I actually tend to agree with the answer from @JanJensen based on electrostatic effects, rather than with the "covalency" argument.

In particular, it has been mentioned in the comments that the $\ce{Al^{III}-OH2}$ is a "far more covalent" bond than one $\ce{Mg^{II}-OH2}$ one. This remark, however, does not actually explain anything, but rather it shifts the question to a new one, i.e. why is it, then, that the $\ce{Al-O}$ bond is more covalent?

Water exchange reactions at the Mg and Al centers have actually been studied extensively, but unfortunately not to the same extent as those for transition metal ions, or for selected lanthanides. The latter in particular are a subject of continuous and very intense interest, because water exchange at e.g. $\ce{Gd^{III}}$ or $\ce{Eu^{II}}$ ions play a pivotal role in the effectiveness of complexes of these ions as MRI contrast agents.

Nonetheless, here are a few very relevant references that will certainly offer some more insight, but probably not the final answer that you are looking for.

An $\ce{^{17}O}$-NMR Study of the Exchange of Water on $\ce{AlOH(H2O)5^{2+}(aq)}$, Inorganic Chemistry, 1998, 37 (19), 4760–4763;
Water exchange on metal ions: experiments and simulations, Coordination Chemistry Reviews, 1999, 187 (1), 151–181.

In fact, when studying exchange reactions at metal ions, the authors of these and other studies tend to consider ions with equal charge as homogeneous series, because the effect of the charge will swamp out any other effect, some times even the relatively large effects of electronic configuration in d-transition metals!

That notwithstanding, the papers above agree that the mechanism for the water exchange reactions for both Al and Mg metal complexes is dissociative in nature, which means that the rate determining step involves the loss of a $\ce{H2O}$ ligand from the inner sphere of the complex.

We can combine this piece of information with the fact that, as mentioned in the second paper above, there is a significant increase in charge density when going from $\ce{Mg^2+}$ to $\ce{Al^3+}$. First, the aluminum cation is smaller (reported values from that reference are: 72 pm for $\ce{Mg^2+}$ and 53 pm for $\ce{Al^3+}$); secondly, it has higher charge. This rather staggering difference can again be explained using the usual electrostatic arguments (i.e. the two ions are isoelectronic, but the Al nucleus contains more protons, so electrons in the latter are held more closely, leading to smaller cations).

Trying to put this all together:

removing a water molecule determines the rate of the overall exchange process according to the experimentally determined dissociative mechanism;
the smaller, more highly charged $\ce{Al^3+}$ exerts a much stronger attraction on each of its bound $\ce{H2O}$ ligands, making said removal much more costly;
in macroscopic terms, the $\ce{Al^3+}$ has a much higher activation energy, which ultimately manifests itself in the much lower rates of exchange that the OP referred to.


Why are antibonding orbitals higher in energy?

You shouldn't stop asking "why," but really explaining why is really, really complicated. It sort is like learning to run. First you learn to waddle around, then you can walk, and then you can run. Starting off such an explanation with quantum mechanics isn't throwing you in the pool it is more like throwing you in the middle of the Pacific Ocean.

Start with this. I assume that you're happy with the shell structure of atoms. Now we want to make molecules.

Consider the reaction: $$\ce{H + H -> H2}$$

Two hydrogen atoms each with a $\ce{1s^1}$ electron bond to form a "molecule" with 2 electrons in some sort of orbital. There is no shell model for a molecule, so how do we conceptualize the two hydrogen orbitals combining? A straightforward approach is the linear combination of atomic orbitals (LCAO). So we have a $\ce{1s}$ orbital from the first hydrogen and a $\ce{1s}$ orbital for the second hydrogen. Remember that each of these orbitals can hold two electrons. We now, with some nativity, just assume that the energy of atomic orbitals must be conserved. So how much in energy the molecular orbital is lower than the hydrogen orbital then the other orbital must exist with that much "antibonding" energy. But for $\ce{H2}$ there are only two electrons, thus the antibonding orbital is empty and the net result is that a $\ce{H2}$ molecule has less energy than two lone H atoms and the molecule will stay together.

Now consider the reaction:
$$\ce{He + He -> He2}$$
Both He atoms have two $\ce{1s}$ electrons. Thus when we try to form a $\ce{He2}$ molecule both the bonding and the antibonding orbitals are occupied. The net result is that the atoms in the $\ce{He2}$ molecule have no net affinity to stay bonded together. Notice also that this would not be true for a $\ce{He2^+}$ ion. With two electrons in the bonding orbital and only one in the antibonding orbital the ion would be "stable" in free space - at least until the ion bumped into some other atom or molecule in interstellar space.

Now using LCAO is naive. However it works well for a lot of chemical reactions. The really nice thing is frankly that the mathematics is fairly simple. Think of Newton's gravity equation. It worked well enough to put men on the moon, but it isn't exact. Now with GPS satellites the GPS clocks must be corrected for relativity. Scientists knew about relativity before we went to the moon of course, but the expanded mathematical treatment just wasn't necessary.

So does that answer your question in enough detail so that you can accept the explanation?

Edit - adding information about quantum mechanics

The easiest treatment of atomic orbitals is to assume hydrogen like orbitals and tweak the value of the orbitals based on the interactions of electrons. The gist here is that the simplest mathematical treatment assumes that the electrons move around the nucleus in defined orbits like the planets move around the sun. It was quickly realized that this sort of modeling was woefully inadequate.

Even with this simple treatment there is a massive fundamental problem that bedevils all such calculations. The three body problem in physics. If our solar system just had the sun and the earth, then it would be possible to use Newton's gravity formula to solve the equation which predicts the movement of earth exactly. However when the rest of the solar system is added, then an exact solution is no longer possible. We can solve the movement numerically by essentially calculating where the planets will be after some time $t$. So based on where all the bodies are in the solar system we calculate a vector for earth and all of the other planets and advance the time by an hour. We repeat the calculation 525,960 times and we can predict where all the planets will be exactly one year from now.

Now skipping several other mathematical models which offered better solutions than the planetary model, we arrive finally at the Schrödinger equation which gives quantum mechanical calculations for orbitals. The high level results are threefold. First the Schrödinger equation is subject to the same three body problem as before so an exact solution is only possible for a system with one electron and the nucleus. Second the electron orbitals are not discrete paths through 3D space but rather smeared out probability functions. Third different types of suborbitals (s,p,d and f) have different shapes.

Now with the more realistic model for atomic structure we can again combine atomic orbitals to get molecular orbitals, with the new constraints introduced by quantum mechanics. But as we did before, when we create a molecular bonding orbital from two hydrogen atoms, then we get an antibonding orbital as well. Thus the same explanations used for hydrogen and helium bonding as before still apply, however the actual values of the bonding energies do change.

I'd be remiss if I didn't point out something else. Going back to the the motions of the planets in the solar system we also stumble into something known as chaos theory in mathematics. Because of all the interactions small errors in the mass or velocities of the planets would result in much different results for the calculations. This is important in chemistry because the ethane molecules in a gas aren't a static entities, but rather dynamic ones. The molecules are rotating and the atoms are vibrating. Molecules are bumping into one another causing rotations and vibrations to change.

Now trying to model all of this in one model is too complex for us to handle yet computationally. So chemists build models for different aspects of the problem and then try to piece together an overall picture.

In the olden days, AAA would print a route map for someone wanting to drive from Augusta, Maine to Tampa, Florida. They didn't draw the whole route on a massive roll, but rather pieced the route together. Each page showed the pertinent decisions relative to that stretch of road. So a page in a large city might be a few miles, but out in the country a page might be a 100 miles.

So rather like AAA, professional chemists don't have one massive bonding model that we keep hidden from new students. Rather we describe atomic orbitals and molecular bonding in a number of different ways depending on the situation. By piecing the overall model of a molecule together we can understand predict overall behavior.


Effect of oxidation aluminium reflectivity

The lifetime of aluminum depends on the environment. If it's near an ocean, where air is humid and often contains salt particles, it may become pitted or look a milk white, rather than shiny, in just months. Anodizing can protect it for years, though.

Because the surface of anodized aluminum is porous (until sealed), it may have a somewhat more or less matte finish depending on treatment. Bright dip anodizing, depending on the aluminum alloy, offers a more shiny finish.

The Alzak process is used for lighting reflectors, and there is information on its specular reflectivity.


As you know you have to remove the others species like $\ce{I2}$ and $\ce{I3-}$. $\ce{I3-}$ I'm quite sure it exists only in solution so simply evaporating the water from Lugol 's solution you should eliminate it. To remove completely $\ce{I2}$ you have to do some others steps.

$\ce{KI}$ is produced industrially by treating iodine with $\ce{KOH}$.(see Wikipedia for reference).

$$ \ce{3I2 + 6KOH -> 5KI + KIO3 + 3H2O}$$

So I think the simple way is to treat Lugol's solution with $\ce{KOH}$ make an ice bath to precipitate $\ce{KIO3}$, filter the solution to remove it and than evaporate water.
You can also try to remove iodine by means of active carbon I know this method is used to purify water see ref and ref.
You can try evaporating the Lugol's solution and re-crystalize it and than eventually  try different solvents to solubilize one of the compounds and leave the other as precipitate but if think is quite difficult to find a better solvent than  water (as you know iodine is very insoluble in water instead $\ce{KI}$ is very soluble).

If you want to make the $\ce{KI}$ from your self here there are two Home experience:

extract potassium iodide from salt
make $\ce{KI}$ from $\ce{I2}$ and $\ce{KOH}$


Educational practice problems/systems/reactions in computational chemistry

There's a free software quantum chemistry package called Psi4. If you're a GNU/Linux user, under Ubuntu or any Debian-based distro, you can install it with a simple terminal command.

sudo apt-get install psi4

They have a well documented site, including a section dedicated to education with simple projects and friendly message boards. As of today, they list the following projects in their education section:

What is the radius of an atom?
What makes a molecule polar or non-polar?
How do we calculate the most accurate energies for a boron atom?
Can we visually predict binding energies?
When does water become $\ce{OH + H}$? A study in molecular orbitals.
What happens to the orbitals as a one-dimensional box gets longer? A study in molecular orbitals.
Is $\ce{C3H+}$ present in the Horsehead nebula?
What is symmetry in chemistry?

The version available in Ubuntu repository is functional, but quite outdated, 1.1.5, as of Mar 2020. If one is serious about learning it, my advice is to download it directlly from their site. For me the only issue with it is that Avogadro, my output viewer of choice, seemingly doesn't provide native support to view Psi4 output files. But I count that as a missing feature in Avogadro, not Psi4.


How do you melt metals with super high melting points?

Tungsten's melting point of 3422 °C is the highest of all metals and second only to carbon (3550 °C) among the elements. This is why tungsten is used in rocket nozzles and reactor linings.  There are refractory ceramics and alloys that have higher melting points, notably $\ce{Ta4HfC5}$ with a melting point of 4215 °C, hafnium carbide at 3900 °C and tantalum carbide at 3800 °C.

Carbon cannot be used to hold molten tungsten because they will react to form tungsten carbide.  Sometimes ladles and crucibles used to prepare or transport high melting point materials like tungsten are lined with the various higher melting ceramics or alloys.  More typically tungsten and other refractory materials are fabricated in a non-molten state.  A process known as powder metallurgy is used.  This process uses 4 basic steps:

powder manufacture - a variety of techniques are available to generate small particles of the material being worked
powder blending - routine procedures are used to blend the constituent particles into a uniform mixture
compacting - the blended powder is placed in a mold and subjected to high pressure
sintering - the compacted material is subjected to high temperature and some level of bonding occurs between particles.


That's an appropriate interpretation of Henry's law, although as you've written it $K$ would be referred to strictly as the solubility constant $H$, see this link to the wikipedia page provided in the comments.

The following shows more formally how to derive the connection between Henry's law, an equilibrium constant and the standard Gibbs free energy for evaporation of the solute.

At equilibrium between the solute in the gas and solution phases, the chemical potential of the solute is equal in both phases, so that

$$\begin{align} \mu(sol)&=\mu(g)\end{align}$$

In the gas phase (assuming ideal behavior)

$$\begin{align} \mu(g)&=\mu^\circ(g)+RT\log\left(\frac{p_i}{p^\circ}\right)\end{align}$$

Now according to Henry's law, the partial pressure and concentration in the solution are related as

$$p_i=K\frac{m_i}{m^\circ}$$

(of course, as you point out, this can already be recast as an equilibrium constant expression, but read on)

so that

$$\begin{align} \mu(sol)&=\mu^\circ(g)+RT\log\left(\frac{K(m_i/m^\circ)}{p^\circ }\right)\end{align}$$

Defining

$$\begin{align} \mu^\circ(sol)&=\mu^\circ(g)+RT\log\left(\frac{K}{p^\circ }\right)\end{align}$$

we can see that

$$\Delta G^\circ_m = \mu^\circ(g)-\mu^\circ(sol)=-RT\log\left(\frac{K}{p^\circ }\right)=-RT\log\left(K_{eq}\right)$$

which is the familiar relation between the standard Gibbs free energy change and an associated equilibrium constant for the equilibrium $\ce{solute <=> gas}$ provided we define the equilibrium constant as

$$K_{eq} = \frac{K}{p^\circ } = \frac{p_i/p^\circ}{m_i/m^\circ }$$


Nucleophilicity Of Amino Acid Side Chains

Aspartic acid < Tyrosine < Threonine < Methionine is the order of nucleophilicity of the amino acid side chains. So, Aspartic acid (neutral form) is the answer. Every side chain has a Oxygen or Sulphur (Methionine) group available for electron pair donation, but in Aspartic acid the Oxygen group has lesser inclination for donation of electron density (due to resonance). Moreover, the solvation layer surrounding the acid group of Aspartic acid makes the job of nucleophilic attachment even more tougher.


Polymerisation of methyl methacrylate - can the C=O group get involved?

C=O bond is strongly polar, and the cleavage of the double bond would rather involve heterolytic breaking with ions formed than homolytic with radicals.


Maybe a numerical example will help you. Let's start from pure water, with the following concentrations :$\ce{[H+] = [OH-] = 10^{-7} M}$. Now we will suppose you add $\ce{10^{-7}} mol$ $\ce{HCl}$ in one liter of this water. Suddenly, the concentration of $\ce{H+}$ should double. This is not possible in the long run. A part $a$ of these supplementary $\ce{H+}$ ions must react with the same amount $a$ of the $\ce{OH-}$ ions to form water. The final molar concentrations of both ions are given by  : $$\ce{[H+] = 2.000·10^{-7}} - a$$  $$\ce{[OH-] = 1.000·10^{-7} } - a$$   This amount $a$ may be obtained by solving the equation : $$\ce{ (2.000·10^{-7}} - a)\ce{(1.000 10^{-7}} - a)= 1.000·10^{-14}$$ The solution is : $a = 0.382·10^{-7} mol/L$. So about $38.2$% of the initial $\ce{OH-}$ ions have been destroyed by the addition of the new acidic ions. And the numerical values of the final concentrations are $$\ce{[H+] = 2.00·10^{-7}}- a = 1.618·10^{-7} mol/L$$$$\ce{[OH-] = 1.00 10^{-7} } - a = 0.618· 10^{-7} mol/L$$ And of course : $\ce{[Cl-] = 1.000·10^{-7}$ $ mol/L}$. If you want you may check the product of the two final concentrations $\ce{[H+]}$ and $\ce{[OH-]}$. It is equal to ${1.000·10^{-14} mol^2L^{-2}}$.


How do I prevent oxygen interfering when I prepare a FeCl2 solution from anhydrous FeCl2 powder?

See if your group or an other (e.g., organic / organic catalysis) has a Schlenk line.  With some training, this allows you to reduce the interference of oxygen greatly, because you may evacuate your reaction flasks and to replace the air by an inert gas such as nitrogen or argon.  Nitrogen is relatively cheap because it equally is used for welding (as gas) or for cooling (think e.g., NMR spectrometers).  Argon gas, on the other hand has the advantage of a density greater than the one of air, thus you may cover and protect your substance under a blanket of argon.
Then, search for Schlenk flasks.  You identify these quickly for their connector to said Schlenk line. Besside round bottom flasks and tubes (a gallery, a second one), there equally are Schlenk flasks to perform a pressure filtration while working under argon enclosing a filter plate (example).
Then, degas your solvent.  In HPLC, bottled solvents for example are placed in the basket of the ultrasound bath while purging them with a steady, slow flow of argon to remove dissolved oxygen.  Do this for your deionized water.
Weigh in $\ce{FeCl2}$ in your dry Schlenk flask, and add the deionized water prepared under the inert gas.


Which of these solution have the lowest vapor pressure in 298 K?

All examples are mixtures of not-volatile (here: salts, but it would be the same for a teaspoon of sugar) and volatile compounds.

The vapour pressure above these mixtures is proportional to the mole fraction of the volatile component. With other words:

The higher the concentation of the salt, the lower the vapour pressure.

But as user80551 pointed out in his comment, we should not look at it this way. Water isn't "just the solvent" here.

It is the volatile compound in the mixture that matters! The higher the mole fraction of the volatile compound, the higher the vapour pressure!

With (again) other words: Raoult's Law ;)


Carbon dioxide concentration from pressure changes

If you know the partial pressures of the gases, then you can calculate the mass fractions.  In the present case, assume you have 1 mole of gas.  Then the number of moles of CO2 is 0.167 moles and the number of moles of air is 0.833.  The mass of CO2 is (44)(0.167)=7.33 grams and the mass of air is (29)(0.833)=24.2 grams.  So the mass fraction of CO2 is $$\frac{7.33}{(7.33+24.2)}=0.233$$
So the ppm CO2 by volume = 167000 and the ppm CO2 by mass = 233000.


Here is a picture of a "classical" carbocation, there is an electron deficient carbon bearing a positive charge.

There are many examples of "non-classical" carbocations, but the 2-norbornyl carbocation is among the best known.

Labeling experiments have shown that the positive charge resides on more than one carbon in the 2-norbornyl ion.  Early on, the data was explained by equilibrating classical ions, but soon another possibility emerged - one involving a single non-classical ion.

The problem comes down to: are the equilibrating classical ions ground state structures with the non-classical ion serving as the transition state, or is the non-classical ion the ground state? This debate went on for a very long period of time, but now most agree that the non-classical structure is the ground state in the 2-norbornyl system.  In fact, a recent, and difficult to obtain, crystal structure for the 2-norbornyl cation has been published proving that the ion exists with the non-classical geometry (thanks to Klaus for finding this reference, see his comment below).

A key difference between classical and non-classical structures is the bonding.  As illustrated above, a classical ion has a carbon with a sextet of electrons and 3 other bonds.  The non-classical ion, on the other hand, involves 3 carbons with 2 electrons spread over them.  This is called a 3-center 2-electron bond (hypercoordinate bonding) and is a clear marker for a non-classical ion.  Notice if you count all of the bonds to any of these 3 carbon atoms (solid and dashed lines) you count 5!  Sounds strange, but such "hypercoordinate bonding" is a permitted consequence of the 3-center 2-electron bond.

Here is a link to a simpler non-classical carbocation discussed here yesterday.


Benzylation and acetylation of amines and alcohols

Phenol is a pretty good leaving group so phenyl acetate is an acetylating agent and fairly labile in aqueous base. The reaction conditions are not clear enough IMO to say whether you would get a or c, but with a basic work-up the OAc can hydrolyse back to phenol. If the amount of Ac2O is limited then any PhOAc formed will acetylate the aniline giving back PhOH.


Reaction of chlorine with hot/cold solution of hydroxide

The chlorate ($\ce{ClO3-}$) is formed by disproportionation of hypochlorite and hypochlorous acid:

$$\ce{ClO- + 2 HClO -> ClO3- + 2 HCl}$$

Higher temperature promotes the formation of hypochlorous acid through hydrolysis of hypochlorite and therefore speeds up the reaction.


Number of water molecules in a box to simulate liquid water

Generally speaking, you use as many molecules in a simulation as you possibly can. There are a couple of caveats. For instance, if you wish to do a simulation using an expensive method (relative to a classical potential) like DFT, you will need to use fewer molecules. This has the obvious drawback that when you use periodic boundary conditions, each molecule will interact with its own reflections much more than we would like (which is not at all). Nowadays, we can certainly do simulations with more than 32 water molecules using DFT.

Also, there is no reason to just keep using larger and larger numbers because it is known that interactions between molecules in the liquid die off reasonably quickly with distance, so at some point you stop gaining anything by adding more molecules. All of the long range effects can be handled by using periodic boundary conditions, which works very well if each molecule is far enough from its own image that a molecule does not have short-range interactions with itself.

Another consideration is that the number of molecules you use depends a lot on what property you are interested in studying. Conventional wisdom about the number of molecules needed is usually associated with the number of molecules that give converged average energies. If you are interested in something like diffusion coefficients, however, you may need quite a few more molecules before convergence is achieved. I have read that up to 512, or even 768, can be needed to get good diffusion coefficients around room temperature for water.

It is true that sometimes a specific number of molecules is chosen to match a specific density you want to do simulations at. This, however, isn't usually too big of a deal as you can always change your box size to match the density you are trying to achieve.

There is no physical reason that people choose powers of 2 for the number of molecules other than that computational scientists tend to use powers of 2. The number 216 is probably used because some seminal paper used this number and got good results, so people tend to just stick with the trend.


How many boiling points does a mixture of liquids have?

I do use my comment above because the meaning of the books is, with little interpretation, clear and the answer is simple and don't require digging in theory of fractional distillation. So it would be pity to let the question unanswered.
A mixture as those discussed boils over a range of T in the sense that its boiling point BP depend on its composition.
The actual mixture, i. e. at a given molar ratio, has one well defined BP (at given P).
If you do not look at a given composition, but you actually conduct a distillation, as the boiling change composition the BP changes all along the process/column/plates, actually spanning a range of T.


A little help understanding some basic algebra(I think) in conversion pathways

Yes, 3.18cm has to be raised to power 3.
$$Density=Mass/Volume$$ or upon rearrangement
$$Mass=Density*Volume$$
The information is provided about edge of the cube made of Osmium and its density. So in order to obtain mass of the cube, it is necessary to find the volume of the cube(since the information about density is supplied). The volume of the cube is edge raised to the power 3,i.e.,
$$Volume of cube=edge^3$$
Since the information regarding edge is provided in inches it has to be converted into cm because the units of density is provided in $g/cm^{3}$.
Once the the volume is also obtained in $cm^{3}$, the mass can be obtained in g.

Follow these steps:

Convert the length into same unit as the the length unit in density(here it is $cm^{3}$). So, 1.25 in=3.18cm
Now to obtain volume use the formula for volume,i.e., to cube the length. So, volume=$$(3.18cm)^{3}\approx 32.16cm^{3}$$
Multiply volume with density to obtain mass. $$Mass=32.16cm^{3}*22.59g/cm^{3}\approx723g$$


Is an allene or a conjugated diene formed?

As the intermediate passes through an allylic carbocation which is fairly stable, it does not matter which H is 'more acidic' - the whole system is in equilibrium. The product you get is the thermodynamically most stable which will be a conjugated diene.


What is the dimension of Avogadro's constant?

The dimension/unit of the Avogadro constant $N_\mathrm{A}$ is actually $1/\mathsf{N}$ or $\pu{mol-1}$. It is shown that way in your first reference as well, i.e.
$$ N_\mathrm{A} = \pu{6.02214076E23 mol-1}. $$
What is "dimensionless" is the Avogadro number or the number of atoms / molecules in a single $\pu{mol}$ of that substance, sometimes written as $N$ or $N_0$.
The numerical value is the same, but the concept is different. Basically the relationship between the two is:
$N_0 = (\pu{1 mol})\times N_\mathrm{A} = \pu{6.02214076E23}$
See also the answer discussed in the comment by Ed V, where the new definition of the constant is explained: Effects of Changing Avogadro's Constant.


When MgCl₂ · nH₂O is heated to the boiling point of H₂O

Wikipedia pretty much has it all already.
There are numerous magnesium chloride hydrates $\ce{MgCl2 · x H2O}$ $(x = 1, 2, 4, 6, 8, 12),$ however in the temperature range from 0 °C to 115 °C the most stable one is hexahydrate $\ce{MgCl2 · 6 H2O}.$
Whether $\ce{MgCl2 · 6 H2O}$ decomposes to hydroxo-salt or only dehydrates depends on the conditions.
When heating takes place in atmosphere of $\ce{HCl},$ anhydrous $\ce{MgCl2}$ is formed:
$$\ce{MgCl2 · 6 H2O (l) ->[>\pu{115 °C}][HCl(g)] MgCl2(s) + H2O(g)}$$
Otherwise, magnesium hydroxochloride is formed [1, p. 522] and $\ce{HCl}$ gas is released:

The product obtained is always the hexahydrate, $\ce{MgCl2 · 6 H2O}.$ It is dehydrated to anhydrous magnesium chloride by spray drying and heating with dry hydrogen chloride gas. In the absence of $\ce{HCl},$ heating hexahydrate yields the basic salt, $\ce{Mg(OH)Cl}:$
$$\ce{MgCl2 · 6 H2O -> Mg(OH)Cl + HCl + 5 H2O}$$

References

Patnaik, P. Handbook of Inorganic Chemicals; McGraw-Hill handbooks; McGraw-Hill: New York, 2003. ISBN 978-0-07-049439-8.


What is resonance, and are resonance structures real?

This answer is intended to clear up some misconceptions about resonance which have come up many times on this site.

Resonance is a part of valence bond theory which is used to describe delocalised electron systems in terms of contributing structures, each only involving 2-centre-2-electron bonds. It is a concept that is very often taught badly and misinterpreted by students. The usual explanation is that it is as if the molecule is flipping back and forth between different structures very rapidly and that what is observed is an average of these structures. This is wrong!  (There are molecules that do this (e.g bullvalene), but the rapidly interconverting structures are not called resonance forms or resonance structures.)

Individual resonance structures do not exist on their own. They are not in some sort of rapid equilibrium. There is only a single structure for a molecule such as benzene, which can be described by resonance. The difference between an equilibrium situation and a resonance situation can be seen on a potential energy diagram.

This diagram shows two possible structures of the 2-norbornyl cation. Structure (a) shows the single delocalised structure, described by resonance whereas structures (b) show the equilibrium option, with the delocalised structure (a) as a transition state. The key point is that resonance hybrids are a single potential energy minimum, whereas equilibrating structures are two energy minima separated by a barrier. In 2013 an X-ray diffraction structure was finally obtained and the correct structure was shown to be (a).

Resonance describes delocalised bonding in terms of contributing structures that give some of their character to the single overall structure. These structures do not have to be equally weighted in their contribution. For example, amides can be described by the following resonance structures:

The left structure is the major contributor but the right structure also contributes and so the structure of an amide has some double bond character in the C-N bond (ie. the bond order is >1) and less double bond character in the C-O bond (bond order <2).

The alternative to valence bond theory and the resonance description of molecules is molecular orbital theory. This explains delocalised bonding as electrons occupying molecular orbitals which extend over more than two atoms.


No, not really.

For a double bond, you'd use the same '\' and '/' characters to indicate cis- and trans- stereochemistry that you use in SMILES. But as you indicate, there isn't a double bond for such stereochemistry.

My personal preference in scripts, code, etc. is to match the double bond, and then use my own code to check the relative positions of substituents around the central double bond. In this case, you could look at a dihedral angle around the C-C-N plane.

For future reference, an excellent resource for testing SMARTS patterns is SMARTSView.


Why does this ligand form 6-coordinate complexes with Transition elements and 8-coordinate complexes with Inner transition elements?

I decided to do some more research for this question as I had faced it before. Turns out the proper name of this ligand is diethylenetriaminepentaacetate ($\ce{DTPA^5-}$). It's an expanded version of EDTA and is primarily used in MRI scanning where it forms complexes with the Gadolinium ion. It forms complexes with a coordination number of six and seven with transition metals simply because these complexes are more stable. Here's Fe(III)-DTPA as an example1:

Experimentally Fe(III)-DTPA has been observed to adopt seven- and six-coordinate
crystal structures. This agrees with our result where the 8-coordinate Fe(III)-DTPA is $\pu{23 kJ mol^-1}$ higher in energy relative to the most stable 7-coordinate structure

With inner transition metals, it forms complexes with a denticity of 8, with Pr-DTPA being an example2:

References:

Sillanpää, Atte J., et al. “Density Functional Complexation Study of Metal Ions with (Amino) Polycarboxylic Acid Ligands.” Phys. Chem. Chem. Phys., vol. 5, no. 16, 2003, pp. 3382–93. doi:10.1039/B303234P.
Fusaro, Luca, et al. “Insight into the Dynamics of Lanthanide-DTPA Complexes As Revealed by Oxygen-17 NMR.” Inorganic Chemistry, vol. 51, no. 15, 2012, pp. 8455–61. doi:10.1021/ic3010085.


Decomposition of acetonide protecting group in chloroform

Deuterochloroform (and indeed 'normal' chloroform) are known to be mildly acidic if left standing for any length of time:

  Chloroform: It reacts slowly with oxygen, or oxidising agents, when exposed to air and light, giving, mainly, phosgene, Cl2 and HCl.

  Source: Purification of Laboratory Chemicals, 7 ed.

To compensate for this, its common to add molecular sieves and a small amount of potassium carbonate to a new bottle of deuterochloroform to keep it dry and to minimise any acid that may be present/may be generated as the bottle ages.

If you look up the entry for acetonides in Greens' Protecting Groups in Organic Chemistry, you'll see that acetonices can come off under quite mild conditions, though personally I've never had an acetonide come off in an NMR tube of chloroform (I assume you didn't leave it standing for months in the tube...)

In order to cleave the acetonide under acidic conditions, you generally also need a nucleophile (methanol is common, water would do the trick), which if present should be very obvious on the NMR, and might go some way to explaining how you lost the acetonide.  You should also consider other explanations, for instance you mentioned concentrating the chloroform- are you sure the rotary evaporator hadn't just been used to take off something wet and acidic?

Hopefully you can just put the acetonide back on without any issues, I suggest picking up a new bottle of chloroform-d and adding some sieves and potassium carbonate (you could try NMR-ing your old bottle to see if theres water/methanol etc in it, and then purify appropriately, but the stuff is sufficiently cheap that I wouldn't waste my time).


Length of NaH bond from the edge length of unit cell

$\ce{NaH}$ has a rock salt type structure meaning each edge has two $\ce{Na-H}$ bonds. When working out bond lengths in a unit cell, you assume the atoms are point charges (i.e. they take up no volume) and so you do not need to know the radius, only the unit cell parameter which as you have said is $\pu{488 pm}$. Since there are two bonds along each edge, the bond length is simply half of the unit cell parameter, therefore $\pu{244 pm}$.

Finally it dawned on me, why you may have been concerned about knowing atomic radii, yes, you need to know the radii to determine whether $T_\mathrm{d}$ or $O_\mathrm{h}$ sites are preferred. However, given the context of the question, there must have been something previously alluding to the octahedral sites being occupied (and hence rock salt structure) rather than the tetrahedral sites (which would give you a ccp structure with half of $T_\mathrm{d}$ sites occupied). However, you still wouldn't need to use the ionic radii in any calculations pertaining to bond length. I have also included a drawn picture of the unit cell of Sodium Hydride to help you visualise how to calculate the bond length.


Why does the proton NMR of ferrocenecarboxaldehyde appear to be four singlets?

I like to address some interesting NMR features of ferrocene derivatives. Try Hard's question about $\mathrm{^{13}C-NMR}$ of ferrocenecarboxaldehyde showing one signal for the substituted cyclopentyl ring prompt me to find some solid evidence. However, it seems like most of ferrocene derivatives give similar results regardless of how strong magnetic field you used. Therefore, Aldrich NMR spectra may not be the best spectra with high definition (e.g., they use $\lt \pu{300 MHz}$ machine for $\mathrm{^{1}H-NMR}$).

Following $\mathrm{^{13}C-NMR}$ values from very recent paper, which published NMR data of few ferrocene derivatives using a $\pu{400 MHz}$ (but $\pu{100 MHz}$ for $\mathrm{^{13}C}$) machine (Ref.1):

The values in left hand side is for $\mathrm{C_p}$ ring (substituted) while single value at the bottom is for all 5 $\ce{C}$s of unsubstituted  $\mathrm{C_{p'}}$ ring. If you look closely, you'd see monosubstituted rings have so close values that they may overlap in spectra taken in low resolution machines.

Reference:

N. D'Antona, R. Morrone, G. Gambera, S. Pedotti, “Enantiorecognition of planar “metallocenic” chirality by a nitrile hydratase/amidase bienzymatic system,” Org. Biomol. Chem. 2016, 14, 4393-4399 (DOI: 10.1039/C6OB00689B).


The relationship between energy and air pressure?

Boiling occurs when the vapor pressure of a liquid reaches ambient pressure.  In the example you asked about, water can be made to boil by removing the air in the chamber in which the vessel holding the water is kept.  When the air pressure gets low enough, there is not enough pressure to keep bubbles of water created by the water's vapor pressure from expanding.  We call this boiling.  At atmospheric pressure the water still exerts vapor pressure but the air keeps the bubbles from expanding.  The only way for water to reach the vapor phase here is for it to evaporate from the surface so we see no bubbles.


Why does liquid water form when we exhale on a mirror?

Why do we have water vapors when our body temperature is also <100°C in the first place?

At normal pressure, water boils at 100°C, meaning that bubbles of pure steam form under water. At lower temperatures, water molecules reversibly move from the liquid to the gas phase and back. The higher the temperature, the higher the vapor pressure, and the higher the equilibrium concentration (partial pressure) of water in air. On the geological scale, there is no equilibrium, and we experience different temperatures, different humidity (related to partial pressure of water in air), and different pressures depending on location.

  Why does ice not exist at T>0°C

Ice, as a pure solid, and water, as a pure liquid, have defined concentration at given pressure and temperature (the equilibrium constant expression for melting includes neither liquid nor solid water, it is simply 1). Above the melting temperature, all the ice melts, there is no equilibrium. This is different from the liquid:gas equilibrium, which exists at temperatures below the boiling point, with lower and lower concentrations of the gas as the temperature drops (here, the equilibrium constant expression included the concentration or partial pressure of water vapor).

  If liquid water can evaporate into gas at T<100°C, then why not ice turns into liquid at T<0°C?

Again, this has to do with pure liquids and pure solids having a constant (or nearly constant) concentration. If you add salt to the liquid, however, ice will turn into liquid below the freezing point (effectively lowering the concentration of water in the liquid). Also, the surface of the ice melts at lower temperature than the bulk, so even for pure water, there can be liquid at temperatures below the bulk freezing point.

  When I exhale on a wall, no water droplets form.

I would do an experiment and check. I think the water droplets are easier to see on a mirror. Try a grand piano (i.e a smooth surface painted with shiny black paint), you might see the water droplets there as well. Or take a long hot shower and check whether water droplets form on surfaces other than a mirror.


Why isn't standard Gibbs energy zero at equilibrium?

METHOD 1
Suppose we were to start off with pure reactants in stoichiometric proportions in the standard state of 1 bar, and end up with pure products in corresponding proportions in the standard state of 1 bar.  The change in Gibbs free energy for this process is $\Delta G^0$, as you have answered in your comment to me.  To bring about this change, we use whatever equipment is necessary, typically using a van't Hoff equilibrium box and constant temperature compressions and expansions in cylinders with piston.  If we had assumed different values of the product and reactant pressures in the initial and final states than 1 bar, then the change in free energy would not be $\Delta G^0$, but rather $\Delta G$, with $$\Delta G=\Delta G^0+RT\ln{Q}$$.  If the values of the pressures in this final state happened to be the same as the partial pressures of the reactants and products in an equilibrium reaction mixture, the $\Delta G$ would be 0, and the Q would be equal to the equilibrium constant K.  So we would have $$0=\Delta G^0+RT\ln{K} $$
METHOD 2
Here, we start out with a mixture of various amounts of reactants and products that are not at equilibrium, and we analyze the variation in the free energy of the mixture as the reaction proceeds.  So, let $n_A$, etc. represent the number of moles of reactants and products at any time as the reaction proceeds, let $G^0_A$ etc. represent the free energies of formation of the reactants and products, and let $p_A$ etc. represent the partial pressures of the reactants and products at any time as the reaction proceeds.  So at any time, the Gibbs free energy of the reaction mixture will be:
$$G=n_A(G^0_A+RT\ln{p_A})+n_B(G^0_B+RT\ln{p_B})+n_C(G^0_C+RT\ln{p_C}+n_D(G^0_D+RT\ln{p_D})$$ with $p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$ , etc., and  $n_A=n_{Ao}-a\xi$, $n_B=n_{Bo}-b\xi$, $n_C=n_{Co}+c\xi$, $n_C=n_{Do}+d\xi$, with $\xi$ representing the extent of the reaction and P representing the total pressure.
If we use this equation to make a plot of G vs $\xi$, the equilibrium of the mixture will be determined when G passes through a minimum.  If we differentiate  this equation with respect to $\xi$ and set the derivative equal to zero, we find algebraically that this happens when $$RT\ln{\left(\frac{p_A^cp_B^d}{p_Ap_B}\right)}=RT\ln{K}=-\Delta G^0$$Work it out yourself and see.
So both very different methods give the same result for the final equilibrium.


Why are drugs mixed with their sodium salts?

A large percentage of pharmaceuticals are sold as salts, and sometimes even as co-crystals, i.e. a drug cation and a drug anion. You are right in the sense that the basic purpose is to make them soluble but this is not the only one. If it is a intravenous drug, salt form may have less pain. A low molecular weight medicine might be in a liquid form yet its salt is a solid. Sometimes you want to make a hydrophobic salt so that its stability is higher in high humidity. Sometimes you wish to improve the taste. There are plenty of factors.

Most likely the reason for a mixture of medicine and its sodium salt is its controlled release and some buffering action. In case of your carmicide which is a mixture of sodium citrate and citric acid, is a buffering system and secondly citric acid alone is very corrosive and very bad tasting (tart). Fucidin is only allowed as a Na-salt.

This review will tell you the principles behind choosing a salt.

Go to Google Scholar and read:
Gupta et al. Salts of Therapeutic Agents: Chemical, Physicochemical, and Biological Considerations, Molecules 2018, 23, 1719; doi:10.3390/molecules23071719

  The choice of a particular salt formulation is based on numerous
  factors such as API chemistry, intended dosage form, pharmacokinetics,
  and pharmacodynamics. The appropriate salt can improve the overall
  therapeutic and pharmaceutical effects of an API. However, the
  incorrect salt form can have the opposite effect, and can be quite
  detrimental for overall drug development.

API= active pharmaceutical ingredient


What happens to iron electrodes (attached to a battery) in copper(II) sulphate solution?

The reaction is spontaneous, you don't need put a power supply.

Let's see the reaction with the potentials (obtained from the table of standard reduction potentials):

$$
\begin{align}
\ce{Fe^0 &→ Fe^2+ + 2 e-} &\quad E^\circ &= \pu{0.44 V} \\
\ce{Cu^2+ + 2 e- &→ Cu^0} &\quad E^\circ &= \pu{0.337 V} \\
\hline
\ce{Fe^0 + Cu^2+ &→ Fe^2+ + Cu^0} &\quad E^\circ &= \pu{0.777 V}
\end{align}
$$

So, the solid formed in both electrodes is copper. The green solution is related to iron sulfate.


Why does a first order reaction only depend on the concentration of a single reactant?

As a purely mathematical definition, a first order reaction is one in which the rate is directly proportional to concentration of reactants. The thing you should look for is physical examples of this. This only happens if you have reactions where one compound is decreasing and the other one is abundance (pseudo - first order) , or , radioactive decay where the decay rate is proportional to number of atoms at a point.
The reaction you have shown with three reactants would most probably me not first order unless two of the three are abundant. So, the solution to the confusion is to think of the physically possible scenarios which the equation models instead of creating hypothetical situations and force the math to model that.


Formation of 3-chloromethylbenzene VS 3-chloromethylbenzene VS 4-chloromethylbenzene

It very much depends on the reagents and reaction type you are using to halogenate the aromatic ring, however, the most common method taught during organic chemistry is electrophilic hologenation, which in this case the methyl will be para directing.

http://en.wikipedia.org/wiki/Electrophilic_halogenation#Scope


Why is the boiling point of hydrogen sulfide higher than that of hydrogen chloride?

Definition of the hydrogen bond

In 2011, a modern and more general IUPAC definition of what constitutes a hydrogen bond was recommended (Arunan et al., 2011). The definition is stated as such:

  The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a molecular fragment $\ce {X-H}$ in which $\ce {X}$ is more electronegative than $\ce {H}$, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation.

It appears that the definition of then hydrogen bond which includes interactions only with $\ce {F, O or N}$ seems too narrow and restrictive, which I personally feel is indeed the case.

Therefore, it is perfectly fine to say that $\ce {H_2S}$ makes hydrogen bonds with other $\ce {H_2S}$ molecules and that $\ce {HCl}$ makes hydrogen bonds with other $\ce {HCl}$ molecules.

Explaining the boiling points

Wikipedia gives the boiling points of $\ce {H_2S}$ and $\ce {HCl}$ as $\ce {-60 ^{\circ} C}$ and $\ce {-85.05 ^{\circ}C}$ respectively. When rationalising boiling point differences, the first consideration is always the strength of the intermolecular forces between the molecules in the liquid.

The Pauling electronegativities of $\ce {S}$ and $\ce {Cl}$ are $\ce {2.58}$ and $\ce {3.16}$ respectively. Evidently, the electron deficiency in the hydrogen atom bonded to $\ce {Cl}$ is more severe compared to that bonded to $\ce {S}$. This suggests that $\ce {HCl}$ molecules are capable of forming stronger hydrogen bonds. However, it is also important to consider the number of hydrogen bonds formed between molecules, as you have already noted in your question. Each $\ce {H_2S}$ can form 4 hydrogen bonds with other $\ce {H_2S}$ molecules, whereas each $\ce {HCl}$ molecule can only form 2 hydrogen bonds with other $\ce {HCl}$ molecules. Therefore, despite the fact that $\ce {HCl}$ molecules can form stronger hydrogen bonds between themselves, the intermolecular attractions between $\ce {H_2S}$ molecules still prevail as they are twice as numerous. Thus, $\ce {H_2S}$ has a higher boiling point than $\ce {HCl}$.

References

Arunan, E., Desiraju, G. R., Klein, R. A., Sadlej, J., Scheiner, S., Alkorta, I., . . . Nesbitt, D. J. (2011). IUPAC definition of the hydrogen bond. Terminology and nomenclature. Pure Applied Chemistry, 83(8), 1619-1636. doi:10.1351/PAC-REP-10-01-01

Electronegativity. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Electronegativity

Hydrogen chloride. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Hydrogen_chloride

Hydrogen sulfide. (n.d.). In Wikipedia. Retrieved January 9, 2018, from https://en.wikipedia.org/wiki/Hydrogen_sulfide


Implicit dependence of reaction rate through reduction potential and activation energy

The rate of a reaction depends on the thermodynamics of its transition state.
It would be better to look at it in terms of Gibbs activation energy. Therefore there two factors affecting the rate of a reaction:

the activation enthalpy $\Delta^‡ H $, that corresponds to the Arrhenius activation energy
the activation entropy $\Delta^‡ S $ that takes into account the probability of a transition state.

The whole dependency of the kinetic constant on these parameters is expressed by the Eyring and Wynne-Jones equation:
$$ k = \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$

Where $T$ is the temperature, $k_\text{B}$ is the Boltzman constant and $h$ is the Planck constant.
Sometimes a reflection coefficient $\kappa$ is added to the equation to take into account some other dependencies. It is near to 1 for most of the simple reactions.
$$ k = \kappa \frac{k_\text{B}T}h \operatorname{e}^{\frac {\Delta^‡ S} R }\operatorname{e}^{-\frac{\Delta^‡ H} {RT} }$$


For the solubility of AB2, Ksp=[A][B]^2=x*(2x)^2. Why is (2x)^2 not considered double dipping?

It's easy, if you consider:

$$\ce{AB2 <=> A + 2B}$$

It's solubility equations would be:

$$K_{\text{sp}}=\ce{[A][B]^2}$$

That equation follows from the law of mass action.

If you consider the amount of A as $x$, wouldn't the amount of B be $2x$ obviously?

If your issue is with why it's squared, it's just the equilibrium law, it follows from law of mass action. You can't really argue why the square is there.


According to Ashley et al, "a rhodizonate spot test kit [for] in-situ testing of lead in paint... can be used for screening (i.e., qualitative) purposes." The test is sensitive not only to metallic (elemental) lead, but also to lead compounds used as pigments. However, the reference cited also states that the sulfide and rhodizonate tests have a high incidence of false positives.

Sodium, potassium or even cesium salts of rhodizonic acid should be equally effective, since the active species is the rhodizonate ion, but be aware that the sodium salt is less stable in water solution and should be used within a few days.


Para-magnetism: Para-magnetism refers to the electronic configuration of an atom having one or more unpaired electron. That means an atom can be called para-magnetic if it has one or more unpaired electron. Now if atom A and B has no unpaired electron they have same para-magnetism (in this case zero). If they both have one unpaired electron still they have same para-magnetism (Here it's one).
Now look at the electronic configuration of the above sated options::
$Mn :: [Ar]3d^54s^2$
$Mn^{2+} :: [Ar] 3d^5$
$Fe :: [Ar]3d^64s^2$
$Fe^{3+} :: [Ar] 3d^5$
So they all have five unpaired electron. That's why they have same paramagnetism.


Wiki on açaí oil

Açaí  oil is obtained by cold pressing.  Steam distillation may be able to extract compounds like vanillic acid but may not get the anthocyanins or the triglycerides.  Many of the health benefits are from the healthy triglycerides and the anthocyanins.


Why are higher acids solid while lower acids are liquid?

Acids with a larger size have greater intermolecular forces than 'smaller' acids, meaning it would take more heat energy to break those bonds in larger acids than in smaller ones. Hence at room temperature the heavier acids would be solid.


What does the pressure in the HPLC(High performance liquid chromatography) do?

The pressure in HPLC systems is not an intended feature, but a consequence of how the system works.

There are several adjustable parameters/properties of a HPLC/UPLC system that will influence on the internal pressure inside the chromatographic column. Some of those are:

Solvent flow:  A greater flow of solvents mean a faster chromatophy. On the other hand, faster runs also mean lower efficiency. An increase in flow leads to an increase in pressure.
Column length: A bigger column will improve the separation efficiency, but also increase the duration of the run. Bigger columns leads to a greater pressure.
Particle diameter: The lower the particle, the greater the separation efficiency, but this also leads to a greater pressure.
Particle pore size: This has the same effect as above, but is not the same thing.

Optmizing this parameters and/or improving the technology related to them is what generally drives the research in the HPLC/UPLC development field, besides of course find the appropriate type of column, solvent gradient, etc. You can't simply make a system where the pressure is as high as possible, or some connectiong will eventually burst. On the other hand, it's not interesting to have a chromatographic run that lasts several hours, or doesn't really separate the molecules, specially when dealing with hundreds or thousands of different analytes.

There is much more to the above in several books and thousands of papers, including very detailed descriptions, experiments and calculations that show a much more precise relation between those and other parameters.


What is the maximum distance over which van der Waals forces act?

There is no strict upper limit you can place on any type of electromagnetic interaction. That being said, van der Waal's forces are due to the formation of instantaneous dipoles which are a result of perturbations of an otherwise spherically symmetric electron density.

The electric field produced by a dipole decays as $1/r^3$. Therefore, it should not be surprising that the interaction between two instantaneous dipoles decays as $1/r^6$. It's actually not completely obvious this should be the case, but it turns out to be true. At very close distances, the van der Waal's forces actually scale as $1/r^7$ due to an effect from quantum electrodynamics, but chemists don't typically worry about this because the effect is negligible in almost all chemical processes.

The actual distance will vary depending on the system, so it's not like one can give actual numbers for the distance. But, for a system like water, the electrostatic force, which decays as $1/r^2$, can be relevant for distances up to, say, 100 angstroms. So, you can try to back out the scale of the van der Waal's interactions from there.

Another way of saying this is that van der Waal's forces can have an important effect on the first solvation shell of a molecule, but they typically do not have a strong effect beyond that distance.


Yeast Fermentation with YPD + Apple Juice

Thank you for an especially clear and detailed question!

  For YPD + apple juice fermentation, I was thinking of doing the same procedure with 50 mL YPD + 50 mL apple juice in each "special flask". Will this work?

I think this would work but it seems like a better comparison to your earlier experiments to keep using 50 mL of total medium, i.e., mix 25 mL of YPD with 25 mL of apple juice.  That way the volume of the fermentation is the same in each experiment.

  Will this be good for the same HPLC setting?

Probably, although it depends on the components of the apple juice you use.  Have you injected just apple juice (unfermented) into your HPLC?  Have you verified that this negative control injection contains no ethanol or other interfering peaks in the region of interest?  If not, those are things you should do to make sure the HPLC method you have developed for YPD medium will still work in apple juice-based media.


Jahn Teller distortion and magnetic moments.

Planar complexes may be affected by this effect- not orthogonal complexes .  ref-(2017)
https://aip.scitation.org/doi/pdf/10.1063/1.4974805

    "The most striking result is that the planar copper(II) (3d9
    ) and nickel(II) (3d8
    )
    chelation afforded very strong ferromagnetic coupling often with 2J ≥ 300 K, owing to the orthogonal orbital
    arrangement between metal ion 3dx2-y2 and oxygen 2pz orbitals. Such a structure-magnetism relationship has
    recently been extended to coordination to rare earth metal ions."


2 brief observations about the answer by Mitchell.

a hexagonal lattice does not exist in 2-dimensions.

By definition,A lattice is an infinite array of points in space, in which each point has identical surroundings to all others. For example consider a 2d cubic lattice; now, for example, choose one of the lattice point (we call it point A) and connect it with an arrow with the point just above and with another arrow with the point on its right. Now move these two arrows as a solid object to another point of the lattice (we call it point B); you will see that the arrow will again connect this other lattice point with other two lattice points, one just above, the other at its right. You find the same thing for both point A and point B, whatever choice you make.
You can do the same job with a hexagonal arrangment of points. Choose again an arbitrary point A in the hexagonal arrangement and plot the two arrows connecting point A to two first neighbouring point. Then again move the two arrows as a solid object from point A to one of the 3 first neighbouring three points. You will see that now arrows will end where no point is present in the space. This is because the 2D hexagonal arrangment of points is actually constituted by two different sublattices, whose symmetries are trigonal.

the different crystal systems are defined by the symmetry operations that are present, not by their metric.

The group of the symmetry operations define the symmetry along the different main crystallographic directions. For example, in the cubic system we have a=b=c and alfa=beta=gamma=90 metric because the symmetry operations that are present impose this.
The tetragonal symmetry simply dictates a=b and alfa=beta=gamma=90; there's no METRIC restriction on the c-axis. The SYMMETRY along the c-axis is different from that along a and b, but in principle it is possible to find a crystal structure that has a tetragonal structure (= symmetry), but a cubic metric, that is a=b=c (and it is still tetragonal!). there are several compounds where this is observed. In conclusion, inequalities strictly hold on symmetry, not on the metric of the crystal structure.


Is there an uncertainty associated with the value 0 K for absolute zero?

As another answer explains, absolute zero is defined in the Kelvin temperature scale as precisely $\pu{0 K}$. But this is not the entire story, as all measurements have an associated uncertainty (all thermometers are imprecise), see e.g. the wikipedia article describing the International Temperature Scale of 1990 and the Provisional Low Temperature Scale of 2000. For instance the second wikipedia article links to a International Committee for Weights and Measures' document describing the uncertainties associated with the calibration of the temperature scale based on measurements of the melting pressure of $\ce{^3He}$.
Recently the SI value of the Kelvin was changed. This is explained in the wikipedia:

The definition of the kelvin underwent a fundamental change. Rather than using the triple point of water to fix the temperature scale, the new definition uses the energy equivalent as given by Boltzmann's equation.
Previous definition: The kelvin, unit of thermodynamic temperature, is 1/273.16 of the thermodynamic temperature of the triple point of water.
2019 definition: The kelvin, symbol K, is the SI unit of thermodynamic temperature. It is defined by taking the fixed numerical value of the Boltzmann constant k to be $1.380649×10^{−23}$ when expressed in the unit $\pu{J⋅K−1}$, which is equal to $\pu{kg⋅m2⋅s−2⋅K−1}$, where the kilogram, metre and second are defined in terms of h, c and $\Delta ν_{Cs}$.
The kelvin may be expressed directly in terms of the defining constants as:
$$\pu{1 K} =  \frac{1.380 649 \times 10^{23}}{(6.626 070 15 \times 10^{-34})(9 192 631 770)} \frac{h \Delta \nu _{Cs}}{k}$$

So what does this mean for the definition of $\pu{0 K}$ ? That doesn't change, that is an absolute, thermodynamic definition. However, evidently the triple point of water is now no longer exactly $\pu{273.16 K}$. You can think of this as shifting the uncertainty from the size of a step to the number of steps that take you from absolute zero to the triple point. But both before and after the redefinition, $\pu{0 K}$ has meant $\pu{0\pm  0 K}$, that hasn't changed.

$^\ast$you can't actually reach absolute zero, so you can only extrapolate to 0 K and provide an uncertainty for this extrapolation, i.e. you could estimate how far your measured temperature is from absolute zero, and provide uncertainties associated with your estimate.


Cobalt may possibly have all those other oxidation states (and others too) for this nomenclature system, we only care about the two most common oxidation states in ionic compounds: $\ce{Co^{2+}}$ and $\ce{Co^{3+}}$. Cobalt(II) compounds would thus be named  cobaltous and cobalt(III) compounds would be cobaltic. Which is yours?


Difference between titanium and titanium dioxide electrode

Used as an anode, a normal untreated piece of titanium metal will "anodize", forming a non-conductive titanium dioxide coating.  In my experience, the current through a large untreated titanium sheet anode drops below the milliamp level in under a minute.

But there are a variety of electrically conductive but oxidation resistant micron-scale coatings that can be formed onto titanium, including platinum (a "platinized titanium anode"); conductive oxides of common metals like Pb or Sn (lead dioxide or tin oxide); or conductive oxides of rare metals like Ir, Ru, or Ta (a "mixed metal oxide anode").  There are some decent electron microscope microstructure photos in Umicore's promotional catalog.

The advantage to using a "valve metal" like titanium or aluminum as the substrate on a coated anode is if part of the fancy conductive coating flakes off, the bare titanium just anodizes and stops conducting, rather than getting eaten away and dissolving into the electrolyte.


Why is the ionization enthalpy of francium greater than that of cesium?

The reason is the same as in the case of barium and radium, a contraction of core electron orbitals with increasing effective nuclear charge. This also changes the shielding (from nuclear charge) of electrons in outer orbitals, and in the end, all $s$ orbitals are contracted and lowered in energy. $s$ orbitals are particularly affected because they have no nodal plane at the nucleus.

This means that the single valence electron in the $7s$ orbital of $\ce{Fr}$ is closer to the nucleus and thus more tightly bound than the $6s$ electron of $\ce{Cs}$. More energy is therefore required to remove the $7s$ electron to form $\ce{Fr+}$ than in the case of $\ce{Cs}$, whose $6s$ electron is more loosely bound.


Why is Vapor Pressure of Liquid Equal to Ambient Pressure at Boiling?

When the vapor pressure equals the applied ambient pressure the liquid can form vapor in the bulk liquid hence "boiling".  To accurately measure this care must be taken to mix the liquid well and to prevent overheating. At lower vapor pressures evaporation is only from the exposed surfaces.


Naming an Ion with respect to its oxidation state

Does the same go for ions? What would the correct way of writing $\ce{Cu^2+}$ in words be?

The most common way of saying $\ce{Cu^{2+}} $ would be "copper(II) ion".  However keep in mind that in aqueous solution, various ligands bind to naked copper(II) ions forming various complexes.  So for example "hexaaquacopper(II)" ions are very common.  In these names the charge is inferred from (a) the oxidation state of the metal and (b) the ligands on the copper.  So copper(II) with four negative charged chloride ligands would be "tetrachlorocopper(II)", although since the latter is an anion, it is often called "tetrachlorocuprate(II)", e.g. on Wikipedia.


What is the difference between radical initiator and radical inhibitor?

The confusing thing about initiators and inhibitors of radical reactions is that both are often radicals.

But it isn't being a radical that is important: it is what happens when that radical reacts in the reaction conditions. If the radical reacts with another radical to give something that is no longer capable of propagating the radical reaction, it is an inhibitor. If it reacts to give more radicals it is an initiator.

Which compounds can do the job of either depends a lot on the specific reaction.


How do I calculate the cell voltage using the following half-reactions and solubility products?

I think you need the Nernst equation. This is something we were just introduced to today in my class, so I’m not so sure about the notation you’re using in your question at the top, but you should be able to correct my answer to fit your needs if you understand the equation I’m about to implement.

$$E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{zF}\ln Q$$

Where $E^\circ_\text{cell}$ = what the $E_\text{cell}$ normally is, under STP with the solutes at a concentration of $1\ \mathrm{mol\ dm^{−3}}$.

$$E^\circ_\text{cell} = E^\circ_\text{cathode} − E^\circ_\text{anode}$$

Since, the reduction potential of copper is lower, it will be the reducing agent in this reaction.

$$E^\circ_\text{cell} = -0.7993\ \mathrm{V} + 0.518\ \mathrm{V}$$

$$E^\circ_\text{cell} = -0.281\ \mathrm{V}$$

Where $R$ is the familiar gas law constant:

$$R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}$$

$T$ is the absolute temperature (it’s always assumed to be $298.15\ \mathrm{K}$):

$$T = 298.15\ \mathrm{K}$$

$z$ is the number of moles of electrons transferred in the half-reaction:

$$z = 1$$

And $F$ is Faraday’s constant, which is measured in coulombs per mole:

$$F = 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}$$

And $Q$ is the also familiar reaction quotient, but as I discovered, it takes some work to calculate, and you have to use the quadratic equation.

$$Q=\frac{[\ce{red}]}{[\ce{ox}]}$$
$$Q=\frac{[\ce{Cu+}]}{[\ce{Ag+}]}$$

$$K_{\text{sp},\ce{CuI}}=10^{-12}=[\ce{Cu+}][\ce{I-}]$$
$$1.0 × 10 ^{-12} = [X][0.090 + X]$$
Do the quadratic formula
$$X = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$
$$[\ce{Cu^+}] = 1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}$$
$$K_{\text{sp},\ce{AgCl}} = 1.8 \times 10^{-10}\ \mathrm{mol\ l^{-1}} = [\ce{Ag+}][\ce{Cl-}]$$
$$1.8 \times 10^{-10} = [X][0.120 + X]$$
Do the quadratic formula
$$X = 1.50 \times 10 ^{-9}\ \mathrm{mol\ l^{-1}}$$
$$[\ce{Ag+}]=1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}$$

$$Q=\frac{1.11 \times 10^{-11}\ \mathrm{mol\ l^{-1}}}{1.50 \times 10^{-9}\ \mathrm{mol\ l^{-1}}}$$

$$Q=0.00740$$

So, plugging this all into the Nernst equation, the answer should be:

$$E_\text{cell}=-0.281\ \mathrm{V}-\frac{8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \times 298.15\ \mathrm{K}}{1 \times 9.649 \times 10^4\ \mathrm{C\ mol^{-1}}}\ln(0.00740)$$

$$E_\text{cell}=-0.155\ \mathrm{V}$$

I should add that since $R$ is a constant, $F$ is a constant, and $T$ is usually something like $298.15\ \mathrm{K}$ or just not measured in elementary chemistry, the equation can just be

$$E_\text{cell}=E^\circ_\text{cell}-\frac{0.0257}{z}\ln Q$$

Which will give the same answer:

$$E_\text{cell}=-0.281\ \mathrm{V}-0.0257\ \mathrm{V}\ln(0.00740)$$
$$E_\text{cell}=-0.155\ \mathrm{V}$$


What would be the intermolecular forces between the molecules of methanetetrol?

Let's try to guess a reasonable structure for $\ce{C(OH)_4}$. The oxygen atoms and the central carbon would probably form a tetrahedron, therefore, those bond dipoles would basically cancel (as in e.g. $\ce{CCl_4}$). However, the carbon-oxygen-hydrogen angle would probably not be 180°. Whether the hydroxyl group would rotate freely about the carbon-oxygen bond or there would be some intramolecular hydrogen bonding preventing that, I would assume that intermolecular hydrogen bonding could still occur. (Which would be the first step towards eliminating water and forming $\ce{CO_2}$.)


Can a reactor be operated at both reflux and above atmospheric pressure?

You could theoretically boil at the higher temperature but why would that be a requirement? Why would you require reflux specifically if you can actually heat the formic acid to higher temperatures under pressure? Heating should work on its own and degassing to drive the equilibrium should work with a back pressure regulator. However,  most lab glassware will not take that pressure. You normally need some stainless steel pressure equipment.
One way might be to use plug flow reactors rather than bombs.


If it doesn't specifically need to be a hydrocarbon, zinc metal melts at 419.5 °C. Could you do an "ice bath" of zinc chunks in molten zinc, maintaining the melt right at its melting point?


Can sodium borohydride reduce hemiacetals?

The major purpose of $\ce{NaBH4}$ in the second step of given reaction sequence is the demercuration, which stops at the hemiacetal as the applied condition is the limit (at $\pu{0 ^\circ C}$ for $\pu{1 min}$).
In general under usual conditions (at $\pu{25 ^\circ C}$ in ethanol for for an hour or two), $\ce{NaBH4}$ reacts with hemiacetals and reduce them to corresponding diols. For instance, $\ce{NaBH4}$ reacts with glucose, which is a hemiacetal in aqueous solutions, to give sorbitol, the corresponding diol (Ref.1). The reaction has been done in water and at room temperature ($\pu{20-25 ^\circ C}$) for $\pu{1-2 h}$. Thus, running the given reduction in $\pu{0 ^\circ C}$ for shorter time make sense.
Also, although lithium aluminum hydride ($\ce{LiAlH4}$) is too vigorous a reagent for reduction, under carefully controlled conditions, it can be used to convert lactone to lactol (cyclic hemiacetal such as monosaccaride) without reacting further to give the corresponding diol. However, excess $\ce{LiAlH4}$ would give the corresponding diol after reacting further with lactol (Ref.2,3):

References:

M. Abdek-Akher, J. K. Hamilton, F. Smith, "The Reduction of Sugars with Sodium Borohydride," J. Am. Chem. Soc. 1951, 73(10), 4691–4692 (DOI: https://doi.org/10.1021/ja01154a061).
2.Glen E. Arth, "Reduction of Lactones to Hydroxyaldehydes with Lithium Aluminum Hydride," J. Am. Chem. Soc. 1953, 75(10), 2413–2415 (DOI: https://doi.org/10.1021/ja01106a040).
Robert A. W. Johnstone, "Chapter 1.11: Reduction of Carboxylic Acids to Aldehydes by Metal Hydrides," In Comprehensive Organic Synthesis - Selectivity, Strategy and Efficiency in Modern Organic Chemistry, Volumes 8: Reduction; Ian Fleming, Volume Editor; Barry M. Trost, Ian Fleming, Editors-in-Chiefs; Pergamon Press: Oxford, United Kingdom, 1991, pp. 259-282 (ISBN: 978-0-08-040599-1).


(In heteronuclear diatomic molecules, atomic orbitals only mix when the electronegativity values are similar.)
(While MOs for homonuclear diatomic molecules contain equal contributions from each interacting atomic orbital, MOs for heteronuclear diatomics contain different atomic orbital contributions.)
(Orbital interactions to produce bonding or antibonding orbitals in heteronuclear diatomics occur if there is sufficient overlap between atomic orbitals, as determined by their symmetries and similarity in orbital energies.)


How do primary alcohols react with dihydropyrans

Please have a closer look at your starting materials and the reaction conditions.

The reaction is performed under protic, ahydrous conditions.
One of the starting materials is a primary alcohol.
The other starting material apparently looks like a cyclohexene where one carbon adjacent to the $\ce{C=C}$ double bond has been replaced by an oxygen atom. This is a dihydropyrane.

Now, let's try a mind game:

Supposed that your second starting material is cyclohexene, what would happen under anhydrous, protic conditions?

   The proton would probably add to the $\ce{C=C}$ double bond to form a cyclohexyl cation. Since the starting material is symmetric, there is only one secondary cation formed. Note that the situation would be different if you would start with methylcyclohexene! Then, a secondary or a tertiary cation could be formed. Remember inductive effects to judge which one is favoured.

How would the primary alcohol come into play here?

   It would probably add to the cyclohexyl cation and yield an ether.

In reality, your second starting material is a dihydropyrane, think cyclic vinyl ether. How will the lone pairs on the oxygen atom effect the protonation?

   If in doubt, draw! Your $\ce{C=C}$ double bond now has two markedly different carbon atoms. Protonation can result in two different cations. I suggest to draw some resonance structures to figure out which one is favoured. Remember the lone pairs ;-)

What happens in the presence of a primary alcohol now?

   Again, it will add to the cation.

How does the final structure look like?

   I'm certain that you have figured this out by drawing and once you have, you will easily identify the structure class :)

Note that the reaction in question is quite useful to reversibly protect alcohols in sequences of chemical transformations.


Is it true that allylic free radical is more stable than allylic carbocation?

This is trickier than it sounds because you have to compare things that are similar or maybe you won't get the comparison you want.

First, you can't compare the energies of two things that don't at least have the same components. In this case, the allylic radical has an extra electron over the allylic cation. Let's say you generated the radical by fragmenting propene via removing a hydrogen atom or a hydride anion, respectively, as the other product. Notice that you there is energy in the other product, and you can't really just compare the radical and cation without taking into account the other product.

Second, you're in a situation where you make one species from the other. In fact, if you ionize the allylic radical, you will make an allylic cation. Notice that since you are ionizing a neutral species, the ionization energy is positive: you're going to need to add energy into the system to remove the electron. This makes sense because that electron is stabilized by the 8 nuclei that are present.

So, while you can't really compare the two species in your question head-to-head, it's clear than you can create the cation from the radical by adding energy to remove an electron. But this doesn't mean the cation is higher in energy; it just means the cation and the electron separated at infinite distance are higher in energy than the radical.

Hope this helps.


Explanation for behavior of acetic acid in aquaponic/hydroponic systems?

The source of the acid shouldn't matter much in adjusting the pH by a small amount.  What is the source of the calcium carbonate?

If it is dissolved calcium carbonate only, as you might find in hard water, then this is strange.  The solution is buffered by the combination of dissolved carbon dioxide and carbonate minerals, so its pH might be difficult to change, but once it has been changed it should stay where you put it, more or less.

If you have solid calcium carbonate (cement, limestone gravel, some pottery (notably not terracotta) or marble chips, perhaps), then what is happening makes a lot more sense.  You add acetic acid, lowering the pH and consuming some of the carbonate from the solution.  There is a large reservoir of carbonate available though, in the form of the carbonate minerals, and that eventually redissolves and the solution becomes more basic with time.


Creating energy profile diagrams for publication

PyEnergyDiagrams

I use my own script. I have posted it on GitHub now.
This is the final result:

It is quite easy to use, on the GitHub readme there is a small tutorial. You have to create an instance of ED class and then add the energy levels to it.

from energydiagram import ED
diagram = ED()
diagram.add_level(0,'Separated Reactants')
diagram.add_level(-5.4,'mlC1')
diagram.add_level(-15.6,'mlC2','last',)
diagram.add_level(28.5,'mTS1',color='g')
diagram.add_level(-9.7,'mCARB1')
diagram.add_level(-19.8,'mCARB2','last')
diagram.add_level(20,'mCARBX','last')

To add the links between the levels you can have a look to the IDs of the level using this the argument of the function plot show_IDs=True:

diagram.plot(show_IDs=True)

And then you can use:

 diagram.add_link(left_ID_levetobelinked,right_ID_levetobelinked)

Export to pgf,pdf and svg

You can export it just saving the plot from the GUI button or programmatically in different formats for publications using matplotlib backend.
 I prefer to export the plot in .pgf   and then I import it in the LaTeX document this allow a full control over fonts and appearance of the figure.

For reaching better quality you have to change a little bit the autogenerated .pgf file.

The first thing you have to do is to find and replace − with -.

Later you can follow the procedure in this answer so that the font of the figure matches with the font size and type of the document.


is dental acrylic dangerous when heated?

What do you mean by acrylic, if it is PMMA. Then be aware that when it is heated that it cracks with unusual ease into the monomer (MMA) which is a strong irritant.

I have (in a fume hood) taken lumps of PMMA and heated them in a test tube with a bunsen, if you distill the product out and then redistill it you can get very pure methyl methacrylate. I have found that if you heat some of this MMA with some AIBN then it will then reform the PMMA.


Which salts can be used to precipitate chloride ions?

The only well known method to check for (the absence of) chloride ions is precipitation as AgCl (solubility 2 mg/l at 20°C).

Mercury(I) chloride is similarly badly soluble, but toxic (and unstable, disproportionation) and therefore bad practice.

Lead(II) chloride is already far more soluble (4 g/l) than AgCl, and practically all other simple chlorides are just very well soluble in water.

Silver nitrate is cheap, the internet sells 10 g high purity $\ce{AgNO3}$ for 25€. Just get some.


Are adamantane crystals covalently bonded?

Adamantane is a molecular compound, much like naphthalene, paraffin, and many, many others. Molecular compounds consist of molecules, which stay pretty much the same in the gas and solid states. A molecule has certain molecular formula. Each carbon in a molecule is linked to a certain number of hydrogens and other carbons. There are no dangling covalent bonds to be formed between molecules, let alone between crystals. Consequently, you can't just take some adamantanes and covalently fuse them; some people would discuss diamond crystal structure in such terms, but only in a metaphorical sense.

Adamantane molecules in a crystal (being non-polar molecules without any specific interactions) are held together only by the weak van der Waals forces. Because of that, yes, the crystal is not mechanically strong at all (again, think of paraffin wax).


I'm sure there are a lot of issues  with my attempt I haven't spotted (or maybe some of the ones I have are in fact lethal), but I saw the question and thought I'd give it a go.
Retro:
Went straight for the Diels-Alder with something resembling Danishefsky's diene. Disconnection to reveal the species I named A and B comes from a Baylis-Hillman reaction, which works reliably with α,β-unsaturated esters (unsure whether the attached enol-looking oxygen would have anything to say about this). Asymmetry in A is provided by a Sharpless epoxidation, and in B by a bit of a longshot centred on MacMillan imidazolidinones.
Forward synthesis:

Notes:
TMS groups are used throughout but could always be beefed up to TBS as they're removed with TBAF at the end anyway.
I'd welcome suggestions of a better way of achieving step IIIa (formation of the enol-ether-like structure in A).
MacMillan catalyst sequence (steps VI to VII):
I'm aware that these catalysts (R = Bn, BnIndole etc.) are designed with one-pot systems using poor nucleophiles and electrophiles in mind, and hence expect this to be a bit of a stretch. I'm wondering if a more robust method might be to go with stoichiometric addition of the imidazolidinone before addition to THF containing the lithium alkyl cuprate shown to the left of step IV, rather than simultaneous addition of a Grignard and copper(I). The cyclisation seems like it would follow naturally and the correct stereochemistry would be provided by the imidazolidinone in accordance with MacMillan's models.
Not sure how big of a concern rearrangement during the Barton McCombie reduction step (IXa) is, but perhaps it would be better to conduct after the Diels-Alder; I don't know.
My hope is that the Diels-Alder step would get some stereocontrol from the 'B' section of the molecule folding down underneath the diene section, but expect this to be the sort of thing that has to be tried out in the lab.


Are there reactions operated at high temperature at industrial level?

The cracking reactions transforming heavy fuel, tar, domestic or heavy oils into lighter molecules like gasoline, kerosene or white spirits, they are all carried out at temperatures higher than 200 °C. The electrolysis of alumina to produce metallic aluminium is done at about 960 °C. The production of iron from ore and charcoal is done in a blast furnace at 1400 °C


Atomic radii cannot really be uniquely nor accurately defined.  However, there is generally a very big difference between covalent, van der Waals, and ionic radii.

So, by most any definition of the respective terms the covalent radius of fluorine will be smaller than the van der Waals radius of neon.  Similarly, the ionic radius of fluoride ion will be larger than the van der Waals radius of neon. The reason for the latter is the higher nuclear charge on neon.

However, most sources list a slightly smaller van der Waals radius for fluorine compared to neon, but a slightly larger covalent radius for fluorine compared neon. So when comparing radii you have to carefully specify what kind of radii you are talking about.


Chemical in (s) mixed with chemical in (aq) vs. chemical in (aq) mixed with chemical in (aq)

Short answer is yes. A lot of times you are trying to control the exothermicity of a reaction by doing a slow addition. This is kind of hard with a powder, and if you dissolve it in a lot of solvent, you can add it slower than you would be able to with a powder.

Without the slow addition, the temperature may increase and decrease your yield or selectivity.


See also: Why is WF6 stable whereas CrF6 is unknown?

Whenever one wants to compare oxidation states, there are a couple main factors to take into account. To reach a higher oxidation state, one obviously has to pay for it in the form of ionisation energy/energies. However, there is a compensatory effect in that elements in higher oxidation states generally get more out of bonding. For example, the lattice energy of $\ce{FeCl3}$ is larger than that of $\ce{FeCl2}$, or the four covalent bonds in $\ce{XeF4}$ are collectively stronger than the two covalent bonds in $\ce{XeF2}$.

The higher ionisation energies of the actinides tend to be smaller than those of the lanthanides, for the same reasons as explained in the linked question (5f orbitals have one radial node; 4f orbitals do not). This generally favours higher oxidation states and disfavours low oxidation states in actinides.


New technology that has replaced the old radioisotope

After searching on Google (now there's an idea)...

If you are asking about what technology was used in medicine before Co-60 therapy was developed:

Co-60 therapy was developed in the early 1950's and was seen as an alternative to other forms of X-ray radiotherapy, or radiation therapy. It is a form of external beam radiation therapy. Radiation sources at the time (mostly derived from radium) required very high energy sources, and proved largely ineffective against deep-set cancers. Co-60 therapy was able to penetrate deeper at lower energies, and therefore was more effective against a range of previously untreatable cancer types. The first Co-60 machine was from the The University of Saskatchewan, and they have an excellent site devoted to the history of Co-60 treatment.

If you asking about what technology is now being used in place of Co-60 therapy:

Conventional X-ray sources were improved with the development the linear accelerator which could produce X-ray energies exceeding 8MV. Nowadays, commercial units can produce photons up to 25MV. Here is a useful link to compare linear accelerator and Co-60 radiation sources for medical applications. There are still many developing areas of use, with some experts claiming there is still very much a place for Co-60 therapy. Co-60 therapy can be applied with high precision with low side-effects and is just one of many approaches available, often used in tandem with other medical treatments.


Graphical representation of reductive powers of NaBH4 and LiAlH4?

This is perhaps more than what you are asking for, but the textbook by Clayden et al.1 has a good overview of reducing agents and their substrates:

Reference

(1) Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; p 534.


Non-zero current flow through an electrolytic cell, when an electric potential lesser than that needed for electrolysis is applied?

As you noticed there is a small amount of current which keeps on flowing when there is no electrolytic decomposition. This "problem" is aptly named as non-Faradaic current. Faradaic current refers to current due to oxidation and reduction. Here is a detailed review: The difference between Faradaic and Nonfaradaic processes in Electrochemistry


What chemical reactions take place when mixing baking soda with citric acid and epsom salt and putting it into a microwave oven?

The solid phase reaction of baking soda and citric acid to form monosodium citrate, water and $\ce{CO2}$ takes place very slowly, so you don't get the visible foaming and bubbling that you do when water is added, or if you were to use baking soda with an acidic juice or vinegar (~95% water).

The molecular formula for Epsom salt is $\ce{MgSO4\dot\ 7H2O}$, meaning it's actually more than 50% water by weight. At $\mathrm{150^oC}$ it starts giving up some of those waters of hydration. Although water is a gas at that temperature, enough of the water is able to wet the baking soda and citric acid to get that reaction going, which then releases more water:

$$\ce{NaHCO3 + C6H8O7 → NaC6H7O7 + CO2 ^ + H2O}$$

Once the above reaction is initiated by the release of enough water from the Epsom salt, the additional water the reaction releases makes it somewhat self-sustaining and the $\ce{CO2}$ provides the foam and bubble show. The fact that additional reaction was observed after adding more water just shows that there was unreacted baking soda and citric acid remaining in the solid "crust".


What is the actual reason for geometrical isomerism?

Geometric isomerism occurs when two structures with the same connectivity are not interconvertible.

Cis-Trans isomerism is common and easy to recognize kind of geometric isomerism. The carbon-carbon double truly has limited rotation. The double bond in the alkene functional group consists of a $\sigma$-bond located within the plane of the molecule and a $\pi$-bond located perpendicular to the plane of the molecule. In the image below (created by David Pilz for Wikipedia), you can see the $\pi$-bond in a perpendicular plane.

The $\pi$-bond is made from $p$-orbitals, so it cannot have the same symmetry as the $\sigma$-bond. You can see the formation of the $\pi$-bond from $p$-orbitals in this picture by Wikipedia user JoJan. In order for the $\pi$-bond to exist the $p$-orbitals must be aligned.

In order for the molecule to rotate from $cis$ to $trans$, the bond axis would need to rotate so that the $p$-orbitals are no longer aligned and can no longer form the $\pi$-bond.  This slide show from James Condon at Roane State(PDF) explains the same concept. Rotation around the double bond requires enough energy input to break the $\pi$-bond, at least $250 \ \text{kJ/mol}$, which is a lot for room temperature.

The clearest demonstration that geometric isomers are clearly different molecules is to look at some examples with drastically different properties. It is important to remember that the experimental observation that these compounds with very similar structures have very different properties predates the explanation that the double bond has limited rotation.

The two isomers of 1,2-dichloroethene have different physical properties due both the difference in polarity and the difference in shape:

  $cis$-1,2-dichloroethane

Dipole moment = $1.9 \ \text{D}$
Boiling point = $60.2 \ ^\circ\text{C}$
Melting point = $-81.47 \ ^\circ\text{C}$
Density = $1.28 \ \text{g/cm}^3$

  $trans$-1,2-dichloroethane

Dipole moment = $0.0 \ \text{D}$
Boiling point = $48.5 \ ^\circ\text{C}$
Melting point = $-49.44 \ ^\circ\text{C}$
Density = $1.26 \ \text{g/cm}^3$

  However, my favorites are maleic acid and fumaric acid. Notice the extreme difference in water solubility as well as the difference in acidity. These molecules behave very different chemically.

Maleic Acid $cis$

Melting point = $135 \ ^\circ\text{C}$ (decomposes)
Density = $1.59 \ \text{g/cm}^3$
Solubility in water $788 \ \text{g/L}$
Acidity $\text{p}K_{a_1} = 1.9$, $\text{p}K_{a_2} = 6.07$

Fumaric Acid $trans$

Melting point = $287 \ ^\circ\text{C}$ (does not decompose)
Density = $1.635 \ \text{g/cm}^3$
Solubility in water $6.3 \ \text{g/L}$
Acidity $\text{p}K_{a_1} = 3.03$, $\text{p}K_{a_2} = 4.44$


Why does the hydrogen atom in HCl (when dissociated in water) bond to the oxygen rather than staying with the chlorine atom?

Look at it from the point of view of the chloride ion. In $\ce{HCl}$, the chloride is attracted to only one hydrogen. But when dissolved in water, the chloride is attracted to the partial positive charges on the hydrogens of several water molecules. Each $\ce{H2O-Cl}$ attraction is less than the $\ce{H-Cl}$ attraction but many of them act together to attract the $\ce{Cl}$ ion away from $\ce{HCl}$. Once the $\ce{Cl}$ is attracted away, that leaves a hydrogen ion to be attracted to the partially negative oxygen of a water molecule.

These forces between multiple waters and an $\ce{HCl}$ are stronger than forces between adjacent $\ce{HCl}$ molecules. Otherwise, $\ce{HCl}$ would be a liquid at room temperature instead of a gas.


How to qualitatively compare bond strength?

You can compare the bond strengths of $\ce{O-O}$,  $\ce{S-S}$,  $\ce{S=S}$ and   $\ce{O=O}$ by considering all possible effects that can alter the bond strengths. The effects which needs to be considered here are extent of π-overlap between the orbitals and inter electronic repulsions between the bonding atoms.

Firstly, it is obvious that double bonds will be relatively stronger than the single bonds. So, it is for sure that  $\ce{O=O} $ and  $\ce{S=S}$ will have higher bond strengths than  $\ce{O-O}$ and  $\ce{S-S}$. Now, we have to compare between two double bonds and two single bonds.

In  $\ce{O-O}$, two similar $\ce{2p}$ orbitals of two oxygen atoms form head-on overlap (σ-bonding) with each other, whereas, in  $\ce{S-S}$, two similar $\ce{3p}$ orbitals overlap. As it is a head on overlap, the extent of overlapping doesn't differ significantly, but for $\ce{2p}$ orbitals being smaller in size, two highly electronically dense $\ce{O}$ atoms comes much close to each other which hugely increase the inter electronic repulsions between the atoms. But in $\ce{S}$, the two bonding atoms remains much far from each other, and the electronic charge density is not also too much on each $\ce{S}$ as the $\ce{3p}$ orbitals are large in size, where the charge can easily diffuse. For this reason, $\ce{S-S}$ bonds become much stronger than  $\ce{O-O}$ bonds. (The reason is similar for the bond dissociation energy order  to decrease as in $\ce{Cl-Cl} >  \ce{Br-Br} > \ce{F-F}$)

But if we compare $\ce{O=O}$ and $\ce{S=S}$, the charge density on oxygen is reduced as lone pairs on oxygen decreases, therefore that effect becomes negligible. In this case, the extent of π-bonding starts to dominate. Being small sized,  $\ce{2p\pi-2p\pi}$ overlap in oxygen becomes much more strong and effective rather than  $\ce{3p\pi-3p\pi}$ overlap (where extent of overlap is not so much) in $\ce{S=S}$. That's why the introduction of double bond becomes much more thermodynamically favourable for oxygen and therefore, $\ce{O=O}$ bond is stronger than  $\ce{S=S}$ bond.

Thus the overall order of bond strength becomes, $$\ce{O=O > S=S > S-S > O-O}$$


Reversibility of the binding of carbon monoxide to haemoglobin

The reaction is not irreversible, but hemoglobin adheres to $\ce{CO}$ about 200 times more tightly than it does to $\ce{O2}$. Thus, part of the treatment for $\ce{CO}$ poisoning is to expose the patient to oxygen, which will slowly displace $\ce{CO}$ from carboxyhemoglobin.

For more on the chemistry involved, see this qestion.


Bond stability is the answer.
The energy required to break a $\ce{C-O}$ bond is $\pu{85.5 kcal/m}$; the energy given off when a $\ce{C=O}$ bond is formed is around $\pu{178 kcal/m}$ (reference).  These bond energies suggest that a carbonyl will generally be a few $\pu{kcal/mol}$ more stable than the corresponding gem-diol $[(2 \cdot 85.5) - 178 = \pu{-7 kcal/mol}]$.
[Entropic effects also play a role (see @Martins comment), but while bond enthalpies will change within a series of compounds the entropic effects should be similar and roughly cancel out when we compare compounds.]
However, the $\pu{7 kcal/mol}$ difference noted above isn't large, so there is a delicate balance in the equilibrium between a carbonyl compound and its corresponding gem-diol. Since $\pu{1.4 kcal/mol}$ will shift an equilibrium by roughly a power of 10, small factors like dipole moments, inductive effects, etc., are enough to stabilize one side of the equilibrium more than the other and significantly shift the equilibrium from one side to the other.
If water is present, then an equilibrium will be established between a carbonyl compound and its corresponding gem-diol.  Below is an interesting comparison of acetone and formaldehyde.  A seemingly small change (changing $\ce{H}$ to $\ce{CH3}$) shifts the equilibrium from one side to the other (by 6 powers of 10!).

Here is a link to a Wikipedia article that has a few more interesting examples of carbonyl-diol equilibria.
In the case of chloral ($\ce{CCl3CHO}$) it is the strong electron withdrawing inductive effect of the $\ce{CCl3}$ group that destabilizes the already positively polarized carbonyl carbon driving the equilibrium to the diol side.
As an aside, if you're not familiar with the history of chloral (or as it is often called, chloral hydrate) google "Mickey Finn" or "knock out drops."

As I've mentioned above a change of just a few $\pu{kcal/mol}$ can markedly shift the equilibrium between a carbonyl and the corresponding gem-diol. Lest the reader think that gem-diols are a rarity, here are a number of examples of compounds that exist predominately in the gem-diol form:

ninhydrin (because of the H-bond)
$\ce{R-(CO)-(CO)-(CO)-R'}$ (middle ketone houses the gem-diol; see detailed explanation)
1,1-dihydroxycyclopropane ($\mathrm{sp^3}$ carbon relieves ring strain)
floral hydrate (similar to chloral hydrate)