text:
How does dimethylsulfoxide serve as a cryoprotectant?
In cell culture, dimethylsulfoxide (DMSO) is added to prevent the formation of ice crystals which may lyse the cells. Exactly how does dimethylsulfoxide act as a cryoprotectant?

I am assuming you are referring to the practice of adding DMSO to prevent lysing while freezing cells for storage. If so, my answer might be helpful.

DMSO/water tends to freeze amorphous glassy with round surfaces rather than crystalline with sharp ends and edges that pierce organelle and cell membranes and segregate dissolved species' compositions during phase transformation.

This question has a simple, but non-obvious answer. DMSO and water are miscible. Mixtures, and in particular homogeneous ones like that formed by water and DMSO, usually have lower melting (or more relevant in this case, freezing) points than either substance does alone. This is the reason salt melts ice: the salt solution has a lower freezing point than pure water. The physics behind this melting point depression are complex, but a simple enough approximation involves disruption of the intermolecular forces holding the solid crystal lattice together.
Water freezes at 0 °C, and DMSO freezes at 19 °C. Mixtures of the two have lower freezing points. More importantly, mixtures of the two likely form less disruptive crystals when frozen. Water is well know for having a solid state that is less dense than its liquid state. Ice crystals take up more room than the liquid water they formed from. Single water crystals can grow quite large, especially as smaller crystals merge. Additionally, water crystals tend to form this hexagonal fractal pattern (snowflakes) which have "pointy bits". DMSO, even at the 10% by weight customarily used, must prevent such large disruptive crystals from forming, especially when the everything is cooled to -80 °C.
DMSO is used for this purpose instead of other water-miscible liquids such as ethanol because it has very low toxicity.

I know this is old but I wanted to provide some additional information in case anyone is using this as an information source. The role in terms of lowering melting temp./preventing ice crystal formation is true. However it is not true that DMSO is substantially less toxic than ethanol. They both become toxic to cells in the single digit percentages in culture media depending on the cell line (e.g. see Timm et al., Cytotechnology, 2013). From personal experience, even hardy cell lines begin to die when cultured for extended periods at 0.5% DMSO or above. When you freeze cells down at 5-10% DMSO you need to be quick because they begin to die within minutes.
A major mechanism of this toxicity is that DMSO makes the cell membrane more permeable. This also contributes to its cryoprotective effects, making the membrane more "flexible" and less liable to being ruptured. Think a rubber ball versus a glass ball.

I'm curious: How do you get the DMSO out again, quickly, after thawing your cells?

Toxicity of DMSO is negligible for short term (<1h) in low (0-10%) DMSO concentrations. While physical the effect on ice crystals is the same (it is amphiphilic.. like soap towards particles in water), it changes effects with regard to the membrane with increasing concentration.
2.5-7.5% DMSO incorporation loosens lipid interaction and membrane stretches and by this gets thinner. 10-25% DMSO leads to increased pore formation and actually dehydration of the membrane (as DMSO shield membrane from water). Beyond that one observes membrane disintegration. (Modulating the Structure and Properties of Cell Membranes: The Molecular Mechanism of Action of Dimethyl Sulfoxide, A Gurtovenko and J Anwar, 2007)
Toxicity within the cell: Same as for crystals, DMSO also shields the soluble molecules from their natural H2O-environment, and inhibits their function. For a shorter period of time, this effect is reversible. We all learned what was written above: "When you freeze cells down at 5-10% DMSO you need to be quick because they begin to die within minutes."
I learned it the same way, but it is not true at least for PBMC. It takes about 60min on 37C for cells to significantly die. Just try and add 5 % to you cell culture, and then GENTLY wash out (180 x g for centrifugation). This is true at least for T cells, a quite sturdy cell type.
Reference: Optimal Thawing of Cryopreserved Peripheral Blood Mononuclear Cells for Use in High-Throughput Human Immune Monitoring Studies, Ramachandran et. al 2012; Cells 2012, 1, 313-324; doi:10.3390/cells1030313


text:
Is enthalpy additive in mixing?
When mixing two fluids of different composition at constant pressure when no heat can escape, is the enthalpy additive?
If not, how to compute the enthalpy of the mixture resulting from mixing two fluids with given composition after equilibrium is re-established?

The enthalpy is not additive under isobar, adiabatic conditions for non-ideal mixtures.
The change in the enthalpy is called the excess enthalpy (enthalpy of mixing). It can be computed with models such as the Margules, van Laar, Wilson, NRTL, and UNIQUAC.
These model give equations for the excess Gibbs energy. The excess enthalpy can be derived from these as described in 6.2 Fundamental Relations of Excess Functions in Molecular Thermodynamics of Fluid-Phase Equilibria (3rd Edition). (F'x already referred to this excellent book.)
One has to be careful with the conflicting sign convetions in the literature.

My question in this case was, how to compute it. That it must be computable is clear. So, given a NRTL model, what is the conceptual algorithm for finding the enthalpy of mixing?

As Ali said, unless the mixture is ideal the enthalpy of mixing (or excess enthalpy) is not zero. This is the basis for a whole field of thermodynamics (and in particular, statistical thermodynamics): solution theories. To quote only a few of the most famous ones:

VST (Vacancy Solution Theory) [I don't have a good reference at hand right now, I'll try to come back and edit it in if noöne beats me to it]
NRTL (Non Random Two Liquid)
SAFT (Statistical Associating Fluid Theory) and its variants: original paper, a sort of review.

In this literature, you'll frequently encounter the name of John M. Prausnitz, whose book are considered classics (though I have not read them myself), including:

Molecular Thermodynamics of Fluid-Phase Equilibria
The Properties of Gases and Liquids

The above are for mixture of molecular fluids… polymer solutions are a bit of a special case of this, and specific models have been developed for them, the best-known being Flory–Huggins.

Also, you did not specify what kind of computation of the mixing enthalpy you were considering, so I have to add that several molecular simulation techniques have been developped to addres this question. For a basic introduction on this topic, I would recommend Singh and Gubbins’ review in Molecular-Based Study of Fluids.

This is intended as an addendum to the other answers, not a full answer in itself.$\newcommand{\b}[2]{\ce{#1\bond{...}#2}}$
Let's say we have a solution of molecule A, and a solution of molecule B. I'm denoting the strength of intramolecular forces by $\b AA,\b BB$ etc.
Whenever we create a real (i.e. non-ideal) solution, we classify the solution on the basis of deviation from Raoult's law:
Positive deviation

$\Delta V_\text{mix}>0$

$\Delta H_\text{mix}>0$

$VP>VP_\text{expected}$

$\b AB<\b AA, \b BB$.  (cohesive forces are stronger than adhesive forces)

Example: water and benzene.

Here, the adhesive forces are weaker. This leads to a net "expansion" of the mixture, and, it is endothermic due to the fact that we are partially replacing strong cohesive "bonds" with weaker, adhesive ones.
Negative deviation

$\Delta V_\text{mix}<0$

$\Delta H_\text{mix}<0$

$VP<VP_\text{expected}$

$\b AB>\b AA, \b BB$.  (cohesive forces are weaker than adhesive forces)

Example: ethanol and chloroform.

Here, the adhesive forces are stronger. This leads to a net "contraction" of the mixture, and, it is exothermic due to the fact that we are partially replacing weak cohesive "bonds" with stronger, adhesive ones.
So yes, the enthalpy is not additive for non-ideal solutions.

Manish's explanation is very nice and clear.  Going back a step, there's an initial confusion that sometimes arises (it did for me).  You can mix two fluids and the first law tells us that the mixture total enthalpy has to equal the sum of the two fluids quantities x their respective specific enthalpies.
To avoid the potential confusion, we should use the term 'isothermal' heat of mixing.  When the mixing of two fluids at the same temperature is non-ideal, there is a change in temperature. If mixing is endothermic, the temperature falls and there is a positive heat required to restore the mixture to the initial temperature and vice versa.  This heat is the Isothermal Heat of Mixing.

source: Stack Exchange

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

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

sort: votes


text:
Is the smell coming out from a portable dehumidifier safe?
A portable dehumidifier requires the consumer to plug it into an outlet to regenerate its crystal. When the crystal is thus "regenerated", the indicator will change from pink color to its original blue color. However, I noticed that there is usually a colorless gas (or smell - it smells differently from the surrounding air) accompanying this change.
So, is the gas (or smell) from the portable dehumidifier (when plugged into a outlet for re-generation of its crystal) harmful?

Can you explain what the crystal is? Based on the common colors you describe and the "Regeneration" I would think that the "crystal" you are describing is just a basic Desiccant (like a silica gel, it's hygroscopic, it absorbs water). Most desiccants are pretty harmless (never heard of any issues with common ones used in consumer goods).
You really need to add more information to your description. What is the make (in case someone wants to look it up)?
The biggest danger I can think of with a portable dehumidifier is bacterial/fungal growth. Both of these can give off a smell, it could be that when "regenerating the crystal" (which if it's a desiccant like I think would mean heating it, is actually "cooking" the growth creating a smell. (But this is me trying to image what could potentially cause that smell, not what is happening)...
Again, not enough information to reach any conclusion. (But I do think you are describing a desiccant).

The dehumidifier that I am referring to is the - Wireless Reusable Portable Mini Air Dehumidifier. (You can google it and it will list out. Basically, it have a indicator that contain blue silica gel to show if it is wet or dry. It contain some crystals which vendors always market out as odorless, non-toxic and safe. It have a 2-pin plug behind that you can connect to the outlet for recharge. The recharge will "heat up" the crystal and the indicator will turn from pink to blue.)

From the MSDS at:
http://www.hrmech.ca/pdf/MSDSSilicaGel(Blue).pdf

Inhalation: Product may be mildly irritating to the nose, throat and respiratory tract and may cause coughing and sneezing. Excessive contact may cause drying of mucous membranes of nose and throat due to absorption of moisture and oils.

Later it also mentions it may cause shortness of breath, lung damage, pneumoconiosis.
This is all related to the dust. It is up to you to interpret their significance, and whether your product is safe.

@Justin This Meta question wasn't a criticism by any means, it was just a good example that I brought up for general discussion.

@jonsca Thank you, I am new to the site and that was helpful. Should I edit the answer?

@Justin, I am going to mark your answer as correct as I think that it could be the smell of the mould/mold in the portable dehumidifier. However, I would hope that you can get yourself one and then you will be able to know what I am trying to emphasis. Thanks.

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


text:
What software is used to generate the PDB molecule of the month images?
What software does the PDB use to generate these "cel-shaded" protein models with outlines around foreground atoms?

(Leptin: May 2012 Molecule of the Month by David Goodsell)

QuteMol is one of the software that can do that…

These illustrations are by David Goodsell, and carry his famous look. Whilst I was not able to determine what specific software he uses, this page suggests that he probably creates his illustrations as a composite of renders and that the outlines and depth cueing are achieved by running a 'find edges' filter over the z-buffer of the scene, which can be output from many renderers.
VMD is a fairly powerful program (that F'x has contributed patches to!) capable of achieving this cel-shaded look with minimal postprocessing by using a GLSL Fresnel shader.
As F'x mentioned, QuteMol can do this style as well, however QuteMol has not been actively developed since about 2007 and is not terribly feature-rich, which is a shame as it's one of very few molecule viewers with competent ambient occlusion.
How to do it with VMD and Blender
In the first instance, I tried using the GLSL shader Goodsell style in VMD on the PDB file 1RON, a neuropeptide. I was not satisfied with the results, but this would probably look decent on a protein SES as illustrated on the aforementioned VMD site.

Note that I have used 2 VMD representations to separate out the style of the backbone and the sidechains, minus the hydrogens. I'm not going for a 1-to-1 clone of the referred style, just something thematically similar.
I have shrunk this VMD output down because as F'x pointed out in chat, GLSL shaded spheres are not antialiased. This said, if you are very tricky, you might be able to force antialiasing using a shader method like FXAA, which can be enabled in arbitrary programs in the very latest nVidia beta drivers (I know...). Alternatively, you can do what I did and manually oversample :D.
Next step - Blender
I output the VdW spheres as two separate VRML files, one for the backbone, one for the sidechains. These can be imported into Blender. I'm going to assume proficiency with Blender, however if you are not familiar, Blender is fairly easy to learn, especially versions 2.5 and later.

I've combined the VdW sphere meshes of each component into a backbone and sidechain object, assigned materials for each (just changing the colours of the default diffuse shader) and have applied one level of Catmull-Clarke subdivision to the meshes, to make the spheres more spherical than the default.
I've also removed the light source and turned on environmental lighting, which lights the scene isotropically and will generate some nice ambient occlusion while we're at it.
I now render this at high resolution, obtaining a standard render and the Z-buffer for the scene. I've scaled them down here. Note that if you want the spheres to be a flat colour, you should set the material to 'shadeless'. I like the subtly shaded look better.

Oh yeah, that looks nice. And here's the Z-buffer. Regrettably, the Z-buffer in blender has no antialiasing which is why I've rendered things huge and scaled down in Photoshop.

I now run the Glowing Edges filter on the Z-buffer and threshold to remove faint spurious edges inside the spheres.

Invert and level, set blendmode to multiply and you're done.

Not perfect, but not bad.
Bonus
Here's the shadeless version, which is closer to Goodsell's ideal.

The Journey Continues!
Poking around with the Toon shaders in Blender, I have found a super easy way to make Goodsell-esque figures!
If you configure your material like so:

You have a shader that will rapidly shade to black at angles far from normal with a lightsource. To make this work, you should disable environment lighting and position a bright point light behind the camera.
You will note that I have adjusted the Toon size to 0.98, which will create a shadow just shy of perpendicular to the light source normal. You will also note that I have set the specularity to zero, which gives an approximately flat finish to the sphere.
Render it up and you will get something like this:


text:
What is Cβ (C-beta) deviation?
Molprobity and some other protein structure validation tools report a Cβ deviation statistic and offer plots for it (example below).  Apparently if the Cβ is greater than 0.25 Å, some attention should be given to the residue.
The statistic is given in units of angstroms, but I'm not sure what it's actually referring to.  With the Rama plot, it's the peptide dihedral angles, but what is this?  And why are the graphs 2-D despite the one dimension given?

This is a protein-specific parameter, or? I'm guessing this from the context, but it might be useful to make this explicit in the question.

@MadScientist yes, I mention that "protein structure validation tools report" this; do you mean edit the title?

I just missed that part somehow, it's fine like it is.

Just browsing through this "Unanswered Question" now (in late 2014). I hope it's not too late?
In protein structure validation, Ramachandran outliers are a strong, but not absolute, indicator that the backbone confirmation of your protein structure deviates from all the high resolution structures known to date, and hence need your attention during the progress of refinement.
If all the phi-psi angles (backbone conformation) all fall into the "allowed/favoured" regions on the Ramachandran plot, your structure model is probably, but not necessarily, esp in the case of low resolution, correct. Hence there are other weaker criteria in protein structure validation, and Cbeta deviation is just one of them. In the light of high resolution structures out there, it turns out that Cbeta also has a favoured range of dihedral angles, suggesting that any misfit between your structure and the "probable" side chain-backbone conformation also deserves your attention during structure refinement.
In MolProbity, the justification and validation procedure are described here.


text:
Hydration of H⁺ ion
I know that $\ce{H+}$ is not possible in water and it is present as $\ce{H3O+}$. But later on I come to know that even $\ce{H3O+}$ is not possible and that it is present as $\ce{H9O4+}$.
Why does this happen? What give that compound so much stability that is not present in $\ce{H3O+}$ or $\ce{H+}$?

I would think that this question is a bit broad… After all, the nature of the proton in water is a large field of study. Have you read your physical chemistry textbook on this topic, or the relevant Wikipedia pages (1, 2, 3). After some introductory reading, you might have a more specific question to ask…

It's basically the same reason why anhydrous $\ce{CuSO4}$ hydrates in water, I would say.. Hydration.

Because by binding to water molecule, you are distributing the positive charge to a wider field--this means stability.

The formation of bonds nearly always leads to a decrease in energy. This is desirable, since less energy $\implies$ stability, so wherever it is possible, bonds tend to form.
$\ce{H+}$ has an empty $s$ orbital. $\ce{H2O{:}}$ has a lone pair. These easily form something akin to a dative bond ($\ce{H2O{:}\bond{->}H+}$), giving $\ce{H3O+}$. Note that there is no single positively charged hydrogen--the charge is distributed over the entire molecule. Distribution of charge leads to more stability.
Now, $\ce{H9O4+}$ (called the "Eigen cation") is even stranger. You can imagine it like this: Three $\ce{H2O}$ molecules approach $\ce{H3O+}$ (remember, this cation is positively charged and water has a lone pair which is attracted to positive charge). Basically, charge-dipole attraction. Of course, the lone pair will "point" towards the hydronium.
Now, these charge-dipole attraction "bonds" resonate. The hydrogens of the central hydronium "detach" from the main molecule and attach to the peripheral ones. In reality, the electron cloud is delocalised and we have partial bonds.

Now, the charge is distributed even more. We can have other species like $\ce{H7O3+}$, $\ce{H15O7+}$, etc. Note that there are crowding issues as well - this is why $\ce{H_{103}O_{51}+}$ probably doesn't exist.

Basically, the "original" $\ce{H+}$ is being solvated. The solvation delocalises the charge and forms extra bonds, reducing the net potential energy and increasing stability. That's basically it. It's similar to what happens when you dissolve any ionic compound in water.

""The solvation delocalises the charge"" This is simply wrong, as well as the mesomery-arrows in the picture above. The movement of the protons along the hydrogen bond is fast (maybe tunneling?) but nevertheless there is a minimum on the energy hypersurface.

@georg hmm, do you have a reference for that? I wasn't too sure of this anyway... But I don't. Think the protons physically oscillate. The bonds may oscillate intead of resonating, though :\

Manishearth, this is elementary physical chemistry. I refuse to name a "reference"! Take any textbook on physical chmistry. Such a mistake is horrible for a "moderator" here.

@georg Sorry, I haven't really understood what you mean to say in the first comment. Solvation leads to the charge being spread over the aggregate, that's what I've seen in textbooks. Some digging confirmed that . I wanted a ref so that I could clearly undersrand what you meant; I don't see how or why the hydrogen moves. Btw, we all can make mistakes. As a moderator, I need not have to be proficient in the topic, and my knowledge need not be perfect.

The protons are closer to one oxygen (solid line) and more distant to the other (dashed line). The thermal motion will cause the protons to jump to the position which has opposite distances, but this is a chemical reaction, although very fast (reason for water having such high polarisation relaxation frequencies and for the high mobility of protons) this is very slow compared to electron "oscillations" named delocalisation. Think of benzene, all bond lengths are identical! Read some basics on delocalisation please, this is very disapointing here.

@georg Oh, I know what delocalization is. I didn't know that this solvation was "jumping" , and not delocalizaton. I mentioned this in my comment as well, though rather confusingly ("the bonds may oscillate"). So we had a sort of misunderstanding. Hmm, I don't see how that adds to stability(since the charge is,still concentrated), but I'll do some research tomorrow and see. If I have time, that is.

The water cluster $\ce{[H9O4]^{+}}$ can be described with resonance well enough as a first approximation, as well as we often use $\ce{H3+O}$ as an approximation to a very complex system of hydrogen bonds. However, to resemble the oscillation better, equilibrium arrows would probably be much more suitable. The most important thing to know about this movement is described by the Grotthuss mechanism as F'x linked that article before.

I think that for this type of question a bit of context as to why you are asking the question, would be useful.
If for instance you are in an Intro Chem class I would say the following:
All chemical systems tend towards their most stable state (in the absence of external energy being applied).
$\ce{H^+}$ is just a single proton, a single positive charge without any negative charge to balance it out. Naked charges are very powerful forces, while it is possible for $\ce{H^+}$ to exist in isolation (even to have a number of them together in a system), the presence of other matter which might be used to balance out the naked positive charge will quickly lead to a reaction.
By forming $\ce{H_3O^+}$ the single proton $\ce{H^+}$ is able to spread its positive charge out over not just itself but the entirety of the bonded $\ce{H_2O}$ molecule. (Note that because Oxygen can be seen in resonance to distribute the negative charge associated with its loan pairs it is particularly good at stabilizing the extra positive charge).
Whats better than  $\ce{H_3O^+}$? Getting even more water molecules to share the burden of the extra positive charge and producing $\ce{H_9O_4^+}$.
Thus the short answer I would give to your question would be:
There are two main factors:

Simply the distribution of that extra positive charge over a larger area/shared by more matter stabilizes the system.
The Oxygen in particular (being able to share the negative charge of its lone pair) can "handle" the extra positive charge better than the lone proton ($\ce{H^+}$)

Hope that helps. Because in my opinion, beyond this explanation the answer gets very complicated and as many have pointed out, broad.

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


text:
Is it possible that diamond can be grown from aqueous solutions?
Assume a cubic diamond crystal with hydrogen-terminated surface bonds. The terminating hydrogen atoms will form pairs that are geometrically close together in alternating diagonal pairs (think of the rectangles on corrugated walking steel).
Next, picture a formaldehyde molecule being pushed oxygen-first into one of these terminating hydrogen pairs. If the oxygen atom can be made to react with the hydrogen pair, a water molecule would be expelled and the remaining $\ce{CH2}$ group would presumably bond to the two bonds of the diamond lattice.
If you repeat this process over the entire cubic face, the result will be a new layer of diamond that is terminated with the same paired hydrogen atoms as the original layer. The entire layer-growth process thus in principle could be repeated indefinitely, resulting in a diamond grown at or near room temperature from an aqueous or other low-temperature solution.
Question: Can anyone think of a fundamental reason why the above process, or some variant of it, could not be used to grow diamond crystals at room temperature? (A "variant" might start with carbohydrate units larger than $\ce{OCH2}$, for example.)
One or more geometrically sophisticated catalyst molecules would be required. Each catalyst molecule would snag, carry, and orient a formaldehyde molecules onto some part of the textured surfaces provided by the cubic diamond face, such as the face itself or along a growth edge.
I did this a long time ago in high school, but my recollection is that the reaction should be mildly exothermic. But even if it is not, energy-source molecules (think of the role of ATP in living systems) could provide energy to enable an endothermic reaction.
So... thoughts, anyone? Is there any deep reason why this approach to room-temperature or near-room-temperature diamond synthesis is clearly not possible?

I'm not saying this is impossible, but this requires to rehybridise the formaldehyde carbon from an sp2 to an sp3, and there's no way I can imagine this taking place in a geometrically sensible way in aqueous media under STP. The formaldehyde carbon does have a free p orbital, but still, the intermediate would have to be twisted, and sufficiently stabilized by water molecules to live long enough for the reaction to complete. I also predict an entropic handicap to this reaction. The only method I know of that can produce diamond at (lower than) 101,3kPa is CVD. Interesting question.

CHM, thanks on both comments. Terminating H atoms would be by construction, e.g. via a thin atomic hydrogen plasma against a cubic face might do it. Also, considerable generalization both of the liquid medium and of STP are permissible, since the underlying constraint is allotrope selection via "conveyance" of the allotrope-distinguishing surface geometry presented by the cubic face (only). So, if a catalyst can both (a) survive and (b) orient the surface reaction with a sufficiently high probability, permissible temperatures could go up substantially. STP as a start supports high orientation.

I would like to say that this is an interesting idea. I don't really think we can say it is impossible, but I don't think anyone right now can rationally design a system to do this. It would be found by dumb luck.
Regarding the C-H terminated surface. These surfaces can certainly have dangling bonds, especially when we're talking about <111>, because something is necessary to stabilize the structure.

@CHM, to be precise it would be a very large hydrocarbon molecule that's a more than a tad heavy on the carbon... :) The current plasma based synthesis methods terminate the surface bonds with something -- I've always assumed hydrogen -- so that part should be fairly easy. I think the harder part is the other issue you mentioned, which is whether simply aligning molecules would really force reactions. I'm going by the very rough (and old) biological enzyme rule of thumb that if you align molecules into roughly the shape of whatever intermediate form is needed, it tends to make them react.

I see no reason that a properly designed protein could not catalyze such a reaction. The problem is designing such a protein.

Thanks @JoeHobbit, I didn't see your comment earlier. I think the real question is whether modern quantum-based modeling can answer this fairly simple forced-proximity question: If oxygen of formaldehyde is pushed firmly into close proximity with those two hydrogens, which of these is most likely?: (a) Nothing; (b) The oxygen reconfigures with the hydrogens to give water; or (c) Some other reaction. My best guess is (a), but I do not really know. Enzymes wreak terrible havoc with "obvious" reaction probabilities all the time, else you and I would not be here reading this, would we?

Regarding diamond growth from liquid phase, the following article by Dr. Robert deVries may be of interest. News & Views: Materials Science, Nature, Vol. 385, p. 485, Feb. 1997.

Non-Innocent Ligands may provide a starting point for designing such a catalyst.

The reaction you propose would be formally equivalent to the protection of aldehydes with 1,3-propanedithiol to form the correspondent dithiane. There are many organic reactions in which aldehydes participate this way, but all of them have something in common: the other species(lets call it donor) must have a free pair of electrons to attack the carbon atom in the aldehyde. In case this pair is not present, but the donor has an acidic hydrogen, we could use a strong base to remove it and form in situ the pair of electrons. (See aldol reactions)
Here's the problem: diamonds have no free pairs of electrons. And terminal hydrogens in a diamond lattice are by no means acidic, hence can't be removed by any base to the best of my knowledge. So there's no way a diamond will react with aldehydes.

Thanks! And yes, that oxygen isn't going to react easily. So, I think my question amounts to whether close, careful alignment of the oxygen near the pair might make possible an otherwise vanishingly unlikely reaction mode. To be frank I doubt it too, but I don't know if modern quantum models of that sort of thing have advanced to the point where they can gives more definitive answers to such oddball questions. (More thoughts welcome, but I'll likely give you the answer nod in a couple of days either way.)

You also have the problem that diamond is not the stable form of carbon at normal temperature and pressure.  So it is very likely that energetics are not favorable for the formation of diamond this way.

That point I'm pretty sure is not a problem, since the energy difference between diamond and graphite is very small, quite trivial in comparison to some of the energetically non-optimal configurations that enzymes produce all the time.


text:
Why is calcium fluoride insoluble in water?
Why is calcium fluoride insoluble in water? $\ce{NaCl}$ is soluble, but $\ce{CaF2}$ is not. Why is this?

Please have a look at the recent topics of solubility and judge and link the answers that concern also this question.

Without specifying the solvent, the meaning is unclear. I guess you meant soluble in water, but please write that out specifically.

This answer to a question about the solubility of silver nitrate and silver chloride will give you a good starting point to answer your question.
In order to dissolve a salt, you have to break apart the ions and hydrate them in solution. You can use the enthalpies of hydration of the ions, and the crystal lattice energy of the solid, to predict which compounds will dissolve. There is quite a nice discussion about how to think about solubilities of salts, and what factors affect solubility using a Born-Haber cycle, in this reference. The values for enthalpies of hydration and crystal lattice energies of sodium chloride and calcium fluoride can be found in standard chemistry texts or with an on-line search.
If you have more questions after looking at this material, come back and ask for clarification.

The calcium cation is much bigger in size compared to the fluorine anion, hence has less polarising power. The polarisation effects are way much less due to the additional small size (relative) of the fluorine anion (low polarisability). This leads me to conclude that calcium fluoride is highly ionic in nature, and should be soluble in water. Where have I gone wrong?


text:
Why do biphasic systems of fluorous and organic solvents form?
It's possible to select a perfluorinated hydrocarbon that is immiscible in another organic solvent, forming a distinct phase boundary. Such systems may become miscible at elevated temperature which is exploited in the design of some catalytic processes1.
Why is it that many fluorous solvents are immiscible with organic solvents? Are their polarities sufficiently different or is there another reason?

(1) Housecroft, C.E.; Sharpe, A.G.; Inorganic Chemistry 2e; Pearson Prentice Hall; 2005, pp. 798-799

Fluorous solvents have the odd property of being both hydrophobic and lipophobic (Ref) and thus are not miscible with either aqueous or many organic solvents as you've noted. Fluorine, as the most electronegative element, does odd things to a molecule (and is often used in pharmaceutical compounds just because of some of these odd properties).
In the perfluorinated hydrocarbons, the carbon-fluorine bonds are quite polar with the electron density higher toward the fluorine. The solvents themselves, however, are not polar due to free rotation about the C-C bonds, so they are not miscible with polar solvents and are hydrophobic.
Because the F is so electronegative, the electrons are quite tightly held and the molecules have a low polarizability and unusually weak London dispersion forces, so they aren't miscible with non-polar solvents either. These weak London forces are illustrated by the fact that the perfluorinated hydrocarbons have nearly the same boiling point as their corresponding normal hydrocarbon, despite having more mass. (Ref)
You mentioned heat as one way to render the organic and fluorous solvents miscible. Adding gaseous $\ce{CO2}$ to the biphasic system also makes the phases miscible in many cases due to the high solubility of $\ce{CO2}$ in both fluorous and organic solvents. (Ref)
This reference suggest that shape may also be a consideration when considering flourous solvent's miscibility with alkanes, also through poor/weak interaction of London dispersion forces in the two liquids.


text:
Why is 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) often considered a prototypical room temperature ionic liquid?
It seems from a brief search of the literature that 1-ethyl-3-methylimidazolium tetrafluoroborate (EMI-BF4) is a prototypical room temperature ionic liquid (RTIL) that has been studied extensively.  However, numerous -- almost limitless -- combinations of cations and anions that give rise to ionic liquids are possible.  Why does it seem that the imidazolium cations are so extensively studied?  Is one possible reason the fact that $\text{EMI}^+$ is aromatic and thus affords the ionic liquid a high degree of stability?
Similarly, why does it seem that $\text{BF}_4^-$ (along with, perhaps, $\text{PF}_6^-$) is the prototypical ionic liquid anion?  I have seen studies of other anions (for example, $\text{CF}_3 \text{SO}_3^-$ and $(\text{CF}_3 \text{SO}_2)_2 \text{N}^-$), but $\text{BF}_4^-$ seems to be one of the most popular.

Here's a link that explains why: https://www.researchgate.net/publication/293958845_SCALABLE_IONIC_LIQUID_ELECTROSPRAY_THRUSTERS_FOR_NANOSATELLITES

This is a very good question. These compounds have been studied for long enough that most people working with them "know" the answers to these questions, but the answers are buried in physical chemistry articles from 20 or more years ago. The imidazolium tetrafluoroborate salts are popular RTILs because 1) they have the "right" properties, and 2) they are easy to make.
Concerning these compounds having the "right" properties:
Most RTILs require two characteristics in the their constituent ions to minimize the intermolecular order which would cause them to solidify. First, they need a cation-anion pair that cannot associate strongly, and second they need to have lower symmetry. Immidazolium tetrafluoroborate (BF4-) salts have both of these. The aromaticity of imidazole is important. The positive charge of the imidazolium cation is distributed through the resonance of the aromatic pi system. The negative charge in BF4- is distributed inductively over the four fluorine atoms. Additionally, imidazolium is planar and BF4- is spherical. Both the diffuse charge distribution and poor shape match prevent these two ions from having a strong association. The two different alkyl groups (ethyl and methyl) on the imidazolium ring are enough to break the symmetry of the system and prevent more efficient packing, which would lead to the substance being solid.
Concerning their ease of synthesis:
The imidazolium salts are easy to prepare from imidazole and various alkyl halides, all commercially available and inexpensive. The predominance of the BF4- anion is also a result of the synthesis. The synthesis of an imidazolium salt RTIL begins with imidazole and and an alkyl halide, say chloromethane. The product of this step is 1-methylimidazolium chloride. This salt is reacted with another alkyl halide, say chloroethane, in the presence of a base, say triethylamine. The products of the this step are 1-ethyl-3-methylimidazolium chloride and triethylammonium chloride. 1-ethyl-3-methylimidazolium chloride is a solid, because the chloride ion is small, with concentrated charge. The tetrafluoroborate salt is prepared by salt metathesis. 1-ethyl-3-methylimidazolium chloride is dissolved in water and an aqueous solution of silver tetrafluoroborate (AgBF4) is added. Silver chloride, AgCl, is virtually insoluble in water (Ksp = 1.77 x 10-10) and precipitates. The solution is filtered, and the filtrate is concentrated to produce 1-ethyl-3-methylimidazolium tetrafluoroborate (EMIM BF4-). AgBF4 and AgPF6 are commercially available, so you will see those ions more frequently. The other ions, while they have good properties for ionic liquids, are not usually available as silver salts, so those reagents have to be prepared. You are already doing three reactions to get to EMIM BF4-. Who wants to do more synthesis just to make their solvents?

Hi Ben, and welcome to Chemistry StackExchange! This is a very nice answer. Maybe you would you have an idea of an introductory book on the topic and/or review article to add?

Nice to be here. I have a book sitting on my desk called Alternative Solvents for Green Chemistry by Francesca M Kerton, 2009 RSC Publishing. The book is introductory on alternative reaction media including ionic liquids, but it assumes that the reader has a strong foundation in undergraduate level chemistry.

I might also suggest these reviews: Thomas Welton, "Room-Temperature Ionic Liquids", Chem. Rev. 1999, p. 2071; J. Ranke, et al, "Design of sustainable chemical products - the example of ionic liquids", Chem. Rev. 2007, p. 2183; and N. V. Plechkova and K. R. Seddon, "Applications of ionic liquids in the chemical industry", Chem. Soc. Rev. 2008, p. 123.

Another answer could be that they're so popular bebcause of their potentional as fuel for ion thrusters to increase the engine's specific impulse.

Could you elaborate on the answer and include a source? I think that others may find this helpful.


text:
Is the specific heat capacity of liquid iron constant?
Does liquid iron have a constant specific heat capacity?  Everywhere I've looked (save one sourceless online reference to $c_p = 611\ \mathrm{J\ kg^{–1}\ K^{–1}}$ here), I find no value listed for liquid iron.  Is this because the heat capacity is temperature-dependent?  Or is this related to a measurement problem?  Or does an accepted constant value exist, but I'm just a poor researcher?

Welcome to chemistry.stackexchange! Please note that while not being explicitly disallowed, such "what's the <numerical value> of <compound>" questions  are discouraged. Please try to ask more conceptual questions.

@Manishearth I understand that.  That's why the question goes beyond "What's the value?" to "If it's not known or constant, why not?"  At this point, I've done the research, and I'm really wondering why it is that this value is hard to track down.  I.e., is it not constant? Is it difficult to measure? Am I not grasping some part of thermochemistry that rules out such a value being well defined?

Hmm, I see. Most probably it's just something that isn't usually needed. Or maybe it's highly dependant on temperature(quite possible). I feel that the question is a bit iffy, but that's just me :).  Maybe you could edit it and make what you want clearer?

Okay, let's answer bit by bit:

Measurement of heat capacity of liquids: this isn't particularly hard to do, and in fact it's much easier to obtain $C_p$ experimentally than $C_V$… however, once you know one, the other can be determined from it and some other thermodynamic properties. Different types of calorimeters are used for liquids, but the best-known is probably the Callender and Barnes apparatus.

“Does an accepted value exist?” and “Is the specific heat of liquid iron temperature-dependent?” — Well, I found a good paper which includes measurements and comparison with previously published values: Thermophysical Properties of Liquid Iron, M. Beutl, G. Pottlacher and H. Jäger, Intl. J. Thermophys., 1994, 15, 1323.

which confirms that considering that heat capacity is roughly constant in the 1800–4000 K range. The authors thus extract the constant pressure specific heat capacity as 825 J/kg/K, apparently in reasonable agreement with previous work that they cite:

“Or am I a poor researcher?” — Well, the above paper is the first result return by Google for a search for "specific heat" and "liquid iron", so I'll let you be the judge of that ;-)

Thanks.  In regard to the third bullet, I don't have access to the article since I'm not affiliated with any subscribing institution.  But hey, I finagled my way into the answer, so that's worth something.

In addition, some freely accessible papers that also give values, either by equation-of-state fitting or direct measurement: here, there

But then, if F'x googles frequently for thermodynamic properties and Excellll doesn't, the results the two of you get up top may be different...

Heat capacities are usually dependent on temperature except for ideal gases. You will generally find the values tabulated at specific temperature and pressure values. For instances for which one is interested in heat capacity as a function of temperature, this is usually approximated through a regression equation. The NIST JANAF Themochemical Tables use the Shomate equation with 5 coefficients, of the form:
Cp° = A + B t + C t2 + D t3 + E / t2
where t is T (in K) / 1000, and A, B, C, D and E are regression coefficients.
If one looks at the Condensed phase thermochemistry data page for iron, one can find the relevant coefficients (A, B, C, D and E in the above equation) and thus calculate heat capacity of liquid iron in the 1809 - 3133 K range. Since all coefficients except A are close to zero, for liquid iron heat capacity is nearly constant in this range.


text:
What is the probability of 3 possible products, between two chemical species?
I am not a chemist. I hope I will be specific enough.
Suppose there are two chemical species $\ce{A}$, $\ce{B}$ with the following properties:

at temperature $t < T_r$, no reaction occurs between $\ce{A}$ and $\ce{B}$ (in any combination).
at $t\ge T_r$, $\ce{A}$ interacts with itself to create $\ce{A_2}$, $\ce{B}$ reacts with itself to create $\ce{B_2}$, and $\ce{A}$ and $\ce{B}$ are reacting to create $\ce{AB}$.
$\ce{A_2}$, $\ce{B_2}$ and $\ce{AB}$ are never reacting.

In experiment, we first mix $\ce{A}$ and $\ce{B}$ in temperature $t<T_r$. Amounts of species mixed are $a$ for $\ce{A}$, $b$ for $\ce{B}$. Then, we add heat to obtain temperature $t\ge T_r$ and start the reaction.

What amounts of $\ce{A_2}$, $\ce{B_2}$, and $\ce{AB}$ can be expected to be produced?
To obtain the amounts, should probability theory be used? E.g., amount of $\ce{AB}$ equals to probability that species $\ce{A}$, $\ce{B}$ will interact ("collide" or similar interpretation).

Assume the rates of the reactions are equal.

I don't think you can get the relative amounts produced without the rates of the various reactions.

Will it help to assume equality of the rates?

You'll also need the initial concentrations. See F'x's answer: you need to solve a system of differential equations.

Well, if you assume the rates are known and the reactions' order follows from stoechiometry (e.g. if they are elementary reactions), you can put the chemical kinetics into simple equations:
$$\frac{\mathrm da}{\mathrm dt} = -k_1 a(t)^2 - k_3 a(t)b(t) $$
$$\frac{\mathrm db}{\mathrm dt} = -k_2 b(t)^2 - k_3 a(t)b(t) $$
($t$ here being time, not temperature).
Knowing initial amounts or concentrations $a_0=a(t=0)$ and $b_0=b(t=0)$, you can pretty much integrate  the system to find out what happens.

Edit — solving this system for $k_1=k_2=k_3$ yields the quantities of AA, AB and BB at infinite time to be as follows:
$$aa = \frac{a_0^2}{a_0+b_0}$$
$$ab = \frac{a_0 b_0}{a_0+b_0}$$
$$bb = \frac{b_0^2}{a_0+b_0}$$

Thank you this is very interesting. I was expecting $ab = \frac{2a_0b_0}{(a_0+b_0)^2} \cdot (a_0+b_0) = \frac{2a_0b_0}{a_0+b_0}$, because this is the probability-approach hypothetical result. If we have a box with balls labeled A and B, then probability of drawing ball A twice, with replacing, is $\frac{a_0^2}{(a_0+b_0)^2}$, while p. of drawing ball A and B in any order is: $\frac{2a_0b_0}{(a_0+b_0)^2}$. Prob. of double-B is $\frac{b_0^2}{(a_0+b_0)^2}$. I wonder what happened with the "2" in $2a_0b_0$, in your result.

Because A and B are conserved, you need to have $aa+ab=a_0$, and also $bb+ab=b_0$, which is one way of seeing that your expressions can't work…

Short test: $a_0 = 10,\space b_0 = 0$. Then, computed values are: $aa = \frac{10^2}{10+0}=10, \space ab=0, \space bb=0$. I think $aa$ should be 5? Five molecules $A_2$, each having two a's – so 10 a's total, like the $a_0$.

@Calikin "this is very interesting ...because this is the probability-approach hypothetical result. " It is not that surprising as you used it as an assumption for the kinetic equations. Off course, in reality it can be (very) different, but since you didn't give any extra info,Fx automatically made this assumption as most reasonable assumption.


text:
Explain the nomenclature of 7-amino-4-methylcoumarin
Can someone explain the nomenclature of 7-amino-4-methylcoumarin, pictured below (from the Sigma-Aldrich website)?

The alpha-carbons of the methyl- and amino-groups here are separated by a minimum of three C's, whereas the numbers 4 and 7 are separated by two digits. What am I missing?

A quick perusal of the internet reveals that there are at least four numbering schemes used for coumarin derivatives. In each case, elision, duplication or out-of-sequence numbering creates a gap that accounts for the odd numbering of substituents.
Skip over the fused ring carbons
Some numbering schemes do not concern themselves with the carbons shared between the rings, as seen here (Fig. 1). This is common in PAH numbering schemes and is recommended by IUPAC1.

4a and 8a
As seen here, 4 is followed by 4a and 8 is followed by 8a.

4' and 8'
Alternatively, the a modifier can be replaced with a prime, as seen here (Fig. 2).

9 and 10
This coumarin-like structure denotes the fused ring carbons 9 and 10.

(1) International Union of Pure and Applied Chemistry; NOMENCLATURE OF FUSED AND BRIDGED FUSED RING SYSTEMS; Pure and Applied Chemistry; 1998, 70, pp. 143-216

This is one of the tricky parts in the nomenclature of fused cyclic systems. Excerpt from page 180 of the 1998 IUPAC Nomenclature of fused and bridged fused ring systems
(I have added bold to the piece of importance to answer your question):

The numbering of the peripheral atoms in a particular orientation starts from the uppermost ring. If there is more than one uppermost ring the ring furthest to the right is chosen. Numbering starts from the non-fusion atom most counterclockwise in the ring selected, and numbering proceeds in a clockwise direction around the system, including fusion heteroatoms. Fusion carbon atoms are not included; each fusion carbon atom is given the same number as the immediately preceding position modified by a roman letter a, b, c, etc. If
  heteroatoms are identified by replacement nomenclature all atoms retain the numbers of the parent
  hydrocarbon.

If one considers coumarin, this is not the IUPAC nomenclature but the numbering rule remains the same. Below is an example of the numbering scheme for coumarin which summarizes why the substituents are numbered by this manner on coumarin.


text:
Why is water ignored in the ionic equilibrium of HF?
While trying to understand the solution of a problem given in my textbook, I realized I'm having some difficulty with the solution. The problem is as follows:

The ionization constant of $\ce{HF}$ is $3.2 \times 10^{-4}$. Calculate the degree of dissociation of $\ce{HF}$ in its $\pu{0.02 M}$ solution. Calculate the concentration of all species present $\ce{H3O+}$, $\ce{F-}$ and $\ce{HF}$ in the solution and its PF.

In the solution of this problem, the equation is given as
$$\ce{HF + H2O <=> H3O+ + F-}$$
The concentration at the time of equilibrium are given as:
\begin{align}
\ce{[HF]}   &= 0.02 - 0.02x, &
\ce{[H3O+]} &= 0.02x, &
\ce{[F- ]}  &= 0.02x
\end{align}
I have the following questions:

Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?
Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?

Write down the dissociation constant equation with and without $\ce{H2O}$.
What changes?
Can you find a formulation of the dissociation constant where it seems natural not to include $\ce{H2O}$?
Read up on "activity".

Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

Write down and solve the equation with $0.02 x$ and with $x$ only. What is the difference?

Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?

$\ce{H3O+}$ actually is $\ce{H+}$ and $\ce{H2O}$ so you can write the equation as just dissociation of $\ce{HF}$ ($\ce{H2O}$ cancels out on both sides):
$$\ce{HF <=> H+ + F-}.$$
Thus we do not take water into consideration.

Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

Because whenever we write an equation
$$\ce{A -> B + C}$$
We write initial concentration of $\ce{A}$ as $c$ while $\ce{B}$ and $\ce{C}$ are zero.
Final concentrations are written as:

for $\ce{A}$: (Initial concentration) - (concentration dissociated)
for $\ce{B}$ and $\ce{C}$:  (concentration dissociated)

Thus re-writing the equations:
\begin{array}{lccccc}
        & \ce{HF} & \ce{<=>} & \ce{H+} & + & \ce{F-} \\
  \text{Initial Amount}:& 0.2      &&    0  &&  0  \\
  \text{Final Amount}:  & 0.2-0.2x &&  0.2x && 0.2x   \\
\end{array}


text:
Silver and milk (or milk products)
In the title I am refering to silver spoons, which, when used to consume milk products (like yogurt), get a weird taste. It could be described as umami/meaty, a little bit like onions. I came to the conclusion that it must be the silver, because this taste is never observed when using steel spoons.
My guess was: the silver (or maybe its oxide?) reacts with an ingredient of the milk product.
My question is: Is this documented/known process/reaction or is this some kind of misperception on my part? If the former is true, what exactly is happening? (By)-products?

You're right: the silver is reacting with sulfur compounds in the food to form a tarnish of silver sulfide. This is most commonly observed, in my experience, using silver teaspoons with boiled eggs, which are pretty rich in sulfur.
There are a number of reactions that can take place depending on the sulfur-containing species - the abstract from this paper from the Journal of Chemical Education (unfortunately pay-walled) has a nice example:

Silver and silver-plated objects react with sulfur and sulfur compounds to produce silver sulfide $\ce{Ag2S}$ or tarnish. Contact with materials that contain sulfur compounds, such as hard boiled eggs, mayonnaise, mustard, and rubber bands can cause tarnish. In the air, a silver object can tarnish owing to the reaction of silver with hydrogen sulfide ($\ce{H2S}$). This is a gas found in the air as a result of some industrial processes and the decomposition of dead plants and animals. The reaction of silver with hydrogen sulfide to form tarnish is as follows:
$\ce{2 Ag(s) + H2S(g) -> Ag2S(s) + H2(g)}$

In Sterling silver specifically, you'll often also get copper sulfide formed.
The paper also mentions that you can remove the tarnish abrasively, or through reacting the tarnish with aluminium (via solution, electrochemically):

$\ce{3 Ag2S(s) + 2 Al(s) -> 6 Ag(s) + Al2S3(s)}$

But a more common approach used by many commercial products is to dissolve off the tarnish, usually in sodium thiosulfate or thiourea. This means you end up with slightly less silver on your item, but is otherwise pretty effective and very mild (it's commonly used for museum pieces).

I'm not totally convinced by this answer: OK it does explain the appearance of tarnish, but is there some evidence that tarnish (or silver sulfide) might have a specific taste, as observed in the initial question? The taste should then be apparent for any foodstuff getting in contact with the tarnish deposit. I would think of some other reaction being involved (but maybe starting from silver sulfide).

Actually, the mentioned boiled eggs also "produce" this taste, I forgot to mention that, so this sounds plausible. What I'd like to know if ingesting silver sulphide is "safe".

Silver sulfide's solubility in water is very low ($8.5 10^{-15}$ g/L), so it should have no taste (and probably low toxicity, though I can't find detailed data on that). Also, it would mean that the spoon would continue to taste that way after the tarnish has formed, not only when eating the milk (because tarnish really forms on the surface and stays there… that's why it's annoying to get rid of it).

And also: milk doesn't actually contain that much sulfur, unless it's condensed milk.

Hmm. Silver is extremely reactive with hydrogen sulphide, as far as I can tell, and given the low solubility it's not unsurprising that it'd hoover up and retain sulphur from any source. If it can't be tasted directly, it's possible that the acidic conditions in various foods could cause the re-release of tiny quantities of hydrogen sulphide (detectable down to 0.02 ppm by some humans, apparently, but more usually ~5 ppm) or related compounds, which would cause the 'taste'.

Oops, misread, ~5 ppm is the maximum detection quantity, and over that the gas poisoning your olfactory nerves means your  ability to smell it diminishes. So it should be detectable in extremely small quantities.

@Aesin That was my guess as well, that the acids in the milk products (never tried it with milk itself, but with yogurt and such) help dissolve the silver sulphide  "layer" and therefor the taste. But...I'm a complete noob in chemistry, so...does this sound reasonable?

I'm not entirely convinced by this answer either, and humbly suggest that what might be going on is some kind of catalytic decomposition of something in the dairy on the spoon surface. That's just wild guessing though.

Are you sure the spoons are made of Silver not Nickel Silver which looks like Silver but contains no Silver.

@Aesin why should silver be oxidized by $\ce{H+}$. Based of the standard redox potentials, I would conclude that such a reaction should not occur.

Yoghurt is an acid product, as it contains lactic acid that originates from the fermentation of lactose by Lactobacillus. This acid may react slightly with silver sulfide and the resulting salt produced the weird taste you mentioned.
Incidentally, silver used to manufacture objects such as a spoon or a piece of jewelry is not pure, soft silver; 7.5% w/w copper is added to it to improve its strength, thus copper might also have a role in the unpleasant taste you felt.


text:
Do noble gasses besides Helium form diatomic molecules at low temperatures?
I know that at extremely low temperatures (mK and lower), Helium can form diatomic molecules. Do the other noble gasses also form molecules at extremely low temperatures?

You might have a hard time to find them named as “diatomic molecules”, unless they are specifically called “van der Waals molecules” or “London molecules”… But searching for “neon (or argon, or …) dimer binding energy (or binding curve)” turns up plenty of both theoretical and experimental papers on the topic. This one is available in full text and includes a good number of references.

PS: I won't have time to write a full answer, but I encourage anyone who wishes to pick up on the paper above and write a full answer based on it and references therein…

Yeah, dimers of noble gas atoms are stable due to dispersion aka London forces. More generally, neutral, non-polar molecules can still form very weak bound states arising entirely from the induction and interaction of instantaneous dipole moments (that is, fluctuations in the electron density around an atom arising from the motion of electrons). My understanding is that physicists would explain this interaction as a special case of the Casimir-Polder interaction. This interaction means that even noble gases are nonideal.
Correctly predicting the negative interaction energy of a noble gas dimer is a pathological test of dispersion-corrected DFT functionals, as F'x's reference1 and the references therein indicate, and many methods fail qualitatively. I recall that Zhao and Truhlar used noble gas dimers to test the efficacy of their various metahybrid functionals2,3. One of their papers2 makes reference to a table of binding energies and equilibrium distances derived4 by Wang and Ogilvy, giving for instance a $\ce{Ne\cdots{}Ne}$ binding energy of 0.35 kJ mol-1 (that's tiny) and less than half that for $\ce{He\cdots{}Ne}$. The $\ce{Ne\cdots{}Ne}$ equilibrium distance is given as 309 pm, which is extreme. As the paper5 you refer to indicates, however, the interaction energy for the $\ce{^4He\cdots{}^4He}$ dimer is much, much smaller than either of these and has an equilibrium distance of 5.2 nm (!!!), which is something else entirely.

(1) Ruzsinszky, A.; Perdew, J.P.; Csonka, G.I.; Binding Energy Curves from Nonempirical Density Functionals II. van der Waals Bonds in Rare-Gas and Alkaline-Earth Diatomics; J. Phys. Chem. A; 2005, 109, pp. 11015-11021
(2) Zhao, Y.; Truhlar, D.G.; Hybrid Meta Density Functional Theory Methods for Thermochemistry, Thermochemical Kinetics, and Noncovalent Interactions: The MPW1B95 and MPWB1K Models and Comparative Assessments for Hydrogen Bonding and van der Waals Interactions; J. Phys. Chem. A; 2004, 108, 6908-6918
(3) Zhao, Y.; Truhlar, D.G.; Comparative DFT Study of van der Waals Complexes: Rare-Gas Dimers, Alkaline-Earth
Dimers, Zinc Dimer, and Zinc-Rare-Gas Dimers; J. Phys. Chem. A; 2006, 110, pp. 5121-5129
(4) Ogilvie, J.F.; Wang, F.Y.H.; Potential-energy functions of diatomic molecules of the noble gases: II. Unlike nuclear species; J. Mol. Struct.; 1993, 291, pp. 313-322
(5) Grisenti, R.E.; Schöllkopf, W.; Toennies,  J.P.; Hegerfeldt, G.C.; Köhler, T.; Stoll, M.; Determination of the Bond Length and Binding Energy of the Helium Dimer by Diffraction from a Transmission Grating; Phys. Rev. Lett.; 2000, 85, pp. 2284-2287


text:
What causes the DNA fragments to stop moving in gel electrophoresis?
I'm currently studying VCE BioChemistry, and we're studying the separation of DNA strings of different lengths via gel electrophoresis.
(This involves having 'clumps' of DNA at one end of a gel medium and applying an electric current, pushing the DNA strands across the gel. Different sized strands move at different speeds through the gel)
At some point, the strands stop moving such that there is a distribution of strands across the gel. What causes the strands to stop?
Surely they'd continue until the end of the gel.
If suppose the electric current could be stopped, thus causing the strands to stop moving, but how would one know when to stop the current?

The DNA does run off the end of the gel. I will suggest that this question be migrated to biology.se

As for when to stop, most loading dyes will contain dyes that will run as a certain molecular weight. Knowing this, you know when to stop your gel. See http://openwetware.org/wiki/Agarose_gel_loading_dye for more details.

@bobthejoe: Electrophoresis, while being widely used in biology, is still a physical-chemistry concept at its root. This question deals with that concept, and thus is fine here, IMO. Also, if you want to suggest migration, use the flag option, that way we are guaranteed to notice it :)

@bobthejoe: I agree with Manishearth. Electrophoresis is definitely a chemistry technique in addition to being a biology technique. I expect the answer to this question has a lot more to do with chemistry, particularly the structure of DNA, than biology.

Also today it is quite common to have the DNA stain already in the gel while the electrophoresis is running (instead of adding a staining solution at the end of the run). This allows to follow the DNA run in "real time".

There are electrophoretic methods where the molecules stop automatically like Isoelectric Focusing, but in the common agarose gel electrophoresis of DNA you mention this is not the case.
The DNA is negatively charged, it will move along the electric field you set up until you stop applying any current. It will happily move out of your gel if you let it run for too long.
To see the progress of your gel some dyes are usually added to the sample like bromophenol blue or xylene cyanol. They don't interact with the DNA, they just provide you a reference on how far your gel has run. You still need to have a rough idea how your samples run compared to those dyes if you want to use the full length of your gel.

As I understand it, the DNA strands don't stop, they just migrate at different rates and you have to stop the process (i.e. remove the electric field) at some point when you have a good pattern.

Oh! So the dyes are added to the DNA before the electrophoresis?

DNA Molecules may stop to move in agarose gel due to high molecular weight , the molecular weight of DNA  affect its mobility in gel , the smaller the molecular weight the higher the speed of migration and long distance . large molecules tend to migrate slowly and they move short distance.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.


text:
What are the differences between the HMQC and HSQC experiments?
This afternoon, I was having some trouble with an HSQC (Heteronuclear Single Quantum Coherence Spectroscopy) spectrum I was trying to acquire, and the spectroscopist recommended I try HMQC (Heteronuclear Multiple Quantum Coherence) spectra, as they are more robust. I did so and got a very nice spectrum in about half the time an equivalent HSQC would have taken, and didn't have to phase it, to boot.
So, what is the downside? Why doesn't everyone use HMQC spectra instead of HSQC? I know they are rather different on a quantum-mechanical level, but as an end-user what, if any, is the difference in the data the two techniques give me, what are the advantages and disadvantages of each, and in what situations might I prefer one over the other?

Most likely, for you the main difference is that HSQC gives better $\ce{^13C}$ or $\ce{^15N}$ resolution than HMQC (as long as the experiment is well set up).

Very interesting; I didn't notice that. I was at 500 MHz, so I'm hoping that won't matter this time, but it is good to keep in mind. Too bad you didn't write this as an answer or I'd upvote it.

The quantum mechanical details matter a lot; usually long-range HSQC is preferable to HMQC or HMBC. See: Mandal, P. K.; Majumdar A. A comprehensive discussion of HSQC and HMQC pulse sequences. Concepts Magn. Reson., Part A 2004, 20A (1), 1–23. DOI: 10.1002/cmr.a.10095.

You get a better resolution in the indirect dimension with an HSQC compared to an HMQC. The HSQC signal is a singlet while the HMQC is usually a multiplet as you see certain couplings on it. So generally you should prefer the HSQC over the HMQC.
One advantage of the HMQC is that it uses significantly fewer pulses than the HSQC. This means that errors in the pulses don't accumulate as much as in the HSQC, you'll lose less signal if your pulses are wrong than in the HSQC.
If you had trouble setting up an HSQC, that could mean that the probe wasn't tuned and matched properly for your sample, or that your pulse lengths were wrong. The proton pulse should be determined on each sample individually; the $\ce{^13C}$ and $\ce{^15N}$ pulses should be set up correctly already. If you're not using a probe with automatic tuning and matching you should also do that manually before the measurement.
There are a lot more HSQC and HMQC variants that are used, e.g. Sofast-HMQC or TROSY sequences that are used for larger molecules like proteins. Another interesting variant you should consider is the multiplicity-edited HSQC, where $\ce{CH2}$ signals will have the opposite sign from $\ce{CH}$ or $\ce{CH3}$ signals.


text:
What is the timescale of disulfide bond formation?
What is the timescale of disulfide bond formation in a given protein? What influences it? Finally, how does one actually measure it?
From what I understand, the reduction and oxidation of bonds happens on the order of 10-8s. However, it appears that the timescale of disulfide bond formation in macromolecules scales with the size of the macromolecule.

I agree that the actual redox process is very fast. The rate limiting issue with proteins and other macromolecues involves the reactive groups finding each other. Thus, from the time of translation to the time of disulfide formation, the peptide chains need to fold properly. Perhaps other subunits need to be assembled and attached. Once everything is in the right place, the disulfide bond should form quickly provided there is an appropriate oxidant present.
You could measure the rate of the entire process by attaching reporters to positions on the protein - say a fluorophore and a quencher. When the process is complete, the quencher and the fluorophore are close in space, and fluorescence stops. This approach will measure the overall rate of the entire process. I am not sure if you can separate out the different components, e.g. specifically the S-S bond forming. I can't point to a specific paper where this is strategy is used, since I am away from my usual computer that likes my journal subscriptions. Hopefully I will be able to update my answer.


text:
How do I calculate the degree of dissociation in equilibrium?
In my textbook, for calculating the percentage dissociation of $\ce{HF}$ for the given equation:
$$\ce{HF + H2O <=> H3O+ + F-}$$
The solution is:
Initial Concentrations
$$[\ce{HF}] = 0.08~\mathrm{M}, \: \ce{[H3O+]} = 0, \:\ce{[F- ]}= 0$$
Equilibrium concentrations
$$[\ce{HF}] = 0.08~\mathrm{M} - x, \: \ce{[H3O+]} = x, \:\ce{[F- ]}= x$$
I am not able to undestand why $x$ is subtracted from 0.08 and not $cx$ [ where $x$ is the degree of dissociation]
I tried solving the same problem taking $cx$ but not able to get the solution, can anybody explain the difference to me? I have tried asking a similar question earlier too but it is really hard for me to get my head around this concept.

Let us first define the terms needed here.

Degree of dissociation (DOD)
Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by $\alpha$.
$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$
Number of moles dissociated
It is defined as the product of the initial concentration of the reactant and the degree of dissociation

Now suppose you have a reaction like this
$$\ce{A->B + C}$$
The initial state of A is always the concentration of A (should be given in the question) while initial moles of B and C are zero (if anything else is not specified).
The final state of A is always defined as $\text{(amount of substance initially present) - (amount of substance dissociated)}$ while for B and C it is just $\text{amount of substance of A dissociated)}$.
Writing our equation again,
\begin{array}{lcccc}
 & \ce{A &-> &B &+ &C}\\
\text{Initial amount of substance}& a && 0 && 0\\
\text{Final amount of substance}&   a - a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD})\\
\end{array}

@BunnyRabbit: In addition to this nice answer from Ashu, I still think (as for the original question) that you should think of your units. As the $x$ in your equation has the unit of a concentration it cannot be the degree of dissociation which is dimenionsless.

@BunnyRabbit: Note that the no of moles here are proportional to the respective concentration: $ c = \frac{n}{V}$ and $V$ is constant in your problem.

x is the extent of the reaction and α is the DoD. if the initial amount of the only reactant is a than
α=x/a
and x=aα
α is useful to dissociation type reactions, where reactant is only one and the coefficient is 1.
such as
PCl5-> PCl3  +  Cl2
N2O4  ->  2NO2
NH3  ->  (1/2)N2  +(3/2)H2

x is subtracted from reactants , because the reaction is forward, so it will change in to ions, or we can say the concentration of reactants will decrease therefore we write minus x, and the acid you mention here is weak acid so it will not dissociate completely, so we donot know directly how much it will dissociate thus we use a variable x ,to find the value of x  we make equation ,eg c-x=x+x then kc =x2/c-x ,then we solve for x , and get the amount which dissociated,


text:
Rock that transfers image
This is a question that has been bugging me for over 15 years!
In high school while studying elements our teacher showed us a clear white rock or element.
It was simply flat on both sides but when placed on a sheet of paper would transfer the letters from the page beneath it to the top surface of the rock.
I remember being amazed by this rock, but could not remember what it was.
Now I am working on a creative project and would love to see if I can incorporate the material in my design.

I used F'x answer as a starting point, and I found the mineral. It's called Ulexite.

The fibers of ulexite act as optical fibers, transmitting light along their lengths by internal reflection. When a piece of ulexite is cut with flat polished faces perpendicular to the orientation of the fibers, a good-quality specimen will display an image of whatever surface is adjacent to its other side (as shown in the photograph).

I'm not sure I understand what you want from the description above “transfer the letters”, but there is a typical high-school demonstration that I can think of that closely matches this description: that of birefringence.

Calcite (featured above) was the first material with which this effect was observed, in 1669, and still an easily obtainable material for demonstrating it.

Very similar to this.  Perhaps my memor is bad.  What i remember is the image from the bottom sheet of paper looking like it was on top of the rock.  So if the rock is one inch thick and sitting on a newspaper, it would be as if the newspaper was on top of the mineral and not beneath.  I think the teacher said they were used for tv screens.   Dont remember it having the double effect.

FX thank you for your answer, helped me find exactly what i was looking for.


text:
Why is arsenous acid denoted H3AsO3?
Inspired by this question, I'm wondering why arsenous acid is frequently denoted $\ce{H3AsO3}$, as opposed to $\ce{As(OH)3}$, which would appear to more accurately reflect its connectivity?
[edit]
I feel like such a doofus, forgetting about sulfuric acid ($\ce{H2SO4}$), phosphoric acid ($\ce{H3PO4}$), and boric acid ($\ce{B(OH)3}$), etc. Great discussion, people!

While Manishearth does the in-depth research, here's a hint: It all depends on which bond in the H-O-X chain, X being arsenic in this case, breaks or disassociates the most easily. Observe for example the contrast between hypochlorous acid, H-O-Cl, and sodium hydroxide, H-O-Na.

@TerryBollinger: Actually no, it's easier than that. Its just that we have two different types of formulae here.

Oh wait, I see what you mean.. Yeah, that is a factor here, just not the main point.

Yep, just a hint, not intended to answer the main point you were researching... which I see is now available below, cool!

As(OH)3 looks like three alcohols, which doesn't match the reactivity of the compounds, despite being technically correct? Or am I totally wrong on this?

Could you have a look at http://meta.chemistry.stackexchange.com/a/182/22--would you mind your question being rewritten to a more general one?

All quoted stuff copied from IUPAC Red Book, IR-4 unless otherwise specified
It's not only $\ce{H3AsO3}$ . What about $\ce{H3BO3,H3PO4,H2SO4,HClO4}$? Or indeed, any other oxyacid?
Basically, we have multiple ways of writing inline formulae of covalent compounds. One is the boring way (molecular formula), and one is the fun way (structural formula) where the formula reveals a bit of the structure. $\ce{As(OH)3}$ reveals the structure, $\ce{H3AsO3}$ doesn't.

For compounds consisting of discrete molecules, the molecular formula, as opposed to the
  empirical formula, may be used to indicate the actual composition of the molecules. For the
  order of citation of symbols in molecular formulae, see Section IR-4.4.

A structural formula gives partial or complete information about the way in which the atoms
  in a molecule are connected and arranged in space. In simple cases, a line formula that is just
  a sequence of atomic symbols gives structural information provided the reader knows that
  the formula represents the order of the atoms in the linear structure.

So, what are the rules for writing the molecular formula?
Well, initially these rules were basically that we just follow electronegativity--the most electropositive element is listed first and so on. Looks like they slightly changed it.
For one, there are actually two ordering standards--one is the extremely boring alphabetical one, which I've never seen used (except in emperical formulae), and one is via electronegativity.
Now, instead of following the absolute electronegativities, the IUPAC has assigned a "pseudo-electronegativity" order (my terminology). The rules are as follows (source):

for two elements in different groups - then the element in the higher numbered group has higher "electronegativity"
for two elements within the same group the element with the lower the atomic number has the higher "electronegativity"
Hydrogen is fitted in to be less electronegative than polonium and more electronegative than nitrogen. Hence the formulae of water and ammonia can be written H2O and NH3 respectively.

This leads to the following "pseudo-electronegativity" order of elements:

(click to enlarge)
The motivation behind this must probably be the erratic and hard-to-quantify nature of "real" enectronegativity.

Now, we can apply these rules in the following way:

If the compound is a binary compound (only two elements, not two atoms), then we just follow the pseudo-electronegativity thingy.

In accordance with established practice, the electronegativity criterion (Section IR-4.4.2.1)
  is most often used in binary species.

If the compound is a coordination compound, we order central atoms by pseudo-electronegativity, and ligands by alphabetical order, as written (eg $\ce{(en)}$ before $\ce{(dmg)}$, but $\ce{C2H4(NH2)2}$ after $\ce{(dmg)}$, even though $\ce{(en)}$ and $\ce{C2H4(NH2)2}$ are the same).

The order of citation of central atoms is based on electronegativity as described in
  Section IR-4.4.2.1. Ligands are cited alphabetically (Section IR-4.4.2.2) according to the
  first symbol of the ligand formula or ligand abbreviation (see Section IR-4.4.4) as written.
  Where possible, the ligand formula should be written in such a way that a/the donor atom
  symbol is closest to the symbol of the central atom to which it is attached.

Finally (barring annoying extra rules for isotopes and hydration), we have what are called "generalized salts". This is for compounds which can be treated as being made up of ions. The ions are ordered by electronegativity, and the elements within each ion are ordered alphabetically{*}.

If the formula of a compound containing three or more elements is not naturally assigned
  using the preceding two sections, the compound can be treated as a generalized salt. This
  term is taken to mean any compound in which it is possible to identify at least one
  constituent which is a positive ion or can be classified as electropositive or more
  electropositive than the other constituents, and at least one constituent which is a negative
  ion or can be classified as electronegative or more electronegative than the rest of the
  constitutents. The ordering principle is then:
(i) all electropositive constituents precede all electronegative constituents;
(ii) within each of the two groups of constituents, alphabetical order is used.

This is the rule we need to write the formula for $\ce{H3AsO3}$. First, we need to clear up what is meant by "treated as being made up of ions". Well, for this, as @Terry noted above, we look for the weaker bond. In $\ce{NaOH}$, we have relative bond strengths $\ce{Na...O-H}$, whereas $\ce{H3AsO3}$ is $\ce{As-(O...H)3}$. Thus, in $\ce{NaOH}$, the "ion" breakup is $\ce{Na+, OH-}$, while in $\ce{H3AsO3}$, the ions are $\ce{H+,AsO3-}$.
So, arsenous acid is broken up into ions, of which $\ce{H+}$ is written first. Within each ion, we order alphabetically, so we get $\ce{AsO3}$, not $\ce{O3As}$. Putting them together, we get $\ce{H3AsO3}$.

*Despite being the IUPAC rule, it seems that, within ions, most people just follow the electronegativity rules again. Selenous acid seems to be commonly named $\ce{H2SeO3}$, when it should be $\ce{H2O3Se}$. But this seems to be a minor offence:

Deviation from alphabetical order of constituents in the same class is allowed to
  emphasize similarities between compounds.

You can write it however you want. Check this out:

Note the presence of formulas such as $\ce{(HO)_2SO_2}$ for sulfuric acid. This information sheet is the same formula sheet I used when I was taught. Serves a good pedagogical purpose - emphasizing structure - and its relation to reactivity, stability, etc.


text:
What does this lambda–sigma notation mean?
I am currently reading the CRC Handbook of Phosphorus-31 Nuclear Magnetic Resonance Data by John C. Tebby (CRC Press, 1991) and on several figures (ex. pages 9 to 14) there is a caption that reads along these lines

$\ce{^{31}P}$ NMR chemical shifts of three coordinate (λ3 σ3) phosphorus compounds. (Tables B to E.)

What does the (λ3 σ3) signify? I've not seen that notation before, and it is on the captions of some of the charts I need.

The notation refers to the valency (number of valence electrons involved in bonds, i.e., either 3 or 5) and coordination number (number of substituents attached) in organophosphorus compounds. For example, it could refer to the tautomeric forms of phosphonate esters:1

So the λ3σ3 refers to a P with 3 bonding valence electrons (the lone pair doesn't count) and 3 substituents attached while λ5σ4 refers to a P with all five of its valence electrons involved in bonds and 4 substituents attached.

Reference

Kraszewski, A.; Stawinski, J. H-Phosphonates: Versatile synthetic precursors to biologically active phosphorus compounds. Pure Appl. Chem. 2009, 79 (12), 2217–2227. DOI: 10.1351/pac200779122217.

To add to the currently accepted answer, this is what I found from IUPAC publications.
In Treatment of Variable Valence in Organic Nomenclature (Lambda Convention) (IUPAC Recommendations 1983):

These recommendations, designated by Lm (lambda), provide a general method for indicating nonstandard valence states of formally neutral skeletal atoms in parent hydrides.
The bonding number, $n$, of a skeletal atom in a parent hydride is the sum of the total number of valence bonds to adjacent skeletal atoms, if any, and the number of attached hydrogen atoms.
A nonstandard bonding number of a neutral skeletal atom in a parent hydride is indicated by the symbol λn. [...] Derivatives formed by substitution of hydrogen atoms of the parent hydride are named by means of prefixes and/or suffixes in the established manner.

Regarding the sigma notation, it seems to have been included in a tentative draft of Nomenclature of Organic Chemistry, dating from 1973. However, it was omitted from the eventual 1979 publication for the following reason (taken from the same source above):

In the first publication of the provisional Section D rules, the symbol σm, where $m$ is the number of σ bonds terminating at the heteroatom, was suggested either as an alternative to or in conjunction with the λn symbol. However, the number of hydrogen atoms attached to the heteroatom would have to be included in this symbolism for a parent hydride, which is not consistent with the basic principles of substitutive nomenclature. In organic nomenclature, implied (or expressed) valence requirements of a skeletal atom in a parent hydride are satisfied with hydrogen atoms after other structural requirements are met. Hence, this symbolism was not included in the revision of the Section D rules for the 1979 edition of the IUPAC Organic Rules.


text:
What organic solvents are suitable to use with potassium permanganate?
I'm attempting to oxidize a substance with potassium permanganate and I was wondering what organic solvents I have at my disposal. The compound is soluble in chloroform and ethanol, but obviously those would not do well with a powerful oxidizing agent. The phase-transfer catalyst I'm going to employ is tetra-n-butylammonium bromide.

Since your compound is soluble in chloroform, is it also soluble in methylene chloride? I found a reference showing good solubility of tetra-n-butylammonium permanganate in that solvent. Other chlorinated solvents, such as 1,1,2,-trichloroethane were mentioned in another article.

It sounds like you need a solvent that 1) does not react with KMnO4 and 2) is immiscible with water.
As @Janice DelMar said in her comment, some of the lower-chlorinated hydrocarbons (methylene chloride, 1,2-dichloroethane, 1,1,2-trichloroethane) should work. If you can live with a high boiling point solvent, you might consider chlorobenzene.
Other solvents that fit these criteria include most alkanes (pentane, hexane, cyclohexane, petroleum ether, etc.), benzene (but not toluene), and diethyl ether.
A quick test to determine compatibility of your solvent with KMnO4 involves a TLC stain. Make (or borrow) a batch of KMnO4 stain as described here. Dip a TLC plate into your solvent and then dip it into the stain with minimal drying between. If the plate turns brown (MnO2), then your solvent was oxidized and is no good. If the plate stays purple, then your solvent is resistant.

Love the KMnO4 stain tip!


text:
How does oobleck work?
Oobleck is basically a mixture of cornstarch and water. It has the interesting property of going hard when pressure is applied. For example, it feels like a liquid when you swirl your fingers through it, but it becomes hard for a second if you punch it.
The common explanation for this is that the starch-water mixture is a suspension, and the  starch particles "lock" when pressure is applied.
What exactly is the mechanism of this "locking"?

Wikipedia already offers a decent summary of two theories of shear thickening behavior - namely the generation of transient 'defects' in a high-density suspension that either resist further growth or are incompressible in the direction of shear.
A potential macroscale analogue of shear thickening is the dramatic increase in viscosity of large particle-air mixtures upon pneumatic removal of the interparticle fluid, as is exploited in jamming skin enabled locomotion. (paper here) This should be roughly comparable to the transient decrease in particle spacing in a shear thickening fluid that is being compressed beyond its relaxation rate.


text:
How do copper sulphate and the like change crystal structure on adding water of hydration?
Copper sulphate, in its hydrated form, is crystalline, whereas the anhydrous form is amorphous.
Gypsum has a similar story-- on heating the crystalline dihydrate we get an amorphous hemihydrate. (Gypsum in fact has two different crystalline forms of the dihydrate, and has an anhydrous form as well).
I've never really grasped how the water makes it crystalline. I suspect it has to do with the relative sizes of the compounds being re-compensated by ligands, but I'm not sure. How does the water affect/create the crystalline properties?

I learned something in trying to answer this question since I've never quite understood what the water was doing in hydrated crystals either. The waters of hydration help to stabilize the crystalline form by holding the crystal together with hydrogen bonding. Water incorporated into the crystal allows the negative dipole of the water to partially diffuse the positive charge and the positive dipole of the water to partially diffuse the negative charge, and would help to minimize repulsive forces between ions of like charge "close" in the crystal. Speaking of $\ce{Cu(SO4)2 .5H2O}$...

Copper is surrounded by six oxygen atoms, provided by two different sulfate groups and four molecules of water. A fifth water resides elsewhere in the framework but does not bind directly to copper.....Water of crystallization is stabilized by electrostatic attractions, consequently hydrates are common for salts that contain +2 and +3 cations as well as -2 anions. 1 (Ref)

A more interesting reference that takes some looking at to get the sense of is given below. $\ce{Fe2(SO4)3.xH2O}$, with iron with a 3+ charge, forms various hydrates with x = 6, 7, 9, 10. (Disclaimer....actually this is the crystal structure for coquimbite with has an occasional Al for Fe substitution thrown in, but the principle is the same)

In this picture, the sulfate ions are shown as tetrahedra and the $\ce{Fe^3+}$ ions are shown as octahedra (presumably because $\ce{Fe^3+}$ ions have 6-coordination sites). The crystal is formed from "clusters of 6 sulfate tetrahedra and 3 iron octahedra which share only corners" and there are two clusters per unit cell. Each of the clusters is linked to the others through hydrogen bond provided by the incorporated water, and parallel chains are held together only through hydrogen bonds.

The individual chain segments are linked through hydrogen bonds only. The geometrical arrangement of the chains gives rise to "channels," ... which are occupied by water molecules linked to the chains by hydrogen bonds.

So, you can see how removing the waters of hydration would cause the crystalline structure to break down and result in an amorphous substance.
Further information from the paper: Coquimbite has $\ce{.9H2O}$. The crystal structure shows that, of the 3 Fe found in a cluster, one Fe is associated with 4 oxygens from sulfate groups, the second Fe is surrounded by 3 oxygens from sulfate groups and 3 water molecules, and the third Fe (or Al) is surrounded by 6 water molecules, so the octahedral representation does make sense.

Hmm, interesting. I'll wait a bit and then accept. Any reason why $\ce{Fe^{3+}}$ ions are shown as octahedra? Is it because there's some coordination compound being formed? (Probably it, the "incorporated water" may be acting as a ligand on the Iron side)

Could you incorporate that into your post? Thanks :)

[Manishearth, I'm not sure I'm even answering the right question below. After reading your question more carefully, I think you were more interested in how water molecules help crystals form, rather than whether or not the anhydrous forms can form crystals also (versus being strictly amorphous). So, oops. Maybe some one else has a good answer on the water molecules part?]
I suspect the answer is an issue of process rather than of fundamentals.
That is, if you could find a non-aqueous medium that readily dissolves the anhydrous form of some salt, my very strong suspicion is that you could grow excellent clear crystals of it that would look almost nothing like the hydrous crystalline forms.
The problem for metallic-anion salts in particular is that finding a liquid that dissolves the salt without simultaneously forming strong coordination complexes with it is going to be tricky, since I'm guessing it's usually the coordination effect that help ionize the units of the material to make it soluble!
Here's a wild guess: Someone, somewhere, has likely done research on how to grow crystals of substances such as anhydrous $\ce{CaSO4}$ (hmm, why does this phrase "dead burnt plaster" come to mind? :) or anhydrous $\ce{CuSO4}$ by using low-temperature molten solutions of... something or other? Not sure what!
And what a delightful crystal that would be to have! It would be way out of the ordinary despite being so very... ordinary?... in one sense in its composition.

Update: Apparently, clear crystals of anhydrous $\ce{CuSO4}$ do indeed exist, and are easier to create than I thought. You just add concentrated sulfuric acid [a] to ordinary copper sulphate pentahydrate solution and let it evaporate slowly. Since $\ce{H2SO4}$ is one of the most aggressive drying agents in existence, my guess is that the $\ce{CuSO4}$ just can't compete for the remaining water molecules and is forced to crystallize without hydration.
[a] Please, never ever attempt to use concentrated $\ce{H2SO4}$anywhere except in a real lab with protections and procedures fully in place. This is a compound that wants water so badly that it willy synthesize water right out of the hydrogen and oxygen bonded into proteins and carbohydrates, leaving only charred carbon. Ironically, the actual acidity of concentrated (vs hydrated) $\ce{H2SO4}$ is quite moderate, comparable to that of vinegar.

Addenda:
-- Another common-exotic crystal may be lithium fluoride, $\ce{LiF}$, although the accuracy of that seems to depend on who you read. Unique how, you say? Even though lithium was an early result of the Big Bang, lithium and fluorine are both consumed rapidly by large hot stars, especially fluorine. Their existence on earth thus appears to depend on processes such as proton irradiation and possibly even intensive neutrino irradiation tht only occur during the explosion of a supernova star, rather than being inherited from before the cataclysmic event. Pretty much all of the elements (except $\ce{H}$) here on Earth are star-forged of course, but the elements in $\ce{LiF}$ crystals thus may be remnants of the supernova event itself. That would... cool!)
-- And speaking of unique, wearable crystals: I've always wondered if there's a market out there for diamond crystals synthesized using carbon that was captured as carbon dioxide during the cremation of a loved one. If there is, feel free -- that is definitely not my bailiwick!

As a possibly useful addition: structures for chalcanthite, the pentahydrate, and chalcocyanite, the anhydrous copper sulphate.
Also, cremation diamonds. (Though, made from ash rather than carbon dioxide.)

Aesin, that cremation diamonds link is... remarkable! I had that idea longer ago than I care to admit. Clearly a missed opportunity. And thanks for all the structures, wow.

Yeah, I was looking for how it forms a crystal. It's OK, I know about conc $\ce{H2SO4}$ (though others may not). Of course, in concentrated form it only goes till $\ce{HSO4-}$, making it not that great, but the water thingy is always there. And the anhydrous crystals concept is interesting--most probably they just retain a brittle structure after dehydration.

Hmm. You have a solid correct answer now -- I'm inclined just to delete this one as overly irrelevant?...

@TerryBollinger  I wouldn't delete it...the suggestion to dehydrate the copper (II) sulfate with sulfuric acid was interesting I thought. I have no idea what would happen, and I would try it if I still had a lab. Multiple answers usually give interesting ways to think about things.


text:
Hanging a spoon on the face
Well, I have seen many comical acts where people balance a spoon (or a fork) on different parts of the face like on the tip of the nose, forehead etc.
The do this by blowing moist air (see below) and immediately putting it on the face

So i wanted to know is what is the force that temporarily binds the skin to the fork. I doubt it's friction or Van Der Walls force.
Can someone please clarify this for me? Is it some sort of chemical bond?

Some addition
Also another method of doing this was to rub the concave part of the spoon on the tip of the nose. This article states how to do it properly. In the 2nd point it states

Eventually you will feel a slight tendency for the spoon to resist on the downstroke.

What is this resisting tendency?

My guess is that the effect is primarily due to suction - the fact that you apply a coat of water to the metal and then press on it can create a pseudo-vacuum.

Sounds more physics to me....

Why would you doubt it is friction?

Thank you to all that have expressed concern about the on-topic nature of this question.  We have discussed it and think that the forces at work are Van der Wals/adhesive forces/induced-dipole/etc interactions which makes it okay for Chem.SE.  We understand that there is much overlap with Physics, but we have elected to retain the question.  Opposing viewpoints are welcome, either on Meta or in chat.

Relevant chatroom discussion. Feel free to drop by and add your views!

I would say the answer on this website is pretty near perfect.
To summarize, you need 2 things:

Enough friction between your nose and the spoon
The center of mass of the spoon should be on a line vertically downwards THROUGH the tip of your nose as illustrated in this image from the same website:

The friction is highest if both your skin and the spoon are free of greasy `lubricants'. Breathing on the spoon will add more friction by depositing a very thin (only a few atomic layers) film of water which, on this scale, behaves solid-like due to ordering effects as shown in this paper.