text:
lindane extraction from aqueous solution using hexane
I am studying the degradation of lindane using iron nanoparticles. I have prepared lindane solution with an acetone and water mixture (1:9 v/v). I tried to extract lindane after reaction with nanoparticles using hexane as a solvent. a viscous third layer has formed between the two layers. I have no idea why this happened. Can anyone please explain why this has happened?

My guess is that the middle layer is an acetone solution of iron chloride with any particulate matter.

According to this publication, lindane has the following properties:
Practically insoluble in water
Specific gravity = 1.85
Melting point = 113 $\mathrm{^oC}$
Solubility in acetone > 200 g/L
Solubility in n-heptane* = 10-14 g/L
* Used as surrogate for hexane solubility for this argument.
Notes: lindane is by far the most soluble in acetone, though only a small amount could have been used in the 1:9 acetone:water solvent.  If we assume that the reaction mixture were extracted with an equal volume of hexane (conservative I think, probably used more hexane), then the solution would be about 1:10:10 acetone:water:hexane. Given the properties of lindane, I'd expect it to make for a gooey, separate, small layer between the water and hexane.
Prediction:

Bottom layer - Water, undissolved iron, dissolved iron chlorides.
  Middle layer - Acetone, lindane, partially reacted lindane, some iron chlorides.
  Top layer - hexane,
  Small amounts of lindane and organic byproducts.

What has happened is that the dreaded "third phase" problem has hit you.
With a very low concentration of lindane in water it will be possible to extract the lindane with hexane. But when the lindane concentration is higher then the organic phase is unable to support the lindane concentration.
I also suspect that the lower phase might be too concentrated in terms of lindane. I think you need to consider the question of "what am I trying to do". If you are trying to study the degradation of traces of lindane then I would suggest you consider working at lower concentrations of lindane which would redcue the loading on both the organic and aqueous phases.
You might need to work using a GCMS or ECD-GC to measure the lindane and the other organics, this would bring down the amount you need to work with.
If you are considering the degradation of large concentrations of lindane then I would suggest that you change to toluene. It is likely that this will be more able to extract lindane than hexane. If you are wanting to study the system with GC then I would suggest that you do the first extraction of the the toluene spiked with an internal standard.
Then transfer the organic phase (200 microlitre) into a new vial containing hexane (1.8 ml) and wash it with water (2 ml) to remove acetone and other hydrophilic substances. You can repeat this process several times before drying the extract with Na2SO4. Then inject into the GC. This method is biased in favour of the lipophilic compounds.
You might well want to try a few method development experiments in which you start with a known mixture of organics such as phenol, lindane, chlorocyclohexane, 1,2-dihydroxycyclohexane and other things which are similar to the things you are considering. Then you can get the recovery yeild of these things in the solvent extraction workup.
I think it is very important to choose solvent extraction conditions that do not give anything other than two clear layers. The reason is that you might lose one or more substances while leaving the other substances in the organic layer. This will cause your experiments to give a misleading impression of what is in the reaction you are sampling.
You might also want to consider the following method. Take a sample out and dilute it with acetone. Next add to it triethyl orthoformate and some tosic acid. Allow it to stand for some time (you will have to work out how long through experiments). Next add triethyl amine (excess) and evapourate it down to dryness. Now redissolve in ethyl acetate and then shake the organic phase with water to remove Et3NH OTs and the other very polar impurities. Now dry with Na2SO4 and filter. Now dilute with acetone for GC.
This is a method which works rather well for sugars and other things with 1,2-diols. It converts the 1,2-diols into acetone acetals which travel a lot better through a GC column. The action of the trethyl orthoformate will dehydrate the mixture and drive the reaction to make it complete.
If you want to make the products even more non polar for GC, the I would suggest that you do the following. After drying the ethyl acetate extract then evapourate it down and then redissolve in acetone. Add silver(I) oxide to this and then some methyl iodide. Allow it to stir in a sealed flask in the dark overnight. The next day add some methanol and stir for a while to destroy the excess methyl iodide. Now filter it and get the sample ready for GC. Now any remaining alcohols which did not react with the acetone will have been converted into methyl ethers.
I have used the acetone acetal and silver oxide / methyl iodide methods to allow me to do GC on sugars and these methods do work very nicely.


text:
Defining a stereospecific reaction
My textbook says that a reaction which gives different stereoisomeric products when the different stereoisomers of starting material are used in a stereospecific reaction.
Is it necessary that all different stereoisomers will give different stereoisomeric products? For example, in bimolecular elimination diastereomers give different geometric isomers while enantiomers don't. So, do we say that this is an example of stereospecific reaction?

Yes, E2 elimination is an example of a diastereospecific reaction.

A stereospecific reaction is one in which there is only one possible stereochemical outcome for the product and that outcome is controlled by the stereochemistry of the reactant.
One of the earliest examples studied by most students of organic chemistry is nucleophilic substitution by the $\mathrm{S_N 2}$ mechanism. This reaction in its simplest form has a concerted mechanism. Reactions with concerted mechanisms often have stereospecific outcomes because the concerted mechanism demands a certain geometric arrangement of the reactants. In the case of the $\mathrm{S_N 2}$ mechanism, the nucleophile must approach from the opposite side of the molecule from the leaving group for steric and electronic reasons.

Likewise, the $\mathrm{E2}$ elimination mechanism leads to a stereospecific outcome. Stereospecificity does not require a one-to-one mapping of stereochemical inputs to outcomes. Rather, it requires a predictable mapping of inputs to outcomes.
For example, consider the $\mathrm{E2}$ elimination of 2-chloro-3-methylpentane with hydroxide. The reactant has four possible stereoisomers and the major Zaitsev product has two possible stereoisomers. Each stereoisomer leads to only one possible stereoisomeric product even though each stereoisomer of the product can be prepared from two stereoisomers of the reactant.

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text:
How can I build the molecular orbital diagram of the complex tetraammincopper(II) (square planar structure)?
I have to build the molecular orbital for the $\ce{[Cu(NH_3)_4]^{2+}}$ complex.
I know I have to use the characters table, the irreducible representations, the rule 5 and, somehow, determine the energy levels of the orbitals sigma and pi.
With this I should be able to build the MO but I don't know how to do it.
PS: I know we usually find this complex as $\ce{[Cu(NH3)4(H2O)2]^{2+}}$, but what I need is a theoretical answer to explain the $\ce{Cu-N}$ bond using MOT.

In principle, it does not matter which central metal you choose and which ligands; if the geometry is identical, the resulting MOs will be identical, too. The problem is as you state that the complex you wish to determine does not have a square-planar geometry (nor is it a tetracoordinated copper) so whatever you calculate will be off. At first, I was thinking of using a nickel(II) analogue (which would be identical save a missing electron) until I realised that $\ce{ls{-}[Ni(NH3)4]^2+}$ likely doesn’t exist and $\ce{hs{-}[Ni(NH3)6]^2+}$ is, of course, octahedral. A little bit of additional searching led me towards the square-planar complex $\ce{[Ni(\textbf{L})](ClO4)2}$, where $\mathbf{L}$ is basically a 3,10-bissubstituted 1,3,5,8,10,12-hexaazacyclotetradecane.[1] We can pretend that this perfectly fits the $D_\mathrm{4h}$ point group (it doesn’t) and that the ‘ammine ligands’ are fully identical (at least they’re symmetry-equivalent) and coordinate exactly from the coordinate axes (well, almost).

The first step is to consider the symmetry of the metal orbitals. We have five of them, and we expect them to group up into one $\mathrm{e}$-representation (two degenerate orbitals) and another three non-degenerate orbitals. A table of the irreducible represenations can be found on meta (soon™) or we can use the one from TU Braunschweig. Thankfully, it already includes the representations of the d orbitals, but for comprehensiveness you can also determine it yourself by thinking ‘does the orbital stay the same ($+1$) or invert lobes ($-1$) if I apply that symmetry operation. Here is the answer for the d orbitals:
\begin{array}{cccccccccccc}\hline
\text{Orb.} & E & 2C_4 & C_2 (z)& 2 C_2' & 2 C_2'' & i & 2 S_4 & \sigma_\mathrm{h} & 2 \sigma_\mathrm{v} & 2 \sigma_\mathrm{d} & \text{Rep.}\\ \hline
\mathrm{d}_{z^2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \mathrm{A_{1g}}\\
\mathrm{d}_{x^2-y^2} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & \mathrm{B_{1g}} \\
\mathrm{d}_{xy} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & \mathrm{B_{2g}} \\
\mathrm{d}_{xz}/\mathrm{d}_{yz} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & \mathrm{E_g} \\
\mathrm{4s} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \mathrm{A_{1g}} \\
\mathrm{4p}_z & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & \mathrm{A_{2u}} \\
\mathrm{4p}_x/\mathrm{4p}_y & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & \mathrm{E_u} \\ \hline
\end{array}
$\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ must be grouped together; only as a group they will be transformed onto each other by $C_4$ or $S_4$. Always if you transform an orbital into empty space with a symmetry operation, find out which orbital was there and include that into a degenerate group.
Thus, the d orbitals transform as $\mathrm{a_{1g} + b_{1g} + b_{2g} + e_g}$. I also included the 4s ($\mathrm{a_{1g}}$) and 4p ($\mathrm{a_{2u} + e_u}$) orbitals for completeness; we’ll need them.

Next, do the same thing for the ligands. Here, we have to make use of group orbitals, i.e. group all the four ammine ligands together. Once we have the represenation of the group orbitals, we will need to reduce that to their corresponding irreducible represenations. Thankfully, we don’t have to consider whether the orbitals are pure or hybrid; we just need to consider σ-type orbitals and π-type orbitals. See the table below.
\begin{array}{cccccccccccc}\hline
\text{Orb.} & E & 2C_4 & C_2 (z)& 2 C_2' & 2 C_2'' & i & 2 S_4 & \sigma_\mathrm{h} & 2 \sigma_\mathrm{v} & 2 \sigma_\mathrm{d} & \text{Rep.}\\ \hline
4 \sigma & 4 & 0 & 0 & 2 & 0 & 0 & 0 & 4 & 2 & 0 & ? \\
8 \pi & 8 & 0 & 0 & -4 & 0 & 0 & 0 & 0 & 0 & 0 & ? \\ \hline
\end{array}
Here is where your ‘rule 5’ — or more accurately, the reduction formula — comes into play. It allows us to determine which irreducible representations a reducible one is made up of. (For smaller point groups you can often just derive it logically, but I wouldn’t do that for a group this size.) The general reduction formula is given in equation $(1)$ ($n$ is the number of times a given irreducible representation $\chi$, whose character with respect to a symmetry operation is $\chi_i$, is part of the reducible representation $\xi$. $h$ is the order of the group.) and the results for the σ and π type orbitals is given in the table below.
$$n = \frac1h \sum_i \chi_i \xi_i $$
\begin{array}{ccccccccccc}\hline
\text{Group} & \mathrm{A_{1g}} & \mathrm{A_{2g}} & \mathrm{B_{1g}} & \mathrm{B_{2g}} & \mathrm{E_g} & \mathrm{A_{1u}} & \mathrm{A_{2u}} & \mathrm{B_{1u}} & \mathrm{B_{2u}} & \mathrm{E_u} \\ \hline
4\sigma & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
8\pi & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 \\
\hline \end{array}
By doing this, we see that the σ-type orbitals transform as $\mathrm{a_{1g} + b_{1g} + e_u}$ while the π-type orbitals transform as $\mathrm{a_{2g} + b_{2g} + e_g + a_{2u} + b_{2u} + e_u}$. (That is all assuming I didn’t miscalculate.)
The next step is to start drawing an orbital diagram. The five d-orbitals should initially be on one level corresponding to the ionisation energy of nickel(II). The virtual 4s and 4p orbitals somewhat higher. The π-type ligand group orbital should be on a level corresponding to the ionisation energy of ammonia, the σ-type ones slightly below. I don’t have values here so I’m going to ballpark the relative energies draw a merely schematical diagram for simplicity. Any orbitals that have the same irreducible representation will mix with each other. If there are only two, these will form a bonding/antibonding pair with the antibonding orbital being relatively higher than the bonding one low. If there are more than two, they will form some kind of ladder with intermediate levels preserving the overall orbital count. Scheme 1 shows what I would end up with.

Scheme 1: Approximate orbital scheme for a fully square-planar, ideal $D_\mathrm{4h}$ $\ce{[Ni(\mathbf{L})]^2+}$ $\mathrm{d^8}$ ion.
Note that there are a number of estimates in there. For example, both $\mathrm{b_{2g}}$ and $\mathrm{e_g}$ are made up solely of d-π interactions; I can’t guess a priori which one would have a higher or lesser energy. Likewise for the relative orderings of other orbitals. The only thing clear is that any interactions stemming from the σ orbitals should be stronger than those stemming from π orbitals due to σ typically displaying better overlap.
To get back to your hypothetical $\mathrm{d^9}$ system, just add the missing electron into the $\mathrm{b_{1g}}$ orbital.

Reference:
[1]: M. P. Suh, B. Y. Shim, T.-S. Yoon, Inorg. Chem. 1994, 33, 5509. DOI: 10.1021/ic00102a026.

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text:
What is the rotational symmetry number used to calculate the rotational entropy from an ORCA frequency calculation?
The quantum chemistry software ORCA gives the following warning on frequency calculations:
CAUTION: The rotational entropy is not quite correctly treated here
         because it includes a symmetry number that is not yet correctly
         implemented in ORCA!
For a nonlinear molecule the correct rotational entropy is:
    S(rot) = R*(ln(qrot/sn)+1.5)
    R    = 8.31441 J/mol/K = 1.987191683e-3 kcal/mol/K
    qrot = 4922043.1563766
    sn is the rotational symmetry number. We have assumed 3 here
       if it is different for your molecule then you should correct
       the printed rotational entropy by manually evaluating the equation
       as given above

For convenience we print out the resulting values for sn=1 - 12:
 sn= 1  qrot/sn= 4922043.1564 T*S(rot)=    10.02 kcal/mol T*S(tot)=    34.72 kcal/mol
 sn= 2  qrot/sn= 2461021.5782 T*S(rot)=     9.61 kcal/mol T*S(tot)=    34.31 kcal/mol
 sn= 3  qrot/sn= 1640681.0521 T*S(rot)=     9.37 kcal/mol T*S(tot)=    34.07 kcal/mol
 sn= 4  qrot/sn= 1230510.7891 T*S(rot)=     9.20 kcal/mol T*S(tot)=    33.90 kcal/mol
 sn= 5  qrot/sn=  984408.6313 T*S(rot)=     9.06 kcal/mol T*S(tot)=    33.77 kcal/mol
 sn= 6  qrot/sn=  820340.5261 T*S(rot)=     8.96 kcal/mol T*S(tot)=    33.66 kcal/mol
 sn= 7  qrot/sn=  703149.0223 T*S(rot)=     8.87 kcal/mol T*S(tot)=    33.57 kcal/mol
 sn= 8  qrot/sn=  615255.3945 T*S(rot)=     8.79 kcal/mol T*S(tot)=    33.49 kcal/mol
 sn= 9  qrot/sn=  546893.6840 T*S(rot)=     8.72 kcal/mol T*S(tot)=    33.42 kcal/mol
 sn=10  qrot/sn=  492204.3156 T*S(rot)=     8.65 kcal/mol T*S(tot)=    33.36 kcal/mol
 sn=11  qrot/sn=  447458.4688 T*S(rot)=     8.60 kcal/mol T*S(tot)=    33.30 kcal/mol
 sn=12  qrot/sn=  410170.2630 T*S(rot)=     8.55 kcal/mol T*S(tot)=    33.25 kcal/mol

Now, I have got the following two questions:

What is the rotational symmetry number $s_n$, i.e. how is it defined?
What symmetry number $s_n$ would a molecule with no symmetry, i.e. $C_1$ point group have? The way I understand it, the rotational symmetry number reflects the rotational symmetry, e.g. $\ce{CHCl3}$ would have $s_n=3$ as there are three equivalent structures superimposed by the $C_3$ symmetry element. So that my guess is that $s_n = 1$ in that case. Is that correct?

ad 1) The rotational symmetry number corrects for the fact that you have to take the Pauli principle into account when deriving the partition function. That is, it accounts for the spin statistical weights of the different rotational levels.

@Paul "It accounts for the spin statistical weights of the different rotational levels" is not a really clear definition to me, sorry. In what way does it do this and what does it have to do with the Pauli principle?

On the link provided in the answer below the symmetry number is defined as: "The "rotational symmetry number or external symmetry number [...] is the number of unique orientations of the rigid molecule that only interchange identical atoms." I think that's a pretty clear-cut definition.

@Paul So given the definition from my comment above, I really don't see the connection to the Pauli principle, maybe you can enlighten me, I never really gotten warm with statistical thermodynamics : )

In the case of molecular hydrogen there are two identical H nuclei and when constructing the partition function or any other statistical thermodynamic quantity, the indistinguishability of these should be taken into account (that is, exchange of identical particles should give $\pm$ in accordance with the generalized Pauli principle). We can construct 4 spin functions: 3 symmetric ones and one antisymmetric one. The symmetric ones combine with odd rotational states and the antisymmetric one with even states.

So that the overall wavefunction is antisymmetric for exchange of the two spin 1/2 nuclei. As a consequence there are 3 times as many molecules with odd rotational states as even ones and this should be taken into account into the degeneracy factors when summing over all energies. The same reasoning can be extended to more complex molecules having identical nuclei.

Ah, so - @Paul, are you essentially saying that treating all nuclei of the same atomic number as identical is not strictly correct in the case where the nuclear spin varies among them? If so, I think it would absolutely be worth writing it up into an answer.

@Paul OK, so in the case of $\mathsf{H_2}$, you end up with $s_n=2$. That is in contrast to the answer of hBy2Py and to 2) of my question. Hmm... Now that I look at it I also see that the definition provided by us at least runs into problems for linear molecules as then $s_n=\infty$, am I right? Or is then $s_n=1$ as it reduces to "basically one rotation".

Only for nonlinear molecules it is the case that the symmetry number is the order of the point group if only pure rotations (including E) are considered.  $s=1$ for linear molecules of $C_{\infty\text{v}}$ and $s=2$ for $D_{\infty\text{h}}$ symmetry.

@Paul I see, so it is actually the number of mappings between atoms that can be achieved by proper rotations, that makes sense...

Your #2 is exactly right, as far as I know, which answers your #1.
A key thing to remember is that the rotational symmetry number doesn't include reflections, so molecules in groups like $C_s$ still have $s_n = 1$ since there are no superimposable orientations achievable only with rotations.
For reference: Table II of the NIST CCCBDB introduction to thermochemistry is a nice, concise summary of the symmetry numbers for the common point groups observed in chemistry.

Thanks for the fast reply : ) The only thing I don't get then is why the ORCA people made $s_n=3$ the default setting, as obviously $s_n=1$ is by far the most common...

@ketbra Agreed, defaulting to $1$ would make more sense to me, too. Perhaps, though, the relative error associated with an incorrect $s_n$ is largest for small molecules, which have a much higher probability of $s_n > 1$, and thus they chose a compromise value.

I see. That makes sense : )


text:
Why is acetic acid stronger than p-nitrophenol?
I know that acetic acid is a stronger acid than phenol because in acetic acid, the charge is delocalized on two oxygen atoms which is better than 1 oxygen and 3 carbon.
But in case of para-nitrophenol, $\ce{NO2}$ reduces the negative charge in the ring, thereby stabilizing the intermediate formed.
So does this mean that the stabilization caused by the presence of $\ce{NO2}$ is less than the delocalization caused by 2 oxygen atoms?
Kindly correct me if I am going in the wrong direction.

A somewhat relevant question

So, this is a pretty complicated topic.  Your reasoning is actually pretty good, but you are missing some key data.  Comparing a phenol to a carboxylic acid is somewhat hard.
The first, simplest, and least satisfying answer is that the $\ce{NO2}$ is farther away from the acidic hydrogen in 4-nitrophenol than the oxygen is from the acidic hydrogen in acetic acid.  Not good enough?  Let's dig further.
We might decide to expand upon your argument that because 4-nitrophenol has one more oxygen in it than acetic acid, there is more potential for electron withdrawal, and hence 4-nitrophenol should be stronger.  Solid reasoning, but consider our first, less satisfying answer.  Both of these effects might be combined, and we could reason that the distance effect is stronger than the increase in electron withdrawing power.  Decent reasoning, but still unsatisfying.
Enter the Hammett Equation:
$$\log \frac{K}{K_0} = \sigma \rho$$
Fully explaining this takes up one or two chapters of a graduate level physical chemistry book.  Quick summary, $K_0$ refers to a reference reaction, which has hydrogen as its substituent.  $\sigma$ is the electron withdrawing(positive) or electron donating (negative) value of a substituent replacing the hydrogen.  $\rho$ refers to a parameter that is descriptive of the reaction itself.  The parameters are empirically determined relative to the acidity of substituted benzoic acids.  When benzene rings are not involved, sometimes an equivalent called the Taft parameter is used.  Run 6 reactions (5 plus reference), draw a line, get your rho value for any reaction.
However, the reference reaction is the dissociation of benzoic acid.  $\rho$ is different for the dissociation of acetic acid and for nitrophenol.  So, even though nitro is a much stronger withdrawing group, its influence is mitigated by a reaction much less subject to such effects.  The Wikipedia article is not terrible on this matter.

hmm.. this sounds interesting but right now, most of it went from over the head. Thanks.

If logarithms and linear regression are not intuitive, it will be superheavy.  But it is pretty cool stuff.

The difference in acidity between acetic acid and phenol can be attributed to the electron-withdrawing properties of a carbonyl function versus a phenyl ring. In both cases, the deprotonated oxygen can be depicted in a resonance structure as forming a double bond to the carbon in question and the charge being delocalised within the ring.
It should immediately strike anybody that a phenyl ring primarily provides a resonance body (i.e. a greater volume into which the charge can be delocalised) while a carbonyl group is also actively withdrawing (both inductively and mesomericly). This is reflected when comparing the $\mathrm{p}K_\mathrm{a}$ values of $\ce{CH3CO-CH3}$ (acetone) and $\ce{Ph-CH3}$ (toluene) — $16.5$ logarithmic units in $\mathrm{DMSO}$ according to the Evans table[1] or approximately $24$ logarithmic units in water ($\mathrm{p}K_\mathrm{a}$ of acetone in water taken from the Brodwell data).[2]
Adding a nitro group to the phenyl ring reduces this difference drastically to $-6.1$ logarithmic units (i.e. para-nitrotoluene is more acidic than acetic acid) — in $\ce{DMSO}$. One should always remember that acidity differences in one solvent can be drastically different in another solvent. For example, the difference between $\ce{MsOH}$ and $\ce{TfOH}$ is $11.4$ logarithmic units in water but only $1.3$ logarthmic units in $\ce{DMSO}$ (that is not a typo). One should also note the difference between acetic acid in water ($\mathrm{p}K_\mathrm{a}\approx4.76$) and $\ce{DMSO}$ ($\mathrm{p}K_\mathrm{a}\approx12.3$).
$$\textbf{Table 1: }\text{Acidity constants of various compounds in various media}\\\text{All data taken from the Evans data unless noted otherwise in the text.}$$
\begin{array}{lrc}\hline
\text{compound} & \mathrm{p}K_\mathrm{a}(\ce{H2O}) & \mathrm{p}K_\mathrm{a}(\ce{DMSO})\\ \hline
\ce{CH3CO-CH3} & 19.3\phantom{000} & 26.5\\
\ce{Ph-CH3} & 31.5\phantom{000} & 30.6\\
\ce{$p${-}NO2-C6H4-CH3} & & 20.4\\
\ce{MsOH} & -2.6\phantom{000} & \phantom{0}1.6 \\
\ce{TsOH} & -14\phantom{.0000} & \phantom{0}0.3 \\
\ce{AcOH} & 4.76\phantom{00} & 12.3\\ \hline \end{array}
Given that a $\mathrm{p}K_\mathrm{a}$ value of para-nitrotoluene has not been reported in water (although it’s $\mathrm{p}K_\mathrm{b}$ value has been),[3] we are not in any position to draw definite conclusions. However, water being the solvent it is, I would still assume the electron-withdrawing ability of a nitro group which relies solely on mesomeric effects to be weaker than the effect of a directly neighbouring carbonyl group. Remember that water is excellent at stabilising charges; $\ce{DMSO}$ not so much.

References:
[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table
[2]: The Brodwell $\mathrm{p}K_\mathrm{a}$ data
[3]: J. C. D. Brand, J. Chem. Soc. 1950, 997. DOI: 10.1039/JR9500000997.

On acetic acid, the carbon in the carboxylic acid group is highly electron-starved: that carbon is bonded to oxygen atoms 3 times! Moreover, that carbon only has one methyl group from which to "take" electrons in order to fill the charge imbalance. This means that the only easy electrons the acidic oxygen on the carboxylic acid group can access is the one on the hydrogen. This makes the O-H bond more ionic in character, and so easy to dissociate in water.
On phenol, the carbon in the hydroxyl group is not electron-starved. Since it is part of an electron-rich aromatic ring, it is a "channel" through which the oxygen atom can access electrons - this lucky oxygen atom can access 6 electrons that are delocalized over 6 carbon atoms in order to satisfy its hunger for negative charges. This means that the O-H bond in question is more covalent in character, and more difficult to dissociate in solution.

Three carbon-oxygen bonds would be orthoacetic acid; I don’t see that as part of the question. Aside from that minor inconsistency, this is a nice and short answer.


text:
What is the relation between universal gas constant R and amount of substance n?
I have found it mentioned in various sources that the value of R depends on the amount of substance $n$ in the ideal gas equation: $PV=nRT$
If this is true, it would imply that $R$ is a function of $n$ and we can replace $R$ by that function.
I want to know what is the relationship between $R$ and $n$. If $R=f(n)$ then what is $f(n)$?
Pradeep's chemistry for class 11 part 1

Chemistry by Arjun

The universal gas constant is given by $R=k_\text{B}N_\text{A}$ where $k_\text{B}$ is the Boltzmann constant and $N_\text{A}$ is Avogadro's constant.

I do not see how R and n being on one side of an equation implies that one is the function of the other. That is simply not true.

@JonCuster I am not saying that them being on the same side of the equation implies that they are related(Read the question before commenting), secondly if you are saying that if two variables on one side of the equation cannot be related, you are again wrong. For example: If someone says that the area of a circle =pir(d/2), then they are not wrong. It may be possible that the value of d=2r may not be substituted in the equation.

The answer is that R is a constant (the universal gas constant) and has no relation to the number of moles. I still do not see anything that would imply that R = f(n) from the ideal gas equation.

Your book states that for a given amount of gas $PV/T$ is a constant which they define as $K$. When you take twice the amount of gas, $K$ becomes two times bigger. $K$ depends linearly on the number of moles $n$, thus you can write $K=nR$ where $R$ is a new constant that defines the increase in the constant $K$ per mole of gas.

I think you are misunderstanding something badly here. Let’s take a sentence that may be the cause of your misunderstanding:

Where $R$ is a constant of proportionality and is found to be independent of the nature of the gas and depends only on the amount of gas taken. For 1 mole of any gas the value of $R$ is the same. (emphasis mine.)

Proportionality constants are the $m$ in a typical linear equation $(1)$. With them and a value $t$, which is potentially present (but physics and chemistry attempt to choose variables in such a way that $t = 0$) it is possible to calculate $x$ from $y$ or vice-versa assuming they are in a linear relationship. As mentioned, it is a constant and thus neither a function of $x$ nor one of $y$.
$$y = mx + t\tag{1}$$
The second sentence I highlighted also includes a potential source of confusion. You might have understood this to mean ‘the value of $R$ is the same for any given gas assuming that exactly $1~\mathrm{mol}$ of it is present.’ However, that is the wrong interpretation. It should be taken to mean: ‘Whichever the chosen gas may be, the $R$ in the equation above will be the same — whether we are calculating with $1~\mathrm{mol}$ or $2~\mathrm{mol}$ or $10~\mathrm{mol}$.’
Indeed, $R$ happens to be a physical constant invariable of the chosen gas. It can be related to other physical constants by equation $(2)$, wherein $k_\mathrm{B}$ is the Boltzmann constant and $N_\mathrm{A}$ is Avogadro’s constant; both of which also invariable with respect to the chosen gas.
$$R = k_\mathrm{B} N_\mathrm{A}\tag{2}$$


text:
What is the difference between the following terms: Incandescence, luminescence, fluorescence, phosphorescence, chemiluminescence and bioluminescence?
What are the differences between  incandescence, luminescence, fluorescence, phosphorescence, chemiluminescence, bioluminescence?

Related: http://chemistry.stackexchange.com/questions/31039/chemiluminescence-vs-bioluminescence/31044#31044

The page on wikipedia gives a rather nice summary on the different types of luminescence: https://en.wikipedia.org/wiki/Luminescence
Incandescence is different from the others in that it is caused by heat. Again,  there's a wiki page for that: https://en.wikipedia.org/wiki/Incandescence

Related: http://chemistry.stackexchange.com/questions/8099/colours-and-transition-metals/8100#8100

That was a 'catchy' question ;)
First a broad division into : Incandescence and Luminescence.
Incandescence is the emission of light from a body that is heated until it glows.
Luminescence is the emission of light by a substance for any reason other than rise in its temperature.
Now the following come under Luminescence:
If the cause of the excitation is due to photons striking atoms, the associated emission of light is called photoluminescence. If it is due to electrons, then the emission of light is referred to as electroluminescence.
If the luminescence results from a chemical reaction, it is chemiluminescence. A form of chemiluminescence, where the emission of light is due to a chemical reaction occurring in vivo, then it's called bioluminescence.
If the luminescence persists significantly (according to my Oxford Science dictionary, a period of time greater than 10 nanoseconds would be "significant") after the exciting cause has been removed, it is called phosphorescence. If it fails to persist for longer than 10 nanoseconds, then it's called fluorescence.
EDIT- @KlausWarzecha answered the Fluorescence-Phosphorescence conundrum better in his answer here : What happens to electrons in metal complexes after excitation by visible light?
Quoting from his answer:

There are however cases, in which the initially formed excited singlet state species first relaxes to its lowest vibrational mode and the emits a photon to reach the ground state.
$$\ce{S_1 -> S_0 + h\nu_{f}}$$
This deactivation via radiation is known as fluorescence. Fluorescence lifetimes are typically rather short (in the ns range) and the emitted photon has a lower energy than the absorbed one.

and on phosphorescence,

In other cases, the excited singlet state undergoes a change in the spin multiplicity (Intersystem crossing) to a triplet state.
$$\ce{S_1 ->[ISC] T_1}$$
These excited triplet states are usually much longer-lived. When they eventually deactivate via emission of a photon, this process is known as phosphorescence.
$$\ce{T_1 -> S_0 + h\nu_{p}}$$

Note that the way I 'defined' fluorescence and phosphorescence has been greatly oversimplified (and possibly wrong), but it should suffice for your purpose. ;)


text:
Do Fermi resonances show in Inelastic Neutron Scattering spectra?
The Raman and FTIR spectra of a urea based mixture shows a strong Fermi resonance in the N-H stretching region, which difficults its interpretation. In an inelastic neutron scattering spectra of the same mixture, would the Fermi resonance be as obvious? Or not show at all?


text:
What is the mechanism for the hydrolysis of the boron-alkoxide complex in NaBH4 reduction?
According to this question, the stoichiometry of the reduction of a ketone is such that sodium borohydride can reduce up to 4 ketones by forming a boron-alkoxide complex.

The next step would involve the hydrolysis of the complex, and I've been searching through textbooks to no avail. What would the mechanism for this hydrolysis look like?

I don’t know whether the mechanism of hydrolysis of the boronic acid esters has been studied. However, the dissociative mechanism immediately suggests itself if only by comparison to similar structures.

Since the workup is typically written $\ce{H+/H2O}$, you may consider some buffered source of external protons. These can protonate any of the alcoholates which can then dissociate. (The $\ce{B=O}$ double bond I drew in should be considered no more than additional stabilisation from oxygen lone pairs; boron is electron-deficient by definition.) Another water molecule can associate: the first bond is cleaved. Repeat three times and end up with four molecules of alcohol and tetrahydroxidoborate $\ce{[B(OH)4]-}$.

3rd species, boron should be negatively charged. Are those arrows dative bonds?

Yes and yes. Which I chose mainly to avoid adding even more charges …

@orthocresol The wrong charge also explains why the chemical warning wouldn’t go away ;)

You could just drop the double bond in the second intermediate...

For an acidic work up:

Protonate an oxygen.
Fragment the tetraalkyl borate into the trialkyl borate and your product alcohol.
Add water to the trialkyl borate and remove the proton.
Repeat.


text:
How to separate alloys?
How do you separate metal alloys into their constituent elements, especially industrially? I don't think you can always use melting points, because alloys, depending on the composition (looking at phase curves) can have one common melting point.
Here's a phase diagram for a copper-nickel alloy, which I think is a system where both components are completely soluble in one another in all ratios.

This composition cannot be separated by cooling it down, as I understand it (even though, as mentioned here, the composition would not be uniform if the molten alloy is not cooled down slowly enough). As described here, the solid that would start to separate (the alpha phase) is itself an alloy.
EDIT: What about electrolysis? Could that work here?

Separate one alloy from another, i.e. distinguish them apart, or separate out the various elements?

@JonCuster separate them into the consituting elements

Maybe it can be interesting I don't know. https://en.wikipedia.org/wiki/Zone_melting

@Hexacoordinate-C I think this is rather used for purifying, don't know if this can really be used to take like 50 tons of steel alloy and separate out the 11% chromium some alloys seem to contain.

When you have a mix of different component you can use the eutectic point(s). If you have only two elements the one which is the more will melt (or the other i don't remember) after the other then you can separate the liquid and the solid and do it again and again as you want. If you have more, you'll need to do it again in other parts.

@Hexacoordinate-C I think this doesn't work for every alloy. I've added the example of Ni-Cu in the question.

I think this depends heavily on the metal. For example, purifying iron for steel via the Bessemer process. Gold can be purified in part by cyanide leaching, but there are other methods of parting gold.

What do you mean by electrolysis?

@IvanNeretin I'm really not sure where I'm going with this, but say we stick two electrodes that are up to the task into the molten alloy and pump current through it. I assume that even though the resistance will be a lot higher for most metals due to higher temp, it will still basically work as a conductor, so it'll just get warm. But what if we were to dump some salt into it? http://web.ornl.gov/info/reports/1963/3445603652660.pdf mentions that that can lead to loss of conductivity, so electrolysis could happen (maybe ionization of metal at anode?). Or would that still only affect the salt?

Essentially, the phases formed from molten metals and molten salts (given that they mix at all, which is not always so) may be either "metal-like" or "salt-like". In the first case it would conduct electric current using electrons, much like electric wires in your house, and would be able to do so for years without any change. In the second case it would be able to undergo electrolysis, like any molten salt, and you might be able to tune the conditions so that the metals separate. Then it is just another recipe starting with "Dissolve both metals..." (dissolve in a molten salt, that is).

there's a series of videos about a chemistry buff who distills platinum from catalytic converter dust from the roadside, and does other fun experiments of that kind.

@IvanNeretin yeah, you make a good point. Thanks)

You seem to approach the separation of alloys much like the separation of water-soluble salts, probably based on the fact that both involve the liquid/solid phase diagrams, which may even look alike. This is wrong; these systems are fundamentally different.
Say, you have a water solution of some well-soluble salt which you need to recrystallize. You cool it down, so the salt precipitates. You filter the crystals out of the solution, then you probably rinse them with something which would remove the traces of solution without damaging the crystals, or just let them dry.
Now suppose you have a metal alloy, maybe not even of the kind you mentioned above; let it have a simple eutectic diagram, like Bi-Cd.

You melt it, then carefully cool it down to a certain point, and end up with a mixture of some (supposedly) pure metal crystals with a still-liquid alloy. What good does it do? You can't really filter it. You can't rinse the crystals. You can't let them "dry". It is useless.
With metal alloys, your best bet is to dissolve both metals and separate them chemically, based on their different properties. In some relatively rare cases, you might be able to melt the alloy and perform chemical conversion of one component without affecting the other; examples include cupellation or Bessemer process referenced earlier in the comments. In yet more exotic cases, you might be lucky enough to chemically etch one component out of solid alloy; this is how Raney nickel is made. Unless you are that lucky, go back to the start of this paragraph.

Thanks for the information on cupellation and Raney nickel, that is helpful. So, you're saying that it generally very much depends on the case? Also, could you elaborate in your answer on how metal composition phase diagrams are different from non-metal ones? As in the sources I linked to, the behaviour of the alloy when heated is shown. The recrystallizing you bring up is valid, but it doesn't have much to do with the issue. Also, where's the problem in decanting molten metal and let it drain unter its weight, separating the residue, using something like a modified Bessemer container?

Generally you have to dissolve both metals and do some wet chemistry. As for the phase diagrams, they are not very different, but the systems are. Molten metal is not quite like water. You decant it, and a good deal of it just remains there (thanks to higher viscosity and surface tension), and you can't rinse it off, nor would it evaporate like water (since its vapor pressure is way lower than that of water).

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text:
Why can entropy be measured in joules per kelvin?
I'm a little confused as to what the measurement associated with entropy represents. It's obviously the joules of energy per kelvin, but how does this measure the disorder of a system?

I'm not 100% sure if the units can be directly related to the interpretation as a measure of disorder. The equation is $S = k \ln W$ and if the system is more "disordered", then $W$ increases, but $W$ and $\ln W$ are both dimensionless; the dimension of J/K arises from the Boltzmann constant $k$. That said, there's probably some explanation that I haven't heard of.

btw, using 'disorder' related to entropy should be avoided as its difficult to distinguish in ones mind between a little and a lot of disorder. The 'number of configurations', ( 'W' or $\Omega$) is better but unclear also, so its better to think about  the number of ways of placing particles into their several energy levels. Classically  entropy  is related to the heat absorbed from surroundings ($T\Delta S$) when an isothermal reaction runs reversibly. If positive the work done is greater than heat of reaction. Only from stat. mech. do we understand entropy fully.

The definition of entropy in 'Classical Thermodynamics' is $$\Delta S = \int\frac{\delta Q}{T}$$
The quantity on the left is the entropy change associated with a physical process. $\delta Q$ is an inexact differential, made exact by an integrating factor $\frac{1}{T}$ This notion was devolped by Kelvin and Cartheodory (I believe). This comes out of the analysis of certain kinds of differential equations called Pfaffian forms.
One notes the dimensions ("units") of this quantity, and observers them to be $\mathrm{J\ K^{-1}}$
It was only later, that the statistical definition of entropy (developed by Boltzman) came to be (and was found to be equivalent to the thermodynamic definition).
$$S = -k_B \sum_{i}p_i\ln(p_i)$$
In this framework entropy is a logarithmic measure of the number of states with significant probability of being occupied. In this expression, the the quantity is changed from dimensionless to dimensionful, by the Boltzmann Constat, which has the units, (you guesed it)  $\mathrm{J\ K^{-1}}$
The measure of "disorder" is in the term $\sum_{i}p_i\ln(p_i)$, which has no units. The units are appended to it by the Boltzmann Constant.

Entropy is a derived quantity. It, therefore, takes on values [and units] from its definition. In this case, the SI base units of J and K (for heat and temperature respectively).
A change in entropy may be defined as being equal to a quantity of heat transferred in a reversible isothermal process divided by the absolute temperature like e.g. in some parts of a Carnot cycle:
$$\Delta{S}=\frac{Q_{rev}}{T}$$
It is simply a relationship that was given its own special symbol for convenience i.e. S is a function that describes the thermodynamic state of a system (which is what makes it important).


text:
How to derive the Gibbs free energy for an ideal gas?

Determine the Gibbs Free Energy for an ideal gas and show that, this energy, and its derivative can be related to the enthalpy of the system via Gibbs-Helmholtz equation
  $$H = -T^2 \left( \frac{\partial (G/T)}{\partial T} \right)$$

So I need to determine the fundamental relation for an ideal gas in the Gibbs free energy representation.
My thoughts:
I know the Gibbs free energy, is defined by
$$G(p,T) = U + pV - TS = H - TS,$$
where $H$ is the enthalpy, then
\begin{align}
\mathrm{d}(U + pV - TS) &=
   V\mathrm{d}p - S\,\mathrm{d}T + \sum{\mu_i\,\mathrm{d}N_i}
   - \sum{X_i\,\mathrm{d}a_i}\\
\mathrm{d}G &=
   V\,\mathrm{d}p - S\,\mathrm{d}T + \sum{\mu_i\,\mathrm{d}N_i
   - \sum{X_i\mathrm{d}a_i}}\\
\therefore \mathrm{d}G &=
   V\,\mathrm{d}p - S\,\mathrm{d}T
\end{align}
Now, if the temperature is constant, and only the pressure change then $\mathrm{d}G = V\,\mathrm{d}p$. It follows that $$G(p)-G(p_0)=\int_{p_0}^{p}{V\,\mathrm{d}p}$$
If $V = \frac{nRT}{p}$, we have
$$G(p) = G(p_0) + nRT \ln\left(\frac{p}{p_0}\right)$$
But, I don't know if this is correct.

I'm not quite sure what this question is asking for ? I don't know what the term "Gibbs Free Energy for an ideal gas" means. What the OP has derived here is the Gibbs Free Energy as a function of pressure, at constant temperature & mole number for an ideal gas. Initially my thoughts were, the OP desired the fundamental relation for an ideal gas in the gibbs free energy representation

Hi, thanks for your answer. The Gibbs free energy is defined as: $G(p.T)=U+pV-TS$. So, I need find the expression $G(p,T)$ for a ideal gas. The other part of the problem is not difficult. Thanks

Okay. I think I get what you want, let me type out an answer for you :)

@getafix This is the first time I have studied these subjects, sorry for my mistake

I wrote a quick answer, to your post. Please edit your question to include the clarification you made in the comments. Ask me if something is unclear, I have to go now. I will fill in some steps later perhaps.

@MathUser Why the two sums are equal? Also why you need the second sum? $$dG=d(U+pV-TS)=-SdT + Vdp + \sum_{i} \mu_{i} n_i$$

The fundamental relation for an Ideal Gas, in the entropy representation is
$$S(U,V,N) =
   NS_0 + NR\ln\left[ \left( \frac{U}{U_0} \right)^c
                      \left( \frac{V}{V_0} \right)
                      \left( \frac{N_0}{N} \right)^{c+1} \right]$$
The Gibbs Free Energy, is the Legendre Transform $G \equiv U[T,p]$
so, $G = U + pV -TS$
First, let's get the equation of state $S(T,V,N) $ from the the fundamental relation. We eliminate $U$ using $U = cNRT$
$$S(T,V,N) =
  NS_0 + NR\ln\left[ \left( \frac{T}{T_0} \right)^c
                     \left( \frac{V}{V_0} \right)
                     \left( \frac{N_0}{N} \right) \right].$$
Also, for an Ideal Gas, $p =\frac{NRT}{V}$
$$G(T,p,N) =
  cNRT - TNS_0
  - NRT\ln\left[ \left( \frac{T}{T_0} \right)^c
                 \left( \frac{p_0}{p} \right)
                 \left( \frac{N_0}{N} \right) \right] + NRT $$
Exercise: Find $G_0 = G(T_0,p_0, N_0)$ and substitute into the fundamental relation derived above to get rid of $S_0$

Yes you are partly on the right track.
 Your final equation is usually written in a more familiar form as
$$G=G^{\ce{o} }+RT\ln\left(\frac{p}{p^{\ce{o}}}\right)$$
and letting $p^{\ce{o}}$ equal to $1$ atm then p is understood to be dimensionless and produces the free energy change with pressure
$$G=G^{\ce{o} }+RT\ln(p)$$
However, this does not help, and to get to your result you quote the basic equation you need which is
$dG=Vdp-SdT$
and at constant pressure
$$\left (\frac {\partial G}{dT} \right)_p =-S $$
(The eqn you need is at constant p although it is not indicated in the equation in your question). Next, as
$$G=H-TS$$
 therefore
$$G=H+T\left(\frac{\partial G}{dT}\right)_p$$
dividing through by $T^2$, working out the derivative $d(G/T)/dT$) and rearranging produces the Gibbs-Helmholtz equation.

I think what they had in mind was for you to start with $$H(T,P)=H(T_0,P_0)+C_p(T-T_0)$$ and $$S(T,P)=S(T_0,P_0)+C_p\ln{\frac{T}{T_0}}-R\ln{\frac{P}{P_0}}$$So,$$G(T,P)=H(T_0,P_0)+C_p(T-T_0)-TS(T_0,P_0)-TC_p\ln{\frac{T}{T_0}}+TR\ln{\frac{P}{P_0}}$$

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text:
Electrode Potential and Effect of Salt Bridge Ions on its Value
My textbook says that for a metal dipped in a solution of one of its electrolytes, the potential difference between the two at equilibrium is the electrode potential and it depends on the nature of the metal, its ions and the concentration of the ions.
However, for a combination of two half cells, as the reaction proceeds, they don't consider the effect of the salt bridge ions that move into the respective compartments but instead consider only the ions of the metal of which the electrode is made up of in the Nernst equation and say that the cell ultimately reaches an equilibrium at which cell potential becomes zero. Why is that so? I get that the salt bridge ions don't react with the initial electrolyte but they would definitely affect the tendency of the anode metal to oxidize right?
Also, consider a solution of ZnSO4 with a Zn electrode. Why can't we just use ZnSO4 in the salt bridge?
Also, in order to measure the potential of a HALF cell why can't we just use a potentiometer? Even if no displacement of charges can occur, a potentiometer can be used to find the potential difference. Granted, it would of no practical use as we would require a reference to generalize the values but it wouldn't it be incorrect to say we couldn't measure the electrode potential of a half cell?

I believe you need to fragment the question. Too many sub questions. For starters, the not-measurable-by-potentiometer part can form a separate question.

But they're all sort of related. The reason I put it together is because the I thought the answer to the first question would give me a sort of intuition about the next. But if that's not the case I will change it when someone gives me an answer. Thanks :)


text:
Reactivity of nitrogen oxygen compounds
I am supposed to find the most reactive compound among the following:
$\ce{NO, NO2, N2O, NO3-}$
Is $\ce{NO3-}$ the most reactive because it is an ion or doesn't it being an ion necessarily mean more reactive? Or is it related to the oxidation state of nitrogen?

There is no such thing as the most reactive compound, unless you specify what is it going to react with. Depending on that, the answer can be almost anything.

@Ivan Isn't stability an index of reactivity? Surely that's independent of what the species are going to react with? [Have I overlooked something?]

Maybe, but stability is not a thing either, unless you specify...

OK, here is one example:
Reactivity with atmospheric oxygen at room temperature. NO reacts spontaneously with O2 to give NO2, whereas the other three are stable to atmospheric oxygen.
Here is another:
Reactivity with water: NO and NO2 react with water at ambient temperatures to form acids (nitrous and nitric) whereas N2O and nitrate ion are stable to water.
Here is another:
Reactivity with itself: NO2 is in a reversible equilibrium with its dimer N2O4, and similarly NO dimerizes to give N2O2 - especially favored at low temperatures. However, N2O and nitrate ion do not react with themselves.


text:
Comparing amines (with respect to basicity)
So, an exercise wants me to compare these three amines:

With respect to their basicity. I know that $\ce{NO2}$ and $\ce{Br}$ groups with "draw" the electric density to themselves, increasing the acidity of the amine. (This is all the information i have. I'm not a chemistry student / chemical engineering student, so i don't have a great background in organic chemistry).
So, i got a few questions:
1 - Is $\ce{HCO}$ a group that will "draw" electric density? Or repel? I'm not sure of that.
2 - Who is stronger (in terms of drawing electric density), $\ce{Br}$ or $\ce{NO2}$? I don't know how to compare these things.

Write mesomeric structures...

@Mithoron i'm sorry, but o don't know how to that in this case..

1 - Is $\ce{HCO}$ a group that will "draw" electric density? Or repel? I'm not sure of that.

It is moderately electron withdrawing. Draw resonance structures

Who is stronger (in terms of drawing electric density), $\ce{Br}$ or $\ce{NO2}$ I don't know how to compare these things.

$\ce{NO2}$ is much stronger.
For 4-nitroaniline, $pK_\text{a} = 1.01^{[1]}$
For 4-bromoaniline, $pK_\text{a} = 3.86^{[2]}$
Between the two, 4-bromoaniline is more basic, since as you noted correctly, stronger electron withdrawing substituents increase the acidity of the amine. Since $\ce{-CHO}$ group moderately electron withdrawing (intermediate between bromo and nitro groups), the order of basicity (most basic to least):
4-bromoaniline > 4-aminobenzaldehyde >4-nitroaniline
[1] https://pubchem.ncbi.nlm.nih.gov/compound/7475#section=pH
[2] https://pubchem.ncbi.nlm.nih.gov/compound/7807#section=LogP

Thanks a lot. Comparing groups such as $\ce{NO_2}$ and $Br$ will always be like: " you have to remember that X is stronger than Y"?


text:
How to interpret intervals on a FT-IR spectrogram of soil samples?
I am studying to analyze the IR spectra of soil samples. And I'm interested in the question - how to interpret the intervals in (1750..1820) and (3020 ... 2820) of the two samples on the graph?
The attached file - 1200..600 spectra of soil samples taken on Nikolet 6700 FT-IR Thermo Scientific in DRIFT mode and particular clippings of (1750..1820) and (3020 ... 2820) spectra.

UPDATE
Scan method – FT-IR DRIFT 1 200 cm-1 .. 6 000 cm-1

Red line – paleosol untreated (only dry the samples and sieve it threw a 2 mm mesh)
Blue line – weathering sandy soil untreated (only dry the samples and sieve it threw a 2 mm mesh)

You might be looking at humic and fulvic acid fractions of organic matter in the soil. Can you give us more details about the soil composition, even at the gross level (clayey, silty, sandy, what you are expecting to find if anything, how was the sample prepared/extracted)?

You are almost certainly looking at bands corresponding to functional groups in the humic acid fraction of your sample.
The band at 1750 - 1820 cm$^{-1}$ is likely the $\ce{C=O}$ stretching mode in $\ce{COOH}$ groups in humic acid, but this band is also used to differentiate the fulvic fraction from the humic fraction (see quoted text below). The band at 2820 - 3020 cm$^{-1}$ is likely aliphatic $\ce{C-H}$ stretching.
The composition of humic and fulvic acid fractions in soils varies quite a bit, and how the samples were prepared can significantly alter the composition of the fractions as well. Keep that in mind when interpreting your results. Your bands appear to be shifted from those in the reference I chose to cite, but that is to be expected depending on the composition of your soil and the organic matter contained therein (and can also indicate greater or lesser degrees of $\ce{H}$-bonding, resulting in red/blue shifts). The soil organic fractions referenced in the study below are sandy loams from Ibri, Oman, in an agricultural region. Your samples will differ in composition based on geography, geology, climate, and other factors.
Reference: Helal, Murad, and Helal. Characterization of different humic materials by various analytical techniques. Arabian Journal of Chemistry 4(1), 51–54 (2011) (link to free PDF from ScienceDirect):

The IR spectra of the three humic fractions are shown in Fig. 2. They have a diversity of bands more or less typical to those distinguishing the humic materials (Stevenson, 1994 and Aiken et al., 1985). Major absorption bands are in the regions of 3400–3300 cm$^{-1}$ ($\ce{H}$-bonded $\ce{OH}$ groups), 2940–2900 cm$^{−1}$ (aliphatic $\ce{C–H}$ stretching), 1750–1720 cm$^{−1}$ ($\ce{C=O}$ stretching of $\ce{COOH}$), 1620 cm$^{−1}$ (aromatic $\ce{C=C}$, $\ce{COO−}$, $\ce{H}$-bonded $\ce{C=O}$), 1280–1230 cm$^{1}$−1 ($\ce{C–O}$ stretching and $\ce{OH}$ deformation of $\ce{COOH}$) and 1040 cm$^{−1}$ ($\ce{C–O}$ stretching of polysaccharide or $\ce{Si–O}$ of silicate impurities). The spectra evidently show predominance of $\ce{OH}$, $\ce{COOH}$ and $\ce{COO−}$ groups which are the most characteristic features of soil humic materials. It is clear from the spectra that fulvic acid is characterized by stronger absorption near 1720 cm$^{−1}$ which implies the high carboxylate capacity, and that of humic acid is stronger than that of humin. This is exactly the case in the results of potentiometric titration (Helal, 2007). The spectrum of fulvic acid is also characterized by the absorption at 1400–1390 ($\ce{OH}$ deformation and $\ce{C–O}$ stretching of phenolic $\ce{OH}$). The results of IR spectra indicate that fulvic acid is more aliphatic, and humic acid and humin are more aromatic. It is obvious that the IR results are in good agreement with the other characterization findings.

Figure 2, from the above-referenced article. FA is fulvic acid and HA is humic acid.

Thank you @Todd Minehard. This time I'm collecting a sample's history and will update this question in Monday, when my colleague from the lab will answer me. After updating the the question I beg to notify you, and I would be grateful for further clarification as this question requires more information on the soil properties and sample extraction/preparation method. I don't understand why in the region (2820 .. 3020), one sample has a lift with two small peaks, and the second has a slight rise with a negative skew.

Today I've updated the question - Red line – paleosoil untreated (only dry the samples and sieve it threw a 2 mm mesh); Blue line – weathering sandy soil untreated (only dry the samples and sieve it threw a 2 mm mesh)

@IlyaRusin - The two peaks in your paleosol are almost exact matches for what are identified as aliphatic $\ce{C-H}$ stretches. The weathered sample spectrum looks to me like it's a little shifted AND indicates that the aliphatic species that would otherwise contribute to those peaks are absent. Have a look at this publication, particularly Figure 1 and the first part of Results and Discussion.


text:
What are disadvantages of using sublimation vs. recrystallization as a method of purification?
I can only think that a small sample can be sublimed while a bigger sample size can be recrystallized. Are there more?

Thermal stability of the material

Sublimation requires specialist equipment

Most chemicals don't sublime under reasonable conditions...

If the resublimation goes wrong, you can find yourself dismantling the vacuum pump and its exhaust pipe in order to scavenge the rest of your material.

The most notable difference is that sublimation requires a volatile solid, whereas most solid can be recrystallized from a solvent system that is not too exotic.
Recrystallizations tend to recover more pure product.
In regards to green chemistry, sublimations are preferred as they generate no solvent waste.

Percent recovery if sublimate is not 100% complete since the fumes may be blown away especially if not executing proper sublimation process.

I think a possible limitation might be that the compound to be purified should be volatile. And also, you don't really know when the sublimation is completely over. That's why we usually wait approximately 5 min after detecting the first crystals and turn over the heater... I mean, it's not a very accurate procedure.


text:
Lipinski's Rule of 5 Oxygen Rules
I would like clarification on the role of oxygen as a hydrogen bond acceptor in Lipinski's Rule of 5 (for drug absorption). I have learned in my medicinal chemistry class that both electron pairs from an oxygen atoms are able to individually form hydrogen bonds.
I was wondering if every sp3 hybridized oxygen would account for 2 possible hydrogen bond acceptors in this rule for drug absorption? If not, what other mechanisms or in vivo findings are not causing oxygen to not count as double?

In Advanced Drug Delivery Reviews, 2001, 46, 3-26, (DOI) Lipinski and coworkers write:

Too many hydrogen bond acceptor groups also hinder permeability across a membrane bi-layer. The sum of $\ce{N}$s and $\ce{O}$s is a rough measure of H bond accepting ability.

So, for the empirical rule of five it's simply the number, nothing more.

In the cited article, the situation for amides is addressed on page 19 in the context of log P calculations and correction factors that increase lipophilicity. Here, the authors write:

PRX, a proximity correction factor for nitrogen
  and oxygen atoms that are close to one another
  topologically. [...] In addition, for each
  carboxamide group, we add an extra 1.0 and for
  each sulfonamide group, we add 2.0

This perspective is problematic when focusing amides; the lone pair from the nitrogen interacts more closely with the carbonyl group and is (mostly) unavailable for hydrogen bonding. The some of N's and O's would be inaccurate in the application of this rule.


text:
Knowing the charge of polyatomic anions
I have an exercise and I have to find the formula of those ions:

Arsenate ion
Perbromate ion
Selenite ion
Hydrogen Tellurate ion

The solution tells: $\ce{BrO4-,\space SeO3^{2-}, AsO4^{3-}, HTeO4^{-}}$.
My question is, how do I get to those results and how do I know their charge to be 1-, 2-, 3-, and 1-?

Can you amend your post to include the work you've done to solve the problem? As it stands, it looks a lot like homework and might be closed by the community here as such.

No. It is not a homework. the book I am using is Chemistry, The central science 13th edition by Theodore L. Brown. It is the exercice number 2.107 on page 79. When looking at the solution at the back of the book, I got the answers I wrote above. I do not understand though, why the respective charges of the ions are like so

I could not understand it though. I am a first year student


text:
Symmetry of trans-1,3,dichlorocyclobutane

Is dis molecule having a plane?or inversion centre or what?and what is it's point group?

Remember dat dis molecule is puckered.

Related: Does trans 1,3-dichlorocyclobutane have zero dipole moment? and Are these two structures of 1,3-dichlorocyclobutane enantiomers?


text:
Molarity of solids and liquids in equilibrium constant expression
Why  is the molarity of pure liquids , solids and excess solvent taken 1 in the equilibrium constant expression?

It's not supposed to be a molarity, it's supposed to be an activity. And, activities of pure liquids and solids are by definition 1.


text:
Is the ideal gas constant not a constant if Celsius is used instead of Kelvin?
For example, Charles' law does not work with Celsius. Why is this? I can only think that the gas constant is not a constant when Celsius is used rather than Kelvin.

I'd say, when Celsius is used instead of Kelvin, there is no such law and no such constant. Physical laws are awfully strict in that they apply to certain things and not to others. You just can't expect them to work after you've changed some physical quantity to another at random, even if the names of the said quantities happen to be kinda similar ("temperature").

Charles law works perfectly well with celsius: the algebra is slightly more complicated but the constant is exactly the same.

I'm voting to close this question as off-topic because this is a physics question with hardly any relevance to chemistry.

I disagree that it is a physics question per se- it is confusion about temperature scales and relative vs absolute.

Charles' Law certainly DOES work with temperature in degrees Centigrade!
If we express it (relative to $T=0$ absolute) as $V/T = k$ and if we convert $T$ (K) to $t$ (°C) as $t = T-273$ then obviously Charles' Law would have the form $V/(t+273) = k$.
More generally if, instead of $T$ we use some linear function of $T$, say $Y= mT + b$, then $$V/T = Vm/(Y-b) = k$$ or $$V/(Y-b) = k/m,$$ where $k$, $b$, and $m$ are constants. We can substitute $k'$ for $k/m$ and get $V/(Y-b) = k'$, but we can't get rid of the $-b$ that way.
The claim that a law (even if it is an approximate one like Charles') "doesn't work" with a different unit of measure confuses a formula with a law. Formulas are great for solving high school level problems, but a deeper understanding is expected in college and beyond. Any strictly increasing function can be used as a temperature scale, for instance $X  = π^T$ could be a temperature scale.
The formula we'd use to express Charles' Law would look very different, but the relationship between volume and $\log(X)/\log(π)$ would be just as constant (and since $\log(π)$ is a constant, we'd have $V/\log(X) = k''$.)

If you consider temperature to be the measure of mean kinetic energy, then to get the correct linear scale, you'd need an absolute temperature. The Kelvin scale (or the Rankine scale) provides that absolute temperature where in the limit of zero temperature, the system is in its lowest energy state.

Why do you need absolute temperature?

Because the third law of thermodynamics should work. You want properties to scale directly with temperature in a linear relationship (which they do with an absolute scale) instead of an affine relationship (linear if you shift the scale). You might as well just shift the scale and make the system linear. Then you just can multiply without that pesky addition.

Charles’ law, for those in need of a refreshment (like me) states that an increase in temperature results in an increase in volume and vice-versa; thus, that temperature and volume are directly proportional.
Typically, this would be an experiment in physics class, measuring the temperature of a gas cylinder and the volume it requires under constant pressure. A number of different temperatures and volumes would be recorded and then filled into a graph. It turns out that that graph gives a linear relationship between the two. Assuming temperature to be on the $x$ axis and volume on the $y$ axis, the equation of the linear relationship would be:
$$V = k\vartheta + V_0\tag{1}$$
Wherein $k$ is a constant (which can be calculated depending on $p, n$ and the gas constant $R$) and $V_0$ is the $y$-axis-bisection of the graph. Since $V_0$ is also a constant, you still observe a linear relationship between $V$ and $\vartheta$. The difficulty is that you need to determine two constants and that you need both for any algebra that you want to perform using this law.
If you extrapolate, you realise that the $x$-axis will be bisected at a temperature of $\vartheta \approx -273~\mathrm{^\circ C}$. You could go ahead and decide to modify your equation so that $V_0 = 0$; that requires modifying what you insert as $\vartheta$ in such a way that you effectively move the graph sideways in the coordinate system. There is a number of general rules for doing so; what you want is to move the graph to the right by $273~\mathrm{^\circ C}$ so the resulting equation becomes:
$$V = k \cdot (\vartheta + 273~\mathrm{^\circ C})\tag{2}$$
And then all some clever people did was to replace $\vartheta + 273~\mathrm{^\circ C}$ with the ‘new’ temperature $T$ using the new unit $\mathrm{K}$; the relationship is such that $0~\mathrm{K}\ \hat{\approx}\  {-273}~\mathrm{^\circ C}$. This leads us to our final equation:
$$V = k \cdot T\tag{3}$$
Equations $(1), (2)$ and $(3)$ are all mathematically equivalent to each other. Thus, they describe the same physical consequence. Thus, unless you learnt equation $(3)$ and equation $(3)$ only as Charles’ law, said law is valid no matter which choice of temperature scale you make. Even Farenheit temperatures would obey the law.

The other answers sort of address your confusion, but not very directly. So let me give it a shot.
For gas laws, the gas constant is always a constant, regardless of which temperature scale you need. It's also a constant regardless of whether you are using it in an ideal gas law or some other law. It's even a constant when it shows up in places that aren't gas laws at all! That's because it's a fundamental constant which relates the statistical properties of molecules to macroscopic phenomena like pressure and temperature.
The catch is, in order to use it, the variables have to be on an absolute scale. The reason for this is a little hard to explain, but I guess an easy way to think about it is that it defines how things change relative to each other. For example, if you double the temperature while holding pressure constant, the volume will also double (Charles's Law). Volume is already on an absolute scale - zero volume means it takes up no space. Pressure is the same way - zero pressure means no force exerted on the walls of the container (on average). Temperature is really the only one that has an arbitrary zero point on a lot of scales, where "zero" doesn't necessarily mean "no kinetic energy."
If you double the temperature as measured in an absolute scale, you are doubling the average kinetic energy of the molecules. If you double the temperature as measured on the (for example) Celsius scale, the kinetic energy doesn't necessarily increase by the same amount. Take a negative temperature. What happens when you double -100? Or look at room temperature. If you go from 20 to 40 degrees Celsius, the absolute temperature only increased by about 7% (20 / 293).
Each combination of pressure, volume and temperature units has its own version of the gas constant, but these are all really just unit conversions of the "fundamental" constant, which is the Boltzmann constant multiplied by Avogadro's number. The key thing to remember is that besides making sure the units match the measurements you have, the temperature scale must be an absolute scale.


text:
Why is the dissolution of AgCl considered endothermic?
The solubility constant of $\ce{AgCl}$ at $25\mathrm{\,^{\circ}C}$ degrees is $1.6\cdot 10^{-10}$.
At $100\mathrm{\,^{\circ}C}$, the $K_{\mathrm{sp}}$ increases to $2.15\cdot 10^{-8}$.
Why is the dissolution of $\ce{AgCl}$ at $100\mathrm{\,^{\circ}C}$ considered endothermic? I would have thought it was exothermic.

The typical way of explaining this is using Le Chatelier's Principle. That is, we could consider the reaction,$$\ce{AgCl(s) + heat<=>Ag+(aq) + Cl^-(aq)}$$This then demonstrates that energy in the form of heat is required to have the reaction proceed. Thus, at a higher temerature, the reaction will shift to the right as predicted by Le Chatelier's principle.
There are obviously more thorough explanations possible, but this is a good place to start.

brief yet understandable. Much appreciated.

Look at the answers to the question I asked about enthalpy change, temperature, and equilibrium.
Basically, the equilibrium constant depends on temperature in a way that is dependent on the sign of the enthalpy change of the reaction. If $\Delta H > 0$, then as temperature increases, the equilibrium constant increases, and vice versa.
You observe the relationship where increasing temperature correlates with increasing equilibrium constant, suggesting the reaction is endothermic.


text:
Entropy change upon mixing cold and hot water

A closed, well-insulated container is filled with $\pu{454 g}$ of water at $\pu{94.4 ^\circ C}$. To the hot water, $\pu{200 g}$ of water ice at exactly $\pu{0 ^\circ C}$ is added. The mixture reaches an equilibrium temperature of $\pu{41.1 ^\circ C}$. Assume the molar heat capacity is constant and all the processes are at constant pressure. The standard enthalpy of fusion for water at $\pu{0^\circ C}$ is $\pu{6.008 kJmol–1}$. The constant-pressure heat capacity for water is $\pu{75.291 JK–1mol–1}$. Water has a molecular weight of $\pu{18.015 gmol^-1}$.
Calculate the entropy change (in $\pu{JK–1}$ ) for the system that happened because of this mixing.

I know the entropy change equals to $q/T$ because $q$ equals to the enthalpy exchange in the system as it is constant pressure, so what I did was:
$$q=(454/18.015)\times 75.291\times (41.1-94.4)+(200/18.015)\times 75.291\times(41.1)+(200/18.015)\times6008 = \pu{172.2 J}$$
Change in entropy = $172.2/(41.4+273) = \pu{0.55 JK-1}$
That is apparently incorrect, what have I done wrong?

The change in entropy doesn't "equals to" q/T.  For a specified closed system, the change in entropy between an initial state and a final state is given by $$\Delta S=\int{\frac{dq_{rev}}{T}}$$ where the subscript "rev" refers to a reversible path between the initial and final states.  The process you described is not reversible.  You need to devise and employ a reversible path to get the change in entropy.
Here is a hint:  For the system comprised of the 454 gm of water initially at 94.4 C, the change in entropy is $$75.291\frac{454}{18.015}\ln{[(273.2+41.1)/(273.2+94.4)]}$$
If you would like a cookbook recipe for how to determine the change in entropy for a system experiencing an irreversible process, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/


text:
Deriving the entropy of mixing of a non-ideal solution
In the earlier posted question named Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics the entropy of mixing was found by assuming an ideal solution. The user Porphyrin mentioned that in case the molecules in the solution do interact with each other, the free energy would have an additional term (n1+n2)x1x2w, with

w≈(2E12−E11−E22) and E12E12 is the interaction energy between molecules of type 11 and 22 when they are adjacent and similarly for
  E11E11 and E22E22 all of which are negative. From a consideration of
  the energies involved ww is expected to be positive.

My question is: how do we go about finding an answer for the entropy of mixing when we do not assume that the solution is ideal?

The calculation of the mixing energy is extremely complicated and only approximate solutions have been obtained as far as I am aware. I shall attempt to outline one of these methods*.
All molecules are assumed to be spherical and of equal size and situated on a lattice with co-ordination number z . Each pair of neighbours AA of type A molecules contributes energy $w_{AA}$, similarly $w_{BB}$ for B molecules and energy $w_{AB}$ for AB pairs. The total energy due to these  interactions
is
$$E_T=N_{AA}w_{AA} + N_{BB}w_{BB}+N_{AB}w_{AB}$$
where $N_{AA}$ is the number of AA pairs and so on. Counting the number of  $A$ interactions and number of B ’s gives
$$zN_A=2N_{AA} + N_{AB}  ~~~\text{and }~~~ zN_B=2N_{BB} + N_{AB} $$
and thus  combining these equations gives
$$E_T=\frac{zN_Aw_{AA}}{2}+\frac{zN_Bw_{BB}}{2}+wN_{AB}=E^T_{AA}+E^T_{BB}+wN_{AB}$$
where $w=w_{AB}-w_{AA}/2-w_{BB}/2$ is the energy gain on mixing per pair of AB neighbours.
In statistical mechanics the partition function is the key to all other thermodynamic functions. In this case partition function is $$
Z= \sum_{N_{AB}}Z_AZ_Bg(N_A,N_B,N_{AB})e^{(-E^T_{AA}-E^T_{BB}-wN_{AB}/KT)}$$
and where $g(..) $ is the number of arrangements of $N_A$  A’s , $N_B$ B ’s and $N_{AB}$  pairs of molecules. Separating out the terms due to pure A and B molecules gives
$$Z=Z’_AZ’_B \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/KT}$$
and as the Helmholtz energy is
$F=-kT\ln(Z)=F_A+F_B+\Delta F$
and for mixing only the $\Delta F$ term is important the others being the energy of each pure component.
Thus
$$\Delta F = -kT\ln\left(\sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT}  \right) $$
The change in energy can be obtained from the Gibbs-Helmholtz equation $(E = F-T(dF/dT))_V$ and is
$$\Delta E= w\frac{\sum_{N_{AB}} N_{AB} g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT} } { \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT} } $$
and this is an equation that seems to have no exact solution and so this is where approximations have to be made.
The difficulty is in the exponential terms, as they contain $N_{AB}$ the aim is therefore to replace these with an average value, and thus the exponential terms become constant wrt $N_{AB}$. It can be shown that
$$\sum_{N_{AB} }g(N_A,N_B,N_{AB}) =\frac{(N_A + N_B)!}{N_A!N_B!}$$
and
$$\bar N_{AB}=w\frac{\sum_{N_{AB}}N_{AB}g(N_A,N_B,N_{AB})}{\sum_{N_{AB}}g(N_A,N_B,N_{AB})} = z\frac{N_AN_B}{N_A + N_B} $$
By replacing $N_{AB}$ by $\bar N_{AB}$ in the partition function produces
$$Z=Z’_AZ’_B \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-w\bar N_{AB}/KT)}
 = \frac{(N_A+N_B)!}{N_A!N_B!}\exp\left(-\frac{zw}{kT}\frac{N_AN_B}{N_A+N_B}\right)$$
the change in energy is then
$$\Delta E = zw\frac{N_AN_B}{N_A+N_B}= w\bar N_{AB}$$
and this can be re-written as molec fractions to produce
$$ \Delta E= z(N_A+N_B)x_1x_2w$$

based on text in Rushbrooke ‘Introduction to Statistical Mechanics’


text:
NMR calculations with NWChem quantum chemistry program
I tried experimenting with some ab initio NMR calculations using the NWChem program. I used quinuclidine as a simple test case. The input file was:
echo
start quinuclidine_b3ylp_opt_nmr_shift

geometry
  c         -6.06576        0.97412       -0.15301
  c         -5.70979        1.98552       -1.28158
  c         -4.36490        2.65888       -0.91745
  h         -6.47615        2.78876       -1.35616
  h         -5.68491        1.45114       -2.25601
  c         -3.37563        1.54341       -0.50314
  c         -3.86410        0.94013        0.84625
  h         -2.33065        1.90335       -0.38494
  h         -3.37977        0.77674       -1.30967
  n         -5.29417        1.27496        1.08291
  h         -3.26800        1.37762        1.67961
  h         -3.67643       -0.15484        0.87417
  h         -5.79345       -0.05430       -0.48418
  h         -7.16227        0.96043        0.02635
  c         -5.40220        2.73070        1.37047
  c         -4.60680        3.55788        0.31857
  h         -5.06760        2.98401        2.39936
  h         -6.47931        3.01065        1.31755
  h         -3.96835        3.24945       -1.77113
  h         -3.61077        3.85719        0.71410
  h         -5.16983        4.48719        0.08363
end

basis
 * library 6-311G
end

dft
 xc b3lyp
end

title "Quinuclidine B3LYP/6-311G geom opt nmr shift"
task dft optimize

basis
 * library 6-311G
end
dft
 xc b3lyp
end

property
 gshift
 shielding
end

task dft property

Next, I calculated the chemical shift of the TMS standard:
echo
start tms_b3lyp_opt_freq_nmr_shift

geometry units angstroms
  si  -5.37335102864362      2.43869139121700     -0.00017138545369
  c   -3.50689688966580      2.43868715601638     -0.00001133876668
  c   -5.99529466705065      4.19430001974021      0.12209335653192
  h   -5.64408778702798      4.67985919142380      1.04819971762123
  h   -7.09781476514675      4.22936546250064      0.12467467866764
  h   -5.64435465416627      4.80382888763710     -0.72767673129762
  h   -3.10677010278341      2.89237170000804      0.92240456642084
  h   -3.10633327621776      3.01074016686707     -0.85387463583008
  h   -3.10622471891726      1.41321522592654     -0.06819713587793
  c   -5.99568159818493      1.45555000740651      1.45925444723856
  c   -5.99571944212990      1.66683194553091     -1.58150765777166
  h   -5.64537419258706      1.88774171172852      2.41177722402156
  h   -5.64408400205021      0.41082962309567      1.41778721376507
  h   -7.09821437068132      1.43547142531899      1.48770835119311
  h   -5.64584598058862      2.22683552653798     -2.46512226305553
  h   -7.09824537965055      1.65038972286304     -1.61218398701376
  h   -5.64368714450789      0.62658083618162     -1.68513442039297
end

basis
 * library 6-311G
end

dft
 xc b3lyp
end

title "TMS VWN5/aug-cc-pVDZ geom opt nmr shift"
task dft optimize

basis
 * library 6-311G
end
dft
 xc b3lyp
end

property
 gshift
 shielding
end

task dft property

However, the resulting chemical shifts are terribly off from the experimental ones. Am I doing an obvious mistake or is the big deviation just the best you can get out of such calculations? What is your experience and approach to ab initio NMR calculations (i.e. chemical shifts and coupling constants)?

I also tried out calculation chemical shifts/coupling constants with ORCA. However, in all the simple test cases I tried out, I always got huge deviations from experiment. This is why I am unsure now if I make an obvious mistake, e.g. not using a solvent model or if those calculations just are that unprecise. I somehow feel that just not handling solvent or using an appropriately sized basis set should completely screw up the results of such calculations.

I know of some cases (such as Pt NMR) where the absolute isotropic shielding constant is EXTREMELY sensitive to the geometry (a deviation of 0.01 Angstroms can lead to a change in this value by a couple hundred ppm or more).  However, you can sometimes get good error cancellation with your reference compound.  Also, try considering solvent in your calculations (COSMO, PCM, etc.) though your mileage may vary.

It is hard to discern what exactly went wrong without a peak at the output files. Did you scale the values output? See this site for more information: http://cheshirenmr.info/index.htm

Can you  upload (or http://pastebin.com/ ) the relevant section of the outputs. I am curious about how they look like. I've never used NWChem nor Orca for chemical shift calculations. Thanks in advance.
Also, notice that it can be important in order to answer your question.

@user1420303 I will do that ASAP.

Before I make an attempt at a full answer, can you post more information about the experimental values, or at least a reference? Also, don't use ORCA for NMR, that part of the program doesn't give the correct answers.

@pentavalentcarbon Don't use ORCA for NMR, that part of the program doesn't give the correct answers. What do you mean by that? Somewhere in ORCAs only documentation I also found the note that ORCA shouldn't be used for NMR calculations as "IGLO is outdated" and "should only be used by expert users". However the ORCA manual doesn't say anything like this. The experimental values can be found for example at the SDBS spectra database.

@deusexmachina As a former ORCA developer, I am interested in (and dispute) the claim about the EPR/NMR module of ORCA. Can you give evidence? It should also be noted that with the recent 4.0 release, a "GIAO implementation for NMR chemical shifts" was made available.

@TAR86 I found this statement in the online ORCA documentation -- so maybe it is outdated. Thanks for pointing out that ORCA 4.0 now includes a GIAO implementation I guess I will try out at some point in time. Really, I tried NMR chemical shift calculations a lot and always got very bad results out of them, as soon as I have some time on my hands I will make a more sophisticated study on how ORCA/NWChem/Gaussian NMR calculations perform on some simple test systems, where reference data are available. But thanks for all the input : )

Those calculations are tricky because some functionals work well, others don't, some basis sets work well while even a better basis might give bad results and so on.
We did some studies in the past using the GIAO Method in G09 with B3LYP and 6-311+G(2d,p) als basis for the NMR calculations and M06-2X/6-311++G(d,p)for geometry optimization. And that worked quite well. However, we always used solvated systems (PCM) which seems to be the common way to do it. Try to reoptimize in solvent and run the NMR calculation in solvent if that's possible in NWChem.

I will perform the calculations with a solvent model and see if it gets better. Thanks : )

While I may lack specific experience in the calculation of chemical shifts, it is immediately apparent that "general purpose" and frankly outdated basis sets of the Pople variety such as 6-311G... are not a good choice. Things may improve if the basis set is (partially) decontracted (to give sufficient variational freedom at the nuclei) - but then one gives up any advantage of the small basis set and may use a better one anyway.

I use Gaussian 09 to obtain NMR $\ce{^{13}C}$ chemical shifts (cs) with great accuracy. The best results are obtain if the calculated cs are scaled. So, I recommend to use this level of theory:

Opt => mPW1PW91/6-31G(d)  with tight criteria (opt=tight)
NNR => mPW1PW91/6-31G(d)  with nmr int=ultafine
To scale: $\text{scaled cs} = \text{(calculated cs)}*1.05 - 1.22$ (See reference)

here I have one example: https://github.com/miroi/open-collection/tree/master/theoretical_chemistry/projects/nmr_diplomovky/derivaty_markerov/derivative_2_Peter/nwchem
where geometry optimization was performed with "classic" basis  6-311++g2d_2p, but NMR parameters were computed with the decontracted NMR-DKH_TZ2P basis set. We used several DFT functionals. Plus cosmo solvent effects. Try this combination.


text:
How would one make a non-toxic pen out of osmium?
Osmium is the densest element in the periodic table. However, it is toxic in its pure form. Is there a way to create a standard pen out of osmium but coat it in a dense material that isn't toxic to have the heaviest pen?
How would this pen be made?
It would be a cap pen (like the Uniball pen pictured below) with no wasted space for plastic mechanical parts.

Osmium alloys with many metals, so it could be covered by an alloy. Osmiridium, an alloy with iridium, is even harder than pure osmium.
It would seem possible to dip the osmium pen blank in an iridium or osmiridium bath (at rather high temperature) to produce a long-lasting casing, rather like dipped chocolate cordials.
If you accept this answer, I'd be glad to receive an award of one of these pens. ;-)

I'm not making these pens, I was just curious. If I was I would give you one ;). Also, osmiridium is nontoxic?

It's considered safe, and is already used for fountain-pen tips: http://pro.tanaka.co.jp/en/library/element/column08/

It turns out that a working pen made out of osmiridium (let's say it uses 45g as 4cm3 with some hollowness) would be about 0.8oz of osmiridium, roughly worth $360. Probably not going to happen anytime soon.

(Note that I have not measured an approximation of the pen I would like it to be modeled after, so my volume measurement is most likely incorrect) [SO app wouldn't let me edit the comment again]


text:
Why isn't CO2 flammable?
Given that $\ce{CO2}$ is comprised of carbon and oxygen, why
isn't it a flammable gas?

What has burned can never burn.

fully oxidized.

If I understand what you are saying, then the issue is that the carbon and oxygen have already reacted. Carbon dioxide is thus not flammable because it is the fully oxidized product of the combustion between oxygen and carbon.
"Made of" has multiple meanings. You're confusing the meaning of physical mixture (flammable) with the chemical composition of a compound (not flammable).


text:
Le Châtelier's Principle and heat
Consider the following reaction at equilibrium.
$$\ce{A->B}, \Delta H < 0 $$
Suppose I increase the temperature. Now, quite a few people would invoke Le Châtelier's Principle and say that since "heat" is a product of this reaction, the equilibrium should shift backwards. This is clearly wrong because "heat" is not a species you can have in a reaction. You can't incorporate it into the reaction quotient.
If you take
$$\Delta G = \Delta H - T\Delta S$$
for $\Delta S < 0$, you could make a claim of the equilibrium shifting backwards at higher $T$. This doesn't have anything to do with the sign of $\Delta H$ however.
Is there an actual theoretical basis for the following claim that does not invoke a principle that does not apply:

For an exothermic reaction at equilibrium, increasing the temperature
  will cause the equilibrium to shift towards reactants.

Then why $K$ the constant of reaction depends on the temperature if "heat" is not a product of the reaction ?

As said already heat does not go into the reaction quotient. The LCP argument (which relates changes in conditions to changes in $Q$) is therefore shaky.

I gave this analogy to a duplicate question that asked this: There are two people. One guy likes chocolates and one guy hates them. They decide to transfer the chocolates. If you start throwing chocolates randomly to both of them, in which side will the reaction go? The guy who hates chocolates would keep giving it to the guy who likes them.
In a similar way, if you provide heat to a reaction, it would move in the direction of the where the heat is absorbed, i.e the endothermic pathway.

Before to show you that what you said is false (I'm sorry ^^) be sure you understand the constant of a reaction depends on the temperature.

Let the same reaction you want $$\mathrm{A \rightleftharpoons B}$$
With a constant $\mathrm{K^{\circ}}$. Imagine you heat your system a little with $\mathrm{d}T>0$, then by Van't Hoff's law we get:
$$\frac{\mathrm{d}\ln(K^{\circ})}{\mathrm{d}T}=\frac{\Delta_rH^{\circ}}{RT^2}$$
Where $R$ is the perfect gas constant which is positive. Then $RT^2>0$.
 So $\mathrm{d}\ln(K^{\circ})$ as the same sign as $\Delta_rH^{\circ}\cdot \mathrm{d}T$.
So if you have an endothermic reaction $\Delta_rH^{\circ}>0$ because $\mathrm{d}T>0$ then, $$\mathrm{d}\ln(K^{\circ})>0$$
Then $K^{\circ}$ will increase with the temperature. If you take $\mathrm{d}T<0$ for an endothermic reaction then the constant of the reaction will decrease with the temperature.
I let you do the same final reasoning for an exothermic reaction. So the "Le Châtelier principle" is still true.
Note: If you know the chemical affinity, you can do the same proof, just a bit longer!

What does K° and ∆H° represent?  Is it the equilibrium constant and enthalpy of a standard temperature. If so then why are you calculating for standard conditions

Firstly, we should add standard symbols to the equation, because only $\Delta G^\circ$ is related to the equilibrium constant $K$, whereas $\Delta G$ is related to the instantaneous reaction quotient $Q$. Therefore any change in $K$ has to be related to $\Delta G^\circ$ and not $\Delta G$. So you have
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
and the argument goes, when $T$ increases, $\Delta G^\circ$ increases (given $\Delta S^\circ < 0$), and therefore the equilibrium goes to the left. The issue is that the equilibrium position is not measured by $\Delta G^\circ$ - it is measured by $K$. And the dependence of $K$ on $\Delta G^\circ$ itself has a temperature component:
$$K = \exp{\left(-\frac{\Delta G^\circ}{RT}\right)}$$
so if you increase $T$, not only does the magnitude of the numerator increase, so does the denominator. That's precisely where this argument fails: you can't necessarily relate a larger $\Delta G^\circ$ to a smaller $K$, if the temperature is changing.
(Note that if you have two different reactions, with given values of $\Delta G^\circ$ at the same temperature $T$, then it is fair to say that the reaction with a larger $\Delta G^\circ$ has a smaller $K$. However this has to be thrown out of the window once your $T$ starts changing.)
So, to relate the change in $\Delta G^\circ$ to the change in $K$ you have to effectively "bring the factor of $T$ over to the other side of the equation".  You just need a bit of algebraic manipulation:
$$\begin{align}
\ln K &= -\frac{\Delta G^\circ}{RT} \\
&= -\frac{\Delta H^\circ - T\Delta S^\circ}{RT} \\
&= -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}
\end{align}$$
From here you can clearly see that if $T$ increases, then the sign of $\Delta H^\circ$ is what controls whether $\ln K$ (and hence $K$) increases or not; $\Delta S^\circ$ has no role to play. This is consistent with the conclusion derived via Le Chatelier's principle.
See also another of my answers on the van 't Hoff equation here.


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What drives intercalation?
From Wikipedia:

intercalation is the reversible inclusion or insertion of a molecule
  (or ion) into compounds with layered structures.

What is the "driving force" / mechanism that's responsible for intercalation? I assume it's not just diffusion.
In particular, why intercalation continues beyond the periphery? i.e. imagine the process has started and the periphery is now occupied by the intercalant: what drives subsequent intercalation to occur if all the periphery is already occupied by low energy host-guest arrangements?

There is no such 'driving force'. Diffusion will cause the two species to collide but while adjacent there has to be an enthalpic (intermolecular interaction energy) or entropic reason for their mutual interaction to cause intercalation so prevent the species  from diffusing apart. Whether energy or entropy or some of both causes the free energy to become favourable vs. separated species may depend on the individual species involved.

@porphyrin if that is the case, wouldn't you expect intercalation to occur only at the periphery?

I would expect that possibly millions of collisions between the host & guest  occur before one that leads to intercalation, clearly the approach direction has to be approx correct but once there is some interaction (even at some odd geometry) as long as it is greater than kT on average  ($\approx 210 \pu{cm^{-1}}$) then intercalating will begin to win out.

@porphyrin sure, but now imagine the periphery is occupied by the intercalant. What drives subsequent intercalation to occur if all the periphery is already occupied by low energy host-guest arrangements?

Things are not static, if this is the case at some point one peripheral guest will leave (due to thermal fluctuations) and so may become fully intercalated or move away altogether. Its just a random thing, but given enough tries,  and there will be millions of them, intercalation will happen if it is the lowest energy state and this is several times kT lower than peripheral energy binding. The intercalated guest may also leave the host, its largely about the energy vs thermal energy although entropy may play a part depending on solvent.


text:
Would combining essential oils and carrier oils be okay in terms of flash point?
With NO scientific, or chemical background, I am making my own beard oils. I found essential oil recipes online to mix with one or many carrier oils to make beard oil. I want to be original by going out on a limb and exploring other scents besides sandalwood, cedar, pine forest, bergamot, patchouli etc etc etc. I get so burned out seeing and smelling the same scents over and over.
With all of that out of the way, I want to combine some essential oils and carrier oils, and then mix in one of the three following oils to make an awesome scent like no other before this. The three following oils are from Crafter's Choice (if that matters.)
The three oils to choose from are

Butterscotch fragrance
Pink Cotton Candy fragrance
Double Bubble Gum (this one is actually a flavor)

I see on the bottles that these all have a flash-point. I'm guessing that has something to do with combustion at a known flash-point temp.
My main worry is, can I combine these flavor and fragrance oils with carrier oils, and or essential oils without worrying about some crazy Frankenstein mad scientist Yosemite Sam mishap on my face at some point in the day.
Would combining these oils be okay?

Personal medical questions are off-topic. We won't speculate on how to formulate health care products that should be formulated with appropriate research and testing, and not formulated by guesswork. This policy is exactly to avoid "some crazy Frankenstein mad scientist Yosemite Sam mishap."


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Relationship between surface area and vapor pressure
How is vapor pressure independent of the surface area and volume of a liquid, although rate of evaporation depends on surface area?
Please someone help me I am totally confused.

The rate of evaporation is proportional to surface area. The rate of condensation is proportional to surface area. The vapor pressure is the equilibrium pressure where the rate of evaporation is equal to the rate of condensation. Since the scaling factor is the same, the vapor pressure is independent of the surface area.

Sorry but why is rate of condensation proportional to surface area?

@samjoe Because it's the microscopic reverse of evaporation.


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Mixing two buffers of the same concentration, but of different pH
Does anyone know how to approach this?
I have two buffers of the same weak acid of the same molarity but at different pH
Let's say the pH of one was 8 and one was 7, how do I go about getting a pH of 7.6?
Thanks in advance

You're going to need that pKa.

What are the values of all pKa of each ones ?


text:
Melting post-1982 minted U.S. pennies and aluminum: Why does the slag turn blue?
After melting aluminum inside of an iron crucible, I tossed in a handful of U.S. pennies that were minted post-1982. The slag began to burn a brilliant blue. What caused this?

Quite possibly copper, but may depend on what country your pennies are from......

In USA. Was thinking it had something to do with the copper.

After 1982, U.S. pennies have been minted as copper-plated zinc - see this Wikipedia entry for information. Apparently 2.5% copper.

Zinc burns in air forming zinc oxide with a bright blue flame. The bit of copper in US cents will add a greenish color to the flame.
N.B. Fumes of burning zinc and some other metals are toxic. Avoid contact or inhaling the smoke.