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Why are noble gases stable
I was recently asked the question "Why are noble gases stable? with the expectation of providing an answer beyond the general explanation of "they have full valence layers" and I couldn't think of one.
I would love to hear a full description of the this stability, I have a solid background in quantum mechanics, so feel free to speak of wave functions or the like if necessary.

Similar, not necessarily a dupe http://chemistry.stackexchange.com/questions/1281/why-are-noble-gases-stable

Actually it is not necessary to dig deep into quantum mechanics. There are several reasons why noble gasses are stable (as gasses at room temperature).
First of all, there is the obvious full valence shell.
Trend in the periodic table make clear that the charge of the nucleus grows from left to right in every period. The attractive force towards the electrons therefore increases. (This holds also from top to bottom.) Electrons in lower orbitals ($n<\text{period}$) now shield the charge of the nucleus. This will be somewhat the same throughout the period. (The orbitals also shrink due to higher charge of the nucleus.) In noble gasses the valence shell is completely filled, providing quite a good shield for the next shell. Also the increase in the main quantum number means a significant increase in the energy level of the next orbital. This makes it unlikely, that a noble gas will accept another electron.
Due to the high charge of the nucleus it is also not easy to remove an electron from an orbital. However, this is possible from (at least) Argon on downwards. For example: $\ce{HArF}$ is stable in a matrix at room temperature. It has a strong covalent $\sigma$ bond ($\ce{H\bond{-}Ar+}$) and a not quite as strong ionic bond ($\ce{[HAr]+\cdots F-}$). Krypton already does some fancy chemistry, that is stable at room temperature. As Uncle Al has pointed out Xenon is well known for its reactivity.
But why the sudden change? It is fairly intuitive, that the maximum electron density for each orbital with increasing main quantum number is also further away from the nucleus. That makes the valence shell very well polarizeable. Also the electrons are better shielded by the previous shells. However, the natural occurrence of these elements are as (homonuclear) gasses.
However, if you bring these elements in close contact with each other it was found out, that they have a very small dissociation energy ($D_e<1~\text{meV}$). This is due to dispersion, and van-der-Waals Forces, which is the main interaction between these elements. However, in $\ce{He2}$ no binding mode was observed (instant dissociation).
But that still does not answer, why those elements are stable as gasses, as pointed out that there are in fact attractive forces. The reason is as simple as obvious: Entropy. If two noble gasses would form a molecule/adduct, the bond/association energy of this molecule would have to compensate for the loss of entropy (Two volume elements would become one, hence the gas would have to expand to cover the room previously occupied, which requires work energy).
The explanation by tschoppi also covers, why from a MO point of view there could not be a bond in $\ce{He2}$. Go ahead and ask yourself the question if this would be true for $\ce{He3}$. We also know that orbital overlap is only one component to the truth. There are many other more. If it was not for such a nice thing like dispersion and other weak chemical interactions we would not exist.

The plural of gas is gases. "Gasses" is a form of the verb "to gas", i.e. to emit gas.

They are stable because the energy is lower. (Ah, the universal answer to every chemistry problem!)
Let me elaborate: If noble gases would occur as diatomic elements, the energy should be lower in comparison to their monoatomic form. But when you combine the atomic orbitals of the bonding partners to the molecular orbitals (MO-LCAO), you then fill in the electrons into all MOs, the bonding as well as the antibonding MOs.
Because antibonding orbitals are more antibonding than bonding orbitals bonding, the overall energy of the compound is raised. So this is a state which the system prefers to avoid, giving you monoatomic elements.

Due to London dispersion the molecule $\ce{He2}$ is slightly bonding, so I believe this answer is not completely correct.

@Martin: Are you talking about intramolecular, next-neighbor London dispersion? Could you give a reference to your statement?

I found it by accident in this marvellous quantum chemistry textbook by Ira N. Levine. However he is referring to Molecular Spectra And Molecular Structure, IV. Constants Of Diatomic Molecules, K. P. Huber G. Herzberg.

http://chemistry.about.com/od/noblegasfacts/a/Noble-Gas-Compounds.htm
Noble gases are reactive. Following are some examples,

I quote the website you link to: "Helium, neon, argon, krypton, xenon, radon have completed valence electron shells, so they are highly stable." They're not as reactive as, say, oxygen. You need high pressures to get these compounds.

Xenon difluoride forms from the elements at low pressure with UV light, J. Am. Chem. Soc., 184(23) 4612 (1962).  Xe reacts with PtF6 like shot, on a vacuum line or at 77 kelvin in liquid SF6,  doi:10.1016/S0010-8545(99)00190-3

It's certainly true that noble gas compounds aren't entirely inert, but it's also certainly true that they are very inert to most conditions. While informative, without context this answer is a bit misleading.

Wrap pipe thread with Teflon tape before screwing together. It seals better, prevents galling, and facilitates dismounting. If it is aluminum pipe, the taped joint often explodes. Look up the /_\H_f of anhydrous AlF3. All the fun is in the footnotes.


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Why does sulfur exist as octasulfur?
Sulfur belongs to the same group as oxygen. Then why oxygen exists as an $\ce{O2}$ molecule while sulfur exists as $\ce{S8}$ molecule? What's the chemistry behind this?

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What is the importance of using a pKa value instead of a pKb value when describing drug chemistry?
Why do we only consider the $\mathrm{p}K_\text{a}$ value of a drug regardless of whether it is an acid or base? Why do we not use the $\mathrm{p}K_\text{b}$ value?

Can you provide an example?  pkb can be obtained readily from pka and pkw.

For easier comparability:
Comparing two pKa values is much easier than comparing a pKa with a pKb value.

The reason we only usually consider the $\mathrm p K_\mathrm a$ is that in a constant medium (in the case of drugs, water) there is a simple relationship between $\mathrm p K_\mathrm b$ and $\mathrm p K_\mathrm a$ so we don't need two scales for acidity and basicity and can define a continuous scale from strong acid to strong base using just one number.
$$\mathrm p K_\mathrm b = \mathrm p K_\mathrm w - \mathrm p K_\mathrm a$$ where $\mathrm p K_\mathrm w = -\log_{10}([\ce{H+}][\ce{OH-}])$, roughly equal to 14 under normal conditions.

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Nomenclature of organic compounds
When I name an organic compound, I follow a preferential order like so: http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/ But I don't understand why this order exists. And intuitively (to me at least), a triple bond has a higher preference than a double bond... then why is the order like in the link, i.e., double bond preferred over triple bond? Is there any chemistry behind this?

Please note that the so-called “Table of Functional Group Priorities For Nomenclature” shown on masterorganicchemistry.com tries to combine the seniority order of suffixes with prefixes and endings. This can be misleading and result in wrong names. In particular, there is no suffix ‘-ane’, ‘-ene’ or ‘-yne’ (these are endings), and multiple bonds (and all groups that follow in that table) are actually not considered in the seniority order. Please do not use such sources.

The naming of organic compounds are governed by IUPAC guidelines, specifically the way the names are formed are by some general principles, which are given on the IUPAC website at their page Guide to Name Construction: General Principles, with naming examples here.
Another rather nice summary is found on the  webpage Nomenclature (Reusch, 1999), states that a organic compound name will have the following features, based on several common chemical characteristics of organic compounds:

•  A root or base indicating a major chain or ring of carbon atoms found in the molecular structure.
•  A suffix or other element(s) designating functional groups that may be present in the compound.
•  Names of substituent groups, other than hydrogen, that complete the molecular structure.

(Specific IUPAC guidelines for different hydrocarbons are included in the link).
Another summary is found on the handout Short Summary of IUPAC Nomenclature of Organic Compounds (Dr. Jan Simek, California Polytechnic State University at San Luis Obispo), which state that the general nomenclature principle is:

IUPAC nomenclature is based on naming a molecule’s longest chain of carbons connected by
single bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or
atoms other than carbon and hydrogen, are indicated by prefixes or suffixes according to a
specific set of priorities.

But this prioritization is more of a standardisation than whether the branch has any chemical properties that lead to an order as determined by IUPAC.

@Artemisia, the 1st quote is the chemical reasons for the prioritisation determined by IUPAC.


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Does my data show quadratic regression in a calibration curve?
This data was taken for my science fair project, and I would like to know the relationship between spectrophotometer absorbance and cell density (taken manually) of algae after a toxicity test. The different levels of toxicity are IC 0 (no chemical), IC 25, IC 50, IC 100 (the most chemical). The IC value is basically the estimated about of the chemical needed to destroy a certain amount of the population. I took the manual density and absorbancy of four samples of different levels of toxicity.
IC value/Cell densities(cells/ml)/absorbancy:
IC 100/28000/0.408
IC 50/42600/0.302
IC 25/70000/0.01
IC 0/138000/0.79
Based off of these points, I would like to make a calibration curve. But when I do this, the highest r^2 is quadratic. Is this an acceptable calibration curve for finding other densities based off of known absorbances? Or would it have to be linear with a very low r^2?

Using higher order interpolations (such as a quadratic) will generally provide a larger correlation coefficient; however this does not mean the quadratic is a better fit unless there is a theoretical reason for why the dependent and independent variables should be related through a quadratic equation.  In your case, absorbance is typically linearly related to concentration.

As you have correctly assumed, the relationship between absorbance and absorbant concentration is linear (law of Lambert-Beer):
$$ A = \varepsilon \, l \times c$$
This should give you a straight line through (0,0) and your data.
Now, for a graphical analysis of data, let's not look at the numbers themselves but on what kind of a graph it gives us. So after 2 minutes with gnuplot I get something like this:

Now, there are two things that I note here:

The absorbance of the $\mathrm{IC_{25}}$ value is very very low (nearly 0) which indicates some wrong measurement. Maybe the cells weren't in the pathway of the light beam, only the medium?
You have four data points.

My advice is the following:

Double check your measurements (the absorbance and the cell density). It would be even better if you did several measurements for the different inhibitor concentrations and then calculate the mean (and standard deviation if you so desire).
Generate more data. It is very wonky to try and justify a linear regression through these 4 data points (although mathematically allowed). However, if all intermittent points are equally chaotically distributed along a straight line the linear relationship still holds (with a fairly high $R^2$). What this does to the interpretation of the data is a completely different story.

Unfortunately, I don't have enough time to take more measurements. What do you mean about the intermittent points? Is there any relationship that I can use here?

With intermittent points I mean more measurements that eventually might give you the linear relationship. Since you don't have enough time to make more measurements, I suggest you point out the problems I wrote above and call it a day. It's not satisfying but the only way I see here. Maybe talk to your supervisor about it?

Your absorbency values seem anomalous compared with the cell density. One would assume that as cell density increased, so would absorbency. Your absorbency decreases first and then leaps up to a maximum. I can't say from your data whether there is any relationship between cell density and absorbency. On the other hand, there seems to be an inverse-square relationship between IC value and cell density. Hope that helps.

What is an inverse-square relationship?

There is an inverse-square relationship between the force of gravity and the distance between two masses. In other words, $F=\frac{k}{r^2}$ This link has more information: http://csep10.phys.utk.edu/astr162/lect/light/intensity.html


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Does the hybridization model gives us any further insight on Molecular Geometry?
It doesn't seem as though the hybridization model adds anything useful to the discussion of molecular geometry as predicted by the VSEPR model. It's just another way of labeling linear, trigonal planar, and tetrahedral geometry (as $sp$, $sp^2$, and $sp^3$, respectively). Considering that orbital hybridization may not be the most accurate view of molecular bonding, what is the utility of having it around?

I find orbital hybridization of very limited use.  Orbital hybridization is a mathematical manipulation of atomic wavefunctions, but that that is the only relationship to a quantum mechanical description of an atom.  Presently, we have no evidence that supports the claim that orbital hybridization happens.  The reason why I am not in favor of using orbital hybridization to describe chemical phenomena is because it is not a predictive tool.  We only know the hybridization of an atom after we know the structure.
VSEPR, however, is a predictive tool, although it has no significant quantum mechanical basis.  The VSEPR model is based upon the idea that pairs of electrons (whether bonding or not) will repel one another.  Geometrical factors are considered to determine how far away pairs of electrons can be from one another (e.g 180 degrees apart for two pairs, and 120 degrees apart for 3 pairs).  The model accurately predicts the H-C-H angle in methane and also predicts that the H-N-H angle in ammonia is closer to 109 degrees than 120 (as is the case in a molecule such as $BF_3$.  Because VSEPR assumes ideal conditions, the model starts to fall apart when considering complex molecules.  For example, while the VSEPR model correctly predicts that $SF_4$ is seesaw shaped, it can only predict that the structure won't be precisely seesaw (with F-S-F angles of 180, 120 and 90 degrees).  It only allows us to suggest that the lone pair on the sulfur will be a "space hog" in a manner of speaking and repel the bonding electrons.
Now, on to the last part of your question why do we still use it?  (This is somewhat speculative and opinion-based.) We have to blame the organic chemists here.  orbital hybridization terminology ($sp^2, sp^3$, ...) provides a lexicon for describing the bonding environments of carbon atoms that is compact and visually constructive for organic molecules.  If I state that there is an $sp^2$ carbon in a molecule, you can begin to visualize what that part of the molecule looks like, that it is a region of planarity, that it cannot be a chiral center, that it likely has a double bond and therefore may undergo certain types of chemical reactions, etc. So while orbital hybridization doesn't provide any insight into molecular geometry, it provides a basis for terminology that chemists can use to transmit meaningful information.

+1. I agree with everything except that hybridization is not predictive. If I've misunderstood what you mean by "predictive," my apologies. Hybridization allows some qualitative predictions, e.g.: we expect H-bearing sp2 atom to be more acidic than an sp3, ceteris paribus, due to additional s-character. Granted, we need to know the structure first. AFAIK, however, that's true for any bonding theory. In MO theory, for example, we still need to know the structure in order to characterize symmetry elements. MO theory is certainly more powerful and more correct, but also more complex.

@GregE. Consider an atom of carbon.  With MO theory, the same 1 2s and 3 2p orbitals are engaging in bonding regardless of whether the ultimate structure contains a single or double bond; however, in the hybridization model, we must first look at the result and then say, "Oh, because there is a double bond, we must have had $sp^2$ orbitals.  Once we can assign a "hybridization", we can make some predictions about the chemistry, I agree.

That's fair. Granted, in MO theory, all atomic orbitals are used in constructing the MOs, not a subset. Still, AFAIK, an assumption (or prior knowledge) of the structure of the molecule must be made. E.g., given three atoms, I can create a plausible molecule which is either linear, bent, or cyclic. The choice will dictate symmetry elements, which in turn determines the valid combinations of AOs. Of course, MO theory will better predict the stability/properties of the resulting structures. In that sense, I would agree. Is there anything I'm missing?

It has to do with the order of events.  In MO theory, there are tools to determine whether or not orbitals will mix (relative energies and symmetry) before creating a structure.  I need to think about a practical example, but that will take a little more coffee..

I would like to add something to this discussion about whether hybridization is predictive. I recently came across a molecule called Propadiene. Looking at the 2D representation may lead one to assume that the two pairs of hydrogen atoms can rotate with respect to each other. In fact, the 2 Hydrogens on one side are fixed relative to the Hydrogens on the other side, due to the issue of availability of p-orbitals for making pi bonds.


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Prediction of products in reaction of dicarbonyl with alpha-haloketone
Following is the question that was asked in my exam. a,b,c,d are the various options, one of which is the answer. I am familiar with the Reformatsky reaction, Aldol condensation, Knoevenagel's reaction and general base catalysed carbonyl reaction. 1 and 2 are my attempts at deriving the mechanism and hence getting at the products. But none of those succeeded to bring about the required products.
I need help in how to proceed through in this case and derive the products.

Just to be clear, your second proposed mechanism begins with the first intermediate formed in the mechanism 1?

yes!! I assumed the first step to be very probable as the active methylene hydrogen are very acidic

I am not sure but possibly, the $\ce{OH-}$ might also attack the carbonyl Carbon, in both the dicarbonyl or the alpha-haloketone

As you've already figured out, cyclohexane-1,3-dione is more acidic than chloroacetaldehyde, the pKa of the diketone in water should be around 5.3.
The enolate of 1 (sorry for the circle in the structure, i'm to lazy to fix it) and 2 undergo an aldol reaction yielding 3. With a bit of charge shifting, we get again a nicely stabilized enolate (4), which dehalogenates (you might call that an O-alkylation as well) to 5. Subsequent elimination under aqueous conditions results in the formation of 6. I'd say this is, more or less, an example of the Feist-Bénary reaction.

Btw, I created the drawings with BKChem.
A conceivable depiction for the "transformation" of 3 to 4 is given below, but please note that this is just a model to please our eyes and minds ;)

This was very helpful. But can you expand on the transformation from 3 to 4. I fail to understand that.

Klaus Warzecha has already provided the correct mechanism, but I'd like to offer a supplementary discussion.
One approach here would be to attempt to eliminate some of the options based simply on first principles. Choices (b) and (c) seem improbable prima facie for reasons of thermodynamics: the carbonyl moiety is not conjugated to the aromatic furan ring, so these should be intrinsically less stable.
Structure (b) has an additional problem, which is that the position of the ether linkage is para- to the remaining carbonyl. A similar problem occurs in structure (d), except in that case the ether linkage is ortho- to the carbonyl moiety. Finding plausible mechanisms that would lead to either of those regiochemical outcomes is difficult. The greater acidity of the 1,3-cyclohexanedione makes it the likelier enolate nucleophile, whereas the 2-chloroacetaldehyde is the likelier electrophile. This means that the oxygens of the original 1,3-cyclohexanedione should remain in place, not undergoing elimination, so they should maintain a meta- relationship.
At the conclusion of that analysis, only (a) survives as the correct choice. The mechanism is fairly rote, and has already been given. The only subtle question, to my mind, is which site on the enolate reacts with which site on the electrophile (i.e., does the carbon of the enolate add to the the carbonyl, or does it substitute the chlorine?); either mechanism can lead to the same product. The enolate carbon is the softer nucleophilic site and should prefer the softer electrophile. Fleming, in his book Molecular Orbitals and Organic Chemical Reactions, provides a table of the absolute hardness, $\eta$, of various Lewis acids. Acetaldehyde is given to be considerably softer than chloromethane. Assuming it's valid to compare those two molecules as a proxy for this reaction, it seems probable that the enolate will attack the carbonyl from the carbon, while the chlorine will be displaced in an SN2 attack by the oxygen. Were the leaving group one of the larger, softer halogens (i.e., $\ce{Br-}$ or $\ce{I-}$), the order might well be the opposite. Aside from the superior leaving group quality of those larger halogens, another mechanistic advantage of the other order is that it results in the hydroxide leaving group of the final intermediate being shifted over by one carbon, so that the E1cB process leading to the final product results from abstraction of a more acidic allylic $\beta$-hydrogen.


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Why is water wet and fire hot?
This may be a silly question but why is water wet?
All the previous questions on water do not explain the reason why is water wet. They assume that its a natural property. However we need to understand why is it wet in the first place. This could apply to other liquids such as milk, juices but the essential element is water.
Similarly another basic property which we take for granted is heat caused by fire. What causes fire to be hot as an inherent property? For the purpose of discussion consider simple fire by burning wood or paper.

What is your definition of 'wet'? By conventional dictionary, something becomes wet when covered/soaked with water or another liquid.

I'll let Richard Feynman answer the fire portion.

Fire is hot because you're absorbing energy from a bunch of extremely high-energy electons, protons, and atoms. Water is wet because it bonds to lots of stuff and is in the liquid state. It feels cool at room temperature because our body has more energy than the water does (water is at room temp, our bodies are at a temp set by thermoregulation).

Two words-surface tension https://en.wikipedia.org/wiki/Wetting BTW, this belongs to physics SE

Are an ancient Greek? Ancient Greeks believe that there are four elements that make up everything: earth, water, air, and fire.

Why is water wet?
Chemically speaking saying that water is wet has not much sense.We could, however, say that "Wetting process" is caused by the presence of a thin liquid particles layer over a material. In our specific case, this depends on the chemistry of water and the chemistry of skin. So the main causes of the fact that water is wet, is its wetting properties determined by a force balance between adhesive and cohesive forces over the skin. Mercury, although is a liquid, will not "wet" you because it has a strong cohesive force that doesn't permit the molecules to get in touch with the sensory cells.  When you touch water or a water wetted material part of the water molecules are transfer to your hands (a hydrophilic material) and so to the receptor cells giving the sensation of wet, this sensation can be caused however in some case of parasthesias even if there are no liquid particles. In fact, there is no "wet receptor" on your skin but the sensation is due to the elaboration of different stimuli from tactile, pressure and temperature receptor present in your skin.
Try to play with plasticine after a while handling it your hands will become hydrophobic you will no more fell the wet sensation when you wash them even if water is still "wet"!
What causes fire to be hot as an inherent property?
Fire is an exothermic process. So energy stored in a chemical bond is released as kinetic energy and electromagnetic wave. The fire has a high temperature because the molecules have great kinetic energies when you get in touch with fire (of course wormed air could be a medium as well for short distance) this kinetic energy is transfer to your skin and so to your thermoreceptors . And you feel what human call hot. In fact, you can feel the sensation of hot even without touching the fire because it even irradiates energy as electromagnetic waves that can be absorbed by your skin and so be feel for his receptors.

"When you touch water or a water wetted material part of the water molecules are transfer to your hands (an hydrophilic material) and so to the receptor cells giving the sensation of wet, this sensation can be cause however in some case of parasthesias even if there are no liquid particles. In fact there are no "wet receptor" over your skin but the sensation is due to the elaboration of different stimuli from tactile, pressure and temperature receptor present in you skin."  Let's see, the reason that water molecules are transferred to your hands is because your hands are made of a 'water-loving' material (i.e., 'hydrophilic').  Sounds like we haven't progressed beyond Moliere (opium induces sleep because of its 'dormative' property), or Nietzsche (by reason of a 'faculty').  Or, perhaps, even Empedocles ('love' and 'strife' rule the universe).  Actually, 'wetness' doesn't seem to be a property at all, but rather the name of a relation that describes a regularity without providing any explanation at all.  Why is water wet?  We have no idea!

Not an answer, but rather an extended comment on another answer.


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Are there any amphoteric substances other than water and ammonia?
I have read about the Brønsted-Lowry and Arrhenius concepts to define a substance as an acid or a base. Our teacher told us that there are substances which behave both as an acid and a base, using water and ammonia as examples. He also asked us to find other substances with the same behavior. Does anyone know any more substances which are amphoteric in nature?

First, it may be worth elucidating the nomenclature a bit and distinguishing between amphiprotic and amphoteric  substances. The former are those that can react as either acids or bases in the Brønsted-Lowry sense, while the latter are those that react in either manner per the broader Lewis definition. As you mention, water and ammonia are amphiprotic (and thus also amphoteric by definition), which is evident even from the fact that both undergo auto-ionization:
$$
\ce{H2O + H2O <=> H3O+ + OH-}\\
\ce{NH3 + NH3 <=> NH4+ + NH2-}
$$
In fact, a large number of protic substances exhibit this type of behavior to an appreciable degree, including concentrated solutions of various acids (e.g., sulfuric, hydrofluoric, etc.). Another canonical example of an extremely common amphiprotic molecule prevalent in nature is bicarbonate, which you find in countless consumer products, as well as within the human body as a physiologic buffering agent.
Under the right conditions, a very wide array of molecules of diverse types can be amphoteric. The two necessary and sufficient criteria for amphoterism are:

The molecule possesses lone pairs, so that it can act as a base.
The molecule can accept electron pairs, so that it can act an acid.

Those are not rare properties, hence amphoterism isn't really an exotic phenomenon.

To drive the point home, every single neutral and anionic species we know exergonically receives a free proton, acting as a base (even many cations should accept protons, but sufficiently small and highly charged cations will likely electrically repel the proton). The pinnacles of unreactivity, helium and neon, will quite gladly react with a free proton to form the hydrohelium and hydroneon cations $\ce{HeH+}$ and $\ce{NeH+}$. I suspect the situation is not analogous for acids; even some neutral species have endergonic electron affinities, such as manganese atoms and benzene.

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Are there any significant industrial processes that use liquid ammonia as a solvent?
There are some interesting laboratory reactions that use liquid ammonia as a solvent (for example, the Birch Reduction where an alkali metal is dissolved in liquid ammonia).
There have also been some high volume industrial processes that use liquid ammonia as a solvent (the weedkiller paraquat has two, quite different, production routes using liquid ammonia producing thousands of tonnes a year). One uses dissolving sodium (like the Birch reduction):

But are there any other significant chemical manufacturing processes that use it?
NB I would include the production of pharmaceuticals or their ingredients as well as any bulk chemicals.

I might be terribly wrong, but I'm  not aware of any other industrial process involving ammonia as a reactant (fertilizers, caprolactam, amino acids) where liquid ammonia is used as a solvent.
As a side note, liquefication at least plays a role over here (Germany) for the transport in rail tank cars.


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Why doesn't EDTA have a double bond?
EDTA (Ethylenediaminetetraacetic acid) has a structure like this

But why, as the name 'Ethylene' suggests, doesn't it should have a double bond between carbons?

A lot of non-systematic chemical names don't describe the composition of the compound, they describe what the compound is synthesised from. That is the case here since EDTA is derived from ethylene by adding two units of the diamine.
It isn't that uncommon in chemical names. Polyethylene, for example, is (simplifying only slightly) a long chain of carbons connected by single bonds. But it is made by polymerising ethylene. Hence has a name that doesn't describe what its structure is but where it came from.

As far as trivial names are concerned, this is one possibility to describe it. Another one is that ethylene is the name for the subunit $\ce{C2H4}$. Just like methylene $\ce{CH2}$ in methylene dichloride ($\ce{CH2Cl2}$) or propylene ($\ce{C3H6}$) in propylene glycol. So it actually describes the structure and not the origin. Names for polymers, however, are always derived from the monomer(s).

@Martin-マーチン, as I understand the matter, your comment is more accurate in this context.


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If atoms are mostly empty space then how are we able to interact with things?
Note: This question has come out of curiosity
Q: We all have read that atoms are mostly empty, they are generally compared to a football stadium in which a pea is kept in the center (nucleus) and flies buzzing in stands (electrons). Then why is it that we can interact with things?
It would be of great help if anyone can provide me with any article because I can't find any.

I suggest this question be moved to physics.SE as the best answer comes from a more physical point of view, though I suspect questions of this nature have been asked there before. I can't seem to find anything that directly addresses your concerns via search, though.

When we think about the structure of atoms and molecules we need to use a range of simplifying analogies as the objects we are talking about are too small to see and behave according to quantum rules we are unfamiliar with in the macroscopic world. None of the analogies we use are perfect and may mislead unless they are brought back to reality by recognising their limitations an checking their implications again the mathematics we know that does describe their behaviour.
That is the problem here. On one analogy atoms are built from nuclei containing protons and neutrons surrounded by electrons orbiting a long way away. This analogy is useful for explaining what happens when, for example, alpha particles are fired at a gold film a few layers of atoms thick. This is the famous Rutherford scattering experiment first done in 1909. This experiment led to the "solar system" analogy for what atoms look like. But really it only explains what they look like from the point of view of fast alpha particles. The mass of atoms is concentrated in the small nucleus and it is the nucleus that alpha particles bounce off, occasionally.
But that is only one, imperfect, way to describe what atoms and molecules look like. It doesn't describe how individual atoms join to form molecules, for example. For that we have to look at how electrons behave. Here a different range of analogies or mathematical descriptions are needed. The solar-system view of electron behaviour is deeply inadequate to describe what electrons actually do. And it is a very poor description of where electrons are. It is better to think of electrons as occupying 'clouds' of probability than being at a single point in space. Those clouds have a range of shapes from simple spheres to multi-lobed, multi-layer things in atoms with many electrons. Many of these have some probability of the electron being anywhere inside the cloud even, in some cases, right into the nucleus. In this analogy atoms are not empty at all but are filled with some probabilistic amounts of electron. No more than two electrons can occupy any orbital which imposes some strong constraints on what can happen when the clouds of two atoms or molecules meet.
Almost all the interactions we experience in normal macroscopic life involve those clouds of electrons interacting with each other. There are a range of different ways they can do this, but at the heart of things they all consist of electrostatic interactions at the atomic or molecular level where electron clouds push (or pull) each other. The low density of electron clouds doesn't matter since the quantum mechanical rules determine what can happen when the clouds meet. Normal interactions in the real world are just the macroscopic impact of these interactions.


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Ozonide v/s Ozonoid
Can anyone please clear the difference between ozonide and ozonoid.
I've already googled it but came up with nothing

I've seen the term ozonoide now and then, both in German and English texts.
It was used for the the primary, as well as for the secondary ozonide. Interestingly, throughout these texts, you'll find this form only once or twice while the same authors mostly use the term ozonide!
While I'm roughly familiar with the Criegee mechanism, I'm not an expert on ozonolysis. Nevertheless, I'd dare to say there's no difference between these terms (at least, there seemingly was none for the authors) and I'd recommend to stay with the established term ozonide.


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Why does road salt reveal this bus's frame?
Our city sprays brine on the roads in the winter to prevent people from sliding around. It works but it makes it impossible to keep vehicles clean. One thing I've noticed is that the brine sticks to the city buses outlining the frame underneath the body panels, like so:

Doing some research, these buses are made out of stainless steel to prevent rusting.
Is there a chemical reason for this happening?

No. the best two explanations don't need any chemistry at all.
If you look closely at the panelling on a bus like this you will find that there is physical  evidence on the outer skin about where the structural frame is. The skin immediately over the frame tends to stick out a little. These physical features are probably enough to ensure that more dirt sticks to those parts of the skin. You can see a similar effect above warm radiators when small bumps on the wall above tend to pick up a disproportionate amount of the dust in the moving air caused by the convection currents from the warm radiator, becoming darker as a result.
Another possible contributor would be the fact that the vibrations in the panel skin of the bus will also be higher at the centre of each panel farthest away from the frame. This might contribute something as well.


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Shape and polarity of the iodate ion
I am trying to figure out the shape of the iodate ion or $\ce{IO3^-}$. No matter how many times I draw it, I keep getting iodine in the centre, and 3 oxygen atoms surrounding it, 2 with single bonds, and 1 with a double bond. This would make it a trigonal planar shape. It would mean that it is not a "bent" shape.
I went to double check on the internet, and I am now confused. On Wikipedia it says it is in fact polar and a pyramidal shape.
Why does iodine have a pair of lone electrons in its shell? Why would it not combine with one of the oxygen atoms to make a double bond?

Is polarity a concept that's even applied to ions? Are there "polar ions" and "non-polar ions"? I've never heard of that before. Also, are you sure you properly counted the electrons (26 valence electrons = 13 electron pairs)? The fact that you drew your structure with two single bonds is strange, as it would imply two oxygen atoms with formal charge -1.

Would you mind detailing all the steps you go through to get to your structure? That way we can point out specifically what went wrong.

I just redrew it for the 6th time and it makes sense. I was trying to pair two pairs of electrons for some ungodly reason.

In case you're having trouble with the concept of formal charge and how to calculate it, perhaps the answers here might be of use. If it was just an oversight, don't worry, that sort of stuff happens often.

let us continue this discussion in chat

You might get more of an understanding of the possible shape (and hence polarity) by looking at other know iodine oxides. For example, periodate ((IO4)-) exists and is tetrahedral.

@NicolauSakerNeto Yes polarity applies to ions. Asymmetric ions will be, in general, polar. Iodate would be non polar if flat but polar if, as it actually is, pyramidal. Also however you draw the bonds to oxygen to calculate formal charge, all the oxygen bonds are actually the same.

@matt_black Yes, figuring out whether a compound has an electric dipole is independent of its liquid charge, so the concept of polarity is easily extended from neutral to charged species. However, I've never heard of anyone explicitly referring to ions as either polar or non-polar, and Google search results are poor because they are swamped with low-level texts on bonding and intermolecular forces. Is there any great use in classifying ions as polar or non-polar? For example, do non-polar ions show substantially increased solubility in low polarity solvents compared to similar polar ions?

Your question is flawed. Ions exhibit charge by the sole fact that they are ions. Iodate is very polar indeed

@pingOfDoom Polarity is a concept that derives from the presence of an electric dipole. Net charges on ions are equivalent to electric monopoles. You can have a monopole without a dipole, and a dipole without a monopole. In the former case, you get non-polar ions. It seems to be rather rarely brought up, possibly because free ions do not exist outside of a vacuum, and when coupled to any other counterion (polar or not), the resulting salt (the smallest unit of which is at least one ion pair) has a dipole itself and is therefore polar.

relevant wiki article http://en.wikipedia.org/wiki/VSEPR_theory

Looking at the formal charges on the atoms, each oxygen has a charge of -2. With the added electron, the iodine has a formal charge of +5. This leaves 2 electrons in the valance shell for the lone pair. So there are 4 items (3 oxygens and a lone pair) to place around the iodine atom, giving it a tetrahedral shape. Since each oxygen has the same charge of -2, there must be 2 bonds to each oxygen. The extra electron is placed on the molecule, not the iodine atom. Since you have a tetrahedral shape with a different corner (lone pair), it has a dipole (like ammonia).

For drawing structures of molecules, I always recommend following these four calculations:

Count the number of electrons we have!
$7 + 3 \cdot 6 + 1 = 26$ (iodine, three oxygens and a negative charge).

How many electrons are required for all atoms to obtain an octet?
$4 \cdot 8 = 32$ (four non-hydrogen atoms, each want eight)

How many electrons are we lacking? We need to share these, i.e. make bonds with them:
$32 - 26 = 6$ (that is: three bonds; either three singles or a single and a double or a triple)

How many electrons can we distribute as lone pairs? It’s those not required for bonding:
$26 - 6 = 20$ (giving us ten lone pairs)

Now we know that we have three bonds and that iodine is the central atom. The only way to do this is to have a single bond to each oxygen. The distribution of lone pairs is trivial since there is only one that will give every atom an octet. Assign formal charges next. Note that each oxygen appears to have seven electrons rather than six, so they get a formal minus. Iodine appears to have five rather than seven so it gets a formal two plus.
Now, consider iodine’s surroundings. There are three bonds and a lone pair. It cannot be planar since the lone pairs always want the largest possible s-character. (The theoretically largest possible s-character would be achieved with $90^\circ$ angles and iodine using only the p-orbitals for bonding. I think I remember a non-zero amount of hybridisation to relax the geometry.)
Since the molecule is non-planar, its point group must be $C_{3\mathrm{v}}$. And since $C_{3\mathrm{v}}$ does not contain an inversion centre or improper rotation, it must be polar, i.e. it must have a non-zero dipole moment.

As an additional note, one single double bond to one of the oxygens makes no sense. There are too many electrons for that to happen. Traditionally, one considered the molecule to have two double bonds to two of the three oxygens and invoke d-orbital hybridisation for iodine. That would lead to ten bonding electrons and 16 lone pair electrons; two oxygens having given up a lone pair each for bonding. However, that view should be considered obsolete, since d-orbitals are energetically too far removed to take part in bonding for non-metal atoms.

In iodate ions there are three double bonds between iodine and oxygen and there is presence of one lone pair.iodine is central atom.the shape of molecule can be seen as pyramidal with corners of base triange be oxygen atoms and iodine at top.And there is no meaning for polarity for ions as they already have a charge and hence polar.

A structure with three double bonds is not a good representation for the iodate anion, as it puts a negative formal charge on the central iodine atom, which is less electronegative than the surrounding oxygen atoms. A better representation would display two double bonds and a single bond.

Also, polarity and having net charges are two distinct things. Polarity is a concept that derives from the presence of an electric dipole. Net charges on ions are equivalent to electric monopoles. You can have a monopole without a dipole, and a dipole without a monopole. In the former case, you get non-polar ions. It seems to be rather rarely brought up, possibly because free ions do not exist outside of a vacuum, and when coupled to any other counterion (polar or not), the resulting salt (the smallest unit of which is at least one ion pair) has a dipole itself and is therefore polar.

This may be of help http://www.chemspider.com/Chemical-Structure.76615.html

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


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Is it realistic to have a natural oxygen source in a cave?
Is it realistic to have a natural oxygen source in a cave? I am writing a story, about people who live underground in a land that is freezing outside. Heat is taken care of by geothermally heated groundwater. How do I make it realistic for these people to breath in the caves?

If you mean is there a mineral that emits oxygen- I don't think so. Plants will easily give them oxygen and equally absorb carbon dioxide but there is problem- plants give off slightly toxic chemicals- a big problem in enclosed space. You need the air to be cleansed by passing through soil- microbes destroy these toxic chemicals- there are underground ecosystems which are totally cut off from the surface.

There are chemical reactions that could make oxygen if you have enough power to drive them. Or you could use plants with enough lighting to drive photosynthesis.

Yes, it is. You can allow outside air to circulate through the cave.

According to this page, the CO2 buildup in the cave might also be a problem.
But, if there was a river that ran on the surface, then oxygen and nitrogen would dissolve into the water. The stream could then pass into the cave and release some of the dissolved oxygen into the oxygen deficient environment in the cave. It would also remove the CO2. Of course, if the surface of the planet is frozen, then the river would have to avoid freezing somehow. But perhaps there is a hot spring on the surface.
My intuition is that the rate of oxygen dissolving out of the water would be very slow and it could be increased by increasing the surface area of the water. In the human lungs, there are many, many tiny branches (Alveoli) that allow the lungs to have a huge surface area so that the slow diffusion of oxygen from the air into our blood can go fast enough for us to survive. Perhaps the cave-people can accomplish something similar if they let their oxygen-rich stream pass over a tall, broad waterfall that would cause the river to break up into tiny droplets, thereby increasing the surface area. This has the added benefit of looking super cool in case your story gets made into a movie :)
Other options:
1) they could generate electricity and use that to make hydrogen and oxygen from water.
2) http://www.mpg.de/621120/pressRelease201003241?filter_order=L

Thank's, that was a lot of help. I like you're first idea the best.


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Viscosity of two different solutions
Suppose that I have 10 kg of plastic/polymer that is non solvable in water. I then divide the plastic/polymer into two equal parts.  Let us name the parts A and B in order to identify them.  I then proceed to make beads from each part. I make $n_A$ beads of radius $r_A$ from part A. I also make $n_B$ beads of radius $r_B$ from part B. Here $r_A>r_B$ and consequently $n_B>n_A$. Suppose that I put the beads from part A into 10 liters of water and I do the same for the beads from part B.  Which one is more viscous, the "solution" from part A or from part B? Note: here $n_A,n_B$ stand for some arbitrary number.

I edited your original question for clarity.  In doing so I removed one statement about "dissolving" the beads.  Since you state that the beads are insoluble, I think this is OK, but please check to make sure I did not change the meaning of your question.

If the polymer is lighter than water then it will form a scum on the top. If denser then it form a layer on the bottom. Though both depend on the size of beads being big enough to not be affected by the water jostling so it can settle.

This is a very interesting question, but I think it would get much more attention, and much better answers, at physics.se. I've tried my best at answering anyway

@Michiel Thanks for your suggestion, but questions older than 60 days can no longer be migrated.

Disclaimer: I will be talking about the shear viscosity. I think that most of what I write is also applicable to extensional viscosity, although with different numerical constants, but I am not fully sure so I will limit myself to shear viscosity.
Most theoretical models predict that the shear viscosity of both solutions will be equal. The reason is that these models describe the viscosity of the suspension $\mu$ as the viscosity of the 'empty' liquid $\mu_0$ $+$ a correction term that depends on the volume concentration of solids, $\phi$.
The most famous one is the Einstein viscosity equation that simply reads:
$$\frac{\mu}{\mu_0}=1+2.5\phi $$
This model works for low solid concentrations when inter-particle interactions can be neglected. Later Batchelor showed that including long-range hydrodynamic interactions an additional correction of $+6.2\phi^2$ should be included, but still only a dependence on the volume concentration is there.
A nice summary of more models can be found in this paper: K.D Danov, J. Colloid Interface Sci., 235, 144–149 (2001). That paper also includes a discussion that shows that the factor 2.5 should be dependent on the mobility of the particle. This would mean that it will be higher for smaller particles, so in your case that the solution with part B will be somewhat more viscosity.
Another overview of models, including models which have an increasing viscosity with increasing particle size can be found here. So that would point to part A giving the more viscous solution.
So to summarize: to first order (i.e. low concentrations) the two viscosities will be the same. At higher orders you have competing effects of the particle radius that should increase the viscosity and the particle number that should also increase it. Which effect is dominant is probably hard to tell theoretically.


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How can I make homemade ink?
I want to make some homemade ink suitable for amateurs to practice calligraphy. I have access to many household chemicals, including acetic acid, sodium hydrogen carbonate, sugar, sodium chloride, and more.
Can someone give me an easy recipe for a colored (any color will be fine) ink that I can make at home? (preferably basic, but slightly acidic is fine too I guess)
Here are some ideas:

Caramelized sugar --- nope, too expensive and smelly.
Something with rust? (iron)
Something like india ink. (pigments are acceptable too)
Cornstarch and iodine.

I don't have a full recipe, but if you're fine with black ink, try to boil some crushed or grounded organic material containing large amounts of gallic acid (or other hydrolyzable tannins) with rusty iron and some vinegar (acetic acid).
Suitable organic materials that come to my mind are oak bark or green tea.
You should end up with an ink that darkens over time. Note that this ink will probably clog fountain pens but should be fine for calligraphic pens.


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Can an aqueous solution conduct electricity forever?
We know that pure water does not conduct electricity, but salt water is a decent conductor. This is commonly explained by saying that “the ions carry the current through the solution”, which is an incomplete explanation, because it does not  indicate what happens when all of the ions have migrated to the electrodes.
More complete explanations of conduction through a salty solution (like this one or this one) rationalize the conduction of electricity in terms of a reduction reaction that takes place at the anode and an oxidation reaction that takes place at the cathode. In the case of salt water, chlorine gas ($\ce{Cl2}$) is formed at the anode and hydrogen gas ($\ce{H2}$) is formed at the cathode.
This explanation seems reasonable, but it implies that the conduction of electricity through a solution is fundamentally different than the conduction of electricity through a wire. A copper wire is (usually) unchanged, even after a large amount of electricity passes through it. In contrast, when electricity passes through salt water, two chemical reactions occur (one at each electrode), which fundamentally change the composition material.
This implies that it is not possible for an aqueous solution to conduct electricity forever. Since we are driving a chemical reaction, we are either consuming our salt (forming $\ce{Cl2}$ gas in the case of an NaCl solution or plating it onto the electrodes in other cases), or we are consuming the water by forming $\ce{H2}$ or $\ce{O2}$ gas.
This is surprising to me. So, I’m asking if my thinking is correct: is it possible for an aqueous salt-containing solution to conduct electricity forever, or it will it always eventually consume the reactants and stop conducting electricity as I have surmised?

No, unless you add more ions, or reverse the reaction for some time(rechargeable batteries).

"Conduct" is a dangerously ambiguous term here. How much of your argument applies if the current is AC not DC?

Good question Matt! I would expect that the usual 60 Hz AC current effectively acts like a DC current on the relevant timescales for the chemistry that takes place, but I am not sure about this.

Pure water does conduct electricity due to presence of equilibrium hydrogen and hydroxide ions. It is a very (very) small current though.

Since nobody has posted a counter example, I will go ahead and say that my claim is true for direct current (which is what I originally had in mind). The current drives a chemical reaction and, at some point, the reactants will run out and the solution will no longer have enough ions to effectively conduct electricity.
Hopefully someone will post a counter-example and make things more interesting :)
In the case of alternating current, it seems possible to plate material onto one electrode and remove it from the other. When the current switches direction, this process would reverse and the electrodes would remain the same mass. In theory, this could go forever, but in reality I wonder if the electrodes would deteriorate and break due to the many cycles of deposition and erosion.

It can be possible, but it depends on the experimental conditions/set up.
I think the following setup could work “forever”:
Two electrodes made of $\ce{Pt}$ in solution of $\ce{CuSO4}$ and $\ce{H2SO4}$ provided with AC source, the bath must be sealed – to avoid evaporation, applying reasonable voltage.
The copper is plating onto one electrode and depleting it to the other electrode.
The reduction: $\ce{Cu^2+ + 2e- -> Cu}$
The oxidation: $\ce{Cu $-$ 2e- -> Cu^2+}$
The electrodes made form $\ce{Pt}$ will avoid change shape of the electrodes during the endless process. The additional $\ce{H2SO4}$ makes the electrolyte more conductive.

So, what reactions are taking place at the electrodes? And how do they reverse to return ions to the solution?

CuSO$_4$ $\Leftrightarrow$ Cu$^+$$^2$ + SO$^-$$^2$$_4$
on both electrodes

That reaction doesn't consume or produce electrons, it just splits CuSO4. So, I don't understand how electrons would be drawn from or returned to an electrode. In the case of salt water, we are taking two Cl- and producing Cl2 gas and 2 electrons at one electrode and taking two electrons from the other electrode to turn 2H+ into H2 gas.

Since the soulutin contains CuSO$_4$, one reaction will be reduction:
Cu$^2$$^+$ + 2e$^-$ $\Rightarrow$ $Cu$
and simultaneously (since the soulutin contains Cu$^2$$^+$) oxidation:
$Cu$ - 2e$^-$ $\Rightarrow$ Cu$^2$$^+$

So, the reaction is basically plating copper onto one electrode and depleting it from another electrode. For DC, one electrode would be consumed and the reaction would stop. But, for AC, the electrodes would alternate depositing and removing materials and end up staying the same size.

yes, thats right

Okay, maybe edit that information into your answer and we can get rid of these comments. I'm also not sure why you need H2SO4. It seems like you are only using the CuSO4. Also, my guess is that this reaction would not really go forever because the electrodes would change shape as the material is re-deposited and eventually small pieces would break off and fall to the bottom of the container.

The plain answer to your question is "no" because the electric current  going through the solution would break  the $\ce{H2O}$ bonds and form $\ce{H2}$  and $\ce{O2}$ gas. Eventually you run out of water due to this process. And don't forget, in a $\ce{H2O + NaCl}$ solution, chlorine gas  is produced which would eventually affect ion concentrations. I believe that Wikipedia has a very detailed description of the electrolysis of water.
And by the way a really simple answer to your question would be to carry out the experiment yourself and see whether it conducts electricity forever. Take a glass of water add sodium chloride,  take a 9 volt battery  connect two wires one at the anode and the other at the cathode and take two pencils sharpened on both ends (graphite is a good conductor of electricity and plus you don't want the wires to be"eaten")  connect each of the wires to one end of each of the pencils. Then put the end of the pencils (the ones without the wires ) into the glass and see it happen for yourself.

I think that Jeroslav demonstrated that the answer is actually "yes" for AC, at least in theory. Nobody here has proposed an example of a solution that can conduct forever for DC. But, I am hesitant to say that it is not possible just because we cannot think of an example. Your proposed experiment will produce gas and demonstrate that a particular solution cannot conduct forever, but it does not prove this generally.

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The science behind fertilizers
Do you know some books or articles that elaborate on the chemistry behind fertilizers?
I have taken organic chemistry so far, and  inorganic chemistry as well. I am also familiar with the Haber-Bosch process.

That is a rather broad and vague question. Do you want to know what they are made of? how they are made? how they work? You need to be more specific to get good answers.

Asides from a general search on Google Scholar, where, unfortunately, many resources are behind paywalls, here are some book references that may be of use for your broad-ranging request:
Methods in Agricultural Chemical Analysis: A Practical Handbook (Google eBook) (Faithfull, 2002) - some of this is available online through Google Books
Fertilizer Manual (UNIDO, 1998) - this, according to its contents has some information about the chemistry of frtilizers.
Chemistry For Agriculture And Ecology (Mido and Satake, 1994), A lot of this book goes back to basics, but chapter 18 is specifically about the chemistry of fertilizers.
Agricultural Chemicals and the Environment (Hester and Harrison, 1996) - once again, has some information about the chemical processes of fertilizers, with a focus on environmental concerns.
I hope these are useful.

Please let me know if you'd like me to add any specific resource.

Thank you. Also, do you do fieldwork regarding fertilizers? If so (and if it is not much trouble), can you send me some journal articles that you used frequently, or any publications that you have contributed to?


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Can I synthesize iron acetate like this?
The oxidation state of the iron doesn't matter too much; I am just trying to make an attractive red-brown ink. Here are some instructions I prescribed to a friend.

Here's an approximately decent iron acetate recipe. I don't know if it'll work, though. Maybe you'll want to heat the acetic acid. The optional first step is to oxidize the iron by hydrogen peroxide or by heating in air. Pour enough heated acid to corrode the iron plenty. Wait. Filter what is now the aqueous solution of iron acetate and acetic acid through some filter paper to remove the undissolved iron. Let the (I hope so) red-brown solution dry: this is because you want to evaporate the acetic acid. When it is dry, you may dissolve what I hope to be a dark red-brown powder in as little distilled water as possible and some alcohol if you so desire. Adjust dilution to suit your hue/shading preference.

Would this synthesize iron (I or preferably II) acetate?

I sincerely doubt you'll get any iron(I) compounds; as far as I know, that's an extremely uncommon oxidation state. IIRC, the initial reaction with hydrogen peroxide will produce iron(II) hydroxide, which will probably go on to react with water and oxygen to form some iron(III) compounds. The hydroxides would ordinarily precipitate out. Adding acetic acid should dissolve that precipitate (the extent of dissolution will depend on concentration) and yield the dissolved acetate. My guess is that you'll mostly obtain iron(II) acetate this way. I could be mistaken, this isn't my strongest area.

Sorry. I meant iron (III).

The oxidation state of the iron should matter. Iron(II) acetate is off-white to tan in color. Iron(III) acetate is red-brown.

Red iron oxide, Fe2O3, from a paint or art supply store, or as polishing rouge (no additives).  dissolve in aqueous acetic acid, filter off any solids, and go on from there.
Dissolving iron in acetic acid will give you Fe(II)acetate.  Air or peroxide oxidation may (probably will) give you Fenton chemistry and muck up your organics.


text:
Why do bubbles form when it rains?
I noticed sometime ago that bubbles formed in puddles/streets whenever it rained. I am wondering if someone could tell me what could cause such bubbles. I examined the bubbles again today more closely and they didn't look "soapy". I know that is not a really valid answer to rule out that leftover detergent/soap could be the cause of such bubble formation. However, the bubbles are short-lived. Furthermore, there aren't "frothy". In other words, there are no clumps of small bubbles.

So by "streets" I'm picturing asphalt. Is that right? The bubble are short-lived; how big do they get?

Yes. They get sort of big. Usually they are roughly between 2 to 4 centimeters in diameter.

I see this too sometimes. I second the idea that there's probably no soap involved, although I always thought that there's probably something else in there.

When you turn on the faucet into a bucket of water, bubbles appear also. It's because the rush of water brings air down and cohesive tension creates a bubble for the air.

I see. But I am not quite convinced that raindrops would be large enough to trap enough air to form a bubble of 2 to 4 centimeters. Maybe the small bubbles coalesce into a large one, but I don't think I've seen such thing happen.

My grandmother was born 1800s.
The farmers had a saying, " bubbles in the puddles, it will rain all day."
Perhaps it has something to do with air pressure or something.
Bubbles form in puddles on low grass and mud too.

Bubbles consist of a gas trapped by a liquid, where the liquid has a surface tension high enough to encapsulate the gas. This gives bubbles three ingredients: the liquid, the gas, and surface tension.
The first ingredient, the liquid, is obviously the rainwater.
The second ingredient, the gas, is (also obviously) air. But what isn't obvious is how the gas becomes entrained in the liquid. Asphalt is porous, meaning the microstructure of asphalt consists of lots and lots of tiny random tubes through the solid material. Thanks to the capillary effect, water is sucked into those tiny tubes, and gas is forced out. This gas being forced out of the porous asphalt material and through the liquid on the surface creates bubbles.
Surface tension is the third ingredient, and controls the bubbles' shape, size, and strength. Surface tension can be enhanced by surfactants (big molecules which, when added to a liquid, agglomerate and increase the surface tension). The surface of a road accumulates lots of surfactants - mostly, oil. Oil consists of many long-chain hydrocarbons, which are perfect surfactants (most detergents are actually long-chain hydrocarbons that are polarized at one end). Surfactants tend to agglomerate (that's why, if you pour a little oil on top of water, it will tend to clump together as it spreads out over the surface of the water). Having oil act as a surfactant increases the surface tension of the rain water on the surface of the road, and makes any bubbles formed in the water stronger. That explains the large bubbles you saw.

The high-level concept here may be correct.  Unfortunately, there are at least three factual problems: (1) surfactants decrease surface tension; related, (2) surface tension must be low in order for bubbles to form; and (3) oils themselves are not surfactants; only the fatty acids derived from them are. There may still be surfactant residues on roadways, but they are definitely not oils.


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Decomposition of iron(III) acetate?
I recently asked about the synthesis of iron (III) acetate. I want a rather concentrated solution to use as a calligraphy ink (which will be dissolved in alcohol/water). Will it be okay to heat the solution of iron(III) acetate, hydrogen peroxide, acetic acid, and water to evaporate all of the other reagents? That is, will heat cause the decomposition of the iron acetate?

I don't think it should be a problem.  Acetate is a stable ion. Iron(III)(or ferric)  ions are also quite stable. Quick wikipedia check shows its stable- it would be no good as a ferric ion test if the ferric ion decomposed to ferrous Fe(II) ions. Wiki lists it as dye.

@user2617804 Would you like to make an answer out of that comment, with links to the sources?

Iron(III) acetate will be stable under these conditions. According to this paper, thermal decomposition to iron oxide only starts at 423 K (150 °C). Heating iron and acetic acid together with an oxidation agent (oxygen from air or, like in your case, $\ce{H2O2}$) is actually a suitable way to prepare iron acetate dyes (source).


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What are the units of P and V when R is expressed in terms of J/(mol K)?
I have learnt the ideal gas equation $PV=nRT$. However, I have some doubts regarding the units of the quantities in this equation.
When we take the gas constant $R=8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}$, what should be the units of $P$ and $V$?

The units for pressure $P$ are pascals ($\mathrm{Pa}$) and volume $V$ is metres cubed ($\mathrm m^3$). The proof can be found by the derivation provided from the University of Waterloo's page The Ideal Gas Law, reproduced below:
Rearrange the formula to make $R$ the subject, hence:
$$R = \frac{PV}{nT}$$
Now rewrite the Ideal Gas Law in terms of its units:
$$\mathrm{J\ mol^{-1}\ K^{-1}} = \frac{\mathrm{Pa\ m^3}}{\mathrm{mol\ K}}$$
or
$$\mathrm{J\ mol^{-1}\ K^{-1}} = \mathrm{Pa\ m^3\ mol^{-1}\ K^{-1}}$$
Given that $\mathrm{Pa} = \mathrm{N\ m^{-2}}$, rewrite that into the equation:
$$\mathrm{J\ mol^{-1}\ K^{-1}} = \mathrm{N\ m^{-2}\ m^3\ mol^{-1}\ K^{-1}}$$
Now the term $\mathrm{N\ m^{-2}\ m^3}$ can be simplified to $\mathrm{N\ m} = \mathrm J$  as work (unit = $\mathrm J$) = force x distance ($\mathrm{N\ m}$)
Hence, the right hand side becomes $\mathrm{J\ mol^{-1}\ K^{-1}}$ as required

Would you agree that pressure is expressed as a force acting on a surface? If so, let's juggle with units and start with the pascal.
$1\ \mathrm{Pa} = 1\ \mathrm{Nm^{-2}}$
will certainly be familiar to you ;)
Would you further agree that we can express the joule as $1\ \mathrm J = 1\ \mathrm{Nm}$?


text:
What are the dimensions of a and b in van der Waal's real gas formula?
In the van der Waal's formula for real gases $\left( p+\frac{a}{V^2}\right)(V-b)=nRT$, what are the dimensions of $a$ and $b$?
I learnt that $a$ is related to the average force of attraction between the molecules and $b$ is related to the total volume of the molecules, but I fail to understand their dimensions.

There seem to be a few ways to write this formula - but this document http://www.roanestate.edu/faculty/chemistry/chemistryslideshows/vanderWaal/VDW.pdf may be of help.

Is there a reason why your other question couldn't be edited to include both the real and ideal gas laws?  It would make the question and answer easier to find for future visitors.

Part of your confusion is in the formula you have written.  According to the Tanner's General Chemistry webpage The van der Waals Equation, if you are considering 1 mole of a substance, then the formula is
$$(P+a/V^2)(V-b) = RT$$
For $n$ moles
$$(P+an^2/V^2)(V-nb) = nRT$$
The Science HQ website Van der waal’s equation, the constants $a$ and $b$ represent pressure and volume corrections respectively.  The units are based on the formula for $n$ moles of substance (the 2nd equation above).
Taking each term on the left hand side individually:
$(P+an^2/V^2)$ is the van der Waals equation equivalent for $P$ in the ideal gas equation, thus this therm must have the same unit, so using $Pa$ as the unit for pressure would require that the units of $a$ to be $Pa L^2 mol^{-2}$  (the website uses the unit $atm$, but the same idea still applies.
In a similar manner, the term $(V-nb)$ is the van der Waals equivalent of $V$ in the ideal gas equation.  So to keep the unit as for volume in the second term, the unit for $b$ must be $Lmol^{-1}$.
One thing to always be careful with, as there are a number of units used for both pressure and volume, it is imperative to keep these constant throughout your calculations.
For example, volume can be expressed as $m^3$ or $L$ and several others (hence the $L^2$ term above would be the same as $(m^3)^2$ or simply $m^6$).
One final document is a master list of $a$ and $b$ constants for several chemicals (once again be careful with the units - unit conversion are provided in the document), van der Waals constants for gases.. Also further explanations of the above can be found on this Purdue University page entitled Deviations from Ideal Gas Law Behavior

Since the van der Waals equation is derived from the ideal gas equation, the empirical constants $a$ and $b$ need to have dimensions so that the units work out.
The ideal gas constant $R$ has dimensions of $\dfrac{\text{pressure}\cdot\text{volume}}{\text{moles}\cdot\text{temperature}}$ since $R=\frac{pV}{nT}$ for an ideal gas. This is one of the definitions of $R$.
For the van der Waals equation, now $R=\dfrac{(p-\frac{a}{V^2})(V-b)}{nT}$
Since we have $(p-\frac{a}{V^2})$, the term $\frac{a}{V^2}$ needs to have units of pressure for this subtraction to not screw up the units.
$$\therefore \dfrac{a}{V^2}\equiv\text{pressure}\implies a\equiv \text{pressure}\cdot(\text{volume})^2$$
Now, we guarantee that $(p-\frac{a}{V^2})$ has units of pressure:
$$(p-\dfrac{a}{V^2}\equiv (\text{pressure}-\dfrac{\text{pressure}\cdot\text{volume}^2}{\text{volume}^2} \equiv \left(\text{pressure}-\text{pressure}\cdot \dfrac{\text{volume}^2}{\text{volume}^2} \right)\equiv (\text{pressure}-\text{pressure}) \equiv \text{pressure}$$
The constant $b$ is easier. $b$ needs to have dimensions of volume so that $V-b$ still has dimensions of volume.


text:
What makes substances explosive
Can someone explain clearly to me why some things are explosive?

It might be better to explain in more detail the type of information you are looking for, and let those who want to answer choose how they would like to answer it.

Look for molecules with very exothermic decomposition.

related: https://chemistry.stackexchange.com/q/4267/102629

I could say explosives are things that blow up, but this would not a helpful answer.
It is also not that helpful to describe explosives as compounds with highly exothermic decomposition. This characteristic alone does not define an explosive.
What matters is rate of the reaction in the bulk compound. This rate defines two key types of explosive. If the reaction, once started, proceeds through the bulk compound faster than the speed of sound we call it detonation. If the reaction is slower than the speed of sound we describe it as deflagration.
This distinction matters because of the different uses explosives can be put to. Propellants, such as the explosives used to drive bullets, tend to deflagrate (as you want a relatively slow, controlled reaction to drive the shell out of the barrel without blowing up the barrel). But if you want something to really go bang and create a mess (as the military do) or if you want something to break up rock (as quarriers and miners do) then you want something that will detonate.
The structures of some well-known explosives are shown below:

Most well known real world explosive compounds are made from these compounds plus some added materials to make them easier to handle. C-4, a well known military explosive is mostly RDX with plasticisers to make it easy to mould like plasticine. The favourite terrorist plastic explosive is Semtex, similar to C-4 but made from a mixture of PETN and RDX. TNT is often used in high-explosive military shells (partially because it can be melted by steam and poured into shell casings). The first modern explosive, dynamite, is based on nitroglycerine mixed with diatomaceous earth (which makes it much easier to handle than the dangerously shock-sensitive pure compound).
The chemical reactions typically involve the production of CO2, H2O and, sometimes, N2.
Another characteristic that matters is how easy the reaction is to set off. Some compounds are very shock sensitive like mercury fulminate or nitroglycerin. This is only useful in detonators (in small quantities) or for chemists who have a death wish. Many bulk explosives require a detonator to trigger an explosion, making them much safer to handle and transport. C-4 in small quantities will actually burn slowly (and was sometimes used by soldiers in the Vietnam war to warm rations).

One thing that the answers so far have not mentioned is that an explosion is a chemical reaction.  So the question is: why does the explosion reaction occur?  And like so many chemical reactions, the answer is that the post-explosion state is lower energy than the pre-explosion state. That would seem to be obvious, given all the energy released in an explosion!  But what's more interesting is that an explosive compound can remain for extended periods in a metastable condition (literally, in an energy state higher than its lowest possible state).  This metastability is precisely why explosive compounds can be useful: they don't just spontaneously blow up.
But this begs the question: why are explosive compounds metastable?  This gets to questions of activation energy or introduction of reactants.  In the first case, it may be required that more energy is required to break the metastability.  This is what is needed for explosive mixtures of hydrogen and oxygen to detonate, for example.  In the second case, the introduction of a reactant may be just what is needed for explosive decomposition.  For example, the explosive $\ce{XeO_3}$ is triggered by mere contact with organic compounds.
So in summary: explosions are particularly vigorous examples of chemical reactions moving to a lower energy state, but the "why" and "how" can be complex.

Chemical compounds possess bond energy. Except at very high temperatures, there are many chemicals which are not the most stable (for that given set of elements). One (simple, crude) way to look at this, these chemicals are "frozen" into shapes which aren't the most stable.
Cellulose is the major ingredient of wood. It is a polymer with the approximate chemical formula $\ce{C6H10O5}$. While wood can last for decades and centuries, wood dust (saw dust) is the same material and is quite explosive when in a dust cloud mixed with air. So, is wood "explosive" or not?
There are many aspects to this: An explosion (a chemical explosion) is a chemical reaction (often a set of related chemical reactions) which occur over a time frame of fractions of a second. This is called the "rate" of the reaction. If I were making an unstable chemical, and as it flowed out of the reaction vessel it reacted, we'd consider that decomposition, rather than explosion. We'd consider the exact same reaction an explosion if it broke some of the equipment. So, we have to be careful to understand what we mean by "explosive". Generally (to simplify), a chemical or chemical mixture is "explosive" if it is mostly stable (decomposing slowly, taking days, weeks, months, years) and if we can add a small amount of energy (a spark, a flame, hitting it with a hammer, etc.) to start a self-propagating and FAST chemical reaction or decomposition.
"Fast": I've already explained why fast is required, but I've not explained what causes "fast". A fast chemical reaction has to be an "easy" rearrangement of the chemical bonds. That means it has to happen between atoms that are close together. I won't go into any more detail here; it would require you to understand quite a bit about stereochemistry and chemical kinetics. Suffice it to say that wood dust allows oxygen in the air and cellulose in the wood to be close together and makes it easy to self-propagate the reaction. "Self-propagation" is a bit of a misnomer. What the term means is that the energy released in the reaction of one molecule (or set of molecules) is more than enough to "unfreeze" another molecule so that it can and will react. We call this the chemical "activation energy", you can learn more about it here.
I look at (chemical) explosives as either:

Mixtures of chemicals (such as black powder) which will react fast enough and with enough energy to result in an incredibly accelerating reaction of more of the same. It feeds on itself.
A single chemical in which its bonds are "strained" and "ready to burst". A very small amount of energy can cause these compounds to detonate, but again the same two requirements of speed and quantity of energy must be met. The same activation energy needs to be overcome and the same propagation needs to happen. But in the case of a single compound, the "ingredients" are naturally "close together" - they are in the same molecule. This is one reason why nitroglycerin is "more dangerous" than black powder, it takes much less to "start".

Note: chemical reactions which don't have "activation energies" happen spontaneously, so you won't be able to make (except fleetingly) a chemical which is both unstable and has no activation energy for reaction/decomposition.
Another way to say this is that all explosives have activation energies (since without activation energy, the reaction would already have happened).

A material that has low reaction rate in bulk are also possible to be explosive when increase its surface area which increase its reaction rate such as dust explosion.


text:
Converting sulfate ion to sulfur trioxide?
I have a solution of only sulfate ions (no other positive ions belonging to the sulfate ions, because the positive ions are attached to the intermediate in a catalytic reaction).
I want to convert the sulfate ions to sulfur trioxide by using $\ce{V^5+}$ ions...
So what if $\ce{V^5+}$ ions could be present and could gain 2 electrons from one sulfate ion to produce $\ce{V^3+}$ and $\ce{SO3}$ (however I'm not even sure if $\ce{SO3}$ would be formed)? Here the two negatively charged oxygen atoms will each lose 1 electron to the $\ce{V^5+}$ ion, so the oxidation state of the central sulfur atom is still $+6$, but the oxidation states of the 2 oxygen atoms are $0$.
Could this be possible? And if so, would it be $\ce{SO3}$ that is formed, or other sulfur oxides?

Please notice that the oxidation of sulfur in the sulfate ion and and sulfur trioxide it the same; the latter is just the anhydride of the corresponding acid of the first.

And how can they then be converted to each other (sulfate->so3)?

A good analogy to your proposed reaction is the formation of carbon dioxide above a solution of carbonate salts.  Like sulfur trioxide, carbon dioxide is an acid anhydride.  The acid for sulfur trioxide is sulfuric acid, $\ce{H2SO4}$.  The acid for carbon dioxide is carbonic acid, $\ce{H2CO3}$.
Dry carbonate salts like sodium carbonate can be made to release $\ce{CO2}$ by heating to high temperatures.  For example, calcium carbonate decomposes to calcium oxide and $\ce{CO2}$ at around 800 °C to 900 °C.  Since such high temperatures are required, this technique does not work with aqueous solutions at ambient temperatures.   (The water would boil off way before gas is released.)
Thus, sulfur trioxide evaporation from sulfate salts would be even more disfavored at room temperature than carbon dioxide evaporation from carbonate salt solutions, because of sulfur trioxide's affinity for water is much higher than $\ce{CO2}$'s.
Also, the reaction mechanism you have drawn is not a dehydration, but a reduction of the sulfate.  That would not form sulfur trioxide, but a reduced sulfur compound, such as sulfite.  Sulfur dioxide gas evaporation from above sulfite salts is probably more favored than sulfur trioxide evaporation, but that isn't saying much. Either one is still highly disfavored under neutral or alkaline pH.


text:
Boiling point of Coca-Cola
Inspired in part by the question boiling point of vodka - given that the exact ingredients of Coca-Cola are kept secret, based on what is known, what would a viable prediction of the boiling point of Coca-Cola be?
Would it be significantly different from related products (e.g. Diet Coke, Coke Zero, etc)?

Your best guess for the boiling point of any version of Coke would be 100 C, the boiling point of water.
Diet Coke is mostly water (the flavourings are a very small amount relative to the amount of water). The largest ingredient will be the sweetener but there will be only a fraction of a gram of that. It is unlikely you will notice any deviation from the properties of water.
Standard Coke has quite a lot of sugar in it. A standard can (~300ml) contains about 40g of sugar. To put it another way, the contents are more than 10% sugar by weight and the solution is about 1/3 mol/L of sucrose (other sugars will be slightly different). A standard calculation using the ebullioscopic constant for water suggests the elevation of the boiling point will be barely 0.2 C, so small you'd struggle to measure it without good instruments and a good experimental setup.

+1 Great answer, so essentially, Coca Cola and just about all soft drinks could be considered as sweetened water,in this respect.

@Amaterasu Well, that is indeed what they are.

Atmosferic preasure will produce a greater difference.

Oh pressure produces diffrence so It depends upon pressure, when the bottle is totally closed it has more pressure etc..

Here's an excellent resource that says the difference between 10% sucrose (aka coke) and water is less than a degree Celsius:
http://kitchenscience.sci-toys.com/boiling_freezing_pressure
Why don't you just measure it? All you need is a bottle of coke, a thermometer and a kitchen. There will be some foam when the CO2 comes out, so heat slowly or be prepared for a sticky mess.

I have a bottle of coke and a few kitchen thermometers. Be right back!

Please do post your results if you get any!


text:
Comparing acidities of substituted and aromatic carboxylic acids
When comparing the acidities of carboxylic acids, we primarily see the electropositivity of the carboxylic acid carbons, i.e. we see how effectively the negative charge on the carboxylate ion is dispersed upon ionization. Electron withdrawing groups increase the acidity by stabilising the anion via electron withdrawal.
Now consider the following two comparisons:-

Formic acid ($\ce{HCOOH}$) and benzoic acid($\ce{C6H5COOH}$)
Benzoic acid and 3-chloropropanoic acid ($\ce{ClCH2CH2COOH}$)

$$\begin{array}{c|c}
\hline
\text{Species} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{HCOOH} & 3.75\mathrm{-}3.77 \\
\ce{PhCOOH} & 4.19 \\
\ce{ClCH2CH2COOH} & 3.98 \\
\hline
\end{array}$$
(source)
My questions:

Why is 3-chloropropanoic acid less acidic than formic acid. The negative inductive effect of chlorine helps in dispersing the negative charge on the carboxylate ion. It is not conjugated and too far to exert positive mesomeric effect. Therefore, the negative inductive effect should increase the acidity by improving the stability of the conjugate base.
Why is benzoic acid weaker than formic acid? Unless I am mistaken, in several other places, the phenylic moiety is treated as slightly electron withdrawing, as in case of alcohol acidity, acidity of hydrogens in base catalysed reactions etc. I am unable to understand the resonance here or any other factor, that might cause this anomaly.

@Martin why edit is rejected? Electron withdrawing groups increases the acidity!

@JM97 Because by saying the opposite, your edit completely deviates from the original reasoning. It may therefore invalidate any answers. You can edit the post to make it more readable, but you should not change the original intention. I believe that was also given as the reason for rejection. In any case, I am not responsible for this post, please direct your requests towards the OP and not towards me.

@Satwik Electron withdrawing groups increases the acidity , do not reduces acidity as mentioned in the question. Please correct it.

In the case of 3-chloropropanoic acid vs. formic acid, I suspect the disparity in expected acidity can mostly be explained by solvation effects. The 3-chloropropanoic acid is much bulkier than formic acid, and hence interacts with solvent molecules differently than formic acid does. There are two essential components to the thermodynamics of solvation:

Due to steric hindrance, it may be the case that the negative charge is less accessible to solvent molecules, and thus more weakly stabilized. I suspect the difference in enthalpy here is actually negligible, but I'm not able to find a reference at this time.
Perhaps more importantly, there is an entropic effect, in that larger molecules tend to require a larger number of solvent molecules to solvate them. These solvent molecules form highly ordered, cage-like structures, which is entropically unfavorable. I believe similar underlying mechanisms are at work in, e.g., the hydrophobic effect, micellization, protein folding, etc.

As Klaus Warzecha points out, if the chlorine were not as far-removed from the carboxylate moiety (since induction rapidly diminishes with growing distance), the strength of the inductive effect would trump the solvent effects.
In the case of benzoic acid, there are probably similar solvent effects occurring. Additionally, the benzene ring is actually electron-donating within the $\pi$-electron system of the molecule, which can only be destabilizing to the carboxylate conjugate base. Unlike in, e.g., phenol or aniline, the excess negative charge cannot actually be delocalized into the ring. Experimentally, the electron-donating effect of the ring (equivalently, the electron-withdrawing effect of the carboxylate) is confirmed by the fact that benzoic acid is severely deactivated in electrophilic aromatic substitution reactions by comparison to ordinary benzene. The benzene ring's electron-withdrawing effect within the $\sigma$-framework (rationalized, according to valence bond theory, by the greater s-character of $\mathrm{sp^2}$ carbons) of the molecule is probably quite weak. There may also be Coulombic destabilization between the strong $\delta^+$ of the carboxylate carbon atom and the $\mathrm{sp^2}$ carbons of the aromatic ring, again due to greater s-character, although that's speculative.

So any substituent showing positive mesomeric effect, bonded to carboxylate carbon, can destabilize the carboxylate anion? I had read that only inductive effect of the substituents has to be considered to determine the order of acidic strength of different carboxylic acids.

Actually, the chlorine atom in 3-chloropropanonic acid does have a significant -I effect; the $\mathrm{p}K_\mathrm{a}$ of propanoic acid is $4.86$. Your expectations were just a bit too high. If you want a stronger effect, you need to get a bit closer:
the $\mathrm{p}K_\mathrm{a}$ of 2-chloropropanoic acid is $2.83$.

But still, even if there is a little negative I effect, does it still not create better charge dispersal than no inductive effect (formic acid).

Well, we should also take into consideration that the alkyl substituent itself has a significant +I effect, just compare formic acid with acetic acid (pKa = 4.76).


text:
Why can we disregard stoichiometric coefficients when computing reduction potentials?
A $\ce{Cr(s) | Cr^3+ || Cd^2+ | Cd(s)}$ galvanic cell has redox half reactions:
$$
\begin{align}
\ce{Cd^2+ + 2e- &-> Cd(s)} &\quad E^\circ &= \pu{-0.4 V}\tag{R1}\\
\ce{Cr^3+ + 3e- &-> Cr(s)} &\quad E^\circ &= \pu{-0.7 V}\tag{R2}
\end{align}
$$
Balanced reaction equation then is
$$\ce{3 Cd^2+ + 2 Cr(s) -> 2 Cr^3+ + 3 Cd(s)}\tag{R3}$$
$$E_\mathrm{cell} = E_\mathrm{red} + E_\mathrm{ox} = \pu{-0.4 V} + \pu{0.7 V} = \pu{0.3 V}$$
Why is $E_\mathrm{cell}^\circ$ independent of molar coefficients $(3,2)?$

You have to be careful of the difference between $E^\circ$ and $E$. $E^\circ$ is the standard electrode potential at defined conditions. Standard thermodynamic conditions are usually

A temperature of 298 K
A pressure of gaseous components of 1 atm (or 1 bar)
A concentration of 1 M

The reference reduction potentials you quote are all standard potentials. These are the contributions to the cell potential when the half-cells meet the criteria above.
$E$ is the cell potential at all other conditions. The correction of $E^\circ$ to $E$ involves temperature, pressure, concentration, and stoichiometry. Behold the Nernst equation:
$$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$

$R$ is the ideal gas constant (this is thermo after all)
$n$ is the number of electrons transferred - here is the stoichiometry!
$F$ is the Faraday constant
$Q$ is the reaction quotient - which accounts for changes in pressure and concentration

Who is the god particle?

What Q should we take? We can scale up the reaction by any factor and Q would be different

@Buraian - You take whatever Q reflects your current situation. The Nernst equation allows you to calculate the cell potential given any starting concentrations in the half cells.

The standard electrode potentials are measured in their standard states when the concentration of the electrolyte solutions are fixed as $1M$ and temperature is $298K$. However, in actual practice electrochemical cells don't have always fixed concentrations of the electrolyte solutions. The electrode potentials depend on the concentration of the electrolyte solutions. Nernst gave a relationship between electrode potentials and the concentration of the electrolyte solutions known as Nernst equation. Even the electrode potential which you have posted (assuming they are not standard potentials) is obtained by Nernst equation, which shows the dependence of electrode potential on concentration (in other words molar coefficients).

Why is E$^\circ_\mathrm{cell}$ independent of molar coefficients (3,2)?

The cell potential is the potential an electron experiences; the coefficients of the equation have no bearing on this. When you use the cell potential to calculate the electrical work a cell does, however, you would multiply by the charge that is flowing through the wire, which would involve $z$, the number of electrons transferred.
So if we take the chemical equation as written by the OP,
$$\ce{3 Cd^2+ + 2 Cr(s) -> 2 Cr^3+ + 3 Cd(s)}\tag{one batch}$$
and multiply it by two, we would have
$$\ce{6 Cd^2+ + 4 Cr(s) -> 4 Cr^3+ + 6 Cd(s)}\tag{two batches}$$
This would not change the voltage observed between the electrodes. However, the molar Gibbs energy for the reaction would be different by a factor of two because the relationship between cell potential and Gibbs energy includes $z$, the number of transferred electrons per reaction:
$$\Delta_r G = -z F E_\mathrm{cell} $$
This is because the molar Gibbs energy of reaction is based on the extent of reaction (which itself is dependent on which coefficients you choose for the reaction, i.e. one batch or two batches in the example).
On the other hand, if we calculate the work available from oxidizing a given mass of elemental chromium in this reaction, we get the same answer no matter which equation we base it on (as it should be). Finally, having more chromium react does not change the voltage of the cell (as long as the solutes are maintained at constant concentration).

When we scale a half reaction in order to balance the number of electrons with the other half reaction, we do not change the intrinsic joules per coulomb (Volts) of the electrons.
So Fe2+ +2e- -------> Fe (s)  E = -0.44V
means that the 2 electrons carry -0.44 Joules per coulomb of energy
If we double the reaction, the reaction energy is doubled, but so is the number of electrons. So the J/C stays the same.  That is the standard reduction potential remains -0.44V
So 2Fe2+ +4e- -------> Fe (s)  E = -0.44V
But when we are trying to combine the 2 reduction reactions as in the case of:
Fe2+ +2e- -------> Fe (s)  E = -0.44V
Fe3+ +e- -------> Fe2+ (s)  E = +0.77V
to get the standard reduction potential for :
Fe3+ +3e- -------> Fe (s)  E = ??
this is a different concept. Here we are not scaling half reactions in order to combine them into a single full reaction.  We are trying to combine 2 half reactions to make a new half reaction (with electrons still present in the equation)
At first glance it's tempting to just add the potentials, since adding the half reactions does give the desired half reaction :

Fe2+ +2e- -------> Fe (s)  E = -0.44V
Fe3+ +e- -------> Fe2+ (s)  E = +0.77V

Fe2+ +2e- Fe3+ +e- -------> Fe (s) Fe2+ (s)  E = ?
(now cancel the Fe2+ on each side)

Fe3+ +3e- -------> Fe (s) E = ?

But we cant just add the potentials.  This is still a half reaction, so the potential is all about the energy per electron in Joules /Coulomb (V)
In reaction 1) the 2 electron carry -0.44 Joules per coulomb, and in reaction 2) the 1 electron carries 0.77 Joules per coulomb
So to get the J/C for reaction 3) we need the weighted average J/C of the 3 electrons
So that 2/3 electrons at -0.44 plus 1/3 electrons at +0.77
2/3 * (-0.44) + 1/3 * (0.77) =  =0.037 Joules per coulomb (volts)
Hope that helps.

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


text:
Could micrograms of polonium really change the viscosity of a whole teapot?
I was reading How radioactive poison became the assassin’s weapon of choice and it opens describing how

The men eventually left, and Andrade cleared the table. As he poured the remaining tea away, he noticed that the consistency of the liquid that tipped into the sink was strange. Gooey. He couldn’t have known it as he puzzled over its weird yellow tinge, but the man who’d been sipping the tea was a 43-year-old Russian dissident called Alexander Litvinenko, and the tea itself, draining away into the London sewers, was lethally radioactive.

Later in the article it mentions the total amount he ingested was about a microgram.
Now, skipping past the part on how you wouldn't notice you are drinking gooey tea, could micrograms of polonium change the viscosity of a whole tea kettle?
It doesn't say how much he drunk, or the size of the teapot, so lets take some guesses. Lets assume he drank a very small amount of tea, say 10 mL to give us a decent upper bound on the concentration of the solution. So that would be 0.1 micrograms/mL or about 0.5 micromolar.
Could a solution in this concentration range change the viscosity of water enough for a casual observer to notice?

Probably the alpha radiation, like light it hits the water molecules and knocks it around so its viscosity increases.

The weird thing is that the bartender who discarded the poisoned tea said it was practically semi-solid: "I scooped it out of the sink and threw it into the bin." I haven't ever heard of polonium being difficult to manipulate in aqueous solution due to increased viscosity. The only thing I can think of is that the decay somehow triggered the polimerization of organic matter present in the tea, and even that is a huge stretch.

@NicolauSakerNeto I'm wondering if it was unevenly distributed, so there was a lot of it in the bottom of the pot, but due to poor stirring, only a little in the top?

I think it is highly unlikely polonium makes water notably more viscous. The concentration is just too low, and radiation doesn't make water more viscous, controlrods are cooled using water and they produce a lot more radiation.
what can happen in the case of tea is that the radiation induces the formation of free radicals that in turn initiate polymerization or other reactions that form larger molecules which would make the solution more viscous, this still sounds highly improbable though.
there are multiple studies on radiation initiated polymerisation:
http://large.stanford.edu/courses/2012/ph241/weil1/


text:
Does chelation take ions out of solution
Does EDTA take $\ce{Mg^{2+}}$ and $\ce{Mn^{2+}}$ out of solution, so that it can be removed by a filter? Or does it just take it out of any chemical reactions going on in the solution?

No- still too small for normal filters. Those ions are in water complexes anyway. It doesn't stop (prevent) chemical reactions rather alter the thermodynamics. EDTA and other organic complexes are the common way to allow animals to absorb ions that would otherwise tend to form oxides when swallowed so its biologically (chemically) available.