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Preparation of iodine solution
How to prepare $\mathrm{0.1M}$ $\ce{I_2}$ solution.
I have found in books that it should be prepared in $\ce{KI}$ solution but none of them explains why? Please explain in detail not just the procedure.

Since $\ce{ I_2}$ is a non-polar covalent molecule, it does not ionize in water. It  cannot also be soluble in the polar water.
$\ce{KI}$, which is a polar, ionic compound, will ionize and dissolve in water. When KI dissolves in water, it ionizes to $\ce{K^+}$ and $\ce{I^-}$.
The $\ce{I^-}$ will react with $\ce{ I_2}$ to form the complex ion $\ce{I_3^-}$. $\ce{I_3^-}$ being negatively charged will dissolve in water.

And in what amount KI should be taken? Does the amount matter in preparing a solution of particular concentration?

Ingredients

Potassium iodide: 10 g
Distilled water: 100 ml
Iodine crystals: 5 g. Preparation: A. Dissolve 10 g potassium iodide in 100 ml of distilled water.
B. Slowly add 5 g iodine crystals, while shaking.
C. Filter and store in a tightly stoppered brown bottle.
This solution has the name of Lugol's iodine solution. As you can see, it has 0.0197 mol of iodine and 0.06 mol of potassium iodide.


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Why is the hydride ion negative?
Recently, I saw a video about sodium hydride and it saw it was made up with $\ce{Na+}$ ions and $\ce{H-}$ ions. My  problem is: aren't all hydrogen ions positive? And in what conditions can $\ce{H-}$ occur?

Sodium and most other metals have a lower electronegativity than hydrogen. Thus, in metal hydrides the metal transfers an electron to hydrogen leading to a negatively charged hydride ion, $\ce{H-}$, and a positively charged metal ion.

It's negative simply because it has more electrons (2) than protons (1), so the main question is how it can occur.

It seems likely that you'd be interested in this; there are actually negative ions of the alkali metals as well, alkalide ions.

And in what conditions can $\ce{H^{−}}$ occur?

In any phase. The hydride ion however doesn't stick around for long in water since it's such a strong base. First, hydrogen isn't that electronegative to start with, so tacking on an extra electron isn't helping with its stability. Second, combination with an hydrogen proton, $\ce{H+}$, is highly entropically favorable because the product, $\ce{H2}$, is a gas.
The $pK_a$ of $\ce{H2}$ has been estimated to be ~35 so that speaks a lot to the strength of the hydride ion as a base.

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How to determine relative size of polyatomic ions?
How would you determine the size of a polyatomic ion?
For example when determining which ion is bigger, $\ce{NH2-}$ or $\ce{I-}$, what would you use to determine the size difference?
I know that $\ce{I-}$ would be much smaller than $\ce{I2-}$ (since the $\ce{e-}$ are not held as tightly in the $\ce{I2-}$ causing the repulsion to increase)
But what basis do we use to determine size of a nonelemental atom like the polyatomic ion or a more complex molecule for that matter?

Well, you'd need atomic/anionic radii, bond lengths and angles.

better late than never i guess haha :P

Well, it's problematic question - how would you compare them - by volume, length, effective volume?

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The effect of hyperconjugation on the stability of alkenes with MO theory
Hyperconjugation stabilizes carbocations and that makes sense because electrons are given to the empty p orbital.
But how does it stabilize alkenes? Can molecular orbital theory be used to explain it?

Since you are familiar with how carbocations are stabilized via hyperconjugation I will keep this brief: With alkenes, it really is the same principle at work. Instead of the empty p orbital, you have to consider the empty $\pi^{*}$ orbitals of the $\ce{C=C}$ double bond. They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\sigma}$ bonds in $\pi$-bond-like manner

and so from those two fragment orbitals you will get two molecular orbitals: one from the in-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be lower in energy then the fragment orbitals, and one from the out-of-phase combination of $\pi^{*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be higher in energy then the fragment orbitals. Since in this interaction there are only two electrons involved, i.e. the ones in $\sigma (\ce{C-X})$, you get a stabilization from the interaction since those two electrons will now occupy the in-phase MO and have thus lowered their energy. The out-of-phase MO is empty and doesn't contribute to the energy of the molecule.

The hydrogen in the pictures should only have an s-orbital, but that's just a minor thing and not really important to the underlying concept. More important is that there is also the interaction of $\ce{\pi(C=C)->\sigma^*(C-H)}$. Strictly speaking this is the Valence Bond Interpretation of hyperconjugation. In Molecular Orbital Theory you just mix all orbitals at the same time and use what works best, i.e. gives the lowest energy.

@Philipp May you please cite a reference or book or sth for this ?

@RaghavSingal A reference (book) for this could be "Valency and Bonding: A Natural Bond Orbital Donor-Acceptor Perspective", by Frank Weinhold and Clark R. Landis.
Section 3.4 talks about hyperconjugation in the s- and p-block elements.

When the sp2 hybridized Carbon atom directly attached to sp3 hybridized Carbon atom having at least 1 hydrogen atom(called α hydrogen), this condition is known as hyperconjugation.
Stability of any hyperconjugative hydrocarbon depends upon no.  of α hydrogen present in compound..
More the no. of α hydrogen, more is the stability of compound

You make it sound that only CH bonds can undergo hyperconjugation, that's not true.

You basically answered nothing. And check up your basics about hyperconjugation.


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Substitute solvent for diethyl ether?
I want to run a reaction that needs diethyl ether as anti solvent in the last step (comes from articles).
But I can't use it due to its harmful effects on the body.
Is there any other solvent?
My reaction is about substitution of $\ce{-OH}$ of lignin with BiBB in THF and in other one in $\ce{EtOAc}$ as solvent.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Please include references to the articles you are referring to in your question and when you use "non-trivial" abbreviations such as BiBB it would also be nice to include a link to some site that shows what hides behind the abbreviation.

I've added a link for BiBB. If its not the correct compound please change it.

It is strange that you are concerned about diethyl ether (which has been used as an anesthetic), but you are not concerned about THF. From the US Pharmacopoeia:

Permitted Daily Exposure
  Tetrahydrofuran 7.2 mg/day
  Diethyl Ether 50 mg/day
  Ethyl Acetate 50 mg/day

If you dry the product thoroughly (above 120°C), you should not have any problems with residual solvents.

Thank you for your answer, but my professor have wanted me to change this solvent

I second this answer. THF is much more hazardous to human health than diethyl ether. I assume that this will be done using proper personal protective equipment, especially fume hood ventilation, then handling ether is not a problem if you can safely handle THF.

However, if you are worried about one of the other ways diethyl ether is more hazardous than THF, then we have a different scenario. Diethyl ether has a lower boiling point (ca. $35\ ^\circ \text{C}$), which means it is much more likely that you will be exposed to the 50 mg/day vapor. Also, in some parts of the world, especially if there is no air conditioning, diethyl ether might be a gas. Diethyl ether also is more flammable (due to its higher volatility), and forms peroxides more quickly (and more explosively).


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Effect on kinetic energy of a body with change in temperature
Why doesn’t the energy of a body doubles when temperature in degrees Celsius is doubled?
I know that it doubles in case of kelvin scale.

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

You might find it interesting to read up on the differences between the different temperature scales. The kelvin scale effectively measures the vibration (average kinetic energy of molecules) so it makes sense that a doubling in the kelvin means a doubling in kinetic energy. On the other hand the Celsius scale is defined by the melting and boiling points of water (quite arbitrary).


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Cu2+ octahedral complexes
$\ce{Cu}^{2+}$ has nine d-electrons, regardless of the ligand field strength it will have one free electron (so it is paramagnetic. Are $\ce{Cu}^{2+}$ octahedral complexes high spin or low spin?

I dont really see what the question whether $\ce{Cu^{2+}}$ complexes are high spin or low spin has to do with the fact that $\ce{Cu^{2+}}$ will have at least one unpaired d electron. Could you explain why you think this is important?

There are not two possibilities for low spin and high spin complex for a d9 metal ion.
The only cases where you can have both is for d4, d5, d6 and d7.
Remember that both configurations arise from the possibility of organizing the d electrons in a more stable way, this is, with the lowest possible energy.
You need extra energy for doing two things:

Promoting one electron to a higher energy level (crystal field stabilization energy, in case of an octahedral compound, $\Delta_O$
Puting another electron on a semioccupied orbital (pairing energy, due to repulsions)

In the complexes where $\Delta_O$ is higher than the pairing energy, the result will be a high spin configuration (it's easier for the electrons to become promoted to a higher level than pairing them with another electron), and if it is smaller, the result will be the opposite.
But this only applies for d4, d5, d7 and d8.
Now think about d1 d2 and d3:
It's nonsense to put the electrons on the upper level, when they can be all by their own in lower energy orbitals, so there are no high spin/low spin possibilities, but only one possible case, the three electrons one on each $T_2g$ orbital.
Now d8 d9 and d10.
There is once again, only one possibility to put the electrons. For your case, d9, you have nine electrons, so 6 of then will be filling the lower energy orbitals for sure, and the other 3, on the upper orbitals.
You may think, well, there is another possibility, promote one of the lower energy electrons to the hole on the higher energy orbital. But this is again nonsense: you are looking for the most energy favorable situation, so, as long as you only have to add to the system the pairing extra energy, why would you also add the promoting energy instead of filling a semioccupied lower energy orbital?
This would result in a excited state of the complex, but this is not what you're looking for when you talk about high/low spin.
The same applies to d8 and d10, there is only one possible situation, and no high/low spin possibilities.


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How to predict the solubility of an organic compound in different kinds of solvents?
How is it possible to know whether a particular solvent will dissolve in a particular kind of solvent, i.e. organic, polar, protic...?
Are there any indicators to look for in molecules or elements that allow one to predict what kind of solvent should be used to dissolve it?

There are some general guesses one can make from looking at the structure but the Abraham solvation equation is commonly used to estimate the solubility of a compound in a given organic solvent:
$$\log P_s = c + e E + s S + a A + b B + v V$$
This equation relies on a set of descriptors to characterize the properties of solvents and solutes to give the water/solvent partition coefficient $P_s$ (the ratio of the solubilities in water and solvent, under certain assumptions):
$p,e,s,a,b,v$ are coefficients that describe the solvent. Abraham and coworkers have collected coefficients for 85 different solvents.
The other variables are descriptors of the solute:
$E$ is the excess molar refractivity, which is a measure of the polarizability of a molecule.
$S$ is the solute dipolarity/polarizability.
$A$ is the overall (summation) hydrogen bond acidity.
$B$ is the overall (summation) hydrogen bond basicity.
$V$ is the McGowan characteristic volume.
You can look at the paper if you want to know exactly what these descriptors are, but their values can be obtained through various experimental and computational means, and combined with the solubility in water, used to estimate the solubility of a given compound in an organic solvent.
Fortunately, you don't have to do all this by hand as several websites, like this, that offer lookup and computation where you only have to specify a compound and pick a solvent from a list.

In general, organic compounds tend to dissolve well in solvents which have similar properties to themselves. So polar molecules tend to dissolve well in polar solvents, apolar molecules in apolar solvents. The solubility is affected strongly by the type of intermolecular forces between the two molecules.
For example benzene and hexane are highly miscible because they are both apolar and interact via Van der Waals forces. In general molecules with large hydrocarbon chains or rings will be soluble in apolar solvents.
A big indicator to look for in molecules is the presence of highly polarised bonds, in particular N-H, F-H and O-H bonds which are able to engage in hydrogen bonding. Protic solvents (those with N-H, F-H and O-H bonds) will form H-bonds with other protic molecules leading to them tending to be very soluble. H-bonding can also occur between protic hydrogens and other highly negatively charged atoms such as the oxygen atom in a carbonyl group and so this leads to protic solvents generally being good at dissolving any significantly polar molecule. For example propanone is soluble in water due to the hydrogen bonding between the water hydrogens and the carbonyl oxygen in propanone.
Polar aprotic solvent such as dimethylsulfoxide (DMSO) are generally miscible with polar protic and many apolar solvents and are similarly able to dissolve a wide range of molecules. After consulting several data sources it seems that DMSO and other similar polar aprotic solvents are miscible with all but the most apolar hydrocarbons and I would assume that they will also dissolve many different organic solids.
In general when looking at molecules to decide which solvent to dissolve them in you need to evaluate what the primary intermolecular interactions will be with that molecule (In particular H-bonding or Van der Waals) and then choose a solvent which has similar properties.
I haven't talked about entropy at all in this answer because I am not entirely confident that I can explain all of the relevant factors that affect it but for dissolving crystalline solids in particular that is also something that you have to take into account but I will leave that to someone more knowledgeable that me to answer.


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Why doesn't limestone thermally decompose instead of turning to marble?
When limestone is buried and put under intense heat and pressure, why does it turn to marble instead of thermally decomposing into calcium oxide and carbon dioxide?

This table relating the vapour pressure of $\ce{CO2}$ above a sample of limestone is important. In air, $\ce{CaCO3}$ beings decomposing slowly at about 600°C via the reaction $\ce{CaCO3\ _{(s)} -> CaO\ _{(s)} + CO2\ _{(g)}}$, but under pressure the equilibrium is shifted to the left and higher temperatures are in turn needed to decompose the limestone.

Short answer:
It's not hot enough, and the pressure is too high.
Long answer:
You say:

intense heat and pressure

The key word is pressure rather than heat. Let's look at the phase diagram of the CaO and CO2 compositional system:

(Calculated and plotted using Perple_X)
Pure limestone and marble are composed of calcium carbonate: CaCO3. In air, the temperature required for calcium carbonate decomposition is 898 °C (considering simple equilibrium and ignoring any vapour pressure effects). As pressure increases, the temperature requirement for this equilibrium decomposition increases very rapidly. It increases so fast that in fact, at above 80 bar, the calcium carbonate will melt into a liquid (call it calcium carbonate magma or lava) instead of decomposing to lime and carbon dioxide.
While 80 bar may seem a lot for us humans (it is higher than the pressure exerted by a 45 kg woman wearing stiletto heels), it is nothing compared to the pressure below us, deep in the Earth. Marble forms at a wide pressure range, but for this discussion let's say that it's somewhere between 2 and 10 kbar. That's 2000 to 10000 bar! Way off the scale of the phase diagram. Also, marble forms in the temperature range of 300 to 800 °C. Again, it's not even enough to melt the calcium carbonate, let alone decompose it to gas.


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Stereo Isomerism - syn/anti
I was studying the types of stereo-isomerism E-Z, Cis-Trans and then I come across syn-anti stereo-isomerism this one is giving me a bit of fight in understanding, especially where lone pairs are taken into consideration.How do classify the molecules with lone pairs in consideration as syn and anti Please help me out here!

Use of the terms "syn" and "anti" to describe the geometry about double bonds is no longer encouraged, but you'll still come across it, particularly in older articles.  Syn and anti are identical to Z(usammen) and E(ntgegen) and were often used to describe the geometry about carbon-nitrogen double bonds. In such cases, the lone pair of electrons is given the lowest priority and the sequence rule applied as usual.  For example, in the following N-methyl imines, the imine on the left has the highest priority group attached to the carbon in the double bond (ethyl) on the same side as the highest priority group attached to the nitrogen in the double bond  (the methyl has higher priority than the lone pair). Therefore the molecule is the syn isomer (but would now be called the Z isomer).

By the same reasoning, the molecule on the right is the anti of E isomer.  Again, the key point is that the lone pair has a lower priority than any atom.

do we have to cut the double bond perpendicularly and then see the priority groups?

@HydrousCaperilla Yes, cut the double bond perpendicular to the $\ce{C-N}$ axis, then figure out which group has the highest priority at the carbon, then do the same at the nitrogen and see if those two highest priority groups are on the same or opposite sides of the double bond.

There is another use of the terms anti and syn that is prevalent among steroid chemists. While the relationships between rings A and B/ B and C are described as trans in tricycles 1 and 2, the relationship between rings A and C in the former is anti while in the latter it is syn.

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Stability of cis vs trans isomers?
Recently, while reading about isomers I learned that cis isomers are more stable than trans isomers.
I referred many books as well as searched around the internet but I didn't find a suitable explanation.
Could someone please explain this to me?

just as like Charges repell each other and the unlike charges attract each other......Keep this rule in mind and think that cis isomers are Same and Near to each other So the they repell each other As well as the trans isomers are far away from each other and Opposite so not to repell and remaiNs Stable.......

This might be helpful: https://en.wikipedia.org/wiki/Steric_effects :)

I'd like to see the scientific basis behind this rule of yours.

As far as I am aware, and from a quick browse though some textbooks, the trans isomer is generally more stable than the cis isomer. This is due to the reduced steric hindrance of the substituents in the trans configuration versus the cis configuration.
For example trans-but-2-ene is more stable the cis-but-2-ene because there is less steric interference between the two methyl groups either side of the double bond. However, there are exceptions to this rule which I couldn't find a quick explanation for. The rule applies to both cyclic and acyclic compounds although there are some obvious cyclic exceptions such as cyclic alkenes, where the cis isomer is more stable, especially for smaller rings, due to the large ring strain caused by the orientation of the trans isomer. This page has a good illustration of this:

What about two OH groups that are cis to each other? Would they be more stable than if they were trans due to hydrogen bonding?

@MelanieShebel Yes, intramolecular H-bonding would have a stabilising effect in cis-1,2-diols.

Contrary to the generally accepted belief, many cis isomers of olefinic compounds are more stable than their trans isomers. The stable cis form of substituted olefins such is 1–fluoro–1, 3–butadiene and l,4–difluoro–1,3–butadiene as being due to intramolecular van der Waals forces between the substituents [ Angew. Chem. , 75 , 793 (1963)  ]. These forces are strong enough to hold the molecules in a cis–preferred configuration despite steric hindrance effects; these effects have been the basis for assuming that the trans form of such molecules is the more stable. But when the substituent atoms are so bulky that they overcome the intramolecular van der Waals forces, then the trans form becomes the more favored structure.
cis cycloalkenes in general are more stable than their trans isomers. The trans double bond causes strong twisting of the ring. Because of the resulting high ring strain small trans cycloalkenes have not been observed and cis isomers show considerable ring strain. However, the latter are sufficiently stable in order to exist. To form a trans isomer the cycloalkene ring must contain at least eight carbons. The energy difference between cis- and trans-cyclooctene is approximately  $38.5\ce{kJ.mol^{-1}}$. Eventually, trans isomers become more stable than cis isomers once the ring contains more than eleven carbons. For more details, please see also: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/12/oc/vlu_organik/stereochemie/cyclische_stereoisomere.vlu/Page/vsc/en/ch/2/oc/stoffklassen/systematik_struktur/cyclische_verbindungen/carbocyclisch/cycloalkene/physikalische_eigenschaften.vscml.html

cis is more stable than trans as long as no. of constituent C atoms is less than 11.
For more than 11 and for acyclic ,  trans is more stable....the reason is that like radicals and charges repel
however for less than11 the repulsion will be balanced best by cis


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Is bicyclo[1.1.0]butane chiral?
Is bicyclo[1.1.0]butane chiral? I think it isn't, but if one of the hydrogen atoms bound to the tertiary or secondary carbon atoms is replaced with e.g. chlorine, the formed compound will be chiral, right?

Bicyclo[1.1.0}butane is an achiral molecule.

Any molecule containing a plane of symmetry must be achiral and bicylco[1.1.0]butane has 2 planes of symmetry.  One plane bisects the molecule and passes through the 2 tertiary carbons, the other plane bisects the molecule and contains the 2 secondary carbons.
Even replacing one of the secondary or tertiary hydrogens with a substituent does not make the molecule chiral.  Such a substitution will remove one plane of symmetry, but not the other one.

In the above bicylco[1.1.0]butane where we have substituted a chlorine for one of the tertiary hydrogens, the plane containing the 2 bridgehead carbons and the chlorine still remains.  However, if you replace one of the tertiary hydrogens and one of the secondary hydrogens, the resultant disubstituted-bicylco[1.1.0]butane will be chiral since both symmetry planes have been destroyed and no other plane or axis of symmetry exists.


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Significance of Proton Tunneling
It has been repeatedly emphasized that proton transfers "involving" three or four membered ring transition states are not favorable and should not be drawn as the primary mechanism for any reaction.
However, with regard to a recent discussion involving $\ce{H3CSH}$ it was noted that the rate of proton tunneling in the anionic form $\ce{H3CS-}$ was likely extremely high.
Carbon-Sulfur Bond Lengths; Resonance Effects (Or Lack Thereof)
So:
a) Is this a valid rule of thumb: avoid intramolecular proton transfers unless the transition state has 5 or 6 members in its ring?
b) What's more significant with regard to proton transfer - proton tunneling or physical proton transfer? Or is the answer an "it depends"?

Was the discussion on Chem SE?  If so, could you reference it?  2) I don't follow, in what sense is the proton in the anion tunneling, what is the proton doing?

@ron http://chemistry.stackexchange.com/questions/18724/carbon-sulfur-bond-lengths-resonance-effects-or-lack-thereof

@ron I'd even guess there's a significant H tunneling rate in the syn form (or an inversion from anti to syn followed by H shift). So I agree it would at least be tough. Unfortunately, my microwave expert retired a few years ago, so I can't ask. –  Geoff Hutchison

a) Is this a valid rule of thumb: avoid intramolecular proton
  transfers unless the transition state has 5 or 6 members in its ring?

Yes

b) What's more significant with regard to proton transfer - proton
  tunneling or physical proton transfer? Or is the answer an "it
  depends"?

It depends, at temperatures around absolute zero, tunneling can become significant.  At ambient temperatures or above, physical proton transfer is much more common.
As to the methyl mercaptan anionic example you cite, to the extent that the various carbon-sulfur double bond resonance structures contribute, I would expect a lowering of the barrier around carbon anion planarization - and consequently a lowering of the barrier to inversion at the carbon atom in the anion.  Therefore, it is not clear to me that tunneling of the sulfur bound proton would be preferred to carbon inversion in order to bring about the syn-anti interconversion, especially since any presumed experiments to generate the anion would not be performed near absolute zero.


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Why isn't washing up liquid dry
Washing up liquid is wet and it is added to water. Why does it not come in a half size package and you add the water yourself? In the same way washing powder is?

There's no technical reason why you couldn't make a powdered dishwashing detergent and I suspect that someone probably does sell it, just as there are liquid, tablet, and powdered automatic dishwasher detergents and laundry detergents. It does seem more convenient to use a liquid detergent for washing dishes in the sink. A liquid doesn't need time to dissolve, is easy to dispense without a scoop and if you're only washing a couple of dishes, you can simply squirt a small amount directly on a dish or use one of those brushes with a detergent reservoir.
I haven't seen any dishwashing detergent that had a full ingredients list, but the one I have says "contains anionic surfactants and enzymes". Anionic surfactant typically implies sodium dodecyl sulfate, which is itself a solid, usually found as a powder. The enzymes, while they could be freeze-dried, could retain more activity if stored in solution. Likewise, the fragrances often added to these detergents may be more suited to a liquid form.


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Concerted mechanism of saponification
The energetics of the mechanism of saponification were previously under debate, but I will admit, I misrepresented a point.
Here is the revised mechanism of saponification proposed by someone other than myself. It is the one on top.
The previously suggested mechanism is in the middle.
The textbook mechanism is the bottom one.
Now, it does seem that if the mechanism were concerted then proton abstraction would be much more plausible from a thermodynamic standpoint. If the second mechanism were akin to scaling Mt. Everest, then the first one looks to me to be more like climbing the rock wall at the gym.
We start with the same reactant - an ester - but in the second mechanism we are going through a dianion intermediate. In the first mechanism however we are simply going through a concerted transition state en route to the stable carboxylate.
So, again, I ask - which is (kinetically) correct?

The reason I prefer the textbook mechanism is that #1 requires bringing 2 anions together, and #2 requires forming a dianionic species - both energetically unfavorable.

Another way of thinking about these three mechanisms is that the top is an E2 mechanism, the middle is an E1-conjugate base mechanism, and the bottom is an E1 mechanism. The middle and bottom represent ends of a mechanistic spectrum and the top represents the mid-point of those two mechanisms.

The question borders on a semantic argument. In the third mechanism (textbook), we would assume that the deprotonation event is very fast. The first mechanism (concerted) is basically saying that the deprotonation is so fast that it cannot be distinguished from the alkoxide leaving.
We know that acid-base reactions are fast. We also know that an anionic oxygen 'kicking out' an alkoxide is fast, at least relative to nucleophilic attack on a carbonyl (the rate determining step for any of these reactions AND the related trans-esterification is nucleophilic attack on a carbonyl to give the tetrahedral intermediate). What we're trying to answer is how fast is fast?
Here is my argument for the third mechanism. While acid-base reaction is fast, the top mechanism is essentially an E2 reaction, which does have a stereoelectronic preference for the hydrogen to be anti-periplanar to the leaving group. I have no doubt that conformation is readily accessible (the C-O bond will be rotating freely). The step is also bimolecular. In the third mechanism, an alkoxide kicking out a leaving group has no stereoelectronic preference. There are no alternate conformations to even consider. The C-O bond may be rotating, but it doesn't matter: there is always an electron pair arranged properly to kick out the alkoxide. Furthermore, the step is unimolecular.
The above is speculation, because I have never seen an alternative to the standard textbook saponification mechanism before. To be honest I never considered the concerted mechanism, and although I think it is a reasonable mechanism, I still prefer the textbook mechanism.

I'm also not sure how we would even design an experiment to address the issue. In both cases, the step in question is after the rate determining step. That makes it difficult to study empirically. It's not necessarily impossible, but definitely beyond my expertise in physical organic chemistry.

good points, and I like the point about the periplanar requirement. Unfortunately we don't cover stereoelectronics at all as the goal of the instructor is for us to do organic "by feel." I also think that while the first mech. is reasonable like you said, the proposer's reasoning behind it is still crappy - there's nothing wrong with a -COOH being formed in basic solution through electron rearrangement.

@Dissenter I feel like if the base approaches from the appropriate trajectory and the proton is lined up properly, the top mechanism will occur. Otherwise, the bottom mechanism will occur. It might also be somewhere in between... If the bond to the leaving group is mostly broken, then the anti-periplanar preference will be diminished. We could call that a concerted, but asynchronous pathway.

you do agree that it's okay to make the -COOH through electron rearrangement though, correct? But not through protonation in basic solution, right?

@Dissenter yes, like we discussed on your previous question, I think it's possible to make the acid through the bottom mechanism. That type of things happens all the time, making an intermediate that's reactive under the conditions. E.g., reaction of a Grignard with an ester, an aldehyde is formed as an intermediate that reacts immediately with another molecule of the Grignard reagent.


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How to balance the disproportionation reaction of manganate anion in basic conditions?
$$\ce{MnO4^2- (aq) -> MnO2(s) + MnO4^-(aq)}$$
How would I balance this in basic conditions? I know we need to separate this into two half reactions but how do we do this if there is only one reactant?

You are referring to the disproportionation reaction of $\ce{Mn(VI)}$ to $\ce{Mn(IV)}$ and $\ce{Mn(VII)}$, which is:
$\ce{MnO4^{2-}(aq) -> MnO2 (s) + MnO4^{-}(aq)}$
Reduction: $\ce{MnO4^{2-} + 2e^{-} + 2H2O -> MnO2 + 4OH-}$
Oxidation: $\ce{2(MnO4^{2-} -> MnO4^{-} + 1e^{-})}$ = $\ce{2MnO4^{2-} -> 2MnO4^{-} + 2e^{-} }$
Combine both, and the final balanced equation is:
$\ce{3MnO4^{2-} + 2H2O -> MnO2 + 2MnO4^{-} + 4OH-}$


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Gas Stoichiometry & Ideal Gas Law
I need a little help with one of the problems below. Any help on the problem below will be appreciated. I don't want to sound like an idiot, I will also provide my work.

Solid $\ce{NaHCO3}$ decomposes to form water vapor, carbon dioxide and sodium carbonate.
  $$\ce{NaHCO3 -> H2O + CO2 + Na2CO3}$$
  If $2.00\ \mathrm g$ of $\ce{NaHCO3}$ decompose at $500\ \mathrm{^\circ C}$ and $700\ \mathrm{mmHg}$, how many liters of $\ce{CO2}$ should be produced?

$$\mathrm{\frac{750\ \mathrm{Torr}}{760\ \mathrm{Torr}}=0.99\ \mathrm{atm}}$$
$$pV=nRT$$
$$2.00\ \mathrm g\ \ce{NaHCO3} \cdot \frac{1\ \mathrm{mol}\ \ce{NaHCO3}}{84.02\ \mathrm g} \cdot \frac{1\ \mathrm{mol}\ \ce{CO2}}{2\ \mathrm{mol}\ \ce{NaHCO3}} \cdot \frac{22.4\ \mathrm L \ce{CO2}}{1\ \mathrm{mol}\ \ce{CO2}} = 0.287\ \mathrm L\ \ce{CO2}$$
$$0.99\ \mathrm{atm}\cdot0.287\ \mathrm L=n\cdot773\ \mathrm K\cdot0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$$
$$n = 0.004159\ \mathrm{mol}\ \ce{CO2}$$
$$0.004159\ \mathrm{mol}\ \ce{CO2} \cdot\frac{22.4\ \mathrm L}{1\ \mathrm{mol}\ \ce{CO2}} = 0.09317\ \mathrm{L}\ \ce{CO2}$$

You can't use 22.4 L of $CO_{2}$ since the system is not at STP.  It is not needed, however, and you can stop your stoichiometry at the moles of gas produced then use the gas law to solve for V, since everything else is provided.

@bobthechemist So what you are saying is that after figuring out the number of moles of CO2 then you just plug it in the the Ideal Gas Law as 'n' to calculate the number of Liters of CO2. Ok now my other question is if you figure out the liters first of CO2 and then plug it in to the 'V', why I am I getting two different answers? One for solving for moles and one for solving for Liters of CO2 and then plugging it in the equation. Is there some sort of mistake I am making? Thanks for your help.

To your first question, yes.  For the second, take a closer look at my comment again; you cannot figure out the volume first.  Your 22.4 L/mol conversion factor is only true at STP, which is not true in this problem.

For dimensional analysis the main function to use is the fraction. Take a look at my edit.

I don't get how dividing 750 Torr by 760 Torr you can obtain a number with pressure dimension unit. This .99 should be unit free to be strictly rigorous.

@Babounet That was just to convert to atms because I wanted to make it easier, I like to deal pressure in Atmospheres. If you do not understand I can do the full dimensional analysis with 750 torr * 1 atm / 760 torr. The torrs cancel out and you have atms left. I just did not feel like to put a dimensional analysis and I thought people would get it. Sorry about that.

I do not like to post an answer for my own question, however I feel the need to clarify my mistakes ton anyone else. So here is the answer:

Calculate the moles of $\ce{NaHCO3}$
Then I need to plug in the moles to $\ce{PV = nRT}$
After finding the volume of $\ce{NaHCO3}$, now I plug it in a dimensional analysis to figure out the volume of $\ce{CO2}$. That is all and the answer would be $\ce{.77\ Liters\ CO2}$

In step 2, do you plug in the moles of $\ce{NaHCO3}$? The law is a gas law and $\ce{NaHCO3}$ is a solid.


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If radium has such a long half-life, how can radon possibly be a threat to us?
If the probability is so low that a radium atom will decay into radon at any given time (the half-life is over 1600 years), then there will be a low amount of radon produced, granted it will be produced for a long time. If this is so, how can radon possibly be dangerous to us?

Isn't chemistry all about electrons? How can this be on-topic?

Radioactivity is a part of nuclear chemistry.

There are two factors here:

Radon is a gas and can thus enter the body very easily, through inhalation.
$\ce{^222Rn}$ and its first 4 decay products are extremely radioactive, with halflives of miliseconds to a few days, releasing alpha and beta radiation.

When the two are combined, we have a problem. Even if the quantity is low, inhaling radon places an extremely radioactive substance at very close proximity to the cells in one's lungs. Normally, alpha and beta radiation are relatively harmless as they cannot penetrate tissue very well, but now they have direct access. All it takes is one mutation in the wrong place to cause lung cancer.
To give you an idea of what a difference these make, the volumetric activity which is one guideline for needing to mitigate radon accumulation in a house is around 800 $\mathrm{Bq/m^3}$, which corresponds to only 141 fg of radon per cubic metre of air. This isn't like a uranium mine or something, but at this level, a certain increased risk of cancer is expected. Apparently some study found a 50% increased risk at a much lower level ~150 $\mathrm{Bq/m^3}$, too.

N.B. a becquerel (Bq) is a reciprocal second—1 Bq means one atom of the radioactive substance disintegrates every second.

This makes sense but I'm wondering if you are able to answer a followup question. Some areas in Montana actually have Radon-therapy mines and things of this sort also exist in Europe. Do you (or someone else) know if these are actually beneficial to health because Radon levels are so much lower, or because the period of exposure is so much shorter than it is for people who come in contact with Radon in their home? Thanks

I found a paper about the Montana mines. As far as I know, it's quackery and it certainly sounds like a bad idea. Certainly nothing new. For a while, early last century, devices for adding radioisotopes to drinking water were a health fad. When a better understanding of radioactivity and its health effects was reached, they were banned and these devices mostly disappeared. Apparently this kind of thing hasn't totally died out.

I think it would be interesting to conduct an official study regarding the potential side effects that these mines have on patients later into life. If the results turned out to match your hypothesis, it could prevent a lot of sufferers of arthritis and bronchitis from worsening their conditions by getting cancer later on.

If I remember right, the decay products of radon are solids instead of gases? So radon atoms that decay in your lungs will deposit radioactive matter in the lungs that can't get back out.

@user137 They are solids, yes but the first 4 major daughter nuclides have half-lives under 30 minutes and the next, Lead-210 has a rather long one so I suspect the effect is more like an increased short-term dose, rather than a cumulative one.

@MichaelD.M.Dryden And while lead is toxic by itself radioactive or not I doubt you'll get enough lead through radon to be a problem.


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Rationalizing enamine stabilities

The solutions manual said that enamine B with its roughly sp2 methyl group can force the nitrogen out of the plane and this messes around with the conjugation.
How about the fact there is a significant degree of resonance donation by the nitrogen, which places a significant partial negative charge on carbon. In enamine A, this carbon is secondary. In enamine B, this carbon is tertiary. Extra inductive donation from the methyl group destabilizes the product.

This is a complex question.  We are dealing with very small energy differences (1-2 kcal/m) and a number of factors each play a subtle role.  Understanding kinetic vs. thermodynamic control and the energy differences between ground states and transition states are probably the two keys.  If isomer distribution was simply based on the effect that steric factors have on conjugation, then why would 2,5-dimethylpyrrolidine produce 90% of isomer B?  If you really want to understand this then see here and

here.  Like most things, when you dig into something, often it is not simple.

Conjugation effects are not very important in this case.
The main reason is steric hindrance. If the more substituted alkene is formed, then both the methyl group and the nitrogen ring will be on the same plane, and repulsions will destabilize the molecule.
In the case of A, the methyl group is more free to move in a pseudo-chair conformation of the cyclohexene ring.
But there are exceptions, for example:

In the second case the more substituted alkene is formed, because is more stable and there is essentially no difference in regard to steric hindrance.

The steric hindrance exemplified in Enamine B is an A-1,3-strain: http://en.wikipedia.org/wiki/Allylic_strain

Yep, in this case the A-1,3 strain is lower than the hypothetical interaction with nitrogen substituents, so trans olefin is formed.

@AlteredState would you say that the inductive and hyperconjugative effect has a role or not?

When it comes to evaluate thermodynamic stability of a more or less substituted alkene, hyperconjugation always has a role. If the destablization caused by the interaction between the amine ring and the sp2 fixed substituent is more important than the hyperconjugative stabilization effect of the more substituted alkene being formed, then you obtain the less substitued enamine, like the example you proposed.

@AlteredState what about inductive effects? Why do you only mention hyperconjugative effects?


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What Causes there to be a Maximum Amount of Salt which can be Dissolved in Water?
When you add salt to water and stir, after a while the salt will no longer dissolve. Why is this?

When a salt dissolves in water, there is a competition between the electrostatic forces that keep the anion and cation together and the ion-dipole forces that solvate the ions.  One could write this equation as:
$$\ce{AB + q H2O \leftrightarrows A(H2O)_{m}+ + B(H2O)_{n}-}$$
where $q=m+n$. Since this is an equilibrium, we can write an equilibrium constant, which is oftentimes referred to as the solubility product: $$\ce{K_{sp}=[A(H2O)_{m}+] * [B(H2O)_{n}^{-}]}$$
Very soluble salts will have a large, but finite, $K_{sp}$.  The magnitude of the solubility product can be rationalized by comparing the energy required to break the electrostatic bonds (the lattice enthalpy) and the energy released by forming solvated ions (solvation enthalpies).  One can use a Born-Haber cycle to predict whether the dissolution of a salt will be endothermic or exothermic, and we would need a similar approach to address entropy changes before we could get a complete, quantitative understanding of the solubility.


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What is the Difference Between the Conjugate Effect and the Inductive Effect?
As the title suggests, what is the difference between the conjugate and the inductive effect in organic chemistry.

Conjugate effect (or delocalization) is an effect in which molecular orbitals (MOs) are conjugated to new molecular orbitals that are more delocalized and therefore generally lower in energy (the amount of MOs stays the same of course). The electrons can move freely in these new extended orbitals. As an example consider this MO diagram of butadiene.

Note that the lowest energy MO is the MO with the least amount of nodes.
Inductive effect on the contrary does not increase delocalization but just pulls electrons towards the inductive (polarized) atom, this, for example, makes lone pairs less available decreasing the nucleophility of a molecule and increase its acidity. A lot of the time this inductive effect will be achieved trough a $\ce{C-X}$ (fluorine and chlorine) bond and you can observe its inductive effect simply by looking at the acidity of a series of more and less substituted molecules. Compare for example trifluoroacetic acid (pKa of 0.23) with regular acetic acid (pKa of 4.76).
I'm not exactly sure though what effect induction has on the size of a MO (but it won't match the effects of conjugation for sure). It is clear however, that the electron distribution will shift toward the spatial direction of the inductive atom.

The inductive effect refers to the shifting of electrons through σ bonds , the conjugations effect through pi bonds


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Ratios of produced gases in water electrolysis
I just did a little water electrolysis experiment at home by immerging two aluminum electrodes in water with a small amount of table salt. The voltage of the battery was 6 V.
What surprised me is that I was able to collect a much bigger volume of gas at the electrode connected to the - than to the + (about a factor 10 by eye). I would have expected a ratio of 1/2 (2 times more hydrogen than oxygen).

Why did I get less gas than expected at the electrode connected to the +?
Was the oxygen consumed by oxydising the electrode? (It appeared quite damaged at the end.)
How can I get a ratio of 1/2?

Maybe you can use vinegar instead of table salt?

We know from stoichiometry that you did produce twice as many moles of $\ce{H}$ at the negative electrode as $\ce{O}$ at the positive electrode. However, we don't know a few important things:

What were the actual volumes of the gases that were finally collected? Volume is notoriously hard to estimate by eye, unless you are comparing the levels in two identical containers. So, we'll take your word for it that it was "more", but we won't place any bets on the factor of 10.
What happened to the oxygen and hydrogen after the water split? Did it react?
What else was in the water that could have reacted? We know there is $\ce{NaCl}$, but maybe there are other dissolved compounds as well?

Question 1 can't really be answered here - the best we can do for now is assume that the volume of hydrogen was 2-10 times larger than that of oxygen. To find out for sure, you would need to conduct another experiment using graduated collection vessels.
Question 2 we can partially answer. If the electrode was corroded, then something reacted - either chlorine, oxygen, or some other more complicated set of reactions. We can get a rough estimate of the volume of oxygen this might consume by assuming it was oxidized by oxygen. If that 0.1 g of the electrode was oxidized, and that it was oxidized into $\ce{Al(OH)3}$, then at 1 atm and 298 K it would consume roughly 137 mL of $\ce{O2}$:
$\ce{Al^3+ + 3OH^- -> Al(OH)3}$
$0.1 \space \rm{g \space Al} * \frac{1\rm{\space mol \space Al}}{27\rm{\space g \space Al}} * \frac{3 \space mol \space O}{1 \space mole \space Al} * \frac{1 \space mol \space \ce{O2}}{2 \space mol \space O} = 0.0056 \space mol \space \ce{O2}$
$PV = nRT$ (oxygen is close enough to ideal under normal lab conditions)
$V = \frac{nRT}{P} = \frac{(0.0056 \space \rm{mol} \space \ce{O2})(0.08206 \space L \space atm \space mol^{-1}K^{-1})(298 \space K)}{1 \space \rm{atm}} = 137 \space \rm{mL}$
$\ce{Al(OH)3}$ is a likely candidate for a reaction product because of the white flakes you observed - aluminum hydroxide is insoluble in water. Aluminum oxide ($\ce{Al2O3}$) is also a potential candidate, and would consume 68 mL of $\ce{O2}$ under the same conditions.
So, assuming 0.1 g is a reasonable estimate for the mass of corroded aluminum, it seems very possible that what happened is the oxygen reacted with the aluminum electrode, reducing your yield of oxygen gas.
How can you get around this? Well, my five minutes of google research did not turn up any good candidates for inert electrodes. Usually graphite electrodes are recommended for other electrolysis setups, but they will react with oxygen as well. I have seen stainless steel recommended, but I would think that gold/platinum/silver are probably your best bets.
Question 3 is also hard to answer. We know that the $\ce{NaCl}$ will react and produce hydroxide ion (further supporting our hypothesis for aluminum hydroxide formation) at low concentrations, and chlorine gas at higher concentrations. If it were producing chlorine gas, you would know - it would be a pale yellow/green and it would make you sick as soon as you smelled it. So it is unlikely that your concentration of table salt was high enough for that to be the competing reaction. However, it would contribute to the aluminum oxidation. The easiest way to rule this out is to not use table salt. I have seen baking soda (sodium bicarbonate) recommended instead, and that would probably work. However, it might still produce hydroxide in solution. What about other unknown substances? If you used tap water, it is likely that there is additional chloride, possibly fluoride, and probably various magnesium and calcium salts. All of these would react during the hydrolysis, and would affect yields. The way to eliminate them is to buy or make distilled or deionized water.
Summary:

Why did I get less gas than expected at the electrode connected to the +?
Was the oxygen consumed by oxidising the electrode? (It appeared quite damaged at the end.)

The answer is most likely yes, the oxygen reacted with the aluminum. I would bet that it formed aluminum hydroxide, which then precipitated into the white flakes you observed.

How can I get a ratio of 1/2?

You might not be able to get it to exactly 1:2, but if you use sodium bicarbonate or a weak acid (vinegar would probably work) instead of salt as an electrolyte, and an inert electrode - like gold, or platinum, you can probably get close. Keep in mind - hydrogen gas is usually the desired product in hydrolysis. Oxygen is cheap if you don't need it to be pure! And even pure oxygen is relatively cheap compared to other gases.

Unfortunately, the previous answers never gives real answers. I will try to explain it now, from the very beginning.
Let's first start with Tony's experiment. He electrolyzed a solution of table salt with two aluminum electrodes at 9 V, and obtained about $10$ times more hydrogen than oxygen. Well ! This is not too difficult to understand, as the usual aluminum metal is always covered by a thin, waterproof, colorless and insulating layer of alumina (aluminum oxide $\ce{Al2O3})$. When dipped into water, aluminum metal is separated from water by this protecting layer. If this layer is rubbed, it will be immediately restored by contact with air. So aluminum pieces cannot be used for electrolyzing water under the usual tensions (about 2 Volts). But Tony used as high a tension as 9 V. This is enough for breaking the protecting alumina layer, and allowing the aluminum atoms to touch water. So now the following reaction occurred : $$\ce{2 Al + 6 H2O -> 2 Al(OH)3 + 3 H2}$$ which has nothing to do with any electrolytic reaction. As a consequence, most of the hydrogen gas obtained at the cathode was mainly due to the previous reaction and not to the electrolytic process. And the precipitate observed must have been $\ce{Al(OH)3}$, as this hydroxide is insoluble, white and flaky. At the anode, the alumina layer was also broken, but water was not oxidized into $\ce{O2}$ as expected. On the contrary, both the chloride ion and the aluminum metal will be discharged in priority according to :
$$\ce{2 Cl^- -> Cl2 + 2 e-}$$ $$\ce{Al -> Al^{3+} + 3 e-}$$ so that practically no Oxygen gas will appear on this electrode. Also chlorine will react with aluminum of the anode to produce  $\ce{Al^{3+}}$ ions $$\ce{2 Al  +  3 Cl2  ->  2 Al^{3+} + 6 Cl^-}$$ This is why the electrode was damaged at the end of the experiment.
At the end, Tony asks how to obtain the ratio $\ce{H2/O2 = 2:1}$. The answer is that with aluminum electrodes, it is simply impossible. He has to use graphite or platinum electrodes, and replace $\ce{NaCl}$ by $\ce{Na2SO4}$ which will not produce chlorine gas at the anode.
Now let's comment Peter Peterson experiment. He does not give as many details as Tony, but he used electrodes of magnesium, aluminum and stainless steel, and pure water. Well ! Pure water is an insulating material. No electrolysis is possible with pure water at low tensions. But he used a rather high tension : $12 V$. As he obtained the same gas at both electrodes, this shows that the main reactions were not due to electrolysis, but to a pure chemical reaction of metal + water. For example $$\ce{Mg + 2 H2O -> Mg(OH)2 + H2}$$ Here too, alumina protective layer on aluminum has probably been destroyed by the high tension, as described in the previous lines. And he should remember that stainless steel resists to oxidation by air, but not to anodic oxidation.
Now let's comment Thomij answer. He has no real explanations to give. He proposed to do more experiments, to work with another electrolyte, and to use platinum electrodes. This is correct, but it does not explain Tony's observations.

It looks like chlorine from table salt, rather than oxygen from water, is created at the positive electrode (I believe the reason is given in http://en.wikipedia.org/wiki/Electrolysis_of_water#Electrolyte_selection ), and the chlorine reacts with aluminum (http://www.chemicalforums.com/index.php?topic=3673.0 ).

This is a likely explanation. Can the reaction between the chlorine and the alunimum lead to a white precipitate? My solution had plenty of white "snowflakes" after use.

@Tony: Aluminum chloride is white (http://en.wikipedia.org/wiki/Aluminium_chloride )

@Tony: Although Aluminum chloride reacts with water, producing aluminum hydroxide (http://en.wikipedia.org/wiki/Aluminium_hydroxide ), which is also white.

True, but the solubility of aluminum chloride in water is huge, so I don't think this is the white compound I obtained.

@Tony: Sorry, looks like it produces some other substance (see the same source), which can also be white:-)

I've done exactly the same electrolytic demonstrations not just with aluminium but also with stainless steel and magnesium materials for both anode and cathode. I had not used any electrolyte because one doesn't need to. On all occasions gases were produced from the process. From doing these demonstrations, both anode and cathode electrodes decomposed over time and what's more interesting is that the same gases were produced at the anode and the cathode - hydrogen. Using magnesium ribbon as the electrodes was more interesting because after doing the electrolytic process on three separate occasions, once using a 12v battery, there was always more gas produced at the anode than the cathode and this again was hydrogen. Another interesting point was that after collecting the gas from the anode and cathode using magnesium ribbon and aluminium, both gases ignited and a clear yellow flame was seen. In my understanding using the same material for both anode and cathode electrodes one will observe them decompose during the electrolytic process. This would happen with any material used for electrodes but given the way some materials are manufactured the decomposition rate could take much longer.


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In the wet medium of an ocean, how does a hard shell form?
In general, if you were to try make something hard in a liquid medium (especially water) it is quite difficult to make the material solid.
Things such as mollusks though, have no problem generating hard shells in a wet medium.
What exactly are the physics behind this formation?

You know cement hardens under water. And your body is a pretty wet place but your bones are still solid. So molluscs aren't particularly unique.

Should the question be rephrased to what is the physics behind making hard substances under water?

I think this would be a relevant video from Minute Earth.

Have you noticed lime deposits? hard stuff sticking to the bath , the kettle etc? They come out of the water. It is called precipitation of dissolved elements, and they form a lattice given a seed crystal and the proper temperatur to do so. This is a chemistry question really. Cells in sea organisms utilize the chemical possibilities of building lattices out of available solvents or food elements.

Carbon dioxide from the atmosphere (or from decaying matter in the ocean) reacts with water to form carbonic acid:
$$ \ce{ CO2 + H2O -> H_2CO_3 }$$
and this reacts with calcium ions to form calcium carbonate:
$$ \ce{H2CO3 + Ca^{2+} -> CaCO3 + 2H+ }$$
The solubility of calcium carbonate is about $13~\mathrm{mg \over L}$, so if the concentration of calcium carbonate is greater than this the excess will precipitate out as solid calcium carbonate.
Shell-forming organisms actively absorb calcium from the water around them, so they are able to increase the concentration of calcium carbonate to above $13~\mathrm{mg\over L}$ and precipitate the excess to form their shells. They can get the carbon dioxide from the water around them or from their own metabolism. The actual details of shell formation is far more complex than this as it's controlled by processes with the cells of the organism rather than just being uncontrolled precipitation. I'm not sure how well the details are understood even today.
Incidentally, this is why shelly fauna are not keen on the acidification of the oceans that results from increased atmospheric carbon dioxide. The solubility of calcium carbonate is strongly $\mathrm{pH}$-dependent and rises sharply as the water gets more acid. Given that the shell is usually intended to stop other animals eating you, having your shell dissolve is generally not good for life expectancy.


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Inductive effect of hydrogen isotopes
Why does the electron-donating inductive effect (+I) of the isotopes of hydrogen decrease in the order $\ce{T} > \ce{D} > \ce{H}$?
(where T is Tritium and D is Deuterium)
Google has nothing to offer. Does it have to do anything with mass, as the order implies?

Where did you see this trend referenced?

Solomons and fryhle

Don't know much but wikipedia says that "Inductive effects can be expressed quantitatively through the Hammett equation...". Can somebody make an answer with this?

Yes, it has a lot to do with mass.  Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus (see "Note" below for more detail on this point). The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.
A shorter bond has less volume to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) over resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.
The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.
Similar arguments can be applied to tritium and it's even shorter $\ce{C-T}$ bond should be even more inductively electron donating towards carbon than deuterium.
Note: Bohr Radius Detail
Most introductory physics texts show the radius of the $n^\text{th}$ Bohr orbit to be given by
$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 m_\mathrm{e}}$$
where $Z$ is the atom's atomic number, $k_\mathrm{c}$ is Coulomb's constant, $e$ is the electron charge, and $m_\mathrm{e}$ is the mass of the electron.  However, in this derivation it is assumed that the electron orbits the nucleus and the nucleus remains stationary.  Given the mass difference between the electron and nucleus, this is generally a reasonable assumption.  However, in reality the nucleus does move too.  It is relatively straightforward to remove this assumption and make the equation more accurate by replacing $m_\mathrm{e}$ with the electron's reduced mass, $\mu_\mathrm{e}$
$$\mu_\mathrm{e} = \frac{m_\mathrm{e}\times m_\text{nucleus}}{m_\mathrm{e} + m_\text{nucleus}}$$
Now the equation for the Bohr radius becomes
$$r_{n} = {n^2\hbar^2\over Zk_\mathrm{c} e^2 \mu_\mathrm{e}}$$
Since the reduced mass of an electron orbiting a heavy nucleus is always larger than the reduced mass of an electron orbiting a lighter nucleus
$$r_\text{heavy} \lt r_\text{light}$$
and consequently an electron will orbit closer to a deuterium nucleus than it will orbit a protium nucleus.

I don't get it. If D is more electronegative than H , then it should show -I effect (as our standard is based on H) ! (because it attract electrons more strongly than H , so acting like electron withdrawing)

The problem is that OP actually meant +I effect : March's advanced chemistry , my class notes , my teacher , textbook , various websites after searching  all have written that D shows +I effect (counterintuitively)  . Also your statement  "Therefore, deuterium is inductively more electron donating than protium towards carbon." implies the same - electron donation means +I effect. What I don't understand is your argument that D donates more electron density due to higher electronegativity , shouldn't it be the other way around ? For eg. F is  electronegative and shows strong -I effect .(not +I)

How big is that effect in numbers (for the Protium/Deutererium/Tritium example)?

This answer is wrong. The reduced mass of the electron in deuterium and tritium differs less than $4 \cdot 10^{-4}$ from the electron mass. This can not account for the observed differences in bond lengths (which e.g. are two orders of magnitude greater in $\ce{H2O / D2O}$).

@aventurin  Sorry, I'm not following. Which bond lengths are 2 orders of magnitude greater?

I wrote about the difference in bond length.

Ron, could you resolve the disagreement here?

A $\ce{C-D}$ bond has a shorter distance, thus can be  in a more favorable position to donate electron density towards an electron deficient center.  The $\ce{D}$ has a greater mass, and would show a shorter bond distance.  This is reflected in the $+ I$  effects noted in the literature.

This answer offers nothing in addition to the existing one.


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how protons(H+) react with matter and cause the solution to be acid?
For example if my hands were put in a HCl solution how does the protons in the solution which in high amounts will interfere/react with my hand(skin->proteins..) and cause it to burn...or how protons react with a plastic bag and cause it to melt.

Well HCl won't desolve all plastics. In case of your hand there are complex proteins involved containing a variety of elements e.g, carbon, nitrogen, oxygen..., when your hand comes in contact of a proton, the proton is a electrophile and seeks negative charge centers, where ever on a protein it finds the negative charge center e.g, on oxygen atom in a hydroxyl(-OH) group in some amino acid in some protein in your hand, it will attach itself to the oxygen and eventually causing it(oxygen) to loose off from the protein as water, because the ion H20^+ isn't stable and the newly formed Oxygen Hydrogen bond isn't easy to break, so the part breaks off as water. millions of such reactions or even trillions happen eventually corroding off the hand or al large part of hand exposed to protons in a mobile medium.
An acid burn isn't actually a burn but corrosion of proteins, and it doesn't either melt things away rather disintegrates them.

it will attach itself to the oxygen and eventually causing it(oxygen) to loose off from the protein as water Do you have a reference to support this claim? I'm no expert, but I'm not sure wether the proteins actually decompose when brought into contact with acid. I would have found it more likely that the change in pH and the accompanying protonation of hydroxyl and amino groups leads to irreversible changes in the conformation of the proteins and thus inactivates them.

The example I gave below is Alanine(An amino acid) reacting with HCl.   _________CH3CH(NH2)COOH + HCl -> H2O + CH3CH(NH2)COCl
As it's a neutralisation reaction (products must be water and a salt).

this reaction is a dehydration reaction that is the reacting molecule looses water molecule

change in pH and the accompanying protonation of hydroxyl and amino groups leads to irreversible changes in the conformation of the proteins and thus inactivates them -  the proteins get decomposed and a part of the protein is detached off as water molecule(in above given reaction of Alanine) which is seen eventually as the protein structure disappearing off. the protein part that still sticks to hand is eventually inactivated as its structure is no longer that of the reacting protein

also when the acid has corroded a part of skin whcih is a protective covering the sensitive proteins inside turn into zwitterions helping the reaction to occur easily.

In the above example I took Alanine found on the inside of our body, a better example would have been Arginine in our skin, but that molecule is a bit large to be shown in the comments.

if you find an answer useful mark it as answered, it helps future users and prevents duplication of questions


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Why is the boiling point of heavy water higher than normal water?
In class, we learned that London forces become stronger as relative molecular mass increases. Not just in organic chemistry but in things like the halogens.
However, as I understand, the London forces are just temporary dipoles forming. This indicates that this would have nothing to do with the molecular mass but instead the number of electrons. My reasoning is that there will be more electrons to form the dipole. This would mean that the trend would be stronger London Forces with more electrons instead of increasing molecular mass.
How can that explain the slightly higher boiling points of heavy water (deuterium oxide ~101˚C) than normal water?

possible duplicate of Why do molecules having a higher $M_r$ have stronger inter-molecular forces?

The linked question only provides an answer with respect to changing number of electrons available to the molecules. That is not the case here, where water is compared to heavy water.

@tschoppi Good point, I had remembered that we already had a question about the relation between mass and inter-molecular forces some time ago but I didn't read through the answer carefully again. I'll retract my close vote.

At atmospheric pressure D2O has the higher boiling point; however, at higher pressure H2O has the higher boiling point. When the pressure is high enough that the boiling point is 220 degrees C, the boiling points are equal.

The fact that D2O is 10% heavier surely counts for something.

The hydrogen bonds in deuterium oxide are slightly stronger than those in water. This is due to a quantum mechanical effect; the bonding interaction has a lower zero point energy due to the greater mass of the deuterium atom. It, therefore, requires more energy to excite the bonding electrons from the ground level to the dissociation point, and a higher boiling point is observed.
You can find a good explanation of the physics behind these effects on this page.

Due to this effect, heavy water is toxic beyond certain concentrations in the human body.

I think in water and D2O the main factor is not Van Der Waals but rather hydrogen bonding. So the boiling point of D2O is higher than H2O not because of the London forces, but because of the difference in electronegativity between the hydrogen and the water, creating an electrostatic attraction.
This begs the more interesting question: why is deuterium's electronegativity different?
Just to clarify, electronegativity is the relative attraction an atom has for an electron.
Deuterium has a lower electronegativity than hydrogen, i.e it wants to give away its electron more. This is because the extra neutron increases the size of the nucleus and I think partially reduces the effect of the positive charge. Since deuterium has a lower electronegativity, there is a greater electronegativity difference between the deuterium and the oxygen, resulting in a stronger hydrogen bond.
Stronger hydrogen bond = stronger intermolecular forces = greater boiling point.
I don't think van der Waals is relevant here.
TL;DR:   In this case, hydrogen bonding is more important than London forces
EDIT
The above explanation is dodgy, I am still not sure to what extent it is true. There are many conflicting sources on the internet.
MUCH MORE INTUITIVE EXPLANATION
Heavy water is heavier, right? So it requires more kinetic energy (i.e more heat) to escape the liquid and vapourise. Hence it has a higher boiling point

You make a good point about the main force being hydrogen bonds. However, I don't see a reason why the neutron will affect electronegativity as the attraction of the electron to the nucleus depends on charge. Since the neutron has a neutral charge, the strength of the pull would not have increase/decrease. The neutron would have no way of reducing the charge, which means no change in electronegativity.

@Jonathanjaya fair point, but this article ought to clear things up. The point is, deuterium's electronegativity IS different,and as you have diligently pointed out though this explanation may not be intuitive (or could be wrong), it was the best out of the others.I will edit my response.
http://en.wikipedia.org/wiki/Kinetic_isotope_effect

I think due to the greater size of deuterium atom, positive charge would be more dispersed, thus more stable. So deuterium's electron can come out with comparatively more ease, thereby reducing deuterium's electronegativity.

The description of the behavior you are trying to describe is not clear/wrong, I think you can afford to use some more technical language to describe the situation.

In D2O ,there would be more extensive hydrogen bonding due to greater electronegativity difference between deuterium and oxygen. Greater the difference more stronger will be the bonding. I just explained why electronegativity difference would be greater in my above comment @J.Ari

I think you should edit your answer to be clearer and include some of this newer information.

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


text:
How to Determine Acid Strength?

The correct answer is E

But I do not understand why?
Shouldn't (v) have the lowest $\mathrm{p}K_\mathrm{a}$ since it does not have a carboxyl group making it a strong acid? How do we determine the strength of acids?

First of all, let's make something crystal clear: The more acidic is a compound, they smaller will be its $pK_a$.
To evaluate the acidity of these compounds you need to analyze the stability of the corresponding conjugate base of each one, in effect, the stability of the same compound but without a proton and a negative charge on an oxygen atom.
First of all, you have two kinds of compounds:

Two alcohols: V and II
Two carboxylic acids: I, III and IV

From those compounds, alcohols will always be less acidic than carboxylic acid. Why? Because the negative charge left by a proton leaving the molecule can be delocalizaed by resonance troughout the carbonyl, but in the case of the alcohols it will be localized on the only oxygen atom, which makes the conjugate base less stable, and therefore, the compound less acidic.
That being said, from the two alcohols, V will be more acidic, because the presence of the chlorine atoms substract charge from the carbon they are bond to (chlorine is more electronegative than carbon), and helps stabilizing the negative charge that is left in the molecule after the proton is substracted.
In regard to the carboxylic acids, you can use a similar reasoning for the carboxylic acid.
The less acidic is the one in which R is just a hydrocarbon chain, which has no effect in contrast to I, which has 1 chlorine atom that stabilizes the negative charge, and to III, which has 3 chlorine atoms that stabilize the negative charge of the conjugate base even more.
Therefore, the order presented in option E is justified.


text:
What would make guanidinium ions and urea more effective denaturants of proteins in some cases, but not in others?
Some proteins do not unfold, or only partially unfold due to chaotropic agents such as guanidinium ions and urea. What are some reasons for this?

Chaotropic agents increase the solubility of nonpolar substances in
  water. Consequently, their effectiveness as denaturing agents stems
  from their ability to disrupt hydrophobic interactions, although the
  manner in which they do so is not well understood. Conversely, those
  substances  that stabilise proteins strengthen hydrophobic forces,
  thus increasing the tendency of water to expel proteins. This accounts
  for the correlation between the abilities of an ion to stabilize
  proteins and to salt them out.
  1

Among other well established chaotropic agents, guanidinium ion ($\ce{Gu+}$) and the nonionic urea, which, in concentrations in the range 5 to 10 M, are the most commonly used protein denaturants. The effect of the various ions on proteins is largely cumulative: $\ce{GuSCN}$ is a much more potent denaturant than the often used $\ce{GuCl}$, whereas $\ce{Gu2SO4}$ stabilizes protein structures.

There are factors which have been investigated to explain why these two chaotropes appear to be effective in different cases (i.e they tend to be effective in some cases but not others). In one experimental investigation, the mechanistic basis of thermodynamic destabilisation due to urea and guanidinium was tested in order to find possibility that they exert their denaturing effect through peptide group hydrogen bonding (done experimentally by measurement of acid and base catalysed hydrogen exchange (HX) in a small peptide model)[2]

Urea

It has been established that urea increases the acid catalysed rate and decreases the base catalysed rate.

The acid catalysed HX rate in urea solution is higher than in the absence of urea. This can be expected when it is understood that acid catalysed HX proceeds dominantly through transient protonation of the peptide carbonyl group and deprotonation of the peptide NH by water or another base like urea and thus the proposed explanation is that urea forms an H-bond to the peptide carbonyl.[2]
In the case of base catalysed HX rate, which strongly decreases upon urea, the peptide NH group transfers its proton to a transiently H-bonded hydroxide. If urea is H-bonded to the peptide N-H, hydroxide can’t access the proton and HX is blocked.[2]

Guanidinium

With gaunidinium, as explained above, the effect of the various ions on proteins varies by some margin as they exert differently on the ion. It doesn’t selectively H-bonds to the peptide group (base catalysis) because guanidinium cannot H-bond to the peptide NH. Instead studies seem to suggest that guanidinium tends to engage in transient interactions with itself.[2]
Guanidium is often found to be approximately twice as efficient as urea in denaturing proteins but this varies with protein target. Helical peptide studies show that gaunidinium can be up to four times more efficient than urea when planar amino acids are major contributors to the helical stability. In contrast, guanidinium is barely more efficient if stabilisation is due to salt bridges. Thus protein targets are major contributing factor to the varying efficiency of guanidinium in denaturing proteins.[2]

Tl;dr
From above discussion, the varying effeciences depend on things like:

acid/base environment
nature of ions (the case of guanidinium)
protein target (e.g planar amino acids in helical peptides)

References

Voet and Voet Biochemistry. Section 7-1. Primary Structure Determination of Proteins.
Urea, but not guanidinium, destabilises proteins by forming hydrogen bonds to the peptide group. Woon Ki Lim 2595-2600 vol 106 no 8.


text:
Phase diagrams (weird region)

On the left, there is a region for $\ce{MgO(ss)}$.
I've been told such regions exist because stuff are not miscible. However, does that mean the composition axis on the bottom is a lie? Namely, because it says ≠100% $\ce{MgO}$ and the region signals pure $\ce{MgO}$.

There is nothing weird in here. What you've been told is wrong, or maybe you have misunderstood it. Regions like this exist because stuff are miscible. (ss) stands for solid solution, meaning that the solid crystal structure can accommodate some amount of the other component (Al2O3 in this case).
The diagram is a bit misleading, because naming something as MgO(ss) is not entirely correct. Adding the (ss) implies that it is not pure MgO. A more correct way to name the fields in the diagram would be to rename MgO to "periclase" and MgAl2O4 to "spinel", which are the mineral names for these compounds. They will not imply pure compositions of the solids, and conceptually allow the solid solutions to occur.
Some applications of the lever rule: For example, take a look at 10% Al2O3 and 2000 degrees. You basically have periclase, which in pure form as the composition MgO. However, it has 10% of Al2O3 in it. It is still periclase though. Let's take a more complicated example: At 40% Al2O3 and 1800 degrees. You have two solid phases, periclase and spinel in roughly equal proportions. The periclase has 10% Al2O3 in it, and the spinel has 65% Al2O3 in it. Note that it's less than the ideal spinel should have!
Also, notice that Al2O3 in the diagram does not have (ss) in it. This means that it's always going to be Al2O3, with no MgO in it at all.
The curves that separate solid solution (ss) fields are appropriately called solvi (singular: solvus). They mark phase changes that are purely in the solid state, without any melting. For example, let's say you are at 70% Al2O3 and 2000 degrees. You have only one phase: a spinel. By cooling down the system, you will cross the solvus line and periclase will be added to the system, on the expense of spinel. You will be able to microscopically see it as patches or stripes or something like that of periclase in the spinel. Wikipedia has an image to demonstrate that (although it's built for eutectic compositions, the morphology is similar).


text:
Reducing agent in preparation of CpRuCl(PPh3)2
Synthesis of cyclopentadienyl-bis-triphenylphosphine-ruthenium chloride $\ce{RuCl(PPh3)2(\eta^5-C5H5)})$:
Dissolved $\ce{PPh3}$, $\ce{RuCl3.3H2O}$ and cyclopentadiene in ethanol and heated under reflux to prepare it.
What is the reducing agent and what is the product formed on oxidation of this reducing agent?
The ruthenium is being reduced but i don't know what is being oxidised.

You'd think it's the ethanol solvent that acts as the reducing agent here, and the byproduct is acetaldehyde.

Could it be possible that triphenylphosphine is the reducing agent here, going to triphenylphosphine oxide?

@jerepierre I believe that is correct.

@jerepierre Couldn't it also be cyclopentadiene, in excess, dimerising? The described synthesis is behind a paywall for me.

In the synthesis of dichlorotris(triphenylphosphine)ruthenium(II), a similar complex, $\ce{PPh3}$ is the reducing agent which is oxidized to triphenylphosphine oxide, $\ce{Ph3PO}$ (references: 1, 2). The reaction is driven by the strength of the $\ce{P=O}$ double bond.
$$\ce{2 RuCl3 \cdot 3H2O + 7 PPh3 ~->~ 2 RuCl2(PPh3)3 + 2 HCl + 5 H2O + OPPh3}$$


text:
What governs the merging of globs of inmiscible liquid?
If I add a small amount of chloroform to a bunch of water, it falls to the bottom as a glob.
If I shake the container vigorously (being carefully not to spill any of the dangerous substance!), the globs of chloroform break apart into many tiny globules. These globules immediately sink to the bottom and collect together by the action of gravity the moment I stop agitating my mixture.
If I watch this pile of globules for some time, I observe that they tend to merge after several seconds, perhaps even a minute or more for some. I suspect that if I reached in with a needle and prodded the globules, they would merge faster. Also, I suspect that if I were to tabulate how long each individual globule survives before merging, I would obtain a half-life curve.
What governs the merging of these globules (I'm guessing surface tension)? Why don't they instantly merge? Why do a few little globules sometimes persist for many minutes, stuck to the side of a much larger globule?

Emulsion chemistry is actually rather complicated, but I'll try to give a conceptual description. Whenever you have a surface dividing two phases, there is an energy associated with creating a certain amount of surface, expressed as a surface energy density with dimensions of energy divided by area. This represents the energy cost of disrupting the intermolecular interactions of each phase at the interface, e.g. for your water/chloroform system, water can't hydrogen bond across the interface so it takes energy to break all the bonds required to make the surface. The larger the surface area, the more energy it takes. Logically, you can see that the lowest energy state is the one in which the interfacial area is minimized, which is why in the absence of external forces like gravity, liquids assume spherical shapes which give the smallest surface area for a given volume.
When you have many small droplets, there is a large surface area to volume ratio which is a high energy state, so it is energetically favourable for the droplets to form larger droplets. However, the reason it doesn't occur instantaneously is that an energy barrier can exist between the two states, for various reasons (electrostatic repulsion is a big one). In most cases, smaller droplets are lost first because the energy barrier to coalescing is surface area dependant. Larger droplets are more stable. Depending on the liquids, different separation mechanisms can occur with different kinetics. (coalescence, Ostwald ripening) I can't give a definite answer, but I'm guessing in your system you would not see an exponential decay-type half-life as that assumes a stochastic process where the probability is always the same. Since the stability depends on surface area, this isn't true in this case.


text:
How to find the mass of the solute given a concentration as a mass fraction?

Formalin is a $37\%~\mathrm{W/W}$ solution of methanal in water. What mass of water must be added to $250~\mathrm{g}$ of methanal to make formalin.

Here's what I've done so far. It's wrong though. According to my book answer is $430~\mathrm{g}$ $\ce{H2O}$.
$$\frac{Cw}{w} = \frac{\mathrm{m\ solute}}{\mathrm{m\ solution}} \cdot 100\%$$
$$37 \% = \mathrm{\frac{m\ solute}{250} \cdot 100\ \%}$$
$$\mathrm{\frac{37}{100} = \frac{m\ solute}{250}}$$
$$\mathrm{0.37 \cdot 250 = m\ solute}$$
$$\mathrm{m\ solute = 92.5\ g}$$
Where have I went wrong? Am I completely off or at least close?

You have been given a mass fraction, $w_{\ce{CH2O}}=37\%~\mathrm{w/w}$, and the mass of the solute, $m_{\ce{CH2O}}=250~\mathrm{g}$. You calculated the mass of the solution wrong.
The mass fraction is defined as
$$w_i = \frac{m_i}{m}\tag1$$
with $m$ being the total mass and therefore
$$m = \sum_i m_i.$$
The latter is also the mass of your solution, the sum of the solute and the solvent.
$$m=m_{\ce{H2O}}+m_{\ce{CH2O}} = m_{\ce{H2O}} + 250~\mathrm{g}$$
Substitute what you know into $(1)$ and transform:
\begin{align}
w_{\ce{CH2O}} &= \frac{m_{\ce{CH2O}}}{m_{\ce{H2O}}+m_{\ce{CH2O}}}\\
0.37 &= \frac{250~\mathrm{g}}{m_{\ce{H2O}} + 250~\mathrm{g}}\\
m_{\ce{H2O}} &= \frac{250~\mathrm{g}}{0.37}-250~\mathrm{g} = 426~\mathrm{g}
\end{align}

You have a couple of problems. In your equation, you've written the mass of the solution is 250 g, but there is 250 g of methanal and the total mass of the solution is 250 g + $m_{\ce{H2O}}$.
Second, the solution is 37% methanal not 37% water so the you've written the equation, you're solving for the wrong thing.
These should help you get on track:
$$\begin{align}\\
f_\mathrm{meth} &= \frac{m_\mathrm{meth}}{m_\mathrm{solution}}\\
m_\mathrm{solution}&=m_\mathrm{meth}+m_\mathrm{\ce{H2O}}\\
\%_\mathrm{meth} &= f_\mathrm{meth}\times100\%
\end{align}$$


text:
Elements of paper
I learnt that everything is made up of atoms even paper! After that, I was thinking what are the elements that made paper? Is there any chemical symbol for paper in general?

There is no chemical symbol for paper since it is not an element but rather a mixture of several different compounds.
Remember that paper is made out of trees. Thus, paper is made mostly out of organic compounds: that is carbon, hydrogen and oxygen (C, H and O). Paper also contains non organic materials to improve its properties. These may be chalk (CaCO3) and kaolin clay (Al2Si2O5(OH)4). Titanium oxide (TiO2) is also commonly used in paper for bleaching it.

Titanium dioxide doesn't bleach paper: it whitens it by making it more opaque via scattering from the small oxide particles. Other bleaches like hypochlorite are used to removed coloured impurities from the wood pulp components.

Correct - I was using bleaching in the optical sense, not the chemical sense.


text:
Coordination numbers of ionic and intermetallic compunds
If $\ce{NaCl}$ crystallizes in an fcc lattice, its coordination number should be 12. Then why do we say that $\ce{NaCl}$ has a coordination number of 6?
Also, if $\ce{CsCl}$ crystallizes in a simple cubic lattice, why is it's coordination number 8 and not 6, as for other simple cubic lattices?

NaCl
NaCl is not monoatomic. When discussing coordination number of cations in ionic compounds, the number refers to the anions the cations is surrounded by (or bonded to, or touching).

The blue Na+ cations are bonded to six other Cl- anions: top, bottom, left, right, forward, backwards (in this image). They are not bonded and do not touch the other Na+ cations in the diagonals.
CsCl
CsCl is not a "simple" simple cubic. Each of the different ions, Cs+ and Cl- by themselves are simple cubic, but they are interpenetrating to form something that looks like a body centered cubic:

A mono atomic simple cubic will have coordination number of six, but this is an ionic compound. Thus, each cation or anion that sits in the center of the cube is bonded to the eight anions or cations (respectively) that sit in the vertices of the cube.

@ron I just read this on http://goo.gl/o28dWh: "Caesium Chloride is a type of unit cell that is commonly mistakened as Body-Centered Cubic. This misconception is easy to make, since there is a center atom in the unit cell, but CsCl is really a Non-closed packed structure type. . . . CsCl crystallize in a primitive cubic lattice which means the cubic unit cell has nodes only at its corners. The structure of CsCl can be seen as two interpenetrating cubes, one of Cs+ and one of Cl-. The ions are not touching one another. Touching would cause repulsion between the anion and cation.

But how is there "repulsion between the anion and cation"? Only like charges repel.

I edited to clarify the body-centered-simple-cubic issue. I'm not sure what they mean about the repulsion.

"Each of the different ions, Cs⁺ and Cl⁻ by themselves are simple cubic . . ." According to Wikipedia, the crystal structure of chlorine is orthorhombic, while that of caesium is bcc. So, how do they combine to form a simple cubic structure for CsCl?

You are confusing the structures of the mono atomic substances with the structures they assume in a compound. These things are not the same.

I thought so; I just wanted to confirm.


text:
Why does a smaller amount of propanol burn slower than a much larger amount?
Why does $30~\mathrm{mL}$ of $\ce{C3H7OH}$ burn in 1 second (inside a water jug) compared to a sixth of the amount: $5~\mathrm{mL}$ of $\ce{C3H7OH}$ (on a plate) burn in 180 seconds? Why does the smaller amount take longer to burn the propanol?

this compound is propanol* the $\ce {-OH}$ is for alcohol, then the terminology is "ol"

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour].|| While editing I took the opportunity to correct the mistake pointed out by @Babounet.

For "burning" you need oxygen because actually when a compound is burning it's what is called an oxidation following this reaction
$$\ce{C3H7OH + 5O2 -> 3CO2 + 4H2O}$$
with propanol called the combustible and oxygen is called the comburant. Both are required as well as a power source to start the combustion process.
So in a plate you have the air surrounding the compound then you have practically an endless amount of oxygen available.
So the reaction will stop when all the Propanol will be consumed.
Hence, in a jug, this time the limiting reactive is the oxygen because you have less oxygen supply. Then the combustion will stop when all the oxygen will be consumed in the jug.
So, to sum up, in the jug the oxygen will be consumed faster than the propanol in the plate, that's why the combustion last longer in the plate even if you have more combustible.

thank you for the edit Martin, I was currently looking for the right way to display the reaction :)
And also for the english correction, in my native language we often use "y" instead of "i" in this kind of words so I'm often confusing :x

If this is the reaction I'm thinking of, seen here, then there is also the difference between a liquid+gas reaction and a gas+gas reaction.  In the former, the reaction is limited to the portion of alcohol that is vaporized, in the latter, virtually all of the material that will combust is already in the proper state for that reaction to occur (the gas phase, that is.)


text:
Identifying an aerogel like material or similar alternatives
A few years ago I saw this ultralight weight hydrophobic material and I have been trying to discover it's identity. I doubt it was aerogel since it was brought in by a high school student, but I cant say for sure.
It was a transparent whitish material that was very light weight but absorbed ridiculous amounts of water, expanding as it absorbed the water. They were in small pellets, approximately 2mm or smaller randomly shaped. When water is poured on it each individual absorbed large amounts of water and expanded, and each pellet became somewhat gel like (in the sense that each individual pellet was kind of squishy). Unless obscene amounts of liquid was added all it took was a quick shake to more or less absorb every drop of water. Even when squished the material generally didn't start releasing the liquid. If you ended add enough water that the material couldn't absorb anymore, it would float on the water. If left out for a few days the water would evaporate, the material would shrink, and the dry material was somewhat squishy (less so than when wet).
If it's off topic for me to ask something like this then how could I go about identifying this with a list of physical properties.

My memory of the material is pretty clear so if more properties for me to add are suggest I'll add them.

This is definitely sodium polyacrylate.


text:
How to check if azobisisobutyronitrile (AIBN) is still active?
I will have some samples of azobisisobutyronitrile and I want to perform Barton-McCombie reaction. The problem is that the reagent is quite old and I need to know whether it has not decomposed.
Do you know any kind of test that I could perform? I was thinking if benzophenone in THF would not become blue after addition of fresh AIBN which would indicate the presence of the radicals. But I am not sure if such a test would be a good idea.

If azobisisobutyronitrile (AIBN) underwent homolytic cleavage over time, do you expect the resulting cyanoprop-2-yl radicals would survive as free radicals for a long period of time? If not, what would be the decomposed products?
Alkali metals reduce benzophenone in THF to form blue diphenylketyl radical anions. I'm not sure azobisisobutyronitrile (AIBN) or cyanoprop-2-yl radical can do that.

Maybe you can test the structural integrity of AIBN by 1H-NMR, IR or Raman spectroscopy?
The -N=N- stretching vibration of a symmetrical trans-azo compound like AIBN is "IR-forbidden" but absorbs in 1576cm-1 region of the Raman spectrum (1). If it decomposes, perhaps you can see the decomposed products in the mass spectrum.
Maybe you can try to activate the subject AIBN as a radical initiator?
a. Perhaps one can observe a minute amount of nitrogen gas upon homolytic cleavage?
b. AIBN can be used as radical initiator in radical polymerization. Once activated as cyanoprop-2-yl radical, it readily attacks olefinic double bond of vinyl monomers. Perhaps you can test such reaction and see if some polymerization occurs using mass spectroscopy?

Reference:

R.M. Silverstein and F.X. Webster: Spectrometric Identification of Organic Compounds (6th ed.)

You'd think that the main decomposition product is tetramethylsuccinyl dinitrile (and perhaps also some peroxidic species.
In 13C NMR the quaternary carbons of both products should have distinct chemical shifts, and assuming relaxation times are similar (they should be) this should give a rough idea about the purity of the product.

I don't think that reaction with benzophenone is good idea - as Wikipedia article about benzophenon says it turns blue when reacted with  alkali metals - becoming ketyl anion. Those anions can trap other radicals but are too aggressive to discern whether your is AIBN good. You could maybe use DPPH - heat it in solution with AIBN and look if it reacts. Or maybe try if it can initialize another, simpler reaction like polimerisation, or maybe analyse it chromatographically - there are lots of options.


text:
How to calculate the activation energy of diffusion of carbon in iron?

At room temperature, $298~\mathrm{K}$, the diffusivity of carbon in iron is
  $9.06\cdot 10^{-26}\frac{m^2}{s}$.
  At $573~\mathrm{K}$ the diffusivity is $1.66\cdot 10^{-15}\frac{m^2}{s}$. What is the activation energy, $Q$?

I am wondering if my solution is correct:
$D=D_0\exp\left\{\frac{-Q}{RT}\right\}$ and $D_0$ is constant (is this correct), thus
$\displaystyle \frac{D_{298}}{\exp\left\{-Q/(RT_{298})\right\}}=\frac{D_{573}}{\exp\left\{-Q/(RT_{573})\right\}} \implies Q\approx 122~\mathrm{\frac{kJ}{mol}}$
I would be very greatful if someone could check my math here as well, if the logic is correct, since this was a test problem.

Maybe I'm wrong but my calculation has given another result:
\begin{align}
D_{298} &= D_0\cdot\exp\left\{\frac{-Q}{298~\mathrm{K}\cdot\mathcal{R}}\right\}\\
D_{573} &= D_0\cdot\exp\left\{\frac{-Q}{573~\mathrm{K}\cdot\mathcal{R}}\right\}\\
\end{align}
Thus, going through the $\ln$ form, you get
\begin{align}
\ln\{D_{298}\} &= \ln\{D_0\} - \frac{Q}{298~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{573}\} &= \ln\{D_0\} - \frac{Q}{573~\mathrm{K}\cdot\mathcal{R}}\\
\ln\{D_{298}\} - \ln\{D_{573}\} &= -\frac{Q}{\mathcal{R}}\cdot\left(\frac{1}{298}-\frac{1}{573}\right)
\end{align}
And then I obtained $Q\approx 38~\mathrm{kJ/mol}$.


text:
Endo rule and stereoselectivity in the Diels-Alder reaction
Do you agree with the stereocenter with the $\ce{-CF3}$ and $\ce{-CN}$ substituents in the product?

source
I thought that since $\ce{-CF3}$ is a better electron withdrawing group than $\ce{-CN}$, it would be placed opposite the diene, giving the product in which $\ce{-CF3}$ points up and $\ce{-CN}$ points down.
However, I believe my proposal above is wrong, because probably the substituent with the highest electron-withdrawing property, the $\ce{-CF3}$ should be placed endo because of secondary orbital interactions. Therefore, I expect the product above should be the correct one.

source
When I try to do the endo-product of the following reaction I came to the other syn product with absolute stereochemistry (S,R) instead of the (R,S)-product shown here. So I'm wondering, why we get endo selectivity in the first reaction and exo selectivity here instead.

Do you agree with the stereocenter with the CF3/CN in the product?

No, I would have expected the other isomer, the one with the $\ce{CN}$ and $\ce{NO2}$ groups pointing down in your drawing.
Rationale:
When exo and endo isomers can form in the Diels-Alder reaction, there is often a preference for formation of the endo isomer.  Although this endo preference is often referred to as the "endo rule", it is not a "rule", just a general, and rather weak, preference for endo product formation.  In fact, in most cases both isomers are formed, just more of the endo isomer (see figure below).
The explanation most often advanced for the endo preference is "secondary orbital interactions." The following diagram illustrates the two possible transition states (TS) in the reaction of cyclopentadiene and an $\alpha,\beta$-unsaturated carbonyl compound.  One TS leads to the exo product, the other TS to endo.  Note how the endo TS has the unsaturated carbonyl bond tucked under the double bonds in the cyclopentadiene ring; the exo TS does not.  It is speculated (1) that the overlap of the carbonyl unsaturation with the diene unsaturation somehow stabilizes the endo TS making endo product preferred.  The accompanying table illustrates how weak the endo preference can be.

image source
In your top example, this endo preference for a TS with overlapping unsaturation would lead me to suspect that the product with the unsaturated nitro and cyano groups (the $\ce{CF3}$ group is not unsaturated) oriented away from the 3-carbon bridge and tucked towards the double bond would be preferred.
Also, keep in mind that the endo preference applies to cases were the Diels-Alder reaction is run under kinetic control.  As the following example shows, the exo product is generally thermodynamically preferred (less steric crowding). So if the reaction is run under conditions (high temperature) where the initially formed kinetic (endo) product reverts to starting material, eventually the thermodynamically favored product (exo) will be formed eliminating the kinetic preference for the endo product.

(1) see here for an alternate explanation for the endo preference

Excellent explanation! The paper is also very helpful to see that the secondary orbital interaction only plays a minor role when sterics is present.

With respect to the second reaction, the endo/exo selectivity is not observable, since there is no substitution at the 1- or 4-positions on the diene. The difference between the product you predicted (S,R) and the shown (R,S) is that they are enantiomers. The product of two achiral reactants must either be achiral or a racemic mixture.