text:
Why do nitro groups confer explosive tendencies?
When suggesting nitration of an aromatic compound in the synthesis of some organic molecule, it was raised that this route should be avoided as to prevent things from going "ka boom."
An explanation was not forthcoming.
So, why do nitro groups tend to make organic molecules explosive? Is it because the $\ce{NO2}$ group really "wants" to be $\ce{N2}$ since a) diatomic nitrogen has a super high bond strength and b) diatomic nitrogen is a gas usually and therefore conversion to diatomic nitrogen would be entropically favorable? Is it because carbon's most oxidized state is also a gas and the $\ce{C=O}$ bond isn't a wimp either - hence the reason many organic compounds are flammable?
Also, what happens first in an explosion? Does carbon become oxidized first, which then provides the activation energy necessary to decompose the nitro group? I ask this because organic carbon compounds by themselves which are flammable - i.e. toluene - aren't considered explosive, but trinitrotoluene is an explosive.

Explosives chemistry is a rather complex topic. I've heard that this book is a good source of information about it (I haven't read it).
In a nutshell, your intuition about the nitro group is accurate. Formation of $\ce{N2}$ is highly energetically favourable. To get an explosive, what we need is a rapid reaction that produces a lot of heat and gas to both cause the obvious effects of an explosion and to propagate the reaction to other molecules of explosive. The rapid part is what separates something like TNT from toluene. The combustion of toluene is also energetically favourable, but in TNT, to create $\ce{N2}$ is not dependant on mass transport of any other species and can thus happen very rapidly, whereas the combustion of toluene is limited by how quickly oxygen is transported to it. If one were to vaporize toluene in the correct concentration in air, this transport problem goes away and an explosion can occur. TNT also has an advantage in this regard because the nitro groups provide a source of oxygen to react with the carbon and nitrogen remaining (not enough for all of it, but it helps). Explosives are often mixed with fuels or oxidizing agents to produce a more oxygen-balanced mixture for a more efficient explosion.
As for what happens first in an explosion, many different reactions can occur during an explosion, but it may be helpful to consider what it takes to actually detonate TNT and think about the timescale of the reactions. TNT is a solid at room temperature and has a flash point of 163 °C making it difficult to even ignite, and while it will burn in a fire, there is no risk of explosion. For an explosion to occur, enough gas and heat has to be produced to propagate the reaction through the bulk of the material. In practice this is done using a much more sensitive explosive (other explosives like lead azide or nitroglycerine are unstable enough to be set off by heat or pressure) to produce a small shockwave that provides the activation energy to initiate a reaction as it travels through the explosive which then sustains the shockwave through the rest of the material. In a normal explosive (not a fuel-air explosive or the like), the reactions that contribute to the bulk of the explosion are limited to what the explosive is made of because the speed of the explosion is too fast for air to play much of a role initially.
In the case of TNT, the experimentally-measured time it takes for a shockwave to pass through is 100–200 fs (no idea how one measures that), so any oxidation of the carbons seem unlikely to contribute much to the the initial explosion, given the only readily available source of oxygen is from the nitro groups which must presumably decompose first. This group proposed a few decomposition pathways for TNT, including homolytic cleavage of the $\ce{C-NO2}$ bond, rearrangement from $\ce{C-NO2}$ to $\ce{C-ONO}$ followed by homolytic $\ce{O-NO}$ cleavage, and $\ce{C-H}$ attack from an adjacent nitro group to the methyl ring substituent, but they found that only the first was fast enough to occur during detonation, the others possible only for lower temperature thermal decomposition. This initial decomposition step is the only thing fast enough to contribute to the shockwave that sets off the rest of the TNT, while the reactions that produce the final products occur (relatively) long after the initial blast has initiated the rest of the explosive.

What may make things go kaboom without true explosion
First of all, whenever you have a mixture of an oxidizer and a fuel, whatever nature, there is always risk of fast combustion.  Common gunpowder is a mixture of solids, and it can go kaboom when confined. Since nitrogroups are oxidizing, while the rest of the molecule is usually reducing, many nitrocompounds are capable of combustion without external oxidizer. The process may proceed without true explosion, for example, TNT in small portions may be safely ignited. It burns with bright yellow flame producing a lot of black thick smoke. Anyway, to produce a kaboom, the compound or mixture must have a way of decomposition producing a lot of energy.
Self-sustaining thermal decomposition, say of hydrogen peroxide, may produce large volumes
of gases, and in case the reaction mixture is confined, the build-up of pressure may eventually lead for the reaction container to... fragment violently.
Detonation wave
A true detonation is described as a process, where instead of burning front, proceeding with subsonic speed, a supersonic detonation front is observed. The speed the front travels with may be above 10 km/s in some solids and over 3 km/s in gases. This implies that  the process does not involve diffusion of active species or thermal energy, that travel usually with subsonic speed. So decomposition in detonation wave is pressure-induced, either from adiabatic heating or from direct mechanical stress. To achieve this, the compound must be able of exothermic decomposition, producing enough energy to sustain the pressure wave.
Nitrogroup role
Commonly utilized nitro-explosives are relatively stable and often require a significant primary explosive charge to induce detonation wave. Nitrogroups, despite having positive energy of formation, are relatively durable groups, with high barrier or bond dissociation energies, double so for nitroaromatics. This means, that in detonation wave decomposition of nitrogroups mostly consumes energy, while formation of other molecules, say, water and carbon monooxide, produces energy. So, the role of nitrogroup is mostly to provide a readily available oxygen for oxidation of the rest of the molecule.
Other cases
However, there is a lot of explosives (most commonly with high amount of nitrogen), that have comparatively low decomposition energy, but are extremely sensitive. Probably, ones of the most infamous would be heavy metal azides, like lead azide. There in addition to readily available exothermic decomposition route, usually some weak bonds are present in the molecule, allowing easy thermal fragmentation, following by rearrangement.
For example, infamous diazomethane have C-N bond energy around 172 kJ/mol, while energy of $\ce{CH2=CH2}$ bond is 611 kJ/mol,  meaning that the process produces roughly twice more energy than consumes - not a lot, but the very low initial energy investment makes diazomethane to explode from a funny look.
TL;DR.
nitrogroups mostly serve as a readily available source of oxygen to oxidize the rest of the molecule


text:
Reuse 20 litre plastic drums for drinking water?
Would you consider it safe to clean and reuse a 20 litre plastic drum that contained 36% Hydrochloric Acid for drinking water? If so how would you recommend cleaning it?

Regardless of what it takes to remove any traces of hydrochloric acid from the drum, the container is probably not certified as food-grade. So in addition to any risks from the acid itself (including things like heavy metals, which are common contaminants in mineral acids), the plastic is not certified to be free of toxic substances left over from manufacturing or that the plastic will not degrade to release such substances over its lifetime.

I imagine the container is made of polyethylene or polypropylene (it should be marked with a numeric code). If that is the case, I have a hard time accepting that there is any downside to using it for drinking water. Chemists don't want heavy metals contaminating their hydrochloric acid, so I think you are OK on that count as well.

Have a look at this—1 ppm of Pb. Acids dissolve a lot of heavy metals and they aren't a problem for most uses and are expensive to remove so you'll find that most mineral acids have a fair bit of heavy metals unless they're the 99.999% assay or semiconductor grade. As far as reusing chemical containers go, it's probably fairly safe. Would I do it in a survival situation without a better option? Yes. Would I take the risk of contaminating drinking water for myself or others so I don't have to buy a new container? No.

Hydrochoric acid is stomach acid.  Any remaining acid can be flushed out with 3 rinses with water to pretty much undetectable levels.  If the acid was intended for food production then the container is totally safe.  If it was intended for cleaning your driveway the quality of the container may be less than ideal.  Remember that the acid will have cleaned the container pretty good and not much dangerous stuff will dissolve later into water.

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text:
Resonance in Propenal
I do not understand why the 3rd canonical form of propenal has the oxygen atom donating an electon.
Oxygen in electronagetive so instead of taking electrons why is it giving?
Maybe this why it is most unstable but how is it even formed ??

What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals.
In the case of propenal the $\pi$ MOs result from the overlap of four $p$ orbitals on the four atoms in the chain. The two occupied $\pi$ MOs show how most of the electron density is between C1 and C2, and C3 and the oxygen but there is also significant electron density between C2 and C3. This reflects the different positions of the double bonds shown in the resonance contributors of propenal.

http://www.chemtube3d.com/orbitalsacrolein.htm

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text:
Draw the four isomers of C2H2Cl2Br2
I have to draw the 4 four isomers for the molecule  $\ce {C2H2Cl2}$  when $\ce {Br2}$ is added to it. At last the four isomers for  $\ce {C2H2Cl2Br2}$.
I have found three of them:

I can't find the last one.

We would need to know what CH2Cl2 is before we can really answer.  Is it geminal, cis or trans?

@Lighthart The first structure is from geminal, the second is from cis and the third is trans (or do I have cis and trans backwards). The only other isomer I can think of is 1,1-dibromo-2,2-dichloroethane, but it is not made from adding Br2 to dichloroethene.

I understand which is which, but the OP does not indicate which molecule is his reactant (or that he has a mix of isomers as reactants

You have only three isomers with the configurations (1R,2R), (1S,2S) and (1R,2S). The one with the configuration (1R,2S) is achiral, as it has a symmetry plan, and it's superimposable on its image by a mirror (i.e. it is identical to its enentiomer (1S,2R)).
This isomer is known as "meso compound".
Meso compounds are a special case of stereoisomers which occur when a molecule has two chiral centers and each chiral center has identical substituants.

The other are too meso compounds.

No, because they don't have a plan of symmetry like this one. Mine is the same as the one you have drawn on the right. We have drawn two conformers of the same molecule.

Like Lighthart already mentioned in the comments, it is important to know which compound was used in the addition reaction. Because of the reaction mechanism, only certain products are formed, this is very specific. LDC3 also explained that in the comments, but let me dissect this a little further.

The first structure is from geminal,

In the scheme below, this is reaction $(1)$ and it forms an achiral product.

the second is from cis

This is depicted in $(2)$ and it forms the meso compound described in Yomen Atassis's answer.

and the third is trans (or do I have cis and trans backwards).

(No you have not.) This is depicted in $(3)$, but in this case a racemic mixture of the (S,S)- and (R,R)-product is formed, depending on which carbon atom in the bromonium ion the nucleophilic attack takes place.

The only other isomer I can think of is 1,1-dibromo-2,2-dichloroethane, but it is not made from adding $\ce{Br2}$ to dichloroethene.

For completeness, this is depicted in $3$.
I could further think of another isomer, which is like the above not formed via addition, it is depicted in $(5)$. Note that these last two reactions also yield achiral products.


text:
What properties of an element determines the maximum number of bonds it can make?
From my conjecture, I think it's a mix between valence electron and the principal energy state, but I'm not sure.
For example, C, N, O and F can only make a maximum of four bonds, as they can only have up to sp3 hybrid orbitals.
For Si, P, S and Cl, they can make a maximum of nine bonds, as they can have sp3d5 hybrid orbitals.
That's what I think, but I'm probably incorrect in some aspect.

You should expand on your statement since it doesn't provide your reason.

LDC3 is right. I can't see how these should get "mixed".

For Si, P, S and Cl, they can make a maximum of nine bonds, as they can have sp3d5 hybrid orbitals. || nope, they can have 4 bonds at most as well.

@permeakra Does $\ce{PCl5}$ have only 4 bonds?

@LDC3 Yep. 3 single bonds and two half-bonds.

What are half bonds?

@LtotheV $\ce{PCl5}$ molecule can be very roughly described as resonance hybrid (i.e. half-sum in this case) of two symmetrical structure. Three Cl atoms are contently bound to P atom, forming a flat triangle with the P atom in the center. one of the remaining electrons of P atom is transferred to one of the remaining Cl atoms, while another electron is used to form a covalent bond with remaining chlorine atom. Another resonance structure is a mirror opposite of described one, with roles of two last Cl atoms reversed.

What would be the molecular geometry of PCl5 in your case? From my textbook, the geometry would be trigonal bipyramidal.

PCl5 has 5 bonds.  Not two half-bonds and three others.

PCl5 exhibits three center two electron bonding

@Dissenter Are you sure it is not 3-center, 4-electron bonding? http://link.springer.com/article/10.1007/s00897010525a

@DavePhD You're correct; I misspoke. Good catch!

Considering the number of other atoms a central atom can be bonded to, "hybridization [is] irrelevant" and "the coordination number ... is not restricted by the number of
hybrid orbitals that the atom can form", quoting On the role of d orbitals in sulfur hexafluoride J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593.
Examples such as $\ce{CLi6}$ and $\ce{TeF8^{2-}}$ are cited and emphasis is placed upon the ionic character of the bonding by the cited reference.
In my opinion, the main limiting factor is how many atoms can simultaneously be close to the central atom, this increasing with increasing radius of the central atom and with decreasing radius of the surrounding atoms.

+1 Agreed. There are plenty of examples with known crystal structures where there appear to be 5, 6, or even 7 interactions between Carbon atoms and neighboring ligands. Certainly 3rd row TM and actinide complexes more readily show large coordination numbers.


text:
Which formula is correct for calculating the heat of dissolution?
I want to calculate the energy change when a solute is dissolved in water. I know that I can achieve this by using the equation $q=mc \Delta T$. My question is does the mass ($m$) change with the amount of solute added to the solvent?
For example, if I had 25g of water ($c=4.18$) as my solvent and 5g of sodium chloride as my solute. Would the energy change equal:
$$q=25 \cdot 4.18 \cdot \Delta T$$
or
$$q=(25+5) \cdot 4.18 \cdot \Delta T$$

Use 30 grams as your mass since the heat capacity of a sodium chloride solution is about the same as the heat capacity of water.  You could use physical chem as a tag.

For a rough estimate you can ignore the mass of the solute if it is relatively small compared to the mass of the solvent (probably not in this case but oh well).

Your energy conservation equation is $$q_{rxn} + q_{soln} = 0$$ so $$q_{rxn} = -q_{soln} = -m_{soln}c_{soln}\Delta T_{soln}$$ Note the minus sign; if $\Delta T$ is negative, the energy will be positive, and vice versa. And yes, you need the mass of the solution here, and the specific heat capacity of the solution too.
If you only need a couple of significant figures and the solution isn't too concentrated, you can assume $c_{soln} \approx c_{\rm H_2O}$.
If you need more precision, though, you'll need to measure or look up $c_{soln}$. And you'll need to include the heat absorbed by the calorimeter in your energy conservation equation, too.

You have to take the mass of the mixture. So, the second expression is true.

but you also have to take into account the different specific heat capacity of the solution to water and its temperature dependence

You are right. But, generally, we assume that the  specific heat capacity of the solution is  the  specific heat capacity of water.


text:
Why do heavier elements have smaller specific heat values.
In the periodic table the trend is that as the mass of each element increase's the specific heat tends to go down. This seems to be counter-intuitive because if I am not mistaken as mass increase's the amount of heat that the substance can hold goes up. So why does the specific heat of these elements go down?

There is no reason to assume specific heat is a nuclear property.

The more particles in a system, the more modes of vibration (and rotation), so one gram of $\ce{H2}$ has about $235 \times$ more particles than one gram of U. The heat is "stored" in the motions of the atoms and electrons, until reaching millions of kelvin, at which point dissociation of the nucleus would be significant in specific heat. See these links and  for more information.

As the amount of a substance increases the Heat capacity of that substance does go up. However, Specific Heat is a correction for this because it is the Heat Capacity divided by the amount of the substance.
For Specific Heat the amount of a substance is irrelevant and its value is largely determined by the degrees of freedom for the constituent particles. The more degree of freedom, other than transnational, the higher the Specific Heat. An example of another degree being rotational energy which does not contribute to Temperature changes.

This doesn't say anything about the trends in the periodic table, which is what the question is about.

This post does not answer the question and it is also incorrect since specific heat capacity c is heat capacity C divided by mass m (not heat capacity C divided by amount of substance n).

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text:
Difference between polypeptides and polyamides
I can't seem to find a very good explanation of the difference between polypeptides and polyamides. So far, I know that polypeptides are a type of polyamides, but that is about it. My guess is that the term polyamides refers both to the polymers made from amino acids and those made from diamines and dicarboxylic acids, but polypeptides are only those made of amino acids, but this is just a guess...
Thank you in advance!

I guess Polypeptides are naturally occurring for eg. Polymers of Amino acids whereas Polyamides are synthesize by condensation process of polymerization for eg. Nylon 6 and nylon 6.6.
Polypeptides are a type of Polyamides. It's only guess

Your guess is correct. Polyamides are polymers where the repeating units are linked by amide bonds. Polypeptides are a specific type of polyamide where the repeating units are amino acids.

Peptides (or polypeptides) are polymers of amino acids; usually created by cell enzymes.
Polyamides usually have a carbon chain between the amides, such as nylon 6 or nylon 66. Peptides could be classified as polyamides, but are not usually included since they have their own group.

Technically, polypeptides are a subset of polyamides.

The idea of a "mer" unit is polymer chemistry and polyamides (e.g. kevlar) are common. They are also liquid crystals. A polyamide is anything that has repeating amide bonds. Any amide (diamide to be precise) can be made into a polyer.

From Organic-ese.com


text:
Why do molecular orbitals in solids merge to bands?
Why do molecular orbitals in solids merge to bands?
For example: In silicon every atom is sp3 hybridised, but when I merge two of these orbitals then it yields a bonding and an antibonding MO. When a third atom binds to one of the $\ce{Si}$ atoms in this configuration, then again we should get two MOs, separated by the same energy. This way we we don't get so called bands. How do the bands form?
I have tried wikipedia and other sites but they answer in brief.

Let's explain this with your example of silicon that every atom is sp3 hybridised. when you merge two of these orbitals, then it yields bonding and antibonding MO. But  when a third atom bounds, it yields  3 MO (not four as you mentioned) bonding, antibonding and non bonding MO. The number of MO equals the number of atomic orbitals constituting them, according to LCAO theory.
The energy of the third MO is in between the nonbonding and anti-bonding MOs. When you add a forth, a fifth and so on atoms, you will have more and more MOs very close in energy and you will form a band.
This is a qualitative description about the formation of a band.

I'll give a quick answer first and come back tomorrow for a more complete version.
In your question, you start with two $\ce{Si}$ atoms and then a third. As you combine these atoms into pairs and molecules, chemistry stresses that the atomic orbitals combine in different ways.
So let's take $\ce{Si2}$. We know that for the $3p$ orbitals, some will become more stable and some will become less stable. There are 3 for each $\ce{Si}$ atom, so a total of 6 molecular orbitals due to combinations of $3p$. (Actually, the $3s$ are also involved, and there's some amount of s-p mixing, but let's ignore that for now.)
Take-home message: when we combine atomic orbitals into molecules, some orbitals become more stable and some orbitals become less stable.
OK, now let's add a third $\ce{Si}$ atom as you said. I'm not going to derive the MO diagram right now. (I'll do that tomorrow and insert the picture.) The main point is that we'll have new combinations. If I take $\ce{H3}$ as another example, I should have 3 molecular orbitals.
As I grow into a linear chain, a 2D sheet, or a 3D solid, I don't have some small number of atoms like 2, 3, 4, 20, etc. Instead, I have 1,000 or 10,000 or many, many more.
Well, those atomic orbitals combine just like they do when you create molecular orbitals. Some become more stable, some become less stable. Yet, rather than having 3 molecular orbitals in $\ce{H3}$ from 3 atomic orbitals, we have 1,000 or 10,000 or many, many more.
Since there are so many orbitals, the energy spacing between them is infinitesimal. These are now called bands.
The type of material (and the atomic composition) will dictate the energy levels of the bands and the so-called "band structure" (i.e., the solid-state equivalent of molecular orbital diagrams). But simply having hundreds and thousands of atoms will give you bands.

This is explained by the Pauli exclusion principle, as applied to conductors: quantum mechanics forbids that fermions have exactly the same spin-state and energy level. In a single atom, that means, for example that H has one s electron in the "up" state and anther in thew "down". In conductors (and semiconductors), electrons from one atom can exchange with those of another, so they must have (slightly) different energy levels, if all spin-states are occupied, effectively splitting the single level into a "band" of energies. See http://en.wikipedia.org/wiki/Pauli_exclusion_principle for more detail.
BTW, this is not true for "solids" in general, e.g. sulfur or window glass, but is true for metals, alloys (e.g. brass) and conductive compounds.


text:
Naming rules and acidity of Cl2CHCH2COOH
How do we name $\ce{Cl2CHCH2COOH}$ ?
Is seeing branched chain our number 1 priority so we name it
1,1-dichloro-3-propanoic acid?
also
Which is more acidic $\ce{Cl2CHCOOH}$ or $\ce{Cl2CHCH2COOH}$ ?
I feel it should be the first one as there is more −I effect but not sure.

Your name is not correct. The carboxylic acid has a higher precedence than chlorine, so your numbering is backwards.

@LDC3 even i thought it should be like that but my book says that more branching is given more priority.

I'm not sure what you mean by more -I effect. How about providing me with more information?

Ok it basically means that it chlorine is electron withdrawing so it is more effective in reducing negative charge on cooh near it so h+ ion is given easily

Since there is an intervening carbon in the second compound, the chlorine atoms have very little effect on the acidity of the acid.

@user166748 It is preferred if you can post separate questions instead of combining your questions into one. That way, it helps the people answering your question and also others hunting for at least one of your questions. Thanks!

1- 3,3-dichloropropanoic acid. Please see http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/
2- The first one  is more acidic, as the inductive effect, due to chlorine atom decrease fast with the distance. (So, your answer is correct!).

With regard to numbering of locants, the principal characteristic group (the characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name, e.g. propanoic acid) has seniority over simple substituent groups (e.g. chloro). On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Note that Rule c takes precedence over Rule f.
Therefore, the preferred IUPAC name for the given structure is 3,3-dichloropropanoic acid (not 1,1-dichloro-3-propanoic acid).

As you have correctly assumed, additional chloro substituents increase the acidity of the carboxylic acid due to their electron-withdrawing inductive effect (−I effect):
  acetic acid: $\mathrm{p}K_\mathrm{a} = 4.756$
  2-chloroacetic acid: $\mathrm{p}K_\mathrm{a} = 2.87$
  2,2-dichloroacetic acid: $\mathrm{p}K_\mathrm{a} = 1.35$
  2,2,2-trichloroacetic acid: $\mathrm{p}K_\mathrm{a} = 0.66$
The acidity of chlorocarboxylic acids is hardly directly affected by the length of the carbon chain:
  2-chloroacetic acid: $\mathrm{p}K_\mathrm{a} = 2.87$
  2-chloropropanoic acid: $\mathrm{p}K_\mathrm{a} = 2.83$
  2-chlorobutanoic acid: $\mathrm{p}K_\mathrm{a} = 2.86$
However, the effect decreases as the distance between the chloro substituent and the carboxyl group increases:
  2-chloroacetic acid: $\mathrm{p}K_\mathrm{a} = 2.87$
  3-chloropropanoic acid: $\mathrm{p}K_\mathrm{a} = 3.98$
  4-chlorobutanoic acid: $\mathrm{p}K_\mathrm{a} = 4.52$
($\mathrm{p}K_\mathrm{a}$ values taken from “Dissociation Constants of Organic Acids and Bases”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)


text:
What is the unit of molar conductivity?
Could you explain what is unit of  molar conductivity along with derivation

Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution.
It's unit in S.I. is: $$\ce{S.m^2.mol^{-1}}$$

What is "S" over here?

S is Simens. It equals $Ohm^{-1}$

Is simens an S.I unit

Yes, as $Ohm$ is an derived S.I. unit.

unit of conductivity of an electrolyte solution =
(unit of length)/(unit of area) * (unit of Conductance)
= (m/m2). S
= S . m-1
unit of molar conductivity = (unit of conductivity of an electrolyte solution  ) / (unit of molar concentration of the electrolyte)
 = ( S m-1) / (mol m-3)
 = S . m . 2 mol-1

This does not look like a full answer yet. Also, please check out what you can find on MathJax in the [help] and this meta-post and the links therein to learn how to format your post with MathJax; it would improve the look of the answer =)

It does look odd that the numerator is expressed in units of conductance times units of area. Conductivity in Siemens per meter divided by concentration in, say, moles per liter should give you something in units of $S \cdot l \cdot mol^{-1} \cdot m^{-1}$ .  But since a liter is simply a volume measurement equal to 1000 $cm^3$ or 0.001 $m^3$, the volume/distance can be cancelled out to just area. So you end up with that $S \cdot m^2 \cdot mol^{-1}$.
Sometimes you see tables that contain molar conductivity values in units of $ 10^{-4} S \cdot m^2 \cdot mol^{-1}$. This makes it handy because the $10^{-4}$ factor gives you the same value as when expressed in units of
$ \mu S/cm $ per $  mmol/l $.
E.g, the specific conductance for NaCl is:
$ 0.0126 \space S \cdot m^2 \cdot mol^{-1}$
$ 126 \times 10^{-4} \space S \cdot m^2 \cdot mol^{-1}$
$126$ $(\mu S / cm) \space \over \space (mmol / l) $


text:
What is "heating under reflux"?
When dealing with organic reactions, heating under reflux is often required, such as the oxidation of Toluene using acidified $\ce{KMnO4}$ and dilute $\ce{H2SO4}$ to Benzoic acid($\ce{C6H6O2})$
What does this mean and how does it differ from "normal" heating?

Many organic reactions are unreasonably slow and can take an extended period of time to achieve any noticeable effect so heating is often used to increase the rate of reaction. However, many organic compounds have low boiling points and will vaporise upon exposure to such high heat, preventing the reaction from proceeding in full.
To address this, heating under reflux is used. This refers to heating a solution with an attached condenser to prevent reagents from escaping.

As seen above, any vapor will condense on the cool surface of the attached condenser and flow back into the flask.
The hot water bath pictured is an optional component of heating under reflux and is usually only used for particularly sensitive reactions. Also, using it limits the reaction temperature to 100 degrees Celsius.

Note: water bath on the picture is optional and usually not used except for particularly sensitive reactions.

You can alwo use an oil bath if you need to reach higher temperatures. Otherwise, a heating mantle is simply used.

the liquid doesnt always condense back into the flask, another type of condenser can be attached so that the now condensed desirable liquid can flow into a seperate beaker or flask. this can be used later in the experiment

@user16616 Isn't that just distillation?

Two comments: 1: In organic labs, not using an oil bath for heating is not an option (you get much better temperature control). 2: Attach cold water to the top of the condenser for a proper counterflow effect!

@Jan For the simple straight-type condenser shown in the picture, the indicated water flow direction from bottom to top is correct in order to ensure that the condenser is always completely filled with cooling water – even at low cooling water flow rates. Certainly, however, this condenser type  is not ideal for heating under reflux, because of the small heat exchange surface area and in particular because of the missing counterflow, which you correctly highlighted.

Use of an air condenser is the preferred option in many labs now for overnight refux. This minimises the risk of the water supply failing and your reaction running dry.

@Jan Bro don't try to be a chemical engineer when you aren't one. You are wrong on so many levels here I am not even gonna try

@PulkitSharma If you have something constructive to say, please say it and don't allude to it hand-wavingly.

@Jan First thing any engineer is taught in the plant is that when a vessel vertical has to be used as a heat transfer media, Gas is fed at the top if used and liquid at the bottom. This way, it is ensured that the heating media is completely filled with the fluid at all times.  Co current and counter current are usually defined in the context when gravity is not at play. In a condenser if you feed water at top, it will drip to bottom along a thin layer and not cause the desired cooling. I said it hand wavingly because anyone who dares think about co/counter current should realize that

@PulkitSharma Thank you. You will find that user7951 already stated essentially that.

Temperature control is important for reactions, especially in organic chemistry. Some reactions are strongly exothermic or have notable side-reactions that can be suppressed at a low temperature. For others, assuming all reactants survive the temperatures in question, van ’t Hoff’s rule dictates that an increase of the temperature by $10~\mathrm{^\circ C}$ increases the reaction rate by a factor of $2$ to $4$. Thus, increasing the temperature is often favourable.
Almost all organic reaction are conducted in a solvent. The choice of solvent dictates the temperature range you can reach; e.g. tetrahydrofurane solidifies at $-108.4~\mathrm{^\circ C}$ and boils at $65.8~\mathrm{^\circ C}$, so any reactions will have to take place at in-between temperatures.
Often, a published reaction will have a set of conditions most likely to work; they typically come with a preferred solvent and a preferred temperature. A Dess-Martin oxidation is typically conducted at $0~\mathrm{^\circ C}$ in dichloromethane. For many reactions, the preferred temperature coincides with the solvent’s boiling point — it means that maximum heating is required to conduct the reaction in that solvent. When heated to the boiling point, the solvent will partially evaporate and recondense on colder surfaces. But since also the concentration of reactants is important, one typically wants to recollect the evaporating solvent.
This is where heating under reflux comes into play. Reflux is the term used to mean ‘letting a solvent boil and collecting its vapour in some kind of condenser to let it drip back into the reaction vessel.’ The most common type of condenser I have encountered for refluxing is the Dimroth condenser as shown in the image below (taken from Wikipedia, where a full list of authors is available).

It is important to connect the cooling water circuit correctly. For some reason, most images found on the internet, including the one in the other answer, suggest a suboptimal cooling. The most optimal cooling efficiency is given in a countercurrent setup. To quote Wikipedia:

The maximum amount of heat or mass transfer that can be obtained is higher with countercurrent than co-current (parallel) exchange because countercurrent maintains a slowly declining difference or gradient (usually temperature or concentration difference). In cocurrent exchange the initial gradient is higher but falls off quickly, leading to wasted potential.

Thus, in the image above, the water supply should be connected to the top connector while the bottom one should be used as water outlet. This allows the strongest cooling efficiency to be at the top of the condensor, which is important, because if vapour manages to get that high up it needs a quick and efficient cooling.

I agree that countercurrent is desirable. However, in some condensers downward flow may mean that the water does not remain in contact with the whole surface. See, for example, the following video at about 1:00 - the cooling water is initially connected downward, and then corrected to an upward flow. https://www.youtube.com/watch?v=h54XyEnYZDA

Where has this downward water flow come from? First chose the proper condenser then adjust the water flow and the heating rate to give condensation where appropriate. The purpose of using reflux is to contain solvents and possibly reactants and to control T; there is no reason to flood a condenser besides it is dangerous.


text:
Why does screening effect decrease due to d-orbital?
In 13th group, atomic radius increases from boron to aluminium. From aluminium to gallium, atomic radii decreases. From gallium to indium, atomic radii increases. And from indium to thallium, atomic radii decreases.
The reason for this irregular trend given is screening effect. While explaining my teacher told that due to d-orbital screening effect decreases.
I can't get why?

The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals.  This is because of something called penetration .
The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals.  In gallium, you have $10$ electrons in the filled $3$d-subshell, and each of these electrons is doing a slightly worse job (relatively speaking) of screening the nuclear charge than the electrons in the s and p orbitals. Therefore, the effective nuclear charge in gallium is slightly higher than that in aluminum, so the increase in the radius is a quite a bit smaller than would be expected based on the difference between boron and aluminum, or gallium and indium.
The trend goes:

$\pu{82 pm}$ ($\ce{B}$)
$\pu{118 pm}$ $\ce{(Al)}$
$\pu{126 pm}$ $\ce{(Ga)}$
$\pu{144 pm \ce{(In)}}$
[covalent radii from www.webelements.com].

This effect is generally known as the d-block contraction.  (It can be more or less pronounced depending on how you define the atomic radii.)
A similar thing happens (in principle) when you go from indium to thallium; except in this case you are now dealing with adding a filled f-subshell to the valence shell.
Electrons in f-orbitals are even worse at screening nuclear charge than those in d-orbitals, therefore again, the effective nuclear charge in thallium is a bit larger than it is in indium, so again the jump in radius is fairly small (from $144$ to $148$ pm).  This effect (of the filled f-subshell) is generally known as the lanthanide contraction.

+1 I was always skeptical of this explanation (I thought relativity may be more important), but I found a reference in support of your answer here: https://en.wikipedia.org/wiki/Lanthanide_contraction. I thought of sharing it here too for others like me!

The exact atomic radius values are:
$$
\begin{array}\
\text{Element}&\text{Atomic radius}\
\ce{B}&\pu{85 pm}\
\ce{Al}&\pu{125 pm}\
\ce{Ga}&\pu{130 pm}\
\ce{In}&\pu{155 pm}\
\ce{Tl}&\pu{190 pm}
\end{array}
$$ The values you mention are covalent radii (https://www.webelements.com/boron/atom_sizes.html), but question is talking about atomic radii instead. If I just change the values, your answer would not make perfect sense. Can you please confirm/update?

Naively, one should expect the atomic radii to increase in a proportional fashion from boron to aluminium, gallium, indium and thallium, because each of these atoms has one more populated shell.
However, the difference is not proportional because there is not a constant proton difference between the elements. From boron to aluminium 8 protons are added to the nucleus; from aluminium to gallium and from gallium to indium 18; and finally, from indium to thallium 32. Wherever there are more protons than expected, orbitals contract; thus, both gallium and thallium are smaller than one naively might assume.
Of course, one should not forget the additional electrons. However, these are added to the d orbitals (and in thallium’s case, f orbitals). These orbitals are much higher in energy than the corresponding shell’s s and p subshells (as it is evident when these orbitals are populated in the periodic table). Furthermore, these orbitals are associated with two (d) or three (f) nodal planes through the nucleus, meaning that the corresponding electrons are further away from the nucleus on average. Thus, their shielding is less effective than that of s and p orbitals.

Hmmm... But the inorganic textbook I read states that nd electrons are usually much more core-like than (n+1)s electrons and then state that it is observed that the (n+1)s and (n+1)p electrons are much more involved in bonding.

Here's an explanation with the help of Slater's rule:

Group 1
Group 2
Rest

s, p
0.35
0.85
1

d, f
0.35
1
1

As you can see from the table, electrons in the same subshell/group contribute only 0.35 screening while the inner electrons do more screening.
The number of electrons that can be accommodated in a subshell increases as s<p<d<f, thus the screening effect decreases as: s>p>d>f.

d and f orbitals experience more  attraction towards nucleus due to lesser shielding effect than s and p orbitals. This is why the ionization energies of d and f block elements are high and so their electronegativities.


text:
IUPAC naming - choosing the main branch
What are the rules for choosing main branch?
Wikipedia states:

Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:

It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.
It should have the maximum number of multiple bonds
It should have the maximum number of single bonds.
It should have the maximum length.

I don't understand how 3 and 4 differ. Which single bonds are we counting? Only carbon-carbon single bonds in the main chain, right?
Also, I heard there is a rule that says if you've two equal length chains, then choose the more substituted one. Why is it not given here?
And now now imagine you've two choices for a chain, like this:

The chemdoodle program gives the name as 2-(1-Hydroxybutyl)-3-methylhexanoic acid .
I simply do not understand why the methyl substituted chain is chosen instead of hydroxy substituted one and how that fits with IUPAC rules given in Wikipedia.

For the given structure, the principal chain could be a 3-hydroxyhexanoic acid or a 3-methylhexanoic acid; i.e. the name of the structure could be 3-hydroxy-2-(pentan-2-yl)hexanoic acid or 2-(1-hydroxybutyl)-3-methylhexanoic acid.

The IUPAC recommendations on this matter have been changed.
According to the old IUPAC recommendations (1979, retained in 1993), the principal chain was the 3-hydroxyhexanoic acid. Thus, the preferred name was 3-hydroxy-2-(pentan-2-yl)hexanoic acid.
However, the relevant rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-45 SELECTION OF PREFERRED IUPAC NAMES
(…)
P-45.5 CRITERIA RELATED TO ALPHANUMERICAL ORDER OF NAMES
P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (…). Alphabetic letters are considered first in the order that they appear in the name; (…)

Therefore, the preferred name is 2-(1-hydroxybutyl)-3-methylhexanoic acid (not 3-hydroxy-2-(pentan-2-yl)hexanoic acid) since ‘hydroxybutyl…’ is earlier alphabetically than ‘hydroxy-pentan-yl…’.

For point 3 carbon-carbon single bonds count. Heteroatoms would build up a functional group. However, I could not think of a scenario where point 4 is needed. It should be a scenario where two chains with an equal amount of multiple bonds and single bonds are present but different in length.
The rule, which you mention, in which the number of functional groups count is point 1.
According to what I have learned the name "2-(1-Hydroxybutyl)-3-methylhexanoic acid" is wrong. There are two equal chains, one has a methyl and one a hydroxyl group. According to IUPAC nomenclature of organic chemistry the group with the higher precedence is hydroxyl. Therefore, chemdoodle gave you the wrong name.
This is a case where this sentence applies which they put above: "Please give it a try and let us know if you encounter any issues."


text:
Are these reaction equations for the formation of the brown ring complex correct?

\begin{align}
\ce{NO3- + 3Fe^2+ + 4H+ &-> NO + 3Fe^3+ + 2H2O}\tag{1}\\
\ce{[Fe(H2O)6]^2+ + NO &-> \underset{\text{(brown)}}{[Fe(H2O)5(NO)]^2+} + H2O}\tag{2}
\end{align}

The above are the reactions given in my book [1] for the qualitative analysis of nitrate ion (formation of the brown ring complex). I think that the reactions above are incorrect. The first one shows formation of $\ce{Fe^3+}$, which should be attacked by excess of $\ce{H2O}$ to give ferrum(III) hexaaquasulphate, not ferrum(II) hexaaquasulphate. Now that should be attacked by $\ce{NO}$ as a ligand and give $\ce{[Fe(H2O)6NO]SO4}$, where the oxidation number of $\ce{Fe}$ is being $3+$.
Here the reactions show that the $\ce{Fe}$ already present gives $\ce{[Fe(H2O)6]^2+}$ and then $\ce{NO}$ acts as a neutral ligand to form the complex. But by the reactions shown in the book, even after formation of the brown ring complex $\ce{Fe^3+}$ is still present in the solution, which according to me shouldn't be because if $\ce{Fe^3+}$ still remains in the solution then water should also have attacked it as a ligand to give $\ce{[Fe(H2O)6NO]SO4}$.
I need some clarity on this. Are the reactions given in the book correct?
Edit: After reading the answers, I do get that the hexaaqua iron(II) will be formed and the $\ce{Fe^3+}$ will be attacked by water to give $\ce{[Fe(H_2O)_6]^3+}$ but why won't $\ce{NO}$ take electrons from the oxidation of iron ($\ce{Fe^2+ -> Fe^3+ + e- }$), make $\ce{NO-}$ and attack the above and substitute one water molecule and give more of the brown ring complex?

NCERT. Chemistry: Textbook for class XII; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-648-1.

Wich book is it ?

I second @Babounet's demand, please add the citation to the book.

It is "Chemistry Part I Textbook for Class XII" published by National Council Of Education Research And Training (India).  http://www.flipkart.com/chemistry-class-xii-part-i-english-01/p/itmdyu9bmzghhwn5

According Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of
Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction
Revisited
The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$
For many years it was thought that iron was reduced to Fe(I) and NO oxidized to NO+, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin Fe(III) (S=5/2) antiferromagnetically couples with NO- (S=1) for an observed spin of S=3/2.
Other than the above, I agree with Dhanajay Gupta that the equations in the book are correct, that some of the Fe2+ is oxidized to Fe3+ in the first equation forming NO, and NO coordinates to remaining Fe2+ according to the second equation, but the second equation doesn't specify the charge on Fe or NO, just the overall charge of the complex.
In the first equation in the book, it should be understood that the ions are aqueous.  For example H+ doesn't mean there are bare protons, it is understood to mean $\ce{H_3O+}$ and Fe2+ is $\ce{[Fe(H2O)6]^{2+}}$ and Fe3+ is $\ce{[Fe(H2O)6]^{3+}}$.
In the second equation in the book, the inner coordination sphere is explicitly shown.
In the article cited above the mechanism of the second equation is explained. Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and "This rate-determining displacement of coordinated water is followed by a rapid intramolecular charge-redistribution process to lead to the final $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+}}$ product".

but why won't NO take electrons from the oxidation of Iron ($Fe^{2+}
 \rightarrow Fe^{3+} + e^-$), make $NO^-$ and attack the above and
  substitute one water molecule and give more of the brown ring complex?

To get the electron from Fe2+, NO would enter the inner coordination sphere, and when the electron transfer occurs, the brown ring complex is formed at that Fe.  If the NO- then left that Fe, it would destroy one instance of the brown ring complex.  Hypothetically, the Fe2+ could reduce NO via an outer sphere electron transfer, and then NO- coordinate Fe3+, but this is not what the article reports occurring.

I see what you did here ;) However, it kind of does not answer the question. (Still upvoted though)

@Martin I added to the answer now

Aah i see. My bad then. Thanks a lot for the correction!

@DavePhD, I have made an edit in my question which made me think why $Fe^{3+}$ ion shouldn't be there. Why isn't that being attacked by the water molecule? And now that you have clarified that Fe is in 3+ and not 1+, why should $Fe^{3+}$ be just $sitting$ there and do nothing?

@DhananjayGupta, please see my edit and the comment above ^

@Robin97 The Fe3+ is hexaaquairon(III) and the Fe2+ is hexaaquairon(II).  $\ce{[Fe(H2O)6NO]^{2+}}$ is a 7-coordinate transition state but it is not stable, one of the waters is displaced and $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$ is formed.

I suggest you also look up the concept of spectator ions. Although not applicable here,  it is perfectly possible for a specie to be doing effectively nothing in the course of a reaction.

@DavePhD, I mean that why is that $Fe^{3+}$ in the solution just sitting there. It could have formed $Fe^{III}(H_2O)_6$ and then form $[Fe^{III}(H_2O)_5(NO^{-})]^{2+}$ by NO attacking and taking the place of $H_2O$.

@Rohinb97 It does form $[Fe^{III}(H_2O)_6]^{3+}$, Fe3+ in the first equation in the book should be construed as short-hand for $[Fe^{III}(H_2O)_6]^{3+}$, just as H+ might be considered as short hand for $H_3O+$.  However, you are not accounting for gain of one electron in proposing that $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$ is formed directly from $[Fe^{III}(H_2O)_6]^{3+}$

This is what I was saying the whole time. The electrons from the oxidation of iron should also help NO to form $NO^-$ and then the reaction which I described above should happen. The reactions given in the book does not account for this part. That is why I feel the reactions given are wrong.

@Rohinb97 For Fe to provide an electron to reduce NO to NO-, Fe needs to be Fe2+.  The book says it is Fe2+ that forms the brown complex.  What's wrong with that? and what alternative are you proposing?

@DavePhD I am saying that the first reaction is correct, and then $both$ the remaining $Fe^{3+}$ and $Fe^{2+}$ should form $[Fe(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{3+}$ and then NO should attack as a neutral ligand and as $NO^-$ respectively to give the final complex: $[Fe^{III}(H_2O)_5(NO^{-})]^{2+}$. Basically I just want to add the attack of $H_2O$ on $Fe^{3+}$ formed in the first reaction, and that too should give me the resulting complex.

@Rohinb97 I agree with "both the remaining $Fe^{3+}$ and $Fe^{2+}$ should form $[Fe(H_2O)_6]^{2+}$ and $[Fe(H_2O)_6]^{3+}$", but I don't follow your reasoning concerning NO- being freely present in the solution, how do you propose the NO- forms?

@DavePhD  It should take the electrons from the formation of $Fe^{2+} \rightarrow Fe^{3+} +e^-$ and form $NO^-$ and then attack $[Fe(H_2O)_6]^{3+}$ and give our brown ring complex.

@Rohinb97 that directly contradicts the mechanism described in the article cited in my answer.  Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and "This rate-determining displacement of coordinated water is
followed by a rapid intramolecular charge-redistribution
process to lead to the final $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$ product".

So you are basically saying that $NO must$ attack as a neutral ligand only? But why? I guess I have convinced you why the hexaaqua ferrum(III) should be formed. Can you edit your answer and tell me why $NO^-$ doesn't attack it and make the ring complex? P.S. The comment section is getting too long. To make this question better and to make it useful please edit the answer to answer my question.

@DavePhD Please see the edit to my question and the comment above ^

@Robin97 yes I added to my answer

Since this reaction takes place in aqueous solution, is is implied, that the reaction equations are involving solvated ions. In general both of the given equations are correct. However, I would prefer the first equation to include the highly charged metal ions as aqua complexes:
$$\ce{NO3- + 3[Fe(H2O)_6]^{2+} + 4H+  -> NO + 3[Fe(H2O)_6]^{3+} + 2H2O}$$
Also note that this is a test for nitrate, and nitrite ions will interfere accordingly.
$$\ce{NO2- + [Fe(H2O)_6]^{2+} + 2H+  -> NO + [Fe(H2O)_6]^{3+} + H2O}$$
It is important to realise, that the Iron (II) solution is given in massive excess for this test to work. Hence the newly reduced nitrogen monoxide will react with the yet unreacted excess of $\ce{Fe^{2+}}$:
$$\ce{NO + [Fe(H2O)_6]^{2+} -> [Fe(NO)(H2O)_5]^{2+} + H2O}$$
Since the iron hexaaqua complexes are nearly colourless, they do not interfere with this visibility of this reaction.

The reactions given are correct. The nitric acid oxidizes the ferrous ions into ferric ions, and itself get's reduced to NO. But this reaction doesn't happen all at once. It's not as if all the ferrous ions are oxidized to ferric in one go. The remaining ferrous ions then form the brown ring complex with NO. Furthermore, the formation of the brown ring complex itself is a redox reaction. Iron is in the rare +1 oxidation state in the complex, whereas NO is actually NO+.
So yes, ferric ions are present in the solution. They were produced as a result of the first reaction, but are not required in the second reaction.

I don't think that $Fe$ is in +1 state,  it's in +3 state with NO in -1.

http://chemistry.stackexchange.com/questions/374/why-is-iron-in-the-brown-ring-compound-in-a-1-oxidation-state
I'm pretty sure it's in +1. This question came in IITJEE once, during the days of the subjective exam. Iron is in +1 state, and NO is also in +1

The accepted answer in the question says iron is in 3+ state. And regarding the answer you have given, I don't think the resultant solution gives the qualitative test of $Fe^{3+}$. Also, can you tell me in which year in IIT JEE?

Oops sorry about that. The question was similar, but the compound was different. Its was K2[Cr(NO)(NH3)(CN)4]
It came in the year 2003.
And i stand corrected about the oxidation state of iron. I was unaware of the latest developments.


text:
Cis-Trans Isomerism
I am confused regarding the fact that in a cis isomer which is less stable as compared to a trans isomer why does it have a higher boiling point ??
According to what i feel trans is more stable so the intermolecular forces are more so melting point is also more as compared to cis isomer but why not the boiling point.
Even generally why is boiling point in inverse proportion with stability?

I think you are confusing some things.
Stability is only regarding to individual molecule.
Instead, boiling point depends on intermolecular bonding. You can have 2 unstable molecules strongly bonded together via hydrogen bonds for instance as you can have 2 stable compounds without hydrogen so they will not have a very high boiling point because of the absence of such hydrogen bonding.
For the melting point, it's another kind of force sticking molecules or atoms together. So there is no space for comparison beteween individual stability and boiling or even melting point of a whole bunch of molecules.
Edit :

Trans on the left and Cis on the right.
Where can you have hydrogen bonds ? You need some Oxygen or chemicaly equivalent atom to make it !

Well even in that case trans can form inter molecular hydrogen bonds whereas cis cant !! So trans should have a higher boiling point @Babounet

Not necessarily. See my edit of my answer


text:
What are the possible concentration pathways for a 3-species system?
Inspired by this question, which asks if a multi-species system could oscillate around its equilibrium, I thought about the possible concentration pathways the following more narrow specified system could take:
$$ \ce{A ->[\text{$k_1$}] B} $$
$$ \ce{B ->[\text{$k_2$}] C} $$
$$ \ce{C ->[\text{$k_3$}] A} $$
What I was interested in are the ratio of those concentrations $A:B:C$. The initial conditions are $A(0)=1$, $B(0)=0$, $C(0)=0$. (For simplicity, $[\ce{A}] \equiv A$, and analogous for $B$ and $C$)

What are the possible concentration pathways for different values of $k_1$, $k_2$ and $k_3$?

We have a cyclic cascade of unimolecular chemical reactions
$$\ce{A} \xrightarrow{\kappa_1} \ce{B} \xrightarrow{\kappa_2} \ce{C} \xrightarrow{\kappa_3} \ce{A}$$
where $\kappa_1, \kappa_2, \kappa_3 > 0$ are the rate constants. Assuming that each chemical reaction in this cascade has mass action kinetics, then we have the following system of linear ODEs
$$\begin{bmatrix} \dot a\\ \dot b\\ \dot c\end{bmatrix} = \underbrace{\begin{bmatrix} - \kappa_1 & 0 & \kappa_3\\ \kappa_1 & - \kappa_2 & 0\\ 0 & \kappa_2 & - \kappa_3\end{bmatrix}}_{=: \mathrm M} \begin{bmatrix} a\\ b\\ c\end{bmatrix}$$
where $a := [\ce{A}]$, $b := [\ce{B}]$ and $c := [\ce{C}]$ are the concentrations of the species. Since $\dot a + \dot b + \dot c = 0$, integrating we obtain the following conservation law
$$a (t) + b (t) + c (t) = a_0 + b_0 + c_0$$
where $a_0, b_0, c_0 \geq 0$ are the initial concentrations.

Stability
The characteristic polynomial of matrix $\mathrm M$ is
$$\det (s \mathrm I_3 - \mathrm M) = s \underbrace{\left( s^2 + (\kappa_1 + \kappa_2 + \kappa_3) s + (\kappa_1 \kappa_2 + \kappa_1 \kappa_3 + \kappa_2 \kappa_3) \right)}_{=: q (s)} = s \, q(s)$$
Since $\kappa_1, \kappa_2, \kappa_3 > 0$, all three coefficients of the (monic) quadratic polynomial $q$ are positive. Hence, both roots of $q$ have negative real parts and, thus, the cyclic cascade is (internally) marginally stable. Given nonzero initial concentrations, the concentrations will not grow unbounded. However, due to the zero eigenvalue, the concentrations will not converge to zero either.

Steady state
If the concentrations of the species do not converge to zero, then to what values do they converge? At steady state, $\dot a = \dot b = \dot c = 0$. Hence, we have the following homogeneous linear system
$$\begin{bmatrix} - \kappa_1 & 0 & \kappa_3\\ \kappa_1 & - \kappa_2 & 0\\ 0 & \kappa_2 & - \kappa_3\end{bmatrix} \begin{bmatrix} \bar a\\ \bar b\\ \bar c\end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}$$
where $\bar a, \bar b, \bar c$ are the steady state concentrations. Using Gaussian elimination, we eventually obtain a line passing through the origin that is parameterized as follows
$$\begin{bmatrix} \bar a\\ \bar b\\ \bar c\end{bmatrix} \in \left\{ \gamma \begin{bmatrix} \frac{1}{\kappa_1}\\ \frac{1}{\kappa_2}\\ \frac{1}{\kappa_3}\end{bmatrix} : \gamma \in \mathbb R \right\}$$
where
$$\tau_1 := \frac{1}{\kappa_1} \qquad\qquad\qquad \tau_2 := \frac{1}{\kappa_2} \qquad\qquad\qquad \tau_3 := \frac{1}{\kappa_3}$$
are time constants. Intersecting this line with the plane $\bar a + \bar b + \bar c = a_0 + b_0 + c_0$, we obtain
$$\gamma = \left( \dfrac{1}{\tau_{1} + \tau_{2} + \tau_{3}}\right) \left( a_0 + b_0 +c_0 \right)$$
and, thus, the steady state concentrations are
$$\boxed{\begin{array}{rl} & \\ \qquad\bar a &= \left( \dfrac{\tau_{1}}{\tau_{1} + \tau_{2} + \tau_{3}}\right) \left( a_0 + b_0 +c_0 \right)  \qquad \\\\ \qquad\bar b &= \left( \dfrac{\tau_{2}}{\tau_{1} + \tau_{2} + \tau_{3}}\right) \left( a_0 + b_0 +c_0 \right) \qquad\\\\ \qquad\bar c &= \left( \dfrac{\tau_{3}}{\tau_{1} + \tau_{2} + \tau_{3}}\right) \left( a_0 + b_0 +c_0 \right)\\ & \end{array}}$$
If, say, $\kappa_1 \gg \kappa_2, \kappa_3$, i.e., if $\tau_1 \ll \tau_2, \tau_3$, then $\bar a \ll \bar b, \bar c$. If $\kappa_1 = \kappa_2 = \kappa_3$, then $\bar a = \bar b = \bar c$.
Assuming that at least one of $a_0, b_0, c_0$ is nonzero, the ratios of the steady state concentrations are
$$\frac{\bar a}{\bar b} = \frac{\kappa_2}{\kappa_1} = \frac{\tau_1}{\tau_2} \qquad\qquad\qquad \frac{\bar a}{\bar c} = \frac{\kappa_3}{\kappa_1} = \frac{\tau_1}{\tau_3} \qquad\qquad\qquad \frac{\bar b}{\bar c} = \frac{\kappa_3}{\kappa_2} = \frac{\tau_2}{\tau_3}$$
Hence, the ratio of the steady state concentrations of two species is the ratio of their time constants.

Oscillations
Let
$$\Delta (\kappa_1, \kappa_2, \kappa_3) := (\kappa_1 + \kappa_2 + \kappa_3)^2 - 4 (\kappa_1 \kappa_2 + \kappa_1 \kappa_3 + \kappa_2 \kappa_3) \\ = (\kappa_1 - \kappa_2)^2 + (\kappa_1 - \kappa_3)^2 + (\kappa_2 - \kappa_3)^2 - (\kappa_1^2 + \kappa_2^2 + \kappa_3^2)$$
be the discriminant of quadratic polynomial $q$. When $\Delta < 0$, the roots of $q$ have nonzero imaginary parts and, thus, the cyclic cascade will exhibit damped oscillations. Hence, when
$$\kappa_1^2 + \kappa_2^2 + \kappa_3^2 > (\kappa_1 - \kappa_2)^2 + (\kappa_1 - \kappa_3)^2 + (\kappa_2 - \kappa_3)^2$$
the cyclic cascade will exhibit oscillatory behavior. Provided that the inequality above is satisfied, the concentrations will oscillate as follows
$$\begin{array}{rl} a (t) &= \bar a + \eta_A \, e^{- \kappa_0 t} \cos( \omega_0 t - \phi_A)\\ b (t) &= \bar b + \eta_B \, e^{- \kappa_0 t} \cos( \omega_0 t - \phi_B)\\ c (t) &= \bar c + \eta_C \, e^{- \kappa_0 t} \cos( \omega_0 t - \phi_C)\end{array}$$
where
$$\begin{array}{rl} \kappa_0 &:= \frac{1}{2} \left( \kappa_1 + \kappa_2 + \kappa_3 \right)\\ \omega_0 &:= \frac 12 \sqrt{\kappa_1^2 + \kappa_2^2 + \kappa_3^2 - (\kappa_1 - \kappa_2)^2 - (\kappa_1 - \kappa_3)^2 - (\kappa_2 - \kappa_3)^2}\end{array}$$
Each species will have its own magnitude and phase lag that are dependent on the initial conditions.

Of course, we can't display every pathway there is. We have to select a wide variety of rate constants to get a "good map" of the parameter-space. The approach I took was this:

Construct the differential equations

$$\frac{\mathrm dA}{\mathrm dt} = -k_1 A + k_3 C$$
$$\frac{\mathrm dB}{\mathrm dt} = -k_2 B + k_1 A$$
$$\frac{\mathrm dC}{\mathrm dt} = -k_3 C + k_2 B$$

Solve them numerically for every possible combination of 0.1, 1 and 10 for each rate constant and the above stated initial conditions
This leads to 27 pathways, which are plotted in a ternary diagram

The title of the ternary diagram is $k_1:k_2:k_3$, which is 0.1 : 0.1 : 0.1 in the first one.

(click for larger image)

What software did you use to generate these diagrams?

@RodrigodeAzevedo Honestly Im not completely sure, either Python / Matplotlib or R. Maybe I'll find the source code when I'm home.


text:
Are all amino acids except glycine chiral compounds?
Specifically, I was thinking of the case where the R group is a carboxylic acid functional group or an amine group. Then there would not be a chiral carbon atom. Is there a restriction on what can be an R group that I am missing?

The alpha carbon is a chiral carbon atom, with the exception of glycine which has two indistinguishable hydrogen atoms on the alpha carbon. Therefore, all alpha amino acids but glycine can exist in either of two enantiomers, called L or D amino acids, which are mirror images of each other.

But if the R group is a carboxylic acid group or amine group, then the alpha carbon atom will not be a chiral carbon atom because it won't have 4 different atoms/groups bonded to it...

@21joanna12 This is true. However, none of the naturally occurring amino acids have these R groups. Synthetic ones could though and so they would not exhibit optical isomerism

In general, there are a variety of amino acids that are not chiral (achiral).  In physiological proteins, the amino acids are generally alpha-amino acids (one carbon between the amino and carboxylic acid groups), but there are also, beta-amino acids, gamma-amino acids, etc.
Just limiting ourselves to alpha-amino acids, other than glycine, the most interesting example of an achiral amino acid is amino malonic acid, which has been given its own three-letter abbreviation (Ama) along with the more-common alpha amino acids.  Ama has being found in the hydrolysates of E. Coli and human proteins.  See Aminomalonic acid: identification in Escherichia coli and atherosclerotic plaque PNAS vol. 81 page 722.
Also, the amino acids of protein orgin generally have an alpha H, but if this H is substituted the compound is still an alpha amino acid.  If the two side chains on the alpha carbon are the same, the compound is achiral. For example alpha-methyl alanine.  An example that occurs naturally in cranberries is 1-aminocyclopropane 1-carboxylic acid. The alpha amino acids with two side chains are sometimes referred to as alpha branched amino acids, and have been found in meteorites.
Another class of achiral alpha amino acids are those with a side chain double-bonded to the alpha carbon such as dehydroalanine.

Plaque material contains bacterial products so Ama from plaques may not be human, after all.


text:
Wade's rules for Zintl ions
How do you apply Wade's rules to Zintl polyhedral cluster anions ?
For example, I believe $[\ce{Bi_3}]^{3-}$ is a triangle. My approach to justify would be:

5 electrons per each Bi atom
3 electrons for the charge
18 total electrons

$18 = 4\times3+6 \therefore$ arachno cluster: trigonal bipyramid ($n+2$ polyhedron) with two missing vertices $\therefore$ linear.
If it was a nido cluster $(4n+4\ \ce{e-})$, then it would be a tetrahedron with one missing vertex $\therefore$ a triangle. Am I mistaken about the observed geometry or is my method incorrect?

Let the number of atoms be n.
If the number of skeletal p electrons is 2n + 2, the structure is closo (a deltahedron with n vertices).
If the number of skeletal p electrons is 2n + 4, the structure is nido (an n+1 vertex deltahedron, with one vertex missing).
If the number of skeletal p electrons is 2n + 6, the structure is arachno (an n+2 vertex deltahedron with two vertices missing).
If the number of skeletal p electrons is 2n + 8, the structure is hypo (an n+3 vertex deltahedron with three vertices missing).
see Excursions Beyond the Zintl Border and Polyanionic Clusters and Networks of the Early p-Element Metals in the Solid State: Beyond the Zintl Boundary
For example:
Neutral Sn, Pb and Ge each contribute 2 valence p electrons, while Bi and Sb each contribute 3 valence p electrons.
$\ce{Sn_5^{2-}}$, $\ce{Pb_5^{2-}}$, $\ce{Ge_5^{2-}}$ and $\ce{Ge_9^{2-}}$ have 2n + 2 skeletal p electrons and are closo (the first three are trigonal bipyrimidal and the last is appoximately a triangular prism with square pyramid caps for the three sides).
$\ce{Sn_9^{4-}}$ and $\ce{Ge_9^{4-}}$ have 2n + 4 skeletal p electrons and are nido. (gyroelongated square bipyramid with one square pyramid apex missing)
$\ce{Bi_4^{2-}}$ and $\ce{Sb_4^{2-}}$ have 2n + 6 skeletal p electrons and are arachno (these two are square planar, an octahedron with two vertices missing).
Also, quoting Farley and Castleman's Observation of Gas-Phase Anionic Bismuth Zintl Ions J. Am. Chem. Soc. 1989, 111, 2734-2735

Since bismuth and antimony are typically trivalent, anionic clusters
of these elements very quickly exceed the maximum 2n + 8 electrons
accountable under Wades rules for “hypo” compounds. In particular, the
previously mentioned $\ce{Bi7^{3-}}$ possesses 2n + 8 valence electrons, and
$\ce{Bi14^{6-}}$ has 2n + 20! Hence, anionic Zintl ions of these species are
particularly intriguing due to the inability of current theories to
account for them and lack of other corresponding metallic compounds
with which they may be compared.

Now going back to the example $\ce{Bi_3^{3-}}$ in the orginal question, the consensus in the peer-reviewed literature is that the sturcture is bent (not a triangular ring and not linear), although the structure of the isolated ion has not been experimentally determined. The reasons for considering it to be bent are:

The structure is isoelectronic with species known to be bent in the ground state including $\ce{O3}$, $\ce{S3}$, $\ce{S2O}$, $\ce{SO2}$ and allyl anion.

Experimental determination published in Stabilization of Ozone-like [Bi3]3- and in a more recent article that it is bent in transition metal complexes. (Similar experiments for isoelectronic [P3]3- and[As3]3-also find bent structures).

Theoretical work in Zintl Anions Analogous to O3.

Though ozone and like species are found to be bent, it is also theorized that cyclic (equilateral triangular ring) ozone represents another higher local potential energy minimum, and this would be the same for $\ce{Bi_3^{3-}}$ too.
$\ce{Bi_3^{3-}}$ has 12 p electrons, which is 2n + 6, which implies arachno with respect to a trigonal bipyramid.  A equilateral triangular ring structure would be consisent with removing two axial vertices or removing one axial vertex and one equatorial vertex, while the bent structure would be consistent with removing two equatorial vertices.

triangular As3 ring was found in organoarsenicum chemistry and in $\ce{[LMAs3ML]^{n+}}; n=1,2; M=Co,Ni$ complexes, implying that triangular structure is acceptable. Bent structure for ozone is preferable because of extra stability of $\pi$-bond for 2nd row elements, however, they are virtually unknown except very special circumstances for 3rd+ rows. Given that, I highly doubt stability of bent geometry. However, relatively high charge density in $\ce{[Bi3]^{3-}}$ makes its formation without external stabilization extremely improbable, so I doubt this consideration are of any use.

I understand that ring As3 is known, but I'm not aware of ring $\ce{[As3]^{3-}}$.  The relative energies of the ring and bent forms of $\ce{[Sb3]^{3-}}$ are calculated in http://pubs.acs.org/doi/pdf/10.1021/jp980004e

$\ce{[Sb3]^{3-}}$ has an optimal angle of 131 degrees, even greater than ozone, and the bent form is more stable by 0.5 eV.

$\ce{As3}$ unit in mentioned complex may be considered as $\ce{[As3]^{3-}}$.

As for anions, highly-charged anions cannot form without solvation stabilization, and in condensed state the stabilization participates greatly in relative stability. For the matter, I recommend to read the link you provided, as it clearly states that for $\ce{[Na4Sb3]^{+}}$ the geometry with triangular structure is preferable. However, the only state I was able to find with 1:1 K/Bi has $\ce{Bi4}$ and $\ce{Bi2}$ units, and $\ce{Ba2Bi3e}$ phase has complex 3d network of Bi atoms.

Yes, for isolated $\ce{[Sb3]^{3-}}$ the bent form is favored by 0.5eV, and for isolated $\ce{Sb3}$ the only minima are the near equilateral structures. As negative charge is taken from $\ce{[Sb3]^{3-}}$ by adding alkali atoms, the near equilateral sturcture becomes more favored, with the ~equilateral slightly favored in $\ce{Na3Sb3}$ and more strongly favored in $\ce{[Na4Sb3]^+}$.  Bent is favored for $\ce{NaSb3}$.

@permearka it is not true that bent three-atom structures fail to exist for third row elements. Trisulfur seems perfectly happy with it.

The wiki article contains a note

To predict the structure of an arachno cluster, the closo polyhedron
  with n + 2 vertices is used as the starting point, and the n+1 vertex
  nido complex is generated by following the rule above; a second vertex
  adjacent to the first is removed if the cluster is composed of mostly
  small atoms, a second vertex not adjacent to the first is removed if
  the cluster is composed mostly of large atoms.

$\ce{Bi}$ atom is definitely a large atom, so we should  remove out non-adjacent vertexes. The only way to choose non-adjacent vertexes in trigonal bipyramid is to take axial vertexes, gaining a triangle.
Also, please not, that removing two adjacent vertexes in trigonal bipyramid may also lead to triangle, if one removes one equatorial and one axial position. removing two equatorial positions will lead not to triangle or line, but to an angle.
However, for $\ce{[Bi3]^{3-}}$ the amount of electrons per atom is exactly six. This means, that 6n rules suite situation better. 6n rule leads to ring in the case.
I recommend to consider Wade's rules as general guidelines, but not a final truth, sometimes they fail spectacularly. For example, consider $\ce{[CH]6}$ cluster. Wade's rule predicts it to be a Trigonal prism. As we know, IRL this molecule prefers to be a ring, with prismane being an esoteric and a quite rare form.

$\ce{[Bi3]^{3-}}$ is isoelectronic with ozone and allyl anion and bent, not a ring. See http://etd.nd.edu/ETD-db/theses/available/etd-07072009-102951/unrestricted/HullMW072009D.pdf and http://pubs.acs.org/doi/pdf/10.1021/ja0038732 $\ce{[P3]^{3-}}$ and $\ce{[As3]^{3-}}$ are also bent.

@DavePhD The links you provided are irrelevant as they do not contain a proof that an isolated structure is bent. The ozone-like bent structure is stabilized by means of covalent interactions with transition metal. In fact the links consider mixed clusters, not $\ce{[Bi3]^{3-}}$

Fig.1.2D of the thesis depicts the isolated ion as bent, so I wouldn't say the link is irrelevant, but I agree that it isn't experiment proof.

@DavePhD Fig. 1.2D depicts hypothetical structure, isoelectronic to allyl-anion. This means, that said structure MAY appear in places where normal allyl-anion may appear, but it does not state anything about the free $\ce[Bi3]^{3-}}$ anion itself. I can draw many similarly isoelectronic structures, it does not mean they would appear IRL.

So-called 2n, 4n, 14n and so forth rules are not necessary at all. Simply, Wade's rule in its original form i.e. 'n' rule is equally applicable to Zintl ions and sufficient to predict their structural type e.g., closo, nido, arachno, hypho and so on. For example, the Skeleton Electron Pair Count becomes six [0.5(3 x 5 - 3)] for [Bi5]3+; indicating (n + 1) formulation and hence it must adopt closo structure. Similarly, the SEPC becomes nine [0.5(3 x 5 + 3)] for [As5]3-, implying its hypho structure from (n + 4) formulation.
Moreover, the contribution of the vertex atom (here it is Bi/As) can be easily calculated from the 'v + x - 2' rule, where v is the # of valance electron(s). The value of x (contribution of exo ligand towards vertex atom) must be zero since there is no exo ligand bonded with the vertex atoms in Zintl ions, Naked Clusters. Jadavpur University, India.


text:
NMR and pseudorotation

I think it has something to do with the fact that at high temperature it interconverts and so all the coupling constants are the same hence quintet but at lower temperatures the frequency of the rotation is less than the sample rate and so there are different coupling constants for the planar and axial fluorine atoms. We have two structures to consider; 3 planar-1 axial and 2 planar-2 axial. If my line of thought it correct the P-31 NMR spectra for each of these should superimpose to give the spectrum shown.

But I'm lost because when I draw my diagrams they don't match the spectrum shown, I end up with a much more complicated splitting pattern.

You are certainly thinking in the right direction.  Fluorine strongly prefers the axial position (for reasons described here), so at low temperature, the isomer that you've drawn on the right (2 axial and 2 equatorial fluorines) is the only isomer present.  As a result we have an $\ce{A2B2X}$ spectrum. The 2 equivalent equatorial fluorines (spin = 1/2) split the phosphorous signal into a triplet.  Each line of this triplet is further split into a triplet by the 2 equivalent axial fluorine's. BTW, the same reasoning that explains fluorines axial preference, also explains why generally $\ce{J(P-F)_{equatorial}}$ > $\ce{J(P-F)_{axial}}$
At room temperature, pseudorotation is rapid (more thermal energy is available and the barrier to pseudorotation can be surmounted) and all fluorines become equivalent on the nmr timescale. Therefore, at room temperature and above a quintet is observed in the $\ce{^{31}P}$ nmr spectrum.

okay I can see that its a triplet of triplets now that you've said! Why does fluorine prefer the axial position?

I've edited my answer to include a link to an earlier answer explaining why fluorine (or other electronegative substituents) prefers the axial position in pentacoordinate phosphorous.

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


text:
Why does superglue ignite cotton?
Cotton is mostly cellulose, a plant saccharide. Superglue refers to a class of cyanoacrylates. What is it about the two that causes ignition?
I'd imagine that the cellulose in cotton is in its linear form and not in its ring form since the ring-linear form interconversion only occurs in solution. Is it that the cellulose attacks the cyanoacrylate?

The three hydroxyl groups, one primary and two secondary, in each repeating cellobiose unit of cellulose are chemically reactive

So, do the hydroxyl groups attack the carbonyl group in superglue?

This might be pure ignorance on my part, but what exactly do you mean by "linear form" and "in solution"? I've never heard of (water-)soluble cellulose, and "linear form in solution" reminds me rather of the mono-sugars, but not the polymer.

I found your post after my son came to me with 2 burnt Q-tips. He used the Q-tips to apply super glue to something, they started to smoke and the ends were black. It would appear that cotton can catch on fire when super glue is applied. He did not put anything else on the Q-tips.

I don't have an answer, but I just recently had an incident with super glue. I was opening a container and a large amount squirted out all over my hands. I reached for the first available thing which was Kleenex and it started to smoke and I could feel it getting warm. I dropped the super glue container to the ground and it ran to the sink. It did not spread any further. Now I know what was happening, I had some inkling based on my ChE degree but I have not practiced in nearly 25 years so I am a bit rusty.

Cyanoacrylates include methyl 2-cyanoacrylatecommonly sold under the trade names "Super Glue".In general, cyanoacrylate (consists of monomers of cyanoacrylate molecules) is an acrylic resin that rapidly polymerises in the presence of water (specifically hydroxide ions), forming long, strong chains, joining the bonded surfaces together. Because the presence of moisture causes the glue to set, exposure to normal levels of humidity in the air causes a thin skin to start to form within seconds, which very greatly slows the reaction. Because of this cyanoacrylate is applied thinly, to ensure that the reaction proceeds rapidly and a strong bond is formed within a reasonable time.

So to sum up, in order to start the reaction off some water is normally needed.. damp things stick better/quicker than dry ones and the glue goes hard faster on a humid day.
In cotton wool, which is made of cellulose, a polymer of sugar molecules, there are lots and lots of hydroxy (-OH or alcohol groups), which can start the reaction in the same way as the water does, only because there are lots of them they can start many more reactions at once.
Since the reaction gives out heat, the cotton bud therefore gets hot (and as it becomes hotter so the reaction goes faster etc), and it may get hot enough to catch fire.

Not exactly. Alcohol is not sufficiently basic (much less than water) to be a really good nucleophile. But cotton is polar, and thus is loaded with water.

Though the monomers in cyanoacrylate glues contain an ester, their polymerization doesn't rely on that ester group directly.  The Wikipedia article for cyanoacrylates shows the polymerization more clearly than I can easily explain in words.  The many hydroxyl groups in cellulose do start polymerization effectively, and the large surface area of cotton wool provides a large number of sites for the glue to cure.
When I read your question I was curious; I had never heard of this before.  I searched the internet and was only able to find one or two examples of people actually trying this successfully, and both of them looked dubious.  I've spilled superglue on cotton shirts and pants any number of times and while it cures almost instantly, it has never caught my shirt on fire.
I decided to try it myself, and I soaked a cotton ball in superglue to see what would happen.  The glue cured extremely rapidly, producing very irritating (colorless) fumes, and the mass of glue and cotton got warm, but it was nowhere near hot enough to ignite - my tap runs hotter than the glue got.  There may be a particular type of glue that cures significantly more exothermically, but I'm inclined to think that this is an internet hoax, and Popular Science agrees.

Maybe you used enough superglue that the thermal mass of the glue itself kept the temperature down?

@Brian It's possible, but the videos that say that superglue will ignite cotton also use lots of it.  To be clear, I'm 100% open to being wrong if given good evidence, but neither Popular Science nor I can make it happen, and there are an awful lot of internet hoax videos of this sort.

Aha, perhaps not, then.  Maybe the 'igniters' presoaked the cotton with rubbing alcohol or something...

But the important point is that the reaction is fairly exothermic because of the presence of many OH groups on the surface and the surface area is high. This means that, given the right conditions, it is possible to cause ignition.

@matt_black No, that's not an accurate statement.  An exothermic reaction does not automatically mean that you can start a fire - it's entirely possible that the energy per reaction is just too low.  Even with a very large surface area, the reaction temperature was only ~315K, and cotton needs to get to ~485K to burn.  That's a substantial difference.  An exothermic reaction is a necessary condition but not sufficient.

The hydroxyl groups don't, it's the adsorbed water on the cellulose. And if it produced fumes, that means it whas just about to catch fire. How on earth did you "measure" those 315K? At that temperature, neither superglue nor cotton produce fumes!

@Karl I would suggest that you try the reaction yourself and watch it do little or nothing.  If your claim is that it bursts into flame, provide evidence.  I indicated that the 315K was an approximation, and the answer I left explains where that approximation came from.  The reaction occurred in my hand.  As soon as it didn't burst into flame, I felt it, and from that estimated the temperature.  Superglue produces noxious fumes at room temperature; it produces more when heated.

Really, you superglued to cotton to your hand, on purpose? Wow. And cyanacrylate does smell rather irritating, but it does not produce fumes at room temperature. Sorry, but what you describe doesn't match up imo.

@Karl Step 1: Add glue to cotton while on sink.  Step 2: Observe lack of fire.  Step 3: Feel cotton and not that glue has not completely set but it still only warm to the touch.  This isn't the place for an argument. If you have a real problem with my answer, feel free to downvote or leave a better answer. Finally, I'm not sure why you're so invested in this myth, but the materials are cheap. Buy some superglue, buy some cotton balls, find some evidence, and prove me wrong.


text:
Why does capillary action occur better in narrow tubes than in wide tubes?
We all have seen the meniscus of a small graduated cylinder in the lab. The reasons for the occurrence of capillary action are clear, but why does it occur more profusely in smaller-rimmed containers?

You have to look which forces act where. You have a force that tries pull the liquid up the capillary, which stems from the gain in energy due to adhesion. At the same time, however, gravity acts against this force. In equilibrium, both forces have the same value:
$$F_\text{g} = F_{\text{surf}}$$
$$ mg=\rho Vg= \rho R^2\pi h g=2\pi R\sigma_s \cos(\varphi)$$
Which leads to the formula presented by Babounet in another answer:
$$ h = \frac{2\sigma_s\cos(\varphi)}{R\rho g}$$
So, to boil it down, if you lower the radius $R$, you lower the force due to gravity $F_g$, but the force due to surface energy $F_\text{surf}$ does not fall as fast, which leads to a higher height $h$.

(source, $W=F_g$ here, naming isn't perfect, $F_{\text{surf}}$ is not drawn into)

A meniscus is formed by the opposing forces of adhesion between the fluid and the walls of the container drawing the fluid up and gravity pulling the fluid down. The adhesive forces are proportional to the diameter of the tube while the gravitational effect due to the liquid's weight is proportional to the square of the diameter. This results in the greater height of the meniscus for narrower tubes.

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].

Remember that it isn't only gravity and adhesion, but also cohesion among the liquid molecules in the container. Now, something I would like an explanation for is why "The adhesive forces are proportional to the diameter of the tube while the gravitational effect due to the liquid's weight is proportional to the square of the diameter." Can you elucidate how these truths were derived? Thanks!

You are correct @user11629, cohesive forces are also important. The difference in strength between adhesion and cohesion for the liquid can also change the shape of the meniscus, for example giving the inverted shape seen for Mercury. Unfortunately I can't find a clear explanation for the different relationships to diameter.

If you look at John's answer, you can see that the weight has to do with the volume of the liquid. For a cylinder, it's the area of the cross section times the height. The area increases with r^2. The force pulling the liquid up (or down) is the surface tension times the length of the three-phase interface (contact line), so it's proportional to the circumference of the tube, which increases linearly with radius. N.B. This force acts in the direction of of the contact angle, which is where the cos() term comes in to extract the vertical component.

According to the Jurin's Law
$ h = \frac{2\gamma\cos(\theta)}{r\rho g} $
With :
$h$ the height of the fluid
$\gamma$ the surface tension
$\theta$ the contact angle with the tube
$r$ the tube radius
$\rho$ the liquid density
$g$ the gravitational constant
So, the $h$ is inversly proportional to $r$. In other words, the larger the tube, the lower the liquid climbs.
Note that this law is only valid if $r < \lambda_\text{c}$ with $\lambda_\text{c} = \sqrt{\frac{\gamma}{\rho g}}  $

This is a nice answer, but really only restates the observations stated in the question; that the meniscus is higher in smaller diameter tubes. Why is this so?

Agreed, long. Thanks for the reference to Jurin's Law, Babounet; I hadn't heard about it before. But I was asking specifically about the chemical explanation about why it is true that this phenomenon occurs.

Capillary is more or less a chemical-physics topic. Not only about "chemistry" so it's hard to give a more "chemical" explanation :/ you can try to look after surface tension concept, it will help you I think. (You'll see why capillary act on the reverse way for Mercury for instance)


text:
Entanglement of large molecules
In many textbooks and chemistry curricula, we can see the description of large molecules (particularly in the discussion of hydrocarbons and other nonpolar compounds) as having "entanglement." However, I think the term is misleading, since the molecules don't really act like tangled shoelaces or two strings pulling against each other while interlocked. Can we try to clarify what actually occurs that we describe as entanglement?
Some things I think may happen:
The molecules are very large, and thus have many sites for IMFs, leading to high viscosity.
By way of this size, the molecules are much harder pressed for movement and so take a while to move (the distinction is that in the first example the increased size begets more IMFs whereas the second causes "entanglement" just by the bulky molecules' shifting past each other).

See also http://chemistry.stackexchange.com/questions/5239/why-do-molecules-having-a-higher-m-r-have-stronger-inter-molecular-forces?rq=1

Well, very long hydrocarbons can become twisted around each other. This is definitely known in the polymer literature. Consider some branched polyethylene with ~1,000 repeat units. It's not too hard to imagine that multiple chains can become entangled in much the same way as shoelaces or strings when you mix a lot of them in a box.
You are also right that with large molecules, there are more sites for intermolecular interactions. Consider that there is a "combination" effect with lots of sites. That is, to pull two very long linear chains apart, you need to break lots of interactions.

Notice that pulling one atom on the bottom away from its neighbor on the top chain will require the movement of all atoms on the bottom chain. So it's slightly harder to separate two long chains, not only because there are so many interactions, but also because of the combination.
Put simply, the difficulty in separating the two chains is slightly higher than simply the sum of the interactions.

I'm extremely sorry, just edit your answer once and I'll remove my downvote. I was actually upvoting the answer. :|


text:
Does the water in hydrates affect final concentration?
Say I have a hydrate, and I am going to dissolve it in water. Does the water in the hydrate affect the final concentration?
Here's an example: If I have $\pu{249.69 g}$ of $\ce{CuSO4·5H2O}$, and dissolve it in $\pu{200 mL}$ of water, will the concentration be $\pu{5 M}$?

Of course it does. The water from crystallohydrate is still water.

Of course, the water in the hydrate affects the final concentration.

Molar mass of  $\ce{CuSO4.5H2O}$ is $\pu{249.5 g/mol}.$
Amount of substance is
$$n = \frac{345.55}{249.5} = \pu{1.385 mol}$$
Concentration:
$$C = \frac{1.385}{0.2} = \pu{6.925 mol/L}$$

But, this is only an illustrative example. It's not real. The solubility of $\ce{CuSO4.5H2O}$ is $\pu{320 g/L}$ at $\pu{20 °C}.$

Just to be clear, I don't think you can actually make a 5M solution of copper sulfate, but I'll ignore solubility for this answer.
When you make a solution of a given concentration from a solid you generally dissolve the solute (your copper sulfate pentahydrate) in a relatively small amount of water and then dilute it to the final volume.  Here, you would dissolve your 249g (1 mole) of copper sulfate pentahydrate in sufficient water to dissolve it but less than 200mL total, then carefully add enough water to reach a solution volume of 200mL, yielding a 5M solution.
The water of hydration would indeed become part of the solution.  If instead you began this process with anhydrous copper sulfate and used 1 mole of that (159.61g) then you would dissolve it in enough water to make the solution but less than 200mL total then carefully add enough to reach your final solution volume and concentration, 200mL at 5M.
The difference here would be that you would have to add 90g of additional water to prepare a solution from the anhydrous compound as compared to the pentahydrate.
If you prepare a solution by adding a fixed mass of solute to a fixed volume of water, you are never going to make the desired concentration and volume.

Firstly, copper would have to be in the 4+ oxidation state for there to be $\ce{Cu(SO4)2.5H2O}$, so this isn't a realistic question.
But, no it won't be 5M, because you need to know the final volume of the solution to calculate molarity.  If you add 345.55 grams of just about anything I can think of to 200 mL of water, the volume will increase substantially.  So not only will the 90 grams of hydrate water add to the volume of the final solution, generally you need to consider how all added substances contribute the final volume.
In some cases, such as magnisium sulfate, a small amount added to water can actually decrease the volume.

I realize this may not be what you are asking, but this is worth pointing out. The most common source for error in solution concentration comes from neglecting to account for the mass of the hydrate. If you weight out an amount of a hydrate, and use the molecular mass of the anhydrous, your final concentration will be erroneous. For example, 5g of $\ce{CuSO4 \dot{} 5 H2O}$ will not yield the same as 5g $\ce{CuSO4}$ (anhydrous). When diluted to a volume of 1 liter:
$$\frac{\ce{5g\,CuSO4.5H2O}}{1} \frac{\ce{1mol\,CuSO4.5H2O}}{249.685\ce{g}}\frac{1}{\ce{1L}}=0.02\ce{M\,CuSO4}$$
$${\frac{\ce{5g\,CuSO4}}{1} \frac{\ce{1mol\,CuSO4}}{159.6086\ce{g}}\frac{1}{\ce{1L}}=0.03\ce{M\,CuSO4}}$$

The molar mass of copper (ii) sulfate penta-hydrate  is $\pu{249.69 g/mol}$ and that for anhydrous copper(ii) sulfate is $\pu{159.61 g/mol}$. Thus, in $1$ mole of the hydrated substance there are $\pu{159.61 g}$ of copper (ii) sulfate and $5 \times \pu{18.00 g}$ (i.e. $\pu{90.00 g}$) of molecules of water of crystallization.
$\pu{5M}$ copper (ii) sulfate solution means $5$ moles of the substance is contained in $1$ liter of solution. The mass of copper (ii) sulfate in $\pu{5M}$ of the substance is $5 \times \pu{159.61 g/L} = \pu{798.05 g/L}$.
This means that to prepare a $\pu{5 M}$ copper sulfate solution, weigh out $\pu{798.05 g}$ of the anhydrous substance and dissolve in $\pu{1 L}$ volumetric flask with distilled or de-ionized water, making up to volume by filling up to the thin ring mark at neck of flask.This can be scaled down by a factor of $5$ by dissolving $\pu{159.61 g}$ of the substance in $\pu{0.2 L}$ ($\pu{200 mL}$) solution, using a $\pu{200 mL}$ volumetric flask.
Now if the penta-hydrate is to be used to achieve a concentration of $\pu{5 M}$ copper sulfate solution, from the previous we see that $\pu{159.61 g}$ copper sulfate is contained in $\pu{249.69 g}$ of the hydrated copper sulfate. Therefore,
$\pu{798.05 g}$ of copper sulfate is contained in $(249.69/159.61) \times \pu{798.05 g} = \pu{1248.45 g}$ of the pent-hydrate, i.e. dissolve $\pu{1248.45 g}$ of the hydrated salt in $\pu{1L}$.
Or if a smaller quantity is desired, say $\pu{200 mL}$ solution: weigh $(249.69/159.61) \times \pu{159.61 g} = \pu{249.69 g}$ of the penta-hydrate and dissolve in $\pu{200 mL}$ solution. Care must be taken to add water in small quantities so as not to exceed the ring graduation at neck of flask and the lower meniscus of solution should align with the mark.

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