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How to deal with solving reactions with not enough background
The book I'm studying with often confronts me with exercises that involve solving reactions such as $\ce{CA(OH)_2 + CO_2 -> ?}$.
Now, most of the time, the questions are just repetitions of material that has been taught, that reaction in particular has been shown many times in the chapter before the questions, $\ce{... -> CaCO_3 + H_2O}$, but not explained.
When I look up such reactions on the Internet to try to understand, I mostly come across people who ask others to balance the equations for them. But that is not really interesting for me, as I find balancing a trivial mathematical task.
I am, however, failing to understand why this reaction (or many other reactions) happen the way they do. My book has also mentioned that you can combine two materials, like $\ce{CO + H2}$, in different ratios, under different temperatures, pressures, and with different catalysts, to get different results (methane, methanol, and many others).
So my question is -- when asked to solve a reaction, and given only the reactants (or possibly also other factors such as mentioned above), is there a mathematical principle (for example, relying on differences in electronegativity) to solve the reaction besides being completely familiarised with molecular orbital theory and quantum mechanics? Because so far, the only alternative I can see is memorising the reactions.
I realise this question is rather broad, and possibly there is no satisfying answer to it or any answer besides to go study quantum mechanics, but to provide some context, I'm a student at high school studying chemistry not as my main subject (so on a rather elementary level), and I'm looking for a perspective how to understand the material I learn and not memorise it.

There are some general principles, so after a while you would begin to make your own generalizations and recognize patterns, but initially there is quite a lot to memorize. E.g. organic chemistry strongly relies on polarity, which is indeed related to electronegativity, but steric effects and much more complex quantum mechanics gets involved at some point, so much so that people often just have to run an experiment and see what happens. In inorganic chemistry, how would you guess that iron (2+) ions react with sulphide to give an insoluble iron sulphide precipitate? It's just experimental...

And in your example, why does calcium hydroxide react with carbon dioxide to give calcium carbonate? We can explain it a posteriori, but you could not safely generalize to similar systems. E.g. I am pretty sure that silicon dioxide and magnesium hydroxide would not react in the same way. So no, it's not like mathematics where an all-encompassing rule can be stated and applied universally: chemistry is still an experimental science. At least, that's the way I see it.

possible duplicate of https://chemistry.stackexchange.com/questions/50684/how-can-i-predict-if-a-reaction-will-occur-between-any-two-or-more-substances

Possible duplicate of How can I predict if a reaction will occur between any two (or more) substances?


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Ultrasound absoprtion by polypropylene beaker for HF acid
In my Ph.D. research, I need to carry out an etching of glass samples in HF acid. I use an ultrasonic bath for the experiment, but to avoid corrosion of steel I put a small plastic (polypropylene) jar with 5% HF with glass samples into the ultrasound bath full of water. Recently, my colleague told me that the plastic jar absorbs most of the ultrasonic energy and my etching process goes the way it would be without the bath at all. So, first of all, is it true that a polypropylene beaker absorbs ultrasound? What kind of beaker should I use instead which wouldn't be affected by HF acid?

Polymers, like other materials as well, attenuate / absorb acoustic waves.  Beside reflection (echo, similar to active sonar), this renders ultrasound suitable for non-destructive inspections.  The extent of absorbing ultrasound depends on several factors (e.g., temperature, density/crystallinity of the sample, deployed frequency) and may be expressed as attenuation and by acoustic impedance.
The following table is part of a lecture about medical diagnostic applications of ultrasound:

(source, slide 5.)
A more detailed listing, specifically addressing polymers, may be found here. From data like
|---------------------------+----------+---------+-----------+-------------|
| Plastic material          | Velocity | Density | Impedance | Attenuation |
|                           |    [m/s] | [g/qcm] |           |     [dB/cm] |
|---------------------------+----------+---------+-----------+-------------|
| Polyethylene, low density |     1950 |    0.92 |      1.79 |         2.4 |
| Polypropylene, Profax     |     2740 |    0.88 |      2.40 |         5.1 |
| Polypropylene, white      |     2660 |    0.89 |      2.36 |        18.2 |
|---------------------------+----------+---------+-----------+-------------|

some differences in polymers of same name, but different brand (and degree of polymerisation, etc.) become apparent. (Interestingly, Teflon/PTFE is not mentioned in their listing, yet according to this study possess a much higher acoustical impedance than PP.)
Hence, beside chemical resistance against HF, you likely have to test the efficiency of the ultrasound waves passing through the walls of your beaker anyway.

Buttonwood, how should I test the efficiency of the ultrasound waves passing through the walls of the beaker?

Fasten the flat bottom beakers immersed in the bath with a clamp with some space just above the resonators to prevent moving / tilting (alternatively, for some baths, there are baskets with holding springs already installed, e.g. http://www.tovatech.com/blog/1504/ultrasonic-cleaner/how-to-use-an-ultrasonic-cleaner-for-usp-method-sample-preparation). Add sample (or sample on sample mount) and HF into a test beaker in question, start the etching for x min; then test with an other sample and new HF solution in a different beaker for the same time, power and soln. temperature; compare the results.

I though it is possible to measure the power of ultrasound inside and outside of the beaker like in optical spectroscopy.

So far I did not witnessed an org.chem. lab with experimental access to record the numbers of dB(A).  This is the reason for the empirical approach "with beaker (material) A it works better, than with B or C". However if your institution has know-how and equipment to record the acoustic spectra, go for it.  An other meta reference found (http://materials.springer.com/lb/docs/sm_nlb_978-0-387-69002-5_60) focusses more on polymers deployed in a medical context; wich may be not applicable here (different frequencies then the ones used in a bath).

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Will an aqueous sodium hydroxide solution degrade over time?
I have a chiller bath, which is attached to a 150 ton cooling unit through 4" PVC pipe. The bath itself holds around 5000 gallons of water, and is made of 304 stainless steel. It constantly circulates through the PVC into the cooling units outside via a 20hp pump. The outside cooling units are made of steel, although I do not know the exact grade of the steel. It was however built for water circulation. I use a sodium hydroxide solution to raise the pH of the water, therefore preventing rust and keeping my water clean. The problem is that I constantly have to add the solution to keep the pH at a level that is acceptable (9-11). I have watched it enough to say that on average the pH drops by about 0.3 a day. I am wondering where all my sodium hydroxide solution is going. Thank you for the responses so far, I apologize for not being more clear on the problem.

Usually, the biggest problems with preventing a sodium hydroxide solution from changing in $\ce{NaOH}$ concentration, or pH, over time are adsorption of $\ce{CO2}$ from the air, and evaporation of water via permeation through the container holding the solution.
If we can really say that it is inside a closed system as you've specified, then any degradation over reasonable time-scales (i.e. months to years) should be negligible.
However, you need to be sure about your "closed system". Is it glass, which could slowly react with the $\ce{NaOH}$? Is it a plastic that could be slightly permeable to water vapor, air (and thus $\ce{CO2}$), etc.? Is it, heaven forbid, a metal that would react and degrade your $\ce{NaOH}$? You need to think about these things at every contact point (vapor or liquid) throughout your system.

Except for a very minute amount, generally solutes do not disappear (Evaporate) when left alone in solution. Your solvent, water, may evaporate over time depending on the conditions, however unless you repeatedly transfer container your are unlikely to lose much of your NaOH.

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Does Ksp depend on the solvent?
E.g. would the solubility and Ksp of  $\ce{Mg(OH)2}$ be the same in water as it would be in HCl? Doesn't solubility depend on the polarity of the solute and solvent, so wouldn't different amounts of a solute dissolve in different solvents. Wouldn't this give different Ksp values? However, I only ever see a single Ksp for anything.
Also, if $\ce{Mg(OH)2}$  is said to completely dissolve in $\ce{HCl}$, is it implied that is completely dissociates into  $\ce{Mg^{2+}}$ and $\ce{OH^-}$ ions, or does it only exist as small solid pieces of $\ce{Mg(OH)2}$  floating around in the  $\ce{HCl}$? I am assuming the latter, as the Ksp of  $\ce{Mg(OH)2}$ is small.
I have a question in my book where $\ce{Mg(OH)2}$ is dissolved in $\ce{HCl}$ and then back-titrated with $\ce{NaOH}$. I can't seem to understand why, if the $\ce{Mg(OH)2}$ dissolves completely, we would need to do a back titration. I thought a back titration was useful when the solute doesn't dissociate completely.

Hydroxide reacts with HCl, you know...

$K_\mathrm{sp}$ for a compound is solvent specific and can change depending on the $\mathrm{pH}$ of solution and on the intrinsic properties of the materials involved.
As $\ce{Mg(OH)2}$ is a base, it will not be in the form of $\ce{OH-}$ in $\ce{HCl}$; it will instead be either $\ce{H3O+}$ or $\ce{H2O}$.
If a salt is said to completely dissolve, it usually means that ions are fully solvated, thus completely surrounded by solvent.
But, remember that if the $\ce{OH-}$ are acidified, the counter ion for the magnesium will become chloride ions.
Lastly, if the solvent changes the nature of the compound, the $K_\mathrm{sp}$ will be that of the different compound it has changed into.
Solubility of:
Magnesium Hydroxide (.0090g / Liter)
Magnesium Chloride (542g / Liter)
Source: Solubility Table

Thanks for posting! For more information about post formatting, see here and here in the [help]. Also, if you're so inclined, stop by and say hello in chat.

"As $\ce{Mg(OH)2}$ is a base, it will not be in the form of $\ce{OH-}$ in $\ce{HCl}$; it will instead be either $\ce{H3O+}$ or $\ce{H2O}$." // WRONG. // $\ce{Mg(OH)2}$ is a solid and thus not in solution.  // There is $\ce{OH-}$ in $\ce{HCl}$, albeit at a very low concentration.

@MaxW yeah like 10 femto molar in a 1M solution. That's totally negligeable and brings very little to the understanding of the problem IMHO. If you think otherwise please explain why.


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Why do we need the identity operator, E
In group theory, in order to assign a point group, one must first identify the symmetry elements present in the molecule.
To give an example, water has:

a rotational axis, labelled C2 since it is a two fold axis, with a π/2 rotation giving back (what appears to be)the same thing
two mirror planes, labelled σv and σv' (with additional labels xy/xz to define what plane the mirror plane lies in)

Symmetry elements present in water
(Note that for the C2v point group, the axis are defined 'arbitrarily', different tables/books might swap x and y)
From this, it would look like water has three unique symmetry elements, however this is not the case, in fact, water has four due to the presence of an additional symmetry element, E, known as the identity operator (sometimes seen as I in certain textbooks).
This E operator corresponds to 'do nothing' (i.e. leave atoms where they are), which is easily seen in the character table for the C2v point group (to which water belongs) in which the characters under E are all '1':
$$\begin{array}{c|cccc|cc} \hline
C_\mathrm{2v} & \color{red}{E} & C_2 & \sigma_\mathrm{v}(xz) & \sigma_\mathrm{v}'(yz) & & \\ \hline
\mathrm{A_1} & \color{red}{1} & 1 & 1 & 1 & z & x^2, y^2, z^2 \\
\mathrm{A_2} & \color{red}{1} & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_1} & \color{red}{1} & -1 & 1 & -1 & x, R_y & xz \\
\mathrm{B_2} & \color{red}{1} & -1 & -1 & 1 & y, R_x & yz \\ \hline
\end{array}$$
$\,$
C2v character table, from orthocresol's Group Theory Tables

The identity, E, consists of doing nothing; the corresponding symmetry element is the entire object. Because every molecule is indistinguishable from itself if nothing is done to it, every object possesses at least the identity element. (Taken from Atkins' Physical Chemistry).

Since transforming a molecule by the identity operator just gives back itself, the use of 'E' in a qualitative sense appears redundant, however it appears in every character table, and as such much have a purpose.
Why do we need the identity operator, E, and what use does it have in chemical group-theory?

Not enough to be a full answer, but: The "why do we need" part has been covered thoroughly. As for "what use it has", to state the obvious, the character of an irrep under $E$ is the dimensionality of the irrep.

@orthocresol, good point, aaide for being needed to complete the  riteria of s group I'd somewhat forgotten that there was a consequence when actually using group theory.

What element of your group is $C_2^4$?  Groups are closed under multiplication, so any product of elements of the group is some element of the group...

@EricTowers, could you expand, not entirely sure what you're saying/what point you're making

You say $C_2$ is an element of your group of symmetries.  This forces $C_2 \cdot C_2 \cdot C_2 \cdot C_2$ to also be a symmetry.  Which element is that product?  (Using the fourth power is a consequence of an error in your description of $C_2$.  Actually, $C_2$ is rotation by $\pi$ radians.  With this correction, we need only ask "which symmetry is given by 'first do $C_2$, then do $C_2$ again'?")

What you are asking arises as a fundamental consequence of the definition of vector spaces and the operations defined for them and, specifically, symmetry groups. The identity operator does not do "nothing" (see below) and it does belong in the formal definition of a group, and it is included to satisfy said definition.
From Byron and Fuller's Mathematics of Classical and Quantum Physics:

...a group is a system consisting of a set of elements and rules for combining the members of this set.

This sets the stage for the definition:

A group is a system $\{G, \cdot\}$, consisting of the set $G$ and a single closed operation $\cdot$, which satisfies the following three axioms:

If $a, b, c \in G$, then $a\cdot(b\cdot c) = (a\cdot b) \cdot c$ (associativity).
There exists an identity element $e\in G$ such that for all $a\in G$, $a\cdot e = e\cdot a = a$.
For every $a\in G$, there exists an inverse element in $G$, denoted $a^{-1}$, such that $a\cdot a^{-1} = a^{-1}\cdot a = e$.

Given this, we can go to more plain English as Wilson, Decius, and Cross do in Molecular Vibrations, where they note that

It is evident that for every symmetry operation possessed by a molecule there is some operation, either the same one or a different one, which undoes the work of the first.

This is an important point, and operationally illustrates what the identity operator does and why it is included in every group:

...if an operation $R$ changes a molecule from the position I to the position II, then there is an operation which shifts the molecule from II to I. Such an operation is called the inverse of the original operation and is written $R^{-1}$. Symbolically, $R^{-1}R = E$, since the symbol $E$ represents the identity operation, which leaves the molecule unaltered.

(For the record, WDC also write that "...$E$ is used for the trivial operation of leaving a molecule in its original position.")

I learnt that there were four axioms (the fourth being that for all $a, b \in G$, $a \cdot b$ must also be in $G$), but it's been a while. Maybe it's already implied by the other three axioms, I'm not sure.

I am sorry, I must have read this in a hurry; as my "fourth axiom" is precisely equivalent to the phrase "closed operation" in Todd's quote. // I assume the ultimate authority on the matter to be a maths textbook. Looking through my notes from Rotman's Abstract Algebra it does list only three axioms. However, like Todd's source, Rotman also separately defines the binary operation, $$, to obey $G * G \to G$. Essentially, that means $$ takes two arguments, both in the set $G$, and outputs one value which is also in the set $G$. [cc @NotEvans.]

@orthocresol I think it ultimately comes down to whether you think of the "closedness"  is a property of the operation or a property of the set. There was an interesting discussion about this on the talk page for the Wikipedia article on (algebraic) magmas. The proponents of the view that it's a property of the operation, pointed out it follows naturally from the definition of a binary operation as being from $G \times G \to G$ (as in Rotman). But I've certainly seen both.

For clarity on why the identity element is so important to group theory in general, it might be worth giving an example of a group and its identity where the identity element isn't so boring and simple as '0' or 'do nothing' such as those in matrix operations and the very complex identity elements in sand pile groups https://plus.google.com/101584889282878921052/posts/QezmLcTCTMJ

It is not that we need the identity operator. It is just that things (character tables, irreducible representations, etc.) work the way they do. As to why they do so, I refer you to any decent textbook on group theory.
I guess one could rebuild the entire theory from scratch (that is, reformulate all definitions, starting from the very definition of a group, reprove all theorems, etc.) without explicitly using $\rm E$. That would be a titanic amount of work, though, and quite pointless at that. Also, many things like "order of a group is divisible by..." would start to sound less natural.
Come to think of it, why do we need the number $0$ in math? It just sits there and does nothing, after all. Well, doing nothing may be quite important: it makes a lot of other things simpler.

♫ Zero is the loneliest number♫

Of course, zero is the identity element in the group consisting of all integers under the operation of addition.

For clarity on why the identity element is so important to group theory in general, it might be worth giving an example of a group and its identity where the identity element isn't so boring and simple as '0' or 'do nothing' such as those in matrix operations and the very complex identity elements in sand pile groups plus.google.com/101584889282878921052/posts/QezmLcTCTMJ

For a set $G$ to be a group, every $a \in G$ must have an inverse $a^{-1}$ which is defined by satisfying $a \times a^{-1} = a^{-1} \times a = e$, where $e$ is the identity of the group.
It is thus impossible to abolish $e$.
Same thing happens for the integers and its zero.

Put another way, if there weren't an identity, we wouldn't be talking about a rotation group; we'd just be talking about a rotation semigroup :)

That's precisely why I said that hypothetical getting rid of E would require a different definition of a group (ditto for pretty much everything else in group theory, for that matter). In the sense of the current definition, the resulting thing would not be a group at all. Perhaps I should have called it differently; how about "shmoup"?

Well, according to this Math StackExchange post, you need the identity element because it is impossible to define an inverse element a if you don't know about the identity element e. The definition of a group includes the identity element (below).
There exists an element e in [Group] G such that, for every element a in G, the equation e • a = a • e = a holds. Such an element is unique (see below), and thus one speaks of the identity element. (Wikipedia)

Symmetry operator $E$ is the lowest symmetry a molecule may possess; and indeed all objects possess this operator.  Yet, among molecules revealing $E$ as the only the operation are the ones exhibiting stereogenic centred chirality, like $\ce{CFClBrI}$.  Hence $E$ is of value, not redundant, and needed.

Actually, as far as I know, electrons do not possess it. Spinor is identical after 720 deg. rotation, not 360.

This is a special case of a phenomenon that occurs throughout mathematics. For example, you could ask why we have zero for similar reasons. The answer is frequently: it's often easier to reason about a collection of things if the thing that represents 'nothing' is considered part of the collection.
For example, if we didn't have zero there'd be two kind of sentences about how much money we have: "I have \$X" or "I don't have any money". If you allow X to be zero then you only need one kind of sentence "I have \$X".
In the case of groups, many theorems are easier to state if you include the identity as part of your collection. For example "the order of a subgroup divides the order of a group" is much easier than "one plus the order of a subgroup divides one plus the order of a group". And "in a group you can always multiply two elements" becomes "you can multiply elements most of the time but there are certain pairs you can't multiply because when you multiply them they cancel out doing nothing but we don't consider doing nothing to be an element of the group". Closer to your example, the statement of the orthogonality relations for character tables would be more complex.
The fact that so many things simplify when you include the identity is good reason for it being there.

It is true that the identity operation belongs to every group, as the other answers point out. One can still ask, why don't we just omit the first column, by convention, since it is all ones anyway?
The simple answer is that it is not always all-ones. It gives important information, namely the dimension of the space of an irreducible representation. In other words, if the first entry is $d$, the group elements are represented by orthogonal $d\times d$ matrices.
(A less serious reason is that the character table would no longer be a square table if we omit the first column.)

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Energy in a Chemical Reaction
In a chemical reaction, is energy always either lost or gained by the reactants? As much as I am concerned, changes in energy can occur during the absorption of heat or the emission of light or heat (these are the most often occurring). Is every chemical reaction accompanied by these conditions? Aren't there any chemical reactions without these conditions? Usually when we observe common chemical reactions (sugar that is melting in the water), we don't see any emissions or absorptions. Are these changes so trifling that we can't feel them (while they actually exist), or there are reactions with no emissions or absorptions?

You might want to narrow the scope of your query: as written, it is rather broad.

Yes, there's enthalpy of mixing if different compounds are mixed.

Your example of "sugar that is melting in the water" likely is an error as your observation relates to dissolution of sugar, rather than to its melting.  In the instance of Sucrose (https://en.wikipedia.org/wiki/Sucrose), there is only a decomposition temperature known, $\pu{186 ^\circ{}C}$, which is much above the temperature of serving a hot tea.  For monitoring even small changes in enthalpy in the course of a reaction, there are (reaction) calorimeters (e.g. DOI: 10.1021/ie070050d, or in biochem, e.g. doi: 10.1074/jbc.M806473200).

your question is good yet lacks the specifics.
there are different so called "energy fields" in which changes occur (entropy, potential ,gibes and so forth).
the thermodynamic field is endless and can be spanned with any variables (using maxwells equations and Legendre transformation)
there is
a sort of an axiom that states that all matter will try to find a way to a lower state of energy.
but that said its not to scientific.
now for the question:
every reaction creates some amounts of electromagnetic waves (as a price for electron "movement") some can be seen with the naked eye as they emit waves in the Uv-vis spectrum some are just infrared radiations of inner states of electron decay to a lower energetic state via vibrational relaxation.
ideally no reaction will occur in the ideal zero as there is no energy to "pay" for any kind of a reaction.
for further read:
quantitive chemical analysis by harris
chapter 17 fundamentals of spectrophotometry.

There are plenty of reactions, e.g., ammonia inversion, where the reactants and products have the same energy.


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Reaction of HBr + 2,4-dimethylpent-1,3-diene yields 3 products?
My understanding is that either pi bond can react with HBr but that you still have to follow Marovnikov's rule or whatever.
I end up with 2 allylic carbocations that are actually the same molecule along with 1 unformed non allylic carbonation. From there there is only one product...

What's wrong with having only one product?

My exam said list all 3 products created.

Maybe they just wanted you to list all the theoretically possible products. But that is a problem with your exam paper, you yourself don't really seem to have a problem.

But what would be the 3 theoretical products? I feel like I did the problem wrong or something.. he definitely made it sound as if there are 3 products formed. Which doesn't make sense because I ended up with 2 equivalent molecules and a 3rd unformed unstable carbocation. Are there other products I'm missing?


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Conversion of an epoxide into an alcohol
I need help changing an epoxide into an alcohol, i.e. the second step in the scheme below. I found a reaction that created two hydroxyl groups but cannot find one that only produces one:

Superhydride (lithium triethylborohydride) does this. JACS 95 (1973)  8486 reports complete selectivity for unsubstituted over disubstituted carbon with very high yield. HC Brown JOC 45 (1980) (1) reports complete selectivity for 2-butanol from 1,2 epoxybutane and 97:3 for the reduction of styrene oxide.

We could add grignard reagent or LiAlH4 (to conserve the carbon skeleton) to the epoxide and then protonate it.

OP is not looking to make any changes to the carbon skeleton.

@Zhe sorry, I read the ques wrongly. Maybe one could use a good H- donor such as LiAlH4

As I can say about this, we can firstly use MCPBA to form an epoxides and then by observing the product we can use Lithium Aluminum Hydride which gives hydride ion(which favours SN2 mechanism) to conserve the number of carbons in the skeleton.
And now, after this the hydride ion attaches to the less hindered site, so after that we have to do the protonation to form this product.


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Synthesis of Anthracene: Mechanism
Recently, I happened to find a problem that involved the synthesis of anthracene from bromobenzene in the presence of tert-butyl lithium, using THF as the solvent, and at room temperature.  I am wondering the mechanism for this transformation.
Taking into account that THF degrades in tert-butyl lithium this is the one that I thought of,
It starts with aryne formation.  Benzyne reacts with ethene in a [2+2] cycloaddition.  This undergoes cycloreversion.  The resulting product reacts with another mole of benzyne to create the dihydro-derivative.  This reacts with another equivalent of tert-butyl lithium to form anthracene.
I have a question regarding this.  I do not think a cycloreversion could occur at room temperature (wiki gives a temperature of 180 degrees and states that the conrotary opening is unfavored).  Is this the case or not?  Am I missing another possible mechanism here?

Bromobenzene is fairly stable towards lithium halogen exchange, without generation of a benyzyne.

@AS_1000 It might represent some work (e.g., barrelene), however a small graphic illustration of your suggested mechanism would ease to follow your argumentation.  As a word of caution, BuLi is known to react with etheral solvents (hence tBuLi is sold in alkanes, and never in THF) to a degree that in conjunction with TMEDA you may use it synthetically (in case of THF, yielding ethene and the Li enolate of acetaldehyde).

Thanks for that, I proposed another mechanism using THF, I also added drawings, but I still have a question regarding it because it involves a cycloreversion that I don't think would be possible at room temperature.

I found a research paper[1] which specifically talks about the reaction you mention. The paper also mentions the same mechanism as you do:

TETRAHYDROFURAN (THF) (1) is known 1 to react with strong bases like
butyl-lithium, and the product (2) is known to decompose by a
retro-cycloaddition to give ethylene (3), and the enolate ion (4) of
acetaldehyde.

Enolate ions (6) are
known 2 to react with benzyne (5) to give dihydrobenzocyclobutenes (7),
and the dihydrobenzocyclobutenes are known to decompose by an
electrocyclic opening to give o-xylylenes (9), which are

trapped by the
benzyne to give dihydroanthracenols (11). By putting these known
reactions together we have found a very simple ' one-pot ' synthesis
of symmetrical 9,lO-unsubstituted anthracenes.

The paper uses warm THF to conduct the reaction which shows that cycloreversion of the side ring actually takes place even at a temperature lower than 180°C.

When bromobenzene (13) is added to a solution of a four-fold excess of
N-lithio-2,2,6,6-tetramethylpiperidine (15) in warm THF, the major
neutral product is anthracene (16) (63% based on bromobenzene).

The paper also talks about what happens when the reaction temperature is changed:

Variations in the Reaction Conditions.-The reaction was carried out as described above, but with the following modifications. (i) When the
butyl-lithium and the tetramethylpiperidine were mixed slowly at 0°
and the reaction carried out at 20°, there was no anthracene formed,
and Nphenyl-2,2,6,6-tetramethylpiperidine was the major product (75%
by n.m.r.). (ii) When the butyl-lithium and the tetramethylpiperidine
were mixed at reflux temperature and then cooled to 0° before adding
the bromobenzene, anthracene was formed (40%, pure). We were not able
to isolate dihydrobenzocyclobutanol. ... (iv) When butyllithium (5
mmol) and tetramethylpiperidine (6 mmol) were mixed at -78° and then
the mixture was brought to reflux temperature before adding the
bromobenzene (1 mmol), the yield of anthracene was the same as in the
original reaction.

Reference:
(1)   Fleming, I.; Mah, T. A Simple Synthesis of Anthracenes. J. Chem. Soc., Perkin Trans. 1 1975, No. 10, 964.


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Discerning Chemical Relatives: Salicyl salicylate and Methyl salicylate
Oil of wintergreen or methyl salicylate is found in Gaultheria procumbens.  The molecular similarity to salicylic acid strikes me as obvious. The hydrogen is replaced by a methyl group.  But am I jumping to conclusion I wonder.  I've seen plenty of molecules which are said to have the same structure, but fail to see it because my knowledge of chemistry is limited.  How do chemists define structural similarity in molecules?


text:
How can I calculate the molar conductivity of a solution with multiple solutes?
I'm running an experiment mixing 2 different solutions, $\pu{5 mL}$ of $\pu{0.1 M}$ $\ce{NaCl}$ with $\pu{5 mL}$ of  $\pu{0.1 M}$ $\ce{KCl}$. I am measuring the conductivity of the total solution in $\pu{mS/cm}$, and also measuring concentration of the electrolytes with electrolyte analyzer, so I will get the individual concentrations of $\ce{Na+, K+, Cl-}$ in the mixed solution.
How do I calculate this mixed solution total molar conductivity given these settings? The literature only shows the calculation for a single solution, and I couldn't find any reference on how to calculate this kind of combination.

If you are interested in the modelling of electrical conductivity of aqueous solutions containing more than a single salt, then there are already some references in literature, for example for the salinity of natural waters (Limnology and Oceanography Methods, doi 10.4319/lom.2008.6.489) or in systems relevant to chemical engineering (Ind. Eng. Chem. Res., doi 10.1021/ie9605903).

(from the latter)


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Factors affecting Kp (pressure constant)
Initially, I found that $K_\mathrm{p}$ depends only on temperature; but later I found that it depends on pressure of $\ce{CO2}$ gas in case of $\ce{CaCO3 (s, aragonite) → CaO (s) + CO2}$.
How do we know whether it depends upon partial pressure of reactants and products or not? Or, this is a special scenario?

If there is a high gas pressure in the environment the reaction is taking place in, $K_\mathrm{p}$ will shift towards the side of the reaction that has fewer gas molecules, as it is "fighting against" the high pressure, in a way. So in your case, since only one side has gas (the product side) if you increase the pressure of $\ce{CO2}$ or any other gas, the equilibrium will shift towards the reactant side.

I am sorry. But does this answer the question?

It seemed to me he was asking if $K_p$ depends on partial pressure of reactants/products, so I was posting about how the equilibrium ($K_p$) shifts due to pressure (which could be partial pressure of a reactant/product, or pressure of some other gas). I hope I didn't misinterpret the question, but I think my post goes some way to answering.

Sir I want to ask that generally we say that K of a reaction depends only on temperature but in this case it is also depending on pressure of CO2 gas.Why is it so?


text:
Simulated Body fluid recipe
I'm working on preparing SBF in my school's lab however a couple of components are missing which are $\ce{NaHCO3}$ (0.350 g) and $\ce{K2HPO4}$ (0.228 g) .
Is it possible to replace these components with $\ce{NaCO3}$  and $\ce{KH2PO4}$ ?
The following link has the ions concentration of SBF:
https://www.irsm.cas.cz/materialy/cs_content/2002/Helebrant_CS_2002_0000.pdf
I'm not a chemist and I would really appreciate your view on this.

No. Or rather, you can do that, but your pH will be way off.

Probably not, sodium carbonate vs sodium bicarbonate are quite different (likewise for the phosphorous containing compounds). It looks like its setting up some kind of buffer, so changing things won't really work. You could just try it but all of the chemicals listed are really, really inexpensive and readily available- just buy what you need

@NotEvans. Thanks...Is there another way to prepare SBF without the chemicals I'm missing? Another recipe :)?

I couldn't say, its very much not my area of chemistry. Google is likely your best bet (but as I say, just buy new chemicals, they're cheap... $\ce{NaHCO3}$ especially, you can find it in the baking aisle of your local supermarket...


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Cold Gas Inflation
What and how is cold gas inflation used as replacement for traditional gas ingredients in airbag deployment? Can we make our own version of cold gas inflation?

you can find answers in this pdf


text:
Can an organic molecule have a triplet ground state?
Can the ground state of an organic molecule be a triplet. This would imply something like a "HOMO" formed by 2 degenerate levels and less than 4 electrons to fill them.
If yes, what are the conditions?

There are molecules like cyclic butadiene having two degenrate level with single eletron in each with parallel spins (triplet) even molecular oxygen which we breathe is triplet oxygen.

Cyclobutadiene has a singlet ground state. And $\ce{O2}$ is definitely triplet, but isn't an organic molecule.

As Martin has mentioned, carbenes are a good starting point if you are looking for organic compounds with a triplet ground state.
In these carbenes, the HOMO is not twofold degenerate, as your question suggests. Instead, there are two singly occupied orbitals with different energy. The energy gap, though, is typically small, and since there is an energy cost associated with pairing electrons in the same orbital, the electrons prefer to stay apart.
$\ce{CH2}$ itself has a triplet ground state, but is very difficult to observe; most methods of generating carbenes actually produce singlet carbenes, and $\ce{CH2}$ is so reactive that singlet $\ce{CH2}$ reacts with other stuff before it decays. Not all that interesting, eh.
There are some examples of persistent triplet carbenes, though.1 This one, reported by Hirai and Tomioka in 1999,2 has a half-life of $\pu{940 s}$ in degassed benzene at $\pu{20 ^\circ C}$:

In general, simple alkyl and aryl carbenes adopt a triplet ground state. These particular aryl groups on the carbene provide a large degree of kinetic stabilisation, which prevents simple dimerisation of the carbene (to form an alkene); the para alkyl groups also hinder reaction at that position. A further computational study of this molecule has been carried out by the Schaefer group, which supports the analysis above.3

References

Hirai, K.; Itoh, T.; Tomioka, H. Persistent Triplet Carbenes. Chem. Rev. 2009, 109 (8), 3275–3332. DOI: 10.1021/cr800518t.
Hirai, K.; Tomioka, H. A Triplet Carbene That Can Almost Be Bottled. J. Am. Chem. Soc. 1999, 121 (43), 10213–10214. DOI: 10.1021/ja991387c.
Woodcock, H. L.; Moran, D.; Schleyer, P. v. R.; Schaefer, H. F., III. The Almost Bottleable Triplet Carbene:  2,6-dibromo-4-tert-butyl-2′,6′-bis(trifluoromethyl)-4′-isopropyldiphenylcarbene. J. Am. Chem. Soc. 2001, 123 (18), 4331–4335. DOI: 10.1021/ja003552q.

There are various examples of molecules in triplet ground state. The smallest (organic) is possibly methylene, $\ce{:\!CH2}$.
Carbenes in general may adopt triplet state as their ground states.
Other possibilities are biradicals (From the IUPAC gold book):

An even-electron molecular entity with two (possibly delocalized) radical centres which act nearly independently of each other, e.g.


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Openbabel: Generate random 3D geometry
Is it possible with openbabel to generate random 3D geometries (i.e. conformers) of a molecule without taking energy into account? Ideally with the property of being truly random, so that after generating conformers like this for a sufficient amount of time, a good coverage of the conformational space is obtained. Because the --conformer options of openbabel takes energy into account, but I want to input the geometries into a qc program that then calculates the energies for the conformers.

I would generate a normal coordination file and write a little script that twists desired bond-angles by different amounts

If you create actual random 3D geometries I expect 99.9% of those being absolute garbage and they won't have anything to do with different confomers. There are techniques which allow for confomer search, for example simulated annealing.

^ this. I don't know what exactly you want to achieve, but maybe some kind of MD simulation could help you get enough snapshots? I mean you have something in the scale of $n_{steps} \cdot N_{atoms}^3$ calculations to do if you want a random, statistical approach to all angles. I don't know, but my boss would ask me if I was crazy if I would start 10k calculations of which 9990 would be garbage...

Well, the thing with the openbabel conformer generator is that it is often not expolaritive enough (sometimes even only a single conformer is returned). So my idea was to generate random conformers and optimize them afterwards to have a more robust method.

99.9% no problem: then I just generate 10000 structures and get 10 conformers that are actually different. because openbabel just gives me basically always the same conformer.

And how are you gonna go through 10000 coords to find the relevant ones?
I would start the other way around. Think about which conformers are important, generate only those coordinates.
If you would tell us what you are aiming towards maybe we could help better...

10000 coordinates with MM is pretty bearable, I guess.

Random conformer search generates random conformers
Open Babel, like many conformer sampling tools, has a set of common dihedral angles to sample. This minimizes the search space, e.g., only 3-4 likely cases for each rotatable bond.
The RandomConformerSearch method indeed generates conformers using a random sampling of dihedral angles. It then (optionally) runs a force field for some number of steps to minimize any steric clashes. No energies are considered.
If you're finding only one conformer output, that's because you're not using the --writeconformers flag:
http://open-babel.readthedocs.io/en/latest/3DStructureGen/multipleconformers.html

PS - I only recently saw this question. There is an Open Babel support forum - I'm not coming to Chem.SE as often as I'd like.

OK, I will keep the open babel support forum in mind. But I tried several combinations of input flags and always got a single conformer. If I have time on my hands, I will try to make a minimal example of the problem I ran into and post it into the open babel forum. By the way, Avogadro and OpenBabel are great, keep up the nice work : ) Anything else in planning?

Doesn't matter what combination of input flags - if you don't use --writeconformers you'll only get the lowest energy one. Alternatively, write a Python script. :-)


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Phenol smell from hands
It is a question which I faced when I was in Senior year in high school, I used my hands to handle phenol which has turned pink due to oxidation, I washed my hands with all kind of soaps but smell in hands didn't go for 3-4 days. Why was it, did it bond to my skin? What I could have done to remove smell?

As an acidic compound, phenol started to degrade the protective layers of your skin, and partially diffused into deeper layers.  This, and because of the moderate solubility in water are the reason the scent of phenol was still recognisable after some time.  While phenol is better soluble in ethanol and methanol, one refrains from these solvents to clean the skin because these actually would help phenol to reach even deeper tissues.
 Your writing that it was recognisable even after 3-4 days, even after extensive rinsing with water and soap, hints to an extended exposure of the skin towards pure phenol, which however should be avoided.
Not visible to human eye, but thanks to antiseptic power, it killed some germs, too.  Earlier in times, it was known and used as carbolic acid.

To note, the use of carbolic acid was discontinued since it inhibits the healing of wounds.

@murmansk Fahlberg found the sweetness of saccharin, Hofman the psychedelic properties of LSD by "contact" with chemistry.  And they didn't die "right on the spot" because they were both careful (dose, duration of exposure towards chemicals) and lucky (pathway of pharmacokinetics).

@paracetamol Well, within rules and reason ... http://www.cafepress.com/+aluminum_license_plate,606152319


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Unknown surface coating
Forgive me if this isn't the best forum for this question, and guide me elsewhere if so.
I have two identical smartphones, both made by Sony (Xperia z2 to be exact). One I purchased new and one second hand. I want to affix an adhesive-backed mounting system to the back of both of them.
With one phone (that I had from new) there's no trouble, everything (e.g. adhesive-backed Velcro) sticks really well. But with the other phone nothing will stick . I've checked that there's no protective overlay sheet and can only assume it's been treated with some sort of chemical coating. Every solvent I've tried to remove this coating just forms a meniscus and runs straight off. Water, detergent, isopropyl, acetone, WD-40 - nothing makes any difference. Any suggestions?

The adhesive I want to use is on a Quadlock (https://www.quadlockcase.co.uk/products/universal-adapter) which is a 'VHB' adhesive. 'Will not adhere to rubber, silicon or TPU'. Just tried the Velcro adhesive on the front of the same phone and it sticks pretty well there. Weird.

Very likely, the title of your question already hits the nail.
Since mobile phone tend to be "smart", filled to the rim with computing power and data, users wish to protect their companions with a coating that repells water.  Often, the devices are not only coated on the front side, which would at least hamper accumulation of residues of gliding fingers, but all around, including the back.  As a result, you might get a cell phone that may drop into a bowl of water, and still remain functional.
Among the different technologies available (a short presentation by CNN, back 2013 is found here), there are polymer coatings, or a mixture of anastase / $\ce{TiO2}$ and PFTE / teflon (like the coating of your frying pan) particles with some perfluorinated liquid (example).  For some, you need a larger installation that appears like a scilab, for others, you may prepare it at home with a spray can (example).  Hence the results in terms of performance, maintenance, and longevity may vary, too.  As a result, if applied both on front and back of the phone, not only water, but the components of "normal glue" (referring to what backs a velcro tape) may be repelled more or less efficiently, too.
Your observation that the coating appears little to non-affected both by water as well as iso-propanol (which for some coatings is a no-go, as it would tear the layer off), as well as for WD 40 (which may be summerised as a blend of apolar hydrocarbons (ref)) allows the speculation that the coating might be either polymer-based, or the triple mix of anastase + teflon + perfluorinated binder.  To discern between the two, remotely, is not possible.

If a mechanical, pocket-shaped mount like

(source)
or a pouch, to be bound around your upper arm while riding a bicycle, like

(source)
is not what you may deploy (or alter a little to fit your needs), and if the coating consists of the "polymer-only" variant, then maybe it is possible to establish a connection between the back plate of the phone, and either the velcro tape or something already connected to the velcro tape with the help of cyan-acrylate based gorrilla super glue.  These glues often bind very strong and to polymers, however partially dissole the polymer a little.  This chemical approach however comes with many "ifs" and uncertain outcome.

As mentioned in my comment above, I want to use a 'VHB' adhesive from 3M. I don't actually need to know what the coating is made of, but any suggestions for other solvents I haven't tried (and can get hold of reasonably easily?)

Ok, you used water, detergent, polar organic (acetone, isopropanol), apolar organic (WD40) which spans quite some range.  Ethanol probably does not offer an effect either.  Acetic acid is sometimes a surprising good solvent.  But if this all has no tangible effect, than the coating is too tight to be removed. Alternatively, exchange of the backside of the housing, if available in a repair shop or by Sony, and by design is foreseen still is the white elephant in the room.

Well, the worst coating would be a perfluorinated one. You're not likely to get much to stick to that. Silicones are more common (in general) and much more likely to be used.  MEK (methylethylketone, aka 2-propanone) works pretty well at swelling silicones, if you allow soaking. Acetone would be similar if you could allow that adequate contact time. The problem is that you should soak the surface for an hour (or more). So how do you slow the solvent's evaporation? In this type of application, using a powder, like flour (finer the better) to slow down evaporation...you may also need to cover it with some sort of film or if that's not feasible keep on replenishing the solvent. I'd definitely soak it for 20 minutes and then try rubbing the area you want cleaned with a solvent moistened rag. Methylene chloride is another good solvent for attacking silicone, but afaik it's no longer available. I don't know what kind of paint strippers you can purchase (it varies by local environmental regulations), but these are also a possibility. Note that it is possible that these could damage the finish and you definitely don't want any of these nasties to leak into your phone or onto the face! An alternate to using flour (or even tissue paper) to "hold" the solvent is to construct a dam around the area (like a circle of toothpaste or cake frosting or sealant - if you've tried a sealant and it doesn't stick very well, then perhaps you could make a dam of than and when done just pull it off. Basically, the dam would be acting as the sides of a pool the solvent goes into. Toluene is another not too bad solvent. Proper safety procedures always when working with chemicals. WD-40 is an awful idea, unless you think an adhesive will stick better to a slippery surface...


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Will mixing methyl 2-cyanoacrylate and cellulose make the cellulose harden quickly?
The other day I was gluing together cotton fabric with methyl 2-cyanoacrylate (superglue) and the fabric started heating up and then eventually began smoking. I have been working on making a 2-part plastic-like substance for personal reasons for some time now, and I know that ionic liquids pull the individual polymer chains that make cellulose apart and that once they are separated they go into solution easily. This in mind, could I get the dissoluted cellulose and add superglue to create a cheap plastic-like material?

The working principle of super glue, based on 2-cyano acrylate, is an anionic polymerisation reaction which in principle triggered by any nucleophile.  Water conveniently may be such a nucleophile (the basic anion of water, hydroxyde $\ce{OH^-}$ were even better) which is why the container's mouth originally is not only closed by a cap, but by a thin metal plate.
Cotton consists mainly of cellulose, and cellulose itself

(source)
posesses many hydroxyl groups which like water may trigger the polymerisation.  Since there are so many that are easily accessible, the process is very fast, and proceeds under liberation of much heat (i.e., an exothermic reaction).  Because of this and because the acrylate may generate a strong 3D polymer, one should not wear cloths made of cotton-based fabric, nor gloves of such.
While it is new for me to learn that ionic liquids are able to dissolve cellulose, it might be that such an unmoderated polymerisation "runs away"; i.e. already proceeds at a high rate and, because of the liberated heat and high local concentration of the nucleophiles and monomers, will become even faster, eventually is no longer manageable.  Beside potential splatter of liquid-as-such (thermal threat), the polymerisation reactions could proceed further even on and with the skin of an operator (chemical threat) if not protected properly.  Be warned, as each of the two are potentially dangerous, in addition to the potential hazard that working with ionic liquids, and chemicals in general, represent.
Wikipedia hints to this potentially relevant Critical Review doi 10.1039/C2CS15311D in Chem. Soc. Rev., 2012, 41, 1519-1537.


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Stabilisation of anions by silicon
In my lecture notes it was stated that in $\ce{Me3SiCH2SMe}$, deprotonation with tBuLi generates an anion in between Si and S. I've been not able to find the $\mathrm pK_\mathrm a$ values (not listed in Reich nor Evans).
Since Si is electropositive it should actively destabilise the anion, meaning that I'd expect to see terminal methyl group deprotonated.
No additional detail about conditions was given, but since it was tBuLi I presume low temperature (kinetic deprotonation).
Is there a reason why this deprotonation takes place? Are the $\mathrm pK_\mathrm a$ values relatively different?

Since Si is electropositive it should actively destabilise the anion

The statement above is largely one based on an inductive effect through the sigma system, in which the polarisation of the C–Si bond is considered. In this sense, Si does destabilise the anion.

Silicon stabilises anions due to negative hyperconjugation (more broadly, σ-conjugation). This is broadly speaking the  pC → σ*Si–C donation, although there is a second, arguably stronger interaction that is also involved in stabilisation of the anion.
The concept of silicon stabilising an anion is easily understood by considering a qualitative molecular orbital diagram:

 The stabilisation of an anion by adjacent sulfur, phosphorus and silicon groups. Taken from Molecular Orbitals and Organic Chemical Reactions (Reference edition), Fleming
In the diagram above the strongest interaction is the one between the p-orbital on carbon and the σSi–C orbital — this results in ψ1 being substantially lowered in energy (qualitatively, E, in the diagram above).
A second (and arguably more important) interaction exists between the low lying σ*Si–C  and the p-orbital on carbon, ensuring that ψ2 remains relatively low energy (if σ*Si–C was higher in energy, as it would be for some other elements, ψ2 would be raised higher, and this would overall counter-act the energy lowering interaction giving rise to E).
As a final point, the stabilisation is now widely accepted not to involve d-orbitals on silicon, as had been previously hypothesised. Although in theory allowing the empty d-orbitals to accept electron density from the anion would lower the overall energy of the system, the d-orbitals aren't of similar enough energies to the orbitals that the anion charge is in, making efficient overlap highly unlikely.

3rd period heteroatoms can stabilize adjacent carbanions charges via hyperconjugation (this is sometimes called negative hyperconjugation).

Consider ylides used for Corey,-Chaykovsky, Peterson, and Wittig olefinations. Granted those have formally positive sulfur, silicon, and phosphorus, but you will not see similar easy deprotonation with tetraalkylammonium for example.
The reactions of silicon-stabilized carbanions are described more fully in Clayden 1st ed., pp 1304–5 (this material seems to have been cut from the 2nd edition).


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Why does the C–C bond have extremely weak absorptions?
My book, Pavia's Introduction to Spectroscopy, tells me this about the infrared spectrum of alkanes:

$\ce{C-C}$ : Stretch not interpretatively useful; many weak peaks.

This is very interesting. What makes the C–C bond so reluctant to absorb infrared wavelengths?
I have been doing a bit of thinking lately. Bonds absorb infrared radiation that match with the harmonic vibrational frequency of the bond. I believe the extent of absorption is based on the tendency of the bond to get vibrated. For some reason, the $\ce{C-C}$ bond is very reluctant to absorb the radiation, and I believe it's because the bond is too "rigid".
If a bond is too rigid, it must show large change in intermolecular force per change in bond distance. In other words, I believe, the value of
$$\frac{\mathrm dF}{\mathrm dr}$$
is quite high. In such a case, the bond wouldn't have a tendency to compress and expand, since the bond has such a high tendency to oppose such changes. Assuming the bond to act as a spring, I have,
$$F=kr$$
where $k$ is the spring constant of the bond-spring. Differentiating this w.r.t $r$, we get:
$$\frac{\mathrm dF}{\mathrm dr}=k$$
So, this makes me conclude that a higher value of $k$, implies a more rigid bond. I also know that
$$U=\frac{1}{2}kr^2$$
Plugging the values of $U$ ($\pu{=346000J}$) and $r$ ($\pu{=1.54 \times 10^{-10}m}$) for the $\ce{C-C}$ bond, I get,
$$k_{\ce{C-C}}\pu{= 2.92Nm^-1}$$
If I compare this with the value of $k$ for the $\ce{C=O}$ bond (one of the most prominent absorptions), I get,
$$k_{\ce{C=O}}\pu{= 1.11Nm^-1}$$
The $k_{\ce{C=O}}$ is just 3 times smaller than $k_{\ce{C-C}}$, but the existence of $\ce{C-C}$ absorption is rather non-existent.
I can conclude one thing for sure, that all that I have thought before is completely wrong.
So here's my question. Why does the C–C bond absorb extremely weakly? And what's wrong with what I have thought about above?

For vibrational spectra the primary transition under investigation is the $v = 1 \leftarrow 0$ excitation (because $\hbar\omega >> k_\mathrm{B}T$, so excited states have negligible thermal population). The intensity of the transition depends on the transition dipole moment:
$$R_{10} = \langle 1 | \hat{\mu} | 0 \rangle$$
in that the intensity $I \propto |R_{10}|^2$. Taking a Taylor expansion of the operator $\hat{\mu}$ with respect to $q$, the vibrational coordinate, at $q = 0$, we have:
$$\hat{\mu} = \mu_{q = 0} + \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} + \cdots$$
and plugging this in we get
$$\begin{align}
R_{10} &= \langle 1 | \mu_{q = 0} | 0\rangle + \left< 1 \middle| \hat{q}\left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0} \middle| 0 \right> + \cdots \\
&= \mu_{q=0}\langle 1 | 0\rangle + \left( \frac{\mathrm d \mu}{\mathrm d q}\right)_{q=0} \langle 1 | \hat{q} | 0 \rangle + \cdots
\end{align}$$
where we took out the values of $\mu$ and $\mathrm d\mu/\mathrm d q$ evaluated at $q = 0$ because those are just ordinary numbers. The first term in the Taylor expansion vanishes because $|1\rangle$ and $|0\rangle$ are orthogonal (a key property of the harmonic oscillator, which is a good model for the low-lying vibrational states at the bottom of the well), and if we neglect the second- and higher-order terms, we have†
$$R_{10} \propto \left( \frac{\mathrm d \mu}{\mathrm d q} \right)_{q=0}$$
The RHS represents how much the dipole moment $\mu$ changes upon vibration of the bond (which is measured by $q$). As such:

for something like a C–C bond, where vibration leads to nearly no change in the dipole moment, this quantity is small, the transition dipole moment is small, and the peak is weak.
On the other hand, the stretching of the C=O unit leads to a large change in dipole moment and the corresponding peak is therefore stronger.
The C=C stretch of alkenes tends to be weak as well, unless it is conjugated to an electron-withdrawing group like a carbonyl group. This can also clearly be linked to the magnitude of $(\mathrm d \mu/\mathrm dq)_{q=0}$.
And incidentally, this is also why molecules like $\ce{H2}$ or $\ce{O2}$ cannot be studied by IR spectroscopy (no matter how much you vibrate, there is no dipole moment), and also why the symmetric stretch of $\ce{CO2}$ is not IR active. In all of these cases $(\mathrm d \mu/\mathrm dq)_{q=0}$ is identically zero.

Unfortunately most of what you've written is kind of off the mark.

It is indeed "reluctant" to absorb the photon, but that is not because of the stiffness of the bond. If the bond is stiffer ($k$ is larger), then it simply needs a photon of shorter wavelength (as the vibrational frequency $\omega = \sqrt{k/\mu}$ will be larger). Instead it is related to the poor interaction of the C–C vibrational mode with the oscillating electric field of the photon. This is going into stuff that I don't know about, so can't explain much more (search "Fermi golden rule" in Atkins MQM or some other quantum book).
$D_\mathrm{e} = kr^2/2$ is not a valid equality here as it simply makes no physical sense. If the energy of the oscillator is at $D_\mathrm{e}$, then physically speaking, the corresponding bond length $r$ should be infinity, as the bond is dissociated. Furthermore, the equilibrium bond length which you used in the calculation ($r_\mathrm{e} = \pu{154 pm}$) refers to the middle of the potential well. If you really wanted to model the C–C bond as a classical oscillator, then your equation should read
$$U = \frac{1}{2}k(r - r_\mathrm{e})^2$$
because the thing that gets squared is the displacement from equilibrium.
In that calculation you used the dissociation energy for one mole of carbon–carbon bonds, and that doesn't make sense if you're trying to calculate the force constant of one carbon–carbon bond.
Even changing no. 3 wouldn't make any difference, because the harmonic oscillator model simply does not work near dissociation (in fact if a bond were to be a harmonic oscillator, it would never dissociate). Therefore bond dissociation energies cannot be treated within a harmonic model; you have to use something else such as a Morse potential.
A typical value of a C–C force constant is on the order of $\pu{400 N m-1}$. A better way of calculating the force constant is with
$$\bar{\nu} = \frac{\omega}{2\pi c} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}} \implies k = \mu(2\pi c\bar{\nu})^2,$$
where $\bar{\nu}$ is the IR wavenumber and $\mu$ is the reduced mass of the oscillator. A very rough calculation is as follows. Plugging in $\mu = \pu{10^-26 kg}$ (for a $\ce{C-C}$ bond), $\bar{\nu} = \pu{1000 cm^-1} = \pu{10^5 m^-1}$, and $c = \pu{3 x 10^8 m s^-1}$, we have $k = \pu{355.3 N m^-1}$. Correct order of magnitude, at least.
It's also nearly impossible that $k_\ce{C=O} < k_\ce{C-C}$, as force constants tend to be correlated with bond order. A quick Google search shows that $k_\ce{C=O}$ is on the order of $\pu{1000-2000 N m^-1}$. (If you go searching for this, you may wish to know that $\pu{1 mdyne/Å} = \pu{100 N m^-1}$, as CGS units are widely used.)

† Of course, there's an additional point that must be made, which is that $\langle 1 | \hat{q} | 0 \rangle \neq 0$, but you'll have to take my word for it as I'm not keen on proving it here. It is most readily shown using ladder operators, and it should be covered in any QM text under the harmonic oscillator.

Wow, lot's of involved QM, that this really should have been over on physics.se. So, as a physicist, I'd mention the chemists' point of view, that C-C bonds are normally very much covalent, which is the reason that C-C vibrations involve very little dipole moment. Your first bullet point should really be the centerpiece of the answer. On top of that, it could be explicitly stated that IR absorption spectroscopy is sensitive to dipole-active excitations, which is so far only implied in the discussion at the top of the answer. +1 ofc

If we assume the simplyfied description, i.e. atoms being tethered together by springs, the spring constant allows describe where (energy wise, along the scale of wavenumbers, the abscissa in the specta plot) the absorption becomes observable.  This however does not say a iota if the absorption band is strong, leading to a low remnant transmittance (the ordinate in the spectra plot).
From a practioner's perspective, C-C bonds often are referred as skeletal vibrations, because they represent vibrations of the backbone of the molecules characterized.  As such, overall the span where to find these absorptions is rather wide ($\pu{1500 - 600 cm^{-1}}$), which contasts with an easy-to-spot absorption band like the one of $\ce{C#N}$ around $\pu{2260 cm^{-1}}$, regardless if you expand the spectral representation, or not.  The better known example are the ones to attribute substitution patterns of arenes

(source)
but as these hints are no longer taught as regularly as in the past, often need an expansion (who routinely looks at the bands at $\pu{900 cm^{-1}}$ and below as well?), and most importantly, since NMR spectroscopy provides this information more rapidly, the focus of attention while inspecting an IR spectrum often shifts to more prominent pattern.

Also in your analysis, you should consider Raman spectroscopy as a complementary method.  Raman is also vibrational, however what show up weakly in the IR tends to show up in the Raman.  In your example c-c bonds are very prominate in the Raman despite being weak in the IR.  Because in the course of a vibration, IR spectroscopy is sensitive to the change of dipolar moments, and Raman spectroscopy is susceptible to the change of polarizabilities of the "electron cloud" involved.


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How to measure distance along solvent-accessible surface of protein complex in PDB file?
I am looking for a straightforward way to measure the distance between two active sites in a protein complex (PDB file).
I need the distance along the solvent-accessible surface of the protein and not the linear distance that would go through the protein, and which the distance wizard will give.
I would prefer a solution using PyMOL, although I would consider alternatives as well.

I would use linear distances between a suitable amount of fixed points as an approximation. You cannot measure exact distances along a rough surface, as the distance will vary depending on minimum measuring size. This is known as the coastline length paradox in math.
In theory the length would approach infinity as the measuring size approaches zero. In practice you can get around the issue if you use a step size compatible with the smallest stable feature.
A suitable measuring size would probably be in the order of ${\sim}\pu{10^{-11} m}$ or $1/10$ the size of an carbon atom.

As Jeppe Nielsen says in the other answer, the distance along the surface is not defined. I would tackle this visually so that the reader can convince themselves that your distance makes sense. Cut the protein in two parts (with a plane that goes through both of the active sites and the "center" of the protein). Then (in 2D) trace a path and measure its length.
Here is an example using xanthine dehydrogenase. Instead of cutting the protein, I used slabbing to remove the top part. This ensures an orientation where the plane is in the plane of the paper.

Using the linear distance shown as a scale bar, you can estimate the path length. This is all approximate, as it should be because the problem is ill-defined. If I choose a different orientation or slab, I will get a slightly different answer. You could also first select some surface atoms to define the path, and then estimate the distance by adding up atom-pair distances. Then, you could visualize the path in a 3D figure and state the distance.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Does cold bluing steel prevent rust?
I've recently purchased a Damascus knife and am patiently awaiting it's arrival in the mail. I have been researching the potential care and maintenance habits for this craftsman blade, hoping to keep it in pristine condition.
I'm wondering what measure(s) could be taken to prevent a Damascus knife from rusting. I will be crafting the handle from my own scales and pins upon receiving the item, so any manipulation done to the steel can be preformed upon the blade in its entirety. The alloys used are a mix of 1095 and 15n20 steels respectively.
One of the potential means that I've entertained the thought of is bluing, AKA a black oxide finish. Bluing is a method commonly used on gun metal in which various gun parts are submerged in caustic hot salt baths for a period of time in-order to attain the coat. This isn't the only method to obtain the oxide layer, it's but one of many. Some sources have stated that bluing provides rust resistance while others are skeptical since most oxide layers as a result of the process are merely 2.5 micrometres (0.0001 inches) in thickness.
The method I'm most interested in potentially trying is cold bluing. Cold bluing can be achieved either by coating the high-carbon steel in mustard, painting it in a commercially sold "perma-blue formula", or submerging it in an acidic at-home substance such as vinegar or lemon juice (boiling vinegar to enhance the effect).
Here is my question Ultimately TL:DR Can a true rust resistant layer be obtained via cold-bluing methods on high-carbon Damascus steel (1095 & 15n20)? Will the acidic substance such as vinegar or "perma-blue formula" simply eat through the steel with time (while I'm letting the bluing process take place before I've neutralized the blade with something such as baking soda), or will it develop a thicker layer of patina oxide layer? Some sources say the bluing layer is rust, therefore rust can't develop on top of rust. Others say bluing is purely cosmetic and can not prevent rust.

I don't think you need to change anything in the blade, just be careful that it won't have occasion to rust. Also Damascus steel has distinct pattern which would be lost if you'd oxidated it.

Store it reasonably dry (=not under the kitchen sink) and wipe the blade with an oily cloth from time to time, clean it after use and not on the next day, and use only water to clean it (or oil it directly afterwards if you used soap), and it should never rust.

i would recommend ezox firearm lubricant, it will leave a dry film of Teflon on the blade till you use it, after use you have to reapply it but that is as easy as wiping oil on it and letting it sit for a few minutes. (its also very good for cleaning off hard to get off things ie tape adhesive/sap/animal blood)

It provides limited corrosion protection. I would suggest just putting oil on it; more corrosion protection and it will not obscure the surface pattern which is  most all that separates Damascus steel from ordinary steels. As I remember regular "bluing" is a complex of ferocyanides ; Oxides of iron are red and black , brown rusts are hydrated oxides. I have now looked at Wiki and see some of the problem.  Molten cyanide salts are use to carbonitride steels;I believe these leave a blue surface- you can't do that. Another source of blue ( and straw and purple) color is different thicknesses of very thin oxide layers. I have a chart showing one hour at 540 F (in air) , or 10 minutes at 590 F, or 20 hours at 485 F will produce a blue. Higher temperature and longer time give thicker oxide. The color of the oxide is determined by the thickness. Assuming the blade is hard like Rockwell C 60, these temperatures will cause some softening. The other methods sound like they will etch the steel . Same conclusion, oil it.

I wouldn’t personally blue a Damascus knife, however I have used commercial blue on high carbon knives and it has added some rust protection a substantial amount more protection I should say than an untreated HC blade.


text:
Transforming AO basis electron repulsion integrals into molecular spin orbital basis?
I am attempting to do project 5 on Daniel Crawford's Group's GitHub. I'm having some trouble reproducing the MP2 energy, which is given in project 4.
In project 4, the MP2 energy is given as:
$$ E_{MP2} = \sum_{ij}\sum_{ab} \frac{(ia|jb)[2(ia|jb)-(ib|ja)]}{e_i+e_j - e_a - e_b}$$
Where i, j are occupied orbitals and a, b are unoccupied orbitals. The ERIs given above are in the MO basis. Crawford's website provides the ERIs in the AO basis. It describes that you can convert the AO basis ERIs into the MO basis ERIs by summing over the MO C coefficients, which I have prepared with SCF, in this fashion:
$$ (pq|rs) = \sum_{\mu\nu\lambda\sigma} C^p_\mu C^q_\nu C^r_\lambda C^s_\sigma (\mu\nu|\lambda\sigma) $$
Where $p, q, r, s$ are  MO basis ERIs and $\mu \nu \lambda \sigma $ are AO basis ERIs. (This is the inefficient formulation of this sum but it looked easier to type in LaTeX; the better version is on the project 4 page which is what I actually coded). I was able to prepare the MO basis ERIs and the MP2 energy correctly  (the website has the right answer) in this method.
On to project 5, where my problem is. Project 5 is to make a CCSD code which uses the spin MO basis as opposed to the non-spin MO basis that MP2 uses, but an intermediate part of that is to show that you can calculate the MP2 energy using these spin MO orbitals. However, I am having trouble doing so. I suspect my problem is that I am not converting my ERIs properly, but I am not 100% sure.
I am trying to use this formula for the MP2 energy (from project 5):
$$E_{MP2} = \frac{1}{4}\sum_{ijab}\left<ij||ab\right> t^{ab}_{ij}$$
I believe it is the same as project 4, where $i,j$ are occupied orbitals, $a,b$ are unoccupied. The formula for $t^{ab}_{ij} $ is given on the project 5 page. When I calculate the MP2 energy with the above formula I get something quite different than what I get from the first MP2 energy formula (from project 4)
I think my problem is that I am misinterpreting what $\left<ij||ab\right>$ means, but I am not sure in what way. The site provides an example code to prepare these (given here), but it's in C++ and I am just doin this in python to practice, so I am making my own. Also, following the advice on the page, I am storing all the ERIs (AO basis, MO basis, spin MO basis) as large 4 index tensors and not using any fancy indexing, just to make it easier to debug. The example code seems simple enough that I think I have reproduced it correctly (except for maybe storing the results), so I think my problem is not a coding one, rather that I am misunderstanding some idea or notation. I am doing three things to obtain it from the given AO ERIs:

Construct the MO ERIs from the AO ERIs, identically to in project 4 (see the four index sum over greek letters I put above). I know I am doing this right as I used these to get the MP2 energy in project 4 correctly.

Construct the spin MO basis ERIs. I use project 5's advice of treating spin as the number of the function (ordered by energy). I have:
$$ \left<pq|rs\right> = (pr|qs) \delta_{r\%2, p\%2} \delta_{s\%2, q\%2} $$
Where $\left<pq|rs\right>$ is the spin MO integral and $(pr|qs)$ is the spatial MO integral. I have the orbitals sorted by energy (the energy given by SCF) and numbered. Say p is #2 and r is #3; then p mod 2 = 0 and r mod 2 = 1, so p would be spin up and r would be spin down, so the spin orbital is killed. This is the method that project 5 is using to simulate spin (at least, that's how I'm interpreting the example code. Could that be my error?).

I anti-symmetrize the spin MO integrals like this:
$$ \left<pq||rs\right>  = (pr|qs) - (ps|qr) $$
I didn't write them here, but the two integrals on the right side have the spin deltas built in to my code as well. Since all the ERIs in the CCSD equations are anti symmetrized ones, I store the 4 index ERI tensor in this form.

I think my problem may be that I am unconsciously mixing up physicists and chemists notation somewhere in my code when I do step 2 in my conversion? In project 5's example code, they use an indexing scheme to read in the spatial MO ERIs, which I have not implemented or really understood, but I think I am emulating correctly by just using a big 4 index tensor to store the ERIs. In my code, I am reading in $(pr|qs)$ and $(ps|qr)$ like this:
prqs = ERIs[p][r][q][s] * (p%2 == r%2) ...
psqr = ERIs[p][s][q][r]

I didn't type all of them but the p%2 == r%2 thing is meant to kill it if the spins are not equal (aka, kill it if the number of the index of the orbitals are of different parity). Here, prqs is in chemists notation (which is what I assume they mean by using parenthesis) and ERIs is my four index tensor (using numpy) of the spatial MO ERIs. Then I antisymmerize them and store them like this:
spin_ERIs[p][q][r][s] = prqs - psqr

by spin_ERIs[p][q][r][s], I mean $\left<pq||rs\right>$. Could my problem be that since I am switching between chemists and physicists notation (again, I assume the <> means physicists notation and () chemists) I need to switch some indices around, or something? It seems like that is built in already, since we have $\left<pq||rs\right> = (pr|qs) - (ps|qr)$, not $\left<pq||rs\right> = \left<pq|rs\right> - \left<qp|rs\right> $or something, but I'm not sure if that's the problem, or something else entirely.
I guess this problem is pretty convoluted but basically, can somebody tell me that I am thinking the right way from my steps 1, 2, 3? Thanks.

Have you tried constraining yourself to closed shell RHF only? It would eliminate the complications due to spin and if you get right results there, you know your problem is in your handling of the spin orbitals.

I am doing so. The test case on the website is water, using STO-3G. So 7 basis functions, 10 electrons. 5 lowest orbitals doubly filled, top 2 are unoccupied.

I am not sure whether this is it, but I think your equation in step 2 should be:
\begin{equation}
\langle pq|rs\rangle = \left(\frac{p}{2}\frac{r}{2}\middle|\frac{q}{2}\frac{s}{2}\right) \delta_{r\%2, p\%2} \delta_{s\%2, q\%2}
\end{equation}
The idea for $(pr|qs)$ is to store $\frac{N}{2}$ spatial orbitals representing $N$ spin orbitals, by remembering that each orbital appears twice (one $\alpha$, one $\beta$). So the indices $p,q,r$ and $s$ go only up to $\frac{N}{2}$ for the spatial ERIs.
Note that we use an integer division here: $\frac{3}{2}=1$ and $\frac{2}{2}=1$. Thus there are always two spin indices mapping to the same spatial index. One might also write the spin indices as compound indices, consisting of one spatial index and the spin state: $0\alpha, 0\beta, 1\alpha, 1\beta, \dots$.
But for the spin ERIs $\langle pq|rs\rangle$ the indices run up to $N$. So with $(pr|qs)$ you would try to access elements outside of it (as soon as the indices become greater then $\frac{N}{2}$). You can also find this in the code of your linked hint.
Furthermore, I think you are mixing up the different notations. Hopefully this will clarify it: According to the book by Szabo and Ostlund the notation for these integrals should be:

$\langle ij|kl\rangle$ is physicists' notation for spin orbitals ($i$ and $j$ are bra-vectors of different electronic coordinates)
$[ij|kl]$ is chemists' notation for spin orbitals ($i$ and $j$ are bra- and ket vector of the same electronic coordinate)
$(ij|kl)$ denotes spatial orbitals following the ordering of the chemists' notation

There is no "physicists'-like" notation for spatial orbitals.
We can relate chemists' and physicists' notation by
\begin{equation}
[ij|kl] = \left<ik|jl\right>
\end{equation}
Also it defines
\begin{equation}
\langle pq||rs\rangle = \langle pq|rs\rangle - \langle qp|rs\rangle
\end{equation}
which means your
\begin{equation}
\langle pq||rs\rangle = (pr|qs) - (ps|qr)
\end{equation}
should be wrong. But maybe you are already using the correct thing in your code there, just with the wrong notation?

Well, I used:
$$ <pq||rs> = (pr|qs) - (ps|qr) $$
Because that seems to be what the "hint script" is indicating; that that I mean, that is what it uses for "value1" and "value2". I suppose I am not 100% sure that the original integrals provided (the spatial AO basis integrals) are in chemist's notation or not; I assume they are because they use (), but maybe that's just the notation on the site?
(reply 1/2)

As to:
$$ <pq|rs> = (\frac{p}{2} \frac{r}{2} | \frac{q}{2} \frac{s}{2}) ... $$
I was assuming they only use p/2, etc in their indexing scheme (which I did not implement). My orbitals (spatial, from scf) are numbered 0, 1, 2... 6, for total of 7 orbitals from 7 basis functions. If p = 1, then what is the 0.5th orbital? Doesn't seem to make sense. Or are you saying 14 total orbitals, such as "0a, 0b, 1a, 1b", both an up/down spin of the original 7 HF orbitals? I did not think of that previously, although it now seems obvious; I will try implementing that and see if it works.
(reply 2/2)

Yes, the spin orbitals would be something like "0a, 0b, 1a, 1b, ..." but since we can only index with integers this has to be translated to "0, 1, 2, 3, ...". It is important here that the $\frac{p}{2}$ division will be an integer division, e.g. 3/2 = 1 and 2/2=1, so there are always two spin indices mapping to the same spatial index.

About your first comment: You're right, the code in the hint already subtracts the exchange integral. Which means it takes care of step 2 AND step 3.

Let me make sure I have a correct thought process (because this is how I coded it):
Since the spatial orbitals were 0 to 6, with 10 electrons, the bottom 5 orbitals (0 to 4) are occupied. I have $\epsilon_0$ though $\epsilon_6$ from scf. Am I right in thinking that, when I change to spin orbitals (which are 0 to 13), it's still the bottom 5 occupied, but those  5 have energies:
$$ E_0 = E_1 = \epsilon_0 $$
$$ E_2 = E_3 = \epsilon_1 $$
$$ E_4 = \epsilon_2 $$
Where latin E is referring to an index of the 14 spin orbitals and greek $\epsilon$ is that of the original spatial orbitals.

Your energy assignments are correct, but keep in mind that spatial orbitals can be doubly occupied and spin orbitals only singly. And we do not change the number of electrons in the system. So 10 electrons can fill 5 spatial orbitals, but we need 10 spin orbitals (0 to 9) for them!

Ah, and with that (remembering that the spin orbitals are 1-slot, not 2-slot), I am getting the right MP2 energy... can't believe I overlooked such obvious things, but thank you.

@iammax Did you manage to reproduce the values given for $\ce{H2O}$? I tried to implement this myself (I already had MP2/3 but no CC), but my equations do diverge to infinity. Maybe we could compare the intermediates?

Sadly, no. I'm still periodically working on project 5. I also get some divergence. I'm probably calculating some of the intermediates wrong, but I'm not sure which ones.
If you are interested, I get these values before my divergence check kicks in:
$$
Iteration:     E(cc)
$$
$$
000:           -2.8203528845609029
$$
$$
001:           -4.3827073354424568
$$
$$
002:           -3.5161279486121662
$$
$$
003:           -6.4768663770805199
$$
$$
004:           -43.4231792529198088
$$
$$
005:           785.2196041556109094
$$
Are you getting the same? Sorry, bad formatting

These are way too large. Mine is now converging at $-0.0726$ instead of $-0.07068$.

I think the answer given for Fedoran is what you are asking for, on a site note, for the implementation you are doing take a look at Coupled Cluster with Singles and Doubles (CCSD) in Python , I think I found a few mistakes in it, but it gives the overall most simple implementation. A tested to work implementation that is very equivalent to the previous link can be found here, python CCSD(T)


text:
Why do compounds form if overall energy of free ion increases according to CFT
According to crystal field theory, a free ion has certain energy and its energy increases when it is symmetrically surrounded by ligands. Now the degenerate d orbitals split into two new set of degenerate orbital, (eg and t2g) of which t2g has lower energy than symmetrically surrounded ion (the Bari centre) but still the energy of t2g is greater than the energy of the FREE ion according to diagram. (And energy of eg is much greater than free ion and Bari centre)

Now, my question is, if the energy of the overall ion afterwards is greater than the energy of the free ion, then why are coordinate compounds considered more stable than the free ion?( according to books)
Or why do they form?


text:
How do scientist obtain the value of enthalpy of formation?
Do they do it experimentally or do/can they use bond enthalpy to figure it out? For example $\ce{C2H4 + H2 -> C2H6}$. The enthalpy of formation, $\Delta_fH$ of formation for the product is listed as -84, however if I use bond enthalpy to calculate the $\Delta_fH$, it gives be a value of -222, which is considerably different. Shouldn't the difference between bonds formed and bonds broken account for the heat given off my the reaction? With this discrepancy I can only conclude that I mixed up the two concept.
Also if $\ce{C2H4}$  enthalpy of formation $\Delta_fH$ equals to 52,  $\ce{H2}$ = 0 and $\ce{C2H6}$ = -84, the difference between the product and reactant is -136. I am confused about the 3 numbers (-84, -222 and -136) and their significance.
Also, how is $\Delta_fH^\circ$ different from $\Delta_RH^\circ$ of reaction? When do I use each one?
I would really appreciate it if someone can clear this up concisely

You seem to have more than one question here. Try to limit your post to one  question that addresses your main area of misunderstanding and, if that still doesn't answer everything, ask a new question.

@Jim Roberts Welcome to ChemSE.  For future reference, be invited to familiarize yourself with mathjx and \mhchem that will help you considerably in formatting questions, as well as answers on ChemSE. To start, have a look at this compilation (https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88); and graduate, at your pace, with this (https://chemistry.meta.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know?noredirect=1&lq=1).

You'll likely get more traction on this question if you provide a table of the relevant values with units. People are less likely to help if they have to look everything up to double check your work.

There is a consent to start with: the standard enthalpies of formation of elements (like $\ce{H2}$), at standard conditions ($\pu{1 bar}$, $\pu{298.15 K}$) are set to zero.  If there are several forms of one element, than the most stable polymorph is chosen as reference (except for phosphor, ref).
More generally, to determine the $\Delta{}_fH^\circ{}$ of other compounds, you may embark experiments by calorimetry. By means of a calorimeter, you are able to record the energetic changes that occur in the course of a reaction.  For example for the exhaustive combustion of carbon with oxygen:
$$\ce{C + O2 -> CO2}$$
all the heat recorded is due to the formation of carbon dioxide. Since this reaction equation was balanced that it equals only one mol of product ($\ce{CO2}$), the more general standard reaction enthalpy ($\Delta_{\text{R}}H^\circ{}$)  equals to the standard enthalpy of formation ($\Delta_fH^\circ{}$).  And by one experiment, after an other, the body of experimentally determined enthalpies grows.
Based on other experiments than combustion, one was able to determine how much energy is necessary to break a bond between atoms which is called bond dissociation energy. Conversely, this value expresses how much energy over all is released if two atoms establish a bond with each other.  In some instances, these energies are (comparatively) small, for example single bonded $\ce{Cl-Cl}$ ($\pu{242 kJ/mol}$), in other instances (comparatively) large, like triple bonded $\ce{N#N}$ ($\pu{945 kJ/mol}$).
If one knows the structure of a compound, which atoms are connected with each other and how (single, double, triple bond), than these bond dissociation energies may be used to estimate the standard enthalpy of formation of the new compound, for example by an increment system. Alternatively, by virtually setting up for example a combustion reaction of exactly one mole of the new compound into $x$ moles of $\ce{H2O}$ and $y$ moles of $\ce{CO2}$.  Then, the calculation to perform is
$$\Delta_{\text{R}}H^\circ{} = \sum_{\text{all products}}{\Delta_fH^\circ{}} - \sum_{\text{all starting materials}}{\Delta_fH^\circ{}}$$
where $\Delta_{\text{R}}H^\circ{}$ again is the overall reaction enthalpy.  Since these estimations are based on some empiric data, they do not a substitute for an experimentally determined value for your newly prepared compound. Often, increment systems leave some of inter- and intramolecular interactions unconsidered (e.g., ring strain, possible existence of polymorphism) which then yields a difference between the estimated, and the actual value of $\Delta_{\text{R}}H^\circ{}$ or $\Delta_fH^\circ{}$.


text:
Interaction of Tris HCl and EDTA with Nylon
I recently learned that we're storing some of our buffers that contain Tris HCl and EDTA in nylon containers.  I can't find anything definitive but I have a suspicion that those components can interact with the nylon.  I was wondering if it is a a good idea to use nylon or if there will be significant reactivity with the components.

Hello and welcome to Chemistry.SE. Please feel free to take the short [tour] or visit the [help] if you have any questions about the site. Best of luck with your question.

$\ce{HCl}$ itself will dissolve any nylon at room temperature, because of strong acidity ($K_\text{a} = \pu{1e6}$). When you have a mixture of $\ce{HCl}$ and EDTA, so the strength of $\ce{HCl}$ will reduced, but it also depends on the concentration of $\ce{HCl}$ in this buffer and room temperature.
If you store this buffer in nylon container, so this container can not be melted anyway, I believe, but for long term use, I recommend that you shall use other stuff to contain it.

It seems you dunno what tris is...


text:
Why is carbonic acid a weaker acid than acetic acid?
Is it because the $\ce{OH}$ group next to $\ce{COOH}$ in carbonic acid donates electron density to the $\ce{COO-}$ by resonance, making it a stronger base, compared to a weaker electron donating inductive effect of $\ce{CH3}$ and thus its conjugate acid is weaker?
What does the resonance contribution look like in this case?
What makes resonance a stronger factor than induction?

Before you ask "why", ask "if".
Carbonic acid is not weaker than acetic. In fact, it is stronger, just as one might expect by looking at that extra electron-withdrawing substituent. The trouble is, carbonic acid is also pretty unstable. At any given moment, most of it exists as $\ce{CO2}$. This leads to much lower apparent acidity.
Think of it this way: you add some acetic acid to a carbonate salt. An equilibrium sets in, much closer to the starting compounds than to the products. In other words, you'll have just a very tiny amount of $\ce{H2CO3}$. But as soon as you have it, it starts decomposing, thus shifting the equilibrium more and more to the right.
Look again at that Wikipedia page, it says:

3.6 ($\rm pK_{a1}$ for $\ce{H2CO3}$ only), 6.3 ($\rm pK_{a1}$ including $\ce{CO2(aq)}$)...

This is something new! Thank you for this information. Although I'd like to ask, doesn't OH DONATE electron density by resonance, or is that not possible in this case?

Even if it does, that doesn't outweigh the electron-withdrawing effect.

@IvanNeretin but isn't inductive effect a weak effect and resonance a strong effect?

@Lalit No. Both can be either weak or strong. In this particular case, the inductive effect is relatively strong, and the resonance relatively weak.

@IvanNeretin Yeah, so it's an exception :-)

The conjugate base of acetic acid is more stronger than carbonic acid because carbonate has $\ce{-OH}$ which shows +R effect but acetate has $\ce{-CH3}$ which shows +I effect.
Therefore, the electron density is little stronger in case of acetate.


text:
What is the most likely reaction between minocycline and tramadol in physiological conditions?
Minocyline (left) and Tramadol (right) are two commonly prescribed medications containing typical organic functional groups.
What would an expert chemist surmise as the first or most-likely reaction involving these drugs in the human body, if any? For example, in the stomach, in a highly acidic environment?
I realize the chemistry here is obviously complex, so an answer excluding biological components is fine. e.g. what would these compounds do in a solution of $\ce{HCl}$ alone?
I'm not an organic chemist, but my guess is that $\ce{HCl}$ can protonate an amine to encourage some kind of reaction.
In any case, some research has shown that the drugs may have a synergistic effect for pain relief: https://www.ncbi.nlm.nih.gov/pubmed/22964800.

You're looking for reactions between these two substances under physiological conditions? Or do you just want to know how they are individually metabolized?

Reactions between.

I would expect that you'd protonate the nitrogens and for ammonium ions. They don't quite have the best functional groups for cross reactivity...

There's no reaction. If there was it would be used purposefully, or dangerous.


text:
Why would the specific heat capacity of water be used as the specific heat capacity of zinc sulfate?
I could not find the specific heat capacity of zinc sulfate and saw that people use water's heat capacity instead. Why is that?

If the substance is dissolved in water, it's very typical (and more correct) to use the heat capacity of water in lieu of the pure salt, acid, base, etc.
The reason is simply that, when in solution with water, there is much more water than the solute. It's an approximation, but a decent one. This is common practice in freshman-chemistry lab courses which I've taught.
It's difficult to find heat capacities for pure salts such as zinc sulfate because the means by which we measure it is typically through a reaction, and typically a salt is dissolved in solvent (e.g. water) in order to perform a reaction. That is to say, reactions of solids are just less common than liquids, gases, and solutions.
That said, the NIST database reports a $c_{p}$ for zinc sulfate of 99.01 Jmol$^{-1}$K$^{-1}$ at 298 K.
Looking at the original paper, I found this plot of $c_p$ as a function of temperature. At room temperature, you can see it is roughly 100 Jmol$^{-1}$K$^{-1}$.
Stuff like this can be easily obtained from citrination.com search engine.