text:
Atomizers in AAS
Electeothermal atomizer ( graphite furnance ) and flame atomizer, are the instruments which help us to convert the analyte into the gas state and they both achieve that. My question is, what's the difference in their way of workig ? I know the flame has its nebulisator which converts the liquid analyte into mist aerosols and then the gas helps the analyte to arrive to the flame. How does the electrothermal atomizer work ? Whats the nebulisator substitution in the graphite furnance ?
Why do we have to turn the analytie in mist aerosol in flame AAS ?

AAS, atom absorption spectroscopy, is a step-wise process to remove a surrounding matrix, and subsequent analysis.
In both variants, the sample is diluted by applying a solvent.  But the solution obtained is destroyed by evaporation, and the mist is desolvatized leading to an aerosol.  The aerosol is then vaporized, yielding individual gaseous molecules (that may be analyzed spectroscopically).  The gaseous molecules may dissociate reversibly into individual atoms (that may be analyzed spectroscopically).  The atoms may ionize (reversibly) into atomic ions (that may be analyzed spectroscopically).
The nebulizer is a typical device for a continuous supply of analyte into the optical path of your instrument, for example between the hollow-cathode lamp and your detector.  You may prepare easily 50 mL of a dilute solution of your sample that is sprayed at constant flow over minutes into the flame observed.
In the instance of the graphite furnace, the supply is discrete, batch-wise.  There is a fixed volume of soluted analyte that is evaporated in a small vessel, and incinerated.  Than -- for a short moment --, the heating is increased at high rate affording the generation of the individual atoms spectroscopically analyzed.  One trick for this is to use a L'vov plate, a little boat of graphite accommodating the analyte solution, that is hold by a surrounding (equally horizontally aligned) tube of graphite.  Because the L'vov plate is heated by the surrounding hotter graphite tube, deposition of analyte on the graphite tube is less likely.  Similar to the graphite furnace, spectrometers based on the electric spark as method of atomization are working discontinuously.

Beside the atom absorption spectroscopy (AAS) are, for example, atom emission spectroscopy (AES), atom fluorescence spectroscopy (AFS), and inductively coupled plasma spectroscopy (ICP) as related means of investigation.

By coincidence (http://chemistry.stackexchange.com/questions/37586/chemical-test-for-tin/37594?noredirect=1#comment58079_37594), I just stumbled over a wikipedia page (https://en.wikipedia.org/wiki/Atomic_absorption_spectroscopy) that shows the graphite tube which helds the above mentioned L'vov plate.


text:
Why are tertiary carbocations the most reactive, if they're already stable?
This may seem silly, but doesn't it seem weird for a compound that's stable (in this context, the tertiary carbocation) to be the most reactive?
I mean, wouldn't it be the least, given that it's already stable and wouldn't want to leave that stable configuration. On the same lines, the methyl carbocation sounds like the most reactive due to its high instability, right? wouldn't it want to react in order to form a stable compound?

A nice analogy I read once in a book." Dolomites  (beautiful group in c the Alps) are very stable (indeed they are there since millions years) but very reactive.  If you do not believe the latter go there with some HCl to see yourself. "

Related (?): Longer the unsaturation, higher the stability and higher the reactivity.

This is an excellent question. Please correct me if I'm wrong, but I think this is what you are grasping at:
First, it is true that tertiary carbocations are generally more stable than primary carbocations (and secondary carbocations) due to having more inductively donating alkyl groups. The hyperconjugative effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations.
Second, transition states involving tertiary carbocations as opposed to primary carbocations are more favorable precisely because tertiary carbocations are more stable than primary carbocations.
Note that this does not mean that tertiary carbocations are more reactive. We don't generally say that something which exists in a transition state is reactive/unreactive. One reason is that the transition state complex cannot be directly captured or observed. Another reason is that it is the transition state ... of course it's going to be reactive - of course it'll change very rapidly!
We may however say that reactants and products are either reactive or unreactive. So if something goes through a transition state involving a tertiary carbocation, it might well be more reactive than something which goes through a transition state involving a primary carbocation.
Of course, before labeling anything as reactive or unreactive, you'd do well to note exactly what you mean by reactivity. To put it lightly, there are myriad reactions. An alkyne such as acetylene is reactive with regard to combustion; acetylene torches are commonplace. However, acetylene isn't going to be reactive with regard to, say, backside attack (for a variety of reasons).
Ultimately I think your issue lies with semantics rather than the chemistry ... still, a good question!

Couldn't a certain carbocation exist on its own or will it only occur as a transition state?

Very interesting question. This also popped up in my head once. And this is how I convinced myself.
The confusing term here is stability. Tertiary carbocations are stable by inductive effect and hyper conjugation, and therefore have the tendency to sustain the positive charge on the carbon atom and stay like this for long. That is why we call it stable: it can stay that way longer.
Kinetic theory illustrates that the reactants present in the container randomly move in a container without any energy loss. Now imagine that there is a carbocation that is not so stable (take n-butane carbocation as an example) and now if a nucleophile would want to attack the cation, how could it attack it if it is not able to hold up by itself? The cation must hold on to itself in order to let the nucleophile come to it and then attack it. Tertiary carbocations have the stability to hold up like that and allow the nucleophile to attack it. Since it can hold up, that is why the reactions involving it favors $S_N1$ mechanisms, whereas a not so stable carbocation (like n-butane) follows $S_N2$, since there is "not enough time" to make a carbocation and proceed the reaction. $S_N2$ is a fast process compared to $S_N1$, since 1$^O$ have to react fast in order to form the product. This could also be supported by the fact that if we want to arrange carbocations in the order of reactivity by $S_N1$ and $S_N2$ mechanisms, they are usually the opposite.
To summarise, it was stable enough to hold on to its configuration and let the attack happen.

You should distinguish between stability — the thermodynamic concept with the adjectives stable and unstable — and reactivity — the kinetic concept with the adjectives reactive and inert.
Also, you should remember that thermodynamic stability is a meaningless concept unless you compare it to something else. Likewise, reactivity is only meaningful if it is compared. Unlike the concept of e.g. orbital energies which have a defined zero-point, there is no good zero-point for either reactivity or stability.
Now what does that mean for tertiary carbocations?
Any carbocation is a reactive species with respect to neutral carbon atoms. So the propan-2-yl cation is more reactive than propan-2-ol and the 2-methylpropan-2-yl cation (tert-butyl cation) is more reactive than 2-methylpropan-2-ol (tert-butanol). I would not want to classify the carbocations according to reactivity within their group. With the exception of very stable cations such as the trityl cation, most carbocations react with a diffusion-limited rate, so determining kinetic parameters is hard.
Within the carbocations, a tertiary carbocation is more stable than a secondary one which in turn is more stable than a primary one. So the tert-butyl cation is more stable than the propan-2-yl one — but remember that both are still very reactive.
However, never forget that with the exception of extreme cases (e.g. trityl), all carbocations are also less stable than any corresponding neutral compounds. In fact, the ‘stability’ of tertiary carbocations is nothing more than a very, very relativised measure of the low stability of different reactive intermediates.

Why stability must be relative unlike orbital energies? We could say that compound A is more stable than compound B if Gibbs free energy of A is less than that of B. This makes perfect sense to me. What is wrong with that?

@adosar You just described a relative stability measure. Gibbs free energy relies on enthalpy and enthalpy has no defined zero-point, so Gibbs free energy also has no clear zero-point. (For the most stable form of an element, this is defined as having zero enthalpy but that leads to us working with negative and positive enthalpies of compounds with an arbitrarily chosen middle point.)

There is misconception related to stability. The actual fact should be whether the stability is before reaching the transition state or at the transition state or after reaching the transition state. If the stability occurs before reaching the transition state, then the reaction may not proceed at high rates.
In the other two instances, the reaction may be facilitated by the stability of the intermediates of the reactions towards product formation.
Imagine you are climbing a hill. If you have more load at the time of stating (due to stability you are pulled down), you may not reach the peak easily. on the other hand if some additional load is given to you after you reach the top, you are more likely to reach the other side quickly (due to stability you are pulled down).
So stability disfavors reactivity, if it is before reaching the transition state and it favors reactivity if it is at or after the transition state.
Of course activation energy (rate/kinetics) and free energy (stability/thermodynamics) are two different terms. So comparing kinetics to thermodynamics is always not straight forward.

Good question.... Tertiary carbonation is more reactive than secondary carbonation which is in turn more reactive than primary carbocation due to stability... Primary carbocation is reactive and unstable so convert again back to a neutral reactant molecule instead of converting into products but tertiary carbocation is stable and there are more chances to form product by reacting with a nucleophile..

Please elaborate.

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text:
Multiple-quantum coherences in NMR
Why are multiple-quantum coherences (coherences with changes in magnetic quantum numbers, $|\Delta m| \neq 1$) unobservable in regular NMR experiments?

I'm not sure what you mean by 'regular' NMR experiments. NMR experiments which observe multi quantum signal are still 'regular' experiments, so I'll assume you asking why you wouldn't see DQ signal in a basic 1D 1H spectrum, for instance. I will attempt to explain without the use of product operator formalism, and scary looking maths.
In its simplest terms, the frequency of a double quantum coherence is about twice the normal observation frequency, and so cannot be observed without transformation back to single quantum coherence. Similarly, zero quantum coherence is typically less than a few kilohertz, and so is also not observable. If you set your receiver to the correct frequency, this transition would be observable, although significantly lower in intensity due to the lower probability of this transition path.

In the case of an isolated, uncoupled spin, there are only two energy states, and the transition between these two states (the single quantum transition) is the only possible transition. For the case of coupled nuclei, the state of spin I is affected by the spins state of nucleus S. The diagram above shows the typical spin transitions for a spin system, IS. The most common transition is a single quantum transition, where only a single spin changes state, and this generates magnetisation in the transverse plane. When both spins change state simultaneously, you can get either zero quantum (the spins flip to opposite states) or double quantum (spins flip the same way) transitions. The frequency of the double quantum transition is roughly twice that of a single quantum transition. In the vector model, double quantum transitions have no magnetisation in the transverse plane (only transverse magnetisation can be observed), but can be transferred to single quantum transition through the process of generation of SQC(first pulse), evolution (linked to couplings) and generation of MQC (second pulse) in a multi quantum experiment. This is typical of a DQF/TDF COSY type experiment, but equally utilised in studies of dipolar couplings of small organics in liquid crystal media (where quantum orders of >5 are typically observed)

This makes sense. I guess what I don't understand now is why within the product operator formalism the DQ/ZQ coherence operators represent transitions and not states/polarizations.

What do you think about @levineds answer?

long's answer (above the figure) is fundamentally incorrect. The most basic thing to recall is that both your detector and amplifier produce dipolar fields and so only dipole-allowed transitions are observable and multiple-quantum transitions are higher order multipoles.
In a liquid state sample, tuning your detector to twice the frequency will not produce a detectable signal. While it is sometimes possible to see such a weak transition in some systems, this is due to an overtone transition (in which mixing between dipole-allowed states makes the transition weakly allowed) not a multiquantum transition. Similarly, zero quantum coherences are also higher order. They cannot be observed because dipoles and higher order multipoles have zero overlap, not because the energy scale is small.
If you had a quadrupolar detector, you could detect multiquantum coherences.
Another way to look at it is that magnetization (like dipoles, a rank-1, vector-like operator) is always 0 for all transitions of higher order at all times and so there is never any signal to pick up in the coil no matter what frequency you are looking.

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text:
Basicity Alteration Due To Ortho Effect In Methoxy Aniline System
I wanted to ask something more in reference to this question.

Suppose instead of toluidine, 2-methoxyaniline and 3-methoxyaniline are present. Now, why is 2-methoxyaniline more basic than 3-methoxyaniline (even though 2-methoxyaniline has reduction of basicity due to ortho-effect but 3-methoxyaniline does not)?
Also what would be the correct order of basicity if nitro group was present in place of methoxy group? Would ortho-effect still take place?

Why would ortho effect reduce basicity? Shouldn't it increase?

Take home message: Just because you have an ortho substituent does not mean you will have an ortho effect.
To have an ortho effect two conditions must be met:

you need an ortho substituent
and, importantly, it must be of sufficient steric bulk to disturb the adjacent substituent.

One measure of steric bulk is the cyclohexane $A$-Value, a table of which can be found here.  From the table we see that methoxy $(A=0.6)$ and nitro $(A=1.1)$ groups have a smaller steric size than that of a methyl $(A=1.7)$ group.  Methoxy and nitro groups generally do not produce an ortho effect.
Therefore, the basicities of the various methoxy anilines (the anisidines) are controlled by a combination of resonance and inductive effects.
\begin{array}{l|r} \hline
\text{Substituent} & \mathrm{p}K_\mathrm{b} \\ \hline
\ce{-H} & 9.13 \\
\ce{o-methoxy} & 9.5 \\
\ce{m-methoxy} & 9.8  \\
\ce{p-methoxy} & 8.7  \\ \hline
\end{array}
(data source)
The basicity of p-methoxy aniline is controlled primarily by resonance effects (too far away for inductive effects to be significant).  Drawing resonance structures we see that the p-methoxy group is pushing electron density into the ring, this tends to keep the nitrogen lone pair more localized on nitrogen (if we also delocalize the nitrogen lone pair into the ring like charges repel leading to destabilization). This results in p-methoxy aniline being more basic than aniline.
The basicity of m-methoxy aniline is controlled purely by inductive effects (resonance is not possible from this position).  The inductively electron withdrawing methoxy group removes electron density from nitrogen, making it less basic than aniline.
The basicity of o-methoxy aniline is controlled by a mix of resonance and inductive effects operating in opposite directions.  Apparently the inductive effect is stronger and wins out (this shouldn't be surprising, note that the inductive effect from the meta position had a 0.7 unit effect on the basicity, and now in the ortho position it will be even stronger, while the resonance effect in the para isomer only changed the basicity by 0.4 units) making o-methoxy aniline less basic than aniline.

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What program could I use to create a protein model from scratch?
I would like to create protein models such as the ones in the Protein Data Bank. I have got GROMACS but I don't know if it can be used to model macromolecules.

Of course it can be, see https://en.wikipedia.org/wiki/GROMACS

You have several options to generate smaller peptides using the peptide builders in

Avogadro or
PyMol

Is that all or should I then get a more advanced software

@PabloXd There are probably some commercial tools for the job, but the two I mentioned are open source, available at no cost, run under Linux, OSX and Windows and you can automate tasks using Python scripts. Btw, if you found my answer helpful on this site, I'd appreciate if you'd accept it. You can always unaccept it later if somebody else provides a better answer ;-)

The free component of the AMBER molecular dynamics package, AmberTools, contains a very nice interface (tleap) for automatically generating coordinates of peptides and proteins from a sequence of three-letter amino acid codes. There's a tutorial on how to build and simulate a peptide using AMBER, but you don't have to run the dynamics. You can use the peptide building functionality on its own. It exports a pretty standard PDB file once you're done.
You could write a Python script that generates the tleap input, or if the peptide is short enough, write a tleap script with the three-letter amino acid codes by hand.


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Acidic strength of para substituted benzoic acids

Comparing the acidic strength, the nitro-substituted group is definitely more acidic. This can be reasoned by fact that a higher delta positive charge which develops on the H due to the strong -M nature of the nitro group.
Another way to check the acidity is to check the stability of the conjugate base.
The conjugate base would be the carboxylate ion which being a -M group creates a lesser electron density at the ortho and para positions. This positive charge is however stabilised by the methoxy group, which is a +M group, and destabilised by the nitro group. So from this viewpoint, the methoxy derivative should have been more acidic.
Why is this not so? What is the fault in my reasoning?

Your chief consideration should be how the negative charge on the conjugate base is stabilised. It is not very important to consider all these little things relating to the multiple resonance structures of the molecular ions.

Anyway, it is important to note that both the methoxy group and nitro group are inductively withdrawing. However, the methoxy group is +M while the nitro group is -M and more importantly, note that the nitro groups is meta-directing in EAS, generating partial positive charges on the ortho and para positions. Even if you claim that the carboxylate group is able to generate positive charges on the ortho and para positions, these are the positions relative to it. Note that the position ortho to the carboxylate group is the position meta to the nitro group.

Thus, the mesomeric withdrawing effect of the nitro group does not destabilise the ion as it does not withdraw via the mesomeric effect from the meta and ipso positions relative to it (i.e. the ortho and para positions relative to the carboxylate group). Thus, your reasoning is flawed.

You are somewhat on the right track. However, consider that acidity depends on the relative stabilization of the acid vs. the conjugate base, since the expression for $\mathrm{p}K_\mathrm{a}$ is as follows: $$\mathrm{p}K_\mathrm{a} = -\log\left(\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}\right)$$
Consider this figure for a visualization:

Electron-withdrawing groups like p-$\ce{NO2}$ (nitro group in the para position) increase acidity by making the carboxylate ion stable relative to the protonated form. However, electron-donating groups like p-methoxy decrease acidity relative to standard benzoic acid by destabilizing the (already negatively charged) carboxylate anion!
TL; DR: An electron-withdrawing group like p-$\ce{NO2}$ will better stabilize the negatively charged $\ce{COO-}$ group than an electron-donating group like p-$\ce{OCH3}$.
I found a neat picture from UC Davis that summarizes what I have said:

How does the carboxylate ion generate a negative charge in the ring? Carboxylate groups are supposed to withdraw electron density from the ring.

Thanks for pointing out 'in the ring'. I've removed that phrase from my answer now. What I mean is that when the COOH is deprotonated, you get a COO- group. The p-NO2 allows a positive charge to develop on the position in the ring at which the COO- group is located. The positive and negative charges are therefore very close together, and this decreased charge separation increases the stability of the conjugate base.

@GnomeSort I think it is the inductive effect of that ring carbon with positive charge. His reasoning is correct when he breaks both bonds, but even in that situation, the +I of both negative oxygens will dominate.

You thought correct but when you are talking about the acidity by taking into account the conjugate basicity then you have to stabilize the negative charge on the acid group which becomes more stable when you introduce an EWG(electron withdrawing group) in the ring which provides -M/-I to the acid group and hence helps it to delocalize its negative charge and become stable.
And there is no concern of the acidity of benzoic acid with the partial positive charge appearing on the ortho positions(here) because the proton is being released by the carboxylic group. Therefore to make it stable we need an EWG.

In Methoxy Benzoic Acid,we have Resonance or mesomeric effect which stabilizes the carboxylate anion.
In Nitro Benzoic Acid, we have Intermolecular Attractions (Some may call it as Hydrogen Bonding) between the Nitrogen(of Nitro group) in one molecule and the Carboxylate Anion in another molecule.
So, since the Intermolecular Attraction is more stronger than the Resonance effect, the Nitro benzoate is more stabilized than Methoxy Benzoate. So, Nitro Benzoic Acid is more acidic.

Those are definitely not hydrogen bonding interactions.

When did i say that they are hydrogen bond?


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Stereochemistry: R and S notation
How do we number the carbon atoms in symmetrical molecules while giving R and S notation. For instance, in 2,3-Dibromobutane, there is an ambiguity over which is carbon C2 and carbon C3. Will we call it (2R,3S) or (2S,3R)?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants
  are assigned to them in the following decreasing order of seniority:
(…)
(j) When there is a choice for lower locants related to the presence of stereogenic centers or stereoisomers, the lower locant is assigned to CIP stereodescriptors Z, R, M, and r (pseudoasymmetry) that are preferred to E, S, P, and s, respectively, which are preferred to the non-CIP stereodescriptors cis, trans, or r (reference), c, and t (…).
(…)

Therefore, the chirality center R receives the lowest locant and thus a systematic name is (2R,3S)-2,3-dibromobutane and not (2S,3R)‑2,3‑dibromobutane.
However, your drawing shows (2R,3R)-2,3-dibromobutane.


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Why cooling an exothermic solution it increases it's solubility?
In my chemistry book writes that cooling an exothermic solution(e.g. $\ce{NaOH + H2O}$) increases it's solubility, and I can't understand why since the cooling should decrease it.

Why do you think cooling should decrease the solubility?

Tables of solubility

Can you expand upon that? Perhaps with a reference

''Solubility of liquids and solids increases as the temperature raises ' - this is from the class book we use in Romania. I think solubility increases with temperature because there is need for heat for hydration of ions. And also, if the temperature is bigger, than the agitation of molecules should be bigger, and the process of diffusion should be faster ... I'm not a native english speaker, sorry.

See the linked question for an answer


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Container to minimise radiation exposure from U-238
Consider a sample of uranium-238 that is undergoing decay to thorium-234. During the transmutation, a gamma ray is emitted from the nucleus. However, the thorium-234 daughter particle does not decay quickly enough to be detected.
What kind of storage container should the uranium sample be stored in to minimise radiation exposure?

Due to its very long half-life, U-238 is not particularly radioactive; the specific activity is only about 12 kBq/g. The emission probability of gamma radiation is less than 0.1 % per disintegration of U-238. Furthermore, the gamma energy is relatively low. Therefore, the direct gamma radiation from pure U-238 (i.e. disregarding Th-234 and further decay products) is usually insignificant.
For example, the gamma dose rate in 1 m distance from 1 kg of pure U-238 (about 12 MBq) is only approximately 0.1 nSv/h – probably less due to self-absorption. By way of comparison, the dose rate from natural background radiation amounts to about 100 nSv/h.
Nearly all of the alpha radiation from U-238 is absorbed within the sample. Only a small fraction originating from the surface of the sample can escape but is stopped by air within a distance of about 3 cm.
Therefore, you do not need any shielding. Just use a suitable closed container to prevent contamination or incorporation.


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Ketene cycloaddition reactivity
Why do ketenes participate in [2+2] cycloadditions with their $\ce{C=C}$ bond and not the $\ce{C=O}$ bond? Can the reactivity be reversed somehow?

Ketenes can participate in [2+2] cycloadditions with their $\ce{C=O}$ double bond.  Ketene dimerization is the best known example of such a cycloaddition.

(image source)
As you add substituents to the ketene $\ce{C=C}$ double bond, you begin to favor cycloadditon across the carbon-carbon double bond at the expense of cycloaddition across the carbonyl double bond.
Some Diels-Alder cycloadditions also involve adding across the ketene carbonyl.

(image source)
With both the [2+2] and [4+2] cycloadditions often the kinetic product involves cycloaddition across the carbonyl, but on mild heating this cycloadduct can rearrange to the product from cycloaddition across the carbon-carbon double bond, the thermodynamic product.
A very nice review of the synthetic and mechanistic aspects of ketene cycloadditions can be found here.

"As you add substituents to the ketene C=C double bond..." Do you mean electronegative, electropositive or any kind of substituents and how does that work?

Any kind, it's not well understood, but the effect is thought to be steric in origin.  Look at the pictures of the two transition states at the top of page 10 in the last link in my  answer ("here") - see how steric factors (repacing H with any kind of R) can start to impact the TS leading to cycloaddition across the carbonyl more than cycloaddition across the $\ce{C=C}$ double bond.


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Target temperature for infused cooking oil
A little context: I make craft chili oils for friends and family. I use a two-step infusion process. First, I infuse the oil for flavor by simmering spices in the oil held around 250 degrees Fahrenheit. Second, I add dried chili flakes into the flavored oil around 250 degrees Fahrenheit. I have used canola and grapeseed oil.
Questions: -Is there an ideal temperature to best extract flavors/fragrances from spices (for ex, from cumin seeds or star anise) into the oil?
-Do the flavor compounds deteriorate with prolonged cook times even if oil temp is kept below smoke point? For example, how does the oil change when kept at 250 degrees Fahrenheit for 1 hour vs 2 hours?

I'm voting to close this question as off-topic because it is better suited at http://cooking.stackexchange.com/

I'm voting to leave the question open because it is about the extraction of fragrances, such as  4-(1-methylethyl)benzaldehyde and (E)-1-methoxy-4-(1-propenyl)benzene, from seeds. This about finding a suitable procedure in the chemistry of natural products and perfectly on topc!

Gowtham - Thank you. I will try posting there as well.

@Klaus Warzecha - Thank you for the vote. I think I'm looking for an answer from a solvent chemistry perspective but I lack the technical vocabulary to build a better question. Your comment was spot on though. Thank you for your time.


text:
What is the product of a copper and vinegar reaction?
I put a copper wire through a hole in the cap of a water bottle, filled the water bottle ~ half way full with vinegar, and left it over night.  The part of the wire inside the bottle that wasn't in vinegar turn black.  Really black.  Why is this?!

UPDATE: I wiped the black stuff off on paper, and it now looks green on the white paper.  Seems like it oxidized over night.  Why so fast?

I think this question is related to the Le Chatelier’s principle:
\begin{align}
\ce{2Cu + O2 &\;<=> 2CuO}\\
\ce{CuO + 2HOAc &-> Cu(OAc)2 + H2O}
\end{align}
You mentioned that half the copper wire was under the acetic acid, so it was not in contact with oxygen. Thus, the lower part did not get oxidized. The upper part of the wire was in contact with oxygen and reacted to form $\ce{Cu(II)}$ oxide. Because there was acetic acid vapor in the atmosphere of the closed bottle, the acid reacted further with the $\ce{Cu(II)}$ oxide, creating $\ce{Cu(II)}$ acetate and water. The removal of $\ce{CuO}$ therefore pressed the oxidation reaction to proceed forward, making more $\ce{CuO}$ and then $\ce{Cu(II)}$ acetate. As a result, the upper part of the copper wire became dark blue/green overnight due to the color of $\ce{Cu(OAc)2}$.

If it looks green/blue on paper, you likely have copper (II) acetate. You likely have the anhydrous version $\ce{Cu(OAc)2}$ which is darker than the hydrous $\ce{Cu(OAc)4(H2O)2}$
Wikipedia Images include:

The rate of formation is a property inherent to the reactants and the products, but it also depends on temperature.

Actually, there is little to no reaction of vinegar on copper metal (hence, the copper wire in solution is clear). Existing $\ce{Cu2O}$ coating will be slowly attacked, forming a cuprous acetate which in air/oxygen creates soluble cupric acetate. The latter serves as a weak electrolyte for a further electrochemical attack of the copper metal with $\ce{O2}$ subject to an inception period delay.
Passing air into the solution, or much more effective, adding $\ce{H2O2}$ promotes the electrochemical reaction as does adding a small amount of a good electrolyte (like $\ce{NaCl}$, which also complexes with insoluble cuprous forming a soluble copper compound). There is a commercial application of the electrochemistry cited in Wikipedia$\ce{^{[1]}}$, which more generally occurs with many transition metals other than copper:
$$\ce{4Cu^+/Fe^{2+}/Co^{2+} (aq) + O2 + 2 H+ -> 4Cu^+/Fe^{3+}/Co^{3+} + 2OH-}$$
Also:
$$\ce{Cu + Cu^{2+} <=> 2 Cu^+}$$ (where cuprous is assumed to be soluble/complexed)
which recycles the cuprous to keep the first reaction active. Interestingly, for those in a hurry, $\ce{Cu}$ /vinegar/3% $\ce{H2O2}$ and $\ce{NaCl}$ as an electrolyte, can be a vigorous reaction with a quick jump start in a microwave! Relatedly, $\ce{Cu/NH3(aq)/O2}$ (or 3% $\ce{H2O2}$ )/$\ce{NaCl}$ is sufficiently energetic to be the basis of commercial copper ore leaching (employs aqueous $\ce{NH3}$ with air and ammonium carbonate, where $\ce{NH3}$ acts as a complexing agent, see $\ce{^{[2]}}$ $\ce{^{[3]}}$ (both sources noting the electrochemical aspects of the reaction).
Now, why the black $\ce{CuO}$ (and basic copper acetate) coating on the part of the copper wire exposed to air, acid fumes, water vapor and dust particles? Because per the reaction above, I surmise that $\ce{CuO}$ and basic copper acetate would be the expected products in low water conditions in the presence of oxygen/air and H+. The dust, I would further speculate, maybe a source of an electrolyte or other metals like iron. This could result in a so-called redox couple favoring also recycling the copper ions, per the reaction:
$$\ce{Fe^{2+} + Cu^{2+} <=> Fe^{3+} + Cu^+}$$
Now, anyone who has heated copper metal in a flame is aware of the fact that even at high temperatures, $\ce{Cu/Cu2O}$ does NOT easily react with oxygen in air to create black $\ce{CuO}$ (and using a methane flame is problematic as $\ce{CH4}$ can reduce any created $\ce{CuO}$ back to $\ce{Cu}$, so a decidedly poor path to cupric oxide).
Many are more likely acquainted with the key electrochemical reaction cited above in the case of metal iron, a process described as the production of RUST. Interestingly, Fe is not readily attacked by cold mineral acids, but is with oxidizing acids as $\ce{HNO3}$ and even feeble $\ce{HOCl}$ (hypochlorous acid). Add oxygen and even weak $\ce{H2CO3}$ (carbonic acid) in the presence of $\ce{NaCl}$ will electrochemically attack iron metal (and, more slowly, copper metal also).
The (original) derivation of the key reaction above, which is apparently a net reaction is based on cited radical chemistry (by H. Liang, Z. M. Chen, D. Huang, Y. Zhao and Z. Y. Li)$\ce{^{[4]}}$  .
For those who trust less in theory and desire experimental results, see my demonstrated electrochemical cell preparation path to $\ce{CuO}$ (with pictures) in one of my threads here$\ce{^{[5]}}$   .
References

https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide
https://www.researchgate.net/publication/259637387_Ammonia_Leaching_A_New_Approach_of_Copper_Industry_in_Hydrometallurgical_Processes
"Kinetics and Mechanism of Copper Dissolution In Aqueous Ammonia", at https://www.academia.edu/292096/Kinetics_and_Mechanism_of_Copper_Dissolution_In_Aqueous_Ammonia
https://www.sciencemadness.org/whisper/viewthread.php?tid=154275#pid625574
https://www.sciencemadness.org/whisper/viewthread.php?tid=84047#pid521659

Just read Wikipedia article on Verdigris again, and it is clear per the preparation path described by Nuygen above, that the product is classic (except for the use of a copper wire in place of a copper plate, with a varying height separation from the 5% vinegar), and accurately detailed by Wikipedia at https://en.wikipedia.org/wiki/Verdigris . To quote, in parts:
"Verdigris is the common name for a green pigment obtained through the application of acetic acid to copper plates[2] or the natural patina formed when copper, brass or bronze is weathered and exposed to air or seawater over time. It is usually a basic copper carbonate (Cu2CO3(OH)2), but near the sea....If acetic acid is present at the time of weathering, it may consist of copper(II) acetate. "
And:
"Verdigris is a variable chemical mixture of compounds, complexes and water.... All the components are in an ever-changing and complex electrochemical reaction equilibrium that is dependent on the ambient environment."
All of which confirms my answer above and challenges the cited 'correct' answer that the product is simply copper(ll) acetate and not too likely, in my opinion, as was claimed on the three-sentence 'correct' answer, to be just dry (anhydrous) cupric acetate.
Actually, I agree with Nuygen on several of claims likely based directly on his observations (namely, a complex mix with an apparent presence of CuO, a shift from light (basic copper acetate per Wiki) to more colored salt, possibly cupric acetate (as also mentioned by Wikipedia).
Here is a video with vertically hung copper sheets at https://www.youtube.com/watch?v=jq5MB1H_hVc . Note the product color and compare to Nuygen's description.
As a side note, vinegar is not pure dilute acetic acid. Depending on the source materials and manner of processing, impurities may be present.

Necro, sorry bud.  10/10 for effort.  0/10 already answered.


text:
How many electrons are in the d orbitals of NiCl2·6H2O?
On Wikipedia, it says that

$\ce{NiCl2·6H2O}$ consists of separated trans-$\ce{[NiCl2(H2O)4]}$ molecules
  linked more weakly to adjacent water molecules. Only four of the six
  water molecules in the formula are bound to the nickel, and the
  remaining two are water of crystallisation.[4] Cobalt(II) chloride
  hexahydrate has a similar structure.

So nickel chloride hexahydrate has an octahedral symmetry and the ligands are two chloride ions and 4 water molecules? Or is $\ce{NiCl2}$ a solid in water, but separates into $\ce{Ni^2+}$ and $\ce{Cl-}$ ions in water since they are ionic?
Does this also indicate that there are 8 electrons in the $d$ orbitals and that it will form a high-spin complex (I know that the low-spin orbitals look the same though) so that there are two unpaired electrons? I think I'm thrown off by the chloride ions because my textbook only explains complex ions like $\ce{[CoCl4]^4-}$ or $\ce{Fe(CN)6}$ which are easier to understand for me.

I think I could answer this, but there is a part of the question I’m not yet understanding. Second paragraph ‘or is $\ce{NiCl2}$ a solid in water, but it probably separates into $\ce{Ni^2+}$ and $\ce{Cl-}$ ions in water since they are ionic, right?’ <- that sentence.

The hexahydrate has octahedral symmetry, and it is very probable that this is the case for the solid and dissolved state of the complex.
The compound will not separate into the ions, because the solvation of three ions vs the solvation of one neutral complex is higher in energy, especially considering the fact that opposite charges attract. So in solution state it is a $\ce{[NiCl2(OH2)4]}$ complex with octahedral symmetry.

The $d$-electron count is 8, as it starts with 10 and has a formal oxidation state of II. This is always the case for $\ce{Ni^{II}}$ in any medium but vacuum.
Whether it's a high-spin or low-spin complex depends on the splitting energy of the $e_\text{g}$ and $t_\text{2g}$ molecular orbitals that arise from the $d$ orbitals in combination with the ligand group orbitals, colloquially known as $\Delta_\text{o}$ or $\Delta_\text{oct}$. Have a look at this Wikipedia article on crystal field theory for more details.
In short, there are several factors that influence the magnitude of the splitting:

Oxidation state of central atom
Geometry
Ligand types

The octahedral geometry usually gives good energy splitting.$^\text{[citation needed]}$ In conjunction with a moderate oxidation state compared to Ni(0), and four mid-range splitting ligands (water), it is a safe bet to say that we will have a low-spin octahedral $d^8$ complex: Three pairs of electrons in the $t_\text{2g}$ orbitals and two unpaired electrons in the $e_\text{g}$ orbitals, that is.
Note at this point that the terms high-spin and low-spin are superfluous. The electrons in the $e_\text{g}$ orbitals will always be in a triplet configuration according to Hund's Rule. So it will always be in a "high-spin" configuration from an experimental standpoint.

Is there a reason for the downvote? I'd be glad to improve this answer.

low-spin octahedral d8 equals square planar. Or, at least, strongly distorted octahedral. Nickel has quite an interesting chemistry here, though AFAIK chloride has not strong enough field to make it low spin. Cianide, on the other hand, does. That said, nickel chloride does dissociate in water, as its solutions have measurable conductance. I suggest you to go for the sources and edit your post accordingly with references and links. I'm not posting the sources here, but they are pretty easily googleable.

@permeakra Quote from the wikipedia page: "In some derivatives, the chloride remains within the coordination sphere, whereas chloride is displaced with highly basic ligands." Water is not highly basic. Distilled water also has measurable conductance. And I don't understand at all what you mean by low-spin octahedral d8 equalling square planar... Maybe check this out for reference?

But your point is noted, I'll scour the interwebs for some information whether the chloride ligands dissociate and update my answer accordingly.

distilled water has low enough conductance to be utilized as dielectric in some applications.

As for "low-spin octahedral d8 equalling square planar"; for ligands with strong field, octahedral coordination for d8 complexes is unstable and the complex breaks down into square one. || as for "Water is not highly basic. " Even wiki states that NiCl2 solutions are acidic because of hydrolysis of Ni2+ cation. Implying that the salt does dissociate in solution.

Let us continue this discussion in chat.

@Jan For reasons pointed out in the spin-off chat, I still consider only octahedral complexes. Nothing square planar going on here.

Edited the answer for more clarity in the last two paragraphs. The discussion of Jahn-Teller distortion and symmetry-breaking effects is beyond scope as I interpret it from the level of knowledge presented about the material in the question by OP.

@permeakra Here's a little (two years) newer publication than the one you cited: DOI
This is all based on EXAFS fitting, which is a highly parametrized process. And seriously, no need to get all riled up about this. Feel free to write your own answer if you don't feel that this one satisfies the requirements of this platform.


text:
Can I get an accurate temperature reading from a flask with a built-in thermometer well?
I recently purchased a flask with a built in thermometer well, but I have discovered that simply setting a rod thermometer in the well does not produce a very accurate reading. I was wondering if there was maybe a specific method for measuring temperature in wells that I am not aware of. Here are some pictures of the flask:

The thermal conductivity of air isn't very high, when compared to organic solvents or silicon oils used for heating baths. Consequently, one doesn't place thermometers in the vapour phase above the liquid when monitoring or controlling temperatures. (That is, of course, except for distillations, where you want to measure exactly that - the temperature of the vapour phase in the head.)
In your case, it might help to fill the well with the same silicon oil that you use for the heating bath. In order to keep the rod thermometer in place (rather than letting it float), you might want to use a Quickfit adapter (similar to this one) that fits the joint of the well.


text:
What is the justification for Hückel's rule?
I know that experimentally, the rule has been verified but I am having trouble understanding why it works. I see no reason why equivalent cyclic structures can undergo resonance and be just as stable. Any clarification would be nice.

Look at this earlier answer, the "Edit: Huckel's Rule: Aromaticity - Antiaromaticity" section may provide what you are looking for.

The short version is that "antiaromatic" compounds would have two unpaired nonbonding elections (see the Frost Circle in ron's link), meaning that they would be diradicals. Other "antiaromatic" compounds have electrons in antibonding orbitals, counteracting the "resonance" stability.

Maybe also the first part of this answer of mine which gives some electronic explanation where the stabilization associated with aromaticity comes from might be of interest to you.

http://onlinelibrary.wiley.com/doi/10.1002/jcc.20470/full

Do have problem with H. Rule specifically or aromacity generally?

The justification comes from molecular orbital theory. I'm going to analyze the MO diagrams for benzene (a molecule with $4n+2$ π electrons) and an MO diagram for cyclobutadiene (a molecule with $4n$ π electrons). These results generalize for all planar cyclic molecules with π orbitals on every atom.

The above π MO diagram of benzene may be obtained through Hückel theory (see also: Pi molecular orbitals of polyenes). Here, the "zero" of energy (referred to as $\alpha$ in Hückel theory) refers to the energy of an isolated carbon 2p orbital that does not participate in bonding. Hence, any orbitals below this energy are bonding in nature, and orbitals above it are antibonding.
In the benzene MO diagram above, you'll notice that all the molecular orbitals that are being filled by the π electrons are more stable than leaving the carbons unbonded. On top of that, it can be shown that the stabilisation derived from delocalised π bonding in benzene is greater than the stabilisation derived from simply forming three localised C–C π bonds, in conjugation with one another, but not delocalised over the entire ring (e.g. in hexa-1,3,5-triene). This is the condition known as aromaticity.
Let's look at the MO diagram for a hypothetical perfectly square cyclobutadiene, for contrast.

As with all molecular orbitals, the most stable orbital is filled first. However, problems start to emerge when going up to next two orbitals. First, unlike the next orbitals in sequence for benzene, these are formally nonbonding within simple Hückel theory.
It can be shown that this situation is less stable than the case for an alternative version of cyclobutadiene, where the double bonds are localised between two adjacent carbons. In practice, cyclobutadiene therefore undergoes a distortion away from a square geometry to a rectangular geometry, which has longer C–C bond lengths for the single bonds and shorter bond lengths for the now localised double bonds (see also: Cyclobutadiene - Jahn–Teller effect or not?). This is the condition known as antiaromaticity.

I have corrected the orbital diagram. (cc @Martin-マーチン) However, there is a serious problem with your statement that a diradical is necessarily unstable and the corresponding implication that cyclobutadiene is antiaromatic because of that. Dioxygen itself $\ce{O2}$ is a diradical and is perfectly stable. The antiaromaticity of square cyclobutadiene is not because it is a diradical: it is because delocalised cyclobutadiene is predicted to be less stable than a localised version. So, delocalisation does not necessarily equate to stability!

In general aromaticity arises from a comparison between fully delocalised π bonding vs conjugated but not fully delocalised π bonding, and I have opted to also edit the answer to reflect this important fact. If you don't like it you may roll back.


text:
How is the "quasi-steady-state" assumption of Michaelis-Menten kinetics different from the steady state assumption of ordinary kinetics?
In Wikipedia's article for Michaelis-Menten kinetics, it titles the "[ES] = constant" section the "Quasi-steady-state approximation". Lehninger's Principles of Biochemistry, 4th Ed. pp. 203ff call this "steady state kinetics".
I know that steady state is an assumption which is used for various kinetic models that have an intermediate, and I believe the term "steady state" is used in other settings as well. Is Wikipedia's use of "quasi" in the title for the enzyme kinetics reflective of $V_0$ only being the measurement of the initial rate, and thus the further assumption that [S] is constant also adds more uncertainty to the steady state assumption? If this isn't the reason (and I'm not certain that it is), why add a further descriptor that denotes uncertainty or inaccuracy when we already know that steady state is an assumption?

Is Wikipedia's use of "quasi" in the title for the enzyme kinetics reflective of [...] the further assumption that $[S]$ is constant[...]?

No I don't think the "quasi-steady state" hypothesis is related to whether $[S]$ varies with time.

[W]hy add a further descriptor that denotes uncertainty or inaccuracy when we already know that steady state is an assumption?

Different fields use different terms for the same concept.  For example, some folks call it the "pseudo-steady state hypothesis" and abbreviate it "PSSH".  Some people just say "steady state assumption".  It's all different vocabulary for the same thing.
I don't think there's a principled reason to prefer one term over another.  If you solve the full differential equations for kinetic models that would normally invoke the PSSH, you'll see one problem.  Before the reaction starts, the concentration of intermediate $I$ is zero.  But when the reaction is going on, we assume the intermediate is at steady state, i.e. that $\frac{dI}{dt}=0$.  But if $I$ starts at zero and has an initial concentration of zero, then it is everywhere zero, which violates the prinicple that intermediates actually occur in real-life chemical reactions.  So there are two phases to such models: the first is the "burst" phase, also known as single-turnover conditions, or unsteady state kinetics.  In that phase $\frac{dI}{dt}$ is not 0 but increases rapidly.  That phase ends quickly and is replaced by the pseudo-steady phase for $I$.  If you want to get precise about the strict definitions of the two phases, you'll need to invoke singular perturbation and asymptotic matching, as described in this thorough book chapter from Mathematical Biology by James D. Murray.
So you could argue that the "pseudo" or "quasi" is good to have in there to remind you of these complexities.  Or you could argue that all those extra letters aren't doing a whole lot of good.  Either way, it's just a semantic issue and is unrelated to differences in enzyme kinetics and more "traditional" chemical reactions.


text:
How can I calculate the charge distribution of a water molecule?
I would like to obtain the 3D charge density of a water molecule for plotting purposes (i.e. not necessarily super-accurate). I know this is a tough problem, but I'd like to give it a try. Since I'm currently using python, I've identified PyQuante and PySCF, and gather that I should learn how to run density functional theory (DFT) calculations. Before I do that, I'd like to find out if there is something else out there I should use instead. Multiple suggestions are welcome.
I've installed PyQuante, which according to it's README.txt may be more accessible to understanding.
For example, the color plots in this answer give a hint of what I'm talking about, though I'll ultimately plot the field strength contours and some electric field lines.
I'd like to obtain a large table of values (or expressions/expansions?), so I can create and work with surfaces and field line plots myself.
Additionally is it possible that there exist supplementary data in published articles. Where I might find that? Although this is about existing data, I actually do plan on putting in the time to have at least a minimal understanding of the calculation itself. If DFT is it, then I will be spending time studying that.

I somehow fail to see what you actually try to accomplish. If you want a large table, what should be in that table? Please be specific about what you need for your plot. What do you mean with field strength contours and electric field lines? Maybe you could add a sketch to illustrate what your goal is. There are many quantum chemistry packages out there, some of them even come with a documentation. Basically, what you need is only a mapping of the electron density or wave function...

How to “get my feet wet” in Density Functional Theory by simulating a water molecule using Python

You need (some) background knowledge and the tools to

generate an input file with the coordinates of your molecule and commands which calculations to perform
run the calculation
postprocess the results

All tools are out there! Unless you have access to the commercially available suites through your institution, I suggest to use those that are free (as in free beer and free speech). Actually, some of these free tools are a good choice in any case!

Input files can be generated using Avogadro, Molden, Gabedit, or wxmacmolplt
You can perform your calculations using NWChem, MPQC (both are in the repositories of every recent Linux distribution) or GAMESS-US, which can be downloaded after registration. In addition, there are pyscf and PyQuante, which you mentioned in your question.
Most of the tools used to generate the input files can be used for postprocessing too.

To quote the slogan of a shoe company:

Just do it!

wow! That's exactly what I needed! This will keep me very busy now. I appreciate you taking the time to give me the "big picture" of what is available.

In pyscf, we are routinely running molden.py to generate molden file and plot the orbital surface with Jmol.  It should not be too hard to write small script to generate density or other charge surface then plot the surface with Jmol.  The basic functions you probably needed should be very close to the functions used by dft module, such as eval_ao, eval_rho.
In the recent Pyscf alpha 2 release, a script pyscf/tools/cubegen.py has been added to generate the Gaussian cube file format.  In my linux box, the density looks fine in Jmol.  You can use it as an example to generate other Jmol formats.

OK I will take a look @qiming-sun. Does it currently do DFT (as well as HF?)  It may be over my head - when I saw mol.light_speed = 137.035989 (1/fine structure)  in the documentation I decided it might be for people who are already experienced.

I should mention that what I want is the charge distribution, (or the fields) so that I can do my own plotting. @qiming-sun, I need the physics part - I want to do the graphics/display part myself. Thanks!

DFT is available.  The package assumes the user has basic knowledge of computational chemistry.  It provides many fundamental functions to achieve basic operations like integration, basis evaluation in real space etc.  The default settings should be enough for most cases, and you don't have to modify anything to do regular computation, eg DFT.  The script cubegen.py is an example about how to calculate charge density on cubic meshgrid with two lines of dft functions.  You still need consider your own data format to record the rest informations, eg the molecular geometry.

Two suggestions: MolCalc: Calculate Properties > Polarity and Solvation gives you a 3D surface plot with the electrostatic potential superimposed.
Another suggestion is Avogadro/GAMESS/MacMolPlt. See for example here

Thanks! I've added some edits to my question. Looks like these won't help.

If you are not looking to actually learn these calculations and just need the contour plots, you can use Arguslab.
This will give you illustrative plots of orbitals as well as total electron density. They won't be super-accurate but they will be quick.

Thanks! I've added some edits to my question. Looks like this won't help.

If you only need the electron density distribution of a water molecule, it doesn't make much sense for you to leanr DFT method. Although I don't think there will be such data out there, it's pretty easy to run such a calculation for people who do DFT calculation regularly. It will take several minutes. My suggestion is that you find someone to help you with the calculation. The calculation could give you the data describing electron density in the 3D space. Then you can play with the data by yourself.

Welcome to chemistry.SE! I think your answer shrinks down to "My suggestion is that you find someone to help you with the calculation." I'm not into DFT, so I leave it to others to decide if this should be a comment or not.

Hi @Pu Zhang, thanks for your comment. I think it always makes sense to learn new things. Though I appreciate your point, and no doubt it takes a serious investment of time to really learn DFT, I'll feel funny using the output without at least a cursory understanding.


text:
Specific gravity of H2O2 bleached viscose
What is the specific gravity of $\ce{H2O2}$ bleached viscose? In literature, I can find that this is $1.52~\mathrm{g~cm^{-3}}$, but for bleached viscose is it not findable. Can I presume $\ce{H2O2}$ treated viscose has the same density?

You can assume that the density remains unchanged.
Hydrogen peroxide simply oxidizes dyes, thereby destroying chromophores (groups that move the absorption of a molecule into the visible spectrum). The viscose structure should remain largely unscathed, and the change in density is so minuscule I would argue it is within the experimental error.


text:
Relative stability of cis and trans cycloalkenes
Why is the stability of (Z)-cyclodecene 2 greater than that of (E)-cyclodecene 1?

I'd think that since 1 contains a trans double bond instead of a cis double bond, it should experience less steric repulsion and hence be more stable.
However, I've read that the cis isomer 2 is more stable than the trans isomer 1 if the ring size is 11 or smaller. For ring sizes larger than 11, the opposite is true.
Why is this so? Also, what would happen if there were more double bonds?

Usually trans-olefins are more stable than their cis isomers for steric reasons, like you suggested.  However in small and medium size rings this is not the case; here the cis-cycloalkene is more stable than the corresponding  trans isomer.
trans-Cyclooctene is the smallest trans-cycloalkene that is stable at room temperature (trans-cyclohexene and trans-cycloheptene have been detected as short-lived intermediates).  In your drawing up above of trans-cyclodecene (a) draw in the two hydrogens on the trans double bond.  One of them is "inside" the ring and causes steric problems with the carbons on the other side of the ring as pictured below.  Because of this destabilizing transannular interaction trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules.  They exist in a conformation as pictured below for trans-cyclooctene.

(image source)
If you can, build a model of trans-cyclooctene.  Notice how you have to pull and stretch the methylene chain in order to attach that last methylene back onto the other end of the double bond - there is a lot of strain in this molecule. Now build a model of cis-cyclooctene; it is easy to loop the methylene chain from one end of the double bond to the other end.
Not until you get to around trans-cyclododecene (12 carbons) do you have enough carbons to easily connect from one end of the trans double bond back to the other end.  trans-Cycloalkenes with less than 12 carbons have additional strain because of this difficulty in spanning the double bond. Once you get to around 12 carbons there is no difficulty spanning the double bond and so once again trans-cycloalkenes become more stable than their cis counterparts.
Since double bond carbons are $\mathrm{sp^2}$ hybridized with 120° angles, if we add more double bonds, the 120° angles make it more difficult to loop the more flexible $\mathrm{sp^3}$ methylene (with their 109.5° angles) chain around the double bonds.

Brilliant answer, thank you, but I was also wondering if there was a way to understand this without making a model (although it would certainly make it easier to understand)?

Just try to imagine how hard it would be to loop the methylene chain from one end of a trans double bond to the other, when the chain is relatively short.

In figure b of the question, wouldn't the 2 sp3 hydrogens from both sides contribute to hindrance?

@Mixcels  Yes, probably some, but the big effect is from having to "stretch" 6 carbons from one end of the double bond to the other.

The problem with figure b here is that it's wrong. The lowest energy conformer is a crown conformation, it's hard to draw but this one isn't one. you really need to look at it in 3D like here http://comporgchem.com/blog/?p=1709 . The strain itself isn't a big problem, those are quite stable too.


text:
Will bleach combined with any of these chemicals create poisonous gas?
I need to know if a combination of bleach and the chemicals below will create poisonous gas. Will it create chloramine, and will there be any visible gas?
1) Sodium laureth sulfate
2) Sodium citrate
3) C12-14 Pareth-3

I need to know if a combination of bleach and the chemicals below will create poisonous gas.
Sodium laureth sulfate

Sodium laureth sulfate is $\ce{CH3(CH2)11(OCH2CH2)_{\bf{3}}OSO3Na}$, where the boldface 3 is an approximation and can actually vary between 1 and 5 or so.  The key point is it contains no nitrogen.

Sodium citrate

Sodium citrate can refer to mono-, di-, or tri-sodium citrate, which have formulae like $\ce{C6H_{7-\it{n}}O7Na_{\it{n}}}$, where $n$ is 1, 2, or 3, respectively.  Again there is no nitrogen.

C12-14 Pareth-3

According to Google, "C12-14 Pareth-3" is a "polyethylene glycol ether of a mixture of synthetic C12-14 alcohols with an average of 3 moles of ethylene oxide."  The most representative compound would presumably be the C13 alcohol with exactly 3 moles of ethylene oxide.  That would have the formula $\ce{CH3(CH2)11CH2O-(CH2CH2-O)3H}$, or, if I did my math right, $\ce{C19H40O4}$.  Again the key point is there is no nitrogen.

Will it create chloramine, and will there be any visible gas?

None of the compounds have nitrogen, so chloramine cannot be created from bleach (chemically known as sodium hypochlorite) and these chemicals.
I think the likely products of bleach treatment would be non-volatile compounds such as inorganic sulfate plus organic compounds like tridecanoic acid, tridecanol, ethylene glycol, as well as oxidized derivatives of all of those organics, possibly including some small amount of $\ce{CO2}$ gas.  But $\ce{CO2}$ gas is not poisonous and not visible.


text:
Is nitrogen of aniline sp² or sp³ hybridized?
I found a thread which discussed this which totally confused me. (refer to the part of the link which talks about $\ce{sp^2/sp^3}$)
The question "How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?" is helpful, but doesn't have aniline specifically. I understand that if the lone pair is in resonance, it becomes $\ce{sp^2}$.
Can someone help me out?

related http://chemistry.stackexchange.com/questions/29168/what-is-the-hybridization-of-the-carbonyl-oxygen-in-a-carboxylic-acid

I marked it as a duplicate anyway, since the premise is (almost) the same, and the answer is general enough to address, that "Atoms that are sp²-hybridized and sp³-hybridized have differing geometries, which is not permitted in the resonance phenomenon."

@Martin I don't understand your generalized statement. What is not permitted in resonance and how is it linked here?

An sp² hybridised atom is trigonal coordinated and an sp³ hybridised atom is tetrahedrally coordinated. When describing a compound with resonance structures, all of these structures have to have the same geometry, i.e. coordination is equal in all structures. Therefore an atom can never be sp² and sp³ hybridised in the same set of resonance structures.

@Martin The resonance hybrid can never be both sp2 and sp3 while the indiviual structures(imaginary ones) are either. Is this true? And by "Therefore an atom can never be sp² and sp³ hybridised in the same set of resonance structures." do you mean to say the answer has to be sp2 or sp3 and  not something in between them?

It can be something in between, but it has to be the same hybridisation in all resonance structures/ configurations.

@Martin Why so? When we donate lone pair to the ring, the N has +ve charge and is planar. So all those resonance structures contribute sp2 while the initial localized lone pair structure is tetrahedral and will contribute sp3. The combined effect of this will have something in between of sp2/sp3.

NO. That is not resonance. You can go ahead and ask this as a new question, I do not have the time to discuss this in the comments. Hybridisation is a helpful concept (not necessarily required for chemistry), resonance is a necessary concept to extend the scope of Lewis structures. Both are just simplifications of the true bonding situation.

@Martin I was talking about these  structures. And considering contributions of different structures on hybridization. Is this wrong? If yes, I will ask another question.

@Wong Did you ask a new question?

The nitrogen in aniline is somewhere between $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized, probably closer to the $\mathrm{sp^2}$ side.  We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal ($\mathrm{sp^3}$ hybridized).  However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, considerable flattening of the nitrogen occurs (if it were completely flat it would be $\mathrm{sp^2}$ hybridized).
We can assess the nitrogen hybridization by measuring its barrier for pyramidal inversion. If a trigonal nitrogen is $\mathrm{sp^2}$ hybridized, the barrier will be zero. On the other hand, in aliphatic amines where the nitrogen is $\mathrm{sp^3}$ hybridized the inversion barrier is typically around $\pu{4-5 kcal/mol}$.

(pyramidal inversion diagram)
In aniline this barrier is very low, somewhere around $\pu{1-2 kcal/mol}$. This indicates that the nitrogen in aniline is not quite planar, but is much closer to being  planar ($\mathrm{sp^2}$) than pyramidal ($\mathrm{sp^3}$).

What is pyramidal inversion?

Why can't it be called completely sp2? If it is in resonance, then it should be sp2. (that question link on my question says so)

See [here] (http://www.expertsmind.com/topic/properties-of-amines/pyramidal-inversion-913906.aspx) for a nice description and picture of pyramidal inversion.

And its contribution to the true representation is lesser as compared to the ones with the lone pair involved in resonance, so sp2 is dominant?

Yes, that's exactly correct.

Hybridisation is a function of geometry. Since it's a requirement for resonance structures to have the same geometry, hybridisation in the different contributing configurations must not change. We know that the amine group in aniline is slightly pyramidal, hence hybridisation cannot be ideal sp². I would rather state, that due to resonance, lesser p character is mixed in the sigma system, since some of the perpendicular p orbital is necessary for the overlap of the pi system.

@Martin So in the end it is more sp2 since it is just slightly pyramidal(sp3)?

@Wong Yes, I assume, that is the case. And this is probably also the reason, why inversion is fast, has a low barrier, respectively.

As the non-bonding electron pair of nitrogen could partially contribute in the resonance with the other  $\pi$ electrons in the cycle. (I will  prove this point indirectly further using an experimental data), the nitrogen atom hybridization is a linear combination of sp2 and sp3.
Now if we consider the nitrogen atom in the pyridine, it  is sp2 hybridized. Because the nitrogen atom has two neighboring atoms (2 carbon atoms) and one non-bonding electron pair, i.e. three neighboring electron pairs. But, the non-bonding electron pair can not contribute in the resonance with the other four $\pi$ electrons in the cycle. In fact, it is  perpendicular to the aromatic $\pi$ system.
The pKa for the pyridine systems is 5.2, for aniline it is 4.6.  So pyridine is a stronger base.  We can link this to the availability of the lone pair in pyridine, as it is not delocalized ( perpendicular to the aromatic $\pi$  system.  While in aniline the nitrogen lone pair can be delocalised using resonance with the adjacent $\pi$ system.

But the lone pair of nitrogen is in resonance with the ring. Since it makes it ortho para directing.

All fine about pyridine. But in aniline, shouldn't it be more sp2-ish?The lone pair is not localised.


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How to convert g/L to mol/L
I know I should divide $\pu{g L-1}$ by the molar mass of the substance, but I don't seem to find the specific answer on Google. So just to be sure:
If I have $10^{-5}~\pu{g L-1}~\ce{Cu^2+}$ solution, do I have $1.57 \times 10^{-7}~\pu{mol L-1}$?
I think I'm missing something.

Dimensional analysis is helpful. g/L divided by g/mol gives mol/L.

Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$
So $$c = \frac{m}{VM}$$
Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).

Grams of $\ce{Cu^{+2}}$ is ill defined.  The $\ce{Cu^{+2}}$ did not get into the solution by itself. It was $\ce{CuCl2,CuSO4}$ or some other salt.
This is probably the reason for the discrepancy.

What discrepancy? The OP's question seems perfectly clear to me.

The OP is mentioning a different answer compared to "the specific answer on google". That is what I meant by discrepancy


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Relation between Temperature and viscosity of Linseed Oil?
I have been trying to get a relationship formula to determine the viscosity of Linseed oil. I have Absolute viscosity of the oil at 33.1 Cp (Source: http://www.engineeringtoolbox.com/absolute-viscosity-liquids-d_1259.html).
The problem is the system where I am using the fluid has a varying temperature, and I want to be able to know what the varying viscosity is. I am aware that the viscosity reduces with increase in temperature, but wasn't able to find a specific formula to relate the two.
Is there a formula to get the relationship between temperature and viscosity?
Thank you

https://chemistry.stackexchange.com/questions/30954/answer/submit

https://chemistry.stackexchange.com/questions/30954/answer/submit

There are many formulas!  This usually means that some of them are OK in general for estimation purposes, but none of them will be especially accurate.  Not easy to choose from the many formulas.
One of the most simple and popular seems to be the Wright equation,
$$\log_{10}\left(\log_{10}\left(\nu + \lambda \right)\right)=A-B\log_{10}(T),$$
where $\nu$ is the kinematic viscosity, $\lambda$ is a parameter, usually taken to be $0.7$, and $A$ and $B$ are compound-specific parameters.  With knowledge of linseed oil viscosity at least one other temperature than the one you provided in your question, you could fit $A$ and $B$ parameters and apply the equation.
However, the best answer to your question is to obtain some empirical data.  A 1977 paper[1] reports on linseed oil viscosity as a function of temperature, for example.  They found that a different equation (not the Wright) gave a reasonable fit to the data.  Specifically, they found
$$\ln(\eta) = -5.73 + \frac{2820}{T}$$
for viscosity $\eta$ in centipoises and temperature $T$ in K.
Another thing to keep in mind is that linseed oil is prone to oxidation.  Oxidized linseed oil may have different properties than undergraded, "pure" linseed oil.
[1]: T. Gallagher; D. E. Kline; J. Am. Oil Chem. Soc. 1977, 54 (2), 68-70. DOI: 10.1007/BF02912392

Thank you very much. "There are many formulas!" was the reason I had to post the question. I think I am okay, for now by ignoring the effect of oxidization of the oil.


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Proton attached to N or O in HNMR spectroscopy
I was wondering why a hydrogen attached to a nitrogen or oxygen is not able to couple with p+ on adjacent carbon atoms and hence why it always appeats as a singlet in HNMR spectroscopy?

It depends on the condition of your experiment if a coupling of a proton of an amino or hydroxyl group with a proton attached to the $\alpha{}$-carbon is observable, or not.  As pointed out by the indicator set by @Mithoron, intermolecular exchange processes need to be taken into account.

If your alcohol (or amine) to be analyzed is very pure, and the deuterated solvent you deploy is both very pure, and dry, and free of acids or bases catalysing proton exchange, than you may observe $^3J$-couplings you expect to see.  For example a freshly opened sample of deuterated DMSO (used as solvent) may contribute to this observation.  See the peak data of the 1H NMR spectrum of ethyl alcohol in DMSO here.
If you aim to analyze an alcohol, not well dried, analyzed in deuterated chloroform or methanol that was frequently exposed to ambient (humid) air, then indeed instead of a triplet of an hydroxyl group  ($\ce{-OH}$)  attached to a methylene group ($\ce{-CH2-{}}$) all what may be spot is a broad singlet signal.  The same reference, 90 MHz 1H NMR in deutero chloroform provides one example.  (Note the chemical displacement of the hydroxyl protons altered, too.)


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How can I purify ethanol from formic / boric / salicylic acid?
I need some ethanol pure and strong enough for technical purposes (fuel, solvent, cleaning agent, dissolving rosin for soldering flux, etc.). Food quality is not required though, denaturated one is fine too. Distilling cheap vodka is possible, but even cheap vodka is expensive because of high tax on food ethanol; also now I don't have a distiller and want to try getting ethanol without building one. Pure denaturated ethanol is sold mostly in huge industrial bulks only. But we have extremely affordable 70% ethanol-based solutions in medicine stores - they are several-percent solutions of boric, formic and salicylic acids respectively. So, my question is: how can I turn these acids into precipitate or gas for easy removal, using not so exotic reagents and conditions?
UPD: I have realized my question may be ambiguous. Every substance I'm speaking about consists of alcohol + ONE of the three mentioned acids, not all of them together. I have a choice which acid I will be getting rid of.

Distillation or freezing should work.

@DrMoishePippik distillation will surely do the job, but as I wrote I don't yet have a distiller. Freezing sounds good, because of course I have a fridge :) Roughly, what temperature do I need, and which acid freezes away better?

Alternatively Sigma Aldrich has 99.5% non-denatured ethanol for sale at 100mL for £17.30 or cheaper if you buy more. http://www.sigmaaldrich.com/catalog/product/sial/459836?lang=en&region=GB

See http://en.wikipedia.org/wiki/Eutectic_system; for ~93% ethanol (remainder water), -120 C (153 K) is needed.

Which country do you live in? In Germany one can buy 96% denatured alcohol in most supermarkets or drug stores. Another option might be outdoor shops, as ethanol is frequently used for camping stoves.

@Gerhard Russia

@Raegdan: cannot help you in that case, unfortunately - never been there. But I am quite surprised that there is no denatured (and hence untaxed) alcohol available in drug stores - you can even get this in the nordic countries, which they tend to be quite strict when it comes to alcohol...


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Mechanisms in kinetic reactions
In my book is given the mechanism of $\ce{HBr}$ forming and I understand every step of it. We did the calculations of the rate of each reaction, which is easy depending on the sign (-,+) of compounds formed or spent, and knowing those rates we were finally able to find the rate of $\ce{HBr}$ which is well defined by the empirical equation of Bodenstein and Lind. If you know all those reactions, its not hard to find the equations you need, but my questions are :
Is there any way we can know which exactly are those reactions (mechanism)  for other examples?
How would we know if any reaction is a branched one ?

As far as I know, the short answer is no; in general, it is not possible to know the reaction mechanism a priori.
In general, the approach is:

Postulate a reaction mechanism/pathway
Use this to calculate the relevant reaction rate(s)
Compare these rate expressions with the results of experiments

If your proposed reaction mechanism matches the empirical results, it is likely (but you can never be sure!) that the mechanism is correct. If not, a new mechanism should be postulated and tested.
That being said, in many examples it may be possible to make an educated guess of a plausible reaction mechanism - based on your own experience or the experience of others (i.e. from articles etc.) - and then investigate whether the proposed mechanism matches the empirical facts of relevant experiments. Often, the proposed mechanism is more or less correct.
In addition, modern advances in computational chemistry may give useful information regarding likely reaction pathways based on the structure and interaction of the involved molecules. To my knowledge, it is not yet possible to use this in a standalone fashion to figure out the detailed mechanism, but it may help out in terms of making a better educated guess.

What about the second question ? Can we know at least if the reaction is a branched one or not ?

Again, as far as I know, the general answer is still no - unfortunately. It might be easier to predict whether or not branching will occur than predicting the reaction mechanism as a whole - but I'm still quite sure that this must be based on experience and (educated) guessing. If anyone has any counter-example I'll be very interested to know :)


text:
What elements and/or substances without water are liquid at room temperature?
I was thinking about liquids, and I started to wonder theses related questions:
1) Besides mercury, what elements are naturally liquid at room temperature?
2) What naturally found family of substances/mixtures that do not contain  $\ce{H2O}$ are naturally liquid at room temperature?

Strange that you didn't think of ethanol ;)  there are also lots of other organic and also inorganic liquids.

duh! of course! I totally forgot about ethanol!

liquefied gases liquid nitrogen, carbon di oxide etc, oil (a natural water hater)

This non-exhaustive list of substances contains 6179 entries which melt at or below 25°C. A few hundred of those are probably gasses at ambient conditions, a few hundred more are repeat entries for a same substance, and a large chunk of the rest is likely represented by some rather exotic compounds. Either way, goes to show there are a lot of possibilities.

(1) is readily answered by looking at most periodic tables. (2) is an extremely long list which SE is not suitable for.

This question is a bit broad in terms of the sheer amount of chemical compounds and mixtures that are liquid at room temperature. Examples include:
Compounds
Acids, bases, many hydrocarbons (e.g. hexane) and many more
Mixtures
Crude oil, aqua regia and many more
In terms of elements, there are only two that are liquid at room temperature (say about 20 °C or 293 K):

Mercury (as you identified).
Bromine

Francium, cesium, gallium and rubidium are close, with melting points at 300 K, 301.59 K, 303.3 K and 312.46 K respectively.
LennTech provides a list The elements of the periodic table sorted by melting point

Sorry, I guess I was thinking on naturally ocurring compounds and mixtures, and also elements. Milk however, gets its liquidness because of the water in it no? It's basically water + other stuff. I guess I meant I meant mixtures that do NOT have h20 in it

The examples I gave are natural, acids, bases, hydrocarbons etc don't have water in them.


text:
Is a microwave oven dangerous?
If we take a look at the electromagnetic specter, microwave wavelength is higher than infrared, which means its energy is very small. To break  chemical bonds, the energy required is 100-1000 kJ/mol, and this kind of energy can be afforded only by UV and VIS. Due to this, we can conclude that microwave radiation has only a vibrational effect on molecules of food and does not break any bonds ( the food remains the same ).
Considering those facts, why is the microwave avoided in concern for health?

No more dangerous than normal oven.

If there is no concern over the quality of the food, I'm guessing the question is more specific to whether the microwave oven is a safe piece of equipment to use.
The U.S. Food and Drug Administration says:

Manufacturers must certify that their microwave ovens comply with strict FDA emission limits. The emission limits are well below the threshold for risk to public health.
  Most injuries related to microwave ovens are the result of serious thermal burns from hot containers, overheated foods, or exploding liquids.
There have been extremely rare instances of radiation injury due to unusual circumstances or improper servicing.

If we're not talking about the oven itself, and rather the nature of chemicals involved with its use, then we could talk about microwaving with plastics with foods. There are sources online that raise concerns over plastics leeching dangerous chemicals into the food.
Harvard School of Medicine says:

If you’re concerned about plastic wraps or containers in the
  microwave, transfer food to glass or ceramic containers labeled for
  use in microwave ovens.
Don’t let plastic wrap touch food during
  microwaving because it may melt.
Wax paper, kitchen parchment paper,
  white paper towels, or a domed container that fits over a plate or
  bowl are better alternatives.
Most takeout containers, water bottles,
  and plastic tubs or jars made to hold margarine, yogurt, whipped
  topping, and foods such as cream cheese, mayonnaise, and mustard are
  not microwave-safe.
Microwavable takeout dinner trays are formulated
  for one-time use only and will say so on the package.
Old, scratched,
  or cracked containers, or those that have been microwaved many times,
  may leach out more plasticizers.
Don’t microwave plastic storage bags
  or plastic bags from the grocery store.
Before microwaving food, be
  sure to vent the container: leave the lid ajar, or lift the edge of
  the cover.

So did I understand right this, if the microwave emits radiation as it should ( with low energy ), then its no concern. But manufactures 'abuse' by selling microwaves with high energy and then it hits us ?

When you say radiation, do you mean inside of the microwave, or outside? The radiation by definition should be low in energy, but could be high in intensity, with a high powered oven. But, even then, manufacturers, at least in the U.S. are required to meet guidelines on emissions (radiation escaping the oven).

Inside of course !

So there's no 'abuse' if it stays inside of the oven, high or low power. Does that answer your question?

Well I was interested on what happens to the food since nowdays Ive read in many articles that it affects the food ( being risky for cancers ), and since it didnt match with chemistry I was wondered about the other possible reasons. If the reason is only in the lunacy of manufactures then the question is answered, if not then I didnt ask for the outside system of microwave.

Well the Cancer Council disagrees with that statement. But the second part of my answer addresses concerns of cancerous chemicals leaching out of plastics.


text:
Why does a hydride like NaH act as a base, but a hydride like NaBH4 act as a nucleophile?
A TA talked about this today in the context of ketones.
If you stick an acetone and $\ce{NaBH4}$ together in a protic solvent, you get 2-propanol.
If you stick acetone and $\ce{NaH}$ together, you get an enolate. What's the difference? They're both hydrides.

First of all, borohydride as a group is not same as hydride. So the mechanisms are very different.
Because a simple $\ce{H-}$ anion is more basic than an enolate $\ce{C=C-O^{-}}$(product of $\alpha$ hydrogen deprotonation of a ketone), it just deprotonate before reduction will happen. Deprotonated ketones (enolates) will no longer be a substrate for reduction.
Borohydride $\ce{BH4-}$ is not as basic as enolate, so deprotonation hardly happen. Thus a reduction will be the predominant reaction.

To complement the answer provided by @Ian Fang and edited by @santiago, sodium hydride's action as a Brønsted base differs from other bases.
"Normal" acid base reactions are reactions of an equilibrium, like the partial deprotonation of malonic esters by 5 m% sodium carbonate in some Knoevenagel reactions:
$\ce{R-H} + \mbox{base} \rightleftharpoons{} \ce{R-} + \mbox{protonated base}$
Deploying sodium hydride as base however often intends a quantitative deprotonation.  The equilibrium is driven to the right hand side, in favour of the protonated base ($\ce{H2}$) immediately leaving the reaction mixture -- so the back reaction can't take place.

The solvents are different for the two reagents. NaBH4 uses protic solvents while NaH requires aprotic solvents. This raises a question. Does NaH reduce benzophenone to the alkoxide or is a radical anion formed which couples along with the liberation of hydrogen? Or do they not react at all?


text:
What does it mean to take reading up to 0.05 mL?
In titration we usually take burette reading upto $0.01~\mathrm{mL}$, but my teacher says that we need to take reading up to $0.05~\mathrm{mL}$. What does that mean? Does it mean if I read $23.56~\mathrm{mL}$ that I should record it as $23.60~\mathrm{mL}$ and if I read $23.44~\mathrm{mL}$ I should record it as $23.40~\mathrm{mL}$?

I believe what your teacher means by this is that the error for your measurement is within $0.05~\mathrm{mL}$ of the estimate, since $0.05~\mathrm{mL}$ is half the value of the smallest measurable volume in a typical $50~\mathrm{mL}$ burette. This is typically how the error of an instrument reading is determined.
The readings you gave would then be: $23.56~\mathrm{mL} \pm 0.05~\mathrm{mL}$ and $23.44~\mathrm{mL} \pm 0.05~\mathrm{mL}$.

Note:Just answering after my practical examination.99% of the time,the meniscus was on XX.05 or XX.00.It was just humanly impossible to even guess the meniscus position further than that without a magnifying glass.I think the burette was made in such a way that the individual drops were about 0.05 cm^3.So my readings were always XX.05 or XX.00. They were not lets say XX.69 or XX.67 or XX.63.Humanly impossible.


text:
How to confirm the type of bonding taking place between two substances?
For example ink on paper: how can I confirm that the bonding is van der Waals and not covalent? Similarly, the same question can be asked about other interaction such as hydrogen bonding, ionic bonding, etc.
Are there analytical or spectroscopic methods for this purpose?

In your example, you mention two mixtures (both the ink and the paper), rather than two pure chemicals. How would you differentiate bonding of the ink (solvent, dyes and pigments) to the cellulose, lignin, sizing and colorant (e.g. TiO2) of the paper? Do you expect one answer be the same for all?

@DrMoishePippik, I was addressing a real-life situation in which one needs to determine "what's going on". If there are several methods one can use to determine the presence of either interaction, that would be helpful.

as written, your question is unanswerable because there are many different components in bot the ink and the paper, and for each pair the answer could be different, and you request one answer..

authors: Денис Забелин


text:
By using the magnetic susceptibility, calculate the number of unpaired elections in the complex
I'm having trouble making sense of this answer. I believe I am doing all the steps right, but my main concern is that I am not arriving at a whole number. If my answer is correct, should I just round up or down?
Question

By using the known values of $\chi$ for $\ce{HgCo(NCS)4}$ calculate the number of unpaired electrons in each of these complexes, and interpret the results in terms of the electronic structures of the complexes.

My Attempt
$$\ce{HgCo(NCS)4}\ 's \   \chi = 16.44 x10^{-6}\  \mathrm{CGS (g^{-1}\cdot cm^3) \ at \ 20°C}$$
$$\mathrm{\chi_M =16.44 \cdot 10^{-6}g^{-1}\cdot cm^3 * \frac{491.8528 g}{1 mol} =8.09\cdot 10^{-3} mol^{-1}cm^3}$$
$$\chi ' _M= \chi_M (\text{metal core electrons}) + \chi_M (\text{ligands}) + \chi_M (\text{other}) - \chi_M$$
Diamagnetic corrections include:
$$\begin{align} \ce{Hg^{2+}}&=40.0\cdot 10^{-6}\\ \ce{Co^{2+}}&=12.8\cdot 10^{-6} \\ \ce{NCS^-} &=26.2\cdot 10^{-6}\end{align}$$
Source
$$\chi ' _M=  (40\cdot 10^{-6})+(12.8\cdot 10^{-6})+(26.2\cdot 10^{-6})(3)-(16.44\cdot 10^{-6}) =-0.02418$$
then using the equation
$$\mathrm{\chi '_MT=\frac{1}{8}n(n+2)}$$
Assuming 298K and setting to zero
$$\mathrm{n^2 +2n-57.6722=0}$$
quadradic formula and i get
$$\mathrm{n=6.65}$$
I figure with all of that arithmetic it would be difficult to arrive at a whole number. So, my intuition is to round up to 7.
Is this approach correct? Which way should I round if so?
And I don't know how I would determine the electronic structure because there's two metal ions in the complex?

Let's start with "what is the species?"
While you write it $\ce{HgCo(NCS)4}$ it's probably better written as $\ce{Hg^{+2} [Co(NCS)4]^{2-}}$.
You can take the two metal ions separately as far as electronic structure.
$\ce{Hg^{+2}: d^{10}}$ So we don't expect any unpaired electrons.
$\ce{[Co(NCS)4]^{2-}}$ Well, $\ce{Co}$ must be $\ce{Co^{+2}}$, so it's $\ce{d^7}$ on paper. Next you'd need to consider the shape of the complex and possibly where the ligand is in the spectrochemical series and the likely orbital occupations.

 Here, $\ce{Co^{+2}}$ is tetrahedral, so it doesn't matter where NCS falls on the series. You expect a high spin complex with 3 unpaired spins.

Note: (NCS) can be a tricky ligand, since it can coordinate with either N or S atoms, but the crystal structure is clearly N-coordination.
I mention this, because while the Hg ion is, on paper, separated from the Co complex, in the crystal structure, there's clearly some Hg-S interactions.
So let's get to your question about the calculation. Yes, if you need an integer, you'd round up. But since you have many, many more than one molecule, your answer implies there are multiple magnetic species.

 This is clear from the electronic structure. Since the calculation above gives three unpaired spins in an isolated $\ce{Co^{+2}}$ complex, there must be some magnetic interactions between multiple species in the solid state. Indeed, this is observed at different temperatures. I'm hardly an expert in magnetic coupling here, but it certainly implies Hg must be involved as well.

Okay so I follow most of that, but if there's 3 expected unpaired spins, why would my calculation show 7?

Well, that certainly implies that it's not just Co d7 involved in the material, doesn't it. :-)

There are multiple academic articles on this compound, depending on the level of detail you want. The article linked in the last section gives some useful information into the temperature-dependent behavior and many relevant citations.


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Determination of mercury in AAS
Its known that the only method to determine mercury in atomic absorbtion spectroscopy is the cold vapor method.
What is the reason that mercury can't be determined with other methods, such as flame, electrothermal AAS or even emission spectroscopy ?

Mercury is an element with poor sensitivity when analyzed by traditional flame atomic absorption. This happens because the element is extremely volatile (remember mercury is a liquid at room temperatures and has significant vapor pressure at room temperature). Cold vapor is a widely used method, that's EPA approved, and shows a significant increase in sensitivity (nearly four fold).
Source
There's many different devices out there that can be used to do elemental analysis, but from a personal experience, if you're working with dry solid material (ex. soil), then X-ray fluorescence (XRF) could be worth looking at. It's relatively cheap and does multi-element analysis. With mercury however, you're looking at a LOD of about 5 ppm, which is... okay. It depends on your experiment.


text:
How long do the bonds in a molecule hold under nominally stable conditions?
Disclaimer: I am not a chemist
I was just drinking my Ice Age mineral water which is 'untouched since the Ice Age' and wondered if the bonds formed in each molecule have remained intact for all those years.
In a liquid that is just lying there underground I'm guessing that there is some, if not very energetic, movement of the molecules. What happens when they bump into each other? Might an atom become dislodged and bind with another molecule?
I have no idea but the thought interested me.

Unless there is nothing to break them, they will be intact. Nothing breaks on its own if it is not disturbed. Once stable, always stable.

It bothers me, seeing so many questions with the disclaimer "I am not a chemist". We all have something new to learn however far we have come along the way toward "being a chemist".

Water with its fluctuating hydrogen bond networks is a special case. If you feel for it, have a look at Water clusters: Untangling the mysteries of the liquid, one molecule at a time, but it doesn't answer your question ;-)
It is more interesting to have a look at paleochemical/geobiological findings.

A complex organic molecule, such as 2$\alpha$-methylhopane was isolated from

2700-million-year-old shales from the Pilbara Craton, Australia
  (Source)

Have look at the website of Roger Summons at MIT and his freely available publications, such as The Exceptional Preservation of Interesting and Informative Biomolecules.
In short, bonds can last a pretty long time!

If you are asking about the water molecules themselves (and not some other materials present such as the paleochemical structure above) the answer is that at least for liquid water, each water molecule exchanges its hydrogen atoms with other water molecules at an extremely rapid pace (millions of times per second). While frozen (a mile deep in Antarctica or Greenland?) for millions of years, this exchange process would be slower, but still I would venture that the water molecules individually are not the same ones they were way back when. Depending on how it was handled after the ice was mined from its glacier, the sample may still be collectively composed of the same atoms that were there millions of years ago.

This is exactly what I was wondering about but am surprised to hear that the atoms move 'millions of time per second'. Now I'm wondering why that happens - in spite of me not being a chemist ;-) Is it due to collisions or to some electrical force... ?

There are at least two factors that contribute to the rapid exchange of hydrogen atoms in water (or when bonded to an oxygen atom in other species like alcohols). Firstly, there is a small amount of ionization of water at any given point in time where two water molecules react to give hydronium ion (H3O+) and hydroxide ion (OH-). Once formed, these species could recombine to give two waters, but are more likely to react with other water molecules to exchange hydrogen atoms.

A second factor is hydrogen bonding - where the hydrogen of one water forms a partial bond to the oxygen of another. This lowers the energy barrier to exchange of hydrogen atoms and makes it a more rapid process than it would be otherwise. Hydrogen is unique in chemistry because it is so small - a hydrogen ion (H+) is a bare proton and can do things other ions cannot do.


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What is the product of a Dieckmann condensation of dimethyl adipate?

My attempt: $\ce{MeO-}$ preferably gives SN2 or E2, being a strong base and good nucleophile. But which reaction proceeds in this case?

I've given a strong hint by changing the title of your question!

Methoxide as a base can abstract a proton from an acidic $\alpha$-position of an ester, that is from $\ce{-C\color\red{H}_2COOMe}$.
The resulting carbanion/enolate can add to the other ester group of your starting material.
The cyclic intermediate eventually loses methoxide to form ...

This Dieckmann condensation is the intramolecular variant of a Claisen condensation.
I suggest to do it on paper!

The only problem here is OMe- is not a good leaving group. So why does it leave? I think it owes to great stability of C=O bond.

OMe is able to leave in this instance because the reaction is under basic conditions, therefore OMe is already in solution (so it is stabilized by these conditions).