -- it is a reactant. So by my definition, the solvent better not have any H's (e.g. tetrachloromethane)

Even if the solvent is water, as it is most (?all) biochemicl processes? I presume you agree that if X is an organic substance and the reaction you describe is done in water (at pH 7, say), water does not supply any electrons to oxygen, merely (it may be envisaged) a pair of protons, and that protonation (by definition) is not an oxidation (or a reduction). Water may be considered part of the reaction, but it does not in any sense combust, surely?

@TomD I really don't care where the matter comes from.  If the matter appears in the product, then it was a reactant (by my own definition). The combustion question itself is a bit "unanswerable" as most species can be oxidized by something.  I have used water as the necessary product because this is what "average" chemists think of when they think of combustion processes (colloquialism), and I would rather not get dragged into some legalistic discussion on precisely what constitutes combustion.

But you say: 'any X that does not contain hydrogen is not combustible'.  What about, for example, oxalate (the dianion): can the carbon-carbon bond not be oxidized? Furthermore, in the (biochemical) combustion of glucose it is electrons 'held' in both carbon-carbon and carbon-hydrogen bonds that are passed (via the Respiratory Redox chain/Krebs Cycle/Glycolysis) to oxygen.  I too don't want to get dragged down into a pedantic discussion, but from a biochemical standpoint, I think your final statement (quoted) is misleading...

...  Again from a biochemical standpoint, in general any X that contains a C-C or a C-H bond is potentially oxidizable (and may be used to 'make' ATP).

@TomD My equation is deliberately set up so that there is no hydrogen source, therefore no hydrogen to make water, and therefore no combustion (colloquial definition) is possible. Perhaps you would like to take a stab at answering the original question. Is there a molecule that does not combust? Can you give an example, or is it your opinion that all molecules can undergo combustion?

let us continue this discussion in chat

@EricBrown can you support your definition of combustion as a reaction that consumes $\ce{O2}$ and produces $\ce{H2O}$?  I think a more realistic definition is an exothermic reaction that involves a fuel and an oxidant.

@bobthechemist I don't think my definition is out of line in a colloquial sense of "combustion."  You are correct that combustion (proper) is anything that can be oxidized. But since anything can be oxidized by something, the legalistic answer to the question is "there are no examples."  But I don't think that e.g., graphite is combustible (in a colloquial sense)--one reason being that there is no driving force by way of hydrogens to form water. BTW I am aware of $\ce{C(s)+ O2 --> CO2}$ but this doesn't seem very much like combustion ala a combustion engine.

Nobody defines combustion that way and for good reason. Charcoal, for example, doesn't contain hydrogen but is quite effectively combusted in air. Otherwise barbeques would not be possible.

@matt_black what is your definition of combustion? Before slamming mine, show me yours. (Mine is an attempt to cleverly answer a question in a way without exceptions.)

Combustion is usefully defined as a sustainable exothermic reaction. This has the benefit of allowing the component substances to be anything you want to include but also of allowing the environment for the combustion reaction to be relevant. Many things we think of as combustible in air (such as paper) cannot sustain combustion if the oxygen level is too low (say <15%). Almost everything is combustible in chlorine trifluoride including concrete, rocket test engineers and asbestos.

Is the exothermic association of O2 with a transition metal a combustion?

@EricBrown Given what I said in the previous comment, it should be obvious why your definition is a poor one. Not every combustion reaction uses oxygen; and there are plenty of oxygen-consuming reactions that do not need the presence of any hydrogen.

It is a constructed one to use to answer a question couched in absolutes.  And it’s actually the definition that is taught from elementary school through university.

@EricBrown Your constructed definition is useless because there are plenty of well-known examples of things we call combustion that clearly don't fit. Finely divided iron (a transition metal) will quite happily burn in air. Why would anyone not call that combustion?

I would not call that combustion in the everyday parlance, I would call that reactivity.

If you were taught this definition, you had poor teachers. And I have already given enough counterexamples for you to judge this for yourself.

You have not written a generalized reaction that describes combustion uniquely.  Please write it out so we can scrutinize it.

And you have not written a definition that covers all the various things normally grouped together as "combustion". Exothermic, self-sustaining reactions that produce light might be both general enough and restrictive enough to cover the common use of term in chemistry and life.

You’re just throwing out ad hominems and not providing a definition. My answer starts out “If you define combustion...”. Your most recent definition of combustion seems to include every single exothermic reaction possible.

PS: http://web.fscj.edu/Milczanowski/psc/lect/Ch11/slide3.htm  (First hit on my Google searching for definition of "combustion") Perhaps you want to write this professor and tell him/her that s/he is poor and his teachers were poor.  (And tell Brown et al. who wrote perhaps the most used textbook in the chemical world that they are poor teachers for using the same definition that I invoked.)


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Can I perform a stable isotope analysis at home?
I'm interested in measuring the stable isotope ratios of some bird feathers. I'm aware that a mass spectrometer is required to measure stable isotope ratios. Is it possible to measure stable isotope ratios at home, using your own 'DIY' spectrometer?

Mass-spectrometer does not have any similarity with normal optical spectrometers, It consist of hi-performance vacuum pump, strong magnet and complicated vacuum chamber with ... additions. It is often big and requires some skill to operate. So you probably should go to nearest lab with said spectrometer. Measuring mass-spectrum is usually quite fast, so the cost should not be an issue.

I am assuming that you wish to measure isotopes of carbon.  It is unknown to me how you will extract sample from a bird feather that is suitable for measurement in a spectrometer, and does not taint the isotope ratios. The spectrometers that can distinguish between isotopes are generally quite sensitive and complicated. So the literal answer to your question is "NO, you can not do it at home with a DIY spectrometer."
But, you might be able to purchase a spectrometer. You might be able to differentiate carbon isotopes in carbonyl compounds with infrared spectroscopy. (But don't count on it, unless you are somehow making a gaseous product such as $\ce{CO}$ or $\ce{CO2}$.)
The usual way of measuring carbon isotopes is via mass spectrometry. You can probably get a crude measurement with a machine that costs $20,000 (on eBay) but then you must get it calibrated and keep it supplied with carrier gas, etc.  In short, it will bankrupt you.
User permeakra has the best idea: send a sample to a lab.  But, I am not sure of any lab will allow you to send samples through their spectrometer, not knowing anything about your reputation or the volume of business that they might expect to get from you.

On 2019-11-29 Ben Krasnow uploaded a YouTube video DIY mass spectrometer measures potassium in dietary salt substitute showing an outstanding construction of a mass-spec apparatus in his home shop based on the articles by Dewdney [1] and Stong [2] with some variations and improvements, like altered electrical scheme and increased emission current.
The main downsides were limited operation time (10–15 min), tedious assembly/disassembly and vacuuming procedures required for each run, low resolution and possible signal artifacts.
However, for the tested probe of potassium chloride, Ben was able to demonstrate the presence of both stable isotopes $\ce{^{39}K}$ and $\ce{^{41}K}$ in the approximate expected ratio of $14:1$.

From the original article [1]:

A mass spectrometer has been designed around simple constituents: the vacuum system is
  made from copper plumbing parts; electrical leads into the vacuum are via darning needles
  pushed through rubber stoppers; the filament of the thermal ionization source is from a small light bulb; object and image slits are made from bits of razor blades sandwiched between brass washers. The magnetic field is supplied by a permanent magnet fitted with homemade cylindrical pole pieces (of such a diameter to give second-order direction focusing). A modest vacuum of $\pu{E-4 mmHg}$ is required. When the instrument is focused (by adjusting the position of the pole pieces) and aligned (by rotating greased rubber stoppers) a resolving power of 50 can be obtained.

Fig. 2. Complete mass spectrometer. (1) 90° elbow, (2) source tee, (3) collector tee, (4) to (9) connections leading to vacuum pump, (10) external part of source, (11) external part of collector, (12) cover on miniature tube socket.

Fig. 3. Cross section of the essentials of the mass spectrometer. The numbered parts are those visible in Fig. 2.

References

Dewdney, J. W. Poor Man’s Mass Spectrometer. American Journal of Physics 1963, 31 (12), 932–937. https://doi.org/10.1119/1.1969211.
Stong, C. L. The Amateur Scientist. Sci Am 1970, 223 (1), 120–128. https://doi.org/10.1038/scientificamerican0770-120.

The small $\Delta{}t$ between the publication of the video, and your post on ChemSE is remarkable.  My interpretation: You know many of ChemSE's nuggets in and out, and seeing the video gave you a «wait a sec, this matches well to one of the 32k questions ...» moment.  Tip to the hat and +1 for your detailed answer.

This is an outstanding answer

@faissaloo Thank you, but it's really Ben Krasnow who deserves all the credit. I subscribed to his channel about five years ago and the content was always highly educational and of top notch quality without any exceptions. Both the levels of preparation and the amount of the background work in his videos is incredible, and yet he manages to make hardcore science accessible to the public. And he drives a DeLorean:)

Not by a homebuilt mass spectrometer, you can't. At least not with a reasonable amount of effort.
A mass spectrometer is a very complicated and sensitive charged-particle optics instrument. A useful mass spectrometer for materials that are not a gas or vapor at near room temperature is far more complicated and expensive than that. A mass spectrometer sensitive enough for distinguishing isotopes in molecules of significant size is even harder. You must be an expert with ultrahigh vacuum, high-speed and ultra-sensitivity electronics, precision high-voltage electronics, particle physics, mechanical fabrication...
It's akin to building your own electron microscope: far from impossible, and hobbyists have done it themselves, but these are people whose main passion and area of knowledge was ultrahigh vacuum and particle physics, not chemistry. I gave up on my diy mass spectrometer project when I concluded that I was not equal to the electronic design.


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Why won't water freeze if you put ice in it, despite ice being frozen?
Just now I got some room temperature water and added some ice to cool it, and it just came to me, like the scientist I wish I was ... why doesn't frozen (freezing point) ice added to higher temperature liquid compound force the water molecules to solidify?

Why won't water freeze if you put ice in it?
It will, even at room temperature. You just need a big enough, cold enough ice cube.
Don't believe it? Add a few drops of water to an ice cube in an ice cube tray (which is the same as adding an ice cube to a few drop of water). Wait a few seconds, turn the tray upside down. No water will fall, presumably because it has frozen. Repeat, but immediately turn the tray upside down. You'll get a wet floor. You can conclude that, given enough time, the water will be frozen by the ice. To the whys and hows:
From the second law of thermodynamics we know that:

An isolated system, if not already in its state of thermodynamic equilibrium, spontaneously evolves towards it.

And

Heat cannot spontaneously flow from a colder location to a hotter location.

So if you put ice and water together in a cup, heat will flow from the water to the ice until they reach thermal equilibrium, which will be a temperature between the water's initial temperature and the ice's initial temperature. If the temperature of the ice was low enough or if there was enough ice for the water to keep giving away heat, then the water will freeze. That's considering our cup is an isolated system.
Why don't you usually see that happening? Well, regular ice cubes are probably around −5 °C, and there usually is more water than ice in a cup, so before the water has a chance to reach 0 °C all the ice has already melted. Besides that, the ice is receiving heat from the room, unless you're in a place where's below 0 °C. In that case, the water WILL eventually freeze, but mostly because of the environment.

Water, like most substances, requires more energy gain or loss to undergo a state change (from solid to liquid, liquid to gas) than to be warmed or cooled by a degree of temperature without changing state. Liquid water requires about 4.2 Joules per milliliter per degree Celsius to raise or lower its temperature. However, the Latent Heat of Fusion (Lf) of water (the energy that must be added to ice to melt it or removed from water to solidify it at the freezing point) is 334 Joules per milliliter.
Therefore, for ice to freeze the surrounding water, the ice must be cold enough that it can afford to gain enough energy to invoke this state change, without melting itself. Now, solid ice has a lower volumetric heat capacity; it only takes about 1.9 J/(mL * K) to change its temperature. So, for an ice cube, let's say 10mL in volume, to freeze 1mL of the surrounding water (at 0°C) without melting itself, the ice must be able to absorb 334J of energy without melting. That means the ice must be -16.7°C, or about 2°F.
That's well within the range of your average freezer (which in the US is normally kept around 0°F to ensure everything stays good and frozen), but that back-of-the-envelope calculation assumes the energy is gained evenly throughout the entire ice cube. This doesn't happen; it's common knowledge that ice melts from the outside in, and that's because it's a relatively poor conductor of heat compared to substances like metals, which sometimes seem to instantly liquefy when they reach their melting point. For the ice to be able to cool that kind of water volume without melting, therefore, it must be much colder; cold enough that the ice can afford to gain enough heat just in a few millimeters of its outer surface layer to freeze the water.
Regardless of this, if you have a fridge with in-door ice and water, and you fill a cup with the ice and then pour the water over it, the ice will clump; the combination of two cubes in close proximity, a small volume of already-cold liquid water in a thin layer between them, and the ice being colder than freezing, results in the ice freezing the water, joining the two cubes. This happens throughout the ice. If you go with crushed ice, which has smaller pieces but more surface area and a larger volume of frozen water to total water, the effect is even more pronounced. But, eventually, the overall thermal energy of the room adds more energy into the glass of water than the ice can absorb without melting. So, it melts.

The latent heat of vaporization (now normally referred to as the enthalpy of vaporization) and the latent heat of fusion (enthalpy of fusion) are different. The former is boiling, the latter is melting. For water, the enthalpy of vaporization (boiling) is 40.7 kJ/mol (2300 J/mL) whereas the enthalpy of fusion (melting) is 6.01 kJ/mol (334 J/mL) - but aside from the "typo", good points.

@R.M. Thanks for the comment. The edited answer fixes the "typo".

At room temperature, water is a liquid. So the ice will gradually turn into liquid.
You may ask your question in a different way:  if you add water to ice, why doesn't the ice turn into liquid?  It does, at room temperature.
If you add water to ice, at freezing temperature (Antarctica), the water turns to ice.
I think it is mostly due to the surrounding air temperature.

Well, yes and no. Ice will freeze cold water in a glass sitting on a table, as the colder ice and warmer water exchange energy to reach a rough equilibrium. But, the overall equilibrium of the larger system, at room temperature, eventually adds more energy than the ice can absorb without melting, so it does.

The question did not specify what the ambient temperature is (it said room temperature water, not system at room temperature). It might be best to first look at the question for an insulated system (no heat transfer), and then perhaps say what happens if you allow heat transfer.

Now if we have a look on values of sp heats of ice and water.
L(ice)=334 kJoule per kg
C(water)=4200 Joule per kg per degree Celsius
C(ice)=2100 Joule per kg per degree celsius
If we want 1 kg of water at 1 degree celsius to first reach zero degree Celsius and then be converted to ice the amount of heat released will be greater than 4200 joules
On the other hand,  If we want 1 kg of ice at minus one degree celsius to first reach zero degree Celsius and then be converted to water the amount of heat required will be around 2100 joules
Therefore in general if we want water to be converted to ice the ratio of ice to water should be more than 2 isto 1.
(Assuming temp of both are at same distance from 0degree celsius)


text:
Why can't you see air but can feel it
Is there a device that can measure the air in the room in lbs?
I would appreciate the anseer

A scale.  Consider an experiment where a container of "air" is submerged into liquid helium (just about the coldest stuff known.)  This will turn the air into solid $\ce{N2}$ and solid $\ce{O2}$ which can then be weighed.
If you knew the area of the room, then you would know the "pounds per square inch" or "p.s.i."
But the above answer, I think, is a little cheeky.  I think it is usually done via the ideal gas law:  $ P V = n R T$, and solving for $n$, the number of moles of gas. Then through Avogadro's number, one can deduce the number of molecules.  The molecules (e.g. $\ce{N2}$) have a known molecular weight, which can be expressed in pounds.
A barometer, a thermometer, and a ruler are the only three devices one would need to get the number of pounds of air.

Thanks for taking the time to answer.  Please, though, consider flagging such "nonsense" (or nearly so) questions for closure.


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Why calcium chloride is used to melt ice over sodium chloride
I'm not quite sure I understand this. My question is:

Calcium chloride is a salt used widely to melt ice on sidewalks and roads.  Explain why one mole of $\ce{CaCl2}$ would be more effective than one mole of $\ce{NaCl}$.

So from everything I've learned, $\ce{NaCl}$ would be more effective for the melting, because calcium chloride dissociates into three ions and sodium chloride dissociates into two, so that would make the boiling point of water with calcium chloride higher.  Therefore, it shouldn't be used to melt ice over sodium chloride, as it would make the water have a higher boiling point so wouldn't melt it as well!

It's a little weird that you're trying to justify changes in the melting point by looking at the effects of dissolved substances on the boiling point. If you have a book around, take a quick look again at freezing point depression (or cryoscopic effect) and compare it with the boiling point elevation (or ebulioscopic effect). In particular, notice how the equations are very similar. What does this similarity imply?

That one mole of $\ce{CaCl2}$ is more effective in melting ice than one mole of $\ce{NaCl}$ is explained by the van 't Hoff factor. The freezing point depression of a solution is calculated by
$$\Delta T=K_fbi$$
where $K_f$ is a cryogenic constant which is specific to the solvent, $b$ is the molal concentration of the solute, and $i$ is the van 't Hoff factor, which indicates number of solute particles. In this case, it is the number of ions produced upon dissociation. Because $K_f$ and $b$ are constant, and $i=3$ for $\ce{CaCl2 \rightarrow Ca^2+ + 2 Cl-}$ and $i = 2$ for $\ce{NaCl \rightarrow Na+ + Cl-}$ the freezing point depression is greater for $\ce{CaCl2}$.
The reason $\ce{CaCl2}$ is widely used as melting agent may have more to do with keeping the concrete intact. Concrete is composed of many calcium-containing species. When water flows over the concrete, the calcium can be leeched out, and the concrete becomes brittle. The higher the calcium content in the water, the less leeching occurs, and the concrete remains structurally sound.

Who cares about moles? One mole of CaCl2 weighs almost as much as two moles of NaCl. So even taking van't Hoff factor into account, CaCl2 should be less effective per kilogram. It is the higher solubility that matters (and makes it more effective, despite the above).

Well, the question cited specifically asks for a mol:mol comparison, for one.

Calcium chloride produces an exothermic reaction when mixed with water or when placed on top of ice, which means that it literally heats up when it comes in contact with ice. Sodium chloride does not produce this exothermic reaction.

More specifically, you're talking about the enthalpy change of solution. For NaCl, this quantity is slightly positive, about 3.87 kJ/mol, meaning for each mole of NaCl solvated in water, 3.87 kJ of heat is absorbed from the surroundings, making the water colder. CaCl2 has a large, negative heat of solution, seemingly in the range of that of NaOH, thus releasing heat that contributes to melting.

Exothermic reaction means next to nothing, when you are using it in the street. All that heat vanishes instantly into the environment.

In addition to the colligative and thermochemical properties, there are practical considerations as to why calcium chloride or sodium chloride would be used to melt ice in a public area. Calcium chloride will make a surface slippery under reasonably cold ($<0~^\circ\mathrm{F}$) conditions presumably due to the hygroscopic nature of the salt.  On the other hand, sodium chloride is fairly corrosive and can damage concrete and vegetation.  I'm not a fan of sending visitors to other Q&A sites, but this answer has a nice table of the pros and cons of several salts used in melting ice.

If you use NaCl on concrete, you will have a water softener where you don't want it. It is one thing to remove calcium from hard water but quite enough to remove calcium from concrete.

The answer is simple. One of the unique properties of any salt is that it completely breaks down when dissolved in water. By that I mean each of the atoms clings to water molecules, preventing them from using their natural ionic bonds to each other. For every NaCl molecule, you can occupy 2 water molecules. For every CaCl2 molecule, you can occupy 3 water molecules. This makes it, Mol for Mol more effective an ice melt.

The solvation sphere is much larger than just one molecule of water per ion.


text:
What is this colorless ferric solution?
I made a solution of iron acetate by placing steel wool in one litre of acetic acid (vinegar $5\%$) and adding a few tablespoons of hydrogen peroxide of questionable age. The solution turned a nice brown almost instantly upon shaking.

I left it overnight and it lost its colour while forming a white precipitate:

I added some more peroxide (new, good peroxide this time), and the moment the peroxide hit the clear an colorless solution it looked like dark red-brown food-coloring! Unfortunately I could only get out my camera fast enough to make the following photo:

I squeezed the air out of the steel wool before the above picture, which caused it to sink (noted this so you don't think something of its sinking). Added one-fourth cup hydrogen peroxide and shook:

With a flash from the top:

Held up to the light:

The result seems to be ferric acetate (III), since ferric acetate (II) is light brown in color (according to the Wikipedia link)--possibly like the light brown in the first/top photo. Ferric acetate (III) is supposed to be a red-brown color (again according to the Wikipedia link).
It is the clear an colourless solution in the second picture that I am not certain as to its composition. Any assistance will be appreciated in identifying that solution.

I would hypothesize that the white precipitate is either acetate or carbonate (more likely carbonate). I would guess that the clear solution occurred as a result of the solution interacting with either the steel wool or the air. I took the steel wool out after making the final dark red solution and it has been 2 days now without it diminishing in colour, so it is likely not the air that was interacting with the solution. Perhaps, if I placed the steel wool back in the solution, it might go clear again. I'll post again in a day or two.

How to make Iron Hydroxide thorugh adding Ammonium hydroxide (household cleaner) to Ferric Acetate: http://faculty.laverne.edu/~nelsonm/chem103/lab9fs.php  See also

According to the wikipedia page on hydrogen peroxide adding H2O2 to a solution of Fe(II) will oxidize it to Fe(III): http://en.wikipedia.org/wiki/Hydrogen_peroxide#Redox_reactions However, Fe(II) was supposed to be brownish not clear I thought.

When you left steel wool in vinegar, iron reacted with acetic acid to produce ferrous acetate and HYDROGEN. The so produced NASCENT hydrogen is very strong reducing agent and it reduced the earlier formed brown ferric acetate (due to the presence of H2O2)  ---> to ferrous acetate. This explains the loss of reddish brown colour.

I don't believe you have any ferric in the colorless picture since you have a metallic iron which would reduce the ferric to a ferrous. I believe you have a ferrous (Iron(II)) acetate characterized as an off-white precipitant.

Since your solution is acidic (from the acetic acid), you cannot have any carbonates; they react with the acid and carbon dioxide bubbles off.
It appears that the brown solution is iron(III) acetate and the white precipitate is iron(II) acetate. The darkness of the solution indicates the concentration of iron(III) acetate.

I would like to add to the previous answers a Pourbaix Diagram, which demonstrates the transition mentioned between ferous and ferric ions as the electrochemical potential or redox properties of the solution change. Source This diagram suggests that the solution does indeed have a pH comparable to that of Vinegar which is pH=2.4 The H2O2 should not affect the pH, but rather the oxidation state (vertical axis below) from Fe+2 to Fe+3.


text:
How Does The Mordant, Ferric Acetate, Interact With Tannins When Ebonizing Wood?
Ferric Acetate (II) and (III) are "mordants," which intensify wood staining in direct proportion to the tannin concentration in the wood. Some wood stainers add tea to increase the tannin content and thereby make the stain darker. How does Ferric Acetate act as a mordant in this case of ebonizing (web archive) wood?

A mordant is a substance used to set dyes on fabrics or tissue
  sections by forming a coordination complex with the dye which then
  attaches to the fabric or tissue (source).

Tannin is a biomolecule consisting of many carbohydrate and polyphenol residues with plenty of hydroxyl and carboxyl groups. Tannic acid is a common type of tannin:
$\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ form colored complexes with these oxygen-containing functional groups. These iron-tannin complexes change the color of the wood. The color intensity increases with tannin content of the wood and can be increased by the addition of tannin-rich tea. Structures with a varying coordination sphere of iron are formed, some of which are shown here (sources: 1 and 2):


text:
In Goodman and Gilman what does R,S-citalopram refer to?
In Goodman & Gilman's The Pharmacological Basis of Therapeutics, 12e R,S-citalopram is referred to. The problem is that from what I am aware of citalopram only has one stereocentre, and hence I am forced to ask were the author having a stroke when they added the comma and really meant the racemic mixture of R and S stereoisomers?

Figure 1: Screenshot of the Relevant Page in Goodman and Gilman

Figure 2: Citalopram
(The latter has the stereocentral carbon highlighted)

It seems fairly clear that the authors intended to refer to the racemic mixture there. You're right that a comma would normally be used to separate the locants of distinct stereocenters when multiple ones are present, but then the individual chiral carbons would have to be numbered, which obviously isn't the case here.


text:
How can diesel fuel produce less carbon monoxide but more carbon particulates?
My guess at why diesel produces less carbon monoxide is that more heat is required for combustion of diesel fuel, so there is less incomplete combustion. Therefore rather than carbon monoxide being formed and released, the ability to completely combust means that carbon dioxide is often formed instead. Correct me if I'm wrong.
However, a risk of diesel fuel is that more carbon particulates are produced. But surely this is just a contradiction, because carbon particulates are a result of a lack of oxygen supply and incomplete combustion. Surely if there is enough complete combustion to form carbon dioxide over carbon monoxide, then there is a high enough oxygen supply and temperature to stop carbon particulates forming?

It seems that the mechanism of particulate formation is not yet well known. This article (preview) states some basics about the composition and formation of Diesel Particulate Matter (DPM):

Diesel particulates form a very complex aerosol system. Despite considerable amount of basic research, neither the formation of PM in the engine cylinder, nor its physical and chemical properties or human health effects are fully understood.
  [...]
  The basic fractions of DPM are elemental carbon, heavy hydrocarbons derived from the fuel and lubricating oil, and hydrated sulfuric acid derived from the fuel sulfur. DPM contains a large portion of the polynuclear aromatic hydrocarbons (PAH) found in diesel exhaust

Also, it seems that the particulates are formed both during combustion and from gas during dilution. During combustion, carbonaceous agglomorates (soot), metallic ash from lubricating oil, and precursors to particle formation during dilution (sulfur oxides and partially burned hydrocarbons from fuel and lubricating oil) are formed. During dilution, materials such as sulfuric acid and other sulfates, as well as heavy hydrocarbons and hydrocarbon derivatives, are for the most part adsorbed onto the carbonaceous agglomorates formed during combustion. Some material may also nucleate by itself, but most nucleation occurs on the agglomorates.
PDF with explanation of ultrafine particle formation mechanisms
Unfortunately, i found it difficult to find any reason why this is such a big problem with diesel and not with gasoline. Perhaps the composition of the diesel is to blame, not temperature or oxygen supply.

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


text:
Chemical methods to extract Oxygen from Rust
Let's say I have a large quantity of rust and want to extract the oxygen from it in a way that allows me to collect the oxygen in a container.
What chemical process can I use to do this and what equipment does it require?
It's preferable if the process is doable at home so I can demonstrate the process but it's not mandatory.

Good question; unfortunately, no good answers. The problem is that oxygen, pure oxygen, doesn't like to exist as such; it will form a diatom like many gases, but it's very electronegative, and so it will prefer to get the extra electrons from something more willing to donate them, like the iron. This is why "oxidation" has that name; until fluorine was isolated in the mid-1800s, oxygen was the strongest known electron acceptor ("oxidizer"). It's easier to run a reaction that gives the oxygen something else to oxidize, producing the raw metal, than to get the iron to complex with something else and release the oxygen. In at least one circumstance, this is very good for humans and other aerobic organisms; oxygen complexes with the iron in your hemoglobin for transport to your cells. In most other circumstances, we could do without this behavior.
Rust is insoluble in water but will dissolve in strong acids. It will react with hydrochloric acid to form iron chloride and water (sometimes producing hydrogen gas if there isn't enough oxygen to go around). The easiest "household" method to obtain oxygen from iron oxide, therefore, is to dissolve the rust in muriatic acid forming iron chloride and water, neutralize any remaining HCl by adding lye to form sodium (and/or potassium) chloride and water, distill the water from the chloride salts by boiling and then cooling the vapor, then electrolyze the water with a battery to produce oxygen and hydrogen. That's cheating, though. First, water's water, and you're adding far more to the party with the acid and base solutions than you would ever get from the rust. Second, muriatic acid and lye, while available in relatively pure form for pool maintenance and for soapmaking respectively, are not exactly everyday household chemicals in those forms (both are components of other household chemicals, such as drain cleaner and lime scale remover, but are somewhat difficult to isolate from them).
Theoretically, if you heat the rust enough, the components will dissociate and you'll be left with pure iron and oxygen. Unfortunately, even with carbon in the mix (producing elemental iron metal and carbon dioxide, which is the reaction behind iron ore refinement and recycling), the minimum temperature required for this reaction to become favorable is 842K = 569*C ~= 1056.2*F. I can't find data on iron-oxygen bond enthalpy or entropy, which is needed to calculate the spontaneous decomposition temperature of the compound, but logic states it's much higher than with added carbon, because you're not getting any energy release by forming the carbon-oxygen bond, so every Joule of energy needed to break the iron-oxygen bonds has to come from outside the material. Theoretically it would cost you about 2.5kWh/kg of heat added to the iron; in practice the amount of heat you'd have to generate (and lose to various inefficiencies) to get the metal that hot is beyond any household device's capabilities, unless your hobby happens to be glassblowing.
The "thermite reaction" between iron oxides and aluminum under heat is another well-known method of separating the iron and oxygen. The rust and aluminum powder, given some initiating heat, will exchange the oxygen to form alumina (Al2O3) and pure iron, and enough heat to keep the reaction going (with plenty to spare; a thermite reaction is too bright to watch without a welder's mask and in fact is used in some situations to weld iron). Two problems with this; first, it's illegal in the US to make thermite without BATFE approval, and second, the reason it works at all is because aluminum oxide is ridiculously stable, more so than iron oxide, and so it takes even more energy to reduce the oxide back down to the raw metal (about 8.2 kWh/kg for aluminum, theoretical, versus 2.25kWh/kg for iron). This is evidenced by the highly exothermic nature of the thermite reaction itself; if creating aluminum oxide produces far more than enough heat to smelt the iron to get the oxygen, think of how much would be needed to go the other way.
Pretty much the only direct route to releasing oxygen gas from rust without a blast furnace is to introduce an even more powerful oxidizer. There are very few known to modern science, and none of them are "household chemicals" to say the least. The easiest to work with would be elemental fluorine gas, and the experienced chemists in the crowd will know that for the extremely relative statement that it is. Fluorine is the most electronegative element in the known universe, and so pretty much any substance that isn't already highly fluorinated (like some polymers, such as Teflon) will react pretty vigorously with it. That includes the water, proteins, enzymes and other compounds in your own body; a fluorine gas leak is a "drop everything and run" type of emergency. As used on rust, the fluorine will simply barge in and displace the oxygen from the rust to form a white to pale green iron fluoride, liberating the oxygen as a gas. You then have to separate the oxygen from the excess fluorine, and one of the standard ways in industry to catch excess fluorine is to pass the exhaust gas over or through a powdered sodium oxide filter, which will form sodium fluoride and liberate more oxygen in a very similar reaction as you got with the iron. Sodium oxide isn't a pleasant chemical to work with either; add water (such as by breathing in any loose powder) and it forms the strong base sodium hydroxide, which eats flesh about as efficiently as anything else you could think of.

You can also go via the same process used for aluminium: electrolytic smelting, either directly (probably), or using a solvent. Not exactly doable at home, though!

A long read but a fantastic answer, seems it wouldn't be too easy to do even given the right industrial equipment!

@Aesin - Electrolytic smelting is pretty much the same basic  iron-for-carbon redox as in a blast furnace, but using electrical arc plasma to generate the necessary heat in a more localized fashion. And no, not exactly the type of thing you could do at home unless you leveraged the main 200A trunk line into a monster DC converter.

Two variations on the acid dissolution:
a. Dissolve the rust ($Fe_2O_3$) in 100% sulfuric acid, producing ferric sulfate with some enneahydrate $Fe_2(SO_4)_3.9H_2O$. When the reaction is done, evaporate the $H_2O$ and electrolyze it.
b. Use 100% phosphoric acid, get $FePO_4.2H_2O$. When the reaction is done, evaporate the water and electrolyze it.
And a third (pure speculation):
c. This way probably won't work, but  organometallics do all sorts of crazy stuff. Ferrocene can be made by heating iron with cyclopentadiene at 300 C (https://en.wikipedia.org/wiki/Ferrocene). I think it might also work with iron oxide, perhaps even better (or not).
$Fe_2O_3$ + 6 $C_5H_6$ --> 2 $FeCp_2$ + 3 $H_2O$ + $(C_5H_5)_2$
or something like that. Collect the water and electrolyze it to get the oxygen. Quite doable in a lab - if it works.


text:
PEG (polyethylene glycol) and ITO (indium tin oxide)
The binding of proteins to an ITO surface can be prevented by coating it with PEG groups (polyethylene glycol). PEG-silane is used to create the coating.
can somebody help with:

Why does PEG-silane bind to ITO (indium tin oxide)? and how? (which bond etc.)
Why does PEG repel proteins?
Are there better coatings to use to repel proteins?

Could you elaborate what you have found/investigated until now?

PEG "silane" is $\ce{HOCH2CH2O-PEG-CH2CH2CH2-Si(OCH3)3}$.  (Pt-catalyzed $\ce{HSi(OCH3)3}$ addition to the terminal allyl monoether.)  A metal oxide surface can be hydrated, (bulk)M-OH.  The TMS end hydrolyzes and dehydrates to tetrahedrally covalently bonded $\ce{HOCH2CH20-PEG-0Si(OM#bulk3)}$.  The alcohol polyethoxy end hydrated waves in the water, and the protein ignores the PEGylated surface.
An especially fine biocompatibilizer rendering nearly any organic surface except silicone rubber invisible to protein deposition and denaturation is poly(N-vinyl pyrrolidone).  It can be linearly radiation grafted into organic surfaces (visualize with Congo Red).  Plexiglass gives a 19 degree average contact angle in saline and implants with neither cellular or humoral reaction.  NVP plasma deposition gives you sludge.  There is no obvious (inexpensive) chemical attachment of its low polymer or oligomer to a surface.  Pity.


text:
Easy-to-do experiments that clearly show outstanding properties of graphene
By chance, we received some monolayer graphene sheets (20 cm x 20 cm) and mixed coper-graphene wires for free at our university. I would like to prepare some very-easy-to-do experiments for the chemistry students that show the awesome properties of graphene, and which could be carried out with very simple lab equipment or even home experiments. What are some good experiments we could do?

You may want to post this in the Physics forum, as many of the outstanding properties of graphene are expected to be physical.  See this Periodic Video: Graphene
Given this, two things to try.  First, graphene should conduct electricity, like graphite but not like carbon.  Second, graphene may dissolve in nonpolar solvents like carbon tetrachloride, similar to buckyballs, to produce colorful solutions.  See this Periodic Video: BuckyBall
Both these experiments should be checked in advance: I'm extrapolating from other experiments, but have not done these myself.

Are you sure that a 20xmX20cm intact film will dissolve? Buckyball is molecular material, graphene is like a polymer: one big thing interconnected with covalent bonds. And why would it be colorful?  Free standing nanotubes and buckyballs can have color. Dissolution only helps separate the coagulates, so show you the color of the freestanding molecules. Graphene has no problem like this as far as I know.


text:
Where did the acac come from in Cr(acac)₃ synthesis by CrCl₃ reagent?
Those are the reactions:
$$\ce{(H2N)2CO + 3 H2O -> CO2 + 2 NH4+ + 2OH-}$$
$$\ce{CH3COCH2COCH3 + OH- -> acac- + H2O}$$
$$\ce{CrCl3.6H2O -> Cr3+ + 3Cl- + 6H2O}$$
$$\ce{Cr3+ + 3acac- -> Cr(acac)}3$$
I don't get from where comes the $\ce{C5H8O2}$ in reaction #2!

$\ce{CH3COCH2COCH3}$ is the condensed formula of pentane-2,4-dione, commonly known as acetylacetone. The $\ce{acac^-}$ is just a shorthand for the acetylacetonate ligand, i.e., $\ce{C5H7O2^-}$, which is just the deprotonated form of the acetylacetone used in reaction two.

The first reaction has no specific purpose in the synthesis of $\ce{Cr(acac)_3}$, right?

@Jerry, actually, the whole sequence of rxns. is bizarre. If the desired end product is $\ce{Cr(acac)3}$, there are infinitely more effective routes. I can only assume the first rxn. is the intended source of $\ce{OH^-}$. The problem is I'm fairly sure it doesn't work. One obvious reason is that the acid-base equilibrium between ammonium and hydroxide should favor the neutral species (water and ammonia) by 5-6 orders of magnitude. AFAIK, the hydrolysis of urea ultimately yields ammonia and carbon dioxide through an unstable carbamic acid intermediate, and, IIRC, enzyme catalysis is required.

Thank you, and yes, those reactions are really weird.

Look like a co-reaction- (1) and (2) happen at the same time- just not in a simple combination so its represented as one reaction. (2) drives (1) towards hydroxyl formation.


text:
Calculating bond disassociation energy and bond energy for ATP
I'd like to calculate Bond Dissociation Energy and Bond Energy for ATP to a similar molecule. What free (at least for academic use) computer programs which natively support and document BDE?
If you don't mind, could you provide a tutorial for using those programs to perform the calculations?

I think your other question goes into this and more, so I'm closing this one for the time being.

An example program is GAMESS.  It is free. But be warned: the calculation of BDEs is an extremely difficult thing to do.  I am presuming that you are a beginner, so you have to be careful for issues surrounding which BDE:
\begin{align}
\ce{A-B &-> A+ + B-}   \tag{ionic}\\
\ce{A-B &-> A. + .B}  \tag{homolytic}
\end{align}
Most electronic structure theory programs assume gas-phase, absence of solvent.  We know that homolytic dissociation is favored in the gas phase, as charge separation into ions is unfavorable in a Coulombic energy.
That being said, the run-of-the-mill electronic structure theory calculations (e.g., HF, DFT, MP2, CCSD, etc. etc.) will not break the atoms apart in a homolytic manner.  They tend to break apart atoms in the ionic fashion.  But this is a problem, as ions are abhorrent in the gas phase!  Basically, simple electronic structure theory methods over-estimate the bond strength, as the product of the dissociation (ionic) is erroneous.
You must use multiconfigurational methods to get it right (e.g. CASSCF, MCSCF). But these are very, very difficult unless you have had a lot of training.  I did them for years, and they were still hard for me.  But they will give you $\ce{A. + .B}$ as a product!  That is the gas-phase BDE.
You may be able to study closed-shell, closed-shell dissociations without much problem, using the HF and DFT methods.  Examples include dimers, host-guest interactions where the products look like good-old fashioned molecules that can stand by themselves. (Think water dimers, etc.)
Also, beware of Solvent Effects (favoring ionic dissociation $\ce{A-B -> A+ + B-}$) and solvation energy.  In fact, the solvation energy is why $\ce{Na-Cl}$ breaks apart into $\ce{Na+}$ and $\ce{Cl-}$.  In the gas phase, it would be $\ce{Na.}$  and $\ce{Cl.}$ radicals.
In summary, BDEs are very difficult things to compute to qualitative accuracy. Quantitative accuracy is even harder.  You may want to have an expert look over your procedure and results before you trust them completely.

Ah yes, bond breaking with electronic structure theory... what a massive headache.


text:
Is energy required to form bonds [phase change]
My question is if any energy is required to form bonds, for instance when there is a phase change?
If I am correct, energy might be required in the beginning, to make the reaction start and then release a bigger amount of energy than it was putted in. However, in terms of molecular behaviour in a phase change, I think there should not be any energy required to make the molecules to form bonds, if they are placed in a system which has a lower temperature than the molecules. Would they not lose the kinetic energy after a specific time?

Bond formation always lowers the energy of the system (or bond formation is a consequence of lower energy, take your pick.)
Indeed, you may have to add energy, because presumably certain bonds must break in order to rearrange the atoms.
Adding energy won't guarantee formation of much higher energy isomers, because the atoms have a large amount of kinetic energy, and so they can just as easily turn around and go back whence they came.
Sometimes the kinetic energy can be dissipated by solvent, collisions with inert gasses, etc. and then you may end up with a measurable quantity of the higher energy species.


text:
On the scale of stuff with a pH
I am a confused (and somewhat anal retentive) non-chemist who is trying to understand what pH is.  I have come to understand it is a measure of acidity, but I have not yet figured out what that means to a precision that I find satisfying.
Please excuse me if these are dumb questions, but they will help me understand the definition:

Does everything have a pH?  What is the scale that pH works upon?  Does a proton have a pH?  A quark?  Does the sun have a pH?  Does the Universe?

""I have come to understand it is a measure of acidity,"" This is wrong! pH is a measure of acidity of aqueous solutions only!

Did you not carry out some research?
Well, you can find information about it in Wikipedia, or even dictionaries like Merriam-Webster or Dictionary Online.
You will see that pH is calculated through:
$$pH = -\log(a_{\ce{H+}})$$
Or which can most of the time be stated as:
$$pH = -\log(\ce{[H+]})$$
$a_{\ce{H+}}$ is the hydrogen ion activity, the 'effective concentration' of hydrogen ions. Without active hydrogen ions, you don't have any pH.
You might find that the pH scale is from 0 to 14, but you can also have circumstances where you have negative pH and pH above 14. It depends on the extent of dissociation of an acid or base to produce active hydrogen ions, or hydroxide ions in the case of bases.
Since I mentioned hydroxide ions, it might also be worth nothing that:
$$14 = pH + pOH$$
And that basically, a pH of 1 is equivalent to a pOH of 13.

pH scales are just a special case of solvonium/solvate scales, which are a way to keep track of the concentration of solvonium/solvate ions. They can be defined for any (usually protic) liquid solvent in macroscopic amounts (the limit of many molecules). Particles don't have pH, and I'm not sure it's possible to define such scales in a useful manner for solids, gasses, plasmas or mixtures of solvents.
Let's look at some cases. Water is capable of self-ionization, according to the equation:
$$\ce{2H_2O_{(l)} -> H_3O^+_{(aq)} + OH^{-}_{(aq})} \ \ \ \ \ K=\frac{a_{H_3O^+}a_{OH^{-}}}{a_{H_2O}^2}\approx \frac{[H_3O^+][OH^-]}{[H_2O]^2}=3.2\times10^{-18}$$
, where the equilibrium constant $K$ is given at 25°C (it varies with temperature). If we consider relatively dilute solutions, then $[H_2O]\approx55.5\ mol.L^{-1}$ and one can obtain:
$$[H_3O^+][OH^-]=K_w^{25°C}=1\times10^{-14}$$
$K_w$ is called the ionic product of water. Taking the logarithm of both sides and multiplying everything by $-1$, we find
$$-log[H_3O^+]-log[OH^-]=-log\ K_w^{25°C}\rightarrow pH+pOH=pK_w^{25°C}=14$$
(At the end I switched $\ce{H_3O^+}$ for $\ce{H^+}$ since they are essentially the same thing. $\ce{H_3O^+}$ is nothing but $\ce{H^+}$ solvated in water)
The $p$ in $pH$ is nothing special; it is merely shorthand for the operator $-log()$, and it can be applied to any measure, though it's not very common to do so. Though at 25°C, the sum of pH and pOH is constant (equal to 14), there is nothing immediately forbidding either value being negative. They just can't be very negative because the approximations in the first two equations breaks down as such solutions are no longer dilute. Also, the sum of pH and pOH is only 14 at 1 bar for a temperature of 25°C. At 0°C, $K_w^{0°C}=1.1\times10^{-15}$, while at 100°C, $K_w^{100°C}=5.1\times10^{-13}$.
Now for other solvents. Water isn't the only substance to self-ionize; another example is is liquid ammonia ($\ce{NH_3}$). Here, we have:
$$\ce{2NH_3\ _{(l)} -> NH_4^+\ _{(am)} + NH_2^{-}\ _{(am})}$$
An ionic product can also be defined for ammonia:
$$[NH_4^+][NH_2^-]=K_{am}^{-34°C}\approx 10^{-30}$$
From the ionic product, we see that ammonia has a smaller tendency to self-ionize compared to water. This is due to several factors, such as the smaller electronegativity of the nitrogen atom and the lower dielectric constant of ammonia, which destabilize the ions formed. In this solvent, the neutral point is around pH 15.
After all this, what's the point of pH? As was mentioned at the start, it's a way to keep track of the concentration (or activity rather) of $\ce{H^+}$ in solvents, and it turns out this species is particularly important to chemists, not only because it's so common and a part of water. Bonding/dissociation/charge transfer happens due to the interaction of electrostatic potentials in molecules which push and pull electrons. Usually, a handful of fundamental units of charge are distributed in volume of a few atoms. However, if you stop to think about it, $\ce{H^+}$ has no electron cloud and is actually a lone proton, characterized by a charge of $+e$ concentrated in the volume of a nucleus. If you recall, an atomic nucleus has about a trillionth the volume of an atom's electron cloud. This means $\ce{H^+}$ has a phenomenally high charge density capable of generating a massive electric potential, and is therefore extremely reactive. Though helium and neon are the most inert substances we know, $\ce{H^+}$ will easily react exothermically with even them to produce $\ce{HeH^+}$ and $\ce{NeH^+}$. In fact, $\ce{H^+}$ cannot exist as a true lone ion in a condensed phase, and it will demand some electron density from any substance. A very large amount of acid-base theory was created solely to understand more about the reactivity of this unique species.

pH stands for "potential hydrogen", originally "power of hydrogen". In simplest terms, it is a logarithmic scale (based on powers of 10, similar to the Richter scale for earthquakes, the decibel scale for sound and signal levels, and others that seek to measure very large differences with smaller numbers) of how many free protons there are in an aqueous solution of a substance, which is in turn a measure of that solution's acidity or basicity.
The number is the absolute value of the exponent of this measurement, so if, for instance, a substance has a pH of 7, that means there's one free proton per 10 million water molecules (1/10,000,000 = 10-7). A pH of one, or less, means there's at least one proton for every 10 water molecules, while a pH of 14 means there is one proton or fewer per 100 trillion water molecules.
Protons, or H+ ions, are attracted to the negative charge region of a water molecule. Water's bent molecular structure, with the two hydrogens closer to one side and oxygen's extra lone pairs on the other, effectively makes each molecule a tiny magnet, which is the foundation for many important characteristics of water, like why ice floats and why some things dissolve in water while others don't. Anyway, the hydrogen loosely "binds" to one of those lone pairs, forming hydronium (H3O+).
There is a natural equilibrium of one hydronium ion per 10 million molecules. At this point, pure water (again being a polar solvent) will actually dissolve itself to a small degree, producing one hydronium and one hydroxide per 10 million. At this pH of 7, the opposite measure, pOH (power of hydroxide) is also 7, balancing the reactivity of the water.
Below a pH of 7, the positively-charged hydronium atoms come to dominate this equilibrium, decreasing the number of hydroxide ions and looking for a substance to which to donate its protons to reach a balance again. This solution, which is now a "proton donor", is one of the classical definitions of an acid (the Lowry-Bronsted definition). Ideally, it will donate its proton to a substance that needs one as badly as the acid needs to get rid of it, but if the acid has enough hydronium, it will stick those hydrogens wherever they'll fit into any other substance's molecules.
Above pH 7, the negatively-charged hydroxide ions reduce the hydronium and dominate equilibrium. The solution is now a "base", a "proton acceptor", and is looking for something from which to take protons. The ideal substance to do so is a proton donor - an acid - but similar to an acid, a base, if strong enough, will take hydrogen atoms from anywhere they're not nailed down.
Now, not all substances which are acidic have a free proton to donate (and so not all of them increase the amount of hydronium ions in water), and not all substances which are basic have a hydroxide anion, nor do they accept free protons into their structure. Therefore, there is another definition, the Lewis definition, stating that an acid is a substance that accepts a "lone pair" of electrons from another substance (a pair of electrons not forming a bond between atoms of the molecule) to equalize its own molecular charge, and a base is a substance that readily donates an electron pair. Regardless of the definition or the exact mechanism, these acids and bases have similar observed properties, and so even if explicit exchange of hydronium atoms isn't involved, the substances still indicate a pH using various testing methods. This prompted the change in name from "power of hydrogen" to "potential hydrogen", in effect making the measurement an "equivalent", applying to both Lewis acids and bases and Bronsted-Lowry acids and bases.


text:
HCl synthesis from sulfuric acid and sodium salt
I'd like know if this is right. The reaction between $\ce{NaCl}$ and $\ce{H_2SO_4}$ should yield $\ce{HCl}$ and sodium sulfate. This is a double substitution, true? No oxidation took place here. The thing is, when I tried the reaction in the lab, the resulting solution (gas bubbled into water) seemed to be sulfuric acid. Simply enough, the solution didn't react with aluminum. I presume that $\ce{SO_2}$ was formed from the oxidation the sulfur (it a little bit like it).
What am I doing wrong? Shouldn't the substitution take place instead?
Process:
$
\ce{NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl}\\
\ce{NaCl + NaHSO_4 \rightarrow Na_2SO_4 + HCl}
$
Adding:
$
\ce{2\space NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2\space HCl }
$
Is this right?

Did you measure the pH of the resulting solution?

Could you include the exact procedures you followed?

@Jerry I added salt to a RBF and then some excess of sulfuric acid. The reaction began almost immediately. There was bubbling and the gas went through a hose to a cylinder with water and crushed ice.

@NicolauSakerNeto I measured the pH and it was 1 or less.

This video suggests another way to identify $\ce{HCl}$. Or you could just wet a pH paper and hold it above the acid solution. A test for $\ce{H_2SO_4}$ would be to add a small amount of a soluble barium salt to the acid. It's a little strange for you solution to be $\ce{H_2SO_4}$ though, since pure sulphuric acid is far from volatile, and chloride probably isn't reducing enough to form volatile $\ce{SO_2}$. How long did you wait for the aluminium to react? It can take several minutes to start breaking through the $\ce{Al_2O_3}$ passivation layer.

@NicolauSakerNeto, your comment would make the beginnings of a good answer.

Ideally, that's the reaction. Realistically, there are a few speed bumps.
First, it occurs stepwise, with the acid salt sodium bisulfate as an intermediate. The first stage happens at room temp between equal proportions of salt and sulfuric acid. The second stage requires temperatures of 200*C (in turn requiring water-free concentrations of acid and dry salt) and an additional equivalent of salt that reacts with the bisulfate.
You did say you were using excess sulfuric acid; if true, you'll react all the salt but what'll be left in the flask is sodium bisulfate, not sodium sulfate. Since you're interested in the HCl gas produced and not the sulfate salt, this isn't a big deal, but I just wanted you to be clear that one mole of sulfuric acid to two moles of salt isn't going to get you the expected 2 moles of HCl unless you're heating a "water-free" combination of the reactants.
Second, if you have too much water in the reaction, the HCl won't bubble out; it'll happily sit in solution. So, dry salt and 10M or higher concentration of sulfuric acid are recommended (as in Nicolau's YouTube video, 18M is plenty strong enough to do the job, and they make and sell it up to 98% concentrations). If you're stuck with 1M, you can heat the flask to boiling, which evaporates some water and increases the acid concentration, liberating the HCl gas (sulfuric acid won't boil below 300*C, so you're fine distilling the HCl like this until you're out of water and the thermometer starts to climb).
Following these steps and bubbling the gas produced through water, what you produce should definitely be hydrochloric acid. Obviously the molar strength of it (and thus its ability to react with test materials as was also indicated by Nicolau) will depend on the relative proportions and amount of reactants, the volume of your trap water, and how much gas is actually absorbed by the water.

Why does the salt react with the bisulfate and not directly with the sulfate? Let me add the reactions to the question.

I'm not sure the bisulphate anion should react with salt, since its acidity is significantly reduced compared to its parent acid (as I mentioned here)

@NicolauSakerNeto But still, $HSO_4^-$ is a strong acid, isn't it? The weak acid would be the sulfate anion.

@NicolauSakerNeto - The bisulfate anion does indeed react with salt; however, because it's a weaker acid in solution, the reaction is endothermic and so requires heating to an equilibrium temperature of 200*C as I mentioned.

@Fiire - The $\ce{HSO_4^-}$ anion in combination with a cation other than hydronium is an acid salt, and acid salts are weak acids in water. The primary industrial use for this chemical is to acidify like a strong mineral acid would, but in applications where a strong mineral acid would be hazardous (such as in the hands of the average individual). These applications include pool water conditioning, cooking/baking, metal pickling such as in jewelrymaking, soil acidification (though typically a nitrogen salt like ammonium nitrate is used), etc.

@KeithS So I should use 1 mol of acid and 2 mol of salt, after heating the acid to 200ºC? Is there be a safer/more effective procedure?

The safer reaction is the "room-temperature" half of the acid/salt reaction. You just need twice the acid to produce the same volume of HCl you'd get if you heated the reaction. Since you're using an excess of acid I assume that's not a problem.

@KeithS Why would double acid produce the same amount of HCl? Isn't the chlorine the limiting reagent? Sorry if I missed something.

No, the chlorine isn't the limiting reagent. In the unheated reaction, all the salt must be reacted with sulfuric acid, producing sodium bisulfate. Therefore, you need equal parts salt and acid. In the heated variant, above 200*C the sodium bisulfate will react with any unreacted sodium chloride to produce sodium sulfate and another equivalent of HCl. This allows the initial quantity of sulfuric acid to react with two equivalents of NaCl; once as sulfuric acid and once as sodium bisulfate.

let us continue this discussion in chat


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Solving carbon black
I want to remove a layer of 2 mm thick carbon black on a copper bowl on the outside from hanging over wood and coal fire.
I have tried different solvents, but the carbon seems to be very resistive.
Which formula solves carbon black without solving copper? The bowl is made from pure copper. It will melt at roughly 1000 degree Celsius and has a high heat capacity.

Carbon in any form is pretty tough to dissolve. What have you tried? And what can the bowl endure? Can it be heated, scratched or sonicated? One possibility is to heat it in an oxygen atmosphere to a high temperature (below the melting point of copper, of course).

I doubt that Jonas knows what "carbon black" is. From his description that crust must be rather rigid, and is not carbon black. He should describe how that crust was deposited.

Well the heating I suggested (if it works) can't be over a regular fire, otherwise the soot from the fire will just add to the layer. It should also oxidise the surface of the copper to $\ce{CuO/Cu_2O}$, though I imagine that's happened by now anyway. Also, at first I assumed the bowl was from a laboratory and that you had access to lab equipment, but you may be asking for a home solution. That complicates things. Scraping the layer off is the solution involving the least complex instrumentation, though it's a lot of work and may damage the surface of the bowl slightly.

Why isn't polishing a viable solution?

By far the safest solution is to physically abrade the surface until you have worn away everything that isn't copper. This is slow and requires some sort of grinding agent but has the upside that you are not likely to injure yourself.
There is a chemical solution that will work but it requires access to some dangerous chemicals and is extremely hazardous even in a well-equipped lab. It is insane to try anywhere else as you won't have adequate safety equipment.
The best way to remove carbon black from anything is to use a mixture of concentrated sulfuric acid and hydrogen peroxide. This is often used (but also often highly discouraged because of the danger involved) in laboratories to clean sintered glass filters contaminated with things like activated carbon. In the case of a copper bowl, the solution will remove the carbon but will also etch the copper surface (it is sometimes used as an etchant to remove copper from printed circuit boards).
The mixture of sulfuric acid and hydrogen peroxide is sometimes known as Piranha solution of reasons that are obvious when you have seen it work.
I would not encourage anyone without good chemical safety training and access to very good safety equipment to work with the mixture. And even then they should work only with small amounts and in a fume hood behind a shield and while wearing eye protection, strong gloves, and a fireproof coat.

I have access to a well equipped lab so it should be possible to stick to the safety rules. But I fear, that the H2O2 will ruin the copper.

Elemental carbon tends not to dissolve in solvents due to its very stable structure, but it will react with oxygen to form carbonates/carbonites, so the heating of the bowl in atmospheric oxygen (as mentioned in a previous comment) could be a possible solution.

All these materials are hazardous by skin contact and inhalation.  The are toxic and evil.  One possibility is soaking overnight in auto supply store generic carburetor cleaner.  Another is overnight soak in automatic transmission fluid.  Gloves, goggles, disposable activated charcoal painter's mask.  If you get anything on a belt or your shoes, you throw them out.  Contaminated clothing must be quickly removed.  Wash skin with soap and water - twice.
http://www.youtube.com/watch?v=bRiZOTTel34
http://www.break-free.com/?location=/products/index.asp
If anything works, wash the cleaned bowl in a good dishwashing detergent solution - twice.  There must be no residual product odor.

When a cooking utensil used over a campfire becomes encrusted on the outside, it's because the contents of the utensil keep it relatively cool.  Simply heating an empty container (below its melting point!) should oxidize off the carbon-based material. (Probably creosote)

I will give it a try as soon I get a large gas torch. The bowl has 60 cm diameter, but it might work.

Hot steam above $\mathrm{250~^\circ C}$ can erode carbon black, $\mathrm{600~^\circ C}$ or higher is better.
Activation/Oxidation: Raw material or carbonized material is exposed to oxidizing atmospheres (oxygen or steam) at temperatures above $\mathrm{250^\circ C}$, usually in the temperature range of $\mathrm{600–1200~^\circ C}$.
See https://en.wikipedia.org/wiki/Activated_carbon

@A.K. please note that the proper formatting for units should be preceded a space.

The situation is not clear.
If your carbon is pure ( why would it be?) then burning down with flame would be the easiest and cleanest solution. Practically nothing solves pure carbon, except melt iron.
If your carbon is not pure, but e.g. you burned some food on the surface, cooking with sodium-carbonate generally help.

I would try chlorinated water from a swimming pool mix. Leaves are broken down somewhat in pool water. I tried it on an electronic circuit with carbon shorts on the substrate and it cleared the carbon current leak between the copper tracks. Be sure to monitor the contact time and wash it off with water.


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Estimating Surface Reactions - Liquid, Turbulent Flow
So, I need a method of estimating the influence of a surface (rate constant would do for a start), the idea is to estimate just Reactant + Surface -> Product in a turbulent liquid flow.
Now I am aware of the book "The properties of gases and liquids" by Poling and Prausnitz which gives me two terms for estimating diffusion in a liquid, one general formula from 1975 and a formula specific to hydrocarbons (applicable in my case) from 1982, namely:
Tyn and Calus, 1975:
$$D(AB) = \frac{7.4\,10^{-4} (\phi M(B))^{1/2}T}{\eta(B) V(A)^{0.6}}$$
Hayduk and Minhas, 1982:
$$D(AB) = \frac{13.3\,10^{-8}T^{1.47} \eta(B)^{\epsilon}}{V(A)^{0.71}}$$
where $\epsilon = \frac{10.2}{V(A)}-0.791$
The other variables are

$M(B)$ = molecular weight of solvent B
$T$ = temperature
$\eta(B)$ = viscosity of solvent B
$V(A)$ = molar volume of solute A
$\phi$ = association factor of solvent B (1 if unassociated)

Given that in  my special case I am looking at a liquid in a turbulent flow, I can use the kinetic theory of gases to estimate a surface collision rate - and include surface efficiency using the work on Baldwin and Howarth from 1982.
For a cylindrical pipe with the solute tending to zero, an upper limit for my rate constant for a surface reaction would be:
$k(s) = \frac{8D}{r^2}$
Now there is one big problem with this: The diffusion estimate is not for a turbulent flow (I might be able to get get a diffusion term from Fluent to get a more accurate estimate).
Secondly, at best this would give me an upper limit.
So my question is, do you know of any better more accurate methods of estimating (at least limits) a rate constant for surface reactions? (Or at least the collision rate with the surface.)

Hi DetlevCM, welcome to chemistry.se. I have edited your post to include Mathjax for your equations. Please take a look at the edit to see how it works.

On the content of your question: I don't exactly understand why you are so interested in a diffusion coefficient. If you have a turbulent flow I would think that species transport is convection dominated, not diffusion dominated

@michielm You are possibly right - diffusion is not the dominant factor in a turbulent flow. It is just that I would need a "diffusion-like" description to allow me to link back to kinetic theory. (Which is the best I have short of trying to go down the long route of setting up a CFD simulation for this aspect of the problem I have.)


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Gaussian and Surface Reactions - Doable?
Independent from my general kinetics question here - Estimating Surface Reactions - Liquid, Turbulent Flow - I am only interested in responses regarding Gaussian in this question.
I would like to know if it is feasible to predict or model the influence of a surface (metal) on a reaction between two hydrocarbons (aromatic and aldehyde for example) in Gaussian.
If yes, how would I go about it? (Building a small metal grid and working with it is very computationally intensive - it is the only approach I can think off...)
(I have access to Gaussian at my university - however I do not have access to Linda, so no shared nodes.)

I'm no expert with surface chemistry and I avoid metals like the plague.  I do use G09 on a daily basis so I'm only speaking from the other side of the fence here but... you may be interested in something like ADF (Amsterdam Density Functional).  Also, I think NWChem may be something to look into as well.  Its freely available unlike ADF.

I doubt my university wants to buy anything - but I came across NWChem once before... but never looked at it in greater detail. Maybe I should - also metal surface effects have only become a focus because they are the only explanation for the reactions I am looking at (liquid phase is what we initially though but Gaussian suggests no...).
Thanks.

Well, do yourself a favor and find an alternative to Gaussian. There are options out there that will be much more efficient and better suited to modeling metals and surfaces.

@LordStryker That is one option - though there is always the issue of support. Heck, I even ended up with chemistry coming from mathematics and originally planning to do CFD - my topic has lead me into chemistry...

I would strongly recommend you to check out Matter Modeling SE where you find people more experienced in this subject (no offense to Chemistry SE). Also, surface catalytic modeling takes a lot of experience, so it might not be the best as your first project. If you do not insist Gaussian (not recommended tool), you can use free or open-source software (Quantum Espresso, CASTEP, etc)

@Greg 7 years late... - Incidentally, in the fuel context, plane wave codes are typically not an good option as the cell would need to be huge to avoid "spillover" across the periodic boundaries. The ORCA developers have been working on point charges for solids that seem to be an interesting approach. Though given that I am no longer active in the field I leave this to others. - And prefer to not deal with metals at all.

@DetlevCM It is never late :) Do not use cluster model/point charges for metal surface simulation...

Well, for spectra in lattice crystals (granted, metal oxides) it has been shown that, point charges provide good results. - There is no reason why a large enough cluster shouldn't provide good results either.
As to "too late": I don't really deal with computational chemistry any more - or at least not at present and not formally. (Just indirectly for an old friend.)

You did go to Gaussian's Documentation on the subject, right?
In principle, it should work.  In practice, you should consider using massive firepower, and the plane-wave codes (NWChem, abinit, VASP, etc. etc.)
Also, I question whether any of these codes properly model bond breaking on a surface, i.e.  metal-catalyzed reactions.  Big simulations don't guarantee correct wave functions.
I should point out that surface reactions with plane waves are also difficult. At least they used to be.  There are/were issues with Ewald summation having to be done in 3D, and so the surface normal dimension used to have to be HUGE to avoid image effects.
This is a tough business.

Thank you for pointing me at this, however I cannot quite see how I can use that. All this tells me is "an option exists"... - Maybe I need to dig around this... (I will also be attending the Gaussian course in Wroclaw in June)
Having said that - when you mention "massive firepower" I suspect I might run into trouble with my university's cluster - I do not have Linda available and the biggest node I can use is 12 cores and 24GB of RAM... (More cores = more nodes, therefore no Gaussian for me)

If you google: Gaussian PBC examples, you get this page: http://scuseria.rice.edu/gau/gdv_pbc/pbc_guide.html

That is a start - lots of fun - I will have a read of it and try to understand as much as possible (Maths background). I was also thinking of trying to get NWChem to run on our computing cluster - being free that should be doable. For now I gave you a +1 on your answer.

Well, I am sure you found the answer (it's been 4 years since you asked the question), but you can use VASP (the Vienna Ab-Initio Simulation Package) or if you don't have or can't afford a licence for VASP, then Quantum Espresso is great too (though it's a pain to get a hang of if you are teaching it to yourself). Also, Schrodinger now comes enabled with Quantum Espresso plane wave code for solid state chemistry calculations.

NWChem is another option for plane waves that I did briefly use - but never went beyond testing (with the resources at the time). But even with a plane wave code, the problem is non-trivial due to the cell size. (Should be a lot better with the computers of today, but I moved on to other details of the same science area.)

I have done some work with reactions on organic surfaces, and hybrid QM/MM calculations helped me solve my problem, particularly using the ONIOM forcefield. I understand that a classical forcefield may not be the best way to treat a metal surface, however.

I actually gave up on that avenue eventually - due to a lack of computational resources. In principle, ONIOM is an option. A colleague suggested freezing bonds which is another option. (But introduces its own issues.)


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Percentage of a conformer in vacuum
I just started with Molecular dynamics simulations. My background is computer science but I'm learning about scientific computing and thus my jump into MD.
Having said that, I was asked to calculate the percentage of each conformer in vacuum at a certain temperature of the Methylcyclohexane molecule. Assuming that entropies and molar volumes are identical.
Can someone point me out to some website or something where I can see how to calculate this?

Page 11699 of this paper has one approach to what you want: basically, you compute the relative Gibbs free energies of the possible stable conformers, assume they can interconvert, and use a Boltzmann distribution to work out what the percentages are.

First, we selected a reference structure $R$ for a given species (neutral or protonated).
Next, for every structure $M$ other than the reference structure $R$ we determined the
  equilibrium constant $K_M$ $$K_M = \frac{[M]}{[R]}$$ from the difference in Gibbs free energies for $M$ and $R$. The fraction of $M$ in the equilibrated sample is given by:
$$x_M = K_M/\left(1 + K_1 + K_2 + ...\right)$$
where the sum in the denominator goes through all structures for a
  given species. The fraction of $R$ in the sample is:
$$x_R = 1/\left(1 + K_1 + K_2 + ...\right)$$

They don't include it in the paper, but the way to calculate an equilibrium constant $K$ from a difference in Gibbs free energy $\Delta G$ is:
$$ \Delta G = RT \ln K $$
where $R$ is the molar gas constant, and $T$ is the temperature (in Kelvin).
The tricky part is then working out what your stable conformers are, generally using some form of either guesswork or conformation sampling algorithm. They use a genetic algorithm of some sort in the paper I linked to, but given that methylcyclohexane isn't that complex a molecule, you might want to just take a look at an energy diagram and discussion of conformers of cyclohexane for some hints.
If you're getting deep into molecular dynamics, you could also use some sort of algorithm designed to use dynamic simulation to sample a wide area on a potential surface, like metadynamics or umbrella sampling, to find your conformers, and probably calculate your free energies at the same time.
You can use a static method to calculate the free energies instead, though -- the method you use to do this will depend on what sort of accuracy you're after, and may vary from a classical forcefield technique through Density Functional Theory to coupled-cluster theory (in ascending order of computational expense).


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Methylcyclohexane chair conformers
I am trying to create the molecule of the methylcyclohexane in the equatorial and axial conformers. I read in this site about the chair conformation. My molecule looks like the following picture (created in Molden)

Can someone point out if this is actually good? I am quite doubtful about the "mid" carbons, should they be push to the center of the molecule like in the drawings in the website?

By the way, if you're just starting out (based on your other question), Avogadro tends to make much better looking graphics than Molden, especially if you combine it with POVRay. If you need lightweight portability, though, you might want to try Jmol instead.

@Aesin There is gmolden, which makes nice pictures and the post script of molden is very good. Vectors ftw!

If you really want to get a feel for what is sensible and what isn't, you should probably look at the "space filling" version of the model (ie where each atom has a shell around it at the van dear Walls radius). This gives a good idea of any important steric interactions of the sort that skeletal models just miss entirely.

It's a little hard to tell from this angle, but that looks like a pretty sensible representation of the axial methyl geometry given on the website you linked to.
I'm not entirely sure what you mean by "push to the center of the molecule", though.

ETA: These might help a little more: this isn't exactly the most accurate geometry, but it's a pretty good representation.

What I meant is, if I see from the top the cyclohexane, does the shape is like a "regular hexagon" or like in the image? From side view I learned that carbons are not all in the same plane.

@BRabbit27: Added some pictures that should help: I think there's probably a camera position which would make the shape a regular hexagon (coinciding with the axis of rotational symmetry of the equivalent cyclohexane).


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How to make artificial diamond?
Carbon can be turned to diamond artificially, but while watching a video I learned that a diamond can be made from the ashes of a man, how is this possible?

First off, take a look at the Wikipedia article: Synthetic diamond. Basically, there are two ways to artificially generate the heat and pressure needed. First is the "conventional" high-temperature, high-pressure method, which uses a press to put a sample of pure carbon graphite under pressure of about 5 gigapascals (1GPa = about 10,000 atmospheres or about 145,000 psi) and about $\pu{1500 ^\circ C}$. This combination encourages the flat hexagonal lattice structure of the graphite to morph into the trihexagonal 3-D lattice of diamond. Second is the chemical vapor deposition method, in which a substrate of diamond crystal is exposed to sublimated carbon plasma (temperatures exceeding $\pu{3000 ^\circ C}$) in a hard vacuum (less than .0001 torr). The sublimated carbon attaches to the crystal substrate, "growing" it.
Both of these can be used to create so-called "memorial diamonds" from the cremated remains of a human. To do that, first the ash is chemically filtered to separate the carbon from "pollutants" naturally occurring in the body that don't combust into gaseous substances during cremation, such as calcium, phosphorous, iron etc. Then, this "amorphous carbon", much like any other carbon, can be used in HTHP or CVD processes, along with additional carbon as may be required to grow the diamond to the proper size. Usually, the HTHP process is used, as it makes the best use of the available carbon (CVD requires sublimating a lot of carbon, more than would be left behind in the ash, and most of which ends up on the chamber walls and not the substrate), and requires less energy (it's easier to maintain mechanical pressure once applied than to maintain temperature).

The diamonds are not actually made solely from the cremated ash of a human. The ashes are just added to the graphite that vis used to synthesis the diamond.
In the diamond synthesis process (look it up in wikipedia) you end up having very high temperatures and pressures. You usually use a mix of graphite and several metals: nickel, iron and or cobalt. Any cremated human residue will react with this, and be dissolved and/or react. When the diamonds are grown from this "soup" there is also a lot of unreacted graphite, as well as metal: with various impurities dissolved in it as it cools from liquid state.
Who is to really know where the cremated remains really end up? It is likely that some of the carbon in the ash, and other elements was taken up by synthesized diamonds, but not all of it will be.

Yes, the diamond can also be created from human ashes. The diamonds which are made from ashes of man is described as the Cremation diamonds. These come under the categories of synthetic diamond or lab created diamonds because they are produced inside laboratories.
These are the methods through which synthetic diamonds are created in the laboratories:-

HPHT (High-Pressure and High-Temperature)
CVD (Chemical Vapor Deposition)
Detonation of carbon-containing explosives.
High-power ultrasound.

If you want to read about the whole process in details then you can read it from the below-mentioned article.
Source:- http://everdear.co/whole-process-create-brilliant-diamonds-ashes/


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Are metals hydrogen bond acceptors or donors?
In a carbonyl or $\ce{-OH}$ group, it's clear the oxygen acts as a hydrogen bond acceptor.
Do metals act as hydrogen bond acceptors or donors and if so, how can I tell as what it's acting?

I dont understand why my question appears unreasonable. I find at least three reviews discussing this topic:
http://pubs.acs.org/doi/abs/10.1021/cr00102a004
http://pubs.acs.org/doi/abs/10.1021/cr00065a002
http://pubs.rsc.org/en/Content/ArticleLanding/2003/DT/B303006G
My hope was somebody could report his/her experience.

Hydrogen bonds are three centre 4 electron interactions that lower the energy of the molecular orbitals involved. We can hypothetically imagine two scenarios,

Metal donor (M$^{\delta ^{-}}-H^{\delta ^{+}}...C$)
Metal acceptor (M$^{\delta ^{-}}...H^{\delta ^{+}}-C$)

(where C is just something else). Metal-Hydrogen bonds where the metal is a donor are scarce since the metal hydride bond is the opposite polarity of the metal hydrogen bond donor, however they do exist.
Metal acceptors are likely to be low oxidation state late row metals with filled $d$ orbitals to donate via $\sigma$ symmetry interaction.
You get these types of interactions all the time in crystal chemistry where we find bridging hydrogens! So in short, it depends where the metal is in the period, if its $dz^2$ is available to donate etc.
If you had a specific example I would be happy to try and work it out, but it looks like you are aware of any literature I would consult!! :)

Only very electronegative atoms— $\ce{O}$, $\ce{N}$ and $\ce{F}$— participate in hydrogen bonding. The donor is the group with the covalently-bonded hydrogen participating in hydrogen-bonding and the acceptor is the atom not covalently bonded.

A $\ce{C}$ atom may also act as a hydrogen bond donor if there are highly electronegative atoms attached to it such as in the case of chloroform $\ce{CHCl3}$.

I'm afraid that TMOTTM is right.


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How do xanthan gum and other hydrocolloids work?
Does anyone know what causes things like xanthan gum to make water into a slimy substance on the molecular level?

What you are essentially asking is: "what is the cause of non-newtonian behaviour in hydrocolloid solutions?" This is a very good question, but it has an extremely complex answer. Maybe the best place to start is to read this article: http://www.sciencemag.org/content/333/6047/1276.full

Also check out pages 21 and 22 of this document: http://www3.imperial.ac.uk/pls/portallive/docs/1/40246.PDF


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What is the highest the boiling point of water can be raised?
As I understand it, the boiling point elevation in a particular solvent (in this case water) is a function of a particular constant of that solvent, and the molality of a dissolved substance. In turn, the maximum molality is a function of the solubility of the compound, as well as the van 't' Hoff factor. Thus it would follow that the highest possible elevation of water's boiling point would be the compound with the greatest ratio of solubility to van 't' Hoff factor. So the question remains: which compound is that?

There are melts of salts with some % of water (eg nitrates used for blueing of steel) with a boiling point about 250 degrees. But all Your reasoning is nonsense, because boiling point elevation equations are limited to DILUTE systems.

The "particular constant" $k$ you refer to in the question isn't an arbitrary constant.  This "law" is usually written as
$\Delta T = km = \frac{M_1 RT_b^2}{1000 \Delta H_1} m$
where k is the "constant you refer to; $M_1$ is the molecular weight of the solvent, $R$ is the gas constant, $T_b$ is the boiling point of the pure solvent, the constant 1000 makes the formula applicable only to water, and $\Delta H_1$ is the heat of vaporization of the solvent, here assumed to be water.  $m$ is the molality of the solute.
This equation is derived in courses in physical chemistry and is itself a simplification to allow slide rule simplification (this is true for a lot of formulas in chemistry).
So what you want to do to make $\Delta T$ as large as possible is to find a solvent with a large molecular weight and a high boiling point and a low heat of vaporization.  The constant 1000, which is the number of grams of solvent in a liter of solvent will also have to be adjusted.  And of course $m$ can be played with.
I don't think anyone has ever worked out what works best for a large $\Delta T$ since boiling point elevation, once used for molecular weight approximation, isn't used for that purpose any more -- and thermometers are a lot more sensitive and accurate than they used to be.

If the 1000 is just the density of the solvent, then why not write it in that way to make the equation more generally applicable? Or are other assumptions leading to the equation also water specific?