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Z-Matrix Generator for Gaussian
I am looking for a way to convert XYZ coordinates into z-matrix for Gaussian input files. I need to specify certain coordinates in the matrix, therefore, the automatic built-in function from gaussview isn't of much help. Do you know of a program that can help me?
Thanks in advance for your help!

molden can write z-matrices, you can load in a xyz file and save it as z-matrix

Chemcraft has a very descent built-in constructor of Z-matrices.

+1 for chemcraft. you can define the coordinates you need and even let the software do the rest.

Try Open Babel, the windows GUI has a lot of features that may be of help depending on what are those specifics you need in your Z-matrix.

Avogadro, nice, clean and very user friendly.
Just draw the molecule, then navigate to menu Extensions → Gaussian.
Here you can format and select z-matrix for Gaussian input.

As mentioned in the comment by Fl.pf, the program molden has a very powerful ZMAT Editor:

You can hand pick the atoms that define bond lengths, angles and dihedrals
You can save as: GamessUS, GamessUK, Gaussian z-mat, mopac, xyz

If you want to convert back to xyz with obabel (for example after replacing a variable in the zmat with a range of values), use mopac (mopin in obabel).
Molden takes a bit to get used to, but this task can be done quite well there.

direct description molden z-mat: http://www.cmbi.ru.nl/molden/zmat/zmat.html ; open babel xyz: http://openbabel.org/wiki/XYZ_(format) ; Redefining a z-matrix from an automatically produced one in molden is a bit of nightmare though

Gaussian provides all you need to do this with the newzmat program.
It literally does all this, no need for 3rd party software.
I found it works best to put the Cartesian coordinates in a file by themselves and save it as xyz, then in the command line just run (Note the atom types need to be in front of the coordinates so
i.e. C   XXXX YYYYY ZZZZZ would be one line in the file if your first atom was carbon)
newzmat -ixyz name_of_xyz_file

It will output name_of_xyz_file.com in z-matrix format.

This is not particularly useful to those that cannot afford Gaussian, and certainly not to those of us who are unable to use it because we are developers of other codes.  It would be nice to add to your answer that this is a commercial program with a restrictive license.

Sorry but the question is implicitly involves Gaussian. This answer is quite useful.

I appreciate that perspective but I stumbled on this looking for a generic solution for z-matrices (other codes accept formats that are trivially difference from Gaussian’s) and the other responses were quite useful to this end.

+1 to them then. I already gave them my +1. I think it is crazy to add clauses to answers. Should I advertise your software as well? Gaussian bad, X good. Let me know what X is.

X is any software that’s accessibly to anyone reading StackOverflow.

Isn't gausview using this exact tool to built the z-matrix in the first place? The one tool that the user didn't find helpful in some cases. Nevertheless, I think you're quite correct for most cases, like generally. / While I agree that it would have been nice to say that Gaussian is a commercial software, that job would have been the OPs job as they tailored the question specifically to it. @JeffHammond


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What physical/chemical reasons could have caused Dorian Gray's picture to degrade?
In the novel 'The Picture of Dorian Gray' by Oscar Wilde, Gray locks away a portrait of his in the attic for 18 years and over that time the picture degrades to an almost unrecognisable version of its former self. The novel explains this as the fulfilment of a wish by Gray to have the portrait age instead of himself but I have been wondering recently, are there be any natural reasons that could explain this transformation if a painting is locked away for a large period of time?

Good thing you didn't ask about the other prominent usage of chemistry in the novel.

Preface: So far, I do not know the novel you refer to, and assume the portrait is a classical oil painting.
Such a painting consists of several layers of products applied.  Similar to a wall in your living room, the canvas is prepared to accomodate the pigments dispersed in linseed oil, what in other fields of application would be called a "primer".  Then there are multiple layers of the painting itself, topped by the varnish.  To mention a few of the potential issues,

Pigments may change their colour upon exposure to air. Pigments like lead white turned over time black, for example.
Pigments may interact with the oil, forming complexes, of different mechanical and complexation strength:

(source)

Components may migrate across the picture and locally, and reactions may occur.  The formation of soaps of lead and zinc are the most prominently known ones, for example based on palmitic, stearic, or azelaic acid.  These processes may alter colors, even rendering them transparent enough to recognise the support of the paint, and they may eventually protrude the surface of the paintings as shown from a topview:

(source, p. 14)
or cross section:

(source)
These may not simply be "washed off" as a normal cleansing would do, as water and humidity in general amplify the problem.  If left untreated, the painting however will age and degrade.
Further reading may be found, for example, here, here, here, here/here.

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making sulfuric acid from pyrite
I recently found a place with an insane amount of pyrite (we are talking about tons) already smashed into smaller pieces, and I'm wondering if there is any way of converting it into sulfuric acid.
I'm aware that this is possible:
$$\ce{4 FeS2 + 11 O2 -> 2 Fe2O3 + 8 SO2}$$
$$\ce{2 SO2 + 2 H2O + O2 -> 2 H2SO4}$$
but it requires to heat up pyrite to quite some temperature :/
Is there any chemical extraction method available?

What's wrong with heating? Yes, chemical methods without heating are technically possible, but insanely costly.

I was an apprentice in a fertilizer manufacturing industry and sulphuric acid was produced on site from originally iron pyrites and then more cheaply from pure sulphur. So look at the costs before venturing forth. I don't know the exact details but nitric acid was also used to 'get the sulphuric acid production going' after releasing the SO2 from the pyrites.
Spent iron was shipped to blast furnace for steel production. But we are talking thousands of tons over a year. There was no water with pure sulphur...

SO2 will not give you sulphuric acid. In a technical process, vanadium pentoxide is used to oxidize it to SO3. Please always take proper safety measures handling acids and gases for you and your environment.

The production of sulfuric acid will be cumbersome. May I suggest that you try to broker the material to an industry that has the equipment to use it. It could be a WIN-WIN for all involved.

There is a direct reaction to convert pyrite to sulfuric acid by treating it with nitric acid.
$$\ce{FeS2 + 18HNO3 → Fe(NO3)3 + 2H2SO4 + 15NO2 + 7H2O}$$

Iron(II) disulfide react with and nitric acid to produce iron(III)
  nitrate, sulfuric acid, nitrogen dioxide and water. Nitric acid -
  concentrated solution. The reaction takes place in a boiling solution.(source)

But the problem is that iron(III) nitrate is soluble and does not precipitate in the solution. You have to distill off the sulfuric acid.
So, in my opinion heating pyrite in oxygen and then hydro-oxidising sulfur dioxide produced to form sulfuric acid is the cheap and efficient way because it does not require any additional chemicals as @Ivan mentioned.
Also, if you have money/curiosity, try using a cheap, oxidizing agent like nitrous or nitric oxide. The advantage in this process is that the products beside sulfuric acid are gases which can leave easily but I recommend you to stick to the plan.

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Mass spectrometry: systematic mathematic transformation for calibration
Given a NaI mass calibration in the m/z 300-2000 range on a QTOF instrument as shown below:

and given any uncalibrated mass spectrum that has been measured with the same instrument settings. What is a general mathematical transformation that calibrates the mass spectrum to the NaI calibration. masses of the NaI clusters are of course known, i.e.

Overall, it seems the masses always deviate by about 0.1 m/z by being always a little bit too high than the reference in this case.
My question now is, how a simple mathematical approach could look like to calibrate mass spectra using the NaI reference. The simplest approach would be to just create a peak list out of the NaI spectrum, i.e.
472.8, 622.7, 772.6, ..., 1971.7
and perform a linear fit against the calculated NaI masses, i.e.
472.7, 622.6, 772.5, ..., 1971.6.
Which looks like this:

Where the slope of the line is 1.00003175282 and the intercept is
-0.153462368663.
Because the overall mass range is high, while the variations in mass are of course small, this gives the ridiculously high $R^2$ of 0.999999999556.
Now I could just apply this linear function
f(m/z) = m/z * slope + intercept
to uncalibrated mass spectra. Is this simple approach viable? Because it does not take into account that the mass deviation might be different depending on the m/z region. From your experience, is it likely for a QTOF-instrument to show significantly different mass deviations in different mass regions? Would it be better to make a "polygonal chain" and divide the overall mass range in separate regions instead?

You already have a tremendously good fit to your data. Try using chi squared instead of $R^2$ (which is useless btw) as then you can work out probability a good fit, see Bevington 'Data reduction and error analysis for the physical science' or Press,  Teukolsky,  Vetterling and  Flannery; 'Numerical Recipes'. You can only really know how TOFms works outside the calibration region by experiment. Otherwise you can only hope that the response is linear.

Thanks for the quick reply. 'You can only know how TOFms works outside the calibration region' I just want to add that I don't intent to measure outside the calibration region (i.e. no extrapolation, only interpolation). Thanks for the references, I will have a look : )

Basically, I am unsure if the normal instrument calibration on a QTOF device is just a rescaling of the m/z scale or if it also includes some other internal adjustments. Because the typical calibration procedure are very tedious and if they are basically just a linear transformation on the m/z scale, Id rather just apply this change during postprocessing ...

The Waters calibration procedure (for Waters Q-TOF) is a polynomial fit (to the fourth or fifth order, I would need to check). The linear fit is quite good, as you show, but if you intend to really measure with ppm accuracy, you might need a better fit of the calibration curve. You should try to plot the residuals (difference between calibrated mass and reference points as a function of m/z) to see if there appears to be a systematic deviation from the linear fit on one edge or the other.

@PLD That's a good idea. I will just try it out and see if I get a good calibration like this and I will plot the difference between calibrated mass and reference points as a function of m/z then.


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Bonding Carbon Fiber to Metal
I have an application for a very thin, strong, electrical conductor that need to act as a (almost) frictionless support "rope". Typically materials like tungsten wire or platinum-iridium alloys have been used. I was wondering about carbon fiber. However, how could I bond it to a metal (brass?) support such that significant tension would not pull it free?
The kind of thickness would be around 100 microns, for a length of maybe 30mm supporting (in the middle) up to 20 grams as it bridges a gap of 20mm.


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Proper disposal of calcium hydride/calcium hydroxide
I had probably 2-3 grams of calcium hydride leftover in a schlenk flask from predrying a distillation of benzyl chloride. For the disposal, I diluted the flask contents in about 30mL isopropanol and then, over a water bath (flask open but under heavy stream of argon), added distilled water every few minutes until well after the bubbling had stopped, which I assumed was exothermic evolution of hydrogen gas. I also thought any leftover benzyl chloride would just make trace toluene, benzyl hydroxide, and calcium chloride.
I then figured the milky-white solution was okay to pour down the sink with lots of water running and removed about half of it in this manner, although I noticed a brief but strong smell from the vapors as I did so and stopped. I checked and noticed some sites place it into a basic waste container. Was it a mistake to pour down the sink? I thought limewater would be harmless with enough water. Were there any other errors in this process? I may have to repeat this in the future with calcium hydride again or phosphorous pentoxide (seen various methods on P$_2$O$_4$, but one mentioned sink disposal as well).

Pfft,  most students wouldn't probably be so careful, I think you done it quite OK.

The safest way to get rid of calcium hydride is to pour over ice. The water quenches the reagent itself, with the ice helping to control the exotherm (if you just add water, a lot of heat and gas is liberated, having a tendency to spray hot chemicals over the whole hood).
Once everythings quenched, it's helpful to add some hydrochloric acid before disposing of down the sink to prevent it clogging the drains.
Note that whilst clean calcium hydride is safe (once quenched) to be disposed of via the normal drainage, you must ensure any chemicals are washed off first (hexane works well), especially odorous things such as amines which you may have been drying with the calcium hydride.

Okay, I guess the isopropanol was too miscible with water and must have been carrying whatever organic material became of the leftover benzyl chloride to cause the smell. Next time I will use a less polar solvent and take care to separate the organic phase into waste, and some acid to neutralize the hydroxide. Thank you for the thorough reply.


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Why is ammonium chloride rather than an alkylammonium chloride salt the product of nucleophilic substitution of a chloroalkane by ammonia?
My textbook said that when an excess of ethanolic ammonia is heated with a haloalkane ($\ce{R-X}$) one should give the products as being $\ce{R-NH2}$ and $\ce{NH4Cl}$ as opposed to $\ce{RNH3Cl}$. I understand that in reality an equilibrium will exist between the alkylammonium and ammonium cations, but I thought amines were stronger bases than ammonia, so I find it odd that I would be asked to give the ammonium salt and an amine rather than the alkylammonium salt as the major product of this reaction. Have I misunderstood something?

You have actually answered the question right in the first sentence of the description.
You are right in saying that $\ce{RNH_3Cl}$ will be formed.
The alkyl halide, $\ce{R-X}$, is the alkylating agent. The reaction medium is basic in nature, so I'd expect the $\ce{S_N2}$ mechanism. So,
$\ce{H3N +RX\to H3NR+X-}$
We know that an excess of ammonia has been used. This simply means that the forward reaction is favoured:
$\ce{H3NR+X- +NH3\rightleftharpoons RNH2 + NH4+X-}$
It's all Le Chatelier's once again - makes extraction simpler, doesn't it?


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Is it possible to retain the stereo-information when using enantiopure compounds?
In a reaction, there is always a probability that the product will be a racemic mixture of the product's enantiomers. Is it possible to use only certain enantiomers in order to achieve just one enantiomer of a product? That is, would using only the D enantiomers of certain compounds ensure getting the D enantiomer of the product?

Yes - as long as the reaction doesn't destroy the existing stereochemistry. See for example https://en.wikipedia.org/wiki/Chiral_pool_synthesis And in all cases you probably meant "enantiomer" not "diastereomer".

Echoing orthocresol's comment, please check your usage of 'diastereomer' vs 'enantiomer'. Right now, it's not clear to me how to respond, but I think it could be a very interesting question and topic for chem.SE.

As is, this question is confusing to me; as my previous commentators, I suspect this is due to misuse of the terms diastereomer and/or enantiomer. Racemate, for example, only makes sense when speaking about enantiomers. I’m voting to close this question as unclear for the time being. Please check your usage of the terms and correct them to the ones you actually want to use to keep the question open/reopen it.

Yes, I meant enantiomer.


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Why are there 3N-6 degrees of freedom in molecules?

Count the degrees of freedom using the minimum number of coordinates required to specify a position.

Let's say there are 3 atoms, 3 coordinates for center of mass, and 3 for an axis. But we need 6 coordinates to specify a line in a three dimensional space: $(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$. Which line do we choose? Even if we have a line, how do we use the remaining three coordinates to specify the atoms?

So the "line"? Are you talking about internal coordinates (Z-matrix)?

N atoms need 3N coordinates to specify their positions as x, y, z.  But in a molecule atoms are attached to one another so 3 coordinates are needed for translational movement of the whole molecule and 3 for whole molecule rotation (about x, y, z axes) leaving 3N-6 degrees of freedom to describe vibrations. Is this what you are asking about?

You have N atoms, each atom can be described by a set of cartesian coordinates (x,y,z). So for each atom we have 3 degrees of freedom which means we end up with 3N degrees of freedom. If we look at an isolated structure without any external field or anything like that then only the relative position of each atom to the others matters. So we can get rid of some degrees of freedom. Translating every atom in the same way into x, y or z direction won't change anything, so we can reduce the degrees of freedom by 3. Rotating it around x, y or z doesn't change anything either, so we are reducing by another 3, so we end up with 3N-6.

Is there any chance you could edit the question to make it clearer what it's asking? Right now I don't really get what it's trying to say.

@orthocresol comment in the wrong place?

Nope, I wanted to ask you because you answered, so I assumed you understood it.

@orthocresol ah got it. I don't think I'm able to do that. All I can "understand" that he is trying to determine the degrees of freedom in a way I don't understand and I don't even know why he's trying this way. I basically answered to the topic and explained how to get to the 3N-6 in (one) correct way.

Heh, fair enough. Hopefully OP clarifies.

What you said is one way to describe it.

@wangge yes, and there are others. But no one here understand what you are trying to do with lines.

I have added a reference in the question about how he describe it.


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Can somebody please explain this statement (about the greenhouse effect)?
"Methane is a potent gas and over a period of 20 years has a 72 times greater effect on warming the atmosphere than carbon dioxide.
Due to my limited english, i am failing to understand how timeline has anything to do with the effect of a gas on the greenhouse effect. I think it has to do with the half life of the gas in the atmoshpere. Maybe methane is broken down more readily than CO2 and the whole thing means that if they remained in the atmosphere for the same amount of time, methane will do a greater damage. Am i right ?

"Methane has a large effect (24 times as strong as carbon dioxide per unit mole) for a brief period (having an estimated lifetime of 8.9±0.6 years in the atmosphere),[12] whereas carbon dioxide has a small effect for a long period (over 100 years). Because of this difference in effect and time period, the global warming potential of methane over a 20-year time period is 72."[ Wikipedia in 2017, quoting Jain, Briegleb et al ].
Simplified, that means:

$\ce{CH4}$ has a shorter half-life in the atmosphere than $\ce{CO2}$, reducing its effect as a green-house gas over time, but

$\ce{CH4}$ is so much more potent in retaining heat than $\ce{CO2}$ that this more than compensates for its short lifetime, making $\ce{CH4}$ a stronger green-house gas over the long term.

And so if I understand correctly even though methane is present as $0.00018$% and $\ce{CO2}$ at 0.04% the 72 times is equivalent to $\approx 0.01$% $\ce{CO2}$, so a slight increase in methane in the future can be very significant.

Lets say that we have 100 cm3 of methane and 100 cm3 of CO2. Suppose that methane decomposes in 5 seconds, CO2 in 60 seconds , that methane is more potent than CO2 but not potent enough to have a greater effect on global warming over the course of 60 seconds. If it remained in the atmosphere for 60 seconds then it would do the greater overall effect on global warming. Is this model correct ? (Assume that the half lives of both methane and CO2 are as stated above and that all the other ir-realistic values are true).

Remember that the time quoted, ~9 yr,  is the half-life of methane. The half-life of CO2, ~27 yr, is given at http://euanmearns.com/the-half-life-of-co2-in-earths-atmosphere-part-1/. You'd need to calculate the exponential decay of each these greenhouse gases and then multiply the CH4 by ~24 to account for its greater effectiveness in retaining IR.

Sir you are talking to someone with fairly basic understanding in this area so it would help if we used a simple model like the one above to discuss this subject. Is my model (and the result which i obtained from it) correct ? Note : My result from that model is that CO2 has a greater overall effect in one minute because : 1- Methane decays more quickly than CO2. 2- Methane isnt potent enough to have a greater overall effect in one minute. But in the first five seconds, methane was causing the greater effect.is this true ?

No, it's wrong. To understand why, you need to learn about gradual, exponential decay. In your model, the methane does not disappear in 5 seconds, it's halved in five seconds.

@DrMoishePippik The factor 24 is confusing because it is based on amount, whereas the 72 is based on mass. The instant (mass-based) global warming potential of methane would be calculated as 66, still lower than 72, which is surprising.

The "Global Warming Potential" is a relative measure of how much one gram of given substance affects infrared absorption compared to one gram of carbon dioxide, given the current composition of the atmosphere.

[OP] i am failing to understand how timeline has anything to do with the effect of a gas on the greenhouse effect. I think it has to do with the half life of the gas in the atmoshpere. Maybe methane is broken down more readily than CO2 and the whole thing means that if they remained in the atmosphere for the same amount of time, methane will do a greater damage.

Yes, that is the gist. Methane has a half-life of about 10 years. If you release 1 g of methane into the atmosphere, the effect over the next 20 years is estimated to be similar to releasing 72 g of carbon dioxide.
On the other hand, if you look at the effect over the next million years, the carbon dioxide will stick around, but the methane is mostly gone after the first hundred years, so the GWP of methane over this longer time span will be much lower (by about a factor of 100,000).
If you look into the literature, you will find that there is no estimate for the half-life of carbon dioxide because it is stable in the current atmosphere. Mechanisms for its removal rely on reactions with solids and liquids on the surface of the earth, and their rates depend on what is going on at the surface and below it (temperature, pH of oceans, living organisms, human activity, etc.). Methane, on the other hand, reacts with oxygen and other species in the atmosphere, turning it into other gases (many of which have their own GWP).
One should note that GWP's are revised continually, with new data available and adjusting to the current composition of the atmosphere.

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Heating diamonds
This is research for a Victorian-era detective story. If a diamond the size of the Koh-i-Noor sat at the bottom of a gaslamp globe for less than an hour, what effect would the flame have? Would the diamond even get hot?

What is the size of the Koh-i-Noor? ;) You might narrow your question down to something specific to chemistry here. For thermal conductivity you might ask in Physics SE. You might also add a diagram or image of a gaslamp globe with the location of the diamond so one could estimate how close to the flame it is sitting, and what it is touching, and of course indicate how big it is.

However it looks like you've already received a great answer so much for my suggestion!

It might get hot (diamond is a very good thermal conductor) and depending on where it's sitting in relation to the flame, it might get some soot deposited on it, but it likely wouldn't burn.
While it is possible to burn diamonds, it takes fairly high temperatures and a ready supply of oxygen to do it, [1] so it generally doesn't happen without an external source of oxygen (or being outright dropped into liquid oxygen).

Great links on burning diamonds - much to be learned there! Actually, being a good thermal conductor is a reason for it not to get as hot as something that is a poor conductor, which could get much hotter on the side exposed to the flame. The conductivity would spread the heat instantly across the whole object, offering more cooling area. That's why some people look into diamond-like carbon films for removal of heat in electronics.


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Proving using ideal gas law that molar volume at NTP is 22.4 litres
The definition of NTP as given by IUPAC is

NTP – Normal Temperature and Pressure – is defined as air
at $20\ ^\circ\mathrm C$ ($\pu{293.15 K}$, $\pu{68^{\circ}F}$) and $\pu{1 atm}$ ($\pu{101.325 kN/m2}$, $101.325\ \mathrm{kPa}$, $14.7\ \mathrm{psia}$, $0\ \mathrm{psig}$, $29.92\ \mathrm{inHg}$, $407\ \mathrm{in}\ce{H2O}$, $760\ \mathrm{Torr}$).
It is also known that volume occupied by $1$ mole of gas at NTP is
$\pu{22.4 L}$.

My attempt at proving the above statement,
For $1$ mole,  $PV=RT$
Substituting $P=\pu{1 atm}$, $T=\pu{293 K}$,  $R=0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$, we get $V=\pu{24.05 L}$ as the molar volume.
On putting $T=\pu{273 K}$ and $P=\pu{1 bar}$ (conditions of STP), we get $V=\pu{22.2 L}$
Therefore, molar volume at STP is $\pu{22.4L}$.
Why is the above calculation in NTP condition giving wrong results? How else to prove that molar volume at NTP is ${22.4\ \mathrm L}$?

see the whole article and the linked section in particular https://en.wikipedia.org/wiki/Standard_conditions_for_temperature_and_pressure#Molar_volume_of_a_gas

The issue is that one mole of a gas at NTP should not be 22.4 L based on the ideal gas. This is only true for a gas at STP. The key difference between the two conditions is they are defined by different temperatures (STP is for 273 K while NTP is for 293 K).
They are also different in that the IUPAC now defines STP with respect to 1 bar while NTP is defined with respect to 1 atm.
See The Engineering ToolBox for more info.

And to add further point, the pressure is one bar, not one atm. So at STP it would be more correctly $\pu{22.7L}$.

@PrittBalagopal good point, I will make note of that explicitly.

You should use 298K for NTP!

@Vaibhav Dixit all the definitions I have seen for NTP define it at 20C or 293K.


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Can we equate nucleophilicity and Lewis basicity?
Please note that it is specifically Lewis basicity, not any other definition of basicity.
Consider the interesting trend in nucleophilicity and basicity of the fluoride and iodide anions. The reason claimed for the weak nucleophilicity of the fluoride ion in polar protic solvents is due to the hydration effects which it experiences as it interacts with solvent molecules. This hydration effect effectively prevents it from reacting with electrophiles. In a sense, the hydration effect severely decreases the availability of the electron pair for donation.
Now, although it is claimed to be a weak nucleophile in aqueous solution, the fluoride ion still behaves as a strong Lewis base. This is logical because the interacting species are different in the two different situations. In the case of nucleophilicty, we are considering its nucleophilicity of the species towards another electrophile (e.g. a haloalkane) placed in solution.
In the case of basicity, we are considering it, in terms of its hydrolysis reactions with the water molecules in the hydration shell, producing the hydroxide ions. Thus, the solution becomes an alkaline one and we say that it is basic.
Now, if we use the same perspective to view nucleophilicity, we can thus say that the water molecule is the electrophile and that the fluoride ion is indeed a strong nucleophile because it has managed to attack the electron-deficient hydrogen atoms on the water molecule.
Consider this: What if I remove all the hydroxide ions produce from the hydrolysis of water from an aqueous solution of fluoride ions and add in an acidic species? Would we still observe a reaction between the fluoride ions and the acid species?
My point is that when we consider Lewis basicity and nucleophilicity of reacting species we are not making a fair comparison and thus, we say that fluoride ions are strong bases but weak nucleophiles in polar protic solvents. This may be true based on the Arrhenius definition of acids and bases. But if we were to look purely at Lewis basicity, it would be certainly be inaccurate.
From the perspective of interactions with water molecules, the fluoride ions are definitely good Lewis bases and also good nucleophiles. From the perspective of interactions with other reacting electrophiles placed into solution, the fluoride ions would be both weak Lewis bases and weak nucleophiles, neglecting the hydroxide ions (because even though the acidic species may react with the hydroxide ions, it is not reacting with the fluoride ions, thus we would see those reactions of the acids with the hydroxide ions as a side reaction).
Therefore, is it appropriate to equate these two terms because they both refer to the electron-donating ability of a chemical species?
If I have overlooked something (i.e. some other differences between the two terms), please inform me via the comments.

I appreciate your perseverance to think so deeply about this!
But...Yes, you have overlooked the main difference between Basicity & Nucleophilicity ,i.e., Basicity (i.e. Lewis basicity) is a Thermodynamic term whereas Nucleophilicity is a Kinetic term. But here you have compared them under the belt of only thermodynamics.
Basicity is measured on the basis of the stability of the species, whereas Nucleophilicity is mainly aimed at the 'donating ability' of the species.
e.g. In polar protic solvents, the order of nucleophilicity in case of halide ions is:

$\ce{F-}$ < $\ce{Cl-}$ < $\ce{Br-}$ < $\ce{I-}$

Which contradicts with the basicity order.
Whereas in polar aprotic solvents, the order is:

$\ce{F-}$ > $\ce{Cl-}$ > $\ce{Br-}$ > $\ce{I-}$

Which follows the basicity order.
I hope this might help you facing your confusion a li'l more logically!

Thanks! This was indeed something I overlooked as I was purely looking at the chemical behaviour of nucleophiles and bases.


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IUPAC name of trimethyl amine
What is the IUPAC name of trimethyl amine? Can we give it an IUPAC name or not? If not then does it mean the IUPAC naming system is limited?

Welcome to StackExchange. Your question shows zero effort and might be closed as such. Edit it to include more info.

Of course there's a IUPAC name for trimethyl amine. If IUPAC wouldn't be able to name such simple molecules it would be completely useless.
The systematic name is: N,N-Dimethylmethanamine

While your answer is correct, we prefer to not spoon-feed the askers. Such an info is right there on Wikipedia.

@PrittBalagopal well, then let us just be consequent and delete such questions right away. Downvoting a correct answer is even worse than answering on a pretty effortless question. (and some language versions of wikipedia don't list IUPAC names...)

When you hover over the downvote button, it says: this post is not useful. I believe this is so, as it doesn't help the community by helping askers who cannot help themselves. We are closing the question. You were apparently quite quick to answer it.


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Is there a shortcut method to calculate the Steric Number?
The steric number is equal to the number of $\sigma$-bonds + the number of lone pairs of electrons on the central atom. It gives us the number of hybridised orbitals.
It is pretty straight-forward to calculate it, but the problem here is that one must always draw the Lewis Structure before one can actually get to calculating the steric number, and then the number and types of hybrid orbitals. Even that is quite simple for a smaller compound, even like XeF$_6$, but when it comes to complex hydrocarbons, it's a little difficult.
My question is that is there any well-known (or not so well-known, but working) shortcut to doing this, so as to save time? It would be great if anyone could share their ideas and help me out.
Thanks in advance.

I’ld like to point out that your first and second sentences contradict themselves. Take for example $\ce{SF4}$. We have two 2-electron-2-centre (2e2c) $\ce{S-F}$ $\sigma$ bonds and one (also 4e3c $\ce{F\bond{...}S\bond{...}F}$ bond. We also have one additional lone pair. The 4e3c bond is also $\sigma$-symmetric. Therefore, we have three or four $\sigma$ bonds — depending on how you count — and thus a steric number of four or five. However, sulphur is $\mathrm{sp^2}$ hybridised, i.e. only three orbitals take part in hybridisation.

This question was posted before I had learned about the concept of banana bonds, and other special bonds, in which multiple centres are present (such as 4e3c, and 2e3c). Hence, I assumed that in all compounds, steric number equals the number of hybridised orbitals.

Interestingly, the steering number is already a gross oversimplification. It always assumes uniform hybridisation patterns, which is likely not true.

All right … I found myself a shortcut, and would like to share this in case it is useful for others. However, this formula is applicable to molecules with only one central atom.
Here is how it goes:

Find $N=\frac{V+M \pm I}{2}$, where $V = n(\ce{e-})$, the number of valence electrons of central atom, which is equal to the group number according to the old IUPAC system, $M = n(\text{atom})$, the number of monovalent atoms directly bonded to it, and $I$ is the number of positive or negative charges present (subtract it if the charge is positive, and add it if the charge is negative). This $N$ is the Steric Number.
Now, find the number of Bond Pairs ($BP$) of electrons, which is equal to the number of atoms surrounding the central atom. However, this is a little difficult for a species like $\ce{H3BO3}$, which is actually $\ce{B(OH)3}$, when written according the IUPAC method of writing the less electronegative atoms first.
Next, find the number of Lone Pairs ($LP$) of electrons, which is equal to $N-BP$.
Now, draw the structure of the atom, using the central atom, drawing the skeleton of the atom using the steric number, and then assigning the Bond Pairs and Lone pairs to the respective bonds/atoms.

That's for an atom with a single central atom.
Now, for a Hydrocarbon, albeit it is not possible to get the shape directly from the molecular formula, it is possible to find its structure and hybridisation if and only if the basic structure of the atom is provided.

For a compound with a single $\sigma$ bond between Carbon atoms, the hybridisation is $sp^3$
For one $\sigma$ and one $\pi$ bond, it is $sp^2$ hybridised, and
For one $\sigma$ and two $\pi$ bonds, it is $sp$ hybridised.

So, essentially, there is no formula for hydrocarbons, but there is a formula for smaller compounds, with a single central atom only.

Short answer: no.

The steric number is a property of an atom, not a compound. You need to know what an atom connected to a given atom to know its steric number. For simple compounds, you can usually determine these connections because the formula suggests a central atom and surrounding groups. For hydrocarbons and other organic compounds, you need to consider isomerism. Given the capability of carbon to form complicated bonding patterns, even simple formulas can produce a fair number of isomers with different bonding patterns and steric numbers.

Let's look at some examples.

$\ce{C4H10}$
This formula corresponds to two compounds with the structures shown:

In this case, both compounds have all four carbon atoms with steric number of 4.
it is not always true that a set of hydrocarbon isomers will always have the same steric number for all carbon atoms or even the same set of steric numbers.
$\ce{C4H8}$
This formula corresponds to six isomers:

Note that four of these structures have two carbon atoms with steric number 4 and two carbon atoms with steric number 3. The other two have all four carbon atoms with steric number 4.
Any method to calculate steric number for carbon atoms in an organic compound using just the formula will fail. You must examine the structure.

Here is an alternate method for one central atom molecules which I found in JD LEE adapted by Sudarshan Guha ed 2021.

Motivation: This method assumes that all the corner atoms have complete octet. Number of $\sigma$ bonds $=$ Number of corner atoms.
Steps

Define $n$ to be: total valence shell electrons of all atoms + number of negative charge(if any)-  number of positive charge (if any).

Write $n$ in the following form: $n = 8j+k $ with $k<8$ and $j>0$ [Remark: this is application of Euclid's division algorithm]

It is now useful to define $Q= 8j$ and $R=  k$, it turns out that $Q$ is number of $\sigma$ bond pairs and $R$ is number of non bonding electrons on central atom.
Important point: To make this method work, we have to consider the valence electron of hydrogen as $7$ (because this method assumes that all the corner atoms have complete octet). Otherwise your calculations will go wrong.

The steric number number is given as: $S = Q + \frac{R}{2}$

Hence, from $n$ , we find the non bonding pairs, bond pairs and finally the steric number.

Example calculation:
Consider methane $\ce{CH4}$, due to absence of charges, $n$  is given as the total valence electrons i.e: $4+ 4 \cdot 7=36= 8 \cdot 4$. We see that $Q=4$ , $R=0$ and $S=4$ hence four bond pairs, no lone pairs and tetrahedral geometry.
Exercise:
Show, using the above method, that the steric number of  $XeF_5^+$ is $6$
Figure out how to extend this method to the molecule $\ce{HNO3}$

I have been teaching my students the same shortcut by AbhigyanC, but expressed a bit differently. Using the same symbols:
LP = (V-M-I)/2
where
LP = No. of lone pairs on central atom
V = No. valence electrons brought by central atom
M = No. of hydrogens or halogens bonded to the central atom
I = Charge of the species
It is a rearrangement of the formal charge formula, and uses the following additional observations:

Hydrogen always makes single bonds
Halogens make single bonds when they are peripheral (at least good enough for General Chemistry)
The net charge can be assigned to the central atom because the allowed peripheral atoms do not take on nonzero formal charges

Of course the steric number is:
N = M + LP
This shortcut allows me (and any student who adopts it) to simply look at a formula and come up with the VSEPR prediction with a simple mental calculation!


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Predicting the vapor pressure of water ammonia solution
I am working on a project that involves the compressors for a vapor compression refrigeration cycle of R717 (ammonia) and came across something that has been boggling my mind a little bit for no good reason. I suspect there is water entering the system and making the compressor work harder than it needs to and thus causing higher energy costs. I was reading an article that stated

For example, pure
  ammonia at -4°F (-20°C) has a saturation pressure of 12.9 psig (0.90 barg). An
  ammonia-water solution of 10% water and 90% ammonia by weight, at the same
  -4°F (-20°C) has a saturation pressure of 10.2 psig (0.72 barg). Therefore to
  maintain an evaporator temperature of -4°F (-20°C) in a system with 10% water
  contamination, the suction pressure would have to be run at 12.9 – 10.2 = 2.7 psi
  (0.18 bar) lower than if the water were not present.

Now I remember from thermodynamics that I can use Antoine's equation to determine the vapor pressure of a substance at a given temperature. I have gone to the NIST and obtained coefficients for water and ammonia. Problem is, at -4F, ammonia is a liquid, but water would be a solid. My issue is, how do I calculate the vapor pressure when, at the given temperature, one of the species is a solid?  Or is that not a concern, because according to the phase diagram, , the solution is a fluid assuming at most 20% water in the solution. If that is the case, the how would vapor pressure be calculated as the antoine's coefficients at only valid for liquid water about 0C? I think I am having a brain fart honestly, but maybe you all could help me out.


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Problem with understanding the transformation of Hessian in primitive internal coordinates to Hessian in delocalized internals
I am currently working out the theory of Hessians in different coordinate systems. Using the Wilson B matrix $\mathbf{B}=B_{ij}=\frac{\partial q_i}{\partial x_j}$, where $q_i$ denotes the $i$th primitive internal coordinate and $x_j$ is the $j$th Cartesian coordinate I can form a $n\times n$ matrix $\mathbf{G}=\mathbf{BB}^\top$. After diagonalizing $\mathbf{G}$ I obtain a matrix $\mathbf{U}$ as the set of nonredundant eigenvectors of $\mathbf{G}$ with eigenvalues $\lambda > 0$. $\mathbf{U}$ spans the nonredundant subspace of the space of primitive internals $\mathbf{q}$. I can use $\mathbf{U}$ to transform the diagonal $n\times n$ Hessian $\mathbf{H}^{\text{prim}}$ in the space of primtive internal coordinates to the space of delocalized internal coordinates using:
$\mathbf{H}^{\text{DLC}}=\mathbf{U}^\top\mathbf{H}^{\text{prim}}\mathbf{U}$.
Can somebody explain this non-linear transformation in a physical and/or mathematical sense? I don't get why I have to use $\mathbf{U}$ two times (apart from deriving the right dimensionality of $\mathbf{H}^{\text{DLC}}$).

Didn't you just explain the mathematical "sense" of it?

I suspect U is orthogonal and thus the last equation would be a similarity transform. Wikipedia will tell you a lot about those, and math.SE the rest.

It does look a bit like a similarity transform these have the general form in your example of $U^{-1}H^{prim}U$ where U are matrices of eigenvectors. In your case they seem to be orthogonal $U^T=U^{-1}$. The transform is equivalent to changing the basis set or equivalently a coordinate ('rotational')  transformation. Usually they are used to diagonalise matrices.

@Porphyrin     It indeed seems to be a similarity transform. Thank you for your anwswer.


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Can thermite be lit while mixed into butane? If not is there a flammable liquid that would work?
I want to create a thermite flamethrower to melt snow since I live in New England and since I am an asthmatic I struggle to shovel and it would be cool, so do you have any idea as to what liquid thermite could be lit in.

Why not just use a gasoline flamethrower? Adding thermite is just going to make life harder (suddenly you can no longer put it out) and make controlled ignition of the fuel stream a lot trickier.

Health considerations aside, you are pursuing a wrong purpose, and doing it wrong, too. You can't mix thermite into butane, for it will just drop to the bottom. Moreover, I don't believe butane flame will ignite thermite. This might be a good thing, for if it would, that would make your other problems look pale in comparison.

Related What-If from XKCD. Basic idea: don't melt snow, move it.

Ordinary propane flamethrowers exist for this purpose, they would be much better than what you are envisioning.  Note that it takes a lot of energy to do and you have to be careful about what's under the snow--it's easy to burn stuff.

There are snow blowers built around jet engines, but they're probably too bulky and expensive for your driveway.

Be aware that thermite flamethrowers tend to kill their operators.

@zwol You can make a pulse jet  though and can make one relatively cheaply.  Colin Furze made one as a snow blower.  It's awfully loud though (your neighbors will be angry)

You remember the boiling point of normal / *n-*butane, under atmospheric pressure, is just about −1 to 1 °C; 30 to 34 °F; 272 to 274 K, and its flashpoint at −60 °C (−76 °F; 213 K) (https://en.wikipedia.org/wiki/Butane)?  Think about safety for you and your environment.

@uhoh Don't forget to watch the thermite launcher  video from Colin Furze as well.  He also has a remarkable number of pulse jet videos.

Thermite is a solid-solid reaction that I think would be greatly inhibited by any sort of intermediate. In either case, the smoke, the dust and the copious amounts of nitrous oxides produced will not make your asthma any better off than if you were to pick up a shovel.
And then there will likely be other safety-, legal-, environment- and relationship-with-neighbours -related issues that alone would make this a very bad idea.

Indeed. The OP should either hire a service or a local kid to shovel the snow.

@JonCuster you need to be paid up on your homeowner's insurance for the latter.

@uhoh - if you aren't paid up on your homeowner's insurance (and an umbrella liability policy), you shouldn't be trying to burn your house down either!

@JonCuster May also be worth looking into a heated driveway if you're already willing to try a thermite flamethrower (though I'm not sure of the effective operating range of heated driveways).

In addition to the points Stian raised, it's important to realize that thermite is very difficult to ignite.  The ignition temperature is very, very high, higher than you can easily get by burning ordinary fuels such as butane, which means that whatever ignition method you use, if it's hot enough to ignite the thermite, would most likely set off the butane first, rather than at the same time.  Depending on your setup, the results could range from "wasteful" to "disastrous."  It's probably best not to bother with a plan like this.

With the proviso 'unless you want to upload a video of the results to YouTube, which will probably pay for the hospital bills and house rebuild resulting'

I'm not at all sure that termite is a useful solution to this problem. The thermite reaction takes place between iron oxide and aluminium, producing a lot of heat and molten iron as a result. This makes it useful for applications like demolition of steel structures and on-site welding of railway rails. In this sort of situation it is a reasonably compact, controllable and portable way of generating very high temperatures.
Having said that the energy density of thermite isn't significant better than any other common fuel and for melting snow it is the opposite of what you want.
Snow (obviously) melts at a fairly low temperature and for this you want a large volume of a moderately warm fluid. Too high a temperature and you are just wasting energy.
A flame thrower isn't a very sensible way to melt snow but even so trying to combine it with thermite makes no engineering sense. Just mixing thermite with a liquid fuel won't really achieve anything except clogging up the jets with metal particles.
Even assuming that melting the snow is the best option something like a kerosene fired space heater (a type of pulse jet) is a vastly more sensible solution as this is a reasonably economical and safe way to create directed stream of hot air).

You can buy radiant tubing that you install in your driveway to have a self-melting driveway.  I'm not sure if it would work in New England winters, but it also uses the principle of "large volume of moderately warm fluid" pretty well.

This will not answer the part of igniting thermite, but addresses the issue of the snow on the driveway, that you want to forget.
Instead of heating from top (by a torch), you may consider heating from the bottom, using a tubing system that is embedded in the ground.  Design-wise, similar to sometimes found installations in a bath room, and may be driven by an extra boiler, surplus heat by a nearby power plant, or geothermic sources. The Federal Highway Administration (here) monitored bridges in several states, updated by such a system where the concern for safety (especially freezing) was in mind.
More recently, in 2014, NYTimes shed some light on it, stating a a yearly cost of about US$ 2000 per 1000 sq ft and year for a hydronic system.

(source)
This technology already finds some application in places where snow occurs regularly like Chicago, or Holland, MI. A similar system has been implemented in a place with snow and geothermal activity - Reykjavik.

+1 for a solution with a much lower probability of resulting in death.

Most liquids don't burn directly: you first must vaporize them into a gas, and then that gas burns.  Butane is no exception.
The real issue* would be that vaporization.  Gases take up more space than liquids, so the result would try to push the particles of thermite apart from each other.  You want them close together to create the usual thermite fire.
So, you'll end up getting lucky: you'll end up with a snowy driveway full of rust and aluminum powder.  It could be worse.  You know, the thermite could have ignited!
* Well, we all know what the real real issue here is, and it has something to do with misapplication of thermite.  We'll let that slide.

SAFETY
Materials mentioned below can easily be dangerous to play with.
Ethyl alcohol is expensive if drinkable and even more poisonous if not.
Do not breathe vapours. Do not drive while melting snow.
Methyl alcohol is NEVER drinkable, is lethally poisonous and best avoided.
Any sprayed combustible fluid that produces impressive flames also can hospitalise or kill you
(or both).
Thermite can also do either or both very easily if you let it.
Hopefully you are familiar enough of the hazardous natures of boiling water and of steam. In snow-lance volumes steam can produce an invisible jet that can kill and maim.
Having noted the kill-joy stuff, lets get down to it ...
________________________________
Thermite: Not really.
Apart from being very very hard to ignite, doing so before it is in contact with snow will waste much of its energy and do a good job of melting the small area per mass used where molten iron drops onto.
ie Apart from what fun you may or may not derive, even if you could deliver thermite to the snow surface at or after combustion time it would not be especially useful per mass used.
Alcohol or other sprayed/misted fluids: For fun effects, ignited sprayed alcohol allows easy production of flames many feet long (ask me how I know) and only MAY kill or maim you. Cost per energy imparted is unlikely to be attractive for snow melting per se - use it if at all as an added special effect.
Water & Steam: Sprayed water at near boiling point will be "somewhat effective".
Melting ice or snow AT 0 degrees C takes 334 kJ/kg and
water cooled from boiling point at 100C to 0C liberate about 420 kJ/kg,
so using 1 unit mass of 100C water will melt 1 unit mass of 0C snow with about 30% extra energy available to bring the snow temperature up from slightly below 0C.
For colder snow you will need more mass of water per mass of snow.
If you produce steam at 100C it has about 5 times as much energy (2260 kJ/kg) as if you cool water from 100C - so you can melt about 5 to 6 kg of snow per kg of steam. You do have of course to provide the energy to either heat the water to 100C or to produce steam, and your butane can be used for that purpose.
For something approaching the visual effects you had in mind, a hot water system with enough excess energy to make SOME steam and an alcohol spray with RELIABLE pilot light ignition would give you a steaming flaming melting snow lance. Sounds like fun.

Like many of the other answers given, "this will not", to quote Buttonwood, "answer the part of igniting thermite, but addresses the issue of the snow on the driveway". As this is Chemistry.SE, I feel that a (physical) chemistry answer – rather than just a physical answer – is in order. And that answer is:
use salt.
Rather than raising the temperature of the snow above its melting point, lower its melting point to below the ambient temperature.

N.B. I'll be coming back to this later and adding a comparison of different salts etc.


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Enthalpy - unit confusion
I think I have trouble interpreting reaction enthalpy in these kind of situations:
I am given that $\Delta _rH^0 = 25,9$kJ/mole in the reaction:
$$\ce{1/2 H2 + 1/2 I2 \rightarrow HI}$$
which I interpret as: if this reaction takes place where 1 mole HI forms, the the produced heat is equal to 25,9kJ.
Now suppose we double the amounts, so that we have the reaction
$$\ce{H2 + I2 \rightarrow 2HI}$$
then clearly the difference in enthalpy should be doubled and be equal to 51,8kJ. Then why do chemists write that  $\Delta _rH^0 = 51,8$kJ/mole and not $\Delta _rH^0 = 25,9$kJ/mole? Because I would think that we can calculate the enthalpy change like this: $2mole\times25,9$kJ/mole.
There is cleary some fundemantel thing I'm missing here or misinterpreting, so I would like some help.

25.9 kJ/mol refers to one mole of $\ce{HI}$ produced while 51.8 kJ/mol corresponds to one mole of $\ce{H2}$ consumed. It is good practice to give the full equation, especially when dealing with reaction enthalpies/energies $\Delta_rX$. For formation quantities $\Delta_fX$, I would always assume one mole of product.


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Do Ligands electrons fill the d-orbitals of the metal ion?
We know that ligands make coordination bonds with d-orbitals of transition metals forming complexes. My question is, if that happened, all d-orbitals will be filled and no electron transition could happen between eg and t2g orbitals to be responsible for magnetism or colour formation.
I need clarification.

You are mistaken. The electron pairs donated in the coordinate bonds by the ligands do not fill the d orbitals of the metal ion.
The basis of crystal field theory is the electrostatic repulsion between the d orbitals of the central metal ion and the electron cloud of the ligands. This repulsion then leads to the breaking of the degeneracy of the orbitals, giving rise to disparity in energy between different sets of d orbitals. This arises due to the different spatial orientations of the 5 d orbitals (i.e. $xy, yz, xz, x^2-y^2, z^2$). Thus, they experience different amounts of repulsion from the ligands when the ligands are oriented in different geometric configurations (i.e. octahedral, square planar, tetrahedral etc.) around the central metal ion.
For more details on crystal field theory and an alternative explanation known as ligand field theory (based on molecular orbital theory), you can refer to the following discussion:
In an octahedral complex, what happens to the electrons donated by the ligand?
Since the d orbitals are not filled up by any ligand electrons, the electronic transitions between the orbitals of different energy levels (such as in the case of $t2g$ and $eg*$) are certainly possible.
Now, you may wonder: Where do the donated electrons go?
Well... based on valence bond theory, they are used in the formation of covalent bonds between the central metal ion and the ligands. These covalent bonding interactions are described as the overlap of the ligands' atomic orbitals with the electron pairs with the hybrid atomic orbitals of the central metal ion.
For example, in the case of a hexaaquairon (III) complex, the iron (III) ion has 5 electrons, each singly occupying one d orbital. As the d orbitals are occupied, the vacant 4s, 4p and 4d orbitals are hybridised to give 6 $sp^3d^2$ orbitals to be used to overlap with the atomic orbitals of the ligands.
Thus, the donated electrons actually are "shared" between the valence atomic orbitals of the ligand and the valence hybrid orbitals of the metal ion.
Hope I have clarified your doubts.

No, hybridisation can and should not be used to describe coordinate bonds.

This is useful. But in case of [Cr(H2O)6]+3. there will be 3 single electrons in 3 orbitals t2g. and all the other orbitals will be hybridized d2sp3. And according to what I understand, there will be coordinate covalent bonds between the lone pair of electrons of the ligands and these hybridized orbitals, meaning that these orbitals will be filled. So, how the electrons will transfer from the t2g orbitals to any higher filled orbital? thus, How could this compound have a color?

@Jan Why not? Although valence bond theory is limited, it suffices in this case right? Is there anything that it fails to provide an explanation for?

@IbramMikhail Sorry if I didn't  make it clear but the orbitals used to make the hybrid orbitals are the 4 d orbitals, not the 3 d orbitals. The 3 d orbitals remain  as they are, in terms of electron population.

Valence bond theory is never a good idea for complexes, full stop. If anything, use crystal field als a low-level theory to describe them. Many of valence bond theory’s assumptions are drastically incorrect for complexes.

Also, the 4d orbitals are energetically so far removed from 4s and 4p (and 3d) that they do not play the slightest of roles in an MO description of a 3d metal’s complexes.


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Homemade "milk of magnesia" (suspension of magnesium hydroxide in water)
I want to make homemade "milk of magnesia" as a suspension containing only magnesium hydroxide and water.
I see that I can purchase a bag of magnesium hydroxide powder. Can I use this to make a suspension or is there a multi-step process?

I guess the first question to ask is why you want to make it at home. The store stuff is dirt cheap...

One gallon (about 4L) of milk of magnesia is about 26 USD. . 500 g (17.6 oz) of Mg(OH)2 is about 17 USD . Since milk of magnesia is 8% by mass, that gallon only has 290 g of Mg(OH)2 in it. The materials are cheaper to make it yourself, but the time and effort might be worth more than 10 USD.

If you plan to ingest it, there may be safety issues involved. Rule 1 of amateur chemistry is don't eat anything you make...

There are some good reasons to make Milk of Magnesia at home. One, I can't presently purchase it the stores (Covid-19 shortage I guess), I use it in a deodorant recipe so I'm not ingesting it and by making it myself I am not shipping water, saving on plastic and I know what is in the concoction. Thank you for helping me.

Milk of magnesia is typically just a suspension of about $\pu{8\%}$ wt/v (i.e. $\pu{8g / 100 mL}$) magnesium hydroxide in water. To get a good suspension you will want to have very fine magnesium hydroxide. Ideally, it should feel like flour when rubbed between you fingers. Regardless, there will be some settling and you will need to shake it thoroughly prior to each usage. No other preparation should be necessary.

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


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why do solids and pure liquids have no effect on yield and equilibrium?
There are three parts to my question.
Part one:

Let's assume we have a reversible reaction at equilibrium  as follows : Calcium carbonate (solid) goes into calcium oxide(solid) and carbon dioxide(gas).
Let's assume that I add more calcium carbonate i.e I add more solid.
Now I understand that there would be no shift in the position of the equilibrium since we added a solid therefore we induced a very negligible change in concentration ,  but what I am confused about is doesn't more reactant mean more product and doesn't more product symbolise a shift to the right?

Part two:

I have read that the equilibrium constant $Kc$ and $Kp$ are really just close approximations of the true thermodynamic constant which uses activities hence the activity of a solid and pure liquid is always 1 .
However upon further research I discovered that for an esterification reaction the equilibrium constant is 4
-First of all, the above reaction involves all pure liquids and hence the thermodynamic equilibrium constant would be one ,however why stick to a value of 4 why not easily say it is one since all activities are 1.

Part three:

If we do take pure liquids in the $Kc$ value how do you measure the concentration of such a species, wouldn't the value be very high.My intution tells me that pure liquids are only taken when a reaction only involves pure liquids only , but they are not taken in heterogeneous reactions or when the liquid involved acts as a solvent as well.However in a case of a heterogeneous reaction only involving solids and liquids what will your equilibrium constant look like .

Lastly I have no knowledge of thermodynamics and I am still a high school student .

https://chemistry.stackexchange.com/questions/5500/how-do-equilibrium-shifts-affect-solids?rq=1 https://chemistry.stackexchange.com/questions/14058/why-are-solids-and-liquids-not-included-in-the-equilibrium-constant-what-about?rq=1

If there are three parts of a question, than it is easier to tackle each of them as an individual question.  Otherwise, the question may be considered as too broad to be answered.

Regarding  your first part, although more reactant means more product,  the percentage yield is the same and thus, there is no shift in the position of equilibrium.


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Alternatives to carbonates?
This isn't strictly 'pure' chemistry but the knowledge to answer it is probably much closer to chemistry than any other subject.
The question is a simple one - are there any other non-toxic aqueous acids that could substitute for carbonic acid ($\ce{CO2}$ dissolved  in water), that could substitute for carbonates as a basis for an experience similar to soda drinks? I'm thinking of chemical suitability rather than economics. Maybe other weak or organic acids? Or non acids even?

If you're talking about $\ce{CO2}$ in the context of beverage carbonation, you may want to read about $\ce{N2}$ being used in some beers.

It's not just mild acidity that gives club soda/seltzer/tonic... its taste, but the bubbles, as well. Citric acid is often used in foods, including powdered fruit-flavored drinks, but it doesn't give the tingly, bubbly sensation of soda. As Cole Porter stated, "I get no kick from champagne" without CO2. C'est la vie.


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Dissolution of Pentahydrate of Copper Sulfate
I'm having some difficulty understanding how the pentahydrate of copper sulfate dissolves in water.  I understand how polar molecules work, and how things dissolve in water, but I can't get my head around this in particular.
What happens to the water molecules in the pentahydrate?  Does the copper sulfate split up at all?  What makes the solution blue, if the solute is split?

When copper(II) sulfate pentahydrade dissolves in water it actually just becomes further hydrated, in what is called an aquo complex (a coordination compound containing only water molecules as ligands). According to this Wikipedia article:

The pentahydrate ($\ce{CuSO4·5H2O}$), the most commonly encountered salt, is bright blue. It exothermically dissolves in water to give the aquo complex $\ce{[Cu(H2O)6]^2+}$, which has octahedral molecular geometry. The structure of the solid pentahydrate reveals a polymeric structure wherein copper is again octahedral but bound to four water ligands. The $\ce{Cu(II)(H2O)4}$ centers are interconnected by sulfate anions to form chains.

The aquo complex maintains the blue coloration that is present in the pentahydrate solid.


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How to compute 2-electron integral for Hartree-Fock code?
I'm working on writing my own code for $\ce{H2}$ in an STO-3G basis set using Hartree-Fock (HF), and I am currently stuck on how to construct the two-electron integral matrix.  I know how to evaluate the primitive two-electron integrals, but I can't figure out how to transform that into the $4\times4$ matrix of possible values for this system.
For my previous integrals, I determined the sum of the primitives by constructing a $6\times6$ matrix of the primitive integrals, and did a transformation using a $6\times2$ matrix of the coefficient values from the fit to the Slater function, for example:
\begin{align}
\mathbf{D} &=
  \begin{bmatrix}
     d_{11} & 0 \\
     d_{12} & 0 \\
     d_{13} & 0 \\
     0 & d_{21} \\
     0 & d_{22} \\
     0 & d_{23}
  \end{bmatrix}\\
\mathbf{S} &=
  \mathbf{D}' * \mathbf{S}_{prim} * \mathbf{D}
\end{align}
I tried doing a similar technique for the two-electron integrals but with a $12\times12$ matrix, but the values didn't come out right.  It seems that I need to make a $4$-dimensional array of all the primitive integrals, but I don't know how to evaluate that in order to determine the sum and hence the final
$4\times4$ matrix.
I've been using Szabo and Ostlund's Modern Quantum Chemistry book as my main reference.

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I am not entirely certain, but transformation from primitive integrals to AO integrals boils down to a 4-index transformation just like the one from AO to MO?! In this case, you will need some sort of 4-index tensor representation, which is precisely what I used back when writing my first exercise SCF (we were provided with the AO integrals through a subroutine).

Do you already have the $12 \times 12$ matrix and are asking how to make the product, or are you asking how to construct the $12 \times 12$ matrix?
Could you please add what you currently have for the $12 \times 12$ matrix?

Are the $d_{ij}$ the contraction coefficients 0.15432897, 0.53532814, 0.44463454?

@pentavalentcarbon: yes

@Martin I only pointed towards the math one because it is a bit more thorough and directly related to the edits I had to make for this question (I couldn't find anywhere on the Chem SE where it was explained how to make matrices with mathjax).

I had already constructed the 12 x 12 matrix, but when I was trying to evaluate the two-electron integral in the same way that I was evaluating the one-electron integral (using matrix multiplication), I wasn't getting the right answer.  I've ascertained at this point (like others have said here) that I need to make a 4-dimensional tensor of 6 x 6 matrices for this, but still trying to figure out how to get the final 4 x 4 matrix form from there.

You can also write this down in a similar way as for the one-electron integral. You already had:
\begin{equation}
\mathbf{S} = \mathbf{D_2}' * \mathbf{S}_{\rm prim} * \mathbf{D_2}
\end{equation}
where $\mathbf{S}$ is $(2\times2)$, $\mathbf{D_2}$ is $(6\times2)$ and $\mathbf{S}_{\rm prim}$ is $(6\times6)$.
For the 2-electron integral we now need
\begin{equation}
\mathbf{V} = \mathbf{D_4}' * \mathbf{V}_{\rm prim} * \mathbf{D_4}
\end{equation}
where $\mathbf{V}$ is $(2\times2\times2\times2)$, $\mathbf{D_4}$ is $(6\times2\times6\times2)$ and $\mathbf{V}_{\rm prim}$ is $(6\times6\times6\times6)$.
You can get $\mathbf{D_4}$ from $\mathbf{D_2}$:
\begin{equation}
\mathbf{D_4} = \mathbf{D_2} * \mathbf{D_2}
\end{equation}
where you do not sum over any index in order to get the 4-index tensor.
The equation for $\mathbf{V}$ is a bit more tricky, since you need to pay attention over which indices you sum (which is actually not specified by the notation). I add an example on how you can do this with Python/Numpy with Einstein summations.
# 1-electron integral
S = numpy.einsum('ba,bc,cd', D2, Sprim, D2)

# 2-electron integral Option 1: form (6x2x6x2) tensor
D4 = numpy.tensordot(D2, D2, 0)
V = numpy.einsum('badc,bedf,egfh->agch', D4, Vprim, D4)

# 2-electron integral Option 2: transform indices one-by-one
V = Vprim.copy()
V = numpy.einsum('ba,bcde->acde', D2, V)    # i
V = numpy.einsum('abcd,be->aecd', V, D2)    # j
V = numpy.einsum('da,bcde->bcae', D2, V)    # k
V = numpy.einsum('abcd,de->abce', V, D2)    # l

You will need to give D2, Sprim and Vprim as input.
(If anyone knows how to write this down in a formal notation, please edit.)
You were asking how to get $\mathbf{V}$ as a 2-index $(4\times4)$ matrix. For this we can simply reshape the above 4-index tensors. So $\mathbf{V}_{\rm prim}$ becomes $(36\times36)$. Now this means $\mathbf{D_4} $ needs to be $(36\times4)$, and not $(12\times12)$. To these reshaped 2-index matrices we can apply again the same transformation equation from above. Here is the code for my example:
# 2-electron integral Option 3: reshape to 2-index matrices
Vprim = Vprim.reshape((Nprim**2, Nprim**2))
D4 = numpy.swapaxes(D4, 1, 2)    # making sure to combine the correct axes
D4 = D4.reshape((Nprim**2, Ncntr**2))
V = numpy.einsum('ba,bc,cd', D4, Vprim, D4)

Now $\mathbf V$ will be:
\begin{bmatrix} (\phi_1\phi_1|\phi_1\phi_1)  & (\phi_1\phi_1|\phi_1\phi_2) &  (\phi_1\phi_1|\phi_2\phi_1) & (\phi_1\phi_1|\phi_2\phi_2)\\      (\phi_1\phi_2|\phi_1\phi_1)  & (\phi_1\phi_2|\phi_1\phi_2) &  (\phi_1\phi_2|\phi_2\phi_1) & (\phi_1\phi_2|\phi_2\phi_2) \\      (\phi_2\phi_1|\phi_1\phi_1)  & (\phi_2\phi_1|\phi_1\phi_2) &  (\phi_2\phi_1|\phi_2\phi_1) & (\phi_2\phi_1|\phi_2\phi_2) \\      (\phi_2\phi_2|\phi_1\phi_1)  & (\phi_2\phi_2|\phi_1\phi_2) &  (\phi_2\phi_2|\phi_2\phi_1) & (\phi_2\phi_2|\phi_2\phi_2)   \end{bmatrix}
And here is the structure of $\mathbf{D_4}$:
\begin{equation}
\mathbf{D_4} = \mathbf{D_2} * \mathbf{D_2} = \begin{bmatrix}
d_{11}*\mathbf{D_2} & 0*\mathbf{D_2} \\
d_{12}*\mathbf{D_2} & 0*\mathbf{D_2} \\
d_{13}*\mathbf{D_2} & 0*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{21}*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{22}*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{23}*\mathbf{D_2} \\
\end{bmatrix}
\end{equation}

Here are some timings for the 3 options. The 2nd one is by far the fastest on my Core i7-3667U, probably because it requires the fewest memory options or the strides are optimal, but I didn't bother figuring out FLOPs/MOPs.

numpy.einsum is not very efficient, since it is pure Python code. But one can easily replace these calls using dot and tensordot, which interface to BLAS. See here.

As was mentioned in the comments, this amounts to the same fundamental operation as the AO-to-MO transformation; they are both 4-index transformations.

(...) each contracted integral $(\mathbf{ab}|\mathbf{cd})$ is expressed as a sum of its component primitive integrals $[\mathbf{ab}|\mathbf{cd}]$ which, in turn, are computed individually, i.e.
$$
(\mathbf{ab}|\mathbf{cd}) = \sum_{i=1}^{K} \sum_{j=1}^{K} \sum_{k=1}^{K} \sum_{l=1}^{K} \mathrm{D}_{\mathrm{a}i} \mathrm{D}_{\mathrm{b}j} \mathrm{D}_{\mathrm{c}k} \mathrm{D}_{\mathrm{d}l} \left[ \mathbf{a}_{i} \mathbf{b}_{j} | \mathbf{c}_{k} \mathbf{d}_{l} \right]
$$

For the naive implementation, assuming you have all the primitives already formed, this is the simplest approach. If you have an AO-to-MO transformation that works, replace the MO coefficients as an argument with the contraction coefficient matrix. To get the naive implementation working, it would probably be easiest to reshape your primitives into a 4-index tensor.

Gill, Peter M. W. Molecular integrals over Gaussian basis functions. Adv. Quantum Chem. 1994, 25, 141--205

From yours and other peoples' comments, I definitely see now that I cannot avoid making a 4-dimensional tensor of the primitive gaussian integrals.  The problem I am having is doing the above summation in a way that condenses all the primitives into a $ 4\times4$ matrix.  This is what I'm trying to accomplish:

$\begin{bmatrix} (\phi_1\phi_1|\phi_1\phi_1)  & (\phi_1\phi_2|\phi_1\phi_1) &  (\phi_2\phi_1|\phi_1\phi_1) & (\phi_2\phi_2|\phi_1\phi_1)\
(\phi_1\phi_1|\phi_1\phi_2)  & (\phi_1\phi_2|\phi_1\phi_2) &  (\phi_2\phi_1|\phi_1\phi_2) & (\phi_2\phi_2|\phi_1\phi_2) \
(\phi_1\phi_1|\phi_2\phi_1)  & (\phi_1\phi_2|\phi_2\phi_1) &  (\phi_2\phi_1|\phi_2\phi_1) & (\phi_2\phi_2|\phi_2\phi_1) \
(\phi_1\phi_1|\phi_2\phi_2)  & (\phi_1\phi_2|\phi_2\phi_2) &  (\phi_2\phi_1|\phi_2\phi_2) & (\phi_2\phi_2|\phi_2\phi_2)
\end{bmatrix}$

I extended my answer to get the $(4\times4)$ matrix. Your mistake was to assume you need a $(12\times12)$ transformation matrix, but it should be $(36\times4)$.


text:
Need help with stoichiometry problem
I'm pretty new to chemistry and I've been stuck on it for hours.
Question: $\pu{10g}$ of the hydroxide of a metal on ignition gave $\pu{8g}$ of oxide. The equivalent weight of the metal is:
a) $\pu{136g}$
b) $\pu{40g}$
c) $\pu{56g}$
d) $\pu{28g}$
I used the law of equivalence but the answer I got was $\pu{3g}$. what am I doing wrong?
How I went about it:
10/E=8/16;
E=20;
E=M(equivalent mass of metal)+16+1;
M=3

I found your numbers confusing, but I think the first equation you wrote is not correct. This is how I'd solve it:
Let's say we have $\mathrm{n}$ equivalents of metal in the sample, and let's call the equivalent weight of the metal $\mathrm{e}$. For the hydroxide sample, we have:
$n(e+17)=10$
where $\mathrm{e+17}$ is the molar mass of the hydroxide $\ce{MOH}$.
For the oxide, we have:
$n/2(2e+16)=8$
where $\mathrm{2e+16}$ is the molar mass of the oxide $\ce{M_2O}$. Note that $\mathrm{n}$ becomes $\mathrm{n/2}$ because of stoichiometry:
$\ce{2MOH->M2O + H2O}$
So you have a system of two equations with two unkowns ($\mathrm{n}$ and $\mathrm{e}$), which you can solve normally. For instance, if you divide the equations:
$\cfrac{n(e+17)}{n/2(2e+16)}=\cfrac{10}{8}$
$\cfrac{e+17}{e+8}=\cfrac{10}{8}$
$8e+136=10e+80$
$2e=56$
$e=28$
So the equivalent weight of the metal is 28g.
Makes sense?

Nothing wrong with Vic's answer. I'd solve it differently, but hopefully you are familiar enough with algebra to understand that there are various ways to correctly come up with the correct answer...
Here's how I did it: There is a loss of 2 grams of what we assume is water.
Now since we know we started from M?(OH)? we assume the reaction was 2OH → H2O + O
meaning ~34 grams of OH would produce 18 g of water and 16 g of oxide oxygen.
Since we know that ~2 grams were lost (assumed to be only water, and all of the water) then we know  2g of water must have come from (2/18)*34 grams of OH (2:18 as X:34), or 34/9 grams of OH. Since we started with 10 grams total, we started with 90/9 grams total and of that 34/9 is OH. Subtract to determine how much metal: (90-34)/9 = 56/9 grams of metal. So far, so good. But what is its oxidation state? Since the question is asking for equivalent weight, we need some way to pin down the equivalency. Well, for every 18 grams of water lost, the residue retained (we assume) 16 grams of oxide. So, 2:18 as X:16 or x= 16/9.
So now, we have the whole enchilada: 90/9ths grams = 56/9 M + 16/9 O + 18/9 OH (as H2O - meaning this treats the OH lost during ignition as H2O rather than OH ) leaving us, after ignition with 56/9 M and 16/9 O. For each equivalent of O= there should be one equivalent of M. This means that since O has an equivalent weight of of 8 that 16/9 ÷ 8 = 2/9 Eq are present (after ignition) so that there must also be 2/9 equivalents of M.  56/9 grams ÷ 2/9 equivalents = 56/2 g/eq = 28 g/eq which is the eq. wt of the Metal.  (A quick search of the periodic table indicates that Fe with an AtWt of 55.845 and assuming we're talking about Fe(OH)2 would produce 2.00 grams of water, theoretically. So, it checks out.)


text:
How do I name this compound
I think I numbered it the correct way but I do not know which is the correct way to name it. Also please give the correct reason.

My answer was:
cyclohexane 1-carboxylic acid 2-oxo 4-carbaldehyde
OR
4-carbaldehyde 2-oxocyclohexane - carboxylic acid

Related: Nomenclature priorities with ethers, acyl groups and ketones

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

(…)
  7. acids
  (…)
  15. aldehydes
  16. ketones
  (…)

Thus, the principal characteristic group of the compound that is given in the question is the carboxylic acid group. This principal characteristic group is expressed at the end of a name by means of a suffix or a class name, i.e. ‘cyclohexanecarboxylic acid’.
The other groups of the given compound cannot be expressed as a suffix. The cyclohexanecarboxylic acid is substituted using the usual principles of substitutive nomenclature.
For the aldehyde group, the suffix ‘carbaldehyde’, which generally can be used when the $\ce{-CHO}$ group is attached to a ring as in cyclohexanecarbaldehyde, cannot be used in this case. Instead, the prefix ‘formyl’ can be used in the presence of a characteristic group having priority to be cited as a suffix or when the $\ce{-CHO}$ group is attached to a side chain.
Likewise, the suffix ‘one’, which generally can be used to form the name of ketones as in cyclohexanone, cannot be used in this case. Instead, the prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix.
Therefore, the correct name for the given compound is 4-formyl-2-oxocyclohexane-1-carboxylic acid.


text:
Naming of an organic compound
How would I determine the systematic name of $\ce{(CH3)2CHCH2CHClCH3}$?
First of all, how do we determine which side we number the carbon atoms from? It has an alkyl group ($\ce{CH3}$) and a functional group ($\ce{Cl}$). I got lucky by going from right to left so I wrote 2-chloro-4-methylpentane but why couldn't we also label this as 4-chloro-2-methylpentane?
Do we take functional groups in preference to alkyl groups when naming them?

Duplicate https://chemistry.stackexchange.com/q/75911/16683

The substituents go alphabetically, and as 'chloro' is before 'methyl' in the alphabet, the correct name is 2-chloro-4-methylpentane.


text:
When are pure liquids included in the equilibrium constant?
For an esterification reaction of pure liquids only, the equilibrium constant will take into account all species, however for an aqueous based esterification reaction, water is not included in the constant
Why are pure liquids used in the first reaction and pure liquids not used in the second reaction, don't pure liquids have an activity of one hence can't we just assume that the equilibrium constant for the first reaction is one, or do we only exclude pure liquids when they act as a solute or they are present in a heterogeneous equilibrium?
Also note that I have a very vague knowledge of thermodynamics.

Water is included, you must have heard of water being removed to progress the esterification in the forward reaction.

Yes but when are pure liquids not included


text:
What is the highest possible effective nuclear charge as a percentage of nuclear charge an electron in any neutral atom can experience?
I was wondering about the following statement "The higher the principal quantum number of the orbital the electron is in, the lower the nuclear charge it experiences." I then argued that this statement is false because when an atom has more electrons, it also has more protons and as the orbitals become more diffuse, the attraction of the protons would significantly outweigh inter-electronic repulsion at higher atomic numbers.
This led me to think about the following question: What is the highest possible (nuclear charge an electron can experience divided by the nuclear charge of the atom)? My guess is that this electron would certainly belong to one of the more diffuse subshells. (i.e. d or f)
Please note that the quantity I am looking at is not effective nuclear charge but it is the effective nuclear charge divided by the nuclear charge.

A high principal quantum number does not necessarily mean more electrons and/or protons. For example Hydrogen can be in an electronic excited state with $n=4$ and higher, as can be observed for example in the Balmer series. So for a specific element your cited statement remains true.
But you want to compare between different elements now. What we are looking for is minimal screening of the nuclear charge by other electrons:
According to  Slater's rules the second $1s$ electron screens with $0.3$ (all other screening constants are larger). Therefore the ratio you are looking for is $\frac{Z_{eff}}{Z}=\frac{Z-0.3}{Z}$ and converges to $1$ for large atomic numbers $Z$.
The only way to get even less screening would be to have no further electrons. As you restricted the question to neutral atoms we are left with Hydrogen. The charge divided by the nuclear charge of the atom, an electron can experience here is $\frac{Z_{eff}}{Z}=1$ for any excited state (principle quantum number) of the Hydrogen atom.

When does it converge to 1? I am still not clear about that part. Could you clarify?

Nuclear charge equals atomic number. I updated it.

Oh ok I think I get it. For a one-electron hydrogen atom since there are no other electrons to shield the nuclear charge, the effective nuclear charge it experiences is certainly +1, which is equivalent to the nuclear charge. How simple! I must have overthought the problem.


text:
Why substitution and elimination reactions are favored respectively at lower and higher temperature?
I read about the competition between nucleophilic substitution and nucleophilic elimination depending on temperature here. Though the webpage clearly says higher temperature favors elimination while lower temperature favors substitution, the explanation on the webpage is unclear.
Need a clear explanation regarding the phenomenon.

@Zhe Good point. You see Mockingbird, elimination is not a "nucleophile" phenomenon, its rather a "base" phenomenon. Regarding your question, consider the fact that eliminations involve more bonds breaking, and hence a higher activational energy. Does that give you any idea why eliminations require higher temperatures?

An even better question is (selectivity aside) what reactions are not favored by heat?

@Pritt But the hyperlinked page also says the reaction 'rate decreases' for substitution decreases as temperature increases. What about that?

@Mockingbird I'm not very sure, but it could be due to the opposing backward reactions.

That's why I hyperlinked. Some points are explained not so clearly in the webpage. Please visit the page.

No. You're comparing relative rates. The absolute rate of both reactions increase.

For the nucleophilic substitutions and eliminations, we can draw up these four generalised schemes:
$$\begin{align}
&\mathrm{S_N1}{:} & \ce{R-X + Nu- &-> R+ + X- + Nu- -> R-Nu + X-}\\
&\mathrm{S_N2}{:} & \ce{R-X + Nu- &-> R-Nu + X-}\\
&\mathrm{E1}{:} & \ce{RCH2-CR2-X + B- &-> RCH2-CR2+ + X- + B- -> RCH=CR2 + X- + HB}\\
&\mathrm{E2}{:} & \ce{RCH2-CH2-X + B- &-> RCH=CH2 + X- + HB}\end{align}$$
Clearly, each substitution keeps the number of freely diffusing particles the same while each elimination increases it by one. Therefore, the entropic term of the eliminations is likely to be higher than that of the substitutions. The rest is given by the Gibbs free energy equation:
$$\Delta G = \Delta H - T\Delta S$$
The entropic factor scales with temperature as Bryce already mentioned so this outweighs any other effects at high temperatures.

When the question is asking about favorability, it is really asking about the magnitude of the Gibbs free energy. The more negative this value, the more favorable a reaction is. Gibbs depends on enthalpy and a temperature scaled entropy: $$\Delta G=\Delta H-T\Delta S$$
When we think of nucleophilic elimination, we know that a hydrogen atom is removed by a nucleophilic group resulting in our final product. Since a hydrogen is removed in elimination, while left intact during substitution, there are more final species formed during an elimination reaction. According to general chemistry principles, this means that there was a larger increase in entropy compared to the substitution. As temperature increases this larger entropy becomes more noticeable as it is now scaled by a larger $T$.
Therefore at low temperature, the enthalpy term dominates Gibbs and the substitution reaction prevails. However as $T$ grows, so does the entropy term and at high $T$ this overrides enthalpy, making the elimination reaction more favorable.

Breaking bonds requires an input of energy. Elimination reactions require the breakage of more bonds than substitution reactions, and therefore require a higher input of energy. This indicates that the activation energy of elimination reactions is higher than that of substitution reactions. Increasing temperature will allow more molecules to reach the required activation energy threshold. Therefore, as temperatures is increased, elimination is favored.


text:
Dimensionality in quantum mechanics
I was reading Introduction to quantum mechanics by David J. Griffiths and came across following paragraph:
" 3. The eigenvectors of a hermitian transformation span the space.
As we have seen, this is equivalent to the statement that any hermitian matrix can be diagonalized. This rather technical fact is, in a sense, the mathematical support on which much of a quantum mechanics leans. It turns out to be a thinner reed then one might have hoped, because the proof does not carry over to infinite-dimensional spaces."
My thoughts:
If much of a quantum mechanics leans on it, but the proof does not carry over to infinite-dimensional spaces, then hermitian transformations with infinite dimensionality are spurious.
But there is infinite set of separable solutions for e.g. particle in a box. So Hamiltionan for that system has spectrum with infinite number of eigenvectors and is of infinite dimensionality.
If we can't prove that this infinite set of eigenvectors span the space then how can we use completness all the time?
Am I missing something here? Any missconceptions?
I'd appriciate any help.

I'm voting to close this question as off-topic because it is so much better suited for Physics.SE.

You may want to read about the spectral theorem.
You need more assumptions for infinite-dimensional spaces (compactness).

The real issue is not whether this belongs on Physics, but instead I would suggest that the Math SEs are better. Beyond that, however, is to note that this really is one of those points where a math proof may not be relevant to physics or chemistry. It is clear that in real physical problems that only a few symmetries tend to occur, and that a readily identifiable set of eigenfunctions (sin/cos, Bessel, Legendre, ...) are known for those.

@FelipeSchneider what kinds of assumptions? I'll check spectral theorem.

@f13 I think the closest you may get to the finite-dimensional case is to ask for compact self-adjoint operators. Those are diagonalisable (in the sense that there is an orthonormal basis consisting of eigenvectors of it).
Kreyszig is a good read on this (specially the last two chapters).


text:
Substances comprised of large molecules are liquids at room temperature?
I am looking for compounds that are liquid at normal temperatures and are physically very large, to be used in an ellipsometer adsorption experiment. I am currently working with decane. The candidates will ideally not be toxic or radioactive (such as $\ce{UF6}$).

What about solutions of proteins? Human serum albumin or something even larger?

@ToddMinehardt Must be a pure liquid for use in vacuum with no secondary components

Maybe someone can comment on the suitability of a room temp. ionic liquid?

You need a molecule with weak intermolecular forces and low symmetry.

https://en.wikipedia.org/wiki/Polydimethylsiloxane ?

Dirk - what are you looking for the molecule to do in your experiment? Provide the largest change per molecule in psi and delta, or what? Presumably a visible ellipsometer, not IR or UV?

Look for liquid crystal molecules, many are large, for example those based on porphyrins.

You may give porphyrins a try.  These dyes are of interest for molecular electronics, and to render them less crystalline / more glassy, alkyl chains are attached to the central motif.  One recent example are the room temperature liquid porphyrins developed in the group around professor Gryko:

(source, Chem Asian J. 2010, 5, 904-909; doi: 10.1002/asia.200900693.)
Besides standard spectroscopic characterisation, the publication contains the results of DSC:

so compound 4d ($\ce{C176H294N4O12}$) may be a good candidate for an organic molecule, without charges and metal coordination, and suitable for ellipsometry / VASE measurments on surfaces.


text:
Combustion of Rocket-Candy with Fe2O3
We know from Wikipedia that the combustion of rocket candy follow this equation :
$\ce{48 KNO3 + 5 C12H22O11 → 55 H2O + 36 CO2 + 24 K2CO3 + 24 N2}$
We can add small quantity of iron (III) oxide  ($\ce{Fe2O3}$) as a catalyst.
The new equation look like this :
$\ce{KNO3 + C12H22O11 + Fe2O3 -> H2O + CO2 + K2CO3 + N2 + ?}$
What product does the combustion of iron (III) oxide  produce ?

I think you may be missing a key point to catalysis; that the catalyst is not net consumed or altered by the reaction. If you were to put the catalyst in the chemical equation as in your second reaction as a reactant, you would technically need to also put it on the right hand side as a product. However, this is not the way a catalyzed reaction is typically indicated. A common way to indicate that a catalyst is used would be as follows:
$$\ce{48KNO3 + 5C12H22O11 ->[Fe2O3] 55H2O + 36CO2 + 24K2CO3 + 24N2}$$


text:
How to separate two aqueous salts?
I have a mixture of two aqueous salts, $\ce{Cr(OH)3}$ and $\ce{NaNO3}$ and I'd like to separate them and purify the $\ce{Cr(OH)3}$ in order to make $\ce{KCr(SO4)12 . H2O}$ (Chromium Alum). Does any one know of a way to synthesize or separate out this compound?

Hydroxide is quite insoluble, so I wonder what's the problem here?

i have read that Cr(OH)_3 is one the few water soluble chromium compounds....

https://en.wikipedia.org/wiki/Chromium(III)_hydroxide No it isn't.

Chromium has quite a lot of water-soluble compounds, but this one is not among them.

Really? Maybe I added excess KOH, and it ended up dissolving. Perhaps i should do i dropwise addition of KOH to $Cr(NO_3)_3$?

Yes, that's the point.

The common way to synthesize chrome alum is the reduction of a sulfuric solution of potassium dichromate with ethanol.

Physical method
According to solubility rules, any nitrate salt is very soluble in water. Hydroxide salt are sparingly soluble in water. In particular, chromium hydroxide is insoluble in water. So, you can easily decant off the nitrate solution leaving behind chromium hydroxide.
Chemical method
Chromium hydroxide is amphoteric i.e. it reacts with both base and acid while sodium nitrate is not.

As base

$$\ce{Cr(OH)3 + NaOH -> NaCrO2 + 2H2O}$$
$$\ce{Cr(OH)3 + 3NaOH → Na3[Cr(OH)6]}$$

As acid

$$\ce{Cr(OH)3 + 3HNO3 → Cr(NO3)3 + 3H2O}$$
You can get back your chromium hydroxide from chromium nitrate by reacting with base and from sodium complex salt by reacting with acid.

Thanks guys. My problem was that i was not adding sufficient KOH to my solution, so it was not precipitating. I added a slight excess and it worked.


text:
Why is triethyl orthoformate stable?
Geminal diols or triols are very unstable and are easily converted into keto or carboxylic compounds. Triethyl orthoformate is a triether of similar structure. I want to know why the former is so unstable and the latter stable.

Why are gem-diols unstable have you thought? Does the reason you have found imply that gem-diethers are unstable as well?

Initially i thought that they are unstable because of stearic hindrance but when I first came across triethyl orthoformate i realised that this is not the case.

No, this has nothing to do with steric stuff. (You can have links to four C all right, and O is smaller than C.) Long story short, this is because H is labile and R is not.

Okay then what exactly is happening?

https://chemistry.stackexchange.com/questions/10722/what-makes-c-o-more-stable-that-coh%e2%82%82

Is this really a duplicate? This question asks for CH(OR)3 while the question in the link asks for CH(OH)3. I don't think this is a duplicate.


text:
How to calculate the specific heat of ethanol?
I am currently stuck with this question. Can somebody please help me with this?

100 g of water was heated from $\pu{15 °C}$ to $\pu{70 °C}$ by burning $\pu{20 g}$ of ethanol. How much energy per mole does this ethanol have?

So far I have calculated that there are $\pu{0.43 mol}$ in $\pu{20 g}$ of ethanol using the formula $n = m/M$.
\begin{align}
m &= \pu{20 g}\\
M &= 12 + 3 + 12 + 2 + 16 + 1 = \pu{46 g/mol}\\
n &= 20/46 = \pu{0.43 mol}\\
\end{align}
From here, I'm not sure if I should divide the energy used to heat up the water by the amount of substance. By using the specific heat formula,
$$Q = c \cdot m \cdot \Delta{}T
= \pu{4.2 J//g \cdot K} \cdot \pu{100 g} \cdot \pu{55 K}
= \pu{23100 J}$$
$$\text{Energy per mole} = \frac{\pu{23100 J}}{\pu{0.43 mol}} = \pu{53720.93 J//mol}.$$
Is this the right answer?

Each gram of water has an specific heat of 4 J/(K.g)*. So heating $\pu{100 g}$ of water from $\pu{15 °C}$ to $\pu{70 °C}$ would take:
Heat = (70°C - 15°C) * 100g * 4 J/(K.g) = 22,000 joules = 22 kJ

$\pu{20 g}$ of ethanol would be:
Ethanol mol = (20g / 46,07 g/mol) = 0.43 mol

So this means that:
0.43 mol = 22 kJ
     mol = 22 / 0.43
     mol = 50677 J = 50.67 kJ

So ethanol has 50.67 kJ/mol vs 53.72 kJ/mol said by you. I would say that you have made fine your homework, there is a very little difference (I have used some decimal to calculate, maybe you not. (Below you can find a more accurate answer)).

* Water specific heat:  water changes it's specific heat when it changes it's heat. For example: (Wikipedia)
Water at  100 °C (steam)  2.080  J/(K.g)
Water at   25 °C          4.1813 J/(K.g)
Water at  100 °C          4.1813 J/(K.g)
Water at  −10 °C (ice)    2.05   J/(K.g)

Normaly it's used:        4.1813 J/(K.g)

A better table can be found in a french page.

Using this table we can do a better aproximation:
Water heat = (70°C - 15°C)        * 100g
Water heat = (293.03 J - 63.04 J) * 100g
Water heat =    229.99 J          * 100g
Water heat = 22,999 J = 22.999 kJ ≈ 22.9 kJ

Ethanol mol = (20g / 46.07 g/mol) = 0.43 mol
Exactly: 0,4341219882787063164749294551769 mol

0.43 mol = 22999 J
     mol = 22999 / 0.43
     mol = 52978.19 J = 52.97819 kJ ≈ 52.97 kJ
Exactly: 52978,196500000000000000000000001 J

Answer: Ethanol has 52.97 kJ/mol. A value closer to your own value.

PD: I am only a normal person who like chemistry, maybe I am wrong. For example in the theory I have no idea what means $Q = c \cdot m \cdot \Delta{}T = \pu{4.2 \frac{J}{g \cdot K}} \cdot \pu{100 g} \cdot \pu{55 K} = \pu{23100 J}$ Well, I have some ideas but I haven't learned it in the school.

Edit:
I notice that you have your own mol of ethanol (46 g/mol instead of my 46.07 g/mol) and your own specific heat of water (4.2 J/(K.g)). You have to do:
Water heat = (70°C - 15°C) * 100g * 4.2 J/(K.g)
Water heat =      55°C     * 100g * 4.2 J/(K.g)
Water heat =
Water heat = 23100 J = 23.1 kJ

Ethanol mol = (20g / 46 g/mol) = 0.43 mol
Exactly: 0.43478260869565217391304347826087 mol

0.43 mol = 23100 J
     mol = 23100 / 0.43
Using mol decimals (0.43478260869565217391304347826087)
     mol = 53130 k = 53.13 kJ
Using only 2 decimals (0.43)
     mol = 53720,93 J ≈ 53.72 kJ
     Exactly: 53720,930232558139534883720930233 J

Answer: Ethanol has 53.13 kJ/mol (With decimals) or 53.72 kJ (with 2 decimal). So yes! You have done right!

Kimchiboy03 assumed a heat capacity of $\pu{0.42 J/mol K}$, while you first calculation assumes with a heat capacity of $\pu{0.4 J/mol K}$ a value that is almost $\pu(5%}$ smaller than the former.  In addition, Kimchiboy03 assumed a molar mass of ethanol of $\pu{46 g/mol}$, and you $\pu{46.07 g/mol}$.  These contribute to numerical differences in the results.  Eventually, I would round $\pu{52978 J}$ up to $\pu{52.98 kJ/mol}$, rather than down to $\pu{52.97 kJ/mol}$.  Even if this may be seen as nit-picking.

@Buttonwood, oh, I used the value of the mol from Wikipedia.


text:
Volume of gas produced in a reaction
Is it correct to say that 'x' litres of a gas were produced in a reaction, calculating x using the ideal gas equation, since the volume occupied by a gas is taken to be the volume of the container it is in?

They usually say "$x$ liters of a gas was produced at STP". At STP, the temperature and pressure are fixed, so the volume depends on the amount of gas produced.

But isn't the volume taken to be the volume of the container, which is constant throughout?

Not if you keep a fixed pressure. Think of pistons that exert constant pressure. More gas means the piston expands more.


text:
Why is nylon durable and how does it have such exceptional strength?
I was wondering how the chemical composition of a synthetic polymer such as nylon results in strength and durability.

Polyamides (such as Nylon) are molecules containing a large number of amide functions:

In those functions, the $\ce{C=O}$ bond is polarized (due to the difference in electronegativity between C and O) : $\ce{^{\delta+}C=O^\delta-}$.
The N-H bond is polarized, too: $\ce{^{\delta-}N-H^\delta+}$.
This leads to H-bonds between the chains:

While H-bonds are weak interactions, their high number (as there are a lot of amides in polyamide chains) leads to strongly bonded chains, which results in increased strength and durability.