text:
How many stereoisomers are possible for the compound in picture?

How many stereoisomers are possible for the compound in picture? I'm getting 8. Is that correct?
My approach was as follows: The left hand methyl groups may be both up or one up and one down. For each of these cases the groups on the right may be in $(E{,}Z'),(E'{,}Z),(E{,}E'),(Z{,}Z')$ configurations. So $4\cdot2=8$.
I think it is quite evident that carbon having 2 π bonds must be linear. I apologize for the non-linearity in the picture. I hope you will be able to understand the question otherwise.

Carbon with two double bonds is linear, not bent.

@IvanNeretin I don't have a better picture...

Good to know you understand that. (Also, drawing it the wrong way may earn you downvotes and close votes.) Now to the point. Your E,Z calculation is correct. As for the "up or down" thing, the right-hand groups may also be either way, so it gets more complicated.

I get your point @IvanNeretin...its really getting complicated..

There are 4 chiral centers in the ring. The diene moiety (upper right) has 2 isomers and the lower right moiety also has 2 stereoisomers. So I counted 36 stereosiomers total.

@SteffX Can you explain your method in detail?

@SteffX 36 sounds kinda superfluous. Are you sure you didn't mean 32?

I'm getting something different.There are 6 stereogenic areas.So I just use $2^6$...is that wrong? @IvanNeretin

@SteffX
I'm getting something different.There are 6 stereogenic areas.So I just use 26...is that wrong?

Half of those are enantiomers. Or rather, all those 64 are pairs of enantiomers.

@IvanNeretin But according to Wikipedia "Enantiomers, also known as optical isomers, are two stereoisomers that are related to each other by a reflection: They are mirror images of each other that are non-superimposable. Human hands are a macroscopic analog of stereoisomerism. Every stereogenic center in one has the opposite configuration in the other." SO ENANTIOMERS ARE ALSO STEREOISOMERS.

https://en.wikipedia.org/wiki/Stereoisomerism

Sorry, it is 64 stereoisomers, aka $2^6$ (I am tired on Friday)

So do you agree with $2^6$ @IvanNeretin i.e $2^5$ pairs of enantiomers?No mistakes in the formulation..right?

OK, then 64 it is.

@IvanNeretin Hi.Can you answer this if possible?Its a similar question. http://chemistry.stackexchange.com/questions/57761/how-many-stereoisomers-are-possible-for-the-compound

The cyclohexane ring contains 4 chiral centers, one for each carbon in the ring bearing a substituent. Also, the ring shows no plane of symmetry.
The diene moiety (upper right) contains 2 double bonds which could each be E or Z, but the terminal double bond has only one substituent, in other words it contains a $\ce{CH2}$ group, so there is no stereoisomer.
Now the fun part. The lower right moiety (which is quite unusual, btw) also has 2 stereoisomers. The $\ce{CH=C=C=C}$ is linear but its bonds cannot rotate freely. In fact, we could consider it as a simple double bond and the terminal methyl group can be E or Z compared to the cyclohexane ring.
There are 6 stereogenic centers so there are $2^6=64$ stereoisomers.

Follow these steps:

Identify all possible sources of stereochemistry. You got pretty far.
Determine how many isomers can derive from each. I think you went wrong here.
Do you have symmetry? Reduce accordingly.
Multiply all the numbers together.

In your molecule:

Double bonds:

the internal double bond in the residue on the top right
the cumulene, which also shows isomerism (E/Z for uneven double bond counts, aR/aS for even ones).

Asymmetric carbons:

the top methyl group
the bottom methyl group
the cumulene-containing side-chain
the other unsaturated side-chain.

Every one of these can be either R/S or E/Z.
The molecule does not have any elements of symmetry, so we cannot reduce.
All we have are factors of 2; two from the double bonds, another four from the asymmetric carbons. This leads us to:
$$n(\text{isomers}) = 2^{(2 + 4)} = 2^6 = 64$$


text:
Why is (NH4)2[CuCl4] square planar complex but Cs2[CuCl4] is tetrahedral?
Why is $\ce{(NH4)2[CuCl4]}$ square planar complex but $\ce{Cs2[CuCl4]}$ is tetrahedral even though both have same oxidation number of copper and same ligands?

In solution, the $\ce{[CuCl4]}$2- is expected to exhibit terahedral, or nearly tetrahedral geometry.
The Jahn-Teller theorem states that degenerate orbitals cannot be unequally occupied. Molecules with unequally occupied degenerate orbitals will distort so as to render the orbitals non-degenerate, with more electrons occupying the lower energy states. Generally, this distortion leads to a reduction in symmetry. In the case of Cu (II), it is a d9 cation with a tetrahedral electronic configuration of (e)4(t2)5, so the three degenerate t2 orbitals are unequally occupied, therefore a distortion towards the square planar geometry of (eg)4(b2g)2(a1g)2(b1g)1, with no unequally occupied degenerate orbitals, is electronically favoured. Nonetheless, experimental evidence (1) indicates that this d-orbital stabilization energy is very small. Since copper is a first-row transition metal (not very big), steric effects dominate electronic effects allowing $\ce{[CuCl4]}$2- to assume a tetrahedral geometry. Most spectroscopic methods I've read involve looking at the molecule in crystal form (because that is where this effect gets interesting), but DFT studies of the $\ce{[CuCl4]}$2- hydrate in concentrated aqueous solution indicate that the anion's tetrahedral structure is more stable (2).
The issue of $\ce{(Cs)2[CuCl4]}$ vs $\ce{(NH4)2[CuCl4]}$ comes up when the complex precipitates and crystals are formed. Due to the borderline electronic effects, the $\ce{[CuCl4]}$2- anion exhibits a great degree of structural flexibility. With large cations ($\ce{Cs+}$ radius is 0.174 nm while $\ce{NH4+}$ radius is 0.143 nm) , it crystallizes as a distorted ("squashed") tetrahedron.
$\hskip1.4in$ (3)
Square-planar co-coordination occurs in such a way that you effectively get octahedra with tetragonal distortion (4). I would imagine that this phenomenon is simply a result of the smaller cation allowing the unit cell to crystallize like that. The general idea with unit cells is that the cation is never allowed to "rattle" around, so the structure will find a way to fill in the space. Given the preexisting electronic favourability of square planar geometry, $\ce{(NH4)2[CuCl4]}$ this distortion favoured by close-packing is made even more favourable.
$\hskip2.1in$ (4)
The canonical reference for $\ce{(NH4)2[CuCl4]}$ 's crystal structure is (5) if you want to take a look at the unit cell.
(1) Emeléus & Sharpe. Advances in Inorganic Chemistry and Radiochemistry, vol. 21, p. 128
(2) https://www.ncbi.nlm.nih.gov/pubmed/21462945
(3) L. Helmholz & R. F. Kruh, J. Am. Chem. Soc., 1952, 74, 1176
(4) Molecular Structure by Diffraction Methods, vol. 1, The Chemical Society, London, 1973, p. 642
(5) R. D. Willet, J. Chem. Phys., 1964, 1964, 41, 2243

thanks for pointing out my typos!
Yes, I was describing the extreme cases; the tetrahedral form in solution and the electronically favourable square planar case. In ammonium chlorocuprate's unit cell, the ion's geometry is the intermediate, non-inversion symmetric case. If anyone has access to (5), we could describe its symmetry. Sadly, this volume seems to have been misplaced in my library.

Haha, my VPN gives me access. The article doesn’t include a picture though. To quote the discussion: ‘This structure contains the first instane of a square-planar $\ce{CuC4-}$ ion. The $\ce{Cl-Cu-Cl}$ bond angle is required by symmetry to be $90^\circ$. However, the $\ce{Cu-Cl}$ bond lengths are not required to be equal. Two are $2.300 \pm 0.005~\mathrm{A}$ long, and two are $2.332 \pm 0.004~\mathrm{A}$ long. Each copper atom completes its normal distorted octahedral configuration by forming two long $\ce{Cu-Cl}$ bonds of $2.793 \pm 0.005~\mathrm{A}$ with adjacent ions and thus forming an …

… infinite two-dimensional network. The $\ce{Cu-Cl-Cu}$ bond angle is linear ($180^\circ \pm 0.2^\circ$) to within experimental error.’

is the bond length distortion due to electronic Jahn-Teller effects (as in http://chemistry.stackexchange.com/questions/43111/jahn-teller-distortions-in-square-planar-complexes), or is it just a consequence of the crystal packing? I have VPN too, but the journal isn't available in electronic format that far back for me

They unfortunately don’t go into detail. I would say that JT doesn’t sound unreasonable, but it could also be some weird packing effects. Again, an image is not included, unfortunately.

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How can we combine all the three gas laws into a single ideal gas equation?
In all texts that I have read, it has been stated that the combined gas law for ideal gases was derived from the individual gas laws proposed by Boyle, Charles and Avogadro.
My confusion is this is that, in each individual law, some variables of the system's state are to be kept constant. However, none of these conditions are considered in the combined gas equation. Also, how is it possible for us to suggest that for two states with variables $P$, $V$ and $T$:
$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$$

Suppose you have two states which have nothing in common (that is, nothing is kept constant): $P_1,T_1,V_1$, and $P_2,T_2,V_2$. Well, you may just move from the first state to $P_1,T_2$ (that would be a constant pressure process which you know how to deal with), then from there to $p_2,T_2$ (that's a constant temperature process).

http://physics.stackexchange.com/questions/99347/deriving-combined-gas-law-from-boyles-and-charles-laws

I think the historical answer is that it was messy. We like to think of science as pushing back the darkness in a defined way but that just isn't what happens. It is usually more like "blind men describing an elephant." Then someone with sight comes along and explains that one has the tail, one has a leg, one has the trunk and one is rubbing the side. Then the pieces fit.

It is indeed the case that the rearrangement of the different gas laws is not very satisfactory as it employs the different variables without any constraints. This issue has been addressed by S. Levine in J. Chem. Educ., 1985, 62 (5), p 399, but in the case you do not have access to this reference, I'll shortly repeat his arguments here.
As $P$, $V$ and $T$ are thermodynamic sate functions, we may write the total differential of $T$ as
$$\mathrm{d}T=\left (\frac{\partial T}{\partial V} \right )_P \mathrm{d}V + \left (\frac{\partial T}{\partial P} \right )_V\mathrm{d}P \equiv X\mathrm{d}V+ Y\mathrm{d}P$$
since this differential must be exact, the following relation must hold
$$\left (\frac{\partial X}{\partial P} \right )_V=\left (\frac{\partial Y}{\partial V} \right )_P$$
We can find simple functions satisfying this condition
$$X=cP\text{ and }Y=cV$$
where $c$ is a constant that we may choose to be $c=1/R$ for later convenience.  Substitution of these functions in the total differential gives
$$\mathrm{d}T=\frac{1}{R}\left (P\mathrm{d}V +  V\mathrm{d}P\right )$$
and integrating on both sides from 0K to T yields
$$T-T_0=\frac{1}{R}\left [(PV)_T-(PV)_{T_0} \right ]$$
and since $T_0=0$ and $PV=0$ at $\pu{0K}$ we arrive at the ideal gas law.

An easier way to combine the simple gas laws is the following. We take Boyle's, Charles' and Avogadro's laws:
$$PV=k_1(n,T),\quad \frac{V}{T}=k_2(n,P),\quad \frac{V}{n}=k_3(T,P),$$
where $k_1,k_2,k_3$ are constants that depends on the fixed value of $n$, $T$, or $P$.
Now, from Boyle's and Charles's laws we have
$$V=\frac{k_1(n,T)}{P}=Tk_2(n,P)\iff \frac{k_1(n,T)}{T}=Pk_2(n,P)$$
In the left hand side of the last equation, all the terms are function of $n$ and $T$ only, and in the right hand side all the terms are function of $n$ are $P$ only. Thus, both terms have to be equal to a constant $C_1(n)$:
$$ C_1(n)=\frac{k_1(n,T)}{T}\iff k_1(n,T)=C_1(n)T\implies V=\frac{C_1(n)T}{P}$$
With the same argument, we can combine this last equation for $V$ with Avogadro's law:
$$V=\frac{C_1(n)T}{P}=nk_3(T,P)\iff \frac{C_1(n)}{n}=\frac{P}{T}k_3(T,P)$$
and both sides of the equation have to be equal to another constant $C_2$:
$$\frac{C_1(n)}{n}=C_2\iff C_1(n)=nC_2\implies V=\frac{nC_2T}{P} $$
Changing the name of $C_2$, we have the magnificent:
$$PV=nRT$$

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Why do these reactions occur?

Identify the products $[A]$ to $[D]$:
$\ce{CH3COONa}+\ce{NaNH2} + \triangle \rightarrow [A]$
$[A] +\ce{CO_2} \rightarrow [B]$
$[B] + \ce{H2O} \rightarrow [C]$
$[C] +\ce{P2O5 } \rightarrow [D]$
$[A],[B],[C],[D]$ are the main product in each step.

According to my book's answer
$[A]=\ce{NaCH2COONa}$
$[B]=\ce{NaOOC-COONa}$
$[C]= \ce{(CH2COOH)}_2$
$[D]= \ce{C_3O_2}$
I can intuitively understand that product $[A]$ is correct.However I never learnt the next three reactions.Can someone explain/give a source from which I can learn why the next three reactions occur ?
Thanks.

You wrote B and C wrong.

That implies that the author of the study material wrote it wrong. BTW according to you what should be the answer? @Mithoron

B, C, and D should all have three carbon atoms. Your B has two, and your C has four.

@aventurin Bad reasoning.The equations may not be balanced.

Abstraction of $\ce{H+}$ by strong base $\ce{NH2^-}$
$$\ce{CH3-COO- + NH2^- -> ^-CH2-COO- + NH3}$$
Carboxylation to form malonate
$$\ce{^-CH2-COO- + CO2 -> ^-OOC-CH2-COO-}$$
Free malonic acid
$$\ce{^-OOC-CH2-COO- + 2 H+ -> HOOC-CH2-COOH}$$
Dehydration of malonic acid with $\ce{P4O10}$ yields carbon suboxide (unbalanced)
$$\ce{HOOC-CH2-COOH + P4O10-> O=C=C=C=O + H3PO4}$$


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Allred-Rochow vs Pauling Electronegativity scale
While observing the trends in electronegativity of the elements, I found a discrepancy between Allred-Rochow and Pauling scales. In Pauling's scale, chlorine is the third most electronegative element ($\ce{F > O > Cl > N}$). In Allred-Rochow's, it is the fourth ($\ce{F > O > N > Cl}$).
While of course the numeric values of the two scales are different, I thought that the rank of the elements would be the same, being electronegativity related to the properties of the atoms (for instance, the partial charge of molecules such as nitrogen trichloride relies on the electronegativity).
So, I was wondering: which scale is more accurate? And, if one is wrong, why are both scales still in use?

Both scales define electronegativity differently, perhaps that's why? Just venturing a guess. I wouldn't say either is wrong.

Yes, they are different things. There's no inherent "electronegativity" just artificial parameters, which people overestimate.

Neither scale is 'more accurate'. As The_Vinz said, each scale may be applied in different areas, and there is a moderately good correlation between the two scales once conversion factors are introduced.
Firstly, let's use the equations for the two scales to explain why the halogen electronegativity orders are different:
Pauling electronegativity is essentially based on the idea that an atom's electronegativity can be defined by the degree to which it polarizes bonds. Pauling's theorem states that the difference between the experimental bond dissociation enthalpy (BDE) of two atoms and the geometric mean of the BDE of each of those atoms' covalent homonuclear bonds is proportional to the two atoms' difference in electronegativity:
$$\Delta=|\chi_{\ce{A}}-\chi_{\ce{B}}|\propto|BDE_{\ce{A-B}}-\sqrt{D_{\ce{A-A}}\cdot D_{\ce{B-B}}}|$$
Essentially, this is a measure of the ionic character of a bond. The geometric mean term in the above equation represents a hypothetical nonpolar heteronuclear bond, so the difference between that BDE and the experimental BDE of the polar heteronuclear bond describes the degree to which the bond is polarized- its ionic character. A scaling factor is introduced and everything is measured with respect to fluorine's Pauling electronegativity value, which is arbitrarily set to 4.0:
$$|\chi_{\ce{A}}-\chi_{\ce{B}}|=0.102 \sqrt {\Delta}$$
So for chlorine (compared to fluorine): (1)
$$\Delta=|250.548-\sqrt{242.58016 \cdot 156.996}$$
$$|\chi_{\ce{F}}-\chi_{\ce{Cl}}|=0.102 \sqrt {55.396}=0.7596$$
$$\chi_{\ce{Cl}} \approx 3.2404>3.04=\chi_{\ce{N}}$$
(mileage may vary with these calculations)
Allred-Rochow (AR) electronegativity, on the other hand, is defined directly by the force resisting removal of an atom's valence electron. This force is equivalent to:
$$F=\frac {(Z^*e)(e)}{4\pi r^2\epsilon_o}$$
where $Z^*$ is the effective nuclear charge, either measured experimentally or calculated using Slater's rules. Introducing some more scaling factors, which make things line up with Pauling electronegativity values, the equation for Allred-Rochow electronegativity is:
$$\chi_\text{AR}=0.359\left(\frac{Z^*}{r^2}\right)+0.744$$
where r is the atom's covalent radius, in Angstroms. $\chi_{\text{AR},\mathrm{Cl}}=2.83 < \chi_{\text{AR},\mathrm{Cl}}=3.07$[2]
The truth is probably somewhere in between. The two scales are often used interchangeably, or a combination of the two values is used. Originally, the advantages had to do with the experimental basis. Bond dissociation energies may be easier to measure than exact effective nuclear charge or ionic radii. Based on their definitions, it seems to me that AR electronegativity would be more useful for looking at physical properties of an atom in a molecule, like ionization. Pauling electronegativity is really a measure of a bond's polarity.
In the case of day to day estimates about polar bonds in organic chemistry, I think Pauling electronegativity is more accurate. Ultimately, though, these are all quantum mechanical phenomena and exact values must be calculated using quantum methods that implicitly "understand" the real phenomenon that is electronegativity.
In the case of nitrogen and chlorine, though, Pauling electronegativity is closer to the "real" situation. For example, B3LYP/6-31G** calculations shown that nitrogen trichloride has a net dipole of -0.584 Debye, pointing towards the trichloride base of the trigonal pyramid[3], so the $\ce{N-Cl}$ bond is indeed polarized in the $\ce{Cl}$ direction as Pauling electronegativity seems to imply.

References:

Properties of Atoms, Radicals, and Bonds: Table 4.11 (https://labs.chem.ucsb.edu/zakarian/armen/11---bonddissociationenergy.pdf)
Gong-Du Zhou, Fundamentals of Structural Chemistry, World Scientific, Singapore, Malaysia, p. 85
Computational Chemistry Comparison and Benchmark DataBase; CCCBDB: "Calculated electric dipole moments for NCl3 (nitrogen trichloride) at B3LYP/6-31G**" (cccbdb.nist.gov)

Pauling's electronegativity scale is determined by considering the strengths of A-A, A-B, B-B bonds.
Allred-Rochow's electronegativity is determined by considering the charge experienced by an electron on the "surface" of the atom.
While the correlation between the two scales is not perfect, it is quite good, since the scales are calculated with different methods:

The correlation between Allred–Rochow electronegativities (x-axis, in Å−2) and Pauling electronegativities (y-axis).
  By Physchim62 GFDL, CC-BY-SA-3.0, from Wikimedia Commons

Also the partial charge is a model, useful in order to make rapid evaluations on real things. Sometimes, when molecules with almost covalent bonds (such as nitrogen trichloride) are modeled with different electronegativity scales, the partial charge may be different, but the actual electron cloud may not agree with the model.
The real electron density can be then calculated with quantomechanical methods, which are much more accurate and yet complicated than the simple model of electronegativity.

still the answer given confuses me, when I look the value for IVA with catenation. Pb is less in cattnation A-A bond strength but is the second by value given in pauling scale.May you explain this. Thank you.

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Number of geometrical isomers
How do I find out the total no of geometrical and stereo isomers of the compound given as $\ce{C2FClBrI}$? Is there any way I can list out all the possible structures mathematically without having to think about each structure in my mind? Any suggestion how you would do it would be wonderful.

Well, do you know what geometry this type of compounds has - you need  to know symmetry of compound.

No but id love to know the more general way to deal with having to tell the total number of isomers of any type of compound

related http://chemistry.stackexchange.com/questions/7975/finding-the-number-of-cyclic-isomers-algorithmically?rq=1 http://chemistry.stackexchange.com/questions/16135/how-to-determine-number-of-structural-isomers http://chemistry.stackexchange.com/questions/33356/is-there-a-formula-to-determine-the-total-number-of-constitutional-isomers

FC(I)=C(Cl)Br,
FC(Br)=C(Cl)I,
FC(Cl)=C(Br)I


text:
Temperature dependence of molarity and molality
My textbook says

Molarity is dependent on temperature, whereas molality is not.

I'm aware of the definitions of both molarity and molality, but consider this scenario:
A solution of $\ce{NaCl}$ in $\ce{H2O}$ is heated. Eventually, all the water will escape and I will be left with only $\ce{NaCl}$ crystals. At the end, there is no solvent to even measure the concentration of what was once a solution. In this case, how does the argument "molality is independent of temperature" still hold true?

In your extreme example, obviously it doesn't. There's no solution to speak of, so you can't really define a concentration, either in terms of molarity or molality. Since, molarity is moles per volume of solution, it is sensitive to temperature (since, volume changes with temperature). No such problems with molality since it is defined as moles over mass of solvent (which doesn't change).

Formally, molality does not change if you take in account the vapor. Otherwise you are just cheating by taking away mass of solvent instead of just changing it's temperature.

You cannot change the composition of a phase and expect that concentrations do not change. Also, applying heat to a solution does not necessarily mean that solvent will evaporate. It does not when the system is closed and consists of the solution phase only.

Both are quantities of solutions. If you evaporate all the solvent, molarity and molality do not mean anything anymore.

Come to think of it, all water will eventually escape even if you don't heat the solution at all. What will happen to molarity?

Recall that molarity $c$ is
$$c=\frac{n_\text{solute}}{V_\text{solution}}$$
whereas for molality $b$ is
$$b=\frac{n_\text{solute}}{m_\text{solvent}}$$
While in the extreme case you mentioned, both quantities are not defined because all the solvent is either boiled away (or for heat-sensitive species, the solvent can even decompose under heating), molarity depends on volume, and the volume of the solution depends on temperature due to expansion/contraction.

Example: 5 moles of solute in 1 liter of solution.
The Molarity at the beginning is (5moles ÷ 1Liter) = 5 M.
When the solution is warmed up and expanded by 1% say, to 1.01L, the new Molarity = (5moles ÷ 1.01Liter) = 4.95 M
P.S. Don't over think this question! The amount of content should not change, all you're talking about is expansion of content when the temperature rises. All matter expands when temperature rises so if you're calculating the molarity of a solution, you may get different readings on a warm day compared to a freezing cold day!

Molality doesn't change with temperature...

You are not answering the OP's query about the temperature dependence of molarity vs. molality.

These two words sound similar but they are not synonyms, even though both of them are used for representing solution concentration. By definition, molarity is the number of moles of solute dissolved per liter of solution. We use capital letter “M” to represent molarity and its formula is M= (# mol SOLUTE)/ (Liters of SOLUTION). Molality is then the number of moles of solute per kilogram of the SOLVENT, NOT solution! We use lower case letter “m” to represent molality and its formula can be represented as: m= (# mole SOLUTE) / (Kilograms of SOLVENT). Most of the time scientists use either molarity or molality to represent solution concentration, but MOLALITY is preferred when the temperature of the solution varies. That is because MOLALITY does not depend on temperature, (Neither number of moles of solute nor mass of solvent will be affected by changes of temperature.) while MOLARITY changes as temperature changes. (Volume of solution in the formula changes as temperature changes, and that is why.)

Of course, I know the definitions properly! My only doubt was in the fact that even though molality is supposed to be independent of temperature, if I increase the temperature of the solution by heating it, some(or all)of the solvent evaporates and not only does the volume of the solvent change, but also the mass of the same and hence, molality will change. What point am I completely missing? From the other answers, I gathered that this argument will not hold good for extreme cases like the one that I mentioned in my question and must not try to apply it to such situations. Am I correct?

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". Also important: We use a capital M for the molecular mass, and a miniscule m for the mass. We refer to molarity in more technical terms as amount concentration and use c for it, while molality uses b as symbol.

Molarity is a temperature-dependent parameter. The volume of a solution increases with an increase in temperature diye to increase in intermolecular spaces and vice versa. Hence, molarity is affected by temperature change.
As
Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax.Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

Okay so no one answered OP properly. Molality(m) is independent of the temperature as long as the temperature change does not actually change the amount of solute. I understand that most resources would not mention this simple fact and expect you to understand it presumptuously. Again, molality(m) is independent of temperature as long as the amount of solvent is not changing.
Example 1: temperature changes --> solution volume changes --> molarity (M) changes --> molality (m) constant
Example 2: temperature changes --> solution volume changes --> amount of solute changes --> molarity (M) changes --> molality (m) changes
See the difference?


text:
Gibbs free energy change at equilibrium
I know that Gibbs free energy change represents the amount of the non-expansionary work that a reaction is capable of doing but what happens to this energy at equilibrium? Why is the system unable to do work? Why is the Gibbs free energy change equal to zero at equilibrium?
Edit: Please don't simply write the derivation that we use to conclude that ∆$G$ must be zero at equilibrium. Please try to explain the physical significance behind this fact.
Please also note that I have already searched this website for an answer to my question, found only the following question but the answers only explained the Math:
Gibbs free energy-zero or minimum
Thanks ever so much in advance :) Regards.

What exactly are you looking for when you say "physical significance"?

At equilibrium, exactly what happens to the energy that had once been present, allowing the reaction to do work? What is the meaning of ∆$G$=0?

It means the chemical potential of the reactants and products is the same

Can you please elaborate that point? I don't quite understand fully.

You would have to allow the use of some mathematics then

Sure, no problem :) I just wanted to get an intuitive sense of this concept, rather than understand the derivation. Please go on.

I detect some inaccuracy in for question. I am going to try to clarify a few things since I perceive genuine curiosity in your question. The answer can not be so short.
First some clarifications:

Why is the Gibbs free energy change equal to zero at equilibrium?

The very common phrase free energy change at equilibrium is inadequate. In thermodynamics the state of a simple system is specified with state parameters.  In the matter at hand temperature ($T$) and pressure ($p$). As the system would be described by $T$ and $p$, any "change" that does not involve changes in $T$ or $p$ is outside of the scope of the theoretical treatment. That is, to say the system is at equilibrium is accurate within a theoretical framework, so change should refers to changes within the same framework. Of course they could be important in uncommon cases but they are not included in the model. So, let use the $^*$ symbol to refer to an equilibrium state. At equilibium $G = G(T^*,p^*)$. Any change that do not involve $T$ and $p$ will let the system in the equilibrium state (although in others way the system can look not the same). So, its change is zero just because is the same function with the same values for its arguments.
At equilibrium the system is in an stationary state. That is, the system (thermodynamic!) state do not change with time. So, if you inspect how $T$ and $p$ evolve in time they do not change anymore. So any spontaneous change won't change the value of $G$. To get a change of $G$ you must make a perturbation on the system by making a change in the environment or the walls that divide them up, but once it is done the system is not in equilibrium state anymore (although it can reach other equilibrium state in the future).
That is why at equilibrium there are not useful work available from the system, because it is in a stationary state in harmony with the environment. So changes won't take place. Notice that to get work from the system it is needed some kind of macroscopic movement (according to the definition of mechanical work).
You said:

I know that Gibbs free energy change represents the amount of the non-expansionary work that a reaction is capable...

It is not exactly true, it is not a requirement that the process involve a reaction.
In the ligh of getafix commentary:
Take in mind that physical significance most times, in the core, is just an inaccurate idea about a physical theory/model/law. Although it may be useful sometimes from a practical perspective (and many times leads to catastrophic conclusions), it has not an add value. There is the nature. There are good formal descriptions of it obtained through years of hard work from experienced people. There are pictorial interpretations of those formal descriptions used when formalism turns too hard/complex (physical significance). I noticed that it is fashion to say that one get the true understanding of phenomena when one get the physical significance. I think that one truly understand the phenomena when feels natural the formal description and is also aware that it is just a description.

I'm really not getting why that post didn't answer OP's query.
Nevertheless, I'm posting the following argument as an answer.

Why is the Gibbs free energy change equal to zero at equilibrium?

Well, it's not 'Gibbs free energy change' $\Delta G$ but rather Reaction Gibbs
Energy $\Delta _r G$ which is zero at equilibrium, if written accurately.
Reaction Gibbs Energy is defined as the rate of change of Gibbs Free Energy per unit change in composition i.e.,
$$\Delta_r G =\frac{\Delta G}{\Delta n} \tag 1$$
where $\Delta n$ is the change in composition of the system.
Now, OP must be aware of the first derivative test for finding local minima; that means for reaching the equilibrium position, the rate of change of Gibbs Free Energy per unit composition must be zero and the rate must change from negative in the neighbourhood of the extrema point of Gibbs Free Energy vs. Composition graph viz.
$$\Delta_r G= 0\;.$$
Now, question might arise as that of the physical implication of association of equilibrium position with $\Delta_r G= 0\;.$
The definition of equilibrium position sheds light on this:
A reaction reaches equilibrium position when it has no further tendency to change; that is, the reaction does remain 'spontaneous in neither direction' .
This happens when the reaction Gibbs energy becomes zero viz. $\Delta_rG=0.$

First of all, that post didn't answer my query because I was looking to get an intuitive feel for this concept and not the mathematical description. What is the definition of ∆$n$ for a system? Yes, I'm aware if the first derivative test, but I'm afraid I didn't understand much of what came after that in your answer. Okay, it's spontaneous in neither direction, but what does all this have to do with energy? Please try to understand what I am asking; the term is "Reaction Gibbs energy" (please elaborate on that too) so what does it have to do with energy and why is this energy becoming zero?

@KaumudiHarikumar: The 'intuitive feel' of the query lies in that very phrase 'spontaneous in neither direction'. Do you know what spontaneous means? Try to correlate it with the Second Law in the light of Gibbs Free Energy.

I don't think I am putting out something truly novel here either; the answer by @MAFIA36790 and the discussion in the link you yourself post it seem quite satisfactory to me, yet I shall try to put a different spin on things so to speak.
Let me define the chemical potential, as the partial molar Gibbs energy of species "$j$":
$$\mu_j = \left ( \frac{\partial G}{\partial n_j} \right)_{p,T,n'}$$
and for a system of components A, B, etc. the fundamental equation of chemical thermodynamics gives us:
$$\mathrm dG= V~\mathrm dp-S ~\mathrm dT+\Sigma \mu_j~\mathrm dn_j$$
Now, I consider a simple equilibria $$R \rightleftharpoons P $$
and additionally, I introduce a parameter " $\xi$ " called the extent of the reaction (it has the dimensions of amount of substance)
for macroscopic changes, the amount of a component $j$ changes by $\nu_j\Delta\xi$ ($\nu_j$ are the "signed" stoichiometric coefficients; positive for products, and negative for reactants)
Now, if we defined $$\Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} $$
This $\Delta _r G$ still holds the meaning you might already be familiar with, namely:
$$\Delta _r G = \Delta _f G\textrm{ (products)} -  \Delta _f G\textrm{ (reactants)}$$
and I have nearly extended it to mean the derivative defined above so that i can relate it to the extent of a reaction (measured in amounts of substance consumed/produced)
and at constant temperature and pressure, from the second equation in this post:
$$ \mathrm dG = (\mu_P - \mu_R)~ \mathrm d\xi$$
so, $$ \Delta _rG = \left ( \frac{\partial G}{\partial \xi} \right)_{p,T} = (\mu_P - \mu_R)$$
Now, as already mentioned, if we were to plot the reaction Gibbs energy as a function of the extent of the reaction, a minimum on the plot would correspond to the first derivative defined above being equal to zero. (or, equivalently in the chemical potential of reactants and products being exactly equal). This is the condition for equilibrium.
Perhaps, stooping down to layman terminology (caution), one can say $\Delta _r G $ is like a "force" that drives a reaction forward, and the system is in equilibrium when this driving force is zero (and moves in the reverse direction when it changes sign).
An appropriate mechanical analogy could perhaps be two weights on either end of a string wrapped across a pulley. If the heavier weight is higher, it would descend raising the lower weight. The system would come to rest when they are in balance.

Thank you, but I have only just graduated high school and I'm not aware of some of the terms that you(and others in the post that I have linked in my question)have used. Nevertheless, I will try to understand what you have said as best I can. At my level, we aren't introduced to many important concepts, which poses a problem for understanding clearly. One last doubt; how can you use the analogy of a seesaw like static mechanism to help explain a dynamic equilibrium? When you say that the driving force is zero, I don't understand this because like I just said, isn't equilibrium dynamic?

It's not a perfect analogy, I admit. Which is have included (a literal) word of caution. If you want to improve the mechanical analogy, then imagine perturbing it equilibrium perhaps? Also, maybe at your level then you could just take Gibbs energy = 0 as a definition of equilibrium?

Yeah, well, this is the part that is very annoying; when I try to understand more but a barrier of some advanced science props up. I only wanted to know why "the ability of the system to do non-expansional work"(for this is how it is defined for us)reduces to zero at equilibrium and in neither of your answers have I found anything useful. Sigh. I guess that I will have to accept the fact that there is no way for me to understand it without understanding the advanced terms and must resort to not knowing for the time being :(


text:
What is active mass?
I'm aware of the fact that active mass is defined as the molarity of a substance, but my textbook states that

"Active masses are dimensionless quantities but for our purposes we generally take them with dimensions of molarity, partial pressures etc."

What point is my textbook trying to get across to me? Why do active masses have no dimensions? Why have we defined this quantity in the first place, if we use it interchangeably with the term "molarity"?
Edit: Please keep in mind that I have only just graduated high school and would be very grateful if you could explain this concept to me in simple terms.

It sounds like the book is focussing on the ratios of molecules. This is the same idea as moles or partial pressures in a gas: they measure the ratios of the compounds present.

Okay, but why? Can you please elaborate?

Well, I want to mention that activity might be intuitively understood by its usage in radioactivity: the activity of the given substance determines how fast it is disintegrating and how much radiation it is generating, analogous to how "reactive" a particular reagent is in the reaction. However, I am not sure if activity and active mass are the same thing...

The term "active mass" is a historical term.
The concept of an equilibrium constant was developed by Cato Maximilian Guldberg and Peter Waage.  The Law of Mass Action has also been referred to as the Law of Guldberg and Waage, historically.
Guldberg and Waage defined the term "active mass" in the 1867 Études sur les affinitès chimiques.

l'on peut choisir un volume arbitraire, par exemple un centimètre cube. Les
   quantités d'un corps qui se trouvent dans lcc du volume total se nomment la
  masse active du corps.

or roughly translated:

One can choose an arbitrary volume, for example a cubic centimeter. The
  amounts of a body located in 1cc of total volume are called the active mass of the body.

And as stated in The Laws of Reaction Rates and
of Equilibrium J. Chem. Ed. 1956, 33 (4), p 178 :

Guldberg and Waage in their study of chemical
  affinity formulated the law of the rate of chemical reactions
  as proportional to the "active mass" of the reagents.
  This they called the law of mass action and
  very clearly defined "active mass" as amount per unit
  volume. (emphasis added)

So, from a historical point of view, it is false that "active mass" is dimensionless.
However, in relating Gibbs energy to the equilibrium constant, it is necessary for the equilibrium constant to be dimensionless, because only the logarithm of a dimensionless quantity is logical.  For this reason, the text mentioned in the OP redefines "active mass" to be dimensionless.

Sir, in my study material, it's written that, active is actually a dimensionless quantity, and for our convenience and purposes we take them with the dimensions of molarity, partial pressure, etc.

@AdeshTamrakar "activity" is dimensionless.  "active mass" is a historical term that professional chemists do not use anymore.  The correct historical meaning of "active mass" is mass per volume, such as mass per cubic centimeter.  The term "active mass" is older than the term "molarity".

Active mass implies that amount of mass which is taking part in a reaction, that it is the effective concentration of a substance.
It's other names are reactive mass or chemical activity.
Units: The units of activity are nominal, rather than real, because chemical activity is formally defined as the ratio of the actual chemical activity of a substance to its chemical activity under some defined standard conditions, and ratios have no units because the units divide out. Or simply it's the ratio of concentration at required conditions to standard condition.The difference between activity and other measures of composition arises because molecules in non-ideal gases or solutions interact with each other, either to attract or to repel each other. The activity of an ion is particularly influenced by its surroundings.
For example : with the hydrogen ion, we use pH = − log10 [H+ ] where [H+ ] is the concentration, i.e. the number of particles (in moles) divided by the volume (in litres).
Once you get a lot of these particles, they bang into each other and take up room (think of billiard balls on a pool table or a room full of tennis balls flying around).
In these circumstances we talk about ‘activity’, which is a way to correct for this. Basically it is what the concentration or amount appears to be if we were using the simplified laws (the effective concentration). For the hydrogen ion, it is also what the concentration appears to be to physiological systems. In other words it uses a correction factor for the simplified laws.
Also Wikipedia says:

In a solution of potassium hydrogen iodate [KH(IO3)2] at 0.02 M the activity is 40% lower than the calculated hydrogen ion concentration, resulting in a much higher pH than expected

I'm afraid I don't understand how taking the ratio functions as a correction factor. Additionally, when using "activity" in the expression for equilibrium constant, the final value does end up having some units. How is this so? Why do we settle for using molarity in place of activity? Can you please further elaborate on all this?

@JM97 I don't think "active mass" and "chemical activity" are interchangeable terms. The idea behind them is the same though, I agree.

@KaumudiHarikumar the equilibrium constant is dimensionless

@KaumudiHarikumar "molarity" is used because it is a readily measurable physical thing, and using it is usually approximately correct.  "activity" is artificially defined mathematically to make the equilibrium constant exactly constant for all concentrations https://www.iupac.org/goldbook/A00115.pdf  http://goldbook.iupac.org/S05915.html It is as if we say, for all people (body weight)/(brain weight) = 27, and when we realize it isn't exactly true, we define something called pseudoweight to make the equation always be true.

Um, OK. I don't know about pseudoweight though.

Part of this answer appears to be a verbatim copy of http://www.3rd1000.com/chem101/chem102q.htm. Please add references if citing external sources.


text:
Can Fe(CO)5 adopt both square pyramidal and trigonal bipyramidal geometries?
I know the hybridization of $\ce{Fe(CO)5}$ is $\mathrm{dsp^3}$. According to my book, coordination compounds with coordination number 5 can interchange between square pyramidal and trigonal bipyramidal geometries. Is it true for this compound? Doesn't seem so according to the picture:

Was my understanding wrong somewhere?

The orbitals used in $\ce{[Fe(CO)5]}$ are 3d, 4s and 4p. If that makes it an $\mathrm{dsp^3}$ complex in your book then so be it. But then consider that orbitals such as 4p and 4d are energetically too far apart to take part in bonding together in a noteworthy manner.

Obligatory link to Why is it wrong to use the concept of "hybridization" while studying complexes?

Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...).
Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, though MO theory and crystal field theory can help understand which might be more favourable.

The structure of iron pentacarbonyl is widely agreed to be trigonal bipyramidal, with three CO ligands occupying equatorial positions, and two CO ligands occupying axial positions.

Trigonal bipyramidal structure of iron pentacarbonyl

Whilst both geometries are possible (and indeed observed, IF5 is square pyramidal), I suspect that your confusion arises from the fact that iron pentacarbonyl is able to undergo a phenomenon known as Berry psuedorotation.

Mechanism of Berry psuedorotation

Berry psuedoroation is the process in which two of the equatorial ligands are switched out for the two axial ligands via a square planar intermediate. This process is sufficiently rapid that if observed by NMR, only one environment for the carbon of the CO ligand is observed (i.e. the psuedorotation is faster than the NMR timescale).

No.Usually dsp3 refers to inner orbital complex while the latter refers to outer orbital complex..Maybe you follow a different convention?

I've never known there to be a difference other than the way people choose to write it, but I'm just an organic chemist, not an inorganiker

@MathJack A hybridisation scheme involving s-, p-, and d-type orbitals is garbage. They are far too much apart to considerably interact with each other.

@NotEvans. As far as i understand, IF5 has sp3d2 hyb. Hence the comparison between IF5 and Fe(CO)5, which I think is dsp3, does not really make sense (as in your answer). The square pyramidal structure of IF5 is very easily explained using VSEPR Theory. Moreover, I don't think that IF5 can show trigonal bipyramidal geometry.

This distinction between "sp3d" and "dsp3" is a relic of an incorrect and flawed theory.

The most recent single crystal structure investigation (high precision, $R_1 = 1.9\%$, $wR_2 = 5.7\%$) of molecular iron pentacarbonyl [1] suggests trigonal bipyramidal coordination environment (also mentioning anti-Berry pseudorotation), as shown by the calculated geometry index $\tau_5$:
\begin{align}
\begin{cases}
\tau_5 &= 0 \qquad &\text{square pyramidal geometry} \\
\tau_5 &= 1 \qquad &\text{trigonal bipyramidal geometry}
\end{cases}
\end{align}
$$\tau_5 = \frac{\beta - \alpha}{60^\circ} = \frac{178.93^\circ - 121.13^\circ}{60^\circ} = 0.96$$
where $\alpha$ and $\beta$ - two greatest valence angles of the iron coordination center calculated from the CIF-file (ICSD #74197).
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#E06633}{\Large\bullet}~\ce{Fe}$.

The magnitude of the $\ce{Fe-C}$ axial vs equatorial band length distortion ($0.44\%$) is notably less than in an alogous p-block trigonal-bipyramidal species (e.g. in $\ce{PF5}$ $\ce{P-F_{ax}} = \pu{1.577 Å}$ and $\ce{P-F_{eq}} = \pu{1.534 Å}$; i.e. distortion $2.8\%$, by gas-phase electron diffraction).
There is a small angular distortion of the trigonal bipyramid in crystalline $\ce{Fe(CO)6}$ with the axial ligands closing on the unique carbonyl ($\ce{C(1)-Fe-C(2)} = 89.47(5)^\circ$, $\ce{C(2)-Fe-C(2a)} = 178.94(10)^\circ$), two equatorial angles opened from their ideal values ($\ce{C(3)-Fe-C(1)} = 121.11(5)^\circ$), and the third slightly closed ($\ce{C(1)-Fe-C(1a)} = 117.8(1)^\circ$). This distortion is of an anti-Berry pseudorotation type, i.e. in the direction away from a square-based pyramidal geometry in which $\ce{C(1)-Fe-C(2)} = \ce{C(1)-Fe-C(3)} = ca. 105^\circ$.

Bibliography

Braga, D.; Grepioni, F.; Orpen, A. G. Organometallics 1993, 12 (4), 1481–1483. DOI 10.1021/om00028a082.

Here the hybridisation is dsp3 in $\ce{Fe(CO)5}$. It is basically an inner orbital complex. So, it is a spin-free orbital. The hybridisation plays a major role in determining the shape in these cases. So, when the hybridisation is dsp3 in most of the cases you get the shape as trigonal bipyramidal, whereas in sp3d we get square pyramidal.


text:
Addition of inert gas at "constant volume"?
My textbook states that

"The addition of inert gases at constant volume does not affect the equilibrium state of the reactants and products contained in that volume."

Although I am not questioning the truth behind this statement, I am very confused by it.
Let us consider a homogeneous gaseous equilibrium system contained within a closed cylinder. The cylinder has a very small hole on its side, through which an inert gas can be pumped in. This is how I visualize the addition of an inert gas to the system. Now, upon introducing a given amount of inert gas into the system, how does the volume remain constant in any way?
While deriving Van der Waals equation for real gases, I learned that the term $V$ in the ideal gas equation stands for the empty space that is available to the gas molecules for movement. Therefore, when we introduce an inert gas into the system, aren't we essentially decreasing this amount of space that is available to the reactant and product molecules for movement?
I am very confused. Please share your insights for it would be tremendously helpful for me. Thanks ever so much in advance :) Regards.

The inert gas that you add does not act like a piston, denying part of the volume to the other molecules in the mixture.  For a constant volume system comprised of an ideal gas mixture, the partial pressures of the reactants and products do not change when you add the inert gas.  It only increases the total pressure, and that change is only because of its own partial pressure.  However, if you are not operating in the ideal gas regime, the addition of the inert gas will affect the equilibrium.

OK, so what you're saying is that according to the assumptions of KTG for ideal gases, the volume of the reaction mixture will remain the volume of the container since the molecules of the inert gas have negligible volumes themselves, yeah?

Additionally, if the reaction mixture is at constant pressure instead of constant volume, these inert gas molecules will exert pressure on the piston, moving it up and hence, increasing the volume, yeah? While studying Le Chatelier's principle for changes in volume, I didn't relate the volume change to pressure change. Instead, I imagined that on decreasing the volume, the equilibrium shifted in that direction which produces lesser number of molecules so that the volume would increase and not so pressure would decrease. I am wrong to think of it like this, right?

I agree to your first comment.  I don't understand your second comment.

You should not :)
"does not affect the equilibrium state" means that the thermodynamical constant is still the same. If you add an inert gas $\ce{G}$ in your system it will not react with other chemical coumponds.
For example at the beginning you have $\ce{A} \rightleftharpoons \ce{B}$ and after you add you inert gas you have $\ce{A + G} \rightleftharpoons \ce{B + G}$. So it not afftects the value of $\ce{K_{eq}}$ because $\ce{K_{eq}}$ depends only of $\ce{T}$ the temperature.
If you use a solid container the volume will not change but the pressure will.
Also we have $$\ce{\mu_i(T,P_i)=\mu_i^°(T)}+\ce{RT\ln\frac{P_i}{P^°}}$$
Then $$\left(\frac{\partial\Delta_r\ce{G}}{\partial n_{inert}}\right)_{\ce{T,V}}=0$$
If you prefer calculus.


text:
The effect of pressure on boiling point?
My textbook states the following:

Qualitatively, I understand why the boiling point of a substance increases when the pressure is increased. However, I learned that if the pressure is increased when the reaction system is at equilibrium, the volume of the system decreases and hence, the reaction will proceed in that direction in which the volume of the system increases. In the above scenario, however, it has been given that water in the gaseous form occupies more volume than its liquid form and so, shouldn't the given reaction proceed in the forward direction, hence reducing the boiling point?
Please note that I am convinced and I believe that it makes sense for the boiling point to increase when the pressure increases but can somebody please explain the flaw in my reasoning with regard to the given equilibrium?
Much thanks in advance :) Regards.

what's the reaction in boiling of water?

I'm afraid I don't understand. What do you mean by "reaction of boiling of water"?

There is no reaction in boiling water, why do you apply le chatliers principle?

There is no chemical reaction but there is a physical reaction, no? An equilibrium does exist between the gas phase and the liquid phase of water(or any liquid, for that matter).

at equilibrium ratio of concentration of products to concentration of reactants remain constant. (K_c) . How would you apply this to a physical change?

In this case, we use Kp=Pressure(of water vapour). At equilibrium, this will be equal to the vapor pressure of the water at that temperature.

Yeah then what, As you decrease volume partial pressure of water increases and to maintain same K_c the partial pressure of water has to decrease this is by its dissolution into liquid water.

The equilibrium will shift in a direction in which the pressure decreases, yes? So it shifts in the direction of the liquid phase for which pressure is...not defined?

Your flaw in the logic lies in that you think that the volume of the system includes the liquid. Le Chatelier's principle on pressure equilibrium only applies to when the compound involved is in gaseous phase, in an ideal sense. Therefore if you increase the pressure (decreasing the volume), then the reaction will progress towards the direction where gas volume DECREASES, which is towards liquid phase. Overall, the entire apparatus increases the vapor pressure on the gas (gas pressure on the external pressure), which increases the boiling point.

OK, but on decreasing volume, why does the reaction progress in that direction in which the volume further decreases?

Le Chatelier's principle states that a reaction that undergoes a change in pressure, temperature, or volume, the equilibrium shifts towards the direction of the reaction that counteracts and subsists with that change. In this case, the change in volume causes an equilibrium shift that subsists by decreasing the volume

The shift in equilibrium should be such that it counteracts the stress(change) applied, yes? So why, on decreasing the volume, will the equilibrium shift in a direction in which the volume further decreases?!

Decreasing the volume increases the pressure. The volume decreases because the pressure increase causes the gas to contract. Look at this link for more info: https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle#Effect_of_change_in_pressure

So if I decrease the pressure, the volume increases and the reaction will shift in that direction in which the volume further increases? Seems very absurd sir/ma'am. I think it should shift in that direction in which the volume decreases, causing the pressure to increase.

I really don't know what to say. Its Le Chateliers principle.

But it's not! Please read my comment again; doesn't it seem like the situation that I'm describing is the exact opposite of the outcome given by the principle?!

Can I get some help here? The volume of the system DECREASES, which means the systems gas' pressure MUST INCREASE. To counteract this, aka reduce the increased pressure, the volume of the GAS mudt decrease

What are you even saying?! "To counteract the increased pressure, the volume of the gas must decrease?!" However are you counteracting the change if you're decreasing the volume even further?! $P$ is inversely proportional to $V$!

Please calm down. The changes in VOLUME is not the actual thing that affects the equilibrium. The change in VOLUME, changes the pressure to be HIGHER, which is what actually shifts the equilibrium. Le Chatelier's principle is about how external forces affect the equilibrium point.

OK, sorry (:P). Since the decreased volume increases the pressure of the gas, according to Le Chatelier's principle, the equilibrium tends to shift in that direction in which its pressure decreases, yes? For a decrease in pressure, the volume increases.  Hence, on decreasing the volume of a gas mixture at equilibrium, it tends to shift in that direction in which its volume increases.

It's because when we increase pressure, we are actually increasing external pressure so according to Le Chatelier's principle, to counteract the effect it should increase the internal pressure. This is the reason that the reaction shifts where internal volume decreases, so that internal pressure increases.

Could you please edit your post so that you are usually complete words? It is difficult to understand the content of your answer with the incomplete words.


text:
Why is the ionic radius of Al(3+) smaller than that of Li+?
I was examining the ionic radii of some ions from this site for a school assignment. I noticed a weird anomaly in the ionic radius of $\ce{Li+}$  as compared to that of $\ce{Al}^{3+}$.
The ionic radius of $\ce{Li+}$ is about 2/3 as large of that of $\ce{Al}^{3+}$. How is this possible, since the electronic configuration of $\ce{Al}^{3+}$ is $[1s^2, 2s^2 2p^6]$ whereas that of $\ce{Li+}$ is $[1s^2]$? It appears that despite the presence of an extra shell in $\ce{Al}^{3+}$, it has a smaller ionic radius than $\ce{Li+}$.
The answers in this post suggests an exactly opposite outcome. All the answers says that $\ce{Mg^{2+}}$ is above $\ce{Ca^{2+}}$, thus it has a smaller radius. By applying  the exactly same logic i can say $\ce{Li+}$ is above $\ce{Al^{3+}}$ so it must have less ionic radius but that is not the case.
As the below picture suggests,(taken from the same post).

I think it is because of diagonal relationship. As $\ce{Mg^{2+}} = \ce{Li+}$ because of diagonal relationship and $\ce{Mg^{2+}} \gt \ce{Al^{3+}}$ because ionic radius decreases for cations along the period by going left to right.
Am I correct?

The ionic radius is not the radius of the particle; it is the radius of its influence.

Al^3+ is shorter because it can attract anions closer.

http://chemistry.stackexchange.com/questions/49863/comparing-radii-in-li-be2-mg2-al3-and-na

@user34388 Why does Li+ has more radius of influence ? Is it because of diagonal relationship ?

Please don’t ‘nag’ us to answer your question. We are all volunteers volunteering our free time. If you want to increase the attention your question gets, you should add a bounty. However, please pay attention to the caveats presented therein.

The size of the 1s orbital in $\ce{Li+}$ and $\ce{Al^3+}$ is not necessarily the same. In fact, they are quite different because of the much larger effective nuclear charge in $\ce{Al^3+}$. One can easily look up the wavefunction for the 1s orbital and see the radial dependence on $Z_\mathrm{eff}$.
Therefore, merely looking at the electronic configuration cannot tell you anything about the size. If that was the case then one would expect $\ce{Cl-}$ and $\ce{K+}$ to have exactly the same size, which is clearly not true.

All the answers says that $\ce{Mg^{2+}}$ is above $\ce{Ca^{2+}}$, thus it has a smaller radius. By applying  the exactly same logic i can say $\ce{Li+}$ is above $\ce{Al^{3+}}$ so it must have less ionic radius but that is not the case

You are missing the point here entirely. Mg and Ca are in the same group which is what makes them directly comparable. Li is not "above" Al any more than F is above Na.

I think it is because of diagonal relationship. As $\ce{Mg^{2+}} = \ce{Li+}$ because of diagonal relationship and $\ce{Mg^{2+}} \gt \ce{Al^{3+}}$ because ionic radius decreases for cations along the period by going left to right.

The diagonal relationship is something that is observed, not a fundamental principle of chemistry. It still has to be rationalised. Saying that it is "because of the diagonal relationship" is akin to saying "electrons move very fast because they have a high velocity".
The diagonal relationship holds because 1) you add an extra shell of electrons going vertically from $\ce{Li+}$ to $\ce{Na+}$, which decreases $Z_\mathrm{eff}$ and increases ionic radius; and 2) you go horizontally from $\ce{Na+}$ to $\ce{Mg^2+}$, which increases $Z_\mathrm{eff}$ (same number of electrons but one more proton) and decreases ionic radius.
Going one more step horizontally to $\ce{Al^3+}$, one can see that $Z_\mathrm{eff}$ should increase again. So your thoughts on the matter are correct, but they just do not get to the crux of the matter.
The origin of the much smaller ionic radius of $\ce{Al^3+}$ is therefore directly attributable to the much larger $Z_\mathrm{eff}$ in $\ce{Al^3+}$ on the 2p valence electrons, compared to that on the 1s valence electrons in $\ce{Li+}$.
Be careful that the ionic radius is not always particularly well-defined and the values can vary from one source to another depending on how they are experimentally obtained.

So, $Z_{\text{eff}}$ is the cause of the problem. There are  many factors deciding physical properties, How would i know where does which apply ? Sorry for so much trouble and thanks for the well detailed answer.

By reading and learning more. There's no way around it - practice makes perfect.

No doubt about that. Can we obtain ionic radii  from just quantum theory equations ?

You would have to have a strict definition of the ionic radius, and the definitions that I am aware of do not involve quantum calculations on free ions, but I assume that if you make up a definition, it would be possible with the computational power we have now to determine ionic radii. Don't quote me on it, though.

Isn't half the distance between a cation and a anion is the ionic radius of cation and anion ? is anything wrong with this definition.

That has nothing to do with quantum theory though. That is the definition one might use when carrying out X-ray crystallography on a solid. Which is why I said "the definitions that I am aware of do not involve quantum calculations on free ions".

The main difference between $\ce{Li+}$ and $\ce{Al^3+}$ is their charge. Aluminium has three times lithium’s charge.
When calculating orbitals, the first step is to set up a Schrödinger equation and then to see what the Hamiltonian operator is made up of. If you didn’t understand that sentence, just rephrase it in your mind to ‘what is the energy of an electron and which forces does it encounter?’ The most important force here is the nuclear charge, which creates a spherical potential energy gradient — the closer to the nucleus an electron is, the lower its potential energy. (You can think of this like a mountain — the lower you are the less potential energy you have.)
The next step is to solve (or approximate — we cannot solve it for systems with more than one electron) the Schrödinger Equation. Its solution will give you orbitals, spherical wave functions. To draw the typical picture of an orbital, we need to define a cutoff value first; this could be ‘the electron has a probability of $95~\%$ of being found within this boundary.’ And only after that can you talk about orbitals and their sizes.
Remember again that the effective nuclear charge is pretty much the most important thing in the life of a (core) electron. Doubling or tripling it increases the forces the electron is subjected to proportionally. Thus, the larger the charge the closer the electrons will come to the nucleus, or the quicker the wave function decays. This is also the reason why atomic radii decrease across a period: a higher effective nuclear charge exercises greater force on the surrounding electrons and draws them closer towards itself.

This part was the theory. What about the practice? Well, ionic radii are even less well defined than covalent radii; it basically comes down to obtaining a crystal structure of a specific salt and measuring distances. So all that the smaller ionic radius tells us is that anions can approach aluminium even closer than they can lithium. Again, this should come as no surprise, as the anions (being charged bodies by themselves) are also more strongly attracted by a triple positive charge.


text:
PIB wave functions and the effects of e- delocalization on UV absorption in conjugated systems
I'll try to keep this short and concise. What made me start thinking about this question was the material we have been covering regarding PIB wave functions in my quantum chemistry class, and material we have been covering regarding ozone absorption in my climate class. I would like to try to understand the relationship between the increased wave function domain that delocalization yields for conjugated pi electrons, and the effect that this has on the wavelengths of light that they can absorb. Like ozone, I understand that most other conjugated pi systems have strong UV absorption, so I would assume there is some correlation there.
Specifically though, I would like to know if this can be explained (at least to some extent) by a simple QM model like the 1D PIB, and if how electrons' wave functions are altered when their spacial domain is extended across the whole molecule by delocalization can explain UV absorption in conjugated pi systems.

You can use the PIB model as a very approximate way of thinking about energy levels in conjugated ($\pi$ electron) molecules. Absorption in triatomics is much more tricky, however, and usually a Walsh diagram is necessary.
The general idea is that as the conjugation gets larger the absorption wavelength for electronic transitions gets smaller and so shifted to the red part of the visible spectrum. (A similar argument applies to particles on a ring for aromatics)
The energy levels of a PIB vary as $E_n \approx (n/L)^2$ where n is the quantum number and L the length of the box, or conjugation. Take this to be sum of all conjugated bond lengths. Thus as L gets larger the energy levels become closer together. Secondly as the number of conjugated bonds increases the number of quantum levels filled also increases, as there are more electrons. The optical transition is from the highest filled level to the next one up (HOMO-LUMO), the energy difference is $E_{n+1}-E_n$ at each value of L. If you take the length of the box to be the number of bonds m of length a, then $L=ma$ it is easy to show that the absorption spectrum shifts to the red as the conjugation increases, although you should definitely not expect accurate values in terms of energies.


text:
Can thionyl chloride convert two equivalents of carboxylic acids to acyl chlorides?
Specifically, I've outlined the arrow pushing heuristic for getting from a carboxylic acid to an acyl chloride in red. Now the mechanism outlined in green is the standard route for what I'm told happens with the release of SO2 gas. However, how stable is this intermediate and is it possible that it undergoes conversion twice as I've drawn in orange?

No. The byproduct of the reaction of thionyl chloride with an acid is unstable and it decomposes very quickly. And even if the compound would be more stable, it would be probably less reactive than thionyl chloride.

Cool, thanks. Why did you put your response as a comment instead of an answer?


text:
Work done when P,V, and T are changing simultaneously
I am given the following P-V graph for an ideal gas:

I have been asked to find the work done in the process from C to A, for which $P$ is directly proportional to $V$. This may be calculated in two ways:

Work done= Area under the line CA, which is the area of the trapezium, given by

$$w = -\frac{1}{2}(P_0+2P_0)V_0 = -\frac{3P_0V_0}{2}$$
(where the negative sign has been introduced since the work done is negative due to decrease in volume.)

Work done = $\int -P\,\mathrm{d}v$. However, in this case, since $P$ is directly proportional to $V$, how to integrate this?

How to calculate the work done in a process where all 3 variables, $P$, $V$, and $T$ are changing simultaneously and $P$ is directly proportional to $V$?

Who said that $\ce{PV}$ is the work done?

The second method $\ce{P}\Delta\ce{V}$ only works when $P$ is constant, otherwise you should use integration.

Ohh, I see. But how to integrate for this case in which I can't substitute $P$=$nRT/V$ since $P$ is directly proportional to $V$?

It does not matter if $\ce{P}$ is directly proportional to $\ce{V}$. The energy is still $\displaystyle\frac23\int\ce{P}\ \mathrm d\ce{V}$.

I'm sorry, but how did you obtain that expression?

The energy is not $\ce{PV}$, but $\dfrac23\ce{PV}$. It is derived from the last formula of this page.

Okay, but how does this help me to integrate this?

Well, $\mbox{work done}=\displaystyle\frac23\int_{\ce{V0}}^{\ce{2V0}}\ce{P}\ \mathrm d\ce{V} = \frac{2\ce{P0}}{3\ce{V0}} \int_{\ce{V0}}^{\ce{2V0}}\ce{V}\ \mathrm d\ce{V} = \frac{2\ce{P0}}{3\ce{V0}} \left(\frac{\ce{V}^2}2\right)_{\ce{V0}}^{\ce{2V0}} = \frac{2\ce{P0}}{3\ce{V0}} \left(\frac{\ce{3V0}^2}2\right)=\ce{P0V0}$

Let us continue this discussion in chat.

I don't understand how the factor of 2/3 is relevant here. The question is asking for work done not the average kinetic energy of the gas molecules.

@orthocresol: That's what I've been thinking too... :/

Your calculation of the area of the trapezoid is the same thing as getting the integral of pdV, aside from the sign  of the result (which has to be taken into account).

As far as I am aware, there is no generic formula for when all three variables are changing at once.
In this case, the easiest way is to simply use the area of a trapezium.
If you insist on integrating, then you need to know what $P$ is as a function of $V$. This derives from basic principles of integration - if you have a function $f(x)$, then $\int_a^b f(x)\,\mathrm{d}x$ gives the area under the curve of $f(x)$ from $x = a$ to $x = b$.
The line $\mathrm{C \rightarrow A}$ can be easily found since the beginning and end points are given. The gradient of the line is
$$m = \frac{2P_0 - P_0}{2V_0 - V_0} = \frac{P_0}{V_0}$$
and the the intercept is clearly 0 from the diagram (you can work it out anyway). In general the equation of a line $y = y(x)$ is given by $y - y_i = m(x - x_i)$ where $(x_i,y_i)$ is any point on the line.
That was perhaps an unnecessarily long explanation but the point is that you now know the relation
$$P = mV$$
holds (where $m$ was determined earlier). Therefore
$$\begin{align}
w &= -\int_{2V_0}^{V_0} P\,\mathrm{d}V \\
&= -\int_{2V_0}^{V_0} -mV\,\mathrm{d}V \\
&= m\left[\frac{1}{2}V^2\right]_{2V_0}^{V_0} \\
&= m\left(\frac{1}{2}V_0^2 - 2V_0^2\right) \\
&= \frac{P_0}{V_0}\left(-\frac{3}{2}V_0^2\right) \\
&= -\frac{3}{2}P_0V_0
\end{align}$$
which is the same as you found via the first method (you also forgot to put the negative sign in the right-hand side of the equation, but I assume that is just a typo).

I thought you need to multiply by $\dfrac23$ :o

@user34388: Bit why?

@user34388 why? The formula you quoted is from the kinetic theory of gases and uses the mean velocity of the gas molecules. It's not relevant to our present question of thermodynamics.

Yes, I realize that it's not relevant! I was asking @user34388, who quoted it in the first place why it is relevant for our present discussion.

@KaumudiHarikumar I wasn't talking to you.

Oh, shucks. Sorry :P


text:
Polar vs. Amphipathic
How can I tell if a molecule is purely polar or amphipathic?
For example, fatty acids are amphipathic because they have a polar carboxyl group and a non polar chain of hydrocarbons. However, sometimes there are molecules that have hydrocarbon chains and side polar groups but are categorized as polar.
What is it that I should look for?
Thank you so much!

The answer is that it is a sliding scale. There is no such thing as "purely polar" or "purely nonpolar." Water is very polar and hydrocarbons are very nonpolar, but there is no clear dividing line between the two categories.
Further confusion in biological molecules such as amino acids can occur, because different individuals categorize them in different ways, considering either just the side chain or the whole molecule. For example the amino acids have the formula $\ce{H2N-CHR-COOH}$ and in the simplest case, Glycine, R=$\ce{H}$. This is clearly a very non polar side chain, but the molecule as a whole is mostly polar, as the side chain is such an insignificantly small part of the molecule. Thus authors who categorize amino acids by side chain (mainly chemists) put Glycine into the non polar category, while authors who categorize amino acids by the whole molecule (mainly biochemists and biologists) put it into the polar category. When it comes to Alanine, R=$\ce{CH3}$ all authors agree it has a non polar side chain and most seem to categorize it as a non polar molecule.
A simple search of the internet shows that there is very little agreement about the amino acid Tyrosine, R=$\ce{CH2-(C6H4)-OH}$. It is classified as polar, non polar, having a polar side chain and having a non polar side chain by different authors. Clearly it has some characteristics of both, having both a benzene ring and a hydroxyl group. The correct definition would be "moderately polar."
You might hope that it is possible to give molecules a number which describes their polarity. This can be done to some extent. Googling "solvents by polarity" will give you lists for common solvents. This is, however, complicated if the molecule of interest happens to be acidic or basic. An acidic molecule will be polar in acidic solution but will become charged (and therefore much more polar) in basic solution, due to loss of $\ce{H+}$. Similarly, a basic molecule will be polar in basic solution but will become charged (and therefore much more polar) in acidic solution due to gain of $\ce{H+}$.
An amphipathic or amphiphilic molecule has a part that is highly polar and a part that is highly non polar. These terms also generally mean not just a molecule that is moderately polar, but one in which there is some distance between the polar and non polar parts, so the molecular weight would be at least into the hundreds.
The example you give of a fatty acid is a good one. (The deprotonated, negatively charged fatty acid anion in basic solution would be an even better one.) Another example would be many cell membrane proteins, which have a non polar, hydrophobic part (that stays in the cell membrane) and polar, hydrophilic parts (that extend out into the aqueous media inside and outside the cell.)
In summary, different substances should not be pigeonholed as polar or non polar as it would require that you draw an arbitrary line, but in general the polarity of two different substances can be compared fairly objectively. And pH may make a difference.

Good thorough answer in my opinion! I made some formatting changes based on the formatting criteria discussed here. It's very well written and makes for a good primer/guideline.


text:
Can carboxylic acids like act as nucleophiles in substitution reactions?

Can carboxylic acids like formic acid act as nucleophile in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ reactions?
For example can formic acid react with an alkyl/aryl halide like benzyl bromide?

I'm asking because I found a very similar question while going through some old organic chemistry test papers. However none of the book's I have read says that carboxylic acids can act as nucleophile in such cases.

Carboxylic acids can act as nucleophiles of course. Consider reactions with isobutene or thionyl chloride (although base certainly helps in this latter case). However, whether it's in $\mathrm{S_N2}$ or $\mathrm{S_N1}$ reaction mechanisms is another matter. If you beast something with enough heat it will usually react. I would not be surprised if a carboxylic acid reacted with triflouromethanesulfonic anhydride without base. It's still $\mathrm{S_N2}$ as mechanistic studies have shown the S=O bonds do not open up like acyl electrophiles.
I suspect you would need to resort to a continuous as opposed to a batch approach to push a reaction between the acid and an alkyl halide without using any base. Expose the reaction to high temperature (I'm thinking 250C and higher)  for 5 seconds or less under continuous liquid flow.


text:
Is the SN1 reaction faster with an axial or equatorial leaving group?

Why does compound 1 undergo the $\mathrm{S_N1}$ reaction faster than 2 even though both proceed via the same carbocation intermediate?

You had the two compounds the wrong way round, by the way. The cis diastereomer reacts faster. Also, please no MathJaX in titles.

@orthocresol do you know what the relative rates of solvolysis are for these two compounds?

@ron I went searching a bit. Can't find anything on the chlorides, but acetolysis of the tosylates proceeds with $k_\mathrm{ax}/k_\mathrm{eq} = 3.24$. JACS 1955, 77, 5562

In compund 2, both substituents can be placed in equatorial positions, whereas in 1 the $\ce{Cl}$ group is forced into an axial position since the bulky t-butyl group has to be placed equatorial. This makes compound 1 more unstable (higher energy), leading to a faster rate of formation of the carbocation.

I think it's highly unlikely that the cyclohexane will adopt a non-chair conformation. The chlorine will just be stuck in an axial position.

Just to add: an alternative way of looking at it is that, with compound 1, upon leaving of the $\ce{Cl-}$, there is relief of 1,3-diaxial interactions. But it is really just a different way of describing the same thing.


text:
Estimate Partition Function from a simulation
I have a system and I've carried out a long molecular dynamics simulation over it. I would like to estimate the partition function $Z.$ Theoretically, one would compute:
$$Z=\dfrac{1}{N!h^{3N}}\int \exp \left(-\beta\frac{-p^2}{2m} \right) \exp\left(-\beta V(r) \right) \ dp \ dr, $$
but if the system I'm dealing with is, say, a protein in water or something like that, this integral is of course absolutely untractable. When thinking of how to estimate the partition function of susch a system from a molecular dynamics simulation, I guess the first thing that comes to mind is the following: Consider all the snapshots obtained during your simulation, label them with the index $j$ and call $E_j$ to the energy of each snapshot. For giant thermalized systems, I guess we can neglect the probability of a concrete configuration repeating, so I believe a nice try would be:
$$\tilde{Z}=\frac{dV_{\text{ps}}}{N!h^{3N}}\sum_{j} \exp\left( -\beta E_j \right),$$
where $dV_{\text{pc}}$ is the volume of small cube in phase space (its volume would have to be adjusted manually depending oh how many snapshots one has taken).
My question, briefly put, is, how can I estimate the error of estimating $Z$ by $\tilde{Z}$?
It seems intuitive that typically $\tilde{Z}$ should be a good aproximation to $Z,$ but one can easilly think of situations where this isn't true: Imagine that my initial configuration lies in, so as to speak, a valley in the energetic landscape, surrounded by a chain of "high mountains" (high-energy configuration) and that at the other side of these "mountains" lie low-energy configuration. It is clear that in such a case there could be a considerable error when estimating $Z$ from $\tilde{Z}.$ But since one does not know the aspect of the energetic landscape beforehand, it seems difficult to deal with this.
The question stops here, but just to provide context I'll explain briefly what I'm thinking about: I want to study a chemical reaction, and to that end I would like to know the free energy profile along a certain reaction coordinate. For that, an Umbrella Sampling technique is used: you drag the reaction coordinate along the reaction path, forcing it to be at certain positions at different steps (otherwhise, the simulation wouldn't take you over the energetic barriers in short simulation times). I am aware of several methods to compute free energies, like FEP or WHAM, which is quite sophisticated, but I was guessing that if the technique I mentioned above to estimate the partition function $Z$ was correct, then the goal of getting the free energy profile would be fairly simple: you just drag along the reaction coordinate, perform molecular dynamics simulations for different restricted values of the reaction coordinate (you force the reaction coordinate to be at a concrete position by means of a harmonic potential), and at each value of the reaction coordinate you estimate the partition function as above. That's the idea and that's why I am interested in having some idea on methods to estimate the error of approximating $Z$ by $\tilde{Z}.$
Note: this question was originally in Physics.stackexchange. It didn't receive any attention at all and I asked for it to be migrated to Chemistry.stackexchange, but the flag was ignored, so I've decided to delete my question on Physics.stacexchange and post it here

What you're saying does make sense. You can, in fact directly estimate the partition function using nested sampling, with error bounds. See for example this paper http://journals.aps.org/prb/abstract/10.1103/PhysRevB.93.174108 . In your case without PBC it's actually even simpler. Also check out Livia's blog here https://liviabpartay.wordpress.com/publications/

There's a deeper problem, in my opinion: your quantity \bar{Z} will not go to the partition function in the limit of infinite sampling. The problem is that your configurations j are already drawn from a Boltzmann distribution (if your simulation is at a defined T), and your expression for \bar{Z} then weights these Boltzmann-distributed energies by a second Boltzmann factor.
Getting the partition function (or configuration integral) from a thermalized simulation is not easy, and hence is not typically attempted.
You might be interested in a discussion of this in Andrew Leach's book, Eqs 6.21-6.28, in the 2001 edition.

Yes, this is precisely what I was worried about. I understand you. Weighting the energies with the Boltzmann factors would make sense if the different snapshots were uniform in the configuration space, but the sampling is already biased towars more probable regions.  So, to estimate the partition function, I would need the density of states... Is the Wand and Landau algorithm the best way to do that?


text:
How to calculate the number of prochiral carbon atoms in a complex organic molecule?

Calculate the number of prochiral carbon in the following compound.

The Wikipedia page on prochirality states:

In stereochemistry, prochiral molecules are those that can be
  converted from achiral to chiral in a single step. An achiral species
  which can be converted to a chiral in two steps is called
  proprochiral.

My approach:
I first considered the sulfur atom to be prochiral as it satisfies the above condition, but it was pointed out, that this is incorrect as the S atom is already chiral.
My doubt:
But should even the double bond containing $\ce{sp^2}$ carbon atoms be considered prochiral?
According to what Martin postend in the comments I conclude that ("all" carbon atoms-4) in the compound are prochiral. I subtracted 4 chiral carbon atoms in the compounds. Tell me if there is still any mistake!
However, I am feeling that a carbon containing two hydrogen cannot be considered prochiral. The Wikipedia article states

If two identical substituents are attached to a sp3-hybridized atom, the descriptors pro-R and pro-S are used to distinguish between the two.

But H atoms are not considered substituents in organic chemistry, see substituent .

The S atom is already chiral.

Well, sulfur is not carbon. You should consider any carbon that could be a chiral centre with one conversion. Hint: Almost all of them are.

@orthocresol Thanks.I get you.But what about the double bonded carbons?

@Martin-マーチン You are correct.S should not be prochiral here.I guess you are referring to the double bonded carbons?

No, I am referring to all carbons. There is not a bit of symmetry in that compound.

@Martin-マーチン So say even a carbon containing two H should be considered prochiral?

Yes. It's in the very same article you linked: "If two identical substituents are attached to a sp3-hybridized atom, the descriptors pro-R and pro-S are used to distinguish between the two."

I consider the concept of prochirality as a property of an achiral compound (or centre), that can be turned into a chiral compound (or centre) in one step (proprochiral: in two steps) by means of substituting or addition. Hence all carbon centres with three different ligands (substituents or hydrogen) can be considered prochiral.
Prochirality is unfortunately a grown concept, which was used in different ways throughout the years and can be confusing at times. I find the definition in the IUPAC gold book quite enlightening and much better than the Wikipedia page that spawned off of it.

prochirality
Also contains definition of: proprochirality
This term is used in different, sometimes contradictory ways; four are listed below.

The geometric property of an achiral object (or spatial arrangement of points or atoms) which is capable of becoming chiral in a single desymmetrization step. An achiral molecular entity, or a part of it considered on its own, is thus called prochiral if it can be made chiral by the replacement of an existing atom (or achiral group) by a different one. An achiral object which is capable of becoming chiral in two desymmetrization steps is sometimes described as proprochiral. For example the proprochiral $\ce{CH3CO2H}$ becomes prochiral as $\ce{CH2DCO2H}$ and chiral as $\ce{CHDTCO2H}$.
The term prochirality also applies to an achiral molecule or entity which contains a trigonal system and which can be made chiral by the addition to the trigonal system of a new atom or achiral group. For example addition of hydrogen to one of the enantiotopic faces of the prochiral ketone $\ce{CH3CH2COCH3}$ gives one of the enantiomers of the chiral alcohol $\ce{CH3CH2CHOHCH3}$; the addition of $\ce{CN−}$ to one of the diastereotopic faces of the chiral aldehyde shown below converts it into one of the diastereoisomers of the cyanohydrin. The two faces of the trigonal system may be described as Re and Si.

The term prochiral also applies to a tetrahedral atom of an achiral or chiral molecule which is bonded to two stereoheterotopic groups. For example, the prochiral molecule $\ce{CH3CH2OH}$ can be converted into the chiral molecule $\ce{CH3CHDOH}$ by the isotopic replacement of one of the two enantiotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The prochiral molecule $\ce{HO2CCH2CHOHCH2CO2H}$ can be converted into a chiral product by esterification of one of the two enantiotopic $\ce{–CH2CO2H}$ groups. The carbon atom of the $\ce{CHOH}$ group is called prochiral. The chiral molecule $\ce{CH3CHOHCH2CH3}$ can be converted into one of the diastereoisomers of $\ce{CH3CHOHCHDCH3}$ by the isotopic replacement of one of the two diastereotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The stereoheterotopic groups in these cases may be described as pro-R or pro-S. Reference to the two stereoheterotopic groups themselves as prochiral, although common, is strongly discouraged.
  See: chirality centre
The term prochirality is also applied to the enantiotopic faces of a trigonal system.

The first definition applies especially to tetrahedral carbon atoms.
You can also work off of another definition which might be a bit easier to comprehend.

prochirality centre
  An atom of a molecule which becomes a chirality centre by replacing one of the two stereoheterotopic ligands attached to it by a different ligand, e.g. C-1 of ethanol; C-3 of butan-2-ol.

stereoheterotopic
    Either enantiotopic or diastereotopic. (In contrast the term constitutionally heterotopic has been used to describe identical groups located in constitutionally non-equivalent positions).

As a rule of thumb you can always substitute a $\ce{-CH2\bond{-}}$ group with a $\ce{-CHD-}$ group and ask your self if it is a chiral centre now.
I personally find it easier to identify stereoheterotopic ligands as this is also important for spectroscopy (esp. NMR).
This is somewhat also mirrored in the third definition.
The second (and fourth) definition applies to doubly bonded carbon atoms especially. If there are three different ligands, then you can assign a Re- and Si-face of that centre. Hence I would also consider these carbons prochiral.
In your case that means that all carbons (except for the ones that are already chiral) are prochiral.

Marked in pink are all chiral elements (the chirality of the nitrogen depends on the inversion barrier).
Marked in dark blue are all carbon centres who contain diastereotopic protons and hence can be considered prochiral according to the first definition. This is also evident from the $\ce{^1H-NMR}$ estimation of ChemDraw.
Marked in green are all carbon centres where you can add a nucleophile to create a stereo centre and are hence prochiral according to definition two.
Marked in orange are all carbons who are part of an aromatic system. Strictly speaking they are also trigonal carbons and technically you can assign a Re/Si to them. Accordingly they fall within the second definition and can be considered prochiral centres.

@glorfindel Thanks for finding all those broken images. I have removed the 'source' statement, as it is completely unnecessary, because the image was already part of the blockquote. (Also I personally dislike using  for these statements; it makes it appear less important, when they are most of the time not.)

@Martin-マーチン thank you for the improvement. The bot does scan the neighbour area of the image for links to the same domain (and if it finds one, it doesn't add attribution), but not as far as necessary to detect this link. I think I can manage to improve it by letting it consider the entire blockquote + neighbour area. In any case, this is one of the benefits of having them submitted as suggested edits, it encourages reviewers to improve the edit beyond what is possible through automated means.


text:
What is the percentage of d-tartaric acid in a mixture of d and l isomers?

Specific rotation for pure tartaric acid is $-12 ^\circ$. A mixture of d-l isomers is prepared which has a specific rotation of $+6^\circ$. What is the percentage of d-tartaric acid in a mixture of $d$ and $l$ isomers?

The pure tartaric acid must have been $l$-tartaric acid.
My attempt:
$$\frac{l-d}{l+d}=\frac{+6}{-12}=\frac{1}{-2}$$ So $d=3l$. Which implies $$\frac{d}{d+l}=\frac{3}{3+1}=0.75 $$
However the answer given in my textbook is $66.67 \%$. Most probably I didn't do any calculation error. So where did I go wrong?

Your calculation is correct.
You can check your result since you know that pure (+)-tartaric acid has a specific rotation of $+12^\circ$ and that pure (−)-tartaric acid has a specific rotation of $-12^\circ$. Therefore, a mixture of $75\ \%$ (+)-tartaric acid and $25\ \%$ (−)-tartaric acid has a specific rotation of
$$0.75\times(+12^\circ)+0.25\times(-12^\circ)=6^\circ$$
A similar exercise can be found in Hart, Hadad, Craine, Hart Organic Chemistry – A Short Course: Google Books
The corresponding solution can be found here: Google Books

And that results in $50~%~\mathrm{e.e.}$


text:
Why is the cis isomer of 1,3-dimethylcyclobutane more stable than the trans isomer?

I expect there to be more steric repulsions between the two methyl groups if they are on the same face as each other. Why is this not the case?

Hint: the four carbon atoms are not coplanar. It is not a flat square.

You should find a realistic stick drawing of the cyclobutane ring, you should then be able to answer your own question.

The four carbon atoms in cyclobutane are not coplanar, as that would lead to a large degree of unfavourable eclipsing interactions. So, it is not a flat square; it adopts a puckered conformation. One of the carbon atoms makes a $25^\circ$ angle with the plane formed by the other three carbons, and this mitigates some of the eclipsing interactions, at the cost of a small increase in angle strain. The conformation is commonly called the "butterfly", and two equivalent puckered conformations interconvert rapidly. Here's a visual reference to help you

Pictured above is an unsubstituted cyclobutane.
Take a closer look at the figure on the left, and notice how there would be some transannular interaction between the two $\ce{H'}$ hydrogens at $\ce{C-1}$ and $\ce{C-3}$, marked in $\color{red}{\text{red}}$. This is bad. In the figure on the right, the $\ce{H'}$ hydrogens are no longer axial, but equatorial, and this is better.
In short, placing substituents in an equatorial position is better than placing them in an axial position.
Now, imagine what the cis and trans isomers would look like for 1,3-dimethylcyclobutane.
In the cis isomer both $\ce{H'}$ on $\ce{C-1}$ and $\ce{C-3}$ would be replaced by $\ce{-CH3}$ groups (since they are on the same side). Of course there will be considerable repulsions if they are arranged in the conformation on the left, which is why ring flip takes place and we will predominantly get the conformation on the right with both the methyl groups in equatorial positions. This is ideal.

The trans isomer would be like replacing one $\color{red}{\text{red}}$ $\ce{H'}$ and one $\color{blue}{\text{blue}}$ $\ce{H}$ at $\ce{C-1}$ and $\ce{C-3}$ with methyl groups. No matter what you do, you can only get one of the methyl groups at the equatorial position, but never both.

[SOURCE]

It is because cyclobutane looks like a square piece of paper with a
  fold along the diagonal. If you add the methyl groups to the corners
  not on the fold, they can be both axial or both equatorial. The axial
  is the less stable conformation than diequatorial. If it is the trans
  compound, then it will have one axial and one equatorial. Just as two
  equatorial groups are more stable, only one equatorial group will be
  less stable.

You need to understand that cyclobutane is a bent structure to reduce angle strain and torsional strain.Read more here http://www.masterorganicchemistry.com/2014/04/03/cycloalkanes-ring-strain-in-cyclopropane-and-cyclobutane/


text:
Why can't one make C60 with a combination of hexagons and heptagons?
$\ce{C60}$ has a pentagon-hexagon system. Why can't one make it from a hexagon-heptagon system, or a pentagon-heptagon system for that matter?

It is possible to build carbon skeletons of other geometric forms (though they usually contain hydrogen as well). But they are not exactly easy to make. Dodecahedrane, for example, has a carbon skeleton with just pentagons, derivatives of cubane (just squares) are known. And they don't have 60 carbons.

Well, let's see. This is not quite a chemical question, though I wouldn't recommend moving it to Math.SE either, because they might have hard time recognizing what's $\ce{C60}$.
Now, let's imagine we're considering locally planar conjugated carbon
structures. Each carbon has to reserve one $p$ orbital for $\pi$-bonding,
which leaves it with 3 bonds. Now, the bonds tend to go in cycles; let's
confine ourselves to 5-, 6-, and 7-membered cycles, because otherwise we'll
have too much of angular strain (not that other cycles are totally impossible, though). So in effect, we are considering graphs
consisting of pentagons, hexagons, and heptagons, with 3 edges meeting at
each vertex. Let's say the numbers of these n-gons are $n_5, n_6$, and $n_7$,
correspondingly.
That's when old Euler kicks in. Polyhedra must meet a certain relation
between the number of faces (F), edges (E), and vertices (V):
$$V - E + F = 2$$
OK, what is the number of faces? Why, that's quite simple: $F=n_5+n_6+n_7$.
Then what is the number of vertices? Apparently, each pentagonal face has
five of them, and so on, so it would be $5n_5+6n_6+7n_7$. But wait, we've
counted each atom three times (for it belongs to three polygons), hence we must divide by 3: $V=(5n_5+6n_6+7n_7)/3$. By similar reasoning, the number of edges would be $5n_5+6n_6+7n_7$, if not for the fact that each edge belongs to two faces and thus is counted twice, hence $E=(5n_5+6n_6+7n_7)/2$.
All in all,
$$
V - E + F = 2\\
{5n_5+6n_6+7n_7\over3} - {5n_5+6n_6+7n_7\over2} + n_5+n_6+n_7 = 2\\
-{5n_5+6n_6+7n_7\over6} + n_5+n_6+n_7 = 2\\
{n_5-n_7\over6} = 2\\
n_5-n_7 = 12
$$
That's pretty much the size of it. You can do more or less whatever you want, as long as you have twelve more pentagons than heptagons. There is no going around it.
Now chemistry kicks in, too, and brings its own limitations. An $sp^2$ carbon wants to have its bonds in one plane, at $120^\circ$ angles. You can convince it to budge, but only so much. If you strain it more, the system becomes unstable. It turns out that two pentagons meeting at one carbon is already too much (not for an $sp^3$ carbon, though; see dodecahedrane). That's why you can't have constructions made of pentagons alone (or pentagons and heptagons, for that matter). Geometry would allow that, but chemistry wouldn't.
Let's see how that translates to actual shapes. Several fused hexagons give us coronene; it's planar.

(It is tilted sideways to give it a 3D feel. There are hydrogens attached on the outer edge; I omitted them for clarity.)
Extend it further, and you'll have an infinite flat sheet; that would be graphene.
Throw in a pentagon, and here's corannulene; it's bowl-shaped.

Extend that pattern, and you'll have fullerene.

Use a heptagon, and get circulene:

See that? It's bent indeed, but it's bent the wrong way. It's a
saddle-shaped bend. An example of negative Gaussian curvature, so to say. You can't wrap it to a sphere, no matter how hard you try.
Can we extend this further? Well, we kinda can (on paper, that is).
Moreover, it can be done in a number of ways, some leading to magnificent structures (purely hypothetical, of course).

Upd. It was brought to my attention that according to some studies, so-called carbon nanofoam might be a real example of structures of the latter kind. I'm not quite convinced (to me, until you have an X-ray diffraction study, everything is just hand-waving), but let's mention that anyway.

I was wondering whether anyone would look at this from a geometrical standpoint, and your answer is even more pleasant to read than I expected! I shall briefly point out Stone-Wales defects and carbon nanofoam as chemically-relevant examples of fused heptagons in carbon structures.

Thought of including the Stone-Wales thing at some point or another, but it didn't quite fit anywhere, and the answer is long enough already (by my standards, that is), so let's save it for another day.

Really, really great answer!