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In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?
s-p mixing in $\ce{N2}$ causes the $1\sigma _g$ orbital to decrease in energy, the $2 \sigma _g$ orbital to increase in energy (to higher than the $1 \pi _u$) yet the $1 \sigma _u$ is left unchanged by the interaction. Why is this?
My lack of understanding may be as a result of the fact that I am unclear as to what exactly it is that is mixing. Clearly they are the s and p orbitals but is this within the atom or is the s orbital on one atom interacting with the p on the other?
Some clarification as to why the $1\sigma _g$ orbital to decreases in energy, the $2 \sigma _g$ orbital to increases in energy would also be much appreciated.

Rob, if you poke around on the internet you can find many sites that show diagrams for how the $\ce{N2}$ molecular orbitals are formed from two sets of nitrogen ($\ce{N1}$) atomic orbitals.  Many of the diagrams are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance.  With that understanding, let's step through your diagram and questions.
It might be useful to take a look at the $\ce{N2}$ MO diagram here.  It's just like your diagram. but they break down the steps and it might be a bit less confusing.
In any case we have a nitrogen atom with 5 valence electrons, 2 in the 2s atomic orbital and 1 each (total of 3) in the 3 degenerate p orbitals.  If we move a second nitrogen atom into the vicinity, these atomic orbitals will hybridize in order to start the bond formation process.
$\ce{N2}$ is analogous to acetylene, they both have a triple bond that is made up of 1 sigma bond and 2 pi bonds.  In acetylene - and nitrogen - the sigma bond is $\ce{sp}$ hybridized.  We form this bond by

Mixing the s and  a p atomic orbitals (the p orbital whose lobe is linear with the s orbital [see the diagrams in the link I supplied above, they call it the $\ce{p_{z}}$ orbital]) on one of the nitrogens creates a bonding and antibonding $\ce{sp}$ orbital on that nitrogen atom.
Same thing happens on the other nitrogen atom
As the 2 nitrogen atoms begin to interact and form bonds, the 2 $\ce{sp}$ orbitals on each nitrogen interact (split further).
The bonding $\ce{sp}$ orbital on each nitrogen interacts to form a lower energy and higher energy pair of bonding molecular orbitals (the $\ce{1\sigma _{g}}$ and $\ce{1\sigma _{u}}$ orbitals in your diagram - BTW, there should be a dashed line from the p orbitals to the $\ce{1\sigma _{g}}$ orbital - it is missing in your diagram, maybe that has led to some of your confusion).
The antibonding $\ce{sp}$ orbital on each nitrogen also interact in the same manner to create 2 antibonding sigma molecular orbitals ($\ce{2\sigma _{g}}$ and $\ce{2\sigma _{u}}$)

Now for pi bonds,

The remaining two p atomic orbitals on each nitrogen are orthogonal to the s and $\ce{p_{z}}$ orbital - they can't mix with those orbitals.
These atomic p orbitals $\ce{p_{x}}$ and $\ce{p_{y}}$ can only mix with their like counterpart on the other nitrogen atom.  This gives rise to a bonding and antibonding $\ce{p_{x}}$ pi molecular orbital and a bonding and antibonding $\ce{p_{y}}$ pi molecular orbital

With that background, let's now look at your questions

s-p mixing in N 2   causes the 1σ g   orbital to decrease in energy,
  the 2σ g   orbital to increase in energy (to higher than the 1π u  )
  yet the 1σ u   is left unchanged by the interaction. Why is this?

I'd say it a bit differently as I did above. Mixing of the 2 bonding $\ce{sp}$ orbitals, one on each nitrogen atom, causes them to split (one is raised and one is lowered).  The same thing happens to the antibonding $\ce{sp}$ orbital on each nitrogen - one is raised, one is lowered.

Clearly they are the s and p orbitals but is this within the atom or
  is the s orbital on one atom interacting with the p on the other?

Hopefully I answered this above.  I tried to distinguish between atomic and molecular orbitals.  If I haven't been clear, post back.

Some clarification as to why the 1σ g   orbital to decreases in
  energy, the 2σ g   orbital to increases in energy would also be much
  appreciated.

Hopefully what I wrote above clarified this as well, let me know if it didn't.


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Is it possible to combust a plasma?
In what we usually consider combustion your fuel and oxidizer are always neutral, but is it possible to have a similar "explosive" combustion process where both the fuel and oxidizer have been singly ionized? If so what might some potential fuels be, and where might more information on the topic be found? A google search doesn't turn up much (other than injecting a small amount of plasma to catalyze an otherwise non-ionized reaction.)

I'm afraid  that this time that the google search doesn't turn up much has good reasons - do you understand what combustion is?

Well chemistry isn't exactly my specialty which is why I'm asking, but as far as I understand combustion is just an interaction between valance electrons which result in new molecules that want to expand.. ionized molecules will feel coulomb repulsion however if they're compressed together I don't see any reason they wouldn't be able to react.

If you specifically mean a hot plasma, I think it might be possible but it's a tough request. To get combustion, you'll need to start off with form of bonding present between atoms in the plasma. This requires finding a molecule with atoms so strongly bound that the bond dissociation energy is higher than the molecule's ionization energy.
Given that the lowest ionization energy known (3.51 eV or about 340 kJ/mol for ditungsten tetra(hpp)) is comparable to relatively sturdy covalent single bonds, there's not much room for negotiation. Ionization energies easily become larger than even strong covalent double bond dissociation energies, so there's a large tendency for any substance to be stripped into its constituent atoms before getting a chance to ionize.
The situation is easier if you allow the usage of "cold" plasmas. Plasma generators are commonly used to scour surfaces completely clean of any organic residues. I don't see why you couldn't just insert oxygen and some organic gas into such a generator, ionize them both with the strong and rapidly varying electric field inside, and cause the organic gas to combust.

In a practical sense, the answer is No.
The reason is that a substance reaching a plasma stage is ionized to a point where there is not opportunity to interact with another substance (in the case of combustion by definition Oxygen to produce water and CO2).
Only after the matter de-ionized follows the returns to a natural substance which can undergo combustion. But by the time this happens, the substance may undergo combustion but is no longer in plasma phase.

The term lighting describes a spontaneous and violent  discharge of electrons from one place to another. (This is what you see in a thunderstorm).  There is no matter undergoing a chemical reaction with Oxygen.  The term combustion applies when a chemical undergoes a oxidation reaction with Oxygen. Therefore, a lighting is not a phenomenon that may undergo a combustion.

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Pressure over a solution
I was wondering something, if I have the same molar fraction for two different substances for a vapor on top of a solution, will this mean that the pressure of each individual substance will be the same ? Since they have the same quantity of molecules ?
This question has a link with Raoult's law : $p_{solvent}=x_{solvent}p^*_{solvent}$, where $p_{solvent}$ is the partial pressure above the solvent, $x_{solvent}$ is the mole fraction of solvent in the mixture, and $p^*_{solvent}$ is the vapor pressure of pure solvent.

If the vapor (i.e. gas) phase above the solution can be treated as an ideal gas, then yes, if two species have the same partial pressure, then they have the same mole fraction.
For an ideal gas mixture, $p_i = y_i P$, where $p_i$ is the partial pressure of species $i$ and $y_i$ is the mole fraction of $i$ in the vapor phase, and $P$ is the total pressure.
The ideal gas assumption will be very accurate in many cases commonly encountered in chemistry, but not always.  Binary mixtures of $\ce{CO2}$ and another compound at high pressure would be one example of vapor phases that are not ideal.  Concentrated vapor of aliphatic carboxylic acids is also not very ideal.

By the way, the way you defined the equation is different from the way Raoult's law is, no? I didn't know you could consider yi for vapor phase also, I thought that it was only valid for the solution

Yes, so (i) if Raoult's law is valid, and (ii) the vapor phase is ideal (those are two separate assumptions), then $y_i P = x_i p^*_i$.

Ok, because I just noticed that I have an exercice that changes a bit the definition of raoult's law the way you did it. I wasn't sure how I was supposed to understand it... Thank you

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Alkali metal enolates
Why do lithium enolates add to an electrophile in direct addition manner, while the sodium and potassium enolates tend to add in conjugated additions? The sodium and potassium enolates are well dissociated, making oxygen a hard base (which is why these enolates participate easily in O-acylations), but still, the alpha carbon atom is made softer, to react well in conjugated additions. This seems to me like a contradiction.

I'll assume we're talking about aldol reactions, ie. the electrophile is a aldehyde, or less commonly, a ketone. Being a more strongly coordinating cation, lithium can actually coordinate to both the enolate oxygen atom AND the electrophile oxygen. This serves to "pre-organise" both reactants for a direct addition via a six-membered, cyclic transition state. The favoured transition states are often chair-like. This is known as the Zimmerman-Traxler model.

The model explains diastereselectivity when that is possible, ie. E enolates give predominantly anti products, while Z enolates give mainly syn products.
Lithium also stabilises the initial aldolate product, by chelation via both oxygens.

The more highly dissociated Na and K cations presumably don't greatly facilitate reaction via a cyclic transition state, nor form a stable chelate of the product. The reaction is also more likely to be reversible, and the products of conjugate addition more stable.
Sent from my Samsung Note 8

Lithium has a very high affinity for oxygen (strong bonding interaction). For the larger (i.e. ion radius) more polarized alkali metals down the periodic table, most notably sodium and potassium, the bonding is more dissociated and hence the reaction of the enolate is more reversible. Usually the conjugated addition product is the thermodynamic product, while the aldol product is the kinetic product. More reversibility will increase reaction time and possibly yield. Therefore sodium and potassium bases are preferred over lithium bases when the conjugated product is needed.
The above was more or less paraphrased from Clayden et Al. Organic Chemistry textbook (p. 752, chapter 29).

Could it be also because oxygen coordinates better to lithium, so the carbonyl group of Michael acceptor will be close for direct addition?

I expanded the question a bit.

@Marko I understand what you are saying. Potassium and sodium will indeed harden the nucleophile and this effect is used sometimes, but reversibility is much more important here, since most direct addition product will be converted back as it is only the kinetic product.

Because the lithium enolates are more strong than sodium or potassium enolates. The lithium enolates addition is under kinetic control, instead the other addition is under thermodinamic control.

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kinetic and thermodynamic parameters of biodiesel production from lard
I want to know if someone has experimental results and calculations of the kinetic and thermodynamic parameters of Base Catalyzed Transesterification of pig fat, if not which is the method used to calculate them?
Thanks!! :)

I don't expect a significant energy difference per ester bond between glyceryl and methyl esters of a fatty acid. The way to drive such a reaction forward is by removing the desired product as the reaction proceeds. Then more product will form as the system seeks a new equilibrium.


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Addition reaction of hydroiodic acid to 2-bromo-3-chloro-2-butene
How can the major product of the addition reaction of 2-bromo-3-chloro-2-butene with hydroiodic acid be predicted, since Markovnikov's rule fails to distinguish it?
$$\ce{H3C-C(Cl)=C(Br)-CH3~+~HI~->~?}$$

TL;DR - Sometimes you need to do the experiment.

Like all questions involving Markovnikov's rule, you should compare the structures of the two carbocations:

At C2 (Bromo)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-CH(Cl)-C+(Br)-CH3}$$
The two issues to consider are resonance and induction.
Resonance
You could draw a resonance structure to show that the bromo group is stabilizing the carbocation:
$$\ce{R2C+-Br <-> R2C=Br+}$$
However, since the bromine atom is much larger than the carbon atom, it cannot form as effective $\ce{p}-\pi$ overlap with carbon.
Induction
The carbocation has two stabilizing alkyl groups. What about that bromo group? Bromine is more electronegative than carbon, but not by much (2.96 vs 2.55 on the Pauling scale). Plus, bromine has a lot of electron density around it to donate, if it could. Bromine thus is a mild electron-withdrawing group by induction.

At C3 (chloro)

$$\ce{H3C-C(Cl)=C(Br)-CH3 + H+ -> H3C-C+(Cl)-CH(Br)-CH3}$$
We once again examine resonance and induction.
Resonance
$$\ce{R2C+-Cl <-> R2C=Cl+}$$
Chlorine is smaller than bromine and closer in size to carbon. Thus, chlorine is a better resonance stabilizer than bromine.
Induction
The same two alkyl groups are present, so we will discount them. Chlorine is more electronegative than bromine (3.16 on Pauling), and it has less electron density around it to begin with. Chlorine is more powerful electron-withdrawer by resonance than bromine.
Compare:
                  Chlorine              Bromine
Resonance         Stronger donor        Weaker donor
Induction         Stronger withdrawer   Weaker withdrawer

In many cases, resonance usually wins, so I might expect carbocation formation preferentially at C3 with the chloro group, but I would expect the other product to form also.
Now... what else might happen.
Regardless of the initial regioselectivity, the carbocation also can be stabilized by forming a halonium ion. Bromine is better at this since it is larger (and less electronegative so it can stabilize the positive charge better).
This halonium would be attacked at the less hindered position (C2), resulting in transfer of the bromine to C3.
So in effect, you can get three products:
$$\ce{CH3-CHCl-CIBr-CH3}$$
$$\ce{CH3-CICl-CHBr-CH3}$$
$$\ce{CH3-CClBr-CHI-CH3}$$
You would then want to actually do the experiment to determine which of these is really the major product.


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Determination of a chemical compound in a non-homogeneous sample
I want to know if it is possible to determine a chemical compound in a non-homogeneous sample. I am asking this because we are aiming to create a device that will detect histamine level in fish without homogenizing the samples. In all the journals we've read (which are a lot), fish samples are homogenized (blended + chemically-treated) before histamine is determined.
I am a computer engineering student with basic knowledge on biochemistry. I'm researching on this because this is our thesis topic (which our adviser has forced recommended to us).
Any help/recommendation regarding this is appreciated. Thank you!

If you analyze non-homogeneous samples, the results can't be interpreted as being representative of the whole organism (or population). So, for example, if histamine is unevenly distributed within the body of a fish, your results would vary widely depending on what part of the fish you analyzed.


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Are there ways to fully dissociate a complex salt?
A complex salt is a salt where the metal and ligands do not dissociate in water.
For eg : $\space\ce{K_4[Fe(CN)_6]}$ when put into $\space\ce{H_2O}$ dissociates as : $\space\ce{4K^+} + \ce{[Fe(CN)_6]^{4-}}$
where $\ce{[Fe(CN)_6]^{4-}}$ is a complex ion which cannot be further dissociated.
So, is there any way to dissociate them, either by heating or by adding something? Or will they dissociate in some other solvent? Or is there no way to dissociate them?

Generally every cation/central atom in water is solvated and usually complexed

Complex ions are in chemical equilibrium with their dissociated forms, and this equilibrium can be shifted just as any other equilibrium. In your example, the complex form is very stable compared to the dissociated forms, but still, e.g. adding strong acid (DO NOT TRY IT!) one can free some HCN.

Let's consider the formation constant of this ion complex at $25^\circ\mathrm{C}$:
$$\ce{Fe^{2+} + 6CN- <=> [Fe(CN)6]^4-}\qquad K_\mathrm{f}=10^{+24}$$
This means that, the equilibrium is highly displaced towards the formation of the  complex ion. The ion complex is very stable.
So, if we want to displace this equilibrium towards the dissociation of the complex ion, we have to add an adequate quantities of  a metallic ion or a ligand that form with $\ce{CN^{-}}$ or $\ce{Fe^{2+}}$, respectively more stable complexes.


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How do I synthesize 3-phenyloxirane-2-carbaldehyde from chloroacetonitrile?
How do I synthesize 3-phenyloxirane-2-carbaldehyde from chloroacetonitrile?

My proposed route involves:

hydrolysis of the nitrile to form a carboxylic acid, and subsequent esterification
then, a Darzens reaction with benzaldehyde
reduction of the ester function with DIBALH and oxidation to an aldehyde with $\ce{H2CrO4}$.

Is there a better method?

The Darzens condensation can be carried out directly from the alpha-chloronitrile. The exact reaction needed here (benzaldehyde + chloroacetonitrile) is listed 15 times in SciFinder using different base catalysts (hydroxide or an alkoxide) and conditions (sometimes with phase-transfer catalysis). The epoxide will be formed as a mixture of stereoisomers, but often chemical yields are high (reportedly upto 100%).
With the epoxide in place, the nitrile can be reduced with DIBAL to give an intermediate imine, which is converted to the aldehyde after hydrolysis. This is pretty much the same strategy you outlined, but with a couple fewer steps.

As for sources of additional synthesis problems, old UC-Berkeley exams are usually very good. Check Chem 3A and 3B at this archive.


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No chlorate esters? Why?
Why are there no chlorate esters? Or are they just undocumented or too unstable? I've seen perchlorate esters, but google "chlorate esters" and only perchlorate esters come up.

perchlorates have no lone pairs of electrons on the chlorine, which leads to a kinetic barrier to any number of reactions.  Chlorates do have such a lone pair, which gives access to various decompositions reactions, even if the absolute oxidation potential is lower.

@Lighthart Together with the fact that alkyl perchlorates aren't just nasty alkylants but rather shock sensitive - I remember some publications from the fromer ASD of the USAAF on the reaction of alkyl iodides with silver perchlorate and subsequent tests - it's not very astonishing that nobody bothered or came close enough to characterize alkyl chlorates.

Your initial statement is wrong. There are organic chlorates. When you want to search compounds, try chemical search engine like PubChem or ChemSpider instead of google. Even google lead you to something if you read carefully. It is no reason that chlorate ester can not exist and they do exist. Below is the link to the PubChem entry of Choric Acid, Methyl Ester, which is Methyl Chlorate.
https://pubchem.ncbi.nlm.nih.gov/compound/18695386#section=Top


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Gibbs Free Energy
My teacher said that

The decrease in Gibbs free energy is the maximum non-mechanical work that can be obtained when the process occurs reversibly at constant temperature and pressure

But for a reversible process always $ΔG = 0$, so doesn't it contradict it? I mean he's just telling that maximum useful work in any process is zero.

You said:

But for a reversible process delta G is always zero.

This is not true; $\Delta G$ is always zero for a reversible process if it is carried out at constant temperature and pressure and the only kind of work involved is the $p-V$ work.
If the only kind of work involved is $p-V$, in the above conditions, it is true that the maximum work obtained is zero. In a general case, try to replace in your notes the work $w$ with $w = w_{pV} + w_{\text{no}\,pV}$. Where $w_{pV}$ is the $p-V$ work and $w_{\text{no}\,pV}$  is the work of any other types, to get the complete history.
In response to your comment:
Teoretically, every thermodynamic problem can be solved by using internal energy function. Many times, in practice is convenient to use others variables. It can be done by obtaining new functions that depends of the variables that we want to. We do that by apling Legendre transformations (http://en.wikipedia.org/wiki/Legendre_transformation) of the internal energy function. One posibility is to get the Gibbs energy.
The energy energy function naturally takes different forms depending on the system. I mean, if there are works others than $p-V$ they must to be included to account for energy changes.
The general form of a kind of work is: $X\cdot dx$, where $X$ is a generlized force (an intensive variable) and $x$ a generalized variable of displacement (an extensive variable), so we have for the change in internal energy ($dU$):
$ dU = \sum _{i=1} ^k X_i\cdot dx_i $
As the energy function change when we add non $p-V$ work, its Legendre transforms corresponding to $G$ also change, so in this case:
$ G \neq H - TS$.
Finally the way for calculating $\Delta G$ depends on the works involved and the specific problem.

Thanks! But then, for reversible process invloving non mechanical work how ΔG is calculated?

You are welcome! I reply you in the answer above.

Reversible process vs reversible reaction
A reversible reaction just means that forward and reverse rate are significant. The reaction will show net changes in concentrations until it reaches a state of equilibrium, where the forward and reverse rate are the same and ΔG = 0.
A reversible process is one where there is no change in entropy, i.e. you carefully choose conditions so that no objects fall off a cliff, no cold objects come in contact with hot objects, and you don't mix solutions of different concentrations.
This is an ideal situation where nothing would happen. You approximate this by imposing very small changes at a time so that the things you are interested in do happen but with very little entropy production.

But for a reversible process always ΔG=0, so doesn't it contradict it?

A reversible process has ΔS(universe) = 0, not ΔG = 0. If ΔG were zero, you would not be able to do any work. The way you can achieve this is, for example, in an electrochemical reaction where you keep the current very low. Under these circumstances, you can measure the electromotive force (emf), and can calculate the work an electron is capable of doing when going from anode to cathode. The chemical reaction is not at equilibrium, but you are running the process in a reversible way (which means you could reverse the direction of the electron by putting in just about the same amount of work as you got out when it went the other direction).
Why does it have to be a reversible process?
A closed system (such as a chemical reaction in a closed reaction vessel) can exchange energy with the surroundings in the form of work or heat. If you run the process in an irreversible way, more energy will dissipate as heat, and the system will do less work. That is why you would not measure the maximal work in this case.
The process is called irreversible because you can't reverse the process. Some of the work you need has not been transferred in the forward process as work, but as heat. So that energy is now unavailable. (In a simpler argument, the entropy of the universe increased in one direction. To get back to the initial state, you would have to decrease the entropy of the universe. This would be a contradiction with the 2nd law of thermodynamics).

Hi Karsten! I am actually confused when original PO said:' The decrease in Gibbs energy is...when the process occurs reversibly at constant p And T.' Is this true? Must the process be reversible for its delta G to be maximum non-expansion work? If we work from G=H-TS until dG=dw(add,max), I see the two constraints imposed are constant T and constant p only, but no constraint on the process must be reversible.

@The99sLearner If the entropy of the universe increases, it means you have "wasted" some of the available Gibbs energy (in the form of heat instead of work). See https://chemistry.stackexchange.com/questions/112900/gibbs-free-energy-and-maximum-work/112958

The way you can achieve this is, for example, in an electrochemical reaction where you keep the current very low. Under these circumstances, you can measure the electromotive force (emf), and can calculate the work an electron is capable of doing when going from anode to cathode. / But it is not reversible process, but spontaneous, irreversible one, in sense it cannot be reversed by reversing the state of neighbourhood.

The term "reversible process" does not mean that $\Delta G = 0$.
Reversible process is an imaginary or idealized process which occurs in infinitesimally small steps, such that the system very close to equilibrium throughout the process.
However, the infinite number of infinitesimally small steps would add up to an finite change.

In a reversible process ΔS total = 0, and ΔG = - TΔS total. So why ΔG is not zero in reversible process? I mean what's wrong in this?

"total" means system + surroundings.  The statement your teacher told you is about how much work the system can do (on the surroundings).

In normal language, a reversible process is one that will undo (and possibly redo) itself, that is, at roughly equal parts equilibrium.  Such a 'reversible process' 'tends' to have a very small DG, largely because of DG= -RT ln K and the closer to DG=0 is the process is, the closer to K=1 is.  However, strictly speaking, 'reversible process' means, chemically, a process that may be reversed.  In this context, there is no explicity requirement that DG = 0.

I disagree with @Lighthart

@DavePhD: About my interpretation of the common language meaning of reversible or something else?

I didn't get you. Yes I know total means system + surrounding, I am just asking why ΔG is not zero for reversible process, as according to the formula ΔG = -TΔStotal since ΔStotal is always zero for a reversible process so ΔG should also always be zero.

@AyushPateria I am saying that even though DeltaS total and DeltaG total are zero under those circumstances, that does not means DeltaG system is zero.

@Lighthart I mean I disagree with the Delta G must be small part of the statement.  For example see http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node34.html.  In that example, the process is reversible if the weight is removed in very small steps, but the change in Gibbs free energy of the system (the gas inside) can be arbitrarily large from the beginning of the process to the end of the process.

What I'm trying (and failing) to say is that at equilibrium with roughly equal proportions of products and reactants (ie the 'common language view') because of DG = -RT lnK, since K is roughly 1 because of roughly equal propotions, DG must necessarily be close to zero because lns of numbers close to one are close to zero.

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Salts that are more hydrophilic than NaCl
What other, rather easy to obtain, salts are more hydrophilic than $\ce{NaCl}$? Is there a hydrophilic scale for substances like the Mohs scale of mineral hardness?

As per DavePhD's answer, the contact angle of NaCl with water is zero, so we need another measure.
I think the word you are looking for is hygroscopic. http://en.wikipedia.org/wiki/Hygroscopy Substances which are so hygroscopic that they dissolve in the water they absorb from the atmosphere are said to be deliquescent, although the property of deliquesence in itself does not directly indicate the strength of a substance as a drying agent.
The strength would best be measured as the equilibrium vapour pressure at a certain temperature, or alternatively, the temperature required to dehydrate: i.e. the temperature at which the vapour pressure of water vapour above the hydrated form of the substance becomes equal to atmospheric.
Cheap substances which are hygroscopic find applications as desiccants
Here is a list of some common ones http://en.wikipedia.org/wiki/List_of_desiccants. It includes the workhorse laboratory desiccants: silica gel for use in desiccators where products are dried under vacuum; molecular sieves, for drying solvents; CaCl2 for drying gases; and MgSO4 and CaSO4 in their anhydrous form for a quick drying of organic solutions. I would assume (though I'm not sure) that all of these would be stronger than NaCl.
Unfortunately the really strong ones are missing, so here's a list of mine. The weaker ones above tend to act by physical absorption, or by changing from anhydrous crystals to hydrated crystals. Stronger ones tend to react chemically.
Concentrated sulfuric acid Strong enough to dehydrate sugar to "charcoal." One of the reasons for the affinity is the energy released when H2SO4 proponates H2O, forming HSO4- + H3O+. More of a curiousity than a desiccant. Not widely used as a desiccant because of its acidity.
Phosphorous pentoxide Reacts with water to form various different polymeric phosphoric acids, ultimately ending up with 2P2O5 + 6H2O= 4H3PO4 though strength is gradually reduced as water is absorbed. Commonly used in the lab when a strong desiccant is required.
Calcium Oxide The reaction CaO + H2O = Ca(OH)2 has been proposed for large scale energy storage (the storage occurs in the reverse direction), but so far has been used only for smaller applications. As the calcium species are basic, exposure to air causes absorption of CO2, which then irreversibly degrades the material to bicarbonate and carbonate (unless heated to a much higher temperature.)
Sodium Metallic sodium, extruded into wire using a press to increase its surface area, is the best way to ensure absolutely dry solvents.

The scale for degree of being hydrophilic is contact angle.
Contact angle ranges from 0 to 180 degrees, 0 being the most hydrophilic and 180 being the most hydrophobic.
180 degrees means that a drop of water on the surface forms a sphere, which only contacts the solid surface at one tangential point.
0 degrees means that the water spreads in a thin, flat layer on the solid surface.
NaCl and NaF have contact angles of 0 degrees according to The Significance of Interfacial Water Structure in Soluble Salt Flotation Systems Journal of Colloid Interface Science vol. 235, page 150, so no salts are more hydrophlic.
KCl has a contact angle of 8 degrees, and KI 25 degrees.

Curious. Is the measurement of contact angle still well-defined when the liquid being measured is able to dissolve the surface it's on?

actually, those values are for salt-saturated solution contacting the solid salt, not pure water

There are salts, like calcium chloride (CaCl2), that are sufficiently hydrophilic to dissolve in water vapor from the surrounding air if not kept in an airtight container. NaCl doesn't do this, so i would have assumed CaCl2 is more hydrophilic than NaCl - but clearly, a contact angle smaller than 0 is impossible. Is there a different word than "hydrophilic" for this, so CaCl2 is more "whatever" than NaCl?

@GuntramBlohm CaCl2 is much more "hygroscopic" than NaCl.

I wonder if the OP is referring to something besides hydrophilicity - maybe solubility, hygroscopy, or something like enthalpy or entropy in solution? For hydrophilicity though, this is a great answer.


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How are pressure vessels heated?
I'm working on a project in which I'll be dealing with pressure vessels. As of now I've drawn/conceptualized everything on theory basis. The diagrams are made on the basis of theory, but then I came through a situation where I'll require to heat water under high pressure at around 300-350 Celsius. So then I decided to use pressure vessels and to use LPG gas for heating it (LPG has flame temperature of around 1100 Celsius, but please note that I want to heat the water kept in vessel at just 300-350 Celsius)
Now, I'm stuck in thinking that how do you heat pressure vessels? Are there any special port/connections provided to heat them/the content inside them? Or are any special type of heaters used to heat them? Would be great if you can tell everything you know about heating it
Optional to read(this could be wrong): Are we supposed to directly heat them on flame? Just like cooking your food on the gas-stove running using LPG? If this is so, then won't it take too much time because the walls of vessels would be very thick?

Sounds extremely dangerous.  At 350 C, pressure will be about 200 atmospheres.  How do you know that the material of the container will be strong enough to hold that kind of pressure, when heated to 1100 C?

No, I don't want to heat it up to 1100C, that's just the flame temperature. I just want to heat the water at around 300-350 (so that it's superheated). And the PSI would be around 1900-2500. I've checked few sites and there are companies that make pressure vessels which can hold that much PSI along with temperature. ;)

I understand the contents will not be 1100C, but the outer surface of the container may be.  Also you are underestimating the flame temperature of LPG which is 1980C http://www.elgas.com.au/blog/453-the-science-a-properties-of-lpg

Yes, it will have to adhere to ASME Pressure Vessel standards (or equivalent outside the US).  However, for inspiration on how to heat water to those temperatures and pressures, you should look at steam locomotive boiler designs.  Combustion takes place in the firebox, and the hot gases are directed down a number of flue tubes that all go through the tank of water, offering large surface area for transfer of the heat to the water.

@DavePhD If that is the case, then think about our normal cooking fry pans. The LPG/flame is same in both the case, so don't you think that if our fry pans can handle 1980C then why can't a specially built vessel won't?

@JonCuster So is it like that no direct heating would take place (putting vessel directly over LPG flame)? And I also agree to your point but isn't there any other way? In schools we're taught that "x" material is heated under high pressure. So over here, how is it heated, and nobody answers it appropriately! If you know anything else about its heating then please let me know. Thanks a lot ;)

@georgemik a frying pan has zero pressure differential.  you are planning for a 2500 psi differential


text:
How can pKas differ across similar protons of the same atom?
For example, citric acid has three acidic protons, all of which are carboxylic acids. Despite being part of the same functional groups, they all have very different pKas. Why (and how?) is this?

The concept of same atom is somewhat misleading here.  They are the 'same type' of atom, not the 'same' atom.

Here is a picture of citric acid.

As you noted, there are 3 carboxylic acid groups in citric acid, so there are 3 acidic protons. A chemist looking at this unionized molecule would say that the carboxylic acid group drawn on the right side of the molecule is equivalent to (the same as) the carboxylic acid group drawn on the left.  Therefore, these 2 equivalent carboxylic acid groups should have the same pKa.  The carboxylic acid group drawn in the middle of the diagram is different from the other two (it is not the same as the other two) and should therefore have a different $\ce{pKa}$.  In chemistry, groups are either the same or different.  If they are the same they must be indistinguishable in all aspects, including physical properties like $\ce{pKa}$
That said, we might expect when we treat citric acid with more and more base to see only 2 different $\ce{pKa}$'s, but that is not the case.  In actuality when we treat citric acid with base one of the two different (right and left or middle) carboxylic acid groups will ionize.  Once that happens the ionized citric acid molecule contains a negative charge.  It will be harder to ionize the molecule again and remove the second proton and place a second negative charge on the molecule (bringing like charges together is destabilizing and costs energy).  The same argument tells us that removing the third and final proton will be even more difficult because we wind up with 3 negative charges on the molecule.  As a consequence of this increasing difficulty to remove protons, we observe 3 different $\ce{pKa}$'s.

Let's begin this discussion with the simpler oxalic acid.
     O
     #
HO-C-C-OH
   #
   O

There are two carboxylic acid functional groups, and there are two different pKa's.  This is because double deprotonation does not happen simulateneously.  One gets deprotonated first, then the other.  After the first deprotonation, the residual molecule has a negative charge.  The result:
     O
     #
HO-C-C-O~
   #
   O

Is much less acidic because the presences of the carboxylate functional group changes the acidity.
Let's continue the discussion with the (still simpler) alpha-chlorosuccinic acid:
    Cl H O
     | | #
HO-C-C-C-C-OH
   # | |
   O H H

The acid closer to the chlorine is more acidic because the chlorine is strongly electron withdrawing, and 'pulls' the electrons out of the carboxylate function, which then 'pulls' electrons away from the hydrogen, making it more acidic.  This effect decreases with distance, so the chlorine pulls on the acid function farther away less strongly.
In a more generalized way, remote groups will modify acidity because of electron withdrawing or donating effects, but the farther away from the acid function, the less effect there is.
For citric acid, there are actually two different flavors of acid function, so one of the kinds will be intrinsically more acidic before deprotonation because of the remove effect.  However, because of the sequential deprotonation discussed above, all three will have different pKa's


text:
Calcination in metallurgical process?
In calcination why is the ore heated in the absence of air in a reverberatory furnace,
does the presence of oxygen affect any thing?

Most of the ore will be oxidized if thermally treated under air. That's why it's usual to heat it under controled atmosphere or even in vacuum.

Yes in calcination, with limited supply carbonates are reduced to oxides and to remove organic matter or to remove water of crystallization or evaporation of volatile components such as arsenic or antimony vapors.
If the oxygen component is more it would actually be then a roasting process where sulfides are converted to oxides since they are easier to reduce.


text:
Which pH indicator will give a dramatic color change with small changes in the amount of carbon dioxide dissolved in water?
I want to do a demonstration for students where they take a capped syringe and change the volume above a solution of seltzer water.  The amount of carbon dioxide dissolved will vary, hopefully changing pH enough to cause an indicator to change color, the more dramatic the color change, the better.
I have tried universal, bromcresol green, and methyl red with little success.  I can add substances to the seltzer water or change the temperature of the water.  I'd like a SAFE system I can create that will change color upon moving the plunger.

Use aqua regia. Safe enough IMO. :)

To get more of an acid shift and thus more dramatic change in the indicator color, create and dissolve more $CO_2$ into solution. Good old baking soda and vinegar will do the trick.

This is a really interesting idea, so props for coming up with it! However, looking at this table, it's evident that you'll need to get a massive pressure difference to shift the pH enough. Any indicator will require a pH shift of 1-2 units for a complete colour change, and it will take a factor of 100 difference in the $\ce{CO2}$ partial pressure to alter the pH by just 1 unit. I don't know exactly what indicator will work, but I think you'll need one with a $pK_a$ around 4.

Try neutral red. It's transition region is from 6.8 to 8.0 > red to yellow
Also check this wikipage for other options.

I once worked in a QC lab where we tested the acidity of organic esters to see if they had any free acid in them. The method was based on titration with strong base in methanol solution, using phenolphtalein.
To do this you would need an aqueous solution of sodium acetate or bicarbonate to get the phenolphthalein into its pink range, but I can tell you from experience that if you have a titration right at its endpoint, atmospheric CO2 and a bit of swirling is all it takes to turn the phenolphtalein back to colourless. Whether you can get the CO2 back out again by sucking with a syringe is another matter. Heating would help.
If you do it this way, you're going to need to change the experiment somewhat, but you should have something that will work. Here's a simple experiment (not exactly what you want but it uses the reagents I mentioned.) http://www.sciencegeek.net/Chemistry/chempdfs/QualitativeAnalysis.pdf
A great thing about phenolpthalein is that when it's in its colourless range it is absolutely colourless (due to the lack of conjugation in the molecule) so you can put lots in for dramatic effect. According to http://en.wikipedia.org/wiki/Phenolphthalein Phenolphtalein is used to test the curing of concrete by CO2, and the reaction and indicator are used in various toys.

Another thought: We used to add solid CO2 pellets to water with indicator in it for fun. That's certainly a way of getting a lot of CO2 into water. If I remember rightly the indicator was methyl orange. The colour changes only while the CO2 pellet is subliming. Once the pellet has completely disappeared, the colour changes back.

I'm going to try pheolphthalein.  I will adjust pH to about 8 and hopefully the pressure change will allow a color change.

Good luck. I had a thought when you mentioned "pressure" because that's important. Assume a student can exert a force of 7lb on a syringe of 1 square inch cross section.  As 1atm=14lb/sq in, that means they can go from 1atm to 1.5atm by pushing, and 1atm to 0.5atm by sucking, not a big difference. And we're only there's only a few grams of gas, so liquid volume needs to be small as possible. I'm sure there's something you can do with CO2-Alkali-Phenolphthalein, but all the examples I have given involve irreversible (at room temp) absorbtion of CO2, not release of CO2. Boiling should reverse!

This may be of interest: http://www.nuffieldfoundation.org/practical-chemistry/blue-bottle-experiment . Here we have a slow reduction reaction which can be temporarily reversed by atmospheric oxygen (by shaking)


text:
Why do people assume 28 oxygens in chlorite mineral formula?
I have been trying to calculate chlorite mineral formulas from microprobe data. However, all of the mineral formula spreadsheets on the internet, including those from the Open University assume 28 oxygens (http://www.open.ac.uk/earth-research/tindle/AGT/AGT_Home_2010/Microprobe-2.html)
However, all the formula I have ever seen for chlorite only have 18 oxygens? (see DHZ or http://www.mindat.org/min-1016.html)
By then calculating my formula assuming 28 oxygens, it means I have easily more cations than the mineral structure would have sites for?
Many thanks for any help

It doesn't mean that there are 28 oxygen atoms.  It means a charge equivalent of 28 oxide anions, considering that each oxide anion has a -2 charge and each hydroxide anion has a -1 charge.
This is based on the chlorite having an formula including $\ce{O20(OH)16}$
For example see $\ce{Mg10Al2(Si6Al2)O20(OH)16}$ in The effect of Cr on chlorite stability and the appearance of a new Al-rich silicate
In otherwords, as explained in OCTAHEDRAL OCCUPANCY AND THE CHEMICAL
COMPOSITION OF DIAGENETIC (LOW-TEMPERATURE) CHLORITES

All chlorite analyses were re-calculated as chlorite structural formulae on the basis of 28 oxygen equivalents (56 negative charges) i.e. assuming an ideal anion framework of $\ce{O20(OH)16}$


text:
Weighted Drilling Fluid Comparison with CaCl₂
I asked a question previously, and the answer was quite illuminating. The answer ushered me forth in a direction that finally made sense. I ended up finding this website, chem.latech.edu/~ramu/cmen522/homework/hw06_04a.pdf, that further explained the dynamic as well. I gathered what was necessary to be done, but wasn't quite sure how to apply it. I am currently a chemist for the oil field and am trying to find an accurate representation of percent water and CaCl2*subscript*, as well as the correct specific gravity of the complete solution. Given a 10.5 Mud weight (10.5 lb/gal), I have struggled to find volumes necessary to add and partial volume actually present.
Further explained - I applied the formula designated $V /cm^3 = 1001.97 + 30.2207(m^.5) − 36.7889m + 37.0838(m^1.5) − 5.7447(m^2)$ for the corrected solution volume and was really close to the correct "loss" of volume from the packing that the answer described. The two following formulas for partial volumes$ −36.7889 + 15.1104(m^−.5) + 55.6257(m^.5) − 11.4894m $(for CaCl2*subscript*) and $18.048 + 0.2722(m^.5) − 0.3341(m^1.5) + 0.1035(m^2) $(for water), don't add up to the total solution loss. I am a bit baffled as to why this is the case. Granted my physical chemistry knowledge is a bit dated, I would assume my excel document is correct. At the 10.5 weighted mud, my molality was 3.586. This resulted in a volume of 1105 with fluid loss and a 1183 of theoretical volume. The excess volume then should be 78.27. The formula for the solute given that molality gave me a volume of 17.63 and the solution was 35.33. The volume together is 52.95. This doesn't equal the excess volume which would suggest there was an error in one of the formulas, or that I have made a significant error on my part.

We have the MathJax plugin installed here, which supports common latex math and chemistry notation. For more information, see this page. Right now, your formulae are hard to follow because of the formatting. For example, is m^3/2 a fractional exponent $m^{3/2}$ or is it $\dfrac{m^3}{2}$? I would normally take a stab at cleaning up a post, but I don't want to make a mistake that will change the meaning.

I tried to fix it. It was intended as a fractional exponent.


text:
HOMO/LUMO and charge distribution
I am a little bit confused between charge distribution and HOMO/LUMO site in determining chemical reactivity. In charge distribution, the sites which are deficient and abundant in electrons are localized (e.g. in carbonyl molecules, we would expect that the carbonyl carbon to be partially positive and we would predict that a nucleophile would attack into that deficient carbon) but in HOMO/LUMO orbitals the site for reactivity differs on the orientation and sign of the orbitals. It may be in HOMO this carbon has an orbital for an overlap but in LUMO a node may appear in that site. My question is where would the nucleophile react in a species, by charge distribution or HOMO/LUMO orbital?

You need both an empty LUMO for the electrophile with an in phase HOMO for the nucleophile at the site of attack AND the proper charge distributions.
The LUMOs are critical; you can't shove electron density into more electron density, and LUMOs represent a lack of electron density.  However, the electron density for a nucleophile will be attracted to the partial positive charges.
Organic chemistry normally requires you make judgments about the relative impacts of various phenomena, and this is another example.


text:
How to represent a 3D model of cocaine?
For my AP Chemistry class we were told to make a molecular diagram that was "better than excellent" and the rest is up to us (we could choose the molecule and how we wanted to represent it).
My group chose benzoylmethylecgonine, or $\ce{C17H21NO4}$, also known as cocaine.
This is proving to be a difficult task as I now realize that cocaine is a pretty complicated molecule and I couldn't find much on the 3D molecular structure.  If anyone could explain its chemical structure, it would be very appreciated.

While I do not think that cocaine is a particularly interesting molecule, the crystal structure has been reported in Ronald J. Hrynchuk, Richard J. Barton, and Beverley E. Robertson, Can. J. Chem., 1983, 61, 481. There is also an open access paper of the XII IUCr Congress available, R. D. Hrynchuk, R. J. Barton and B. E. Robertson, Acta Cryst A, 1981, A37, C72. The crystal structure file has been deposited to ccdc.cam.ac.uk and can be downloaded free of charge. However, as far as I can see, it does only contain the information about the space group, etc, but no atomic co-ordinates.
However, the hydrochloride structure has also been reported in E. J. Gabe and W. H. Barnes, Acta Cryst., 1963, 16, 796-801. And the cif file can be also obtained at ccdc.cam.ac.uk.
I did a quick calculation on this geometry on the DF-BP86/def2-SVP level of theory, the plain xyz coordinates are attached at the end of this post. You can open them with molecular viewers such as Avogadro to create your own model.
The following, however, has been produced with the commercial ChemCraft program.

I do not know how far you want to go with your presentation, there are some pictures on wikipedia. I can also recommend checking the record of it on ChemSpider.

Geometry of cocaine hydrochloride:
symmetry c1
Cl      -2.051693000     -0.493009000      4.153090000
C       -2.451519000     -3.565091000      6.602382000
C       -6.214292000      1.164927000      8.772468000
C       -7.187531000      2.140573000      9.037661000
C       -8.160694000      1.917991000     10.029303000
C       -8.167271000      0.711328000     10.754343000
C       -7.200746000     -0.269358000     10.489848000
C       -4.535361000     -2.199809000      5.904903000
C       -4.965503000     -2.410764000      3.577976000
C       -0.472992000     -3.311748000      5.082266000
C       -3.681531000     -2.746076000      7.063993000
C       -3.208615000     -1.681350000      8.100725000
C       -1.959949000     -0.929634000      7.627840000
C       -0.886578000     -1.913758000      7.125382000
C       -0.580604000     -3.058883000      8.128726000
C       -1.633721000     -4.157723000      7.786625000
C       -5.209882000     -1.120147000      9.257751000
C       -6.215633000     -0.043947000      9.503197000
H       -2.776079000     -4.325870000      5.870251000
H       -1.141638000     -5.101415000      7.477122000
H       -5.458897000      1.318599000      7.989093000
H       -7.190344000      3.080452000      8.463274000
H       -8.921920000      2.687588000     10.235464000
H       -8.932096000      0.535630000     11.527409000
H       -7.186414000     -1.224606000     11.036739000
H       -4.915209000     -3.233215000      2.840991000
H       -4.408859000     -1.522439000      3.213266000
H       -6.015572000     -2.140911000      3.803492000
H        0.202195000     -3.979332000      5.655364000
H       -4.361351000     -3.437105000      7.610613000
H       -1.009855000     -3.900141000      4.313786000
H        0.111577000     -2.526413000      4.564425000
H       -2.986886000     -2.247300000      9.031392000
H       -2.216795000     -0.241161000      6.793930000
H       -1.551394000     -0.325210000      8.463897000
H        0.011514000     -1.355303000      6.795935000
H        0.453167000     -3.437387000      7.998115000
H       -0.655253000     -2.701279000      9.175171000
H       -2.290875000     -4.410651000      8.642544000
N       -1.471722000     -2.634200000      5.937688000
O       -4.202728000     -0.704253000      8.439562000
O       -5.256578000     -2.245544000      9.737495000
O       -5.375360000     -1.329144000      6.012186000
O       -4.299429000     -2.894414000      4.763398000
H       -1.903686000     -1.736871000      5.153339000

This was beyond amazing thank you.  You mention that you don't find it an interesting molecule, what would you consider interesting?  Either way, thanks so much.  I really do appreciate it.

@JohnSmith You are welcome. I consider small molecules like carbon monoxide way more interesting. Also small rings like cyclopropane, beta-lactames/lactones and boron compounds... cocaine's biological role is interesting, the molecule itself is just another organic compound.

You can also use this interactive image from chemapps.
For cocaine, go to https://chemapps.stolaf.edu/jmol/jmol.php?model=CN1C2CCC1C(C(C2)OC(=O)C3=CC=CC=C3)C(=O)OC
If you want any other molecule, replace "model=(SMILES of the molecule you want)"


text:
How to calculate to actual calcium content in a pudding mix
How can I determine the calcium (Ca2+ ion) content in pudding mix given the amount of milk, number of servings made with one package, the percent of recommended daily dose of calcium per serving, and the recommended daily dose of calcium?

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text:
How can I calculate molality from molarity?
How can we determine the boiling point of 1 molar aqueous solution (density 1.06 g per ml) of KBr? If I use the colligative property formula than I require molality instead of molarity. Here the density given is the one of the solution not the solvent.

The key here is to find the mass of the solvent. If you take one mL of the solution, it will contain $10^{-3} \mathrm{mol}$ of KBr.
The mass of $10^{-3} \mathrm{mol}$ of KBr is  $10^{-3}\times(39+79,9)=10^{-3}\times 118.9=0.1189 \mathrm{g}$.
Or, the mass of one mL of the solution is $1.06 \mathrm{g}$. This means that the mass of the solvent in one mL of the solution is  $1.06 -0.1189=0.9411 \mathrm{g}=0.9411\times 10^{-3}\mathrm{kg} $.
The molality is:
$$m=\frac{10^{-3} \mathrm{mol}}{0.9411\times 10^{-3}\mathrm{kg}}=1.063\,\, \mathrm{mol/kg} $$


text:
Why is the volume in this compress gas cylinder constant throughout?
I am stuck on this gas question and its really stressing me out.

A compressed cylinder of gas contains $2.25 \cdot 10^3~\mathrm{g}$
  of $\ce{N2}$ gas at a pressure of $4.25 \cdot 10^7~\mathrm{Pa}$ and a temperature of
  $19.4~^\circ\mathrm{C}$. What volume of gas has been released into the atmosphere
  if the final pressure in the cylinder is $1.80 \cdot 10^5~\mathrm{Pa}$?
  Assume ideal behaviour and that the gas temperature remains unchanged.

I know I have to use the ideal gas law for this.
I've seen the solution for this problem, but something still doesn't make sense to me.
In the solution they used $$\frac{nRT}{p} = \frac{n_\mathrm{f}RT}{p_\mathrm{f}},$$ to find the final amount of nitrogen gas in moles, $n_\mathrm{f}$, remaining in the gas cylinder. But why is the initial volume and final volume the same? Since gas is being released into the atmosphere, shouldn't the final volume change?

It is about the volume of the cylinder, which of course does not change. You do not also consider the volume of the escaped nitrogen gas.

Why don't we consider the volume of the gas released? Isn't, V, the volume of gas there is in the container and since gas is being released, shouldn't there be a change in the final volume of gas?

No, V is the volume of the container, because as per definition, an ideal gas cannot occupy any volume by itself.

Gases always occupy whole volume of the container, and it's not to going to change  throughout.


text:
Aromatic ring substitution: how are MNT and DNT further nitrated if nitro groups deactivate benzene rings?
This is probably a common question because it involves TNT precursors, but if an $\ce{NO2}$ group deactivates benzene rings toward electrophilic aromatic substitution reactions, then how can toluene be tri-nitrated? I understand that the methyl group will donate electron density and activate the ring toward the first substitution, but after two nitrations I'm sure the nitro groups slow the third substitution. So what's the trick? How does toluene get three nitro groups?

Deactivated does not mean does not react but rather *makes it harder to react. So you just need more ‘pressure’ as shown in Jori’s answer.

On a industrial scale by using highly concentrated nitric acid known as fuming nitric acid (usually > 90%, but in the last step from DNT to TNT may also be anhydrous) along with oleum, which is a mixture of $\ce{SO_3}$ and sulfuric acid. Preparation in the lab may be achieved by more or less ordinary nitration at elevated temperatures. In a micro scale experiment [1] they used two different types of nitrating acid:

Nitrating acid I contains concentrated nitric acid 19% v/v, concentrated sulfuric acid 58% v/v and water 23% v/v.
Nitrating acid II is a mixture of concentrated nitric acid 25% v/v and oleum (30% $\ce{SO_3}$) 75% v/v.

Nitrating acid I was used for $\ce{Toluene -> MNT}$ and $\ce{MNT -> DNT}$ (65 °C). Nitrating acid II was used for $\ce{DNT -> TNT}$ (110 °C).
1 The Stepwise Nitration of Toluene: A Multistep Microscale Synthesis Based on an Industrial Process
Richard A. Russell, Robert W. Swltzer, and Robert W. Langmwe
University College, University of N.S.W., Northcotl Drive, Campbell ACT. 2600, Australia


text:
Solid and Liquid Equilibrium
I understand why solids and liquids are not included in the $K_{eq}$ expressions. However, I'm wondering what the $K$ value for a reaction involving only solid or liquid reactants and products looks like. Theoretically, it'd be 1 no matter what since $1^n = 1$. Is this right? Also, do all such reactions go to completion?

No, it definitly isn't.

So then what is it?

Related: Why do liquids and solids have constant concentrations?, Why are solids and liquids not included in the equilibrium constant? What about in a reaction rate calculation?

These do not address my question. I get that liquids and solids have non-variable concentrations and thus can be considered contributors to $K_{eq}$ via rearrangement, and thus they are not part of the changing expression. What is the K value for A + B --> C + D where A, B, C, and D are solids or liquids?

If they concentrations don't change, they aren't reacting and indeed you have K=1.

Mithoron - this is not a duplicate. That question has nothing that addresses what I'm asking.

If something is reacting it's concentration will change, although it can be neutralised by regeneration, or external addition

You could calculate an equilibrium constant at a given temperature for a reaction from the standard free energy change ($\Delta G^{\circ}$) for any reaction using:
$\Delta G^{\circ} = -RT \ln K$
$\Delta G^{\circ}$ can be calculated for reactions using tabulated standard free energies of formation ($\Delta G_{f}^{\circ}$) using the following equation:
$\Delta G^{\circ} = \sum{\Delta G_{f}^{\circ} \mbox{(Products)}} - \sum{\Delta G_{f}^{\circ} \mbox{(Reactants)}}$
If the reaction only contains solids and liquids, then K will still depend on the ratio between the product and reactant concentrations.  However, these concentrations will not change as a reaction moves towards equilibrium, which implies that the reaction quotient ($Q$) remains constant


text:
Is the subtrahend in the Nernst equation related to ohmic drop?
Let us consider the Nernst equation $$E= E^\circ -\frac{2.303RT}{nF} \log Q$$ which once at $\pu{298.15 K}$, is the equivalent  of $$E= E^\circ - \frac{0.059}{n} \log Q .$$
Does the part of the Nernst equation $-\frac{0.059}{n} \log Q$ address Ohmic drop? In other words, is $-\frac{0.059}{n} \log Q$ related to solution (electrolyte) resistance of an electrochemical cell?

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In short: No.
The Nernst equation describes the thermodynamic equilibrium state. By definition, there is no current flowing through the cell. Therefore, there is no ohmic drop in any term within the Nernst equation.
The Q term relates to the activities of the reactants and products at equilibrium.

authors: Дана Касымбаева


text:
Why does temperature affect cell potential?
The Nernst equation describes the relationship between cell potential and temperature. But why does temperature affect cell potential?
My understanding is that the collision model of kinetics is not relevant to electrochemical cells, plus an increase in temperature decreases cell potential.
So why does the potential change?

Depending on how you write the Nernst equation the temperature might increase or decrease the potential of the cell. This depends more on the reaction quotient than on anything else.
The temperature comes in to the equation as a scaling factor where RT/nF has units Volt. This essentially defines how much the voltage changes per a decade change in the reaction quotient.
The temperature dependence here is just a matter of the scaling factor.
Going back to the collision model, that is more about kinetics than thermodynamics and it is not at play for the Nernst equation.
Nernst equation describes the thermodynamics of the electrochemical system.
The temperature dependence can be explained by the temperature dependence of work-functions.
Any electrochemical reaction by definition is an electron transfer. The electron goes from the highest occupied MO of one species to the lowest unoccupied MO of the other. Both these levels are temperature dependent.

Thank you. Could you please expand on the role of the work function? Is it analogous to the first ionization energy?

I did add a couple lines on that. It is essentially the interplay between first ionization energy and the electron affinity.

I would like to ask one thing. Does temperature affect standard cell potential?

Suppose the log term is 1 thus we must have standard cell then will we say that emf is independent of temperature

To a first approximation, NO. However, there are side effects in a lot of the reactions that do cause variations. Battery voltage is a great example.

$$\Delta G^\circ = -nF\Delta E^\circ$$
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
$$\Delta E^\circ = \left(\frac{\Delta S^\circ}{nF}\right)T + \frac{\Delta H^\circ}{nF}$$
This is where the temperature dependence comes from. Of course, a requirement is that the the reaction quotient remains at 1, because as soon as it differs, the $\Delta E^\circ$ no longer represents the standard electrical potential of the directly cell, and so it must be adjusted for using the Nernst equation.


text:
Ozonolysis of nitronate
What is the mechanism for the ozonolysis of a nitronate anion? What gas is produced in this reaction? Is there any chemoselectivity in the presence of a double C=C bond in the molecule?

What is the mechanism for the ozonolysis of a nitronate anion?

Probably very analogous to ozonolysis of a double bond.

A typical synthetic procedure is detailed in OrgSyn.  Note the addition of dimethylsulfide during work-up; this is common in ozonolysis reactions as it helps reduce the ozonide and also reduces any other dangerous, highly oxidized by-products.  The discussion section of the OrgSyn article says,

"There are various methods available for transforming a nitro group
  into a carbonyl group, including the Nef reaction (strongly acidic),²
  permanganate oxidation of nitronate anions (basic, oxidative),³
  persulfate oxidation of nitronates (basic, oxidative),⁴ treatment with
  a mixture of organic and inorganic nitrite (neutral, oxidative),⁵ and
  treatment of either a free nitro compound or a nitronate anion with
  aqueous titanium(III) (neutral, reductive).⁶ Each method is limited,
  however, by poor yield, inconvenience, or lack of generality.
With the proviso that the substrate not contain a reactive
  carbon-carbon double bond, the present ozonolysis procedure⁷ appears
  to provide a convenient and efficient method for carrying out the
  desired "

The Nef Reaction is mentioned above, it is worth taking a look at the linked article since the Nef reaction is complementary to the ozonolysis reaction of nitronates.  You may also find the "Reaction mechanism" section interesting.

What gas is produced in this reaction?

According to the above mechanism, no gas would be generated, rather $\ce{NO3^{-}}$ is produced

Is there any chemoselectivity in the presence of a double C=C bond in the molecule?

No, as mentioned in the OrgSyn reference, this approach can't be used when other double bonds are present.

what do you mean by DMS helping reduce other highly oxidized by-products? What could they be, and why would they be dangerous?

@Dissenter  Ozonolysis is not a delicate reaction, lots of things get oxidized and chewed up.  Hydroperoxides, peroxides and tetroxanes can be formed in the process and they are all explosive.  A dialkyl sulfide can reduce them to less dangerous by-products.


text:
Reaction balance
I got this redox reaction on my exam that I didn't know how to balance.
$$Sodium~iodate+sulphur~dioxide+water\ce{->}sodium~sulphate+sulphuric~acid+iodine$$, thus it would be:
$$\ce{NaIO3 +SO2 +H2O->Na2SO4 +H2SO4 +I2}$$
How does one go about balancing this using the oxidation number method?
I understand that S goes from +4 to +6 losing 2 electrons and I from +5 to 0 gaining 5 electrons.
Do we need to consider that there's 2 S atoms on the products (thus each atom loses 1 electron)? So I guess we  need to only multiply the reactants' side sulphur with 5/2, while in other cases we need to multiply sulphur both in the reactants and the products side.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Firstly, sodium periodate is NaIO4, not NaIO3.

@MARamezani I think you're editing to aggressively.  If his/her exam said periodate, and he/she doesn't understand what periodate is, that could be addressed in the answer.  If you change "periodate" to "iodate" it totally changes the question.   It's easier to give a good answer if someone can see what the person is actually asking with out trying to dig through the edits and find the Dead Sea scrolls.

@DavePhD Sorry, I thought that was too minor and was a typo rather than a misbelief. "Inexperiencedness" causes trouble.

@user14537 sent a comment asking if we can solve the exercise using only oxidation number method:
$$\ce{NaIO4  + SO2 +H2O ->Na2SO4  +H2SO4 +I2}$$

The change in oxidation number between periodate ion and iodine molecule is $7\times 2=14$ (as we have two atoms in iodine molecule).
So, the first step is to multiply $\ce{NaIO4 }$ by $2$.
The change in oxidation number between $\ce{SO2}$ and sulfate ion is $2$.
As the changes must be equal, we have to multiply $\ce{SO2}$ by $7$ and $\ce{H2SO4}$ by
$6$. (As you have already a sulfate ion in   $\ce{Na2SO4}$.
Now, as you have $12$ hydrogen atoms in the right side, you add $6$ molecules of water to the left side:
$$\ce{2NaIO4  + 7SO2 +6H2O ->Na2SO4  +6H2SO4 +I2}$$

The equation is balanced.
I hope it's clear now.

I'm trying to find an algorithm for this kind of problems, so the first step is equating the iodine atoms, shouldn't we be doing the same for sulphur (multiplying SO2 by 2)? And then why do we raise the coefficient only of H2SO4 to get the 7 ions? I suppose that is a combination of charge and inspection type balancing?

The first step is to calculate the "change of oxidation number" for the element iodine. It's from 7 to zero, i.e. 7 . As we have two atoms of iodines in the left side, the change is 14.

You mean right side, yes I understand that.

The first step is to calculate the "change of oxidation number" for the element iodine. It's from 7 to zero, i.e. 7 . As we have two atoms of iodines in the left side, the change is 14. So, we multiply the periodate by two, to balance the element iodine. On the other hand, the change of oxidation number" for the element sulfur is 2, and as the changes must be equal for the two elements (sulfur and iodine), we multiply SO2 by 7 (so the change will be 14). To balance the element sulfur, we multiply sulfuric acid by 6 as we have already the element sulfur in Na2SO4. Then we add water to balance

Yes I understand that, but why don't we e.g. multiply Na2SO4 by 3 and H2SO4 by 4?

Because you unbalance the element sodium (which is related to element iodine in periodate and already balanced)

Let's try to solve the question starting by iodate ion and then by periodate ion (as the post changes several times while I am answering the question).

Iodate ion:
$$\ce{2IO3^-  + 10e^- + 6H2O ->12OH^- +I2}$$

$$\ce{ SO2 +4OH^- ->SO4^2- +2e^- +2H2O}$$
If we multiply the second equation by $5$:
$$\ce{ 5SO2 +20OH^- ->5SO4^2- +10e^- +10H2O}$$
 Now, we add the the first and the last equation:
$$\ce{2IO3^-  + 5SO2 +8OH^- ->5SO4^2- +4H2O +I2}$$
Now, we can rearrange this equation:
Let's begin by adding $8\ce{H+}$ to the two sides of the equation ( and combining $\ce{H+}$ and $\ce{OH-}$) to make water:
$$\ce{2IO3^-  + 5SO2 +4H2O ->5SO4^2- +8H+ +I2}$$
We combine $\ce{4SO4^2- +8H+ }$ to make sulfuric acid:
$$\ce{2IO3^-  + 5SO2 +4H2O ->SO4^2- +4H2SO4 +I2}$$
By adding $2\ce{Na+}$ to the two sides of the equation, we find:
$$\ce{2NaIO3  + 5SO2 +4H2O ->Na2SO4  +4H2SO4 +I2}$$

Periodate ion:

Now, if we try to answer the question with  periodate ion:
$$\ce{2IO4^-  + 14e^- + 8H2O ->16OH^- +I2}$$
$$\ce{ SO2 +4OH^- ->SO4^2- +2e^- +2H2O}$$
If we multiply the second equation by $7$:
$$\ce{ 7SO2 +28OH^- ->7SO4^2- +14e^- +14H2O}$$
Now, we add the the first and the last equation:
$$\ce{2IO4^-  + 7SO2 + 12OH^- -> 6H2O + I2 + 7SO4^2-}$$
Now, we can rearrange this equation:
Let's begin by adding $8\ce{H+}$ to the two sides of the equation ( and combining $\ce{H+}$ and $\ce{OH-}$) to make water:
$$\ce{2IO4^-  + 7SO2 + 6H2O -> 6H2SO4 + I2 + SO4^2-}$$
By adding $2\ce{Na+}$ to the two sides of the equation, we find:
 $$\ce{2NaIO4  + 7SO2 +6H2O ->Na2SO4  +6H2SO4 +I2}$$
Conclusion: I think the question must be asked with periodate ion.

Can't it be solved directly by the oxidation number method then? I am not really familiar with this method with the hydroxide ions. By the way I meant to write Sodium iodate, I just got confused and wrote periodate the first time.

@user14537 please see my second answer. I hope it's clear now.

$\ce{2NaIO3 + 10e^- + 12H+ ->2Na+ +I2 + 6H2O}$
$\ce{5SO2 +10H2O ->5SO4^2- +10e^- + 20H+}$
$\ce{2NaIO3 + 5SO2 + 4H2O ->Na2SO4 + 4H2SO4 + I2}$


text:
Two liquids that react to produce a gas less dense than air at room temperature
Does anybody know two liquids (or a solid and a liquid) that can react to produce a gas less dense than air (or similar) at room temperature?
This question is intended to make a chemical induced reaction when a lever is pulled to produce gas and decrease an object's mass.

If the other products aren't significant, I'll recommend the all famous alkali metals' reaction with water:
$$\ce{2Li(s) + 2H2O(l) -> 2LiOH(aq) + H2(g)}$$
Hydrogen gas is less dense than air, both at RTP and at STP, because the main components of air (by mole percent) are $\ce{N2}~(78\%)$, $\ce{O2}~(20.95\%)$ and $\ce{Ar}~(0.934\%)$ - all of which have a molar mass larger than hydrogen.

As a side note, basically any reaction that produces Hydrogen gas is what you need. I wrote this example because it's cheap enough to obtain.

lithium isn't exactly cheap

@permeakra FWIW, cheap is relative. They never mentioned what's cheap supposed to mean.'

In case you want a safe and reliable method, consider magnesium granules/foil and acetic acid. The reaction is quite exothermic, but the reagents are reasonably safe. Do not use magnesium powder for the reaction, it is likely to react violently.
In case you want a cheap and reliable method, consider aluminium granules and sodium hydroxide. The reaction is, again, quite exothermic, and again, I recommend to avoid aluminium powder. Note: sodium hydroxide is unstable on air.
The military option is ferrosilicon with sodium hydroxide, but ferrosilicon is not readily available and the reaction is, again, quite exothermic.
Standard lab option is to use zinc granules with hydrochloric acid, but zinc is quite costly.

The other answers are very good, so purely in the interest of diversity I'll mention the methide (methanide?) carbides such as aluminum carbide or magnesium carbide.  With aluminum, the reaction is:
$\ce{Al4C3(s) + 12 H2O(l) → 4 Al(OH)3(aq) + 3 CH4(g)}$
Methane ($\ce{CH4}$) is less dense than air, but not by nearly as much as $\ce{H2}$ which is generated by reactions mentioned in other answers.  I've never tried the reaction myself but I believe it is comparatively less vigorous than hydrogen generation from alkali metals or alkali earths.  But on the up-side, the relative change in mass of the system will be higher with methane evolution since it weighs more than hydrogen.
I'm not sure why you even need a gas less dense than air.  If you're using a closed container (n.b. this is a bad, extremely dangerous idea unless it is a balloon or something), the mass of the system won't change.  If you're using an open container, the mass of the system will change as the gas is evolved, whether or not it is less dense than air.
Calcium carbide is not a methanide, but an acetylide, so when mixed with water it generates acetylene gas.  Acetylene is very nearly the same density as air (its molecular weight is 26 g/mol, compared to an average of 30 g/mol for air.  The reaction of calcium carbide with water was formerly used widely in carbide lamps, so presumably it is relatively easy to control.
Calcium carbide is also cheap compared to alkali metals such as lithium, and acetylene evolution will cause an even higher relative change in mass of the system than methane evolution.
$\ce{CaC2(s) + 2H2O(l) -> HC#CH(g) + Ca(OH)2(s)}%edit$


text:
Entropy change of system in adiabatic irreversible process
In an adiabatic irreversible process we know that q = 0. Now change in entropy is defined as q/T. So when q is zero, why is change in entropy not zero ?

also: http://chemistry.stackexchange.com/questions/10706/entropy-change-in-adiabatic-process?rq=1

The change in entropy is the reversible heat flow divided by $T$.  So if the process is irreversible, you can't use the irreversible process's heat flow to find the entropy change.

So, how do you find entropy change in an adiabatic irreversible process ?

To find the entropy change in an adiabatic irreversible process, you need to figure out (i.e., identify) a reversible path between the same pair of initial- and final equilibrium states of the system (any reversible path will suffice) and calculate the integral of dq/T for that path.  For more on this, see the following link:  https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/


text:
Which has higher priority (Cahn-Ingold-Prelog)?
Which group gets higher priority and why?
$\ce{R-CH_2OCH_3}$
$\ce{R-CH=O}$

Aldehyde over ether, oxygen in the $\ce{C=O}$ double bond counts twice.

As mentioned by Yomen , the double bond counts for two atoms with a single bond between them . They are called ghost atoms . This makes aldehyde group at higher priority than the ether .
Wikedia has a very nice explanation on it :

If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a ghost atom that has the same atomic number as B but is not attached to anything except A. In turn, when B is replaced with a list of attached atoms, A itself is excluded in accordance with the general principle of not doubling back along a bond that has just been followed, but a ghost atom for A is included so that the double bond is properly represented from both ends.

A triple bond is handled in a similar manner .
Wikipedia link : Cahn–Ingold–Prelog priority rules

$\ce{R-CH=O}$ has higher priority over
$\ce{R-CH_2OCH_3}$. Because, if $\ce{R}$ group is the same, then we have to compare between the next substituents: For $\ce{R-CH=O}$  you have one  carbon atom connected to 2 oxygen atoms (a double bond is equivalent to two simple bonds) and one hydrogen atom.
For $\ce{R-CH_2OCH_3}$, you have one  carbon atom connected to one oxygen atom and two hydrogen atoms. So, $\ce{R-CH=O}$ has higher priority.


text:
How can enthalpy change of a system be negative while entropy change is positive?
$$\Delta G = \Delta H_\text{system}-T\Delta S_\text{system}$$
$$
\begin{array}{ccc}
\hline
\text{Sign of}~ΔH & \text{Sign of}~ΔS & \text{Spontaneity} \\
\hline
+ & + & \text{The reaction is spontaneous at high temperature} \\
+ & - & \text{The reaction is never spontaneous} \\
- & - & \text{The reaction is spontaneous at low temperature} \\
- & + & \text{The reaction is always spontaneous} \\
\hline
\end{array}
$$
When enthalpy change is negative, the reaction is exothermic, which means it releases energy into the surroundings. If the system is losing energy, shouldn’t the entropy of the system always decrease? I understand mathematically that $\mathrm{d}S = \mathrm{d}Q/T$, and if there is heat exchange, then entropy change can be positive.
But intuitively, if energy is taken away from the system, shouldn’t the entropy of that same system decrease? For some reason, I always thought that more energy means more entropy.

For example, suppose you have a solid block of TNT.
It explodes and releases much energy. $\Delta H$ is negative.
Gaseous products like nitrogen, carbon dioxide and water vapor are formed. The system has become more disordered, so entropy has increased.

"The system has become more disordered, so entropy has increased" I understand this statement intuitively, but how to actually express this mathematically? Does this mean that reversible conversion of TNT to gases requires $\triangle H$ to be positive?

@Eisenstein Mathematically, $\Delta H$ is negative and $\Delta S$ is positive and $\Delta G$ is negative.  Reversible would be much different than an explosion.    Reversible means the change is slow.   $\Delta H$ , $\Delta S$  and $\Delta G$ would all be the same, but more work on surroundings would be obtainable if it was a reversible process.


text:
How to calculate the oxidation state and the number of equivalents oxidized with YBa₂Cu₃O₇
I've hit some trouble with these calculations and need help sorting out confusion.
In the lab we synthesized $\ce{YBa2Cu3O7}$, per the lab manual $\ce{Y, Ba, O}$ have the usual charges of +3, +2, and -2 respectively. Spectroscopic studies show that no copper (III) centers are present in the material, but rather there are missing electrons from the copper-oxygen bonds. For the purpose of the titration though, the missing electrons are thought of as coming from the copper center and we're assuming there are copper III centers present.
We used idodine to tritrate, and there is the assumption that copper is doing the oxidizing. The text states, "Copper(I) is not an oxidizing equivalent: It cannot oxidize iodide to iodine."
The relevant equations are
$$\begin{align}
\ce{Cu^{3+} + 2I- &-> Cu+ +I2}\\
\ce{Cu^{2+} +I- &-> Cu+ + \frac{1}{2}I2}
\end{align}$$
I've calculated the theoretical OS to be
$$\ce{Y} (3^+ )+\ce{Ba} (2^+ * 2)+\ce{O} (2^-*7)= \frac{7}{3 \ \ce{Cu}}\ce{Cu}^{2.3+}$$
We have to calulate the theoretical number of equivalents of $\ce{I-}$ oxidized by 1 g of the sample.
Would this be correct for the oxidation state? If so would that mean that the total equivalents of Iodine are an addition of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ (the two equations)? Also I'm feeling brain dead, because I can't figure out the fraction of $\ce{Cu(II)}$ and $\ce{Cu(III)}$ that would average to 2.3.

You forgot one relevant equation:
$\ce{Cu+ + I- -> CuI}$ which uses up some iodide as copper precipitates with it.
Getting the fractions is easy. You know that they must add to 3 and that that charge balance must apply. I marked the fractions as $x$ for copper(II) and $y$ for copper (III) and set up this system of equations which is easy to solve:
$$\begin{align}
x+y&=3\\
2x+3y&=3\times\text{average oxidation state}
\end{align}$$
You multiply the fraction with the charge of the species and the average oxidation state by three because in one mole of the compound you have three moles of copper, so that the equation represents a charge balance.

For the purpose of this question, I'm just providing the equations from the experiment and the textbook which says, "Copper(I) is not an oxidizing equivalent: It cannot oxidize iodide to iodine." As for the math that you provided, what was the reasoning for multiplying II by 2, III by 3, and the avg OS by 3?

It won't oxidize iodide, but it will make a precipitate  with it. You multiple the fraction with the charge of the species and the average oxidation state by three because in one mole of the compound you have three moles of copper, so that the equation represents a charge balance.

if you edit your question to include that information, I can give you the upvote.

Also, just checking here, I wound up doing the math and found the formula to be $\frac{2x+3y}{x+y}=2.3$. I had trouble solving your equation

Why replacing a number with variables? Express one variable from the first equation and replace it with the expression you got in the second equation.

Because both x and y are fractions whose denominator is the sum of the two; they're both fractions. I found their LCD to be 10. If you're confident in your math, could you show the work? Also, you've still ignored that my question does not ask for the reaction of $\ce{Cu^{+}}$.

x+y=3, then x=3-y and 2(3-y)+3y=3*2.3. y=0.9 x=2.1

Your fundamental assumption is that all of the oxygen is $\ce{O^2-}$. In oxide ceramics, however, it is possible for oxygen atoms to dissolve into the matrix. zirconia,  titania, and bismuth oxide can all easily accumulate non-stoichiometric oxygen. In fact this ability to "dissolve" oxygen into a solid matrix is how fuel cells are able to operate.
You are correct that the yttrium is $\ce{Y^{3+}}$, barium is $\ce{Ba^{2+}}$, and copper is $\ce{Cu^{2+}}$, but in YBCO the oxygen can form $\ce{O^{(2x-2)-}_x}$ sites as part of $\ce{Cu2O_x}$ ($1\leq x \leq 4$) clusters. Note: for $x = 2\ \ce{O^{2-}2}$ is not inherently a peroxide ion, you need cooper pairs of electrons to have superconductivity (see image below). Thus another way to present the structure of YBCO may be as $\ce{YBa2Cu3O5(O2)}$ or rather $\ce{YBa2Cu3O_{5+x}(O2)_{1-x}}$ or even $\ce{YBa2Cu3O_{3+3x}\nu^0_x(O4)_{1-x}}$ $(0 \leq x \leq 1)$ as the ceramic can be non-stoichiometric and produce oxygen vacancies which affect the superconductivity.

@OscarLanzi yes I suppose you could have O− radicals, especially at room temperature but this is not likely to provide cooper pairs below $T_C$.  My point was that oxygen does not have to be 2- and thus copper does not have to be 3+ to balance. I have edited things.


text:
Can all water soluble ionic compounds conduct electricity?
I'm taking a first year university chemistry course, and I am reading a lot of contradicting things. Some people say that ionic compounds that are soluble in water are always electrically conductive, others say that it maybe conductive.  Which one is it?
I have this problem to think about:

Suppose that an unknown chemical compound exhibits the following properties:

it is crystalline but shows no electric conductivity in the solid state
it melts at $\mathrm{300^\circ C}$ and decomposes at $\mathrm{400^\circ C}$
the compound is soluble in water, the solution shows no electrical conductivity.

What kind of chemical bond would you expect in the
  given compound? Try to describe this type of bond in detail.

Everything screams "ionic" (high melting point, solid at room temperature, crystalline structure, non-conducting in solid form) besides the fact that it doesn't conduct in water.

It's probably stable molecular crystal.

hmm, never even heard of those in the course! are all inonic compounds then conductors in water?

Covalent molecular solids can also exhibit high melting points if the molecules are large enough and have high symmetry (they pack well into a crystalline structure). See for example adamantane. Furthermore, simple ionic compounds usually don't decompose at such low temperatures. Lastly, I'd be quite interested to hear if there actually is such a thing as an ionic compound which dissolves in water but does not dissociate into electrolytes.

Melting points of 300 °C and above aren't unsual for organic compounds, think in terephthalic acid or trimesic acid. Both are soluble in water, but I'd expect some conductivity here.

DavePhD is right! Your material is an organic compound.
In order to be soluble in water, it should have some polar substituents. I'd however exclude carboxylic acids or phenolic $\ce{OH}$ (except maybe in the proximity of a carbonyl group) since these will partly dissociate and yield to minor conductivity.
If the melting point wouldn't be that high, inositol would be a candidate.
The nucleobases thymine and uracil do show melting points in the 300 °C range and are both soluble in water.

All soluble ionic compound conduct according to Debye-Hückel-Onsager theory.
Think of something molecular.

Oops I was only thinking of the melting point

I wouldn't say "always". But yes, and by default, ionic compounds are usually going to conduct electricity. I would study electrolytes before trying to decide these things, if you know electrolytes, the answer would be obvious to you

I don't know if you should have posted something like this as an answer. It would be really nice if you bring examples as to be the exceptions.


text:
do candles burn wood varnish fumes?
A friend of mine bought a piece of furniture that had a super smelly wood varnish, so I read around online and someone said to light candles to burn the chemical fumes.  Does anybody know if there is a chemical reaction between the wood varnish fumes and the candles? they are unscented candles and I assume they can break down the hydrocarbons in the wood varnish fumes.  Are there any other harmful chemical in the wood varnish fumes that are not oxidized by the candles that cannot be smelled by humans?

He should let it stay to dry. Simply keep it on sun or in a dry, warm well-ventilated room for some time (days to weeks).

Smelly compounds can be destroyed by free radicals wich are formed by a burning candle. But you need a lot of candles for your apartment.
Candles were also used in bathrooms (after toilet visits) to destroy smelly compounds like $\ce{H2S, CH3-SH, CH3-S-CH3}$ ...
A better and safer way is it too use cyclodextrin. This compound absorbs molecules and destroys them as well. Cyclodextrin is the working stuff inside the air-improvement sprays.


text:
Which is more basic: aniline or pyridine?
Please explain why, thank you very much.

You can look at the pKb values.  It turns out pyridine is more basic.  This is difficult to predict from looking at the Lewis structures of the molecules.  What you should see is that the lone pair on pyridine is on an sp2 hybridized N.  This makes it less basic than an sp3 N.  The N in aniline is sp3 but the lone pair is in resonance with the pi electrons of the phenyl ring, lowering the basicity.


text:
Why do NMR coupling constants change with temperature?
Why do coupling constants change with temperature in NMR?

Could we have some additional background information? For example, how do coupling constants respond to temperature? Many folks (myself included) may have only passing knowledge of this phenomenon. I know that they are temperature dependent (and I can probably posit why they are temperature dependent). However, I do not know if the coupling constants get larger or smaller as temperature increases (or maybe both depending on structural elements).

As it stands, your question is hard to answer without this information. Do you want to know why coupling constants change with temperature? Do you want to be able to predict the change given the structure of a molecule? Do you want to use the temperature-dependent change to learn something about structure?

Alright. May we change it to "Why do coupling constants in NMR spectroscopy change with temperature?"

Let’s use vicinal 3-bond $\ce{H-C-C-H}$ proton coupling constants as an example.  Coupling is a through bond phenomenon and as such the coupling constant will change depending if we have better (the coupling constant will increase) or worse (the coupling constant will decrease) overlap between the two $\ce{C-H}$ bonds involved.  Overlap is maximized when the two, vicinal $\ce{C-H}$ bonds are aligned in an eclipsed conformation ($\ce{H-C-C-H }$ dihedral angle =0°) or anti ($\ce{H-C-C-H }$ dihedral angle =180°)  conformation.  When the dihedral angle is ~90° the overlap of the two vicinal $\ce{C-H}$ bonds is minimized.  This thinking is embodied in the Karplus equation which relates vicinal coupling constants to dihedral angle.  The following graph illustrates this dependence of the vicinal coupling constant on the dihedral angle.

image source
All of this tells us that an experimentally observed vicinal coupling constant will be dependent upon the relative populations of the eclipsed, staggered and anti conformations. Said differently, the vicinal coupling constant will be an average of each of the individual coupling constants for each conformation, weighted by the relative population of that conformation. Since the relative conformer populations change with temperature, the coupling constant will also change with temperature.

One reason that coupling constants vary with temperature is because the alignment of the coupled atoms changes with temperature.
For example, in 1H NMR, proton-proton coupling between vicinal protons (HCCH) can be estimated by the Karplus equation, in which the coupling is mainly a function of dihedral angle between the two separate C-H bonds.  However, the degree to which the molecule can rotate freely around the C-C bond is a strong function of temperature.  Thus at high temperatures the observed coupling should be an average of all possible dihedral angles, while at lower temperatures where particular conformations are stable on NMR time scales, the Karplus equation can predict coupling directly.
This is just one reason for the T sensitivity.  I hope other chem.se users with more experience in NMR can provide better, more detailed answers.


text:
Which method can be applied to separate ferrofluid liquid from oil?
I want to know if there is any substance, liquid or any other stuff to separate ferrofluid from oil after mixing them.

Typical ferrofluids are suspension of nano-sized magnetic particles in oil. If you add more of the same oil and mix thoroughly, separation is impossible.
If you add a chemically different oil, you might be able to remove this compound from the mixture again by distillation in vacuum.

You might be able to recover the magnetic particles by extraction with a suitable solvent (thinking of petroleum ether or hexanes) in order to get rid of the oil. A frit might come handy.