text:
Macgyvering a Spectroscope
I am currently playing a dungeons and dragons campaign in which there are some new elements. My party, being the nerds that we are, is looking to experiment with them and attempt to discover new ones using a spectroscope. The problem is that we need a solvent that will dissolve our sample. I was wondering if there were any suggestions as for what to use, or how to obtain it, as the game takes place in a medieval time period. Acetone would be best, but I'm not sure that it's possible to extract that from nature, although it does occur naturally. Thanks!

Are the samples minerals from which the elements have to be refined?

No, we either have, or can refine the materials ourselves.

I was asking because strong acids, such as sulfuric acid or nitric acid would have been possible solvents. For these, historical methods of  preparation, that would fit in the medieval time frame, are known.

That's a good point, someone had mentioned aqua regia, but I didn't know how to find the medieval preparation methods, nor whether that actually existed. Originally, I was thinking perhaps melting it with something similar to thermite, do you think that would work?

Btw, welcome to Chemistry.SE! I always like questions on chemistry in fiction :) If you like to know a bit more on how this site works, please have a look and take the tour :)

Acetone can be easily produced by heating calcium acetate, which is also easily produced using table acetic acid. Acetone is second to easiest organic solvent to come with, the first being ethanol.

Is magic allowed?

The medieval "nerd" in a D&D campaign might use acids to get minerals into solution.
Sulfuric acid ($\ce{H2SO4}$, vitriol oil) was initially prepared by heating iron(II)-sulfate ($\ce{FeSO4 · 7 H2O}$, iron vitriol) in a retort. The emerging sulfur trioxide was collected an led into water.
Nitric acid ($\ce{HNO3}$) was prepared by heating mixtures of potassium nitrate ($\ce{KNO3}$, saltpetre) with iron vitriol.
Hydrogen chloride ($\ce{HCl}$) was available by the reaction of vitriol oil with halite ($\ce{NaCl}$).
Mixing hydrochloric acid with nitric acid would then give aqua regia.
Another method to get the cations of minerals with low solubility in water into solution is the so called Aufschluß.
Here, one would either melt the mineral with

a mixture of soda and potash, or
a mixture of sulfur and potash

in a crucible and eventually leach out the now soluble material with water.

Admittedly, our D&D alchemist still doesn't have a spectroscope :\
In order to characterize the metals in the samples, he could use his Lötrohr, a kind of blow pipe and blow the flame of his oil lamp onto the sample place on a piece of charcoal or a gypsum plate. From the colour of the flame and/or the colour of the residue formed, he could draw some conclusions on the metal.
If our medieval science hero is a bit ahead of his epoch, he would also carry some borax in a bag and create borax pearls for a bead test.


text:
What is a baseline?
I am currently doing a lab write up for biology, in which the purpose of the lab is to "Develop a Method for Measuring Peroxidase in Plant Material, and to Determine a Baseline." This is supposed to be accomplished by finding the relationship between time elapsed since the beginning of the reaction (in minutes) and the $\mathrm{pH}$ of the solution. However, I'm not exactly sure what is meant by a baseline.
The lab packet, which contains instructions and background for the experiment, has this description of a baseline:

Baseline is a universal term for most chemical reactions. In this investigation, the term is used to establish a standard for a reaction. Thus, when manipulating components of a reaction (in this case, substrate or enzyme), you have a reference to help understand what occurred in the reaction. The baseline may vary with different scenarios pertinant to the design of the experiment, such as altering the environment in which the reaction occurs. In this scenario, different conditions can be compared, and the effect of changing an environmental variable (e.g., pH) can be determined.

However, whether due to an ambiguous description, or due to an error in my thinking, I am still a bit confused as to (a) what a baseline is, and (b) how baselines are used.
If asked, I would guess that a baseline is either a specific set of conditions (like what you'd find in a control group of an experiment), or  that it's the results you'd obtain from a specific set of conditions, but I'm really not sure. I don't understand it well enough to apply it in the next part of the lab, in which we are supposed to compare some part of the first section of the experiment (the baseline, I think) to some part of the second experiment, when the pH of the solution is changed.

I am not sure how applicable my opinion would be. However, in spectrometry the baseline is a way to reduce white noise. In spectrometry, we take a spectrum of the conditions before analyzing the sample. We then substract this spectrum from the spectrum of the analyzed sample. I suppose you could do the same for your situation. Measure the pH of the solution for x seconds. Then add the reagents and measure for x seconds as well.  Or the baseline in the lab could refer to the calibration of the probe.

I think it means a control (standard) sample that could be used for reaction purposes. Its definetly strange the way theyve put it tho.

http://www.bgs.ac.uk/research/groundwater/quality/BaselineUK/baselineMore.html
Maybe this can help

In the context of your lab, I think that the term "starting point for the experimental conditions" could fairly correctly be inserted in place of "baseline".  Once you have measured the parameters of interest for the starting point of your experiment, then you have something, a baseline, for comparison to the measured parameters after you perturb your system or the conditions of the reaction or measurement.  Without this baseline you cannot quantify the effects or interpret the meaning of measurements made under arbitrary conditions.
One comment to your question was "I think it means a control (standard) sample...". I'm not sure whether this is where the commenter was headed, but this is not a bad analogy. Here, your established baseline is like the standard against which you compare your future results.

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What reaction happens when carboxylic acids are treated with calcium hydroxide and distilled?
What reaction takes place when $\ce{Ca(OH)2}$ reacts with $\ce{R-COOH}$? What is obtained after distillation?

Acetic acid (as an example of a typical carboxylic acid) reacts with calcium hydroxide to form calcium acetate. This is a simple acid-base reaction which forms a salt and water:
$$\ce{2CH3COOH + Ca(OH)2 -> (CH3COO)2Ca + 2H2O}$$
Upon heating, calcium acetate undergoes decomposition to acetone and calcium carbonate. The acetone can be distilled off from the mixture, leaving behind solid calcium carbonate:
$$\ce{(CH3COO)2Ca ->[\Delta] CaCO3 + (CH3)2C=O}$$
At even higher temperatures calcium carbonate itself can be decomposed to calcium oxide and carbon dioxide:
$$\ce{CaCO3 ->[\Delta] CaO + CO2}$$
References

http://www.docbrown.info/page04/OilProducts10.htm

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Mechanistic Explanation of Circular Dichroism
I'm wondering if someone can give me a physically motivated reason for why chiral molecules interact with left and right circularly polarized light (LCP) (RCP) differently. Particularly, a quantum mechanical view of what's happening here.
I understand why it is that a molecule has to be chiral in order for it to manifest circular dichroism (my use of terminology may not be perfect here so I apologize). That is, both the electric and magnetic dipole operators must be non-zero and there's an argument here on the basis of point groups.
Nothing about that, however, is particularly illuminating when it comes to understanding why I should expect a molecule to interact differently with LCP and RCP light.
I don't know what else I can even say. A good answer would address the different behavior of both the light and the molecule in these different situations, as I imagine that's a necessary piece to the puzzle.
Also, if one has some references (a review or textbook chapter etc.) that would be appreciated.

see http://chemistry.stackexchange.com/questions/1012/is-there-a-simple-way-to-get-the-circular-dichroism-of-a-molecule-from-its-struc

also http://chemistry.stackexchange.com/questions/6271/molecular-chirality-and-optical-rotation

Chiral molecules interact differently with chiral light, just like a screw (chiral) interacts differently if you twist it clockwise or counterclockwise with a screwdriver.

Hmm... Perhaps I need to do some reading before I can really give words to the thing that's bothering me cause it might just be based on simple misunderstandings. @jjgoings do you know of any references that discuss this on a simple level I could look into?

@jheindel Check out "Molecular light scattering and optical activity"(2004) by Lawrence Barron. Mechanistically, it will help if you consider chiral molecules like the helicenes (https://en.wikipedia.org/wiki/Helicene) rather than a 'typical' chiral molecule with a stereocenter. Simplistically, think about how it will be easier to polarize the electron density if the circular polarization matches the contour of the helicene rather than polarizing with the opposite 'handedness'. If, say, LCP light polarizes the molecule more strongly than RCP, there will be a difference in absorption intensity.

I think this page from Carnegie Mellon gets at what you are asking (despite being presented in a pretty ugly format).

see also the answer here https://chemistry.stackexchange.com/questions/73665/dependence-of-the-angle-of-rotation-on-the-wavelength-of-plane-polarized-light

Some part of a mechanical macroscopic explanation, acoustical activity where mechanical transverse oscillation direction is being rotated, is described in

Frenzel, T., Köpfler, J., Jung, E. et al. Ultrasound experiments on acoustical activity in chiral mechanical metamaterials. Nat Commun 10, 3384 (2019).

The notion “activity” refers the rotation of the linear polarization axis of a transversely polarized mechanical wave. The rotation angle is proportional to the propagation distance and does not depend on the orientation of the incident linear polarization. It has nanometer precision under time-harmonic excitation at ultrasound frequencies in the range from 20 to 180 kHz. Polarization rotations are as large as 22° per unit cell.
Add a net neutral electrical charge distribution to the components of the metamaterial where the negative charge distribution has a different elastic or inertial nature than the positive charge distribution to get a circularly polarized electromagnetic wave.
A quantum mechanical explanation was requested. Just add stochastic bias to the mechanical input so that only inputs of at least a minimum value get statistical significance. This makes the setup equivalent to quantum mechanics.

Not a singel vote. I have done what I can with this answer. What is missing with this answer?

Perhaps this is more like what the OP was looking for: Physical Principles of Circular Dichroism, Steven S. Andrews* and James Tretton,  J. Chem. Educ. 2020, 97, 4370−4376. Anyway, as I type this, you have 1 upvote and 1 downvote. I will give an upvote for an interesting take on the matter.

I would prefer if downvotes and upvotes were shown separately. As it is shown now I can not tell that someone found something wrong with the answer. I just guess that they mean that there can be no twist of the polarization plane, just a probability of a larger  one. Is this the issue? Effectively there is no difference since there are so many molecules involved.

I have no idea what others voted or what reason(s) they may have had. Right now, you have 3 upvotes, one being mine, and 1 downvote. Best not to get too discouraged: my most upvoted answer is about fill gas in potato chip bags! Some other of my answers, with some thoughtful effort put into them, have one or a couple of upvotes. And in a couple of cases, clearly wrong answers blew my correct answer away, rep wise. Annoying, but nothing much to do about it.

Another thing that might cause negative reactions is that the meta material has an elastic constant most commonly considered to be 0, and impossible. This is the case only in static stress/strain. I will eventually add something on that.

I do no longer support this answer and I will remove in in a while. This answer could eventually explain just a piece of the phenomenon.

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How to determine the equivalent weight of carbon dioxide correctly?
I just started learning about equivalent weight and encountered this solved example.

Problem Statement:
$4.215~\mathrm{g}$ of a metallic carbonate was heated in a hard glass tube and the $\ce{CO2}$ evolved was found to measure $1336~\mathrm{mL}$ at $27~^\circ\mathrm{C}$ and $700~\mathrm{mmHg}$ of pressure. What is the equivalent weight of the metal?

My attempt:
First of all lets find the volume occupied by $\ce{CO2}$ at NTP.
We know,
$$\frac{P_1 V_1}{T_1}=\frac{P_2V_2}{T_2}$$
Now, as it is given that
$$\begin{align}
&P_1 = 700~\mathrm{mmHg}\\
&V_1 = 1336~\mathrm{mL} = 1.336~\mathrm{L}\\
&T_1 = 27~^\circ\mathrm{C} = 300~\mathrm{K}\\
&P_2 = 760~\mathrm{mmHg}\\
&T_2 = 273~\mathrm{K}
\end{align}$$
Hence, $$V_2=\frac{P_1\times V_1\times T_2}{T_1\times P_2}=\frac{700\times 1.336\times 273}{300\times 760}=1.12~\mathrm{L}$$
Also, according to the law of equivalence, we get
$$\text{equivalents of metal carbonate} = \text{equivalents of }\ce{CO2}$$
Now comes the part where I am really stuck, i.e. how to determine the volume occupied by $1$ equivalent of $\ce{CO2}$.
I started with finding the equivalent weight of $\ce{CO2}$ by considering the reaction of carbonation of water which produces $\ce{H2CO3}$
$$\ce{CO2 + H2O -> H2CO3}$$
As we know the equivalent weight of $\ce{H2CO3}$ is $\frac{M}{2}$ because it has a basicity of $2$.
Hence, according to the law of equivalence,
$$\text{Equivalent of }\ce{CO2} = \text{Equivalent of }\ce{H2CO3}$$
which gives us the equivalent weight of $\ce{CO2}$ to be $22~\mathrm{g/equivalent}$. Hence, the equivalent volume of $\ce{CO2}$ at NTP is $11.2~\mathrm{L}$
Now consider this another method.
The equivalent weight of $\ce{C}$ in $\ce{CO2}$ is $3~\mathrm{g/equivalent}$ as $1 \ce{C}$ combines with $4$ equivalents of $\ce{O}$ to form $\ce{CO2}$, hence we get the equivalent weight of $\ce{CO2}$ as $3+2\times8=19\text{ g/equivalent}$ which is super weird.

My deal with the question:
What am I doing wrong in the second method for finding the equivalent weight of $\ce{CO2}$ and if you wanna know why I added the equivalent weights of the elements was because I had read that for finding the equivalent weight of an electrolyte we add the equivalent weights of the cations and anions. Hence, I tested this method of finding the equivalent weight on $\ce{SO4^2-}$.
Here is my attempt of finding the equivalent weight of $\ce{SO4^2-}$ using the method for electrolytes. We know that equivalent weight of $\ce{S}$ is $\frac{32}{2}=16$ and that of $\ce{O}$ is $8$, so we have
$$\text{Eq. } \ce{SO4^2-} = \text{Eq. }\ce{S} + 4 \times \text{Eq. }\ce{O} = 16 + 32 = 48~\mathrm{g/equivalent}$$

For those who are attempting to solve the problem:-
The solution provided by the book is as follows:-

 Volume of $\ce{CO2}$ at NTP=$\dfrac{1336\times273}{300}\times\dfrac{700}{760}=1120~\mathrm{mL}$
 Suppose the equivalent weight of the metal is $\ce{E}$.
So, Equivalent Weight of metal Carbonate $=\ce{E}+30$
Because, Eq. wt. of ${\ce{CO3}}^{2-}=\frac{60}{2}=30$
Now,
 $$\begin{align}\text{equivalents of metal carbonate} &= \text{equivalents of }\ce{CO2}\\\frac{4.215}{E+30}&=\frac{1120\text{(vol. at NTP)}}{11200\text{(vol. of 1 eq. at NTP)}}\\\text{So, } \text{E}&=12.15~\mathrm{g/equivalent}\end{align}$$

The whole concept of equivalent weight is somewhat obsolete. It is much easier to use molar mass instead as that is independent of the actual chemistry that is happening. A note on you MathJax: It is better to not use \dfrac{}{} as it does not render on mobile devices, it is also redundant, since $$ ... $$ sets fractions already in display mode.

@Martin-マーチン-Yeah, after reading all the other posts regarding equivalent weight on Chemistry.SE it seems that everyone is saying that, but isn't the concept of equivalent mass makes it easier to understand electrochemistry and redox reaction and solve them with a lot of speed and accuracy.

@Martin-マーチン-I would still appreciate it if you could tell me what is wrong in my thinking in the above post because I am a student who is doing self study and it has left me in tears after searching for it through all the books that I have and material regarding it on the internet but nothing is there that answers my doubt that I have posted here.

Metallic Carbonate means $\ce{M^_x(CO3)_y}$ .
For example there are $\ce{Li^_2(CO3)}$, $\ce{Ca(CO3)}$, and $\ce{Al^_2(CO3)_3}$
and $\ce{M^_x(CO3)_y -> M^_xO_y + yCO2}$
$PV=nRT$
n= 0.0500 moles carbon dioxide evolved
mass of the evolved carbon dioxide is 0.0500 moles $\times$ 44.0 g/mol = 2.20 grams
mass of the remaining metal oxide is 4.215 g - 2.20 g = 2.015  grams.
Of the 2.015 grams of the remaining metal oxide, 0.0500 moles $\times$ 16.00 g/mol = 0.800 grams are oxide, while the mass of metal is 1.215 grams.
There are 24.3 grams of metal per mole of carbon dioxide evolved.
Because the ratio x:y is unknown, it is unknown whether 1.215 grams corresponds to 0.05 moles, 0.1 moles or some other number of moles.
However, since there is one mole of carbon dioxide evolved per carbonate ion, and carbonate has a charge of -2, the molar mass of metal per charge of the metal ion will be 24.3/2 = 12.15 grams.  This is what is meant by equivalent weight of the metal in the question.

I have edited my post to include the answer provided by the book you might wanna have a look.

@user350331 so you have 4.125g in the question, but they have 4.215g in answer. Is the typo yours or the book's?

@DavePhD-I am extremely sorry in the question too it is $4.215~\mathrm{g}$. I apologise again for the inconvenience.

@user350331 so then the mass of metal is 1.215 grams.  There are 24.3 grams of metal per mole of CO2 evolved.  (This is consistent with the metal being Mg.)

Although your solution does provide the correct equivalent weight of the metal but it doesn't answer the concern that I have written in the last section of my post. Can you shed some light on what is wrong in the second method by which I determined the equivalent weight of $\ce{CO2}$

@user350331 The question wants you to find the molar mass of the metal per charge of the metal.  That is what they mean by equivalent weight of the metal.  So the important fact is the -2 charge of the carbonate ion.  I think you go wrong where you are adding equivalent weights of atoms to get an equivalent weight of CO2.  I never heard of doing that.  There are multiple definitions of "equivalent weight" depending upon the purpose, but here it is (total molar mass of the ion)/(absolute value of total charge of the ion).

@DavePhD-Yeah, I did have the feeling that I was definitely wrong in adding the equivalent weights of the atoms to determine the equivalent weight of the molecule but as mentioned in the post we do add the equivalent weights of the cation and anion to get the equivalent weight of the electrolyte, so to conclude I was not really sure what was the conceptual problem in  doing so.

You began with
4.125 g of $\ce{M^_x(CO3)_y}$
And you finished with
1.925 g of $\ce{M^_xO_y}$
The subscripts on the general formulas are the same because the charge on oxygen and carbonate are the same.
Since 0.5 mol * 44 g/mol = 2.2 g of $\ce{CO2}$ evolved then 16/44 of that mass (0.8 g) of oxygen remains. (16/44 being the mass ratio of the one oxygen atom remaining for each one molecule of carbon dioxide evolved).
1.925 g metal oxide minus 0.8 g oxygen leaves 1.125 grams of metal

There is no such metal oxide with a ratio of 1.125 masses of metal per 1.925 masses of metal oxide.  The metal in the problem would have the an atomic mass corresponding to charge of:

+1=11.25 amu
+2=22.5 amu
+3=33.75 amu
+4=45 amu
+5=56.25 amu
+6=67.5 amu
+7=78.75 amu
It is possible, however that the sample contained a mixture of a given metal element with some proportion of atoms in different oxidation states

I see that the values in the original question changed, indicating that the metal is Mg


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What is a laboratory set-up for the synthesis of ethanol from ethene?
I am doing a chemistry investigation and one component of that requires me to synthesize ethanol from ethene. Ethene itself is gaseous at room temperature and pressure so I was wondering if someone could suggest how I can carry out this synthesis with the use of acids to obtain ethanol.
To give you an idea of what I am actually doing for this investigation, I am trying to find the yield of ethanol when ethene is hydrated using different acid catalysts. However, because ethene is gaseous, I don't know how I am going to actually do this experiment.
To further clarify, I know what reagents I need to use and I know the mechanism of the reaction, I just don't know how to do it practically within a laboratory setting.

Addition of water to ethene requires high temperatures (~300ºC) and possibly high pressures too. I'm afraid that common lab glassware won't cut it. You will need to keep a pressurized environment of a potentially explosive gas, so safety might be an issue.

Well, kudos to @xyz123 who has found a good laboratory process of preparing ethanol from ethene. To my dismay, I couldn't find any proper laboratory preparation but I would like to point out that making ethanol from ethene is actually an industrial process.

Of course , of course. They would use high pressures and relatively high temperatures for that process (Oh alright, 300 isn't really that high but whatever). I was looking for a way to do it in the laboratory. Thank you for your help :)

The lack of literature for a laboratory setting should be an indicator, that this is probably not that easy to achieve and maybe also a bit too dangerous.

I don't think the lack of literature says that, @Martin-マーチン I think the real issue is no chemist would want to synthesize in a laboratory a very cheap, widely available product from a gaseous reagent. It's probably and likely like you could achieve this via a hydroboration oxidation, but heck, even in that case, it's easier just to start from triethylborane. Industrially, it's a different story altogether as the economics may be very different.

You should try bubbling the ethene gas through dilute sulfuric acid. It should form ethyl sulfate and get hydrated with water forming ethanol. As the gas will bubble through the solution, you should have an ample amount of liquid so that most of the gas can maximally react with the sulfuric acid before bubbling out as gas. You should do the reaction in a fume, hood too. Then I think that you could separate the ethanol from the water by distillation of the ethanol from the mixture, or you could separate it by exploiting the great difference in the freezing points of ethanol and water.
If you could connect tubes in such a way with special chemistry glassware, then maybe you could make some kind of reflux apparatus like an acid reflux apparatus where the gas funnels back into the bottom of the liquid and cycles through several times then you could increase your yield. Just if you use some cyclic mechanism, don't release the gas too quickly or you could break your apparatus.
I'm not exactly sure what the experimental design is. But due to the concentration of gas in the reaction chamber, it's important to have an escape valve for the gas for safety purposes of equalizing pressure. If you know what you want, an experienced glass-blower can make your apparatus.

Thanks for the reply. I just need to check two main things;

What should my set up look like?

Currently, my idea is to use a set up where I synthesize ethylene from ethanol by heating the ethanol in a boiling tube and letting the gas flow from the boiling tube, through a glass tube into a container which contains my acid solution and another test tube, which is where I'll collect my ethene and where it will be bubbled into the acid solution. Is that fine?
Below that container will be a bunsen burner to provide the heat.

Should I use organic acids as well as inorganic acids?

Could you backup this method with some literature or a general procedure? How effective is that, given that those reactions are equilibria and without removing the ethanol, will not be very effective. How do you get rid of the sulfuric acid? If you would go ahead and distil the mixture, it will just cleave ethanol again. Most importantly you should perform chemical reactions always in a fume hood.

Patent US 2,683,731 gives some technical details. An older patent US 2,497,959 describes the process with 60-65% sulfuric acid. During the  process, diethyl sulfate might be formed too. The GHS statements H340 and  H350 add an interesting twist to the procedure. Well, you can't live forever...


text:
Why is gaining hydrogen called reduction when gaining electrons is called reduction? Aren't they opposites

Reduction is gain of hydrogen  (Source)
Reduction is gain of electrons (Source)

Is it because a hydrogen has an electron so gaining hydrogen is technically gaining electrons? But that doesn't seem right as oxygens also have electrons and gaining oxygens is oxidation.
What is confusing me is when we usually talk about hydrogens, its a hydrogen ion. And hydrogen ions and electrons are opposite. I realise this situation is not a hydrogen ion but a full hydrogen, but it's still doing my head in.

Indeed, we're talking about "full" hydrogen. Gain of hydrogen ions would have pretty much nothing to do with reduction.

Perhaps its easier to understand reduction and oxidation by defining what happens as (a) an oxidising agent is an electron acceptor and (b) a reducing agent is an electron donor.

@porphyrin Yeah, until we move on to study the Krebs cycle, where you'd literally die of old age before you assign all the oxidation states and figure out where the electrons are going, yet everybody is talking about redox processes, and that quite confidently.

@Ivan Neretin ? don't understand what the Krebs cycle has to do with at, its even a pain just to look at :) I've worked on electron transfer extensively and thinking of electron donors and acceptors is by far the easiest way of dealing with these reactions, also you get $\Delta G$ directly as donor-acceptor redox.

@porphyrin Krebs cycle is just an example of redox system where thinking of electron donors and acceptors is probably not the easiest way of dealing with it.

Related: Difficulty understanding redox in terms of hydrogen and oxygen transfer?

As a blanket statement the gain of a hydrogen atom cannot be considered a reduction. The IUPAC gold book defines it as follows:

reduction
The complete transfer of one or more electrons to a molecular entity (also called 'electronation'), and, more generally, the reverse of the processes described under oxidation (2) and (3).
oxidation

The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').
An increase in the oxidation number of any atom within any substrate.
Gain of oxygen and/or loss of hydrogen of an organic substrate.

All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation of an organic substrate that can be rationally dissected into steps or primitive changes. The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons or hydroxide ions, or by nucleophilic substitution by water or its reverse and/or by an intramolecular molecular rearrangement. This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1. For example the oxidation of methane to chloromethane may be considered as follows:

As you can see, the reverse of (3) is your first statement with one significant addition: organic substrate.
To understand this one only has to look at the electronegativities. With an electronegativity of 2.20 (Pauling) for hydrogen, it is less electronegative than most other non-metals, or in general, elements that organic substrates are made out of.[1] If you follow the electronegativity scheme of assigning oxidation states, then adding a hydrogen atom (one proton and one electron) results in an decrease of the oxidation number of the element that the hydrogen atom was added to.[2]
On the other hand, adding oxygen would increase the oxidation number of the element (except fluorine) it was added to, since its electronegativity is the second highest, i.e. 3.44 (Pauling).[3]
For example, adding dihydrogen to ethene, the carbons are reduced, while hydrogen is oxidised.
$$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4
  + \overset{0}{H}_2
  -> \overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6}$$
Adding dioxygen to ethane to form ethane-1,2-diol, the carbons are oxidised, while oxygen is reduced.
$$\ce{\overset{\color{orange}{-3}}{C}_2\overset{+1}{H}_6
  + \overset{0}{O}_2
  -> \overset{\color{orange}{-1}}{C}_2\overset{+1}{H}_6\overset{-2}{O}_2}$$
You can go as far as looking at the addition of water to ethane to form ethanol. One carbon will be oxidised and one will be reduced.
$$\ce{\overset{\color{orange}{-2}}{C}_2\overset{+1}{H}_4
  + \overset{+1}{H}_2\overset{-2}{O}
  -> (\overset{+1}{H}\overset{-2}{O})H2\overset{\color{orange}{-1}}{C}-\overset{\color{orange}{-3}}{C}\overset{+1}{H}_3}$$
Just keep in mind that your first statement is only true when hydrogen is added to to more electronegative elements. The reversal is the case when adding it to a less electronegative element, like a metal.
A second point to keep in mind is that oxidation states are bookkeeping tools only. The world of bonding is not strictly ionic or covalent; it is often somewhere in between. Therefore electrons are almost never completely transferred and the actual charge distribution might be quite different.
References:

Hydrogen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/hydrogen/electronegativity.html
Electronegativity Considerations in Assigning Oxidation States
Oxygen: electronegativity. WebElements [http://www.webelements.com/]. [Online]; Dated 30th Nov. 2016; https://www.webelements.com/oxygen/electronegativity.html

The hydrogen definition is a simplified form which works well with organic compounds or other electronegative non-metals. It fails for metals. Specifically, if you react a metal with hydrogen, the metal will be oxidised in spite of additional hydrogen being in the compound.
Or in other words, $\ce{NaH}$ is the oxidised form, $\ce{Na}$ the reduced one.


text:
Is it possible to melt a diamond?
I am school student. In my school text book there is a fact about diamond. That is

Each carbon atom is covalently bonded to four other carbon atoms. A lot of energy is needed to separate the atoms in diamond. This is because covalent bonds are strong, and diamond contains many covalent bonds. This makes diamond's melting point and boiling point very high.

If a diamond is a single molecule, how does a single molecule melt? Is that fact correct? Please tell me the answer!

Related: http://chemistry.stackexchange.com/questions/8590/what-is-the-melting-point-of-diamond

The melting point of diamond is certainly an exotic topic. Diamond is slightly less stabile than its allotrope, graphite, and will in most circumstances turn to graphite way before its melting point. If you have enough pressure it will stay diamondlike and it will at some point melt.
A diamond is not a single molecule unless you stretch the definition of molecule. It is usually defined as the smallest fundamental unit, and for diamond I'd call the tetragonal carbon unit cell the smallest fundamental unit, or maybe even a carbon atom. Diamond, having one of the highest values for heat transfer among substances would certainly melt almost like a single molecule though. But in theory, one and one carbon decides to leave behind it's friends bound by the diamond structure boundaries behind to seek their fortune in the hippie camp that is called liquid.

It's true that the high melting temperature of diamond is caused by its structure. Usually a substance melts when you overcome the intermolecular attraction forces, whereas for diamond the forces involved are interatomic, which are several orders of magnitude stronger. However, one could argue that fusion is a physical process, and in this case we're breaking chemical bonds so it's more of a chemical process which would make it, technically, not fusion.


text:
How to specify a different isotope in the ORCA quantum chemistry program?
I would like to carry out a calculation using deuterium instead of hydrogen.
I tried the following ORCA input file:
!cc-pVDZ B3LYP opt freq

*xyzfile 0 1 bi_guess.xyz

The contents of bi_guess.xyz:
34
Coordinates from ORCA-job kie_computation_bi_b3lyp
  N   -1.36522940768289      1.43090120637739     -1.29989163642319
  C   -0.44442228001060      0.38103756632797     -1.36699056491989
  S   1.11938175575132      0.89828742472636     -0.68162779041360
  C   0.39936643411072      2.45263744108356     -0.17598894082966
  C   -0.88737920921729      2.55508413237739     -0.58575891095156
  C   -1.81988687350823      3.70483170564282     -0.34149534660368
  H   -2.02914244079129      4.28099946722255     -1.25957521107675
  H   -1.38928363000205      4.40377070305222      0.38811759209947
  H   -2.78814652651823      3.35404865266229      0.05306751853438
  C   -2.42958273785510      1.55203191145485     -2.29150912214068
  H   -2.13103494174077      1.01505106288885     -3.20503185208371
  H   -2.58921987988251      2.60821448709764     -2.54320899415084
  H   -3.38528969660347      1.12099748064241     -1.94940612103283
  C   1.24350647485456      3.43182218478396      0.59010704636463
  C   2.34455035926304      4.09444738435286     -0.25628872453066
  H   1.70917002774078      2.91928017185310      1.45151071835754
  H   0.59149967441242      4.21059012151118      1.01775365078364
  H   3.02263338128754      3.34129107817330     -0.68945118759167
  H   2.94960251527178      4.78111271092864      0.35865283243980
  H   1.90986274364991      4.66846424028840     -1.09098042345273
  C   -0.58896270899179     -0.87335406540114     -1.89161187230330
  O   0.62483817548640     -1.53234971307262     -2.12144915093809
  C   -1.81093059712914     -1.61627205233074     -2.20313985412716
  C   -1.78295926336086     -2.62665829989993     -3.19412484968867
  C   -2.91185414985907     -3.39920051566432     -3.46720312726732
  C   -4.10190122740441     -3.19660857990940     -2.75883040599387
  C   -4.14025228929028     -2.21734676116007     -1.75913366985694
  C   -3.01353021224531     -1.44485219171837     -1.47688481122363
  H   -3.04567986529299     -0.72442989166239     -0.65687581329096
  H   -0.86205885646938     -2.78256837851931     -3.75966518262305
  H   -4.98552287920138     -3.80160445008789     -2.97478885669734
  H   -5.05406866849322     -2.06721957203355     -1.17765661592474
  H   -2.86371065052297     -4.16497387835216     -4.24648540168365
  H   0.55630745024474     -2.41808278363585     -1.72738492075924 M = 2

However, the output shows me that there are only normal protons, no deuterium atoms. What am I doing wrong here?

Important correction: It seems that at least with the Orca installation I am using (seems to be ORCA 3.0.1)  you have to specify the mass without the equals sign, e.g. "H 0.556307 -2.4180827 -1.72738492 M 2".

Indeed, sometimes ORCA wants an equals sign, sometimes it insists it not be there ... and sometimes it just doesn't care.

ORCA cannot read isotope information from XYZ files -- the OpenBabel specification does not include it and so ORCA doesn't look for it in the geometry file.
You need to copy-paste the geometry into your input file:
* xyz 0 1
  ...
  H   0.55630745024474     -2.41808278363585     -1.72738492075924 M 2
*

Alternatively, if you are only interested in the frequencies, normal modes, and/or IR spectrum, you can edit the atomic masses in the .hess file after the run is completed and use the orca_vib utility program to re-run the analysis with the substituted isotopes. See pp. 201-202 of the ORCA v3.0.3 manual for more details.

Editing the .hess file is mentioned in the ORCA manual, as you pointed out. I intend to compare the gibbs free enthalpy G of the deuterated compound with the protio isotopomer (to compute the kinetic isotope effect by comparison with the transition state). I was unsure if the free enthalpy G will also be available via orca_vib.

@ketbra Ah, yeah, I think you have to include the isotope mass in the input file in order to get the correct thermochemical values.


text:
How does the inverted Marcus region explain chemiluminescence?
The invert Marcus region offers one explanation of chemiluminescence, the process by which light is one of the products of a chemical reaction.
I tried reading one of Marcus's original papers on the subject [1], but I don't really get his argument. I get confused about which of the potential energy surfaces in his Figure (copied here, hopefully that's okay) that refer to the ground state reactant, ground state product, excited reactant, and excited product.

[1] R.J. Marcus (1965), "On the Theory of Chemiluminescent Electron-Transfer Reactions", The Journal of Chemical Physics 43(8):2654-2657

Marcus theory of electron transfer shows that it is possible for a reaction with a large negative free energy ($-\Delta G$ ) to be slower than one with a less negative free energy. The region where this happens is called the 'inverted' region and is shown in figure (c) in your diagram. You can see that the energy of the products is far below that of the reactants but that there is still a substantial activation energy, that is the energy from the minimum of the R curve to the crossing point of the R-P curve on the left.
In fig (A) this is called the 'normal' region and the free energy is close to zero and there is still an activation barrier, in (b) the reaction is slightly exothermic has an activation energy and is still in the 'normal' region. What is missing is the case when the reaction is activationless, this lies between figs (b) and (c) and occurs when the P curve crosses the R curve at its minimum. In this situation the reaction has its maximum rate constant.
The figure shows some data illustrating the rate constnats for electron transfer in the normal(left side) activationless (top) and inverted inverted (right) regions vs $-\Delta G$. The data is for electron transfer from bases in DNA to different intercalated dyes. The fitted Marcus curve is the solid line one from classical Marcus theory the other from a quantum calculation.
Hopefully this will description allow you to follow the paper you have been reading. (You may also want to  look at BBA v811, p265, 1985)
It was not until late 1980's that Marcus theory was confirmed experimentally, first in the solid state and then with chemically linked donor and acceptor molecules as this prevents diffusion, which had hitherto greatly confused interpreting data, and also meant that the inverted region could not be observed.

The energetics for an electron transfer reaction
$$\ce{D^* +A \rightarrow D^+ + A^-  }$$ is
$$\Delta G = E^o_{D^+/D} -E^o_{A/A^-}-E_{s_1} +\epsilon_r$$
for an excited state $\ce{D^*}$ with energy $E_{s_1}$ and electrode potentials $E^o$. Marcus in the paper also includes a solvation term in a solvent, of dielectric constant $\epsilon$ to account for change in redox. This term can often be neglected but is
$$\epsilon_r = \frac{\Delta e^2}{4\pi \epsilon_o}\left (\left(\frac{1}{r}-\frac{1}{2a}\right)\frac{1}{\epsilon}-\frac{1}{\epsilon_ar}\right)$$
where $\epsilon_a$ is the dielectric constant of acetonitrile, $2a$ is radius of donor + acceptor and r their separation.
The rate constant is calculated from
$$ k_\mathrm{et} = k^0\exp\left(- \frac{(\Delta G +\lambda)^2}{4\lambda k_\mathrm{B}T}\right)$$
where $\lambda$ is the reorganisation energy and is the vertical (positive) energy between the minimum of the acceptor potential and the donor potential. Its is determine experimentally from the maximum in the rate vs $-\Delta G$ plot as is $k^0$. In the plot $\lambda \approx 1~\mathrm{eV}$ and $k^0 \approx 10^{13} \pu{s^{-1}}$.
The chemiluminescent reaction can be considered as the reverse of the example given.

Well explained the basics of Marcus theory, but I'm still confused about how it relates to chemiluminescence. As far as I know, the barrier for chemiluminescence is lower than the barrier for crossing to the ground state via a non-radiative transition, but I don't see it in the figures.

I'm not sure that the Marcus paper does actually explain this; his papers are often very hard to follow. It mainly explains the eqns I give but in an old notation. In a chemiluminescence reaction the product is formed in the excited state so this would be R in the Marcus figure, so the process is backwards as it were. however,  now R is an excited state that emits to its ground state. This is below R and has v. approx the same shape but is not shown in the figure. I'm not sure even if Marcus theory does explain chemiluminescence. You will have to look for more recent papers.

Researchers were of the opinion that, chemiluminescence could be a prime suspect to justify why it was so hard to observe Marcus inverted region in electron transfer reactions. To explain why so, let us assume that the equilibrium reactant and product curves have a large free energy gap relative to the reorganization energy, which would mean that the ET reaction occurs in the inverted region. In other words, the two curves would intersect on the conventional left side of the reactant parabola.
Now, because of the large free energy gap between the equilibrium ground states of the reactant (R) and product (P), there is a possibility that a lower-lying excited state of the product (P*) can be energetically favorably placed to cross the R potential at its minimum. If such a situation is prevalent, then a barrierless electron transfer reaction between R and P* would prefer to populate the excited state of the product, P.
As a result, an electron transfer reaction would end up populating the product excited state P* more than the ground state. That excited population can radiatively relax to the product ground state and result in a luminescent signal, which in this particular situation is known as chemiluminescence.

The reason the inverted region was not observed was that reactants were free to diffuse. As they come closer the electron transfer rate constant increases hugely and so all that is measured is a diffusion controlled rate constant. Keeping reactants in fixed positions allowed the inverted region to be observed as shown in the figure in my answer below.


text:
Tallow and polymer pellets
Bank of England has issued a new £5 note which contains Animal Tallow. I was wondering if you can explain to me as to how the animal tallow is used in this process. Also would you be able to speculate as to tallow of which animal could potentially be used in this production process?
Does the animal tallow completely transform or are its original characteristics retained?
Bank of England official Press Release is as follows:

We are aware of some people’s concerns about traces of tallow in our new five pound note. We respect those concerns and are treating them with the utmost seriousness.
This issue has only just come to light, and the Bank did not know about it when the contract was signed.
Information recently provided by our supplier, Innovia, and its supply chain shows that an extremely small amount of tallow is used in an early stage of the production process of polymer pellets, which are then used to create the base substrate for the five pound note.
Innovia is now working intensively with its supply chain and will keep the Bank informed on progress towards potential solutions.

It's likely used in the production of some base product of a processing additive, i guess. Likely because it has no big other uses and is therefore cheap . How did some neurotics even find out about it?

Animal tallow is used early on in the creation of various plastics. The molecules that make up the tallow are converted to various different compounds.

According to the UK news outlet The Sun:

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

Suet is mainly used to make tallow, though it is also a widely used cooking ingredient.
Stearic acid is a waxy material who's ionic form (stearate salts like sodium stearate) is widely used in detergents. This property gives it the desired smooth, slippery property desired in the manufacture of the polymer notes.

The answer from the B of E is not an answer at all. The animal derivative is more than likely to be a stearate from fat. Long chain fatty acids and their derivatives act as lubricants in the process of making film and make the product easily handleable. The actual substance is likely to be calcium or magnesium stearate at a level of 0.01% to 0.05%.


text:
Is the equation pH + pOH = 14 valid at different temperatures?

$\mathrm{pH} + \mathrm{pOH} = 14$

For water at 60 degrees Celsius,
$K_\mathrm{w} = 1 \times 10^{-13} = [\ce{H+}] \times [\ce{OH-}]$
Hence, $[\ce{H+}] = 1 \times 10^{-6.5} = [\ce{OH-}]$
So $\mathrm{pH} = -\log(1\times 10^{-6.5}) = 6.5 = \mathrm{pOH}$
which gives $\mathrm{pH} + \mathrm{pOH} = 6.5+6.5 = 13$, not $14$
So is my math or chemistry wrong, or are there limitations on $\mathrm{pH} + \mathrm{pOH} = 14$?

$\mathrm{pH + pOH = 14}$

This equation only holds true at around $25~^\circ\mathrm{C}$, where the water autodissociation constant $K_\mathrm{w} = 10^{-14}$.
Mathematically, you can write
$$\begin{align}
K_\mathrm{w} = [\ce{H+}][\ce{OH-}] &= 10^{-14} & &\\
\log([\ce{H+}][\ce{OH-}]) &= \log(10^{-14}) & &\text{(logarithms of both sides)} \\
\log[\ce{H+}] + \log[\ce{OH-}] &= -14 & &(\log(ab) = \log a + \log b) \\
-\log[\ce{H+}] - \log[\ce{OH-}] &= 14 & & \\
\mathrm{pH + pOH} &= 14 & & \text{(definitions of pH and pOH)}
\end{align}$$
It's fairly clear from this that if $K_\mathrm{w}$ is $10^{-x}$ at some other temperature, then you will obtain $\mathrm{pH + pOH} = x$.

IOW, pH+pOH=-log(Kw).


text:
Matching a Slater-type wavefunction with a minimal (STO-nG) Gaussian basis set
I have on hand a high-precision wavefunction expressed in Slater orbitals.  I need to express it as accurately as possible in a minimal Gaussian basis set.
For background: I am currently working with STO-6G, but it is not good enough to describe the cusp condition with the accuracy I need. How can I get to a basis like STO-7G, or even beyond to something like STO-10G, preferably in a form that is compatible with QChem?

TL;DR
The procedure to represent a Slater-type orbital (STO) as a linear combination of Gaussian-type orbitals (GTO) is outlined in W. J. Hehre, R. F. Stewart and J. A. Pople, J. Chem. Phys. 1969, 51, 2657 and is a quite straight forward least square fitting.
Derivation
I'll be quoting/ rephrasing quite liberally and take the 1s orbital as an example. The STO is defined as
\begin{align}
\phi_\mathrm{1s}(\zeta_1, \mathbf{r}) &=
  (\zeta_1^3/\pi)^{1/2}\exp(-\zeta_1 \mathbf{r}).
  \tag1\label{STO-1s}
\end{align}
The general form of a Gaussian-type 1s orbital is defined as
\begin{align}
g_\mathrm{1s}(\alpha, \mathbf{r}) &=
  (2\alpha/\pi)^{3/4}\exp(-\alpha \mathbf{r}^2).
  \tag2\label{GTO-1s-general}
\end{align}
The STO is expressed as a sum of $K$ GTOs; the coefficients and exponents are obtained for a STO with $\zeta=1.0$ and the scaled uniformly. Hence we can write
\begin{align}
\phi_\mathrm{1s}'(\zeta, \mathbf{r}) &=
  \zeta^{3/2}\phi_\mathrm{1s}'(1.0,\zeta\mathbf{r}),
  \tag3\label{GTO-1s-zeta1}
\end{align}
where
\begin{align}
\phi_\mathrm{1s}'(1.0, \mathbf{r}) &=
  \sum_k^K d_{\mathrm{1s},k}\,g_\mathrm{1s}(\alpha_k,\mathbf{r}).
  \tag4\label{GTO-1s-lc}
\end{align}
In order to obtain the coefficients $d$ and exponents $\alpha$ the contracted GTO is fitted to the form of the STO via a least square method. Note that this is a quite unique approach, since most other basis sets are optimised to produce the minimal energy. It all comes down to minimising the integral
\begin{align}
I &=
  \int \left(
         \phi_\mathrm{1s}(1.0,\mathbf{r})-\phi_\mathrm{1s}'(1.0,\mathbf{r})
       \right)^2 \mathrm{d}\mathbf{r}.
  \tag5\label{min-int}
\end{align}
For $K=3$, i.e. STO-3G, the following values are obtained:
\begin{array}{llll}\hline
k & \alpha_{\mathrm{1s},k} & d_{\mathrm{1s},k}      & I\\\hline
  &                        &                        & 3.31\times10^{-4}\\
1 & 1.09818 \times10^{-1}  & 4.44635 \times10^{-1}  &\\
2 & 4.05771 \times10^{-1}  & 5.35328 \times10^{-1}  &\\
3 & 2.22766                & 1.54329 \times10^{-1}  &\\\hline
\end{array}
Implementation
Finally you need to scale the exponents to use the right exponent. From $\eqref{GTO-1s-zeta1}$ we see
$$\mathbf{r}\mapsto\zeta\mathbf{r}_{\zeta=1.0}$$
and from $\eqref{GTO-1s-general}$ we know $\mathbf{r}\mapsto\mathbf{r}^2$ and therefore the exponent
$$\alpha_\zeta\mapsto\alpha_{\zeta=1.0}\zeta^2.$$
Hydrogen for example has $\zeta=1.24$ (Tab. IX) and therefore the exponents that need to be implemented are:
\begin{array}{lll}\hline
k & \alpha_{\mathrm{1s},k} & d_{\mathrm{1s},k}      \\\hline
1 &       0.16885540       &      0.44463454        \\
2 &       0.62391373       &      0.53532814        \\
3 &       3.42525091       &      0.15432897        \\\hline
\end{array}
For Q-Chem this is explained in the manual under 7.4.3 and needs to have the form
\begin{array}{llll}
\text{\$basis}\\
\ce{X} & 0\\
L & K & \text{scale}\\
α_1  & C_1^{L_\mathrm{min}} & C_1^{L_\mathrm{min}+1} &\dots& C_1^{L_\mathrm{max}}&\\
α_2  & C_2^{L_\mathrm{min}} & C_2^{L_\mathrm{min}+1} &\dots& C_2^{L_\mathrm{max}}&\\
\vdots    & \vdots & \vdots &\ddots & \vdots \\
α_K  & C_K^{L_\mathrm{min}} & C_K^{L_\mathrm{min}+1} &\dots& C_K^{L_\mathrm{max}}&\\
\text{****}\\
\text{\$end}\\
\end{array}
where $\ce{X}$ is the atomic symbol, $L$ is the angular momentum, $K$ is the degree of contraction, $\text{scale}$ is the scaling applied to all exponents, $\alpha_i$ are the exponents, and $C_i^L$ are the contraction coefficients.
As a practical example, the STO-3G basis set shall look like this:

$basis
   H   0
   S   3   1.00
      3.42525091             0.15432897
      0.62391373             0.53532814
      0.16885540             0.44463454
****
$end

For the actual fitting procedure, i.e. minimising $\eqref{min-int}$ you should consult a more technical oriented network page. It might be on topic at Mathematics, Computational Science, or even Mathematica depending on the question that troubles you. Please check their guidelines before asking.

General remarks
The most principle shortcoming of GTOs is that they cannot describe the cusp correctly. It's a trade-off you just make for faster integral evaluation. With STO-6G you are getting the orbital quite right already and there is a reason why we usually use STO-3G and not some higher contraction. It just offers the best cost : accuracy ratio.
This can be quite neatly seen in the graphical representation. The black curve  is the original STO, purple is STO-6G, cyan is STO-4G, and red is STO-2G plotted online at the wonderful FooPlot. Hack it here (I hope that works).

Going from STO-6G to STO-7G won't give you much more accuracy, maybe not even going to STO-10G. You probably have to go more into the direction of STO-20/30G. At that point it is probably more economic to switch to a program that can handle STO natively. From Wikipedia I see two, that handle those: ADF, and MolDS. The latter seems young, but is free.
Addendum:
If the FooPlot link does not work, here are the used formulae:

(1/pi)^(1/2)*exp(-abs(x))

0.679*(2*0.152/pi)^(3/4)*exp(-0.152*x^2)+
0.430*(2*0.852/pi)^(3/4)*exp(-0.852*x^2)

0.292*(2*0.088/pi)^(3/4)*exp(-0.088*x^2)+
0.533*(2*0.265/pi)^(3/4)*exp(-0.265*x^2)+
0.260*(2*0.955/pi)^(3/4)*exp(-0.955*x^2)+
0.0568*(2*5.22/pi)^(3/4)*exp(-5.22*x^2)

0.130*(2*0.0651/pi)^(3/4)*exp(-0.0651*x^2)+
0.416*(2*0.158/pi)^(3/4)*exp(-0.158*x^2)+
0.371*(2*0.407/pi)^(3/4)*exp(-0.407*x^2)+
0.169*(2*1.19/pi)^(3/4)*exp(-1.19*x^2)+
0.0494*(2*4.24/pi)^(3/4)*exp(-4.24*x^2)+
0.00916*(2*23.1/pi)^(3/4)*exp(-23.1*x^2)

Thanks a lot for your contribution! Yes, we can find the STO-6G from textbooks or the paper you mentioned. After searching for a couple of days, I really cannot find the coefficients and exponents for STO-NG with N>6. I will see if I can work it out and share here.


text:
How can I determine the structure of complexes with ethylenediamine as a ligand?
I am a bit confused about how to solve this problem.  I can interpret coordination complexes like $\ce{[CoCl6]^4-}$ but I just came across a coordination complex that I am unsure of how to solve.
It is $\ce{[Mn(en)3]^3+}$ where en stands for ethylenediamine.  The problem first asks that I draw the structure (not sure how to do this as ethylenediamine is a large molecule).  I would also like to know whether this complex would be tetrahedral, octahedral, trigonal planar, etc.  I'm just kind of confused as to how to solve this problem when its a large molecule as the ligand rather than a molecule like $\ce{Cl6}$.

Point of detail: The chloride ligands are not one monolithic counter-ligand. Each ion complexes ~independently with the central metal.

I was about to contest hexachloridocobaltate(II) when I realised it was cobalt(III) that preferred tetracoordination of chlorido ligands due to its small size.

In principle, there is no difference between having monodentate ligands such as chlorido ligand or more complex ligands such as ethylenediamine.
The first step is always to draw out the structure of the ligand by itself. Then, identify how many coordinating entities it contains. Ethylenediamine is $\ce{H2N-CH2-CH2-NH2}$ and both nitrogens can form coordinate bonds.
In the next step, identify how many coordinating entities you have in total and thus what the maximum coordination number is. As mentioned, each $\ce{en}$ has two potentially coordinating nitrogens for a total of six dative bonds. Thus, a hexacoordinated octahedron is the maximum possible.
Continue by looking at the metal. What do you know about its properties in complexes? Does it prefer any type of complex? In this case, manganese(III) is absolutely fine with having an octahedral geometry, so that’s taken care of.
Once you’ve done all this, you will notice that the metal enjoys octahedral coordination and there are enough Lewis bases for octahedral coordination to occur — the complex will likely be octahedral. If the complex had been $\ce{[Ni(en)2]^2+}$, we would have arrived at ‘square planar’ by the same methodology.
Next, draw the complex. Start with the metal and the octahedral geometry and put a nitrogen at the end of each line. Then, try to connect the nitrogens with $\ce{CH2-CH2}$ fragments. There should be two possible, enantiomeric outcomes.

Hey thanks a lot! One thing that I am curious about is "What do you know about its properties in complexes? Does it prefer any type of complex?" are there some atoms that prefer octahedral/tetrahedral/square planar? Is there a way to tell which prefers which, or is it mostly memorization?

@Andrew The most obvious case is nickel(II)’s preference for square planar complexes. And then maybe some small ions that don’t prefer octahedral but are fine with tetrahedral. That’s just an exception phrase to catch those before you start using octahedral for everything ;)

Basically $\ce{en}$ is a strong field ligand and is a neutral one so the $\ce{Mn}$ in $+3$ state would have $3\mathrm{d^4}$ configuration and $\ce{en}$ would pair them up in the first two orbitals so there would be an inner orbital complex formation with octahedral geometry having $\mathrm{d^2sp^3}$ hybridization.

Please note that it is practically always a very bad idea to extend hybridisation to transition metal complexes. See my other answers on that topic.


text:
Finding electron configuration by following the Aufbau principle
In the question, "Electron Configuration of Tellurium", there is mention of the 'follow yellow brick road' method of finding electron configuration.
What I'd learned in the past was to find the nearest noble gas (in the row above the element of interest) and count from there.
I searched in Google for this method and see that it appears to also called the Aufbau principle & diagonal method. However I'm not getting a good grasp on how it works.
Could someone explain this method to me?
Also, would this method work for something with a weird electron configuration, like copper?

I think the more common name is to follow the aufbau principle.

Related https://chemistry.stackexchange.com/questions/73325/while-filling-electrons-we-follow-aufbau-principle-but-not-while-removing-why

The so-called rule is to fill the subshells in increasing order of $n+l$. For subshells with the same value of $n+l$, the subshell with lowest $n$ is filled first.
$$\begin{array}{cc} \hline
n+l & \text{Orbitals} \\ \hline
1 & \mathrm{1s} \\
2 & \mathrm{2s} \\
3 & \mathrm{2p,3s} \\
4 & \mathrm{3p,4s} \\
5 & \mathrm{3d,4p,5s} \\
\vdots & \vdots \\ \hline
\end{array}$$
For all intents and purposes, this is exactly equivalent to your heuristic of "go from the nearest noble gas", since to get the configuration of the noble gas itself you have to go through this same ordering of subshells. You can easily verify that the table above reproduces the order of filling of the Periodic Table.
Chromium and copper are exceptions to the rule (obviously there are many more).


text:
Smoke Bomb with Potassium Nitrate and Sucrose (Or is it caramel?)
I've scoured the internet for a proper chemical reaction between caramelized sucrose and Potassium Nitrate for the past few days. So far, every chemical reaction I see drawn out is between pure sucrose and potassium nitrate, while the smoke bomb in question requires caramelized sucrose, an entirely different thing altogether that probably will give you a different reaction.
Am I over-thinking this, or is Caramelen/Caramelan the reactant, instead of sucrose? If someone could be so kind as to draw out the chemical equation between the two, that'd be awesome, too. (Heat is part of this equation, as the substance is lit to produce the smoke)

http://chemistry.stackexchange.com/questions/22404/what-substance-in-a-gummy-bear-reacts-with-heated-potassium-chlorate-to-create-a

Caramel is, depending on how badly caramelised it was, either mostly sucrose, mostly carbon, or some badly-defined intermediate. And please don't try building a bomb if you don'tknow what you're doing. Or on second thought, just don't build a bomb.


text:
Total oxidation of toluene free enthalpy
I have trouble finding whether this reaction is spontaneous or not :
$$\ce{C7H8 + 9O2 -> 7CO2 + 4 H2O}$$
I consider (from the CRC Handbook of Physics and Chemistry) :

And so I find :
$$\Delta_r H° = 7\cdot (-393.5)+4\cdot(-241.5) - 9\cdot0 - 1\cdot 50.5 = -3670 \;\text{kJ/mol}  $$
and
$$\Delta_rS°= 7\cdot213.8+4\cdot188.8-9\cdot205-1\cdot0= 406.8\;\text{J/mol/K}$$
So at T=298 K I have
$$\Delta_rG°=\Delta_rH° - T\Delta_rS° = -3670\cdot 10^3-298\cdot406.8= -3.8\; \text{MJ/mol}$$
This would mean the reaction is very spontaneous, but according to experimental data, it should not. (The reaction only occurs when heating and with a catalyst).
What did I get wrong ?

Your problem is that you are trying to calculate a kinetic property (spontaneous versus nonspontaneous; on a compound level that’s inert versus reactive) with thermodynamic data (exergonic versus endergonic; on a compound level that’s stable versus unstable).
Thermodynamically, we should all be carbon dioxide and water, as should any other organic compound. However, the activation energy of combustion of organic compounds is rather large. Thus, while the reaction is thermodynamically favourable, it is not spontaneous in the absence of activation energy.

I think another way of saying this is that the kinetics of the reaction is very slow.


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Microwaving a glass of water, what happens?
Most of us here would already know the simplified idea behind microwaving food: Microwave radiation hits the water molecules present in food, which excites them and causes 'em to vibrate rapidly in situ resulting in the heating up of food. Very straightforward.
But what if you decide to microwave a glass of water?
I can't seem to find any reliable literature that deals with this. A lot of highly unreliable sites spout nonsense from "The water becomes poisonous" to "The water becomes radioactive", so I've immediately dismissed those. However, some sources suggest that the "water will explode".
Now I was under the impression that the water would perhaps come to a boil a bit faster than it would by simply lighting a fire underneath it.
So after reading the last few sources it wasn't really difficult to believe that an explosion might occur. At first I simply extrapolated my theory of 'rapid boiling' to the theory of 'Very rapid boiling' and was content with that. It made sense to me since I presumed that, just as in the conventional method of boiling water, bubbles of water would be formed at the bottom of the glass albeit at a much faster rate, build up, rise, thus resulting in an explosion. But after some reflection, I noticed a problem with this 'extrapolation'.
Why do the bubbles even form in the first place when you conventionally heat water? Simply lighting a fire below a beaker of water does not ensure that the entire content o the beaker is uniformly heated at the same time. Obviously the layer of water molecules in direct contact with the bottom of the beaker will get heated rather quickly. They rise up, colder water descends, etc etc, you know the deal: convection. It's the sizable temperate difference between the bottom region and the rest of the water, that enables the water at the bottom to get vaporized at some point forming bubbles while the water immediately above it is still in the liquid state.
The problem I see that prevents me from extending this theory to microwaving water is that the mode of heating is different in the case of the microwave: Radiation
Now since radiation is faster at heating a given volume than convection is; for all practical purposes, I think it'd be safe to accept that the water in the beaker will be heated uniformly (The beaker's pretty normal sized...it's dimensions aren't of an order of magnitude greater than that of the speed of light/electromagnetic radiation, so it'll be alright to take the apparatus as being 'uniformly heated'. Had to say this before someone inevitably points out in the comments that all the water being heated to constant temperatures like this at the exact same time is impossible).
So if the beaker is heated uniformly, then I see no reason for bubbles to form...in other words, I don't see an explosion coming.
So what really happens if you microwave a glass of water?
Will it quickly and steadily vaporize? Will it explode? Or does something else happen? And why does it happen?

I do this all the time, However for the good reasons in the answers above I put a single ceramic Raschig ring in there whenever I use clear glass. I have one at home, stolen from the university 10 years ago for that very purpose. But then again, I am a nerd.

Microwave radiation is too low in energy to cause molecular vibrations. It causes molecular rotations.

The glass is quite irrelevant. The water gets heated by the microwave and the glass gets heated by the hot water. The rest is all nonsense.

I would've considered "explosion" a valid answer with justification. It's reasonably easy to superheat water in a glass container that is very smooth. Superheated water is above the boiling point but not boiling and will boil very violently given a shock or seed of some kind.

https://www.youtube.com/watch?v=2FcwRYfUBLM

@MaxW That's what I thought for a long time, too, but this Wikipedia article cites this paper as part of an argument that it's actually excitation of the broad vibrational absorption of the hydrogen bonding network in water.

I'm not too experienced on this site, but wouldn't this question be more suitable for [physics.se]?

Actually it is neither :)  Microwave ovens (usually 2.4 GHz) are too low a frequency to effectively excite molecular vibrations or rotations.  Heating of water in a microwave oven comes from [dielectric loss] (https://en.wikipedia.org/wiki/Dielectric_loss). This is when the microwave electric field pushes the electrons back and forth (rather than the atomic nuclei, as in molecular vibrations and rotations).  The moving electrons exhibit some loss similar to ohmic heating in conductors.

A glass of water won't be heated uniformly; the outer bit of the water absorbs some of the microwaves by converting them to heat, so the inside bit sees less microwaves. (Handwavy explanation)

Reminds me of a Mythbusters episode I watched recently.  They tested the theory that watering plants with water that was first boiled in a microwave would kill the plants.  In fact they found the exact opposite.  The plant that was watered with microwaved water actually grew taller than the control plants.

As luck would have it, they also did an episode that addressed superheated microwaved water too.  https://www.youtube.com/watch?v=1_OXM4mr_i0

@ByteCommander (Di-)electric properties of materials are a subject in chemistry as much as in physics. In the sense that many chemists understand a lot more about it than many physicists. (Except for the standing wave pattern in the microwave oven, which we all gladly leave to you electrical engineers. ;-))

@Evan Dielectric loss is just another word for "inelastic interaction with an electric field". That can be the electron shell, but in this case it is forced (as opposed to resonant) rotation/vibration  of polar molecules.

The heating effect in a microwave heater I think is due to covectional currents of rotation of water molecules to the substance may be microwaving of the same substance (water) may had caused something unexpected rather than just getting heated up

Long story short,It explodes.:)

Heating water on a hot plate is safe, because the hottest point is at the bottom of the pot. A lot of relatively small bubbles appear there without much overheating of the water, because there is a lot of nucleation at the uneven phase boundary steel-water.
In a microwave, the hottest place is IN the water. The glass does not get heated by microwave (at least not much), and radiates off some heat to the surrounding.
Problem: In clean water, there are few good nucleation points to form bubbles, only some dust particles perhaps. So the water gets overheated rather strongly, and a first bubble that appears can grow a lot before it has cooled its surroundings down to 100 °C. That one huge bubble can throw most of the water out of the glass. It boils over, violently.
Btw. a microwave does not heat its content uniformly. It forms a standing electromagnetic wave (that's not radiation, strictly speaking) in the oven, like a rope swung quickly between two people, or a guitar string. The wave pattern has knots at a distance of $0.5c/f\approx6\ \mathrm{cm}$ (with $f=2.45\ \mathrm{GHz}$ and speed of light $c$ which of course is a bit less in your chicken), where there is very little heating. That's why the microwave oven has the rotating plate, to generate some uniform heating. For pure water, that doesn't matter, because convection sets in anyway and distributes the heat.

I microwave glasses of water all the time; I'm just careful how long I microwave it, and I stir the water briefly every 30 seconds or so. By ensuring that I don't wind up with any superheated water, all that ends up happening is that I get hot water :)

@DoktorJ The most common way to keep the water from boiling over is putting a glas rod or similar in the cup. Opening the oven every few seconds seems a bit cumbersome to me.

? A said, a glass of water in a microwave oven will overheat without much visible change (gas bubbles), and then suddenly boil over. With some bad luck, your impatient hand is just in the oven at that time.

The mode of heating of a water glass in a microwave and on a stove is actually very similar. While it's true that microwave radiation penetrates somewhat into the body of water, the penetration depth is rather small.
The main problem is that on a stove, you get uniform heating from the bottom, with temperature usually far higher than the boiling point of water. "Boiling" occurs when the water at the bottom of the container is hot enough to turn to vapour regardless of the surrounding pressure - it's actually a bit hotter than the boiling point. At the same time, the rest of the water is significantly colder, which is why you see the bubbles long before we tend to consider the water at "boiling temperature".
In a microwave, the water is heated in a bit different fashion. In the simplest model, you're heating it from the sides - similar to heating a glass of water in a normal oven. This already makes a huge difference - since you're heating the water from the sides, the bulk of water isn't heated through convection. The main effect is that the heating appears a lot more uniform, and the heat is distributed mostly through diffusion - when the water gets close to boiling point, a large bulk of the water is close to the boiling point, and convection can't carry that heat anywhere.
And this point is the tricky part. The water isn't capable of boiling, because to vapourize, it has to overcome surface tension - and the heat near the boiling point isn't powerful enough for that yet. Boiling is easiest on contact with e.g. sides of the glass, but that's not where the hottest water is, unlike with a stove. And now, you disrupt the water - bump into it, put a coffee been into it, a spoon... you form a large surface area in water that's already at boiling point, and remove the only thing that's preventing it from boiling. Across the newly formed surfaces, water starts vapourizing, which increases the pressure, which disrupts the surfaces further, causing further vapourization... and you get a splash of boiling-hot water.
Now, the actual way stuff is heated in a microwave oven is even more complex than this - but that's not really necessary to explain why superheating is a lot bigger problem in an oven than on a stove, so I'll leave it at that :)
In a way, this is similar to the coke & menthos effect (though instead of vapourization, the carbon dioxide in that experiment comes out of a water solution). You have a liquid with dissolved carbon dioxide at an equilibrium - in fact, the liquid is already supersaturated when you open the bottle due to the drop in pressure. Drop the menthos inside, and you create a great nucleation source for the dissolved carbon dioxide - the surface of the candy is rather rough, and it gets rougher as it dissolves, so the surface tension is much smaller than it would usually be, causing the carbon dioxide to quickly come out of the solution, causing foaming and expansion -> boom.
So, what would happen if you could actually heat the water uniformly? The water would start boiling along points of contact with the glass, so you'd see bubbles of boiled water carrying the heat away quickly. The "sweetspot" for "exploding water" is exactly having pockets of boiling water apart from any nucleation sources, and then introducing a nucleation source. This is pretty hard to do even in a microwave oven, but it is possible - and the same hazard applies any time you heat a container in a way that doesn't leave the most heat in places in direct contact with the container :)

And how deep does the microwave field penetrate into the water?

@Karl If I remember correctly, it's one or two centimeters for 2/3rds absorption in cold water. As the water gets hotter, the penetration is easier, so close to the boiling point, most of the radiation actually makes it through a typical glass of water (to be reflected on the other side of the microwave). I couldn't find a good model that takes everything into account at once - it really is quite complicated. But this should be close enough for a typical microwave. This is also why unfreezing is so slow (and pointless) in a microwave - the penetration depth in ice is about ten meters :)

"you get a splash of boiling-hot water" Most likely, a splash of above boiling-hot water.

@DavidRicherby I don't think so. When the water starts boiling, it is suddenly full of nucleation points where water evaporates, and the temperature almost instantly drops down to 100°C.

@Luaan There is actually no real "radiation" in a microwave oven. A standing wave pattern is formed, which is distorted by the glass of water. The penetation depth will depend a lot on the salt/carbonate content. Two centimetres penetation depth in a circular, 5cm dia glass will make for quite even heat distibution. Afaik, penetation depth is largely dependent on electrical conductivity (inversely), while heat transfer depends on the dielectric properties. So there's no direct connection, while you are right, ice does not heat in the microwave.

@Karl It's not that simple. The dielectric heating depends on the dielectric properties of the substance, heat transfer doesn't (though it's often proportional, it's not a rule). As water gets hotter, this effect diminishes - the penetration depth increases, so less energy is absorbed. And the absorption isn't even over those 2cm - think about the absorption curve with depth, there's barely any absorption at the edges. That's the core issue here - the "skin" of the liquid doesn't get warm enough to boil, while the inside does - heat transfer between the two is very slow.

@Karl The temperature quickly dropping to 100C sounds very plausible -- good point. But 5cm would be a very narrow glass.

@DavidRicherby To your hand, it makes of course no difference if it gets hit by superheated water or an equivalent amount of condensing steam. Autsch! ;-)

A microwaved glass of water will 'bump' if the glassware is clean and the microwave heating is uniform.   The water has some tensile strength, so a bubble will not form at the exact boiling temperature without some nucleus (low surface tension due to a gas void in a boiling stone, for instance), so the liquid can become superheated.   On reaching in for the hot liquid, it may be the case that moving the liquid in the container causes an eruption as the superheated liquid boils.
It is true, too, that even without a nucleus to form a bubble, the air/hot-water surface will evaporate water vapor rapidly, but that has the effect of  cooling the water surface and if the water is not in motion, a cool-water layer
on the surface will form even as the bulk of the water becomes superheated.
Superheating is dangerous, of course; if your hand is holding the cup
when it bumps, a burn is likely.   A spoon lowered into superheated water
will usually make it erupt.   Eventually, a cosmic ray will make it erupt.

The water has no here relevant "tensile strenght", it's the surface tension of the bubble-to-be that keeps it from growing, until the water is hot enough to overcome the surface tension. The eruption can and w ill occur spontaneously.

Tensile strength is a way to express the difficulty of forming a small bubble (when surface tension is proportional to 1/r, small bubbles that form thermally must grow past a critical size or they collapse).  A small bubble in liquid is inhibited just as a small crack in a solid is, by "tensile strength".

Chemists should agree calling surface tension for tensile strenght of liquid under these circumstances. I understood what you meant and liked the expression. I support you Whit3rd. Chemists make imprecision a science! Your answer is totally within +/-1 logarithm of the probable answer, which is good enough :-P

The phenomenon has some similarities with https://en.wikipedia.org/wiki/Ultimate_tensile_strength, but is just something different. Why do you use a bad analogon when you know better  (critical size of bubble) and there is a perfect, direct explanation? I'm a chemist, and find that highly unscientific, @StianYttervik !

Because, the philosophical difference between physics, materials science and chemistry is the models we use to explain different phenomena. While it is not rigorous to explain surface tension as resistance like tensile strength, it is exactly something a chemist could use as explanation for an observed phenomenon. And then we are indeed ultimately in agreement that the best model overall would be surface tension. No argument against it, and I would have been cross to see it used formally. But here it is used informally, and in my view appropriately. For this phenomenon - the model works.

Of course the other danger of superheated water is that, not only is it liable to erupt all over you but, also, when it does it is, by definition, hotter than boiling water so it scalds you harder than boiling water would.

Water is a polar molecule as you know. This means it has positive and negative ends. The microwave is a electromagnetic wave. These waves are the right frequency to cause the molecules to get vibrating and cause tension. As you know, temperature is defined as the kinetic energy of a particle. This means that the water would heat up. But the directional vibration of the water depends on where it is located in the microwave. This it is almost random. The random movements can collide with impurities and jagged features of the container or water to nucleate the bubbles. This means that the bubble would float out of solution. This results in boiling. If the water is pure and the container is smooth, then the water will not have a nucleation point and will not bubble. What will happen is that some of the water molecules will just steam out but the rest of the molecules have no way to nucleate until something goes in, like a spoon, or sugar.

Always stay clear of the center of the plate, rather put your cup or glass on a side so it travels the most while heating.
Microwaves have "fixed waves" and should there be a region of water always under a specific wave, the water will explode just like a droplet of water in OIL.  It wants to evaporate, with no air around to do so, so BOOM it needs to go somewhere, and it does.
On the other hand, make it go around the platter, avoiding dead center, the waves will just heat up the water little by little, eventually boiling and creating movement and bubbles, hence space for the vapor to go.
A try to avoid the center of the revolving plate whenever I put food in the microwave too, even leaving a space in food if the plate is bigger than half the diameter of the platter.
I read somewhere "make a hole in the middle of the food it'll heat more evenly".  Well, ONLY if that hole is aligned with the middle of the rotating platter, then
yes.


text:
Variations in protonation in reaction mechanisms
English and German Wikipedia give slightly different mechanisms on Fischer indole synthesis. The thing bugging me is the protonation of N and the source of the proton.
Look at the intermediates 8 (Ger) and 5 (Eng). In the German Wikipedia imines are not protonated, in the English they are. Or look at intermediates 10/11 (Ger) and 7 (Eng) - protonation occurs but on different N-atoms of the aminoacetal. Ammonia won't leave the molecule unless it is previously protonated, but how important is protonation in other steps? Aren't those molecules in an equilibrium with all their protonation variations anyway?
And so does the reaction mechanism even specify the source of the proton? Do specific steps take a proton from the molecule, whereas other steps take it from the surroundings? I always thought the molecule simply grabs the most available proton (the closest and "loosest" one).
German Wikipedia:

English Wikipedia:

English Wikipedia is more correct. Under acidic conditions, carbonyl and imine additions should happen via the oxocarbenium and the iminium ions, respectively.
Also, the in German version, the step between 5 and 6 is not a step; that's a resonance structure. There's also sloppiness in 3 and 10. Those attacks should come from the lone pair. Technically, those steps should've happened before the addition anyway, as I've already pointed out. Bottom line is that, again, the English version looks better.

There's actually something happening from 5 to 6, namely a deprotonation. I agree the German one is sloppy, 5 shouldn't even be there if you use the N lone pair to expel water in 4.


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Does every solute spend at least $t_M$ in the mobile phase?
In 'Fundamentals of Analytical Chemistry' 9th edition by Skoog, West, Holler and Crouch, there is the following passage (page 865).

The dead time (void time), $t_M$, is the time it takes for an unretained species to pass through a chromatographic column. All components spend at least this amount of time in the mobile phase.

Apologies if it seems a silly question, but how does that make sense? Wouldn't all components spend exactly that amount in the mobile phase? That is, of course, assuming a constant flow rate for the mobile phase. Is this where I have gone wrong?

What this passage means is that all components will spend at least that time inside the column.

@Variax Well that's a given but for most components $$\text{time spent in mobile phase} \neq \text{time spent in the column}.$$ Furthermore, this isn't limited to just this sidenote. It is noted elsewhere in the book as well.

@ToddMinehardt So by lower bound you're essentially saying that different unretained species spend a different amount in the mobile phase? For me, that goes against the definition of unretained. Furthermore, the book does later use dead time $t_M$ as a measure of speed for the mobile phase (i.e., pure eluent).

@LinearChristmas - Thanks for pointing out a problem with what I wrote. I'd revise my comment to say that dead time is simply the rate of the eluent through the column. Thus, it is the lower bound on time for what can travel through the column, as it represents unimpeded flow (say, 1 mL/min, isocratic), and therefore anything that interacts with the immobile phase will elute at a time greater than the dead time.

@ToddMinehardt ;) I think I finally got it, too. Retained does not necessarily mean entering the stationary phase (hence coming to a "complete stop"). It can also simply slow down which should have been obvious to me from the start... If you write a short answer, I'd be happy to tick it!

Dead time ($t_{\bf M}$, also called holdup time) is the time it takes for the mobile phase (eluent) to traverse one length of the column, for a given flow rate. Because moieties in the eluent do not interact chemically with the stationary phase, $t_{\bf M}$ is therefore a lower bound on retention times for the species that do interact (chemically) with the stationary phase.
Simply put, anything that does interact with the immobile phase cannot elute faster that the pure mobile phase: retention times will be greater than $t_{\bf M}$.

Also called void time! :)

Yes, it turns out that the world of chromatography is rife with terminology that can be contradictory/incomprehensible/etc. :)

@ToddMinehardt How we measure dead time if the column is always filled with mobile phase? The detector would always get signal from the mobile phase.


text:
How does a diatomic molecule vibrate in response to heat?
Do you know of any applet/animation that shows how a diatomic molecule like $\ce{O2}$ or $\ce{N2}$ vibrate in response to/ to produce heat?

Can it vibrate only on one axis, the one joining the two atoms? What happens to the electrons during the vibration?

That's the only degree of freedom available. Otherwise, you're probably coupling rotation with the vibration.

It can vibrate only by extending and contracting the bond. The more energy that is added the bigger the bond excursion. (More vibrational levels are excited).  The bond only extends by a few % of its length in lower vib levels so its pretty hard to see. If heaps & heaps of energy are added (temp say $1000$ K)  then the bond can break as it is  actually anharmonic in its motion. When this happens the atoms fly apart.  Additionally the molecule can also rotate while it is vibrating. Rotation is slower than vibration so many vibrational periods occur within a $360$ degree rotation.

This post actually contains three questions, each of which would probably have benefited from being asked in individual posts.

The diatomic can vibrate only by extending and contracting the bond. The more energy that is added the bigger the bond excursion. (More vibrational levels are excited and have bigger excursions). The bond only extends by a few % of its length in lower vibrational levels so its pretty hard to see. If heaps & heaps of energy are added (temp say $1000$ K) then in a real molecule the bond can break as it is actually anharmonic in its motion. When this happens the atoms fly apart.
Additionally the molecule can also rotate while it is vibrating. Rotation is a slower motion than vibration so many vibrational periods occur within a $360$ degree rotation.
At any given constant temperature in a sample of molecules the energy moves between vibrations, rotations and translations so that each type of energy is in equilibrium.
Some of the energy can  also be transferred as  radiation, in the infra-red (for vibrations hundreds to a few thousand $\pu{cm^{-1}}$) and microwave region ( in the range few $\pu{cm^{-1}}$) for rotational transitions but only if the diatomic molecule has a permanent dipole, i.e. is not homonuclear. If these molecules are in equilibrium with surrounding then these surroundings must also emit/absorb radiation to keep the total energy constant. If you were to suddenly isolate your block of (heteronuclear diatomic) molecules, say in space, then the radiation emitted would over a few hours remove energy so that the molecules will end up in their zero point levels.
The figure shows the potential energy profile of HI. The the average length for the for the lowest level (quantum number $0$) is approx $106.4 $ pm and the chnage due to vibration is about $\pm 10 $ pm and clearly greater for higher vibrational levels. The vibrational period is $\approx 14 $ fs.

In a hetero-nuclear diatomic molecule microwave radiation can start a molecule rotating by causing it to move from one rotational state to another, it can also loose energy by emitting photons. But in homo-nuclear diatomic molecules there is no dipole and so radiation cannot influence it (other than to electronically excite it if the radiation has enough energy, i.e. visible/uv photons). Rotation  and translational energy levels can be populated or depopulated in all types of molecules by energy transfer during a collision.


text:
Is my teacher correct about organic compounds?
So in my chem notes its written that CCl4 , CH2Cl2 are both organic compounds as they are derived from CH4 which is also an organic compound.
It says that because CCl4 and CH2Cl2 are "derivatives" of CH4
Well...then lets say we involve CH4 in a reaction and take the product and involve the product in another reaction and then take the product of that reaction too and involve IT in another reaction and keep on doing this 10 times and then end up with SO2 does that mean that SO2 is an organic compound too?
I mean if this is the definition then ANY element/compound can be termed as an organic compound right?
Please help.

sigh... https://en.wikipedia.org/wiki/Organic_compound

maybe she means that it only goes till 1 reaction. Like if the product of the first reaction of CH4 is another organic compound and not the consecutive reactions?

You should really check what derivative means...


text:
What is the major hydrolysis product of 2,2,2-trifluoroethyl 2-acetamido-3-phenylprop-2-enoate?
In the below reaction we have to find the major hydrolysis product of 2,2,2-trifluoroethyl 2-acetamido-3-phenylprop-2-enoate.

My doubt is why the major hydrolysis is not when $\ce{OH-}$ at the $\ce{C}$ of $\ce{CF3}$ as it is more electropostive. However the major product is given by the following reaction, why?

That's a not a hydrolysis product. Are you just asking for the "hydrolysis product" because of the reaction conditions?

@Zhe which one is not hydrolysis product . The one which I thought or the other one

The other one. How is that hydrolysis?

Also, C-F bonds are notoriously hard to hydrolyze. You should have supposed that hydrolysis would simply break the ester, that would be at least logical.

@IvanNeretin why it is hard to hydrolyse

Plain hydrolysis would just replace trifluorexthoxide with hydroxide.
But you might want to consider that your staring material in equilibrium with another tautomer:

Deprotonation of the $\ce{OH}$ in the right structure, followed by an intramolecular reaction (transesterification/lactonization) in which $\ce{CF3CH2O-}$ is kicked out should give rise to an oxazolone.

How can we know which reaction can produce major product

Does the second reaction is also hydrolysis

@J.Doe Hydrolysis is an extremely broad term. I'd rather call this a cyclization, which is still broad but describes the outcome of the reaction much better. On a second look, this looks a bit an intramolecular variant of the Erlenmeyer-Plöchl reaction.

That reaction is not hydrolysis, it's a transesterification.

@Variax Yes, just as I wrote in my answer ;)


text:
How to differentiate diastereotopic protons by NMR in flexible groups?
In (1R,2S)-2-(hydroxymethyl)cyclohexan-1-ol shown below, Ha and Hb are diastereotopic (they will have different chemical shifts), but because of fast rotation, NOESY should show that both couple with Hc, right? How to differentiate diastereotopic protons by NMR in flexible groups?

Hmm... I am wondering if the rotation really will be that fast. I guess there will be some preferred position of the hydroxyl group. If you would then perform low-temperature NMR you could maybe be able to freeze the molecule in that preferred conformation. The only way I could imagine you would be able to differentiate between $\mathsf{H_a}$ and $\mathsf{H_b}$ though would be to perform ab initio calculations and compare these to the experimental chemical shifts. Why are you interested in that anyway?

I don’t think you can distinguish which is which in a freely rotatable group like the one you drew. They will be different, both signals will split up as dd (potentially ddd, if coupling to the hydroxy proton is observed). There will be significant roofing. Careful analysis of the spectrum should reveal two different shifts coupling to each other but very close.
Rotation around the $\ce{C-C}$ axis should be rapid on the NMR timescale resulting in the couplings $^3J_{\ce{HaHc}}$ and $^3J_{\ce{HbHc}}$ averaging out to one common value and the signal of $\ce{Hc}$ having a dddt-type pattern (three doublets to the neighbouring ring protons, a triplet to the side-chain protons).
All NMR methods I know to determine relative stereochemistry (e.g. NOESY, Murata, …) rely on rotation being slow on the NMR timescale.
Potentially cooling your sample down to very low temperatures ($-50~\mathrm{^\circ C}$ or less) could inhibit the rotation and allow for intermittent resolution of the protons. However, that should also produce a more distinct chemical shift difference, so it may actually not help.

In the frozen-in state, shouldn't the coupling constants to Hc give a good idea which is which (Karplus equation)?

@Karl Yes, imho. However, I’m not sure if the chemical shifts in frozen state can be mapped back to those of the non-frozen state. (I never did low-temp NMR, so … ^^')

Right. The average chemical shifts in the freely rotating hydroxymethyl group might easily reverse the peak order in the spectrum.

There are a number of methods that have proven successful in a number of cases, but no universally applicable method exists. Some thoughts to throw into the mix, to add to Jan's answer.
The greater the rate of rotation, or conformational exchange, the more difficult to implement a method. That is, if no preferred conformation exists, it becomes very difficult. The methods developed by Murata (J. Org. Chem., 1999, 64 (3), pp 866–876 DOI: 10.1021/jo981810k) make use of the 2 and 3 bond couplings to carbon, as well as the 3J H-H coupling. A very good reference that covers much of the recent work is by Bifulco, G. et al. (Chem. Rev. 2007, 107, 3744-3779 DOI: 10.1021/cr030733c).
For complicated natural product systems, often the difficulty is accurately determining the H-C coupling constants, and a lot of work is being done in this area of NMR methodology.  The review by Parella et al (Prog in Nucl Magn Res Spec 73 (2013) 17–55 DOI:10.1016/j.pnmrs.2013.07.001) provides some good insights here, and the very recent review by Tormena (Prog in Nucl Magn Res Spec 96 (2016) 73-88 DOI:10.1016/j.pnmrs.2016.04.001) discusses some complementary methods of analysis using modelling and theoretical calculations.
Freely rotating chains of course exist as a staggered average of rotational conformers. Sometimes, even in acyclic systems, specific conformers are preferred, and the coupling constants reflect this. However, in fast rotating systems, the couplings average across the lifetimes of all conformers - if you are seeing 3JH-H coupling of about 7 Hz, you have a very fast averaging system, with no strongly preferred conformation. If you have couplings that are more toward 3-4 (gauche) or 10-12 (anti), then it is likely that a preferred conformation exists. As Jan suggests, cooling can help, but even simply changing solvent can have benefits. Going to a non-polar solvent will promote intramolecular H-bonding, which can help stabilise a preferred conformer. Even going to a higher magnetic field can help favour the one population.