Understanding a chemical reaction from kegg

The $n$ of the polyphosphate should be decreased by 1 on the reactant side relative to the product side:

$\ce{(P)_{n} + Glc -> G 6P + (P)_{n-1} }$

Polyphosphate, like ATP, contains several phosphoanhydride bonds.  These are "high energy" bonds that can be used for biosynthesis.  But to do so one of these bonds needs to be broken, and that is exactly what happens when one of the end phosphates on the phosphoanhydride chain is peeled off and trasferred to glucose.


Probably you are having problems with Le Chatelier's Principle.

Suppose you have an equilibrium established between four substances $\ce{A}$, $\ce{B}$, $\ce{C}$ and $\ce{D}$.

  What would happen if you changed the conditions by increasing the
  concentration of $\ce{A}$?

According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of $\ce{A}$ decreases again - by reacting it with $\ce{B}$ and turning it into $\ce{C + D}$. The position of equilibrium moves to the right. Of course, this assumes that there is still some amount of $\ce{B}$ left in the reaction vessel.

For better understanding, refer this.


Formation of hydrogen bromide: Does the concept of "order" apply to this reaction?

We don't assign orders. An order either simply is there, or it isn't. In particular, this expression doesn't look like $\ce{\underbrace{(something)}_{\substack{\text{may include }\ce{[H2]},\\ \text{but not }\ce{[Br2]}}}\cdot[Br2]^a}$ for any $a$, hence there is no order.

Alternatively, if we look at the reaction at its initial moments when $\ce{[HBr]}\approx0$, the denominator simplifies to just $1$, and we may say that the reaction has order $1\over2$ with respect to bromine.

Having or not having an order doesn't have any particularly far-reaching physical consequences. Moreover, it is not even an intrinsic property of a reaction; it may depend on the approximation used, as you saw above.

The scary rate equation comes from the so-called steady state approximation and the detailed reaction mechanism with intermediates and stuff, which is about two times longer than my answer, so I'd better refrain from including it. Well, here is the first line, just to give you the taste of the whole thing:
$$\ce{Br2}\underset{k_{-1}}{\stackrel{k_1}{\longleftrightarrows}}\ce{2Br^.}$$

To tell the truth, most real reactions are like that. The well-known $r=k\ce{[A]^a[B]^b}$ expression with stoichiometric coefficients for exponents is but a great oversimplification. It applies to reactions with no intermediates, which are quite rare.


How do chemists measure purity in high purity metals and other reagents?

As the comments alluded, to analyze purity at high purity, you analyze the impurities. I'll use XRF for this example, but ICP, MS, HPLC, or other methods could work in the same way.

With XRF's $1-2\%$ precision directly measuring the gold would give you a maximum supportable purity of $98-99 \%$. To verify a metal as $99.99\%$ the impurities are measured which are usually in the parts per million range, added up and subtracted from $\pu{100\%}$.

For example, let's say you have a gold bar with $\pu{35ppm}$ silver, $\pu{20ppm}$ copper, $\pu{10ppm}$ lead, and $\pu{5ppm}$ mercury with all other elements below a , $\pu{0.5ppm}$ detection limit (also called a threshold limit). The total impurity would be the total of the measured impurities and the detection limit of the unmeasured impurities. Note: most analysis is by metal basis which excludes the other elements that may be present.

$$\mathrm{Impurity = 35 + 20. + 10. + 5 + [0.5 \times (59-5)] = \pu{97 ppm} }$$

The $59$ is for the $59$ naturally occurring metal elements and the $5$ is for the five elements we measured ($\ce{Au, Ag, Cu, Pb, Hg}$). $\pu 2 \pu\%$ of $\pu{97ppm}$ is $\pu{1.94ppm}$ and the measurement plus the maximum error is $\pu{99\!.ppm}$. Taking $\pu{99\!.ppm} = \pu{0.0099\%}$ and subtracting from $\pu{100.00\%}$, we get an inferred purity of $\pu{99.991\%}$ Gold ($\mathrm{100\% - 0.01\% = 99.991\%}$). The gold would still be sold at $\pu{99.99\%}$ regardless of it being purer than needed as selling the gold by lots would be unnecessarily complicated and time intensive.


Simple practical, non-toxic, non-flammable reaction for slow release of CO2

I would recommend not touching or coming near calcium carbide at any rate. The acetylene it creates is explosive in certain concentrations (with a flame to ignite it) and notorious for extremely loud bangs. And it also is rather a diversion for what you want to do.

Instead, consider carbonates. Baking soda is one of the prime household sources of carbonates, and you just need to add an acid to release carbon dioxide; and nicely, the only by-product is water. You should use rather dilute acids so that you can slow down the reaction rate. Vinegar might not be the acid of choice (it has a distinct smell) but either essence of vinegar diluted or citric acid (which you should be able to buy from baking suppliers, too) should work. Or lemon juice.

To make the reaction sufficiently slow, you would need to (as I said) dilute the acid let it be poured into the baking soda solution slowly. Experiment a little! For a heat source, you could likely just put your average heating cushion underneath.

(Personally, though, I would use the steam cleaner because it sounds most kick-arse!)


Why are there no edge dislocations in a face-centered cubic (FCC) lattice?

I have to stretch my memory to remember how this goes (and read a bit from here).

First, how do you define burgers vector? You take the end point of the dislocation and make a circle around it as it were a perfect lattice, the extra (or missing step) from the full loop is the burgers vector.

Now, is the burgers vector also a lattice vector? If yes, then it is (edge)dislocation. If no, then it is only a partial (edge)dislocation (partial dislocations have their own name).

Now, how does this work on a fcc-lattice case. On quick read for link I pasted, it seems that fcc-lattice cannot have edge dislocations, because the burgers vector is not a lattice vector.

You may want to go this through by yourself and check that I didn't make any mistakes.


Specific Heats for Non-Constant Processes

Cp of an ideal gas doesn't depend on pressure.  Cp of any substance is defined as the partial derivative of enthalpy respect to temperature at constant pressure.  But, if the enthalpy of the substance is independent of pressure, then it doesn't matter if the pressure is constant.  However, determining Cp by measuring the amount of heat required to change the temperature by a certain amount of the substance at constant pressure is still a standard method of measuring Cp.


Density of water increases from 273 to 277 K due to increase in entropy?

At $0°$C, the water is not yet made of independent $H_2O$ molecules. It is still a mixture  of hexagones containing 6 molecules $H_2O$. And these hexagones are separated by independent $H_2O$ molecules. There is still a certain order in the water at 0°C. These hexagones are destroyed between $0°$C and $4$°C. These hexagones are more cumbersome than 6 molecules $H_2O$, because there is a hole in the center of the hexagones. That is why the density of water at $0$°C is lower than at $4°$C.


Textbook chemistry vs Real life chemistry and impurities

Well, scientists usually purchase the purest grade of chemical available if they want to use it for a reaction. (Leading to the absurd case that the less pure reagent grade is often dismissed for the purer analytical grade.) Usually, special care is taken for these chemicals to remove especially those impurities which could hurt most — you really don’t want any palladium in your sample of sodium bromide; that could do some really bad things.

That directly leads to a second point: No chemist would ever use ‘table salt’ or tap water unless in dire need. As you state, the impurities are basically incontrollable.

With that in hand, usually chemists have a good idea of what they just added and how that influenced the reactions. But often some things still go wrong. One of the most famous examples in organic chemistry is the Nozaki-Hiyama-Kishi coupling, which was initially reported as requiring only a $\ce{Cr^{II}}$ salt — usually $\ce{CrCl2}$. However, yields of other groups were much much worse than those originally reported by Nozaki and Hiyama. It wasn’t until eleven years after initial publication that — independently of each other — the original authors and Kishi both discovered that $2~\text{mol-}\%$ $\ce{Ni^{II}}$ are required as a co-catalyst for the success of the reaction. The original chromium salt batches of Nozaki and Hiyama were obviously contaminated by enough nickel(II).

An inorganic example I once heard of in a solid state chemistry course was a crystal structure published with experimentals in a solid state reaction. (I can’t remember the exact compound.) Other authors tried to reproduce the reaction but got much worse yields than the $100~\%$ that are expected in solid state reactions. The solution was that the square unit cell contained a hydride anion in its centre, almost invisible due to its low electron density. This hydride was introduced to the reaction from the silicon grease the authors used for the glass joints.


The shape of $\ce{XeF4}$ is indeed square planar as you mentioned it is so because:

The lone pairs are kept in anti position to minimise repulsion between them and also from the electrons in fluorine atoms. Keeping the lone pair anti results in 4 empty positions which are filled with fluorine, giving rise to a square planar shape and octahedral geometry.

Please fell free to comment any doubt, and give the tick if you are satisfied.


What is the difference between the esterification and transesterification?

You are getting too specific as though these are biological processes. They do not have to be... ANY ester made is an esterification reaction.

So any ester can be transesterified into a different alkyl alkanoate and any carboxylic acid or derivative (even nitriles) can be esterified. See the diagram:


As far as I am aware, and from a quick browse though some textbooks, the trans isomer is generally more stable than the cis isomer. This is due to the reduced steric hindrance of the substituents in the trans configuration versus the cis configuration.
For example trans-but-2-ene is more stable the cis-but-2-ene because there is less steric interference between the two methyl groups either side of the double bond. However, there are exceptions to this rule which I couldn't find a quick explanation for. The rule applies to both cyclic and acyclic compounds although there are some obvious cyclic exceptions such as cyclic alkenes, where the cis isomer is more stable, especially for smaller rings, due to the large ring strain caused by the orientation of the trans isomer. This page has a good illustration of this:


How to choose the level of theory for modelling reactions of polymers?

The question is by far not an elementary one. Short answer is: unfortunately, there is no silver bullet.

For a molecule of around 65 atoms I would go with DFT as the methodology which provides probably the best quality/cost ratio for molecules of this size.

Now, you need to choose functional and basis set, and this is a tricky business. But to start from:

Do not blindly use B3LYP.
Try to search literature for the same types of calculations (energetics of ring opening reactions, optical properties) for similar compounds. Maybe someone already did the so-called benchmark calculations, i.e. compared on how good different functionals in combination with different basis sets are on reproducing the properties (say, specific rotation) either measured experimentally or calculated at some high level of theory (say, CCSD(T) with reasonably sized basis set).
Do your own benchmark calculations with different functionals and basis sets. First, benchmark different functionals with the same basis set (say, Dunning correlation-consistent basis sets, cc-pVTZ or better cc-pVQZ), vary the functional, compare results with experimental values, choose the best one. Once method is chosen, try to decrease and also increase (if possible) the basis set to see the effect of basis set size (with the above mentioned bases it is quite simple: if cc-pVTZ was used to benchmark different functionals, try the best functional in combination with cc-pVDZ and cc-pVQZ). Choose the smallest basis set that reproduces experimental results with desired accuracy.

P.S. You could also ask Google, there are some resources of information here and there on how to choose method and basis set, though, as I said, this question is not elementary, so only general advice can be given.


It's not really a molecule, so you don't really have a Lewis structure consisting of one magnesium and two boron atoms.  For a lattice structure see Wikipedia.  The lattice structure is hexagonal with alternating magnesium and boron layers, the boron layers having twice as many atoms per unit area as the magnesium layers.

The referenced picture is given here  and consists of green balls (magnesium) plus pink balls (boron), the latter covalently linked with each other.  If we model the magnesium as $2+$ ions (yes, the electronegativity difference is only 0.8, but de la Mora at al. indicates that the magnesium-boron bonding is predominantly ionic), then each boron atom has one negative charge and so becomes isoelectronic with carbon.  Then each of the hexagonal, covalently bonded boron layers is isoelectronic and isostructural with a graphite layer.


Variational coupled cluster (vCC)
First, you can make a variational CC with a non-unitary T operator in the following way.
Recall the variational principle (taken from Wikipedia):

$$\varepsilon[\Psi] = \frac{\langle\Psi| \hat{H} |\Psi\rangle}{\langle\Psi|\Psi\rangle}$$
The variational principle states that

$\varepsilon \geq E_{0}$, where $E_{0}$ is the lowest energy eigenstate (ground state) of the hamiltonian
$\varepsilon = E_{0}$ if and only if $\Psi$ is exactly equal to the wave function of the ground state of the studied system.

So all you have to do to make a variational CC is set $|\Psi\rangle = e^{\hat{T}}|0\rangle$ in the above expression. This is called "variational coupled cluster" or "vCC", because it literally uses the "variational principle" shown above.
Unitary coupled cluster (uCC)
Unitary transformations preserve eigenvalues, so if $\hat{U}$ is a unitary transformation (unlike $e^{\hat{T}}$), then $UHU^\dagger$ has the same eigenvalues as $H$. Since the energies are eigenvalues of $H$, unitary coupled cluster can recover the correct energies with the unitary operator $U = e^{\hat{T} - \hat{T}^{\dagger}}$. For a long time people thought vCC and uCC were just two ways of writing the same thing, but this was shown to be wrong very recently (G. Harsha et al. On the difference between variational and unitary coupled cluster theories. arXiv:1711.00579 [cond-mat.str-el]; J. Chem. Phys. 2018, 148 (4), 044107.).
Something that a lot of people don't know
Regular CC is actually variational in the full correlation limit. This means that for $\ce{Li2}$ for example, CCSDTQPH is variational. This is because $\ce{Li2}$ has $6$ electrons, so CCSDTQPH is FCI, and FCI is variational. CCSDTQP and CCSDTQ and even CCSDT(Q) are not likely to overshoot the variational limit either since they are very good approximations of CCSDTQPH for $\ce{Li2}$.


finding cation of unknown salt in solution from precipitate

Among the options for the second salt in solution, one is less likely to be found in the presence of $\ce{K_2SO_4}$.

These solutions for analysis are prepared by adding decent amounts of two salts with good solubility to water. Under these conditions, a double displacement reaction may occur and in one case, a precipitate will be formed:

$\ce{K_2SO_4 + CaCl_2 -> 2KCl + CaSO_4}$

I doubt that the person preparing the solutions for analysis used calcium chloride, since

this would have required to remove the precipitate by filtration or decanting off the supernatant solution
the concentration of $\ce{Ca^{2+}}$ in the solution would be rather low. Remember that these qualitative tests are not necessary suitable for trace analysis.

The second salt therefore is most likely sodium chloride ($\ce{NaCl}$), potassium chloride ($\ce{KCl}$) or magnesium chloride ($\ce{MgCl_2}$).

I would do a flame test first. If you see a yellow coloured flame, the second cation is the sodium cation ($\ce{Na+}$). Similarly if you see a lilac coloured flame, the second cation is the potassium cation ($\ce{K+})$ and if you initially see a colourless flame and finally see an intense white coloured flame,the second cation is the magnesium cation ($\ce{Mg^{2+}}$).
If you want to test for the magnesium cation ($\ce{Mg^{2+}}$), place a drop of the solution on a spot plate, make sure it's alkaline by adding aqueous $\ce{NaOH}$ if necessary, and add a bit of a titan yellow solution. This azo dye with its misleading name (there is no titanium in it) will form a red precipitate with the magnesium cation ($\ce{Mg^{2+}}$).


There is no direct connection between having unpaired electrons and having a colour. In fact, many colourful compounds have no unpaired electrons and yet display a bright colour: potassium permanganate, potassium dichromate, azobenzene, sudan red, phenolphthalein, iodine and many, many more. Arguably, there are more coloured compounds known in the singlet state (which means no unpaired spins in the vast majority of cases) than in the dublet, triplet or any other state.

So what causes the colour? It is the possibility of electron transitions from a lower to a higher orbital. If that transition is equivalent to a photon with a visible wavelength, we observe a colour. Examples:

Most phenyl groups absorb light at c. $\pu{250 nm}$ — to short a wavelength to be visible; they are ‘ultravioletly coloured’.
The electronic transitions of most transition metal complexes (except $\mathrm{d^0}$, $\mathrm{d^5}$ and $\mathrm{d^{10}}$ complexes) are well within the visible range of $400$ to $\pu{700 nm}$.
In rare cases, extensive electronic systems may even allow the absorption maximum to be shifted outside of the visible range into the infrared range. Chlorophyll comes rather close to this, absorbing red light (the longest visible wavelengths).

For potassium hexacyanidoferrate(II), the colour we observe is yellow so the absorption must be the complementary blue/violet, indicating a relatively large energy difference.


Why is sulfuric acid used in the Jones oxidation of alcohols?

It protonates the chromic acid to form an oxonium ion, increasing positive charge at the chromium center.

Using $\ce{H^+}$ typically protonates the electrophile, making it more electrophilic (more electron deficient = more electrophilic), just like with acid-catalyzed esterification. In fact, you're making a chromate ester in this reaction.

See the first step here in this image (and the rest is related, also):
Image source


Help with absolute configuration of a substituted cyclohexane?

The most helpful way that I was taught to identify the configuration of stereocenters is a right-hand/left-hand rule. First, assign priority numbers to each of the substituents according to atomic mass, or the atomic mass of their first point of differentiation. Next, identify the direction that the substituent with the lowest priority is pointing in (into the plane/paper, out of the plane/paper, or parallel with the plane/paper). Then imagine that your thumb is the lowest priority substituent. If the fingers of your right hand can curl in the direction of the 1-2-3 substituents, then the configuration is R. If the fingers of your left hand curl in the direction of the 1-2-3 substituents, then the configuration is S.

I found this method to be extremely helpful because no matter the orientation of the stereocenter, this method will work because you can adjust your hands to match the orientation of the stereocenter and identify it as R or S.


what would happen if you got a full package of wet silica gel on your hand after it tore open?

Like user137 has already written in the comment section it should not be harmful unless exposed for a long time. The main charateristic is its hygroscopy, so skin may feel dry.

To cite from a safety data sheet (source):

4 FIRST AID MEASURES
4.1 Description of first aid measures
If inhaled
    If breathed in, move person into fresh air. If not breathing, give artificial respiration.
In case of skin contact
    Wash off with soap and plenty of water.
In case of eye contact
    Flush eyes with water as a precaution.
If swallowed
    Never give anything by mouth to an unconscious person. Rinse mouth with water.

Nevertheless: Always use the safety data sheets your distributor provided and read them before dealing with the chemicals, if the safety hazards are not clear.


Why does moving from left to right on the periodic table result in higher conductivity?

Actually, this only applies for metals. This trend does not apply for non-metals: non-metals form covalent bonds, which are due to the sharing of electrons - thus, there are no free electrons that can carry charge. Additionally, atoms are held tightly by covalent bonds and thus cannot dissociate into ions that carry electric charge.

With regards to metals, metallic bonding consists of metal ions surrounded by a sea of delocalized electrons - and it is these delocalized electrons that can carry charge. Each atom in a metal has the same number of electrons as the charge on the ion. For example, Na+ has one delocalized electron for one atom, Mg2+ has two, Al3+ has three, and so forth. So as the charge increases, the number of delocalized electrons per atom increases - and so the number of delocalized electrons in the metal overall also increases. This allows it to carry more charge per unit time, leading to a higher rate of conductivity.

Similarly with regards to ionic lattices, a greater charge means there will be more ions formed per molecule that is made molten. So one mol of NaCl will have 2 x 6.022 x 10^23 ions total, while 1 mole of MgCl2 will have 3 x 6.022 x 10^23 ions. So, MgCl2 has more ions to conduct electricity and thus shows greater conductivity.

I'm not sure if it has much to do with the actual 'ease' of breaking bonds. Both NaCl and MgCl2 would have to be in a molten state to be able to conduct electricity anyway, so I don't think it plays a role for ionic compounds. For metallic compounds, again, as they are simply ions surrounded by electrons there isn't any bond-breaking to be done.


entropy of vaporization vs entropy of fusion

Here I'm assuming you are interested in comparing the magnitude of $\Delta S^{\circ}_{fus}$ at $T_{fus}$ with that of $\Delta S^{\circ}_{vap}$ at $T_{vap}$, for the elements in their standard states (so, for instance, hydrogen would be $\ce{H_2}$ rather than $\ce{H}$).
Like you, I was unable to find a comparative tabulation of these values on the internet.  Fortunately, one can readily generate such a table using Wolfram Mathematica's chemical database.
For each element:
$$\Delta S^{\circ}_{fus} \text{ at } T_{fus} = \frac{\Delta H^{\circ}_{fus}}{T_{fus}}$$
$$\Delta S^{\circ}_{vap} \text{ at } T_{vap} = \frac{\Delta H^{\circ}_{vap}}{T_{vas}}$$
Wolfram has the above data for all of elements 1–93 (hydrogen through neptunium), except for helium (which can't be solidified at standard pressure, which is 1 bar), astatine, and francium.*
Here I've plotted $|\Delta S^{\circ}_{vap}|$ vs. $|\Delta S^{\circ}_{fus}|$* for these 90 elements, and added a y = x line. From the placement of the points relative to this line, you can see that all except one of the elements have $|\Delta S^{\circ}_{vap}| > |\Delta S^{\circ}_{fus}|$.
That single exception is hydrogen, for which:
$$|\Delta S^{\circ}_{fus}| \text{ at } T_{fus} = 39.8 \frac{J}{mol K}$$
$$|\Delta S^{\circ}_{vap}| \text{ at } T_{vap} = 22.3 \frac{J}{mol K}$$

*Note, however, this complication:  Most, but not all, of these measurements were done at standard pressure (1 bar). For instance:
"When heated at standard atmospheric pressure, arsenic changes directly from a solid to a gas, or sublimates, at a temperature of 887 K. In order to form liquid arsenic, the atmospheric pressure must be increased. At 28 times standard atmospheric pressure, arsenic melts at a temperature of 1090 K. If it were also measured at a pressure of 28 atmospheres, arsenic's boiling point would be higher than its melting point, as you would expect."
https://education.jlab.org/itselemental/ele033.html


What does Pauli’s exclusion principle mean in atomic or fundamental way?

Pauli's exclusion principle is a consequence of electrons being indistinguishable fermions. Fundamental particles are indistinguishable in that two of the same type differ only in a few properties but otherwise behave identically (compare that say to two apples - they are always in principle distinguishable because they differ in so many ways).
Fermions exhibit an associated statistical property when you have a collection of the otherwise indistinguishable particles. The statistical property follows from how the wavefunction of the collection behaves under symmetry (exchange) operations. You have two choices when you swap particles in such a wavefunction: retain the sign or switch signs. That is one fundamental distinguishing property between fermions and bosons (all particles can be classified as one or the other). It happens that nature allows both cases. We chose to call one fermions (to honor Enrico Fermi who co-discovered them), and the particles we call electrons are fermions (you can look into the details of the Standard Model to work out exactly how changes to the properties of the electron would change the universe). Fermions have half-integer values of the spin quantum number. Bosons have integer values. If you have multiple indistinguishable fermions, the QM wavefunction changes sign (is antisymmetric) under exchange of any two of the particles. For the exchange to alter the sign the two exchanged fermions must differ in some property, making the wavefunction antisymmetric, for instance if intrinsic angular momenta (spins) have opposing quantization (quantum numbers $\pm \frac12$).
The consequence of this is explained well in the Wikipedia

If two fermions were in the same state (for example the same orbital with the same spin in the same atom), interchanging them would change nothing and the total wave function would be unchanged. The only way the total wave function can both change sign as required for fermions and also remain unchanged is that this function must be zero everywhere, which means that the state cannot exist. This reasoning does not apply to bosons because the sign does not change.


What is the reasoning for higher pH values in this scene?

Branched amines are more basic than their linear counterparts because inductive effect of alkanes places more electron density in the adjacent carbon atom. The higher the electron density near/at the amine group, the higher its pH will be.


Near IR spectroscopy vs. IR spectroscopy

This is more or less speculation, since they haven't released detailed technical specifications, but given the size of the instrument and the need to avoid moving parts to keep it durable, I would wager that it uses an IR LED for illumination, a diffraction grating for dispersion (although from the picture, it doesn't look like there's room for a conventional reflective diffraction grating setup so they might be doing something else), and probably some kind of silicon CCD or InGaAs diode array for detection. If this is the case, it's likely impossible to convert for use in conventional IR spectroscopy—not as a matter of programming, but simply that the hardware is not equipped to do it. Normal mid-IR spectroscopy (below about $4000\mathrm{\ cm^{-1}}$) involves much longer wavelengths beyond what LEDs can do, which generally means some kind of incandescent lamp, optics that don't involve glass (which blocks mid-IR), and a detector that can deal with such low energy photons (there are a few semiconductor-based sensors that looks promising, but most spectrometers use thermal detectors). There is also the matter of dealing with all the water in the air—many IR spectrometers have gas purge systems in their sample chambers.

There has been a lot of interest in standoff IR spectroscopy for things like explosives detection and advances in mid-IR optics and sensors may make this kind of small instrument possible, but we're not quite there yet. Certainly an instrument designed for NIR is not likely to have any usefulness for mid-IR, without being designed with it in mind.

A related technique that is much better suited to small handheld instruments is raman spectroscopy, since the light involved is actually usually in the NIR. I doubt this thing is sensitive enough to do it well, but it might be possible to add filters to the illumination and measurement ports to give it a shot. It's definitely possible to do raman with a handheld instrument though—I saw one of these at a conference recently. Quite expensive, but very nifty.


The term polyfuranoside is a combination of

the Greek prefix poly, which means many and
the term furanoside, which indicates that the structure is somewhat related to furan, a five-membered (aromatic) heterocycle that contains an oxygen atom. Forget about the aromaticity here, what counts is

five-membered ring
contains an oxygen atom

If you take a look at the large diagonal of the structure, you'll recognize a large number (12) of these structure elements connected by glycosidic bonds, hence the other description, docosanasaccharide.


Interpretation of Coloumb operator in Hartree Fock equation

From the labels ($i,j$), I am guessing that $\chi$ is a molecular orbital (MO). When formulating the equations for molecular orbitals, one assumes that there are as many (spin-)orbitals as there are electrons. In any event, when writing down the Coulomb operator in the MO representation for the ground state, only occupied orbitals appear.
Therefore, I believe that the equations you are showing here are for the general problem, not just $\ce{H2}$. In some sense, the equations are the same for either, it is just a question of how many summands there are. Finally, note that the last equation you showed is valid for any system, but that it is actually part of a set of such equations, one for each MO.


Ring chain tautomerization of hydroxyacetone

At first, even though this is not related to the question, I want to say that you're confused between double-headed arrow between the two resonance forms and double harpoon in reversible reactions. (The harpoons should be written as below.)

Your incomplete mechanism has two errors:

The hydrogen atom bonded to carbon is far less reactive than the one bonded to oxygen.
The positive carbon, which is bonded to oxygen, a very electronegative element, is very, very unstable, which is not likely to happen.

This is the right mechanism for the reaction:

Step 1: Oxygen is a good nucleophile, so it is most likely to take the most acidic hydrogen.
Step 2: Then, the negative charged oxygen will attack the electrophile carbon atom, forming epoxy ring.


What makes heat resistant proteins heat resistant?

Szilágyi and Závodszky published an article in the journal Structure which analyses a number of different structural parameters of proteins of moderately thermophilic ($45~\mathrm{^\circ C} < \vartheta_\mathrm{opt} < 80~\mathrm{^\circ C}$) and hyperthermophilic ($\vartheta_\mathrm{opt} \approx 100~\mathrm{^\circ C}$) organisms compared to homologous mesophilic organisms.[1] I shall summarise their results in this answer.

They analysed the parameters number of cavities, volume of cavities, surface area of cavities (all relating to cavities on the surface of a given protein), number of hydrogen bonds, number of unsatisfied potential hydrogen-bonding candidates, number of ionic pairs (the ionic pairs were analysed for distances of $4~\mathrm{Å}, 6~\mathrm{Å}$ and $8~\mathrm{Å}$) number of α helices, number of β strands, number of unordered loops and the ratio of polar to apolar amino acids (both of exposed and of buried residues). These were all assigned a $P$-value; this defining the probability of the observed effect occuring if the null hypothesis ($H_0$, ‘this feature does not contribute to heat stability’) were true.

The single most important influence on the stability of both moderately thermophilic and hyperthermophilic proteins proved to be additional ions pairs as shown in table 1 below.

$$\textbf{Table 1: }\text{$P$-values of ion pair distances in both}\\\text{moderately thermophilic and hyperthermophilic proteins}\\
\begin{array}{lcc}\hline
d/\mathrm{Å} & P_{45{-}80} & P_{100}\\ \hline
4 & 0.178\phantom{0} & 0.026\\
6 & 0.024\phantom{0} & 0.016\\
8 & 0.0048 & 0.014\\ \hline \end{array}$$

While close-contact ion pairing is not as important for moderately thermophilic proteins ($P_{45{-}80}$), this quickly changes for less-close contact ion pairings; in hyperthermophiles, all types of ion pairing are significantly important. Other influences were:

polarity of buried areas; while the difference was not as significant for only moderately thermophilic ($P_{45{-}80} = 0.156$) or hyperthermophilic proteins ($P_{100} = 0.101$), over the full data set there was a greater proportion of polar amino acids buried (i.e. not superficial) in thermophilic proteins when compared to mesophilic proteins ($P_\mathrm{full} = 0.047$).
polarity of exposed ares; this mainly stabilised moderately thermophilic proteins ($P_{45{-}80} = 0.024$).
number of cavities; this was mainly a feature of hyperthermophilic proteins ($P_{100} = 0.033$).

Furthermore, the authors point out that the choice of amino acid is somewhat influenced; however, the effects are not as strong. Moderately thermophilic proteins tend to favour arginine over lysine and use less lysine than mesophiles; hyperthermophiles incorporate both more lysine and arginine residues. Tryptophan, glutamate and aspartate are significantly increased in hyperthermophiles; moderate thermophiles incorporate less tryptophan and slightly more aspartate and glutamate than mesophiles. Also, hyperthermophiles exhibit a preference for large unpolar amino acids (leucine and isoleucine) and and have a reduced methionine and asparagine content. Since both latter amino acids decompose more readily at elevated temperatures, their reduction was expected. Cysteine content is either similar or reduced, showing that increased disulphide bridging does not contribute to thermic stability.

Previous studies and theoretical models[2] have shown that ion pairing is expected to have a marked effect on stability. At room temperature, the formation of a salt bridge leads to an unfavourable reduction of solvation which overcompensates any enthalpy gained. At higher temperatures however, this desolvation penalty is strongly reduced and the stability gain from salt bridges thus increased.

The authors do point out one major caveat concerning hyperthermophilic proteins they studied:[1]

  It should be noted, however, that the number of known structures of proteins from organisms with $T_\mathrm{opt} \approx 100~\mathrm{^\circ C}$ is very small (there are only five such structures in our data set). This means that our conclusions pertaining to proteins from extremely thermophilic organisms might be of limited validity; in fact, it is possible that much of the observed deviations are because of ‘sample bias’ (i.e., the peculiarities of the available protein structures).

References:

[1]: A. Szilágyi, P. Závodszky, Structure 2000, 8, 493. DOI: 10.1016/S0969-2126(00)00133-7.

[2]: A. H. Elcock, J. Mol. Biol. 1998, 284, 489. DOI: 10.1006/jmbi.1998.2159. Reference [42] in [1].


According to the UK news outlet The Sun:

  The new polymer note uses beef tallow made from suet, which is hard fat found around the animal’s kidneys, stomach and other organs.
  [...]
  The animal product is used as a source of stearic acid.
  This allows the note to be lubricated and to ensure it is smooth inside your wallet.

Suet is mainly used to make tallow, though it is also a widely used cooking ingredient.

Stearic acid is a waxy material who's ionic form (stearate salts like sodium stearate) is widely used in detergents. This property gives it the desired smooth, slippery property desired in the manufacture of the polymer notes.


How can I draw alpha helices in ChemDraw?

I use ChemDraw Professional 19.1.1.32. If you follow  File>Open Templates>Advanced BioDraw, you will find the black helix that I have reoriented from horizontal to vertical. The red helix was enlarged and colored red. I hope this is of help.


I think that the link that Ron mentioned answered the question adequately, although let me add my own take.  (What makes carbon special and versatile?)

I heard that the reason Carbon is able to make such stable multiple bonds is that its orbitals are close enough to its nucleus that they form extremely effective overlap when bonding.  It is therefore the best atom because it is the smallest atom that can form four bonds.  When you have 5 valence electrons already and can only make 3 bonds with 1 lone pair, that usually leads to weird bent shapes (although Nitrogen is also pretty common in organic compounds).

It's kind of like, which hotel is more versatile in booking large families—a hotel with four open rooms right next to a breakfast bar (Carbon), a hotel with three open rooms and a full room right next to a breakfast bar (Nitrogen), or a hotel with any amount of open rooms, except you need to walk a mile for breakfast (any other element)?


How to prepare and store anisaldehyde tlc staining solution? What will be the color of this solution? What is its shelf life?

p-Anisaldehyde (PAA) stain

This posting from www.ochemonline.com (which is available under a CC-By-SA-3.0 licence) provides the following information about the stain and its behavior. The information below is consistent with my personal experience using PAA stains.

Color: Range of colors on a light pink background (my experience was mostly blue spots)
Medium shelf life

The post on stains goes on to say:

  This stain is light and oxidation sensitive and will gradually turn pink/orange. The stain should be kept in an aluminum-covered jar while in use, and the excess should be kept cold and in the dark. Once the stain turns dark red, it should be discarded and made fresh again.

Now you have your answer - the stain should turn pink over time, it should produce (mainly) blue spots on a pink background, and you have a visual indicator of its shelf-life. Once it is dark red, it is done. To keep the stain around longer, wrap the jar in foil.

Brown spots on white background

The iodine stain is the only stain I know that does this for most compounds.

Not all the spots are visible

PAA is principally an electrophile and stains nucleophilic compounds more strongly than other types of compounds.

Personal recommendation

I grew frustrated with the PAA stain because it degraded quickly and switched to the vanillin stain. The recipe is in the link posted, and reproduced below scaled down. Make small batches to avoid waste. You want to use the stain faster than it decomposes.

  Dissolve 3 g of vanillin in 50 mL of ethanol. Slowly add 0.5 mL of concentrated sulfuric acid.

This stain is very sensitive to functional groups and produces a wider range of functional group specific colors than PAA. Vanillin (the structure of vanillin is simiular to PAA, but with an additional OH on the ring) is both an electrophile and a nucleophile and so reacts with a wider range of compounds.

Ochemonline indicates that it has the same "medium shelf-life" as PAA, but in my experience, PAA decomposed more quickly. I left my vanillin stain in an amber glass jar without foil wrapping in my hood at room temperature, and since I was doing lots of TLC every day, I used it faster than it degraded. PAA stain degraded much more quickly for me under these conditions, and I was throwing half of it out despite the number of TLC plates I was doing. If you are not doing TLC every day, then wrap in foil and refrigerate.

  Best reason to use vanillin: The awesome smell!

Poor spot quality

This main not be a problem with your stain.

Big streaky spots are indicative of too high concentration in your solutions.
Alternatively, you may need a more optimal solvent. For example, when doing amines on silica, you may want to add 1% triethylamine to your TLC solvent since silica is acidic. Amines and other basic compounds will streak less in these conditions.
Big streaky spots may also mean poorly resolved ocmpounds
Faint spots may mean poor compatibility with stain or that the concentration was too low.
If spots run together, perhaps your lanes are too close, the bottom of the TLC plate is not level, or your plate touched the side of the developing chamber.


Trying to understand the necessity of orbital approximation or the independent electron model

Setting the stage

Once the Born-Oppenheimer approximation is introduced, we have (first) to solve the so-called electronic Schrödinger equation that could be written as follows,
\begin{equation}
    \newcommand{\el}{_{\mathrm{e}}}
    \newcommand{\elel}{_{\mathrm{ee}}}
    \newcommand{\elnuc}{_{\mathrm{en}}}
    \newcommand{\nucnuc}{_{\mathrm{nn}}}
    \newcommand{\op}[1]{\hat{#1}}
    \op{H}\el \psi\el(\vec{r}\el) = E\el \psi\el(\vec{r}\el) \, . \tag{1}
\end{equation}
Here $\op{H}\el = \op{T}\el + \op{V}\elel + \op{V}\elnuc$ is the electronic Hamiltonian, $E\el = U - V\nucnuc$ is the electronic energy, and $\psi\el(\vec{r}\el)$ is the electronic wave function. The electronic Schrödinger equation, which is a partial differential equation, is fairly complicated. One of the most frequently used methods of simplification of such equations is technique known as the separation of variables.

Separation of variables technique

The separation of variables is aimed to broke a partial differential equation of many variables into a set of simpler partial differential equations of fewer variables. Note though, that the separation of variables is by no means a universal approach: for some equations a solution of the form mentioned above make the separation of variables possible, for others it is of no use.

The most common way of separation of variables for a partial differential equation in a function $f(x_1, x_2, \dotsc, x_n)$ of $n$ variables $x_{i}$ is by writing its solution as a product of $n$ functions $f_{i}(x_{i})$ each of which is a function of one variable only
\begin{equation}
    f(x_1, x_2, \dotsc, x_n) = f_1(x_1) f_2(x_2) \dotsm f_n(x_n) = \prod_{i=1}^{n} f_i(x_i) \, , \tag{2}
\end{equation}
and then substituting solution of this form into the equation.
Using algebraic manipulation one could try to break the resulting equation into a set of $n$ independent ordinary differential equations for each function $f_{i}(x_{i})$ and if succeed, one could solve these independent equations and by plugging all $f_{i}(x_{i})$ back into (2) obtain the solution of the starting equation.

Separation of particles coordinates

It is fairly well known that an attempt to separate the coordinates of individual particles in the time-independent Schrödinger equation
\begin{equation}
    \op{H} \psi(\vec{r}) = E \psi(\vec{r}) \, ,
\end{equation}
by making a substitution $\psi(\vec{r}) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i})$
will be successful only if the potential energy function can be expressed as the sum of the separate potentials for each individual particle
\begin{equation}
    \op{V}(\vec{r}) = \sum_{i=1}^{n} \op{v}(\vec{r}_{i}) \, ,
\end{equation}
which implies that the particles do not interact with each other. For such a system of non-interacting particles the time-independent Schrödinger equation, partial differential equation of a function of $n$ variables, breaks up into a system of $n$ independent partial differential equations of functions of $1$ variable only,
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\op{H} = \sum\limits_{i=1}^{n} \op{h}_{i}$ and $E = \sum\limits_{i=1}^{n} \varepsilon_{i}$. All this is relatively easy to establish, and besides, this derivation is done in many textbooks in great details, so I leave at as an educational exercise to OP.

Separation of electrons coordinates

Electrons, on the contrary, interact with each other, so that a simplification of the electronic Schrödinger equation by means of separation of coordinates of individual electrons won't work. Mathematically, it is the interaction potential $\op{V}\elel$ that will prevent the separation of the electronic coordinates,
\begin{equation}
    \psi\el(\vec{r}\el) \neq \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}

But what if we can approximate $\op{V}\elel$ with some model potential which will allow the separation of electronic coordinates? Obviously, such an approximate potential should have the form of the sum of one-electron potentials each of which depends only on the coordinates of $i$-th electron as we have already seen. Physically it would mean that electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. Hence the name of the approximation, the mean field approximation. The corresponding approximate potential can be formally written as follows
\begin{equation}
    \op{V}_{\mathrm{MF}}
    =
    \sum\limits_{i=1}^{n} \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
The electronic Hamiltonian $\op{H}\el$ is then approximated with the sum of effective one-electron Hamiltonian operators $\op{h}(i)$
\begin{equation}
    \op{H}_{\mathrm{MF}}
    =
    \sum_{i=1}^{n} \op{h}(\vec{r}_{i}) \, ,
\end{equation}
where
\begin{equation}
    \op{h}(\vec{r}_{i})
    =
    - \frac{1}{2} \nabla_{i}^2
    - \sum\limits_{α=1}^{ν} \frac{Z_{α}}{ r_{α i} }
    + \op{v}_{\mathrm{MF}}(\vec{r}_{i}) \, .
\end{equation}
And in contrast to eigenfunctions $\psi\el(\vec{r}\el)$ of the true electronic Hamiltonian $\op{H}\el$, eigenfunctions $\psi_{\mathrm{MF}}(\vec{r}\el)$ of the mean-field Hamiltonian $\op{H}_{\mathrm{MF}}$ defined by the corresponding eigenvalue equation
\begin{equation}
    \op{H}_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el)
    =
    E_{\mathrm{MF}} \psi_{\mathrm{MF}}(\vec{r}\el) \, ,
\end{equation}
can be written as products of one-electron wave functions
\begin{equation}
    \psi_{\mathrm{MF}}(\vec{r}\el) = \prod\limits_{i=1}^{n} \psi_{i}(\vec{r}_{i}) \, ,
\end{equation}
where $\psi_{i}(\vec{r}_{i})$ are solutions of the corresponding one-electron Schrödinger equation
\begin{equation}
    \op{h}_{i} \psi_{i}(\vec{r}_{i}) = \varepsilon_{i} \psi_{i}(\vec{r}_{i}) \, .
\end{equation}

The main problem with the mean field approach described above is that so far we have just assumed that $\op{V}_{\mathrm{MF}}$ exist in one form or another, but we have no idea of how it actually might look like.
There is a well-known workaround, but its a different story.

Note that I ignored spin completely in the discussion above which is, of course, fundamentally wrong, but spin won't make a big difference with respect to the key concept discussed, the mean-field approximation.


Why does solid phase peptide synthesis (SPPS) always proceed from the C to the N terminus?

The reasons are

The use of an acylating agent in excess (e. g. acetic anhydride or N-acetyl imidazole) drives the reaction forward.
UNCAs (urethane-type protected  N-carboxy anhydrides (e. g. Fmoc, Z, Boc, ...)) can be used. They prevent racemization of the amino acid, that it is attached to.

But there are (despite of ribosomes themselves) also instances, where N -> C directionality was applied:

http://onlinelibrary.wiley.com/doi/10.1002/anie.199710991/abstract
http://www.ncbi.nlm.nih.gov/pubmed/10891165


Why was Nobel Prize given to Baeyer for his "faulty" Baeyer's Strain Theory?

Scientists are human beings after all. Nobody gets a Nobel Prize for inventing a "formula". It is the idea that makes an impact and new ideas are respected. Scientists like normal human beings make honest mistakes, but try their best to be correct. The guy who got the Nobel Prize in polarography, Hervosky, in the 1950s made errors in other electrochemistry papers, does this mean the prize be snatched back from him as well? Max Planck when he derived his blackbody radiation law, he did not believe much in atomic/molecular theory (later on he changed his views). He also received a Nobel prize.

If you read the citation/biographer by the Nobel Prize committee, the award was not due to some cycloalkane strain angle but the prize motivation was "in recognition of his services in the advancement of organic chemistry and the chemical industry, through his work on organic dyes and hydroaromatic compounds."

The Nobel Prize clearly says "Von Baeyer’s work was at once pioneering and many-sided. With admirable penetration and extraordinary experimental skill he combined dogged perseverance and, even at 70 years old, a youthful buoyancy in his work. He was careful never to overestimate the value of a theory. While Kekulé sometimes approached Nature with preconceived opinions, von Baeyer would say: “I have never set up an experiment to see whether I was right, but to see how the materials behave”. Even in old age his views did not become fixed, and his mind remained open to new developments in chemical science." (https://www.nobelprize.org/prizes/chemistry/1905/baeyer/biographical/)

You can see here, (good) scientists are open to change and are willing to listen to others. I found very good scientists to be very humble (with a small fraction of outliers), but only the mediocre ones are at times rigid and arrogant.


How can I solve a non-cubic crystal structure?

The lattice constants and the symmetry of the lattice determine the position of the diffraction peaks observed along the $2\theta$-scale.  The intensity of the diffracton peaks, leaving out increase by accidental or systematic superposition of them, is due to the arrangement within the asymmetric unit, i.e. the motif which -- after application of the symmetry operators (centres of inversion, proper axes, rotoinversion axes, glide planes, mirror planes) -- builds the unit cell.

So you face the phase problem (as single crystal diffraction analysis does, too). If you do not want to rely on already existing software (like DASH, MAUD, Jana, or FOX to mention a few), you will look up the Rietveld refinement.  Beside by the comment by @porphyrin, as shown in this lecture briefly providing some background, the numerous properties of sample and experiment (including their statistics) equally lead me to the recommendation to use software already proven to work.  The International Union of Crystallography maintains a software directory, along with -- under their educational umbrella -- the notes of a dedicated workshop about powder diffraction.

Perhaps contacting a local crystallographer may be helpful, too; like via the American Crystallographic Association ACA.


Effect of magnetization on oxidation (rusting) of iron

OK, those who say magnetic fields have no effect can buy a relatively inexpensive Magnetizer (as I once did) and try performing a controlled experiment (as I attempted) with the rusting of iron. There is some apparent evidence that alignment of magnetic fields can promote radical activity.
This UK study on the effect of a magnetizer in reducing the amount of chlorinating pool chemicals required, is an example of a successful application. Reference, see Magnetic treatment of swimming pool water for enhanced chemical oxidation and disinfecting.
However, with respect to possible applications and significant results, apparently electrochemistry applications are at the top. For example, see Zou, P.; Leddy, J., Magnetized Nickel Electrodes for Improved Charge and Discharge Rates in Nickel Metal Hydride and Nickel Cadmium Batteries, in Electrochemical and Solid-State Letters 2006, 9, A43-A45. To quote from the article:

...magnetized electrodes yielded higher currents than the corresponding nonmagnetic electrodes by 70 and 230 %, respectively.

Source material for background includes Magnetic field induced motion behavior of gas bubbles in liquid, by Keliang Wang, Pucheng Pei, Yu Pei, Ze Ma, Huachi Xu & Dongfang Chen, in Scientific Reports 6, Article number:21068 (2016), doi:10.1038/srep21068 ), interesting article comments include:

the kinetics of oxygen bubbles is mainly proportional to intensity of the electromagnetic field. The magnetic-field induced rotational motion of oxygen bubbles in a square electrolyzer can increase liquid hydrodynamics, thus solve the problems of oxygen bubbles coalescence, and uneven distribution of electrolyte composition and temperature. These types of oxygen bubbles movement will not only improve energy saving and metal deposition for energy storage and metal refinery, but also propel object motion in application to medical and martial fields.

Note, with my experiment reported above with iron rusting, I observed significant more frothing/air bubbling activity around iron in the presence of the Magnetizer. One could look at my experiment as a crude rendition of an iron metal/air battery cell with the Magnetizer forming a magnetic iron electrode.
Finally, a link to a prior StackExchange comment on the topic here.


Given volume of reagents find the chemical formula of nitrogen oxide

As a hint, you could write balanced equations for oxidation to N2O, NO, NO2 and N2O5.  (Did I forget any nitrogen oxide?)  Then you can compare moles (or volumes, since we are talking gases) of starting materials and products.


If you were to perform Full CI (within a finite basis) then I don't think that it matters whether you start with SCF orbitals or some other choice.  This solution will have the lowest conceivable energy no matter what.

But if you are doing truncated (less than full) CI, then you are very much at the mercy of your starting orbitals. In general, one would want to start with orbitals that resemble those of the actual state that you are trying to describe via CI.

Otherwise, your precious excitations (singles and doubles) may be consumed correcting the "slop" that you provided, rather than moving toward the full CI limit.


How do breaking chemical bonds turn into kinetic energy

Breaking chemical bonds does not turn their potential energy into kinetic energy. All bond dissociation enthalpies are positive (for stable bonds that actually exist), which means that breaking them is an endothermic or heat absorbing process, and therefore they absorb kinetic energy (heat is a transfer of molecular kinetic energy).

What you might be thinking of is exothermic reactions, where the enthalpy of the reactants is lower than the products, and so the reaction releases heat, or increases the kinetic energy of the surroundings.

The short answer to your question is "because energy is conserved, it has to go somewhere, and heat and light are the only places it can go."

I'll try to think about a better way to explain it that isn't too far away from reality, but at the same time isn't too complicated. If I can, I'll edit this answer later to include it.


Usually the dot is put there to emphasize that the nitric oxide is a free radical that includes an unpaired electron.  This is especially notable by comparison with $\ce{NO^+}$, which does not have the unpaired electron.

Note that the nomenclature $\ce{·NO}$ should not be rendered as showing the unpaired electron on nitrogen.  The unpaired electron is actually incorporated into the molecular orbital structure of the nitric oxide molecule, covering both atoms.


How are point group character tables typeset correctly?

Notations and conventions used for the description of symmetry in rigid molecules are established in Notations and conventions in molecular spectroscopy: Part 2. Symmetry notation (IUPAC Recommendations 1997):

  Symbols for symmetry operators are printed italic, with sub- and super-scripts that are upright except for the variables $k$ and $n$ which are replaced by numbers for a specific operator.

For example:
$E$ – identity operator
$\sigma$ – reflection operator for reflection across a general plane
$\sigma_\mathrm{h}$ – reflection operator for reflection across a horizontal plane
$\sigma_\mathrm{v}$ – reflection operator for reflection across a vertical plane
$C_n{}^k$ – $n$-fold rotation operator for $k$ successive rotations through an angle of $2\pi/n$ about an $n$-fold rotation axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$
$S_n{}^k$ – $n$-fold rotation-reflection operator for $k$ successive rotation-reflections about an $n$-fold rotation-reflection axis for a rotation through an angle of $2\pi/n$ followed
by a reflection in a plane perpendicular to the axis, where $n = 2, 3, \ldots$; $k = 1, 2, \ldots, (n-1)$
$i$ – inversion operator through the centre of symmetry

Likewise,

  Symbols for symmetry groups are printed italic with upright subscripts, following common usage.

For example:
$C_n$, $S_{2n}$, $D_n$, $D_{n\mathrm{h}}$, $D_{n\mathrm{d}}$, $C_{n\mathrm{v}}$, $C_{n\mathrm{h}}$, $T$, $T_\mathrm{h}$, $T_\mathrm{d}$, $O$, $O_\mathrm{h}$, $I_\mathrm{h}$, $C_{\infty\mathrm{v}}$, $D_{\infty\mathrm{h}}$

However,

  Symbols for irreducible representations of point groups, which are usually called symmetry species in spectroscopy, are printed upright.

For example:
$\mathrm{A}$, $\mathrm{B}$, $\mathrm{E}$, $\mathrm{A}_1$, $\mathrm{A}_2$

The example table given in the question should thus be typeset as follows:


Finding formula of reduced scandium oxide using its mass and mass of obtained metal

Since scandium oxide is an ionic compound, the method is exactly the same as obtaining the empirical formula, from the ratio of the amounts of elements $n(\ce{Sc})$ and $n(\ce{O}).$
$$n(\ce{Sc})=\frac{m(\ce{Sc})}{M(\ce{Sc})}=\frac{\pu{0.929 g}}{\pu{44.96 g mol-1}}=\pu{2.07E-2 mol}$$
The mass difference comes from the loss of oxygen atoms:
$$n(\ce{O})=\frac{m(\ce{Sc_xO_y})-m(\ce{Sc})}{M(\ce{O})}=\frac{\pu{1.423 g}-\pu{0.929 g}}{\pu{16.00 g mol-1}}=\pu{3.09E-2 mol}$$
The ratio between the amounts of elements is close to $2:3$, therefore the formula is $\ce{Sc2O3}$. You can check that using oxidation states.


Confusion about definition of Bond Enthalpy

Bond energies are usually given for each bond-type. You can see this table of values, for example (that's a great site). In order to clasify them we could say that particular type means clasify the bond according to the atoms and to bond-order. So you have C=O, C-O, N=C, N-C, and so on.

If you check the site, look that those values are positive, meaning that the energy -as you said- is the necessary to break the bond. If you need to estimate the difference in enthalpy of formation of:

$$\ce{A<=>B}$$

you will need to change sign of bond-enthalpy values (because usually entalphy is of formation).

The energy is calculated per mol, as you say. And it is a very big value, bigger than enthalpy of vibration, rotation or translation.


How could a scientist from the 1800s to the 1920s run spectroscopy of chemicals?

Light from a flame, the sun, or a star, was passed though a narrow slit, then directed through a prism (F) and observed.  The prism is rotatable about a vertical axis to adjust the wavelength range being observed.  Bright lines in the spectrum are emission transitions and dark lines are absorption transitions.

Further progress was made using diffraction gratings instead of prisms and using photography instead of naked eye observation.  Photography permitted observation of infrared and ultraviolet lines.

By these techniques, known elements were detected in the sun and other stars, and helium was discovered in the sun before it was discovered on Earth, all before 1900.

For further information see Spectroscopy History.


What kind of filter can remove zeolite dust from ethanol?

Since this never got a proper answer I thought I'd come back with a follow-up.
Essentially the root cause was low quality sieves (I had just picked the cheapest I could find on the internet). There are probably other things that work but all of the following captured the dust when used alone:

Whatman #6 filters do a decent job. I have #597 too but I don't recall if they worked here. (Whatman filter grade chart).
A bed of sodium bentonite over a #597, which  I was using to filter other stuff, also incidentally captured the sieve dust.
I never actually tried celite; but it was recommended in the comments so that could be a solution, too.

And the following also helped with the issue:

Simply letting the solutions sit for a day then pouring off the top left a significant portion of the dust settled at the bottom. I'd do this anyways before running it through a filter, to lighten the filter load.
More importantly, they were just cheap sieves. The second time I used them after baking them there was significantly less dust, and by the third time the issue was basically solved.

The next time I get cheap sieves I think I might wash them in the solvent first just to get the initial dust load out, recharge them, and then use them for their intended purpose. Although I suspect that if I balance the time, cost, and energy of doing this with the extra cost of better sieves, it's probably a good idea just to pay for the better sieves.
Pretty obvious in retrospect but maybe a fellow non-chemist will find it useful some day.


How do I calculate the hydronium ion concentration if I know the strontium hydroxide concentration?

Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use

$$K_\mathrm{w} = \ce{[OH-][H3O+]}$$

$$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$


Why does nitration of N,N-dimethylaniline occur at the meta position?

In the presence of these strong acids the $\ce{-NMe2}$ group is protonated, and the protonated form is electron-withdrawing via the inductive effect. This discourages attack at the electron-poor ortho position.

Under the conditions I know for that experiment, you get a mixture of para- and meta-product, but no ortho-product due to steric hindrance.


The white stuff is aluminum hydroxide. The water is fine, filter the solids with a coffee filter if it doesn't settle to the bottom. And remove the ladder.


What chemical properties make some grasses usable for producing textile fabrics?

A large part is the agriculture behind the fabric and what the industry and customer demands. Cotton and bamboo are easy to farm because we have a lot of experience and techniques for raising and harvesting and processing the plants and fibers ( especially cotton, which is available as GMO round-up ready to simplify weed control, and centuries of breeding for better quality and quantity of fiber ). Additionally, the textile industry has machines that are calibrated for cotton, other plant fibers may not work as well and re-designing the machines would be expensive. Customer demand is important, and customers are used to cotton clothing.

As far as chemical reasons go I don't have a lot of specifics. However, different plants have different compositions of fibers. Though all plant fibers are mostly cellulose/hemicellulose, lignin is also important, and would make for woodier, stiffer, tougher fibers ( though bamboo gets pretty woody, I don't know much about bamboo textiles ). Other plant fibers may contain pigments that would have to be bleached out before it could dyed with other colors. Length of fibers is also important. Shorter fibers would be harder to work into threads and textiles.

And keep in mind that the textile industry is not limited to cotton and bamboo. For centuries, cotton was a luxury material because picking the seeds out of the cotton boll was very hard work, cotton was far more expensive than silk. The widespread adaption of the cotton gin ( and slavery ) made cotton cheap and easily available. Prior to this, fabric would have been made from flax to make linin, or from hemp. Flax and hemp are still commonly grown for fabric. ( and we haven't even mentioned animal sources of fiber such as silk or wool )


How can I tell if a molecule is polar or not?

Every sufficiently asymmetric molecule will be polar, but some more than others

The polarity of molecules is related to the polarity of bonds within the molecule, but just having polar bonds is not enough to create a polar molecule.
 Consider, for example, CCl4 and CHCl3. Carbon tetrachloride has 4 fairly polar bonds but they form a regular tetrahedron and the polarity of the individual bonds cancel each other out to leave a molecule which has no net dipole moment and is non-polar. But the chloroform molecule (which swaps a hydrogen for a chlorine) is fairly polar (~1.1 Debye–the unit used to measure dipole moments–usually abbreviated to D) because the symmetry is lower and the three C-Cl bonds now all point away from the C-H bond and their dipoles don't cancel out.

Similarly carbon dioxide is a linear molecule so the dipoles of the (fairly polar) C=O bonds point in exactly the opposite direction and so cancel out. Water, however is bent so the dipoles of the polar O-H bonds  do not cancel out leaving a polar molecule with a significant dipole moment (~1.8 D).

Saturated hydrocarbons are usually thought to be non-polar for most meaningful purposes. But those that lack symmetry still have measurable (but tiny) dipole moments. Propane $\ce {CH3CH2CH3} $ and isobutane $\ce {CH3CH(CH3)CH3} $, for example, both have small but measurable dipole moments (0.081 D and 0.132 D) so are, strictly, polar. In practice, though, these molecules have such low polarity that the effects of their net dipole are dominated by other non-polarity related intermolecular forces and can be though of as non-polar.

But the critical message is that the most fundamental feature that determines polarity is symmetry or the lack of it.

Also note that we are talking about discrete molecules here. One of your initial examples, $\ce {SiO2}$, is a network solid not a discrete molecule and the ideas won't apply.


Use RT/F = 0.059 or 0.06
It will simplify your calculations.

Note:- I have converted ln into log.
So formula becomes:-
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{0.059}{n}\log\frac{[\text{Products}]}{[\text{Reactants}]}$$

Just try this in problems. Calculations would become easier.


This is possible, just use the density equation for a crystal lattice
$$ d = \frac{z\times M}{V_c \times N_\mathrm A}$$
Where $d$ is density; $z$ is the number of atoms per unit cell; $M$ ist he molar mass of the material (isotope in this case); $V_c$ is the volume of a unit cell; and $N_\mathrm A$f is avogadro's number.  Since silver is FCC, $z= 4$ ($\frac{1}{8}\times 8 + \frac{1}{2}\times 6 = 4$) and $V_c$ is equal to the lattice parameter squared ($a^3; a = \pu{0.409nm}$).

This method will give you a very close approximate value but is limited in accuracy to the experimentally observed density as it does not account for free space in the crystal due to vacancies in the lattice or voids at grain boundaries. For example, if you calculated the densities of iridium and osmium you find that iridium should be denser, but experimentally you will not observe this result because iridium has a sufficiently high concentration of vacancies that osmium is actually denser.


What is the structure of $\ce{FeSO4 \cdot NO}$ that is formed when $\ce{NO}$ is passed through ferrous sulphate solution?

The structure is octahedral.

The Fe ion is at the center of the octahedron.

Five water molecules and the NO molecule occupy the vertices of the octahedron.

Sulfate is a separate spectator ion.

The overall charge of the iron coordination complex is 2+.

  If it is brown ring complex, why does the complex sphere break on heating to give $\ce{NO}$? In general, when do the complex spheres break ?

"Brown ring" refers to the macroscopic appearance of the product in a tube.  This is unrelated to the chemical structure.  The binding of NO to Fe is an exothermic reaction.  In accordance with the Van't Hoff Equation, the equillibrium constant shifts in favor of the reverse reaction as temperature increases.

For detailed infomation see Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of Nitric Oxide to Aquated Iron(II)

Interestingly, for many years it was thought the iron was reduced to Fe(I), the NO being oxidized to NO+; however, according to the above reference, the current thinking is that the iron is oxidized to Fe(III), and the NO is reduced to NO-.

In other words, the complex is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$


Tin nitrate can be prepared by mixing tin and nitric acid at various proportions.

$$\ce{5Sn + 12HNO3 -> 5Sn(NO3)2 + N2 + 6H2O}$$

  Tin react with nitric acid to produce nitrate tin(II), nitrogen and
  water. Nitric acid - diluted solution. This reaction takes place
  slowly. Impurities: nitrogen(II) oxide (chemiday 1)

$$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$

  Tin react with nitric acid to produce nitrate tin(II), ammonium
  nitrate and water. Nitric acid - very dilute solution. This reaction
  takes place slowly.(chemiday 2)

$$\ce{3Sn + 8HNO3 -> 3Sn(NO3)2 + 2NO + 4H2O}$$

  Tin react with nitric acid to produce nitrate tin (II), oxide nitrogen
  (II) and water. Nitric acid - diluted solution.(chemiday 3)

A brief information on tin nitrate can be found here*

  Stannous Nitrate is obtained in white leaflets having the
  composition $\ce{Sn(NO3)2.20H2O}$ by cooling to -20° C. a solution of
  stannous oxide in nitric acid of density 1.2. It is also formed by the
  action of highly diluted nitric acid on the metal, some of the acid
  thus suffering maximum reduction to ammonia:

  $$\ce{4Sn + 10HNO3 -> 4Sn(NO3)2 + NH4NO3 + 3H2O}$$

  According to C. H. H. Walker, tin dissolves in nitric acid forming
  stannous and stannic nitrates, the relative proportion of these two
  salts produced depending on the temperature and the strength of the
  acid; moreover, the yellowish white precipitate which separates when
  the somewhat concentrated acid is employed is said to be a hydrated,
  ill-defined, stannic nitrate.

*http://tin.atomistry.com/stannous_nitrate.html


In the case of $\ce{HNO3}$ and $\ce{H2SO4}$, the proton from sulphuric acid protonates the $\ce{-OH}$ group of $\ce{HNO3}$ , which gives $\ce{NO2+}$ , $\ce{HSO4-}$ and water.
This mixture of nitric acid and sulphuric acid is widely used as a nitrating mixture for nitration of aromatic compounds.

In case of other acid mixtures, no reaction happens, but there is an equilibrium. The stronger acids dissociates almost completely , decreasing the dissociation of the weaker one by common ion effect.


Your final statement:

  in truth, whatever the $K_a$ it's not actually COMPLETELY dissociated

is exactly right.

Strong acid is very often stated to be "completely dissociated", but that's just wrong.

For example, nitric acid has a $K_a$ of 24, so 1M nitric acid about 96% dissociated. If the concentration is higher, it will be less dissociated and if the concentration is lower it will be more dissociated.

I would think of the solvent leveling effect this way:

Suppose you have 1 liter of 0.1M ethanol in water and you add 0.1 moles of sodium amide thinking that it will deprotonate the ethanol.  That's not going to happen because most of the protons that protonate the amide will come from the solvent (water) rather than the ethanol.


Why does Chem3D minimize nitrogen dioxide (radical) so badly?

By the looks of it, Chem3D uses generalised classical force fields to describe the molecular interactions.

Such force fields are optimised to provide a reasonable fit to multiple properties for lots of different molecules. For example, they reproduce reasonable bond lengths not only for $\ce{NO2}$ but also for the various bonds in organic molecules.

They are also constructed to reasonably reproduce solvation properties which depend on the molecule-specific partial charges, which in turn alter bond lengths and angles.

So it is unlikely that you have done anything wrong.

If you require a more accurate model, either apply more accurate force fields (e.g. those fitted specifically to $\ce{}$) or switch to quantum chemistry methods such as DFT.


What does it mean to say that a strong base is only slightly soluble?

Solubility is different from dissolution, which is itself different from dissociation.

Solubility tells you the extent to which a substance dissolves. Dissolution is the kinetic process of solvation, when solvent molecules interact with solute molecules or ions. Dissociation is, from IUPAC,

  The separation of a molecular entity into two or more molecular entities (or any similar separation within a polyatomic molecular entity). Examples include unimolecular heterolysis and homolysis, and the separation of the constituents of an ion pair into free ions.
  The separation of the constituents of any aggregate of molecular entities.

Dissociation is the mode of solvation for ionic compounds.

I think your confusion stems from associating solubility and dissolution improperly. ChemGuide directly answers your question here:

  Some strong bases like calcium hydroxide aren't very soluble in water. That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions. Calcium hydroxide still counts as a strong base because of that 100% ionisation.


How do you find the enthalpy change for a given reaction?

Step 1 Calculate $q_\text{solution}$. To do so, you have to have in mind that the mass of the solution is aproximatly equals to $100\ \mathrm{ml}$ of $\ce{HCl}$ plus the mass of $\ce{Mg}$.

\begin{align*}
   q_\text{solution}&=100.5\ \mathrm g\cdot4.20\ \frac{\mathrm J}{\mathrm{mol\cdot K}}\cdot\left(318.0\ \mathrm K-295.4\ \mathrm K\right)\\
   &=9.54\cdot 10^3\ \mathrm J
  \end{align*}
Step 2 Calculate $q_\mathrm{rxn}$

\begin{align*}
 q_\mathrm{rxn} + q_\text{solution} &=0\\
 q_\mathrm{rxn} + 9.54\cdot 10^3\ \mathrm J &= 0
\end{align*}
Step 3 Calculate the value of $\Delta H$ per mole.

\begin{align*}
 \Delta H&= \left(\frac{-9.54\cdot 10^3\ \mathrm J}{0.500\ \mathrm g\ \text{of}\ \ce{Mg}}\right)\left(\frac{24.31\ \mathrm g\ \text{of}\ \ce{Mg}}{1\ \mathrm{mol}\ \text{of}\ \ce{Mg}}\right)\\
 &= -4.64 \cdot 10^5\ \frac{\mathrm J}{\mathrm{mol}}\ \text{of}\ \ce{Mg}
\end{align*}


Effect of temperature and Le Chatelier's principle

Your teacher is right, let's see why. First off, here's our reaction $$\ce{2 NO + H2 -> N2O + H2O}\qquad \Delta H = -351\ \mathrm{kJ/mol}$$
So the reaction to the right is exothermic, it gives off heat.  That tells us the products are more stable than the reactants and at equilibrium $\mathrm{\frac{[Products]}{[Reactants]}~>1}$.  Here we could guess that if we heat the reaction (put heat back into the reaction), we will tend to level out the equilibrium thereby pushing the reaction back to the left.

To examine this guess let's work an example.  You were given an enthalpy of reaction, but for a "back of the envelope" calculation, let's run a comparison and assume that the change in entropy will be roughly the same for the two conditions (temperatures) we're going to compare.  With this assumption, we can let $$\Delta G= -RT\cdot\ln K_\text{eq}= -351\ \mathrm{kJ/mol}$$  We know $\Delta G$, $R=8.31\times10^{-3}$ and let's say the data is given for room temperature, so $T=298\ \mathrm K$.  Solving for $K_\text{eq}$ we find $K_\text{eq}=3.6\times10^{61}$.  Now let's let our reaction equilibrate $100^\circ$ above room temperature (now $T=398\ \mathrm K$).  $\Delta G$ and $R$ are fixed, so lets resolve the equation and we now find $K_\text{eq}=1.2\times10^{46}$, indeed heat has pushed our reaction to the left.  The teacher was right.


In a nucleophilic substitution reaction, does the other atom to which the nucleophile is bonded matter?

When undergoing an SN rxn, does the other atom to which the
  nucleophile is ionically bonded matter?

It can matter.  If the counterion shifts the covalent-ionic nature of the bond connecting it to the nucleophile, then the nucleophilicity of the nucleophile can change (increase or decrease).  However in your example, the counterions don't make a significant difference.

  Which is more valid, and why?

On paper, they are both valid but

$\ce{H2S}$ is very toxic and corrosive, so the first method using sodium hydrosulfide would likely be preferred.  However, in industry, where raw material cost is very important, both of these methods are successfully used.

However both methods suffer the same drawback - once some of the desired thiol is prepared, it can react further with another molecule of alkyl bromide to produce a thioether; usually a mixture of thiol and thioether is obtained.

In the lab, neither method is used

In the chem lab, the preferred method to prepare simple thiols is through the use of thiourea.  Note the resonance structures for thiourea (just disregard the lower half of the diagram) that place negative charge on the sulfur, making it a good nucleophile.

(image source)

In the first step the thiourea reacts with the alkyl halide via an $\mathrm{S_{N}2}$ reaction to yield an isothiouronium salt, which is subsequently hydrolyzed by sodium hydroxide to generate the thiol. This synthesis can be carried as a one pot, two step reaction.

(image source)


Strange green and gold coloured chemical

I worked for a time with squaraine dyes that were often greenish in the crystalline state, but ink-blue when in solution.

However, I would suggest that you take an analytical approach to identifying this compound.  Have you done any NMR, IR or even UV/Vis spectroscopy?  Elemental analysis?

These would be a much safer way to go about identifying the mystery compound.

Good luck, and I'd love to know what you discover!


To determine the C-rate, you should charge and discharge two cycles very slowly say at 0.1 mA, between say 1.5 V to 4.3 V for your Li-ion (unless it's a very old laptop battery it should be LiCoO2), and measure how long it'll take. The C-rate would be your capacity divided by one hour.

Don't know of a way to measure it using heat or internal resistance.

Batteries generally have an adverse effect when heating up, so unless it is much below freezing, they should be ok.


From our friends at Periodic Videos, the following come to mind:

Gun cotton, more precisely known as nitrocellulose, which is a very simple nitration that you could show prior to the pyrotechnics.

Sodium and Potassium are also good, though you may not always get an explosion, depending upon the available hydrogen above the metal.  Which suggests...

Hydrogen! This is the classic explosion in oxygen. In fact, on a small scale, the electrolysis of water into hydrogen and oxygen can be used to show that the mixture of these gases is very explosive. But keep it small!

All of these experiments avoid outright explosives, which seems prudent in a high school setting.


I am sorry to see that they are still teaching these concepts of the 19th century as facts. The concept of primary valence and secondary valence is a translation of Werner's terminology (Nobel) as Hauptvalenz and Nebenvalenz. Haupt means something "main" and neben means something "extra, supplementary". What is valence in old school terminology? It is the combining power of an element.

Suppose you have $\ce{Co^{3+}}$, how many counter anions do you expect? $\ce{Cl^-}$? $\ce{CoCl3}$. This is the primary valence of 3. Standard electrostatics, that is, the number of positive charges must equal negative charges. Charged particles cannot stay alone under normal daily conditions.

However, Werner found that compounds of definite composition can also be found such as $\ce{[Co(NH3)6]Cl3}$. But cobalt could only take three chlorides with it? How come there is additional ammonia, that is a part of the molecular "Komplex".   He invoked the idea of Nebenvalenz, i.e. supplementary valence to explain the additional ammonia molecules built into this molecule. Indeed this was a complex idea for him and he named it as such Komplexverbindung (complex compound).

  I have read that primary valencies are non directional (this
  description suggests primary valencies are a physical thing).

Delete this from your mind. This does not mean anything today.


Entropy of Activation and Temperature Dependence

The equation
$$\ln\left(\frac{k}{T}\right) = \frac{-\Delta H^{\ddagger}}{RT} + \frac{\Delta S^{\ddagger}}{R} + \ln\left(\frac{k_\mathrm{B}}{h}\right)$$
does not assume that $\Delta S^{\ddagger}$ is temperature independent. To evaluate its T dependence you might proceed as follows:
$$\left(\frac{\partial \Delta G^{\ddagger} }{\partial T}\right)_p = -\Delta S^{\ddagger}$$
$$\rightarrow\left(\frac{\partial [RT\ln K^{\ddagger}] }{\partial T}\right)_p =  \Delta S^{\ddagger}$$
$$\rightarrow \Delta S^{\ddagger}=\left(\frac{ \partial [RT \ln (hk/k_\mathrm{B}T)] }{\partial T}\right)_p $$
It follows that you can evaluate $\Delta S^{\ddagger}(T)$, $\Delta S^{\ddagger}$  at different temperatures, from the change of $[RT \ln (hk/k_\mathrm{B}T)]$ with respect to temperature.


How could we figure out the recipe of Coca Cola?

I'd say some combination of a GCMS / LCMS. The chromatography separates the components & can often be relied upon to ppm level identification.

Then compare the MS peaks for hits against one of the standard databases e.g. NIST.

I suppose that strategy would identify most of the peaks. The left over peaks, which don't give a good match against a GCMS database, would be more challenging to identify. One would need to screen potential raw materials one suspects may have gone into the blend to see if one generates a hit.

Of course, this identifies individual chemical species. Any practical recipe on the other hand is unlikely to be based on pure species amounts but would use available ingredients which are each complex mixtures of pure species.

The GCMS  / LCMS approach won't tell you what pseudo-components (e.g. lemon oil  ) went in & in what proportions. So you figured out the chemical composition but not really the recipe.

Sometimes, rarely, two compounds can have very similar MS peak patterns. In those cases (or to reinforce a GCMS analysis) one could also compute the Kovats Index (KI) of each peak on the GC based on retention times alone (and the injection of a standard Hydrocarbon mix). There's a good amount of info. and databases on the KI flavor / natural product ingredients.

That would be another strategy to confirm peak identity.

On the molecular biology side of things I'm sure there are techniques using PCR etc. that could amplify and characterize traces of DNA segments that came via ingredients of plant origin. e.g. One might be able to say a certain plant oil extract was used. Although at high dilutions & after extensive processing it would be a challenge to get enough starting matter for these techniques to work.


What does 'forces' mean in the output of Gaussian calculations?

The forces you have listed are the derivatives of the energy with respect to a displacement of a particular atom in each Cartesian direction. These forces are useful for finding stationary points on the potential energy surface, which occur when the force on each atom is zero. Computationally, it's not possible or worthwhile to try to get all the forces to be exactly zero, so stationary points are determined by a combination of criteria involving the forces and the displacement between optimization steps.

The vibrations in the system are related to the second derivatives of the energy with respect to the atomic coordinates, which are contained in the Hessian matrix.


Precedence of 1,2 carbocation rearrangement

The obvious order is via the stability of the carbocation of the group.

I think you meant the migrating groups stability?
This is not what textbooks say.
Reactions are ruled by the $ΔG$ of the activated complex.
This may be close  to educts, or to products, or some halfway
state or a real intermediate minimum is on the reaction path.
So, in some cases the model of a separate carbenium ion
(in a ionic pair maybe) is useful, sometimes you need to
think of a three-center bond state.


Conjugate addition vs Electrophilic addition

Grignards unless modified with Cu(I) salts add 1,2 not 1,4 so the EtMgBr will add to the ketone to give the t-alcohol and leaving the double bond intact.

Consider then the HCl step - if it is strong enough to protonate the double bond leading to Cl addition then it will protonate the t-alcohol in preference leading to elimination or t-Cl formation. So the answer is to do the reaction on the double bond first then the Grignard addition. Protonation will occcur at the carbonyl oxygen but you have to remember that the double bond is in conjugation with the carbonyl system. This polarises the double bond, leading to Cl- attack at the end of the system.

The Grignard is more nucleophile than base so you will get majority addition to the ketone though some deprotonation/elimination may occur depending on the reaction conditions.


Why doesn't a fractured salt crystal stick back together?

You would have to get the two surfaces (or rather their crystallographic axes) together at the atomically perfect original angle to each other. Otherwise no overall (ionic) binding energy will result: Even if the surface is perfectly even, half the ions will be at places where they feel repulsion rather than attraction by the other side.

With polished metal surfaces, this works much better, also because metal atoms are (in "average") more mobile on a surface than ions.

Another point is humidity. Many surfaces, and certainly ionic substances, adsorb a layer of water from the air, practically instantly. It'll take quite a while before that has diffused out again when you press the surfaces against each other.


Selection of parent carbon chain for IUPAC naming

When assigning numbers to carbon atoms in an aliphatic skeleton, your first priority must be the longest carbon chain. The chain with red numbers is one carbon atom smaller than the one with blue numbers. Blue - 9, red - 8.
It's also helpful to know that when you have two carbon chains of the same length, the one that's the most branched must be assigned the highest priority. Here we have exactly this case, where two chains of the same length can be found. One of them has 4 branches (2 ethyl radicals, 1 alcohol and 1 bromide) and the other one has 3 branches (1 hydroxyl, 1 ethyl and 1-(1-bromoethyl)).


Under acidic conditions, a blue colour is observed. This results from the formation of N,N'-diphenylbenzidine. In the course of the reaction, diphenylamine is oxidized to a quinoid radical which subsequently adds to another molecule of diphenylamine.
Under slightly acidic conditions, bleaching of the solution is observed. Sulfite is oxidized to sulfate while $\ce{MnO4-}$ is reduced to $\ce{Mn^{2+}}$.


Should the liquid be included in Kc expression?

In the equation that you stated initially, water would be omitted from the equilibrium constant expression.

Your second question (about esterification that you posted in a comment to one of the answers) is extremely interesting.

I have given it some thought: in the standard case of esterification water should be included  because it is responsible for causing the reverse process ( hydrolysis of ester formed).

But in the case of zeolites, which by virtue of their structure, probably act as shape selective catalysts..ester hydrolysis by the water would be affected. (in some cases, I have read that selectivity towards ester formation is nearly 100%) In a case such as this, one can safely remove water from the equilibrium constant expression.

(I am not be a 100% correct, please excuse me, since I am just a student myself. Perhaps, someone else can provide a better/clearer explanation)


Boiling point of alkynes from n = 2 to 10

I would like to try to give a more detailed answer, but this is all I can give right now. I am fairly confident this is what is going on.
The dominant intermolecular force which determines the boiling point is dispersion. Because dispersion is the result of instantaneous dipoles, the strength of the interaction is proportional to the "surface area" of the molecule. There is no well-defined measure of the surface area of a molecule, but the most common way to describe this would be by using a van der Waals' radius for the atoms/groups.
The important point, however, is that when we add another carbon with its hydrogens, to a hydrocarbon, the mass increases linearly, while the surface area increases proportional to the square of the van der Waals' radius.
So, if we believe that dispersion forces are truly proportional to surface area and hence truly proportional to the boiling temperature, then a change of mass proportional to $\sqrt m$ should make a linear change in the surface area of the molecule, which should make a linear change the boiling temperature.
Notice, however, that this argument should be exactly the same for a linear alkane, and not just an alkyne. I made this same plot for the first fifteen n-alkanes, and the trend is very linear.

As to why there isn't a similar trend for the melting points, melting temperature is much more sensitive to symmetry than the boiling point, so how the molecules stack will make quite a difference. I would expect that hydrocarbons do not form very good crystals, so the trends may be highly dependent on the specific molecule.
Also, notice that the more branched these molecules are the less this argument should apply to the boiling point because each groups added at branching points will add less surface area than for the straight alkane because some of the surface area will be "redundant" so to speak.
Great observation! I hope my answer is right. I'm sure this has been noticed before, so it would be interesting if someone could find references.

Another way of thinking about this which I like is this:
Imagine that we can add mass to the carbon chain continuously rather than in lumps of atoms. Because adding mass also adds length to the chain, adding mass adds surface area. From this perspective, the relationship between $\sqrt m$ and surface area is the same as the relationship between length and surface area. This follows because these molecules are shaped a lot like cylinders.
So, the only jump left is from surface area to boiling point, which is already understood to be related through dispersion forces.


If the question you were asked to answer was literally "what can you predict about the $\ce{^31P}$ NMR spectrum of $\ce{TiP2O7}$ (from the crystal structure), then I believe you have vastly overcomplicated the problem.

This is a question about solid-state NMR (as $\ce{TiP2O7}$ is a solid).

In solid-state NMR of crystalline compounds, the chemical shift depends on the exact arrangement of atoms within the crystal unit cell. The number of signals (for each element) depends on how many non-equivalent atoms of a particular element there are within one unit cell.

From the screenshot, it says that "formula units per cell" = 4. As the formula is $\ce{TiP2O7}$, this indicates there are 4*2 = 8 different phosphorus atoms in the unit cell.

So the answer is:
You expect to observe 8 phosphorus signals in the $\ce{^31P}$ NMR spectrum.

There is no need to simulate the NMR spectrum with simpson. In fact, it is not possible to simulate the spectrum using simpson with only the crystal unit cell parameters. In order to set up a simulation in simpson, you actually have to supply the chemical shift and the interactions between the sites. You can calculate the interaction between the sites from the distances specified in the unit cell, but there is no information about chemical shifts!


What is the best way to remove/break carbon carbon bonds?

A possible pathway is as follows:

At high temperatures, formation of the more substituted alkene will be favored. Catalytic amounts of base provide a kinetic pathway for this rearrangement. $\ce{OsO4}$ creates a vicinal diol, while $\ce{NaIO4}$ cleaves the $\ce{C-C}$ bond and oxidizes the alcohols to carbonyls. Additionally, $\ce{NaIO4}$ oxidizes $\ce{(HO)2OsO2}$ back to $\ce{OsO4}$, allowing the reaction cycle to continue.


Electrolytic rust removal appears more like a semi-hoax to me. I think it is more like a mechanical removal of rust layer. As we dip a rusty object in sodium bicarb solution and you pass cathodic currents, we would see bubbling of hydrogen gas from the rusty object. These bubbles apparently help to loosen up the tight rust layer (bottom up). The cathodic current may also reduce some iron oxide.

The anode, which most videos suggest should be made of iron, is also useless in a sodium bicarbonate solution. Any inert conductor would have worked like a graphite anode. The reason is that the iron from the anode must go into the solution, if it were a true electroplating. However, in most videos iron oxides from the anod float as a precipitate of iron hydroxides. The whole solution turns rusty. Finally, people suggest a mechanical brushing or polishing. I am sure if a rusted object is polished with an abrasive, the results would be the same as electrochemical "rust" removal.

Now some over-the-counter rust stain removers are quite effective. They contain powerful reducing and perhaps some chelating agents. No, it is not oxalic acid, if that is coming to mind. These reducing agents immediately remove the rust by chemical means, but the iron surface is no longer shiny (greyish black). One has to polish it again to look shiny.


Why do real gases deviate from ideal gas behavior?

Real gases deviate from ideal behaviour because their particles (atoms for inert gases or molecules) occupy some finite space and do exert interactive forces among them. Completely ideal behaviour is hypothetical because of the reasons above. At low pressure and high temperature, real gases behave approximately as ideal gases. In ideal behaviour, gas particles don't occupy space and do not have any interaction, as assumed in the kinetic theory of gases. But in reality this is not the case: we get errors by applying the ideal gas law.
That's why van der Waals corrected it by introducing suitable constants.


What determines the strength of reducing and oxidizing agents?

Because compounds can be oxidizing agents like potassium permangenate (KMnO4) and reducing agents LiH4, what makes a compound an oxidizing or reducing agent is oxidation & reduction tables. As oxidation is the giving off of electrons and reduction is the acceptance of electrons, like plus vs minus relationship, if you have an oxidation table you can turn it into a reduction table by putting the flipping the table, changing the signs and reversing the equations. Anyway, reduction tables are more standard, where the strongest oxidizing agents have the most positive/largest standard reduction potentials and the strongest reducing agents have the most negative/smallest standard reduction potentials.

The standard reduction potentials are determined with a volt-meter connecting two cells together as electrons pass through the salt-bridge.

This site explains it  rather nicely

But there are some patterns such that

$\ce{2M(s) + 2H2O(l) -> 2M+(aq) + OH^{-}(aq)}$ $\ce{ M = Li, Na, K, Rb, Cs}$

Cesium reacts more violently (in an explosive reaction which occurs as hydrogen gas is ignited from the heat of the strongly exothermic reaction) than all the metals above it because it is a stronger reducing agent reducing agent being oxidized itself more than the metals above it because it has the the lowest ionization energy due to shielding as the many shells of electrons around the nucleus of Cesium reduce the pull of the positive charged nucleus on the electrons due to the electrons in the electron shells repelling the valence electrons further from the nucleus by like charge repulsion. Ionization energy is simply the measurement of the heat energy required to cause an atom to lose an electron in the gas phase.

This makes sense, but there are anomalies to consider not going by the electrochemical series from experiments and just trying to rationalize off of stuff like ionization energies, electronegativities, etc that I started off my answer with. The reason that you might find anomalies is because the reduction potentials in voltages of elements are calculated in aqueous solution, while ionization energies are calculated in the gas phase, though many oxidation-reduction reactions occur in liquid solutions, one potential difference being the failure to consider enthalpy of solvation when considering ionization energies. Furthermore, Pauling's electronegativites can be calculated from running some equations through elements considering the physics of ionization energies, so ionization energies and electronegativites give you a similar perspective.


How to determine the configuration of an propargylic alcohol?

I think your working is correct, and the molecule should be assigned as S, with the alkyne taking priority over the disubstituted alkene such that the priority goes:

alcohol
alkyne
alkene
hydrogen

This leads to the S configuration, as you propose.

There is a chance that your supervisor is confused because in IUPAC naming nomenclature, the alkene often (and indeed, in this case) takes priority over the alkyne, which would of course lead to assignment of the molecule as R.


Choosing fish-oil that won't oxidize in long travel?

First off, a "fishy" smell and rancidity are two different things.  A fishy smell is likely due to impurities, primarily small volatile amines such as trimethylamine.  These impurities are unrelated to oxidative degradation of the fish oil.

Rancidity in fish oil is the result of oxidation.  Oxidation leads to the formation of lipid hydroperoxides and short-chain aldehydes, which can have "off" tastes and give rise to unpleasant odors, but in a very different way than the classic fishy smell.

Fish oil's primary components are docosahexaenoic acid and eicosapentaenoic acid.  They are more sensitive to oxidation than many other fats because they contain many double bonds.

The best bet is to look for oil that is freshly prepared and highly purified.  Some fish oils are esterified with ethanol to form ethyl esters right "on the boat".  These ethyl esters can be distilled to remove most impurities.  Optionally, the purified esters can be converted back to triacylglycerols to give "molecularly distilled" fish oil in the triglyceride form.