text:
Rationale for assigning antibonding MOs
How do people decide if a molecular orbital (MO) is bonding or antibonding? From Wikipedia article on antibonding MO:

[...] if the antibonding interactions outnumber the bonding interactions, the molecular orbital is said to be antibonding

and vice versa.
How do we define what an "antibonding interaction" is for cyclic molecules? Does one nodal plane count as 1 or 2 antibonding interactions? For example, the π2 molecular orbital of benzene may have 1 nodal plane, but how many antibonding interactions would that MO have?

Similarly, how do we define "bonding interactions" for cyclic molecules?

There are a couple ways that people define bonding and antibonding orbitals. One of the simplest is to just look at the energy of the MOs formed compared to the AOs they were formed from: if the MO is lower in energy, it is bonding and if the MO is higher in energy, it is antibonding.
Like I said, this way of doing thing is relatively simplistic, but you will see it pretty commonly in textbooks on the matter. The method of course isn't perfect, in particular because it essentially assumes that all local bonds are affected equally by a particular orbital.
We can see this is not the case by looking at the $\pi_2$ orbital you show. While the orbital is listed as bonding, the interactions of the p-orbitals where a positive lobe is next to a negative love will locally decrease the bonding between those two atoms (ie the MO is antibonding with respect to those local bonds).
So, for a simple first approximation, you can use the energy of the MO compared to the AO, but for a more complete sense, you should look at the symmetry of the MO and see what effect it has on a given local bond.
In terms of the ordering of the MOs themselves, assuming you don't have experimental way to determine the ordering like ionization energies, you can heuristically use an approach like you suggest where you compare the number of nodes and less nodes will mean lower in energy, more nodes will be higher.
In terms of comparing to the original AOs, it can be tricky without experimental data, however there are again general qualitative ways of ordering the MOs with respect to their original AOs. For example, if you have 2 MOs formed from two overlapping AOs, one will have to be higher in energy and the other will have to be lower.
When you get more orbitals overlapping, the arrangements can vary, but the sum of the energies has to remain the same. So, for example, if you have 3 overlapping AOs, you could form 2 bonding and 1 antibonding MOs or vis versa so long as their combined energy is the same as before.
Beyond this, you really just have to read up on MO theory to get a sense of how qualitative energy ordering is done. There is a lot more to it than what I've written (or could write) and it's probably better to build your understanding of the theory from the ground up rather than getting it piecemeal.

Hi Tyberius, I'm not sure what you mean by "look(ing) at the symmetry of the MO and see what effect it has on a given local bond".

Also, if we were to "examine the energy of the MOs formed compared to the AOs they were formed from", and compare their relative energies in order to reach a conclusion whether the MO is bonding or antibonding, this begs the question: how do we know if a particular MO is higher or lower in energy compared to the AOs it was formed from?

@JonathanSmith So that is what I'm referring to in the 2nd paragraph of my answer when I talk about the π2
orbital. If you look at the interaction between atoms that have a positive lobe next to a negative lobe, that is an antibonding interaction because those atoms won't be drawn towards each other.


text:
How does HCl dissolve in water if the intramolecular force is stronger?
So what I know now is that the H in HCl is more positive and the Cl is more negative, while the H in H2O is more positive and the O is more negative. When HCl enters water, its more positive atoms (H) interact with the more negative atoms of H2O (Oxygen), and this attraction breaks the molecules apart.
However, wouldn't this mean that the dipole-dipole intermolecular force is greater than the intramolecular force? But isn't intramolecular forces stronger than intermolecular forces?

I can’t really answer this because I want to back this with thermodynamic data that I can’t find, but in general your assumption is wrong; the intramolecular covalent bonds are orders of magnitude stronger than dipole dipole interactions. The dissociation of $\ce{HCl}$ in water is a reaction in which new bonds are formed.

Yes, intramolecular forces are stronger than intermolecular force. But notice that here the amount of $\ce{H2O}$ is much greater than $\ce{HCl}$, and not 1 molecule of $\ce{H2O}$ is trying to break 1 molecule of $\ce{HCl}$. Rather every molecule of $\ce{HCl}$ gets surrounded by the molecules of $\ce{H2O}$ and so the attraction by the intermolecular force gets multiplied by far and it helps to overcome the intramolecular force of $\ce{HCl}$.
Again, another noticeable fact is between $\ce{H2O}$ and $\ce{HCl}$, not only dipole dipole attraction is present but  hydrogen bonding is possible too which is another pretty strong intermolecular force compared to the others.

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text:
Why do gases need to be cooled to liquefy/solidify?
My book states:

Gases do not liquify on compression only, although molecules come close to each other and Weak forces operate at a maximum.

Why is it that we cannot just keep on compressing  gas till it solidified or liquified? Why do we need to reduce its temperature?
Also, if that is true, what would happen if I kept compressing a gas, without siphoning out the excess energy?

What your book states is not generally true. Two counter-examples:

Ammonia ($\ce{NH3}$) will liquefy at room temperature and a pressure of approximately $\pu{10bar}$. (CRC Handbook of Chemistry and Physics 44th ed; information as cited on Wikipedia’s data page).
Butane ($\ce{C4H10}$) will liquefy at room temperature and a pressure of just over $\pu{2bar}$. (Same source as cited on Wikipedia’s data page).

Especially the second is well-known to all those who have a butane lighter of transparent material. If it is half-full, a clear distinction between the liquid and gaseous phases can be seen inside at room temperature. Naturally, the inside of the lighter must be under positive pressure for gas to flow out if a flame is required.
On the other hand, there are a great many gases that cannot be liquefied at room temperature by applying pressure. One of these is nitrogen ($\ce{N2}$). In this case, applying enough pressure will make the sample transform directly from the gaseous to a solid phase.
The difference between nitrogen and ammonia/butane is their thermodynamic data, most importantly their critical point. The critical point defines the last position of the liquid/solid separation of a phase diagram. If the temperature is higher than the critical temperature, no liquid phase will be observed under any compression. Likewise, if the pressure is higher than the critical pressure, a substance will not liquefy even under extreme cooling (it will go straight to the solid phase).
The critical points of the compounds mentioned herein are:
$$\begin{array}{lrr}\hline
\text{Compound} & \vartheta_\text{crit} [\mathrm{^\circ C}] & p_\text{crit} [bar] \\
\hline
\ce{NH3} & 132.4 & 112.8\phantom{0} \\
\ce{C4H10} & 152\phantom{.0} & 37.96\\
\ce{N2} & -146.9 & 33.90\\ \hline \end{array}$$
As you can see, the critical pressures of all compounds are well above standard pressure (hence they can be liquefied under reduced temperature). Nitrogen’s critical temperature is below room temperature, so it cannot be liquefied by compression; the others can.

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Byproducts of LiAlH4 reduction of amides
I know that $\ce{LiAlH4}$ is able to reduce amides into amines, e.g. benzamide into benzylamine. However, I am unsure what other by-products are formed as a result.
$$\ce{LiAlH4 + C6H5CONH2 -> C6H5CH2NH2 + H2O + ??}$$

Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required:  e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]:

In the workup conditions, $\ce{H3O+}$ is followed by $\ce{^–OH}$. An aqueous acidic solution is often used to carry out the protonolysis step that follows the $\ce{LiAlH4}$ reduction (as shown in the following mechanism). The excess of acid that is typically used converts the amine, which is a base, into its conjugate-acid ammonium ion. Hydroxide is then required to neutralize this ammonium salt and thus give the neutral amine.

Although water itself rather than acid can be used in the protonolysis step, for practical reasons the acidic workup is more convenient. Thus, the extra neutralization step is required.

The latter $\ce{H2O}$-quenching step is the reason why $\ce{AlH3}$ is not remained among reactive species and, according to Brown, the brutto-formula of the metal-containing side-product is $\ce{LiAlO2}$ which is in agreement with stoichiometrical equation for reactants (amide : LAH = 1 : 0.5)[2]:
$$\ce{2RCONR'R'' + LiAlH4 -> 2RCH2NR'R'' + LiAlO2}$$
It is also worth noticing that if there is moisture, LAH readily reacts with water producing $\ce{LiOH}$ and $\ce{Al(OH)3}$. Also, there is a chance of side-reaction with the solvent, such as formation of lithium amide complexes with THF.
Recent review of catalytic and non-catalytic methods of amide reduction [3] also covers few more possibilities:

When less than stoichiometric quantities of $\ce{LiAlH4}$ are used for the reduction of primary amides, the corresponding nitrile may be formed:

Moreover, high selectivity toward the nitrile is also observed with amides containing electron-rich quinoline systems. If traces of water are present,
partial reduction of amides to aldehydes over $\ce{LiAlH4}$ can also comprise an undesirable competing reaction. Hydrolysis to the alcohol product has also been reported for secondary amide substrates.

GC analysis also shows presence of $\ce{LiOAlH3}$, $\ce{Li+AlH_xR_{4-x}-}$ in the system [4], though I would doubt that for benzamide $x$ can be less that 3.
References

Loudon, G. M.; Parise, J. Organic chemistry, 6th Ed.; Roberts and Company Publishers: Greenwood Village, Colorado, 2016. ISBN 978-1-936221-34-9.
Nystrom, R. F.; Brown, W. G. Journal of the American Chemical Society 1948, 70 (11), 3738–3740. DOI 10.1021/ja01191a057.
Smith, A. M.; Whyman, R. Chemical Reviews 2014, 114 (10), 5477–5510. DOI 10.1021/cr400609m.
Thiedemann, B.; Schmitz, C. M. L.; Staubitz, A. The Journal of Organic Chemistry 2014, 79 (21), 10284–10295. DOI 10.1021/jo501907v.


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Etymology of alanine
Oxford dictionary online gives etymology of alanine as:

Coined in German as Alanin, from aldehyde + -an (for ease of pronunciation) + -ine.

But I see no resemblance to the aldehyde structure in alanine. Is this etymology incorrect, or am I mistaken about the resemblance?

In the original German paper [1] Adolf Strecker used Aldehyd-Ammoniak or aldehyde-ammonia as a precursor, that's where the name derives from:

Vor einigen Jahren habe ich gezeigt, daſs Aldehyd-Ammoniak und Blausäure beim Erwärmen mit verdünnter Chlorwasserstoffsäure sich zu einer schwachen Basis, Alanin genannt, vereinigen [...]:
  $$\ce{\underset{\text{Aldehyd-Ammoniak}}{C4H4O2 * NH3} + HCl + \underset{Blausäure}{C2NH} +2 HO = \underset{Alanin}{C6H7NO4} + NH4Cl}$$

As David Richerby mentioned in the comments, Strecker's brutto-formula ($\ce{C6H7NO4}$) deviates from the moden one ($\ce{C3H7NO2}$), also the reaction scheme is a bit different.

Strecker, A. Annalen der Chemie und Pharmacie 1854, 91 (3), 349–351. DOI 10.1002/jlac.18540910309.

@andselisk Easy-to-understand answers to easy-to-understand questions always get lots of votes. Think of it as compensation for the times when you've spent an hour writing a detailed answer to a difficult question and it scores +2 because nobody can be bothered to read it.

Having said that, I'm a bit confused. Isn't alanine $\ce{C3H7NO2}$?

@DavidRicherby I'm not 100% sure, but most likely it's brutto-formula was established incorrectly at that time. Many other sources around that period of time (mid 19th century) refer to alanine as to $\ce{C6H7NO4}$, or, to meet the era's spirit, $\ce{C^6H^7NO^4}$.

Ah, I see. Might be worth adding a note about that to the answer.

User364914 may be confused by the "odd" formulas in Strecker's 1854 paper. At that time there was no universal agreement on the masses of atoms. Was carbon 6 or 12? Was oxygen 8 or 16? Strecker used 6 for carbon and 8 for oxygen.
The aldehyde is acetaldehyde, $\ce{NH3}$ is OK, blausaure is hydrogen cyanide and $\ce{OH}$ is water. The corrected structures should have half the number of carbons and oxygens. The modern equation should read:
$$\ce{C2H4O.NH3 + HCl + HCN + H2O -> C3H7NO2 + NH4Cl}$$
You may want to read about the Karlsruhe Congress of 1860.

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Why is density of water maximum at 4 degree celsius?
Why is density of water maximum at 4 degree celsius?
We were having a discussion to this question in class and I could gather the following points.
1) $H_2O$ exists in a cage like structure in ice form, due to extensive hydrogen bonding.
2) density=mass/ volume : We are not changing the mass here, therefore increase in volume will directly correspond to decrease in density and vice versa.
2) This cage like structure can trap gases and result in increase in volume. Therefore, forming of this cage like structure will result in increase in volume and breaking of this structure will result in decrease in volume.
3) At 273K , this cage like structure of ice exists as shown. On heating, we find that molecules gain kinetic energy and this cage like structure breaks. Due to this breaking, the volume decreases. At 277 K or (4 degree celsius), it so happens that the entire cage like structure is broken which results in decrease in volume resulting in maximum density.
Therefore, density of water is maximum at this temperature.
4) At this stage, the cage has been entirely broken. Now, the molecules form this cage like structure again due to which volume again starts to increase as a result of which density decreases.
Here's what I couldn't understand:

1) I cannot understand why trapping of gases results in increase of
  volume. The gases are trapped in the free space inside the cage.
  Whether that free space is occupied by gases or whether it remains
  unoccupied shouldn't result in increase of volume. Then why it's said
  that trapping of gases inside the cage like structure formed by ice
  results in increase in volume?
2) I am not sure I heard the 4th point right because it clearly
  doesn't make sense. If at 277K, the cage like structure broke, why it
  will reform on further heating. On heating, we are giving the
  molecules more kinetic energy.They should roam about and not get tied
  down to a cage like structure. Why that cage like structure is being
  formed again after 4 degree celsius?

https://chemistry.stackexchange.com/questions/73440/maximum-density-of-heavy-water-at-11-6-degree-celsius/73442?s=2|0.1249#73442

The occurrence of a maximum indicates two competing factors in different temperature regimes. At low temperatures, directed hydrogen bonding favors a low-density ordered framework; at high temperatures, the framework is disrupted by thermal motion, but this is compensated for by thermal expansion, which (I speculate) we might attribute to entropic effects. Hence a maximum density must exist. I don't know what to think about your point (4).

Trapping of gases isn't relevant to the basic properties of water and ice.

The answer appears to be fairly complicated, but fundamentally it results from a competition between two different options for water molecules interacting.
One place to start would be The structure of water; from ambient to deeply supercooled, Lars G.M. Pettersson and Anders Nilsson, J. Non-Crystalline Solids 407 399-417 (2015) where they discuss the 'inhomogeneous structure hypothesis'. At its heart they define high density liquid (HDL) and low density liquid (LDL) (not connected to cholesterol!) structures. LDL is tetrahedral, HDL is more distorted. LDL is directional H-bonded, while HDL is more van der Waals influenced. LDL minimized enthaly, HDL maximizes entropy. These are all consistent with a variety of experimental and simulation work given in the paper.
A recent paper applying DFT to the problem, Interstitial Voids and Resultant Density of Liquid Water: A First-Principles Molecular Dynamics Study, Sohag Biswas et al., ACS Omega 3 2010-2017 (2018), suggests a similar general picture. Again, the tradeoff of hydrogen vs van der Waals bonding appears to be important. Different exchange/correlation terms provide somewhat different answers. One suggested reason for the HDL structure is that, by incorporating an 'interstitial' water molecule into the LDL, one disrupts the directional hydrogen bonds and enhances the breaking and making of them, blurring out the structure (while being higher density).
So, it isn't a simple one-phase picture, but still seems to be a tradeoff of entropy and enthalpy of different configurations in the water.

" LDL minimized enthaly, HDL maximizes entropy. " should probably be " LDL maximized enthalpy, HDL maximizes entropy. "?


text:
C-F bond lengths of octafluoropropane using Bent's rule
I was trying to predict different $\ce{C-F}$ bond lengths of octafluoropropane,  $\ce{CF3-CF2-CF3}$ :

Applying Bent's rule here, we see that

For $\ce{C^2}$, we have $\ce{C^3}$ and $\ce{C^1}$ towards which s-character of $\ce{sp^3}$ orbitals is richer. Hence $\ce{sp^3}$ orbital of $\ce{C^2}$ has lesser s-character directed towards $\ce{F}$.
For $\ce{C^1}$ (or $\ce{C^3}$), we have only  $\ce{C^2}$ toward which  s-character of $\ce{sp^3}$ orbitals is richer. Hence $\ce{sp^3}$ orbital of $\ce{C^1}$ has greater s-character directed towards $\ce{F}$.

Using this I think $\ce{C^2 - F}$ bond length is greater than $\ce{C^1 - F}$.
Q. Is this explanation reasonable?
Q. Can we use Bent's rule to predict bond-lengths here or we should consider steric effects of $\ce{CF3}$ (trifluoromethyl) groups ? (here both lead to same result)
I mean in general on which rule should we rely on? On wikipedia article on Bent's rule its written that often times Bent's rule prediction would contradict predictions based on other classical effects alone (steric effects etc). In such cases how can we justify any prediction?
(Wikipedia) Bent's Rule - Bond lengths

A prediction based on sterics alone would lead to the opposite trend, as the large chlorine substituents would be more favorable far apart. As the steric explanation contradicts the experimental result, Bent’s rule is likely playing a primary role in structure determination.

Just for fun I've checked several crystal structures of various fluoroalkyls and I can not see the trend in $\ce{C-F}$ bond lengths for the terminal and internal carbons, and it can be inconsistent even within a single structure with more than one fluoroalkyl chain. But still there is a chance that octafluoropropane is different as I didn't find crystal structure of this particular molecule.

@andselisk thank you very much for the information! Actually this question was on our test and our teacher explained this using steric crowding. It might be a combination of various factors making predictions like mine irrelevant.

Fluorine and hydrogen have roughly the same size - in fact, fluorine is widely used as an isostere of hydrogen in medicinal chemistry, meaning that people replace hydrogen with fluorine to make new drugs or better drugs that have the same shape. I'd therefore be cautious about explaining this on the basis of supposed steric differences.

Could you explain why there would be "even lesser s-character"? The link is not very clear. Shouldn't there be more s-character, in my opinion?

@TanYongBoon Yes you are correct its a typing mistake! Thanks!

AFAIK, $\ce{CF3SO3H}$ is a stronger acid than $\ce{FSO3H}$ Does it affect your considerations?

@permeakra Hmm so I should consider group electronegativity instead of just considering just the directly bonded atom. That's very helpful! Thanks!

It is very much non-trivial to attempt estimations on a perfluorated alkane using Bent’s rule. In Bent’s rule, one should not only consider the effects of the electronegative substituents on the substituted atom itself, but also onto the next atom. Consider for example the following table taken from Bent’s review article:[1]
\begin{array}{lr}\hline
\text{Compound} & \angle(\ce{R^1-O-R^2}) \\ \hline
\ce{Me-O-Me} & 111^\circ \\
\ce{Me-O-H} & 107{-}109^\circ \\
\ce{H-O-H} & 104.5^\circ \\
\ce{F-O-F} & 103.8^\circ \\ \hline\end{array}
This very nicely corresponds to increasing group (!) electronegativities[2] in the series $\ce{Me} > \ce{H} > \ce{F}$. $\ce{CF3}$ should have a higher electronegativity than both $\ce{Me}$ and $\ce{H}$ (compare the acidities of formic acid, acetic acid and trifluoroacetic acid). Yet $\angle(\ce{C-O-C}) = 120^\circ$ in $\ce{F3C-O-CF3}$.[3] This corresponds to a higher $\mathrm s$ contribution than for the previous groups indicating other, more important contributions (here the anomeric effect of oxygen acting on the $\ce{CF3}$ group).[4]
Luckily, there are no such secondary effects in perfluoropropane. However, if you want to estimate all bond lengths compared to propane or 2,2-difluoropropane, these must be considered. Specifically, going from propane to 2,2-difluoropropane should not only reduce the $\ce{C-C}$ bond lengths due to increased $\mathrm s$ contribution but also due to partial double bond character due to a gauche-type effect which can be rationalised in Lewis formalism by the following resonance structure:
$$\ce{H-C-C-F <-> \overset{+}{H}\bond{...}C=C\bond{...}\overset{-}{F}}$$
Of course, the partial $\ce{C-C}$ double bond and the less-than-single $\ce{C-F}$ bonds also mess with the observed bond angles in 2,2-difluoropropane; they don’t only influence bond lengths.

So to attempt an estimation of the differences in bond length of the two $\ce{C-F}$ bonds in perfluoropropane, we would need to know how the group electronegativites of the $\ce{CF3}$ and $\ce{C2F5}$ groups compare to the electronegativity of fluorine and to each other. I don’t see any simple way to do this.
Mack et al. have published gas phase structures of 2,2-difluoropropane and perfluoropropane. These are their results (all values in picometres):[5]
\begin{array}{lcc}\hline
\text{compound} & d(\ce{C-C}) & d(\ce{F2RC-F}) & d(\ce{R2FC-F}) \\ \hline
\ce{Me2CF2} & 151.2 & \text{–/–} & 137.0 \\
\ce{(CF3)2CF2} & 154.6 & 132 & 134 \\
\ce{Me2CH2} & 153.1 & \text{–/–} & \text{–/–} \\ \hline \end{array}
As you can see, the experimental values just confirm your prediction of the central $\ce{CF2}$ having longer bond lengths than the terminal $\ce{CF3}$ groups. As expected, all $\ce{C-F}$ bonds in perfluoropropane are shorter than the bond lengths of 2,2-difluoropropane. And the $\ce{C-C}$ bond lengths increase from 2,2-difluoropropane to propane to perfluoropropane.
The conclusion is that indeed $\ce{CF3}$ is slightly less electronegative than fluorine, but the effect seems minimal.
References and Notes:
[1]: H. A. Bent, Chem. Rev. 1961, 61, 275–311. DOI: 10.1021/cr60211a005.
[2]: Bent explicitly considers group electronegativities. Only by that is the group electronegativity of a methyl group lower than the electronegativity of hydrogen.
[3]: A. H. Lowrey, C. George, P. D'Antanio, J. Karle, J. Mol. Struct. 1980, 63, 243–248. DOI: 10.1016/0022-2860(80)80330-9.
[4]: F. R. Leroux, B. Manteau, J.-P. Vors, S. Pazenok, Beilstein J. Org. Chem. 2008, 4, 13. DOI: 10.3762/bjoc.4.13.
[5]: H.-G. Mack, M. Dakkouri, H. Oberhammer, J. Phys. Chem. 1991, 95, 3136–3138. DOI: 10.1021/j100161a034.


text:
Why the tendency of ClO3 to form dimer is high?
I read that tendency of $\ce{ClO2}$ to form dimer is less and that of $\ce{ClO3}$ to form dimer is more.
I can understand the reason for the former because the unpaired electron is more delocalized in this case. So, if an electron is localized, i.e. it is at rest then the tendency of the molecule to form dimer will be high. Whereas if the electrons are delocalized, i.e. it is moving, then the tendency of the molecule to form dimer will be less.
I don't understand why the tendency to form dimer in the case of $\ce{ClO3}$ is high. There are vacant $\mathrm{d}$ orbitals with chlorine. Then why is that?

Strongly related: https://chemistry.stackexchange.com/questions/20083/what-is-the-structure-of-cl₂o₆

Yes ClO2 dimer is less stable and very much speculated. See this  and this.


text:
Why do stabilised ylids lead to trans alkenes in the Wittig reaction?
Why do stabilised ylids (such as $\ce{Ph3P=CHCO2Me}$) lead to (E)-alkenes in the Wittig reaction, in contrast to unstabilised ylids which are (Z)-selective?
The (E)-alkene is formed from stereospecific collapse of the more stable trans oxaphosphetane. Does the observed preference have anything to do with thermodynamic control over formation of the intermediate oxaphosphetane?

TL;DR: With stabilised ylids, the trans oxaphosphetane is formed, under kinetic control. This is due to minimisation of dipole-dipole repulsions in the transition state for oxaphosphetane formation:

There is no equilibrium established between cis and trans oxaphosphetanes. The formation of the trans oxaphosphetane is irreversible, and its collapse leads to the formation of the (E)-alkene.

Long answer
Part 1: Oxaphosphetane reversibility?
The mechanism of the Wittig reaction has historically been controversial. Pioneering research, most notably by Edwin Vedejs and his group, has led to our current understanding of the mechanism as being a concerted [2+2] cycloaddition to form the oxaphosphetane directly. For more information, see: Which is the currently accepted mechanism of a Wittig reaction?

Within this framework, the formation of (Z)-alkenes from unstabilised ylids is understood to be because of a minimisation of steric repulsion in the cycloaddition transition state, as was proposed by Vedejs in 1988.1 For more information, see: Why does the unstabilised Wittig reaction selectively form cis alkenes?
For a long time, it was believed that oxaphosphetane formation from stabilised ylids was reversible. Consequently, the trans oxaphosphetane was preferentially formed (due to minimisation of steric repulsions), under thermodynamic control. The collapse of this then led to (E)-alkene formation.
However, this has been proven to be all but incorrect (and is sadly still propagated on widely used websites such as Wikipedia). A major problem with this theory was that oxaphosphetanes formed from stabilised ylids decompose so fast that they are not detectable via NMR spectroscopy.2 It is therefore implausible that the reversal of oxaphosphetane to starting materials can occur before forward reaction to form products.
The nail in the coffin was provided by Vedejs.2 If oxaphosphetane formation (from stabilised ylids) is indeed reversible, then it would mean that if the cis oxaphosphetane is synthesised, it should decompose to form the (E)-alkene via isomerisation to the trans oxaphosphetane. While the oxaphosphetanes themselves cannot be synthesised (they simply decompose to products), the corresponding betaines can be synthesised by deprotonation of β-hydroxyphosphonium salts (their conjugate acids).
The work by Vedejs proved that the syn betaine - which necessarily cyclises to form the cis oxaphosphetane - decomposed stereospecifically to the (Z)-alkene, hence disproving the idea of reversibility in the Wittig reaction. First, the relevant β-hydroxyphosphonium salts were synthesised:

Deprotonation of this formed a betaine, which would immediately cyclise to form a cis oxaphosphetane (this was well-established by then; see the question linked earlier on the mechanism).

If any of the (E)-alkene was formed, then it would mean that there was a mechanism by which either the betaines, or the oxaphosphetanes could interconvert. This would therefore not be conclusive evidence as to whether oxaphosphetane formation was reversible. However, if none of the (E)-alkene was formed, then it would necessarily mean that betaine and oxaphosphetane formation were both irreversible.
In their experiments, an E:Z ratio of > 98:2 was observed for all cases except for R = Ph. This indicated that in fact oxaphosphetane formation was not reversible, and that the stereoselectivity of the Wittig reaction cannot be based on a thermodynamic argument.
For R = Ph, 8% of the (E)-isomer of ethyl cinnamate was observed. Experiments with added 4-chlorobenzaldehyde ruled out the possibility of reversal to the ylid + carbonyl, as any ylid formed would preferentially react with the more electrophilic 4-chlorobenzaldehyde, leading to ethyl 4'-chlorocinnamate - but this was not seen. Furthermore, ethyl cinnamate was shown to be configurationally stable under the experimental conditions. It was finally found out that this occurred because, instead of deprotonating on oxygen, DBU deprotonated the carbon atom α to the ester, which led to loss of stereochemical integrity. This was further supported by deuterium labelling studies.

As a consequence of this, Vedejs concluded that oxaphosphetane formation was irreversible, and noted that for reactions with a stabilised ylid, the Hammond postulate implied a late transition state, which would "feel" stabilising or destabilising influences of the oxaphosphetane. Vedejs therefore proposed an almost-planar transition state, with the stereoselectivity rationalised by steric repulsions between the $\ce{CO2Me}$ and $\ce{R}$ substituents in the TS:

Part 2: Steric repulsions or dipole alignment?
In 2003, Varinder Aggarwal's research group managed to synthesise stabilised ylids with a $\ce{P(OMe)3}$ group instead of the usual $\ce{PPh3}$, and discovered that these ylids led to substantially lower (E)-selectivity.3 Vedejs' model could not be used to rationalise these observations:

To explain these, Aggarwal and Harvey put forth a different model, based on dipole-dipole interactions and supported by DFT studies:4

For the usual $\ce{PPh3}$ ylids, there is a strong dipole pointing from ylid carbon to ester group; in the trans TS, this dipole points opposite to the dipole of the C=O group in the carbonyl component. Hence, there is electrostatic attraction in, and stabilisation of, the trans TS. This explains the usual high (E)-selectivity.
However, for the new $\ce{P(OMe)3}$ ylids, the dipole moment of the ylid is much smaller. This is because the negative charge on the ylid carbon is delocalised into not only the ester group, but also the σ* orbital of the P–O bonds, which are energetically low-lying. This accounts for the lower (E)-selectivity of the phosphite-derived ylid.
As far as I know, this is the currently accepted model for why stabilised ylids lead to (E)-alkenes. Our mechanistic understanding of the Wittig reaction is very nicely summed up by Aggarwal and Harvey,5 in a grand computational study of all types of ylids. In it, they provided a full potential energy surface for the reaction of $\ce{Ph3P=CHCO2Me}$ with benzaldehyde (note that there is a typo in the ylid starting material in the diagram):

Ending with a quote from their paper, which perfectly summarises this answer:

In the case of the triphenylphosphonium stabilized ylide, TS2 is very low in energy, showing that reversal cannot occur in these reactions, which is in agreement with experiment [...] E/Z selectivity is thereby determined solely by the relative energy of cis and trans TS1s. These energies, as well as the geometry of the respective TS1s, are strongly affected by a dipole-dipole interaction between the two reactants [...] The outcome for stabilized ylides is thus a high E selectivity.

References

Vedejs, E.; Marth, C.F. Mechanism of the Wittig reaction: the role of substituents at phosphorus. J. Am. Chem. Soc. 1988, 110 (12), 3948–3958. DOI: 10.1021/ja00220a037.
Vedejs, E.; Fleck, T. J. Kinetic (not equilibrium) factors are dominant in Wittig reactions of conjugated ylides. J. Am. Chem. Soc. 1989, 111 (15), 5861–5871. DOI: 10.1021/ja00197a055.
Aggarwal, V. K.; Fulton, J. R.; Sheldon, C. G.; de Vicente, J. Generation of Phosphoranes Derived from Phosphites. A New Class of Phosphorus Ylides Leading to High E Selectivity with Semi-stabilizing Groups in Wittig Olefinations. J. Am. Chem. Soc. 2003, 125 (20), 6034–6035. DOI: 10.1021/ja029573x.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. On the Origin of High E Selectivity in the Wittig Reaction of Stabilized Ylides: Importance of Dipole−Dipole Interactions. J. Am. Chem. Soc. 2005, 127 (39), 13468–13469. DOI: 10.1021/ja0539589.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. Reactivity and Selectivity in the Wittig Reaction: A Computational Study. J. Am. Chem. Soc. 2006, 128 (7), 2394–2409. DOI: 10.1021/ja056650q.


text:
Diastereomers and H-NMR
What effect does stereochemistry have on the number of signals in an H-NMR?
For example, if you consider the two structures below, would the number of signals and splitting differ due to the restricted rotation? If so, which hydrogens are being affected?
EDIT: This is what I think the splitting would look like.

Enantiomers have identical spectra. What you show are two different diastereomers. Different boiling point, different chemical behaviour, different spectrum.

How about you predict what the spectrum of each looks like, and I correct you if necessary? Not absolute chemical shifts, just the number of peaks and their splitting?

The methylene protons are equivalent in the first structure but not the second.

Note that cyclobutane rings are typically not planar but puckered. If I am not missing anything, the left-hand compound has $C_\mathrm{s}$ point group at most while the right-hand one has at least $C_\mathrm{2v}$.

The first thing you need to take into account is that cyclobutane rings are not planar, as they are drawn on paper, but puckered. Thus, a three-dimensional structure of the two isomers would look something like this:

I expect the structure on the left to be rather stable as it is, in the way I drew it, minimising steric interactions. Any other puckering would can either be traced back to that one by rotation or mirroring or put the two methyl groups into axial position which is sterically undesirable. Thus, we can expect only one significant species in solution.
This species should have $C_\mathrm{2v}$ symmetry if I didn’t miss anything. Using the principal axis — the one through the centre of the cyclobutane ring — we can identify four sets of homotopic hydrogens. We therefore expect four signals in the proton NMR spectrum.
Coupling constants, especially those along the ring, should be clearly visible and identifyable (if not confounded due to signal overlap).
The species on the right is less rigid. It can interconvert easily between different puckered states and in each one there will be one methyl group in axial position and one in equatorial position. Still, all structures can be traced back to the one I’ve drawn through mirroring and rotation unless I missed something. So here again, we have only one set of signals at all temperatures.
This species has a lower symmetry: $C_\mathrm s$. There is no rotation symmetry axis, so none of the ring protons are actually homotopic. Transforming one onto another is only possible by mirroring, hence they are enantiotopic. Since the methyl groups and the tertiary hydrogens are also non-identical, I expect a total of six signals at low temperature. However, repuckering would transform the axial into the equatorial methyl group and vice-versa. Therefore, I expect only a set of three signals to be visible at room temperature due to rapid interconversion.
These three signals would be:

the methyl group
the tertiary proton
the set of secondary protons

Since the protons rapidly exchange positions, I would expect the coupling constants at room temperature to be averaged. At very low temperatures, after freezing the rearrangements, these should become distinct.
So yes, rigidness of a ring due to stereochemistry does affect the number and type of signals. cis-1,3-dimethylcyclobutane is very rigid and unlikely to transform from one arrangement to another. This leads to a clean, full set of signals. trans-1,3-dimethylcyclobutane, on the other hand, interconverts rapidly between different orientations, mapping the otherwise non-identical methyl groups and protons onto each other. It therefore shows less signals than it should at low temperature.
A similar effect can be observed in cyclohexane and derivatives, where cooling to low temperatures also freezes the ring flip and causes the signals of axial and equatorial protons to be distinct.


text:
Titration with Ca(OH)2

In the laboratory, $20.0$ milliliters of an aqueous solution of calcium hydroxide, $\ce{Ca(OH)2}$, was used in a titration. A drop of phenolphthalein was added to it to indicate the end point. The solution turned colorless after $20.0$ milliliters of a standard solution of $\pu{0.050 M}$ $\ce{HCl}$ solution was added. What was the molarity of the $\ce{Ca(OH)2}$?
  (A) $\pu{0.010 M}$
  (B) $\pu{0.025 M}$
  (C) $\pu{0.50 M}$
  (D) $\pu{0.75 M}$
  (E) $\pu{1.0 M}$

I did not know how to answer such kind of question using formulas, that is why I applied logic. At first, I wrote such a reaction:
$$\ce{\underset{\pu{20 mL}}{Ca(OH)2} + \underset{\pu{0.05 M}}{\underset{\pu{20 mL}}{HCl}} = end point}$$
Then, I understood that if the volume of both reactants is equal, the molarity of $\ce{HCl}$ must be higher than that of $\ce{Ca(OH)2}$, as $\ce{HCl}$ changes the nature of $\ce{Ca(OH)2}$.
As I knew that the molarity of $\ce{Ca(OH)2}$ is just a bit less than that of $\ce{HCl}$, I chose B, and this variant turned out to be the right one.
In the book, I have such explanation:

In the titration, the reaction is:
  $$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$
  The acid to base ratio is $2 : 1$, or (moles acid used) = 2(moles base used), so $M_\mathrm{a}V_\mathrm{a} = 2M_\mathrm{b}V_\mathrm{b}$, where $M$ is the molarity and $V$ is the volume expressed in liters. Then
  $$M_\mathrm{b} = \frac{M_\mathrm{a}V_\mathrm{a}}{2V_\mathrm{b}}$$
  $$M_\mathrm{b} = \frac{\pu{0.05 M} \times \pu{0.02 L}}{2}$$
  $$\pu{0.02 L} = \pu{0.025 M}$$

I do not understand the explanation. Although it comes up with the same variant as I did, if I had such formula I would use it this way $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ as we have 2 moles of acid.
Then, the answer would be $0.1$. The same formula is used on this forum. Can you help me understand what is wrong here please?

Try reading about the terms 'molarity', 'normality' and 'valency factor' and  how they are related.

(1) Meaning of the textbook solution
$$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$
This reaction clearly tells us that
$$\frac{moles_{acid}}{moles_{base}}=2$$
moles=molarity* volume(in litres)
Therefore, $$\frac{moles_{acid}}{moles_{base}}=\frac{M_{a}V_{a}}{M_{b}V_{b}}$$
(2) What is wrong with your interpretation?
You wrote $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$.
$2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ actually implies that more number of moles of base need to be taken. For neutralisation, for every one mole of base, we require two moles of acid since each calcium hydroxide will release 2 hydroxide ions (for which two H+ are required).

You can also look at this in the aspect of calculating the number of mmols of HCl used to titrate the Ca(OH)2.  Once you have this, you divide that by 2 to get the number of mmol of Ca(OH)2 that were titrated.  Once you have that you divide that number of mmol by the number mL, and you get the molar concentration.
mmol of HCl = 20*0.05=1,
mmol of Ca(OH)2 = 1/2 = 0.5
M Ca(OH)2 = 0.5mmol/20mL = 0.025 M

Please do not write "equalities" between expressions which are not equal, like in your last sentence, which is wrong, and which is repeated here in italics :  1/2 = 0.5 M Ca(OH)2 = 0.5 mmol/20 mL = 0.025 M


text:
Transiesta Electrode Run
I am using a (1,1,1) gold electrode for electron transport experiments using Transiesta. I have never used it before. I am wondering if when I do the electrode run, do I build a system where it is the left and right electrodes connected to each other? or do I use the entire device (molecule AND electrodes) and just set a flag in the file?
(In response to andselisk's request for more information about Transiesta: it is a computational chemistry program that is a part of the Siesta Suite. It is used mainly for electron transport calculations.)

It looks like spam...

@Mithoron Actually I asked OP to provide a description of what Transiesta is (I'm that illiterate, yep:) ), if this is what made you think so. I don't think this is spam, more on a par with other software-related (e.g. Gaussian) questions, I believe.

Yeah, I agree with @andselisk - see https://departments.icmab.es/leem/siesta/.

I don't see any evidence for this being spam - @Mithoron, you can retract your flag if you change your mind.

How is this spam???? I do undergrad research in computational chemistry and molecular mechanics. I use software that sometimes has a lack of documentation. Some of the programs I use are Gabedit, Jmol, Gaussian, and now siesta. I am extremely confused as to how this question could be deemed as spam???

I'm not a pro myself, and am still experimenting/learning about SIESTA and TranSIESTA. What me and my colleagues usually do is run the electrode files, i.e., the left, then the right, then the device (which includes both the left and the right electrodes).
In response to your question, your electrodes should be part of your device, but in their respective .fdf files, they should have lattice parameters such that the electrodes are periodic and semi-infinite. Then in the main device, the left electrode coordinates should be listed first, followed by the device, then the right electrode. The electrodes need not be directly connected (I don't think you should, but then again, I've never tried that).
An example might help.
Say I have a rectangular sheet of atoms arranged in a $m$ by $n$ grid = $mn$ number of atoms (in the $x$ and $y$ direction respectively), and a unit cell of this sheet is $p$ by $q$, where $p$ = $m$ and $q$ = $n$/10. And the transport direction in this case is the $y$ direction (in the manual it must always be in the $z$ direction, but for the sake of simplicity we use $y$ here).
For an electrode of 3 unit cells (you can have any number, but the manual suggests at least 3 last I remembered), my left electrode .fdf file would include the coordinates of the first $p$ times 3$q$ atoms, with the lattice parameter block containing a cell size that would ensure the 3 unit cell electrode is periodic (so that it's semi-infinite to the side not connected to the device). I do the same for the the right electrode file, but this time with the last $p$ times 3$q$ atoms.
Moving on to the device, the .fdf file would have coordinates of all $xy$ atoms, arranged in the order of left ($y$ = 0 to 3$q$), middle (non-electrode, so coordinates of $y$ =
 3$q$ to $m$-3$q$), and the right ($y$ = $m$-3$q$ to $m$). The lattice parameter in this full device file doesn't quite matter as the electrodes I don't think, and typically I use a huge box so that there are no effects from periodic boundary conditions.
And that's how I usually do it. You can play around with the parameters and flags as well to suit your computation needs. After TranSIESTA I run tbtrans (which is a post-processor they provided), but I'm getting weird computation times with that. Also, this is SIESTA 4.0 I'm using, so double check the manual if your version is different.
Feel free to comment or provide suggestions on my methods, as I would appreciate if I could get more insight on how to use the software.

Ok so there are no specific flags that need to be put in the FDF when I am running transiesta on the unit cell for the electrode? I am using a 1,1,1 gold molecule as it is the most stable crystalline structure. I am using 3-7-3-7 as the unit cell. So that is what I put in the calculation for the electrode, but is there any thing i need to add so that it knows its a electrode? Or just when I do the entire scattered region the unit cell output goes in where the right and left electrodes are?

Not on the electrodes no. By executing TranSIESTA on the left and right .fdf files you'll get .TSHS files, which you then specify in your scattering region .fdf file where it can find the left and right electrode .TSHS files. The only place you need to 'tell' TranSIESTA the electrodes would be the scattering region .fdf file - that is, you set the paths to your left and right electrodes in that scat. region file.
Maybe go through the examples that in your SIESTA folder (they come with the compilation), and maybe some internet examples as well. You'll understand better with familiarity.

There is a manual available that states (page 111).

TranSiesta
  defines the Left Electrode to be the first atoms specified in the SR
  .fdf
  file,
  and the Right Electrode to be the last ones.  The transport direction has to be considered
  to be the the third cartesian axis, the
  z
  axis.  The Left Electrode atoms must have smaller
  z
  components than the Right Electrode atoms.  It is also crucial that the atomic positions
  specified at the left (right) EL calculation must be equivalent to the left (right) electrode
  part of the SR setup.  Here, equivalent means that they can be made equal by a simple
  translation  in  space.   It  is  also  possible  to  use  buffer  atoms.

There is more information that I think may address further doubts in the manual as well. I am not very familiar with the program. So, this is as much as I can help you. You could try asking the mailing list, though SIESTA's mailing list does not get many responses to be honest.


text:
How is ATP generated by cellular respiration?
I learned in a high school level Biology class that the chemical equation for cellular respiration is
$$\ce{6O2 + C6H12O6 -> 6CO2 + 6H2O + ATP}$$
When I looked up the chemical formula for ATP I found that it is
$$\ce{C10H16N5O13P3}$$
Knowing that, that must mean that the first equation becomes
$$\ce{6O2 + C6H12O6 -> 6CO2 + 6H2O + C10H16N5O13P3}$$
One of the products contains nitrogen and phosphorus, but none of the reactants contain those two elements, but there should be in order to produce a product that contains nitrogen and phosphorus. With my high school knowledge, I cannot figure this out. Where do the nitrogen and the phosphorus atoms come from?

sigh Insert "+energy which harvested  and then magazined in ATP" instead of simply write ATP if you wish, but it doesn't exactly fit into equation, doesn't it?

But isn't ATP one of the products in the chemical equation?

No... it isn't. Very simply speaking, that equation, without ATP inside it, produces energy. The body takes that energy and channels it into a second reaction: ADP + phosphate + energy -> water + ATP. Thus, ATP is not formed directly in that reaction; it is an indirect product. What you're learning in your biology class is a further simplification of what I've just said.

Oh ok, now I understand. So the chemical equation produces energy which is then used in another reaction. Thank you.

One note at the beginning: please do not be scared! You can scroll right down to the bottom of this answer to read the conclusion which in itself should sufficiently answer the question. The bits in between are merely elaboration and need not be known in full detail.

The reaction one typically receives in high school biology classes to explain cellular respiration is already a simplification on many levels. It is presented as $(0)$, but in reality that is not one single reaction. Whether or not $\ce{ATP}$ has already been put in is irrelevant.
$$\ce{C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O}\tag{0}$$
In fact, this reaction is a series of steps, each performed by a single, specific enzyme. The first reaction, that basically immediately happens when glucose enters an animal cell is phosphorylation as shown in $(1)$.
$$\ce{\underset{{glucose}}{C6H12O6} + \underset{{ATP}}{C10H12N5O13P3^4-} ->[enzyme] \underset{{glucose-6-phosphate}}{C6H11O9P^2-} + \underset{{ADP}}{C10H12N5O10P2^3-} + H+}\tag{1}$$
I have written $\ce{ATP}$ and will continue to do so for all products in the form they are in at physiological $\mathrm{pH}$
Glucose-6-phosphate is isomerised to fructose-6-phosphate, which cannot be distinguished by sum formulae alone, so I’m leaving this step out. Another enzyme comes along which phosphorylates fructose-6-phosphate again to give fructose-1,6-bisphosphate; see $(2)$. So far, all we have done is use up $\ce{ATP}$ at no benefit for the cell.
$$\begin{multline}\ce{\underset{{fructose-6-phosphate}}{C6H11O9P^2-} + \underset{{ATP}}{C10H12N5O13P3^4- ->[enzyme] }} \\ \ce{\underset{{fructose-1,6-bisphosphate}}{C6H10O12P2^4-} + \underset{{ADP}}{C10H12N5O10P2^3-} + H+}\end{multline}\tag{2}$$
In a retro-aldol reaction, fructose-1,6-bisphosphate is split into glycerinaldehyde-3-phosphate (commonly abbreviated GAP) and dihydroxyacetonephosphate (commonly abbreviated DHAP). These once again are structural isomers that bear the same sum formula and they are interconverted. Hence I am pretending in equation $(3)$ that only GAP is a product. This assumption is close enough to valid in sum formulae.
$$\ce{\underset{{fructose-1,6-bisphosphate}}{C6H10O12P2^4-} ->[enzyme] 2 \underset{{GAP}}{C3H6O5P^2-}}\tag{3}$$
We have now reached the phase of glycolysis (the consumption of glucose) called the pay-off phase. Previously, as you may have noticed, all we did was consume ATP. It is now almost time to get some back. But first, we need to prepare GAP for its prominent role in reaction $(4)$:
$$\begin{multline}\ce{2 \underset{{GAP}}{C3H5O6P^2-} + 2 HPO4^2- + 2 \underset{NAD+}{C21H26N7O14P2-}} \\ \ce{->[enzyme] 2 \underset{{1,3-bisphosphoglycerate}}{C3H4O10P2^4-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{4}$$
It may seem confusing that I labelled an anion as $\ce{NAD+}$ (its common name). This is because there are two phosphate diesters that can each be protonated once. Fully protonated, $\ce{NAD+}$ is indeed a cation — only not at physiological $\mathrm{pH}$.
In the previous step, we synthesised a phosphoric carboxylic mixed acid anhydride. These are very energetic and consequently its hydrolysis is now used in reaction $(5)$ to synthesise $\ce{ATP}$. We are breaking even.
$$\begin{multline}\ce{2 \underset{{1,3-bisphosphoglycerate}}{C3H4O10P2^4-} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} ->[enzyme]} \\ \ce{2 \underset{{3-phosphoglycerate}}{C3H4O7P^3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{5}$$
After this step, we again have an isomerisation that cannot be shown in sum formulae. The phosphate group is moved from position 3 to position 2 giving 2-phosphoglycerate. This can then be dehydrated as shown in $(6)$ to give phosphoenolpyruvate: a very important cellular intermediate. We have also generated our first two water molecules.
$$\ce{2 \underset{{2-phosphoglycerate}}{C3H4O7P^3-} ->[enzyme] 2 \underset{{phosphoenolpyruvate}}{C3H2O6P^3-} + 2 H2O}\tag{6}$$
Finally, a second set of two $\ce{ATP}$ molecules is formed by dephosphorylating phosphoenolpyruvate to give free pyruvate as shown in reaction $(7)$.
$$\begin{multline}\ce{2 \underset{{phosphoenolpyruvate}}{C3H2O6P^3-} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 H+} \\ \ce{->[enzyme] 2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{7}$$
We can draw an intermediate line here, since the process typically known as glycolysis has reached its close. We can sum it up in reaction $\text{(1–7)}$. We still have not generated any $\ce{CO2}$ and only $\ce{2 H2O}$. We also already have $\ce{2 ATP}$ per molecule of glucose and 2 mysterious molecules called $\ce{NADH}$.
$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 HPO4- + 2 \underset{NAD+}{C21H26N7O14P2-} ->[steps]} \\ \ce{2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+ + 2 \underset{{ATP}}{C10H12N5O13P3^4-} + 2 H2O}\end{multline}\tag{1–7}$$
The next step is still very easy to explain. It is called oxidative decarboxylation and adds a cofactor known as coenzyme A of $\ce{CoA}$ covalently to what was pyruvate before the reaction. See equation $(8)$.
$$\begin{multline}\ce{2 \underset{{pyruvate}}{C3H3O3-} + 2 \underset{NAD+}{C21H26N7O14P2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} ->[enzyme]} \\ \ce{2 \underset{{acetyl-CoA}}{C23H35N7O17P3S^3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 CO2}\end{multline}\tag{8}$$
Acetyl-CoA now enters into a magic known as the Krebs cycle or citric acid cycle. As the name implies, this is constantly going around. At the beginning, the two carbon atoms of acetyl-CoA are added to oxaloacetate. After the ten cycle steps, we will have lost two molecules of $\ce{CO2}$, effectively reforming the oxaloacetate substrate. And once again some of the steps are merely isomerisations which do not show up in some formulae. Are you ready? Here goes. Reaction $(9)$ has already been described in the text above:
$$\begin{multline}\ce{2 \underset{{acetyl-CoA}}{C23H35N7O17P3S^3-} + 2 \underset{{oxaloacetate}}{C4H2O5^2-} + 2 H2O} \\ \ce{->[enzyme] 2 \underset{{citrate}}{C6H5O7^3-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-}}\end{multline}\tag{9}$$
After isomerisation of citrate to give isocitrate, this is oxidised to give oxalosuccinate as shown in $(10)$.
$$\begin{multline}\ce{2 \underset{{isocitrate}}{C6H5O7^3-} + 2 \underset{NAD+}{C21H26N7O14P2-} ->[enzyme]} \\ \ce{2 \underset{{oxalosuccinate}}{C6H3O7^3-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{10}$$
For each oxalosuccinate, one molecule of $\ce{CO2}$ can be removed to give α-ketoglutarate as shown in $(11)$.
$$\ce{2 \underset{{oxalosuccinate}}{C6H3O7^3-} + 2 H+ ->[enzyme] 2 \underset{\alpha{-ketoglutarate}}{C5H4O5^2-} + 2 CO2}\tag{11}$$
α-ketoglutarate is then once again attacked by coenzyme A in the same manner as in step $(9)$. This can happen because both pyruvate and α-ketoglutarate have the same structural feature. Refer to equation $(12)$.
$$\begin{multline}\ce{2 \underset{\alpha{-ketoglutarate}}{C5H4O5^2-} + 2 \underset{NAD+}{C21H26N7O14P2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} ->[enzyme]} \\ \ce{2 \underset{{succinyl-CoA}}{C25H36N7O19P3S^4-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 CO2}\end{multline}\tag{12}$$
We have now lost all $\ce{CO2}$. What we need to do is regenerate the oxaloacetate from the beginning of the Krebs cycle. The first step of this is to generate succinate from succinate-CoA by hydrolysis as given in $(13)$.
$$\begin{multline}\ce{2 \underset{{succinyl-CoA}}{C25H36N7O19P3S^4-} + 2 H2O + 2 \underset{{ADP}}{C10H12N5O10P2^3-} + 2 HPO4^2-  ->[enzyme]} \\ \ce{2 \underset{{succinate}}{C4H4O4^2-} + 2 \underset{{coenzyme A}}{C21H33N7O16P3S^3-} + 2 \underset{{ATP}}{C10H12N5O13P3^4-}}\end{multline}\tag{13}$$
Next, we oxidise the saturated succinate molecule to an unsaturated fumarate. This is typically perfomed by yet another cofactor called flavin-adenine-dinucleotide or $\ce{FAD}$; see reaction $(14)$.
$$\ce{2 \underset{{succinate}}{C4H4O4^2-} + 2 \underset{FAD}{C27H31N9O15P2^2-} ->[enzyme] 2 \underset{{fumarate}}{C4H2O4^2-} + 2 \underset{FADH2}{C27H33N9O15P2^2-}}\tag{14}$$
Water is added to the double bond to give malate (reaction $(15)$):
$$\ce{2 \underset{{fumarate}}{C4H2O4^2-} + 2 H2O ->[enzyme] 2 \underset{{malate}}{C4H4O5^2-}}\tag{15}$$
And malate is finally oxidised by using $\ce{NAD+}$ as shown in $(16)$ to give the starting material of the Krebs cycle, oxaloacetate.
$$\begin{multline}\ce{2 \underset{{malate}}{C4H4O5^2-} + 2 \underset{NAD+}{C21H26N7O14P2-} ->[enzyme]} \\ \ce{2 \underset{{oxaloacetate}}{C4H2O5^2-} + 2 \underset{NADH}{C21H27N7O14P2^2-} + 2 H+}\end{multline}\tag{16}$$
Technically, we could end it here. Let’s look again at the final products of reactions $(1–16)$. Thankfully, a lot cancels out.
$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 4 \underset{{ADP}}{C10H12N5O10P2^3-} + 4 HPO4^2- + 10 \underset{NAD+}{C21H26N7O14P2-} +} \\ \ce{2 \underset{FAD}{C27H31N9O15P2^2-} ->[steps] 6 CO2 + 6 H+ +} \\ \ce{4 \underset{{ATP}}{C10H12N5O13P3^4-} + 10 \underset{NADH}{C21H27N7O14P2^2-} + 2 \underset{FADH2}{C27H33N9O15P2^2-}}\end{multline}\tag{1–16}$$
We’re still not there yet. We have generated a lot of $\ce{NADH}$ and $\ce{FADH2}$ which somehow needs to be transformed back into $\ce{NAD+}$ and $\ce{FAD}$ for the next iteration to be performed — remember that cellular energy is, as you noted in your question, typically equated to $\ce{ATP}$. And we still haven’t used oxygen at all and are not producing enough water.
The next set of reactions to bring us closer to our goal is the electron transport chain. By a series of reactions that is not easily described (but basically involve electrons being moved in redox processes from one compound to another), oxygen molecules acquire four electrons and four protons each to become water. The electron transport chain that performs this will, at the same time, use the energy liberated by the generation of water to set up a proton potential: protons ($\ce{H+}$) get pumped from one side of an intracellular membrane to another.
This proton potential of course is a driving force since there is a lack of protons on the other side. Protons may travel back through special trans-membrane proteins. These proteins can be imagined like a watermill and start rotating as protons pass through. This rotation allows them to literally slap a phosphate onto an $\ce{ADP}$ molecule to give $\ce{ATP}$.
As a theoretical maximum, the regeneration of $\ce{NAD+}$ from $\ce{NADH}$ gives around 2.5 equivalents of $\ce{ATP}$ while $\ce{FADH2}$ can only be used to generate around 1.5 equivalents of $\ce{ATP}$ upon the reverse reaction to $\ce{FAD}$. These numbers are, however, generalised, since some energy is also lost in transport and other processes.
Therefore, our theoretical maximum yield of $\ce{ATP}$ is 25 from $\ce{NADH}$, a further 3 from $\ce{FADH2}$ and a further 4 that are directly produced in the reaction sequences. Each of these $\ce{ATP}$s is generated by the reaction of a phosphate anion onto an $\ce{ADP}$ molecule, technically involving the loss of water or an equivalent.

As you may see, writing up a complete reaction for the entire cellular respiration process is very difficult. We can, however, arrive approximately there by a few basic principles:

there are up to $\ce{32 ATP}$ generated over the course of the digestion of one molecule of glucose
each $\ce{ATP}$ is generally synthesised by some form of $\ce{HPO4- + ADP}$ — therefore, the phosphorus part is balanced.
the exact process is not necessarily important; a lot cancels out and the end result counts.

This in total may lead us to question why exactly the reaction taught at schools is often the one presented in equation $(0)$. Well, we can see that exactly six molecules of carbon dioxide are formed in the overall process per molecule of glucose. This makes sense, as it means each glucose carbon atom effectively leaves the body as $\ce{CO2}$, the carbon compound with the lowest remaining energy. The number of oxygen and water molecules cannot, unfortunately, be accurately established for the single process. However, schools want to teach the concept of a balanced equation. And if all carbon leaves the body as $\ce{CO2}$, then all hydrogen must eventually leave it as $\ce{H2O}$. To accomplish that, a certain set of oxygen molecules is needed — exactly 6.
The reaction $(0)$ thus never actually happens in our body. That is a good thing. You can see what it looks like by heating suger to very high temperatures and then adding a match — actually don’t. It’s the same combustion as any other combustion. Part of it can be adequately summarised of many partial reactions, part of it is just there so the summary fulfills the required stoichiometry. And even the $\ce{32 ATP}$ are only an approximation based on a lot of averaging.
What we can definitely say, however, is that each of the $\ce{32 ATP}$s is produced by the reaction of phosphate and $\ce{ADP}$. And therefore, at least consistent with your high-school level, the reaction should better be written as:
$$\begin{multline}\ce{\underset{{glucose}}{C6H12O6} + 6 O2 + 32 HPO4- + 32 \underset{ADP}{C10H12N5O10P2^3−}} \\ \ce{-> 6 CO2 + 6 H2O + 32 \underset{ATP}{C10H12N5O13P3^4−}}\end{multline}\tag{full}$$

Oh @Jan didn't you know that the best way to scare someone is by telling them "Don't be scared"? I suggest you add a tl;dr at the beginning.

@PrittBalagopal if you feel you can write a good one, be my guest!

@NilayGhosh I was called out for it on [meta.se] ;)

can you describe that why school text book  write the combustion reaction of glucose to represent respiration,,in actual respiration 10 H20  are formed and not 6,,

@Jan i am a high school student ,,can you please describe me in brief that why school text book write the combustion reaction of glucose to represent respiration in actual respiration 10 H20 are formed and not 6

@ArunBhardwaj I can’t. It doesn’t make sense to me. A balanced reaction with six oxygen molecules and one glucose molecule would yield six water molecules.

@Jan, I am also very confused, also in respiration the oxygen are needed just for the last step after the ETS and not anywhere in between , but we know in aerobic respiration 6 CO2 molecules are released for 1 glucose, Ok so we have all of those 6 carbons from glucose but along with them we have 12 oxygen also, but glucose only have 6 oxygen from where do they come from? I don't even know? combustion is way different from respiration it should not be represented in the same manner, isn't it?

@ArunBhardwaj No, thermodynamically they’re the same. The route a process takes does not matter as long as starting and ending points are the same. So a total of 12 extra oxygen atoms must be added in some way or another.

@Jan but the ending point is not same, we have 10 H2O and not 6 ??

@ArunBhardwaj Well, except you don’t have ten. There must be an error because thermodynamics requires you to get six water molecules from the equation. Unless you’re putting in extra hydrogens and oxygens.

This is a very good question imo, and I had this doubt about a year ago. The fact is, that ATP isn't produced in the reaction, as orthocresol had mentioned, but rather the energy obtained from the respiration is used to synthesize ATP from ADP. The ATP serves as an "energy currency". When the energy is required, the ATP is converted back to ADP and the released energy is used for whatever biological purpose.
For further reading, head over to Wikipedia's article about Adenosine Triphosphate.


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Why does iron have an abnormally high ionization energy?
Along a period the ionization energy should increase because the atomic number is increasing, but there is negligible increase in shielding. However, $\mathrm{IE}_\ce{Mn} < \mathrm{IE}_\ce{Fe} > \mathrm{IE}_\ce{Co} > \mathrm{IE}_\ce{Cu}$, so why does iron have such a high ionization energy?
I've considered that iron has a $\mathrm{4s^2 3d^6}$ configuration that goes to a $\mathrm{4s^1 3d^6}$ configuration upon ionization, while $\ce{Co}$ has a $\mathrm{4s^2 3d^7}$ that goes to $\mathrm{4s^0 3d^8}$ configuration, $\ce{Ni}$ has a $\mathrm{4s^2 3d^8}$ that goes to a $\mathrm{4s^0 3d^9}$ configuration and $\ce{Cu}$ has a $\mathrm{4s^2 3d^9}$ that goes to a $\mathrm{4s^0 3d^{10}}$ configuration (Lang, J. Chem. Ed., 2003).  In the Long paper, the authors explain the increase of IE at iron is due to "electronic structure" and I would like to know exactly what about the electronic structure it is that causes iron to have a high IE than expected based on a general periodic trend.
I was thinking that adding an electron to the $\mathrm{3d}$ orbitals might be energetically favorable enough to lower the ionization energy of $\ce{Ni}$, $\ce{Co}$ and $\ce{Cu}$, so rather than iron having a high ionization energy the surround metals just have a low ionization energy because the ion is stabilized by having another $\mathrm{d}$-electron added? Does that sound reasonable?
Another thing I was thinking was that iron is the first transition metal to have a paired $\mathrm{d}$-electron, could that effect the shielding and increase the ionization energy?

The first ionization energy of cobalt is only 0.3% lower than that of iron. So I wouldn't say that iron has an abnormal high ionization energy.

@aventurin I mean "abnormal" in that it defies a general periodic trend, the magnitude itself is not abnormal

"I mean "abnormal" in that it defines a general trend" makes no sense to me at all. Did you mean "defies"?  Take a look at the chart here: https://en.wikipedia.org/wiki/Ionization_energy.
By my count, for the s-block elements the "trend" always holds.  For the p-block elements, it holds in 5/6ths of the cases. And for the d-block, I can't tell from the chart, but claiming it is increasing for them seems agross exaggeration to me; I'd say they might have a very slight tendency to increase, but they're so close to being flat-lined, that it seems to be not a useful claim. If you count the exceptions to the rule, and you expect your "rules" to hold at least 90% to 95% of the time (this isn't the Social Sciences, after all), then it is a just plain lousy rule. Nothing unusual about that: I think the Periodic Table is best when used to "predict" Atomic Number (they increase moving right and down, without exception), and so untrustworthy with the other periodic trends as to not be something in my "mental toolbox". Use it at your own peril. (The at. no. "trend" was a joke, of course)

I did mean "defies", I apologize for the confusion and I fixed my mistake.  I know the trend is not a rule, like all other trends, but this is still something that people research and have published papers about.  I'm unclear on an aspect  of a peer-reviewed paper, and I would like an answer, I don't see why that's a problem.

Could you please add some formatting, especially paragraph breaks to your answer? As it is, it is too time-consuming for me to read thus I am abstaining from voting.


text:
Order of acidity in aromatic molecules with various functional groups
What factors need to be considered in order to rank the acidity of aromatic molecules with various substituents?

Given benzene with 1 $\ce{OH}$ group. How does adding the following functional groups affect acidity, and what are the rules to determine this?

an additional $\ce{OH}$ group
$\ce{Cl}$
$\ce{NO2}$
$\ce{NO2 + Cl}$
$2\times\ce{NO2}$ groups
$3\times\ce{NO2}$ groups

Based on what I know about acids, the following increase acidity:

charge - the less electrons, the more acidic (more able to grab hydrogen)
electronegativy increases.
compounds with more resonance (more double and triple bonds) - more stability.
more electron withdrawing groups.
lower orbitals - the more s character (less p,d,f), the closer the electrons are to the nucleus, and the lower the energy, which means more stability.

The following decreases acidity

steric hindrance - where 3D structure of a molecule make solvent less accessible, and thus more weakly stabilised. (The difference may be small)
larger molecules tend to require a larger number of solvent molecules to solvate them.

Draw the resonance structures for each of the molecules.


text:
Physical meaning of thermal parameters dimension (Å²)
In single-crystal x-ray crystallography both isotropic and anisotropic displacement parameters $U_{ij}$ of thermal ellipsoids have dimension of square  angstrom ($Å^2$) as follows from the definition of Debye–Waller factor ($T$ is dimensionless):

For isotropic approximation (conditions of Bragg's law):
$$T_\text{iso} = 8\pi^2U\left(\frac{\sin \Theta}{\lambda}\right)^2 \to [U] = Å^2 \tag{1}$$
For anisotropic displacement parameters (ADPs):
$$T_\text{ani} = 2\pi^2 \sum_{i=1\\j=1}^3{H[i]H[j]U_{ij}e^\star[i]e^\star[j]} \to [U] = Å^2 \tag{2}$$
where $H$ is reflex's index; $e^\star$ is length of the reciprocal lattice basis vector.

In both cases $U_1$, $U_2$ and $U_3$ ($U_{ij}$ in general) are basically showing contribution of oscillation along three principal (orthogonal) axes and are measured in $Å^2$. So, is there any physical meaning that can be attributed to square angstrom in this case? At first glance, it seems rather confusing that a parameter that seems to represent a linear quantity is measured in square units of length.

Take a square root and be done with it.

The link you provided for the definition of Debye-Waller factor gives a different equation for $T_{iso}$ then you have included here. In the link, the U does seem to be squared. Also, it should be that the exponential of what you wrote is $T_{iso}$

@Tyberius I'm not sure which equation you are pointing to, and I mostly provided the link for DWF ($B$) as a quick reference what this thing is. In general $B=8\pi^2U=8\pi^2\bar{u^2}$, if that is what confused you ($\bar{u}$ differs from $U$ as it is a root-mean-square amplitude of the atomic vibration).

I'm referring to the first equation in that link. "However, as can be seen from the basic expression for the isotropic Debye-Waller factor, $T=\exp(-8\pi^2\langle u^2\rangle \sin^2(\theta)/\lambda^2)$, the larger the second moment, the more rapidly the scattering from the atomic center in question falls off with increase in the scattering angle." So it seems as though the factor to consider is $\langle u^2\rangle$, the variance, which should have units of angstroms squared.

As given by @Tyberius the $\langle u^2 \rangle $ term is the mean square displacement of the lattice points perpendicular to the scattering plane. As the scattering also depends on $\sin^2(\theta)/\lambda^2$ for a given  set of displacements the diminution in intensity becomes more important for large scattering angles.

..and some more points I forgot about. (a) Debye assumed that the lattice points move independently from one another, (b) the average is over a longer time than lattice vibrations, probably only important for fs duration experiments and (c) as the calculation producing the Debye-Waller factor involves averaging displacements from lattice positions the sharpness of peaks is not affected only their intensity .

@porphyrin it seems you're most of the way towards an answer...

This looks like it has an answer by @Tyberius disguised as a comment...

@NightWriter I figured a combination of mine and porphyrin's comments would answer the question. The main reason I didn't answer at the time was that I don't know much about thermal ellipsoids, so I wasn't certain it addressed his concern.

Model of a crystal and deriving the expected diffraction image
In single-crystal X-ray crystallography, the diffraction image is due to scattering of electrons in a crystalline sample. A convenient way of describing the electron density is to first specify crystal symmetry and mean atomic positions (coordinates), and then to describe in more or less detail how atoms deviate from these mean positions (modeled by atomic displacement factors). Combining this with a spherical description of the electrons "belonging" to each atom (modeled by atomic form factors), you arrive at an electron density.
This description is useful because in the model, the electron density is the convolution of the crystal lattice points with the coordinates in the unit cell with the distribution of atoms around mean positions with the distribution of electrons around the atoms. The convolution theory states that the Fourier transform of a convolution is the product of the Fourier transforms. So in this model, you can separately Fourier-transform the different levels of the model and then multiply them to arrive at the expected diffraction image (via the structure factors).
Atomic displacement and Debye-Waller factor
The atomic displacement around a mean position may be described through a probability density function $p_k(\mathbf{u})$, where $\mathbf{u}$ is the position (3D-vector) with the mean position as origin. The Fourier transform of this function defines the Debye-Waller factor (source: equation 1.4.8):
$$T_k(\mathbf{h}) = \int p_k(\mathbf{u}) e^{2 \pi i \mathbf{h} \cdot \mathbf{u}}d^3 \mathbf{u}\tag{3}$$
Here, $\mathbf{h}$ represents coordinates in reciprocal space (just as Fourier transform of a time series gives you information in frequency space, the Fourier transform connects real space - coordinate $\mathbf{u}$ - with reciprocal space - coordinate $\mathbf{h}$).
If the displacement is modeled by a Gaussian, the integral can be determined as (source: equation 1.4.10)
$$T_k(\mathbf{h}) = e^{-2 \pi^2 \langle(\mathbf{h} \cdot \mathbf{u})^2\rangle}\tag{4}$$
This reflects that the Fourier transform of a Gaussian is a Gaussian again, and the width of one Gaussian is the inverse of the other (i.e. if the atoms are widely distributed, the signal will diminish a lot with increasing resolution). Again using the time series as an example, if a sound is played for a very short time, the pitch will be ill-defined (or if an NMR FID signal decays rapidly, the peaks will be broad).
Displacement parameters $\mathbf{U}$
The displacement parameters $\mathbf{U}$ or $\mathbf{U^{jl}}$ in the OP's equation (1) and (2) are derived from the $\langle(\mathbf{h} \cdot \mathbf{u})^2\rangle$ expression in equation (3), either direction-independent in the case of an isotropic approximation (spherical distribution around the mean) or separately along the three lattice axes (leading to six anisotropic displacement factors). I am not explicitly showing this - it gets a bit complicated because the derivation requires transformation into the sometimes non-orthogonal coordinate system based on the crystal lattice (see here). In any case, the expression is a mean square displacement (analogous to a variance of a one-dimensional function). For that reason, the dimensions are length squared and the typical units are $Å^2$. As stated in the IUCR nomenclature document:

the elements of the tensor U have dimension (length)2 and can be directly associated with the mean-square displacements of the atom considered in the corresponding directions

and

If the atomic pdf is assumed to be a trivariate Gaussian, the characteristic function corresponding to this pdf - by definition, its Fourier transform - can be described by the second moments of the pdf, which in the present context are called anisotropic mean-square displacements.

Second moments (variance) have dimensions that are the square of the dimension of the quantity described by the probability distribution function (pdf). In this case, we are looking at the distribution of the position (dimensions length), and the second moment will have dimensions of length squared.

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


text:
Can you model intermolecular interactions by using the Born-Oppenheimer approximation to get a potential function for nuclei?
Here is a claim about the modeling of chemical systems. I can’t remember if it’s true or not:
Suppose we have a bunch of molecules and we want to know how they interact. Eg we want to look at the physics of two $\ce{CO_2}$ molecules bumping into each other. We can do this by treating the nuclei as classical particles moving in a potential function determined by the ground state electronic wavefunction for a given configuration of nuclei.
Eg, we want to know how hydrogen molecules bounce off each other. We do this by modeling it as two protons with momentums and positions, and we iterate forward through time using a Hamiltonian where the potential energy operator uses the Hartree-Fock method or something to calculate the electronic potential.
This approximation obviously ignores some important things, for example the momentum of electrons. But that’s small compared to the momentum of the nuclei and plausibly doesn’t matter.
Is this approximation good? Is it commonly used?

The Born-Oppenheimer approximation is very used in ab initio molecular dynamics (AIMD). When the Born-Oppenheimer approximation is used, it is more specifically called Born-Oppenheimer molecular dynamics (BOMD). Various software packages have implemented this method, here is a list of a few of them:

CP2K
CPMD
Gaussian
Orca

Other types of AIMD is also used, including Car-Parrinello molecular dymanics (CPMD) and Ehrenfest molecular dynamics.
As always it is important to remember that the Born-Oppenheimer approximation is only a good approximation if the relaxation time of the electronic structure is much faster than that of the movement of the nucleus.

the Born-Oppenheimer approxiation is only a good approxmation if the relaxation time of the electronic structure is much faster than that of the movement of the nucleis.

In what kind of situations is this condition true?

An example I can think of on the top of my head is a scattering experiment. If two nuclei are sent towards each other at "high" speed, then the electrons will NOT relax fast compared to the motion of the nuclei, thus the Born-Oppenheimer approximation cannot be used.


text:
Reactions which are impossible theoretically
Take an example of reaction like :
Burning of carbon in air to form Carbon dioxide, say heat required is 100 joule (assume so)
And we provide the system say 20 joule heat
Theoretically it would be not possible. But the alone C atoms present near heat source will absorb their required heat (which would be very very small indeed) and get converted into co2
Hence, there will be small value of reactants which are converted into products then how can we say reaction is not possible. Thus at room temperature is it true that there is some concentration of products in any reaction.

If your said reaction requires 100J to proceed under given conditions, and if you supply only 20J at the same conditions... the reaction wouldn't proceed at all. ;)

But as i have written few molecules which near burner will get their required energy (which may be in order of minus  15) from burner. Like atoms at bottom of breaker lying just close to flame of burner

If the $20$ J you mention is the average energy this will correspond to a certain temperature. However, molecules have a distribution of energies, caused by their random motion in solution or in the gas phase, so that some have far more energy than the average and some far less. The relative populations of vibrational and rotational energy levels a molecule has are determined by the Boltzmann (exponential) distribution. This means that at a given temperature some molecules are more likely to react than others. If the $100$ J you mention corresponds to a temperature at which the reaction proceeds with a reasonable rate constant then if you only supply $20$ J the rate constant will be $e^{-5} \approx 7.10^{-3}$ times smaller. This may be a small enough rate constant to suspect that the reaction is not proceeding, but its just a matter of waiting !
(If you have a temperature gradient across your sample then of course the reaction rate constant will also vary across the sample being greatest where the temperature is greatest.)


text:
Why is dimethylmercury more toxic (to animal life) than methylmercury or ethylmercury?
It's obviously a two step process to reduce dimethylmercury to the less harmful inorganic form.
Is neuronal uptake also enhanced due to lipophilicity? Is there some other mechanism of toxicity I'm overlooking?

Hmm what are your sources? There are studies that show that this is not the case.


text:
Rate order of decomposition of hydrogen peroxide
I have been having some troubles finding a source for my investigation. I am inverstigating the decomposition of hydrogen peroxide, which (according to my teacher/supervisor) is a second order reaction with the use of catalase, and looks like this:
$$\ce{Rate = [H2O2][catalase]}$$
My investigation requires the rate order, but I am not supposed to calculate it myself due to "the high chance of errors in the experiment". I thought this wouldn't be an issue and was willing to investigate said rate equation myself, though it is not an option at my school and the teacher. I have been advised to find a source which states proves this, though I have been having hard time doing so myself.
Are there any articles or research papers I could purchase/read which would allow me to justify my statement for the previously mentioned rate order?


text:
Reaction between magnesium metal and sodium hydroxide: Does this even happen?
Came across this question in a book:

Among aluminum, zinc, tin and magnesium, which metals (if any) will react with an aqueous solution of sodium hydroxide to displace dihydrogen?

The answer to the question is : Aluminum, zinc and tin.
Their reasoning?

Under normal conditions, magnesium metal does not react with an aqueous solution of sodium hydroxide.

Given that the book does not go on to explain the rationale behind this one-liner, naturally my question would be: "Why not?"

QUESTION:
Why doesn't magnesium react with an aqueous solution of sodium hydroxide and release dihydrogen?

[@paracetamol thinks]
Initially, I was under the impression that they were talking about the oxide layer that forms over magnesium metal (the very reason we buff the surface of magnesium ribbons with sandpaper before igniting it), which would give the reaction a slow start. Then, the statement could've been: "Magnesium metal does not immediately react with aqueous sodium hydroxide (owing to the presence of an oxide layer)".
But earlier on in the chapter, there is explicit mention of the reaction between aluminum metal ( which also naturally come with an oxide coat) and aqueous sodium hydroxide, to form sodium aluminate and dihydrogen. There was no mention of an oxide layer hampering the reaction.

[@paracetamol thinks harder]
I then (cursorily) looked at the Electrochemistry aspect of the reaction; because whenever we're asked if a metal reacts with a mineral acid (to liberate dihydrogen), we're expected to flip open our table of standard electrode reduction potentials and say:

Metals having a higher electrode reduction potential than that for the reaction $$\ce{2H^{+} -> H_{2}}$$ (which is arbitrarily taken as zero volts) will be able to liberate dihydrogen from the solution. However, metals such as copper, silver, mercury, etc which have a lower standard electrode reduction potential than that for $H^+$ will not liberate dihydrogen.

When I compared the standard electrode reduction potentials of aluminum, zinc, tin and magnesium, I've observed that all of them have electrode reduction potentials that are higher than the electrode reduction potential for $H^+$. Meaning, that all those metals (magnesium included) should be able to displace dihydrogen from the solution, at least, from the high-school electrochemistry perspective.
 (Yes, I know the textbook mentions an alkali in the question, but what I'm trying to point out is, that since the four metals have higher electrode reduction potentials that $H^+$, then they ought to behave the same way. So if aluminum, zinc and tin can liberate dihydrogen (as the textbook claims) then why not magnesium too?)

[@paracetamol Googles]
I Googled my question, and arrived at this Quora post. Same question as mine, but it's generated such lackluster answers, that they would only be worthy of...well...Quora. O:)

[@paracetamol bamboozled]
I can't for the world figure out why magnesium doesn't react with aqueous sodium hydroxide (if it doesn't). Anyone?

NaOH lowers the concentration of H+ in solution, so there is less H+ to begin with and considering that magnesium slowly reacts with water, I'm not astonished that it doesn't react with basic water.

Magnesium does react with hot water (not so much with cold water). I see no reason for it to not react with hot sodium hydroxide (though it might happen slower).  Also, magnesium is a very active metal and in mixture of fine powders reacts (actually, burns) with things like sand, many metal oxides, and for example, hydrate of barium chloride. Pretty sure it should burn with powder of solid sodium hydroxide.

You need to use the Nernst equation to adjust the potentials accordingly to the much higher pH value. The potential of hydrogen gas and protons equalling zero is only true at pH zero.

The reason why aluminium is so reactive to NaOH is because it destroys the oxide layer (the oxide is acidic). The oxide on the Mg might well not have this character. Worth a look.

For a metal to react with sodium hydroxide solution it's oxide must be amphoteric, not just basic.  Alkaline earth metals are called that for a reason; except for beryllium the oxides are purely basic (and in fact moderate to strong in basic strength) in water, so sodium hydroxide won't make those metals react.
Magnesium, as a typical alkaline earth metal, can be made to react with acids displacing hydrogen, even when the "acid" is a transition metal salt that hydrolyzes slightly like cupric sulfate.

But we are looking at the properties of the metal. How are the properties of their oxides relevant to the discussion?

Metals cannot react with aqueous hydroxide ions.  Their oxides/hydroxides can, if these are amphoteric.


text:
How many tetrahedral and octahedral holes are there in a bcc unit cell?
When I tried to count, I found 24 tetrahedral holes (each formed by 2 body centers of 2 unit cells sharing a face and 2 vertices of the same edge of that face). As each hole is being shared by 2 unit cells, there are 12 holes per unit cell.
Also,I found 6 octahedral holes (6 at face centers being shard by 2 unit cells each and 12 at every edge center being shared by 4 unit cells each).
Generally I was taught that for $n$ packed spheres, there should be $2n$ tetrahedral holes. Does the analysis above mean that this rule doesn't apply here?

Bcc isn't a close packed structure, so I don't think one should expect that the usual 1:1:2 ratio of sphere:octahedral hole:tetrahedral hole is fulfilled. As far as I can tell, your analysis leading to 6 octahedral and 12 tetrahedral per unit cell is correct (textbooks state this when describing the structure of α-AgI). However, these holes are slightly distorted from perfect octahedral/tetrahedral symmetry.

Exactly what you are asking :  https://youtu.be/J8gsZ50Vcqs

It's a partial answer :
The diagram shows two places where irregular octahedral voids exist in a BCC lattice.
By symmetry you'll be able to say that there exist:
6 octahedral voids on each surface and 12 octahedral voids on each edge length which gives us $6$ octahedral voids per unit cell.
A professor goes on for an hour to explain it here:
https://youtu.be/J8gsZ50Vcqs

BCC doesn't have any tetrahedral voids, however we have octahedral holes at face centres. Four adjacent corners and two closest body centres. In a homogeneous lattice the octahedral is compressed along the body-centre-atom axis. However I believe we can form a hetrogenous lattice and form a close approximation to octahedral.
We have thus $6\times1/2 = 3$ distorted octahedral voids.

BCC does have distorted tetrahedral voids

A BCC has 6 octahedral holes and 12 tetrahedral holes.
At each face of the bcc, there is one octahedral hole.
There is also an octahedral hole on each edge.
We divide the holes found on the faces by 2 as they lay between two unit cells, and we divide the holes found on the edges by 4 because they likewise lay between 4 unit cells. There are 6 faces and 12 edges in a cubic cell.
6/2 + 12/4 = 6 octahedral voids
Likewise, there is a total of 12 tetrahedral voids found within a bbc.
If you have trouble visualizing tetrahedral holes, use a dice (or any cube) and point one corner towards you so that you are staring down the C3 axis of rotation. You should be able to see that the edges connected from the corner closest to you to the corners further from you clearly form a tetrahedron.


text:
Mixing solvents of different polarity
Suppose a non-polar solid is dissolved in ethanol. Water is added to the solution. Will the solid separate, if it isn't soluble in water?

https://chemistry.stackexchange.com/questions/75100/is-it-possible-for-a-compound-to-be-soluble-in-a-mixture-of-solvents-but-not-the https://chemistry.stackexchange.com/questions/32025/how-do-i-predict-the-solubility-of-a-compound-in-a-mixture-of-water-and-an-organ

Sorry, I worded it wrong. Will edit.

Yes. How much water you have to add before seeing separation will depend on the specific system (solute, conc., T).