Intrinsic Reaction Coordinate calculation-- both sides of TS lead to same minimum

Be sure you have really obtained TS, that means stationary point, i.e. all gradients are zero. Than, you should have one imaginary frequency corresponding to the motion along the reaction coordinate. If this is true and you still get both IRCs going to same minimum, your assumptions regarding the shape of Potential Energy Surface (PES) could be wrong.

It is easier to find the minima first, so be sure you have well optimized reactant, product, and if by any means reasonable, also intermediates. It is possible that there exist some high energy intermediates, which complicate the whole story (if you are not aware of them).


Calculate the molarity of Na+ ions in 20 mL of a .130 M solution of Na2Cr2O7?

$0.0052$ is the number of moles of $\ce{Na+}$ ions in solution, but to find the molarity, you need to divide by $\pu{0.02 L}$. Therefore, the final answer is $\ce{2\cdot 0.130 \cdot 0.02 \cdot \frac{1}{0.02} = 0.260 M}$


You might find this table of Wood Densities helpful. For example, note the density of balsa and that of Lignum Vitae. In fact, Lignum Vitae is so dense that it has been used for bearings on ships and even nuclear sub!


Basically there are only 3 ways of increasing the pressure.

1. By decreasing the volume of the gas (i.e. decreasing V)

2. By adding more amount of gas (i.e. increasing number of moles "n")

3. By increasing the temperature of the gas (i.e. increasing T)

Now your question being, would increase in pressure "P" keeping "V" and "n" constant result in the rise in temperature "T". So at the first place you need to increase the pressure.

By keeping V and n constant point 1 and point 2 discussed above is ruled out.

Now you want to observe T from changing P that means T, i.e., temperature over here cannot be used to increase P because you want to see how temperature varies with pressure, i.e. temperature here is to be observed.

So temperature cannot be used here to rise pressure. How would you increase pressure now? Out of the three factors which could have increased the pressure, two of the factors are constant and the last factor, i.e., temperature factor is to be observed and hence cannot be used to rise the pressure. So how will you now increase the pressure? It's not possible to increase the pressure without using any of the above factors.

As you are not able to increase the pressure in the first place, then how are you expecting to see the variation in temperature?


How to determine optical activity of a compound without asymmetric carbons theoretically?

A substance will be optically active if it is not identical to its mirror image. Check out this great answer to see why. So the dullest and brute-force-est way to find out whether a molecule is chiral is to draw it, to mirror it, and then to attempt to overlay image and mirror image using only rotation. (Remember to draw molecules three-dimensionally!)

But there are a few shortcuts. You can look at the molecule as a whole and check out what symmetry features it has — determine its point group. If a molecule has a centre of symmetry (i.e. inversion results in the identical molecule), a plane of symmetry or improper rotation (which is a rotation followed by mirroring along a plane orthogonal to the rotation axis) it is achiral. If you cannot find any of these symmetry elements, the molecule is chiral.


Is the polarisation of halogens of higher atomic number easier because the electrons are further away from its core?

Short answer: yes.
The Wikipedia article on Polarizability has a nice section on this, which I would like to supplement with further quotations from its cited source.$^1$
pg 24

Since electrons are charged particles they respond to electric fields.
...
The electron cloud is said to become polarized in response to the electric field. The ability of the electron cloud to distort in response to an external field is known as its polarizability.

pg 25

We define the polarizability of a molecule as the magnitude of the dipole induced by one unit of field gradient, which works out to be in units of volume. Often, the larger the volume occupied by the electrons, the more polarizable those electrons.

My understanding is that it's less that the electrons are farther away from the nucleus in higher elements, but rather that they occupy a larger volume and are therefore more greatly affected by the electric field.

There is probably a nice analytical expression for the relationship of the polarizability to the terms we're used to seeing in the Schrodinger equation, but I'm not aware of it. In my experience, it is typically calculated numerically using perturbation theory. I did find an example where they worked through the perturbation solution for a particle in a box being subjected to an electric field here, section 4.1.1.

$^1$ Anslyn, Eric V. and Dougherty, Dennis A. Modern Physical Organic Chemistry. University Science Books, California, 2006.


What is the significance of "magic numbers" of atoms in nanoparticles?

As indicated in the comment above, "magic numbers" (i.e., atom counts with much higher stability) have been known in atomic clusters of many types for a long time.

Each magic number has a specific geometry. Some are non-spherical, so it's hard to make generalizations about surface area or lattice structure.

A very famous non-gold example is buckminsterfullerene ($\ce{C60}$). Multiple groups had seen the signature of 720 Daltons in mass spectra of soot for years before the official discovery. There are other magic numbers for $\ce{C_n}$ clusters.

In nanoparticles and nanoclusters, this is an active area of research, but the main difference is stability. The so-called magic-numbers are simply more stable. (Case in point, the thiol-termated gold nanoparticles were created by heating up the samples until only the most stable ones survived.)

While I'm not an expert, I suspect the increased stability likely means less reactivity, lower catalytic activity (i.e., the species is more stable, less reactive so less likely to activate).

I'll give a comparison with Hückel's 4n+2 $\pi$ electron rule. We know that benzene is aromatic and has increased stability relative to the 5-membered and 7-membered conjugated hydrocarbon rings. It's less reactive and thus requires specific reaction chemistry.


Where did the rest of the elements come from?

By the "original periodic table" I assume you're talking about the "elements ocurring in measurable quantities in nature".--everything till Uranium sans Technetium and Promethium.

The method of preparation of these elements is pretty much the same -- bombardment. Just the projectile and substrate differ.For example, the (accidental) neutron bombardment of Molybdenum in old cyclotron parts led to the discovery of Technetium.

There are some other sources as well. Analyzing fission products can get you new elements--that's how Promethium was discovered. The process of fission creates smaller nuclei--but there is a shower of neutrons created as well (which can bombard everything in sight). Most of the discoveries have been due to the "shower of neutrinos", because generally all "smaller nuclei" are of already-known elements. Promethium was synthesized by the "formation of smaller nuclei", though.

Since fission occurs in nature (Usually it's just a little bit, though we sometimes have natural fission reactors), we actually have all the elements up till californium existing in minute quantities in nature (source).


There have been a number of interesting papers just this year on this. Simon Duckett from York is leading the charge, and has recently demonstrated successful hyperpolarisation of a number of amines, including ammonia, as well as a number amides. Previously, a limiting factor in the success of the SABRE method has been the ability of the substrate to coordinate to the catalyst centre. Recent results show that this can be overcome using a relayed polarisation transfer method using a co-ligand.

For a couple of examples, check out:

Chem. Sci. 2018, 9 (15), 3677–3684
Sci. Adv. 2018, 4 (1), eaao6250
anything else by Simon Duckett in recent years


This is the a peak from one of the vinyl protons on an allyl group. This splitting pattern is an ABX pattern from the bold hydrogen below. It is technically a triplet of doublets of doublets.

$$\ce{R-CH2-C}\textbf{H}\ce{=CH2}$$

Check out the following simulated NMR spectrum of 3-chloro-1-propene (allyl chloride) from nmrdb.org

Now image if the coupling constants are a little different and the structure collapses into what you have.


How to separate two liquids with both having the same properties?

Typical vegetable oils are not just one chemical compound, but a mixture (after removing everything that is not) of glyceride triesters. The chance that there is some overlap between your two oils is big, so the answer to your immediate questions likely is no, completely impossible.

Even if there is no overlap in the chemical composition, you would probably still have to separate the oil mixture into all its components, and then know how to mix them together again to get the original two oils.

The latter is possible, of course, with a reasonable analytical effort, the former is close to impossible to perform on a macroscopic sample. A GC(-MS) can easily separate (and identify) the compounds, but that's a few billion molecules or so. Doing liquid column chromatography would be very tricky, with any reasonable apparative effort would give you a few milligrams of each compound, and perhaps there are still a few you can't separate.

Distillation is impossible (boiling points are high and too similar), fractionation by crystallisation would only work well for some of the triglycerides, etc. All in all it's a bad mess. Don't mix two oils and try to separate them afterwards. ;-)


This elimination is unfavorable because C--C bonds are generally quite stable. Decarboxylation would involve the formation of an unstable carbanion before proceeding to the elimination as you have outlined. (Note, this is more favorable under different functional group conditions, namely when a neighboring carbonyl can stabilize the carbanion through the formation of an enol  https://www.masterorganicchemistry.com/tips/decarboxylation/ ) So, the formation of the alkyne is favored here over the elimination of CO2.


The application of MO theory to (electron-deficient) cluster compounds such as boranes is generally summarised in a set of rules, known as Wade's rules. We can look at the closo structure $\ce{[B6H6]^2-}$ in more detail to see where the two negative charges come from.
In closo boranes, each BH fragment points outwards from a vertex of a polyhedron. For $\ce{[B6H6]^2-}$, it turns out that this polyhedron is the familiar octahedron, which has six vertices.
What we must do now is to count how many electrons and how many orbitals each BH fragment contributes to the entire cluster. This can be done fairly easily:

Here we have neglected the boron 2s orbitals. One 2p orbital on boron has the correct symmetry to overlap with the hydrogen 1s orbital, forming bonding (blue) and antibonding (red) orbitals; the other two 2p orbitals remain non-bonding (gold).
So, each BH fragment contributes two valence electrons to the cluster. Since there are six BH fragments, the total valence electron count is (so far) 12. Saying this implies that the boron 2s electrons aren't valence electrons, which is, in a sense, true; they do not participate much in the delocalised bonding in the entire cluster.
If we use the three bonding and nonbonding orbitals to form properly symmetrised MOs across the whole cluster, it turns out we get seven bonding MOs across the entire cluster. I'm not keen to ChemDraw these MOs, so I've taken a picture from Housecroft's Inorganic Chemistry (4th ed.):

The lowest energy MO ($\mathrm{a_{1g}}$) comes from an in-phase combination of all six bonding fragment MOs. The three next lowest ($\mathrm{t_{1u}}$) come from overlap of two bonding fragment MOs with four nonbonding 2p orbitals on the other boron atoms. Three more ($\mathrm{t_{2g}}$) come from "sideways", or "tangential", overlap between four boron 2p orbitals.
In general, it turns out that for a polyhedron with $n$ vertices (at least, where $n \geq 6$ — see comments), there will be $n+1$ bonding MOs which can be made using the fragment MOs at each vertex.
Now, in order to get a stable cluster, we need to be able to fill all these $n+1$ bonding MOs with electrons. In the case of $n = 6$, we have 7 MOs to fill, and so we need 14 electrons. However, as established previously, we only have 12 valence electrons!
The solution is therefore to tack on two extra electrons, and that's why the closo borane requires a (2−) negative charge. If it didn't have this negative charge, it wouldn't be able to fill all of the bonding MOs in an octahedron, and it's quite likely that it would choose to adopt a different geometry that is no longer closo.


You have multiple options.  All of them assume that the distillation setup is empty, at ambient pressure.

Of course, like mentioned in the comment by @Waylander, put a little of (silicon) grease on the glass joints prior to the distillation, just enough that they become transparent.  There are multiple kinds of grease, suitable for different temperatures / pressures during the application.
If grease is not an option, there are sleeves made of PFTE / teflon to be put on the joints.  They are available in different diameters, and forms, e.g.

(source, source)

Especially for vacuum distillations, but not limited to this: the best remedy for this issue is to dismount the distillation right after the recovery of the product fraction(s), while the parts (e.g., column, head) are still considerably warm.  Otherwise, once cooled to rt, the setup then often will stuck much better than ever intended.  Depending on size of the distillation and remaining temperature of these pieces, wearing gloves (to protect your hands against the heat) may be helpful.  Then, it is better to allow the warm pieces to cool to rt while being put e.g. on cork rings.
«Forgotton» and already stuck glass joints may be detached from each other if you immerse them in the basket of a ultrasound bath.  Addition of some of detergent to the water bath helps while occasionally gently pulling the pieces apart.
If you have the dexterity (some develop this over time / by experience), wear both gloves to protect your hands against heat and goggles: you may heat the circumference of the outer wall of the glass joint with a heat blower or even a heat gun / pocket burner.  The aim here is that the outer part of the joint dilates quicker than the inner part.  It is the most risky variant because uneven heating may shatter your distillation into pieces; especially if the parts joined together are made of different types of glass with different expansion coefficients.  It this is not for you, just ask your local glass blower for assistance.


Reduction of conjugated double bond by sodium borohydride

Scifinder shows lots of reactions such as you describe. Often (but not always) copper is used to help favour the 1,4-addition (alkene reduction) process.

The simplest example I can see is the reduction of 3-penten-2-one[[2]] - when treated with NaBH4 the majority (89%) is reduced at the ketone only, while 11% of the doubly-reduced product is formed:

Obviously different substrates would give different product ratios, but certainly this is a plausible (and precedented) reaction.


Role of substituents in radical initialized polymerization

The compound IV is a great compound for cationic polymerization , the explanation is the oxygen which is giving electron with resonance : The form where the oxygen has a plus charge and the end carbon a minus is the reason why this compound will be able to start a radical polymerization but with a really low constant of polymerization (kinetic constant), and also the reason why the cationic polymerization will have high constant.

Compounds with strong (electron donor and tiny) resonance will not be good for radical mechanism, but if you put a phenyl group behind the ether function it will be better because the resonance effect will be "diluted".


How can we tell if an ion is amphiprotic

To be amphiprotic means that the chemical species can donate or accept H+ ions.

$$\ce{H2CO3<=>[H+] HCO3^- <=>[OH-] CO3^{2-}}$$
$$\ce{H3PO4 <=>[H+] H2PO4^- <=>[OH-] HPO4^{2-}}$$
$$\ce{CH3COOH <=>[H+] CH3COO^- <=>[OH-] \text{No Change}}$$

Thus since the acetate anion can't donate a proton, it is not amphiprotic.


How to calculate percent error for absolute zero?

It sounds like this is one of those canned labs that ask you to do standard things, even when they do not apply.

As you noticed, the percent-error formula needs you to use a value in the denominator.  If you use 0K in the denominator, you get a divide by zero.  If you use your 15K extrapolation, the percent error will always be 100%, no matter what the error actually was!

In such cases, I would look at the percent error with respect to your data.  Why?  Well, because somebody said you need to calculate a percent error and you might as well use a real datapoint rather than divide by 0 or some other silly result.  If you choose your coldest datapoint, you can say something about the percent error of the extrapolation process at the least.  Thus, if your coldest temperature recorded was 2C, you could say "My extrapolation suggests that absolute zero is 260C below my coldest recorded value.  The actual absolute zero is 275C below my coldest recorded value.  This gives me a 5.7% error on my measure of absolute zero relative to my coldest recorded value."


What is insoluble in drinking alcohol, but edible?

Ethanol does not dissolve chocolate, which consists mostly of butter and sugar. These mini alcohol filled chocolates are very common party candies. You could try inserting one of these into your cupcakes after you have baked them.

You could also try making a mold for a gelatin-based shot cup, as gelatin is also insoluble in ethanol. Gelatin is capable of absorbing alcohol, however, so I do not recommend making these too far in advance, lest they turn soggy.


In TOF Mass Spectrometry how does the detector 'know' the difference between ions of different elements?

I suppose that you are referring to a Wiley–McLaren time‐of‐flight mass spectrometer [Rev. Sci. Instr. 65, 3344 (1994)] which is basically a tube containing an electrode stack for accelerating the ions toward a detector. As mentioned in the comments, one can discriminate different masses if one uses a pulsed experiment, for instance using pulsed lasers or pulsed electric fields to extract the ions.

Another option is to add a magnetic field to the setup in order to bend the trajectory of the ions. The radius of the trajectory depends on the mass-to-charge ratio of the ion and one can record spectra as a functoin of the magnetic field to get a certain mass selected or use a position-sensitive detector to measure several masses simultaneously.

Yet another solution is to use quadrupole–time-of-flight mass
spectrometry, where radio frequencies are applied to four electrode rods placed on the circumference of a circle. Depending on the frequency that is applied to the rods, only ions of a certain mass-to-charge ratio make it out of the quadrupole and the qaudrupole thus serves as a mass filter.

Note that all these techniques are sensitive to the mass-to-charge ratio only and one cannot distinguish between, e.g., O$^+$ and S$^{2+}$.


How do oxide formations affect voltaic cells?

There are several questions in your text.
First, you imagine that "a metal's ability to form an oxide layer is related to its reactivity". No. It is not. Aluminum is rather reactive with water. But sodium is more. And metallic sodium is not protected by any oxide layer. On the contrary. Sodium reacts dangerously with water. Aluminum is a strange metal, as it forms an oxide that has the same density as the metal. So this oxide can perfectly cover the surface of the metal.
Then you ask about the difference of reactivity between zinc and nickel. I cannot answer this question, as I don't know enough of the nickel chemistry.
Then you say that aluminum oxide reacts with acids. It is right, but this reaction is extremely slow. If some aluminum piece or foil is put into a solution of $\ce{HCl}$ made by mixing the concentrated solution ($36$%) by an equal amount of water, you will have to wait a couple of minutes before the oxide layer is dissolved according to : $$\ce{Al2O3 + 6 HCl -> 2 AlCl3 + 3 H2}$$ And this last reaction is so slow that the expected reaction producing hydrogen gas happens only two or three minutes later according to $$\ce{2 Al  +  6 HCl  ->  AlCl3  +  3 H2}$$ In this case, the production of gas is violent and highly exothermic, but it only happens with delay.
At the end, you ask why "I even got a voltage in the aluminum cell if the oxide layer does not allow any aluminum ions to be transferred". This is a good question, for which I don't have a real answer. It may be due to some default of the alumina layer.


Why do strong acid-weak base reactions "essentially go to completion"?

This is a matter of equilibrium. Let's consider the  acid-base reaction of $\ce{HCl}$ and $\ce{NH3}$:

$$\ce{NH3 + HCl <=> NH4+ + Cl-}$$

Which can be broken apart into:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

$$\ce{HCl + H2O <=> H3O+ + Cl-}$$

For these reactions at equilibrium:

$$K_\mathrm{eq}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \hspace{1cm} K_\mathrm{b}(\ce{NH3})=\ce{\frac{[NH4+][OH-]}{[NH3]}} \hspace{1cm} K_\mathrm{a}(\ce{HCl})=\ce{\frac{[H3O+][Cl-]}{[HCl]}} $$

$$K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) = \ce{\frac{[H3O+][Cl-]}{[HCl]}} \cdot \ce{\frac{[NH4+][OH-]}{[NH3]}}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \cdot \ce{[H3O+][OH-]} = K_\mathrm{w}K_\mathrm{eq}$$

$$K_\mathrm{w}K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) \Rightarrow$$

$$K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot \frac{K_\mathrm{b}(\ce{NH3})}{K_\mathrm{w}}=\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \Rightarrow$$

$$\mathrm{p}K_\mathrm{eq} = -\log \left(\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \right) = \log \left(\frac{K_\mathrm{a}(\ce{NH4+})}{K_\mathrm{a}(\ce{HCl})} \right) = \mathrm{p}K_\mathrm{a}(\ce{NH4+}) - \mathrm{p}K_\mathrm{a}(\ce{HCl})=9.24-(-8)=17.24$$

Just from the $\mathrm{p}K_\mathrm{a}$ difference of the acid and conjugate acid of our reaction, we figured out that at equilibrium the products are present in a $10^{17.24}:1$ ratio to the reactants, meaning the reaction has essentially gone to completion.


How to calculate the temperature from heat of neutralisation?

First, you need to recognize what the limiting reagent is.  Because NaOH and HCl react in a 1:1 ratio, and there is less HCl ($\pu{25mL}$ of $\pu{1.79M}$ is fewer moles than $\pu{50 mL}$ of $\pu{1.6M}$), HCl is the limiting reagent.
Second, given the amount of limiting reagent, how much heat will be released (or absorbed).  $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$
Third, you need to approximate that the solution has the heat capacity of water, which is $\pu{4.18 kJ/K L}$.
If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average.  So here: $$(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$$
So finally calculate the amount the temperature of $\pu{75 ml}$ of "water" will increase when $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$ of heat is added, and add this to $\pu{25.55 ^\circ C}$.


Pectin is a heteropolysaccharide rich in galacturonic acid:

The carboxylic acid in galacturonic acid gives pectin the ionizable character. Pectin can react with bases to form carboxylate anions along the polysaccharide chain.

There exists a sugar-derivative and polysaccharide with opposite behavior - they react with acids to produce cations along the polysaccharide chain. The modified sugar is glucosamine, and the polysaccharides are chitin and chitosan(http://en.wikipedia.org/wiki/Chitosan). The amine ($\ce{NH2}$) group on glucosamine is basic and will react with acids to become cationic.

Glucosamine is used as a dietary supplement, and chitin and chitosan are found in various foods, like fungi and shellfish.


Modelling planar molecules and their evolutions

One-dimensional Fabre and Bechgaard salts, TTF-TCNQ,

http://www.chem.tamu.edu/rgroup/dunbar/Publications/2002-147.pdf
platinum blues

DOI:10.1002/asia.200700370
hexaphenylbenzene discotic liquid crystals
http://www.crcpress.com/product/isbn/9781439811436
"Chemistry of Discotic Liquid Crystals: From Monomers to Polymers," Sandeep Kumar (2010)

http://link.springer.com/chapter/10.1007/978-94-017-2319-0_16
middle
http://physics.aps.org/synopsis-for/10.1103/PhysRevB.82.180401
  Organic magnets

http://pubs.acs.org/cen/topstory/8201/8201notw8.html
Ladderanes

stacks
http://www.google.com/patents/WO2011017711A2?cl=en
http://www.halcyon.com/nanojbl/Images/ConPoly3.gif
and a central hexacoordinate silicon with oxygen bridges


Reference for electronegativities of different metal oxidation states

Pearson conveniently lists cumulative experimental data in the 1988 paper [1], referrring to the earlier work of Moore [2].
Selected values of $I$ (ionization potential), $A$ (electron affinity), $χ$ (absolute electronegativity – probably, that's what you are looking for) and $η$ (absolute hardness) for iron and cobalt cations are:

Table I. Experimental Parameters for Monatomic Cations (eV)
$$
\begin{array}{lcccc}
\hline
\text{ion} & I & A & χ & η \\
\hline
\ce{Fe^2+} & 30.65 & 16.18 & 23.42 & 7.24 \\
\ce{Fe^3+} & 54.8 & 30.65 & 42.73 & 12.08 \\
\ce{Co^2+} & 33.50 & 17.06 & 25.28 & 8.22 \\
\ce{Co^3+} & 51.3 & 33.50 & 42.4 & 8.9 \\
\hline
\end{array}
$$

So it looks like the relation is a bit different:
$$\ce{Fe^2+} < \ce{Co^2+} < \ce{Co^3+} < \ce{Fe^3+}$$
Complete table as a screenshot:

References

Pearson, R. G. Absolute Electronegativity and Hardness: Application to Inorganic Chemistry. Inorganic Chemistry 1988, 27 (4), 734–740. https://doi.org/10.1021/ic00277a030.
Moore, C. E. "Ionization Potentials and Ionization Limits"; Natl. Stand. Ref. Data Ser. (U.S. Natl. Bur. Stand.); 1970, NSRDS-NBS 34. (NIST - PDF)


The proper and straightforward way to solve nomenclature problems is to construct the systematic name for the given structure according to the current IUPAC recommendations. However, for such multiple-choice questions, it may be easier to check the given answers and try to draw a structure for each proposed name as follows – especially when the particular nomenclature rules that are considered in the exercise are not exactly known.

The name 1,1-diformyl propanal (A) can be ruled out since the atom ‘1’ (i.e. the $\ce{C}$ atom of the aldehyde group) of propanal has only one $\ce{H}$ atom that could be substituted; therefore it is impossible to place two formyl groups at atom ‘1’. Furthermore, after substitution of the $\ce{H}$ atom of the aldehyde group, the compound would no longer be an aldehyde.

The name 1,1,3-ethane tricarbaldehyde (D) does not make sense since ethane has only two $\ce{C}$ atoms; therefore it is impossible to place a carbaldehyde group at atom ‘3’ of ethane.

It is possible to draw structures for the names 3-formyl butanedial (B) and 2-formyl butanedial (C) based on the structure of butanedial:

Note that the resulting structures for 3-formyl butanedial and 2-formyl butanedial are actually identical, and both structures correspond to the structure that is given in the question. The correct choice is 2-formyl butanedial (C) rather than 3-formyl butanedial (B) since the locant ‘2’ is lower than ‘3’. (Note that the correct spelling of the name is actually 2-formylbutanedial, i.e. without space.)

Anyway, even the name 2-formylbutanedial is actually not in accordance with current IUPAC recommendations. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), aldehydes can be systematically named in various ways:

  P-66.6.1 Systematic names of aldehydes

  Aldehydes are systematically named in three ways:

  (1) substitutively, using the suffixes ‘al’ for $\ce{-(C)HO}$ and ‘carbaldehyde’ for $\ce{-CHO}$;

  (2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;

  (3) by using the prefixes ‘oxo’, denoting $\ce{=O}$, or ‘formyl-’, denoting the substituent group $\ce{-CHO}$.

In particular,

  P-66.6.1.1.2 The suffix ‘carbaldehyde’ is used when more than two $\ce{-CHO}$ groups are attached to an alkane.

Therefore, the preferred IUPAC name for the compound given in the question is ethane-1,1,2-tricarbaldehyde.


The transparency of a plastic does not depend on its nature but on how it is made.

To get a transparent plastic it must be cooled very fast when it is in molten state.
If the cooling is too slow, the polymer begins to order it self in more or less crystalline domains.

When it is not isotropic anymore, it is also not transparent.


Why we make pyrogallol alkaline to absorb oxygen?

Because that makes them $\ce{\dots-O-}$, which is even better for oxidation. See, this thing is just ready to give away two electrons and become the corresponding quinone without any structural rearrangement.


Extent of dehydration by concentrated sulphuric acid

The most stable Alkene is the most highly substituted Alkene. (d) gives a tri-substituted alkene, all others give the less disubstituted alkenes. Most stable Alkene (d) therefore more of it is formed! (Edited thanks to Bon for highlighting my error)

Also a factor to keep in mind is conjugation. (a), (b) and (d) will have conjugated double-bonds, while (c) will not. This conjugation also adds stability to the system. But the major stability contributor is the higher substituted alkene in (d).


For resorcinol, the naming is covered in a footnote to the Wikipedia article:

  "Wir nennen den neuen Körper, da wir gefunden haben, dass er auch aus dem Ammoniakgummiharz erhalten werden kann, Resorcin, um an seine Entstehung aus Harzen und seine Beziehung zum Orcin zu erinnern." (We name the new substance, since we've found that it can be obtained from ammoniated resin gum, resorcin, in order to recall its creation from resin and its relation to orcin [i.e., orcinol].)

Which is itself a quote from  H. Hlasiwetz and L. Barth (1864) "Ueber einen neuen, dem Orcin homologen Körper", Annalen der Chemie, 130 (3) : 354-359.

At some point the ending -ol got added, to reflect the alcohol moieties in the compound.

Then name orcinol itself is derived from its association with orcein, a dye substance derived from orchil lichens.


The advantages of OFET's over traditional CMOS transistors all derive from how OFET's can be manufactured.  Traditional transistors are manufactured using vacuum deposition techniques.  These are slow, time consuming, batch processes, hence the higher cost for electronic components produced by such methods.

OFET's can be manufactured using standard coating technologies that have been widely used in industry for a long time, and are well understood.  Therefore, the manufacturing equipment needed is readily available. With this coating equipment uniform layers ranging from sub-micron to tens of microns (you don't need thicker layers for an OFET, but the coating technology would allow you to coat much thicker layers if you wanted to) can be coated at high speed on thin, flexible, web supports such as aluminized mylar. Multiple layers can be coated by making multiple passes down the coating line, into a drying oven and then on to the next coater.  Thousands of OFETS can be obtained from a small, coated, mylar spool. Since the process is continuous and high speed the manufacturing cost of the device is much reduced compared to similar CMOS devices.  Each layer contains an active organic compound dissolved in a low-cost plastic (for example, polycarbonate) binder, so material costs are reduced as well.  Another key advantage is that since organic materials, plastic binders and flexible web materials are used, the final product is flexible and can be easily cut to any desired shape.  This allows the manufacturer to use the product in new applications where flexibility (newspapers) and shape (electronic clocks) might preclude the use of rigid CMOS devices.


Why is it wrong to calculate the amount of oxygen directly from the products of a combustion reaction?

Let's first write the chemical equation for the reaction (I will leave it unbalanced because the stoichiometry is irrelevant):
$$\ce{C_xH_yO_z + O2 -> CO2 + H2O}$$
To do this question, we need to find the mole ratio of carbon, hydrogen and oxygen in $\ce{C_xH_yO_z}$, then we can find its empirical formula.

From the above equation, it is clear that all the carbon in $\ce{CO2}$ originated from the compound and that all the hydrogen $\ce{H2O}$ originated from the compound. However, all the oxygen in $\ce{CO2}$ and $\ce{H2O}$ didn't just come from the compound, it came from the compound and gaseous oxygen. Therefore, from your method, you are just finding the total amount of oxygen in the reactants, not the number of moles of oxygen in the compound.


Syntactic vs Semantic validity of molecular graphs

Generative machine learning methods in chemistry are typically tasked with generating new molecular structures. Think of it as 'dreaming up molecules.'

We see these kinds of ML models with predictive text programs.. if I'm typing, my phone attempts to guess the word or complete my sentence. (NB this doesn't work so well with molecule names or scientific terms.)

There are a wide variety of ways to represent a molecule for such a generative ML method. One of the common ways is via SMILES in which the molecular graph is encoded as a text string:

c1ccccc1  benzene
CC(=O)O   acetic acid

In this case, syntax can be a problem. SMILES uses parenthesis to represent branching, and they must match. Numbers are used to represent ring closures, which must match as well.

Consequently, many generative ML method generate syntactically incorrect molecules (ignoring questions of chemical valence, etc.)

c1ccccc2 error
CC(=OO)) error

There are solutions. Noel O'Boyle and Andrew Dalke suggested DeepSMILES which encodes a new syntax that's resistant to such issues.

As you mention in your answer errors in connectivity and/or valence represent semantics - that is what makes a chemically viable molecule (i.e., carbon does not take 5 bonds, isolated oxygen atoms aren't stable, etc.)


What qualifies to be considered as a lone pair-lone pair repulsion?

When applying VSEPR, imagine that the various electron pairs occupy "orbitals", in essence volumes shaped like balloons, protruding from the central atom (in this case chlorine). The electrons ("orbitals") repel each other.
Which repulsive interaction is strongest?
One assumes that, placed in equivalent relative positions (angles), LP-LP (LP=lone pair, BP=bonding pair) repulsion is strongest, followed by LP-BP and then BP-BP. The idea is that bonding electrons are more distant from the central atom, whereas nonbonding (LP) electrons are held tightly by the central atom and therefore represent a more "concentrated" charge, and therefore repel other electrons most strongly. The repulsion is also stronger the smaller the angle between the electron pairs.
The strength of repulsion determines the order in which to prioritize the interactions, but this can be a little tricky, as your example shows, because it is a "non-linear" problem. When more than one pair of LP is involved, the geometry should minimize LP-LP repulsions followed by LP-BP repulsions, but there is a tradeoff. While (c) places both LPs on opposite ends ($180^\circ$) and would seem to minimize the repulsion, there is a bit of a tradeoff between LP-LP and LP-BP repulsions. Geometry (c) results in a lot (6) of LP-BP repulsions at right angles ($90^\circ$), whereas geometry (a) has 4 LP-BP repulsions at right angles. This leads to positioning of the LPs in equatorial (!) positions, even though LP-LP distance is not maximized.
Such arguments are not transparent, they are based on a crude method that only approximates the "real" electron distribution. The experimental geometry is in fact a distorted version of (a), with the LP-BP angles slightly greater than right angles ($>90^\circ$).
My practical advice is to simply remember that according to VSEPR lone pairs are always placed in the equatorial plane in trigonal bipyramidal electron pair geometries (those with 5 pairs). See for instance the following table:


Why is fluorine deactivating while OH is not?

F is the most electronegative element of the periodic table.Due to its -I effect, it shifts the electron density of any alkyl group towards itself, resulting in less density of electron cloud over the alkyl or phenyl group.This deactivates the ring for electrophilic substitution reactions.However , oxygen also being an electronegative element , has capacity to donate its lone pair electrons forming a back bond in case their is a deficiency in alkyl group.In case of cyclic rings that is in aromatic structures , the lone of oxygen is in  resonance with the ring during attack of the reagent , that is during formation of sigma complex.However this is not shown by fluorine due to its high electronegativity.Do not get confused with other halogens.Cl ,Br and I are ring activating groups as they also can form the same back bond as oxygen.


How can I calculate the standard reaction enthalpy if I lack one formation enthalpy?

The standard enthalpy of formation of $\ce{CuS}\,\mathrm{(s)}$ is -48.4 $\mathrm{kJ}\cdot\mathrm{mol}^{-1}$. See this reference  for your requested data and more.


Is there a name for the opposite reaction to the dissolution?

Effervescence if there are bubbles and

Degasification more generally speaking.

There is also the term outgassing, but that term is broader than just gas coming out of liquid solution.


I'm not an expert in Fenton chemistry but it appears that difficulty in getting the Fenton reagent to reduce COD to below 500 ppm is not unusual.

Does your particular wastewater have  chelants such as phosphates, EDTA, formaldehyde, and/or citric/oxalic acids?
Perhaps at long reaction times all the iron you add is being precipitated as iron (hydr)oxides.  Could you lower the pH further?

  Reaction rates with Fenton’s Reagent are generally limited by the rate
  of $\ce{.OH}$ generation (i.e., concentration of iron catalyst) and less so
  by the specific wastewater being treated. Typical $\ce{Fe:H2O2}$ ratios are
  1:5-10 wt/wt, though iron levels < 25-50 mg/L can require excessive
  reaction times (10-24 hours). This is particularly true where the
  oxidation products (organic acids) sequester the iron and remove it
  from the catalytic cycle. Fenton’s Reagent is most effective as a
  pretreatment tool, where CODs are >500 mg/L. This is due to the loss
  in selectivity as pollutant levels decrease...In addition to free radical scavengers,
  the process is inhibited by (iron) chelants such as phosphates, EDTA,
  formaldehyde, and citric/oxalic acids. Because of the sensitivity of
  Fenton’s Reagent to different wastewaters, it is recommended that the
  reaction always be characterized through laboratory treatability tests
  before proceeding to plant scale.

My tip would be to increase the concentration of both $\ce{H2O2}$ and $\ce{FeSO4}$, and, if possible, to make sure the pH is 3 (or maybe even lower).  If that doesn't work, do you think iron chelators could be present in your particular wastewater sample(s)?


In configuration interaction (and in coupled cluster) we typically think of excited determinant being caused by single particle excitations of the form $\hat{t}^a_i = \hat{a}^\dagger_a\hat{a}_i$ where $i$ labels a virtual orbital and $a$ labels an occupied orbital, causing an excitation of the Hartree Fock reference determinant $\Psi_0$.
The resulting CI expansion would go:
$$
\Phi_\text{CI} = \Psi_0 + \sum_{i\in occ\\ a \in vir} \hat{t}^a_i \Psi_0 + \sum_{i,j\in occ\\ a,b \in vir} \hat{t}^{ab}_{ij} \Psi_0 + \sum_{i,j,k\in occ\\ a,b,c \in vir} \hat{t}^{abc}_{ijk} \Psi_0 + \: ...
$$
Hence, in the calculation on Li mentioned above, we would not consider the excitation $\hat{t}^2_1 = \hat{a}^\dagger_2\hat{a}_1$ in the first place as it would annihilate the Hartree Fock determinant $|12>$ containing a particle in orbital 2.
By this logic the determinant $|23>$ is a singly excited determinant caused by $\hat{t}^3_1= \hat{a}^\dagger_3\hat{a}_1$, or class 1 as described by Helgaker.
Note:
As determinants $|23>$ and $|32>$ only differ by a multiplicative constant (-1) only one needs to be considered and the resulting coefficient/cluster amplitude will take care of the relative sign.


How quickly does sulfur hexafluoride mix with the atmosphere?

Up front: this depends on what you mean by "stay in". It would diffuse out of the container and, given the surroundings are closed, obtain a small equilibrium concentration.

Not that I could provide an exact answer to your question (it is rather a physics question), but this paper (paywalled) deals with the diffusion of the $\ce{N2-SF6}$ pair. The binary diffusion coefficient is reported to be 0.0945 cm2/s. In their experiments they used a container of d = 0.5 m, l = 1 m, introduced a small amount (2-5%) of $\ce{SF6}$ at the bottom of the vessel and measured the time it took to obtain the equilibrium concentration of the mixture. For this setup, this was ~15 h, with a reported time constant of 4-5 hours (I believe this is the time it takes to reach a value of [local $\ce{SF6}$ concentration]/[equilibrium concentration] of 0.5).

They used a lengthy formula derived from a one-dimensional model for the comparison of calculated and measured concentrations and in my opinion the measured data agreed quite well with the calculated ones.


What metals have been used in photography?

I wanted to contribute what I know so far:

Most photography processes are based on silver halides, which are light sensitive, decomposing into metallic silver and halogens. Here is a claim that silver salts are the most light-sensitive, but it is not well-supported. Silver processes include black and white, as well as more complex color processes like E-6 and C-41 which use dye couplers and color-sensitive layers. I don't know what makes the layers color sensitive, but the reaction is like $$\ce{AgCl + photon -> Ag(s) + {1\over 2}Cl_2(g)}$$ Apparently, after a photon is received, freeing a silver atom, this atom must combine with another free silver atom through diffusion over the lattice of the silver halide crystal before it can become part of the image. There may be a better reference, but this fact is mentioned in the photographic developer Wikipedia article.
Platinum and palladium photography exists. These are also based on metal halides, but they usually include ferric oxalate as a sensitizer. Also, they appear to be based on alkali salts like sodium tetrachloropalladate. Palladium is above platinum on the periodic table, and was used as a cheaper alternative. I saw some old black-and-white photographs in a museum recently, which were labelled as platinum or platinum-palladium prints.
Other metals with light-sensitive salts include mercury and gold, but I am not aware of these being used in photographic paper (although mercury saw early use as a developer in the Daguerreotype process). These undergo a different reaction than silver halides, at least in the case of mercury - undergoing disproportionation rather than releasing free halogen.
The Cyanotype process is based on iron, using both ferric ammonium citrate and potassium ferricyanide. It is the source of the name "blueprint". The light-sensitive compound is ferric ammonium citrate; the citrate reduces the ferric ion under exposure to UV light. The resulting ferrous ion then reacts with potassium ferricyanide to form insoluble ferric ferrocyanide (Prussian blue).

Light-sensitive compounds outside of chemical photography:

In biological systems, chlorophyll is mostly organic, but contains magnesium in a key role. This contrasts with the opsins such as rhodopsin which are responsible for vision. They are proteins bound to cofactors, I think the cofactor is always retinal which is organic - so no metals in vision as far as I know.
Light-sensitive synthetic organic molecules exist, such as aryl phthalate esters. However, I am aware of no photographic process based on organic molecules.
Glow-in-the-dark toys are usually based on metallic phosphors such as zinc sulfide and strontium aluminate.
Wikipedia has a category of light-sensitive chemicals which seems under-populated, but includes many inorganic compounds, as well as hydrogen peroxide and LSD.
Digital photography of course uses metals, with CCDs containing capacitors formed from metal-oxide-semiconductors. Whether this can be usefully related to the chemistry of chemical photography I do not know.

I'm not sure what the overall conclusion is. Why metals are important but not necessary; why silver seems to be uniquely sensitive to light, and how to classify and understand this uniqueness...


As @Waylander pointed out, it appears this reaction has not been performed and/or recorded in any literature, so it is quite dangerous to speculate.

But keeping that aside, A 3D perspective reveals that abstraction of protons from the methyl group in quite unhindered.

Hence, the triiodo intermediate is well anticipated.

However, a quick glance at spatial orientation of iodine atoms reveals the reaction may be dead slow in the next step.

Notice that the Burgi-Dunitz trajectory, which we may assume the incoming nucleophile to take, is hindered by the large iodine atoms and the methyl group.

It is quite safe to assume that the attack at the carbonyl carbon is unfavoured, preventing the release of the $\ce{CI3-}$, and ultimately $\ce{CHI3}$ never appears.

EDIT: Apparently there is some relevant literature available for similar compounds, as mentioned in this answer. Thanks to Mathew for searching and pointing it out.


Why would the following reaction not proceed through SN2 reaction?

The substrate in this reaction is an Acyl Chloride, which is highly susceptible to nucleophilic acyl substitution, more so than other carboxylic acid derivatives, so SN2 attack (as mentioned in the comments) is highly improbable. The mechanism for this nucleophilic acyl substitution is as follows:

With equilibrium constant $K_c >> 1$ favouring the products on the right.


Use a solid or liquid stationary phase in gas chromatography

Solid adsorbents in gas chromatography were used about 30-40 years ago. They are obsolete now. The efficiency of solids (or packed column GC) was much lower than coated columns. Most modern GC stationary phases are coated or sometimes chemically bonded  to the silica surface inside the capillary.

  how does the stationary phase not get carried away with the mobile
  phase? since a high-pressure gas is used doesn't the stationary phase
  just get carried along with the mobile phase, especially if it is
  liquid?

The viscosity of the coating is so high that it does not get carried away. Imagine a very viscous wax like substance stuck on a glassy surface. Smear it very thinly. The coating is now a few microns thick. Try blowing away thick waxy substance from a glass surface, can you ever get rid of it?

You may think that the pressure is high, but recall that Pressure= Force/Area. If the capillary area (i.e. cross section) is so small, the force is very small too. Capillaries diameters are on the order of microns. That is why you can use a silica capillary in GC. If we had used a wide bore capillary say 4 mm i.d., and used the same pressure with the same wall thickness, it would have shattered.


Acid chlorides and reduction by hydrogenation

Hydrogen gas and platinum metal simply work here because the C=O part
  of the acid chloride is flat.

It's more than just that.  Acid chlorides are significantly more reactive towards catalytic hydrogenation than other carbonyl compounds.  They are reactive enough that a poisoned catalyst must be used along with mild reaction conditions to achieve reduction to the corresponding aldehyde, and have the reaction stop there (Rosenmund reaction).

So the question becomes, "why are acid chlorides so reactive towards catalytic hydrogenation."  It is thought that because the acid chloride $\ce{C-Cl}$ bond is so reactive that the palladium inserts into this bond to form intermediates such the one pictured below.

Other typical carbonyl compounds can't form such an intermediate; consequently acid chlorides are noticeably more reactive towards catalytic hydrogenation. Anhydrides, where the carbonyl is roughly as reactive as an acid chloride's carbonyl towards nucleophilic attack, have not been reported (to my knowledge) to produce aldehydes under Rosenmund conditions supporting this mechanism involving a palladium intermediate.

Historical Note: Prior to the application of metal hydrides to the reduction of carbonyl groups, the Rosenmund reduction was a major synthetic tool for chemists.  It was one of the few methods available for the straightforward transformation of acids, esters, amides, etc, to aldehydes.


Make very accurate solutions without weighing

Are there any other ideas how to get exact concentrations using small amounts of the desired substance?

Every mass spectrometry lab should have microbalances to prepare samples and standards. A lab which can afford an LC-MS can also afford a semimicro balance or a microbalance. Such balances are meant to weigh 0.01 mg to 0.001 mg (with extreme caution).
Two simple options:
(i) Since you are apparently not the first user of LC-MS system in your group, ask how other senior researchers prepared standards in your lab?
(ii) Another (crude) option is that you try weighing by difference. You will tare your analytical balance to zero. By an analytical balance I mean a balance which reads to four decimal places in grams, this is the most commonly used in research labs. Place your standard container on the balance and read the mass. Carefully transfer approximate amounts to your desired container in which the standard will be prepared, and note the weight of the standard container again. The change in mass corresponds to the amount transferred. Practice it with some commonly available substance to see if this works and if your balance is sensitive enough or not. Balances age with time.


problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate,

There are several things which may be problematic in your approach.

For the stated problem you do not need to invoke Nernst equation. Simply use the balanced equation.
When potentiometric titration curve are to be estimated, it is far more convenient would use the half-cell instead of a full reaction. I am assuming the redox electrode is dipping in iron solution.

\begin{align*}
    E=E^{0} -\frac{0.0592}{5}\log{[Fe^{2+}]/[Fe^{3+}]}
\end{align*}

where $E^{0}$ is the iron half cell value. You should not calculate equilibrium constants from potentiometric titrations.

See page 17 for a full example Potentometric Titration clculation

I do not know when people will stop teaching normality. It is obsolete (18th century concept). Your normality conversions are incorrect.

  The problem mentions the following concentrations for the titration:
  Fe2+:0.1 N (I assumed it was 0.1/5=0.02 M)
  H2SO4: 1N (probably 0.1M for H+)
  KMnO4: 1N (0.2M MnO4- following the same steps)

1 N sulfuric acid will be half the molarity =0.5 M

1 N permanganate will be 1/5th the molarity = 1/5 M

Check the concept of normality in your textbook.


Where does the residue accumulate in the boiling point composition curve?

Let's start with the mixture $x_A$, which is made of 85% A + 15 % B. When heating this liquid, the first bubble vaporized appeared at a temperature just a little above $T'_A$, as given by the horizontal line. And the composition of the vapor is nearly pure $A$, with $1$% $B$. As a consequence, the liquid is loosing some $A$. its volume decreases, but as it does not loose $B$, its composition changes from $15$% $B$ to maybe $20$% $B$. As it has more B, it is necessary to heat the liquid to a higher temperature. The composition of the liquid is a point following the lower curve upwards to the right hand side of the graph. Meanwhile the composition of the vapor slightly changes and contains a little bit of $B$.
When the volume of the liquid has decreased by more than one half, the remaining liquid contains more $B$ than $A$. Its composition is $20$% $A$ and $80$% $B$. The boiling point of this mixture has increased : it is now about half way between the boiling temperatures of the pure $A$ and $B$ stuff. And the vapor gettin out of this liquid can be obtained in drawing a horizontal line from the point $0.8$ on the lower curve. This line cuts the upper line $y$ at a point $y_A$ = $85$%, and $y_B$ = $15$%. It is possible to continue heating the rest of the liquid. The amount of liquid is smaller and smaller, and the vapor has a composition following the upper $y$ curve.
Now, if you have a hot vapor containing the same proportion ($15$% B , $85$% A), and if you coli it down, the first drop of liquid happening at the beginning of the condensation has a composition corresponding the th horizontal line passing at $y_A, y_B$. So this liquid droplet contains $80$% $B$ and $20$% $A$.


In Diazonium Ion why does N contain 4 bonds?

One way to think about this structure, the benzenediazonium cation, is to consider the structure of benzonitrile - C6H5CN - just replace the positive nitrogen with a neutral carbon atom. This has the same basic electronic structure as the diazonium ion, but differs in that the carbon nucleus has one less positive charge than the nitrogen nucleus.

I don't think the concept of "coordinate bonds" is very helpful here. There are many different species with positively charged nitrogen, such as tetramethylammonium salts, that have four full covalent bonds to nitrogen.


What reaction is expectable from the atomized reactants?

The rate for a reaction is dependent upon the rate constant and, depending upon what type of reaction it is, the concentrations of the reactants.  The rate constant can be further broken down into an entropic factor (how likely are the reactants to collide in an orientation that will produce a reaction) and an energy factor (the activation energy).

Assuming for the moment that the reactant concentrations are such that they will not control the direction of reaction, and that the entropic factors don't differ dramatically for one pathway over another (that's usually the case for similar reactions), then the first reaction to happen will be the one involving the reaction with the lowest activation energy.  This will lead to the kinetically controlled product.

If the back (reverse) reaction from the kinetically controlled product also has a low enough activation energy such that it happens rapidly under our reaction conditions, then essentially we are setting up an equilibrium.  In this case, thermodynamics will eventually control the product distribution and we will eventually wind up with the product that has the lowest free energy, the thermodynamically controlled product.

Many reactions first produce the kinetically controlled product, but with time, the thermodynamic product becomes the major product.


How do we know if a compound is amphoteric or not?

Sorry to tell you but your teacher's opinion is not right, just because the word "react" is not enough. Usually you label oxides as "amphoteric". You may say that if an oxide reacts with an acid and forms a salt or if an oxide reacts with a base forms a salt then you can say that it is an amphoteric oxide. Take the example of aluminum oxide, when dissolved in HCl it will form aluminium chloride, but we can also dissolve it in NaOH, where it forms sodium aluminate.

However, we can broaden this definition further (good for advanced classes). One formal definition is "A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: H2SO4 is an acid when studied in water, but becomes amphoteric in superacids."

See: https://goldbook.iupac.org/html/A/A00306.html


"The problem I face is I have been taught that the pure liquids aren't included in the expression of $\mathrm{K}_{c}$ since they have a concentration of 1."

Liquids have a constant concentration (really activity), so in general, it is not helpful to include them in the equilibrium expression. However, that does not mean that the value is 1. Pure water has a concentration of roughly $55.5\ \mathrm{M}$. You can absorb this concentration into the equilibrium constant, since it is a constant. Therefore,

$$\mathrm{K}_{w} = \ce{[H+][OH-]} = K_{\mathrm{c}}\ce{[H2O]}$$

EDIT:

Just for the record, note that the autoionization of water is generally written:

$$\ce{H2O + H2O <=> H3O+ + OH-}$$

since we don't have naked protons in water.

This changes the equilibrium constant slightly:

$$\mathrm{K} = \frac{\ce{[H3O+][OH-]}}{\ce{[H2O]}^{2}}$$

but $\mathrm{K}_{w}$ is still the same.


How to measure quantities at standard state when this state is a hypothetical one?

The standard states are limiting states. In these limits you can write the chemical potential of the component as

$\mu = \mu^\circ+RT\log(x)$

where $x$ is a measure of the concentration of solute (for Henry's law, where the standard state is a hypothetical state related to the behavior as $x \rightarrow  0$) or solvent (for Raoult's law, where the standard state is approached as $x \rightarrow  1$). In other words, in the limiting states the activity goes to 1. The meaning of $\mu^\circ$ depends on the limiting law (hypothetical with properties of an infinitely dilute solution or pure solvent).

A standard state can be hypothetical (as may happen when Henry's law is invoked), but it can also be real (as in Raoult's law, for a pure solvent). It is a state in which the sample behaves ideally as defined by the above equation, in the limit of infinite dilution or of a pure solvent. This limiting behavior provides a convenient and predictable reference, like setting the potential energy to zero when particles are separated by an infinite distance. In the case of Raoult's law, the standard state is the (not particularly mysterious) pure substance. In the case of Henry's law, the reference state is found by extrapolating behaviour at infinite dilution to the standard concentration (1 molal or 1 M, conventionally). This may be a hypothetical state.

Using reference standard states (at 1 bar of course) can simplify combining or comparing data for different substances or conditions. Of course when considering real solutions, activities may diverge from concentrations and activity coefficients deviate from unity.

For a mixture of substances, for each component you can describe individual thermodynamic properties relative to the ideal standard states. By ideal it is also meant that individual solutes do not interact.


Formation of zwitterion from amino acids

The reason amino acids exist largely in their zwitterionic form at biological $\mathrm{pH}$ (usually around 7) is due to the $\mathrm{p}K_\mathrm{a}$ of the constituent groups. It's not that the oxygen 'wants' to lose a proton, but more that at that $\mathrm{pH}$ the equilibrium lies towards the deprotonated state (things are rarely 100% protonated/deprotonated).

In general, carboxylic acids easily lose their proton to form carboxylates (the anionic charge is delocalised over the three atoms making the proton loss 'favourable'), and amines act as bases, to pick up a proton.

At a $\mathrm{pH}$ of 7, the medium is basic enough to cause the carboxylic acid to be deprotonated, and the amine to be protonated. If you look at the diagram below, you can see that increasing or decreasing the pecan shift the equilibrium towards fully protonated (at low $\mathrm{pH}$) or fully deprotonated (at high $\mathrm{pH}$) amino acid.

Ionisation state of amino acids over 0-14 $\mathrm{pH}$ range, taken from Biochemistry, Berg (7 ed.)


Solubility of salt in heavy water vs. "regular" water

Well, consider a salt as $\ce{AB_s}$ in water you have :

$$\ce{AB}_s\rightleftharpoons \ce{A_{aq}^{q+} + B_{aq}^{q-}}$$

And $\Delta H_{s}=-\ce{U +\Delta H_{aq}}$ where $\ce{U}$ is the lattice energy.

When the salt needs to be solvate, $\ce{O-H}$-bonds of water molecules around this salt need to be rearranged. More, the "$\ce{DO}$-bonds" in $\ce{D_2O}$ are stronger than "$\ce{HO}$-bonds".$^1$ Then it will be harder to rearranged hence it will cost more energy to solvate the salt which will be then less soluble in heavy water than in water.

I found an article in the JACS and as you can see on the image below the maximum solubility of $\ce{KCl}$ in heavy water is a little less than in water (you can make a linear extrapolation) the difference is not quite importante but it looks it exists.

Look at the second reference $^2$.

$^1$ : https://arxiv.org/ftp/arxiv/papers/0706/0706.1355.pdf, p-5

   Nuclear quantum effects are particularly important in the different
  properties of light (H2O) and heavy water (D2O) where the more restricted atomic vibrations in D2O
  reduce the negative effect of its van der Waals repulsive core so increasing its overall hydrogen bond
  energy.

$^2$ : For the image, http://pubs.acs.org/doi/abs/10.1021/je60070a011


For a molecule to be chiral it must have non-superimposable mirror images.  Here is a drawing of the two mirror images for 2-bromobutane.

The chiral carbon atom is denoted by an asterisk.  In the case of 2-bromobutane there are 4 different substituents attached to it. The molecule is chiral, you can't pick up one of the mirror images and superimpose it on the other.

Note to think about:

4 different substituents may make an atom chiral, but the molecule as a whole may still be achiral

For example meso compounds

image source

some molecules without 4 different substituents attached to an atom may still be chiral .

For example, compounds with axial chirality such as chiral allenes (1,3-dimethylallene), chiral biphenyls, chiral helicenes, etc.

image source

image source

image source


(partial answer)

Interesting question.  Silver also forms an unusual "subfluoride", Ag2F.  My understanding is the Ag is NOT in dimers.  But the structure has sorts of "sheets" of silver in that all of the octagonal holes are filled but only in alternating layers.  This allows silver atoms to be very close to each other, similar to the metal.  Presumably there is a lot of silver-silver bonding going on.  But more like sheets of metal than dimers.

The explanation of Frenkel defects you usually hear is that the cation is small enough versus the structure to allow dislodging by radiation (i.e. not a thermodynamics distribution of defects but a provoked one). See the Wikipedia page for instance.  However defect models are idealizations.

It would further be interesting to know how often the Frenkel defect is associated with some higher conductivity.  A purely billiard ball dislodged model would be less prone to conductivity than one where silver-silver bonding is the driver.


How do amino acids look like at the isoelectric point?

By Henderson relation we have

$$\mathrm{pH=pI=p}K_\mathrm{a}+\log\left(\frac{[\ce{A-}]}{[\ce{AH}]}\right)$$

So if $\mathrm{pH - p}K_\mathrm{a} < 0$ then $[\ce{A-}] < [\ce{AH}]$

And if $\mathrm{pH - p}K_\mathrm{a} > 0$ then $[\ce{A-}] > [\ce{AH}]$

If you want then to know the exact proportion you need to know the concentration of your amino-acid.


Example from biochemistry

The Michaelis-Menten equation for enzyme kinetics is an example.

$v=\frac{k_{cat}E_0S}{K_S+S}$ where $S$ is substrate concentration and $E$ in enzyme concentration.

At low $S$ ($S\ll K_S$), the reaction is 2nd order with $v=\frac{k_{cat}}{K_S}E_0S$, i.e. first order in both enzyme and in substrate.  However at high $S$ ($S\gg K_S$), the reaction is first order only in enzyme with $v=k_{cat}E_0$, and is zero-order in substrate.  The substrate concentration doesn't have to be in molar excess to other reactants.

Example from process control

In chemical engineering, nonlinear systems such as higher-order reactions occuring in a continuous process at steady-state are often modeled using linear (which you can think of as first-order reactions) systems as an approximation for developing process control techniques.  A MATLAB documentation page has some good information.

Another example, somewhat contrived

As a last example, consider an unstable molecule $A$.  It can decompose to $B$ by two entirely separate elementary mechanisms.  The first is $\ce{A + A -> B + B}$.  The second is $\ce{A -> B}$.  At high concentrations of $A$, the second-order kinetics will dominate and the decomposition of $A$ will be second order.  But lower the concentration enough, and the reaction will become first order in $A$ because the rate the bimolecular reaction will be dwarfed by the unimolecular route at low $A$.


It seems to have the 1:1 $\ce{Br}$ ratio spaced by 1.

I think a reasonable assignment would be that it is $\ce{[BrCCOH]^2+}$

m = 120 and 122

m/z = 60 and 61


Yes, Aaron.

It will.

Because Iodide is very good leaving group. In iodoform test, we use base... which will convert 2-iodopropane into propan-2-ol. The latter will give haloform the same way as ethyl alcohol and any other methyl ketone does!


The curly brackets denote "activity of" the species therein.

See the Wikipedia section: Basic definitions and properties of Equilibrium constant


Sublimation and recrystallization of caffeine

This sounds like a nice project.

For the purification of caffeine, sublimation in vacuum and recrystallization from ethanol are both conceivable.

But first you have to separate it from the other ingredients, such as carbohydrates.

According to the manufacturer, 100 mL of Red Bull, a popular brand, contain 11 g sugars and 32 mg caffeine.

In order to isolate the caffeine, two methods are possible:

extraction with $\ce{CH2Cl2}$
solid phase extraction on a C18 column

The latter method might be interesting. Imagine that your pour the energy drink through a preconditioned column of this adsorbent. The caffeine will  be retained, whereas everything else passes through.

In a second step, you wash the caffeine off the adsorbent using ethyl acetate. Then, dry the organic solvent with sodium sulfate, filter off, evaporate the ethyl acetate, dry the crude material and check the melting point. If necessary, use any of the above two methods for purification.

If you'd like to read this in more detail, have a look at a procedure provided at Open BU, an open access initiative of the Boston University.

UPDATE

I'd probably use just one of the methods for purification, depending on the equipment available. But if you insist on using both, I'd recrystallize first.


What is the correct discharge reaction in nickel-cadmium cell?

Note that the three equations are different only in whether the nickel and cadmium oxides are hydrated (and thus where water appears in the equation). Also, the Chemwiki equation is not balanced because nickel oxide is $\ce{Ni2O3}$ and not $\ce{NiO2}$.

The correct equation for the ChemWiki version is:

$$\ce{Ni2O3 + Cd + 3H2O -> 2Ni(OH)2 + Cd(OH)2}$$

Many metal oxides react with water to form hydroxides. This kind of behavior is probably most frequently taught with the alkali metals and alkaline earth metals:

$$\begin{align}
\ce{2NaO + H2O} &\ce{-> 2NaOH}\\
\ce{CaO + H2O} &\ce{-> Ca(OH)2}\end{align}
$$

However, other metals undergo these kinds of reactions. For example, the chemistry of rust involves the equilibrium between iron oxides and iron oxide hydroxides:
$$\begin{align}
\ce{FeO + H2O} &\ce{<=> Fe(OH)2}\\
\ce{Fe2O3 + 3H2O} &\ce{<=> 2Fe(OH)3}\end{align}
$$

  Nickel

The three ionic nickel species in the reactants are different only on the degree of hydration:

$$\begin{align}
\ce{Ni2O3 + H2O} &\ce{<=> 2NiO(OH)}\\
\ce{NiO(OH) + H2O} &\ce{<=> Ni(OH)3}\end{align}
$$

  Cadium

The two ionic cadmium species in the products differ only by degree of hydration:

$$
\ce{CdO2 + H2O<=> Cd(OH)2}
$$


The first part of your schematic basically describes a dessicant dryer or deliquescent dryer. These are common industrial air drying processes that produce very low dew point properties. Generally air (or other inert gas) is passed through a column of desiccant, under increased pressure.

Desiccant dryers use common desiccants that include activated alumina, molecular sieves and silica gel. In all cases, there is a maximum amount of water that can be adsorbed/absorbed before the desiccant needs to be regenerated; typically regeneration is done by heating the desiccant column, although other methods, such as purging can be used. Industrial desiccant dryers have the heater element installed as part of the construction, although any bench chemist can probably tell you about heating red silica gel beads from desiccators in the oven until they turn blue.

Deliquescent dryers are similar, but involve a highly hygroscopic agent that slowly dissolves, or deliquesces, as it absorbs more water. Usually the agent cannot be regenerated, and the waste liquid is drained and discarded. Fresh drying agent then needs to be added. This is most likely the scenario for NaOH.

So, this would need to be a major consideration for your proposed system; how to safely and effectively dry your NaOH. This is a very caustic material to have to try to continually regenerate or replace.

Yes, I would be concerned that you would transfer NaOH from the first chamber. To check this, you could easily test the gas outlet with some litmus.

Can you use dry air to remove excess water from other materials (such as acids)? In theory, this may be possible and will depend on the relative hygroscopicity of the drying agent and the material to be dried; you need to be sure that the air you are supplying your wet material with is dry enough to receive water and not simply supply more. Purge drying (passing dry air over a desiccant) is another form of regenerating industrial desiccants, but is not as effective as heating. Sulfuric acid is itself used as a desiccating agent as it is very hygroscopic. I don't know the relative hygroscopicity of NaOH and H2SO4/CH3COOH, but there is no way I would ever consider passing air from one container directly into the other chamber. This is a very hazardous setup, and is fraught with much danger. Heating an acid solution in a suitably ventilated environment under controlled conditions is still, to my knowledge, the most effective way to increase the concentration of the solution.

You should also consider the energy efficiencies of trying to 'dry' a sulfuric acid solution by passing dry air over or through the solution versus heating it. The amount of time and materials required to produce sufficient dry air to effectively concentrate a sulfuric acid solution is likely to be far in excess than the time/material investment required to simply heating it.


David Hamric did a four year Rare-Earth Metal Long Term Air Exposure Test

and has a wonderful set of time-lapsed photographs on his site.

Europium is the most reactive to oxidation.

Several sources, such as Avalon Advanced Materials: Europium make statements like:

  Samples of the metal, even when coated with a protective layer of mineral oil, are rarely shiny and it readily ignites in air above 150°C.

So rather than reacting with the mineral oil, I would say that europium and possibly others are reacting with oxygen despite being covered with the mineral oil.

THE SOLUBILITY OF O2, CO, AND N2 IN MINERAL OIL AND THE TRANSFER OF CARBON DIOXIDE FROM OIL TO AIR. The Journal of Biological Chemistry (1927)
 vol. 72,  pages 545-548 gives quantitative values for the solubility of oxygen in mineral oil, but I can't understand what units are being used. It says solubility is 0.134 +/- 0.004 at 24 degrees C.  I think this means 0.134 volumes of gas phase oxygen at standard temperature and pressure per volume of oil. Whatever the units are, it does show that significant oxygen can be present in mineral oil.

Also, I'll quote your own answer to another question:

  Even when these [alkali] metals are stored under mineral oil, oxygen can dissolve in the mineral oil and react.

Original Source: Management of time-sensitive
chemicals (II): Their
identification, chemistry
and management


Arrangement of elements in the reactivity series

The reason that lithium is on the top of the activity series is due to its behavior in aqueous reactions with water and acid. Due to the lithium atom's very small size (only two energy levels), the lithium ion interacts well with water forming ion-dipole interactions.  The lithium atom's small size compared to sodium and potassium is more entropically favorable in the formation of ion-dipole interactions with water and consequently so are reactions involving lithium ion formation favored to to other alkali metals.

The reasoning behind this is that reactions favor a greater state of entropy (disorder) in the universe. With larger cations, more water molecules have to become ordered around the cation to form the appropriate ion-dipole interaction.  However, due to the lithium cation's small size, less water molecules are required to form the ion-dipole interaction, thus less order, being more favorable in terms of entropy.


The method you mentioned is known as directional solvent extraction and seems quite straight forward and efficient. It is a relatively new method of desalination. Soybean oil, hexanoic acid, decanoic acid, and octanoic acid can be used (the last two are most effective) [ source: https://dspace.mit.edu/handle/1721.1/78539 ]. This video shows the setup used.


The biggest problem you will have is getting a high enough temperature, not the oxidation of the iron.

Putting the crucible in a charcoal furnace won't produce a high enough temperature to cause the reaction or the liquefy the iron: you need to have a crucible that can be driven to a much higher temperature. Normally this is done by blowing air (or, if you have it, oxygen) into the furnace. In practice what is often done is to mix the iron oxide and an excess of carbon in a single vessel: the carbon burns to create the temperature and also acts as the reducing agent to create the iron. This still requires a source to force air into the mixture. This appears to be how ancient craftsmen did the reaction, kicking off the iron age.

For an account of how to do this see this blog describing a modern recreation of some of those ancient techniques.


Calculating the mass of water removed by air conditioning condensation

Step 0) Realize that sometimes the answer given for a homework problem is wrong.

In this case it appears that the correct answer is $\pu{6.11 kg}$ of water, not $\pu{6.00 kg}$ of water.

The approach to the problem given below is to simply calculate the total water mass in the house before and after cooling, then subtract the latter from the former to get the mass of water removed by condensation.

Step I) Assemble parameters applicable before and after cooling:

$R =$ gas constant $= \pu{8.31m^3~bar~mol^{-1}~K^{-1}}$

$V = \pu{245m^3}$

$MW_{water}=\pu{18.0g~mol^{-1}}$

${n=\frac{PV}{RT}}$

Step II) Calculate water mass before cooling:

$T = \pu{33^oC = 306~K}$

$P_{saturation} = \mathrm{50.3~mbar^{(1)}}$

$P = 80\%$ of $P_{saturation} = \mathrm{40.2~mbar}$

${n=\mathrm{\frac{0.0402~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 306~K}=387~mol~\ce{H2O}}}$

$\mathrm{387~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=6.97~kg~\ce{H2O}}$

Step III) Calculate water mass after cooling:

$T = \mathrm{25^oC=298~K}$

$P_{saturation} = \mathrm{31.7~mbar^{(1)}}$

$P = 15\%$ of $P_{saturation} = \mathrm{4.76~mbar}$

${n=\mathrm{\frac{0.00476~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 298~K}=47.9~mol~\ce{H2O}}}$

$\mathrm{47.9~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=0.862~kg~\ce{H2O}}$

Step IV) Calculate total water removed from the house:

$$\mathrm{6.97~kg~\ce{H2O} - 0.862~kg~\ce{H2O} = 6.11~kg~\ce{H2O}}$$

Reference:
1) Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.
....


There are two reasons.

As mentioned by Ivan Neretin in the comments, carbonate is the conjugate base of a much weaker acid than chloride is the conjugate base of.  In other words, carbonic acid is less acidic than hydrochloric acid, and carbonate is more basic than chloride.  This means that carbonate solutions will have higher pHs than chloride solutions, in general.  At higher pH, the equilibrium between $\ce{NH4^{+}}$ and $\ce{NH3}$ is shifted more towards $\ce{NH3}$, which is volatile and smelly.
Carbonate is also potentially volatile.  If you leave ammonium carbonate, even in solution, around forever in an open container, it will eventually completely disappear.
$$\ce{(NH4)2CO3(s) -> H2O(g) + 2 NH3(g) + 2CO2(g) }$$This also favors more vaporization of ammonia.  In contrast, in ammonium chloride, chloride is essentially completely non-volatile.  Thus ammonium could not evaporate without a buildup of negative charge in the solution, highly unfavorable.


Neutralizations of any acid with oxides, hydroxides and carbonates always produce a salt and water.
The only neutralizations that do not produce water are reactions of an acid $\ce{HX}$ with ammonia $\ce{NH3}$ or with organic derivates of ammonia, like amines. These reactions produce an ammonium salt $\ce{NH4X}$ and no water
$$\ce{NH3 + HX -> NH4X}$$ If the acid is $\ce{H2Y}$ or $\ce{H3Z}$, the neutralization reactions are $$\ce{2  NH3 + H2Y -> (NH4)2Y}$$ $$\ce{3 NH3 + H3Z -> (NH4)3Z}$$ It never produces $\ce{H2O}$. If one atom $\ce{H}$ from $\ce{NH3}$ is replaced by an organic radical $\ce{R}$, a primary amine $\ce{RNH2}$ is obtained and the neutralization reaction produces also an ammonium salt $\ce{RNH3X}$ $$\ce{RNH2 + HX -> RNH3X}$$ Similar reactions can be obtained with secondary and tertiary amines. All these reactions produce salts and no water.


Still taught to reverse oxidation half cells in electrochemistry?

I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and USA. The Ostwald school of thought, wrote the electrode potentials in accordance with the electrostatic sign of the electrode with reference to the standard hydrogen electrode. This was the so-called European convention. Keep in mind that electrostatic signs of the electrodes are invariant. It does not matter how you write them. A silver electrode in Ag+/Ag half-cell is always positively charged with respect to the hydrogen electrode under standard conditions.

The American school of thought, with its own influence such as Gibbs, Lewis and Randall, Ligane, Latimer, had a thermodynamic view of the signs relating electrode potentials to the Gibbs free energy. Latimer wrote a very famous book "Oxidation Potentials" which is available from Internet archives. Yes, under those conditions we can flip the signs back and forth while keeping Gibbs free energy in mind. The tug of war between European convention vs. American convention went on until the 50s-70s. Textbooks are usually 20 years behind current research...these issues lingered on.

Finally, electrochemists decided, including Allan J. Bard (from US who wrote a very influential electrochemistry book taught all over the world), that let us go back to the original notion- keep the electrostatic signs associated with the electrode potentials which are invariant with respect to the way we write them. This has "almost" terminated the sign convention issue of US vs. European signs of electrode potentials. The plus and minuses took almost a century to resolve. It is amazing to see the insight of those early scientists such as Ostwald, that world finally accepted their views.


As has already been said in the comments, this is a free radical-based reaction. The peroxide $\ce{R1-O-O-R2}$ acts as a radical initiator( where $\ce{R1}$ and $\ce{R2}$ represent alkyl groups). These peroxide radicals generate $\ce{^.CCl3}$ which attacks the double bond and completes the reaction as shown.

An important point to note is that $\ce{^.CCl3}$ is regenerated at the end of the reaction, so it can be described as a catalytic intermediate for this mechanism


What prompts sodium to give up an electron?

To asses the possibility of a chemical change, we need to look into the spontaneity as given by the thermodynamic Gibbs' Free energy change at constant T and P. For your particular case, the change occurs when an electron accepting ion is present to form the crystalline compound after the electron exchange, in solution where the hydration releases energy or in certain solutions such as ammonia, where solvation of free electrons also take part.

Case (1). The changes involved are $$\ce{Na_{(s)}->Na_{(g)}^+ +e^{-}}$$, $$\ce{M_{(g)} +e^{-}->M^{-}}$$ $$\ce{Na_{(g)}^+ + M_{(g)}^{-}->NaM_{(s)}}$$

Where $M$ is an electronegative atom.The ionization energy ($\Delta H_{I}$) and the sublimation energy ($\Delta H_S$) required in the first transformation need to be compensated for by the electron gain enthalpy($\Delta H_{eg}$) released in the second transformation and the lattice enthalpy($\Delta H_{F}\text{lattice}$) released in the final transformation. If it is compensated, the change will be spontaneous. Moreover, the enthalpy released (negative) must also compensate for the lost entropy since gaseous $M$ is converted to an ordered crystal. $\Delta G=\Delta H-T\Delta S$ tells you the compensation required for $\Delta G$ to be negative.
($|\Delta H_{eg}+\Delta H_F|-|\Delta H_I+\Delta H_S|>|T\Delta S|$)

Case (2) Elemental sodium in water might lose electrons to the hydrogen evolving hydrogen gas and forming dissolved sodium hydroxide.
$$\ce{Na_{(s)}->Na_{(aq)}^+ + e^{-}}$$
$$\ce{H2O +e^{-}->\frac 12H2 +OH-}$$
here, the relesed hydration energy compensates the required sublimation energy in the first step and the negative electrode potential in the second step. There is net positive entropy change which favours the reaction, making $\Delta G$ negative.  ($|\Delta H_{hyd}(\ce{OH^-,Na^+})|+|T\Delta S|=|\Delta G|$)

Case (3) very similar to case (2) but here the solvation is both of free electrons and that of metal ions by ammonia (liquid). The equations are similar except that now instead of the hydration enthalpy, solvation enthalpy will be released. the entropy change is also positive here, supporting the change.


Major product of mononitration of 3-methylphenol

The product you are suggesting is very unstable as compared to the ones given as answers due to steric factors. Thus it would not be formed.

  Since −OH and −CH3 are o,p-directing groups, shouldn’t the major product be 3-methyl-2-nitrophenol? Why not?

The reasoning that hydroxyl and methyl are o,p directing groups is not at supportive as to why 3-methyl-2-nitrophenol should be formed in preference to the ones given as answers. The o,p directing effect just rules out the meta product and steric factor decides among the remaining possibilities here.


Actually not all of the last three are aromatic.  Look carefully at the third compound; the heteroatom is boron, not nitrogen, and so does not contribute the electron pair that would be needed to make six pi electrons in an aromatic ring.  That compound  is instead antiaromatic.

So JAM 2017 is apparently rendering both of the first two compounds as aromatic.  This, however, is dubious.  The first compound in the list is a pyrone, which not everyone would call aromatic.  To make an aromatic contributing structure in pyrone we would have to render the carbonyl group as its zwitterionic structure ($\ce{C^+-O^-}$) and then distribute the positive charge around the ring which has an electronegative oxygen atom.  We'd have to weigh the favorable effect of aromaticity against the unfavorable effect of putting a positive charge into an electronegative ring.

As for the second compound, we would need to "polarize" two carbonyl groups and concentrate two positive charges into the small ring, and that is not happening even with aromaticity.  There is a saving grace, however: if the $\ce{N-H}$ protons are tautomerized onto the oxygen atoms, converting the "keto" structure to its "enol" form, then the tautomerized molecule is 2,4-dihydroxy-1,3-diazine, which would be unambiguously aromatic.

What it boils down to is that aromaticity is often not the simple yes/no question often suggested in introductory chemistry texts.  Given this ambiguity, one could potentially defend an answer of 2, 3, or 4 (but not the last three for the two larger numbers) to the JAM 2017 question.


What is the highest possible effective nuclear charge as a percentage of nuclear charge an electron in any neutral atom can experience?

A high principal quantum number does not necessarily mean more electrons and/or protons. For example Hydrogen can be in an electronic excited state with $n=4$ and higher, as can be observed for example in the Balmer series. So for a specific element your cited statement remains true.

But you want to compare between different elements now. What we are looking for is minimal screening of the nuclear charge by other electrons:

According to  Slater's rules the second $1s$ electron screens with $0.3$ (all other screening constants are larger). Therefore the ratio you are looking for is $\frac{Z_{eff}}{Z}=\frac{Z-0.3}{Z}$ and converges to $1$ for large atomic numbers $Z$.

The only way to get even less screening would be to have no further electrons. As you restricted the question to neutral atoms we are left with Hydrogen. The charge divided by the nuclear charge of the atom, an electron can experience here is $\frac{Z_{eff}}{Z}=1$ for any excited state (principle quantum number) of the Hydrogen atom.


This has a lot to do with nucleation.  Just like crystallization and initial rain drop formation, for boiling to occur an initial site within the solution for a bubble of vapor to form on is required. So if you have very clean water that has been deaerated in a highly polished vessel there is an increased chance for superheating to occur due to the reduction in potential nucleation sites.  Tap water can superheat too, it is just less likely to do so because it typically contains more impurities that can serve as nucleation sites. This why boiling chips are commonly used in lab distillations.  They provide a site for vapor nucleation and usually prevent superheating of the material to be distilled.  Introducing other sites for nucleation, such as dissolved bubbles, solid contaminants, rough spots on the wall of the container all serve to lessen the probability of superheating.  I'm not sure what ions would have to do with the superheating process other than they are often representative of microscopic solid impurities that can provide that surface for the initial nucleation event.


Why does eutectic point have three phases, but consolute point have only two phases?

I just realised, the 2 phases present in a liquid-liquid solution system with temperature below its solution temperature is a) conjugate solution enriched in A
b) conjugate solution enriched in B

No pure liquid A or pure liquid B phase exists.


Is the current situation possible for recreaction in a lab?

25MPa is just north of 3600psi for those of us that haven't converted to SI units; 320*C ~= 608*F. Your average lecture bottle is rated to about 3000psi, while compressed gas cylinders can go as high as 6000psi. Trouble is, most of them don't stand up well to being heated by an open flame. Temperature is another beast; your Bunsen burner can easily achieve a flame temperature well in excess of 600*F (and a simple variation, the Meker burner, can hit more than twice that), but your Pyrex glassware is only rated to 500*F, and that for short periods. Bunsen burners are often used for glassworking on the lab bench, creating custom tubing rigs by softening straight tubing and then bending or stretching to create bends and bottlenecks. If the glass can be molded over a Bunsen burner, it's not going to be able to sustain the temperatures you're talking about.

The short of it is that both the temperature and pressure are possible, but neither are achievable with a benchtop apparatus in a fume hood; you'll need an HTHP reactor vessel rated for your experimental conditions. They exist, but whether you can get access to one is another question only your superior can answer. If you were planning on visually studying the change in the cellulose, you'll need a sapphire viewport, and these get very expensive very quickly.

Once you have the right tool for the job, it becomes a rather academic task of calculating how much volume of reactants are required to achieve the desired pressure in the vessel when heated to temperature, then filling the vessel and evenly heating it. Only two of the vessels in the linked column hold any appreciable amount of material at the proper temp and pressure, and one of them will just barely meet your needs; if you unevenly heat any of these vessels, by the time the far end reaches the proper temperature the spot being heated will have failed.


The statement is actually valid.If you go from S to Te, the compounds of Se and Te are stable at lower oxidation states due to inert pair effect.
So, the tendency of these group 16 elements to be in higher oxidation states decreases down the group.That means, tendency of the elements getting oxidised decreases down the group and thus their reducing property(reducing others, getting oxidised itself) decreases frm S to Te.


Transition state theory or activated complex theory is a very advanced topic and deals with chemical reaction dynamics. It goes beyond the scope of my knowledge, however, I will attempt my best to explain with what I know.

As stated, the activated complex is a temporary unstable product. As two particles with sufficient kinetic energy collide; along the reaction path, they distort the bonds and rearrange into new atoms. The location of the transition state for the activated complex is at the highest point; or peak, of an potential energy diagram.

As stated in my general chem textbook Principles of Modern Chemistry by Oxtoby et. al, p. 685:

  The activated complex is assumed to exist as if it were in equilibrium
  with the reactants and the theory focuses on calculating the rate at which the
  activated complex passes through the transition state to form products.

Statistical thermodynamics is required to calculate all motions of particles, including both rotational and vibrational energy.

In the transition state, change from products is mediated when bonds between particles or molecules become out of phase (or antisymmetrically stretched) by vibrational energies. The frequency of the molecules becoming out of phase with one another decreases as the reaction proceeds forward and the new product is transformed. The energy transferred from vibrational energy to achieve the activated complex is turned into kinetic energy.

The Eyring formula makes it possible to calculate reaction rates using transition state theory:

$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}K^†$$

The kappa term is referred to as the transmission coefficent and refers to the probability that the particles will create the transition state. $K_\mathrm{B}T/h$ measures the rate at which the activated complex dissociates to form into it's products.

This equation is unwieldy to us; as you stated previously, it is extremely hard to determine equilibrium constants for the activated complex since the instability achieved at the height of the potential energy for the reaction exists for only a brief period of time.

Instead, we can relate the Eyring formula to thermodynamics and substitute $\exp (-\Delta G^†/RT)$ for $K^†$.

$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta G^†/RT)$$
$$ k_r = \kappa \frac{K_\mathrm{B}T}{h}\exp (-\Delta H^†/RT)\exp(\Delta S^†/R) $$

The equation that Arrhenius derived is:

$$k_r = A\exp (-E_a/RT)$$

Activation energy is expressed as activation enthalpy and activation entropy. The pre-exponential factor $A$ is substituted with $K_\mathrm{B}T/h$. This equation and the use of enthalpy and entropy as driving forces is used extensively in enzyme catalyzed reactions.


Will this beta-branched secondary alkyl halide undergo SN1 or SN2?

Secondary alkyl halides are on the borderline of $\ce{S_{N}2}$ and $\ce{S_{N}1}$, so either could be operating. Branching at the position beta to the halide will further hinder the backside attack necessary for $\ce{S_{N}2}$. Furthermore, the solvent is polar protic, favoring $\ce{S_{N}1}$. In this case, the secondary carbocation can rearrange to a more stable tertiary carbocation, which leads to the observed product.

Full mechanism:

I am of the opinion that it's a difficult question given the conflicting factors, and I don't like it as an exam question because the relative influence of those factors isn't usually taught or easily predicted.
I imagine that the solvent plays a large role here. Cyanide is a typical nucleophile for $\ce{S_{N}2}$, but in water its nucleophilicity will be reduced by hydrogen bonding. In order for the $\ce{S_{N}2}$ mechanism to take place for this hindered electrophile, the nucleophile will need to be as strong as possible.


Understanding the mechanism of Blanc chloromethylation

The result is the same no matter which resonance form you use. Resonance forms are only formal representations, extreme forms of the true structure which is somewhere in between.

Note that to get from the blue resonance form to the red resonance form, you'd need to push an arrow:

So there is no real difference which resonance form you use - in the end you would draw exactly the same arrows. In the blue reaction, you draw two arrows. In the red reaction, you only draw one arrow. But, to get from the blue resonance form to the red resonance form, you have to draw one arrow anyway.

Don't read too much into the number of arrows or whatever - the only point I am trying to make is that from a mechanism-drawing perspective, it does not matter which resonance form you use, because all roads lead to the same product anyway. All those arrows are just formal bookkeeping.


Why do significant figure rules contradict each other?

Bear in mind the significant digit rules are just  simplifications of the rules of absolute (+, -) and relative (*,/) variance propagation.

The rule of significant digit numbers for summing is for 2 statistically independent variables. Their  correlation coefficient = 0 and the error is $$  s_\mathrm{C}= \sqrt{(s_\mathrm{A})^2+(s_\mathrm{B})^2}$$
and therefore about at the same order as for less precisely known variable.

Multiplying a variable by number means summing the same variable $N$ times, with the correlation coefficient 1 and the error is growing linearly by $N$. As here applies the multiplicative  rule for propagation of relative variance.

$$ s_\mathrm{C,r}= \sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{B,r})^2}\\=\sqrt{(s_\mathrm{A,r})^2+(s_\mathrm{3})^2}=s_\mathrm{A,r}$$

I agree with the comment to follow the rule of significant digits on the final result only, to avoid cummulative rounding errors.

On the other side, it can be hard to track the digits but for simple formulas. Applying the variance propagation rules may be easier way to go.


How do I calculate the grams of a product after a reaction?

The concentration $c$ of a solution is the molar quantity of the solute $n$ devided by the volume of the solution $V$:

$\begin{equation}
c = \frac{n}{V}
\end{equation}$

as you state, for example for HCl 2.5 mol/L (or, equivalent, 2.5 M).  Inversely, if you know volume and concentration of your solution, you may access the molar quantity of your solute.  Actually, there is such an equation for each of the solutions, indepent from each other; you may index the variables accordingly, too.

Yet watch, there is a catch: mixing the two solutions together increases the total volume, and simultaneously decreases the concentration of the specis in question -- without altering the molar quantity of $\ce{Na+}$, nor $\ce{Cl-}$ ions present.  Hence "2.5 M of NaCl" (as measure of concentration) in the solution in common is incorrect.

Because of the initially asked question, one of the solutions contains the limiting reagent.  Figure out which one -- this includes the stoichiometry in the equation of chemical reaction. Knowing the molecular amount $n$ of it, your are able to determine the amount of sodium chloride formed; initially as moles, later (via molecular mass) in grams.


The sum of the oxidation states of all of the atoms needs to equal the total charge on the ion.

For example, for $\ce{NH4+}$, if each hydrogen atom has an oxidation state of +1, and the overall charge is +1, then we can solve for the oxidation state of nitrogen:

$$\begin{aligned} 4(+1) + OS_\ce{N} &= +1\\
OS_\ce{N} &= +1 -4(+1) \\
OS_\ce{N} &= -3
\end{aligned}$$

Similarly, we can do $\ce{SO4^2-}$. If each oxygen atom has an oxidation state of -2, and the overall charge is -2, we can solve for the oxidation state of sulfur:

$$\begin{aligned} 4(-2) + OS_\ce{S} &= -2\\
OS_\ce{S} &= -2 -4(-2) \\
OS_\ce{S} &= +6
\end{aligned}$$


How can one tell from the MO diagram of the cyanide ion that the HOMO is carbon-centred?

The 2s and 2pz atomic orbitals are of the same symmetry in the molecule. That means that the 2 sigma* and 3 sigma are actually hybrids. The former is a hybrid were most of the electron density points to the bond (making it partially bonding), while in the latter most of the electron density points outside, i.e. a lone pair in the carbon.