Why is there a discrepancy among sources on the atomic radius of some elements?

Atomic radii are not really well-defined. There is a great deal of uncertainty associated with them, mainly due to the somewhat philosophical question ‘where does an atom end?’ Orbitals, in theory, extend into infinity for all (mathematical) intents and purposes, yet the probability of locating an electron in Far, Far Away is zero by all sensible rounding methods. However, the decrease is exponential, so you need to define a cutoff value if you want to know where an orbital ends to draw the beautiful pictures you probably have seen before.

This only gets worse when considering entire atoms rather than orbitals. We typically consider atoms as spherical (hence radius). But for every element except hydrogen, helium, lithium and beryllium (possibly including boron, too) there is at least one additional non-spherical orbital occupied. Can we still truly define an atom as a sphere? And even if we can, consider the case of transition metals; should we be talking about some cutoff value for the outermost s orbital or about one for the lower shell’s d subshell, whose electrons are considered valence electrons?

Now in theory, you could ignore all those philosophical discussions and just take the solid-state structure of an element, measure the shortest distance between two nuclei, divide it by 2 and call that the atomic radius. Well, that is what is done. However, many elements have different structures that they adopt or can adopt at room temperature and standard pressure. For other, especially non-metals, the shortest interatomic distance is not an atom-atom contact but an interatom bond. In a select few, the interatomic bond can have different bond order values; sixtytrees noted carbon whose most stable phase is graphite (bond order 1.5) but which also features a diamond allotrope with a bond order of 1.

The covalence problem can be overcome with a set of correction factors that introduce further uncertainties and are added differently by different sources.

All these issues mean that there is not one ‘true’ way to determining atomic radii; and since the methods differ, so will the results. It is only natural for some elements to be larger than others in one scale but smaller in another.

The whole problem gets infinitely worse with ionic radii. Ionic compounds exist as crystals with anions and cations — but where to the cations end and the anions begin? How do we deal with the fact that no bond is truly ionic and no bond between two different elements truly covalent?

Even if you determine e.g. the ionic radii of $\ce{Na+}$ and $\ce{Cl-}$ by various methods, (e.g. from comparing sodium fluoride, sodium carbonate, magnesium chloride, ammonium chloride …) and then add the two values together to arrive at an interionic distance in the $\ce{NaCl}$ crystal, it is very likely that your result is off.


Why is charcoal useful as a fuel compared to wood? Don't we lose energy by making it?

As I understand it, charcoal --mostly carbon, plus some amount of inorganic ash-- burns cleaner - and hotter - than wood, and has more energy content per unit of weight than wood. The process of making charcoal involves heating the wood in the absence of air or with limited contact with air - driving off water and low molecular weight oils and odor components that produce a lot of the smoke from burning wood, and converting cellulose (a major component of wood) into carbon plus water.


What is a "tri-salt polymer" as described by this product that eliminates and prevents mold?

Beside the three-step deprotonation  of (ortho) phosphoric acid, $\ce{H3PO4}$, along

$\ce{H3PO4 <=> H2PO4^- + H+ <=> HPO4^{2-} + 2H+  <=> PO4^{3-} + 3H+}$

where each of the oxygen-bound hydrogen atoms may be substituted by a cation, phosphoric acid may dehydrate partially to yield polyphosphates.  Which seems, reading a related Canadian patent CA2504014 (A1) (deposit in English), the working principle here:

(loc. cit., p. 6; cations generalized)

The know-how includes tuning the proccessing as such that the composition of these blend polyphosphates is obtained reliably, since the complete dehydration simply would yield the waxy polyphosphoric acid used as acidic catalyst in the lab.


Home heating oil and motor diesel are very similar in terms of density and chain length (basically same stuff) with minor differences in sulfur content (at different eras in different countries) and with colorant for tax purposes.  [Similar to comment above.]

In general the middle distillates would have a bit more sulfur in them if just based on distillation cut.  In order to meet lower sulfur requirements for diesel, additional processing is needed in the refinery to get the sulfur out.  As the US (and other countries) started putting in stricter and stricter sulfur requirements for diesel (for pollution reasons), it has had the impact of driving diesel prices up to be more than gasoline--traditionally it was cheaper.

[Note also that, at least in the US, it is possible to buy untaxed, undyed diesel for use in farm machinery.]

Note that "RFO" or residual fuel oil is a very different beast from home heating oil.  It is much heavier and higher boiling point fraction in the column.  It also typically has MUCH more sulfur in it (from concentration, the sulfur containing compounds tend to be heavier).  RFO is used in large onshore power plants (declining in the US but still some remain and many overseas) and also in ship power plants ("bunker fuel").  Ship use is both in steam boilers and in "marine diesels".  Larger shipboard diesels have evolved to become larger and larger, competing with steam turbines, and they are designed specially so that they can use a more crappy fuel (don't need motor diesel). RFO is also often burned (for heat exchangers and/or cogen) in island refineries (that lack access to natural gas).

Note that shipboard use of RFO is expected to decline after 2020 because of new restrictions on open ocean fuels (the "IMO law").  RFO burning is already prohibited in close proximity to North American and European coastal waters. The issue is sulfur content.  Ships can continue to burn the RFO if they install onboard sulfur scrubbers, but current indications are most will just use a lower sulfur fuel.  The easiest way to do this is to use something close to a motor diesel (rather than desulfurize RFO itself).  Will have a little more sulfur than most land diesels, but still somewhat desulfurized...and just because it is a middle distillate "cut" inherently has much less sulfur than RFO (which collects the nasties when distilling).  Also there are still questions about enforcement (i.e. cheating) or even legality of regulating open ocean operations.  But overall impact will be a big drop in prices for RFO as market goes away and more pressure on middle distillate prices.  [Ships can also shift to LNG but this is even trickier than scrubbers because, in addition to engine modifications, the infrastructure to bunker LNG does not exist in most ports.  But perhaps over time, we will see migration to LNG, a zero sulfur fuel.]

A decent reference is here:  https://www.amazon.com/Petroleum-Refining-Nontechnical-Language-William/dp/1593701586 It is a few years old so won't have the IMO discussion but will give you a good overview on the whole situation and is VERY friendly in terms of plain language.  I bought it before going into the industry and worked through the whole book in a few days...totally helps with some initial grounding in things.  After that you can fill in the blanks with whatever you are doing (marketing, buying, refining) by just asking for specs.

See also: https://www.eia.gov/tools/glossary/index.php?id=petroleum EIA glossary--you may also find some other info if you scratch around that website...a lot of pages and FAQs are a little buried.


This seems like more of an engineering problem than "chemistry problem." I'm not sure how long you want the process to run, nor how critical the 10% salt concentration is.

I'd try adding a pure water bubbler in front of the salt solution to saturate the air being blown through with water.


Physical reason for why delocalisation leads to stability?

A simple QM explanation I can think of would invoke Hückel theory. It would go this way:

Suppose you have a two-state system (analogous to having two orbitals) with eigenvalues of the Hamiltonian $\lambda_1$ and $\lambda_2$. These correspond to the orbital energies. So just for argument's sake, let's say that

$$\begin{equation}
\mathbf{H} = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}
\end{equation}$$

then obviously you have $\lambda_1 = 1, \lambda_2 = 2$. Let's say this system has two electrons. They would both go into the orbital with the lower energy, i.e. $\lambda_1$. In the context of organic chemistry, this would typically mean that the first orbital would be bonding in nature (hence filled), and the second orbital would be antibonding (hence empty), although that's not always the case.

Now, you perturb the system by making the non-diagonal elements of the Hamiltonian negative (analogous to allowing the orbitals to overlap, i.e. allowing delocalisation of electron density between the two). Just for argument's sake, let's change $\mathbf{H}_{12}$ and $\mathbf{H}_{21}$ to both be $-0.1$. If you recalculate the eigenvalues you get

$$\lambda_1 = 0.99, \lambda_2 = 2.01$$

The new eigenvectors will be linear combinations of the old eigenvectors – one in-phase and one out-of-phase combination. The "bonding" or in-phase combination of the two orbitals will have decreased in energy, whereas the out-of-phase combination of the two orbitals will have an increased energy. If you put both electrons into the bonding orbital, you get a net stabilisation.

In general, interactions between filled and unfilled orbitals are stabilising in nature, since both electrons occupy a lower-energy bonding orbital.

On the other hand, interactions between two filled orbitals are destabilising, because you would have four electrons and you would have to fill both of the new orbitals. If you add up the two new eigenvalues, the total energy is greater than it was before when $\mathbf{H}$ was unperturbed.

This is not evident in my example above. The reason is that I made the assumption that $\langle i | j \rangle = \mathbf{S}_{ij} = \delta_{ij}$. In extended Huckel theory you would solve the generalised eigenvalue equation, $\mathbf{Hc} = E\mathbf{Sc}$. However with my assumption, $\mathbf{S}$ is simply the identity matrix and this reduces to $\mathbf{Hc} = E\mathbf{c}$. A theorem from linear algebra states that the sum of the eigenvalues of $\mathbf{H}$ is simply equal to the trace of $\mathbf{H}$. Therefore, as long as we assume $\mathbf{S}_{ij} = \delta_{ij}$, merely changing the off-diagonal values will not change the sum of the orbital energies.

In general, the basis vectors are not orthogonal. If you solved the full equation, $\mathbf{Hc} = E\mathbf{Sc}$ then you would find that the sum of the new eigenvalues is greater than the sum of the old, unperturbed eigenvalues.

Obviously this is ignoring other important aspects, e.g. electron-electron repulsions (that is a problem intrinsic to Huckel theory), but it's a start.


Insoluble doesn't mean unreactive. The filler must be something which would be oxidized by permanganate in acidic solution.


Difference between probability and probability density of finding a particle in space

The latter is the limit of probability per volume, as volume approaches zero, at a particular point in space.


How did Millikan measure the mass of these oil droplets?

By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser)

The rate of fall here means terminal velocity of oil, $v_\mathrm{t}.$ Oil was made to drop at constant rate so no acceleration. Drag force on oil drop is:
$$F_\mathrm{d} = 6\pi r\eta v_\mathrm{t},\tag{1}$$
where $r$ is the radius of oil drop and $\eta$ is the viscosity of air.
Assume oil droplet is spherical. Apparent weight of oil droplet in air is the true weight minus the upthrust (which equals the weight of air displaced by the oil drop). So, it can be written as:
$$w = \frac{4\pi r^3(\rho - \rho_\mathrm{air})g}{3}.\tag{2}$$
No acceleration is taking place so the total force acting on it must be zero and thus the two forces must cancel each other (i.e $F_\mathrm{d} = w).$ Solving for $r$ gives
$$r = \sqrt{\frac{9\eta v_\mathrm{t}}{2g(\rho - \rho_\mathrm{air})}}.\tag{3}$$
The density of oil is known so by known so by putting $r,$ you can calculate the mass of oil droplet as
$$m = \rho V = \rho\frac{4\pi r^3}{3}.\tag{4}$$


In an octahedral complex, what happens to the electrons donated by the ligand?

This is a very good question, I must say. It requires the understanding of the very fundamentals.

You're right, if the electrons from the ligand pair up with the electrons of the metal, the electrons cannot undergo $d$-$d$ transitions, and no color would be seen.

But do they pair up? Think carefully. Recall one of the most basic assumptions of the Crystal Field Theory: The attraction between the ligand and the metal is assumed to be purely electrostatic in nature. This very important assumption is the sole reason for the explanation of colors of these complexes. As the ligands approach the metal atom, they create a crystal field. These fields tend to repel the electrons of the metal and increases their energy. A perfectly symmetric spherical  crystal field would increase the energy level of all the orbitals equally and to the same level. But this isn't the case in actual coordination complexes. An octahedral crystal field would approach along the axes of the $d_{x^2+y^2}$ and $d_{z^2}$ and tend to repel them to a greater extent than the other orbitals ($d_{xy}$, $d_{yz}$, $d_{zx}$). This difference in energies of the orbitals is known as crystal field splitting. We can now group these orbitals into two groups, three low energy $t_{2g}$, and two high energy $e_g$ orbitals.

Keep in mind that the ligands do not pair with the electrons of the metal. They simply repel the orbitals electrostatically and increase their potential energy in the process.

So, now how do the complexes actually get their color then? When the orbitals split, the difference in their energies is called Crystal Field Stabilization Energy (CFSE) and is denoted by $\Delta_{\text{o}}$. When photons of light are incident on the complex, it absorbs the photons which possess the energy equal to that of the value of $\Delta _{\text{o}}$. From quantum theory of electromagnetic waves, it's known that the energy of a photon is given by:

$$U = \frac{hc}{\lambda}$$

Where $h$ is known as the Planck's Constant, with a value of $\pu{6.626×10^{-34}Js}$, and $\lambda$ is the wavelength of light. If the value of the wavelength lies in the visible light spectrum, you can find the color of the light absorbed by the complex.

How do you find the color emitted out? There's a very easy and fun way to find that out. Take a look at this color wheel:

Find where the color of absorption lies. Then the color of the complex lies on the opposite side of the color of absorption.

This is a brief discussion about the colors of complexes. Let me know if you need more clarification.


Chemical reaction notation in differential equations

In general, there is no way to know exactly how a net reaction should be turned into a differential equation. Or, perhaps it's more accurate to say that you can't know which differential equation should describe the reaction.

In your case, since you're talking about a decay process, this will always be first-order with respect to concentration and should thus be written as,

$$\frac{dA}{dt} = -k[A].$$

Some other examples are:

$\textbf{Elementary Reactions:}$

If a reaction consists of only a single step, or the rate is known to be affected by only a single step, then the rate can be written down similar to how you guess above. One example are decay processes, which are always first-order. Consider, $$\ce{H2O2(l) ->[k]H2O +\frac{1}{2}O2}$$
Because we know this is a decay process, we can write down,$$\frac{-d[\ce{H2O2}]}{dt}=k[\ce{H2O2}]$$
$\textbf{Reactions with Simple Physical Models:}$

I'm thinking specifically of the derivation of the Michaelis-Menten Equation. I'll provide a sketch of this.

The goal is to describe the kinetics of enzyme catalysis. Because the reactant is relatively localized (i.e. it's trapped in an enzyme), there can't be that many different states. At most, we can have this sequence of steps:
$$\ce{E +S<=>[k_1][k_{-1}]ES<=>[k_2][k_{-2}]EP<=>[k_3][k_{-3}]E +P}$$where $\ce{E}$ is the enzyme, $\ce{ES}$ is the enzyme-substrate complex, and $\ce{EP}$ is the enzyme-product complex.

On nothing but physical intuition, we make two assumptions:

The rate of change of $\ce{ES}$ is zero (i.e. constant concentration)
The reverse step involving $\ce{EP}$ can be ignored, assuming $k_3\gg k_{-3}$

From these assumptions, and equilibrium reactions, we can write down a number of differential equations which give us a very useful and quite accurate equation.

The differential equations one can write down abide by the law of mass-action, which basically just says if we write down all the places some mass can go, then we can know the rate of change for a particular step.

For instance,
$$\frac{d[\ce{ES}]}{dt}=k_1[\ce{E}]+k_1[\ce{S}]+k_{-2}[\ce{EP}]-k_{-1}[\ce{ES}]-k_2[\ce{ES}]$$

As you can see, you basically just pick a step and tally up all the ways things are reacting to form a molecule at this step and all the ways things are leaving this step.

$\textbf{Takeaway:}$

Pretty much, in order to write a differential equation describing rate of changes of reactions, we must have some physical intuition about what is going on. Once we have this intuition, we can almost always get somewhere meaningful.


Why do some nucleophiles attack at the carbonyl group while others at an alpha-carbon in ketones?

For starters, you shouldn’t call it a nucleophilic attack at the α carbon; it is a deprotonation (something quite different since the latter doesn’t access the carbon atom in any way).

The second thing to note is that both do indeed happen. However, the nucleophilic attack on the ketone is not productive: nothing will happen that creates a new product or a different pathway. The only productive result of hydroxide’s attack onto the carbonyl carbon in the long run is the re-elimintion of either of the geminal hydroxy groups to reform the carbonyl and we’re back at the start.

Please don’t stop thinking like this (‘what happens when $\ce{X}$ attacks $\ce{Y}$ rather than $\ce{Z}$?’) — it’s a great way to spot side reactions, alternative pathways, problems with functional groups etc.


What's the point of solubility products?

Solubility products figure prominently in equilibria involving precipitated species.  Following is an example that may arise in a practical situation.

Problem:  We have 0.01 M ferrous ion in water and we propose adding a base to drive out the iron as hydroxide.  We do not want excess base dissolving into the water so we will try magnesium hydroxide.  How well will it work?

The proposed reaction is then

$\ce{Fe^{2+} + Mg(OH)2(s) <=> Mg^{2+} + Fe(OH)2(s)}$

$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}$

Compare this equilibrium constant with:

$K_{sp}(\ce{Mg(OH)2})=[\ce{Mg^{2+}}][\ce{OH^-}]^2=5.61×10^{-12}$ (source)

$K_{sp}(\ce{Fe(OH)2})=[\ce{Fe^{2+}}][\ce{OH^-}]^2=4.87×10^{-17}$ (Ibid.)

So then

$K=\frac{[\ce{Mg^{2+}}]}{[\ce{Fe^{2+}}]}=\frac{K_{sp}(\ce{Mg(OH)2})}{K_{sp}(\ce{Fe(OH)2})}=1.15×10^5$

Since the reaction stoichionmetry implies that the sum $[\ce{Mg^{2+}}]+[\ce{Fe^{2+}}]$ will be constant at the original iron concentration of 0.01 M, we find that the iron level remaining in the water may be reduced by five orders of magnitude, the equilibrium concentration is below $10^{-7}$ molar!

Magnesium hydroxide, despite being supposedly "weak" because of its limited solubility, is actually like a strong base here, displacing the weak iron hydroxide base almost quantitatively -- but, in effect, self-regulating because of that limited intrinsic solubility.  The same property applies with respect to most heavy metals commonly found in water; their hydroxide solubilities, like that of iron, are much less than magnesium's.  Thus magnesium hydroxide (or, in practice, magnesium oxide which becomes the hydroxide in situ) is a good agent for treating water to remove heavy metals.  See here, under "Applications".


Why does baking soda and vinegar remove stains?

You might try a comparison test to see if baking soda and vinegar is any more effective than plain water, or water and one of the ingredients. There are some possible reasons why each might be effective:

Some dyes and stains are set by exposure to acid, such as vinegar, which makes them harder to remove. It is possible that alkaline baking soda and water have the opposite effect, removing the stain.
Some stains, such as from red cabbage and red wine, are indicators, and change color when exposed to alkali or acid. The natural dye alizarin changes from dark red to lighter yellow under acid conditions, as in vinegar.
A powder such as baking soda may act as a sponge or adsorbent because of its large surface area. Talcum powder, baking soda or fuller's earth are also used for that purpose.
There might be a physical effect from the bubbles, as you state. For that reason, perhaps, some use plain seltzer (carbonated water) for stain removal.

Note that there are many conflicting statements on the efficacy of these stain removal techniques. It would be of benefit to humanity if someone conducted controlled experiments on removing a variety of stains with these chemicals! In the meantime, I just put a bit of HE laundry detergent on a stain, or a commercial cleaner such Lestoil, on the spot, before laundering.


You can remove tin from bronze (copper-tin alloys) and leave behind the copper. However, doing so may not leave you with copper in an easy to use form.

Tin is a less reactive metal than copper. Tin reacts with all acids, like $\ce{HCl}$, while copper only reacts with oxidizing acids, like $\ce{HNO3}$.

$$\ce{Sn(s) +2H+(aq) -> Sn^{2+}(aq) + H2 (g)^ }$$
$$\ce{Cu(s) + 2H+ (aq) -> No \ Reaction}$$
$$\ce{Cu(s) + 4H+(aq) + 2NO3- (aq) -> Cu^{2+} (aq) + 2H2O (l) + 2NO2 (g) ^}$$

Soaking your bronze in acid (even vinegar might work) will eventually dissolve the tin away from the copper. Heating, or using a stronger acid like the $\ce{HCl}$ in muriatic acid, will likely increase the rate. However, since the composition of the alloy is uniform throughout, the copper will likely come out of solution as a fine powder.


Why is the inversion barrier larger in PH3 than it is in NH3?

Ammonia is the classic system for $\ce{sp^3}$ hybridisation save methane. The lone pair (and each of the $\unicode[Times]{x3C3}$-bonds) has almost $25~\%$ s-character which corresponds nicely to $\ce{sp^3}$. However, the whole system can also swing around, changing its hybridisation to $\ce{sp^2}$ and back; a process during which the lone pair is temporarily in a p-type orbital and the s-character of the bonding orbitals increases.

Elements outside of the second period show a much smaller tendency to involve the s-orbital in bonding. The bonding orbitals only have an s-character of approximately $16~\%$. This also means that the phosphorous lone pair has a much higher s-character of approximately $50~\%$, while the bonding $\unicode[Times]{x3C3}$-orbitals have a larger p-character. (This also explains the much smaller $\ce{H-P-H}$ bonding angle of approximately $90^\circ$.) An orbital of high s-character has a long way to go to turn into a p-type orbital, and the three largely p-type $\unicode[Times]{x3C3}$-orbitals have an equally long way to go to give $\ce{sp^2}$-type orbitals so the interconversion and thereby the inversion of $\ce{PH3}$ is strongly hindered.


According to David R. Lide, ed., CRC Handbook of Chemistry and
Physics, 90th Edition (CD-ROM Version 2010), CRC Press/Taylor and Francis, Boca Raton, FL, the C-H bond dissociation energies for increasing chlorination of methane are as follows:

$D\left(\ce{H-CH3}\right) = (439.3 \pm 0.4) \; \text{kJ/mol}$

$D\left(\ce{H-CH2Cl}\right) = (419.0 \pm 2.3) \; \text{kJ/mol}$

$D\left(\ce{H-CHCl2}\right) = (400.6 \pm 2.0) \; \text{kJ/mol}$

$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$

The effect of different halogens in H-CX3 is shown below:

$D\left(\ce{H-CF3}\right) = (445.2 \pm 2.9) \; \text{kJ/mol}$

$D\left(\ce{H-CCl3}\right) = (392.5 \pm 2.5) \; \text{kJ/mol}$

$D\left(\ce{H-CBr3}\right) = (399.2 \pm 8.4) \; \text{kJ/mol}$

$D\left(\ce{H-CI3}\right) = (423 \pm 29) \; \text{kJ/mol}$


Well, XPS analysis should detect atoms as light as Carbon, but not atoms as light as Hydrogen.

Far be it from me to claim any expertise on microscope sliders, but I have used one or two in my career, and I am hazarding a guess. It is likely a pure glass ($\ce{SiO2}$) sheet laminated with some acrylic glass. The hydrogen in the acrylic polymer are too light to appear on the XPS.

There is a slight possibility, although I consider it very unlikely, that it is glass with a substrate of moissanite on top, which would in theory make it scratch proof, due to the hardness of $\ce{SiC}$.

If you would include whether the %'s are weight percent or atomic percent, it would be easier. If it is atomic, then the case for $\ce{SiC}$ is more likely, since the ratio ($\ce{Si:O} > \ce{1:2}$) and the extra Si would be in SiC. If it is weight percent then there is some math to be done. I still think it is just glass and acrylic glass.


How do you calculate a half cell potential Copper(II) and Copper Sulfate?

Here is a figure of the famed Daniell cell:

Note the spelling of the name: it was invented in 1836 by John Frederic Daniell. At $\pu{25 ^\circ C}$, and with $n = \pu{2 equivalents/mole}$, as correct here, $(\ln 10)\frac{RT}{nF} = \pu{0.02958 V} = \pu{29.58 mV}$.


I tried to do something similar when I had a humid car.

In chemistry, we often use $\ce{CaSO4}$ or anhydrite to serve as a desiccant. You could probably get cost-effective plaster for similar purposes. You could also get gypsum and heat it to generate plaster of Paris yourself.

As you're probably aware, people also use silica gel in products (and food) to minimize water content.

Either material should be much better as a desiccant than regular salt.

One key criteria is to increase the surface area - make sure the material is a fine powder and enclose it in something porous. I've seen people use various cloth materials, paper towels, etc.

As to how much you need, you can work out how much it will eventually absorb. If I take plaster of Paris to gypsum, each mole of $\ce{CaSO4}$ will take ~1.5 moles of water. So for every 136.14 g of anhydrite or ~145.15 g of plaster, you can absorb ~27 g of water. (I use ~ here, since I'm not convinced that regular plaster of Paris will be completely $\ce{CaSO4*0.5H2O}$.. some will likely already have extra water in it.)

The trick is that humidity will continually increase in the room if there's exchange of outside air.

Personally... I'd go with an electric dehumidifier.


Regioselectivity in electrophilic aromatic substitution of benzofuran and indole

In electrophilic substitution of aromatic systems sigma complexes are typically formed as reaction intermediates.  Let's look at the sigma complexes formed in the reaction of benzofuran and indole with an electrophile ($\ce{NO2+}$) at the 2 and 3 positions and compare their stability.

With both benzofuran and indole attack at the 2 position produces a sigma complex where the positive charge is nicely stabilized by the benzene ring (1a and 1b) . The stabilization is analogous to that found in a benzyl carbocation.

Attack at the 3 position produces a sigma complex where the positive charge can be stabilized by the lone pair of electrons on the adjacent heteroatom (resonance structures 3a and 3b), but because nitrogen is less electronegative than oxygen we would expect resonance structure 3b to be a bigger contributor than resonance structure 3a. Therefore we would expect the sigma complex produced upon attack at the 3 position to be more stable in the case of indole than benzofuran.

The overall result would then be to produce more of the 3 isomer in the case of indole as compared to benzofuran.


Determining the coupling on a substituted benzene ring

Benzene rings follows the same general principles as other spin systems for coupling.

A symmetrically para-substituted benzene, such as 1,4-dichlorobenzene, will have a 1H spectrum which will contain just a singlet. All four proton environments are identical. An XY para-disubstituted benzene will have the aromatic peaks that appear somewhat like two sets of doublets. The appearance is actually incredibly useful diagnostically,as it gives a useful indication of the type of ring substitution you have. They are, however, not true doublets, and consist of second order coupling arising from coupling between magnetically non-equivalent sites. That is, the splitting of the apparent doublets is not a true reflection of the coupling constant.

The para proton on a mono-substituted benzene ring does appear as a triplet. The ortho coupling giving rise to the ~7Hz splitting comes from coupling from two chemically equivalent protons at the meta position. Coupling to 2 equivalent protons will give rise to a 3 line splitting (from your 2nI+1 rule). Strictly speaking, a mono-substituted benzene ring is just a very special subset of a para-disubstituted benzene ring system, but does not typically show the classic splitting expected in a para system, because the 5J coupling is essentially zero.


Is there a quantitative test/analysis technique that can determine if some chloroform has degraded into phosgene?

Not a quantitative test but a good proof of phosgene. You could use a pH paper strip. Chloroform should show a neutral pH, but if there is phosgene/HCl the paper will show acidity.


Mechanism for reaction of NOCl (Tilden reagent) with primary amines

Why would the chloride ion attack the electron rich amine? The amine is the best nucleophile is the system and will attack the NO+. This gives a RNH-NO nitroso species which is then protonated on Oxygen, loses water and gives the diazonium cation. This adds loses nitrogen rapidly and adds Chloride to give the primary alkyl chloride.

More can be read about this process here (which I where I sourced the reaction scheme) or by searching the term "nitrosation of primary amine"


Why is the specific gravity of 0.9 % saline 1.0046?

0.9% medical saline is defined to contain $\pu{0.90 g}$ sodium chloride per $\pu{1000 ml}$ solution at $\pu{22 °C}$. Its density is $\pu{1.0046 g/cm^3}$ at $\pu{22 °C}$.

The density of water is $\pu{0.9978 g/cm^3}$ at $\pu{22 °C}$.

If the volume would stay constant when you add $\pu{0.90 g}$ sodium chloride to $\pu{1000 ml}$ water, the resulting density would be

$$\pu{0.9978 g/cm^3} + \pu{0.0090 g/cm^3} = \pu{1.0068 g/cm^3}$$

The density of the 0.9% saline shows us that $\pu{1000 ml}$ of it contain

$$\pu{1004.6 g} - \pu{9.0 g} = \pu{995.6 g}$$

or

$$\frac{\pu{995.6 g}}{\pu{0.9978 g/cm^3}} = \pu{997.8 ml}$$

water.

This means that the volume of the liquid phase increases when we add sodium chloride to water, what explains the difference in density from $\pu{1.0068 g/cm^3}$ to $\pu{1.0046 g/cm^3}$.


In solution, the iron(III) will be hexaaqua iron(III), 6 water molecules coordinating the iron ion, with the oxygen atoms being at the vertices of an octahedron.

Hexaaqua iron(III) has a high spin d5 electron configuration.

All the d-d transitions are spin forbidden.

Hexaaqua iron(III) is nearly colorless.


Non-trivial cases where the frozen core approximation is bad?

The frozen core approximation (FCA) falls apart whenever you need orbital relaxation of the core to accurately describe your system. From a theory-based argument, there are multiple electronic structure methods which rely on this (CCSD, orbital optimized methods) to better account for correlation effects. However, given that the core orbitals do not change appreciably during most processes (ionization, binding, standard chemical reactions), the error introduced by the FCA is all but completely cancelled (you can still get sub kcal/mol accuracy with the FCA).

However, one case where it rears its head is when you're trying to calculate the single-point energy of a system. In this case, the FCA will get in the way of recovering (for example) the full CI result in the full basis set limit.


Is it possible for a reaction to have an entropy change of zero?

Yes, entropy change of reversible process is zero.

This will also help


What are the conditions for making amines from alkyl halides?

Ethanol is just a solvent often used for this reaction because 1) the reactants and products are soluble in it and 2) the reaction proceeds at a reasonable rate in boiling ethanol.  If your trying to prepare the corresponding primary amine, then a large excess of ammonia should be used.  Otherwise, if the concentration of the initial product (the primary amine) becomes comparable to the concentration of ammonia, then the primary amine could react with the starting alkyl halide and a complicated mixture of products could ensue (see this article for a nice description of all the possible products that could be formed if the ammonia concentration is not high enough).

The reaction is typically run at reflux (boiling ethanol).  Since ammonia is a gas at room temperature it would quickly escape from the reaction and we wouldn't have a high concentration of ammonia for long.  Therefore, the ammonia is condensed into a solution of the ethanol and alkyl halide (usually an alkyl bromide) in a glass tube at low temperature.  When the desired concentration of ammonia is obtained, the tube is sealed and then heated to allow the reaction to take place.


I will assume you are asking for percent mass/volume, $\mathrm{g/ml}$, since to calculate percent mass/mass you need the density of the liquid (unless you assume it is water which is $1~\mathrm{g/ml}$ so will give the same answer).

We have $500~\mu\mathrm{M}$ of hydrogen peroxide, which means we have $500~\mu\mathrm{mol}$ of hydrogen peroxide per 1 litre of solution. $500~\mu\mathrm{mol} = 0.0005~\mathrm{mol}$. Now you can work out the molar mass of hydrogen peroxide and multiply it by the number of moles to get the mass of hydrogen peroxide per litre of solution. This gives $0.017~\mathrm{g}$. There are $1000~\mathrm{ml}$ in 1 litre. So you have $0.017~\mathrm{g}/1000~\mathrm{ml} = 0.000017 = 0.0017\%$.

For your question about whether hydrogen peroxide is toxic to the eyes, you can search the internet for the Materials Safety Data Sheet (MSDS) of hydrogen peroxide (Sigma-Aldrich is a good site to search). This contains all the hazards associated with this chemical, and when you do laboratory work you always have to look these up for each chemical before you start a reaction. The MSDS states that hydrogen peroxide can cause severe eye damage, among other hazards.


What is a 'triplet biradical transition state'?

A triplet biradical state simply implies that the bonding pair will be cleaved homolytically .  For example the triplet biradical transition state for

Will be


Stubborn CuAAC reaction as derivatization for ESI-MS

How to get a better turnover while maintaining a reasonably short reaction time?

To speed up the reaction, you should look at changing (1) the reagents, (2) the catalysts, or (3) the reaction conditions.  In particular you might be dealing with poor solubility for your alkyne substrate if you are doing the reaction in aqueous conditions.
1. Reagents
You could try the picolyl azides I mentioned in the comments in lieu of the 4-acetophenyl azide you originally tried.  These azide reagents chelate copper significantly, thus increasing reaction rates since proximity of two of the three components of the reaction (\ce{Cu+} and aizde) is enhanced by the chelation effect.
2. Catalyst
There are a variety of exotic $\ce{Cu+}$ organometallic complexes that work better than the "standard" copper salts.  An example publication is Sebest et al. 2018, which reports that the N-heterocyclic carbene complex $\ce{[CuI(Mes‐6)]}$ (Mes-6 being an NHC ligand pictured below) worked well at loadings down to 0.05 mol% in their trial reaction.

This might be overkill if you are not using the ligand THPTA used in your original protocol.  Maybe you could just use that?
3. Conditions
You an always heat more, but I think a more important variable is your solvent.  Is your reaction solvent one that dissolves all the species, including alkyne, azide, and copper(I) complex?  Instead of aqueous conditions, can you try acetonitrile or 100% DMF?  Or n-butanol?
Additionally, the Presolski reference you give mentions that high chloride concentrations (~0.2 mol/L) can compete with ligand for copper ion.  Plasma has about a 0.1 mol/L chloride concentration, so maybe it's an issue?  Probably speaks to the need to use a good copper ligand like THPTA.
Bottom line
If it were me, I'd try getting some THPTA or other ligand to improve $\ce{Cu+}$ solubility in organic solvents, and trying the reaction in non-aqueous solvents where your steroid is likely to be soluble.  I'd also get the picolyl azide if my lab could afford it.  If that doesn't work, and I still wanted to improve things, I guess I'd look at increasing temperature or using more exotic NHC ligands for the copper.


Why do only some metals make protective oxide layers?

The simple answer is that metals which form sufficiently insoluble oxide layers stifle the very process which caused the oxide layer in the first place.
It's not just a matter of energy of formation of the oxide, but whether the first reaction product formed will be soluble or insoluble, then whether it will be adherent or non-adherent. In the case of the alkali metals, the oxide may be adherent (reaction with oxygen produces a layer that sticks), but is so soluble that it dissolves in water and refreshes the metallic surface. Likewise in the case of the alkaline earth metals - even $\ce{CaO}$, which is almost insoluble, is soluble enough to give a pH of 12.5, and soluble enough to be non-protective.
Now, magnesium oxide is quite insoluble, but probably is on the surface as $\ce{Mg(OH)2}$, a secondary product. You might consider that $\ce{Mg(OH)2}$ is $\ce{MgO}$'s failed attempt to dissolve and go away from the surface. Instead, it precipitates, and not only that, it adheres quite well. So further reaction with water is inhibited. However, if you heat the water, the $\ce{Mg(OH)2}$ becomes more soluble, and corrosion proceeds, although not as fast as sodium reacts. In flowing hot water, where the $\ce{Mg(OH)2}$ could be carried away, corrosion would be faster; in a flask, stirring would exert a similar rate increase.
The effectiveness of $\ce{Mg(OH)2}$ as a stifling passivator may be related to its layer structure:
 (File from Commons, by User:Smokefoot)

Like a clay layer, the hydroxide ions fuzz out of each layer of close-packed magnesium ions, and the metal surface could be like a half-layer of $\ce{Mg(OH)2}$, with $\ce{OH-}$ ions sticking out and bonding to separate layers of $\ce{Mg(OH)2}$, or perhaps the metal itself just hold onto a layer of $\ce{Mg(OH)2}$ - or maybe a little of both. This adherence is necessary to stifle corrosion. The clay layer-like characteristics of $\ce{Mg(OH)2}$ are also evident in milk of magnesia, which is a stable suspension - if the suspension were not stabilized by clay-like interactions, it would settle fairly rapidly and need good shaking before use. (I shake it anyway.)
Corrosion on aluminium is stifled by an adherent, quite insoluble oxide, but not completely: the oxide layer continues to thicken, and corrosion continues, diminishing, but never stopping completely. Nature never sleeps.
Iron rusts with $\ce{H2O}$. Iron oxide could be adherent (it is on stainless steels, when mixed with some chromium), but unfortunately, pure iron oxide tends to form a hydrate which has a greater volume than the oxide, and as it forms, it peels away from the metal surface, leaving bare metal or a very thin layer of oxide, which can still corrode. Chloride ion is especially effective in promoting the continuation of corrosion on iron and steel (even stainless) and aluminium.


Reduction of nitrobenzene with LiAlH4 and NaBH4

The structure you provided is known as azobenzene . It is a parent compound in a well-known family of aromatic azo compounds. The compounds usually are not very thermally stable, but often have intense color and are used as light duty dies (or as pH or other indicators) Aliphatic azo compounds are not stable at all and very prone to fragmentation with formation of molecular nitrogen and free radicals and some of them are used exactly for that.

  I couldn't find the mechanism for this reaction anywhere

The reduction of the nitro-group might procede through attack of hydride onto the nitrogen atom or electron transfer onto the group with subsequent reaction of the anion with proton (the exact mechanism depends on conditions and reagents). It can produce a wide amount of monomeric products, in particular: nitrosocompounds, hydroxilamines and amines. It is known, that amines can react with nitrosocompounds forming the $\ce{N=N}$ bond. It looks like the first step in the reaction is a nucleoplilic attack of the amine nitrogen onto nitroso nitrogene with subsequent elimination of water.


Add another amine. It is better to choose an amine with just few more carbons.


How do I do stoichiometry for 2 limiting reactants

Your balance equation is correct except for not simplifying it. Simplified equation should be:
 $$\ce{2Al + 2NaOH + 6H2O ⟶ 2Na[Al(OH)4] + 3H2}$$

I also think you're getting a little bit confused about your equation. You probably think reacting $\ce{Al}$ and $\ce{NaOH}$ with same mole ratio (2:2) make them two limiting reagents. If that is the case, you are absolutely wrong. As M. Farooq commented elsewhere, the one reactant, which is present in the least amount in a stoichiometric sense, is the limiting reactant. Let's see your question in hand:

The amount of $\ce{NaOH}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{35 g}\; \ce{NaOH}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{40 g}\; \ce{NaOH}}  = \pu{4.38 mol}$

The amount of $\ce{Al}$ in a typical container of Drano $ = \pu{500 g} \times \frac{\pu{2.5 g}\; \ce{Al}}{\pu{100 g}}\times \frac{\pu{1 mol}}{\pu{27 g}\; \ce{Al}}  = \pu{0.463 mol}$

Since $\ce{NaOH}$ and $\ce{Al}$ react with 1:1 ratio according to the equation, $\ce{Al}$ is clearly the limiting reagent.

Hence, $\pu{0.463 mol}$ of $\ce{Al}$ reacts with $\pu{0.463 mol}$ of $\ce{NaOH}$ if you added enough water to $\pu{500 g}$ of Drano container. The leftover amount of $\ce{NaOH}$ would be: $\pu{(4.38 - 0.463) mol} = \pu{3.92 mol} = \pu{3.92 mol} \times \frac{\pu{40 g}\; \ce{NaOH}}{\pu{1 mol}} = \pu{157 g}$

Now, how much water you need? You need more than $\pu{0.463 mol}\times \frac{6}{2} = \pu{1.39 mol}$ of water, which is more than $\pu{25.0 g}$ (to be excess).

Note: Similar calculations would appied to find out amount of $\ce{H2}$ released by the reaction.


What does a number in brackets after another number mean? I.e. 21(1)cm-1

The results of measurements and other numerical values of quantities are often given with an associated standard uncertainty. A numerical value and the associated uncertainty may be expressed as shown in the question:

$$\begin{align}
y&=21(1)\ \mathrm{cm^{-1}}\\[6pt]
&=a(b)\ \mathrm{cm^{-1}}
\end{align}$$

where
$y$ is the estimate of the measurand (e.g. the result of a measurement) expressed in the unit $\mathrm{cm^{-1}}$,
$a$ is the numerical value, and
$b$ denotes a standard uncertainty expressed in terms of the least significant digit(s) in $a$.

It is important to note that the given uncertainty refers to the least significant digits of the given numerical value. For example, in the expression

$$l=23.4782(32)\ \mathrm m$$

the $(32)$ represents a standard uncertainty equal to

$$u(l)=0.0032\ \mathrm m$$

Many physical constants are also given in this form. For example, the previously recommended value for the molar gas constant $R$ as given by NIST from 25 June 2015 until the new value became available on 20 May 2019:

$$R=8.3144598(48)\ \mathrm{J\ mol^{-1}\ K^{-1}}$$

where the $(48)$ represents a standard uncertainty of

$$u(R)=0.0000048\ \mathrm{J\ mol^{-1}\ K^{-1}}$$

This form is in accordance with various current standards, in particular

JCGM 100:2008 Evaluation of measurement data – Guide to the expression of uncertainty in measurement
ISO/IEC Guide 98-3:2008 Uncertainty of measurement   Part 3: Guide to the expression of uncertainty in measurement (GUM:1995)
ISO 80000-1:2009 Quantities and units – Part 1: General

The Guide to the Expression of Uncertainty in Measurement (GUM) also shows other permissible forms:

  7.2.2 When the measure of uncertainty is $u_\mathrm c(y)$, it is preferable to state the numerical result of the measurement in one of the following four ways in order to prevent misunderstanding. (The quantity whose value is being reported is assumed to be a nominally 100 g standard of mass $m_\mathrm S$; the words in parentheses may be omitted for brevity if $u_\mathrm c$ is defined elsewhere in the document reporting the result.)

  1) “$m_\mathrm S=100{,}021\,47\ \mathrm g$ with (a combined standard uncertainty) $u_\mathrm c = 0{,}35\ \mathrm{mg}$.”

  2) “$m_\mathrm S=100{,}021\,47(35)\ \mathrm g$, where the number in parentheses is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ referred to the corresponding last digits of the quoted result.”

  3) “$m_\mathrm S=100{,}021\,47(0{,}000\,35)\ \mathrm g$, where the number in parentheses is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ expressed in the unit of the quoted result.”

  4) “$m_\mathrm S=(100{,}021\,47\pm0{,}000\,35)\ \mathrm g$, where the number following the symbol $\pm$ is the numerical value of (the combined standard uncertainty) $u_\mathrm c$ and not a confidence interval.”

Note that item 2) corresponds to the form given in the question.

Concerning item 4), however, the GUM notes

  The ± format should be avoided whenever possible because it has traditionally been used to indicate an interval corresponding to a high level of confidence and thus may be confused with expanded uncertainty (…).

Furthermore, ISO 80000 notes

  Uncertainties are often expressed in the following manner: $(23{,}478\,2\pm0{,}003\,2)\ \mathrm m$. This is, however, wrong from a mathematical point of view. $23{,}478\,2\pm0{,}003\,2$ means $23{,}481\,4$ or $23{,}475\,0$, but not all values between these two values. (…)


Calculated 13C NMR Shifts of brominated Carbons

The authors note in their review article* that such effects are expected as "heavy atom effects" due to electron-correlation effects and spin-orbit coupling: (references stripped for clarity)

  This being said, one way to improve the computed chemical shifts of heavy-atom-bound carbon atoms is to utilize a method that better captures electron correlation effects, such as MP2 or higher-level methods along with correlation-consistent basis sets (see section 4.3). This will likely result in significant improvement for carbon atoms attached to sulfur or phosphorus as well as to halogens, although the improvement is expected to be greater for sulfur or phosphorus. Further improvement requires the use of methods that determine relativistic spin–orbit corrections.
  [...]

  Another option for improving chemical shift calculations with heavy-atom errors is to utilize a linear regression method, as discussed in section 3.6. Due to the systematic nature of heavy-atom effects, they are amenable to improvement via this approach, provided that the linear regression data consists of nuclei similar to the nucleus of interest. That is, a linear correlation correction to carbon nuclei attached to a given number of bromine atoms, for example, should be derived from empirical data for other carbon nuclei attached to the same number and type of halogen atoms. In fact, this approach has been demonstrated specifically for such cases.

The latter section cites two relevant articles:

"Theoretical investigation on $\ce{^1H}$ and $\ce{^13C}$ NMR chemical shifts of small alkanes and chloroalkanes" J. Chem. Phys. (2006) 125, 144309.
"A hybrid quantum mechanical and empirical model for the prediction of isotropic $\ce{^13C}$ shielding constants of organic molecules" Phys. Chem. Chem. Phys., 2002, 4, 5498-5507.

Both articles suggest ways to use DFT methods with empirical corrections that work fairly well.

* Michael W. Lodewyk, Matthew R. Siebert, and Dean J. Tantillo, Chem. Rev., 2012, 112 (3), 1839–1862.


Selecting between redox and neutralization reactions for a mixture of HCl, H2SO4 and NaOH

As Klaus and I mentioned, the first reactions seem very strange from a chemical perspective, even if they are correctly balanced. This is where a chemist needs to exercise their intuition and experience.

Balancing reactions is nothing but solving a linear system of equations, so it is entirely possible to propose reactions which balance perfectly, but which are based on questionable chemistry. For example, in all but the last reaction, the sulphur atoms in $\ce{SO_4^2-}$ anions are being reduced to lower oxidation states. While this is possible in some conditions, it takes quite strong reducing agents, and none of the reagents shown are known to possess significant reducing character. Reactions which release gaseous oxygen are also not particularly common. Furthermore, two reactions propose the formation of sulphur chlorides, which would actually hydrolyse in contact with water.

Disregarding intuition, it should be possible to pick out which reaction is the most favourable by simply comparing the Gibbs free energy of reaction in each case; the most exergonic reaction, with the most negative value of $\boldsymbol{\Delta_\mathbf{r}G}$, should dominate (assuming no kinetic factors bar it from happening, which is unlikely in this situation). I did a very quick and dirty calculation, out of curiosity, by taking the standard Gibbs free energy of formation $\Delta_\mathrm{f}G^\circ$ for each compound and summing them together to obtain the standard Gibbs free energy of reaction:

$$\Delta_\mathrm{r}G^\circ = \sum\limits_\text{products}\Delta_\mathrm{f}G^\circ - \sum\limits_\text{reagents}\Delta_\text{f}G^\circ$$

This is only meant as a qualitative way to compare the reactions, as I am assuming standard state conditions for the reactions, which are not necessarily true. Furthermore, for a few species, I didn’t find the correct values in these two sources [1] [2]. $\ce{SCl2}$ and $\ce{S2Cl2}$ probably can’t be found tabulated for the aqueous phase because of their propensity to hydrolysis, so I used the free energies of formation for the gas and solid phases, respectively. I also didn’t find a source for aqueous $\ce{Na2SO4}$, so instead I used the data for the solid decahydrate $\ce{Na2SO4 . 10 H2O (s)}$. The results I got were:

$$ \small
\begin{array}{lcc}
\hline
\text{Reaction} & \Delta_\mathrm{r}G^\circ / \mathrm{kJ\ mol^{-1}}\\
\hline
\text{A} & +440  \\
\text{B} & +1940  \\
\text{C} & +590  \\
\text{D} &  \mathbf{-2610}  \\
\hline
\end{array}
$$

As you can see, even though the calculation was very rough, the huge disparity in free energy values suggests that the first three reactions are completely overwhelmed by the last one, so that of the four possibilities listed, only the last has any chance of occurring. Strictly speaking this doesn’t mean the last reaction happens, only that it’s much more favourable than the others presented; there could be an unlisted reaction which is even more favourable than all four mentioned in the question. However, empirical data confirms that the last reaction, a simple neutralisation reaction, is the major event when mixing $\ce{HCl}$, $\ce{H2SO4}$ and $\ce{NaOH}$ in practically any conditions.


Electrocatalysis of Hydrogen Evolution Reaction with metal oxides

You can use a similar, DFT-based approach to estimating the electrocatalytic activity of oxides, as has been done with the oxygen evolution reaction, but since most oxides are not stable at reducing conditions present in HER, there's not as much literature on that particular material-reaction pairing.  As an aside, a popular theory of metal-adsorbate bond formation, the d-band model was formulated specifically for metals, so the question as to whether or not there's a model with comparable simplicity that includes oxides (or other binary compounds) is still somewhat of an open question.


Free API to view 2D representation of Molecules

Open Babel can convert all kinds of input into pictures with 2D structures.


I think you are confusing the wavefunction $\Psi$ with the probability density function $\Psi^2$. The former is a state function describing the behavior of an electron. The latter describes the "probability of finding an electron".

Wavefunctions are usually complex functions, but even the simplest sine wave functions can be negative in the algebraic sense: the value of $\Psi=f(x)<0$. These regions where $Psi=f(x)$ has different algebraic signs are the regions corresponding to the orbital lobes of different phase, like in a p-orbital. However, $\Psi^2$ is always $\Psi^2 =f^2(x)\ge 0$. Information about the "location" of the electron is gained, information about its phase is lost.

Take as an example the two wave functions representing two atomic orbitals $AO_1$ and $AO_2$ on the graph below. $\Psi_{AO_1}=\cos{(\frac{\pi x}{2})}$. $\Psi_{AO_2}=\cos{(\frac{3\pi x}{2})}$. $\Psi_{AO_2}$ has regions where $\Psi_{AO_2}<0$ when $\frac{1}{3}<|x|<1$.

Now let's look at the probability density functions $\Psi^2_{AO_1}=\cos^2 {(\frac{\pi x}{2})}$ and $\Psi^2_{AO_2}=\cos^2 {(\frac{3\pi x}{2})}$. Notice that both functions are always $\ge0$.

Okay... now, when atomic orbitals combine to form molecular orbitals, one of those conservation laws (angular momentum) gets involved, and the number of orbitals has to remain constant. Thus, molecular orbitals are linear combinations of atomic orbitals. For example, if two atomic orbitals combine, two molecular orbitals are produced.

Let's go back to our two sample sine wavefunctions above and pretend they are atomic orbitals overlapping. One of the molecular orbitals could then be: $\Psi_{MO_1}=\frac{1}{2}\Psi_{AO_1}+\frac{1}{2}\Psi_{AO_2}$. This is the orbital representing constructive overlap between these two wavefunctions. To get the other orbital, we can't just add again, or we would get the same orbital. If we subtract, we get a different orbital - the one representing destructive overlap. The graphs of these two "molecular orbitals" and their probability density functions are below.

Again note the difference between the MO wavefunctions and their density functions. If these were atomic nuclei at $x=\frac{1}{3}$ and $x=-\frac{1}{3}$, then $\Psi_{MO_1}$ would be bonding and $\Psi_{MO2}$ would be antibonding (or perhaps nonbonding).


Why allene cannot be described with an allyl system?

Why isn't allene planar?

Let's start by looking at the structure of allene and seeing what happens if we make it planar.

(image source)

As the above image shows, the 3 carbons in allene are linear (form a straight line) and the terminal hydrogens lie in orthogonal planes (e.g. the planes containing the hydrogens are rotated 90° one from the other as shown in the bottom half of the drawing).  Allene has two orthogonal pi bonds (e.g. the planes of the two pi bonds are also rotated 90° one from the other).  Each one of those pi bonds contains 2 electrons.  The terminal carbons are $\ce{sp^2}$ hybridized, while the central carbon is $\ce{sp}$ hybridized.

Now let's twist one of the terminal methylenes by 90°.  It doesn't matter which one, so let's just say we twist the one on the right, and twist it so that its red p-orbital lobe lines up with the red lobes on the other double bond.  Now our molecule still has the 3 carbons arranged in a line, but all of the atoms - including the hydrogens - now lie in the same plane; we've made our molecule planar.

To accomplish this we broke one pi bond.  In our planar molecule we have 3 pi electrons (2 from the original pi bond on the left and 1 from the terminal methylene p-orbital that we rotated) in the 3 aligned p orbitals that exist in the plane of the screen.  We are also left with 1 electron in the p-orbital on the central carbon that is perpendicular to the screen (the p-orbital that was part of the double bond on the right side of the molecule).  So we took a molecule with two stable double bonds, rotated a p-orbital 90° and wind up with a planar molecule that has 3 aligned p-orbitals with 3 pi electrons and one electron by itself in a single p-orbital.  This is a biradical, an allyl-like (it's just allyl-like because the 3 carbons in our molecule are still linear, whereas the 3 carbons in the allyl system are bent) radical and a radical in a p-orbital.  Converting 2 double bonds to a biradical is energetically uphill, it does not happen at room temperature. Allene prefers to keep its 2 stable double bond structure rather than be a planar biradical molecule because the former geometry is much more stable.

  Why does $\ce{CS2}$ have another geometry than $\ce{H2CCCH2}$?

It doesn't. Carbon disulfide and carbon dioxide have the same structure and hybridization as allene.  It is just a little harder to notice because we don't have hydrogens on the ends of those molecules to "tip us off" that the two double bonds are orthogonal.

Here is a picture of the orbitals in carbon dioxide (it's exactly the same in carbon disulfide)

(image source)

Just like in allene, we have 3 atoms arranged in a linear fashion and the central carbon is $\ce{sp}$ hybridized; there are two pi bonds that are orthogonal to one another.


Does the multiplication and division significant figures rule apply to unit conversions as well?

In regards to the calcium situation, it depends on whether you measured 40g of calcium (which would be 2 sig. fig.), or 40g was the stated value (1 sig. fig.). Yes, the number of sig figs in your answer should equal the number of sig figs in the least accurate piece of data.

When converting units, the 1000 is taken as the exact value, i.e. it has infinite accuracy. Here, it is not considered, so the answer should have 3 sig fig from 12.4 (which is also why scientific notation is used to decrease ambiguity).


Finding 'UV index' at a point might be difficult. UV index is an index created by I think was WHO, and is calculated using
 - Thickness of ozone above
 - Cloud density of the area
 - Altitude of location
 - Sun's height
more about it here - https://www.who.int/uv/publications/en/UVIGuide.pdf
So I think you are better off finding the UV light intensity instead, using some measurement tools. There's few different options like photo reactive dyes, and digital scales, but probably the most cheap and quantitative data can be obtained using arduino or other similar development board and creating a small module yourself. Good Luck!


Understanding how to calculate the pH of a buffer with ice tables

The answer given is totally wrong. With 0.1 molar HCl, then the pH would be 1.0 from the HCl and virtually none of the acetic acid would be dissociated.

REVISED PROBLEM

  I made that exercice up. The original exercice has $0.1$ molar $\ce{HCl}$ and $0.1$ molar $\ce{CH3COOH}$.

This makes no sense.

The work shown in the second image is for 0.1 molar $\ce{CH3COOH}$ and 0.1 molar $\ce{CH3COO^{-}}$. That work gets the right answer, but the wrong way. At equilibrium: $$[\ce{CH3COOH}] = 0.1 - x$$
but $x$ is so small that $$[\ce{CH3COOH}] \approx 0.1$$ and $$[\ce{CH3COO^{-}}] \approx 0.1$$ which simplifies the mathematics considerably.


Approximation of electron density of a small molecule

You are looking for density functional theory. The theory is somewhat involved and you may be interested in A Chemist's Guide to Density Functional Theory or similar text book. A full description of the variety of ways to compute the (ground state) electron density is beyond the scope of an answer and better suited to a textbook.

The general gist is to optimize the energy with respect to the electron density numerically.


precipitation reaction - cellulose in schweizer reagent reacts with acid

It's possible to regenerate cellulose from Schweizer's reagent by dilution in plain water, though neutralization of the high pH+ cuprammonium solution with an acid is more efficient. $\ce{HCl}$ works as well as $\ce{H2SO4}$. Economics probably drives the choice.

The chemistry is based on breaking cellulose, a polysaccharide (polymer)  β(1→4) linked D-glucose, into smaller, soluble, segments, and reconstituting the longer polysaccharide through neutralization.


You have an error somewhere in your math.
Assuming you were supposed to interpolate $K_C$ from both experiment 1 and 2, you would do the following:
Experiment1 yields $K_C=3.31$. Experiement two yields $K_C=3.40$. Interpolation yields $K_C=3.35$.
This would give you $$B=\sqrt{\frac{C^2}{A\cdot K_C}}=\sqrt{\frac{1.49^2}{1.4\cdot 3.35}}=0.69$$


Despite being less reactive,alkanes are highly flammable

In chemistry, reactive and flammable are two different term. For example, Diethyl ether is a rather non-reactive material but it is highly flammable.
In your case, propane or butane are mostly unreactive to common chemicals, but it does react with oxygen at an elevated temperature- that is they are flammable. Under similar condition, propene or butene are more flammable.


What (simple) household chemicals react with oxygen?

You don't have to reinvent the wheel.
Food industry already uses quite simple oxygen scavengers for decades, and one of the most affordable formulations include iron powder mixed with sodium chloride.

Typically, an oxygen scavenger is packaged in small packets (sometimes labeled as "oxygen absorber" or "antioxidant") that are added with the product inside the pact and sealed.
The absorber itself consists of iron powder and an inert absorbent (sodium chloride, activated carbon, food-grade clay or a composite such as nylon 6/nanoclay).
When placed inside a package with a product intended for storage, it begins to actively absorb oxygen and completely removes it within 12–96 hours.
The drawback of using Fe is that it promotes rapid lipid oxidation, if it's leaked and gets in contact with the product.
However, since you are already using stainless steel container, going with iron-based quencher seems like an optimal choice to me.

More historical background and interesting tech details can be found in Wiley's Encyclopedia of Packaging Technology [1, p. 845]:

  […] The iron-based oxygen scavengers are the most prevalent and they are examined
  first.

  The iron technology is based on rapid oxidation of iron powders to prevent headspace oxygen from reacting with foodstuff components or to reduce oxygen permeation rates through container walls. Reduced iron powders (zero-valent iron $\ce{Fe^0}$) are highly reactive and possess a higher capacity to react with oxygen than partially oxidized ferrous iron ($\ce{Fe^2+}$) compounds. Large exposed surface area of iron powders per unit weight, which increases with reduction in mean particle size and especially when using highly porous "sponge" iron morphologies, promotes simultaneous oxidation at multiple reactive sites, resulting in faster rates of oxygen scavenging. Presence of electrolytes and an aqueous interface to facilitate electron transfer during oxidation is known to accelerate oxidation reactions of transition metals. This is exempliﬁed by rapid rusting of ferrous metals and steels exposed to road salt and salted seawater, in contrast with slow rates of corrosion in the dry climates. Ferrous and ferric iron oxides and hydroxides produced by oxidative reactions are basic. As a result, acidic environments promote oxygen scavenging by shifting equilibrium to iron oxides. At low pH values, oxidation of reduced iron proceeds in humid environments even in the absence of oxygen, by deoxygenizing water molecules and resulting in hydrogen gas formation.

  Since low pH values and aqueous interfaces require the presence of water, water vapor diffusion and variation in relative humidity of the environment can be used to control activation of oxygen scavenging capacity of iron powders. The speciﬁc formulations are designed to trigger the activation at desired RH levels, control the reaction rate, achieve the full oxidative potential of the iron powder, maintain compatibility with various products, and ensure food safety. The reaction formula for complete oxidation of reduced iron to eventually form ferric oxide trihydrate complex (commonly known as rust) is

  $$\ce{4 Fe^0 + 3 O2 + 6 H2O → 4 Fe(OH)3 → 2 (Fe2O3* 3 H2O)}$$

  Many transition metals are recognized as being sufficiently oxidative to become candidates for an oxygen-scavenger formulation; however, iron does offer unique advantages that have driven the industry to general use of this medium. Iron has a relatively high afﬁnity to combine with oxygen on a per unit weight basis compared to most alternatives. Complete oxidation of 1 g of reduced iron removes 300 cm³ of oxygen at STP conditions. Elemental candidates that exceed it have drawbacks such as the odor problems with the use of sulfur and the propensity of aluminum to form an almost impermeable oxidized skin layer that limits further oxidation. Iron rust, on the other hand, ﬂakes off as it forms and it is highly permeable to further duffusion of water and oxygen. Of great importance is the food safety of iron powders, which are recognized as nutrients important for healthy blood and commonly used as food enrichments in milled flour, breads, breakfast cereals, and baby foods. The same cannot be said about copper, zinc, aluminum, and many other transition metals. The abundance and relatively low cost of producing elemental iron are also important, especially in comparison to choices such as palladium and platinum metals and catalysts based on them. Then there is the ready ability to manipulate reactive capability of iron powder formulations to adapt to a wide variety of applications for both the rate of oxygen scavenging and the activation by the moisture in a package headspace. One of the few drawbacks is that the oxidation of iron is a temperature-dependent reaction and normally slows dramatically as the temperature approaches freezing.

  Electrolytes and acidifying agents in a scavenging formulation are commonly supplied by metallic salts […]. The practical considerations of food safety have dictated that common table salt ($\ce{NaCl}$) is often the best choice for the electrolyte, while various metal and organic salts of strong acids are often used to increase the acidity of the formulation. […] When the packaged product does not have a required water activity to fully activate an iron-based scavenger and provide the desired reaction rate, premoisturized silicas, zeolites, and activated carbons are used as self-contained water carriers in various proprietary formulations to provide a reliable activation.

References

The Wiley Encyclopedia of Packaging Technology, 3rd ed.; Yam, K. L., Ed.; John Wiley & Sons: Hoboken, N.J, 2009. ISBN 978-0-470-08704-6.


Enol content in 3 is greater then enol content of 2

Greater the acidic nature of alpha Hydrogen,greater is the tendency for Tautomeric behavior. H at alpha position in ketone is more acidic then H at alpha position in an ester.Pka values are shown below.The withdrawing power of ester is less compared to carbonyl group in ketone. Since lone pair on oxygen is in conjugation with C=O in ester,its withdrawing power decreases(as shown below).A more acidic α-hydrogen implies a weaker C−H bond. Summarizing,the enol content increases with the acidity of the enolic hydrogen.

$$\ce{carbonyl <=> enol}$$
$$\mathrm{K_{eq}=\frac{[enol]}{[carbonyl]}}$$

The equilibrium between keto and enol form is tipped to enol if H at alpha carbon is more acidic.

Hence enol content in 3 is greater then enol content of 2.

[]

  Pka of cyclohexanone is as per this link: https://www.drugbank.ca/drugs/DB02060

Enol of 4th compound is aromatic and should be most stable.

Therefore extent of enol content must be 4>1>3>2.
As a general guide line the following can be used for predicting enol content.

The enol form of cyclohexanone  has bond angles around double bond other then 120${^o}$ as shown below using avagadro.


How did Mendeleev know elements from compounds or mixtures?

Please do not underestimate the scientists of 19th century. They were as creative, intelligent and perhaps more genuinely dedicated to science than the scientists of the 21st century. Spectroscopy was the tool of the trade to identify and verify that a given substance is not a mixture.
The original reference which established that Didymium was a mixture is in German [1] (English title: "On the decomposition of didymium in its elements").
I don't know enough German to read it conveniently, but it is clear that the authors did fractional crystallization of a lot of salts, made measurements absorption spectra, did emission spectroscopy and finally led to a beautiful result. The original paper has plenty of spectra! From [1, p. 490]:

Da sonach die exacte Zerlegung des Didyms in mehrere Elemente realisirt ist, so schlage ich vor, die Bezeichnung Didym nunmehr ganz zu streichen und beantrage, für das erste Element, entsprechendder Grünfärbung seiner Salze und seiner Abstammung die Benennung:
Praseodym mit dem Zeichen $\ce{Pr}$
und für das zweite, als das „neue Didym“, die Benennung:
Neodym mit dem Zeichen $\ce{Nd}.$
Aus den Atomgewichtsbestimmungen, die in allen Einzelnheiten nach Bunsen ausgeführt wurden, ergaben sich als vorläufige  Werthe für Praseodym = 143·6, für Neodym = 140·8, wobei dem entsprechenden Oxyd die allgemeine Formel $\ce{M2O3}$ zukommt.

English translation with assistance of DeepL:

Since, therefore, the exact division of the didymium into several elements has been carried out, I propose that the name 'didymium' be deleted altogether and that the first element be given a name corresponding to the green colour of its salts and its origin:
Praseodymium with the sign $\ce{Pr}$
and for the second, as the "new didymium", the naming:
Neodymium with the character $\ce{Nd}.$
From the atomic weight determinations, which were carried out in all individual units according to Bunsen, it resulted as provisional values for praseodymium = 143·6, for neodymium = 140·8, whereby the corresponding oxide is assigned the general formula $\ce{M2O3}.$

Anyway, historically, it is not Mendeleev who had to decide whether the elements were actually elements. You may ask the right question, how did the element discoverer decided that it was an element in Mendeleev's time? The answer is that each element has a story to tell and that is why Isaac Asimov wrote a beautiful book called "The Search for the Elements". Go ahead and get hold of it.
Reference

v. Welsbach, C. A. Die Zerlegung des Didyms in seine Elemente. Monatshefte für Chemie 1885, 6 (1), 477–491. DOI: 10.1007/BF01554643.


Why is the active space for the dinitrogen cation defined this large in my example?

The active space is a truncation of the full CI space. Hence, including more virtual orbitals will lower the energy and eventually approach the FCI limit. The larger the active space, the more accurate the results.

The smaller active space will give you a good qualitative description.
But if your results turn out to be quantitatively wrong, increasing the active space is one way to improve them.

To get an idea about the higher orbitals, take a close look to the Molpro output. You might want to check the Print options, to get all desired energies and orbital coefficients. Furthermore, you can use for example put to create molden files and create 3D isosurface plots of all MOs.

Possible active spaces

The smallest meaningful would be to include the singly occupied $3\sigma_g$ orbital, together with its anti-bonding pendant $3\sigma_u$. Furthermore all the $1\pi$ orbitals should be included. This makes a CAS(5,6):

occ,   3,1,1,0,3,1,1,0
closed,2,0,0,0,2,0,0,0

Your occ,3,1,1,0,2,0,0,0 card actually excludes the anti-bonding $3\sigma_u$ and $1\pi_g$ orbitals, which corresponds to a ROHF calculation.

The next larger space would be the full valence active space, covering the whole $2s2p$ AO shell:

occ,   3,1,1,0,3,1,1,0
closed,1,0,0,0,1,0,0,0

Addtionally one could also include the $1s$ core orbitals with closed,0

The next thing to include would be the next higher lying virtual orbitals coming from the $3s3p$ AO shell: occ,5,2,2,0,5,2,2,0. After that one could go on with the $3d$ and $4s4p$ shells.

The occ,3,2,2,0,3,2,2,0 card would include the bonding $2\pi_u (3p_{x/y})$ and anti-bonding $2\pi_g (3p_{x/y})$ orbitals, but not the $4\sigma_{g/u} (3s)$ and $5\sigma_{g/u} (3p_z)$ which are similar in energy.


Why isn't standard Gibbs free energy always zero?

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

It isn't directly $G$, but formula for change in free energy. Also, in standard conditions $\Delta G = \Delta G^\circ$, so it isn't necessarily 0.

Relevant reading on a website


I think you have got the calculation right. However, I would also consider the coupling between F and H because coupling through 3 bonds is also usually visible. So, each of your 8 lines will be split into four by the H-atoms.
That will be the major part of the spectrum. You would also get weak signals with a different splitting pattern due to the other isotope of boron ($\ce{^10B, I=3}$, abundance=20%).


No cylinder regulator that I know of will provide fine control of flow rates on the order you require.  The purpose of a regulator is to control the line pressure, not the line flow rate.  In theory you could stick a flowmeter in-line and try to finesse the regulator pressure and shutoff valve (if there even is one) to reach the flow you want, but it would probably be a maddening exercise.

What you need is a mass flow controller (MFC) of some kind.  Depending on your budget and your needs for adjustability and precision of the flow rate, there is a variety of options:

If you know you will only require a specific, set flow rate (or, a small number of set flow rates) you might do okay with a flow orifice: set the upstream pressure appropriately and the flow should match the spec.
If you need to be able to vary the flow rate and an approximate value is acceptable, something like these dial-knob flow control valves would probably work okay.
If you need precise control of the flow rate, you're probably up against something like these Porter mass flow controllers. They're quite pricey (a few thousand USD) but they work extremely well.  You'd need both the in-line MFC unit and the separate control box to make it work.

IMPORTANT: One thing to be aware of is that all of these technologies are gas-specific to a greater or lesser extent.  The orifices and dial-knob controllers depend at least on the viscosity and density of the gas, and the Porter MFCs depend also on the specific heat of the gas.  The differences will be minor, so I wouldn't worry about it for the first two, but if you need highly precise flow rates, you will have to compensate for the differences if you should need to switch gases.  Porter actually calibrates each MFC to a specific gas, so if you only need argon flows, you'll be all set.


Why should the redox reaction happen between the copper ions and iron atoms?

This is a good question, but you may have gotten off on the wrong track by thinking about the electronic shell structure (although what you have written seems correct, if a bit old-fashioned).  The relevant phenomenon here is the relative abilities of the metal ions to oxidize (or remove electrons from) each other.  For your specific example, the copper ion is a better oxidizing agent than the iron atom.  Therefore it is able to take two electrons away from the iron atom, oxidizing it and leaving the iron 2+ ion behind.

As for why this happens, at your current stage of education the best answer is probably "because iron is easier to oxidize than copper".  You can look up the electrochemical activity series to find out how various metals relate to each other in terms of how easy they are to oxidize.  Copper is one of the harder ones to oxidize, but if you were to mix copper metal with silver ions, you would end up with silver metal and copper ions, because silver is even harder to oxidize than copper.

Does that help?


In a nutshell, water is very much not like semiconductors. Its conductivity is due to altogether different mechanism, and follows different trends.

Typical semiconductor is a rigid network of atoms held together with bonds. Nearly any "wrong" atom would provide either extra electrons or holes (that is, missing electrons). Holes and electrons are the charge carriers here. Being very small, they move fast, so you only need a tiny amount of those impurity atoms (say, a few ppm) to create significant conductivity.

Water is quite different. You can't have free electrons in water, to begin with; they will instantly react with something and become ions. Ditto for holes. Ions are the only allowed charge carriers in water. Ions are as big as atoms, and each carries a thick coat of water molecules stuck to it, so effectively they are even bigger than atoms. An ion would move in an electric field, but with great difficulty, much like a man pushing his way through a busy crowd. (Compared to it, the electrons would be like little sparrows darting overhead.) Therefore you need a lot of ions to create some real conductivity.

Things that produce ions in water (dissociate, as we say) are called electrolytes, and there is plenty of them around. For example, table salt dissociates into $\ce{Na+}$ and $\ce{Cl-}$; sulphuric acid dissociates into $\ce{H+}$ and $\ce{SO4^2-}$. Elemental chlorine ($\ce{Cl2}$), being a neutral molecule, does not dissociate; also, you can't add much of it, for it is a gas, and its solubility is not that great. Over time, it would react with water to produce certain compounds which do dissociate, so you'll have some conductivity after all, but that's a weird and inefficient way of achieving it. Using any soluble salt is much more straightforward.

The choice of particular electrolyte depends on your ultimate purpose.


The product of a reaction series involving 1,3-dithiane, 1-chloro-3-iodopropane and butyllithium

Some right ideas, but unfortunately, not correct. The correct answer is (C).
The protons in between the two sulfurs are more acidic because of hyperconjugative stabilization into the sulfur. Addition of 1-iodo-3-chloropropane results in the substitution product as you guessed. Another equivalent of butyllithium does the same thing again, except, this time, we do an intramolecular substitution for ring closure. Mercury has high affinity for sulfur, so that deprotects what is analogous to a ketal to give the ketone.

The key for these kinds of problems is to approach it from both ends. For example, the products are all carbonyl groups. Where would those groups come from? They can only be from the last step because otherwise butyllithium would add into a carbonyl. This gives you some idea for what 'C' would look, and so on.
For the forward direction, it's really important that you not kill all the reactivity by substituting off the heteroatoms. If you do that, you'll end up with an alkane, and there's no good chemistry unless you want $\ce{C-H}$ activation magic. Since butyllithium is not adding into anything, consider acid-base chemistry because it's also a very powerful base.


What are the products of alkene oxymercuration with diols?

There are so many structures here that it is at times difficult to keep things straight. The best way to approach your question is to take advantage of what is known about oxymercuration. If you enter "oxymercuration" at Search on Chemistry at the top of the page you will find a wealth of information at your disposal right here on ChemSE.
To start, the alkene is the (E)-propenylbenzene 1 and the diol is butan-2,3-diol 3, devoid of stereochemistry. Formation of mercurinium ion 4 with mercuric trifluoroacetate (2) is drawn as unsymmetrical because in the transition state for addition the carbon bearing the greater positive charge is the site of SN2, C-O bond formation. Mercurial 5 is a racemate, ignoring any stereochemistry of 3, which is what is formed after the first reaction. The second reaction, reduction with NaBH4, proceeds via radical 6 to form alcohol 7. A minor product may be bis-ether 9 arising from alkene 1 and mercurial 5. For mercurial 5 to form 1,4-dioxane 8 requires an oxidation of the carbon bound to mercury and a reduction of Hg+2 to Hgo as shown by the red arrows in structure 5. How do we know that this process does not occur? Because if it did, NaBH4 would not be required! Consider the oxymercuration of alkene 1 in the presence of water. If what you are suggesting were true, then 1 would form a diol rather than the observed mono-ol prior to the participation of NaBH4.


Why exactly does atomic radius increase down a group?

The description of $Z_\mathrm{eff}$ you gave is a bit too simplistic. As you correctly said, that would predict the same value of $Z_\mathrm{eff}$ for all elements in a group, which is not true.

In general we have $Z_\mathrm{eff} = Z - \sigma$ where $Z$ is the nuclear charge (that solely depends on the number of protons) and $\sigma$ is the shielding constant, which reflects the electron-electron repulsions but is not a simple function of the number of core electrons.

One of the earliest models to determine $\sigma$ was described by Slater in 1930.1 These are easily found online as "Slater's rules". These are quite simplistic, and so people tried to find better ways to calculate $\sigma$. Nowadays, one very popular source for values of $Z_\mathrm{eff}$ are the Clementi values.2 That's still quite long ago, so you can imagine that since then, we have come up with even more complicated ways to calculate it.

A peculiarity is that the values of $Z_\mathrm{eff}$ actually increase down the group. At the very least, that should dispel the myth that $Z_\mathrm{eff}$ only depends on the number of core electrons.

(Source: Keeler & Wothers, Chemical Structure & Reactivity: An Integrated Approach, p 264)

"Surely that should mean that the atomic radii decrease down the group!" you say.

I don't fault you for thinking that at all. However, that's not what we see: the atomic radii clearly increase going down the group. The second factor is the principal quantum number, $n$.

In the hydrogen atom, an electron in the 1s orbital has a much smaller radius3 than an electron in the 2s orbital. That's something that comes right out of the quantum mechanical description of the hydrogen atom.4 The same can be said for other atoms: a 2s orbital in lithium is smaller than a 3s orbital in lithium.

Note that I didn't compare a 2s orbital in lithium with a 3s orbital in sodium. Why? That's because sodium has a larger effective nuclear charge, and this serves to contract all the orbitals in sodium relative to those in lithium. Essentially, the orbitals are pulled inwards by the larger $Z_\mathrm{eff}$.

So, as you can see, going from lithium to sodium, there are two competing factors:

Larger $n$ suggests that the outermost electron should be further away from the nucleus.
Larger $Z_\mathrm{eff}$ suggests that the outermost electron should be nearer to the nucleus.

Unsurprisingly, it turns out that the variation of atomic radius is dependent on a combination of the two. We know that the atomic radius increases going down the group, so that must mean that the increase in $n$ (going from 2 to 3) outweighs the increase in $Z_\mathrm{eff}$ (going from 1.28 to 2.51), although it's sadly difficult to find an exact mathematical formulation in any textbook. It's likely that the exact dependency is very complicated and nowadays, such atomic properties are mostly calculated computationally anyway, which removes the need for such a mathematical formula.

Notes and references

1 Phys. Rev. 1930, 36 (1), 57

2 J. Chem. Phys. 1963, 38, 2686

3 I'm talking about $\langle r \rangle$, not the radius of a Bohr orbit, although that does also increase going from 1s to 2s.

4 I'm avoiding the circular logic that "higher energy means further away". It is further away because that is just the form of the radial wavefunction obtained by solving the Schrodinger equation - period.


Rate of aromatic nucleophilic substitution reaction on ortho and para fluoronitrobenzene

Reaction mechanism of aromatic nucleophilic substitution is as given -

Source : https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/
The rate determining step is the formation of the C-Nu bond, which causes the development of a negative charge on the carbon atom at ortho position to leaving group (fluorine). Presence of electron-withdrawing groups helps to stabilize the negative charge, and hence increase the rate of reaction.
The -NO2 substituent exerts not only a -M effect (resonance) but also a strong -I effect (inductive effect, which pulls the negative charge away from the ring). Inductive effect decreases rapidly as the distance increases, so its effect is much more pronounced for ortho than for para (Note that the initial negative charge is developed at the ortho position). As a result orthofluoronitrobenzene has a higher rate of reaction than parafluoronitrobenzene.


Why does the test tube face downwards in gas production?

If you mean the test tube containing the solids (reaction tube), then that tube faces downwards to promote the flow of gas into the receiving tube.  The idea is to let the gas sink away from the reaction zone and towards the receiving tube as it cools and becomes more dense.

The receiving tube, of course, faces down so that the ammonia gas, lighter than air, can float to the top after it meets the air.  So you have the ammonia flowing down with cooling through the reaction tube and then flowing up with buoyancy into the receiving tube.

Now for my bonus question:  Had you been producing carbon dioxide by heating magnesium carbonate, would you change the reaction tube?  How about the receiving tube?


At the eutectic point, what is the effect on the melting point of adding impurities?

Lets state my assumption that we can represent the free energies of the mixtures in fairly regular thermodynamics of phases ways (as I will get to here quickly). I will also assume fairly limited solid solubilities (think something like the Au-Si system).

OK, just take a system of A and B. The Gibbs free energy of a binary mixture (either solid or liquid) is:

$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + RT(x_{A}\ln x_A + x_{B} \ln x_{B})$, with $x_{A} + x_{B} = 1$, $G_{A,B}^{sol/liq}$ being the Gibbs free energy of pure A (or B) in the solid (or liquid) at that temperature, and $H_{AB}^{sol/liq}$ being the enthalpy of mixing of A and B.

For a eutectic, the free energy of the liquid drops below the tie-line connecting the free energies of the solid phases, and it clearly does this in such a way that the eutectic point has a composition that is intermediate to the compositions defining the solid tie-line. In a perfectly symmetric and regular system, this puts the eutectic at $x_{A} = x_{B} = 0.5$, but at least keep in mind that liquid composition has reasonably large values of $x_{A}$ or $x_{B}$, certainly larger than the minority composition in either solid end-member.

Now, we add a third element. The Gibbs free energy formula becomes longer, since it now takes in to account the binary and ternary interactions.  So, one gets:

$G(T) = x_{A}G_{A}^{sol/liq} + x_{B}G_{B}^{sol/liq} + x_{C}G_{C}^{sol/liq} + x_{A}x_{B}H_{AB}^{sol/liq} + x_{A}x_{C}H_{AC}^{sol/liq} +  x_{B}x_{C}H_{BC}^{sol/liq} + x_{A}x_{B}x_{C}H_{ABC}^{sol/liq} +  RT(x_{A}\ln x_A + x_{B} \ln x_{B} + x_{C} \ln x_{C})$, with $x_{A}+ x_{B} + x_{C} = 1$. This is pretty ugly, but we just added a little bit of C, so $x_{C}$ is small.

Now, back to the assumptions on the system being considered. For the liquid, $x_{A}$ and $x_{B}$ are reasonably large ($\sim 0.5$). In solid A, $x_{A}$ is near 1, $x_{B}$ is small(ish), and $x_{C}$ is really small.  In solid B, $x_{B}$ is near 1, $x_A$ is small(ish), and $x_{C}$ is small. Really small.

What has really changed? Well, for $x_{C}$ being really small, you can throw out all terms linear in C. The only thing left (!) is the entropy of mixing terms, where adding a little C results in a change of  $RT x_{C} \ln x_{C}$. But, this gets applied to both solids (A and B), and also to the liquid. All the free energy curves shift down by the same amount on an absolute scale. By geometry this means that the tie line connecting the two solid phases and the liquid at the eutectic point does not shift to first order in $x_{C}$.

The eutectic point will not shift.


Generally, you don't make pH of $7\pm2$ by dilution alone. The required dilution is so great that it would leave very little of your compound, and then any tiny impurity would affect the result greatly. Think of a droplet on the pH meter probe left from the previous measurement. Distilled water does not help much. All the same, it contains $\ce{CO2}$ from the air, so its pH is usually about 5.5, not 7 as one might expect.

When people want to have certain pH level more or less reliably, they use buffer solutions. Come to think of it, that's what nature itself does, because our blood is a buffer solution as well.


Here is a specific example from biochemistry. In the hydrolysis of ATP, the first step is the attack of a hydroxide anion at one of the phosphates. The triphosphate portion of ATP has multiple negative charges at neutral pH, so this reaction is very slow.

When catalyzed by an enzyme, the enzyme often binds to a magnesium cation in such location that it shields some of the negative charge. This makes the reaction proceed faster.

  Do electric charges increase activation barriers?

Yes, they do. And there are various strategies to lower the activation barrier in these cases. One is providing a counter ion as described above. Another is to change the charge via acid or base catalysis, i.e. protonating or deprotonating one or the other reactant to decrease electrostatic repulsion.

  Would the former have a lower activation energy since there is no repulsion between the molecules?

Yes, and it would be sufficient to remove one of the charges in the latter to lower the activation barrier because electrostatic repulsion requires both particles to be charged.


Defining equilibrium position vs equilbrium constant

You're using the word "pressure" too casually. There are partial pressures of species and total pressure (which is the sum of all partial pressures of all species, regardless of if they're participating in the equilibrium or not).

Partial pressures of reactants and products factor into the equilibrium constant $K_{\mathrm{p}}$. If you increase the partial pressure of a gas that is not participating in the equilibrium, none of the other partial pressures change, so the equilibrium is unaffected, but the total pressure will change. Thus, depending on the circumstances, changing the total pressure may or may not change the position of equilibrium.

The key is you should focus only on the partial pressures of interest. All other pressures are irrelevant, unless they somehow tell you about a change in a partial pressure of interest.


What will be the possible conditions to react bromo alkane with diethyl phenylmalonate?

You have tried a number of sensible approaches and nothing is working well. I think the problem here is that the anion you are creating is very stable and hindered and no amount of tweaking will significantly improve it. You either accept the yield as it is and scale the reaction accordingly, or change your strategy.

Your next step is hydrolysis and decarboxylation to give a substituted phenylacetic acid so I suggest you start with ethyl phenylacetate. You could alkylate it with NaH (1eq) in DMF with KI then 1-bromopentane. That should alkylate well, but there will be a percentage of di-alkylation so I'm going to suggest you condense ethyl phenylacetate with pentanal and then hydrogenate (or otherwise reduce the double bond). I think this will give you a cleaner reaction mixture and higher yield.


What does it mean for chemical reaction network to be in equilibrium?

Ignoring the math, "in equilibrium" means the forward and reverse reactions are proceeding at the same rate, so there is no gross change in the amount of reactants. In a closed system, if initially one reaction was faster, then it's product will start to dominate, and, by the law of mass action, the reverse action will "catch up" because there is now a larger supply of those reactants.

A simple physical analog:

Balance a ruler on two fingers, one finger near the end of the ruler, the other finger inset from the end towards the middle.
Gradually move your fingers together.

No matter where you start, the fingers will come to rest near the middle of the ruler.


First order, parallel reaction integration

Start by solving the rate equation for $\ce{[A]}$:
\begin{align}
\frac{\mathrm{d}\ce{[A]}}{\mathrm{d}t}
&= -k_\mathrm{B}\ce{[A]} - k_\mathrm{C}\ce{[A]}&
\Longleftrightarrow&&
\ce{[A]}
&= \ce{[A]_0}\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t).
\end{align}
We can then solve the rate equation for $\ce{[B]}$:
\begin{align}
\frac{\mathrm{d}\ce{[B]}}{\mathrm{d}t} &= k_\mathrm{B}\ce{[A]} \\
&= k_\mathrm{B}\ce{[A]_0}\exp\left(-(k_\mathrm{B} + k_\mathrm{C}\right)t)\\
\ce{[B]}-\ce{[B]_0}
&= \int_{\ce{[B]_0}}^{\ce{[B]}}\mathrm{d}\ce{[B]}' \\
&= k_\mathrm{B}\ce{[A]_0}\int_0^t\mathrm{d}t'\,\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t')\\
 &= \frac{k_\mathrm{B}\ce{[A]_0}}{k_\mathrm{B} + k_\mathrm{C}}[1-\exp\left(-(k_\mathrm{B}+k_\mathrm{C}\right)t)].
\end{align}
I use primes to distinguish the variable of integration from the limits of integration.


Nucleophilic substitution with chloroethane: hydrogen cyanide vs. potassium cyanide

Because hydrogen cyanide is not a very strong acid.

Potassium cyanide exists as potassium and cyanide ions, so here the cyanide ions can act as strong nucleophiles. Hydrogen cyanide is only a weak acid and so it will not dissociate to produce a significant amount of the cyanide ion nucleophiles. So it is a better idea to use potassium cyanide.


If the orbitals of the metal core is distorted. the amount of splitting is not high enough to promote the pairing of energy. therefore the 2 electrons will go to two separate eg orbitals. let the splitting energy due to this distortion be β1. the electron in the lowers orbital will have an energy loss of β1/2 and the electron in the higher orbital will have an energy gain of β1/2 and therefore the net gain in the energy is 0. which does not give any additional stability to the complex(as there is no release of energy).
hence the distortion does not have a strong effect.


Basic Intro to Quantum Chemistry. Two Slit Experiment and Bohr Model

As some commenters have stated, this is a complicated topic, and if you really want to understand it you would probably be best to study quantum mechanics itself. Although it's detailed it's also interesting!

However, as to your question, the first thing to ask is what does it mean for an electron to be "observed"? It can't mean just looking at them with your eyes, because then any matter we looked at would suddenly go crazy when all of its electrons change their behaviour. It also doesn't mean just looking in a normal light microscope, because electrons are too small to see in a light microscope. In fact to really properly "see" an electron, we would need to use very powerful high energy equipment. So high energy in fact, it would give the electron enough energy to fly out of the atom it was attached to. After that it perhaps could well fly off like a "normal particle" electron with a Newtonian path.

Now the obvious question everybody then asks is, is there a way to "look at" the electron bound to an atom without giving it so much energy? And the answer according to physics is no, it is actually impossible, to really "look at" a bound electron enough to make it act like a particle, it always takes a big amount of energy. You could "look at" it just a little bit, just enough to find what energy level it is in for example - but as you probably know this isn't enough to stop it behaving like a spread out wave (atomic orbital).

Now as to your question about why it doesn't fall in to the nucleus, you have actually already answered it. Why would it? The earth doesn't fall in to the sun. You mentioned that electrons are very light, but that doesn't really matter. If we hollowed out the earth to make it much lighter that wouldn't affect its orbit. Now the real reason is a kind of "quantum mechanical" version of this argument that I could explain if you like, but it is complicated and really this is the key reason.


Why is cyclooctatetraene non planar but the cyclooctatetraenide anion planar?

Fulfilling Hückel's rule means that you get to be aromatic.

The reason cyclooctotetraene is non-planar is because its planar arrangement would be antiaromatic. It twists to avoid it, because antiaromatic compounds are unstable with respect to localisation of the π-electrons (see here: What is the justification for Hückel's Rule?)

The veritable nanosecond that a molecule has the potential to become aromatic though, it will leap at the opportunity. Any angle strain from being planar is so quickly overtaken by the huge jump in stability that aromaticity brings that the molecule will twist to form an aromatic structure in a blink.


Which is more soluble in water, HF or SO2?

There are a couple different forces at play here. One big difference between the compounds is that $\ce{SO2}$ is much more volatile, with a boiling point of $\pu{-10 ^\circ C}$; in other words it's a gas. It does strongly hydrogen bond with water however, and is pretty soluble at $\pu{94 g/L}$.

$\ce{HF}$ is also pretty volatile, but much less so than $\ce{SO2}$ with a boiling point of $\pu{20 ^\circ C}$. It also is a more polar molecule than $\ce{SO2}$ and hydrogen bonds even stronger with water. In fact, $\ce{HF}$ is fully miscible with water, so clearly that would be the more water soluble compound.

To summarize, the water solubility of $\ce{SO2}$ is $\pu{94 g/L}$, whereas $\ce{HF}$ is fully miscible, thus $\ce{HF}$ is the more soluble compound in water.

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