OK, this was a new one to me too, but I found a few resources which might be useful:

A Computational Organic Chemistry blog post
Michael Scott Long blog post
Nature Chemistry review

The basic idea is this. Most bonds are ionic (i.e., both atoms are charged, but electrostatically attracted) or covalent (i.e., the electron pair is shared).

(This is largely cribbed from the Comp. Org. Chem. blog post. Thanks to Steven Bachrach)

If we were writing a multi-configurational wavefunction to describe the bond, we might have something like:

$$\psi(\mathrm{VB}) = c_1\psi_\mathrm{covalent} + c_2\psi_{\ce{A+B-}} + c_3\psi_{\ce{A-B+}}$$

So normally, we'd look at which is the dominant configuration - maybe it's the covalent one ($c_1$) or one of the ionic ones ($c_2$ or $c_3$). Instead, sometimes the covalent form isn't even stabilizing!

From my perspective, the classic example given for this charge-shift bond is $\ce{F2}$ which of course has many $\pi^*$ anti-bonding electrons, so the classical bond order is 1, but ignores that there's a lot of antibonding character. But you can't really consider $\ce{F2}$ an ionic compound!

I think the big question is the future - can this new type of bonding show general predictability and utility. Clearly most chemists have a general intuition (if sometimes wrong) of covalent and ionic bonding. Besides the authors, I'm not sure if anyone else has an intuition of charge-shift systems.


What does an electron's spin of 0.5 and minus 0.5 signify?

It is very tempting (and often also very useful!) to picture electron spin as an angular momentum vector, similar to a spinning top. Using this analogy, there are two properties (or numbers) of this angular momentum vector that we need in order to describe the electron spin. The first one is the spin itself and this is often designated the symbol $s$. The second number is the projection of the spin on a particular axis (often the $z$ axis) and this is assigned the symbol $m_s$.
You probably learned already that not all values of $s$ and $m_s$ are allowed in quantum mechanics. From measurements we know that an electron has a spin of $s=1/2$ (always positive) and its projections can be $m_s=+1/2$ or $m_s=-1/2$. The quantum numbers $s$ and $m_s$ are just numbers we use to label a particular state of the electron, but when we do measurements we also measure units. The units of angular momentum are J s (Joule second) and from comparing the classical definition of angular momentum with the rules of quantum mechanics, we can derive that one unit of angular momentum in quantum mechanics has a value of $\hbar$. This means that the angular momentum of the electron has a value which is basically half of what you would expect based on classical arguments and sometimes people like to emphasize the weirdness of this by saying things like "an electron has to rotate twice around its axis to come back to its original position". Of course this not true because the angular momentum of the electron is not a consequence of a rotation at all (the electron is a point particle). In addition, the sign of the projection of $s$ has nothing to do with this hypothetical rotation either. The sign of $m_s$ does influence the interaction of the electron with an inhomogenous magnetic field. If you have a magnetic field gradient along the $z$ direction, electrons in one state move up while electrons in the other state move down and you will see two separated spots on a position sensitive detector behind the magnet. This is the basis of the famous Stern-Gerlach experiment. Note that you have to perform these kind of experiments with neutral systems that behave like a single electron because a charged electron would experience a much stronger interaction between its charge and the magnetic field (Lorentz force) than between its spin and the magnetic field (Zeeman interaction).
What I find very interesting about half-integer spin is that there is no orientation possible for which the electron spin does not interact with the magnetic field (which would correspond to $m_s=0$ or a perpendicular orientation of the electron spin with respect to the magnetic field).


Why is WF6 stable whereas CrF6 is unknown?

The answer simply has to do with the accessibility of the high +6 oxidation state.

In Cr, the 3d electrons drop in energy extremely rapidly as you remove electrons. So, it is much harder to remove multiple electrons one after another; the only Cr(VI) compounds that we know of are paired with extremely hard bases like the oxide ion, viz. CrO3, CrO42−, Cr2O72−.

Apparently, fluoride isn't quite hard enough to stabilise Cr(VI). Another factor is sterics - squeezing six fluorines around Cr is relatively difficult as compared to having three or four oxygens.

In W, the 5d orbitals have two inner radial nodes. The two pockets of electron density close to the nucleus shield the bulk of the electron density, which is far away from the nucleus. So, when you continuously remove electrons from W, the remaining 5d electrons don't drop in energy so rapidly.

Here are Latimer diagrams for Cr, Mo and W (from Shriver et al., Inorganic Chemistry 6ed.); you can see that the stability of the higher oxidation states increases going from Cr to Mo to W (more obvious if you construct a Frost diagram, but I'm too lazy to do so):

This trend is hardly limited to the Group 6 elements. OsO4 features Os(VIII), whereas the highest oxidation state of iron is Fe(VI), which is already extremely difficult to achieve. Likewise Au(III) is pretty normal but Cu hardly goes above Cu(II).


Chemical Equilibria - Fraction Bound equation based on Kd, [R], [L], [RL], is actually an average?

When a reaction is at equilibrium, there will be stochastic fluctuations. These are usually not measurable (for example, if we have more than a mole of products and reactants in our reaction mixtures).

In this specific case, the fluctuations are easy to measure because the total number of antigen molecules per cell is countable, and you are separating cells (I presume) before analyzing the number of antibodies bound.

If you expect a fluctuation of $\sqrt{N}$ for N objects, that is significant for $N= 150$ but not for $N = \pu{6.022e23}$ when expressed relative to $N$.


Why is sulphuric acid added in the ripper titration?

There are 3 forms:

sulfur dioxide (molecular) $$\ce{SO2}$$  ,
Bisulphite $$\ce{HSO3-}$$
Sulphite $$\ce{SO3^2-}$$
(Additionally, bisulphite can become oxidized with water to form sulphuric acid)

These species exist in equilibrium, and shifts in pH with shift the concentration of each species.

The addition of sulphuric acid lowers pH and favors the molecular form. The molecular form can react with iodine and water. When there is no longer any of the molecular form to react with the iodine, the end point its reached and the color changes.

$$\ce{SO2 + I2 + 2H2O → H2SO4 + 2HI}$$


What is the crystal structure of Calcium Acetate?

Too long for a comment, hence an answer:

What you are asking about is not crystal crystal structure, it's called crystal morphology. Crystal structure is a model describing atomic positions and overall symmetry within a unit cell, whereas crystal morphology is the appearance of the crystal on a macro-scale. And yes, crystal morphology is also affected by the conditions crystals are growing under (concentration and evaporation rate, in your case), so that the same substance can have crystals of different shape.
The formula for calcium acetate and hence the equation is wrong, the correct one would be $(\ce{OAc-} = \ce{CH3COO-})$
$$\ce{CaCO3 + 2HOAc → Ca(OAc)2 + H2O + CO2}$$
and, since your reaction's stoichiometry suggested 1:1 ratio between calcium carbonate and acetic acid, I suspect you didn't use enough acid leaving $\ce{CO3^2-}$ in the solution. Then, as the primary product – calcium acetate $\ce{Ca(OAc)2}$ – has crystallized (white edgy crystals on the bottom), the remaining calcium carbonate $\ce{CaCO3}$ started to precipitate or co-crystallize on the walls of the beaker (what you call a "styrofoam look-a-like" phase), resembling how calcite grows in nature (see this and other images of natural calcite on MineralMiner's FTP server):

Also note that there is no such thing as $\ce{CO3}$. It's either carbonate $\ce{CO3^2-}$ (solution) or carbon dioxide $\ce{CO2}$ (gas).
Regarding the the impurities from the chalk, usually there are parts of shells of marine animals and silicates (can be eliminated by filtering), also minor amounts of $\ce{MgCO3}$. However, if you took a chalk bar from the classroom, this is a different story as it's often nearly 50% $\ce{CaSO4}$ (I hope you didn't do that).


What's the Acid Ionization Constant of the Hydrogen Proton?

The equillibrium between a proton plus water molecule and a hydronium ion is better considered using the concept of proton affinity.

\begin{aligned}
\ce{H^+(g) &-> H^+(aq)} \\
\Delta E &= −1530~\mathrm{kJ/mol}
\end{aligned}

You could roughly convert to an equillibrium constant using

$$\Delta G = -RT \ln(K)$$


I will start off by addressing Jon's comment above. Yes, the Bohr model is flawed. I think it is still worth learning about it just from a historical standpoint, to see how we discovered the quantum mechanical description of the electron, but when you are studying it you absolutely have to remember that all of what he said was effectively rubbish.

Now on to your questions.

  Is there a relationship between an electron's energy and its distance from the nucleus?

The first thing that should be said is that in QM, unlike in the Bohr model, there is no well-defined "distance from the nucleus". Electrons exist everywhere (except at nodes) and the probability density is given by $\lvert\psi\rvert^2$. However, it is possible to calculate the "most probable distance from the nucleus" using the radial distribution function, $r^2 R^2$, where $R(r)$ is the radial part of the wavefunction. Essentially, you take the derivative of this with respect to $r$ and set it to 0 - you can read more here. For a 1s electron in hydrogen, you will find that the most probable radius is approximately $52.9 \text{ pm}$, a quantity called the Bohr radius and denoted $a_0$.

(As an aside: oddly enough, this correlates with the flawed Bohr model, which said that the radius of the 1s electron in hydrogen was $52.9 \text{ pm}$. However, what Bohr said was that the electron was always at this distance, which is wrong. QM says that the electron is most likely to be found at this distance.)

Anyway, you will find that there is some degree of correlation between the most probable radius and the energy of the electron. If we consider the 2s orbital of hydrogen, you can perform exactly the same calculation as you did for the 1s orbital, and you will find that the most probable radius is $4a_0$.

  Also, it is often mentioned that an atom is 99.99% empty space etc...

That's just a simplified view presented in popular science (or elementary chemistry classes) that ignores the quantum mechanical model. No serious chemistry book would write something like that. I don't know exactly how it's calculated, but I can make a good guess. You take any atom you like, let's say hydrogen. You look up the atomic radius - Google tells you it's $52.9 \text{ pm}$. (No surprises there.) So you can think of the atom as a sphere with that radius, and calculate the volume of the sphere: $V = \frac{4}{3}\pi r^3$. Then you go and look up the volume of an electron and a proton (a physicist would get a headache from just reading the phrase "volume of an electron"), and add it up, and you find that the "atom" is "99.99% empty space".

As Ivan succinctly said - with the QM description of the atom, there is no longer such a thing as free space since the electron "exists everywhere". Apart from that, just like how the distance from the nucleus is no longer well-defined in QM, the "volume" of an atom is also no longer well-defined. We pretty much threw that view out of the window.


When an acid is added to water, why does the hydroxide ion concentration decrease?

As long as you stay in the dilute regime, that is if water molecules have most of the time to interact just with each other, $K_w$ won't change. Add 20% of anything soluble (salt, acid, alcohol, acetone, ...) and things look different.


Can one reduce solubility of NaCl by adding CO2 under pressure? Will the salt fall out of solution?

NaCl is vastly more soluble than carbon dioxide, so this method won't work.  You can get some NaCl to precipitate (assuming you have a concentrated solution) by bubbling HCl in, thanks to the common-ion effect, so you could for example get pure NaCl from rock salt this way.  But somehow I don't think that's the situation you have in mind :-)


I think you have some typo or flaw in your question.  "E = 89.9 pm" doesn't make sense.  You are saying energy equals a length. The edited version "E = 89.9 eV" has energy units, but the theory never predicted such as large value.  And "r = -1/4 pm" doesn't make sense either.  Distance can't be negative. Also, even it were positive 1/4 pm, the theory never predicted such a small magnitude value.

The initial calculations of Heitler and London were:

r = 80pm
E = -2.4 eV

A major source of error was inability to compute integrals.  Additionally, the calculations neglect ionic contributions.

See Computational Imperatives in Quantum Chemistry for a historical overview.


There is no such convention explicitly telling what comes out first.
Here, for a traditional formula of phosphoric(V) acid $\ce{H3PO4}$ writing a proton first serves only a didactic purpose.
It's visually easier for students to keep a track on dissociation as the order of the elements both in formula and among the products is preserved.

On the other hand, shall one use a coordination formula of phosphoric(V) acid $\ce{[PO(OH)3]}$, it probably would make more sense to use a reversed order and put $\ce{H+}$ at the end:

$$
\begin{align}
\ce{[PO(OH)3] &<=> [PO2(OH)2]- + H+}\\
\ce{[PO2(OH)2]- &<=> [PO3(OH)]^2- + H+}\\
\ce{[PO3(OH)]^2- &<=> [PO4]^{3-} + H+}
\end{align}
$$

One more minor thing.
Equations are usually aligned about arrows, and a tabular array of products and reactants emerges when the same repeating compound ($\ce{H+}$) is shown first.
Note how easier it is to grasp the information when more compounds are aligned:

$$
\begin{align}
&\color{green}{\text{aligned}} & &\color{red}{\text{misaligned}}\\
\ce{H3PO4 &<=> H+ + H_2PO4^{-}(aq)} &\quad \ce{H3PO4 &<=> H_2PO4^{-}(aq) + H+} \\
\ce{H2PO4- &<=>H+ + HPO4^{2-}(aq)} &\quad \ce{H2PO4- &<=> HPO4^{2-}(aq) + H+} \\
\ce{HPO4^2- &<=> H+ + PO4^{3-}(aq)} &\quad \ce{HPO4^2- &<=> PO4^{3-}(aq) + H+}
\end{align}
$$


Why is the manufacture of sulfuric acid known as the contact process?

According to this site:

  The Hot gases( $\ce{SO2}$
  ) evolved from burning of sulfur ore comes in
  contact with catalyst bed,  So the name of this process is called
  contact process.


Are there different types of melamine resin?

Certainly the are many different types of melamine resin.

Is there a way to make a melamine resin without using formaldehyde?

Yes, a different aldehyde could be used such as acetylaldehyde.  Glucose, benzaldehyde, acrolein and other alternatives to formaldehyde have been studied also (see page 33 of Polymer Synthesis by  Sandler et al.).

Is there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer?

Good sources of information about formaldehyde-melamine-sodium bisulfite copolymer and possible alternatives are the patent Resilient foam based on a melamine-formaldehyde condensate and related patents.

First the introductory portion of the patent explains:

  It is true that foams based on melamine-formaldehyde condensates have been described in various patent publications, but they have not hitherto been employed industrially for heat or sound insulation in the construction industry. They are produced by stirring air into an aqueous melamine resin solution which contains an emulsifier and a curing agent. Such as described, for example, in German Pat. No. 870,027, have the serious disadvantage that they are very hard and brittle, and easily break or crumble when handled. It is alleged in French Pat. No. 1,073,642 that melamine resin foams can be produced by heating a resin powder in a mold under reduced pressure. This process, however, does not give useful resilient foams. U.S. Pat. No. 3,093,600 describes melamine resin foams which are said to have improved resilience and resistance to crazing as a result of the incorporation of triols, e.g. trimethylolpropane. However, it has been found that the resilience, and especially the recovery after compression, of such foams is inadequate for many applications.

Then in describing what is considered new about the invention, many different alternatives are included. Such as adding aldehydes other than formaldehyde, urea, and many other substances.


What are the steps to creating synthetic glucose?

A relatively simple process is described by Kobayashi et. al where they use cellulose, a structural component of plants, which is ball milled, mixed with activated carbons and HCl and hydrolysed to produce glucose in high yield (88%).

It requires only easily available ingredients and the ability to heat (and preferably stir) to 453K. Reference is given below.

High-Yielding One-Pot Synthesis of Glucose from Cellulose Using Simple Activated Carbons and Trace Hydrochloric Acid
Hirokazu Kobayashi, Mizuho Yabushita, Tasuku Komanoya, Kenji Hara, Ichiro Fujita, and Atsushi Fukuoka
ACS Catalysis 2013 3 (4), 581-587
DOI: 10.1021/cs300845f


How toxic chemically is plutonium (Pu), neglecting the radioactive damage?

The toxicity is primarily due to radioactivity and to absorption by the body, where that radioactivity can act internally. There is, "significant deposition of plutonium in the liver and in the "actively metabolizing" portion of bone," according to Miner and Schonfeld.
Many $\ce{Pu}$ isotopes are primarily alpha-emitters, with "high energy radiation with low penetration and thereby requires minimal shielding. A sheet of paper can be used to shield against the alpha particles emitted by plutonium-238," from Wikipedia. The risk of holding a small amount of plutonium, if it is sealed inside almost any container, is negligible, because the the heavy, positively-charged alpha particles are stopped by the container in a very short distance.
By that same token, when absorbed, the plutonium migrates to rapidly dividing liver tissue and bone's hematopoietic cells. The alpha particles are stopped in a very short distance, depositing all their energy in the few nearest cells and massively interfering with cellular reproduction. Small amounts can cause various types of cancer, such as leukemia, a cancer of the previously-mentioned blood cell producing tissues.
That said, plutonium is not quite as toxic as one might think: "The U.S. Department of Energy estimates that the lifetime cancer risk from inhaling 5,000 plutonium particles, each about 3 µm wide, to be 1% over the background U.S. average... no human being is known to have died because of inhaling or ingesting plutonium, and many people have measurable amounts of plutonium in their bodies."
Perhaps what Rhodes means by "not that radioactive" is that the radiation from plutonium is not very penetrating. The isotope used in nuclear weapons, 239Pu, has a half-life of about 24,000 years, comparable to radium, or to 231Pa. If I've done my math  correctly, one mole of 239Pu, about 6e23 atoms, with  half-life of 8e11 seconds, should emit about 400,000,000,000 alpha particles per second. Of course, if it were a sphere, most of those particles would be absorbed inside the mass of metal, releasing energy as heat. However, spread through a human body, each particle has a much greater chance to cause damage. And 238Pu, used in radioisotope thermoelectric generators, is far more radioactive!


What does it mean that an electron pair "spends more time" nearer an atom?

When you come to the quantum level (looking at one electron) things don't behave anymore like we normally think. To make it anyway understandable those expressions were invented such as "spending more time". Electrons are less described as a dot but as a smearing (electron density).
Chlorine and fluorine have higher electronegativities than hydrogen. The electron density therefore is not equally distributed between the two bonding partners but drawn towards fluorine/chlorine. To illustrate this, you can say "the electron spends more time at F/Cl".

This thinking was established as molecular orbital theory (s orbitals, p orbitals, ...). There have been other models before, e.g. you might know the shell model (k shell, l shell, ...). For experimental evidence you might want to take at look at this -- but most of all the citations therein: http://en.wikipedia.org/wiki/User:Chem507f10grp4/sandbox#Experimental_Evidence_Supporting_Molecular_Orbital_Theory
Basically, the shell model ran into a series of properties it did not calculate correctly or precisely. Molecular orbital theory was found to predict many properties and came to well calculated values.


Comparing O−O bond length in hydrogen peroxide and dioxide(2−) ion

The way to tackle this is to look at the lone pair repulsions between the 2 molecules. It is known that $\ce{N-N};$ $\ce{O-O};$ $\ce{F-F}$ single bonds are quite unstable due to lone pair (lp) repulsions.
An evidence of this is that $\ce{N}$ exists in molecular state as $\ce{N2}$ using multiple bonds, so that it's lone pairs do not repel each other, however $\ce{P}$ doesn't do that. It is capable of forming $\ce{P4}$ because in the third period the size of the atoms are large enough to properly distance the lone pairs.
One more evidence of this act is that among diatomic halogen molecules the bond length is in the order $\ce{Cl2} < \ce{Br2} < \ce{F2} < \ce{I2}$ the bond length of $\ce{F2}$ should have been lower (2p–2p overlapping is much stronger hence the bond should have been the shortest). However, due to lp–lp repulsions it is now much longer bond.
With this we can now see that in $\ce{H2O2}$ there is only two lone pairs on each oxygen atom while in $\ce{O2^2-}$ it has three lone pairs on each atom, hence it has increased lp–lp interaction.
As a result of this $\ce{O2^2-}$ tends to elongate (relatively more than $\ce{H2O2})$ its bond to decrease this intense lp–lp interaction, hence the longer bond.


What is non toxic to fish, has a low freezing point, and floats on water?

There are oils sold for swimming pools to slow or stop evaporative cooling . That would be a problem for air breathers like mosquito larva. I don't know anything about it because I went with 1" thick Styrofoam panels. That is what I currently have on my pond ( 10' X 5 '), works very well .I cheat a little and have 150 watt aquarium heater in the 700 gallons. My pond is currently 59 F while nearby open water is 49 F. For your tiny area the Styrofoam is fine leave about 1/2" on all sides for air . On the other hand why not do the obvious and put in a 25 watt ( if they make them that small) aquarium thermostat/heater ?


Why is nicotinoyl chloride available as its hydrochloride salt? How can we separate them?

Unfortunately, i don't have access to the original reference from 1928 Über die Einwirkung von Thionylchlorid auf die Pyridin-monocarbonsäuren by Meyer and Graf, but it's cited in The Preparation of Nicotinyl Chloride from 1953.

In this later note, where dried and grounded potassium nicotinate was reacted with oxalyl chloride in anhydrous benzene and the resulting nicotinyl chloride eventually destilled off under reduced pressure (75-90 °C, 10-12 mm), the authors state:

  On exposure to the air, the oily acyl chloride soon solidified to nicotinic acid hydrochloride.

With other words, the combination of moisture and the basicity of the pyridine moiety results in the decomposition of the product. This is apparently prevented by deactivation of the pyridine upon conversion to its hydrochloride.


Have there been no advances in the determination of effective nuclear charges since Clementi and Raimondi in the 60s?

I would say the most important reason why no development for effective nuclear charges is that the concept of effective nuclear charges has fallen to disuse in the recent years.  I can think of two important reasons:

Effective nuclear charges is intended to be used to explain electronic structure of isolated atoms.  To this day some atoms can still pose a challenge, but those are the atoms that experience heavy relativistic effects and would need other special treatments than shielding effect along.  However, for the most part, chemists are not interested in atoms anymore.
Its theoretical framework is too simple.  Chemist need their results to be extremely accurate.  A 0.1 eV change can easily be a game changer in an organic reaction.  That, unfortunately, cannot be achieved by a model that has only one adjustable constant.

However, the idea still lives on.  For example, pseudo-potential or Effective Core Potentials can be considered as more general and modern version of effective nuclear charge concept.  Instead of simply changing the charge but keeping the Coulomb potential unchanged, ECP provides the entire effective potential of core electrons. Valence electrons and sometimes even sub-valence electrons has to be excluded from such treatment to be able to produce results with sufficient accuracy. These pseudo-potentials are actively developed and widely used today, can provide chemically relevant accuracy in complex molecular systems and may qualify as "improvement to effective nuclear charge", although they are evolved into different species already.


Why does strong Lewis acid-strong Lewis base interactions prevail over hard-soft acid-base interactions?

The Klopman-Salem equation accounts for the energy gain (or loss) when the orbitals of one reactant overlap with those of another. The equation is

$$\Delta E=-\sum_{ab}(p_a+p_b)\beta_{ab}S_{ab}+\sum_{k<l}\frac{Q_kQ_l}{\varepsilon R_{kl}}+\sum_{r}^{\text{occ}}\sum_{s}^{\text{unocc}}-\sum_{s}^{\text{occ}}\sum_{r}^{\text{unocc}}\frac{2(\sum_{ab}c_{ra}c_{sb}\beta_{ab})^2}{E_r-E_s}$$
where $p_a$ and $p_b$ are the electron population in the atomic orbitals $a$ and $b$, $\beta$ and $S$ are the resonance and overlap integrals, $Q_k$ and $Q_l$ are the local charges on atoms $k$ and $l$, $c_{ra}$ is the coefficient of atomic orbital $a$ in molecular orbital $r$, where $r$ refers to the molecular orbitals on one molecule and $s$ refers to those on the other, $E_r$ is the energy of molecular orbital $r$ and $E_s$ is the energy of molecular orbital $s$.

The first term is a closed-shell repulsion term, the second term is the Coulombic repulsion or attraction, and the third term is the interaction of all the filled orbitals with all the unfilled of correct symmetry. This equation was derived using perturbation theory being the first term is a first-order perturbation term and the third the second-order perturbation term.

Now, the HSAB theory is affected by charge and orbital interactions. The h-h interactions is mainly electrostatic in nature, while in a s-s interaction the orbitals involved are close in energy. With the Klopman-Salem equation we can see that the second term is the hard contribution and the third is the soft contribution. There is a simplified form, which accounts only for frontier orbitals, that is
$$\Delta E=\frac{Q_{\text{Nu}}Q_{\text{El}}}{\varepsilon R}+\frac{2(c_{\text{Nu}}c_{\text{El}}\beta)^2}{E_{\text{HOMO}(\text{Nu})}\pm E_{\text{LUMO}(\text{El})}}$$
So, back to the question, a hard molecule or ion is characterized by a large HOMO-LUMO gap (definition of hardness) and the second term, in the simplified version, is very small, so the contribution is mainly electrostatic, i.e., the more charged the sites in the molecule/ion the greater will be the stabilization (or repulsion, in case of like charges). The electric field of the two parts in the molecules/ions would diminish $R$ and the term would be very large. In the case of soft-soft interactions, the orbital term is the dominant because the HOMO-LUMO gap is smaller.

But sure, the interaction has to be between to molecules or ions, not a single specie. You can apply this to nucleophiles and electrophiles as well. A hard nucleophile (negative charge localized and low-energy HOMO) would tend to react with a hard electrophile (positive charge localized and high-energy LUMO), leading to a great stabilization by the electrostatic term. Meanwhile a soft nucleophile (high-energy HOMO) would react with a soft electrophile (low-energy LUMO), stabilized by the orbital term, because $E_{\text{HOMO}(\text{Nu})}-E_{\text{LUMO}(\text{El})}$ is small.

If it is necessary, I can edit with a MO diagram to see this more visually.


How does NaCl (or any inorganic salt) increase surface tension?

I hope this picture helps.

The interaction between a molecule of water and an ion is stronger than the hydrogen bonding that occurs between two water molecules.


You could vacuform heated thin PET film over it.  That would not stop biology.


What are quantifier and qualifier ions in mass spectrometry data for GC-MS or LC-MS?

This is a general "identification" rule which should not be restricted to mass spectrometry.
In chromatography you may be analyzing very closely related analytes whose masses are very close or even identical. Think of structural isomers (low resolution MS is blind to enantiomers, isotopomers, isotopologues etc.). Structural isomers may very closely related mass spectra and their molecular ion peaks would be identical in mass, but their fragmentation patterns might be slightly different. So the "good" advice is pick not one but more than one line in order to positively identify a target analyte.
Think of the legal consequences of false positives. A sportsman or women may loose their career if the tests show a false positive or a criminal may go scot free. Therefore analytical chemists have a huge responsibility on the society in terms of having a very high degree of confidence in identifying or quantifying anything.
Although most of the people think that mass spectrometer is a "perfect" detector in chromatography, it is nowhere close to being one. Many compounds do not ionize well and hence go undetected. It is extremely useful though.


Any substance too poisonous to measure an LD50?

Of course no. Botulotoxin is probably the strongest known, and still its $\rm LD_{50}$ is counted in nanograms per kilogram, which is pretty manageable. Sure, working with such tiny amounts requires some special measures, but still, it is way greater than one molecule. You can divide it again, and again, and again.

So it goes.


What do the horizontal lines in a P-V phase diagram mean?

the graph is trying to show three things (pressure, density and temperature)  in 2D.

The yellow portion shows the super critical fluid of $\ce{CO2}$.

The blue portion liquid $\ce{CO2}$

The light green portion shows the gas phase for $\ce{CO2}$.

The horizontal lines are called "tie lines." The two phases at the ends of the tie lines are in equilibrium.

If you follow the line for say for 12 $^\circ$C, from the degree notation there is at first one phases, a gas phase.

Now when you reach the dotted curve there are two phases a gas phase and a liquid phase which are joined by a horizontal tie line.

The whole thing is a lot easier to figure out with a more traditional phase diagram.

Found above on page:
https://www.engineeringtoolbox.com/CO2-carbon-dioxide-properties-d_2017.html


Hill's system alphabetizes elements symbol or name?

Go straight to the original source, ON A SYSTEM OF INDEXING CHEMICAL LITERATURE; ADOPTED BY THE CLASSIFICATION DIVISION OF THE U. S. PATENT OFFICE, and you can see that Hill uses the element symbol to determine alphabetical order.

In your example $\ce{C5H8NNaO4}$ is correct.


I believe this may be a duplicate of this question:
Why is BCl3 a monomer whereas AlCl3 exists as a dimer?
From what I remember, the stronger the backbond, the smaller the tendency to undergo dimerisation, however, way back in the dark ages when I was a student and asked my second-year chem prof, his answer was that the simple reason why none of the tri-halides of boron undergo dimerisation is steric hinderance - i.e. the boron atom is just too small to accommodate the fourth larger Cl/Br/I atom. He then pointed me to several sets of further reading.
However, you will probably find a better answer at one of the following links:

Mulliken J. Chem. Phys. 3, 635 (1935); https://doi.org/10.1063/1.1749567
Lappert, et.al. Chem. Soc., Faraday Trans. 2, 1976,72, 539-55 https://doi.org/10.1039/F29767200539


What chemical substance is obtained by dissolving C60 in olive oil?

There is no chemical reaction between fullerene and olive oil, that is to say that the two molecules do not break or make chemical bonds to produce a new chemical species.  However, fullerene is soluble in olive oil and so there are typical solvent-solute interactions.  A homogenous solution of fullerene in olive oil produces a light purple color typical of many fullerene solutions.

Since Moussa's 2012 experiment where he fed fullerene in olive oil to rats and found that their life span roughly doubled, many companies have been marketing a dark-red liquid comprised of olive oil and fullerene.  It is likely that in many of these highly colored liquids, the fullerene is not completely dissolved and light scattering by the fine particulates are producing the observed color.


Finding the kp with density, temperature, and pressure

You can use the formula for the ideal gas
$$\rho=\frac{p\cdot M_\mathrm{mean}}{R\cdot T}$$

As pressure, temperature and density are given, you get the mean molar mass.

$$M_\mathrm{mean}=\frac{\rho RT}{p}$$

From that, you get molar fractions.

$$x_{\ce{O2}}=\frac{ M_{\ce{O3}}-M_\mathrm{mean}}{ M_{\ce{O3}}- M_{\ce{O2}}}$$

$$x_{\ce{O3}}=1-x_{\ce{O2}}$$

From that, you get partial pressures.

$$p_{\ce{O2}}=p\cdot x_{\ce{O2}}$$

$$p_{\ce{O3}}=p-p_{\ce{O2}}$$

From that, you get $$K_\mathrm{p}=\frac{p_{\ce{O3}}^2}{p_{\ce{O2}}^3}$$


Assuming that the reservoir does not absorb any heat and there is no loss of heat from the setup by:-

Radiation of heat from the reservoir,
Convection currents in the air/surrounding fluid (if any),
The entire water is at the same temperature so that heat absorbed is not used for increase in temperature,
Conduction loss from the reservoir to the ground/support it is on and air(negligible but thought I should mention)

Fulfilling these conditions ensures that the energy is only used by the water for vaporization.
So, yes, these conditions are fulfilled. But, that is never the case when experiments are conducted in labs.
Edit: The above conditions are more than just an isolated system.


I don't understand your question completely, but If you want to understand what's happening in an organic reaction you need to know which type of reaction is in question and you have to look at the mechanism. This is a radical reaction with the following mechanism (excluding the termination step):


Molecular orbital energies prediction with ML algorithms

As several comments have mentioned, interpreting a machine learning method can often be difficult. Outside the question of machine learning, I think your question centers around "what would I want to know to roughly predict HOMO / LUMO energies?"

Unfortunately QM9 isn't very diverse chemically, but if we think from a general qualitative molecular orbital picture (e.g., building up diatomic molecules), you want to know:

Number of valence electrons (i.e., which orbitals are filled)
Atom type (which tells you valence electrons and atomic orbital energies)
Type of bond (e.g., single sigma bonds clearly will yield different MO energies than pi bonds)
Conjugation - conjugated and aromatic systems will have different MO energies than plain sigma-bonded species
Bond distances (from the Coulomb repulsion which encodes a distance $1/r$)

Given the above information, it's fairly easy to sketch out MO diagrams for $\ce{O2}$ or $\ce{HOH}$ so one can imagine that they're important for a machine learning algorithm too.

Now how do you know if they actually matter to the ML model?

One easy way is to drop one of the descriptors and retrain the model. Does it get worse? How much?

There are statistical techniques to give you some insight as well. For example, one can use LASSO to drop factors from the input representation that don't have much effect. You can also look at simple statistical correlation between input factors. I'd guess that "Atom Type" and "Atomic Number" are probably highly correlated with each other and only one is truly necessary.. but that's just a guess.


For group 7 elements, are anions a lower energy state than being neutral atoms?

Yes, the electron affinity for group 17 elements (group VII by older IUPAC nomenclature) is generally exothermic, so the uptake of an electron should be favourable.

But as Ivan mentioned in a comment, there is no such thing as a free or lonesome electron. Each electron must first be displaced from another atom — ionised; the corresponding thermodynamic property is the ionisation enthalpy. For every element, the ionisation enthalpy is positive, thus it is always unfavourable to displace an electron from a neutral atom. No exceptions.


Radioactivity and Kinetics of disintegration

No, as hey answered, neutrons are made of two down quarks and an up quark.  Protons are made of two up quarks and a down quark.  It is the conversion of a down quark into an up quark that results in the neutron --> proton conversion.  Read here for more information about this reaction.
The confusion with this probably comes from the fact that net reaction  in a neutron to proton conversion is $n^{0} \rightarrow p^{+} + e^{-} + \overline{\nu_e}$, so it looks like a neutron is falling apart into three sub-particles.
No, the atom will not become ionized from this event itself.  Consider what is happening in this reaction:
$p^{+} + e^{-} \rightarrow n^{0} + \nu_e$
Yes, an electron is removed from the atom, but at the same time a proton is as well.  We should know that the atoms remains neutral because of conservation of charge. If the atom has no net charge at the beginning of the reaction, and no charged particles are ejected, it must be neutral at the end of the reaction as well.
Jon Custer's response is spot on here.  Gamma ray emission represents the movement of the nucleus from a metastable state to its ground state via the release of electromagnetic radiation.


What reaction happens between glucose and ammonium chloride when drying?

When a sugar solution turns brown (or black), it's generally a sign of polymerization of the sugar molecules triggered by dehydration reactions that create more reactive molecules. A classic example of this is caramelization, in which sucrose is dehydrated by heating and forms many types of furans and pyrans and the oligomers caramelan, caramelen and caramelin. The latter three all contribute to the brown color.
Although caramelization typically requires heat, a similar effect can also be achieved by use of an acid catalyst. A simple example is treatment of sucrose with concentrated sulfuric acid, which very quickly results in a solid black mess.
In your case, you've started with a nearly neutral solution of glucose in ammonium chloride. The solution is nearly neutral because it is a 1:1 mix of a moderate base (ammonia) with a strong acid (HCl). As you evaporate off the water, however, some ammonia evaporates as well, resulting in a slight excess of HCl and a lowering of pH. The combination of mild heat and acid is likely causing a caramelization-type reaction at a lower temp than is usually required. The ammonia might also enhance the color via a Maillard-type reaction.
A good review of carmelization is: Sengar, G.; Sharma, H. K. Food caramels: a review. J. Food Sci. Technol. 2014, 51 (9), 1686–1696. DOI: 10.1007/s13197-012-0633-z


Concentrating mixtures of volatile compounds

The simplest method would be fractional distillation: if you have a mixture of miscible liquids with different boiling points, as is the case in your question, then you can heat them slowly. When the boiling point of the most volatile compound in the mixture is reached, it begins to boil and the temperature does not increase until it has all boiled off. The vapour formed is cooled with a condenser and collected as a liquid. As soon as the temperature begins to increase again, all of this compound has evaporated, so it can be removed. Provided the boiling points aren't too close, you could use this to separate out all the compounds in the mixture. A diagram of the set-up (from Wikipedia) is below.

This would be easier with water than ethanol, since it has a higher boiling point so is less likely to boil at a similar temperature to your desired compound.
(The most common way to separate out mixtures in chemistry is chromatography - for example, HPLC or GC - but I'm not sure you'd want to put the results in food! HPLC, which stands for high performance liquid chromatography, involves the use of a solvent mixture which normally includes chemicals like acetonitrile or methanol, neither of which are safe to ingest.)


Given that a 2-place balance is no more accurate than $\pu{+/- 0.01 g}$, you can say that the two measurements were within $\pu{0.01 g}$ of the average, $\pu{12.53g}$, then just use this average for your calculations.


What is the simplest approach to balance a complex reaction equation?

$\ce{ a K4Fe(CN)6 + b Ce(NO3)4 + c KOH ->\\ d Ce(OH)3 + e Fe(OH)3 + f H2O + g K2CO3 + h KNO3}$

Because there is one $\ce{Fe}$ in $a$ and one in $e$, we know that $a=e$.

To better see things, I rewrite the formulae.

For $\ce{Fe}$: $a=e$
For $\ce{Ce}$: $b=d$
For $\ce{C}$: $6a=g$
For $\ce{N}$: $6a+4b=h$
For $\ce{K}$: $4a+c=2g+h$
For $\ce{H}$: $c=3d+3e+2f$
For $\ce{O}$: $12b+c=3d+3e+f+3g+3h$

Now I replace $e, d, g, h$ with their counterpart, leaving just $a, b, c, f$. I get

For $\ce{K}$: $4a+c=2(6a)+(6a+4b) \Rightarrow c=14a+4b $
For $\ce{H}$: $c=3b+3a+2f$
For $\ce{O}$: $12b+c=3b+3a+f+3(6a)+3(6a+4b) \Rightarrow c=39a+3b+f$

Replacing $c$.

For $\ce{H}$: $14a+4b=3b+3a+2f \Rightarrow 11a+b=2f$
For $\ce{O}$: $14a+4b=39a+3b+f \Rightarrow -25a+b=f$

Combining at $f$.

$11a+b=-50a+2b \Rightarrow 61a = b$

Now I assume $a=1$ and get

$a=1$
$b=61a=61$
$c=14a+4b=258$
$d=b=61$
$e=a=1$
$f=-25a+b=36$
$g=6a=6$
$h=6a+4b=250$

And, finally

$a+b+c+d+e+f+g+h=674$

The last step could also be achieved by taking $a+b+c+d+e+f+g+h$ and replacing $c, d, e, f, g, h$, then replacing $b$, which should eventually lead to $674a$.


How is the trend of acidity in the sequence: phosphinic, phosphonic, phosphoric acid explained?

The stability of conjugate bases is indeed a good way of estimating the relative acidities of different compounds.

However, in the case of the three compounds you've brought up, it is not sufficient. Taking the conjugate bases of each acid, one sees that each has a doubly bonded oxygen (oxo) and a deprotonated hydroxy. These are the only relevant groups as far as resonance is concerned.

Note that phosphorus-bound hydrogens cannot delocalize the negative charge, and neither can phosphorus-bound hydroxyl groups.

Given this, all three acids actually have the same resonance stabilization. Any resonance structure where a hydroxyl $\ce{O}$ donates an electron pair to the phosphorus is insignificant because it creates increased charge, likewise any resonance structure where the hydroxyl $\ce{O}$ receives an electron pair is insignificant because oxygen already has a complete octet in its hydroxyl form.

Since the resonance argument only tells us that their $\mathrm{p}K_{\mathrm{a}}$'s are similar (looking only at first $\mathrm{p}K_{\mathrm{a}}$) we can use the argument that the additional hydroxyl groups reduce the polarization of the $\ce{O-H}$ bond. This can be demonstrated by looking at the resonance structures of the acids (in neutral form). The hydroxyl $\ce{O}$'s in the neutral form can donate electron density to the phosphorus; however when there is only one hydroxyl group, all the positive charge is localized there, while when there are two or three, the positive charge is delocalized across three hydroxyl groups.

Note that in the case of the neutral acid, it is ok to have resonance where $\ce{O}$ donates to $\ce{P}$ because $\ce{P}$ does not gain a $2-$ charge as it would with the conjugate base.


How to find the oxidation number of Co and Cr in [Co(NH3)6] [Cr(C2O4)3]?

$\ce{Co}$ is known to have two possible oxidation numbers in solution : +II and +III. Both make complexes with 6 $\ce{NH3}$. If $\ce{Co}$ is at +II, the charge of the complex is $2$+, as $\ce{NH3}$ is not charged.
If $\ce{Co}$ is at +III, the charge of the complex is $3$+ for the same reason.
For chromium, it is similar, as $\ce{Cr}$ may have oxidation number +II and +III. But the ligand oxalate is charged $-2$. The complex of $\ce{Cr}$ with three oxalate ions must compensate the charges of the cobalt complex (+$2$ or +$3$). So the charge of the chromium complex has to be -$2$ or -$3$. In other words, the complex ion is either $\ce{[Cr(C2O4)3]^{2-}}$ or $\ce{[Cr(C2O4)3]^{3-}}$. Let's compare the two possibilities, and the oxidation numbers of $\ce{Cr}$ in these two structures.
To get a complex $\ce{[Cr(C2O4)3]^{2-}}$, $\ce{Cr}$ oxidation number should be equal to $x$, with $x + 3(-2) = -2$. So $x$ = $+4$ or +IV. And unfortunately, this oxidation number IV does not exist for $\ce{Cr}$ in solution. This structure has to be abandoned,
On the other hand, to get a complex $\ce{[Cr(C2O4)3]^{3−}, Cr}$ must have an the oxidation number $y$ given by : $y +3(-2) = +3$. So $y$ = $+3$ = +III, and this is possible and even welcome. It corresponds to one of the requirements developed previously. As a final result, the complex ion $\ce{[Cr(C2O4)3]^{3-}}$ has a charge $3-$ that can be coupled with $\ce{Co(NH3)6^{3+}}$.
As a consequence, both Co and Cr are at oxidation number +III in the original structure $\ce{[Co(NH3)6][Cr(C2O4)3]}$


What is the mechanism for thermal decarboxylation of a generic carboxylic acid (not special cases like beta-ketocarboxylic acid or malonic acids)?

Under sufficiently alkaline conditions, i.e. when the acid is deprotonated, the carboxylate can undergo an electron transfer reaction with a suitable partner (oxidant). In the course of this process, the carboxylate is oxidized to an acyloxy radical, which subsequently fragments to yield an alkyl (or alkylaryl) radical and carbon dioxide.

$$\ce{R-COO- ->[-e^-] R-COO\cdot -> R\cdot + CO2}$$

A photochemical variant of this reaction was intensively examined in the group of Axel Griesbeck at the University of Cologne, Germany. For further reading, have a look at

Acc. Chem. Res., 2007, 40, 128-140 (DOI)
J. Phys. Chem. A, 2006, 110, 3356-3363 (DOI)
Synlett, 2004, 2347-2350  (DOI)


Can phenol be detected with a ceric ammonium nitrate test?

With unsubstituted phenol ($\ce{PhOH}$) dark-brown precipitation is formed:

$$\ce{(NH4)2[Ce(NO3)6] + 2 PhOH -> [Ce(NO3)4(PhOH)2] + 2 NH4NO3}$$

Other phenols should produce wine-red solution as expected [1, p. 478]. Also, this and other phenol tests are discussed in [2].

Pavia, D. L. A small-scale approach to organic laboratory techniques, 3rd Ed.; Brooks/Cole, Cengage Learning: Belmont, CA, 2011. ISBN 978-1-4390-4932-7.
Atul, S. The Pearson Guide to Objective Chemistry for the AIEEE; Pearson Education India, 2010. ISBN 978-81-317-3381-3.


Two different factors play a role when considering a reaction. The energy of the bonds etc. (enthalpy) and the entropy. Together, these two components make up what is called the 'Gibbs Free Energy'. (Wikipedia)

The change in Gibbs free energy ($\Delta{}G$) of a system must be $< 0$ for a reaction to be spontaneous. From this it is possible to see that a reaction where the entropy increases drastically, but the enthalpy increases. In this case a reaction can be endothermic.

A case of terminology: The 'more correct' term (my teacher always said) is exergonic or endergonic. What this means is that the reaction has a $\Delta{}G < 0$ or $\Delta{}G > 0$ respectively.

As @Cohorent mentioned before, it is also possible to have a system that is larger than just the chemical reaction (adding a battery for example). Meaning that the reaction is maybe endergonic, but a different energy source is used to drive the reaction. Note then, that the $\Delta{}G$ for the entire system is still below zero.


How to calculate the pH of a buffer after dilution

Adding water means diluting the buffer.

Use the known concentration, the volume of the buffer and the volume of water added to calculate the new concentrations.


How to determine whether inversion takes place?

A) is $\mathrm{S_N2}$ due to the secondary $\ce{C}$ and the sufficiently good nucleophile $\ce{I-}$. So inversion takes place.
B) is $\mathrm{S_N1}$ due to the tertiary $\ce{C}$. The solvent is also more polar than in A). So the stereochemical information would be lost.


Regenerating the imaginary data of a 2D NMR spectrum with a hilbert transform

If you have the raw data (the FIDs in the ser file) still available, there is no reason to use a Hilbert transformation, you can just process with xfb instead of xfb n again. xfb n also doesn't really discard the imaginary part, it processes the spectrum from the raw data again, but doesn't save the imaginary processed data.

The ser file contains all the recorded FIDs, that data is sufficient (together with the parameter files) to recreate the full spectrum. That your Topspin processes 2D spectra automatically is just some specific configuration, it doesn't do that by default.

As far as I understand it, if you zero-fill your data at least once, all the information content of the imaginary part is then present in the real part of the processed spectrum. So I'd guess that under those conditions, the Hilbert transformation shouldn't have any downside, but I'm not totally sure about that.


Why is there no salt bridge for hydrogen–silver bromide cell?

The function of a salt bridge or a diaphragm is to avoid mixing of 2 different electrolytes or electrolysis products from the respective half-cells, while allowing ion electro-migration between them.
This cell does not need a salt bridge as it uses the only electrolyte $\ce{HBr(aq)}$ and the is no need to separate the $\mathrm{AgBr}$ electrode from gaseous hydrogen.


Is palladium on carbon acting as an oxidising or reducing agent?

Palladium on carbon ($\ce{Pd/C}$) is neither an oxidant nor a reductant. The reaction is the dehydrogenation of piperidine: $\ce{C5H11N -> C5H5N + 3H2}$ and here, $\ce{Pd/C}$ is merely a catalyst.

Even in standard hydrogenation reactions using $\ce{H2(g) + Pd/C}$, the $\ce{Pd/C}$ is not a reducing agent. The $\ce{H2}$ is the reducing agent and Pd/C is again only a catalyst.


A better way of thinking about it, without getting too bogged down in details, is that the plant absorbs the energy of photons and uses it to drive chemical processes - ultimately the synthesis of sugar out of carbon dioxide and water - that are energetically "uphill", and would not happen spontaneously without the energy from the photons.  The energy from the photons is used by the plant to move electrons from one place to another, but no electrons are actually generated or consumed in the process ... photosynthesis is a cycle, so you always end up back where you started, with some molecules of $\ce{CO2}$ and $\ce{H2O}$ having been consumed (along with a lot of light energy), and some extra molecules of sugar and $\ce{O2}$ having been produced.

The places where the photonic energy is used to mobilize the electrons are protein complexes called photosystem I and photosystem II, which are concentrated in the chloroplasts of green plants.  The details of how this happen are still not completely understood, but a LOT of progress has been made fairly recently using fancy equipment like laser spectrometers that can measure physicals processes that occur on timescales shorter than a nanosecond.  This has allowed researchers to map the progress of the electron as it moves from one part of the protein complex to another, following absorption of the photonic energy.

Does that help at all?


What is the area that 1 litre of gasoline would spread above the sea?

The hint here assumes a layer thickness of (crude) oil of about 1 mm. If I remember the pictures in the News around Deepwater Horizon correctly, and compare what I saw with own eyes at the local petrol station, petrol has a lower viscosity; likely spreads out wider than crude oil.

From here we learn the layer thickness is much lesser than 1 mm (millimetre), namely 168 nm (nanometre); hence all these colourful interference patterns in visible light.

$$ 1 \text{ L} = 1 \text{ dm}^3 =  10^{24} \text{ nm}^3$$

Deviding by 168 nm provides with googles built-in calculator an area of about $$5.95 \times 10^{21} \text{ nm}^2 = 5952 \text{ m}^2$$

To set this into a context, UEFA and FIFA agreed on the size of European Football / Association (soccer) field ($ 105 \times 68 \text{ m}$) which equals $7140 \text{ m}^2$.

Addendum:  According to this, an American Football field measures "$120.0 \text{ yd } (109.73 \text{ m}) \times 53.3 \text{ yd } (48.74 \text{ m})$ or $6400 \text{ yd}^2 (5351.2 \text{ m}^2$)".


Why is it important that glucose’s third OH group points to the left?

This is the concept called stereochemistry.

Each carbon with 4 bonds has an approximately tetrahedral geometry.  We assign a configuration to the situation by numbering the substituents in terms of priority. Hydrogen is lowest.  Oxygen is highest.  The top carbon is second, and the other carbon is third.  We then trace a route from 1 to 2 to 3.  This would be a clockwise circle, and we would call this configuration S.  If we switched sides between the H and the O we would follow this circle widdershins and call the configuration R.  So the substituents can't be exchanged because the configuration would change.

So what?

When there is only one of these tetrahedral centers (called stereocenters) in play for a situation, it does not matter at all which one you have.  But the moment a second stereocenter comes into the picture, chemical properties begin to change.  A fairly tractable analogy is feet and shoes.  When you aren't wearing shoes, your left foot and your right foot are about the same.  But when you try to put shoes on, putting the right shoe on the left foot is slightly more difficult than putting the right shoe on the right foot.  In this loose analogy, the 'reaction' of putting your shoes on is slowed based on how the stereocenters of foot and shoe interact.

Onto glucose ...  Because there are 4 stereocenters in glucose, it has a distinct shape from its isomers.  The other chemicals in your body also have their own shapes/configurations.  Glucose reacts slightly differently than other sugars in the body because of the way these shapes fit together.  Therefore you need to keep the configurations known and well marked.

Most people (about 94% of the rest of you) don't intuitively see stereochemistry concepts and need to make models.  However, pragmatically, if you can take this explanation, and remember that 'it just matters' to keep the notation consistent, you can probably get through most of biochemistry.

If you want to make models, at your next wine and cheese party get some toothpicks, and two pieces each of four different kinds of cheese.  Put four of the cheese chunks on the end of the toothpicks, and then stick the tooth picks in a sausage or something.  Then make an exact copy.  After that is done, exchange two of the cheese chunks on one model and investigate what you have.  There should be no way you can twist or turn or rotate one model to have its kinds of cheeses in the same orientation as the other.  Once you have done this, the following picture will make much more sense:


Intramolecular Cannizzaro reaction product

Your reasoning is correct and the problem is with the publisher as they made  a mistake while compiling these previous year questions.
I found out the original 2003 question paper(pdf). The question you mentioned had the following options:

It is clear form the options that the answer is C(which is the product you expected) which can also be confirmed by the key attached at the end of the pdf.


What are the most important resonance structures of 5-aminocycloheaxa-2,4-dienone?

The handwriting of your TA probably stems from ancient egyptian script, I had a hard time reading it. In the end I agree with jerepierre, interpreting it as "6E" or "8E" short for 6/8 electrons. The last statement then translates to "but all have 8E."
Before continuing to answer your question, I want to make a statement about resonance structures in general. Resonance structures are used to express a complicated bonding situation in terms of simpler, Lewis type structures. None of these structures exist individually, only a mixture of all possible resonance structures at the same time exists. I am not a fan of declaring single resonance structure as having more or less impact on the overall structure. Of course one can go ahead and calculate the total wave function in terms of resonance structures, generally known as valence bond theory, but this is overkill for such simple molecules. Then you get a mathematical representation of the electronic structure which you can decompose into contributions of the individual resonance structures.
A simpler approach is to analyse the molecule with Natural Bond Orbital (NBO) Analysis. I have done this for you, but only as an educational effort. Maybe after understanding this concept it is easier to grasp why certain structures are make a smaller contribution.
Let me go ahead and explain a little NBO to you.  The electronic structure of a molecule is in general highly delocalised. All orbitals are stretched over the whole molecule. Without changing the electron density, it is possible to localise (transform) these orbitals to represent two centre interactions, that can be interpreted as bonds. This gives an approximate picture in terms of a Lewis structure. There will be electron density which cannot be assigned to a bond. The smaller this density is, the better is the fit as a Lewis structure.
For this educational approach I picked the 5 resonance structures you have posted plus an additional ionic form. (I would have detracted a point for the last structure.) There are many more ionic forms, but these will not have big effects, e.g. cleavage of a carbon carbon bond.

The following is based on NBO analyses from DF-BP86/def2-SVP.
\begin{array}{crr}\hline
\text{Structure} & \text{% Lewis} & \text{% non-Lewis}\\\hline
\mathbf{1}   & 96.4  &  3.6  \\
\mathbf{2}   & 96.8  &  3.2  \\
\mathbf{3}   & 96.2  &  3.8  \\
\mathbf{4}   & 95.8  &  4.2  \\
\mathbf{5}   & 96.3  &  3.7  \\
\mathbf{6}   & 95.2  &  4.7  \\\hline
\end{array}
The first statement that should be made is that this molecule is quite well described by Lewis resonance structures. However, delocalised bonding would describe the molecule much better. All structures are reasonably close to each other, as one would expect for a very delocalised system.
The ionic structure has the smallest impact, which is feasible because nitrogen-hydrogen bonds are better described as covalent. However, it proves that there can (in the right medium) be a proton shift from nitrogen to oxygen, and is therefore worth consideration.
The structure with the second largest error (% non-Lewis) is $\mathbf{4}$. A positively charged carbon would be by far more electronegative than nitrogen. And many imine-type structures are known, which makes $\mathbf{5}$ more important. (In this sense I would disagree with your explanation, which could also justify the detraction of a point.)
In general: Carbocation structures that are not in close proximity to (much) more electronegative elements should be considered as having a lesser influence on the electronic structure of a molecule.


$\ce{H_{2}O}$ is the water molecule, two hydrogen atoms attached to a central oxygen atom, $\mathrm{C}_{2v}$ symmetry, thermodynamically minimal structure of these atoms, Adam's ale, etc.

HHO is a poorly defined term often bandied around by 'water powered car'... enthusiasts. I'm not sure that it's supposed to represent a molecule so much as a state of hydrogen and oxygen that conventional chemistry has no concept of.

Here's a post on skeptics.SE about it
And here's an acerbic article on rationalwiki that mentions 'HHO'

I think we're supposed to believe that it's a gas of hydrogen and hydroxyl radicals, or of atomic hydrogen and oxygen, or protons and hydroxide ions, or something, that is metastable at room temperature and that also happens to violate the first law of thermodynamics when produced and combusted. Why this glaringly obvious anomaly in quantum electrodynamics and thermodynamics has never been observed anywhere in nature before is anyone's guess (probably part of the conspiracy).

These guys should write up a few papers, get published in Nature, collect the Nobel chemistry, physics and peace prizes...


No. When the gas - let's assume it is an ideal gas - rises up it will expand because the pressure $p$ sinks the higher it comes. If it is allowed to expand quasi-statically under adiabatic conditions ($\delta Q = 0$) then it does work $W$ (pressure-volume work $W = - p \mathrm{d}V$ to be more precise) on its environment, and, hence, its internal energy $U$ is reduced, and its temperature $T$ changes because according to the first law of thermodynamics

\begin{align}
  \mathrm{d} U &= \delta Q + \delta W \\
  &\overset{\delta Q = 0}{=} \mathrm{d} W = - p \mathrm{d}V
\end{align}

and also for an ideal gas you have

\begin{align}
  \mathrm{d} U &= C_{V} \mathrm{d}T
\end{align}

where $C_{V}$ is the heat capacity. Hence the temperature changes according to

\begin{align}
  \mathrm{d}T = - \frac{p}{C_{V}} \mathrm{d}V \ .
\end{align}


How to remove formaldehyde (smell) from clothes/jeans?

Formaldehyde is indeed used for disinfection, but usually there is a subsequent procedure involving formaldehyde neutralization with $10\%$ to $25\%$ aqueous ammonia solution introduced (as a spray) at elevated temperatures. Ammonia reacts with formaldehyde producing water-soluble hexamine which, even though also has unpleasant smell of rotten fish, is then easily washed out from the fabric:

$$\ce{6 CH2O + 4 NH3 -> C6H12N4 + 6 H2O}$$

I'd suggest to wash the clothes at higher temperature ($\pu{40 .. 60 ^\circ C}$) as you would normally do, but adding some ammonia; afterwards, soak the clothes in cold water for a while (solubility of hexamine increases in cold water), wring out and air dry.

In some countries there are also ozone-chambers (basically a sealed wardrobe with an $\ce{O3}$ generator inside, offered by dry cleaning companies) where organic pollutants are oxidized; formaldehyde is no exception:

$$\ce{CH2O + 2 O3 -> CO2 + H2O + 2 O2}$$

Both vinegar and soda can also be used, though I'd think their effectiveness is going to be lower.


Letting Copper sit in NaOH for 5 min (for monolayer assembly)

Such steps are usually cleaning steps to remove deposited organic materials. $\ce{NaOH}$ is also known as lye and will dissolve basically all organics. In this case, it should leave the $\ce{Cu}$ and copper oxides alone, leaving a clean surface.

I don't have as much experience with copper monolayers, but similar cleaning steps are used in my lab, where you might treat with ozone or some other step to remove any deposited organic "gunk" from the metal surface.

In this case, I believe you don't want to use an acid, because the thiols react with the copper oxide surface, not strictly the copper itself.


Can a small amount of scandium be added to gold, platinum or palladium to create a wedding ring?

The go-to place for alloy information is ASM International's Alloy Phase Diagram Database (providing your institution has access).

Scandium-Gold: as noted in the comment by @MaxW, you can get several atomic percent (but less than 10 at.%) Sc in Au. There exist 6 compounds across the diagram which you would want to avoid most likely.

Scandium-Platinum: likely up to about 10 at.% of Sc in Pt. At least several compounds to avoid.

Scandium-Palladium: also order of 10 at.% of Sc. 5 compounds to avoid.

All the compounds should be avoided because (1) they are higher in scandium content, and (2) they are mostly line compounds and unlikely to be malleable.

So, yes, you can get scandium in to a gold-platinum-palladium mix. Find a good jeweler and they can melt it in for you. Scandium is not all that expensive (at least compared to the cost of getting the jeweler to do a custom alloy).


How would one compare the magnitude of covalent character between SnCl4 and SnF2 using Fajan's Rules?

Fajan's rules for the prediction of bond character can be summarized in two points:

Large cation with low positive charge + small anion $\rightarrow$ ionic
Small cation with high positive charge + large anion $\rightarrow$ covalent

In $\ce{SnF2}$, we have a large cation with low positive charge. $\ce{Sn^2+}$ is larger than $\ce{Sn^4+}$ because it has 2 more electrons in the $5s$ orbital. The anion $\ce{F^-}$ is smaller than its heavier group homolog $\ce{Cl^-}$. For this combination large cation + small anion, the bond character is predicted to be more ionic and less covalent.

On the other hand, $\ce{Sn^4+}$ is a small cation with high positive charge, and $\ce{Cl^-}$ is the large anion, as already stated. Therefore, according to the rules, $\ce{SnCl4}$ is the tin halide with more covalent bonds, and indeed, tin(IV) chloride consists of tetrahedral $\ce{SnCl4}$ molecules in the solid, liquid and gaseous state.

However, the predictions made with these rules do not always agree with the real bonding situation, and $\ce{SnF2}$ is an example. There is experimental evidence for covalent $\ce{Sn-F}$ bonds in tin(II) fluoride (references: 1, 2).


What is the difference between a mixture and a solution?

Let's try some examples for mixtures and solutions. Think about the answers yourself before you uncover them ;-)

We cut carrots, celery, and green beans into pieces and throw them into boiling water to prepare a vegetable soup. Mixture or solution?

  It's a mixture, because the pieces of the different vegetables can be retrieved by a physical method, such as picking them with a fork.

Throw a spoonful of sugar into a cup of water and stir until there's no sugar left on the ground of the cup. Mixture or solution?

   This is a solution. It it homogenous; even the tiniest part of the water contains the same amount of sugar and there is no way to get back the sugar (unless you boil the water off).

Take a handful of razorblades and a handful of ants. Throw everything in a bowl. Wear gloves! Shake. Mixture or solution?

   This is a mixture. You can separate the components with a magnet - or by luring the ants out of the bowl with a trace of sugar water.

Melt iron. Add cobalt. Don't mix with bare hands ;-) Let it cool down until it is a solid again. Mixture or solution?

   This is a solution. Solutions do not need to be liquid. They can be solids as well. You can cut the metal block into pieces and mill every piece down to a fine powder and the amount of iron and cobalt in every grain of powder will be the same. Again, you cannot fiddle the components apart by any physical method.


Why thermal decomposition of iron(II) nitrate and sodium nitrate yields different products?

Iron(II) cation is moderately reducing- an iron(II) cation has a valence electron configuration of $\mathrm{3d^5 4s^1}$, and by giving up an electron in its $\mathrm{4s}$ orbital the cation will transform into an iron(III) cation with $\mathrm{3d^5}$ valence electron configuration, which, with every $\mathrm{d}$ orbital half filled, has a higher degree of symmetry over $\mathrm{3d^5 4s^1}$ electron configuration and hence is more stable (thermodynamically favored).
Sodium(I) cation, on the contrary, has a highly stable valence electron configuration of $\mathrm{2s^2 2p^6}$, with all its $\mathrm{2s}$ and $\mathrm{2p}$ orbitals filled, identical to the electron configuration of the chemically inert noble gas neon, and either removing or adding one electron to the sodium(I) ion will lead to an electron configuration much less stable and hence thermodynamically unfavored.
You may notice that in the decomposition reaction, the iron(II) cation is oxidized into iron (III), which is favored, and the sodium cation remain intact.
Put it in another way, we may imagine a first step thermal decomposition reaction of iron(II) nitrate resembling decomposition of sodium nitrate and yields iron(II) nitrite and oxygen gas as products. Then we easily notice that iron(II) nitrite is reducing, and oxygen gas is a strongly oxidant, hence the two products of the first step decomposition can further react with each other to produce iron(III) oxide and nitrogen dioxide.


How to calculate the mass of hydrogen in a borane sample?

The problem does not need any difficult approach, not even relative atomic masses, if you trust the one who made this exercise. Since borane (whatever type it is)* consists (by exercise's hypothesis) only of hydrogen and boron and since boron is $85.63\%$ by mass, hydrogen's mass percentage would be the remaining $100\% - 85.63\% = 14.37\%$. Therefore, we have $\pu{6.60 g}$ of hydrogen in this sample.

The compound borane, $\ce{BH3}$, is about $78.1\%$ boron per mass. However, there is a class of compounds called boranes which also only consist of boron and hydrogen. From this class pentaborane(9), $\ce{B5H9}$, with about $85.7\%$  boron by mass would fit the exercise.


Why do principal energy levels in an atom get closer together as n increases?

The energy levels for an atom are similar to a "particle in a box", except that the walls have a different shape. When you analyzed the "particle in a box" situation, you used a rectangular box. An atom is spherical, so the walls of the atom open outwards at the same rate as the sphere increases in surface area. You should have noticed that the energy levels for a "particle in a box" decreased in distance at higher energy levels. The surface area of the sphere brings the energy levels even closer.


Determining structure factors for crystals from lattice and motif

Q1) A crystal is a convolution of a lattice and motif, so by the convolution theorem, the fourier transform of the crystal (i.e. your structure factor) is the product of the fourier transform of the lattice and the fourier transform of the motif. That is, this is general.
Q2) If the crystal is centrosymmetric* then the structure factor is real - this is a property of the fourier transform (the imaginary sine terms have opposite signs and equal other and are therefore zero). If your lattice and motif generate a centrosymmetric crystal then your fourier transform will be real. The motif doesn't need to be centrosymmetric to generate a centrosymmetric crystal.
*and the atomic form factors are real, but this isn't relevant.


Why are rates of reaction faster for ionic compounds than for covalent compounds?

Not all ionic reactions are not faster, i.e. have a larger rate constant than those between neutral species, however, at similar activation energies there are effects that only ionic reactions suffer. There are two considerations, the dielectric constant of the solvent used, water, hexane, ethanol, acetone etc., and the relative charges on the ions, similar or opposite as well as their value, $\pm1,\pm 2 $ etc.

For slow reactions, that is not diffusion controlled, (see below) the 'primary kinetic salt effect' is important. This shows that the rate constant can increase or decrease depending on the ionic strength of the solution and the charge on the ions. It is based on the Debye-Huckel theory of ions in solution which relates activity coefficients in dilute solution to the ionic strength.

The rate constant for a bimolecular reaction  has the form
$$ \log_{10}\left(\frac{k_2}{k_2^0}\right) = 2\alpha z_az_b\sqrt I$$
where I is the ionic strength, the z the charges on the two ions, $(-1,+1),(-1,-1) $ etc. and $\alpha \approx 1.6 \cdot 10^{-2} \pu{ m^{3/2} mol^{-1/2}}$ is a constant originating in Debye-Huckel theory. At low ionic strength ($ \lt 0.5 \pu{mol dm^{-3}}$ ) the rate constants increase or decrease linearly with $\sqrt I$.

The effect of increasing the dielectric constant (relative permittivity) is to lower the rate constant, thus rate constants are lower in water ($\epsilon = 78$) than in cyclohexane ($ \epsilon = 2.2$) for the same reaction. This is due to the fact that, at a given distance, in a high dielectric solvent the electric field of the ion is greatly reduced compared to its value in a low dielectric solvent. Thus in a high dielectric the ions are far more isolated from one another and so rate constants are only affected when ions come in relatively close contact. Conversely in low dielectric solvents the ions 'feel' the presence of one another at a large separation, this can considerably increase or decrease the rate constant depending on relative charges.

The rate constant is given by
$$ \ln(k_2) =\ln(k_2^{\infty}) - \frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}k_BT}$$
where $k_2^{\infty}$ is the limiting rate constant for a solvent of infinite dielectric constant. The permittivity of free space is $\epsilon _0$, the relative permittivity $\epsilon$, q is the charge on the electron, $d_{ab}$ the separation of the ions at reaction and $k_B$ the Boltzmann constant. A plot of the rate constant vs. $1/\epsilon$ yields a reasonably straight line  for all but the lowest $\epsilon$ values.

When a reaction's activation energy  is low, less than $\approx 10 \pu{kJmol^{-1}}$ the reactants react so rapidly on contact that the rate constant in determined by how quickly they can approach one another in solution; the rate constant then becomes diffusion controlled and thus depends on the solution viscosity.

Using the Stokes-Einstein equation and assuming equal sized molecules the rate constant has the simple form
$$ k_2= \frac{8RT}{3\eta}$$
where $\eta$ is the solvent viscosity. Water has a value $\approx 1\cdot 10^{-3} $ Pa s or $1$ centipoise (cP) at room temperature. In fluid solution such as hexane the rate diffusion controlled constant is typically $10^{10}$ to $10^{11} \pu{dm^3mol^{-1}s^{-1}}$.

The effects of ions changes the diffusion controlled rate constant for the reasons described above and is now
$$ k_2= \frac{8RT}{3\eta}\left(\frac{x}{e^x -1}\right)$$
where
$$ x=\frac{z_az_bq^2 }{4\pi \epsilon _0 \epsilon d_{ab}kT}$$
and is the term used before.

When the ionic charges are the same and the dielectric constant small the rate constant is greatly reduced, and conversely it is enhanced when the charges are opposite in sign to one another. In high dielectric solvents these changes are greatly reduced. These effects are similar to those described above.


Hydrogen venting: Why no flame arrestors?

Hydrogen has a very broad range for combustion in air, with a LEL (lower explosive limit) of about 4% H2 in air and a UEL (upper explosive limit ) of about 96% H2 in air (or 4% air in H2 if you prefer) - well, actually a more recent reference from Matheson indicates 4% to 75% - I've learned something new today, thanks! It also has a very low ignition energy (energy input required to get it going). So, your exhaust system can fill up with lots of H2 and then get set off easily. This can quickly go from a small flame zone to deflagration in the exhaust system (sub-sonic) to detonation (supersonic reaction front), and the reaction will continue until you are outside of the LEL/UEL range.

Look over the Matheson reference, and you will see that there are very few other gases with such a broad LEL/UEL range. For example, methane (or natural gas) is only 5% to 15%, propane is 2.1% to 9.5%. So, if you fill up a space with methane or propane, only so much of it will burn, then it will quench as the mix goes out of the LEL-UEL limits.

The implication is that a hydrogen-air mixture can sustain detonation/deflagration across a much broader range, so it can (and by Murphy will) make a lot more flame and shock and awe. Bottom line, don't have anything in the way of all that excitement exiting the facility as quickly and completely as possible.

An article in Chemical Engineering discusses the issues briefly, although doesn't go into the engineering details (a bit odd for an engineering magazine?).


Can a mixture of water miscible solvent and water be separated by fractional freezing on a dry ice bath?

The short answer is

  Yes

This is exactly how acetic acid is purified: Acetic acid freezing distilation

In this case, acetic acid (freezing point $16\ ^\circ \text{C}$) freezes first, leaving a water-enriched fluid behind. Conceivably you could remove water from methanol or acetone in the same way (except that water would freeze first). In practice, fractional distillation can be done more economically on a large scale. Fractional distillation requires 1 step (with lots of theoretical plates), while you need to repeat fractional freezing multiple times to get to high purity. Acetic acid is purified by fractional freezing perhaps because it may azeotrope with water.


How do atomic radius and nuclear charge influence ionisation energies?

When we are considering atomic radius:
The atomic radius of noble gases are considered the highest in a period since we only consider their Van der Waals radius (distance between two similar atoms in their closest approach). Although technically they should have lower atomic radius due to increased $Z_\text{eff}$.

As for ionisation energy:
The ionisation energy of noble gases are way higher than the other elements of the same period due to stable configuration


What is the "‡" symbol meaning in a reaction mechanism?

The symbol is called "double dagger" (sometimes also "double cross") and is used to denote transition state (a maximum in an energy diagram; also often denoted with "*" or "TS") or a related physical property.
Note, however, that a transition state and an intermediate are two different terms.

The symbol has the peculiar origins: as written by H. Eyring, it's entirely the merit of the secretary of the department, Miss Lucy D’Arcy [1, p. 9]:

  The symbol, ‡, first appeared in the second paper dealing with the activated complex.
  In that manuscript a star was used to designate the activated state but
  Miss Lucy D'Arcy, the departmental secretary, lacking a star used a plus and minus sign.
  Believing the typesetters would interpret this as a star it was left and by
  their option the cross, ‡, has become the almost universally used sign for the activated state.

Also, note that according to IUPAC recommendations [2, p. 1080]:

  In accordance with previous IUPAC recommendations (IUPAC QUANTITIES (1988)) the symbol ‡ to indicate transition states ("double dagger") is used as a prefix to the appropriate quantities, e.g. $Δ^‡G$ rather than the more often used $ΔG^‡.$

In the future, I suggest to use services like Shapecatcher or Detexify to find the name of the symbol.
Finding its usage in chemistry afterwards is normally a trivial task.

References

Eyring, H. Models in Research. International Journal of Quantum Chemistry 1969, 3 (S3A), 5–15. DOI: 10.1002/qua.560030705.
Muller, P. Glossary of Terms Used in Physical Organic Chemistry (IUPAC Recommendations 1994). Pure and Applied Chemistry 1994, 66 (5), 1077–1184. DOI: 10.1351/pac199466051077. (Free Access)


When does nucleophilic attack and proton transfer happen?

Because you are a beginner to organic chemistry, I'll tell you this: Both esterification (e.g., Fischer esterification) and de-esterification reactions are reversible. The question you had at the end is why the last step of base catalyzed de-esterification reaction is only forward reaction. As you have learned in general chemistry class, acid-base reaction is one of the fastest reactions in chemistry (not counting nuclear reactions). Thus, the last step is governed by preference for acid base reactions:

$$\ce{R-COOH + OH- -> R-CO2- + H2O}$$

The $\mathrm{p}K_\mathrm{a}$ of $\ce{R-COOH}$ is about $4.7$ ($\mathrm{p}K_\mathrm{a}$ of $\ce{CH3-COOH}$ is $4.75$) while $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O}$ is $15.7$. Therefore, forward reaction is about $10^{10}$ times faster than backward reaction.

Also remember, this reaction is in aqueous medium, therefore there are much more $\ce{H2O}$ molecule available than $\ce{R-OH}$ molecules. Thus, going back to ester reaction is much slower than becoming acid (governed by Le Chatelier principle).


The sharp, intense band at 1716 $\mathrm{cm^{-1}}$ is indeed characteristic for a carbonyl group ($\ce{C=O}$ stretch, in this case aldehyde or ketone). The medium and broad bands between 2732-2957 $\mathrm{cm^{-1}}$ suggest that it is more likely an aldehyde. In this case, they would be assigned to the $\ce{=C-H}$ stretch, and the intensity of the broad band fits better to a saturated aldehyde, i.e. there is no double or triple bond in $\alpha, \beta$-position to the $\ce{CHO}$ group.

Your assignment of the peaks at 3624 and 3533 $\mathrm{cm^{-1}}$ to a primary amine is reasonable, since primary amines have two bands, while a secondary amine would have only one, weaker band at lower wavenumbers. One of the bands in the area of 700-940 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{N-H}$ wagging vibration mode.

The sharp band at 3416 $\mathrm{cm^{-1}}$ could indicate a terminal alkyne ($\ce{C \equiv C-H}$ stretch), the $\ce{C-H}$ stretch band of this group at approximately 2900 $\mathrm{cm^{-1}}$ would then be included in the broad band at 2957 $\mathrm{cm^{-1}}$. The weak band in the area between 2000-2250 $\mathrm{cm^{-1}}$ can then be assigned to the $\ce{C \equiv C}$ stretch.

Sources: 1, 2, 3, 4


Why are potassium organic salts ever preferable to the sodium equivalents?

The differences in the properties of the alkali cations are subtle, and the rationale for choosing one over the other will depend heavily on the application. I suspect you're right, that to some extent natural abundance of, e.g., potassium salts relative to sodium ones in certain mineral contexts explains the industrial preference for one or the other -- but, really, variations in abundance are just further examples of these subtle differences in the properties of the cations!

One specific example that I recently came across is the Hofmeister series (see also the Wikipedia article), where the choice of cation makes a demonstrable difference in the solubility properties of proteins.  Speaking also from personal experience, the choice of, e.g., $\ce{Na2SO4}$ versus $\ce{K2SO4}$ versus $\ce{(NH4)2SO4}$ as a component of the supporting electrolyte in electrodeposition of copper from acidic $\ce{CuSO4}$ matters significantly to the properties of the deposit obtained.


Why is it that less useful energy can be extracted in explosive reactions?

I think there are two relevant points here.

1) explosive reactions can give significant useful energy - e.g. internal combustion engine

2) step-wise processes may go slower and release less heat / light other energy that is not so useful. This is a bit like in thermodynamics you get maximum useful energy from slow reversible processes.


Reaction of Cu in hydrochloric acid bath with exposure to air over several months?

In the presence of excess hydrochloric acid, copper oxides are not the immediate reaction product of copper metal and oxygen (e.g. from the air).  The air does oxidize the copper...

$$\ce{Cu + 1/2 O2  + 2 H+ -> Cu^{2+} + H2O}$$

... but the final oxidation product is copper(II) ion, not copper(I).  (Copper(I) is formed initially but is more prone to air oxidation than copper metal is, so it is relatively quickly converted to copper(II).)  Additionally, in the presence of chloride anions, the copper(II) ions are preferentially complexed by chloride.  The low concentration of hydroxide means that hydroxide complexes are not favored, and chloride is a better ligand than water for copper.

The actual complexes formed vary in their stoichiometry and their color.  $\ce{CuCl2(aq)}$, $\ce{CuCl3^{-}(aq)}$ and $\ce{CuCl4^{2-}(aq)}$ are all possible:

$$\ce{Cu^{2+} + n Cl- -> CuCl_n^{(n-2)-}(aq)}$$

$$n \in \{2, 3, 4\}$$

Nurdrage made a nice video showing how the reaction of $\ce{HCl}$, air, and copper metal leads to copper chloride.  It's the first of the three methods he presents for making copper chloride.  As he notes, the reaction is very slow.
$%edit$


To answer this question I would first assign the R-S configuration to the original molecule and the molecule proposed by you and show you that both are identical. At the end I will also show you a simpler way to tell that both are identical.
Original molecule:

Proposed molecule:

As you can see, both the molecules have the same R-S configuration and thus are identical.
Now to see this easily and intuitively just assume you are looking at the original molecule from inside the screen. What do you see? You will observe that the original molecule will look just as same as the proposed molecule which means that both are identical. This method works because our frame of reference doesn't change the optical properties of a molecule until and unless we change the configuration of the molecule by doing some reflections.


You have, for each compound, $\mu_i = \mu^\circ_i + RT\ln(P_i/P^\circ)$.
So, replacing each potential in $\Delta G$ by its expression, assuming $P^\circ = 1$ bar, we get :
$$\Delta G = c(\mu^\circ_C + RT\ln(P_C)) + d(\mu^\circ_D + RT\ln(P_D)) - a((\mu^\circ_A + RT\ln(P_A)) - b(\mu^\circ_B + RT\ln(P_B))$$
Then we develop and group standard potentials and logs:
$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT(c\ln(P_C) + d\ln(P_D) - a\ln(P_A)-b\ln(P_B))$$
Then, using calculation rules on logarithms:
$$\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\left(\ln({P_C}^c)+\ln({P_D}^d)-\ln({P_A}^a)-\ln({P_B}^b)\right)$$
And finally, since $-\ln(x) = \ln(1/x)$ and $\ln(x)+\ln(y)=\ln(xy)$,
$$\boxed{\Delta G = (c\mu^\circ_C + d\mu^\circ_D - a\mu^\circ_A - b\mu^\circ_B) + RT\ln\left(\frac{{P_C}^c{P_D}^d}{{P_A}^a{P_B}^b}\right)}$$
Then, try expressing the quotient of the reaction in terms of partial pressures, assuming standard pressure is 1 bar and you'll find out it gives the inside of the log!


Energy of photons of given frequency and number

Analogy: you have 5 bottles of coke 1 L each; what is the total volume?
The formula you shown $(E = hf)$ is derived for exactly one photon.
Now, since you have 50 billion of photons $(N = \pu{5.00e10}),$ guess what the total energy would be?

 $$E_\mathrm{tot} = Nhf = (\pu{5.00E10})(\pu{6.63E-34 J s})(\pu{1.00E9 s-1}) = \pu{3.32E-14 J}$$


Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state?

In the steady-state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. So in this case only if k2 >> k1 and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P.

\begin{align}
\frac{d[\ce{S1E}]}{dt} &= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}] - \ce{k2}[\ce{S1E}] = 0\\
\end{align}

Therfore:

\begin{align}
[\ce{S1E}] &= \frac{\ce{k1}}{k_{-1}+\ce{k2}}[\ce{S1}][\ce{E}] = K_a[\ce{S1}][\ce{E}]
\\
\end{align}

Similarly for the second reaction:

\begin{align}
[\ce{S2E}] &= \frac{\ce{k3}}{k_{-3}+\ce{k4}}[\ce{S2}][\ce{E}] = K_b[\ce{S2}][\ce{E}]
\\
\end{align}

The enzyme E is involved in both reactions and its total concentration(bound and unbound) is constant. This total concentration of enzyme $[E]_{0}$ is equivalent to the concentration of the free enzyme before adding the substrates. The concentration of the free enzyme at a certain time t is [E]:

\begin{align}
[E]_{0} &= [E] + [\ce{S1E}] + [\ce{S2E}]
\\
\end{align}

if you substitute $[\ce{S1E}]$ and $[\ce{S2E}]$ with the previous expressions then:

\begin{align}
[E] &= \frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}

Eventually:
\begin{align}
\frac{d[P_{1}]}{dt} = k_2[\ce{S1E}] = \ce{k2}(K_a[\ce{S1}][\ce{E}]) = K_p[\ce{S1}][\ce{E}] = K_p[\ce{S1}]\frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}

and similarly for the rate of production of $P_2$


Explanation of terminology in equation used for energy calculation using Hartree–Fock method

For every $x$ and $t,$ $\Psi(x,t)$ is a complex number. $\Psi^*$ is the conjugate of that number, no more, no less. The reason it seems like sometimes it's only the $t$ part that gets conjugated is simply that often it is the only part of the wavefunction that is complex. Let's use an example:
$$\Psi = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{-\mathrm iE_n t}.$$
We want to calculate $\Psi^*$. Well, since the conjugate of the product of two numbers is the product of their conjugates (that is, $(zw)^* = z^* w^*),$ let's do it step by step.
First we need to conjugate $\displaystyle\sqrt{\frac{2}{a}}$, but since it's a real number, it is equal to its conjugate. So we leave it alone and move on. Now we need to conjugate $\displaystyle\sin\left(\frac{n\pi x}{a}\right),$ but again, this is a real number, because $\sin x$ is real whenever $x$ is real. The last part is $\mathrm e^{-\mathrm iE_n t}$. This is actually complex, so we need to conjugate it, and its conjugate is $\mathrm e^{\mathrm iE_n t}.$ So putting it all together, we have
$$\Psi^* = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{\mathrm iE_n t}.$$


How does cooling increase the selectivity of a reaction producing the kinetic vs the thermodynamic product?

First, there is no rule that "the" kinetic product will be different than the thermodynamic product.  It depends on the particular reaction you are interested in.

Thus, I think you are really asking "in reactions where the kinetic product is different than the thermodynamic product, what does cooling select for?"

Let's consider two reactions, the first with a reactant R to a thermodynamic product T and a second, separate reaction where R goes to the kinetic product K.  The rate constants for these reactions are $k_1$ and $k_2$, respectively.

$$\ce{R ->[k_1] T}$$
$$\ce{R ->[k_2] K}$$

Let's suppose both reactions follow classical transition state theory, which is a very good approximation to many everyday reactions and is a great starting place even for the weird cases that do vary.

TST says (if you assume that the $\kappa$ transmission coefficient is 1):

$k_1 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_1/RT
}$

$k_2 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_2/RT
}$

The "kinetic" product is the one with the higher rate constant; this is the definition of kinetic.  Doing some algebra,

$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$

...we see that for $k_2$ to be higher than $k_1$, i.e. for $\ln{\frac{k_2}{k_1}}$ to be greater than 0, then $\Delta G^{\ddagger}_2$ must be smaller than $\Delta G^{\ddagger}_1$ (note the negative sign).

The effect of temperature is still in the equation:

$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$

The assumption of TST is that $\Delta G^{\ddagger}$ is not a function of temperature.  As temperature is lowered, then the $RT$ divisor in the equation gets smaller, which means that the difference $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$ affects the ratio of rate constants more.  Thus, under the assumptions we have made, cooling will favor the kinetic product more. As temperature is raised, eventually $RT$ becomes much greater than $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$, so selectivity for the kinetic product vanishes because $\ln{\frac{k_2}{k_1}} \rightarrow 0$, i.e. $\frac{k_2}{k_1} \rightarrow 1$.

You might ask, how can the ratio go to one; doesn't that mean that we have assumed the equilibrium constant between T and K is also 1?  The answer is no, we haven't assumed that, but we have made a number of other assumptions that mean we we won't be able to predict the net rates of T and K formation.  The most important one is that we have neglected the reverse reactions $\ce{T -> R}$ and $\ce{K -> R}$; introducing them would give the same qualitative effect of cooling the temperature but would complicate the math considerably (but would let us wind up with net rates of T and K instead of the gross single-directional rates we have calcualted here.  Second we have assumed that $\kappa = 1$, which might not always be true.


How would one construct a 3D model of a molecule based on the crystal structure of its salts?

In general, the crystal structure of a molecule is a highly accurate geometry. So yes, you can take a salt and modify as needed. (The caveat in the comment is useful - keep in mind that different groups may be protonated or deprotonated depending on solvent and pH.)

There are two other caveats:

1) The conformation in the crystal structure is not necessarily the lowest energy conformer in solution or the gas phase. You may wish to perform a conformer search to find a low-energy minima.

2) If you are performing further calculations (e.g., DFT for electronic structure, force fields for MD, etc.) the geometry of the crystal structure will not likely be a minima of the computational method, so geometry optimization or equilibration (for MD) is needed before further simulation.

That said, the crystal geometry is a good place to start for many uses. Certainly if a crystal structure is available, I'd suggest you use it rather than build a geometry by hand.


What are the products of the reaction of NH4Br (aq) + Hg2(NO3)2 (aq)?

According to this table, mercury (I) bromide ($\ce{Hg_2Br_2}$) is somewhat insoluble (among other mercury (I) salts), so it gets precipitated as $\ce{Hg_2Br_2(s)}$ instead of $\ce{Hg_2Br_2(aq)}$, making the overall equation this:

$$\ce{2NH4Br(aq) + Hg2(NO3)2(aq) -> 2NH4NO3(aq) + Hg2Br2(s)}$$

Mercury (I) ions are insoluble because every orbital is full, making it unable to form ligands with water.


The problem is that we were taught incorrectly in freshman physics.  They told us that $Q=nC\Delta T$, where C is called the heat capacity.  However, they neglected to tell us that this only applies if no work is done (i.e., the volume is essentially constant, as for a solid or liquid).  Otherwise, it gives the wrong answer for the heat Q.

"Heat capacity" is a bit of a misnomer for C.  From what we then learn in thermodynamics, a better term would be the "internal energy capacity," because that is how Cv is more properly defined.  For an ideal gas,  $$nC_v=\frac{\Delta U}{\Delta T}$$ or $$\Delta U=nC_v\Delta T$$This has nothing to do with the heat added unless the volume is constant and no work is done.  In that case, $$Q=\Delta U=nC_v\Delta T$$So the freshman form of the equation is recovered only if no work is done.


However: $\Delta S_{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

No. $\Delta S_{env} = \frac{q}{T}$, assuming the process is reversible (as usual). So you get $\Delta S_{universe} = -\frac{\Delta G}{T}$ for a process where the heat exchange with the environment is reversible (and there is no non-PV work).

For more background, see https://chemistry.stackexchange.com/a/114323


It depends on what degree of accuracy you need and what order of magnitude the final concentrations will be.

If you need high accuracy and concentrations are in the range you are describing, I would put a class A volumetric flask on a balance, add the appropriate weight of solute, add water to not quite the full volume, dissolve the solute, and add water to the volumetric line, seperately for each solution.

If you need concentrations much lower than what you are saying, when weighing becomes your largest source of error, then it is more accurate to make a concentrated solution, and the dilute the concentrated solution.  For example, you would use a class A TD (to deliver) volumetric pipette to measure the concentrated solution into a class A volumetric flask.


Molecules which are made up of exactly the same elements but differ in atomic arrangement are called isomers.

Since the two isomers have the same mass and charge, any seperation techniques which rely on a difference in mass and/or density would be very difficult.

Mass spectrometry is an analytical technique in which components of different mass are separated and detected, typically by ionizing (charging), accelerating through an electric field to a given velocity and finally, 'bending' through a magnetic field. The final position of the is detected as a function of its charge-to-mass ratio.

Any two isomers of $\small\ce{(C3H3)^+}$ have the same charge-to-mass ratio and therefore would be very difficult to separate and detect with mass spectrometry.


If you can smell something does that mean that this thing loses molecules?

Yes, the thing you smell surely does lose some molecules, and consequently some weight. It can be said that it slowly evaporates or sublimates. But the strength of smell varies greatly in different compounds, as does the rate of evaporation (sublimation). Some compounds may be gone completely in a few minutes, and some would stay for years, or even decades.


Examples of electronically conductive foods?

I wonder why nobody mentioned the famous vodka with gold flakes. Admittedly, this hardly qualifies as food, and the amount of gold is minuscule, but in the absence of any solid examples with true metallic character I think it may deserve a curious look.


What happens to a greenhouse gas after it absorbs infrared radiation?

Actually a greenhouse gas absorbs light in the infrared wavelengths of light. One of two things will happen after the gas absorbs the IR radiation: 1) the molecules to move around and vibrate faster, which is really the definition of temperature, so the air gets warmer, and 2) the gas molecules can re-emit the IR radiation. In the second case, the radiation is emitted isotropically, so that some of it just goes out into space as it would have had the gas not been there in the first place, but some of it will be directed back towards the surface of the earth, warming the surface by absorbtion of the IR (heat).

Note that most of the heat being absorbed by greenhouse gases is coming from the ground after the ground absorbs sunlight and warms up, not from the sun shining through the air and being absorbed by the greenhouse gases directly.