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Seashell and grape juice

My daughter is working on a 5th grade science fair experiment.  She soaked white seashells in various drinkable liquids to test the effects of those liquids on teeth (teeth are represented by seashells). We were expecting to find loss of mass and discoloration of the seashells, but we were completely surprised by the reaction observed in the grape juice seashell.  The pictures attached are not great at showing what I mean, but besides the discoloration, the seashell shimmers!  There seem to be crystals all over it. It appears like it is covered in glitter.  Has anybody heard of such chemical reaction?  I researched this, but cannot seem to find anything to explain it to my daughter.  We don't know if the juice simply ate away at the outer protective layer of the seashell to expose the sparkly inside (not likely) or are we looking at a chemical reaction (more likely).

Off the top of my head I suspect calcium tartrate. Calcium from the shell and tartaric acid from the grape juice.

Hello and welcome to Chemistry.SE! My first take was that your "not likely" and "more likely" comments should be reversed, but I think MaxW might be onto a good idea too. Did you try other acidic juices?  The reason I ask is that tartaric acid, which would make the calcium tartrate, is found in high quantities in grapes. If you didn't see this effect with citric acid containing juices for example, then that gives some evidence that this may be due to the tartaric acid found in grape juice. Very interesting! Also, if you are interested, you can learn more about this site via the short [tour].

Thank you.  The only other juice she's testing is orange juice. So far, the experiment is three days in, but it is the first time we took the seashells out of their liquids.  Other than loss of mass, there is no change in the appearance of the OJ seashell, not even color change.  If the acid found in grapes could be the cause of the shimmery seashell, then maybe I should have her try a side experiment and test red grape juice, versus white grape juice, versus juice from fresh grapes.  That might prove to be even more interesting.

We also tried Pepsi and Diet Pepsi, both acidic, and the only observations there was the mass loss and color change.  No shimmer.

Absolutely. I assume that these are basically the same type and size of shells? About the most acidic juices you might have around are lemon and lime juice, which are around 5% citric acid and have a pH of around 2.2 (grape juices vary from pH of 3 to 4).

I just saw your cola posts, those are just as acidic as  lemon/lime juices. I still like your idea of trying different types of grape juice.  If multiple types give the shimmer, I think you've established that it's something in the grapes, probably tartaric acid as @MaxW suggested.

Yes, the shells are the same type and weight. They all started at about 3 grams each and we are using a scale with a sensitivity of .01 grams.

Thank you for all the help and interesting ideas. I think I'll be taking my daughter to the grocery store tomorrow and we'll take this experiment further.

Please let us know your results, and someone will likely write up a good summary answer for you from all this back-and-forth!

It might be interesting to try this with vinegar and with vinegar spiked with cream of tartar  (Potassium bitartrate).

I warn you that I've virtually no knowledge of the composition of grape juice - I didn't even remember that it contained tartaric acid, for example. My admimission of my ignorance is to guard you from letting my technical vocabulary  fool you into thinking that just because I might sound knowledgeable, that what I say is correct concerning grape juice and sea shells.
Calcium tartrate, when pure, is colorless (its crystals are "white"). When made from grape juice, I would expect that the crystals are colored. This would be due to impurities which co-precipitate & are incorporated into the crystal structure. I was able to find some pictures of crystals of calcium tartrate using Google which were colored (purplish red in one case, orange-red in another) and which might be consistent with what you saw. See also https://en.wikipedia.org/wiki/Tartrate.
On the other hand, grape juice contains other organic acids which might also form colored crystals. Two families in particular I am thinking about: tannins and anthrocyanins. Both are polyphenols, which means they have two or three -OH (hydroxyl) groups attached to one aromatic ring (the simplest example of an aromatic ring is benzene).
Sheen is generally not a property of a chemical compound's pure crystal. The prototypical example is the gemstone opal, which is composed of microscopic silica spheres packed into a regular array. The point here is that "sheen" is more likely due to microscopic particles included in the crystal rather than an inherent property of a pure crystal. There are exceptions to this, however.
The reason I mention the polyphenols is that they typically form very highly colored complexes (salts) with iron. Often they are colored brown, purple or black (and the that their crystals might have a sheen.) It appears to me that the shell has turned purple-black due to the polyphenols reacting with trace quantities of iron. A way to check this would be to take an iron nail (ungalvanized, unplated, uncoated) and let it rust in some grape juice and see what you get - if its just the typical orange-red rust, then there probably aren't many polyphenols present. But, note that in anaerobic (low oxygen,eg. in unmixed water) conditions, iron will rust to a black color, which will turn slowly red-brown when allowed to air dry. Wine is well known to contain tannins, which come from grape skins.
Anthrocyanins are known to change color with changes in acidity (pH). So, you could check to see if the crystals change color in water with added acid (vinegar, say) and base (ammonia, say). With supervision, Draino in water is also an option; it contains a very strong base, sodium hydroxide, but it definitely shouldn't be used by a child - it can cause severe burns to the skin and worse to the eyes. Most powdered detergents also are basic.
Grape juice contains large amounts of tartaric, malic, and citric acids. Neither of the latter two are likely to be the source of your crystals, imho. Two other experiments you could try are: 1. Add some rubbing alcohol (40% or even better 70%) to a bit of the juice. You might get some crystals (sediment).
2. Put a shell into some Cream of tartar in water. (The Cr. of T. will be more soluble with some vinegar or lemon juice added, making it more acidic). If the crystals you saw were indeed calcium tartrate, then this should give you the same thing. When I saw the picture of the black seashell, I immediately assumed tannins (and iron) were present, fwiw.


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Does the reaction of 1-bromobutane with cyanide ion follow an SN1 or SN2 mechanism?

In my textbook it's mentioned that the reaction undergoes $\mathrm{S_N2}$ mechanism. I understand that the carbon attached to the bromine is primary, but both solvents here are polar protic solvents, which favour the $\mathrm{S_N1}$ mechanism. Can anybody explain how to reconcile these?

Agreed, $ \ce{Br-} $ is a good leaving group, so one would expect that an SN1 mechanism is not too difficult to have in a polar protic solvent. But do not forget that the mechanism of any reaction is determined by not only the nature of the leaving group, but also the carbon skeleton. A primary carbocation is extremely difficult to obtain, and will not, in general, be formed unless the corresponding anionic complex is extremely stable. For example, the equilibrium $$ \ce{R-CH2-F + SbF5 <=> R-CH2+ + SbF6-    }$$ lies predominantly towards the products side because the anion $\ce{SbF6-}$ is an extremely stable system (which is why the mixture $\ce{HF + SbF5}$ is considered a super acid.). Had $\ce{AlBr3}$ been used before adding $\ce{KCN}$ to the solution, I would've gone for SN1 mechanism.
So, firstly, the primary carbocation is a difficulty if you consider an SN1 mechanism.
Moreover, the formation of a carbocation would immediately facilitate rearrangements. Since a primary carbocation is unstable, Wagner-Meerwein rearrangements will take place and your end product would have been a substitution at the second carbon instead of the first (terminal) one.
As for the solvent, it is a mixture of $\ce{Et-OH}$ and $\ce{H2O}$ to facilitate the dissolution of both reactants. Because one of them is an ionic compound, and the other an organic compound, getting them both in the same state can only be achieved by the mixture of an organic solvent and water; the organic solvent being such that it is, itself, miscible with water. A polar protic solvent weakens the ionic bonds by screening, and favours ionisation. This doesn't imply that a polar protic solvent will always completely ionise your solute.

Would it have been SN1 or SN2 if there was a secondary haloalkane?

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Nomenclature of Electronic states
We often see electronic states are named like nπ* or ππ* states. What is the reason behind this kind of naming ?

Hello and welcome to Chemistry.SE!  At this point I suggest that you take the short [tour] of the site.  Best of luck...

Can you give specific examples? I'm not sure what your question is.

Are you sure you are not looking at the labels of electronic transitions? For example $\mathrm{n-\pi^*}$ would be the the transition of an electron from a nonbonding orbital to a $\pi$ antibonding orbital.

As @Ben Norris writes the $\pi\pi^*$  refer to electronic transitions between states. The notation is just a simplification/shorthand to indicate the type of molecular orbitals the HOMO /LUMO electrons are in, rather than using proper symmetry labels.

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Selectivity of Zinin Reduction
What is it, that makes the Zinin Reduction a selective reduction, and not a complete reduction, in the case of dinitrobenzenes? Also, is the reaction a selective one even when it is done for aliphatic dinitro compounds, or is it, in that case,  a complete reduction to diamine? What about polynitro-aromatic compounds?
Do the relative positions of the two nitro groups in dinitrobenzene affect the reaction and its selectivity?

Related: http://chemistry.stackexchange.com/questions/34066/selective-nitro-reduction-of-poly-nitro-compounds/53698#53698

To whomever keeps flagging the comment, the question cited is related to this one, the commenter is not purporting that this question is a duplicate of that one or that the linked question answers this one completely.


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Why can UV light initiate a reaction between hydrogen and chlorine gas?
Can someone explain me how does UV light help combine chloride gas and hydrogen to produce hydrochloric acid?
$$\ce{Cl2(g) + H2(g) -> 2HCl(g)}$$

This is a standard (photochemically induced) chain reaction involving the H and Cl  radicals that occurs after the photon breaks the chlorine bond. Similar reactions to this are described in many phys. chem. textbooks where the standard (generic) scheme used describes the hydrogen/bromine reaction.

Before going into the mechanism of this reaction, I suggest you look up free radical mechanism, as this reaction takes place through that.
$\ce{Cl-Cl}$ bond in $\ce{Cl2}$ is weak enough to be broken by mere UV rays (present in sunlight), and hence they undergo homolytic cleavage (the resultant products are $\ce{Cl}$ atoms, not ions) to form two $\ce{Cl}$ free radicals. Now these $\ce{Cl}$ free radicals are extremely reactive (due to one free electron that it can share in a covalent bond) and hence attack hydrogen gas.
Now is the time you consider energy lowering. $\ce{H-Cl}$ bond is lot more stronger than $\ce{H2}$, and hence the formation of $\ce{H-Cl}$ bond formation will be favored instead of $\ce{H-H}$. Therefore, $\ce{Cl}$ free radical attacks $\ce{H2}$ gas to break the bond and form $\ce{HCl}$. The other $\ce{H}$ atom hence liberated forms a bond with the other $\ce{Cl}$ free radical.

The HCl bond is actually slightly weaker ( $35728 \pu{cm^{-1}}$ ) than the $\ce{H2}$ bond ( $36099 \pu{cm^{-1}}$ ) . The H atoms are formed with chain  $\ce{Cl + H2 <=> HCl + H}$. The H reacts rapidly in chain steps $\ce{H + Cl2 <=> HCl + Cl }$ and $\ce{H + HCl <=> H2 + Cl}$ so its concentration is always low. The termination step is $\ce{Cl + Cl +M <=> Cl2 +M}$,  M being any un-reactive gas. [H] is too low for its equivalent termination step to be important. The photochemical chain reaction can be explosive and makes a good demonstration.


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What kind of glass glue can replace silicone?
I tried to put the 'Ferrofluid' and the clear liquid in a large glass tank.
My glass tank was bonded with silicone. However, silicon reacted with 'Ferrofluid' ('Ferrofluid' became sticky and hardened).
Perhaps 'Ferrofluid' seems to have chemically reacted with silicon because 'heptane' is the base.
What kind of glass glue can replace silicone?
Perhaps the epoxy is good, but it is not elastic.

how big a tank do you need? 10x17 inches ok?

picture is just a sample.
Actually I want to make 65 x 65 x 3 inch.

Wow. I want to see this when it's done ;) So, your requirements are basically a material that will maintain the structural integrity of this glass tank filled with heptane, but that will not interact with your ferrrofluid? What is the ferrofluid material?

https://youtu.be/RtDdCWEuDwo
This video is a prototype.
I will only expand its size.
Ferrofluid is composed of nanoparticle size iron powder, surfactant, heptane and so on.
The transparent liquid consists of water, alcohol and so on.


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Relationship between a temperature increase of an exothermic reaction, Keq, and kf (forward rate constant)
I'm currently stuck on problem #30 for the Chemistry Olympiad 2015 local exam which reads:

For a reversible exothermic reaction, what is the effect of increasing temperature on the equilibrium constant ($K_\mathrm{eq}$) and on the forward rate constant ($k_\mathrm{f}$)?
(A) $K_\mathrm{eq}$ and $k_\mathrm{f}$ both increase
(B) $K_\mathrm{eq}$ and $k_\mathrm{f}$ both decrease
(C) $K_\mathrm{eq}$ increases and $k_\mathrm{f}$ decreases
(D) $K_\mathrm{eq}$ decreases and $k_\mathrm{f}$ increases

The answer for this problem is D.
My approach:
To find the relationship between $K_\mathrm{eq}$and $k_\mathrm{f}$, I first used this theoretical reaction of
$$\ce{A -> B + \mathrm{heat}}$$
The forward rate can be defined as
$\mathrm{rate(f)} = k_\mathrm{f}[\ce{A}]$
and the backwards rate as
$\mathrm{rate(b)} = k_\mathrm{b}[\ce{B}]$
Since we're finding the relationship between these rates with $K_\mathrm{eq}$, rate(f) = rate(b) because the reaction is at equilibrium. This allows the equation
$k_\mathrm{f}\ce{[A]} = k_\mathrm{b}\ce{[B]} $
to be set up. Dividing both sides by $k_\mathrm{b}$ and [A] gives us
$k_\mathrm{f}/k_\mathrm{b} = \ce{[B]/[A]}$
And
$K_\mathrm{eq} = \mathrm{ products/reactants }= \ce{[B]/[A]} = k_\mathrm{f}/k_\mathrm{b}$
Due to the fact that the temperature is raised with an exothermic reaction, the reaction is reactant favored and more A will be formed. Since
$K_\mathrm{eq} = \mathrm{ products/reactants }= \ce{[B]/[A]}$
$K_\mathrm{eq}$will decrease. However, why doesn't $k_\mathrm{f}$ decrease? Since
$K_\mathrm{eq}= k_\mathrm{f}/k_\mathrm{b}$,
how could $k_\mathrm{f}$ increase while $K_\mathrm{eq}$ decrease if they're equal to each other in that equation? Shouldn't $k_\mathrm{b}$increase or $k_\mathrm{f}$ decrease to "=" to the decreased value of $K_\mathrm{eq}$?

$\Delta\text{K'}f  < \Delta\text{K'}r$ so $\text{K'}{eq} < \text{K}{eq}$

You know that both the forwards $k_f$ and reverse $k_r$ rate constants will increase with temperature, simply because there is more energy present and so it takes less energy to surmount the activation barrier, both from the product and reactant sides.
The question is then which is changed more by changing the temperature. The equilibrium constant is as you write $K_Q=k_f/k_r$ and now using the Arrhenius equation we get
$$K_Q=\frac{A_f\exp(-\Delta E/(RT))}{A_r\exp(-(\Delta E+E_0)/(RT))}$$
where the A's are pre-exponential factors (and are constant vs temperature) and $\Delta E$ the activation barrier for the forwards reaction and $E_0$ the extra energy for the back reaction and  is positive since the reaction is exothermic. (If you sketch the barriers hopefully this should be clear).
Simplifying gives
$$K_Q=\frac{A_f}{A_r}\exp(E_0/(RT))$$
As T  is increased $K_Q$ becomes smaller. Thus the answer is (D).

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How does haloform reaction take place if the reactants NaOH and X2 react with each other?
If $\ce{NaOH}$ reacts with halogen to produce $\ce{NaX}$ and $\ce{NaXO3}$, then how does haloform reaction take place?

Take a look at the mechanism here.

The disproportionation of halogens $(1)$ is an equilibrium reaction; depending on the concentrations of reactants and products it can go both ways. Since hydroxide (or more generally: a base) is among the reactants, increasing pH will drive the reaction towards the product side.
$$\ce{3 Br2 + 6 OH- <=>> 5 Br- + BrO3- + 3 H2O}\tag{1}$$
The first step of the haloform reaction is to enolise a ketone to the corresponding enole $(2)$. This is also an equilibrium lying on the reactants’ side this time.
$$\ce{RC(=O)-CH3 + OH- <<=> RC(-O^-)=CH2 + H2O}\tag{2}$$
So you are correct, both equilibria should be away from what is required for the haloform reaction to proceed. However, both are equilibria. For both, we can measure an equilibrium constant and use it to calculate the concentrations of both the enolate and the free dihalogen. I don’t have all the numbers here, but the result will be that both reactants are present in sufficient concentrations to drive the reaction forward. Remember that the full haloform reaction is no longer an equilibrium process; each bromination and the final separation is a non-reversible step; see $(3)$.
$$\begin{gathered}\ce{RC(-O^-)=CX2 + Br2 -> RC(=O)-CX2Br + Br-}\\[0.8em]
\ce{RC(=O)-CBr3 + OH- <<=> RC(-O^-)(-OH)-CBr3} \\[0.4em] \ce{-> RC(=O)-OH + CBr3- -> RCOO- + HCBr3}\end{gathered}\tag{3}$$
The mere presence of non-reversible steps will drive any reaction forwards.


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What happens when a body of water and a pice of ice are placed in a pricessly controlled zero degree environment
My understanding is, if you put ice into a >0 degrees environment, it will be 0 degrees as it melts and if you put water into a <0 degree environment, it will 0 degrees as it freezes. but what happens if you put water and ice in to an environment maintained exactly at 0 degrees. I'm inclined to think it should be said water freezes below 0 and ice melts above 0, I'd go so far, pending a satisfactory answer of course, is to suggest, what happens to water and ice at 0 degrees is nothing at all. Please put me out of my misery on this :-D

My understanding is, if you put ice into a >0 degree environment it
  will be 0 degrees as it melts and if you put water into a <0 degree
  environment it will 0 degrees as it freezes.

Correct. If you place ice in a container where the temperature is above  $0\ \mathrm{^\circ C}$, the ice will first heat up to $0\ \mathrm{^\circ C}$, and then undergo phase transition to a liquid. The same works the other way around, as you stated. Placing water in a sub-zero temperature will cause the water to cool down to $0\ \mathrm{^\circ C}$ and then it will freeze (form a solid). See picture below:

But want happens if you put water and ice in an environment maintained exactly at 0 degrees.

This really depends on the other qualities of water. If you add salt in to the water, it will remain as a liquid at $0\ \mathrm{^\circ C}$. This is due to freezing-point depression, which is described reasonably well in Wikipedia.
If you are considering pure water (no salt or other solvents mixed), then you will get an equilibrium of water and ice. It might be easier to understand, if you keep in mind that although the overall temperature is $0\ \mathrm{^\circ C}$, there will be some variation between single molecules, some having a lower amount of thermal energy and some having a higher amount.

If you add salt to an ice/water mixture the temp of the water will drop below 0 C. The ice too must be lowered in temperature. This melts some of the ice. Then mixture now extracts heat from the environment to warm up to 0 C thus melting all the ice.

Thanks but wouldn't the equilibrium of water and ice be, the ice parts being below zero and the water parts above zero? or would this equilibrium have a different state?


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Lennard-jones cut-off
I am using a Lennard-Jones potential in a computer simulation and I was wondering if i should place a cut-off for low distances since the potential blows up quickly   making the calculations quite difficult.
I am trying to simulate basic molecules ( few atoms ) , calculating bonds stretching , bond angle bending and lennard-jones interactions and trying to minimize the total energy to get an optimal structure prediction .

Please describe the system you are trying to simulate. What exactly does 'blow up' refer to? If it is uncontrollably high energies, then you probably will not benefit at all from having a low value for the cut-off, since the energy from LJ-potential grows only when the intermolecular distances are small.

Still need more info:
Which forcefield are you using? If you are not using any general FF, then please point to the source where you are getting the parameter values.
Which simulator are you using, or are you doing the calculations yourself?
Which molecules are you trying to simulate? Are you simulating them in a solvent, or in vacuum? Do all of the particles have LJ-sites?
Also, please elaborate on the 'blowing up'. If you are using any standard MD software (GROMACS, DL_POLY etc.) you should be able to see which energy term(s) grow unphysically high.

I am building my own simple force field in python, using parameters from : https://github.com/labello/amberff/blob/master/gaffparams.py . I am doing all the calculations myself .I hope to simulate simple molecules in vacuum like hydrocarbons . I am programming everything from scratch including the optimization method , the Lennard-Jones term keeps getting very large

LJ potentials should get really large only if distances are small.
Maybe that's what you got?

It's hard/impossible to pinpoint the reason for your problems without seeing the  code and knowing which terms you use. If you feel like sharing it, then you should probably head over to Computational Science and make a post there.
A few general things come to mind:

Check that you have $\sigma$ and $\epsilon$ values inserted in the correct order.
Check the correctness of the combination rules you are using. It seems the force field you linked previously contains noncovalent terms from OPLS/AA. Make sure that the rest of the FF is compatible with OPLS/AA.
Check the units you are using, and make sure they are all the same. Probably won't make your simulation blow up, but it's still worth to take a look, especially if you are using parameters from different FFs.


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What is the difference between cofactors and coenzymes?
I've just been wondering what are the differences between the two and I can't seem to find a formal delineation.

Enzyme cofactors are a collective name of all the chemical compunds or elements associated with the enzyme to increase its efficiency. There are mainly two types- inorganic ions and organic compounds. Inorganic ions are also known as enzyme activators (Cl ion in salivary amylase). Organic compunds are again divided into two- co-enzymes and prosthetic groups. Co-enzymes are loosely associated with the enzyme. They can be removed from the enzyme and undergo cyclic reactions to again combine with enzyme (EX: NAD, NADP). But prosthetic groups are tightly associated with enzyme molecules(EX : FAD). Hope this helps.

A  cofactor is a non-protein chemical compound. It is bound to the protein
Coenzymes are cofactors that are bound to an enzyme loosely.
Coenzymes are a special case of cofactors. (With Prosthetic groups)

These terms, according to this paper, have not been defined consistently in textbooks. The same paper suggests the following definitions:

Prosthetic group, coenzyme and cofactor are all generally defined. Whilst a cofactor is any factor essentially required for enzyme activity or protein function, a coenzyme is the cofactor which is directly involved in enzyme catalysed reaction. A cofactor which is not directly involved in enzyme catalysis, or is associated with the function of a non-catalytic protein is not a coenzyme. A prosthetic group (covalently associated non-protein constituent required for a particular function), on the other hand, may also be termed a coenzyme if it is directly involved in catalytic reaction. A prosthetic group which is not involved in enzyme catalysed reaction but functionally essential to the enzyme or a non-catalytic protein is also called a cofactor. Coenzyme and cofactor weakly bound to enzyme or protein are however, not classified as prosthetic group.

So the prosthetic group is defined via its binding properties, the coenzyme is defined via its reaction, and a cofactor is defined via its chemical makeup. This results in lots of overlap and gray areas. A three-dimensional Venn diagram might be necessary to decide in a specific case which term or terms are applicable.
It does not help that hemoglobin, often the first example of a protein with a prosthetic group presented to students, binds to a metal via an organic molecule and is sometimes called an enzyme (but it is just an honorary enzyme).

Cofactor is a non protein chemical compound that bound to the protein while a coenzyme is a Cofactor that bound to the enzyme

Your answer adds nothing new to previous answers.


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Calculate hardness of water
A sample of water has its hardness due to only $\ce{CaSO4}$. When this water is passed through on anion exchange resin, $\ce{SO4^2-}$, the $\ce{SO}$ ions are replaced by $\ce{OH-}$. A 25.0 mL sample of water so treated requires 21.58 mL of 0.001 M sulfuric acid for its titration. What is the hardness of water expressed in terms of $\ce{CaCO3}$ in ppm? Assume the density of water is 1.0 g/mL.
I do not understand where does $\ce{CaCO3}$ come from if we started with $\ce{CaSO4}$at the beginning.

The problem isn't going through gyrations just to drive you crazy. It is really the way that water analysis works. The "overall" hardness is expressed in terms of the "equivalent ppm of $\ce{CaCO3}$".
The two biggest contributors to water hardness are typically $\ce{CaCO3}$ and $\ce{MgCO3}$. Now let's say that you want to buy a water softener to replace cations $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ with $\ce{Na^+}$ and the $\ce{CO3^{2-}}$ with $\ce{Cl^-}$. Figuring how many gallons of water you'd use in a week and knowing your water's hardness as "equivalent ppm of $\ce{CaCO3}$" you could figure out how much ion exchange resin you'd need. So there is a functional reason that "equivalent ppm of $\ce{CaCO3}$" is helpful.

But I am interested is calcium carbonate present in sample above or I just have to calculate possible amount of calcium carbonate that can be formed from amount of calcium from calcium phosphate.

You're calculating as if the reaction were $\ce{CaCO3 + H2SO4 -> CaSO4 + H2O + CO2}$

I calculated molarity of sulphuric acid and looking at reaction Ca(OH)2 + H2SO4 -----> CaSO4</ sub> + H< sub>2 O AND then found moles of calcium phosphate as it is equal to moles of sulphuric acid. Moles of calcium carbonate is equal to moles of calcium phosphate soi found its mass and found hardness in ppm . I wonder If It is ok.

@chemistry - Yes, moles of calcium carbonate is equal to moles of calcium phosphate. But did you find the mass of calcium carbonate or that of calcium phosphate to convert to ppm?

I found mass of calcium carbonate and divaded it by mass of water.

Reporting hardness in terms of "mg/L as $\ce{CaCO_3}$" is common in water treatment practice. The whole point is to get all of your compounds' concentrations in relatable terms so that you can subtract/add them easily (similar to normality). This is called equivalence. In the case of hardness (or titrations), you are concerned with the equivalent amount of protons a compound can give or release. In the case of calcium sulfate, you have two equivalents because the calcium and sulfate both have a charge of 2 (they give/take two protons). The first thing you need to do is convert calcium sulfate into an equivalent weight (EW), which is the molecular weight divided by the number of equivalents, so $\frac{136.14 g/mol}{2 eq/mole}$ = 68.07 g/eq. Note: the English of your question was a bit difficult to understand; read the problem carefully as you may need the EW of calcium hydroxide, not sulfate. The problem approach is the same though, only the number would change.
Now, to get this in terms of mg/L as $\ce{CaCO_3}$ you use the following equation: (mg/L as $\ce{CaCO_3}$)=(mg/L of the species)($\frac{EW_{CaCO_3}}{EW_{species}}$) Similar to calcium sulfate, calcium carbonate has an equivalence of 2 (+2/-2 charge), so the EW is (100/2=50). The problem you are working on is very simple because you only need to convert one compound - often we need to convert many compounds to find, e.g., the alkalinity of a water.
Hint: To get mg/L species you use the titration information - recall that the equivalent definition we are using refers to the amount of protons given/received, so the titration info tells you just that!


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How can I know if there is a phospholipid (phosphatidylcholine) in a solution?
I have a solution which probably contains a phospholipid (phosphatydylcholine), I would like to know if there is an easy method to show the presence of the phospholipid.
I came across TLC, but I couldn't perform it since it's too difficult and expensive to buy chloroform.
So, If you guys know an easy method to show the presence of phospholipids please reply.

Your question kind of asks two different things:
1) Whether there is an easy method to show the presence of hosphatydylcholine
and
2) Whether there is an easy method to show the presence of phospholipids in general.
My first comment is that if TLC with chloroform is too expensive, then you are probably out of luck. This is not a simple thing to do, particulary the specific analysis of a single particular phospholipid in the presence of others. This Wikipedia article gives some insight on this:

There are no simple methods available for analysis of phospholipids
  since the close range of polarity between different phospholipid
  species makes detection difficult. Oil chemists often use
  spectroscopy to determine total Phosphorus abundance and then
  calculate approximate mass of phospholipids based on molecular weight
  of expected fatty acid species. Modern lipid profiling employs more
  absolute methods of analysis, with nuclear magnetic resonance
  spectroscopy (NMR), particularly $\ce{{^{31}P}}$-NMR, while HPLC-ELSD
  provides relative values.

Unfortunately, these methods are going to be orders of magnitude more expensive to do yourself, as compared to TLC, though outsourcing may or may not be a viable alternative in your case.
The same article discusses a phospholipid characterization technique based on refractive index measurements that, again, may or may not be a viable type of analysis for your specific use case.


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Wilkinson's catalyst: why this structure?
I wonder how the structure of Wilkinson's catalyst contributes to its catalytic properties. For example, what role do the phosphine ligands play? What properties make Rh the best TM centre?
Thank you!

Wilkinson's catalyst is a very famous molecule...it has its own Wikipedia page for example. From your readings of readily available resources such as this, which part are you having trouble with, in relation to the role of the phosphine ligands? What makes you say Rh is the best transition metal centre? Best for what?

@long Thanks for your comment! I was just wondering whether the phosphine ligands act to stabilise Rh in a low oxidation state, provide the bulk to prevent the binding of larger ligands that may be contaminants, force H and the alkene to be cis to one another to allow the alkene insertion to occur or something else? I am just interested in the whole process of discovering new catalysts: surely, chemists don't just screen thousands of complexes, and every single component / ligand they add has to play a role. Wilkinson's catalysts is just an example, but I want to understand general principles.


text:
Why is CF3COOH exceptionally acidic?
The pKa values of almost all carboxylic acids lie much above 0. But this is violated by trifluoroacetic acid(-0.25). How can this be justified?

There's nothing strange in this, normal consequence of inductive effect.

Related: Naming rules and acidity of Cl2CHCH2COOH

Try drawing the structure of the conjugate base $\ce{CF_3COO^-}$. In general, the more stabilized the negative charge of the conjugate base is, the more the equilibrium favors that form, thus the more the acid dissociates, thus the "stronger" the acid is.
So for a stronger acid (lower $\ce{pK_a}$), the negative charge must be more stabilized. When comparing $\ce{CF_3COOH}$ (TFA) to $\ce{CH_3COOH}$ (we'll use acetic acid to represent a "typical" carboxylic acid), TFA is stronger than acetic acid because its conjugate base can better stabilize the negative charge. If we draw the structures of the conjugate bases (I encourage you to do so), we will see that they both can stabilize the negative charge via resonance in the carboxylate functional group. However, TFA also has three highly electronegative fluorine atoms which withdraw electron density "through the single bonds" via induction.
As stated above, the $\ce{-CF_3}$ moiety is an electron withdrawing group, by the inductive effect. So our negative charge (rather, the electron density it represents) will be slightly drawn through the bonds toward the fluorines, thus it is further delocalized and thus more stable.
Note that we do not have convenient structural diagrams to show induction like we do resonance, but you should be able to imagine the electrons "sloshing" more towards one side of a bond/functional group. It's the same concept as a good old polar bond like $\ce{H-Cl}$; we say Cl "hogs" the electron density and pulls it away from H. In the same way $\ce{-CF_3}$ will draw electron density toward itself.
So to summarise, in general:
Stronger acid ~ more stable conjugate base. Whenever you want to qualitatively rationalize the relative strengths of two acids, draw their conjugate bases and determine which is more stable. (If your acid was neutral, its conjugate will be negative, and it will be easier to use this method than for a positive acid/neutral conjugate.) Stabilizing factors are size and electronegativity of the atom bearing the charge, resonance delocalization of the charge, hybridization of the orbital holding the charge, induction delocalization of the charge, and hyperconjugation which donates electron density and may destabilized a negative charge. SERHIH, roughly in that order of significance.
In this example, TFA and acetic acid are tied at size & electronegativity because both place the charge on an oxygen atom, tied at resonance because both may use the carboxylate functionality, tied at hybridization because the atoms in a carboxylate are necessarily $\mathrm{sp^2}$, but now induction is present in TFA and not acetic acid, so we expect TFA will be a stronger acid.

If $\ce{CF3COOH}$ loses a hydrogen atom, the $\ce{CF3COO-}$ ion is highly stabilized by the electron withdrawing power (-I) of the fluorine atoms.
The acidic strength of a acid is directly proportional to the stability of its conjugate base.

Your second sentence sounds a lot like a mathematical statement, is there an equation I'm not aware of?

No..It's just a statement..More stable is the conjugate base more strong the acid will be...That's what directly proportional means..

Directly proportional is a mathematical statement with a specific meaning. Just because Y increases when X increases does not mean they are proportional, the relationship may not even be linear at all. For example, when temperature increases the equilibrium constant Ka increases, but the the relationship is quite nonlinear: K = Ce^(-a/T). Your statement literally implies there is some equation pKa = (constant)*∆H[base]. This would be very nice, and I hope it's true, but personally I'm not aware of such a model.

@electronpusher for that matter, recall the definition of $K_a$ for the reaction $$\ce{HA<=>H+ + A-}$$ would be $$K_a=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$ You can say that $K_a$ is directly proportional to concentration of conjugate base, which is $\ce{[A-]}$


text:
How does a 5th hydrogen bind to carbon atom in CH5[+]?
Can we bind a fifth hydrogen to a carbon atom if carbon is not in excited state?
We have a big discussion in class and cannot clearly answer the question. If carbon atom excites an electron from $2\text{s}$ orbital to $2\mathrm{p^3}$ orbital, it can covalently bind four hydrogens each with one electron. In this case the $\ce{CH5^+}$ couldn't be created.
But what if carbon atom stays in its ground state so that there will be $2$ electrons in its s-orbital and $2$ electrons in its p-orbital? We discuss that then four hydrogens (each with one electron) could bind to carbon covalently, filling its p-orbitals, and the fifth hydrogen nucleus (without any electron) would bind to carbon s-orbital via coordinate bond.
Is it possible that two hydrogen atoms (each with one electron) would fill one vacant (plain) orbital of other element, e.g. carbon, and thus bind to it?

Then I guess you wasted quite a bit of time because you didn't know about three-center two-electron bonds.

related http://chemistry.stackexchange.com/questions/4027/carbon-with-5-bonds

also http://chemistry.stackexchange.com/questions/32319/hypervalency-in-elements-in-the-second-period

Actually the first diagram is fairly close to the truth. Two of the hydrogens sharing one of carbon's p orbital - that's precisely what a 3c2e bond is.


text:
Is liquid hydrogen rocket propeller pollution free?
If one searches for the different fuel types used in rockets, one can find that over the years, NASA and other space agencies have used both solid fuels and liquid fuels. I can see that more and more rockets are propelled by using a combination of Liquid Hydrogen (LH) - Liquid Oxygen (LOX). Then, I assumed that the reaction would be:
$$\ce{2H2(l) + O2(l) ->[heat] 2H_2O(v)}\tag{1}$$
By looking at this reaction, one can see that the product of this reaction is water (which most probably will be in a vapor state). Now, I have three questions:

Is this reaction correct?
If the product is water, does that mean that such a combustion reaction is pollution free?
I assume that I am wrong thinking that is pollution free. What would be the polluting agent in this situation and how bad is it, compared to let's say the burning of the same amount of Kerosene?

Thank you!

I'll address you questions in order:
1) Yes, that is basically the correct reaction, although not quite balanced;  here is how I would write it:
$$\ce{2H2(l) + O2(l) -> 2H_2O(g)}$$
2) Ideally, if there were no side reactions, then yes, this particular reaction is "polution free", and the only product is water.
3) You are correct in your thinking that in the end this does not represent a net pollution-free energy source.
The reason for this doesn't lie in the reaction taking place at the rocket, but in how they produced the hydrogen in the first place. Unfortunately, there are no significant deposits of $\ce{H2}$ on our planet waiting to be pumped out and used as fuel. So we  have to make it, which requires engergy.
This means that the $\ce{H2}$ is not a net energy source, rather it is essentially acting as a battery, storing the energy from some other source. So, you can think of the actual pollutants of the $\ce{H2}$ rocket fuel as being the products from whatever process went into generating it in the first place.  This largely depends on what part of the world the hydrogen generation took place. They could have burned coal or other fossil fuel to produce the energy, or they could have built a dam to generate hydroelectric power, etc.  Any of the negative aspects of producing a particular type of energy would be considered a negative product of the $\ce{H2}$ production.

Same for $\ce{O2}$ as well. Plenty of it in the atmosphere, but energy has to be used to separate and liquefy it.

Absolutely; far from an energetically-downhill task.

Thank you very much for your answers. So, if you would get H2 from dams or wind mills, that would reduce the pollution, which is actually nice. But what about the answer from Oscar Lanzi. He states that the hot exhaust would form NOx. Do you also agree with this statement?

I do. My answer is really in the perfect extreme of getting exactly the desired reaction. Oscar's basically correct statement is an example of my #2 about side reactions. The same thing happens in an automobile. Ideally, the chemical equation is hydrocarbon plus O2 gives CO2 and water. But, under the high temperatures of combustion, particularly the case of an auto where you have more N2 than O2, you get nitrogen oxides, partially burned hydrocarbons, etc. That said, you are not using atmospheric O2 in the case of rocket fuel, so that wouldn't be nearly the relative problem it is in an auto.

Hydrogen is not quite pollution-free when it burns in air; some of the nitrogen will form NOx in the high-temperature environment.  In a rocket with hydrogen and purified oxygen we get pure water in the combustion reaction but the hot exhaust can still form some NOx in air.  Still, this is a lot less polluting (and more efficient for the rocket) than carbon-bearing fuels.

Thank you Oscar. This is great and interesting information. So, would the NOx form because of the Nitrogen in the air, or? And is there any chance that this NOx would contribute to the formation of Ozone up in the stratosphere? As far as I know, NOx can react with VOC's, forming ozone, which would actually be beneficial for the stratosphere.

Actually NOx is detrimental to ozone.  Like other free radicals such as atomic chlorine, it sets off a free radical chain reaction that causes the ozone to break down to diatomic oxygen.  See https://www.learner.org/courses/envsci/unit/text.php?unit=11&secNum=10.

I definitely agree with what you said in your answer. My only further comment (not a disagreement) on the NOx-ozone cycle is that in the troposphere, during the daytime and in the presence of VOC's, NOx will promote photochemical ozone production down low where we don't want it. As you said, this case is not as bad as the case of carbon-bearing fuels, which also produce their own VOC's, among other things.


text:
How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?
How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure? Is there some kind of mathematical proof for it? Is there some way I can understand this intuitively?
I have been trying to think about it using this picture but I cannot seem to get anywhere.

I'd start with counting how many neighboring T and O each sphere has.

So, each sphere has three T and three O voids. Then?

Should be much more than that. Don't you miss the voids formed by the next layer (the one on top of this)?

If we place the next layer on the T's, then all the T's will get filled, and we will only be left with O's. So, any sphere in the second layer will have 3O's.

No, the T's won't get filled. If anything, they will get completed.

Yeah, that's what I meant, they will get "completed"

OK, then how many of them will be there?

To be precise, if in a close packed structure (ccp or fcc) there are $n$ atoms or ions then the number of octahedral voids and tetrahedral voids will be $n$ and $2n$ respectively.
For example, there are 8 tetrahedral voids per unit cell of fcc structure $(Z_{\text{eff}}=4)$. If you divide the FCC unit cell into 8 small cubes then each small cube has 1 tetrahedral void located at its own body center. Thus, total number of TVs in a unit cell $= 8 = 2  \times  Z_\text{eff}$

Now, again consider a CCP or FCC unit cell. The body centre of the cube is not occupied but it is surrounded by 6 atoms (4 in same plane, 1 above and 1 below). On joining these face centers, an octahedral void is formed. Thus, number of OV at body centre of the cube is 1.

Besides the body center, there is one OV at the center of each of the 12 edges. It is surrounded by 6 atoms, three belonging to the same cell unit (two on the corners and once on face center), and the three belonging to two adjacent unit cells.

Since, each edge of the cube is shared between four adjacent unit cells only one fourth of each void belongs to a particular unit.
Therefore, OV at body centre = 1
OV at the 12 edges of the cube and shared between four unit cells  $= 12\times\frac{1}{4}=3$
Total number of OVs $= 1 + 3 = 4 = Z_\text{eff}$

How many other  octahedral voids surround a n octahedral void ...I tried to do this but I am getting confused

Try to count the voids around the central octahedral void. Its much easier to visualize.

Well the octahedral void is surrounded by 12 other octahedral voids according to the diagram in it's own unit cell

One more thing.Considering the spheres at one  face centre which forms a square with the other face centre in it's plane,will only adjacent  face centres touch it and the  face centre at "a" distance from it won't touch  the face centre opposite to it?

Can you please answer my question as I am confused here.......will help me explain the number of neatest neighbours in FCC.......

Let $E$ and $F$ be the planes that contain the centers of the spheres of the first and second layer, respectively. We notice that neither the centers of tetrahedral or octahedral voids lie in these planes.
Thus, when counting the tetrahedral and octahedral voids in a cubic or hexagonal close packaging of equal spheres, it is sufficient to count the voids that are centered in the space between $E$ and $F$.
If we regard a fixed sphere, we find four tetrahedral and three octahedral voids next to it and between $E$ and $F$.
Since each tetrahedral void is adjacent to four spheres, while each octahedral void is adjacent to six spheres, we get $4 \cdot \frac{1}{4} = 1$ tetrahedral and $3 \cdot \frac{1}{6} = \frac{1}{2}$ octahedreal void per sphere.


text:
Detection of Tungsten Hexafluoride
I'll be doing some work (as a software engineer) with a vacuum process chamber that consumes various nasty gases.  There are numerous safeguards in the system, but as a cautious type, I decided to read the MSDS of the chemicals involved.  The one that makes me the most nervous is tungsten hexafluoride; one mole apparently reacts with water to form 6 moles of HF, which I distantly remember from chemistry class as being bad news.  Being largely composed of water, I find this troubling.  If there were failures in the multiple safeguards and a leak did somehow occur, how could it be detected?

First off, +1 for reading the MSDS then being concerned about the potential formation of HF...really, nice going there. It looks like you've found a very serious concern, as your memories of the nastiness of HF are all too true.  Personally, I think this is such a serious issue that it is a question for a safety engineer or some such person who is thoroughly familiar with your system. You can maybe get ideas to pass on from here, but you absolutely don't want to take an answer from here as a solution to your very specific, very serious issue.  Regardless, good luck!

@airhuff, I'm sure the engineers in charge of safety are taking the proper safeguards, but I was posing this more as a question of how I could notice if something was wrong.  I.e. what would be the warning signs of a leak?  You're right that I should probably just ask the engineers, though, as they likely have plans in place for detecting and responding to serious accidents.

Yea, I didn't mean to discourage the question, it's just that chemists eyes pop out when they see "leak" and "HF" in the same paragraph ;) What is the chamber made of?

@airhuff, I don't know the precise details, but it's a heavy duty vacuum chamber and it's being built by folks who have experience working with these kinds of things.  There are several layers of safeguards (all valves are inside a  sealed box), so obviously the process guys take this stuff seriously; maybe I'm just overly paranoid.  I was thinking more: if you're working with HCN and smell almonds, you know you have a problem; not sure what you can detect if you're exposed to HF.

Your nose is very sensitive to F and Cl (and HF and HCl), at the parts per billion level. The threshold vales for harming you are higher. But, you smelling the intensely sharp acrid odor of a halogen (or halogen acid) is not, of course, calibrated. Usually continuous air monitoring is used, along with appropriate respirator protection when opening up the system.

Sorry to jump back on my safety horse one more time, but, if you can smell it, it's too late. From the Wikipedia entry for hydrofluoric acid:"Hydrogen fluoride gas is an acute poison that may immediately and permanently damage lungs and the corneas of the eyes. Aqueous hydrofluoric acid is a contact-poison with the potential for deep, initially painless burns and ensuing tissue death. By interfering with body calcium metabolism, the concentrated acid may also cause systemic toxicity and eventual cardiac arrest and fatality, after contact with as little as 160 cm2 (25 square inches) of skin.".

$\ce{WF6}$ is not an uncommon gas used in industrial manufacturing of semi-conductors. Commercial gas detectors are readily available, and will detect $\ce{WF6}$ down to low ppm range, which is about an order of magnitude less than the $\mathrm{LC}_{50}$ for rats, apparently.
You could google tungsten hexafluoride detector, but some examples include (no affiliation or endorsement):

http://www.newcosmos-europe.com/gas-by-name/144-tungsten-hexafluoride-detection.html
http://www.toxicgasdetection.co.uk/gas/?gas=Tungsten%20hexafluoride&search=Tungsten%20hexafluoride&form=gas&id=416
https://www.gas-find.com/hexafluorure-de-tungstene.html

The most common type of detector is a diffusion monitor with electrochemical sensor.
Some sensors actually use conversion of $\ce{WF6}$ to $\ce{HF}$, and then detect the $\ce{HF}$. Commercial $\ce{HF}$ detectors at low ppm range are also common.


text:
Relation between activation energy and temperature sensitivity
In my reaction engineering textbook (Levenspiel), it is stated that reactions with high activation energies are very temperature sensitive, while reactions with low activation energies are relatively temperature-insensitive. This principle was then applied to a homework problem of ours where for two reaction pathways in competition, at high temperature, the reaction rate of the higher activation energy pathway would dominate, with the opposite being true at low temperatures.
Could someone explain why the high activation pathway dominates at high temperatures? In my mind, I still expect the rate of the low activation energy pathway to dominate due to path of least resistance principle.

Were you given a graphic along with that question? If so, attaching a picture to your question would give a good starting point for discussion, though this is not absolutely necessary.

You are correct, the higher activation energy reaction will never dominate, it will only become more competitive at higher temperatures.

In chemicals reactions that have more than one product, being one of them the kinetic product and the other the thermodynamic product, the temperature dependency is a relevant factor. For example, the kinetic vs. thermodynamic control can manipulate the regiochemistry of the alkylation of enolates. In other words, you have a competition between reactivity and stability.
Reactions are usefully to describe it in terms of potential energy diagrams, $\Delta G$ vs. reaction coordinate ($\xi$), where a one-step reaction have a global maximum corresponding to a transition state leading to the product and the free energy of this transition state, relative to the reactant, determine only the rate of reaction. The enthalpy part of $\Delta G$ deal with the stability of the molecule or intermediate but this doesn't give information about the rate of reaction.
Absolute rate theory (or transition state theory), by Eyring, have an expression for the rate constant $k$:
$$k=\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}e^{-\Delta H^{\ddagger}/RT}$$
where $\kappa$ is the transmission coefficient, $k_{B}$ the Boltzmann's constant, $h$ the Planck's constant, $R$ is the gas constant and $T$ is the absolute temperature. Here, the term $\frac{\kappa k_{B}T}{h}e^{\Delta S^{\ddagger}/R}$ varies slightly with temperature compared to the enthalpic term. Arrhenius equation also reflects the temperature dependence of reactions:
$$k=Ae^{-E_{a}/RT}$$
For remark, both relationships are temperature dependent.
So, if you have a reaction under kinetic control, the ratio of the products is determined by the relative energies of the transitions states. Whereas under thermodynamic control, the ratio of the products is determined by the relative energies of the products. This leads to the question if there are any relationship between $\Delta G$ or $\Delta G^{\ddagger}$, where you might search about Bell-Evans-Polanyi formula, Hammond's postulate and the Marcus equation.
Suppose you have a reaction
$$\text{A}\leftarrow\text{R}\rightarrow\text{B}$$
with the energy profile:

The product A have a lower $E_{a}$ or $\Delta G^{\ddagger}$ (kinetic product), whereas B have a higher $E_{a}$ but a lower $\Delta G$. The potential energy profile generally is temperature independent, i.e. the barrier and the heat of reaction are determined by the molecular structure. When you raise the temperature, the kinetic energy of the molecules is raised and this helps the molecules to pass the potential energy barrier more easily (you can check it by a Maxwell-Boltzmann distribution for a molecule changing the temperature and fixing an energy corresponding to $E_{a}$), being obvious that if the barrier is smaller will be more easy to pass.
If the temperature is high enough to make the molecules pass over all the barriers on the energy profile, the molecules will choose the pathway with the most stable product. All of this is determined by whether the products are or aren't able to equilibrate under the reaction conditions (search for the principle of microscopic reversibility). So, at lowers temperature you will disfavor the equilibrium and only get the kinetic product.


text:
Why are electrons shared equally when calculating formal charge, but unequally when calculating oxidation state?

When calculating formal charge - electrons are shared equally between
the atoms in the bond.

When calculating oxidation state - electrons are both given to the
most electronegative atom.

Why is this different?

I am assuming it is something to do with the intended use of calculating formal charge and oxidation state, which I believe are:

Formal charge can be used to find the most stable structure.
Oxidation numbers can be used to determine what will become oxidised
in a redox reaction.

However, even based on their uses, I am unsure why they are calculated differently.
*Images from Wikipedia

Are talking about a complex, for instance like $\ce{FeCl4^−}$ ?

I am unsure what a the implications of something being a complex are (I just googled what a complex itself referred to). But my question was for a compound such as $\ce{CO2}$. I have added some images to clarify. Thank  you.

The oxidation state is a form to distribute the charge of atoms in molecules considering to have only ionic bonds. In contrast, the formal charge is the distribution of the electrons considering to have only 100% electron-pair covalent bonds.
Actually, the formal charge is not very realistic and is more or less a bookkeeping tool. Now there exists theoretical and computational methods to calculate more reasonably charges on atoms from electronic structure calculations like Hartree-Fock, post-HF, semi-empirical and density-functional methods.
One example, taken from Anslyn's Modern Physical Organic Chemistry, is the case of the tetramethylammonium. One would place a positive formal charge on the nitrogen atom, but this is more electronegative than carbon, so there is a contradiction. The calculations mentioned before, indicate that the nitrogen is essentially neutral and the positive charge resides on the methyl groups. So the formal charge is more adequate for indicate the charge on the molecule and not on the atoms.

Nice explanation to cut through Gordian knot. I'd start off second paragraph with "No bond is purely ionic or purely covalent. Thus neither model is absolutely realistic and both are more or less ..."

@MaxW you're right. I was going to write it but I forgot.

Thanks for the answer Andres. I am a bit confused when you say formal charge is for charge on the molecule. We still calculate the formal charge for each atom right (and then maybe sum them up to find the charge on the molecule). But in calculating the charge of each atom, why is 1 electron distributed to each atom in the covalent bond for formal charge, whereas both electrons are taken by the more electronegative atom for oxidation state? Why isn't the same process used for both formal charge and oxidation state? Thank you.

@K-Feldspar the formal charge is only a convenient form to visualize an electron distribution in a Lewis structure. In the basics, the formal charge is what you know, I mean, the sum of the atoms in a neutral molecule is zero and so on. The thing is that charge is actually not what you "see" in, for example, electronic structure calculations.

@K-Feldspar Think as you have two types of bonds, the ionic bond and the covalent bond. In covalent bonding, two atoms share one electron pair per bond. In ionic bonding, the electronegative atom take the electron from the electropositive. In formal charge, you only account for 100% covalent bonding, i.e. the electronegativity stuff is fully neglected. In oxidation states, you only account for asymmetrical electron distribution, associated with a 100% ionic bond, for example in Na$^{+}$Cl$^{-}$. By definition, formal charge and oxidation state accounts for different types of bonds.


text:
Can you combine elemental gold and boron?
Is it possible to combine gold and boron to form compounds? A simple example being:
$$
\ce{Au + B -> AuB}
$$

yes. http://link.springer.com/chapter/10.1007%2F978-3-540-45280-5_27

@MaxW Thank ! is my equation correct  or am i missing something ? Also is the reaction made with gaseous gold and bore ?

That is the overall reaction but making AuB isn't as easy as the equation would make it seem. Gold is a solid at room temperature of course.

@MaxW yeah of course, but heating the reactants at 2075° (so the bore and gold should be liquid) (bore fusion point is 2075 and gold is 1064)  and then freezing down the solution would work ?

I don't know how to do it. No point in me speculating and telling you wrong. But at a temperature of 2000 degrees how do you protect the material from atmospheric oxygen? What container material do you use? And so on...

AuB does exist, at least in gas phase.

http://aip.scitation.org/doi/10.1063/1.1675226

Yes, gold and boron can bond with each other to form alloy clusters. They are very important in semiconductor industry and used in fabrication process. Here(1) is a short note on gold-boron alloy:

Photoelectron spectroscopy and density-functional theory are combined to investigate the electronic and structural properties of a
  series of $\ce{B-Au}$ alloy clusters: $\ce{B6Au_{n}^{−}}$ and
  $\ce{B6Au_{n}}$ ($\ce{n = 1−3}$). Rich spectral features are
  observed for each species, and vibrational structures are resolved for
  numerous detachment transitions of $\ce{B6Au^−}$ and
  $\ce{B6Au^{2−}}$. Electron affinities of $\ce{B6Au_{n} (n = 1−3)}$ are
  evaluated to be 2.70 ± 0.03, 2.91 ± 0.02, and 3.21 ± 0.05 eV,
  respectively. Global structural searches are performed for both the
  anions and their neutrals. The calculated electronic binding energies
  are compared with experimental measurements to establish the anion
  global-minimum structures and their corresponding neutral states.
  The ground-state structures of these clusters are shown to be planar
  or quasi-planar with an elongated B6core, to which the first and
  second Au atoms are bonded terminally and the third Au in a bridging
  position. All three anion clusters are π antiaromatic.
  Natural bond orbital analyses show that the $\ce{B-Au}$ bonding is
  highly covalent, providing new examples for the $\ce{Au/H}$ analogy
  in Au alloy clusters.

You can find more information about gold-boron alloy clusters in the reference below.
Reference

The Journal of Chemical Physics 138, 084306 (2016); doi: http://dx.doi.org/10.1063/1.4792501
https://www.google.com/patents/US3211550
https://www.ncbi.nlm.nih.gov/pubmed/23901981
http://aip.scitation.org/doi/10.1063/1.4816010
https://www.researchgate.net/publication/235886590_On_the_structures_and_bonding_in_boron-gold_alloy_clusters_B6Aun-_and_B6Aun_n_1-3
http://www.google.co.in/patents/US3137595

In a rather different vein is boronizing gold for surface hardening.  Here boron combines not with the gold but with a base metal alloyed into the gold.  According to the abstract from*  (full article behind  a paywall):

Surface hardening of gold containing Ti, Zr, Hf, V, Nb, Ta, Cr, Mn,
  Fe, Co or Ni was tried by means of boronizing technique with powder
  boron under argon atmospheric furnace at 900 to 950 degree C for 6
  hrs. The obvious surface hardening was possible for the gold alloy
  which contains more than 5%Ti, 5%V, 3%Cr, 15%Mn, 10%Fe, 5%Co or 5%Ni.
  Surface hardness and thickness of the hardened layer strongly depended
  on the alloying element added. The phases formed on the hardened
  surface were investigated by X-ray diffraction analysis.

Reference
*https://www.researchgate.net/publication/29677483_Surface_Hardening_of_Gold_with_Boronizing_TechniqueMaterials_Metallurgy_Weldability


text:
How to determine whether two hydrogen are homotopic, diastereotopic, emantiotopic?
I have been learning about nmr and I am confused about how to identify whether two hydrogen on different carbon atoms are homotopic, diastereotopic or emantiotopic. I can do identify this when two hydrogens are attached to same carbon atoms, but when it is different I am lost.I have searched lot on web, but they only include examples of two hydrogens on same atom, not on different ones. So can anyone help determine what the two hydrogen are in these compounds and why ? Thanks.

Honestly I think that it is hard to understand 3D symmetry by looking at 2D drawings on paper. A ball and stick model set really helps you understand better. Once you really have manipulated the 3D models then it is much easier to understand the 2D drawings.

If two H atoms are related by a C2 axis they are homotopic, if they are related by an Sn improper axis (including plane and center of symmetry) they are enantiopotic. All the others are diastereotopic. Another approach is to change one hydrogen for a different group (say X) then you get compound A, then start with the original compound and change the other hydrogen for X and you get compound B. If A and B represent the same compound the hydrogens are homotopic, if A and B are enantiomers the H atoms are enantiotopic and if A and B are diastereomers the H atoms are diastereotopic.

You determine by labeling. Take the two hydrogens in question and label them one at a time to get two molecules. What's the relationship between these two molecules? If the molecules are the same, then the hydrogens were homotopic. If they are enantiomers, the hydrogens are enantiotopic. Otherwise, if they are diastereomeric molecules, the hydrogens are diastereomeric.


text:
Uncertainty Principle in the Ground State of a Rigid Rotor
This is a fairly basic question which has bothered me for some time and I haven't been able to satisfactorily resolve.
In the ground state of the rigid rotor the energy is zero. That is, there is no zero point energy for this system. This means we definitely need to make sure we aren't violating the uncertainty principle for position and momentum.
So, the first place to look at is the wavefunction of the ground state. The wavefunction of the rigid rotor are the associated Legendre polynomials, and for the ground state (ignoring normalization) this is:$$P_0^0(x)=1.$$The point being, this is just a constant.
So, my first intuition is to say that we aren't violating position-momentum uncertainty because the wavefunction is only a constant so we know nothing about its location. This, however, I think is wrong because we solve this in spherical coordinates and the angle $\phi$ is constrained to be within $[0,2\pi]$. That is, our particle is somewhere on a sphere. Thus, this explanation doesn't work because the particle is somewhere on the sphere so we have some certainty of its position.
My next thought was that the uncertainty principle I'm thinking of is written as follows:$$\Delta p_x\Delta x \ge \frac{\hbar}{2}$$This, however, is wrong because it means that when the ground state energy is zero, the angular momentum is zero (that is, this is a closed spherically symmetric wave).
Thus, the next explanation I thought is that if I solved this in Cartesian coordinates rather than spherical coordinates, I would still get the same answer that the ground state energy is zero, but the x, y, and z position and x, y, and z momenta would all have positive, finite values which do not violate the uncertainty principle.
So, hopefully I've convinced you that this requires a slightly more nuanced answer than one might expect.
Is the last explanation correct in that when solving this in spherical coordinates, we should actually think about some other uncertainty principle than the one depending on Cartesian coordinates written above? Something like $$\Delta L \Delta r \ge \frac{\hbar}{2}$$
So, let me know if my reasoning has gone astray and where the solution lies. I feel a bit embarrassed that this isn't obvious, but I think my last point is where the answer lies.
Specifically, I'd like to see if anyone can address the question of the uncertainty principle in the ground state when this problem is solved in Cartesian Coordinates.

I think the uncertainty relation you want is between the angle $\phi$ and the angular momentum $l_z = -\mathrm{i}\hbar(\mathrm d/\mathrm d\phi)$. Using the generalised relations here you should find that $\Delta\phi \Delta l_z \geq \hbar/2$ Not sure about Cartesian coordinates though.

Ya, I think you're right. Or maybe the angle $\phi$ and the total angular momentum?

Your intuition is correct; because you know the energy is zero you cannot know in what direction the rotor is positioned if uncertainty is to be obeyed.

I have chanced upon a book that discusses angle-angular momentum uncertainty, and addresses the exact contradiction you raised about $\theta$ being restricted to $[0,2\pi)$, but I don't think I am qualified to explain it: Fayngold & Fayngold, Quantum Mechanics and Quantum Information, pp 384-388. It is to do with the domain over which the operators involved are Hermitian. I've uploaded the relevant pages here.

@orthocresol Wow, thanks for posting that here. That turns out to be a much more subtle problem than I expected. Maybe once I get the time to fully understand what's going on there I'll write up the answer. I get the general idea though. That's pretty interesting.

As far as the uncertainty relation between $L_{z} = -i\hbar \partial/\partial\phi$ and $\phi$ is concerned, there is a simple way to understand why we cannot use the commutation relation $[\partial/\partial\phi,\, \phi] = 1$, which leads to the problematic inequality $\Delta L_z \Delta\phi \ge \hbar/2$.
Remember that the variable $\phi$ is defined on a circle, i.e., $\phi = 0$ and $\phi=2\pi$ represent the same point. A function of the form $\phi\psi(\phi)$, where $\psi(\phi)$ is $2\pi$-periodic, has a discontinuous jump by the amount $-2\pi \psi(0)$ at $\phi = 0$ (or $2\pi$). Hence, when applying $\partial/\partial\phi$ to $\phi\psi(\phi)$, one should make sure to differentiate the discontinuity as well, which leads to an additional term: $-2\pi \psi(0)$ times a Dirac delta function. Then, the correct commutation relation should be $[\partial/\partial\phi,\, \phi]\,\psi(\phi) = \psi(\phi) - 2\pi\psi(0)\delta(\phi)$, and we have the modified uncertainty principle
$$
\Delta L_z \Delta\phi \ge \frac{1}{2}\big|\langle[L_{z}, \phi]\rangle\big| = \frac{\hbar}{2}\Big|\langle\psi|[\partial/\partial\phi,\, \phi]|\psi\rangle\Big| = \frac{\hbar}{2}[ 1- 2\pi|\psi(0)|^2].
$$
For a normalized angular momentum eigenfunction $\psi_{m}(\phi) = e^{im\phi}/\sqrt{2\pi}$, the right-hand side of the above yieids zero. This is exactly what should be the case when the uncertainties in the angular momentum and the angle are respectively zero and $2\pi$.

Very well put! This is almost exactly what the reference that orthocresol linked above says, but you put it in a much more understandable way. I see where the problem arises now, and it will be good to keep this in mind as this is an artifact which could arise in lots of different periodic systems like this.

@jheindel I'm glad because you seem to like this way of understanding the anomalous term in the uncertainty relation :) Compared to what's found in the standard literature, this approach is less formal (i.e., doesn't involve discussions on domains of operators and all that) but, I believe, is much more intuitive.


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Total number of stereoisomers of cyclobutane-1,2,3,4-tetracarboxylic acid
Question

Find all stereoisomers of:

My attempt
I got four isomers, and none of them seem to be optically active as they either have a plane of symmetry or a point of symmetry:

However, my sir keeps insisting there are five isomers. Could anybody help me identify the fifth one?

I don't believe there is a fifth.

I agree with you, there are just 4 stereo isomers. If anyone of them would have been optically active then we could have said that there are 5 isomers, but as you correctly pointed out none of the stereo isomers are optically active.

Sometimes the teacher is wrong.

Related: Total number of stereoisomers of truxillic acid

You can view 1,2,3,4-tetrasomethingcyclobutane as a square with black and/or white corners (black representing e.g. up, then white representing down configuration of the substituent). Two squares represent the same object when after applying some of the following operations on the first you get the second one:

rotation in the paper plane (of course)
mirroring AND color inversion

By drawing all 24 = 16 possible squares and using these two rules, you quickly find that they represent only 4 different objects, as seen on the following sketch.

Note that the second "controversial" operation mirroring AND color inversion is in our scenario, because of D2 symmetry of all squares, equivalent with more proper, "mirror-free" operation - rotation around some in-plane axis, considering that obverse of the square has inverse colors. (Because of the symmetry, the operation can even be reduced to color inversion only.)


text:
silicon nitride and sCO2
I want to use silicon nitride ball bearings with a reciprocating piston that will be powered by $\ce{CO2}$ liquid heated to  $\mathrm{288^oC}$ becoming super critical. That temperature is the maximum and the cylinder will not exceed that staying well below the $\mathrm{768^oC}$ rating of the bearings.
I can not use lubrication and the 1/4 bearings will be stacked 3 high with 4 of them surrounding the rod for a total of 12 bearings in direct contact with the rod that could have traces of supercritical $\ce{CO2}$ upon it. The bearings will be in contact with each other to provide both a seal and to align the piston and if there is any contamination it will spread.
Can I expect any type of reaction between the supercritical $\ce{CO2}$ and silicon nitride bearings?

I wouldn't expect anything.

Each of these materials is largely inert and there is no reason to expect that they would react with each other. Regarding supercritical $\ce{CO2}$, this Wikipedia article states the following:

Supercritical $\ce{CO2}$ is becoming an important commercial and industrial solvent due to its role in chemical extraction in addition to its low toxicity and environmental impact. The relatively low temperature of the process and the stability of $\ce{CO2}$ also allows most compounds to be extracted with little damage or denaturing.

And regarding silicon nitride:

Silicon nitride is a chemical compound of the elements silicon and nitrogen, with the formula $\ce{Si3N4}$. It is a white, high-melting-point solid that is relatively chemically inert, being attacked by dilute $\ce{HF}$ and hot $\ce{H2SO4}$. It is very hard (8.5 on the mohs scale). It is the most thermodynamically stable of the silicon nitrides. Hence, $\ce{Si3N4}$ is the most commercially important of the silicon nitrides and is generally understood as what is being referred to where the term "silicon nitride" is used.

Additionally, silicone nitride is used in the most critical, high temperature, high stress of applications, such as the bearings for NASA's (now defunct) Space Shuttle engine bearings.


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Understanding how electrons are counted for lone pairs vs. bonds especially when understanding resonance?
This is taken from the book Organic Chemistry I as a Second Language 3E: Translating the Basic Concepts[1] from page 35.
In the following image:

It is seen that when the electrons from the double bond are pushed into the lone pair electrons, we get 2 electrons as a lone pair.  And when a lone pair electrons are pushed to from a double bond we remove 2 electrons from the oxygen.
The part that confuses me is how the electrons are being counted (especially for octet rule). Let's take the example of oxygen that starts out double bonded (the molecule left of the double headed arrows) and then the double bond electrons get pushed into a lone pair (Oxygen on the right of the double headed arrow with a formal charge of 1).  If I count the first oxygen (double bonded oxygen to the left of the double headed arrow), I get 6 electrons where the double bond counts as 1 electron for each bond, now if I go to the right of the arrow I see that same oxygen I count 7 electrons.  How can a lone pair which is made up of 2 electrons result as being counted as 1 electron when it becomes part of a bond or vice versa?
Reference:
[1]: Klein, D. R. Organic Chemistry I as a Second Language 3E: Translating the Basic Concepts; John Wiley & Sons: Nashville, TN, 2011.

I'm glad you re-posted this question. I tried to post a comment suggesting that you not delete the question altogether, but you had already deleted it! This question is much better stated and easier to read than your original post; nice job.

Yea, I did it because I became unsure as to what I was trying to ask, and I just wanted to have a fresh start. Deleting wasn't necessary but I did it anyways.  Thanks for the advice!

this image is wrong, you are trying to draw acetic acid and that is why you are confused. there is a hydrogen on the second oxygen you have drawn the 6 valence electrons of oxygen but have magically bonded it the carbon? 6 electrons on oxygen 1 goes to an hydrogen the other 1 goes to the carbon bond.

Count 2 electrons per bond and you'll be fine. The partners of a bond usually share two electrons in their mutual orbitals. While you are correct in thinking 2 divided by 2 is 1, thats not how the octet rule works - count them all.
Left w/ dbl bond: 4 lone pair electrons + 4 bond electrons = 8.
Left w/o dbl bond: 6 lone pairs electrons + 2 bond electrons = 8.
Octet rule satisfied.

Then what about formal charge? If octet rule is being satisfied how do I figure out the formal charge for an atom? How can I know that the oxygen on the left with no double bond has a formal charge of 1?

Formal charge can be tricky sometimes, with ambiguous atoms. H, S, etc. Oxygen is usually easier. Count the lone pairs as 2, count bonds as 1. Then compare with the state of the atom. Oxygen has 6 electrons to start with. So, 4 lone pairs + 4 bond electrons (/2) gives 6 and net formal charge 0.  It helps "knowing" that Oxygen needs 2 bonds to be 0 formal charge, "knowing" that N needs 3, C needs 4, all these things which you learn in a while.


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Why is the compound 3,4-dihydroxy-3-cyclobutene-1,2-dione more acidic than acetic acid?
Why is the compound 3,4-dihydroxy-3-cyclobutene-1,2-dione more acidic than acetic acid?

I don't get how its resonance stabilization is far greater than that of acetic acid despite the fact that the conjugate base has high angular strain. How is it more stable than acetate ion? In the consideration of acidity, why is only the resonance dispersion of charge taken into account and not the angular strain?
For stability comparisons of the anions so formed I am stuck at the resonance structure too:

Finally, I want to know whether it would be more acidic that formic acid?

related http://chemistry.stackexchange.com/questions/32832/condition-for-nahco3-to-react-to-form-co2-with-organic-acids

This molecule is commonly known as Squaric acid. The Wikipedia entry gives the possible resonance structures of the conjugate base (added here for convenience):

You can see that there are four possible resonance structures for the conjugate base, resulting from the proximity of two ketone groups and two hydroxyl groups around the ring. This proximity allows movement of electrons around the square ring and between the ring and each oxygen atom.
On the other hand, acetic acid (or rather, the conjugate acetate anion) has only two resonance structures, as seen here:

Since the principle is that a molecule with more resonance structures is more stable than a molecule with less resonance structures, squaric acid anion is more stable than acetate anion. Since a more stable conjugate base makes an acid stronger, squaric acid is considered to be stronger than acetic acid.
This conclusion is also evident by the Pka values. This table provides Pka values for many acids, including squaric, acetic and formic acid. You can see that squaric acid has two values for two acidic protons - 1.5 for the first proton and 3.5 for the second proton. acetic acid has a Pka value of 4.75 and formic acid has a Pka value of 3.75.
This means that the strongest acid of the three is squaric acid, followed by formic acid, and the least acidic of them is acetic acid.
Squaric acid > Formic acid > Acetic acid

Can you help in the resonating structures of first anion formed when only one proton is extrascted

It is as you drew it in your notebook. Keep in mind protons are constantly associating and dissociating from charged oxygen atoms (any of the four oxygen atoms) when equilibrium has been reached. If we imagine those brief moments when the molecule is resonating in a monoprotic form, it is as you drew it.

then it has got only two resonating forms for the first ionisation then why are those more stable than those of acetate ion


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What can be added to hydrochloric acid to make a gel or paste?
I have a bottle of brick cleaner containing 18-20% hydrochloric acid. What can I add to it to make a gel or paste without unwanted reactions? The application is still just cleaning bricks as the product is intended, I just don't want it to run like a liquid.

wall paper paste would be a simple choice

You don't want the acid sticking to the brick because washing it off will be a chore. As long as the acid is there it will keep eating at the cement.

@MaxW The reason for wanting it to stick to the brick faces is to stop it running into the mortar line, as the acid also eats into the lime mortar which is undesirable.

There is another consideration too. Reacting an acid with carbonate yields carbon dioxide gas. So the gas would bubble and splatter the gel outward -- at you. The HCl solution is "thin enough" that it would splatter a lot less.

great question (also to clean WC) it's a shame it has been downvoted

Xanthan Gum can handle hydrochloric / muriatic acid (HCl) for a while, depending on how concentrated the acid is.  Here is how I made an HCl gel to clean an old toilet that was heavily coated with lime scale...
I first dissolved a level teaspoon of xanthan gum into 1/4 cup of rapidly stirring water.  After a few hours of letting it sit, with occasional stirring, I had a very thick, viscous goo.  To make the HCl gel, I mixed one part of this goo with 1.5 parts of 31% hydrocholoric acid, by volume.  This gave a gel of a suitable viscosity to get the lime dissolved off the toilet.  I don't know how long this gel would last before the HCl broke down the gum, but my gel lasted several hours, which was enough to get the toilet clean.
The following URL gives more hints on how to make mineral acid gels:
https://dowac.custhelp.com/app/answers/detail/a_id/4951

your link is dead

Ground flax seeds and Corn starch are cheaper alternatives to Xanthan Gum. Remains to know if they could handle the hydrochloric acid long enough to get the toilet clean...


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Canonical MOs vs. Localized MOs: Do both represent reality in the same way?
In my understanding localized molecular orbitals (LMOs) are equivalent to "standard" molecular orbitals, often called canonical orbitals (CMOs—by the way, why are they called canonical?). We can produce LMOs by linear combination of CMOs and they represent the same physical state.
I have two questions:

Is this just a mathematical gimmick or does this represent the reality? Is it possible for a molecule to be in such a state where every electron is in such a localized orbital? Or is this just a thing we do because it makes it easier to understand?
If LMOs are equivalent and have the same energy then why do calculations always end up with canonical molecular orbitals and never with LMOs? And wouldn't there be an unlimited amount of equivalent solutions?

Thanks @Jan . After reading the answers in the other question several times I might have gotten answers to most of my questions but not all. You say there that we lose information when going from canonical MOs to LMOs? This would mean the MOs are better at describing reality? Orbitals aren't "real" in the first place so this is something which is quite hard to understand for me.
Is this also the reason why a calculation ends up with canonical MOs, because those are the ones with all the information? Can we calculate the energy of a molecule from LMOs?

This is absolutely not a duplicate. Answer pending. (Also, DSVA, this isn't two questions -- it's something like eight questions. But, it's all good. :-)

Canonical MO is produced from the solution of eigenequation, e.g. f|i> = e_i |i>

NOTE: In the below, I'm implicitly discussing a ground-state, closed-shell wavefunction, where all occupied orbitals are doubly occupied.  The discussion would be similar for open-shell wavefunctions, but there are complexities that I won't address here. Also, once one starts noodling at excited states, things get complicated pretty quickly, and (AFAIK) some manipulations that are valid for a ground state wavefunction are invalid for excited states.

In my understanding localized molecular orbitals (LMO) are equivalent to "standard" molecular orbitals, often called canonical orbitals (btw. why are they called canonical?). We can produce LMOs by linear combination of MOs and they represent the same physical state.

The "canonical" MOs (CMOs) are called such because they are in a certain sense the "default" MOs for a wavefunction. The sense in which they are the "default" is that they diagonalize the Fock matrix that defines the energy of the system and is used in the course of the self-consistent field (SCF) calculation of, e.g., the Hartree-Fock method for solving the approximate many-electron problem.
Occupied localized MOs (LMOs), when all considered together as a unified wavefunction, in total contain the exact same information as the wavefunction composed of the occupied CMOs.  This is because the LMOs are obtained from a unitary transformation of the occupied CMOs, and the linear algebra indicates that the overall wavefunction is invariant to such a transformation.

If LMOs are equivalent and have the same energy then why do calculations always end up with canonical molecular orbitals and never with LMOs?

The italicized phrase of this question is false. LMOs individually are not equivalent to individual CMOs, and they do not have the same energy. Strictly speaking, the energies of individual LMOs are actually undefined. Only the total wavefunction composed of LMOs is equivalent to the total wavefunction composed of CMOs.
As noted above, the reason why we always get CMOs out of an SCF energy calculation is that they are the set of orbitals naturally associated with the solution of the SCF problem.

And wouldn't there be an unlimited amount of equivalent solutions?

Yes. Every possible unitary transformation among the occupied canonical orbitals, of which there is indeed an unlimited number, would provide a set of orbitals that gives an equivalent total wavefunction.  Not all of these transformed orbital sets are especially useful, however. Specific examples of useful ones include the various localization transforms that are the topic of the question, the series of variously defined "natural orbitals", and (somewhat related) the Löwdin symmetric orthogonalization for atomic orbitals.

Is this just a mathematical gimmick or does this represent the reality? ... Or is this just a thing we do because it makes it easier to understand?

The transformed MOs do represent reality, and often provide a very powerful and/or convenient means for analysis and understanding of the overall wavefunction. You do have to know what is or is not a valid use of any particular set of transformed orbitals, however.
One specific application of transformed orbitals I know about that seems very powerful is the family of local pair natural orbital (LPNO) methods for accelerating correlated calculations. The v4.0 release of ORCA includes implementations of their advanced domain-based (DLPNO) version of the method for several calculation methods (MP2, CASSCF-NEVPT2, etc.; see various of the 2016/2017 Neese citations listed here).  The DLPNO-CCSD method has already proven to be extraordinarily powerful in enabling correlated calculations of large systems—the below figure is a reproduction of Figure 5.5 from the ORCA v3.0.3 manual showing the enhanced performance of (D)LPNO-CCSD methods for a series of benchmark systems (click to enlarge):

Is it possible for a molecule to be in such a state where every electron is in such a localized orbital?

Yes, this is always true, as long as the transformation that is applied to the occupied orbitals includes no unoccupied orbitals. No matter what set of orbitals you use, whether canonical or transformed, every occupied orbital in the wavefunction is always fully (here, doubly) occupied. Transformation among the occupied orbitals doesn't change the occupancy.
Inclusion of some un-occupied orbitals (often called virtual orbitals) in the transformation, though, will lead to orbital occupations between zero and two. This is what happens in the classic natural bond orbital analysis of Weinhold & co., as can be seen in the sample natural population analysis output (pdf link; "Occupancy" column) provided on the GENNBO website .


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How can isotope labelling be used to determine the mechanism of ether cleavage with hydrogen iodide?
When methyl tert-butyl ether (MTBE) is treated with one equivalent of hydrogen iodide, the ether C–O bond is cleaved.
I read that if the oxygen in MTBE is isotopically labelled, it is possible to work out the mechanism, but I am not able to understand the function of $\ce{^{18}O}$.
$$\ce{CH3-^{18}O-C(CH3)3 + HI ->[S_N1/S_N2] products}$$
How does isotopic labelling help determining the mechanism?

If your products are simply an alcohol plus an alkyl iodide, then you really don't need the isotopic labelling. If you have an $\mathrm{S_N1}$ mechanism operating, then iodide ion attacks the tert-butyl cation, and you get the products
$$\ce{Me-O-^tBu + HI ->[S_N1] MeOH + ^tBuI}$$
If you have an $\mathrm{S_N2}$ mechanism, then iodide ion attacks the methyl group, displacing t-butanol:
$$\ce{Me-O-^tBu + HI ->[S_N2] MeI + ^tBuOH}$$
In both cases it should be trivial to take a proton NMR of the crude reaction mixture and determine which set of products are present.

The isotope label is more pertinent if you are forming two alcohols as the products. In that case, if the $\ce{^18O}$ remains attached to the methyl group (for example), it means that the $\ce{^tBu-O}$ bond was cleaved: this indicates an $\mathrm{S_N1}$ mechanism. In other words:
$$\ce{Me-^{18}O-^tBu ->[S_N1] Me^{18}OH + HO-^tBu}$$
Likewise
$$\ce{Me-^{18}O-^tBu ->[S_N2] MeOH + H^{18}O-^tBu}$$
However, you would have to be careful when running this experiment: if you leave it too long, especially under acidic conditions, the labelled oxygen will simply be exchanged with unlabelled oxygen from water by a degenerate nucleophilic substitution reaction, destroying any precious evidence that you had obtained:
$$\ce{R^{18}OH + H3^{16}O+ <=>> R^{16}OH + H3^{18}O+}; \quad \ce{R} = \ce{Me}\text{ or }\ce{^tBu}$$
where the equilibrium position lies to the right because of the larger concentration of $\ce{H3^{16}O+}$. Equilibrium tells us nothing about kinetics, though, so as stated earlier, you will have to stop the reaction (or monitor it using e.g. mass spectrometry, since NMR won't tell you the difference between oxygen isotopes) before this happens.

The function of the oxygen isotope is to determine what mechanism the reaction undergo. The mechanism of ether cleavage is SN1 or SN2 depending on the nature of the carbons bonded to oxygen. If both carbons are primary, cleavage involves a SN2 reaction, the halide ion being the nucleophile. With secondary or tertiary carbons, the $\ce{C-O}$ bond is cleaved by an SN1 mechanism.
In your case you have primary and tertiary carbons bonded to oxygen, but the tertiary carbons are particularly susceptible to cleavage by acid. So your reaction with one equivalent of $\ce{HI}$ will undergo cleavage with SN1 mechanism, forming $\ce{(CH3)3CI}$ and $\ce{H3COH}$.
The SN1 or SN2 mechanism starts with the formation of the oxonium cation:

And then, the characteristics of the carbon atoms bonded to the oxygen select between the SN1 or SN2. If the mechanism is SN2 the nature of the nucleophile also is important, otherwise not. The solvation energy, the strength of the bond with carbon, electronegativity, polarizability and sterics involving the nucleophile are considering factors.
Taking the case if the reaction undergo SN2; the transition state would have a partial $\ce{H3C-I}$ bond and a partial $\ce{H3C-O}$ bond. The free energy of this transition state is higher in energy because the $\ce{C-I}$ bond is weak ($\pu{213 kJ/mol}$), compared to $\ce{C-O}$ ($\pu{336 kJ/mol}$). In general the $\ce{C-I}$ bond is the weaker of the 14-17 groups general elements in organic chemistry (the source is a table from the organic chemistry class that I had). So, the activation energy of the SN2 mechanism is high.
However, if the reaction is SN1, the relative nucleophilicity has no effect on the rate of the reaction. Moreover, the substitution proceeds by rate-determining heterolytic dissociation of the reactant to a carbocation and the leaving group. So, the stability of the carbocation, the nature of the leaving group and the solvent's ionizing power is important. The electronic effect that stabilizes is the electron release, the accepting ability of the leaving group and the capacity of the solvent to stabilize the charge separation in the transition state. The mechanism would be:

where, like I said in the comment, the $\ce{-O(H+)CH3}$ part of the oxonium cation is a good leaving group. For the other hand, the carbocation intermediate is tertiary, i.e. the most stable carbocation. This stability is due to hyperconjugation.

That does not at all mention why the oxygen needs to be labelled.


text:
How does density of alkali metals increase down the group?
Size of alkali metals increases down the group, so volume also shows increment, and since volume is inversely proportional to density, how does density also increase down the group?

Because mass is also increasing, presumably faster than volume.

Do you mean "why"? "How" suggests you may want to know whether the increase is linear or some other mathematical function.

The molar volume increases from 13.10 cm$^{3}$/mol for Li to 71.07 cm$^{3}$/mol for Cs. Mass increases from 6.9 gm/mol for Li to 132.9 gm/mol for Cs. So, the mass per atom increases faster than the atomic density decreases.

$Density=mass/volume$
1) If mass is increasing and volume is decreasing, then density (mass/volume) will increase.
2)If mass is decreasing and volume is increasing simultaneously, then the density (mass/volume) will decrease.
*3)If both mass and volume are increasing, then we need to check which one of them is increasing at a faster rate(since both are contradictory factors)
a) If mass is increasing at a faster rate than volume, then density will increase.
b) If volume is increasing at a faster rate than mass (i.e denominator in mass/volume is increasing making the overall fraction smaller), then density decreases.
You can apply same logic when both volume and mass are decreasing simultaneously.
Generally, we see that in alkali metals the rate of increase of mass is greater than rate of increase of volume, therefore the density increases down the group.
Also, note that there is no definite reason for why mass is increasing at a faster rate than volume. So asking "why" does it happen isn't a great question.

Its just because density is directly proportional to mass. And the mass increases faster than that of volume expect for the case of potassium which is lighter than sodium.