Preferred Lewis structure for sulphate anion and XeO4

You are reasoning as if the ion $\ce{SO4^2-}$ is planar. It is not the case. It has a tetrahedral geometry. And what you see in Atkins' book is a projection of the tetrahedron on a horizontal plane. So the angles between the bonds are not 90°, but larger. It is the same for $\ce{XeO4}$. This reality will solve all your questions simultaneously.


Process to separate Lauric Acid from Coconut Oil

With respect to economy, waste management and regulations concerning food processing, your current approach, a solvent-free crystallization, is probably the most reasonable approach.

There's a Unilever patent which claims that the crystallization, both stirred and quiescent, is significantly improved by addition of sucrose laureates.

The patent holders attribute this to the observation that the additive leads to the formation of

  large and non-porous spherulites, which highly increases separation efficiency


There are a lot of questions in this one post which is probably why there has not yet been an attempt to answer this post.

1.) What is the Schrödinger wave equation?

The wave equation (i.e. wave function) is a function which contains all the information about a chemical system.  With an appropriate operator, one can extract this information and gain insight into the system.  For example, the Hamiltonian operator will extract the energy of a system from the wave function.  The wave function is an eigenfunction while the energy (in this example) is the eigenvalue of the eigenfunction.  For example, in the Schrödinger equation

$$ \mathcal{H}\Psi = E\Psi$$

$\mathcal{H}$ is the Hamiltonian operator, $\Psi$ is the wave function (eigenfunction) and $E$ is the energy (eigenvalue).  Notice that by operating on $\Psi$ with $\mathcal{H}$ returns $E$ AND $\Psi$, which by definition, makes this an eigenfunction-eigenvalue equation.

2.) What are radial probability functions?

These are probability distributions for the probability of an electron residing some distance from the nucleus of an atom such as the hydrogen atom.  Consider this plot of various radial probability distribution functions of the hydrogen atom.

Here we see that the probability of an electron (in the hydrogen atom) in a 1s orbital is zero at the nucleus (at 0,0) and a maximum a hair shy of $2 a_0$.  Interestingly an electron in a 2p orbital has a small probability of residing close to the nucleus ($< 1a_0$) but is more likely to reside further away from the nucleus than an electron in a 1s orbital.  This makes sense given that a 2p orbital is higher in energy because 'it' is further away from the nucleus than a 1s orbital.

3.) What are nodes in radial and angular functions?

A node is where the function is zero at a particular point.  This simply means in a physical sense that at this point (at the node), the probability of an electron existing at that point is ZERO.  Nodes are a result of the separation of variables technique when solving the appropriate equation(s). When any one component (or variable) is zero then the resulting function will be zero at that point.

4.) What is electron cloud?

Electron 'cloud' is just a nicer away of saying electron density.  Electrons are not strictly particles.  Electrons do not just 'orbit' a nucleus.  Electrons also exhibit wavelike behavior.  (see: double-slit experiment - wave-like behavior and J. J. Thompsons cathode ray tube experiments - particle-like behavior).  The 'cloud' is obtained simply by 'squaring' the wave function.  More technically, its taking the product of the wave function and its complex conjugate.  The 'cloud' is just a probability distribution of where an electron is likely to be (usually given in 3-dimensions).

5.) What is the Hamiltonian operator?

The Hamiltonian operator is the energy operator to the Schrodinger equation.  Because it is Hermitian, it will return a 'real' result (as opposed to a complex result).  There are different versions of the Hamiltonian depending upon what approximations you want to make.  The full Hamiltonian operator can be written as (in terms of atomic units)

$$\mathcal{H} = -\sum\limits_{i=1}^{N}\dfrac{1}{2}\nabla_i^2
              -\sum\limits_{A=1}^{M}\dfrac{1}{2M_A}\nabla_A^2
              -\sum\limits_{i=1}^{N}\sum\limits_{A=1}^{M}\dfrac{Z_A}{r_{iA}}
              +\sum\limits_{i=1}^{N}\sum\limits_{j>i}^{N}\dfrac{1}{r_{ij}}
              +\sum\limits_{A=1}^{M}\sum\limits_{B>A}^{M}\dfrac{Z_A Z_B}{R_{AB}}
$$

Each term on the right hand side of the equation (going from left to right) is defined as

the kinetic energy of the electrons
the kinetic energy of the nuclei
the potential energy of the electron-nuclear attractions
the potential energy of the electron-electron repulsions
the potential energy of the nuclear-nuclear repulsions

We can write the Hamiltonian using a bit more compact notation such as

$$
\mathcal{H} = \hat{T}_{e} + \hat{T}_{N} + \hat{V}_{eN} + \hat{V}_{ee} + \hat{V}_{NN}
$$

where negative signs are assumed for each $\hat{T}$ operator.  The terms denoted here correspond to the respective terms in the equation before this one.

If we take into account the Born-Oppenheimer approximation (the B.O. approximation assumes that the electron motion is orders of magnitude larger than the nuclei they belong to... so much faster that the nuclei are, relatively, standing still with respect to the electrons), we can make two changes to the full Hamiltonian giving us the electronic Hamiltonian.

$$
\mathcal{H} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee} +\ \mathrm{constant}
$$

The nuclei are treated as frozen in space when the B.O. approximation is applied for reasons explained above.  Therefore the kinetic energy of the nuclei goes to zero.  The nuclear-nuclear repulsion is now a constant because the nuclei are fixed.  The repulsion will never change.

6.) What is a Laplace operator?

The Laplace operater (or Laplacian) is simply a differential operator of the form

$$
\nabla^2 = \dfrac{\partial^2}{\partial x^2} +  \dfrac{\partial^2}{\partial y^2} +  \dfrac{\partial^2}{\partial z^2}
$$

if you are working with Cartesian coordinates. This is simply a set of second-order derivative operators within the kinetic energy operator.

While I may have addressed each question, there is a lot more depth to these answers that lie well beyond the scope of this post.  I encourage you to find more detailed and insightful answers in well-established texts.  I highly recommend Szabo & Ostlund for a solid introduction to quantum mechanics.  You can snag a paperback version of this excellent work for about $10.


Is the caffeine in coffee an oxidation agent?

No, caffeine is not known as an oxidizing agent.


Cyclohexane derivative formation of carambolaflavone using a Lewis acid catalyst

@Waylander provided a correct written explanation for your question. I'll expand on his comments by considering the effect of temperature and provide diagrams. Using M+ = Sc+3 to save some space, the Lewis acid  coordinates with the more basic of the two acetate oxygens in 1 to provide the reactive oxonium species 2. [The oxonium species is planar about the double bond but I have left it in the chair conformation as a space-saving device.] This step may be reversible. At room temperature there is not sufficient activation energy to permit substitution of the aromatic ring. Following the red arrows of phenolic species 3, axial attack of the phenol oxygen occurs on the oxonium ion 2 to form, kinetically, compound 4. At 50 oC this reaction is reversible catalyzed by Sc+3 to reform 2 and 3 via the blue arrows. Following the green arrows, now there is sufficient energy to disrupt the aromaticity of 3 and allow for axial attack by carbon on the oxonium ion 2 to form, kinetically, structure 5. An increase of temperature to 70 oC permits coordination of the tetrahydropyran oxygen with the Lewis acid permitting ring opening, bond rotation about the red bond in 6 and ring closure of conformation 7 forming the thermodynamically stable, all equatorial tetrahydropyran ring of 8. [Structures 6 and 7 may also be drawn as benzyl cations with the carbonyl oxygen protonated.]


How to obtain the Raman spectrum along every coordinate of a scan in Gaussian?

No, there is no automation for this, and there should not be.

Using compound jobs is unfortunately very common among computational chemists, but they should always be viewed critically. If you are using any kind of IOp, then the probability that the second part of your job is doing what you want is low. Additionally, you do not really have any control about what happens if something happens to fail. In such cases, you also give up control over the validity of your checkpoint-file, which you might want to reuse.
Another issue is that these are basically two jobs running one after another and in most cases need different memory requirements, are more effective with another setup. It's the dream of every administrator (and co-worker) when you block resources you don't need for an extended period of time just to avoid a little more waiting.
Lastly, with the introduction of %oldchk= in in revision D.01, it is completely superfluous. Every well organised queuing system should be able to run jobs one after the other when specified (i.e. let one job wait for another to finish).

Only run compound jobs if you are 100% sure they cannot fail (and even then it's questionable).

From a theoretical standpoint, you should consider how much sense your approach makes. As TAR86 pointed out, Raman intensities are expensive. Additionally frequency calculations are only meaningful at well converged local minima structures. You are employing the harmonic oscillator approximation, which will break down if you are not using a local minimum (or at least a stationary point).

If you really need this approach, then you need to do it manually by extracting the geometries (and possibly the MO) and perform the frequency job on top of it.


Which is a better Lewis structure?

There is no such thing as a better or a worse Lewis structure. They should be properly referred to as major or minor resonance contributors/forms.

I looked up a textbook to check the rules of determining which resonance structure contributes more. The order is:

Structures that satisfy the octet rule are greater contributors than structures which do not satisfy the octet rule.
As long as the octet rule is not exceeded for second-row elements, the contributing structure with the greater number of covalent bonds contributes more.
When two or more structures satisfy the octet rule with the same number of covalent bonds, the major contributor is the one with the smallest separation of oppositely charged atoms.
Among structural formulas that satisfy the octet rule with the same number of covalent bonds, and in which one or more atoms bear a formal charge, the major contributor is the one in which the negative charge resides on the most electronegative atom.

In the first structure, all atoms possess a complete octet. But in the second structure, nitrogen only has six electrons around it: consequently, by rule 1, the first resonance structure is the major contributor.

  Are they resonating structures of $\ce{NO+}$?

Not exactly.

The actual structure of the nitrosonium cation is somewhere in between the two forms. This means that the N–O bond is somewhere between a triple bond and a double bond. However, since we have established that the first form contributes more to the actual structure, it is going to be closer to a triple bond than a double bond. You could think of it as a 2.9-bond. Molecular orbital theory tells us that the actual bond order in $\ce{NO+}$ is 3, but that's an entirely separate story for another day.

The cation does not resonate between two distinct structures. Let me cite another example. Benzene also has two different resonance structures, both of which contribute equally (since they are identical to each other). Does that mean that the C-C bonds flip between being a single bond and being a double bond? No. All the C-C bonds in benzene are 1.5-bonds, at all times. The common usage of alternating single and double bonds (the Kekule structure) is just us being lazy since many mechanisms are easier to draw that way. Similarly, the $\ce{NO+}$ cation does not flip between the two resonance structures, which represent two extreme ends of a spectrum of possible structures. The actual structure is always in the middle of the spectrum. Which side the actual structure leans towards, is determined by which resonance form contributes more.

At risk of sounding like a broken record - The actual structure is somewhere in between the two resonance forms. It does not oscillate between two different resonance forms. This is a very important concept to understand!

  I think the structure in which the less electronegative atom has a positive formal charge is better, so the first structure is better. Is the above reasoning correct?

Yes and no. There are several methods of measuring electronegativity, and in all of them, nitrogen is less electronegative than oxygen. By your reasoning, this means it would be more favourable to place the positive formal charge on nitrogen, i.e. resonance form 2. So your reasoning is a little bit off there.

On the other hand, you would be correct if there wasn't a difference in the number of bonds between the two resonance forms. As I mentioned at the top of the post, the number of bonds takes precedence over the electronegativity issue.


Separation and qualitative identification of cations

Well there is a pretty simple way, that you can do in any laboratory by standard reagents too.
React the unknown mixture with $\ce{NH4OH + NH4Cl}$
(ammonium chloride is put as a standard to prevent the precipitation of Zn, Mn, Cu and Ni hydroxides, however, you don't have those in your sample, so you can proceed with only ammonium hydroxide too)
now you will observe precipitate as

Green- confirms $\ce{Cr^{+3}}$

Reddish-brown- confirms $\ce{Fe^{+3}}$

White- confirms $\ce{Al^{+3}}$

(These are colours of the corresponding hydroxides)

Now if you have multiple salts, you carry out the following conformatory tests

For $\ce{Cr^{+3}}$, add a few dorps of strong oxidising agent in an acidified medium ($\ce{KMnO4}$/$\ce{H+}$ would do), this oxidises $\ce{Cr^{+3}}$ to $\ce{Cr2O7^{-2}}$ which gives the super famous chromyl chloride test.

For $\ce{Fe^{+3}}$, add a few drops of $\ce{CH3COONa}$ and you will initially observe a dark-red solution of $\ce{Fe3(OH)2(CH3COO)6}$, which on heating precipitates as blood red $\ce{Fe(OH)2CH3COO}$

For $\ce{Fe^{+2}}$ simply add KI, very distinct black $\ce{FeI2}$ precipitates out

For $\ce{Al^{+3}}$ add ammonium phosphate, you will see a cloudy white precipitate

You can read more here, Vogel is pretty much the most solid and trustworthy book for practical inorganic chemistry
You are welcome!


Why isn't just a fully occupied valence s orbital stable?

Elements with just 1 or 2 electrons in the most outer $\ce{s}$ orbitals, like $\ce{K}$ or $\ce{Ca}$, have very well screened off the central charge of the kernel. As consequence, the effective charge of the kernel and related energy needed to ionize the atom are relatively low.
OTOH, elements with just few missing electrons in the most outer $\ce{p}$ orbital, like $\ce{O}$ or $\ce{Cl}$, do not have well screened off the central kernel charge, which is relatively high, what leads to considerable electron affinity to some extra electrons.
Combining these 2 kinds of elements leads to energy release, as energy spent by ionization of the former is  compensated by energy release by capturing electrons by the latter and by a covalent contribution to the element bond.
In case of solid ionic compounds, there is additional large energy release du the ionic lattice energy.


Decomposition of lead(IV) oxide at different temperatures forms different types of oxides. At around $\pu{600 ^\circ C}$, lead(II) oxide forms. Wikipedia adopts the scheme of lead(IV) oxide decomposition in air summarized by White and Roy [1]:

$$\ce{PbO2 ->[\pu{293 ^\circ C}] Pb12O19 ->[\pu{351 ^\circ C}] Pb12O17 ->[\pu{375 ^\circ C}] Pb3O4 ->[\pu{605 ^\circ C}] PbO}$$

You can then reduce lead(II) oxide to lead using carbon or carbon monoxide at around $\pu{1200 ^\circ C}$. The effect of dissolved inorganic carbon is discussed by Wang et al. [2]
$$
\begin{align}
\ce{PbO + CO &->[\pu{1200 ^\circ C}] Pb + CO2}\\
\ce{2 PbO + C &->[\Delta] 2 Pb + CO2}
\end{align}
$$
Note that lead(II) oxide can also decompose to release toxic fumes of lead which can be harmful and also doesn't serves your purpose of getting solid lead.
References

White, W. B.; Roy, R. Phase Relations in the System Lead-Oxygen. J American Ceramic Society 1964, 47 (5), 242–249. DOI: 10.1111/j.1151-2916.1964.tb14404.x.
Wang, Y.; Xie, Y.; Li, W.; Wang, Z.; Giammar, D. E. Formation of Lead(IV) Oxides from Lead(II) Compounds. Environ. Sci. Technol. 2010, 44 (23), 8950–8956. DOI: 10.1021/es102318z.


Searching for a solvent with a low carbon dioxide solubility

To find a solvent with low solubility for non polar gases like CO2 you have to compare the polarity of the solvents. A very polar solvent like water is the best choice.
There's a related question here: Dissolving of non polar gases in water (liquid)

At the end of this answer I have added two links. The first one gives you the solubility of CO2 in water. According to this source, at room temperature and around atmospheric pressure of Co2, the molar fraction in water is 0.00061.
The second link compares many other solvents. At room temperature and 1atm of CO2 the molar fraction is 0.012 in hexane that is the lowest for straight chain alkenes(Fig.1). Overall, among alkanes and cyclic alkanes, cyclohexane is the only one that at room temperature has a CO2 solubility below 0.01(fig.2).
Among aromatic hydrocarbons, you can reach a molar fraction as low as 0.006 if you use 1-Methylnaphthalene(fig.3) and this value is interestingly similar to what you can reach using methanol(you can find a graph comparing alcohols there as well(pag.129). Apparently the CO2 solubility with aromatic compounds is comparable to alcohols(polar).

With all this in mind, water is the best option and if you add salts the solubility of CO2 decreases further as you mentioned.

Solubility of Carbon Dioxide in Water:
https://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/solubility%20of%20carbon%20dioxide%20in%20water.pdf

CARBON DIOXIDE IN NON-AQUEOUS SOLVENTS:
https://srdata.nist.gov/solubility/IUPAC/SDS-50/SDS-50.pdf


How to trace an energy versus time diagram?

Assuming you want to plot heat exchanged by reaction versus time, you need at least to know what reaction you want to study. Then you will have to assess two different quantities: enthalpy if the process is isobaric (thermodynamic) and reaction rate (kinetic). This kind of data can be found on NIST Database.

If you do not find reaction enthalpy $\Delta_{R} H$, you can assess it by using enthalpies of formation:

$$ \Delta_{R} H = \sum\limits_i \nu_i h_{\mathrm{f},i} $$

Where $i$ indexes reactants and products, $\nu_i$ are stoichiometric coefficients (note: they are negative for reactant) and $h_{\mathrm{f},i}$ are enthalpies of formation.

Then you need to find out which is the kinetic $v$ of your reaction and solve the related differential equation. If you do not find any information you might try the activated complex model:

$$ v = \frac{1}{V}\frac{\mathrm{d}\xi}{\mathrm{d}t} = k(T)\prod_j x_j^{\alpha_j} $$

Where $\xi$ is the molar coordinate of your reaction (De Donder variable), $k(T)$ the kinetic constant, $x_j$ are reactant concentrations and $\alpha_j$ are partial orders of reaction (they can be taken equals to $|\nu_j|$). Solving this differential equation - for a given set of initial conditions - will provide you a relation, maybe explicit, between $\xi$ and $t$.

Merging both step results will lead to enthalpy-time expression which can be plotted. Of course those calculation might be perform numerically.

If you were looking for an hand-made and pedagogic Energy Plot like this, you have not searched so far.

Method exposed above does not create this kind of plot which shows internal energy of sum of compounds versus coordinate of a molecular transformation. Gap between the initial and final state is the energy relaxed/absorbed by the reaction (often pure heat). Such plot are arbitrarily plotted because the activated state is often unknown and definitely non measurable. Furthermore everything between initial and final state is also not-well-known because we just have access on macroscopic data so we ignore microscopic details. Last hint: activation energy (maximum) is always higher than the higher state, you need extra energy to initiate the transformation.


How much ozone is produced by ionization of air and how turn ozone into oxygen

Even if the UV light idea doesn't work out very well I still want a way to do this

With some photochemical background, I suggest to forget about running such a UV lamp

Remember that your plasma will be formed in air. For every molecule of ozone formed, there are much more harm- and odourless $\ce{O2}$ molecules around. If you cleave $\ce{O3}$

\[\ce{O3 + O2->[\mathrm{h}\nu][\lambda\ \lt\ \mathrm{300\ nm}] [O2 + O* + O2] -> O2 + O3}\]

the resulting oxygen atom will be captured by the next $\ce{O2}$ to form ozone again.

The overall process will not reduce the number of ozone molecules, unless $\ce{O*}$ reacts with $\ce{N2}$, which might lead to some other unpleasant species ;)

Note that these processes are fortunately effective in the ozone to protect us from short wave UV radiation, which brings us to the next aspect of the setup.

In order to photolyze ozone, you'll neeed a light source that emits UV at $\lambda\lt300\,\mathrm{nm}$. It's not that these light sources would not be available, but this UV range is not that of a tanning booth lamp! Remember that the ozone layer protects our DNA from running havoc! The least you can expect from short-wave irradtion is the sun burn of your life! Consequently, you will have to shield the whole setup properly!

In summary, I absolutely like the idea of building a plasma speaker yourself and there even seem to be some DIY kits around, but I suggest to forget about the UV part, which can be much more dangerous than the ozone formed.

UPDATE

Burak Ulgut brought up a very interesting point. Back in the days, plasma speakers using other gases (or gas mixtures) were tried, and even went commerical.

This site in German describes a helium driven plasma speaker system :)


Calculate thr unit cell length of perovskite

Experimental X-ray crystallography data yields a value of 4.182 for the lattice constant for $\ce{KMnF3}$. See this reference for further details. Given that the paper was published in 1961, there are likely more recent refinements experimentally. That said, your number is within the error of observed values.

Reference: K. Knox. Perovskite-like Fluorides. I. Structures of $\ce{KMnF3}$, $\ce{KFeF3}$, $\ce{KCoF3}$, $\ce{KNiF3}$, and $\ce{KZnF3}$. Crystal Field Effects in the Series and in $\ce{KCrF3}$ and $\ce{KCuF3}$. $\it{Acta\,Cryst.}$ (1961) $\bf{14}$, 583.


How to estimate the equilibrium constant of the reaction between NaF and HCl

We can entirely ignore the sodium cation; it is only a spectator ion. This means, the reaction we are observing is:

$$\ce{HCl + F- <=> Cl- + HF}\tag{1}$$

The equilibrium constant for this reaction is:

$$K = \frac{[\ce{Cl-}][\ce{HF}]}{[\ce{HCl}][\ce{F-}]}\tag{2}$$

We can expand this equation:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]}\tag{2'}$$

And then we realise that that is nothing else than:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]} = \frac{[\ce{Cl-}][\ce{H+}]}{[\ce{HCl}]} \cdot \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]} = \frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{HF})} \tag{2''}$$

So the equilibrium constant is the fraction of the acidity constants of $\ce{HCl}$ and $\ce{HF}$. We know that $\ce{HCl}$ is a much stronger acid than $\ce{HF}$. This is reflected by the inequation:

$$K_\mathrm{a}(\ce{HCl}) > K_\mathrm{a}(\ce{HF}) \tag{3}$$

Since the numerator is larger than the denominator, the value of the fraction must be larger than $1$. Therefore, the product side is preferred.

If your book arrives at any other conclusion, it is disregarding experimental results and should be replaced.


When I dissolve sugar in my cup of tea/coffee, does it become a liquid?

When the sugar dissolves, it is broken up into individual particles; each molecule is surrounded by water molecules. Essentially, you are trying to decide if a single molecule is a solid or liquid when dissolve.

The best fit is a liquid since the molecules are free to move independently and will take on the shape of the container.


Unusual oxidation states of alkali metals

You are talking about alkalides, salts where the anion is an alkali metal. There is a very brief overview on Wikipedia which also provides a couple of examples, including $\ce{[Na(\text{cryptand[2.2.2]})]+Na-}$. A good university level inorganic text book such as Greenwood and Earnshaw or Housecroft and Sharpe will provide more detail.

If you are interested in unusual oxidation states of alkali metals you might also like to know that in electrochemical experiments there is some evidence for $\ce{Cs^3+}$, which is isoelectronic with $\ce{Xe^2+}$. Again I think Greenwood and Earnshaw discusses this, but I don't have it to hand at the moment to confirm.


Why do dianions (such as malonate) bind cations more strongly than anions?

This is known as the chelate effect.

The main reason why you observe this is that cations in solution have an ordered solvent shell around them, especially in polar solvents where there will be defined solvent geometry around the shell; often octahedral for metal ions in water. The formation of a complex with a bidentate ligand such as malonate is more favourable than that with a monodentate ligand such as acetate because the same number of solvent ligands are lost with half the number of ligands complexed. This results in a greater increase in entropy - you can see this in terms of the degrees of freedom gained by the water molecules more than compensating for those lost by the new ligands.

  Is it simply because of the proximal availability of another $\ce{O-}$ group that can bind to cations ?

This is likely also a factor, although in most cases less important. In the case of the monodentate ligands there is a small energetic cost associated with bringing the anionic oxygen centres close to complex the metal centres. In the case of the bidentate ligand, some of this cost is "already paid" in the enthalpy of formation of the ligand - it is preorganized. This contribution becomes dominant for macrocyclic ligands.


Is possible to figure out if a battery is NiCd or NiMH?

NiCd cells are much more dense then NiMH, so if you have two cells of the same size (e.g. AA), it's easy to determine by "hefting". However, with a sealed battery, this may not be so obvious. You could determine the density by Archimedes' method, but you probably don't want to get the inside of the battery pack wet (unless you can thoroughly dry it), to prevent electrolytic corrosion.

BTW, NiCd cells are much more resistant to damage from over-charge and from complete discharge than NiMH, and are still widely used in portable power tools and other high-drain applications. NiMH provide more total energy by mass, and more environmentally "friendly'.


All complex anions you noted that include only non-metal atoms are always considered covalent. Except for a few extreme examples including fluorine and hydrogen, non-metals only form predominantly covalent bonds between themselves.

It gets a little more tricky when considering those anions whose central atom is a transition metal. In this case at first approximation, we can consider the anion a coordination compound and the coordinate bonds within coordination compounds are typically considered more covalent. Moving on to a second approximation clarifies the picture: the formal oxido-ligands are in fact very strongly attached to the metal and do not undergo ligand substitution. This means that they are even more covalent than other coordination complexes.

Therefore, it is typically safe to consider the bonds within complex anions to be predominantly covalent.


In my own work, I've never used the term polymorph for anything but molecular crystals in a highly but differently ordered solid state.

  In the case of crystal solids, it seems clear that allotropy is a particular case of polymorphism.

Yes, allotropy is the polymorphism of elements in the same state. This does not mean that each solid allotrope has to be highly ordered: amorphous forms of carbon, phosphorus or selenium are considered as allotropes too.

The definition of allotropy isn't confined to the solid state: ozone is an allotrope of dioxygen.


Unusual hydrolysis of methyl 2,4,6‐trimethylbenzoate

In the typical reaction of hydroxide ion with a methyl ester, the hydroxide attacks the carbonyl carbon of the ester, resulting ultimately in elimination of methoxide. Given the basic conditions, the carboxylic acid formed immediately deprotonates to a carboxylate, and the methoxide is protonated to methanol. In older texts particularly, this reaction mechanism is referred to as BAc2, indicating "bimolecular base-catalyzed acyl-oxygen cleavage" (*) because the bond that is broken is between the acyl group and the oxygen of the methoxyl group [see for example Ingold, C.K. (1953) Structure and Mechanism in Organic Chemistry, Cornell Univ. Press].
An alternative mechanism is the BAl2 mechanism, designating "bimolecular base-catalyzed alkyl-oxygen cleavage". As the name suggests, in this mechanism, the bond between the alkyl group and the oxygen is broken. To do this, the hydroxide attacks the methyl carbon in an SN2 reaction, directly yielding the carboxylate and methanol products. This mechanism is also described in detail in Ingold's book and many other sources.
In the years since this mechanism was proposed, there has been some debate about how often it really occurs. Certainly for most esters, BAc2 is the dominant mechanism, and even when reaction by that mechanism is very slow, only a fraction of molecules, if any, react by the BAl2 option.
There seem to be two requirements for the BAl2 type reaction. First, the alkyl group must be methyl or a very ring-strained lactone. Other alkyl groups react much more slowly, as is observed in any SN2 reaction. Second, the carbonyl carbon must be sterically hindered to the point that the BAc2 mechanism is essentially impossible.
I am not certain that the tri-methyl benzoic acid in your question is hindered enough to react exclusively by the BAl2 mechanism, but it is certainly hindered enough that BAc2 would be substantially diminished.
You can find a more thorough discussion of these mechanisms in many textbooks, including the Ingold reference above. Another paper that may be useful is Mueller, P. and Siegfried, B. (1974) SN2 Reactions with Carboxylic Esters. Selective Cleavage of Methyl Esters. Helv Chim Acta 57:987.
*Note that since the net reaction in my example is $\ce{R-CO2Me + HO- -> R-CO2- + MeOH}$, the base is a stoichiometric reactant rather than a catalyst, so "base-catalyzed" is a misnomer. That terminology is used because the mechanism is the same if a different nucelophile is used, and the product is a non-acidic acyl derivative rather than a carboxylic acid. In that case, the base is indeed a catalyst.


Amount concentration versus number concentration

The "amount" of a substance is the proper term for the "number of moles" (Wikipedia; Gold Book). So, the two quantities are related by a factor of the Avogadro constant, $\pu{6.022\times 10^23 mol-1}$.

If you have $\pu{1 mol}$ of helium in a $\pu{1 m3}$ box, the amount concentration is $\pu{1 mol m-3}$, or $\pu{0.001 mol dm-3}$ if you prefer those units.

The corresponding number concentration is $\pu{6.022\times 10^23 m-3}$.


Molarity is defined as "moles of solute per volume of solution", which implies that the system is in the liquid state.

As such, molarity is undefined for anything in a gaseous state -- there is no solute/solvent distinction in the gas phase.

You can define the concentration of a gas, though, and that calculation would be exactly the one you've described: placing $X$ moles of a gas in a container of volume $V$ yields a concentration $C=X/V$, with units of, e.g., $\pu{mol\over dm^3}$.

Note that while these units might look the same as the units of molarity, they are different. In this concentration value, the volume is the volume of the container; in a molarity value, the volume is the volume of the solution.


Respect to other chemicals in reduction/oxidation potentials

You cannot define ‘vacuum’ oxidation potentials, or otherwise derive them de novo without having some zero-point. There is, technically, no self-suggesting zero-point around — unlike for example temperature, where $0\,\mathrm{K}$ is equivalent to ‘molecular movement frozen’.

Part of the reason is, that the electron(s) lost or gained during the redox reaction must go or come from somewhere. Therefore, one has chosen the hydrogen cell as a centre of the scale. Why?

Hydrogen is the easiest element with the least electrons.
It’s simple to build.
It actually also comes with a good further distinction: The one between noble and non-noble metals.

You can actually easily recalculate the standard potentials with respect to another half-cell like magnesium: Just add/subtract that cell’s potential with respect to hydrogen to all other potentials. It’s a bit like moving the temperature scale’s zero point from $0\,\mathrm{K}$ to $273\,\mathrm{K} = 0\,\mathrm{°C}$.


I could be wrong but I think this reaction works by forming benzyne, which then undergoes enolate attack:

I don't think hydroxide is basic enough to do this, so no I don't believe your proposed reaction conditions will work. Is there any reason you can't just use NaNH2, or BuLi, LDA etc.?


I found something very similar to what I sought in Dose, Nature 179, 734 (1957) though it involves aldimines and a decarboxylation step. The author gives a very well explained mechanism for the process and reports varying transamination ratios of valine based on the substitution of the aldehyde. An example:


What is meant by "fairly" when used as a quantifier for the solubility of a chemical?

It seems that you understand the concept very well and that this is just a minor confusion about word fairly.

If you would characterize the solubility of a compound (solid, liquid, or gas) by words, rather than by giving the exact solubility in $\pu{mg\cdot mL^{-1}}$, you would probably come up with terms like

highly soluble
soluble
slightly soluble
insoluble

So, fairly means that the solubility isn't excellent, but it's not that bad either. It is acceptable, which of course isn't very precise.

  Is it that fairly soluble in water means that a part of it dissolves while another part doesn't dissolve?

That one part does not dissolve will happen in all aqueous solutions when you add to much of the compound that you want to dissolve. Just think in sodium chloride, which has a solubility of about $\pu{390 g\cdot L^{-1}}$ in water. When your solution is saturated, it can't take any more. If you add another spoon of the salt, it will not dissolve and precipitate.


Halogens and the free energy of activation in a resonant benzene system

Resonance between a carbon and a halogen atom decreases as we move down the series $\ce{F, Cl, Br, I}$.  This is because overlap is most effective between orbitals of the same size.  Therefor, carbon-fluorine overlap, both involving $\ce{2p-2p}$ orbitals should be much stronger than carbon $\ce{2p}$ orbital overlap with the $\ce{3p, 4p, etc_{.}}$ orbitals in the larger halogens.  Consequently carbon-halogen resonance becomes less significant as we move down the halogen series.  If we now consider that 1) the resonance and inductive effects for the para-halogens are in opposition to each other (tend to counteract each other) and 2) the resonance effect decreases as we move down the halogen series, then the order of halogens in the Table you mentioned makes sense.


From "Statistical Mechanics: A Concise Introduction for Chemists" by Benjamin Widom (pp. 35-36):

  For the diatom of ordinary hydrogen ($\ce{{}^{1}_{1} H}$), where each
  nucleus is just a proton of spin quantum number 1/2, there are three
  possible symmetric states of the pair of spins (nuclear spin triplet)
  and only one possible antisymmetric nuclear spin state (spin
  singlet). [...]

  At high temperatures, which generally means $T \gg
> \theta_{\mathrm{rot}}$ (where $\theta_{\mathrm{rot}}$ is the
  characteristic rotational temperature) and so for $\ce{H2}$ means $T$
  well above 88 K, the three ortho nuclear-spin states and one para
  state are equally populated, so equilibrium hydrogen at high
  temperatures is a mixture of ortho- and para-hydrogen in the ratio
  3:1. In pure hydrogen the magnetic interactions between molecules that
  could cause transitions between nuclear spin states are so weak that
  such hydrogen may be cooled to arbitrarily low temperatures with the
  ortho:para ratio remaining 3:1. That is normal hydrogen. Paramagnetic
  impurities such as $\ce{O2}$, or a catalyst such as $\ce{Pt}$ on the
  surface of which $\ce{H2}$ molecules may dissociate, allowing the
  atoms to recombine with their nuclear spins in a different relative
  orientation, catalyze the ortho-para conversion. Cooled in the
  presence of such a paramagnetic impurity or such a catalyst, then, the
  hydrogen may have its ortho:para ratio continually adjust to that
  which would correspond to thermal equilibrium at the given low
  temperature. [...]

  Cooled in such a way, then, hydrogen becomes pure para-$\ce{H2}$ at
  low temperatures, in contrast to "normal" hydrogen, which remains a
  mixture of ortho and para in the ratio 3:1. If the hydrogen is cooled
  in the presence of a catalyst that catalyzes the ortho-para conversion
  and thus allows it to become pure para-$\ce{H2}$ at low T, and if the
  catalyst is then removed and the hydrogen warmed back up to 298 K,
  say, the result is a sample of pure para-$\ce{H2}$ at normal
  laboratory temperatures. Pure ortho-$\ce{H2}$ cannot be prepared in
  this way.


The electron gain enthalpy you've mentioned is measured in the gas phase, not in aqueous solution like the order of basicity.

When you go down on a group, two important properties are increased: nuclear charge and atomic radius. However, the increase in atomic radius is much more pronounced than of the nuclear charge. Considering this, as you go down on the halogens group the electrons added feel less attracted to the nucleus and, therefore, the electron gain enthalpy become less negative.

The order is only apparently changed for $\ce{Cl-}$ and $\ce{F-}$ due to fluorine small size. The additional electron would be added to a "small" $\ce{2p}$ orbital in fluorine (and there would be more electron-electron repulsion) while in the chlorine atom it would be added to a "big" $\ce{3p}$ orbital that can better accommodate the additional charge, so the enthalpy is less negative for the electron gain in $\ce{F}$ then for $\ce{Cl}$.


The HOH bond angle for an isolated water molecule is accurately known to be 104.5°  (104.52°±0.05° from Rotation‐Vibration Spectra of Deuterated Water Vapor J. Chem. Phys. 24, 1139).

In the liquid phase, values of 105.5° (calculated) and 106° (experimental) are reported in Structural, electronic, and bonding properties of liquid water from first principles J. Chem. Phys. 111, 3572.

The gas phase water dimer (two otherwise isolated water molecules hydrogen bonded to each other) has also been considered and an angle of 104.7° is reported in the above reference.

For ice Ih (common form of ice), values ranging from 106° to 108° have been reported.

Attempting to explain the trend, the intermolecular hydrogen bonding in the liquid and solid phases weakens the intramolecular covalent O-H bonds, and from a VSPER approach causes the electrons of the O-H bond to be more like the lone pair electrons.


Which of the following elements forms a water-soluble, alkaline oxide?

solid oxide which dissolves in water

Magnesium oxide $\ce{MgO}$ is (practically) insoluble in water.

You might be interested to know that burning of magnesium was the classical flash light in the early days of photography:  Wikipedia.


Meaning of Flammability Limits and reaction in Air of Ammonia

Chemical reaction would proceed in an air/ammonia mix, even outside the concentration range at which the reaction is self-sustaining, particularly if there were a catalyst to help. It is just that at the specified concentration range, when it reaches $651^{\circ}\mathrm{C}$ at one point in the mixture, the reaction proceeds fast enough to heat the rest of the mixture to that ignition temperature.

This is similar to an Otto cycle spark-ignition engine: with the correct mixture, the flame front propagates from the spark plug throughout the cylinder.

In the Ostwald process, a $1:10$ ration of $\ce{NH3}$ and air can sustain the reaction at the surface of a catalyst.

In a similar fashion, hot copper can catalyze the oxidation of acetone, methanol or ethanol, providing a surface where the reactants can combine fast enough to keep the surface hot.


In a suggested lab, the RSC recommends using propan-2-ol:

  If there is any unused potassium remaining at the end of the experiment, remove it with the tweezers and place it in a beaker containing about 100 cm3 of propan-2-ol to dissolve away – it will fizz, giving off bubbles of hydrogen. When the fizzing has stopped, dispose of the resulting alkaline solution down the sink flushing it away with plenty of water. The acidic solution remaining from the experiment may also be disposed of down the sink flushing it away with plenty of water.

The same is recommended for sodium.


What functional groups can 900-2500nm IR spectrometers detect?

You recall vibrations IR spectrometers detect in the range of $4000\dots\pu{800 cm^{-1}}$ typically are either fundamental vibrations and combinations.  Tuning for shorter wavelengths / higher wavenumbers however renders recording absorptions about overtones instead of fundamentals more likely.  On occasion, you may encounter this part of the electromagnetic spectrum accessed by the larger UV-Vis-NIR spectrometers (e.g., PerkinElmer Lambda series) with a spectral range from $175\dots\pu{3300 nm}$); here, NIR is the label about near infrared.
The typical overtone is less probable than the fundamental, thus -- in parlance of Beer-Lambert's law, $\mbox{Abs} = \varepsilon \cdot c \cdot d$ -- the molecular absorption coefficient $\varepsilon$ is smaller.  While this may appear on first sight as a drawback (at same analyte concentration $c$ and  pathlength $d$, this lowers $\mbox{Abs}$ recorded), it very well may become an advantage if you want to record absorption bands about matter in high concentration, e.g., to check the quality of grain and crops in general  where NIR spectroscopy is used as such.
IR spectroscopy has a long tradition of chemometrics, e.g. in the on-line process control of petroleum fractions where whole spectra are used to determine concentrations of components (e.g., keyword principal component analysis) allowing to map a sample for its chemical composition as tool in histology (a review).  Again, tuning the senors to record NIR instead of the range of $4000\dots\pu{800 cm^{-1}}$ allows to see some things literally in a different light (e.g., Landsat satellites, farming).

(credit)


How should I identify my leaving group in SN1 reactions if there might be more than one option?

This case (and similar cases) you need to think about the following:

When a leaving group leaves, only one bond is broken. By this logic, $\ce{OH}$ is a leaving group - it can leaving by the breaking of the $\ce{C} -\ce{O}$ bond. The alcohols you mention, $\ce{CH3OH}$ and $\ce{CH3CH2OH}$ require breaking two bonds to leave: the the $\ce{C} -\ce{O}$ bond and a $\ce{C} -\ce{C}$ bond.
What is that acid $\ce{H-Br}$ doing? When you are given a strong acid as a reagent, very commonly the first step in the reaction's mechanism is a proton transfer (an acid-base reaction). Proton transfers are very rapid; they often happen before anything else has a chance to. In this case, the proton transfer is also exothermic (or exergonic if you prefer) - the equilibrium constant is on the order of $10^8$ in favor of the products. What does the product of the initial proton transfer look like? Is there a better looking leaving group on that structure?


Benzyl and allyl cations are stabilised by the neighbouring $\unicode[Times]{x3C0}$-systems. The electrons in the $\unicode[Times]{x3C0}$-system are actually required for (formally) form double bonds to ensure all-octet structures as well as possible. The cationic structures will contain an electron sextet somewhere (most often drawn on the benzylic carbon for the benzyl cation).

The $\unicode[Times]{x3B1}$-oxygen and $\unicode[Times]{x3B1}$-nitrogen systems on the other hand contain free lone pairs on the heteroatoms. These are not used for double bonds, so they can be shared with the cation to form a new double bond. The resulting mesomeric structure is an all-octet structure which are usually much more stable than non-octet structures (at least for main group elements).

The nitrogen-containing carbocation could even, after the double bond has been formed, loose its proton to create a neutral all-octet structure, the formal $\ce{HCl}$ elimination product.


The short answer: no. We do not know how to solve the Schrodinger equation analytically, even for systems as simple as the He atom.

The much longer answer:

The Schrodinger equation is the basis for almost all theoretical chemistry: chemistry that attempts to predict properties of compounds without an actual experiment. The problem is that the Schrodinger equation is unsolvable for any compound with more than two electrons, and so we must make some approximations to solve it. This leads to methods such as Hartree-Fock and Density Functional Theory.

While these approximations are usually pretty good, they are still only approximations, and sometimes stop working correctly. There is an added problem that the Schrodinger equation is only an approximation, and the proper theory to use to describe light-matter interaction is Quantum Electrodynamics, which is incredibly difficult to solve for a single photon reflecting off of a mirror.

Long story short: there may or may not be a "unified theory" for describing exactly how all bonds behave in every situation, but there are currently methods that are good enough to predict a large portion of the behavior that we observe.


Electrophilic substitution in meta-nitrophenol

Electrophilic substitution can happen at only that place where electrons are present which can be donated/shared with some other electron-deficient species. In the resonance structures of m-nitrophenol shown above , there is a negative charge on hydrogen or carbon(because charge keeps on shifting between carbon and Hydrogen) and they are not that electronegative to handle the negative charge . So they want to stabilize themselves by giving off their electrons to electron deficient species by letting electrophilic substitution happen. You can see lone pair (negative charge ) of OH group  is not delocalised by NO2 group if it is present in meta position . Although NO2 group is still in Conjugation with benzene ring but for OH group, NO2 can show nothing but only -Inductive Effect.
So , now after making changes in my answer , In my opinion , electrophilic substitution will happen on left side of the molecule at ortho position. On the right side at ortho position , electron density is less as compared to the left side because NO2 is showing -I effect making conditions less suitable for electrophilic substitution to happen.


Does CN4 with the following structure exist?

All kinds of weird molecules exist in space. The molecule is formally known as bis(diazo)methane according to chemdraw. By looking at the structure it contains a Diazomethane moiety. Resonance would also take place to introduce the charges at both nitrogen atoms.  The structure doesn't seem to exist on any chemical database nor any official papers have been released relating to the exact structure. It may exist considering the fact that all atoms have a total of 8 surrounding electrons.
I hope this provides some clarity


Sensitivity of Carbon NMR Compared to Proton NMR

The receptivity ratio is ca. $5800:1$, as listed in the Bruker tables (under the column Receptivity > Natural rel. $\ce{^13C}$). Receptivity is a product of the natural abundance and an "inherent NMR sensitivity", which scales as $\gamma^3 I(I+1)$. For proton and carbon, $I = 1/2$, so the $I(I+1)$ term cancels out and we are left with

$$\begin{align}
\frac{\text{receptivity }\ce{^1H}}{\text{receptivity }\ce{^13C}} &= \frac{\text{abundance }\ce{^1H}}{\text{abundance }\ce{^13C}}\cdot \left(\frac{\gamma_\ce{H}}{\gamma_{C}}\right)^3 \\
&= \frac{100\%}{1.1\%} \cdot (4)^3 \\
&\approx 5800.
\end{align}$$

Why $\gamma^3$? One must bear in mind that at the end of the day, the signal that is detected is not the magnetisation directly. The precessing magnetisation that is excited in an NMR experiment induces a voltage in the coil that surrounds the sample. This voltage is the actual quantity that is detected.

By Faraday's Law, the voltage generated is proportional to the rate of precession, i.e. the Larmor frequency $\omega_0$, as well as the actual magnitude of the magnetisation $M$. The Larmor frequency is itself proportional to $\gamma$ (it is defined by $\omega_0 = -\gamma B_0$), so that accounts for one out of three.

Any magnetisation that we can measure must be derived from the equilibrium magnetisation $M_0$. Since the equilibrium magnetisation points along the $z$-axis, $M_0$ is the expectation value of $z$-magnetisation $\langle M_z \rangle$ at equilibrium. $\langle M_z \rangle$ itself can be related to the populations of the spin-up and spin-down states, conventionally denoted as $n_\alpha$ and $n_\beta$:*

$$\begin{align}
\langle M_z \rangle &= \gamma \langle I_z \rangle \\
&= \gamma \left[ \left(\frac{\hbar}{2}\right)n_\alpha + \left(-\frac{\hbar}{2}\right)n_\beta \right] \\
&= \frac{\hbar\gamma}{2}(n_\alpha - n_\beta)
\end{align}$$

So that accounts for one more occurrence of $\gamma$. The third and final one comes from the population difference $(n_\alpha - n_\beta)$. According to the Boltzmann distribution, we have

$$\begin{align}
\frac{n_\alpha}{n_\beta} &= \exp\left(\frac{\Delta E}{k_\mathrm{B}T}\right) \\
&\approx 1 + \frac{\Delta E}{k_\mathrm{B}T}
\end{align}$$

where $\Delta E = E_\beta - E_\alpha$ is the (positive) energy difference between the $\alpha$ and $\beta$ states. In the second step we have expanded the exponential as a Taylor series and truncated at first-order, making the (in this context, very valid) assumption that $\Delta E \ll k_\mathrm{B}T$. So now:

$$\begin{align}
n_\alpha - n_\beta &= \left(1 + \frac{\Delta E}{k_\mathrm{B}T}\right) n_\beta - n_\beta \\
&= \frac{\Delta E}{k_\mathrm{B}T} \cdot n_\beta
\end{align}$$

The energy of the spin states in an external magnetic field (the Zeeman splitting) will depend on the gyromagnetic ratio:

$$\Delta E = \gamma\hbar B$$

So, working backwards: a nuclide with a smaller gyromagnetic ratio, when placed in an external magnetic field,

has a smaller difference between the population of spin-up and spin-down states (by a factor of $\gamma$);
which leads to a smaller equilibrium magnetisation (by a factor of $\gamma$);
which, when excited, generates a smaller voltage in the detection coil (by a factor of $\gamma$).

Finally, it should be noted that this does not take into account any noise in the spectrum; it is just a measure of the signal that is available. As far as I can tell, the (root-mean-square) noise also scales with the frequency, and this leads to a theoretical signal-to-noise ratio that scales as either $\gamma^{3/2}$ or $\gamma^{11/4}$ depending on the analysis made (essentially, the exponent is slightly reduced from the nominal value of 3). I don't have the ability to explain this properly, but I can provide references that are probably helpful:

Abragam, A. The Principles of Nuclear Magnetism; Clarendon Press: 1961, especially pp 78 onwards.
Hill, H. D. W.; Richards, R. E. Limits of measurement in magnetic resonance. J. Phys. E: Sci. Instrum. 1968, 1 (10), 977–983
Hoult, D. I.; Richards, R. E. The signal-to-noise ratio of the nuclear magnetic resonance experiment. J. Magn. Reson. 1976, 24 (1), 71–85

* Populations should not be interpreted as the actual number of particles in a specific eigenstate, but rather the ensemble average of the coefficient corresponding to the specific eigenstate in the wavefunction. In other words, if each spin has a state $|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, then we have $n_\alpha = \overline{c_\alpha^* c_\alpha}$ where the bar indicates an average over all spins in the ensemble. For more information please consult a suitable resource on density matrices.


How long does it take to replicate Friedrich Wohler's synthesis of urea?

OK, I am not answering the question but the author Kyros might be interested in how Professor Benjamin Silliman, Sr. (1779-1864) described Wohler's synthesis in his Elements of Chemistry (1831) (Vol II, pg.600) just 3 years after the landmark experiment.
Although Silliman conceded that the properties of the "artificial" urea had the properties of urea, he appears to have been skeptical. Wohler wrote to his mentor J. J. Berzelius in Sweden about his discovery on how he, Wohler, could make urea without the need of kidney, be they animal, human or dog. Berzelius was delighted because the reaction was an example of isomerism, a term Berzelius had coined.

Preparation. — By double decomposition, by muriate of ammonia and cyanate of silver; or by single decomposition, by cyanate of lead and liquid ammonia. In this second process, the oxide of lead is liberated and a compound is obtained, in colorless rectangular prisms. In their properties, they are identical with urea, and their composition is exactly the same; they give the same results in analysis as cyanate of ammonia, provided 1 equiv. of water is supposed to exist in that salt.
Still the artificial urea, although from the mode of its formation it would appear that it contains only cyanic acid and ammonia, yields neither, by chemical agents. Potassa does not evolve ammonia, nor do the stronger acids give carbonic acid and evolve cyanic acid, with the peculiar odor which appears when the cyanates are decomposed by a stronger acid. The artificial urea gives no precipitate with salts of lead or silver.*


What are good representatives for the concepts of endergonic and exergonic?

As you say you are starting with Gibbs free energy, just pay attention to the $\Delta G$ values. Remember that $\Delta G= \Delta H -T\Delta S$ so it doesn't depends only on enthalpy but also on entropy. A process where $\Delta G<0$ is exergonic and in the opposite case it is endergonic.

About the examples we can say that:

Salts that disolve in water are exergonic processes, $\Delta G_\mathrm{disol}<0$.
Sodium bicarbonate will disolve in vinegar to produce $\ce{CO2}$ (because of the high concentration of acetic acid), it is also exergonic.


If temperature and volume are same, the pressure of gas obeying van der Waal's equation is smaller than that of an ideal gas"

Your statement is false. There are cases where the pressure from the Van der Waals pressure is higher than the pressure from the Ideal Gas Law.

The Van der Waals equation introduces two corrections to the Ideal Gas Law:

It accounts for attractive forces between gas particles. This means that the Van der Waals pressure can be lower than the Ideal Gas pressure.
It accounts for the small but nonzero volume of the particles. This means that the Van der Waals pressure can be higher than the Ideal Gas pressure.

Depending on the amount and nature of your gas, the volume and the temperature, either one of these effects can dominate.

Qualitatively:

In a situation where gas molecules are squished together very closely, the Van der Waals equation yields:

Higher pressures for molecules that are big but not sticky, like isobutane, and
Lower pressures for molecules that are small but sticky, like hydrogen chloride.

Quantitatively:

I'm attaching a spreadsheet with a bunch of examples; pardon the German. For a number of situations described by:

substance, column A
volume (liters), column B
amount (moles), column C
temperature (°C), column D

the pressure is given according to the Ideal Gas Law (column I) and the Van der Waals equation (column J).

There are examples that follow your 'rule', and the Van der Waals pressure is lower than the Ideal Gas Law pressure; e.g., row 6.

There are examples that do not follow your 'rule', and the Van der Waals pressure is lower than the Ideal Gas Law pressure; e.g., row 11 or 22.

Keep in mind that both equations only apply to gases; look up phase diagrams and make sure the gases in question don't condense under the conditions given. Rows 12, 13, 23 are supercritical fluids.

You'll find that the Ideal Gas Law is quite precise in most cases, and the cases where it's not, are those where the gas is close to not being a gas anymore.


Preference of locants in organic nomenclature

As pointed out by Loong, the name you prefer is the correct name: 4-ethyl-5-methyloctane.

Generally we try to do the following:

Assign the smallest locant possible to the carbon bearing the highest priority functional gorup.
Assign locants to generate the lowest possible sum of the locants.

Often it is straightforward. In this case, both assignments of generate locants of 4 and 5, which sum to 9. Both place an alkyl substituent at the lower locant.

In this case, the locants are assigned so that the alphabetically first of the two substituents of equivalent priorty receives the smallest locant.

  Correct - 4-ethyl-5-methyloxctane: ethyl is alphabetically before methyl and should get a smaller locant

  Incorrent - 5-ethyl-4-methyloctane: ethyl is alphabetically before methyl and should get a smaller locant


Is identifying a chemical equivalent to be able to produce it?

No, the capability to identify a compound does not imply capability to synthesize that compound. Indeed, "analysis" and "synthesis" represent two branches of chemistry, and many chemists spend years trying to synthesize compounds of known structure.

Modern structure determination relies on roughly two steps. First is separation of sample mixtures into constituent parts, using various forms of chromatography. Next is analysis of chemical composition and bonding patterns using techniques such as IR spectroscopy and nuclear magnetic resonance spectroscopy, as well as mass spectrometry (MS), which weighs molecules (and their destructively ionized fragments) with extraordinary precision and dynamic range.

Advanced MS instruments can analyze very small amounts of collected samples, without special preparation, and at atmospheric pressure. Pure samples are not required for the instruments typically used in forensics and product testing (DART). When a pure sample is desired, one or more separation devices can feed the MS instrument in-line (LC-MS, GC-MS, MS-MS).

MS can measure molecular weight so precisely that the chemical formula can be deduced directly based on the fractional weights of the elements due to varying isotopic abundance. Often, a chemical formula is almost sufficient for unique identification. For example, the NIST Chemical WebBook contains only 4 entries with the same formula as the VX agent, all for VX or nearly identical compounds. The PubChem database contains only 1 (VX itself).

It is also possible to identify compounds using known chemical or physical interactions linked to a readout such as a color change. This principle is used in devices such as home pregnancy tests and lead test kits. The US military even has custom field test kits for some chemical weapons.

It should be noted that compounds introduced into an organism may be identified indirectly via metabolites found in blood or urine, even if the original compound is long gone. Again, VX agent is a suitable example.


Trying to displace iron from ferrous sulphate heptahydrate using aluminium foil but nothing has happened

You are correct: Commercially available metal foil is typically made from aluminum, even if it's commonly called 'tinfoil' sometimes. As you say, technically, tin foil is made of tin; aluminum foil is made of aluminum.

Jannis Andreska is also correct: The passivating oxide on the aluminum is likely preventing the iron ions from reaching the $\ce{Al^0}$ and swiping their electrons.

So, you are correct again: The yellow/orange color you're observing is probably oxidation of the $\ce{Fe^{2+}}$ to $\ce{Fe^{3+}}$, by $\ce{O2}$ in the air.

One possible approach to breaking the oxide would be to connect the foil to the negative terminal of, say, a AA battery, with the positive terminal connected to a junk stainless steel spoon, also submerged.  That might not be enough voltage to break down the oxide film, though; on the other hand, a higher voltage might lead to corrosion of the spoon and the release of iron and other metals into your solution.  May or may not be a problem for your experiment?

Regardless, I would recommend adding some table salt $\left(\ce{NaCl}\right)$ to your solution, both to increase its conductivity and also to introduce chloride, which may help to attack the oxide on the aluminum, in tandem with the applied potential.


Comparing adamantane carbon types for accepting OH using MM ForceField

One thing to try would be just to build the two isomers and compare their total energies .. that would at least show which position was preferred thermodynamically (i.e. more stable).  It sounds like you are more interested in the kinetic consideration though (i.e. which reacts faster) ... in that case, looking at the relative energies of the transition states will give you some basis for rationalizing the preference.  I am not sure how good MM is at dealing with transition states, however ... you would need to rely rather heavily on your chemical intuition to find the correct structures, and even then, the bond angles and bond lengths may be distorted to such an extent that your forcefield is not well parameterized to deal with them.


Kinetic vs thermodynamic control of HBr to 1,3-Butadiene

A transition state does not ‘represent the energies of […] reactants and […] products’. A transition state is a certain arrangement of atoms somewhere between the most stable atom arrangement the reactants can have and the most stable atom arrangement the products can have. The key characteristic of a transition state is that it is directly en route from reactant to product and that it is the arrangement with the highest internal energy.

For any arrangement of atoms you can calculate a single point energy: reactants, products and the transition state. The energy difference between reactants and the transition state is colloquially known as activiation energy although that may not always be correct. It is, however, important to realise that each arrangement has its own energy that cannot a priori be deduced from any other arrangement. Thus, a transition state’s energy is not directly related to the energies of reactants or products (with the exception that by definition it must be higher in energy than both).

While it is a general rule that the transition state leading to the thermodynamically more stable products has a lower energy than any other transition state, this is not true for all cases. Indeed, for any reaction in which one can discuss kinetic (lower energy transition state) versus thermodynamic (lower energy products) control, the transition state energies must be ‘reversed’ with respect to the product energies.


Why does a carbonyl group form instead of alkene in alcohol oxidation?

Technically, it would be possible for an entire chromate group $\ce{CrO4^2-}$ to be eliminated along with a β-hydrogen to give a double bond. However, that elimination would have to proceed via either an $\mathrm{E2}$ or an $\mathrm{E1_{cb}}$ mechanism as the chromium(VI) centre by itself is much more happy being reduced to chromate(IV) (the intermediate chromium species) in the electrocyclic breakdown of the complex you have been taught.

The entire reaction proceeds in acidic conditions, so any conjugate base mechanism is basically impossible. And likewise, there is no sufficiently strong base present that would allow an $\mathrm{E2}$ mechanism to occur, either.


How much CO₂ should one expect in an IR spectrum?

From what I've read (and personal experience with IRs I've run), a relatively intense $\ce{CO2}$ asymmetric bond stretching signal at approximately $\ce{2350cm^{-1}}$ is an extremely common background artifact in IR spectra (along with the broad, weak signals downfield of $\ce{3000cm^{-1}}$ attributable to $\ce{O-H}$ bonds from ambient moisture). Depending on your software, it's often possible to run a "dry" IR reading with no sample present in order to establish a baseline spectrum of the surrounding air, which can then be checked against by your software (so that overlapping signals from the baseline air spectrum are deleted from the spectrum of your sample). Alternatively, you can take measures to dry the optics (I've once used $\ce{N2}$ for that purpose; alternatively, air directed through $\ce{CaCl2}$ drying tubes could possibly be used, but I can't attest to the efficacy of that). That manual drying procedure takes some effort, and might be futile in any case depending on the conditions of the lab. My one experience with that approach was disappointing.


Why would two isomers have different solubilities in water?

In the most general terms: Molecules consist of nuclei and a surrounding electron cloud. That cloud is dependent on molecular charge and geometry - so it will be different for two different isomers. It follows that its interaction with the nuclei and electron cloud of a third molecule, say, water, will be different.

More specifically, one will find that properties such as

polarizability (related to van-der-Waals forces)
bond dipoles (related to H-bonds and static dipole interactions)
effective electronegativity (related to H-bonds and static dipole interactions)

On a very basic level, any solvation involves the formation of solvent shells. When varying the shape and surface of the solute, those shells will change as well. One will then find shells that work well and others that work less well, influencing the solubility.


How to choose proper active space in method such as CASPT2 and CASSCF?

Choosing the active space in any calculation with a multiconfigurational reference (CASSCF, RASSCF, GASSCF, CASPT2, NEVPT2, MRCI, the list goes on) requires more chemical intuition than more commonly-used "black box" quantum chemistry methods, such as DFT. Here, chemical intuition means preexisting knowledge about 1. the system under investigation, and 2. the properties being calculated. For example, a minimal active space may be sufficient to qualitatively capture static correlation in a system, but the active space might change when targeting excitation energies, and might be different again when calculating spin-orbit coupling constants. More specifically, it is understanding what orbitals are most important in composing your wavefunction and have the largest influence on molecular properties of interest. Unfortunately, this can also involve quite a bit of trial and error.

Some very general recommendations on the workflow for performing multiconfigurational calculations are in the numbered list in this answer (yes, a shameless plug).

Here is a more complete list of references I've found useful in my own research:

[1], [2] are slides from the group that develops GAMESS, one of the premier packages for multiconfigurational calculations. The second set of slides in particular has examples that highlight the chemical motivation for different active spaces depending on the molecule and the property (internal rotation in ethylene, internal rotation in $\ce{H2C=NH}$, dissociation of $\ce{H2C=O} \rightarrow \ce{CH2} + \ce{O}$, isomerization of bicyclobutane to 1,3-butadiene, and others).
[3] (paywalled) is a more formal overview of the above. Excellent.
[4] is a set of very detailed and theory-focused slides from the original authors of DALTON. I link to them only for completeness; there aren't many practical examples here.
[5] (paywalled) is a comprehensive review article from the authors of COLUMBUS, with many examples. Excellent.
[6] (paywalled) is an article from the authors of COLUMBUS on the application of multiconfigurational methods to geometries of small molecules. It has good explanations on the choice of active space based on electronic configurations.

In addition, there are many other online resources that are more tutorial oriented: [7], [8], [9], [10]. These may appear to be program-specific, but the chemistry is transferable between different programs; only small technical details in how to set up calculations should be different.

Finally, some programs have nice examples or helpful hints in their documentation. DALTON (PDF), Molcas, and ORCA (PDF) come to mind.

References

[3]: Schmidt, M. W.; Gordon, M. S. The construction and interpretation of MCSCF wavefunctions. Annual Review Of Physical Chemistry 1998, 49, 233-266. DOI: 10.1146/annurev.physchem.49.1.233
[5]: Szalay, P. G.; Müller, T.; Gidofalvi, G.; Lischka, H.; Shepard, R. Multiconfiguration Self-Consistent Field and Multireference Configuration Interaction Methods and Applications. Chemical Reviews 2012, 112, 108-181. DOI: 10.1021/cr200137a
[6]: Shepard, R.; Kedziora, G. S.; Lischka, H.; Shavitt, I.; Müller, T.; Szalay, P. G.; Kállay, M.; Seth, M. The accuracy of molecular bond lengths computed by multireference electronic structure methods. Chemical Physics 2008, 349, 37-57. DOI: 10.1016/j.chemphys.2008.03.009


Red Cabbage Indicator colour change with Sodium Metabisulphite

http://mylespower.co.uk/2012/04/06/homemade-ph-indicator/

Now we add the reducing agent (model compound).

That lessens conjugation and bleaches the anthocyanine.  "But the colour it currently is is very pale" If the new compounds' residual optical absorbances are in the green, the solution will be iodine purple by transmission.


Flipping a molecule vs. mirror behind compound

If you place a mirror behind the compound and look into the mirror,
  you get the second drawing

"If you place a mirror behind compound 1 and look into the mirror" - If by that you mean, generate the mirror image of compound 1, well OK, but that does not produce compound 2. It generates the molecule labeled "mirror image" in the following diagram, and it is a non-superimposable mirror image.


Will acetone damage/attack anti UV coating on glasses?

Without knowing the material your glasses are made of, I would discourage cleaning glasses with acetone.

Often, glasses use no longer use glass as lens material -- polymers like polycarbonate represent an alternative, especially if a high degree of correction is intended, or simply because of weight concerns.  Polycarbonates however withstand ketones (and acetone is an example of ketones) poorly (listing).  As a result, contact with this solvent may blind your glasses permanently.  Happens to safety goggles worn in the chem lab, too -- yet better the goggles, than damage to the eyes.

In addition, what is the material of the frame?  Hence, I recommend a mild detergent; and if this is not working, consult your optician -- they have an ultrasound bath.


Regioselectivity in nitration of a 1,2-disubstituted aromatic compound

The molecule is symmetrical so the two positions on the far side of the alkyl groups are exactly the same. Similarly, the two positions closest to the alkyl groups are the same. For the far positions, each one is para with respect to one alkyl group and meta with respect to the other. This is not a contradiction because ortho, meta and para are purely relative terms.

I suspect that you are right that substitution will mostly occur at the far sites. The directing effect of the alkyl groups is essentially the same for every site because they are all ortho or para to one group and meta to another. Therefore, the steric effect of the bulky methyl groups adjacent to the near positions will disfavour substitution there.


How and why are all the substituents in methyl cyclopropane eclipsed?

The geometry of the molecule is highly restricted by the presence of the ring.

The presence of the ring caused the $\ce{C-C-C}$ ring to be planar, because every triangle is planar.

Since there is only one carbon atom directly to the right to the circle, it must be eclipsed, so are the hydrogen atoms by VSEPR.


Zinc in a single displacement reaction with hydrochloric acid

The zinc replaces the hydrogen in that is reduces (gives electrons to) $\ce{H^+}$, forming $\ce{H2(g)}$ and $\ce{Zn^2+(aq)}$ according to this reaction: $$\ce{Zn(s) + 2H^+(aq) -> Zn^2+ + H2(g)}$$ This reaction relies on the prescence of $\ce{H^+}$.

The hydrogen in acids exists as $\ce{H^+}$. Acids are also ionic compounds that involve hydrogen as the donor of electrons. For example, hydrogen gas and chlorine gas react to make hydrogen chloride: $$\ce{H2 -> 2H^+ + 2e^-}$$ $$\ce{Cl2 + 2e^- -> 2Cl^-}$$ $$\ce{2H^+ + Cl^- -> 2HCl}$$  The reverse of this happens when the hydrogen chloride is dissolved into water.  this forms hydrochloric acid.The compound breaks apart into hydrogen and chlorine ions:$$\ce{H2O + HCl -> H+ + Cl-}$$ Hydrochloric acid completely dissociates, meaning that every $\ce{HCl}$ molecule yields one $\ce{H^+}$ ion.

Essentialy, acids increase the amount of $\ce{H^+}$ in solution. Normally, in pure water, concentrations of $\ce{H^+}$ are very low. The hydrogen in $\ce{H2O}$ is covalently bonded to the oxygen, so it is difficult for the hydrogen to react. In order to liberate a hydrogen ion from $\ce{H2O}$, this must occur: $$\ce{H2O -> H^+ + OH^-}$$
This reaction happens infrequently in water, so concentrations of $\ce{H^+}$ are very low. Therefore, zinc barely replaces hydrogen in water.


I believe the concept assessed here is the idea of aromaticity. You must first be able to recognise that the lone pair held by each carbon atom with the negative formal charge is in a different type of orbital. In the carbon atom of the 6-membered ring, it is in an $\ce {sp^2}$ hybrid orbital that is in a plane perpendicular to the $\pi$ electron cloud of the indene molecule. This lone pair is not delocalised into the indene $\pi$ system. On the other hand, in the carbon atom of the 5-membered ring, it is in an unhybridised p orbital that is parallel to the other p orbitals of the indene molecule. This lone pair is, in fact, delocalised into the indene $\pi$ system. As such, it is much less available for donation to  an acid.

Protonation at sites C and D would also be unfavourable due to destruction of aromatic character. Between sites A and B, I have explained above that the lone pair at A would be much more available for donation due to lack of delocalisation in to the $\pi$ system. Hence, the preferred site of protonation would be site A.


Electolysis of H2O to produce O2(g), in a small, portable unit

In scuba diving the main issue is pressure. To be able to breathe, you need to counteract the pressure of the column of water on your rib cage. This is why divers breathe from compressed gas tanks. Otherwise their rib cage and lungs would be crushed by the pressure. At higher pressures, air becomes too viscous to be breathed without too much effort, and the build-up of dissolved nitrogen in the tissues leads to narcosis ("being drunk"), so the oxygen has to be mixed with helium, that is much less viscous. So, with theses 'artificial gills' or with electrolysis, you still need a tank of helium, and a powerful compressor.

Overall, it is much more convenient to carry tanks of compressed mixtures of gases. Compressing gases is a form of energy storage (that could even power cars). But if you are in a nuclear-powered submarine, then you would be better off doing electrolysis.


The two compound names that you are asking about refer to different compounds. Phenol is numbered as follows:

The structures of 2-chloro-5-methylphenol and 5-chloro-2-methyl phenol are below.

Perhaps you were confused about the 2,6 or 3,5 isomers? In this case, the chloro- substituent should get the lower locant number because it is alphabetically first.


Why is dimethyloxonium cation stronger acid than hydronium?

A powerful but often overlooked factor in determining a substance's acidity and basicity is the effect of the solvent (or lack thereof) in which it is dissolved.

You are not wrong; in the gas phase, $\ce{H_3O^{+}}$ is indeed a stronger acid than $\ce{(CH_3)_2OH^{+}}$, by a factor of about $5\times 10^{18}$. This can be seen by comparing the proton affinities of the conjugated bases, dimethyl ether ($\mathrm{804\ kJ\ mol^{-1}}$) and water ($\mathrm{697\ kJ\ mol^{-1}}$), respectively.

In other words, a mole of gaseous dimethyl ether reacting with a mole of gaseous protons releases $\mathrm{804\ kJ}$, while a mole of gaseous water releases only $\mathrm{697\ kJ}$. This means that in the gas phase, water is a weaker base than dimethyl ether, which implies that protonated water is a stronger acid than protonated dimethyl ether.

Why does the tendency invert in aqueous solution? Water can form hydrogen bonds, a very strong type of molecular interaction. $\ce{H_3O^{+}}$ has three protons available for hydrogen bond donation, whereas $\ce{(CH_3)_2OH^{+}}$ only has one. This means that the solvating water stabilises $\ce{H_3O^{+}}$ more than $\ce{(CH_3)_2OH^{+}}$. Higher stability means reduced acidity, and the difference happens to be enough that the relative order of acidity in water is the opposite of that in the gas phase.


Does a zinc metal coating react with coffee in a coffee grinder?

If you plate the pieces with zinc or silver, the main source of zinc/silver in the ground coffee will be from abrasion between the parts and the coffee beans and not from chemical attack.

A few milligrams of zinc or silver (or iron for that matter) won't to do any harm to your health, specially if diluted in many coffee cups along several months or years.

In fact, Zinc was used therapeutically in dosages upwards of hundreds of milligrams per day for months, or more, with no permanent health issues.

Silver, on the other hand, has much lower acceptable exposure limits and while not particularly toxic, I'd not put it in parts of the device that are guaranteed to put metallic silver into the ground beans.


When you manipulate the equation to solve for t you'll want to subtract 1/[A]0 from 1/[A] and then divide by k.


Comparison of basic strength of benzylamine and ammonia

Everything you said is correct, except that it is more relevant to consider the inductive effect of methyl on nitrogen: electron density is donated to nitrogen.

The inductive effect of phenyl is on the methyl, decreasing its ability to donate electron density to nitrogen, but apparently not completely.

I suppose that if you increased the negative inductive effect of the phenyl group with a couple of nitro groups, the basicity of the compound would decrease.


Is elastomeric better with more polymer, or with added calcium carbonate?

Stuff like calcium carbonate is generally used as a cost cutter, as it is far less expensive than even the cheapest elastomer (polymer). It allows the producer of the formulation to make a profit/make more profit, essentially.

Generally speaking, adding calcium carbonate to an elastomeric formulation will decrease the useful properties of said formulation, including its resistance to weathering. It's resistance to wear and tear (as represented by mechanical properties like tensile strength and abrasion resistance) will also suffer. Calcium carbonate containing formulations, all other things being equal, will also be less elastic and somewhat stiffer.

A polymer that has poor resistance to weathering (like SBR rubber) will generally also not benefit from calcium carbonate as a formulation ingredient. Instead, special anti-oxydant/anti-ozonants will improve this cheap polymer's fight against the elements.

But if a composition does contain significant amounts of calcium carbonate it is not automatically disqualified as suitable: that will largely depend on the end-user's expectations. In other words: cheap can work well too.


As Mithoron says in the comments, tert-butyllithium is one choice.  Like pure organolithium compounds generally, this is pyrophoric and reactive to many common atmospheric gases or vapors. An article by Schwindeman et al. [1] recommends using a nitrogen or argon atmosphere.

In the case of alkaline earth metals, magnesocene provides a relatively volatile magnesium compound, subliming at $\pu{100 °C}$.  This is also air- and moisture-sensitive and again, should be handled under nitrogen or argon.

References

Schwindeman, J. A.; Woltermann, C. J.; Letchford, R. J. Safe Handling of Organolithium Compounds in the Laboratory. Chemical Health and Safety 2002, 9 (3), 6–11. https://doi.org/10/ctj2sj. (PDF mirror: chemistry.ucla.edu)


Database for experimental diatomic ionization energies

NIST Atomic Spectra Database has the atom part. Not sure about molecules.
https://physics.nist.gov/PhysRefData/ASD/ionEnergy.html


There is no anomaly in here. Fluorine’s oxidation state in $\ce{HOF}$ is $\mathrm{-I}$ as the theory says. Hydrogen’s is $\mathrm{+I}$. This leaves oxygen with an oxidation state of $\mathrm{\pm 0}$.

That final fact is where the oxidative power comes from. Oxygen, the second-most electronegative element in the periodic table, has the same oxidation state as in molecular form. It is also bonded to an even more electronegative fluorine. Even in hydrogen peroxide oxygen would me more reduced. Thus, any reaction of $\ce{HOF}$ involves oxygen gaining electrons, not fluorine.

Remember that many sources on the internet can be wrong.


Gay-Lussac's Law, in its most basic form, says that, assuming constant volume (V) and amount/moles (n), the pressure (P) of a gas is directly proportional to the absolute temperature (T) of a gas.  We can write this mathematically in a couple different ways:

$$ P = kT$$
 (assuming constant V and n), where k is some constant.  More often, we use it in the form below, where we know the pressure and temperature under one set of conditions, and then change those conditions (again, keeping V and n constant):

$$
\frac{P_1}{T_1} = \frac{P_2}{T_2}
$$

As for the next statement, "all gases have the same amount of particles", as stated this makes no sense.  Perhaps what you meant to say was that, under identical conditions, equal volumes of an ideal gas contain equal number of moles (and therefore equal numbers of gas particles).  For example, it can easily be shown using the Ideal Gas Law that one mole of an ideal gas occupies 22.4 L at STP.


What is "stuffy"/"stale" air composed of?

It may not be a complete answer but I can think of a few distinguishable features that may or may not be chemical reasons:

chemicals created by bacteria and organisms that are present in everyday households (mold, light activating some production from bacteria and micro-organisms, $NH_3, S^{2-}$ related compounds (they smell bad), etc).
Compound this with the acknowledged lack of circulation and you have funky smelling air.
high humidity: how one can "feel" "stale, muggy" air.
high carbon dioxide concentration can make the air feel heavier (from interior design sources).


Why does copper sulphate (CuSO4) in water increase surface tension?

From Sayok#6589 on Discord:

  When dissolving CuSO4 in H2O, it does not remain as CuSO4. It's a soluble salt, so you make [Cu(H2O)6]2+ and [SO4]2-. The first coordination shell of either ions is filled with water, forming electrostatic/polar bonds between the water molecules and the ions.
  Because the water molecule are forming bonds with charged ions, their polarity is slightly increased compared to a regular water molecule.
  Surface tension of H2O is formed because of the polarity of the water molecules. They like each other and want to stick with each other. In general, the more polar a solution is, the more surface tension it can have.
  Increasing the polarity of the water molecules because of the presence of the charged ions in solution thus results in a small increase in the surface tension.


What would be an appropriate way to clean up equipment contaminated with ethidium bromide?

As you stated in the question, there is some controversy surrounding the mutagenicity of ethidium bromide. However, a recently published opinion piece titled "The Myth of Ethidium Bromide" quickly dispels most of the uncertainty:

  Ethidium bromide, as far as can be told from the data, is not a human mutagen. It’s not a mouse mutagen or rat mutagen either. Nor apparently a mutagen in cows and other farm animals, where it’s used in veterinary medicine at concentrations one thousand times higher than the red solutions that are so feared in biology labs, seemingly with no bad effects.

This knowledge does however not give us clear directions as to how the clean-up and decontamination should proceed. For this we can turn to the Material Safety Data Sheet (MSDS).
The relevant sections are Hazards Identification, Handling and Storage, and Exposure Controls / Personal Protection, numbered 2, 7, and 8, respectively.

Hazards Identification tells us that it's fatal when inhaled (in its solid form). This means that when it's not in solution, you should be wearing a respiratory mask or have a fume hood panel separate you from the dust. The fume hood should be operational, of course.

Handling and Storage states that you should:

Avoid contact with skin and eyes.
Avoid formation of dust and aerosols.
Provide appropriate exhaust ventilation at places where dust is formed.

Finally, Personal Protection informs us that you should be wearing a lab coat and glasses (duh), but also gives us some data on the breakthrough time of the substance through nitrile gloves, which is 480 minutes. There is also a section on respiratory protection, but it is prefaced with "Where risk assessment shows [respirators] are appropriate," which is something that you must decide for yourself.

Thus, the answer is, as almost always, it depends:
In any case you will want to avoid brushing and dusting off the contaminated surface to reduce the formation of ethidium bromide dust. Personally, I would recommend to wipe all surfaces first with water (according to Wikipedia the solubility is 40 g / L) until clean using paper towels (and gloves, of course). Following that, one final wipe-down with ethanol may be appropriate, but that is left for you to decide.


Turning the amino group into an amide serves two purposes:

We are introducing a sterically bulky group. We expect the amide bond to be as complanar as possible with respect to the benzene ring to maximise electronic interactions. That should block one of the two ortho positions sterically.
We are turning an electron-rich, $+M$ aromatic system into an electron-poor, $-I$ one.

Especially the second transformation will greatly reduce the reaction rate. Furthermore, a bromine atom is mildly deactivating, too, due to its $-I$ effect (the weak $+M$ effect is neglegible and only responsible for the ortho/para directing abilities). Thus, it is very likely that we can selectively stop the reaction after monobromination.

The question still arises why we only obtain the para-product. Obviously the amide behaves much like a bromide and displays a weak $+M$ effect in spite of the amide resonance. Yet we can imagine the the proximity to the bulky amide group inhibits substitution at both ortho-protons sufficiently to allow isolation of the para-product.


Why does heating transform the aquacomples of slightly acidified cobalt(II) chloride solution to the chlorido complex?

You may know the formula for Gibbs Free Energy $\Delta G$. This is taught as:
$$\Delta G = \Delta H - T\Delta S\tag{1}$$

In this case, we would be observing $\Delta G$ for equilibrium $(2)$.

$$\ce{[Co(H2O)6]^2+ + 4 Cl- <=> [CoCl4]^2- + 6 H2O}\tag{2}$$

Within the temperature range between $\pu{25^\circ C}$ and $\pu{100^\circ C}$, we should not expect the enthalpic term $\Delta H$ to change significantly. In fact, it is usually correct at first or even second approximation to consider it constant. If we do, the only thing in $\Delta G$ that changes with temperature in equation $(1)$ is the $-T\Delta S$ term. It basically implies that a higher temperature will increase the entropic contribution.

You already noted that the product side is favoured entropically since five free-moving particles are turned into seven in the course of the reaction. Higher temperatures should thus favour the product side.

One more thing: since we have established that $\Delta S$ is positive, we can immediately see that $\Delta H$ must be positive, too; i.e. the reaction must be endothermic. This is because $T$ is positive by definition and therefore $-T\Delta S$ is always negative. However, the only way to tip an equilibrium to the opposite side is to change the sign of $\Delta G$ — which can only work if a positive $\Delta H$ can offset a negative $-T\Delta S$.


Stability of carbocation is directly proportional to +M effect .
So ,
 carbocation will more stable on A then on B and then C as NO2 has high -M. Effect it's last..


How do we predict the properties of the materials we make through chemical reactions?

Well, you've described one of the goals of computational chemistry. It turns out that this problem is harder than intuition might predict. I recently read an article on the state of quantum computing that happens to summarize the issue for chemists nicely.

  We do not have enough computing power to do this well.

Computational quantum chemical models have become more and more sophisticated to take advantage of increasing computing power, but there still is not enough computing power. Part of the issue is that at the center of what we are trying to do is a multibody problem, which means that any attempts to optimize the electronic state for one electron in a molecule perturbs the state for other electrons. Then we go and adjust those other electrons, which perturbs our first electrons. This is a certified Really Hard ProblemTM.

At the end of the day, we currently cannot determine exact solutions to multielectron wavefunctions. So approximations get made, and the approximations are getting better. In fact for smallish molecules, we are getting quite close to a variety of known experimental properties. However, for larger molecules, the algorithms still need trained and calibrated on real experimental data. As the molecules get bigger, or the materials being modeled become more complex, our ability to throw quantum chemical calculations at them diminishes. Modeling of proteins mostly uses molecular mechanics and statistical dynamics.


How to interpret a singlet term symbol for an s¹p¹ configuration?

The two states you have drawn correspond to the (antisymmetrised) wavefunctions

$$\begin{align}
\Psi_i &= \mathrm{s}(1)\alpha(1) \cdot \mathrm{p}(2)\beta(2) \\
\Psi_j &= \mathrm{s}(1)\beta(1) \cdot \mathrm{p}(2)\alpha(2)
\end{align}$$

Neither of the two states above are valid, as they do not obey the requirement for indistinguishability. For fermions such as electrons (spin-1/2), this is achieved by making the wavefunction antisymmetric upon permutation of any two electron labels. However, neither $\Psi_i$ nor $\Psi_j$ obey this:

$$\begin{align}
\mathcal{P}(\Psi_i) &= \mathrm{s}(2)\alpha(2) \cdot \mathrm{p}(1)\beta(1)  \neq -\Psi_i \\
\mathcal{P}(\Psi_j) &= \mathrm{s}(2)\beta(2) \cdot \mathrm{p}(1)\alpha(1)  \neq -\Psi_j
\end{align}$$

The appropriate wavefunction is a linear combination of these states (and a couple more):

$$\Psi_\text{singlet} = \frac{1}{\sqrt 2}[\mathrm{s}(1)\mathrm{p}(2) + \mathrm{p}(1)\mathrm{s}(2)] \cdot \frac{1}{\sqrt 2}[\alpha(1)\beta(2) - \beta(1)\alpha(2)]$$

If you expand these you can see the constituent wavefunctions

$$\begin{align}
\Psi_\text{singlet} &= \frac{1}{2} [\mathrm{s}(1)\alpha(1) \cdot \mathrm{p}(2)\beta(2) \\
&\qquad + \mathrm{p}(1)\alpha(1) \cdot \mathrm{s}(2)\beta(2) \\
&\qquad - \mathrm{s}(1)\beta(1) \cdot \mathrm{p}(2)\alpha(2) \\
&\qquad - \mathrm{p}(1)\beta(1) \cdot \mathrm{s}(2)\alpha(2)] \\
\end{align}$$

You need four "simple" wavefunctions in total because you have to account for the spatial indistinguishability of the two electrons (hence need a linear combination of two spatial wavefunctions) as well as the spin indistinguishability (linear combination of two spin wavefunctions).