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Why don't we see these lanthanide species?
For most lanthanide metals$^{[1]}$, the stable oxidation state is III. The general electronic structure$^{[2]}$ is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}.$$
Elements that have the d-electron are La, Ce, Gd, and Lu. Furthermore, the f-subshell is considered relatively stable in states $$f^0, f^7, \text{and} f^{14}.$$
We can conclude that La, Gd, and also Lu easily form $\ce{E^3+}$ ions. Yet, as would be predicted by this easy approach, we would also see $$\ce{Sm+, Tm+ (f^7 \ and \ f^14 ), \\ Pr^5+, Dy^5+ (f^0 \ and \ f^7)}.$$
Why is this not the case?

$^{[1]}$ Ce, Pr, and Tb also have the oxidation state IV. Eu and Tm have the additional state II.
$^{[2]}$ Ordering (relative energy) changes with the number of electrons.

I strongly recommend having a look at these questions:

Howcome orbitals become 'core-like' when electrons are removed?
What is meant by 'electrons of like/unlike rotation'?
What's up with this quarter / three-quarter rule?

Original topic: Predominance of III oxidation state for lanthanides

This is really the exact same question but just in a different context of transition metals: Cr(II) and Mn(III) - their oxidizing and reducing properties?
The answer is because exchange energy (which is the reason behind the stability of half-filled and fully-filled shells) is not the main factor that decides the predominance of an oxidation state.
Whether an oxidation state can be reached depends on the balance between two things:

An energetic cost of having to ionise the element, i.e. the ionisation energy
The energetic payback obtained by forming covalent bonds / ionic bonds / solvation in solution / etc.

Exchange energy effects only give rise to minor variations in ionisation energies. The increase going from the nth to (n+1)th ionisation energy is far greater than any exchange effects.
Just like the magnesium example I gave in the other question, the lanthanides have very small second ionisation energies, and these can also be easily compensated for by bonding or ion-dipole interactions in solution. Take your example of $\ce{Sm+}$. The fact that it has a $\mathrm{f^7}$ configuration means that the second IE is marginally larger than would be expected. However, this marginal increase hardly changes the fact that the cost of ionising Sm a second time can be easily recouped.
Likewise, your example of $\ce{Pr^5+}$ is totally unrealistic because those 5th IEs are so huge that there is absolutely zero chance that they will give it up, no matter what exchange stabilisation there is.
The predominance of the +3 ionisation state for the lanthanides is therefore simply a consequence of IE1 + IE2 + IE3 being sufficiently small, and IE4 being way too large. That's all there is to it. Note how the variations in the graph below (due to exchange energy effects) are so small that all the IE graphs never cross over each other. IE5 is so huge that they didn't even bother plotting it!

(source: Inorganic Chemistry 6ed, Weller et al., p 630)

If possible in terms of ~simple mathematics, why is the fourth ionisation energy so high with the respect to the third one (in the bulk, about $1500 \ kJ/mol$ bigger) while third $IE$ is usually less than $1000\ kJ/mol$ bigger than $IE_2$? This isn't exactly what I asked, but it would still be interesting to know :) (just a reference is enough, since you have gone above and beyond in satisfying my curiosity)

I agree it would be interesting. The explanation in terms of ionization energies is enlightening and certainly explains a lot but it does, to some extent, beg the question. A little semi-quantitative bookkeeping showing the broad-brush origin of the ionization energies for example of the ubiquity of +3 in the lanthanides (such as something like the semi-empirical nuclear model) would be interesting to see.

(The greatest mystery for me is the magnitude of the compensatory effects, the energy cost is reasonably easy to grasp).

@DanSheppard: Glad you agree, and it would indeed be useful. It could be possible to expand the answer here, but I think a better option is to reply under Howcome orbitals become 'core-like' when electrons are removed?. That is basically what the question is about.


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Howcome orbitals become 'core-like' when electrons are removed?
It seems to me that f-orbitals for lanthanide metals are treated as 'core-like' when a certain number of electrons have been removed.
Or, as Radiochemistry puts it,

The 4f binding energy is so great that the remaining 4f-electrons are regarded as "core-like". [Italic and emphasis added]

How is this justified exactly?
Effort
I am aware that

each electron will be harder to remove as Culonic attraction increases with positive ionic charge,

and probably

lathanide contraction is a part of the explanation.

You have to check out some of these other threads:

Why don't we see these lanthanide species?
What is meant by 'electrons of like/unlike rotation'?
What's up with this quarter / three-quarter rule?

Original topic: Predominance of III oxidation state for lanthanides

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What is meant by 'electrons of like/unlike rotation'?
Explaining the reasons behind electron repulsion,
Radiochemistry asserts that,

Interelectronic repulsion in related not just to electron pairing, but also to [the] angular momentum of the electrons. [---]
e.g., in $\ce{Pr^2+ (4f^3)->Pr^3+ (4f^2)}$ ionisation removes repulsion between [electrons] of like rotation [---]
$\ce{Pm^2+ (4f^4)->Pm^3+ (4f^3)}$ ionisation removes repulsion between [electrons] of unlike rotation [---] [emphasis added]

What exactly is meant by 'electrons of like/unlike rotation'?

You may glance at these links, but only if you want to.

Why don't we see these lanthanide species?
Howcome orbitals become 'core-like' when electrons are removed?
What's up with this quarter / three-quarter rule?

Original topic: Predominance of III oxidation state for lanthanides

It simply refers to electrons in orbitals with the same sign of $m_l$.
The quick and easy way to determine the ground-state term symbol of a lanthanide is to fill the 4f orbitals from most positive $m_l$ to most negative. This is in order to get the largest possible value of $M_L$:
$$M_L = \sum m_l$$
This is not valid by quantum mechanical arguments due to the indistinguishability of the electrons. However, it always leads to the correct ground-state term symbol because of Hund's second rule, which essentially states that the term with the largest value of $L$ is the most stable (after the first rule has been taken into account). Since the value of the projection quantum number $M_L$ ranges from $-L, -L+1, \ldots, +L$, a ${}^{2S+1}L$ term necessarily contains a term with $M_L = +L$. Therefore, by maximising the value of $M_L$, you will automatically generate the term with the largest possible $L$.
For example, consider the $\mathrm{f^3}$ ion $\ce{Nd^3+}$. You would want to "fill the orbitals" in this way:

where you have first and foremost put the electrons all with the same spin in order to satisfy Hund's first rule, and then filled them in decreasing order of $m_l$ in order to satisfy Hund's second rule.
The classical argument behind Hund's second rule is that electrons that have the same sign of angular momentum "collide" into each other less and as a consequence repel each other less. It is like two people running around a track; if they ran in the same direction, they would bump into each other less than if they ran in opposite directions.
Therefore, removal of an electron with "like rotation" (as you would if you were to ionise $\ce{Nd^3+}$ to $\ce{Nd^4+}$) is marginally more difficult than expected. Do not imagine it as a huge effect. It is very small and only leads to minor deviations from the ionisation energy trends (quarter-shell and three-quarter-shell effects).

Great answer! Many things that I'm hearing for the first time. If I understand correctly, removing the electrons with the same sign of $m_l$ is the same as removing electrons of like rotation. This is harder than removing electrons of unlike rotation. Is removing an electron from $m_l = 0$ and any other $m_l$ also considered 'unlike rotation'? To answer your comment from the other thread ... If you like, you can expand your answer here or give a short explanation there.  In the former case, I might close/delete the other one. Well done to you!

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Avagadro's number
Ive been reading about moles and I can't seem to understand avagadro's number. I get that the idea is 1g of hydrogen, 12g of carbon, and 16g of oxygen would all have the same number of atoms, but why? Is it just a coincidence, what's the reason behind this?

A mole is a measurement of the number of atoms in a substance. Specifically, it is equal to the number of atoms in the substance divided by some constant $x$. We define the molecular weight of a compound to be the number of grams of that substance that are required to make up a mole of that substance. Now, all that we need is to find some value for $x$ that makes it easy to discover the molecular weight of an atom/compound. It turns out that protons and neutrons have around the same mass, and electrons have negligible mass compared to the two of them. Because of this, we can define one unit of "mass" to be the mass of a neutron or proton, allowing us to find the molecular weight of some compound just by summing up all of the protons and neutrons in that compound. Since this is our molecular weight, it will be measured in units of $\dfrac{g}{mol}$. Now, we just need to find our $x$ such that our conception of a mole corresponds to a correct number of atoms. It turns out by experimental results that this number must be $6.022 * 10^{23} \dfrac{atoms}{mol}$, or avogardo's number.


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What is the mechanism for the reaction of an organic hydroperoxide with sulfuric acid?
The question asked to react the organic hydroperoxide substrate, $\ce{R-OOH}$, with dilute sulfuric acid, $\ce{H2SO4}$.
In the last step, I removed a proton, making it act like a catalyst, along with the removal of $\ce{H2O2}$. But I am very unsure about its correctness.
Please confirm or correct if this is a wrong answer.

See, that peroxide was there for a reason. I don't believe it would walk away just like that.

I really don't get. Please explain.

I would propose something like the Hock rearrangement. It should be especially feasible in this case as double bond participation from the anisole ring is expected.

As Marko has commented it is like the Hock rearrangement, which is the migratory step in the cumene process.
It is somewhat similar to the Beckmann rearrangement except that there is an oxygen instead of a nitrogen.

The acid protonates the peroxide, generating a great leaving group on electron-deficient oxygen.
Migration towards electron-deficient oxygen occurs with loss of water to give an oxonium ion.
The migratory aptitude here is determined in the same way as for the Baeyer-Villiger reaction. There is a buildup of positive charge on the migrating carbon during the migratory step, and therefore the group that can best stabilise the charge preferentially migrates. The methoxy-substituted ring is electron-rich and the positive charge is more stabilised on it than on the unsubstituted phenyl group. The methyl group is even worse at stabilising positive charge.
Under the dilute acid conditions this oxonium ion is hydrolysed.

Well done! I should have written it on paper before posting my answer to see the whole process.

The magical appearance and disappearance of protons... $$\ce{RR'MeC-OOH ->T[magic] RR'MeC-OOH2+ ->...->T[more magic] RMeCO + R'-OH + H2O}$$

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How Does This Reaction Proceed?
$$
\ce{CH3OH + CH3CCNa+ ->  ?}
$$
The answer is $\ce{CH3CCH +CH3ONa+}$ but I don't understand why.

@R.M. http://meta.chemistry.stackexchange.com/a/3052

$\ce{CH_{3}C#CNa}$ exists in a solution as a cation and anion shown below:
$\ce{CH_{3}C#CNa  <=> CH_{3}C#C- + Na+} $
$\ce{CH_{3}OH}$ has very slightly acidic hydrogen (thanks to electron pulling ability of oxygen), that can be removed by a strong base.
When put in the same solution, $\ce{CH_{3}C#C-}$ will pull out $\ce{H+}$ from $\ce{CH_{3}OH}$ like so:
$\ce{CH_{3}C#C- + CH_{3}OH  <=> CH_{3}C#CH + CH_{3}O-}$
$\ce{CH_{3}C#CH}$ is a weaker acid than $\ce{CH_{3}OH}$, which means that $\ce{CH_{3}C#C-}$ is a stronger base than $\ce{CH_{3}O-}$. Therefore, $\ce{CH_{3}C#C-}$ will completely pull out hydrogen from methanol.
And overall reaction looks like the following:
$\ce{CH_{3}C#CNa + CH_{3}OH  <=> CH_{3}C#CH + CH_{3}ONa}$

Propyne is much weaker acid so reaction will be complete

@Mithoron You're right. pKa of propyne is much bigger than pKa of methanol. I updated the answer.


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How do I arrive at correct furanose structures when starting from a Fischer projection?
While numbering the furanose ring from the left are we supposed to place the atoms on right of the Fischer projection to the top? Isn't it just the opposite to what we do while numbering from the right?

In the image the atoms/groups lying on the right of the 2nd Fischer structure is present to the bottom of every furanose carbon of the 3rd image (chronologically). So aren't they supposed to be present at the bottom even if we start from the left (referring to structure A)? But I find Wikipedia says they are supposed to be at the top.

Tl;dr: It doesn’t matter how you do it, as long as it is consistent and correct.

@Jan I don't get you.

http://chemistry.stackexchange.com/questions/44332/the-proper-formula-of-sucrose-why-both-forms-are-correct-and-how-one-is-transf

Also, why are you linking to sucrose rather than to fructose? I thought, your question was about fructose? (I know it is present in sucrose, but still.)

It was the only wikipedia page where the structure of Beta-D-fuctose is provided where it is numbered from the left.

A Fischer structure is one way of depicting structures such as sugars or amino acids to quickly draw them and have a definite way of seeing stereochemistry on every carbon. The rule behind Fischer structures is that the highest oxidised atom should be at the top or as close as possible to the top. The following convention is used to turn tetrahedral carbons into a square environment:

This allows you to assign absolute configurations to every carbon atom.
When instead drawing the same structure as a ring like in your lower images, you just need to take care that every carbon has the same configuration. (Note that your drawing of β-D-fructose is wrong; it should be drawn with a very wavy line like in my image below, because of the constraints of Fischer projections.)

Once you have gotten there, you are free to rotate the ring around as much as you wish while retaining the absolute configuration at every stereocentre. Note especially, that a rotation of $180^\circ$ of my structure of β-D-fructose will lead to you one on the left. If you don’t believe me, print it out, turn the paper around and then nudge the forwards-pointing part of the ring downwards below the tetrahydrofuryl-oxygen.


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Why convert to moles for stoichiometry calculations?
Why do you have to convert the reactants and products of an equation to moles when doing stoichiometry problems? Why can't you just convert straight to grams to solve the problem?

You should use more descriptive titles, it is obvious that you have a question/problem to pose on a Q&A site.

Possible duplicate of Why do people still use the mole (unit) in chemistry?

Equations tell you how much of each molecule you need in relation to others. If you used grams the numbers would change because not every molecule weighs the same.
E.g. if you have $200$ hydrogen and $100$ oxygen molecules you could make $100$ water molecules. But if you had $200\ \mathrm{g}$ hydrogen and $100\ \mathrm{g}$ oxygen, you don't know how many grams of water that gives. You need to convert using the weight.

Grams depends on the molecule itself, oxygen weighs more than hydrogen. But using moles gets rid of the mass concern, because a mole represents a specific amount of molecules, independant of weight. Since you want to convert from reactants to products, you need to know how many molecules react with other molecules, not mass. Mass doesn’t affect the outcome of a reaction, atom counts do.
Looking at a balanced equation, you’re looking at atoms. For example, $\ce{H2 + O -> H2O}$ (incorrect equation but for demonstration purposes) means that for every 2 atoms of hydrogen, and every 1 atom of oxygen, you get one molecule of $\ce{H2O}$. Now because the atoms weight different amounts, this is not the same as saying 2 grams of hydrogen and 1 gram of oxygen gives 1 gram of water. First of all, you can’t react 3 grams and end up with 1 gram, and also if you do the stoich, you have far more of one atom that the other.

You have to use moles as the coefficients in a chemical equation refer to the molar quantity. If the coefficients were given in units of mass instead, you could use mass ratios. However, this is not the convention of chemical equations. Thus, you must use molar ratios.

The answer is because in chemical reactions there is a conservation of moles, rather than mass. Moles are the total amount of molecules in the system, those remain constant. Naturally using numbers like $6.022 \times 10^{23}$ is inconvenient, so moles help with keeping track. Mass on the other hand, which is equivalent to energy in the system, is not conserved in systems that are not isolated (as most in real life are not).

Not quite right. Overall mass is conserved in chemical reactions. In fact moles are not conserved.

jiggunjer is right that chemical equations tell you how much of each molecule you need in relation to others. "How much" in the sense of an integer count.
So consider the simple reaction $$\ce{2H2 + O2 -> 2H2O}\tag{1}$$ Thus two molecules of hydrogen and one molecule of oxygen yields two molecules of water. Now we can create any integer multiple of that reaction. So the following are also true. $$\ce{200H2 + 100O2 -> 200H2O}\tag{2}$$
$$\ce{2\times10^6*H2 + 1\times10^6O2 -> 2\times10^6H2O}\tag{3}$$
$$\ce{2billion H2 + 1 billion O2 -> 2 billion H2O}\tag{4}$$
So in the first equation above the multiple is 100, the second uses $1 \times 10^6$, and the third one billion.
The problem with atoms or molecules is that they have an exceedingly small mass. So for things on an atomic scale it is convenient to count in terms of moles. One mole of atoms or molecules, which is $6.022 \times 10^{23}$ of them, gives a mass large enough for us humans to weigh and manipulate. So if we let $\ce{N_A}$ be equal to Avogadro's number, then we can say that $$\ce{2N_AH2 + N_AO2 -> 2N_AH2O}\tag{5}$$ or in plain English two moles of hydrogen and one mole of oxygen react to produce two moles of water.
Now lets use equation 1 and multiply each side by "1 gram(s)" as we can do in algebra. We get: $$\ce{2~grams~H2 + 1~gram~O2 -> 2~grams~H2O}\tag{6}$$ In chemistry mass is conserved. This equation has a very, very large problem. There are three grams of matter on the left but only two grams on the right. The missing gram of matter isn't the energy in a balloon pop, but rather the energy of a very sizeable nuclear bomb.
Now we could try writing the equation in terms of the mass of the reactants and products, which is sometimes done. $$\ce{4~grams~H2 + 32~grams~O2 -> 36~grams~H2O}\tag{7}$$
Now if we try to use our algebra trick again we can divide both sides by "4 grams" to get: $$\ce{H2 + 8O2 -> 9H2O}\tag{8}$$ but we can see that the stoichiometry is wrong for this equation.
Now let's digress in another direction. From Wikipedia "The gram, 1/1000 of a kilogram, was provisionally defined in 1795 as the mass of one cubic centimeter of water at the melting point of ice. The final kilogram, manufactured as a prototype in 1799 and from which the International Prototype of the Kilogram (IPK) was derived in 1875, had a mass equal to the mass of 1 dm3 of water at its maximum density, approximately 4 °C." So the notion of a kilogram was created by humans as some arbitrary mass long before the notion of an atom or molecule was truly understood by scientists. Thus chemistry was tied into the existing system of measures.
Just as a mass of 1 gram is arbitrary, so too is Avogadro's number arbitrary. It has been defined now for a long time as the number of atoms of carbon in 12 grams of $\ce{^{12}C}$. But some other element/isotope could have been chosen. So 12 grams of $\ce{^{12}C}$ and 16 grams of $\ce{^{16}O}$ do not have the same number of atoms because of the binding energy in the nucleus.

You could use mass ratios if you wanted but the convention among chemists is to use molar ratios since they are easier to work with.


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Reaction between lye and water?
I am planning on making some homemade soap soon and have read online that breathing in the fume generated from the reaction between lye and water is not recommended because it is "hazardous."
I am a little bit confused because I thought the reaction between lye and water is simply a highly exothermic dissolution reaction and therefore the "fumes" would consist of only steam?
                           NaOH + H2O -> OH- + Na+ + H2O + HEAT

If this fume is corrosive, does that mean there is some gaseous hydroxide as well generated from the heat? Why else would this fume be dangerous?

The fumes will contain trace amounts of NaOH. The fumes are microscopic water droplets, and these droplets will still be caustic, as they are essentially a solution of sodium hydroxide.
There is no real chemical reaction going on here, but rather a dissociation of the sodium and hydroxide ions, which is an exothermic process. Solid sodium hydroxide is 'hazardous', the resulting solution is 'hazardous' and the fumes above the dissolving lye are 'hazardous'.

Add the water to the lye - the slow reaction with slow diffusion means there are no fumes.  Misinformation about this (reference Soap Queen's alarmist and inaccurate instructions about mixing water and lye) has been pervasive on soap making websites

Do you have any reference for this?

Perhaps an analogy would be NaCl in the air around the seashore, where unprotected iron will rust rapidly.
NaOH is dangerously corrosive, and even minor amounts inhaled or on skin are to be avoided.
Localized heating from mixing "lye" with water, or using products such as oven cleaners at elevated temperatures (where they work best), does carry the risk of exposure to caustic fumes, particularly if done on a daily basis long term.
Fortunately, most oven cleaning is more on the order of once a year.

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Drying solvents while taking care of purity
I need to dry some solvents like $\text{MeOH, EtOH, and  DMSO} $ and I have to take care of purity of the solvents. These solvents are HPLC grade.
I am interested in freezing point and boiling point of these solvents  with maximum error of $ \pm  0.5  C$, hence, I do not prefer to add silica gel (or other dehydrating agent) in the flask containing solvent.
What are other ways to dry solvents without putting something in them.

Drying is a form of purification. Why do suspect that 'putting something in' your solvent is going to disturb the purity beyond the current water content? Have you researched (ie search engine) purification methods for organic solvents?

Then still add silica gel. It is not what you think it is.

My concern is, adding say silica gel may cause entry of other impurities in the solvent (some impurity present in the silica gel).

Trust me, you are not the first chemist out there concerned with purity.

I agree, however I am just being cautious about the freezing and boiling points. These temperatures will serve as calibration points for other devices, so it is very important to take care of this small thing in my experiment.

You should distillate the solvents anyway, so adding something to them is not such a problem. There is a plenty of lab manual about solvent purification.

Even distillation is becoming a yesteryear technology. Modern labs are moving towards low-hazard, efficient purification systems like Pure-Solv. A lot depends on the volume you need, but the key is to purify, via whatever route, immediately prior to use.

Forget about the silica, use established protocols. Destill your alkanols. Boil methanol over magnesium, ethanol over sodium. Pre-dry DMSO over calcium sulfate or barium oxide, then decant and reflux/destill with calcium hydride at reduced pressure.

There is actually a book on preparing laboratory chemicals you can look up:

"Purification of Laboratory Chemicals" by WLF Armarego and CLL Chai

However, given that all the chemicals you want to dry are extremely hygroscopic, I'm not sure how long they would stay super dry for calibration purposes (and I'm not sure you need super dry solvents for calibration either). If you need it to last, you may need a large dry box.

I would suggest refractional destilation as a method to isolate your solvents from the impurities. You can also repeat this procedure until you have obtained a desired purity.


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Concentration of NaOH to dissolve glass
I am doing a physics experiment in which it is required to etch the outermost glass layer from a coverslip (regular lab apparatus). What concentration of NaOH should be dissolved in water, and how long should the coverslip be placed in the base?
Any idea or information that can help me figure out myself is welcome.

The thickness of the attacked layer (or the corresponding loss of weight per surface area) due to alkali attack on glass mainly depends on the following parameters:

type of glass (most laboratory glass is borosilicate glass 3.3)
concentration of the alkaline solution (or the corresponding pH)
temperature
exposure time

Note that the density of typical borosilicate glass is 2.23 g/cm³ (at 25 °C). Thus, an attacked thickness of 1.00 µm corresponds to a loss of 22.3 mg/100 cm².
The following figures are taken from product information for Schott DURAN® glass:

Alkali attack on borosilicate glass as a function of pH, exposure time: 3 h

Alkali attack on borosilicate glass as a function of temperature, calculated from the weight losses, c(NaOH) = 1 mol/l, exposure time: 1 h

Alkali attack on borosilicate glass as a function of time
For example, in order to remove a layer of about 1 µm, you can expose the glass for a period of approximately 40 min to an alkaline solution with c(NaOH) = 1 mol/l at 100 °C.


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Tunneling corrections in reaction rates
I've studied Transition State Theory since I asked this a while ago, but so far I have not been able to find a detailed explanation of how tunneling corrections are implemented into reaction rates computations.
I am thinking specifically in reactions involving proton transfer along a hydrogen bond. Electrons don't pose any trouble, because they are treated quantically. But for protons, which are usually beheld as classical particles, semiclassical corrections accounting for tunneling must be made, since a full quantum treatment of protons seems undoable.
First idea: If one performs a molecular dynamics simulation, one could at each step think of the proton "just for a moment" as a quantum particle, find an approximated wavefunction thanks to the WKB method and perform a Monte Carlo simulation to decide "where" will we find the proton at the next step. This seems doable, but it's not what is done, as far as I know (sometimes performing molecular dynamics simulations is not possible, or not useful).
Second idea/ problem: Imagine we have a Free Energy surface (I am concerned with two-dimensional problems, where there are two relevant reaction coordinates to be tracked), which informs us about the precise changes in free energy along infinitely many reaction pathways. For each of those, a classical reaction rate can be computed by means of Eyring equation and then we could take averages over some sets of paths via path integrals or something like that. Even better, one can employ Variational Transition State Theory to derive estimations for the reaction rate, once the Free Energy surface is known. I expect this will not be hard. But I don't have a clue as to how to make the tunneling corrections for the reaction rate in this scenario, although I suspect is in this setting where most tunneling corrections are made.
I've certainly found a lot of bibliography dealing with Quantum Transition State Theory and Tunneling Corrections, but absolutely nowhere do they explain with all the mathematical and physical details the concrete formulas they are employing. Instead, they write a lot (without much equations), give you other references, where you can find still some more references -and so on.
So, in brief, my question is: could anyone provide me with a reference containing a thorough, complete and detailed explanation, not lacking the mathematical details, of how does one compute the tunneling corrections for reaction rate constants?

Have you checked the original papers by C. Eckart, R. P. Bell, H. Shin, J. O. Hirschfelder and E. Wigner? These contain a fair degree of mathematical and physical treatment.
There is also a nice paper from S. G. Christov called "Some aspects of the theory of proton-transfer processes" published in J. Res. Inst. Catalysis 1968, 16 (1), 169-194 (pdf).
Other possible references:
Johnston, H. S.; Heicklen, J. Tunnelling Corrections for Unsymmetrical Eckart Potential Energy Barriers. J. Phys. Chem. 1962, 66 (3), 532–533. DOI: 10.1021/j100809a040.
Some papers published by D. G. Truhlar together with T. N. Truong, B. C. Garret, R. T. Skodje.


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Explain why alumina-containing zeolites possess chemically active surfaces
I think zeolites possess acid sites that are catalytically active in many hydrocarbon reactions. The pore system only allows molecules that are small enough to enter, hence it affects the selectivity of reactions by excluding both the participation and formation of molecules that are too large for the pores. Is that right or not ?

Welcome to Chemistry! This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Alumina containing zeolites have composition that can be represented as

where M is metal ion with m charge, and so on, respectively.
Meaning you have $[SiO_4]^{4-}$ and $[AlO_4]^{5-}$ tetrahedra which make up negatively charged framework. Positive ions compensate for this negative charge. There is a rule from 1954. called Löwenstein rule, which states that x/y has to be less than 1, which in turn implies there are no Al-O-Al bonds, as two tetrahedra have Al atoms in their tetrahedral positions. This gives really great internal specific area and resistance to high temperature, which makes this type of zeolite a great candidate for catalysis (heterogenous).
The following picture nicely represents the active sites in zeolites - they act as molecular sieves, as only the specific molecules fit in the pores:


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What happens when potassium iodide is added to a copper (II) salt?

$$\ce{Cu^2+ + KI -> ??}$$

I thought of this reaction:
$$\ce{Cu^2+ + 2KI -> CuI2 + 2K+}$$
But since $\ce{CuI2}$ doesn't exist, the $\ce{I-}$ ions must reduce $\ce{Cu^2+}$ to $\ce{CuI}$ and simultaneously get oxidised to $\ce{I2}$ and thus it will proceed the following reaction:-
$$\ce{2Cu^2+ + 4KI → 2CuI + I2 + 2K+}$$
I gave this reaction to my teacher and he gave me an another version of the reaction:
$$\ce{2Cu^2+ + 5KI -> 2CuI + KI3}$$
He says that he performed the reaction experimentally and obtained a white salt ($\ce{CuI}$) in a brown solution ($\ce{KI3}$).
Now, I started wondering what will happen if I alter the reaction and add sodium/lithium iodide instead of potassium iodide. Will it form sodium/lithium triiodide. Definitely not/ Or maybe, it will form.

Two questions:

What is the correct reaction?
What will happen if we add sodium/lithium iodide instead of potassium iodide?

Related questions:

Why is NaCl3 possible?
What reaction takes place when potassium iodide is added to a tetraammine copper(II) complex?
Why triiodide ion does not form ionic bond with cation having +2 charge?
Why Cu+ is unstable in aqueous medium?

When writing a reaction, you should take care to keep charges and atoms balanced. So the very first reaction you postulated is good (it has charges and atoms balanced) but then it goes South.
If you are only interested in the redox process, then it makes sense to think only about those reactions that form part of the redox process.
$$\begin{align}\ce{Cu^2+ + e- & <=> Cu+}\tag{red}\\
\ce{2I- & <=> I2 + 2 e-}\tag{ox}\\
\ce{2Cu^2+ + 2 I- & <=> 2 Cu+ + I2}\tag{redox}
\end{align}$$
The next step, however, is that the $\ce{Cu+}$ ions will combine with iodide to form insoluble $\ce{CuI}$. Thus, we need excess iodide.
$$\ce{2 Cu^2+ + 4 I- -> 2 CuI v + I2}\tag{precipitation}$$
At any of these points, you could decide to include counterions. For example, your iodide could be $\ce{KI}$ and your copper salt could be $\ce{CuSO4}$. That would give:
$$\ce{2 CuSO4 + 4 KI -> 2 CuI v + I2 + 2 K2SO4}\tag{counterions}$$
Or maybe you decide that only $\ce{KI}$ is important, then of course this becomes:
$$\ce{2 Cu^2+ + 4 KI -> 2 CuI v + I2 + 4 K+}\tag{only KI}$$
Finally, you will typically have an excess of $\ce{KI}$, so that that can further react with the iodine generated to redissolve it. (Iodine is poorly soluble in water, $\ce{KI3}$ is well soluble, hence Lugol’s iodine.)
$$\begin{align}\ce{I2 + KI & -> KI3}\tag{Lugol, balanced}\\
\ce{I2 + I- & -> I3-}\tag{Lugol, ionic}\end{align}$$
Thus, a full equation will look like any of the following:
$$\begin{align}\ce{2 Cu^2+ + 5 I- & -> 2 CuI v + I3-}\tag{full, ionic}\\
\ce{2 Cu^2+ + 5 KI & -> 2 CuI v + KI3 + 4 K+}\tag{full, semi-ionic}\\
\ce{2 CuSO4 + 5 KI & -> 2 CuI v + KI3 + 2 K2SO4}\tag{full, balanced}\end{align}$$
Since the alkaline halides are generally very similar to each other, I think that the reaction should work pretty much identically if potassium is replaced with sodium or lithium. Personally, I didn’t have a chance to perform it, since my inorganic student labs only had pre-mixed potassium iodide solutions.

So, if replaced with sodium or lithium iodide, will it form sodium/lithium triiodide? I don't think the compound exist.

@NilayGhosh We are talking about aquaeous solutions. And I have no reason to believe that a sodium cation or a lithium one should cause the breakdown of the triiodide anion. It would be a simple experiment; add $\ce{NaCl}$ to a $\ce{KI3}$ solution and see what happens; likely nothing.

I've tried this reaction which proceeds immediately, but I am stuck on the redox potentials. I don't understand why I can't calculate a positive electrode potential.
Cupric ions reduced to cuprous ions +0.15V
Iodide oxidized to Iodine -0.54V
It seems to me this reaction should NOT be spontaneous. But it does happen, quickly and readily.

@124c41 Two immediate possibilities: 1) the standard potential of $\ce{Cu^2+/Cu+}$ might be different from $\ce{Cu^2+/CuI}$; or 2) the equilibrium might be driven to the product side through solubility (CuI is very poorly soluble).


text:
Boiling point of: Halogenoalkanes vs alkanes vs alcohols
We know that Boiling point (alcohol) > Boiling point (halogenoalkane) > Boiling point(alkane) with same number of $\ce C$ atoms, but is the following, good explanation of that?:
Since e.g. $\ce{CH_3CH_2-O^{\delta_1 \textbf{-}}-H^{\delta_1}+}$, similar with $\ce{ClCH2CH3}$ and with $\delta_2\textbf{-}(\text{at}\ \ce{Cl})\Longrightarrow \delta_1 \textbf{-}\stackrel{\text{bigger, electronegative}}{>}\delta_2 \textbf{-}$, thus electron pair is closer to oxygen atom, so more energy is needed to move that electron in alcohol than in halogenoalkane, so higher temperature is needed to reach boiling point.
Does this have any sence, if not then could somebody explain why does it not have sence?

Oxygen is more electronegative, hence alcohols are more polar. Even more importantly, they also are capable of forming hydrogen bonds, which is not the case with haloalcanes.

The cause of this is the hydrogen bond that forms in the liquid state between the molecules, and debye bonds that form between polar molecules.
A hydrogen bond can only occur between an electronegative atom and a hydrogen atom, himself linked to an electronegative atom  (F, O or N).
In order to evaporate the liquid, you have to provide enough energy to break the hydrogen bond. So the more bonds there are, the higher the boiling temperature. Soft debye bonds can also occur between two polar molecules and have to be broken too.

The alkanes can not participate in any bonds, thus have the lowest boiling temperature.
The halogen alkanes may not form hydrogen bonds, but they are polar, thanks to the very electronegative halogens. They form debye links, that are not very strong but sufficient to raise the boiling temperature.
The alcohols have -OH extremities, very likely to associate themselves in hydrogen bonds, that are stronger than debye links, and thus alcohols have the highest boiling temperature.


text:
Reduction of a disulfide bond
I'm trying to reduce a disulfide bond, I've tried 2-mercapto-ethanol, because I've seen in literature that using a thiol is the most common procedure. I carry out the reaction at 80o C but so far I have no positive results

Reveal your reagents and expected products please.

@permeakra Is very simple indeed I'm trying to reduce the dsulfide bond in cystamine I've used 2-mercaptoaceticacid and 2-mercaptoethanol.

Here is a good source to look into http://cen.acs.org/articles/90/web/2012/02/Better-Disulfide-Reducing-Agent.html.


text:
How to find out unpaired electron in S2 molecule?
Theoretically electronic configuration and number of unpaired electron can be found out by using molecular orbital theory. But, practically I have seen only up to 20. For example, for Ne2
But for S2 (32 electrons) how to find out the electronic configuration and number of unpaired electrons.

The molecular orbital energy-level diagram of $\ce{S2}$ is given below:-
(source)
We can see  how close the energy the 3s and 3p atomic orbitals are because their energy separation will determine whether the  $\ce{π_3_{p}_{x,y}}$,  or the $\ce{σ_{3}_{p}_{z}}$ molecular orbital is higher in energy. Because the ns–np energy gap increases as the nuclear charge increases , the $\ce{σ_{3}_{p}_{z}}$ molecular orbital will be lower in energy than the $\ce{π_3_{p}_{x,y}}$ pair.
Each sulfur atom contributes 6 valence electrons, for a total of 12 valence electrons. Ten valence electrons are used to fill the orbitals through $\ce{π_{3}_{p}_{z}}$and $\ce{π_3_{p}_{y}}$, leaving 2 electrons to occupy the degenerate $\ce{π^{⋆}_{3}_{p}_{z}}$ and $\ce{π^{⋆}_{3}_{p}_{y}}$ pair. From Hund’s rule, the remaining 2 electrons must occupy these orbitals separately with their spins aligned. With the numbers of electrons written as superscripts, the electron configuration of $\ce{S2}$ is $\ce{(σ_{3}_{s})^{2}(σ^{⋆}_{3}_{s})^{2}(σ_{3}_{p}_{z})^{2}(π_{3}_{p}_{z,y})^{4}(π_{3}_{p}^{⋆}_{x,y})^{2}}$ with 2 unpaired electrons at the antibonding $\ce{π^{⋆}}$ orbital and thus is paramagnetic. The bond order is (8 − 4) ÷ 2 = 2, so we predict an $\ce{S=S}$ double bond.


text:
Why is my sugar paste becoming a brick?
I know this question may sound to be not inherent with chemistry, but since making the sugar wax paste involves acids and heating, I thought this would be the right place.
Preparing the paste is quite simple:
1 cup of sugar, 2 tbs of lemon juice and 2 tbs of water. Then the process expects to heat the compound until it becomes abra colored. The the mixture is moved away to cool down.The result would be a malleable and sticky paste at ambient temperature. My problem is that my paste becomes really hard and impossible to shape.
What could I do to avoid the compound to become a brick?

It may be more appropriate for Seasoned Advice (Cooking SE)

My main concern is how can I avoid sugar to crystalize?

Are you making caramel candy or something else? If yes, this receipe sounds really weird (the sugar should be caramelized before adding fluid, then you probably add way more fluid).

You don't have enough liquid in the recipe. Double the lemon and water to 4 tablespoons each, and you should have better luck. Adding more liquid will bring the sugar/liquid ratio more in line with what is needed to make the paste pliable. Add more liquid if you still have issues.

According to this video you seem to be using (much) less water and lemon juice than you should. To your solace, a bit of google-ing shows that a lot of people tried to prepare sugar wax without obtaining the expected results.


text:
Why is the discovery of chiral molecules in space a big accomplishment?
Today, I came across the discovery of a chiral pair of molecules in interstellar space. I know that a chiral molecule is a molecule that is not superimposable on its mirror image. But why is this discovery such a big deal?  I think that the existence of chiral pairs should be obvious where multiple molecules exist. So why is this discovery significant?

The story is a big deal because it might help explain why molecules in living creatures have a single chirality
One of the great mysteries of living systems is that, when they use chiral molecules, they only use molecules of one chirality. There is no obvious mechanism to kick-start this preference and, while there are many esoteric theories, there is very little evidence that explains why one chirality was chosen over another.
Proteins are, for example, made from amino acids which are almost all chiral. And they are all left-handed (technically they are all levorotatory, see wikipedia). Most sugars used in biological systems are right-handed. There are simply no convincing theories that explain the origin of this preference.
One modern theory suggests that there is a subtle energetic preference for one chirality caused by the interaction of molecules with the electrons emitted in certain types of radioactive decay (the weak nuclear force involved in radioactive decay is one of the few fundamental forces to have a mirror-image preference). But other experts think the preference is arbitrary.
Resolving this would be big news for biology.
The reason why the observation of chiral molecules in space (the news story only reports the observation of a molecule that is chiral not that we can observe the chiral preference) is that it creates hope than we can observe directly whether the chiral preferences of life are inherent in the universe or are arbitrary.
We don't yet know whether propylene oxide (the molecule in the news story) exists with more of one chirality than the other, but now it has been observed, we can expect to find out as instruments improve. So the big story is that we now have some hope that we can test some of the theories about why life has a handedness preference.

I believe that this is the news article in question (link).
The fact that there are left and right versions in space makes sense. The ability to see the chirality in space might be a first... the technology/ability to see chirality might lead to finding life out there because life is preferential toward one chirality or another DNA and many sugars are right handed molecules in life (on earth); proteins are left handed. the "reason" for selective chirality of life is basically unknown.


text:
Why does a larger DNA gel box require a higher voltage to move DNA at the same speed?
If I run a small agarose DNA gel and use a voltage of 90V, the DNA migrates the length of the gel in about an hour. If I use a medium size gel I have to use a voltage of 140V to move the same DNA the same distance (the equivalent of the small gel) in the same amount of time. For a large gel it takes about 170V. What is it exactly that requires a higher voltage to maintain the same speed of DNA movement?
I think this is more of an electrochemistry question than a biology questions so I put it here.

It’s not an electrochemistry question, more like an electrophysics question.

It's actually a physics question. The speed of the particle in the gel is proportional to the force that acts on it. The force that acts on the charged particle is proportional to the strength of the electric field $E$. The strength of the electric field is proportional to the voltage $U$ and inverse proportional to the length of the gel $d$.
$$E = \frac{U}{d}$$
So if you want to have the same strength of the electric field you have to double the voltage if you double the length of the gel.


text:
How can I keep homemade laundry detergent from disassociating?
We've been using this recipe to make laundry detergent. Works fine, but after a few days it disassociates into a clear bottom layer and a pink top layer. It's mildly annoying to have to re-stir it every time. Can we do something about it?

You should use about 0.25 - 0.35% xanthan gum (thickening agent) per the total weight of the mixture. Using xanthan gum is kind of tricky. You want to thoroughly mix the gum with the baking soda and washing soda (maybe in a bag), and slowly add the sodas to the water while mixing. If you don't do this, the xanthan gum will form lumps that are difficult to manage.
Hydroxyethyl cellulose might also work as a thickener, but it could increase the sudsing effect. Also, I would try adding a dash of dish soap; only use about 1% of the total weight of the mixture though, or else the sudsing effect will cause the washing machine to foam over.

It sounds like your emulsion is breaking down. Borax is used an emulsifier to get the soap molecules to mix more effectively in water (soap itself is an emulsifier, which is why it is able to remove grease and other hydrophobic molecules in water wash. In fact, most of the active ingredients in detergent are emulsifiers or surfactants of some sort). There are a number of factors that could be contributing to this, including the hardness of your water, and the actual ingredients you use.
There are a couple of things you could do to try to form a more permanent emulsion:

increase the amount of borax in the recipe.
make sure you dissolve your borax fully in warm water before it is mixed in.
Mix well. By well, I mean get a paint mixer on the end of a power drill and have a cuppa while you mix it.
As you are making an oil-in-water emulsion, you might consider combining (and dissolving) all ingredients EXCEPT the melted soap first, and then add the melted soap slowly as you mix well (see above for definition of mix well).
try adding an additional emulsifier to assist stabilising the emulsion. You could cheat and add a little commercial liquid detergent as a starter.

How much is "a little"? And, he or regular?

There's no precise formula. For a bucket of detergent, I'd start with a capful of additive. If that doesn't hold, add a little more after 24 hours and remix. If it does hold, use less next time! Not sure whether HE or regular formulation would make a difference to the emulsion stability.

The laundry soap recipe contains 3 classes of reagents: 1) Pure Soap (sodium stearate),  2) dispersant (borax adsorbs on clay and other solid particles, conferring a negative charge, and borax is also alkaline: pH~9.2) and 3) alkalizers (Na2CO3, pH~11.6) and sodium bicarbonate (pH~8.3).
The CRC Handbook lists sodium stearate as "soluble", but that needs qualification. A 2% solution is clear or nearly clear (but has a strong Tyndall effect); a 10% "solution" is quite turbid; and higher concentrations may separate quickly into layers. This separation is favored by a high ionic strength. NaCl is used commercially to "salt out" sodium stearate from the hydrolysis of tristearin (beef fat).
Thorough mixing of the warm sodium stearate mixture will help homogenize the mixture, but borax, sodium carbonate and sodium bicarbonate all tend to favor salting out the sodium stearate.
A cleaner can be optimized for the dirt it is supposed to remove. If your dirt is more oily/greasy, you may not need so much borax or sodium bicarbonate. Less salt will give a more stable solution. If your dirt is more clayey, dusty, or muddy, then less soap and more dispersant and alkalizers will work better (and the lower concentration of stearate would be stable longer).
Another possibility is to start with a soft soap (made from oleic acid). These tend to be less prone to separation because they form more viscous associations and are less solid when they begin to separate. The foaming could vary and be more significant for HE washing machines however.


text:
Percent difference: What to place on the denominator?
I'm genuinely confused on what to put on the denominator for the formula $$\%~\mathrm{difference} = \mathrm{\left|theoretical~-~experimental\right|\over theoretical}$$ or some other similar formula with different variable names that mean the same thing.
Specifically, I have a reading on the iron content per pill written on the bottle, and I have an amount that I calculated from doing some math with the experimental data. The problem is, I know that the calculated value is experimental, but the iron content per pill isn't really theoretical because it's known. Anyway, which should I put in the denominator?

Also, welcome to Chem.SE! :-)

If it's %difference, shouldn't you multiply the RHS of your equation by 100?

The experimental value is the one you measured/calculated yourself, from your own experiments. The theoretical value is the one that you obtained from some third party, to which you are comparing your experimental value.
So, the number on the bottle is the theoretical, and the number from your data and calculations is the experimental.

It seems that you are really asking about the errors in the measurement otherwise its trivial and you plug in the numbers and multiply by 100 to get percentage. You should then place the decimal so that your result is to the same number of decimal places as the smallest number of places you have in either number used in the calculation. If you are asking about errors you can use the method below. You have two readings from different sources with different error values. Before you can decide what the overall error is you must have some estimate of the respective errors, even if one is very small or zero. Next you should use an error propagating formula to estimate the error in % difference.
Let your equation be y = f( x, z ) where f is the function in x and z then the variance (std deviation squared) is

thus in your example x could be theoretical and z experimental values thus you have y=(x-z)/x = 1-z/x. The sigma's are the respective measured standard deviations. Once you have the total error you can decide how to write  %difference +- error. (The absolute value does not matter here as the formula squares values)

While this type of uncertainty propagation calculation is important to know how to carry out, I don't think it answers the OP's question.

I added a comment that should answer this at the end of the first paragraph.  If the OP is doing the calculation and is worried about precision then its not much more effort to calculate the errors properly.


text:
Decreasing Powder Size
I have a small amount of a powder that is roughly 100 microns but I need to get it down to less than 10 microns. It is a small amount so I can't send it to any factory specializing in that as far as I know. In your opinion, what's the best way (cheapest) way to decrease the particle size (so far I've only been able to find coffeee grinders but it is unlikely they'll be able to get the particle size down to 10 microns) to less than 10 microns?

A ball mill can get you to sub-micron particle size with a minimal capital investment - less than 500 USD on eBay.


text:
Does anyone know how to work out the amount of sodium sulfate present in water?
I need to find out 2 ways to determine the amount of sodium sulfate in an aqueous solution. I have searched the internet for assays.

I would use barium nitrate and weigh the precipitate, assuming that the solution does not contain any other sulfates.  Barium sulfate solubility is only 0.0002448 g/100 mL at room temperature.  An experiment is here http://web.iitd.ac.in/~arunku/files/CEL212_Y14/CEL%20212%20Lab%204%20Sulfate.pdf. You could also try something like this http://www.thomassci.com/Laboratory-Supplies/Water-Quality-Test-Kits/_/SULFATE-TEST-KIT.


text:
Energy level of hybrid orbitals in the molecular orbital energy diagram
I would like to know what is the level of energy of a hybrid orbital?
For instance, lets consider the $\ce{N2}$ molecule. According to its geometry, we know that there is an orbital 2p and 2s that are going to form 2 sp orbitals.
 In the molecular orbital energy diagram, where should we place the these 2 sp orbitals? Would they be in the middle of a 2p and a 2s? And which p orbital do we take to take the middle ($p_x$,$p_y$,$p_z$, $\pi$ or $\sigma$?) Will the sp* be higher or lower that the $2p_x$ and $2p_y$?

Hi and welcome to chemistry.stackexchange.com. Since you already took the tour, all that remains is to point you towards the [help], where any unclear aspects of how the site works will be explained =)

Note that molecular orbital theory does not require hybrid orbitals.

Right, I did not know that molecular orbital theory does not require the use of hybrid orbitals, cheers.

@Bbruyne If you decide to use hybrid orbitals then, yes, your guess is correct: Since an $\mathrm{sp}$ orbital is an equal mixture of an $\mathrm{s}$ and a $\mathrm{p}$ orbital it would be energetically in the middle between those orbitals. For the mixing you would usually use the $\mathrm{p}$ orbital that points in the direction of the bonding axis which conventially is taken to be the $\mathrm{p}_{z}$ orbital.

@Bbruyne What do you mean by $\mathrm{sp}^{}$? There is no such thing as an anti-bonding $\mathrm{sp}$ orbital if that is what you mean. Mixing orbitals to form hybrid orbitals is conceptually totally different from letting, say, two $\mathrm{s}$ orbitals interact to form a $\sigma$ and a $\sigma^{}$ orbital.

I assume that each sp orbital of the two N will interact to form a  σ and a σ* orbital.

@Bbruyne Yes, the two $\mathrm{sp}$ orbitals pointing towards each other would form form a $\sigma$ and a $\sigma^{}$ orbital. The $\mathrm{sp}$ orbitals pointing away from each other would interact as well but much weaker due to the low orbital overlap. Thus the splitting between the resulting $\sigma$ and $\sigma^{}$ orbitals would be quite small and so those $\mathrm{sp}$ orbitals could essentially be considered as non-bonding orbitals (holding the $\ce{N2}$ lone pairs).

When we have an sp hybrid orbital, it is usually made of the s orbital and the p orbital that points in the bounding axis ($p_z$). It's energy will be the mean of the energy of the initial orbitals.
In the example of $\ce{N2}$, it is essential to bear in mind that each sp won't form an $\sigma$ and an $\sigma^*$ orbital. Only the one pair of the sp orbitals, the one that overlap the most, will do. The other pair will contribute to the $\ce{N2}$ lone pairs because there is nearly no overlapping. The 2 others bonds that form the triple bonds are made with the 2 left orbitals $p_x$ and $p_y$.
By the way, it is not necessary to use hybrid orbitals for the molecular orbital theory.
Credits to Philipp and Ben Norris.

"Its energy will be the mean of the energy of the initial orbitals" How does this align with the answer posted here? https://chemistry.stackexchange.com/questions/41097/why-does-the-mixing-of-sigma-2s-and-2p-orbitals-lower-the-energy-of-the-sigma-2s?rq=1


text:
Back titration of excess HCl used for dissolving limestone

$15.0~\mathrm{mL}$ of $1.4~\mathrm{M}\ \ce{HCl}$ was mixed with $1.00~\mathrm{g}$ of limestone (impure $\ce{CaCO3}$) until all the solid had dissolved. The solution was then transferred to a conical flask and made up to $200~\mathrm{mL}$ with water. A $20.0~\mathrm{mL}$ portion was then neutralised by $8.50~\mathrm{mL}$ of a $0.1~\mathrm{M}\ \ce{NaOH}$ solution.
Calculate:

Amount of excess $\ce{HCl}$ in the $20.0~\mathrm{mL}$ portion
Amount of excess $\ce{HCl}$ in the $200~\mathrm{mL}$ portion
Amount of $\ce{HCl}$ which reacted with $\ce{CaCO3}$

My attempt:
When the limestone is dissolved,
$$\ce{2 HCl(aq) + CaCO3(s) -> CaCl2(aq) + CO2(g) + H2O(l)}$$
$$n(\ce{HCl}) = c \times V = (15 \times 10^{-3}~\mathrm{L}) (1.4~\mathrm{M}) = 0.021~\mathrm{mol}$$
$$n(\ce{CaCO3}) = x~\mathrm{mol}$$
When the solution is diluted to $200~\mathrm{mL}$ the molarity of $\ce{HCl}$ can be found by,
\begin{align}
n_\mathrm i &= n_\mathrm f\\
c_\mathrm i V_\mathrm i &= c_\mathrm f V_\mathrm f\\
c_\mathrm f &= \frac{c_\mathrm i V_\mathrm i}{V_\mathrm f}\\
\therefore c_\mathrm f (\ce{HCl}) &= \frac{(15\times 10^{-3}~\mathrm{L})(1.4~\mathrm{M})}{(200\times 10^{-3}~\mathrm{L})} = 0.105~\mathrm{M}
\end{align}
I'm having doubts about my working up until this stage. Am I correct so far?

It is generally a good idea to approach problems in the order that sub-questions are asked in. There is often a paedagogic reasoning in that order that can help you approach the answer.
In this case, the key point is that the limestone is impure $\ce{CaCO3}$, and that thus we do not know how much of the acid was initially neutralised with $\ce{CO2}$ evolution. Approaching the question from this side, as you did, is not very helpful, since we are missing an important value to calculate from.
What we do know, however, is the final end of the experiment: that $8.50~\mathrm{ml}$ of $0.1~\mathrm{M}\ \ce{NaOH}$ were able to neutralise a certain portion of the remaining hydrochloric acid. You should think of this as being a titration.
Once you have realised that, it makes sense why the first question asks for the amount of substance in the $20.0~\mathrm{ml}$ aliquot — it is the one you can actually calculate quickly, and it will lead you to the entire solution.
Thus, your plan of action is:

find the concentration or amount of $\ce{HCl}$ in the $20.0~\mathrm{ml}$ portion by known formulae used for titration problems (your work so far shows you can do this);

 We have $\ce{HCl + NaOH -> NaCl + H2O}$, so
 $$\begin{align} n_\ce{HCl} = n_\ce{NaOH} &= (\pu{8.50 mL})(\pu{0.1 M}) \\ &= \pu{0.00085 mol} \end{align}$$

thencefrom calculate the concentration or amount of $\ce{HCl}$ in the entire $200~\mathrm{ml}$ sample from similar formulae;

 A $\pu{20 mL}$ portion of the $\ce{HCl}$ simply has one-tenth of the original amount of $\ce{HCl}$, so in the entire $\pu{200 mL}$, we have:
 $$n_\ce{HCl} = (\pu{0.00085 mol}) \times \frac{\pu{200 mL}}{\pu{20 mL}} = \pu{0.0085 mol}$$

Next, decide how much hydrochloric acid was lost in the dissolution of the limestone.

 At the start, we had a grand total of
 $$\begin{align} n_{\ce{HCl},\mathrm{tot}} &= (\pu{15.0 mL})(\pu{1.4 M}) \\ &= \pu{0.021 mol} \end{align}$$
 which you already calculated. After reacting with the limestone, we were left with only $\pu{0.0085 mol}$ of that. So, the amount that reacted is simply:
 $$n_{\ce{HCl},\mathrm{reacted}} = \pu{0.021 mol} - \pu{0.0085 mol} = \pu{0.0125 mol}.$$


text:
Why is the pOH of 1M HCl 14?
Why is the pOH of 1 molar HCl 14 in an aqueous solution?
I was learning about strong bases and strong acids in chemistry within this context:
We have 1L of pure water at 25°C, and the pH of water is 7 and the pOH of water is 7, so it is a neutral solution.
Then we add 1 mole of HCl into the water. HCl, because it is such a strong acid, disassociates completely in the water. After some basic calculations, we found that the pH of HCl was 0 and the pOH was 14, but the way he showed this was using the water's pH and pOH - does this make sense?
So we found that the pH of water is 0, and then he did something funny and got a pOH of 14. Why is the pOH of this solution 14?
Here is the video, video mark, 10:02: Khan Academy - pH & pOH of Strong Acids and Bases

pH + pOH = 14. This follows from $k_w = 10^{-14}$.

@aventurin keep in mind this is at STP; at higher temperatures $pK_w$ will be lower.

In the moment from the video that you referred to, Sal was using the equation $\rm pK_w=pH+pOH$ to find the $\rm pOH$ of the solution. $\rm pK_w$ is equal to 14, since $K_w$ is equal to $1\times 10^{-14}$, giving us the new equation $14=\rm pH+pOH$. We know that the $\rm pH$ of a 1M solution of $\ce{HCl}$ is zero from earlier on in the video. This gives us that $14=0+\rm pOH$, or in words, that the $\rm pOH$ must be equal to 14.

In case the $\mathrm{p}$'s are confusing, why don't we take a step back and think about what $K_\mathrm{w}$ means?
The reaction of water's autoionisation is $\ce{H2O(l)->H+(aq) + OH-(aq)}.$ This gives the following expression for $K$, the equilibrium constant:
$$K_\mathrm{w}=[\ce{H+}][\ce{OH-}]$$
Remember that $K_\mathrm{w}$ will be a constant at standard conditions (actually, it is $1.0\times10^{-14}$ at STP). But let's try to write this equation in terms of $\mathrm{p}K_\mathrm{w}$, remembering our log rules!
$$-\log K_\mathrm{w}=-\log ([\ce{H+}][\ce{OH-}])$$
$$\mathrm{p}K_\mathrm{w}=-\log [\ce{H+}] -\log[\ce{OH-}]$$
$$\mathrm{p}K_\mathrm{w}=\ce{pH}+\ce{pOH}=14$$
Now you should understand where this magical equation comes from. Since water is approximately neutral, it will have $\ce{pH}=\ce{pOH}=7$ at standard temperatures.
You got that the $\ce{pH}$ of $\ce{HCl}$ was $0$ from taking the negative log of $\pu{1M}$. That means that $\ce{pOH}=pK_w-\ce{pH}=14-0=\boxed{14}$.


text:
Is it safe to use lab-grade calcium nitrate as a fertilizer?
I want to use lab-grade calcium nitrate in my hydroponic system (as a fertilizer) to grow lettuce, is it safe?

Usually you can buy chemicals in different purity grades.
If you buy something that has the grade purum >95% will be your desired ingredient. However there is not usually any information about what the other <5% will consist of.
Personally I would never ingest (or use in food production) anything that is not food grade. Food grade does not necessarily mean very pure, it just means that there is nothing in the other x% that is potentially harmful. Look out for an FCC number, that is the Food Chemical Codex, when you order it and you should be fine.

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The problem is by now I do not have access to Calcium-Nitrate from stores as a fertilizer but I have access to lab-grade ones, if it has some bad thing as impurity (like heavy metals) in it does the plants absorb that or they ignore that?

You can't relay on the plant to filter out potentially harmful substances. You have no way of knowing what else is in the calcium nitrate. You could always call the supplier and ask them for more information.


text:
Which is the most reactive element in the periodic table?
Which is the most reactive element in the periodic table? Is it francium, caesium, lithium or fluorine?

What about radioactive elements?

Fluorine is the most reactive non radioactive element on periodic table.
It exists in gaseous form at room temperature and even reacts with glass. So it is almost impossible to store it in pure form.and caesium is the most reactive metal

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text:
Reactant energy distribution and spontaneity of reactions
I wonder about the implications of substances having an energy distribution around a mean, as opposed to them all having the same energy level, on the spontaneity of reactions.
2 somewhat separate questions:

Can a reaction with a high activation energy, but which in the end has a negative Gibbs free energy change at certain temperature, occur spontaneously even though at this temperature the energy level of the reactants is lower than the activation energy? If I understand correctly, the answer would be yes because the energy of the reactants is just an average of a broader distribution, and so some small fraction of the reactants would attain the $E_a$ even if on average $E_a > E_{\text{reactants}}$?
Wouldn't it also be possible that some non-spontaneous reactions (with a positive change in Gibbs free energy) occur spontaneously? I mean we know that if we invest $\Delta G$ energy, we will force the non-spontaneous reaction to occur. But what if some small fraction of the reactants get to the energy level they would have had if $\Delta G$ energy was invested in them (without it being actually invested). Wouldn't that mean that for that fraction of the reactants, in the high end of the energy distribution around some average, the reaction would occur (while not occurring for the rest and vast majority of the reactants who fail by themselves to attain the same energy level they would have had if ΔG energy were invested in them)? Is there a flaw in this understanding?

I guess the question here is whether the idea that $\Delta G > 0$ reactions are non-spontaneous is a statement of probability that says that such reactions are just very unlikely or is it an hard and fast rule like the conservation of energy?

Yes; you are on the right path in your second paragraph, but distinguish between activation energy and free energy as they are not connected. I'm sorry but I think that your third paragraph is not all correct as it stands. If the free energy change is positive the reaction is not spontaneous by definition. Perhaps you mean, if the overall enthalpy change is positive can the free energy now be negative if the overall entropy change is also positive? The answer is yes.

The thermodynamic quantities only give us information about initial and final states at equilibrium. The activation energy is obtained from kinetic, that is rate constant measurements usually vs, temperature and not from thermodynamic quantities. In the vast majority of reactions the activation energy is greater than thermal energy (approx kBT, where kB is Boltzmann's constant). These reactions are generally slow because, by the decreasing exponential nature of the Boltzmann distribution with energy, not many reactants have enough energy to react in any given time period, say 1 sec. (This is lucky otherwise we would all have reacted eons ago!) In some special types of unimolecular reactions, e.g. electron transfer, bond dissociation & isomerization, with zero or very small activation energy then clearly these will be very fast possibly with rate constants $ >10^{12}\ /sec $.
In the case where the products are of higher energy than reactants, reaction still occurs but the equilibrium lies on the side of the reactants. Again by the Boltzmann distribution argument products react faster to reactants than the other way round.

Thank you for your answer. I wonder, though, if, for example, a reaction needs 50kJ/mol of work done on it to occur (ΔG = +50), then wouldn't a tiny fraction of the reactants achieve that level of energy even while the average reactant would have 50kJ/mol less energy than what's needed for the reaction? And if so, it seems that that a tiny fraction would be able to react even though the 50kJ of energy weren't invested in it. If this scenario is possible in principle, isn't that a case of a non-spontaneous reaction occurring spontaneously?

You ask what happens when products are at a higher energy than reactants. In this case some reactants do still go to products but by the Boltzmann distribution it is more likely that products return to reactants so the equilibrium is on the side of the reactants. This is because in this case the activation barrier is smaller for products. At equilibrium, products and reactants are always interconverting. If you completely removed product the reaction would proceed until all reactant was used up. Thermodynamics tells us nothing about how long this might take.

I think you are confusing $\Delta G$ and $\Delta G^\circ$, where $\Delta G = \Delta G^\circ + RT \ln (Q)$.  At $\Delta G$ = 0, $Q$ is equal to the equilibrium constant $K$, and $\Delta G^\circ = -RT \ln (K) $.
But $\Delta G$ and spontaneity and all that is about being away from equilibrium and getting back to equilibrium: $\Delta G = RT \ln (Q/K)$.  So if you're not at equilibrium $Q$ < $K$ then $\Delta G $ < 0 and more product will be made spontaneously to get back to equilibrium whether the sign of $ \Delta G^\circ$ is positive or negative.


text:
Ignoring the difference of concentrations of reactants. Equilibrium
Recently at school, while studying equilibrium we learned that in problems where the initial concentrations of reactants and the equilibrium constant are given, there is a special formula to determine wether the change in reactants can be ignored where C is the concentration of the reactant with the smallest concentration and K is the equilibrium constant: L=C/K. We learned that if L is bigger than 500, we can consider the change in the reactants negligible. However this is where my problem lies: i do not see why the C is included, doesn't the size of K by itself tell us wether a little or a lot of reactants is transformed? Not to mention, some cases turn out flat out wrong. For example:
A+B=(reversible) 2C.
Initial concentrations: A=10^-20M
B=10^-20M.  C=0 M
K=10^-8
By the equation we learned, L=10^-12 and thus the variation in the reactants should not be neglected.
However, after doing the math, x~10^-24 which is 4 orders of magnitude below the beginning concentration of the reactants and completely negligible, is something wrong with my logic? Or did i discover a flaw in my textbook :)
Thanks

You can't use it that way because technically C is the smallest, being 0. So L=0. This rule of thumb was developed for systems already in equilibrium, if I remember correctly.

Let's look at your example and set up the ICE (intial-change-equilibrium) problem. Then I'll give you more inclusive of thumb.
Let's say reaction equation is
$\ce{A + B <=> 2C}$$
The law of mass action is:
$$K = \dfrac{[\ce{C}]^2}{[\ce{A}][\ce{B}]}=10^{-8}$$
Let's set up the ICE matrix. The change in concentration $x$, will be subtracted from $[\ce{A}]$ and $[\ce{B}]$ and added to $[\ce{C}]$.
$$\begin{array}{|c|c|c|c|}\hline
\ & [\ce{A}]\ (\mathrm{M}) & [\ce{B}]\ (\mathrm{M}) & [\ce{C}]\ (\mathrm{M}) \\ \hline
I & 10^{-20} & 10^{-20} & 0 \\
C & -x & -x & +2x \\
E & 10^{-20} - x & 10^{-20} -x & 2x \\ \hline
\end{array}$$
Now we pug in to the law of mass action:
$$10^{-8} = \dfrac{x^2}{\left(10^{-20} -x \right)^2}$$
Since all numbers have to be positive, let's take the square root of both sides:
$$10^{-4} = \dfrac{x}{10^{-20} -x }$$
Now we have a ratio that says that $x$ is four orders of magnitude smaller than $[\ce{a}]_i -x$. I could have easily used the initial concentration of $\ce{B}$ Put another way:
$$x \approx  K\cdot [\ce{A}]_i$$
If this value is much less than $[\ce{A}]_i$, then we can simplify.


text:
Why exactly does molten NaCl explode, when it is poured into water?
Why does molten $\ce{NaCl}$ explode, when it is poured into water?
$\ce{NaCl}$ has a high melting point, $1074\ \mathrm{K}$ ($801~\mathrm{^\circ C}$). $\ce{NaCl}$ has a molar mass of $58.44\ \mathrm{g/mol}$, it has specific heat capacity of $36.79\ \mathrm{J/(K\cdot mol)} = 629.53\ \mathrm{J/(K \cdot kg)}$. therefore the $\ce{NaCl}$ at melting point temperature has $491,033\ \mathrm{kJ/kg}$ more thermal energy than $\ce{NaCl}$ in STP conditions ($1\ \mathrm{atm}, 20~\mathrm{^\circ C}$, temperature difference $= 780\ \mathrm{K}$)
According to one amateur video in youtube, the molten salt explodes when it is poured into water (c. 2:15)
What is the exact cause of the explosion?
The author of the video reasons that the phenomenon is purely physical and it is caused by that water heats up, vaporises and expands as a gas inside the glimp of very hot $\ce{NaCl}$ salt. But is it really everything that happens there?
Other possible processes present in such occasion are (this is just a list what comes into my mind):

rapid crystallisation of the $\ce{NaCl}$
chemical reaction between $\ce{Na}$ and water: $$\ce{2 Na (s) + 2 H2O (l) -> 2 NaOH (aq) + H2 (g)}$$ (this causes explosion if $\ce{Na}$ is inserted into $\ce{H2O}$)
reaction between $\ce{Cl}$ and water: $$\ce{Cl2 + H2O -> HOCl + HCl}$$
the rapid solubility of hot $\ce{NaCl}$ into water
thermal decomposition (thermolysis) of $\ce{H2O}$ into either monoatomic or diatomic hydrogen and oxygen, and reactions that follow this.

Also, the author fails to explain why no explosion occurs with sodium tetraborate nor sodium carbonate when each of them, in molten state, was poured into the water.

The author is completely right, and all other processes you mention are either totally negligible or downright impossible.

Also, Water is not $\ce{H20}$, but $\ce{H2O}$ (‘H twenty’ versus ‘H two Oh’).

@IvanNeretin That comment is not very helpful. Cool that you know the answer, but it doesn't really help OP.
The most interesting part is why this doesn't occur for sodium tetraborate and sodium carbonate. Perhaps this is why OP is sceptical of the explanation given by the author.

@Brian You comment sounds quite puzzling to me, as OP never said a thing about sodium tetraborate or sodium carbonate.

@Brian I’ll agree with Ivan that that was not part of the question asked or I would have had a more difficult time answering it.

Sodium Chloride hydrolise at high temperatures,btw

@IvanNeretin I only meant to say that it would be nice if you took the time to explain why the author is right, and why all of OP's suggestions are negligible or impossible. :-) That's what I was talking about when I mentioned "it doesn't really help OP". I tried to put a line break afterwards, but they get removed. The latter part was to bring to people's attention that different behavior is observed for sodium tetraborate and sodium carbonate (as it was shown in the linked video). If you believe this simple explanation for NaCl, wouldn't you expect the same for the other two?

@Jan He is asking if the author has the correct explanation. He links the experimental basis for the discussion, although he doesn't specifically bring up the discrepant behavior of sodium tetraborate and sodium carbonate... That's why I did... Because the authors explanation sounds very plausible, but since it doesn't occur for sodium tetraborate and sodium carbonate, we're missing part of the explanation. Which is what we're after, right :-)?

@Brian I'll modify the question to ask about why the same behavior is not observed with sodium tetraborate and sodium carbonate.

Now that makes the question quite different and a lot harder.

I’m not sure I approve of the moving question after it has been answered. I think I would prefer a second question asking for the interesting observation with borate and carbonate. CC @Brian. $$%$$Oh yeah. Protip for Chem.SE: $$%$$ will fit in a display maths block which includes vertical space ;) Doesn’t work on SEs that don’t have MathJax.

I am fairly certain the author has never had his hands on molten sodium carbonate; it decomposes.

Understand that sodium chloride is not made up of sodium metal and chlorine gas but of sodium ions and chloride ions, held together by ionic interactions. Under these thermal conditions (liquefication), the compound will not decompose into its elements and therefore all reactions you suggested which include elemental chlorine or sodium cannot occur.
Dissolution of sodium chloride in water is neither strongly exothermic nor strongly endothermic, so any effects stemming from the dissolution are neglegible.
Also note that you neglected an important variable in your calculations. As far as I can tell, you only calculate how to arrive at melting-point hot solid sodium chloride. To liquefy, additional melting enthalpy has to be applied, a further reservoir of energy to draw from. This melting enthalpy is of course released upon rapid crystallisation, but you should really subsume it into the heat energy difference altogether.
Finally, thermal decomposition of water is not exactly a process that will happen quickly, as exemplified by (water) steam temperatures that can be huge. Check out this unrelated answer of mine about how much energy is required to heat water to $100~\mathrm{^\circ C}$ and boil it.
So all things considered: The principle contribution to the explosion is the rapid heating of water including boiling, leading to tightly compressed gaseous $\ce{H2O}$ which expands explosively. This concurs with the recrystallisation of $\ce{NaCl}$ which accounts for a non-neglegible portion of the energy added, but which is technically already present in molten $\ce{NaCl}$. All other processes are minor or impossible.

Yes. Your answer makes sense. But when the experimenter tried with another salt the same didn't happen. What is unique about NaCl for this to happen?

@shre_sudh_97 That is a different question that you didn’t ask.

The question has been modified to include this question as well.

Also really interested why it does not work with other types of salt

$\ce{NaCl}$:

Molar mass: $58.44~\mathrm{g~mol^{−1}}$
Melting point: $801\mathrm{°C}$
Specific heat capacity: $36.79~\mathrm{J~K^{-1}~mol^{−1}}$
Std enthalpy change of fusion: $27.95~\mathrm{kJ~mol^{−1}}$

The energy required to melt $58~\mathrm{g}$ $\ce{NaCl}$ ($20\mathrm{°C} - 801\mathrm{°C}$) is roughly $56.6~\mathrm{kJ}$.
$\ce{H2O}$:

Molar mass: $18.02~\mathrm{g~mol^{−1}}$
Specific heat capacity: $75.375~\mathrm{J~K^{-1}~mol^{−1}}$
Heat of vaporization: $40.657~\mathrm{kJ~mol^{−1}}$

The energy required to vaporize $18~\mathrm{g}$ $\ce{H2O}$ ($20\mathrm{°C} - 100\mathrm{°C}$) is roughly $46.7~\mathrm{kJ}$.
So $58~\mathrm{g}$ $\ce{NaCl}$ of $801\mathrm{°C}$ could vaporize $21.8~\mathrm{g}$ water of $20\mathrm{°C}$ giving roughly $27$ liters of vapor. This suffices to make some kind of explosion. However, if a explosion really occurs depends also on how fast the heat is transferred from the molten salt to water. A high viscosity of the molten salt e.g. will prevent fast heat transfer.

Nice quantitive answer. As you say its a sort of explosion, a physical one due to pressure not due to chemical reaction. Incidentally  I don't see why a viscous solid will prevent heat transfer since  there has to be  lots of interaction between molecules; if anything would have thought that the opposite would be true.

When the molten salt has low viscosity it is more likely to form smaller parts (drops) when poured into water. The greater surface would then enable a faster heat transfer and thus a more vigorous reaction. A liquid with high viscosity, such as a molten glass,  has less tnendeny to do so. This can be seen in the video where the borate retains the shape of a single drop that slowly cools down from the surface to the middle.

Having looked at the video, it is clear that in the last experiment a small region of water around the very hot molten NaCl boule is turned to steam that rapidly expands and the shockwave causes the fish tank to explode. This must occur as there was sufficient energy left in the molten salt to completely vaporise a layer water next to it even though thermal diffusion (thermal conduction) in the water is carrying heat away and any bubbles of steam will try to rise out of the water. The steam produced is probably also super heated by the > 800 C molten salt and will continue to expand after the water boils. The reason that the cooler but still liquid salt did not vaporise enough water to crack the tank is probably that thermal conduction cools the melt and also removes heat from the water surrounding the boule and this wins out over heating enough of the water to vaporisation do do any damage.  Incidentally the  molten salt is probably molecular in nature; there is no water in the melt to make ions. Molecular NaCl (as a diatomic molecule) is well known in the vapour phase and its spectroscopy has been well studied as has NaI the iodine equivalent.

Yes, I think you nailed it. What I saw: As the molten salt entered the water, it was sheathed by a bed of superheated steam that keeps water away. Perhaps because of splashing, a small amount of superheated steam near the surface exploded. This caused a) a venting of the steam near the surface upwards and b) a sudden unsheathing of the molten salt from its steam bed as the surface shock propagated downwards and disrupted the hitherto organized molten salt flow. Once it became turbulent, vastly more energy transferred to the water, water instantaneously superheated and an explosion resulted.

I suspect rather strongly that this is a coulombic explosion.
This phenomenon was only recently described to explain the violent explosion of alkali metals in water. The explosion of alkali metals in water share some properties with molten salt explosions, specifically:

it is extremely fast and violent - much faster than expected
a similar set of alternate explanations (steam pressure, hydrogen combustion) have been attempted but have not been shown to work

Since molten salt is also conductive, a similar thing could be happening here.
One important observation that supports this is that the molten salt explosion only occurs when the salt is heated above a certain temperature. It is not enough for the salt to be simply molten. If the salt is below this temperature, there is almost no reaction (apart from steam). But when above this temperature, there is a completely different and explosive reaction. This can readily be seen in the different attempts in the OP's video. The third attempt (in which he made the salt 'really hot') is the only one that explodes.
An explosion will occur if the repulsion forces of the built-up charge within the salt exceeds the surface tension of the liquid salt. And it seems likely to me that both the electron affinity of the salt and its surface tension depend on temperature. If the force is below the surface tension, the salt will hold together while eventually cooling. If the force is above the surface tension, the salt will explode. This is consistent with observation.
This also provides a potential explanation for why the same explosion does not occur for sodium tetraborate or sodium carbonate. The electron affinity and surface tension are both likely to be different from those of NaCl.
The key to achieving an explosive effect is to bring out the energy within the salt (not just the energy at its surface). A coulombic explosion explains this because it predicts that the salt itself blasts apart, instantly exposing every internal portion in a violent chain reaction. This makes that internal thermal energy available almost instantly.