Testing for the presence of Hexamethylenetetramine / Formaldehyde

I don't believe the premise that E239 (the E number designated to hexamethylene tetramine) is used to preserve fish within the EU is correct - in January, 2014, the European Food Safety Authority published an article in their journal about this preservative where it states quite clearly that the only permitted use for this additive is in Provolone cheese. There are concerns that over-consumers of this particular cheese may accrete more levels of E239 within the body than are healthy.

E239 does appear to be used in the UK for preserving fish, or at least, it is listed as one of its intended uses. Whether its actually used or not I haven't been able to determine, but given we're in the EU, technically, it shouldn't be. If I source any other research regarding this, I'll add it later, but currently, it seems it depends where you're buying your fish from in the first place, that is, the country of origin.


There are a number of things that exceed the speed of light:

Particles in a medium such as water can enter faster than light [in that medium] but are slowed and release energy as Cherenkov radiation.
The phase velocity of a wave can exceed that of light [in that medium].
An arbitrary point, such as the point on a pair of scissors where blades,
meet, can move faster than light.
Most puzzling to me: entangled particles. Einstein called this "spooky action at a distance". Although each particle moves at C or less, information might be shared between them in an indefinitely short time; see superluminal communication. Entanglement does apply to chemistry, as do other odd quantum effects; e.g. the detection of single photons with silver halide photography as they form an interference pattern in the double-slit experiment.


Nucleophilicity/basicity in protic and aprotic solvents

Why is an atom with high polarizability a strong nucleophile?

An atom with a higher polarizability will share his electrons more easier since they are less attracted by the nucleus. The more the electrons are near the nucleus, the more it's hard the share them with an electrophile. $\mathrm{I^-}$ is thus more nucleophile than $\mathrm{F^-}$.

As you'll se below, the solvent can affect the nucleophilicity and reverse what I just said.

In polar aprotic solvent, why is $\mathrm{F^-}$ a stronger nucleophile than $\mathrm{I^-}$?

When talking about nucleophilicity, why does the polarizability matter in a polar protic solvent, but not in a polar aprotic solvent?

First, basicity and nucleophilicity are not the same at all. For instance, a strong base can be non-nucleophile, such as LDA for instance.

Then, let's talk about the polarizability of the $\mathrm{F^-}$ and $\mathrm{I^-}$ anions. The fluoride has a small polarizability: its charge is dense, very localized. It's called an hard anion. Iodide has a great polarizability with its dispersed charge. It's called a soft anion.

Using a protic polar solvent, the anions will be well solvated, even better if they are hard anions because the hydrogen bondings will be more efficient. The anion $\mathrm{F^-}$ is thus more solvated than $\mathrm{I^-}$ and less reactive.

In the case of an aprotic polar solvent, the anion is not so well solvated and the fluoride will be more reactive.


What is the proper laboratory technique to place a solid in a beaker that will be weighed?

There is no correct or universal weighing protocol per se. The key question is what is the desired level of accuracy? Your teacher's approach is quite cautious but this is not the universal way, as what you did is also common. He/she is apparently avoiding a spatula as well. I personally dislike paper and plastic weighing boats. If you ever had a chance to weigh dry salts or organic crystals on a dry day, these materials charge like crazy and literally all particles fly apart the moment you touch the plastic boat with gloves.

Pharma companies, where a high level of accuracy is required and legal formalities are important, chemists often use weighing funnels. They are not popular in academia (pricey!). In the picture, you see a latest type of balance which avoids static issues. On the top you see a weighing funnel. This one is plastic coated with anti-static material but glass is more common. After weighing, the funnel containing the weighed substance is placed on the mouth of a volumetric flask and washed very carefully. You can't do this with paper.

https://www.laboratorynetwork.com/doc/new-smartprep-weighing-funnel-weigh-in-samples-more-accurately-0001


It sulfonates on N first and then rearranges on heating to give sulfonation in the 4 position. JACS 68 (1946) 969


Dissociation of ions

Water, as you probably know, is highly polar and has a very high dielectric constant. When ionic compounds are added to the water, the electrostatic force of attraction between the ions get very weak and hence will result in dissociation of the compound.

The percentage of dissociation of a compound depends on various factors such as solvent, strength of ionic bond, etc.

$\ce{HCl}$ is a polar molecule and hence it dissociates into its constituent ions, namely, $\ce{H+}$ and $\ce{Cl-}$ ions.

Your initial question "what is the pH of HCl" is not quite right. More about it in the next section.

What is pH?$\space$ Can pH be negative?

pH by definition is a measure of hydrogen ion concentration.

$\mathrm{pH} = -\log[\ce{H+}]$ where $[\ce{H+}]$ denotes the active hydrogen ion concentration.

Hence, $\mathrm{pH}$ of $\ce{HCl}$ in an arbitary context won't make any sense since the $\mathrm{pH}$ of $\ce{HCl}$ would depend on its concentration.

The $\mathrm{pH}$ of a substance can be negative. $10\ \mathrm M$ solution of $\ce{HCl}$ would have a $\mathrm{pH}$ of $-1$.

The stronger the acid, the smaller is the value of $\mathrm{pH}$. $\ce{HCl}$ is a strong acid because it ungergoes almost 100 % dissociation.

Theory of acids and bases and their relation with pH

The theory of acids which you are refering to were developed to classify acids and bases rather not to measure the strength of the acids and bases.

However, one can arrive at the strength of the acid/base according to the theory by logical reasoning. For example, if someone were to ask you which one of the two compounds is an Lewis acid, you would investigate both the molecules and find out which one has a higher tendency to accept electrons to find the stronger acid.

Therefore, the theory of acids proposed by Arrhenius and Brønsted–Lowry are in no way related to $\mathrm{pH}$.

pH on the other hand, is a measure of strength of an acid and is indepdent of the above theories.

It is quite easy to compare $\mathrm{pH}$ with Arrhenius/Brønsted–Lowry theory of acids since according to the theory, the molecule which is willing to donate free $\ce{H+}$ ions will be considered an acid. A claim that higher the concentration of $\ce{H+}$ will be a stronger acid is reasonable, understandable and acceptable.

Bonus:
In case of bases, we have a strength measurement scale similar to the pH. We make use of hydroxide ion concentration in place of hydrogen ion concentration to find the strength of the base. We call it $\mathrm{pOH}$.

These two are related by the following formula when the acid/base are dissolved in water at room temperature.

$14 = \mathrm{pH} + \mathrm{pOH}$


Determining the stereochemistry of the product(s) when an alkyl shift in a carbocation generates a chiral carbon

Since this problem is about the stereochemistry, I'd try to address it according to sophomore undergraduate Organic Chemistry text books used in the USA.

Briefly, any nucleophilic substitution reaction at primary carbon predominantly undergoes with SN2 mechanism while those reactions at tertiary carbon undergo SN1 mechanism exclusively. Meanwhile, reactions at secondary carbon would undergo either mechanism, based on reaction conditions, solvents, etc.

The "carbocation" given is a primary carbon even though it is a chiral carbon due to the presence of deuterium atom in place of hydrogen. Note that the stereo-effect caused by deuterium can be considered similar to that of hydrogen. Nonetheless, the bond energy difference (the $\ce {C–D}$ bond dissociation energy listed as $\pu{341.4 kJ/mol}$ while that of $\ce {C–H}$ as $\pu {338.4 kJ/mol}$) may cause the isotope-effects in kinetic of reaction mechanisms (that is a different avenue than what in discussion). Since $\ce{-CH2-Cl}$ and $\ce {-CHD-Cl}$ are similar and $\ce {C}$ atom in both cases is primary, the mechanism of the reaction should be SN2 exclusively. Therefore, the stereochemistry will be $\ce {100\%}$ inverted.

For reading for substitution reactions involving alkyl halides: article. Please also note that, one of example given on that website, the compound $\ce {Ph-CH(CH3)CH2-Cl}$ does not undergo methyl group migration to give benzylic carbocation during its nucleophilic substitution reaction.


Equilibrium :. Why does a system at equilibrium try to undo the effect of external stress applied on it?

It is not that a system follows Le Chatelier's Principle, as it is not a law such as Newton's laws of motion,  but rather that this is a shorthand way of working out what happens; a 'rule of thumb' if you wish.

If a system is in stable equilibrium it must have the lowest energy possible under the prevailing conditions. If now a perturbation is made, say by changing temperature or pressure, the system is no longer at equilibrium because some energy has been added or removed. It will now change to lower its energy and again reach equilibrium under the new conditions. Le Chatelier's Principle is then a statement of this.

For example, suppose that as a result of molecular motion, a small part, A, of a larger volume of a gas spontaneously increases by  $\Delta V$. Thus the pressure  decreases, $\Delta p < 0$, and so the net force on A exerted by the surroundings is in a direction that tends to restore equilibrium.


This reaction is considered a modified Friedel-crafts acylation. The reaction is formally a series of electrophilic aromatic substitutions, with hydroxyl leaving groups picking up protons (1), however the Lewis acid catalysis obviously plays a role in the mechanism. I think it increases the electrophilicity of the phthalic anhydride in the first step, and facilitates the elimination of the hydroxyl groups. For the mechanism without the involvement of the zinc(II) chloride, see (1). Here is a mechanism emphasizing the role the Lewis acid might play in catalyzing the reaction:

This reaction works in a similar way to the Nencki reaction (2)
In order to get the form of the molecule shown as the product of your reaction, the $\mathrm{pH}$ needs to be increased in order to open the lactone ring.(3) This is fluorescein-A, the form of the molecule that actually fluoresces.

References:

Jag Mohan, In Organic Analytical Chemistry: Theory and Practice; Alpha Science International Ltd.: Pangbourne, England, 2003, p. 497 (ISBN: 1-84265-119-6).
M. Nencki, N. Sieber, "Ueber die Verbindungen der ein‐ und zweibasischen Fettsäuren mit Phenolen," Journal für Praktische Chemie 1881, 23(1), 147-156 (https://doi.org/10.1002/prac.18810230111).
Clifton J. Stephenson, Ken D. Shimizu, "A fluorescent diastereoselective molecular sensor for 1,2-aminoalcohols based on the rhodamine B lactone–zwitterion equilibrium," Org. Biomol. Chem 2010, 8(5), 1027-1032 (https://doi.org/10.1039/B918823A).


How can both kinetic and potential energy increase simultaneously in a open substance?

You are adding energy by heating, this generally does not break chemical bonds (unless the temperature reaches hundreds of degrees) but instead the weak intermolecular potential energy between molecules (sometimes call van-der Waals interaction) and so the solid will melt, then the liquid will boil, obviously :)

The kinetic and potential energy of the sample has increased. Kinetic energy in translational motion of the molecules and internal potential and kinetic energy of the molecules has also increased as they have absorbed some of the heat as vibrational and rotational energy.  (A non-linear molecule with N atoms has $3N-6$ different ways of vibrating, e.g ethane has 18 ways).

The reason that the temperature increases rapidly is due to the fact that if you heat a pan on the stove, or a boiling tube with a Bunsen burner  is that convection of heat away into the air is very poor. So your boiling tube is effectively insulated. If it were completely insulated then less added heat would be needed to reach the same temperature. As an admittedly odd example, you will know that a plumber can join Cu pipes with solder using a blow torch, but only if they are dry. If they contain a even small amount of water more of the heat is carried away (and heats the water) and soldering becomes impossible.

If you heat a sealed tube, it is likely to explode violently due to increase in pressure; so don't do it! (To prevent explosion is why pressure cookers have a pressure release value).


I'd say it works by coming out as a liquid. Then it diffuses through the porous rust layer or travels along the steel edge and expands to gas- as the mixture has a highly suppressed vaporisation point to push the rust from the inside. Steel has a high surface affinity for hydrocarbons as demonstrated by high difficult it is to clean. Partial vaporisation is plenty powerful enough to knock off rust.


What is more reactive towards a cyanohydrin formation an aldehyde or an Ketone?

Aldehydes will generally form cyanohydrins more readily than ketones, but some ketones such as trifluoromethyl or acyl ketones may behave more like aldehydes due to the electron deficiency at carbonyl carbon.


Carboxylic acid strength due to the inductive effect

The strength of the carboxylic acid (and any other Bronsted acid, for that matter), is related to the 'stability' of its conjugate base, the carboxylate anion. The carboxylic acid and the carboxylate anion are in equilibrium with one another, and the relative acidity of carboxylic acids depends upon the position of this equilibrium.

One of the main reasons why carboxylic acids are acidic is due to the ability of the charge to be delocalised around the pi-system:

However, in addition to this delocalisation about the pi system, extra stability can be gained by the presence of electron withdrawing groups adjacent to the carboxylate group (i.e. -I groups attached to the R group in the diagram above).

These electron withdrawing groups draw electron density away from the carboxylate anion via the sigma system (the single bonds). This is advantageous as the stability increases with the number of atoms it can be spread across (i.e. how diffuse the charge is).

Electron releasing groups (+I as you call them) destabilise, by increasing the electron density around the carboxylate group, which of-course makes it less favourable for the carboxylic acid to lose a proton and hence makes it less acidic.

The +I groups aren't involved in breaking the O-H bond to any great extent, as you allude to. This is because when the proton leaves, it transfers its electron density to the oxygen it was attached to, cleaving the bond and allowing it to leave as H+. If the +I groups were donating into this bond, you would have to have the H leaving as H-, which is rarely observed.


To quote Sime's textbook on Physical Chemistry, experiment 3 on the measurement of heats of combustion of hydrocarbons using bomb calorimetry: "we will assume that $\Delta U = \Delta U_r^o$; that is, we will assume that $\Delta U$ is independent of pressure". There is no further justification of the statement. The Parr Instrument Company provides an introductory guide to bomb calorimetry (document 483M) which has the following to say:

Users interested in high precision calorimetry
will understand that test results obtained in a bomb
calorimeter express calorific values obtained at constant
volume. This differs from the usual conditions
under which fuels are burned at constant pressure
rather than constant volume. If the sample contains
elements which normally would increase in volume
after combustion, a correction must be applied to
account for the heat equivalent of this work. This
value is so small as to be negligible in routine work,
and it usually is disregarded.

The required pressure correction can be computed from thermal expansion and isothermal compressibility coefficients. For instance, for water one estimates that if the bomb pressure rises by $\pu{2 GPa}$ over standard pressure then the energy will increase by $\approx \pu{80 J/mol}$. Compare this correction with the heat of combustion of benzoic acid ($\approx \pu{3221 kJ/mol}$) and (assuming the coefficients for benzoic acid are of the same order of magnitude) it's clear that for condensed substances the correction is (often) negligible.
For gases, based on the assumption of ideal gas behavior, an isothermal PV change for the products will leave $\Delta U$ unchanged, so that in fact $\Delta U_r = \Delta U_r^o$.


How to change the direction of evaporation and thus location of condensation?

The simple answer is that water evaporate where the concentration in the air is lower than the vapor pressure some liquid water at a given temperature. Then it will condense at a location where the temperature is lower that the saturation temperature of water at that partial pressure.

Industrially, though, one widely used technique to avoid condensation when the surface temperature is low enough for it to happen is to use hydrophobic coatings to prevent the water vapor from nucleating on certain surfaces. A quick google search will reveal lots of these for different applications.


Does anisole react with concentrated nitric or sulfuric acid

Anisole (methoxybenzene) is an electron-rich arene that will react with sulfuric acid or nitric acid in an electrophilic aromatic substitution.

I suggest that you look the reaction up. Then grab a pencil, some sheets of paper and start to draw. Draw the arenium ions (Wheland intermediates) for the addition of the electrophiles at the different positions of anisole, have a look at the possible resonance structures and enjoy chemistry.

UPDATE

Btw, you can do the same reaction with aniline or phenol without causing an explosion.

My comment on aniline and phenol only works on paper but not in real life and Mithoron's comment on aniline and phenol is correct!

Nitroanilines and nitrophenols are synthesized in a different way.
In principle, nitration of aniline is possible with a mixture of nitric and nitrous acid, or after protecting the aniline as its formanilide.

Technically, 4-nitroaniline is produced by reacting 4-chloronitrobenzene with aqueous ammonia at elevated temperatures in an autoclave.

Details can be found in Ullmann's Encyclopedia of Industrial Chemistry:

  In the alternative continuous process, molten 4-chloronitrobenzene and 40 % ammonia liquor in a molar ratio of 1 : 17 are pumped at 20 MPa through a steel reaction coil held at 237-240 °C by oil bath heating. Inflow is controlled
  to give a contact time of ca. 10 min, and outflow is directed into an expansion chamber with a tangential flow of hot water to give, in quantitative yield, a suspension of product that is readily isolated and washed.

The same is true for the formation of 4-nitrophenol. Again, the starting material is 4-chloronitrobenzene, which is reacted with strong aqueous $\ce{NaOH}$ to yield the phenol.

These two reactions are examples for nucleophilic aromatic substitutions!


Polymerizations expelling water, equally known as polycondensation (PET is a typical product), are a subset of polymerizations. The radical polymerization here is more likely to proceed differently with three reaction types running concurrently: initiation, propagation, and termination.

During the initiation, light splits the phenylketone into two fragments with one unpaired electron each. This is symbolized by the dot next to the oxygen atoms, denoting that these chemicals are radicals:

These radicals may react with one electron of the double bond of a neutral acrylate in the propagation reaction. Symbolized by blue is the newly formed single bond between the two reagents (first line).  Simultaneously, the other electron of the former double bond represents a new site where again a neutral acrylate may react (second line).  This growth of the polymer chain may continue again and again, or propagate (third line).

As soon as there are radicals formed, there may be a number of reactions where these carbon-centered radicals are consumed.  This cancels the further growth of the polymer chains.  Among these termination reactions, hydrogen-transfer / redox reactions as well as dimerizations are the most prominent ones.  Some which seem likely to occur are depicted below:

Structure
Your starting material consists of a starter, and a blend of bifunctional acrylates.  Thus the curation will yield polymer chains that are cross-linked with each other.  The illustration below (for a different material) symbolized this by the blue zig-zag lines:

(Konuray et al., Polymers 2018, 10, 178 doi 10.3390/polym10020178, open access)
Because you started with multiple different monomers, it would be an example of branched copolymers.


Neither Raman nor IR Active vibrational modes

Yes it is possible. While no normal modes can be both IR and Raman active in molecules with a centre of symmetry, as mentioned in a comment molecules with silent modes exist.

The easiest example is  probably the HCH out-of-plane twisting of ethylene:

Source: http://www.shodor.org/succeed-1.0/compchem/labs/vibrations/ethylene.jpg

Due to the symmetry of the molecule, when the HCH twist out of plane the dipole and polarizability of molecule is unchanged.

Other examples can be found here

https://www.chem.purdue.edu/jmol/vibs/c6h6.html

where the modes of benzene are sketched. The silent modes easily seen are kind of distorted breathing of the ring, that due to the C6 axis of symmetry again let the molecule unchanged both for the electric dipole moment and its polarisability.


Reaction of carboxylic acids with primary amines

After this step the $\ce{OH-}$ which was eliminated …

No. $\ce{OH-}$ is never eliminated but formally. To eliminate the oxygen required, it needs to first accept a proton from somewhere leading to $\ce{R-C(O^-)(OH2+)-NHR}$. The hydroxide ion is simply much too bad a leaving group for any $\mathrm{S_N}$ reaction; it must always be transformed into a better one.

This leads to a severe problem in the reaction. Basic conditions hinder the reaction because they decrease the degree of protonation on that oxygen. However, under acidic conditions the amine will be protonated in solution ($\ce{RNH3+}$) and a protonated amine no longer has any lone pairs that can attack nucleophilicly. You’ll want neutral to mildly basic conditions for the reaction to work out.

Another thing to consider: the ‘elimination of a proton’ is a simple acid-base reaction. Proton displacements are typically much more rapid than any carbocation eliminations because protons, being as small as they are, can tunnel. Additionally, you should never compare the stability of a naked proton against that of a naked carbocation in these eliminations. A naked proton will not be generated, it will attach to a base immediately — and there are enough bases around: $\ce{OH-, H2O, RNH2, \dots}$ Furthermore, carbon-heteroatom bonds are pretty stable if the heteroatom side does not equal a good leaving group (e.g. $\ce{NHTs}$). Thus, these are hardly ever cleaved.

In practice, you would not use a particularly acidic or basic medium in this reaction. Rather, you would use acid activation techniques such as the Steglich conditions ($\ce{DCC, DMAP}$).


Why is the addition of methanol during the prepartion of the liquid crystal cholesteryl benzoate necesary

Pyridinium hydrochloride and methyl benzoate are both soluble in methanol and thus will be washed through the filter leaving behind clean cholesteryl benzoate. If you don't use the methanol the solid you produce will be impure and sticky; the methyl benzoate and pyridinium hydrochloride will not pass through the filter but will bind to the product crystals.. Methanol also removes any excess of benzoyl chloride by reacting with it.


Relative Energies between LGOs and 2s orbital of oxygen in H2O MO Diagram

You probably already understand that the oxygen 2s orbital is lower in energy than the hydrogen 1s to begin with. For the unconvinced reader, consider the fact that the ionisation energies of oxygen and hydrogen (1314 and 1312 kJ/mol respectively) are quite similar, which implies that O2p and H1s have (somewhat) similar energies.

The question is then about how much lower in energy the H1s + H1s bonding combination is, relative to the H1s orbital itself. However, the hydrogen atoms are quite far apart: using the H–O–H bond angle (104.45°) and H–O bond length (95.84 pm) quoted on Wikipedia, the H–H distance can be calculated as being 151.5 pm, over twice as long as in dihydrogen itself (74 pm). This means a very poor overlap between the 1s orbitals, so you should not have an in-phase combination that is significantly lower in energy than the original hydrogen 1s orbital.

For the same reason, the "antibonding" LGO should also have roughly the same energy as the original 1s orbitals. The only reason why we are looking at bonding & antibonding LGOs is because of symmetry; it doesn't have anything to do whether they are actually bonding or antibonding to an appreciable extent.


In the paper that you've mentioned$^{\ast1}$, there is something described, which is called effective bond order ($\mathrm{EBO}$).
$$\mathrm{EBO} = \frac{1}{2}\left(\sum_{i=1}^N \eta_{b,i}-\eta_{a,i}\right)$$
... with $\eta_b = 2 − x$ as the occupation number of an occupied orbital and $\eta_a = x$ as the occupation number of the corresponding antibonding orbital. (In multireference quantum chemistry methods, the resulting averaged orbitals can have occupation values of $0 \leq x \leq 2$)
In 2013 Ruipérez, et al. calculated the effective bond order for several homo- and heteroatomic dimers.$^{\ast2}$ Based on their very high quality calculations, the

[...] results show that the effective bond order ($\mathrm{EBO}$) of the $\ce{MoU}$ dimer (5.5) is higher than that for the tungsten dimer (5.2), known to date as the molecule with the highest $\mathrm{EBO}$.

So somewhat arbitrarily, $\ce{MoU}$ seems to be one of the best candidates for the highest bond order.

References:
$^{\ast1}$. Björn O. Roos, Antonio C. Borin, Laura Gagliardi, "Reaching the Maximum Multiplicity of the Covalent Chemical Bond," Angew. Chem. Int. Ed. 2007, 46(9), 1469–1472 (https://doi.org/10.1002/anie.200603600).
$^{\ast2}$. Fernando Ruipérez, Gabriel Merino, Jesus M. Ugalde, Ivan Infante, "Molecules with High Bond Orders and Ultrashort Bond Lengths: $\ce{CrU, MoU,}$ and $\ce{WU}$," Inorg. Chem. 2013, 52(6), 2838–2843 (https://doi.org/10.1021/ic301657c).


Does mercury chloride (ii) evaporate at room temperature?

According to this 1911 Journal of the American Chemical Society article (Johnson, F. M. G, JACS 1911, 33, 777), the vapor pressure of $\ce{HgCl2}$ is 3 mmHg at 152 degrees Celsius. Since the article is behind a paywall, I will reproduce the data below:

$$\begin{array}{|c|c|}
\hline
T/ \mathrm{^\circ C} & P/ \mathrm{mmHg} \\ \hline
152 & 3\\
195 & 20 \\
231 & 82 \\
256 & 198\\
262 & 237 \\
275 & 325\\
278 & 421 \\
283 & 481\\
302 & 754\\
\hline \end{array}$$

The authors did not report values for lower temperatures since their method focused on setting a pressure in the system and then measuring the temperature of sublimation/boiling point. However, using the data and some help from the Clausius-Clapeyron relationship, an estimate for the room temperature vapor pressure is $2.4\times 10^{-4}\ \mathrm{mmHg}$, which assuming you are at 1 atm, is 0.32 parts per million when the atmosphere is saturated. Consult a safety data sheet or other source for toxicity considerations at that exposure level.


In spectroscopy, is it possible for the sample to be excited multiple times?

Very technically? Yes.

Realistically? The probability is small enough that even if it does happen, the peaks for the multiple transitions are going to be small enough that we cannot really observe them on the spectrum. The lifetime of a given excited state is so small compared to the analogous time in the ground state that it can basically be considered zero for the situation you're talking about. I'm sure it has happened, but compared to the one at a time transitions, your spectrum isn't really going to show it.


Hessian Matrix and Physical directions: Potential Energy Surface

I think you have the causality backwards here. It's not as if we have the vibrations to start with and they happen to be matched by the eigenvectors of the Hessian. We choose to look at these particular vibrations precisely because they are the eigenvectors of the Hessian.

Consider the example of water. We know that it has it has $9$ degrees of freedom, because each of its $3$ atoms are described by $3$ Cartesian coordinates. However, if we are interested in the energy of a configuration, we know that some of these degrees of freedom don't matter. For example, if we shift the x coordinate of each atom by the same amount, we haven't changed the molecule at all, only its location. This leads us to seek a set of coordinates that describe the potential energy surface without considering parts that don't matter. For water, we can remove $6$ as shifting along or rotating about any axis doesn't change the energy of the molecule.

This leaves us with $3$ remaining degrees of freedom, internal motions of the molecule that do affect its potential energy. What do these look like? I could come up with all sorts of valid variations on this. Off the bat, I might think that a good representation would be individual stretches of each $\ce{O-H}$. While there isn't anything strictly wrong with this representation, it does have the issue that motions along these coordinates are coupled, meaning changes along a given mode affect some or all of the others.

What can I do to avoid to avoid this coupling? Well, within the harmonic approximation, all the coupling between modes is described by off-diagonal elements of the Hessian matrix. So, if I can find a representation where the Hessian is diagonal, I will have a nice set of uncoupled internal motions. This is exactly what we get from the eigenvectors. Now, instead of two coupled $\ce{O-H}$ bond stretches, the normal modes give two decoupled anti/symm vibrations of both $\ce{O-H}$ bond.

The normal modes aren't any more physically correct than my previous example, but they are easier to analyze because they are decoupled. In this way, the "neatness" of the normal modes is really a fundamental mathematical property rather than a fundamental physical one.


I think your answer is correct as long as you can justify it. (IV) is clearly absurd but why do you eliminate (III) ? However in my opinion at $\ce{20 ^{\circ} C}$ there would be no reaction from a practical point of view, i.e. one with reasonable yield and reaction time.


Why is the IUPAC name of glycerol expressed in the way that it is?

Glycerol and trihydroxypropane are common names. The IUPAC name is propane-1,2,3-triol. The common name glycerol comes from the root glyco- which means sweet. Glucose, glycogen, and glycerin (another name for glycerol) have the same etymology.

Trihydroxypropane makes sense as a common name because there are only 3 locations for the hydroxyl groups on the 3 carbon chain, so the name is basically unambiguous - multiple hydroxyl groups on the same carbon are not stable (see geminal diols).


I'm assuming you mean "do the alcohol groups in carbohydrates such as sugars cause inebriation".

If that is what you mean then the answer is no. Inebriation is a function of some biological effects that are very specific to small alcohols not just to the OH group in a molecule (I suspect that methanol causes inebriation as well as ethanol, but given that it also kills you this is not something i'm willing to test).

Some of the biological effects are based on specific interactions with proteins in nerves, others are gross physical effects (e.g. dizziness is a product of changes to the density of fluid present in the balance organs in your inner ear when the alcohol concentration in the blood is high). The neuromodulation effects are very specific to the size and shape of the whole molecule not just the OH group and the dizziness is a function of the density of ethanol and its ability to circulate in the blood (larger alcohols and other OH-containing chemicals won't do this).

The body has a whole bunch of mechanisms to handle and manage sugars (as they are a primary way of providing fuel to cells). It would be surprising indeed if animals were inebriated by such compounds as they have evolved sophisticated ways to use them as a key part of their metabolism.


Mechanism of reaction of benzenediazonium chloride with ethanol to form benzene and acetaldehyde

In reality, benzene and acetaldehyde are the minor product of the reaction of benzenediazonium chloride with ethanol. The major product is ethoxybemzene (phenetole). Peter Griess, who had discovered diazonium salts in 1858, had reported benzenediazonium salt (nitrate or sulfate) with ethanol undergoes aforementioned redox reaction in 1864 (Ref.1). However, by 1901, it was confirmed that the products from the redox reaction is minor (about ~9%) when compared to major product, phenetole (about ~64%) when benzenediazonium chloride was used. By 1956, Kelley (Ref.2) found that the yield is even low as 3% and his student, Miller (Ref.1) later confirmed this result.
Now, it is believed that the reaction follows the scheme $\bf{A}$ and $\bf{B}$ (Ref.1):
$$\ce{Ar-^+N#NX- + RCH2-OH -> Ar-O-CH2R + N2 + HX} \tag{$\bf{A}$}$$
$$\ce{Ar-^+N#NX- + RCH2-OH <=> Ar-N=N-O-CH2R HX \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -> ArH + RCHO + N2 + HX} \tag{$\bf{B}$}$$
where process $\bf{A}$ proceeding through a $\mathrm{S_N2}$ displacement of the diazonium groups while process $\bf{B}$ involves a homolytic fission of the bonds to produce radicals which lead ultimately to aromatic hydrocarbon and aldehyde.
Proposed mechanism:
$$\ce{Ar-N=N-X -> Ar^. + N2 + X^.} $$
$$\ce{Ar^. + R-CH2-OH -> ArH + R-^.CH-OH} $$
$$\ce{AR-^.CH-OH + X^. -> R-CHO + HX} $$
It was found that when oxygen is added to the system rate of reaction has not changed significantly. The propagation steps with $\ce{O2}$ has been suggested as:
$$\ce{AR-^.CH-OH + O2 -> R-CHO + H-O-O^.} $$
$$\ce{H-O-O^. + AR-CH2-OH  -> AR-^.CH-OH + H-O-O-H} $$
Ref.1 and Ref.2 have reported the first order kinetics of this reaction with $k = 0.96 \times 10^{-4}$ at $\pu{25 ^\circ C}$ and $k = 2.03 \times 10^{-4}$ at $\pu{30 ^\circ C}$, which are consistent with other published data.
Note: More insight on mechanism, read Ref.3.
References:

Robert Warren Miller, "The stoichiometry of the reaction of benzenediazonium chloride with ethanol," MS Thesis 1957, University of Arizona, Arizona (PDF).
A. E. Kelley, PhD Dissertation 1956, Purdue University, Indiana.
DeLos F. DeTar, Takuo Kosuge, "Mechanisms of Diazonium Salt Reactions. VI. The Reactions of Diazonium Salts with Alcohols under Acidic Conditions; Evidence for Hydride Transfer," J. Am. Chem. Soc. 1958, 80(22), 6072–6077 (https://doi.org/10.1021/ja01555a044).


Why does the liquid in an Oralmedic stick flow uphill after I turn it upside-down?

The most likely explanation is a combination of two capillary effects, where a combination of surface tension and adhesion of the liquid to surrounding solid surfaces causes macroscopic flow to occur.

For the first, the internal diameter of the hollow stick is small enough that if one were to pull off the cotton buds at the top and bottom, capillary forces would likely cause the fluid to flow only very slowly through it. This provides some resistance to flow, sufficient to keep the liquid from gushing up to the 'snapped bud' end when you invert the stick. The surface tension associated with this capillary effect also results in the liquid behaving like a "plug" in the hollow tube: gas cannot "slide" around the liquid meniscus, which is also an important component to the physics.

The other role of capillary action is active in the fine threads of the lower cotton bud itself.  Once the top bud is snapped, the liquid starts to slowly flow down toward that unsnapped bud. Once the liquid reaches it, it wicks aggressively into the finely structured cotton.  Since no gas can get past the liquid in the tube, conservation of mass requires that the rest of the liquid continue to flow down toward the unsnapped bud.

The reason why flow continues toward the unsnapped bud when you invert the stick is a combination of all three of the above factors: the capillary action drawing the liquid into the unsnapped bud is essentially unaffected by the inversion of the stick, and since no gas can slip around the remaining liquid in the tube, that remaining liquid has to continue flowing up toward the unsnapped bud.

Another way to think about it is this: as each microscopic parcel of liquid wicks into the unsnapped bud, it leaves a region of slightly lower pressure behind it.  Since the end of the liquid column facing the snapped bud is open to atmosphere, that results in a pressure differential against the snapped-bud side of the liquid column that pushes the liquid up toward the unsnapped bud.


In a weak acid-strong base titration why is [H+] no longer equal to [A-] after addition of the base? (A2 level chemistry)

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$

The base removes the hydrogen ions from solution, and the acid further dissociates to maintain LeChatelier's equilibrium. The $\ce{H3O+}$ ion concentration will decrease as it is continually removed whereas the $\ce{A-}$ concentration will increase until the acid is fully neutralized into a salty solution.

I can't make much sense of your chemical equations, but it looks to me like they are incorrect. There is no reason for formation of the ionic compounds in solution. They should actually be written in reverse. And you have no strong base to react with the hydronium ions.


Is there an opposite to shielding effect?

Slater's rules are an attempt to lump the effect of all other electrons on the wavefunction, and thereby other properties such as energy, of an electron (described by a hydrogen-like wavefunction). The effect of electron-electron repulsion is modeled indirectly by saying that inner shell electrons effectively screen the attractive nuclear charge sensed by electrons further away from the nucleus. For simplicity screening is modeled as a one-sided effect: electrons further from the nucleus are not expected to significantly shield those closer to it, since an outer electron is close to the nucleus less frequently than an inner one (or, using more accurate language, the density near the nucleus of the outer electron is lower).
Slater's rules amount to a method of estimating appropriate exponents in a hydrogenic approximation of the electron wavefunction. They are semi-empirical, useful as a guide to explain why certain trends are observed, and fit data because they contain "fudge-factors" (parameters obtained from fits to data, not from fundamental theory). Still, that they work at all indicates what amazing insight Slater had in selecting the shape of the function and using the method to predict a number of atomic properties across the periodic table. If you haven't you should look at his original paper (Ref. 1), it's impressive and accessible. This is also fairly well explained in a Wikipedia article.
Reference

Slater, J.C. Phys. Rev. 36 (1): 57–64. doi:10.1103/PhysRev.36.57.


Calculating volume of gases in different conditions

I haven't found any flaws in your math either, and I can only join those who recommended to include units in your calculations. By googling around I found this problem appeared in 2012 Australian Science Olympiad Exam, and $\pu{13.1 mL}$ was one of the options there. Maybe your source of the answer key is incorrect?

Now to the fun part. I'm pretty sure you were expected to use ideal gas law, but come on, it's Venus and no way under the pressure of $\pu{92 atm}$ any gas would behave even remotely as the ideal one. In fact, $\ce{CO2}$ on Venus with its critical temperature of $\pu{304 K}$ and critical pressure of $\pu{7.39 MPa}$ is going to be in supercritical state!

You have to use Van der Waals' equation:

$$
pV_\mathrm{m} = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2},
$$

where $p$ – pressure; $V_\mathrm{m}$ – molar volume; T – temperature; R – molar gas constant; $a$ and $b$ – characteristic parameters related to the critical temperature $T_\mathrm{cr}$ and pressure $p_\mathrm{cr}$ (reduced form):

$$
a = \frac{27R^2T_\mathrm{cr}^2}{64p_\mathrm{cr}};
\qquad
b = \frac{RT_\mathrm{cr}}{8p_\mathrm{cr}}
$$

Using the above-mentioned values coefficients $a$ and $b$ can be found:

$$
a = \frac{27\cdot (\pu{8.314 J mol-1 K-1})^2\cdot (\pu{304 K})^2}{64\cdot\pu{7.39e6 Pa}} = \pu{3.65e-1 J2 mol-2 Pa-1}
$$

$$
b = \frac{\pu{8.314 J mol-1 K-1}\cdot \pu{304 K}}{8\cdot \pu{7.39e6 Pa}} = \pu{4.28e-5 J mol-1 Pa-1}
$$

Now you need to plug both values in the first equation and solve quadratic equation for $V_\mathrm{m}$. Wolfram Alpha claims $V_\mathrm{m} = \pu{2.56e-2 m3 mol-1}$, so

$$
V(\ce{CO2}) = \pu{2.56e-2 m3 mol-1}\cdot\pu{2e-2 mol} = \pu{5.12e-1 L},
$$

which, of course, deviates quite a lot from the value found using ideal gas law.


What is a food-safe level of exposure of plastic containers to heat?

You ask:

  I'd like to know if the information passed around the internet regarding the release of BPAs into the contents of plastic containers is grossly over-exaggerated or if there is a grain of truth to it that I should be concerned about to a degree.

In the United States, the Food and Drug Administration (FDA) has published  Bisphenol A (BPA): Use in Food Contact Application, in which they state:

  FDA’s current perspective, based on its most recent safety assessment, is that BPA is safe at the current levels occurring in foods. Based on FDA’s ongoing safety review of scientific evidence, the available information continues to support the safety of BPA for the currently approved uses in food containers and packaging.

From the link (above), you can go to the 2014 Updated safety assessment of Bisphenol A (BPA) for use in food contact applications, where they conclude that the "no observed adverse effect level (NOAEL)" for BPA is 5000 $\mu\mathrm{g}$ per kg of body weight per day ($\mu\mathrm{g}$/kg-bw/day) and that exposure estimates are "0.2 $\mu\mathrm{g}$/kg-bw/day (mean) and 0.5
$\mu\mathrm{g}$/kg-bw/day (90$^{\rm th}$ percentile) for the adult US population aged 2 years and older." The estimated dietary intake (EDI) is given as 1.1 $\mu\mathrm{g}$/kg-bw/day for population under 2 years old and 0.5 $\mu\mathrm{g}$/kg-bw/day for population greater than 2 years old. Thus, the exposure levels are on the order of a thousand times less than the NOAEL.

The European Food Safety Authority (EFSA) has recently (January, 2015) revised the tolerable daily intake (TDI) of bisphenol-A (BPA) from 50 $\mu\mathrm{g}$/kg-bw/day to 4 $\mu\mathrm{g}$/kg-bw/day. See this press release and this FAQ for more details.

The summary from the press release is:

  EFSA’s comprehensive re-evaluation of bisphenol A (BPA) exposure and toxicity concludes that BPA poses no health risk to consumers of any age group (including unborn children, infants and adolescents) at current exposure levels. Exposure from the diet or from a combination of sources (diet, dust, cosmetics and thermal paper) is considerably under the safe level (the “tolerable daily intake” or TDI).

The scientific opinion on the risks to public health related to the presence of bisphenol A (BPA) in foodstuffs, used by the EFSA to bolster the down-revision of the TDI for BPA, has been summarized in a very nice form suitable for the lay public, available here.

What about heating?

In one publication (Assessment of Bisphenol A Released from Reusable Plastic, Aluminium and Stainless Steel Water Bottles. J. E. Cooper, E. L. Kendig, and S. M. Belcher. Chemosphere 85(6), 943-947 (2011)), the authors state:

  At room temperature the concentration of BPA migrating from polycarbonate bottles ranged from 0.2–0.3 mg/L. Under identical conditions BPA migration from aluminium bottles lined with epoxy-based resins was variable depending on manufacturer ranging from 0.08 to 1.9 mg/L. Boiling water significantly increased migration of BPA from the epoxy lined bottles. No detectable BPA contamination was observed in water stored in bottles made from Tritan™ copolyester plastic, uncoated stainless steel, or aluminium lined with EcoCare™. The results from this study demonstrate that when used according to manufactures’ recommendations reusable water bottles constructed from “BPA-free” alternative materials are suitable for consumption of beverages free of BPA contamination.

and they conclude

  As shown previously by our results and those of many others, BPA migrates into water stored in polycarbonate plastic, especially when heated to high temperatures. Relatively high levels of BPA are found to be released into drinking water from reusable aluminium bottles with an epoxy-based liner. The amount of BPA released from the epoxy coating was greatly increased by heating. No detectable BPA was observed in water to migrate from unlined stainless steel, EcoCare™ copolyester lined aluminium, or copolyester Tritan™ plastic water bottles. To avoid exposures to BPA unlined stainless steel, copolyester lined aluminium or copolyester plastic drinking bottles should be used.

So, should you be worried? In my opinion, that's largely a function of whether or not you use plastic that contains BPA. A few years ago, it might have been difficult to find BPA-free plastics, but there is no shortage of BPA-free items available now. It's easy enough these days to remove any doubt from the equation by simply using containers that don't contain BPA.


Chemiluminescence - Determining Temperature from Luminol Imaging

However, given the intensity of the light and with the knowledge that luminol is being used, is it possible to determine the approximate temperature of the point based on its intensity and the intensity of the surrounding region?

Probably not! The energy released from the excited state of the 3-aminophthalate dianion formed in the oxidation of luminol (or rather its dianion) by hydrogen peroxide in alkaline is released in form of light. There's little to no heat released in this chemoluminescent reaction.


What happens to the "leftover" Ions in the Mass Analyzer

Most would end up being sucked out though the vacuum pump with the other residual gases. I assume that a lot of them collide with the sides of the vacuum chamber and are neutralized before sublimating in the vacuum.

The quads can become contaminated and require cleaning after several years of heavy use. Because there are such small quantities at that stage of the process, it's not uncommon for an MS to reach the end of it's service life before the quads need to be cleaned.


To what degree an N99 active coal filter can protect against products of burning paraffin oil?

Activated carbon's forte is vapor adsorption of hydrocarbons due to the activated carbon's high surface area, and this mask is probably very good at adsorbing fumes from hot parafin.

However, activated carbon does not tend to make a good material for filtering out the most harmful components of that kind of smoke, namely particulates smaller than 2.5 microns. In fact, the specification on the mask for particulate removal is that "that it blocks dust of size 0.1 mm and more" (meaning 100 microns and larger).  The World Health Organization has classified particulate matter of 10 micron (one tenth of the size claimed to be removed by the mask) to be a Group 1 carcinogen $\mathrm{^{(1)}}$.

Your filters are likely very good at filtering the particulate size most relevant to a particular application. However, for your application (production of very fine smoke particulates), I suggest a mask that is rated to remove 95% of $\pu PM_{2.5}$.

1) Outdoor Particulate Matter Exposure and Lung Cancer: A Systematic Review and Meta-Analysis, Environ Health Perspect; DOI:10.1289/ehp.1408092


Is it possible to design molecules of desired qualities?

Your title and the question itself arise two different issues, and I'll try to address both.

Yes it is possible to design molecules of desired qualities, only it's not all that simple. It requires a great deal of calculations, practice, empirical laws, heuristics, experimentation, guesswork, and magic. Eventually this body of knowledge grew so wide that people started to count it a science in its own right. It became known as chemistry.
Yes we can take into account every property of every atom and compute the properties of any compound they constitute. The problem, though, quickly becomes prohibitively expensive in terms of computation time. The reverse problem (to come up with a molecule of desired properties via calculations alone) is even more costly.

So it goes.


First I'd locate the bottle which causes the problem. Usually HCl is #1 suspect, but to be sure you can put a vial with smelling salts (aqueous solution of $\ce{(NH4)2CO3}$) or ammonia in the box with acids; white coating of $\ce{NH4Cl}$ on the bottle signifies the leak.

It's also a good practice to store acids in glass bottles with a proper joint (teflon ring) and a screw cap (e.g. Merck's SafetyCap). I would strongly recommend to get the proper bottles as soon as possible. Plastic bottles are only used to reduce the production and transportation costs, they are a poor choice for a long-term storing of chemicals. Even thick plastic is prone to diffusion, whereas glass is not; also plastic is, well, plastic, and is prone to mechanical deformations, so it's also tricky to maintain an impermeability of a bottleneck-cap joint over time.

In the meantime I'd wrap the necks of the bottles with parafilm, and/or pour some baking soda (sodium bicarbonate) on the bottom of the container where you store the bottles. This should help to neutralize the vapors before they reach the furniture and your nose, but it's a temporary solution.


What is the difference between nuclear charge and ionic charge?

Nuclear charge is the charge present inside a nucleus, due to the protons. For example, a carbon atom has $+6e$ nuclear charge.

Ionic charge is the charge due to the gain or loss of electrons in the valence shell. So, the sodium ion has an ionic charge of $+1e$ (due to loss of one valence electron). That ion's nuclear charge is still $+11e$ (there are still 11 protons only).

The reason because nuclear charge as an answer to this question is unaccetable is because nuclear charge - being inside the nucleus - has largely no affect on the intermolecular forces between the magnesium atoms. Those magnesium atoms have intermolecular forces governed only by the valence electrons only.


How does the Lennard-Jones potential relate to energy?

We have some nomenclature issues here.

In physics, the electric potential and gravitational potential refer to energy per unit something. In the gravitational potential, it's usually energy per unit mass, e.g. kJ/kg. In the electric potential, it's usually J/C (more commonly known as a Volt).

Here, though, we're not talking about a potential in that sense. Rather, I suspect (though I may be wrong) that potential was once a shorthand for potential energy surface, which is a way to think about the potential energy of a system with respect to various parameters. In the case of an L-J potential, there is only one parameter, the distance between the two particles of the system.

As mentioned in my comment though, I don't understand what you mean by two systems. The Lennard-Jones potential describes the energy of one system. If you put two of these systems together, the potential becomes ill-defined.

Added later: I found today that the Wiki page on thermodynamic potential says the same thing, namely that the "thermodynamic potential" is actually an energy instead of a potential. It quotes an ISO/IEC document that I don't have access to. I'm still not sure of the origin of this nomenclature.


Reactions of differents halohydrins with tert-butoxide

Chair conformation RR-1 (shown as its alkoxides) is in equilibrium with conformation RR-2, which is the unfavorable partner owing to the strong 1,3-diaxial Me/alkoxide interaction. It is this conformation that leads to the epoxide 1. This route is possible if the rate of formation is much greater than the rate constant for RR-1 --> aldehyde 2. RR-1 has concentration on its side and anti-periplanar alignment of bonds for ring contraction. I prefer ring contraction but this does not preclude epoxide formation.
The RS isomer has two chair conformations, RS-1 and RS-2.  Again, RS-2 is an unfavorable conformation (vide supra) but has proper bond alignment for ring contraction. Conformation RS-1 has the necessary bond alignment for E2 elimination to form cyclohexanone 4 via the enol 3.
A priori, it is not easy to predict with certainty what the product distribution will be. The epoxide can only arise from the RR isomer while aldehyde 2 can arise from either bromohydrin. Organic chemistry is an experimental science.


Why do different elements form different types of carbides?

According to wikipedia,

  Carbides can be generally classified by chemical bonding type as
  follows:

  salt-like carbides
  covalent carbides
  interstitial carbides
  "intermediate" transition metal carbides.

This type of classification is based on the electronegativity of the element to which the carbon is bonded.

According to this site:-

  The most electropositive metals form ionic or
  saltlike carbides, the transition metals in the middle of the periodic
  table tend to form what are called interstitial carbides, and the
  nonmetals of electronegativity similar to that of carbon form covalent
  or molecular carbides.

Actually credits goes to @jatin  who gave me an important hint:-

  You might wanna consider diagonal relationship which accounts for
  similarities between those particular metals.

As you can see that Li and Mg, Na and Ca and Be and Al all form diagonal element pairs. So, they are expected to have similar properties. So, considering their electronegativity, the diagonal element pairs have almost same value (though Li and Mg electronegativity is quite different). So they form similar type of carbide.

(original picture)

As for the intestitial carbides, @bon said that

  Insterstitial carbides only form with large cations (Ti and W), as you might
  expect if the anion (very small) to fit in the spaces in the cation crystal
  structure.


Why does the cathode in an OLED require a low work function and the anode require a high work function?

The operation of light emitting diodes involves the electrochemical creation of radical - cations and -anions that upon recombination lead to molecules in their emitting excited states.
The formation of the above charged species happen when the applied forward potential is positive enough to inject (reduction) electrons from the cathode to the LUMO of the molecule and extract electrons (oxidation) from the HOMO to the anode.
Therefore, the electrodes must have work functions about the levels you want to operate onto.
This means that the cathode must have oxidation potential / work function near the LUMO energy level. Opposite, to work on the extraction of electrons from the HOMO, a metal lining down at higher oxidation potential is needed.
Note that this design not only reduces the potential needed for the functioning of the LED, but minimise energy barriers (if any) at the emitting material / electrode interfaces.
Also note that a reverse order can still lead to light emission, but the ON potential to be applied to this "unnatural" configuration is inevitably higher.


How do valence shell configurations affect attraction and repulsion forces between atoms?

Firstly, why should covalently bonded atoms be incapable of bonding? Bond changes are the very basis of quite a huge amount of chemical reactions. Oxygen occurs in the $\ce{O2}$ form, which is covalently bonded, but readily breaks its O–O bond in order to form other bonds.

Now, to your question. Why are elements in group 18 (generally: think of the xenon fluorides) inert? Why is sodium so willing to give up its electron, while neon doesn’t form any compounds at all, when they only differ by one proton in the nucleus?

The answer lies in the orbital energies. You can find a table listing the outer shell orbital energies page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.). Let us compare the outer orbital energies of Ne and Na. Neon has two 2s electrons with an energy of $-48.47\ \mathrm{eV}$ and four 2p electrons with an energy of $-21.59\ \mathrm{eV}$, while the 3s electron in the sodium atom lies at $-5.14\ \mathrm{eV}$. The difference is overwhelming. When a ionic compound is formed, an electron is removed from the less electronegative atom and added to the more electronegative atom, forming a cation and an anion. Now, the Na electron is much easier to remove than the Ne electron – you just need $-5.14\ \mathrm{eV}$ to do the task, while eliminating the Ne electron is much more difficult. The electron is then placed into an atom such as fluorine, whose outer 2p subshell lies at $-18.65\ \mathrm{eV}$: you needed energy to remove the Na electron, but energy is released when you place it in the F atom (since it is put into a low energy orbital). Placing an electron from F to Na wouldn’t release energy because it is much more stable in the fluorine atom, due to their orbital energies (if you want to simplify things my anthropomorphizing the electrons, you could say they’re “happier” in the fluorine atom). Removing an electron from the neon valence shell is very difficult because it has very low energy; adding one doesn’t bring much because since the second shell is full, you would have to add it to the third shell – its energy is similar to the Na 3s shell, so it’s not convenient.

I have to note here that while the first ionization energy corresponds more or less to the orbital energy (Na: $495.8\ \mathrm{kJ/mol}$ which corresponds to $5.139\ \mathrm{eV/atom}$), the electron affinity doesn’t (F: $-3.399\ \mathrm{eV/atom}$), because of electron-electron repulsion in the fluorine p subshell. That’s why an isolated $\ce{Na+ - F-}$ compound cannot exist: the additional energy released when forming the lattice is needed to form a stable entity.

Having said this, we can move on. You may ask: Yes, but why are the F electrons more stable than the Na ones, when Na has more protons? The reason is that the outer Na electrons are in the third shell, and thus they have higher energies. In a hydrogen-like atom (an atom with only one electron, which is clearly an oversimplification, since it doesn’t involve screening) the orbitals have the following energies:

$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$

You can see that raising the $Z$ (atomic number) value from 9 (F) to 11 (Na) isn’t enough for lowering the energy, since $n$ (the principal quantum number, i.e. the shell) changes from 2 to 3.

So, this is why sodium forms positive ions, fluorine forms negative ions, and neon doesn’t form any ions at all (chemically). Nota bene: as I said, the formula above doesn’t take screening into account – when going down a group, it (usually: see d-block contraction) prevails over the growing nucleus. That’s why Ar is less inert than Ne, even though the formula above would suggest the contrary.

But what about covalent compounds? According to VB theory, bonding occurs through orbital overlap. Lower energy orbitals mean smaller overlaps, which is inconvenient if you want a stable bond: Carbon forms bonds while neon doesn’t because its orbitals are less stable (see table) and can overlap with other orbitals. Its orbitals are less stable because neon has more protons with the same principal quantum number – see the equation above to understand the concept.

In short: noble gases, especially those positioned higher in the periodic table, don’t bond because their outer orbitals have the lowest energies. Ultimately, this is the result of having the largest number of protons possible for a given outer shell in the ground state.


Simultaneous solubility of CaF2 and SrF2

Calculate the simultaneous solubility of $\ce{CaF_2}$ and $\ce{SrF_2}$ given that:

$\mathrm {K_{sp}(\ce CaF_2)= 3.9 \times 10^{-11}}$ and $\mathrm {K_{sp}(\ce SrF_2)= 2.9 \times 10^{-9}}$

Let $\ce{[Ca^{2+}]} = x$ and $\ce{[Sr^{2+}]} = y$
So the dissolving $\ce{CaF_2}$ contributes $2x$ of $\ce{F^-}$ and the dissolving $\ce{SrF_2}$ contributes $2y$ of $\ce{F^-}$. So the total $\ce{[F^-]} = 2x + 2y = 2(x+y)$
Substituting into the two $\text{K}_{sp}$ equations leads to "messy" cubic equations which could be solved numerically since there are two equations with two unknowns.
$3.9\times10^{-11} = x(2(x+y))^2 = 4x(x+y)^2$
$2.9 \times 10^{-9} = y(2(x+y))^2 = 4y(x+y)^2$
But going back to the original $\text{K}_{sp}$ equations we have:
$\ce{[Ca^{2+}][F^-]^2 = 3.9 \times 10^{-11}}$
$\ce{[Sr^{2+}][F^-]^2 = 2.9 \times 10^{-9}}$
and by dividing the two we get $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$ or $x = 0.0134y$. Substituting this into the second equation we get:
$2.9 \times 10^{-9} = 4y(x+y)^2 = 4y(1.0134y)^2 = 4.108y^3$ or $y=8.9\times10^{-4}$
So:
$\ce{[Sr^{2+}]} = 8.9\times10^{-4}$
$\ce{[Ca^{2+}]} = (0.0134)(8.9\times10^{-4}) = 1.2\times10^{-5} $
$\ce{[F^-]} = 2(8.9\times10^{-4} + 1.2\times10^{-5}) = 1.8\times10^{-3}$
Note that this can also be simplified using significant figures a different way. Since the two $\mathrm {K_{sp}}$ values only have two significant figures and  $\ce{[Ca^{2+}] = 0.0134 [Sr^{2+}]}$, we can assume that for all practical purposes that all the $\ce{F^-}$ comes from the dissolution of the $\ce{SrF_2}$. So:
$\ce{4[Sr^{2+}]^3} = 2.9\times10^{-9}$
Thus
$\ce{[Sr^{2+}]} = 9.0 \times10^{-4}$
$\ce{[F^-]} = \sqrt{\frac{2.9\times10^{-9}}{9.0 \times10^{-4}}} = 1.8\times10^{-3}$
$\ce{[Ca^{2+}]} = \dfrac{3.9\times10^{-11}}{(1.8\times10^{-3})^2}=1.2\times10^{-5}$


Concentrating Oxalic Acid by Evaporation

Depends on how hot the plate is I'd say. Oxalic acid is hygroscopic, so it will be hard to completely dry it by just subjecting it to heat as you might just cause it to break down.

There's a paper and procedure on Organic Synthesis on how to obtain the anhydrous form, so you should check that out to see if that meets your requirements.


Why does steel wool catch fire when rubbed with a battery?

Steel wool is a mat of very fine steel fibres. Because of their small size, it's easy to heat them up quickly to the point of burning and their relatively large surface area to volume ratio lets them oxidize quickly. If you try it with a flame, you'll find that it's quite easy to burn steel wool. When you short the terminals a battery across steel wool, a lot of current passes producing heat (think about incandescent light bulbs), and especially if you touch the battery such that only a few fibres are touching, enough heat is produced to cause them to burn.


Look at dipole moment as a vector with the head of vector pointing towards the more electronegative atom. Ccl4 has 4 vectors pointing outwards but these vectors cancel, hence 0 dipole. Chcl3, 1 vector from H towards c, 3 vectors from c to each cl atom. Note they have tetrahedral geometry. The net vector would be along one of cl atoms but as other cl atoms have components of their vectors opposite to the victorious cl atom, dipole would be reduced. Ch2cl2, net vector would be somewhere between the 2 c==>cl vectors but there are not much opposite components of vectors as compared to chcl3, it has greater dipole than the former. Ch3cl, net dipole is along the cl atom and further each H==>C dipole is increasing the magnitude of the vector. These problems are based on visualization and vectors help you visualize better.


How can potassium hydroxide be prepared from available potassium chloride (KCl) and Quick Lime (CaO)

Mixing $\ce{KCl + CaO}$, or $\ce{KCl + Ca(OH)2}$, will never produce pure $\ce{KOH}$ without $\ce{Ca(OH)2}$. And this Ca(OH)2 will prevent soap from being synthesized out of oil, as $\ce{(Ca(OH)2}$s destroy soap in case a little bit of soap has been synthesized. The only way of producing $\ce{KOH}$ out of $\ce{CaO}$ or $\ce{Ca(OH)2}$ is to mix it with potassium carbonate $\ce{K2CO3}$. Of course $\ce{Ca(OH)2}$ is not very soluble in water. But $\ce{CaCO3}$ is still more insoluble. So the following reaction is possible :$$\ce{Ca(OH)2 + K2CO3 -> 2 KOH + CaCO3(s)}$$ And the insoluble $\ce{CaCO3}$ can be filtrated to produce a KOH solution


How do you calculate the dipole moment of larger molecules?

The total dipole of a molecule can be thought of as the sum of dipoles of individual functional groups:

$$
\vec{\mu}_{\text{total}} = \sum_{i}^{N_\text{groups}} \vec{\mu}_{i}
$$

Because each dipole is represented as a vector with both magnitude and direction, this amounts to vector addition.

In the case of water, consider the dipole of each $\ce{-OH}$ group. The above reference gives it as 1.53 Debye, and the $\ce{H-O-H}$ angle is $105^{\circ}$. Because coordinates aren't given, the angle has to be used instead:

$$
\vec{P} + \vec{Q} = \sqrt{\lVert \vec{P} \rVert^{2} + \lVert \vec{Q} \rVert^{2} + 2\lVert\vec{P}\rVert\lVert\vec{Q}\rVert\cos{\theta_{PQ}}}
$$

which gives an estimate of 1.64 Debye. Not very good compared to 1.85 Debye! The group addition approximation completely neglects how the electronic structure of the molecule changes upon going from (using water as an example)

$$
\ce{2OH^{.}} \rightarrow \ce{H2O}
$$

The issue is probably that the two individual $\ce{-OH}$ groups interact strongly and favorably when combined together in water, compared to when they are apart; that is, the electron density changes quite a bit due to the bonding.

I've also neglected to explain how these groups can be defined; there is no unique definition of how to partition a molecule. For example, take a carboxyl group:

Are all four atoms treated together as a single group, or are the $\ce{C=O}$ and $\ce{C-O}$ and $\ce{-OH}$ treated separately? Maybe this is a simple case but it becomes more unclear for something like a protein or a coordination complex.

The molecular dipole can also be calculated using electronic structure theory, which removes the ambiguity surrounding how to partition a molecule and incorporates those "non-additive" effects.

As Geoff mentioned in the comments, the definition of the dipole moment created by a set of charges can also be used:

$$
\vec{\mu} = \sum_{a}^{\text{charges}} \vec{r}_a \times q_a
$$

where $q_a$ is the charge, and

$$
\vec{r}_a = \vec{R}_a - \vec{O} = (R_{ax} - O_x, R_{ay} - O_y, R_{az} - O_z),
$$

where $\vec{R}$ is the position of the point charge, and $\vec{O}$ is a common origin, usually taken to be either $(0,0,0)$, the center of mass, or the center of nuclear charge. It doesn't matter what this point is for uncharged (neutral) systems, but there is an origin dependence for anything with a non-zero total charge. The $\{\vec{R}\}$ are taken to be atomic positions, and the charges $\{q\}$ can come from simpler Mulliken or Löwdin population analyses, or more complicated schemes of which many are designed to reproduce total dipole moments. Care must be taken to deal with units properly, specifically when converting: $0.393430307 \, ea_0 = 1 \, \mathrm{Debye}$.


Way to establish the acid/base properties of a compound from its chemical formula?

$\hspace{4.3cm}$

Consider the two compounds above. Both have the formula $\ce{C5H10O2}$, yet one is about 20 orders of magnitude stronger an acid than the other. Without knowing the general structure of a molecule (what functional groups it is comprised of), suffice it to say no, you cannot know how acidic the molecule is.


Nomenclature of Coordination compounds containing nitrogen

As per rules, double m is used when the ligand is just $\ce{NH3}$, ammonia:

Cite from IUPAC Nomenclature of Inorganic Chemistry, Recommendations 2005:

  IR-7.1.3 Representing ligands in additive names

  Additive names are constructed by placing (sometimes modified) ligand names as prefixes
  to the name of the central atom. For anionic ligands, the anion endings ‘ide’, ‘ate’ and ‘ite’
  (see Section IR-5.3.3) are changed to ‘ido’, ‘ato’ and ‘ito’, respectively,
  when generating
  these prefixes. Names of neutral and cationic ligands are used unchanged, except in a few
  special cases, most notably water (prefix ‘aqua’), ammonia (prefix ‘ammine’), carbon
  monoxide bound through carbon (prefix ‘carbonyl’), and nitrogen monoxide bound through
  nitrogen (prefix ‘nitrosyl’) (cf. Section IR-9.2.4.1).

E.g. in $\ce{[CoCl(NH3)5]^{2+}}$ pentaamminechloridocobalt(2+).

For amines, the ligand name is unchanged, e.g.
$\ce{[Co(en)3]^{3+}}$ tris(ethane-1,2-diamine)cobalt(3+).


Why a low conductance ion such as lithium is used in batteries?

The conductivity values you cite are for ions in water. there is no water in a lithium battery otherwise, it would explode and burn before leaving the factory. Beside low conductivity can be compensated by increasing concentration of the ions or a variety of other architectures.

Further, there's a few reasons lithium is optimal in batteries.  It's not environmentally toxic (like beryllium), has a higher oxidation potential than the other alkali metals, but most importantly is its high energy density in both Joule per weight and per volume, as well as power per weight, which no other metal can beat.  The only disadvantages for lithium are its flammability and cost.

(source)


Are there any non-Jahn-Teller type distortions?

Apparently there is some theorem e.g. to be found in an article from Isaak B Bersuker with title "SPONTANEOUS SYMMETRY BREAKING IN MATTER INDUCED BY DEGENERACIES AND PSEUDODEGENERACIES" in Advances in Chemical Physics, Volume 160, First Edition. Stuart A. Rice and Aaron R. Dinner. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc; p 187-

The theorem reads like:

"DEGENERACY AND PSEUDODEGENERACY ARE THE ONLY SOURCE OF SPONTANEOUS SYMMETRY BREAKING (SSB) IN ATOMIC SYSTEMS"

And can be reformulated by:

" [...] can we get structural instability of high‐symmetry configurations in any electronic state without involving interaction with other states [...]? The answer to these questions is no: any instability and SSB in polyatomic systems leading to their formation, deformation, and transformation involves at least two electronic states, degenerate or pseudodegenerate. This is the two‐state paradigm in structural instability."

So in essence that means all types of molecular distortions can be traced back to some JT effect.


Potential energy of electron density at the nucleus

I think it's useful to expand slightly on the comments because I think the language here should be as clear as possible.

You're concerned about the convergence of the integral of a function when some piece of the function is unbounded, i.e., for $f(x)$ on domain $[a, b]$, $\exists c \in [a,b]$ such that $f(c) = \infty$. Frequently, integrals of this type do not converge.

Your function certainly looks like that but the resolution of this problem is actually hidden in the $dV$. Typically, we treat the integration of the wave function as a separable problem, i.e, we integrate the radial and spherical components separately. When you look at the integral over the distance from the nucleus, you get a factor $4\pi r^{2}$ because you are integrating shells and $dV = 4\pi r^{2} dr$. So even though $\frac{1}{r}$ diverges at zero, $\frac{r^{2}}{r}$, while not defined at zero, does converge.

$$\lim_{r\rightarrow 0}\frac{r^{2}}{r} = 0$$

From this perspective, no piece of the integral is infinite, and the integral converges.


How to verify the accuracy of a pH meter

Your school should have a purchased buffer solutions that have been calibrated against a NIST standard.  If they don't, have them purchase several in order to span the entire pH range, here's one from Sigma-Aldrich.  For your calibration, pick a standard solution that is close to the range you plan to work in.  Here's a couple of links with some more "tips" on how best to actually perform the calibration.

tips #1
tips #2


Entropy as a state function independent of the path

Entropy is indeed a state function, and thus depends only on the state of the system.  Hence it doesn't matter how you get from state A to state B, the entropy change will be the same.  The analogy would be that it doesn't matter which path you use to get from the base of a mountain to the summit, your elevation change will be the same.  This is because altitude is a state function: your altitude depends only on how high you are (the state of your system), not how you got there.

But, since $dS= \frac{\text{đ}q_{rev}}{T}$, one can calculate the entropy change by integrating along a path connecting the two states, but that path must be reversible.  I.e., it doesn't matter if the way the system changed from A to B was reversible or irreversible.  $\Delta S$ will be the same.  However, to calculate $\Delta S$ for that change from A to B, you need to integrate along some reversible path that connects A to B.

Also, it is possible to sum the effects of an infinite number of infinitely small steps.  That's what happens when we do an integration.  [More precisely, in an integration we calculate the sum in the limit as the step size goes to zero and the number of steps goes to infinity.]  For a graphical representation of this, take a look at the graphics in Wikipedia's entry on Reimann sums (https://en.wikipedia.org/wiki/Riemann_sum), and imagine what happens in the limit as the bar width goes to zero.

The comments by Andrew, and Chet Miller, are both useful additions.  Further expanding on Andrew's comment: The system's net change isn't affected by the path.  This means that, to distinguish between the result of making a change in a system by a reversible vs. irreversible path, we need to look at differences in effects upon the surroundings.


The mechanism that I have submitted in the edit of my post is correct. The reason both products are formed, is due to the the enol intermediate being planar with respect to the oxygen, allowing it to be attacked from either plane. This allows for the formation of a the so-called racemic mixture.


What is the best way to calculate the mass percentage of Calcium Carbonate from water?

Water contains negligible amount of $\ce{CaCO3}$ or $\ce{MgCO3}$, as these are almost insoluble in water.
What you are interested in are bicarbonates $\ce{Ca(HCO3)2 + Mg(HCO3)2}$. Their content is known as the temporary or carbonate water hardness that can be expressed as $\ce{CaCO3}$ equivalent.
The complement to the carbonate hardness is the permanent hardness, formed mostly by $\ce{CaSO4, MgSO4, CaCl2, MgCl2}$. The sum of the temporary and permanent hardness is called the total hardness.
Bicarbonates ($\ce{HCO3-}$) are not threat to health themselves, as they convert to $\ce{CO2}$ in the stomach and are present in blood as the component of the blood $\ce{pH}$ buffer.
What can be a subject of health concern is the total content of $\ce{Ca^2+}$ (and in lesser extent $\ce{Mg^2+}$). It is again expressible as $\ce{CaCO3}$ equivalent.
Calcium/magnesium content is not measurable by resources usually found in households or local shop center.
The best way is to ask the water provider for the data.
The 2nd best way, close to home-like determination, is to obtain the set for aquarists from a pet shop for determination of carbonate and total water hardness. I am a former aquarist, and this was available even in 1980s in the former communist Czechoslovakia.
Determination is done by counting  the drops of the colored solutions, mixed with the tested volume of water, until the color changes.
Carbonate (in fact bicarbonate) hardness determination is based of reaction of an acid with bicarbonates: $$\ce{HCO3- + H+ -> H2O + CO2 ^}$$
Drop counting ends when yellow color (methyl orange dye indicator) changes to red. The dye turns red by the first excess of the added acid, as has red color in acidic solutions.
Total hardness is based on reaction of $\ce{Ca^2+ + Mg^2+}$ with ethylendiamintetraacetic acid (EDTA) in alkaline solution, enriched by the addition of the dye indicator Eriochrome black T,
$$\ce{Ca^2+ + H2EDTA^2- -> 2 H+ + CaEDTA^2-}$$ The EDTA first reacts with free $\ce{Ca^2+ + Mg^2+}$. The drop count ends when red turns to blue, as EDTA captures $\ce{Ca^2+ + Mg^2+}$  bonded to the Eriochrome black T.
As result, you obtain the carbonate (temporary) and total (temporary + permanent) hardness in 2 possible units: $\pu{mmol/l}$ or $\pu{dGH}$ (the degree of General Hardness).
$\pu{1 mmol/l = 5.6 dGH}$.
$\pu{1 mmol/l}=\pu{100 mg/l}= 0.01\%$, as molar mass of $\ce{CaCO3}$ is cca $\pu{100 g/mol}$.
The typical total hardness of drinking water is $\pu{1\!-\! 3 mmol/l}$,  i.e. the equivalent of $\pu{100\!-\! 300 mg/l}$, resp. $0.01\!-\! 0.03\%$ of $\ce{CaCO3}$.


Calculating bond disassociation energy and bond energy for ATP

An example program is GAMESS.  It is free. But be warned: the calculation of BDEs is an extremely difficult thing to do.  I am presuming that you are a beginner, so you have to be careful for issues surrounding which BDE:
\begin{align}
\ce{A-B &-> A+ + B-}   \tag{ionic}\\
\ce{A-B &-> A. + .B}  \tag{homolytic}
\end{align}

Most electronic structure theory programs assume gas-phase, absence of solvent.  We know that homolytic dissociation is favored in the gas phase, as charge separation into ions is unfavorable in a Coulombic energy.

That being said, the run-of-the-mill electronic structure theory calculations (e.g., HF, DFT, MP2, CCSD, etc. etc.) will not break the atoms apart in a homolytic manner.  They tend to break apart atoms in the ionic fashion.  But this is a problem, as ions are abhorrent in the gas phase!  Basically, simple electronic structure theory methods over-estimate the bond strength, as the product of the dissociation (ionic) is erroneous.

You must use multiconfigurational methods to get it right (e.g. CASSCF, MCSCF). But these are very, very difficult unless you have had a lot of training.  I did them for years, and they were still hard for me.  But they will give you $\ce{A. + .B}$ as a product!  That is the gas-phase BDE.

You may be able to study closed-shell, closed-shell dissociations without much problem, using the HF and DFT methods.  Examples include dimers, host-guest interactions where the products look like good-old fashioned molecules that can stand by themselves. (Think water dimers, etc.)

Also, beware of Solvent Effects (favoring ionic dissociation $\ce{A-B -> A+ + B-}$) and solvation energy.  In fact, the solvation energy is why $\ce{Na-Cl}$ breaks apart into $\ce{Na+}$ and $\ce{Cl-}$.  In the gas phase, it would be $\ce{Na.}$  and $\ce{Cl.}$ radicals.

In summary, BDEs are very difficult things to compute to qualitative accuracy. Quantitative accuracy is even harder.  You may want to have an expert look over your procedure and results before you trust them completely.


Why does radium have a higher first ionisation energy than barium?

I think it's also important to mention relativistic effects here.  They already start becoming quite visible after $Z=70$, and $\ce{Ra}$ lies a good bit after that.

In very heavy atoms, the electrons of the $\ce{1s}$ orbital (actually, all orbitals with some electron density close to the nucleus, but the $\ce{1s}$ orbital happens to be the closest and therefore most affected) are subjected to very high effective nuclear charges, compressing the orbitals into a very small region of space. This in turn forces the innermost electrons' momenta to be very high, via the uncertainty principle (or in a classical picture, the electrons need to orbit the nucleus very quickly in order to avoid falling in). The momenta are so high, in fact, that special relativity corrections become appreciable, so that the actual, relativistically corrected momenta, ($p_{\text{relativistic}}=\gamma p_{\text{classical}}$) are somewhat higher than the approximate classical momenta. Again via the uncertainty principle, this causes a relativistic contraction of the $\ce{1s}$ orbital (and other orbitals with electron density close to the nucleus, especially $\ce{ns}$ and $\ce{np}$ orbitals).

The relativistic contraction of the innermost orbitals creates a cascade of electron shielding changes among the rest of the orbitals. The final result is that all $\ce{ns}$ orbitals are contracted, getting closer to the nucleus and becoming shifted down in energy. This is relevant to the question because the $\ce{7s}$ valence electrons in $\ce{Ra}$ are more attracted to the nucleus than one would expect from a simple trend analysis, since they rarely take into account the increase of relativistic effects as one goes down the periodic table.

Thus, the first (and second) ionization energy of $\ce{Ra}$ becomes higher than expected, to the point that there's actually a upward blip in the downward trend. Eka-radium ($Z=120$) would have far stronger relativistic effects, and can be expected to have a significantly higher ionization energy compared to $\ce{Ra}$. In fact, relativistic effects will conspire to make the group 2 metals slightly more noble! Though the periodic table becomes such a mess near the super heavy elements that it's hard to say whether it'll be a clearly visible trend, or just one effect to be combined with several others.


Writing the net reaction for reaction of sodium hypochlorite with hydrochloric acid and potassium bromide

I'd like to add some things to what Tanith Rosenbaun said.
First, the point of net ionic equations is not just to write out the equations in terms of ions, but to focus on only those ions that experience a change in charge or valence.  Ions which don't change in charge don't really participate in the reaction and called spectator ions.
Here's the last equation from Rosenbaum's answer:
$$\ce{2H3O+ + 2Cl- + 2Na+ + CO3^2- -> 2H2O + 2Cl- + 2Na+ + CO2}$$
Notice that the sodium and chloride ions are unchanged in this reaction.  So we drop them, and just write out the ions that change:
$$\ce{2H3O+ + CO3^2- -> 2H2O + CO2}$$
This is the net ionic equation: it focuses precisely on what is interesting, which in this case is the neutralization reaction.
Now let's go back to your problem:  People have been asking what you know about it.  You seem to know that you are transferring a solution of chlorine dissolved in cyclohexane.  You got that chlorine from the following reaction:
$$\ce{2HCl + NaClO -> Cl2 + NaCl + H2O}$$
and the chlorine preferentially dissolves in the cyclohexane.  Now it's important to understand that the cyclohexane doesn't play any role in the second part of your experiment, it's just how you transport the chlorine.  (And it's why you need to be careful with acids around bleach: chlorine gas is poisonous!)
The key question is: what happens when you put elemental chlorine in the presence of dissolved potassium bromide?
$\ce{Cl2 + KBr -> ?}$  (Note: this is not balanced yet!)
In particular, do you know what you produced when that solution turned dark red?  Hint: this is a very important property that halogens like chlorine and bromine have towards each other, which is indeed seen best via a net ionic equation.  I'll happily provide the answer if you've tried some stuff and are still stuck.


This structure has a phosphonic and a carboxylic acid moieties with a minimum number of carbon atoms. As such, it should be soluble in basic water to deprotonate at least one site


Is HPLC the best way to detect residues from anti-coagulant rodenticides

UV(/VIS) spectroscopy is the most common HPLC detection method, it's quite cheap and powerful but you can only detect substances which absorb light in the range you are using for detection. You also don't get any real structural information, most of the time identification is only possible using reference material. But most HPLC systems will be equipped with such a detector.

Fluorescence detectors can be used for fluorescent molecules and usually you got a much lower limit of detection. Like for UV/VIS you don't get structural information, but identification works quite well, since you can selectively only detect molecules with the used excitation and emission wavelengths.

Other detectors would be for example refractive index detector, which only really work well if HPLC is run isokratic or MS, which gives you structural information.


There's no structural formula because it's what we call an amorphous solid. That is, it's like glass. It has no clearly defined crystal structure, and is likely to be extremely hard, but also extremely brittle.

As to why it's ferromagnetic or fluorescent, I'm unsure. It's likely they don't quite know, either. I suspect it has something to do with the curious bond structure of Q-carbon, which is a mishmash of both 3-way and 4-way bonds. In comparison, graphene is uniformly 3-way bonds, while diamond is a fairly uniform composition of 4-way bonds.

I speculate (don't quote me on this, I can't find a reliable source) that the way the electrons interact across the carbon atoms has something to do with why it illuminates (as photon emission is not an uncommon effect of electrons changing energy levels) and with why it's magnetic (as ferromagnetism explicitly refers to the quantum mechanical properties of electrons).

Either way. Please take this explanation with a grain of salt, but, if you're able to find more reliable sources, feel free to tack it on.


How many vibrations are expected for butatriene and cyclobutadiene?

Perhaps an explanation of where these mysterious equations come from will help you understand how to apply them properly.

I only need you to make one, single (correct) assumption: all atoms in a molecule are capable of moving in each of the three Cartesian coordinate directions.

That is, each of those little hydrogens can wiggle around in the x, y, and z coordinates as much as it pleases.

That means that we have a total $3N$ ways that the whole molecule can move because each of $N$ atoms can move in three directions.

Now, let's consider how molecules move. If every one of those atoms happened to be moving in the same direction, what would that look like?

Well, that would be translation of the molecule (i.e. moving). Any molecule can do this in three directions.

So, excluding translation we have $3N-3$ ways of moving around.

Now, another thing we know molecules do is rotate. By the same reasoning, we can say that a molecule can rotate in all three coordinate directions.

This leaves us with $3N-6$ ways to move around.

But, what if that molecule was linear? Well, then that molecule can only rotate in two unique directions because spinning around like a cylinder is known to be a zero energy process. Another way of saying that is that it doesn't happen.

So for a linear molecule there $3N-5$ ways of moving around still because we only subtracted out two rotational modes.

Of course, now we see that the only other way our molecules can move is by vibrating.

Thus, for a linear molecule, $$Number\ of\ normal\ modes=3N-5$$and for a nonlinear polyatomic molecule,$$Number\ of\ normal\ modes=3N-6$$

So, using these equations which now hopefully make some more sense, it's quite obvious that we do indeed include the hydrogens in our $N$ and find there are 18 normal modes of vibration for both molecules.

And, if there were an isomer of the formula $\ce{C4H4}$ which was linear then there would be 19 normal modes, but there is no such molecule so, yes all molecules with the molecular formula $\ce{C4H4}$ have 18 vibrational modes.


Amending soil pH to change the color of hydrangea flowers

Although it is true that soil $\mathrm{pH}$ plays a role in the color of hydrangeas, the presence of aluminium is actually what determines the flower color. Aluminium is a naturally occurring mineral that is found in most  soils, however, just because a mineral is in the soil doesn’t mean the plant is absorbing this mineral. This is where soil $\mathrm{pH}$ comes into play – plants are able to absorb aluminium when the soil $\mathrm{pH}$ is between 4.5 and 5.8 with the best range being between 5.0 and 5.5. When the $\mathrm{pH}$ moves to 6.0 and above then the plants will not be able to absorb the aluminium despite it being present in the soil.

Adding either sulfur or magnesium sulfate (epsom salt) to the soil will lower the $\mathrm{pH}$, but this will not help with the presence of aluminium in the soil, which is why many aluminium sulfate is so highly recommended.

This is not to say that magnesium sulfate would not be helpful in other ways, however, as magnesium is the central atom of pigment chlorophyll, which will help to improve the foliage quality of your hydrangea.


Calculate water hardness from grams of CaCO3

Unless I'm missing something, I don't see the problem with your working either. Perhaps check whether you got the original concentration of $\ce{CaCl2}$ given in the problem right (the volume of the solution does not affect your final answer). The given answer may well be wrong, it's irritating but it does happen.

There is one thing I want to comment on though: never round off your answer too early. Since you obtained the intermediate answer of $801 \text{ mg}$, I am assuming you used a precise molar mass for $\ce{CaCO3}$ - I will take this to be $100.09 \text{ g mol}^{-1}$. (If you used the value $100$ you would have obtained $800 \text{ mg}$.) Your calculations should be:

Number of moles of $\ce{CaCO3}$ formed upon addition of excess $\ce{CO3^2-}$

$$\eta_{\ce{CaCO3}} = (0.400 \text{ mol dm}^{-3})(0.02 \text{ dm}^{3}) = 0.008 \text{ mol}$$

Mass of said $\ce{CaCO3}$

$$m_{\ce{CaCO3}} = (0.008 \text{ mol})(100.09 \text{ g mol}^{-1}) = 0.80072 \text{ g} = 800.72 \text{ mg}$$

Hardness of water in $\text{mg L}^{-1}$

$$[\ce{CaCO3}] = \frac{800.72 \text{ mg}}{0.02 \text{ L}} = 40036 \text{ mg L}^{-1}$$

This is your final answer and if you want to round to the appropriate number of significant figures, you should do so only at this point. Your problem is that you rounded too early (after step 2) and this means you obtained an imprecise final answer, $40050 \text{ mg L}^{-1}$.

Here, the appropriate number of sf is technically 3, since the original concentration of $\ce{CaCl2}$ is given to you in 3 sf. Since you obtained the imprecise answer $40050 \text{ mg L}^{-1}$, you would have rounded this to $40100$. But if you stuck to the precise value at step 2 and obtained the answer $40036 \text{ mg L}^{-1}$, you would round this to $40000$.


Is there a good sol gel method for carbon nanotube thin film deposition?

I think it may depend on what you mean by "effectively." I imagine that it would be difficult to get a thin film of oriented CNTs using a sol-gel method, since the gel would necessarily have randomly oriented tubes.

The other potential issue is making the sol-gel in the first place. Unless they are functionalized, carbon nanotubes won't gel in hydrophilic solvents. It might be possible to create gels in hydrophobic solvents, but a quick search didn't turn anything like that up.

This link has a brief description of a typical process - you can see that the idea is to make the sol-gel with something that will gel in aqueous solution first (silica, for example), and then add the nanotubes.

So, if you want pure CNT thin films, that would probably be a lot more difficult.


What is the substituent on this benzaldehyde? (MS attached)

The molecular ion $\ce{[M]^{\bullet +}}$ peak is likely at 136 m/z. The peak at 137 m/z is due to the 1.1% natural abundance of $\ce{^{13}C}$.

The peak at 135 m/z is the $\ce{[M-1]^{+}}$ peak typical of aldehydes:

$$\ce{ [R-C(=O)-H]^{\bullet +} -> [R-C+=O <-> R-C\equiv O+] +H ^{\bullet} }$$

Another typical fragmentation of aldehydes is the loss of the CHO for $\ce{[M-29]^{+}}$, which you see at 107 m/z.

$$\ce{ [R-C(=O)-H]^{\bullet +} -> R+ + [HC=O] ^{\bullet} }$$

Since the main contributor to the $\ce{[M+1]^{\bullet +}}$ peak is $^{13}\ce{C}$ we can estimate the number of carbon atoms ($C$) in the molecule from the ratio of intensities of the $\ce{[M]^{\bullet +}}$ to the $\ce{[M +1]^{\bullet +}}$ peak:

$$\frac{I_{M+1}}{I_{M}}=\frac{1.1}{98.9}C$$
$$C=\frac{98.9I_{M+1}}{1.1I_{M}}=89.9\frac{I_{M+1}}{I_{M}}$$

The intensities appear to be $I_M \approx75$ and $I_{M+1} \approx 7$.
$$\therefore C=89.9\frac{7}{75}=8.4$$

Since the number of carbon atoms needs to be an integer, round down to $C=8$. We can now work backward to get a formula. Eight carbon atoms have a mass of $8*12=96$ g/mol, leaving 40 g/mol to be the rest of the formula. We know one is an oxygen (aldehyde) (16 g/mol), leaving 24 g/mol. The most number of hydrogen atoms that can be associated with eight carbon atoms is 18 (2n+2), so there has to be some other atom, maybe an oxygen or nitrogen?

I will leave rest to you:

Maybe there is a second oxygen?
How many hydrogens do you have?
What is the final formula?
How is it different from benzaldehyde?
How could those extra atoms be placed on the ring? (there are many possibilities)
Do you have any other evidence that suggests which of the possibilties from #5 is correct (IR, NMR, etc.)?


Does the speed of an ion affect the current that can flow through a solution?

There is a strong dependence on temperature in conductivity of ionic solutions; see http://www.eutechinst.com/techtips/tech-tips44.htm, for example. However, thermal energy, i.e. molecular motion, is random, and does not favor current flow in the forward (or backward) direction. There are a number of factors that temperature influences:

Viscosity usually changes  inversely with temperature, so ions move more freely at higher temperature.
Ionic solutions have other species than OH- and H+; for example OH3+ [water + hydroxyl radical] is the hydronium ion; see http://en.wikipedia.org/wiki/Hydronium. The amount of each species varies with temperature, as shown in that reference. See also http://en.wikipedia.org/wiki/Self-ionization_of_water. Depending on mass, hydrogen bonding etc. different species have different conductivity.


Melting and boiling point trend in Group II

Its goes down because the attraction between electrons in sea of electrons and cations in the metal lattice becomes weaker therefore less energy is required to interrupt the lattice and easier to melt .


It's a little hard to tell from your second visualization, but to me, it simply seems as if the molecules have shifted somewhat with respect to the periodic box.

A unit cell or periodic boundary "box" is simply a frame of reference. There's no walls. The molecules don't "see" it.

It's merely an indication that there's translational symmetry. In short, in the left-hand view of the second image, the molecule "breaking the box" is also wrapped around into the box by translational symmetry.

I don't know what visualization tool you're using, but if you display more than one unit cell, you'll easily see that the molecules "wrap" around.


I will not attempt to define "proximity effects" in my answer. If you are looking for a definition, this is probably not the best answer. I think the term itself is quite vague and rather qualitative if you will. I've personally seen the concept of "proximity effects" being used more in chemical biology / enzymes chemistry context, than in organic chemistry.

In the first example, I am not even sure if there is indeed any proximity effect in operation. There could be; but it really depends on the solvent effects. How does one really separate the different factors in play? Doing gas phase simulations could give you an answer. The first example you have cited, can indeed be more easily be explained in terms of relative stabilities of the "carbocation" resonance forms: it is more favourable to have the positive charge at position 2 (allylic, secondary), than to have the positive charge at position 4 (allylic, primary). In terms of resonance or MO theory, this picture translates into position 2 being more electron deficient than position 4. A nucleophile would attack at the most electron deficient position (ceteris paribus), and this would be the fastest reaction path. Attack at position 2 leads to the kinetic product, as shown in the book you've cited. Whether there is any additional "proximity effects" - such as the nucleophile being "closer" to position 2 than position 4, is up for debate and cannot be concluded from this result. On the grounds of carbocation stabilisation alone, the observed product can be explained. Note however, that under the conditions, this reaction could be reversible and eventually generate the more stable thermodynamic product.

The second example, is perhaps a more appropriate application of 'proximity effects'. However, I'm not sure if it is appropriate to call this an 'enthalpic phenomenon'. I would have guessed a more probabilistic/entropic origin of this phenomenon. In this example, I would interpret proximity effect as simply the fact that if you are doing a 1-3 ring closure (to form a 3 membered ring), your two reaction centres are closer in space, and hence the reaction is probabilistically more likely than a 1-6 ring closure. I don't know how using the term 'proximity effect' helps give a better explanation, though. If you're interested, you should read up on rates of ring closures in a book of your choice, that discusses both enthalpic and entropic contributions to analyse the relative rates of ring closure.


How is hybridisation of a carbon atom determined by looking at the molecule's structure and observing the lone pair and/or sigma bond present?

The answer you are looking for is probably this:

Count the number of $\sigma$ bonds made by that atom, and add to that the number of lone pairs present on that atom. This number is called the steric number.

If the steric number is $2$, hybridisation is $sp$.
Similarly, $3 \Rightarrow sp^{2}$, and $4 \Rightarrow sp^{3}$ hybridization.