text:
Lowering the melting point of water with salt
I was trying to calculate amount of salt needed to lower the melting point of water by 5 °C. According to my solutions, I need to make sure $$n\mu(\ce{H2O(s)}) =n\mu(\ce{H2O(l)})+ RT\log(x)$$where $ x=\frac{n}{n+2m}$
I, however, thought that I need to make sure that
$$n\mu(\ce{H2O(s)})+m\mu(\ce{NaCl(s)}) =n\mu(\ce{H2O(l)})+m\mu(\ce{NaCl(aq)})$$
After all, the solid ice still has a certain contribution. And I would have assumed that water can be seen as the solved according to the first convention? So why don't we need to consider the solid ice, and what is wrong with my way of thinking?

You should write independent equilibrium expressions for the chemical potential of each species.
In a freezing point depression problem the solvent in frozen form is at equilibrium with the solvent in liquid form, so you can write that
$$\mu(\ce{H2O(s))}=\mu(\ce{H2O(sol))}$$
Next, by Raoult's law (assuming it holds), the chemical potential of solvent in the solution (the right-hand term) differs from that of the pure liquid by the term $RT\log(\chi_1)$ where $\chi_1$ is the mole fraction of solvent, so that
$$\mu(\ce{H2O(s))}=\mu(\ce{H2O(l))} + RT\log(\chi_1)$$
This is just the beginning. Next you need to recognize that the standard Gibbs free energy change for melting is just $\Delta_m G^\circ=\mu(\ce{H2O(l))}-\mu(\ce{H2O(s))}$ (assuming the solution is at $\pu{1 bar}$ or approx normal pressure) and invoke the Gibbs-Helmholtz equation in differential form:
$$\left(\frac{\partial(\Delta G /T)}{\partial (1/T)}\right)_p=\Delta H$$
so that
$$\left(\frac{\partial\log(\chi_1)}{\partial (1/T)}\right)_p=-\frac{\Delta_m H^\circ}{R}$$
Now you integrate the van't Hoff equation from $\chi_1 = 1$ (pure solvent) to $\chi_1 = 1-\chi_2<1$, while assuming $\Delta_m H^\circ$ to be constant over this range of $T$. Finally, assuming $T\approx T_m$, expressing $\chi_1$ in terms of solute molality, and assuming the solute concentration is low, results in the freezing point equation
$$\Delta T = K_m m_2 $$
where $K_m$ is the cryoscopic constant
$$K_m=\frac{RT_m^2M_2}{1000 \Delta_m H^\circ}$$
Here $M_2$ is the molar mass of the solute.
You can use this last expression to compute the molality of solute $m_2$ required to depress the freezing temperature by $\Delta T$.
The process of dissolving the solute is not important (whatever happens before it is fully dissolved doesn't matter as long as it is soluble). Also, it is assumed an ideal solution forms, which means interactions of the solvent with solute, and between solvent molecules are equally strong. And the salt remains in solution (none is incorporated into the solid phase). Note also that liquid solvent remains in equilibrium with solid solvent at the melting point $T_m$, and that we assume that the solute concentration is dilute (this is important to derive the last expression, but Raoult's law may hold even for substantial solute concentrations).

Thank you! Two questions: 1. So the salt we use doesn't matter, we might as well have used KI if the amount is the same? 2. Why doesn't the enthalpy and entropy of dissolving our salt matter at all? After all, by freezing our water, we force the salt to become solid again, right? So why don't we have to consider that in our equations?

The process of dissolving the solute is not important (whatever happens before it is fully dissolved doesn't matter as long as it is soluble). Also, it is assumed an ideal solution forms, which means interactions of the solvent with solute, and between solvent molecules are equally strong. And no, the salt remains in solution. And remember that the liquid solvent remains in equilibrium with solid (at a constant T) and we assume the solute concentration is dilute.

@MathewMahindaratne thanks for your edit - that was a weird oversight of mine.


text:
Why do we use hydrochloric in the test for sulfate ions but nitric acid for halide ions?
Recently, I have learnt about testing for ions in qualitative analysis (BBC Bitesize).
The link above which I used for reference mentions that for the test for sulfate anions, hydrochloric acid $(\ce{HCl})$ is used to ensure that there is no presence of carbonate ions.  For the test for halide ions $(\ce{F-},$ $\ce{Cl-},$ $\ce{Br-},$ $\ce{I-},$ etc), nitric acid $(\ce{HNO3})$ is used instead, for the same purpose.
I would like to know why the acids used are different, since they serve the same purpose.
Update: $HCl$ cannot be used for removal of carbonates in halides since $HCl$ contains the $Cl-$ halide.  However, as to why nitric acid cannot be universally applied to all, I am still unsure.

Wouldn't adding $\ce{HCl}$ interfere with a test for $\ce{Cl-}$?

@MaxW how about adding nitric acid in the test for sulfate ions?

Quoting the linked page: "Barium nitrate solution can be used instead of barium chloride solution. However, nitric acid is added first to acidify the test solution. Sulfuric acid cannot be used because it contains sulfate ions - these would interfere with the second part of the test."

The link above which I used for reference mentions that for the test
  for sulfate anions, hydrochloric acid (HCl) is used to ensure that
  there is no presence of carbonate ions. For the test for halide ions
  (F−, Cl−, Br−, I−, etc), nitric acid (HNO3) is used instead, for the
  same purpose.

Good question, but this is not a universal practice. In reality, the choice of acid depends on the actual sample. Qualitative tests, like these, have become merely academic exercises due the to advances in instrumental analysis these days. As you said, one cannot add HCl to destroy carbonates in a sample which is to be tested for halide ions. Dilute nitric is the only choice, because you would not use sulfuric acid in the case of halides because (i) your test sample may contain ions which will precipitate with sulfate (Ba, Ag, Pb, Ca etc.). Concentrated nitric acid would be a poor choice as well because it will oxidize iodide to free iodine.
An excellent book, which is far better than any online resource, is Vogel's Qualitative Inorganic Analysis. It should be available in any good library and it will give an in depth view of qualitative analysis.

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Usage of TFA in Pfitzner-Moffat oxidation and Collins vs Jones oxidation
Is the addition of TFA in Pfitzner-Moffat oxidation to protonate the DCC to form a positive carbocation for the alcohol to attack? Or is there more to it? Why is TFA used and not other acids? Are other acids to weak for this?
Lastly for the Collins vs Jones oxidation, the only discernible differences are the solvent: $\ce{H2SO4}$ vs DCM, and that the Collins oxidation makes use of a pyridin chlorochromate, does this complex dampen the oxidative effect of chromium? So much so that the Collins oxidation can not directly oxidise a alcohol to a carboxylic acid, whereas Jones reagent with $\ce{H2SO4}$ and $\ce{CrO3}$ can?

Actually, other acids have been employed (in the total synthesis of Camptothecin
by Rapoport, phosphoric acid was used. The reaction took 30 hours)

https://chemistry.meta.stackexchange.com/questions/524/i-question-the-use-of-question-in-question-titles

The Collins oxidation employs the bis-pyridine complex of CrO3 in DCM. Pyridinium chlorochromate in DCM is the Corey-Suggs oxidation.

In the original communication of 1963 (JACS, 85, 3027), Moffatt and Pfitzner report that TFA serves poorly and HCl and H2SO4 do not work at all.  All three work well as their pyridinium salts. H3PO4 was first one their list.

Moffat & Pfitzner have accidentally found their famous oxidation while working on mono-nucleotides (acually 5'-phosphates). Breifly, when pyromidium thymidine-5'-O-phosphate was reacted with dicyclohexylcaebodiimide  in anhydrous DMSO at room temperature resulted a violent reaction to give 90% of 5'-aldehyde product (Ref.1). They further pursued this finding and published their finding as specific reaction (Ref.2). The acid they have used in first finding was anhydrous orthophosphoric acid, which have given 90% yield. Hence, it is not necessary to have TFA as the acid. Following is the suggested mechanism for Pfitzner-Moffat oxidation:

About Collins vs Jones oxidation, both of them used chromium-based oxidants. Oxidations with chromium oxidants proceed through a similar mechanism. Different between Collins and Jones are conditions: Anhydrous (in pyridine/$\ce{CH2Cl2}$) vs aqueous (use of $\ce{CH2Cl2}$). Following is the suggested mechanism for chromium-based oxidants:

After first oxidation to aldehyde, chromium oxidant needs $\ce{CH-OH}$ part available to further oxidize to carboxylic acid (see in both mechanism, how oxidation initiated and finished). In anhydrous conditions, this feature is lacked. So, oxidation stops at aldehyde moiety. However, trace of water make aldehyde in equlibrium with its gem-diol moiety, which gives the necessary requirement of $\ce{CH-OH}$ feature. So, in Jones oxidation condition, oxidation will further continue to give carboxylic acids.
References:

K. E. Pfitzner, J. G. Moffatt, “The Synthesis of Nucleoside-5' Aldehydes,” J. Am. Chem. Soc. 1963, 85(19), 3027-3027 (https://doi.org/10.1021/ja00902a035).
K. E. Pfitzner, J. G. Moffatt, “A New and Selective Oxidation of Alcohols,” J. Am. Chem. Soc. 1963, 85(19), 3027-3028 (https://doi.org/10.1021/ja00902a036).

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text:
Looking for a radio-transparent epoxy/resin
I have much hobby experience with 2-part resins, but this time I'm looking to
make a repair part for the inside of a microwave oven (everything I've tried so far heats up, but the original part does not).  I need the result to be hard enough (shore durometer D ~ 75) to rotate the glass turntable tray within the oven, withstand ~ 250 degrees Fahrenheit (it will not be in contact with food) and be radio-transparent at the frequency of 2.4 GHz.
I've reached out to over a dozen epoxy/resin manufacturers, but am having trouble finding such a product.  Thank you in advance.

You might not be able to find a home-usable thermosetting resins such as epoxy or RTV silicone rubber that meets your needs: low dielectric loss (dissipation factor), moderate hardness and stability at microwave cooking temperatures.
Bakelite (urea-formaldehyde thermosetting resin) might be usable, as it's hard and temperature resistant, though its dissipation factor is ~0.005, a tiny bit higher than ideal. However, one-off molding of Bakelite is non trivial, requiring pressurized and temperature controlled molds.
A table of electrical properties of some plastics shows polystyrene, polypropylene and some fluorinated plastics such as polytetrafluoroethylene with dissipation factors <0.001, but these are less hard than Bakelite.
Glass-filled plastics such as glass-epoxy laminate (printed circuit board, PCB, material) and RO4000® hydrocarbon ceramic laminates might be better choices. However, instead of casting a part, you'd need to machine the part from one or more PCB's laminated together. This might actually be easier, for a one-off part, than casting.

He needs a plastic that won't absorb the microwaves emitted within a convention microwave oven.

@MaxW, that's what the dissipation factor shows The lower the factor, the less absorbed. That said, the factors are usually given for 1 MHz, rather than 2.4 GHz, but the glass-filled PCB material cited is ~0.003 at 10 GHz.

@DrMoishe_Pippik Thank you for your answer.  The FR4 approach is interesting!  I wonder what modern microwave oven turntable tray couplers are made of.  (I don't know whether one can tell by looking at it in the oven under the glass tray.)

@DrMoishe_Pippik If you're still around, I have a side question: How would one laminate/glue PCBs together; wouldn't I have the same heating problem since the glue is not microwave transparent?

Possibly, but the amount of energy absorbed depends on the geometry and size of the material. You could use a non-carbon based cement, such as a ceramic cement. See http://www.sauereisen.com/wp-content/uploads/HIGHTEMP.pdf


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Miller indices: How does one translates the orientation to the origin of the unit cell?
In a section discussing Miller indices, my textbook says the following:

The rules for determining the Miller indices of a direction or an orientation in a crystal are as follows: translate the orientation to the origin of the unit cell, and take the normalized coordinates of its other vertex.

What I'm confused about is how one translates the orientation to the origin of the unit cell?
Since the textbook just mentions this without elaboration or illustration, it is not clear precisely what this means or how one would go about it. I would appreciate it if people could please take the time to explain and demonstrate how this is (mathematically) done. Sources would be appreciated, if you have them.

When you reach a confusing point like this, consult another book or source. I usually found it beneficial  to read the same material from 2-3 books. Multiple sources gave me a better understanding of the topic being studied.

@MaxW I attempted to find clarification using a Google search before I posted this question. See the slide beginning Determine the Miller indices for the plane: https://www3.nd.edu/~amoukasi/CBE30361/Lecture__crystallography_B.pdf It's the same problem: The author mentions the process, but I don't see any explanation or illustration of how it is actually done.

try reading this https://en.wikipedia.org/wiki/Miller_index

@MaxW I did, but there isn't even any mention of translations or "origin".

Just to clarify: It isn't the basics of Miller indices that I'm confused about, it's the specific aforementioned concept that I quoted from the textbook. I did attempt to research this myself before bothering people with the question.

Is it translation you don't understand? // Translation is a mathematical concept. Instead of x=0, y=0, and z=0 being the origin, you can translate the origin to a new point and create a new set of coordinates.  So If I want the origin at 1,2,3 the x'=x-1, y'=y-2, and z'=z-3. You can also rotate coordinates in all three axes.

@MaxW I'm familiar with the mathematical concept of translation. What I'm confused about is what is meant by "translating the orientation" to the origin of the unit cell. I understand the concept of an "origin of the unit cell". But what does it mean to "translate the orientation" to the origin of the unit cell (say, a simple cubic)?

@ThePointer Note that you may train yourself a bit, e.g. on https://www.doitpoms.ac.uk/tlplib/miller_indices/index.php

Don't focus too much on the paragraph that you cited, it's not the best description of how to find Miller indices. The word orientation there just means direction. To rephrase your textbook: if you start a vector in the origin of the unit cell, the Miller indices of the direction of this vector are the same as the coordinates of the vector's terminal point. Assuming that the vector ends on a lattice point so that the coordinates are integral. (not sure if my wording is clearer than the original).

Let's say your structure contains a short piece of helical DNA, and it (the helix axis) starts at point P1 and goes to point P2. You want to know the Miller index associated with that direction. So you move the entire helix (without rotating) so that P1 is now at the origin. The new position of P2, lets call it P2(prime), will be at:
$$ P_2' = P_2 - P_1 $$
Just checking that is is a translation (I have to subtract the same vector from the entire structure, otherwise I'm rotating or warping it):
$$ P_1' = P_1 - P_1 = (0,0,0)$$
Again, the pair of two points still points in the same direction, but now $P_1$ is at the origin. Taking the fractional coordinates (i.e. along the unit cell axes) of $P_2'$, you can turn those into Miller indices $h$, $k$, and $l$ by multiplying with a large number and rounding to the next integer. Generally, the direction will be off because of the rounding. However, in special cases, e.g. if your DNA is on some crystallographic symmetry axis, the Miller indices will match your direction exactly.

Could you clarify "and you have a Miller index. h, k, and l will not typically be integers unless your DNA is on some crystallographic symmetry axis.". Miller indices, by definition, must be integers. The direction doesn't need to be a symmetry axis to have (integer) Miller indices.

@marcin I edited the answer. Here is an example: Let's say your fractional coordinates are (0.3147, 0.5324, 0.4143). Multiplying by 10000 tells me that the direction with Miller indices (3147, 5324, 4143) will point in just about the same direction (within sigfigs). You are unlikely to measure that reflection though. Multiplying by 32 gives (10.0704, 17.0368, 13.2576) so the Miller indices (10,17,13) have a similar direction to our direction of interest, and you might actually measure that reflection.


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Radial Probability Distribution Curve versus ψ² versus r curve for 1s orbitals
My question is basic, but I have already referred to a couple of books.
This is an excerpt from Linus Pauling's book:

the most probable distance of the electron from the nucleus, which is the value of $r$ at which $D(r)$ has its maximum value, is seen from Figure 21-1 to be $a=\pu{0.529 \mathring{A}}$, which is just the radius of the normal Bohr orbit for hydrogen.
  [Figure 21-1 is the Radial Probability Distribution Curve (RPDC) for the 1s orbital, see below.]

And on the next page he states that:

The function ${\psi^2_{100}}$ has its maximum value at $r=0$, showing that the most probable position for the electron is in the immediate neighborhood of the nucleus; that is, the chance that the electron lie in a small volume element very near the nucleus is larger than the chance that it lie in a volume element of the same size at a greater distance from the nucleus.

Don't these two statements contradict each other? $\psi^2$ versus $r$ says that the probability density is highest at $r=0$. RPDC says that the probability is highest at $r=\pu{0.529 \mathring{A}}$. Why is there such a stark difference between probability density and probability? When $r$ tends to $0+$ we observe high electron probability density (wrt the $\psi^2$ curve) but minimized electron probability (wrt the RPDF curve)
I do understand, however, that Radial Probability Distribution Curve represents probability and $\psi^2$ versus $r$ represents probability density. But why is there a difference in these two?
For reference:

This is almost certainly fully discussed in (and quite possibly a duplicate of) https://chemistry.stackexchange.com/questions/104430/why-are-the-radial-wavefunction-and-radial-distribution-function-different; https://chemistry.stackexchange.com/questions/92244/what-is-the-difference-between-%cf%88-%cf%88%c2%b2-radial-probability-and-radial-distribut; https://chemistry.stackexchange.com/questions/107097/what-is-the-exact-definition-of-the-radial-distribution-function

I do kind of get the language in the answers but the math is kinda advanced for me as a high schooler to understand. Could you please try and word it down for me by reducing the math a bit? That would help a lot. Thanks in advance @orthocresol

And also I agree to all these answers completely with respect to the math. After a while of reading I've almost understood the math entirely but I still have a doubt as to why doesn't chemistry adhere to the mathematics of the equation. Math states that the RPDC curve and psi^2 are same but the same does not hold true when it comes to the arrangement of electrons whereby the electrons are more concentrated closer to r tends to 0+ which is chemically acceptable in the RPDC but mathematically unacceptable. Why?

Correction: "Math states that the RPDC curve and psi^2 are same". What I meant to say was Math says that the graphs are acceptable just the way they are but the chemistry behind the arrangement does not.

Does this answer your question? What is the difference between ψ, |ψ|², radial probability, and radial distribution of electrons?

It is very important to note here that $R(r)$ is just the radial part of wavefunction. Wavefunction does not describe any observable. The Born interpretation says that the probability density of finding electron between any two points $x_1$ and $x_2$ is given by
$$\rho=\psi\psi^*.$$
While the probability is
$$P =\int_{x_1}^{x_2}{\psi\psi^* \mathrm{d}x}.$$
(This is only in one dimension. In spherical coordinates you must integrate over the volume).
Why is there a difference between these two quantities? I think it will be easier to understand using an analogy. Consider a rod of unit length in which the charge varies as a function of $x$. Say the function is
$$q(x)=x\mathrm{e}^x-1$$
The average charge of the rod is $0$. While the average charge density of this rod is $\mathrm{e}-2$. Where average is given by
$$\langle f \rangle=\frac{\int^{b}_{a}{f \mathrm{d}x}}{b-a}.$$
I hope you have understood the difference between function density and the function itself. The radial probability and probability density are something similar. By seeing the graphs you can to a certain extent predict the average of the probability density and probability. In the second graph it is $0$ while in the second graph it is close to the peak which is $a_0$.

Probability is basically probability density* volume.
Now we're taking some concentric shells around the nucleus and discussing the probability of finding the electron in those region.
So when you are too close to the nucleus, the volume of the shell is very small. So even if the probability density is high, the value of probability is low.
When r increases, volume of the shells increase, resulting in an increase in probability.


text:
What lubricant can be used for mechanical parts in a system containing diethyl ether?
I make a thermal actuator that works with the expansion of diethyl ether at its boiling point of around 34 °C. For sealing and lubrication of the polypropylene piston and barrel, I need a lubricant that is immiscible with diethyl ether and does not react with it.
I found that diethyl ether is slightly soluble in water, and that it dissolves oils. I suppose this means that many of the everyday oil- and water-based lubricants are not fit for this application (as diethyl ether will dissolve in the lubricant)?
Any other suggestions for a lubricant that can be used here?

Let me ask: can't you consider another liquid with low boiling point, other than ether? Diethyl ether is a great solvent, and anything that comes to my mind (oils, silicone, etc) are soluble in it.

Thanks. An alternative would be N-pentane (boiling point 36 °C), which apparently is insoluble in water, but which is a little bit harder to obtain. Would it be possible to super-saturate a water-based lubricant to minimize dissolution of diethyl ether in it?

Solid lubricants, such as graphite and molybdenum disulfide might reduce friction, and are not soluble, but might not help sealing... you'd need to test.


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Balance Repeatability Sample or Population for SD
A question for the masses about standard deviation (SD), selection of population vs sample in balance measurements.
Analytical balances are typically checked for repeatability by taking 10 weight measurements and finding the uncertainty by calculating standard deviation.
There are two forms of standard deviation: population and sample. Population divides the sum of differences squared by $n$; where as, sample divides sum of differences squared by $n-1$.
If I take 10 measurements of a standard weight, is this not the entire population? Or does the fact that I will repeat this test in the future make it a sample? I would typically select population in this case; however, most of the balance manufacturers I am reading are quoting $n-1$. What is the correct SD?

Use the sample standard deviation: your samples are not the putative entire population.

The "population of measurements" are the infinite number of measurements that could be made with the balance in the exact state that it is in at the moment. Of course that isn't practical. If you measure a sample for years there would be wear on the sample, electronic drift of the balance, weights on the balance would corrode and so forth. The notion is that your ten measurements would be a "sample" from the infinite number of possible measurements. // You have to be careful with "repeatability." The repeatability  of individual measurements is different than the repeatability of the mean.

First, please see my two comments following @Mathew Mahindaratne's answer. Then you might be interested in the pros and cons of the two sample standard deviations that are applicable when the population mean is unknown (which is typically the case): https://stats.stackexchange.com/a/416102/247352 . Bottom line: us chemists generally use the formula with the Bessel correction, i.e., the factor of 1/(N-1) before the summation sign, because it is less negatively biased than the other one and its square is an unbiased estimate of the population variance.

Standard Deviation Formulas in Maths is fun is a nice article to review. It explains why we have use two different formulas to find standard deviation for the population and sample. However, I'd like to point out that your explanation of calculation is a bit off. The correct formula to calculate standard deviation is:
$$ \bbox[yellow,5px,border:2px solid red]
{
\sigma = \sqrt{\left(\frac{1}{N} \sum_{i=1}^N (x_i-\mu)^2\right)}\qquad (1)}$$
where $\sigma$ is the standard deviation of the population and $\mu$ is the Mean (the simple average) of the population of numbers:
$$\mu = \frac{1}{N} \sum_{i=1}^N x_i$$
According to the Math is Fun website:

To calculate the standard deviation of those numbers:

Work out the Mean (the simple average of the numbers)
Then for each number: subtract the Mean and square the result
Then work out the mean of those squared differences.
Take the square root of that.

The correct formula to calculate standard deviation of a sample is:
$$ \bbox[white,5px,border:2px solid blue]
{
s = \sqrt{\left(\frac{1}{n-1} \sum_{i=1}^n (x_i-\chi)^2\right)}\qquad (2)}$$
where $s$ is the standard deviation of the sample (say you choose $n$ numbers out of total of $N$ numbers population) and $\chi$ is the mean (the simple average) of the chosen numbers, hence: $$\chi = \frac{1}{n} \sum_{i=1}^n x_i$$.
Note: In our old days when I was taking analytical chemistry, we were instructed to use equation (1), if the $N \gt 10$, but to use equation (2) whenever we have to deal with data where $N \lt 10$.

Several points. First, equation 2, with n instead of the Bessel correction n-1, is also a correct sample standard deviation. It is more negatively biased than equation 2, but has some statistically nice properties. Second, if the population is infinite, N i.i.d. samples are taken and $\mu$ happens to be known (which is unlikely) and $\sigma$ is unknown, then equation 1 can be used to give yet another sample standard deviation, i.e., the left hand side is "sigma hat". Finally, the 2012 QUAM is recommended for lots more info: https://www.eurachem.org/images/stories/Guides/pdf/QUAM2012_P1.pdf

@Ed V: Thanks for the valuable insight.

Three of the sample standard deviation definitions (and their pdfs and expectation values) are given here: https://stats.stackexchange.com/a/410311/247352 . Note that @Mathew Mahindaratne's equation 1 gives $\sigma$ if N is finite and each of the N elements is sampled. His equation 1 also gives an estimate of $\sigma$ if the conditions are as stated in my previous comment. Bottom line: use equation 2 unless there is good reason to do otherwise. And for large N, it won't make much difference anyway.

Thank you folks for the clarification.  Very good display of the equations and explanation of usage


text:
In situ preparation of Potassium Tert-Butoxide
I am attempting to prepare potassium tert-butoxide. I was going to suspend anhydrous toluene and anhydrous tert-butanol, dissolve (reflux) the potassium metal (pure), then distill the excess tert-butanol as a benzene azeotrope.
2 questions.

The metal comes stored in mineral oil. Should I rinse it in toluene to get the oil off?
How will the presence of oxides effect the reaction (if at all). If so, how can I mitigate this?

tBuOK is supposed to be well soluble in THF, and badly in toluene.Where does the benzene come from? Don't use benzene if at all avoidable, everything else is very bad practice. You should filter the K-butoxide solution to remove any residual oxide etc. That way also the solvents dry themselves during the reaction without any further step.


text:
How to calculate the number of atoms in a gas molecule?

$\pu{20 mL}$ of sulfur vapour at $\pu{1000 °C}$ reacts with $\pu{40 mL}$ of oxygen gas to form $\pu{40 mL}$ of sulfur dioxide gas. Determine the number of atoms in the molecule of sulfur in the vapour state at this temperature.

My working out is as follows:
$$\ce{\underset{\pu{20 mL}}{S(g)} + \underset{\pu{40 mL}}{O2(g)} -> \underset{\pu{40 mL}}{SO2(g)}}$$
I then used the formula $n = m/M$ to find the amount of substance.
$$n = m/M = 20/32.07 = x$$
Then, to find the number of atoms, I used the formula $n = N/N_\mathrm{A}$.
$$N = nN_\mathrm{A} = x\cdot\pu{6.022e23} = \pu{3.80e23}~\text{(atoms)}$$
However, I am uncertain as to whether this is correct as I did not utilize any of the gas laws. For this question, which of the gas laws should I have used?
Any answers would be appreciated.

Welcome to our site! BTW I noticed that you used the volume $(20\pu{mL})$ of $\ce{S}$ in the place of mass "$m$". Are you sure you want it that way?

@WilliamR.Ebenezer Should I be using one of the gas law equations to solve for "n" and then use the formula N=N/Na to solve for the number of atoms?

Instead of using any 'gas laws', why don't you just remember the equation $pV = nRT$?

I'm afraid I misread the question, further fueled by seeing your usage of Avogadro's number.

I have rewritten the answer, sorry for the confusion caused earlier.

@WilliamR.Ebenezer That's alright.

The question isn't about counting all the atoms of sulphur but the number of atoms of sulphur in a single molecule of sulphur vapour. This can be done with simple gas-law reasoning.

Reason for rewrite: refer to edit summary.

From Wikipedia on Disulfur ($\ce{S2}$)

This violet gas is commonly generated by heating sulfur above 720 °C

Since the sulfur is well above $720^\circ\pu{C}$, it would most likely be in the diatomic state, more like an analogous structure to dioxygen ($\ce{O2}$).
We can also judge this from the molar ratios too, as rose22 has pointed out. Since it is given that $40~\pu{mL}$ of $\ce{SO2}$ is formed from $20~\pu{mL}$ of $\ce{S_x}$ (assuming the atomicity of $\ce{S}$ to be $x$ at this temperature) and $20~\pu{mL}$ of $\ce{O2}$, they must come to terms with a simple whole number ratio.
$$\ce{\underset{20~\pu{mL}}{S_x}+\underset{40~\pu{mL}}{xO2}->\underset{40~\pu{mL}}{xSO2}}$$
Statement 1: Since the ratio of volumes of gases at a particular temperature and pressure is equal to the ratio of amount of substance in moles, it is clear that:
$$1:x:x=20:40:40$$
That is to say, $x=2$, so the atomicity of sulfur here is $2$.
In the question, the OP said:

However, I am uncertain as to whether this is correct as I did not utilize any of the gas laws. For this question, which of the gas laws should I have used?

Statement 1 is a  result from the universal gas law.
$$PV=n\text{R}T$$
When $P$ and $T$ are constant,
$$\frac{V_1}{V_2}=\frac{n_1}{n_2}$$
Some may argue that this relationship is called Avogadro law, but again, this law too is a gas law, evidently.

@KarstenTheis If you're referring to the line "dead wrong  to say that there are no gas laws involved", I was replying to question itself, which was(and is still) asking if gas laws were required or not. Not referring to the answer, sire. Anyway, I respect you for explaining your downvote.

I did not realize until now that the question and the other answer were by the same person. You answer makes more sense to me know. I also appreciate the bit about properties of $\ce{S2}$ you included.

The question is (my emphasis):

20 mL  of sulfur vapour at $\pu{1000 ^\circ C}$ reacts with 40 mL of oxygen gas to form 40 mL of sulfur dioxide gas. Determine the number of atoms in the molecule of sulfur in the vapour state at this temperature.

For this question, we must understand that the number of sulfur atoms in the sulfur molecule (the subscript in its chemical formula) is actually an integer ($x$). The volume ratios determine the mole ratios and thus, are able to determine the stoichiometry of the equation. From this stochiometry, the value of ($x$) can be determined.
The working out is as follows:
Let $\ce{S_x}$ be the formula of the molecule making up the sulfur vapour.
Volume ratios (Sx:O2:SO2) = 20:40:40 = 1:2:2 = mole ratios, so the chemical equation with stoichiometric coefficients is:
$$\ce{S_x + 2O2 -> 2SO2}$$
Thus $x = 2$ (to have a balanced equation).
Therefore, the molecule making up the sulfur vapour is $\ce{S2}$.
To conclude, there are 2 atoms of sulfur in this particular molecule of sulfur.

I agree with the corrected statements made in your answer - however, what I meant was that no gas laws equations are required to solve this problem. Also, thank you very much for your additional information.

@rose22 Without using gas laws you wouldn't be able to use direct proportionality between gas volumes and the amounts of substances. It doesn't matter whether you write those down or not — they are the basis of your assumption. Also, note that this doesn't really answers your the question (where is the answer for the number of atoms?) and is more justified as an edit or a shortened comment.

@andselisk Okay, thank you for your input. I'll ensure that the answer is more coherent and distinguishable next time.

I am up-voting because the answer is correct, and the incorrect statement about not using the gas laws have been removed.

@andselisk There was some misunderstanding about the wording of the question, which asks for the number of sulfur atoms in one sulfur molecule, if I understand correctly. I hope folks who downvoted for that reason (or for the now removed statement about not needing any gas laws) reconsider their downvote.

@KarstenTheis I noticed that, but since OP initially attempted to find the total number of sulfur atoms, I thought there might be some incorrect translation of the question. Note that the first revision of William's answer also addresses this number, using an assumption about the composition of sulfur molecule (which, as it's turned out, should be the answer) as an aid to solve the problem.

Also, this solution only makes sense if the question would explicitly say that the reaction was complete, which it doesn't. I think the question is still badly formulated (not OP's fault, I agree).


text:
Choice of buffer pH to avoid aggregation
I want to purify IgG which has a theoretical pI of 6.4-9.0 with an anion exchange column. I also want to keep the pH of my buffers as low as possible and "outside" the range of the pI to prevent aggregation.
Still, I will need to increase the pI so that IgG will have negative net charge and bind to the column. What pH do you think I should choose in order to have the IgG binding to the column and to prevent aggregation as much as possible?
Thanks in advance!

This is an extraordinarily wide pI range, perhaps its because you a purifying an IgG with very many subunits (?). In that case, it might not be good idea to purify this by the means of anion exchange. 'Common' globular proteins (monomers, dimer, trimers or tetramers etc) have one theoretical pI, and you keep the pH of the buffer at approx. 1.0 pH unit above or below the theoretical pI value - preferably towards physiological pH values (to retain proper activity). In the case of anion exchange you need to think of the surfaces charges.


text:
Does energy come from fuel or oxygen? (surprise?)
Question: Is it oxygen or fuel that releases energy during burning?
The answer from this article:
Why Combustions Are Always Exothermic, Yielding About 418 kJ per Mole of O2 (Journal of Chemical Education, 2015)

Our analysis provides simple answers to two related, important
  questions: “What makes fire hot?” and “How much heat is produced?” in
  the combustion of a certain fuel. One can now explain that the
  double bond in O2 is unusually weak, and therefore the formation of
  the stronger bonds in CO2 and H2O results in the release of heat or
  an increase in thermal motion.
Generally, our analysis indicates that it is not the organic fuel
  but rather O2 that is “energy-rich”.

It is usually said that fuel contains energy and that oxygen only enables the release of energy in the sense like enzymes enable reactions. Does anyone want to comment this? Because, from this article it seems that it is just the opposite: fuel enables the release of energy from oxygen.
(This question was initially nutrition-related: Does energy come from nutrients or oxygen?)

Surprised that it makes its way to an ACS journal. What is pointed out it is obvious and not a surprise at all. It might be still necessary to bring it to the attention of a class, but I would say in general and rather elementary education. And surely (now?)adays teachers might spread misconceptions. That a reactant behaves as a catalyst might be one of them, as seen in your question when you mentioned oxygen as comburent treated as an enzyme.

Not to be pedantic. I think is worth pointing to this fact, and might be indeed necessary. But already the title is at least curious, as for exothermic reactions that are not combustions exist. Or their mentioning that fire is always hot. I just comment because not a real answer comes to my mind. Except that on elemental basis yes, most of the energy comes from oxygen changes - tge rest being necessary to order fuels for the energy that they release/allow to be released.

Alchimista, this is an educational article in the Journal of Chemical Education. Why are you surprised :-)? It is an educational article.

Because as it is is not highly educational starting from the title. @M. Farooq because as I said it might be important to point out  facts which are probably obscured by bad practices and /or colloquial wording. But those facts shouldn't be presented as a sort of findings nor a title like this makes sense. Plus a lot of considerations -  more substantial or just technical and semantic - that are rooted in thermohenistry. The entire idea of where the energy comes from is exaggerated as for we are dealing with reaction hentalpy not with the formation hentalpy of reactants.

It seems that the statement "combustions are always exothermic" is simply based on the defintion of combustion, rather than something fundamental about the chemistry. For instance, the formation of nitric oxide, $\ce{1/2N2(g) + 1/2O2(g) →NO(g)}$, is endothermic, with $\Delta _f H^{\circ}[NO]=91.3 kJ/mol$. However, combustion is typically defined as an exothermic reaction.  [See, e.g., https://en.wikipedia.org/wiki/Combustion] Thus while redox reactions exist in which a substance combines with oxygen or some other oxidant endothermically, these aren't combustion reactions only by definition.

I find this somewhat misleading. Energy/enthalpy is much more assigned to the whole reaction or pairs of reactants. One can argue that in general the X-O bonds are very strong, or the C-C, H-H bonds are comparably weak. Also, oxidation reaction in eg F2 gas are highly exothermic yet they have no double bond.

@Greg The quoted JCE article says that the X-O bonds are no stronger than the bonds in the fuel, and that it is the weakness of the O=O double bond/ two bonds that is the source of the energy.

@theorist "combustions are exothermic" did seem circular to me also. The examples seemed cherry picked. The number of reactions that fit the theory means nothing if the reactions are cherry picked, and there was no mention of what proportion of all reactions this large number reactions was, not that they were randomly selected. It makes sense to me that the energy is not in the fuel so much as in the oxygen but I do get the feeling that some of the author's arguments are (self?)-deceptive (those that I could understand, most of it was way over my head).

[OP:] It is usually said that fuel contains energy and that oxygen only enables the release of energy in the sense like enzymes enable reactions.

An enzyme is a catalyst, so it does not change the enthalpy of a reaction. That part is correct. "Oxygen only enables the release of energy" is incorrect. Oxygen is one of the reactants, and the oxygen atoms also appear as products.

[Dr. Schmidt-Rohr in JChemEd paper] Generally, our analysis indicates that it is not the organic fuel but rather O2 that is “energy-rich”.

This might be true in the context the combustion reaction with the set of organic molecules he chose to include. The organic fuels exclusively had C-H, C-C, C=C, O-H, C-O and C=O bonds, if I understand correctly from the abstract and the figure. He is saying that the bond energy of most bonds in reactants and products are similar (e.g. C-H vs O-H, C=C vs C=O, 2 C-C vs C=O), and that the O=O "double bond" is an outlier.
I think it is worthwhile to state the amazing fact that we have a biradical (see https://chemistry.stackexchange.com/a/15061) in our atmosphere and are breathing in a very reactive substance. The only reason that our red blood cells aren't constantly undergoing mutations from oxidative damage is that they lack DNA once they are released into the bloodstream.
If I were to study the combustion reactions with elemental metals, I might come to a different conclusion than the paper. Try comparing $\ce{Na + O2}$ with $\ce{Hg + O2}$.
Final remark (quoting myself):

When a raw egg rolls off a countertop and falls on the floor, does it crash because the countertop is so high or because the floor is so low?

Excellent answer (+1) and I loved the egg thing! I have always believed that it is the chemical reaction, in all its (often very complex and/or obscure) glory that matters. So, for example, powdered PTFE is the oxidizer of powdered Mg 'fuel' in MTV countermeasure flares (https://en.wikipedia.org/wiki/Flare_(countermeasure) ). Of course, PTFE is just the generic version of Dupont's trademarked Teflon and, aside from run-ins with NaK and such like, I have never considered my Teflon/PTFE beakers and such to be particularly high energy! There are also 'cold flame' combustions that are hand safe.

and does the egg crash because its shell is weak or because the floor is hard?

Energy usually comes from reactions not one component of the reactions
Describing oxygen as "energy rich" doesn't make sense outside of the context of reactions oxygen can participate in. And, in that context, it doesn't make sense to partition the "energy* between the things that are reacting.
Chemical reactions can release energy. This happens because the bonds in the results of the reaction are in a lower energy state than those in the starting materials. Oxygen often participates in exothermic energy-releasing reactions because the resulting bonds (eg the bonds between carbon and oxygen in carbon dioxide) are stronger (or in a lower energy state) than the sum of all the bonds in the starting materials.
It is the strength of the bonds in the product that is significant here and not the "energy stored" in the starting materials (eg a hydrocarbon and oxygen gas). It isn't that dioxygen molecules "store" energy, it is that many oxygen compounds in the products are in low energy states with strong bonds. The release of energy makes sense only in the context of a chemical reaction and not in terms of one of the starting materials. Many compounds with strong bonds to oxygen (even silicon dioxide (silica) can still release lots of energy when reacted with the right compound (which is why they don't let schools do experiments with chlorine trifluoride which sets sand on fire).
And the idea that oxygen is the "store" of energy will confuse us when talking about compounds which, by themselves, are effective stores of energy. Many explosives can be persuaded to undergo strongly exothermic reactions without having to do something as gross as reacting with another compound but just by decomposing. Describing these as "stores" of energy makes sense. Getting compounds like nitroglycerine or nitrogen triiodide to release that energy sometimes takes little more than a bad glance or the touch of a feather. But try persuading pure oxygen to do anything similar and you won't get very far.
In short, describing one component of a reaction as a "store" of energy distracts form the key issue in chemistry which is thinking about reactions and confuses reactions involving multiple components with decompositions that release energy all by themselves without having to engage with other molecules.


text:
Identifying Name/Purpose of piece of ellipsoid glassware with internal tube?
I recently went to a market in Southern France and found this little piece of equipment for sale along with a whole bunch of other chemistry things. the guy had no idea what it was for or where he'd gotten it from, but for €1 I decided I had to have it because I've never seen anything quite like it (and I'm mildly infatuated with old chemistry equipment...). I'm wondering if anyone here would be able to identify it?
An image with a 10 pence coin for size comparison.

My first thought was some sort of condenser, because the design is a quite similar due to the external jacket with an internal tube. However on closer inspection the internal tube has a cut out section which would mean any coolant would mix with the substance being condensed.
this diagram shows a little bit more detail about the internal construction of the piece (cross-section):

There is also some text on the body of the tube which says:

EX 2ml COLD E - MIL LINE

Thanks for any information someone could give me about what this piece of equipment is and/or what it was used for.

Could it be a liquid/vapor separator? If you invert the drawing, liquid entering either of the bottom tubes would exit the other, but vapor could exit the large-diameter tube. The side-exit of the inner tube could direct liquid spray away from the vapor tube.

It could be a kind of "atomizer" giving a sprayed flow out of the little "outlet"?

https://www.reddit.com/r/chemistry/comments/rqiaij/is_anyone_able_to_identify_this_glassware/ - related

What you got there is a tilting head dispenser/tilting pipette dispenser. You are lucky that you found one for €1 because it usually costs around $20.
It is small in size because it is to be fitted on the mouth of a container and so called by another name "bottle-top dispenser".
Its is to be attached to a bottle or reservoir, its main function is to dispense the liquid from the container in precise volumes. It is usually used in the fields of analytical chemistry, molecular biology, and pharmaceutical research where accuracy and precision of volumes matter.

Flask tilts back to fill to desired volume, forward to dispense.

Source: https://www.coleparmer.com/p/kimax-tilting-dispenser/43264


text:
What is the reason for phase separation?
I have seen phase separation as a phenomenon taking place. However I do not understand why this phenomenon occurs as a counter example to the natural laws of diffusion along the concentration gradient. During phase separation the richer side becomes richer with the atoms which it already has and becomes more and more deficient with the poorer atoms and vice-versa. Similar things will occur in the opposite case as well. How can phase separation be explained in detail, for example with factors like the chemical potential?How do you quantify whether 2 liquid interactions will be unfavorable or not?

The 2 liquids' interactions are unfavourable, so they simply go for minimal contact policy.

@WilliamR.Ebenezer Yes but how do you quantify whether 2 liquid interactions will be unfavourable or not?

What do you mean by "in detail?" That makes this question very broad and unspecific, and therefore unsuitable for the site. It would help if you could narrow down the question, perhaps by including in your post the last question in your comment, "how do you quantify whether 2 liquid interactions will be unfavourable or not? "

@BuckThorn edited

The change in Gibbs energy has to be negative for a process to happen (at constant pressure in the absence of work). Gibbs energy is a combination of entropy and enthalpy.

I do not understand why this phenomenon occurs as a counter example to the natural laws of diffusion along the concentration gradient.

Diffusion is driven by entropy changes, specifically the entropy of mixing (in an ideal solution, the enthalpy of mixing is zero, so the entropy term is the only consideration). The entropy goes up when a solute moves from high to low concentration.
Phase separation occurs when the enthalpy term dominates. If substances A and B are initially mixed and then undergo phase separation, some change occurred (change of temperature, added third component, etc.) so that now phase separation more exothermic (lower energy for [A to interact with A and B interact with B] than [A interacting with B]). Under these circumstances, phase separation can happen.

What will be the typical shape of the G vs X curve for a single phase two component system?

How will the G vs X curve look like in such circumstances?There is a typical "common tangent" method that we can apply to find out the equilibrium condition for a multi-phase system.Now that works pretty well when I have two concave upwards shapes.What if I had concave downwards?How would I find the minimum free energy?Common tangents could not be drawn then isn't it?

@user586228 See https://chemistry.stackexchange.com/a/113061/72973 for the equilibrium.

@user586228 http://web.mit.edu/hujuejun/www/Teaching/MSEG%20803/12%20--%20Solution%20Theory.pptx


text:
How is the 'rate law' and 'order' for the ozone decomposition reaction can be defined?
How is the 'rate law' and 'order' for the following reaction can be defined?
$$\ce{2O3 -> 3O2}$$
"The expression that correlates the rate of reaction with concentration of reactants is known as rate law for that reaction."
So the rate law only include the concentration of reactants, not products.
But the rate law for this reaction is:
$$r = k[\ce{O3}]^2[\ce{O2}]^{-1}$$
where $k$ is rate constant.
"Order of a reaction is the number of reactant molecules whose concentration determines the rate expression."
Order for this reaction w.r.t. ozone is 2 and that w.r.t. oxygen is -1. But why we are considering oxygen, being a product of this irreversible reaction. This also means that the rate of this reaction will decrease with the increase in concentration of oxygen.
Now it seems that, this is a reversible reaction. But chemical kinetics is studied for only irreversible reaction?

Chemical kinetic describes both reversible and irreversible reactions.
The reaction scheme and rate expression do not refer to the same thing. The reaction scheme is an overall one whereas the rate equation refers to a particular multi-step reaction. Two examples are given below:
Ozone can be destroyed by UV light in the $240$ to $\pu{320 nm}$ region with the scheme $(\ce{M}$ is an inert gas needed to take away excess energy):
$$
\begin{align}
\ce{O3 &->[$hν$] O2 + O} \\
\ce{O3 + O + M &-> O2 + O2 + M} \\
\hline
\ce{2 O3 &-> 3 O2}
\end{align}
$$
Similarly, oxygen molecules can react with short wavelength UV as
$$
\begin{align}
\ce{O2 &->[$hν$][$(<\pu{242 nm})$] O + O} & \\
\ce{O + O2 + M &-> O3 + M} &|\cdot 2 \\
\hline
\ce{3 O2 &-> 2 O3}
\end{align}
$$
which also gives a similar net reaction. This is the Chapman mechanism and is very important in atmospheric chemistry.


text:
Why does a tetrahedral molecule like methane have a dipole moment of zero?
I was studying chemical bonding when I noticed something odd.
We say compounds like $\ce{CCl4}$ and $\ce{CH4}$ have a tetrahedral geometry (which is a 3D structure) but when we talk about their dipole moments, we say they have no dipole moment. We give the reason that as the H atoms are opposite each other (hence assuming it to be a 2D structure), they cancel out their bond moments.
But why? In the beginning we saw that they are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero.
But certainly I am wrong as these values have been calculated scientifically. So can someone please point out where i am going wrong in my understanding?

Devil's advocate addon question: Does a single crystalline piece of methane also have no dipole moment? ;-)

@Karl Based on your very helpful answer to another question by uhoh, I will make a wild guess: dielectric relaxation spectroscopy? ;-)

@EdV Well, some substances have a finite dipole moment in the terminal (--> zero frequency) regime. Another word for frequency dependent dipole moment is complex permittivity. ;-) Your guess can tell the answer ...

Actually, I'm not 100% sure if the answer to my question is "of course not, silly, it cancels out", or "obviously, stupid". If someone would explain that, I'd throw a 100pt bounty into the audience.

Karsten's answer is excellent, but here is a figure that shows the mathematics involved:

The central atom (green) is at the center of the cube, the four other atoms (purple) are at alternating vertices and the geometry should be clear.
Alternatively, if you orient the molecule so one peripheral (purple) atom is directly "above" the central (green) atom, then each of the other three atoms is just $1 - \theta$ ($\approx 70.52877940 ^\circ$) away from being directly "under" the central atom, so each contributes $\cos(1 - \theta)$ times the bond dipole. This is the downward component of the bond dipole from one of the lower atoms.
But $\cos(1 - \theta) = 1/3$, so this is simply (bond dipole)/3, and there are three of these lower atoms, so the three downward components exactly balance the one upward component.

Nice figure and geometry treatment. I'm so happy someone actually wants to know the math!

If you put the carbon at the origin (0 0 0) and the hydrogens on (1 1 1), (1 -1 -1), (-1 1 -1) and (-1 -1 1), it is also easy to see the dipole moments add up to zero (corresponds to orientation 4 in my answer).

Indeed! I was a bit reluctant to post this, given your excellent answer. But I did the figure years ago and thought it might be helpful. The cube also directly relates to your first figure and the symmetry is clear: the two equal upward components equal the two equal downward components.

(cc @Karsten) Even though both answers address this particular question, it seems to me the forest vanished behind a tree (a tetrahedral one, apparently). OP might come up with the similar question for the symmetry resembling every platonic solid out there, and trying to explain each and every case with 3D images and geometric constructions seems counterproductive. IMO it should be explicitly stated that centrosymmetric molecules have zero dipole moment (and are also non-chiral — as a bonus), probably add a proof why it is so. Scientific approach is a unified approach, after all:)

@andselisk I think there is a balance between answering the actual question and the question that should have been asked. Also, there are cases where the general rule does not erase the misconception of the OP - in this case why methane looked "unbalanced" when one bond is pointing straight up.  I started on the symmetry argument in my answer, but I did not mention the symmetry of inversion. I appreciate your comment and insight, and I think there is space for one more answer and many more questions (e.g. is a center of inversion necessary and sufficient for the dipole moment to vanish?)

There is no difficulty in doing the cases of 2 through 8 non-central atoms around the central atom, and the 7 attached atoms case even involves the equatorial pentagram's Golden Ratio, while the 8 attached atoms case uses the cubic or twisted cubic symmetry. But I was under the impression that answers should have a "laser-like focus" on answering the asked question. The Help information says generalizing the answer is desirable, so the answer would be "for the ages". I am new here, so I do not yet have a finely calibrated sense of balance. Thoughts?

@andselisk Simply stating that "centrosymmetric molecules have zero dipole moment" would probably just produce the response "but why?" from the OP. If he/she could "see" that was true without any help, the question about a particular geometry would never have been asked. In fact asking "what exactly does centrosymmetric mean" is not a silly question in its own right.

Orientation 1
Consider the orientation of methane below. The partial positive charges are distributed equally around the central atom. Comparing the left with the right side of the molecule, the top left and top right cancel each other out, and the other two are in the middle with the carbon. Comparing the top with the bottom side of the molecule, the two top hydrogen positions cancel out with the two bottom hydrogen positions. Comparing the front with the back side of the molecule, the bottom front and the bottom back hydrogen positions cancel each other out and the other two are in the plane of the paper.
For a polar molecule, the positive partial charges have to be separated from the negative ones along a direction (along the overall dipole moment, which is a vector). Here, the positive partial charges are on the outside and the negative partial charges are on the inside.

If you like vectors, you could also say the two top bond dipole moments added up point straight upwards, while the two bottom dipole moments added up point straight downwards. If you add those two up, you get a net dipole of zero.
If you like symmetry arguments, there are six mirror planes along the H-C-H planes (the left-right and the front-back are easy to see), so there can't be a dipole. The argument is as follows: If there were a dipole moment, for example left to right, and I apply the left-right mirror plane, it would have to switch directions, but the molecule is still the same. Because of this contradiction, the dipole moment has to be zero in that direction. The same argument goes for the three-fold and two-fold axis in the molecule.
Orientation 2

Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to 0.

The upper H atom is straight up, while the downward H atoms go at an angle, with the component down being 1/3 a bond length. So in that orientation, it also cancels out, but is more difficult to believe.

Orientation 3
In this orientation (used for Fisher projections in sugar chemistry), you can see the symmetry nicely as well one up, one down; one left, one right; two front, two back.

Orientation 4
Maybe my favorite: two up, two down, two left, two right, two front, two back - that is symmetric.

"In the beginning we saw that [methane molecules] are 3D structures with one H being at the top and other 3 at the bottom, with the bond dipole moment directed from each H towards C. Due to this the component of the bond moment of the 3 downward H atoms along the line of the upper H atom would cause a net upwards bond moment which is not equal to zero."

So what happens if you turn the molecule around so that one of the other H atoms is now at the top?
Remember that a tetrahedron is symmetric, so that it looks the same when viewed from any of its four corners.  So if the tetrahedral methane molecule had a non-zero dipole moment pointing towards one of the hydrogens, by symmetry it would also have to have an equal dipole moment pointing towards all of them.  But then it wouldn't be a dipole, but an octupole — a dipole can, by definition, only point in one direction.

Nice symmetry argument! This could be generalized to "any molecule with centrosymmetry has a dipole moment of zero", but the proof did not fit into this margin (@andselisk)

But is the symmetry perfect? What if we account for tiny aberrations (thinking of spins)?


text:
Why doesn't the Pilling-Bedworth ratio of silicon cause issues in silicon technology?
According to Wikipedia, silicon has a Pilling-Bedworth ratio of 2.15. In theory this should mean an SiO2 film on silicon should have a fairly large amount of compressive stress and high probability of fracturing. After all, 2.15 is even higher than the 2.14 PB ratio of iron, which flakes quite readily as rust. However in practice this is obviously not the case, and in fact the growth of oxide films on silicon is usually performed quite simply by thermal growth without any further technique. Is this simply due to the low thickness of these films or is there another reason silicon oxide behaves nicely?

The film is amorphous for one.


text:
Conditional kinetic equations
From a biological system, I got the following reaction:
$$A+B+D\rightarrow C+C_1D$$ ($C_1D$ is a certain complex)
However, if at any given time there are no $D$s to react, and only then, instead the following reaction does happen (until a new $D$ appears, at which point the first reaction switches in instead):
$$A+B\rightarrow C+C_2$$
So, the second reaction happens if and only if there are no $D$s.
Therefore, this system is like a "conditional" system, where under existence of $D$ the first reaction happens, until all $D$s are used, and then . At any given time, more $D$s can appear and then the whole reaction system acts again as described above.
In the above system, what are the kinetic equations?
(Note: of course, this is not the complete reaction scheme, but the minimum one to extract equations for. However, it should be noted that I consider $[D]=g(t)$, i.e. not constant equal to the initial concentration of $D$. Of course $g(t)$ also depends on the aforementioned reactions too, but on others also)

Without a mechanism you can't get the rate equation (i.e. rate vs concentration of species). And the mechanism will be complicated because the way you set it up, reaction 2 is shut off completely as soon there is a little bit of D. That is very unusual.

You will only be able to solve these equations numerically if D varies in time externally to the scheme given, ie. as some arbitrary g(t). If you want exclusive reactions (or nearly so ) then the rate constant for reaction 1 should be vast compared to that for 2 such that when D is present  1 is favoured, but you will always have some of reaction 2.


text:
How to calculate molecular emission spectra?
Is there free software that can calculate the fluorescence emission spectra of organic molecules?
Let's say, for example, rubrene
There is a bunch of quantum free quantum chemistry software:
https://en.wikipedia.org/wiki/List_of_quantum_chemistry_and_solid-state_physics_software
However, which one would be the most suitable for this purpose? Could this be done on a reasonable up-to-date laptop? Of course, computing times will depend on the number of atoms in the molecule and the desired precision. What would be approximate calculation times? In the order of hours, days or weeks?
I can imagine that this is most probably not a simple "draw molecule and push calculate-button"-task, but any hints where to further dig into are appreciated.
Addition after the question was put on hold:
This question certainly can fill a semester of a lecture on quantum or computational chemistry. In that sense, yes, it is broad.
However, instead of detailed instructions or input scripts, I was hoping for some rough guidance about the essential steps and the available software which can be recommended for that purpose (and equally important: which software can be excluded or cannot be recommended to avoid investing a lot of time and find out later that there are much better options). And of course, whether you need a super-computer or whether you can do it on a laptop within a reasonable amount of time.

As a start you could assume harmonic potentials and with the displacement between excited and ground state potentials calculate the Franck-Condon factors for displaced and different frequency oscillators, see Chem. Phys. Letts, v 18 p 481, 1973 for the equations. V simple calculation. Most aromatics only have a couple of FC active modes that dominate the spectrum which is just as well as there are 3N-6 modes and using all combinations/overtones   in a full calculation involves a truly vast number of pairs of levels.

Thank you for the reference, I will check. Although, I my goal is not to implement the equations myself. Therefore, I'm hoping for established software to take the geometry of the molecule (i.e. position and type of atoms) as input, which I guess need to be calculated first. So, in short, ideally SMILES (or others) code in --> spectrum out.

You could use the Avogadro software to create input files from SMILES for a number of different codes (Gaussian, Orca, Psi4, ...). But you will still have to select the actual method and basis set and tell it run a geometry optimization and frequency analysis. This would give you an IR/Raman spectrum. Fluorescence is more complicated than that, as it includes couplings between different electronic and vibrational excited states.

Could you edit your question to include the additional information from your comment. Your question has been flagged as too broad.

@Feodoran, thank you, Avogadro seems to be already one step further from 2D to 3D. Gaussian is commercial, Orca academic/commercial, so you can recommend Psi4 to look at further?!
Karsten Theis, the input is focussed on SMILES, the output is focussed on emission spectrum. The way from input to output (as you say) is apparently broad. How could I possibly narrow down my question if I cannot judge the large variety of different software packages? That's why I'm asking experts for guidance.

What I was trying to say: you can get something half-way black box for IR spectra. But since fluorescence is more complicated than that ... I don't think what you are looking for exists.

If you are working as a non-profit, try ORCA. This is particularly good for spectroscopy. Here is the link to the download and forum. They have a lot of users and community forum is also good.

As you mentioned, Gaussian can do it but is proprietary.
I believe ezSpectrum (https://github.com/iopenshell/ezSpectrum) might be able to do what you are looking at but I have not used it myself so far.
There used to be a program available called FCClasses which would run the Franck-Condon computations for you based on (I think) Gaussian output, but it seems like the author took it down, I can only find a front-end to it right now: https://github.com/jcerezochem/fcc_tools The FCClasses homepage is mentioned as http://www.pi.iccom.cnr.it/fcclasses on that page but, as mentioned above, no longer works.


text:
Equilibrium Constant dependence
The equilibrium constant (EC) of a certain reaction is only dependent on temperature. For instance, suppose at a certain temperature, the EC of $\ce{A + 3B <=> 2C}$ is $K$. Now if I divide the whole equation by 2 to get $\ce{(1/2)A + (3/2)B <=> C}$, why will my new EC $K'$ be the square root of the previous one,namely $K$?

K depends on the temperature and the chemical equation. This is also true of reaction enthalpy and reaction Gibbs energy. It might be surprising because these quantities look  like intensive quantities (units are joule per mole), but they double if you double the coefficients of the chemical equation. They are per amount of substance, but specifically per extent of reaction. Once you know $\Delta_r G$ changes when you double or half all coefficients, it is clear that K changes too via $\Delta_r G^\circ = R T \ln K$

There is a simple proof behind the result.
Let's take
$$\ce{aA + bB \rightleftharpoons cC +dD}$$
Then the equilibrium constant of this reaction (K) will be
$$\ce{K= \frac{[A]^a [B]^b}{[C]^c[D]^d}}$$
Now if you divide the whole equation by a  number say n then
$$\ce{\frac{a}{n}A + \frac{b}{n}B \rightleftharpoons \frac{c}{n}C +\frac{d}{n}D}$$
For this equation you have the equilibrium constant as
$$\ce{ K'= \frac{[A]^{ \frac{a}{n}} [B]^{ \frac{b}{n}}}{[C]^{ \frac{c}{n}}[D]^{ \frac{d}{n}}}}$$
So
$$\ce K' =(K)^\frac{1}{n}$$


text:
If Beer's Law appears to apply to a high concentration (>0.01M) of a solute, is it valid to use for concentration calculation?
Internet consensus seems to define "high concentration" for Beer's Law as >0.01M. If one has a solution at higher concentrations than that, but the absorbance vs. concentration is still linear, is it valid to use Beer's Law to calculate concentration from absorbance data (assuming that the concentration is less than the highest point for which one has experimental data)?

I have never encountered this "internet consensus". There are a number of underlying assumptions for Beer's law to be valid: I have seen references to as many as 13 conditions. And you would not want the analyte to form dimers, as concentration increased, because this would change the molar absorptivity. But there is nothing intrinsically problematic about, say, 0.1 M analyte. You appear to have a handle on this, so best of success!

@EdV, do you remember the reference which listed 13 conditions? Since you are an expert on the limit of detection, we should start a post on this concept. Please share your thoughts, in your experience which is the "best" way to determine LOD.

@M.Farooq I just e-mailed two papers stating 11 conditions (not 13: my mistake). As for detection limits, take a look at my short list of references on the topic: https://stats.stackexchange.com/a/410372/247352 . This includes my book, Joel Tellinghuisen's Analytical Chemistry paper (published last month), and lots more. What I recommend on detection limits is in my book and multiple papers on the topic, and at my book's associated web site (URL at the bottom of my e-mail). Frankly, I prefer not to get into that at this Q & A site: it is way too complicated.

Thank you Ed. Most people take LOD as a fixed definition. Nice to see that feature in Analytical Chemistry. Will have a look. The acknowledgement is interesting "I have profited greatly from frequent exchanges with EdV on matters concerning detection limits. I cannot thank him enough."

@M.Farooq We actually exchanged 185 e-mails, over about 4.5 months, and more than 150 files: sims, spreadsheets, pics, papers, etc. It was an enjoyable intellectual adventure.

Internet consensus seems to define "high concentration" for Beer's Law
  as >0.01M

Keep in mind that Beer's law is an approximation. Look at the more advanced version. Beer's law-advanced version. The statement that a concentration of 0.01 M is the upper limit of Beer's law is incorrect due to several reasons. It is the absorbance value which we should worry about rather than the concentration. One can have a solution which is 0.01 M yet its extinction coefficient is not that large at that particular wavelength, we would still be fine.
This trick is used in atomic absorption spectroscopy, where you choose wavelength, which is not a resonance line of that particular atom. As a result, one can use relatively high concentration without getting a non-linear curve.
Secondly, chemists tend to infatuate little bit with linear calibrations. It is okay to have a non-linear curve. Quadratic equations can be used to fit the empirical curve and one can get accurate analytical results.

... moreover, what a chemist would call a nonlinear curve may still be a linear from a statistics/data analysis/curve fitting point of view: as a chemist I understand linear calibration to refer to a function that is linear in the concentration. As chemometrician/data analyst, I can also understand linear model refering to a calibration function that is linear in the coefficient: in curve fitting/statistical modeling the coefficient is what needs to be calculated, and the fitting methods that can be used depend on this type of linearity, nonlinear terms in concentration are "harmless" here.

Looking at the actual transmission points to a 2nd way of adjusting: choosing appropriate optical path length. OTOH, concentration will actually matter if at too high concentration new chemical species appear which have different absorption characteristics (dimerization, changes in solvation).

@cbeleites Your point about linearity in coefficients is one that should be far better known! And the distinction between y - a = bx and y = a + bx, in terms of mathematical linearity, also should be better known.

Dear EdV, Could you elaborate more on linear in coefficients and your distinction b/w y-a=bx vs. y=a+bx. What is a test for that?

@M.Farooq So, exactly as cbeleites said, "fitting methods that can be used depend on this type of linearity", i.e., whether the equation is linear or nonlinear in the coefficients. This paper, which I will e-mail to you right after this comment is posted, explains things very well: J. Tellinghuisen, "Statistical Error Propagation", J. Phys. Chem. A 105, (2001) 3917-3921. See the bottom of the first page and left half of the second page, in particular.

@M.Farooq The other matter concerns the superposition principle (https://en.wikipedia.org/wiki/Superposition_principle ). So y = a + bx is an affine transformation, i.e., y is a scaled and shifted x, but this "linear" relationship between y and x is not mathematically linear because it does not satisfy the superposition principle. Indeed, y consists of a linear term plus a constant. In contrast, (y-a) = bx is linear. Of course, we all know this: we are always subtracting the background or offset or baseline!

The concentration is abitrary; take copper sulfate (CuSO4), with a molar extinction coefficient of ~50 dm3 mol-1 cm-1; its absorbance will be relatively low (A $\approx$ 0.5 at 0.01 M in 1 cm cell, so T $\approx$ 32 %).
Compare this with KMnO4, $\epsilon \approx 220$ dm3 mol-1 cm-1. At 0.01 M, A $\approx$ 2.2 (T $\approx$ 0.6%).
Even more striking are organic dyes with extinction coefficients upwards of $50\,000$ dm3 mol-1 cm-1 and suddenly you're at a point where a concentration of 0.01 is sufficiently high that only one photon in 10500 is actually getting through to your detector!
It's the absorbance that counts; a general rule of thumb is to configure your concentrations so that you keep your absorbance in the range 0.1-0.6 as this corresponds to 80% to 25% transmission, well within the dynamic range of most instruments.  If you have a shiny instrument, you may be able to work outside this range.
As for linearity, this is an obsession of physical scientists.  The linearity of BL is only an approximation. If you have done an investigation with your analyte and looked at how absorbance varies with concentration, and you are satisified with its approximation to linearity, then run with it.  If however you notice a significant curve to your Abs/Conc data, then you may wish to revisit the BL law and its logarithmic derivation.
HTH.

Concentration alone is not a criterion here, and I would refrain even more from stipulating one threshold concentration applicable for all samples.  True, from perspective of the sample (possible unwanted association of the analyte's molecule) and linearity of the signal, you are in a better situation if your analyte is not too concentrated.
From the perspective of the spectrometer's detector, the old-school approach is to record the absorption bands around $\mathrm{Abs} \approx 1$.  For a cell of $\pu{1cm}$, given $\mathrm{Abs} = \varepsilon \cdot c \cdot l$, once you know about your strongest absoption $\varepsilon_{\mathrm{max}} = 10^n$ in your region of interest, you then would record at a concentration not higher than $c \approx 10^{-n}$.
Do not forget that the simultaneous presence of (very) strong and weak absorption bands does not hinder you to split the range to record into multiple sections -- each recorded at a different (diluted) concentration.  Instead of $\mathrm{Abs}$ on the ordinate, taking into account the sample's concentration, you would trace $\log \varepsilon$ as function of wavelength ($\mathrm{nm}$), or proportional to energy (e.g., $\pu{cm^{-1}}$) on the abcissa.

(reference)


text:
Is it possible to produce sodium bicarbonate from potassium bicarbonate?
I balanced the chemical reaction which is as follows:
NaCl + KHCO3 = NaHCO3 + KCl
I've also gathered solubility values which are the following:

KHCO3
22.4 g/100 mL (20 °C)
NaCl
35,9 g/100 mL (20°)
KCl
34.2 g/100ml (20°C)
NaHCO3
9.6 g/100ml (20 °C)

Since NaHCO3 is less soluble than the other compounds, I could separate it by boiling off some of the water up to a certain point, but I'm not sure how to predict which way the reaction will be going and if it will proceed completely.
How do I find out which way the reaction will proceed, and to which extent it will be complete?
I'd be glad for any help.

You show the bicarbonates, i.e., the hydrogen carbonates. Is that correct?

Sorry, I corrected the title! Thank you Ed. That being said and done I'd also be interested about the case of the carbonates. But I think I could extrapolate myself for them from the information I'll receive when this question is answered.

You could run it over an kation exchanger column that is loaded with Na⁺. If we're talking "household" chemistry, such ion exchanger may be harvested from an old dishwasher, and it can be loaded with Na⁺ using NaCl (that's the salt the dishwasher asks for).

Boiling off some water is a decent plan. Perhaps it should be done at reduced pressure so as to keep the temperature low. Magnesium and calcium bicarbonates decompose at temperatures above 60C in water to yield insoluble carbonates, and several data sheets for sodium bicarbonate suggest the same. One said decomposition was complete at 100C. I suppose you could consider that any H2CO3 in equilibrium with NaHCO3 could decompose to CO2 + H2O. Heat would just drive the equilibrium to completion. I suppose that's why I could find no solubility data for NaHCO3 or KHCO3 at temperatures higher than 60C.
One way to go about the preparation is to dissolve 58.5 grams (1 mole) of NaCl in 260 grams H2O, and 100 grams of KHCO3 (1 mole) in 260 grams H2O. Heat both to 60C; they will completely dissolve. Then pour one into the other (it may make a difference which way you do it) and stir. The amount of water is sufficient to dissolve 1 mole (84 grams) of NaHCO3 at 60C (but the other salts could and probably will affect this). So the mixed solution will be clear; all salts will be soluble.
Cooling to 20C will (should!) ppt 34.1 grams of NaHCO3, leaving 49.9 grams in solution (again, ignoring the effect of the KCl). Cooling to OC should ppt another 14 grams of NaHCO3, leaving 35.9 grams in solution. You could then boil off about half of the water before the solubility limit of the KCl is reached, but because of the possible loss of CO2, you may not get much more NaHCO3.


text:
Antibonding, Nonbonding and Bonding orbitals
Recently I came across nonbonding orbitals. I know something about bonding and antibonding orbitals and their formation, but I have no idea about nonbonding orbitals. Could someone explain their formation and the difference when compared with the other two?

Could you edit your question to include where you came across nonbonding orbitals? Could you also describe your understanding of bonding and antibonding orbitals to give an indication what level of theory would be helpful as an answer?

In basic MO theory, when two atomic orbitals mix, they form an in-phase (bonding) and out-of-phase (antibonding) molecular orbital. The primary criteria for this mixing are having the same symmetry and being near in energy. If you mix your atomic orbitals together, but find that some of the atomic orbitals have no symmetry match with a similar enough energy, then these unmatched atomic orbitals do not mix and become non-bonding MOs.
Consider hydrogen fluoride. The 1s orbital on hydrogen has the same symmetry as the $p_z$ orbital on fluorine. These mix into a bonding and antibonding orbital. Hydrogen however does not offer any other occupied atomic orbitals with matching symmetry for the $p_x$ and $p_y$ atomic orbitals on fluorine. Therefore, these orbitals remain as non-bonding. Furthermore, the 2s fluorine atomic orbitals are much lower in energy than hydrogen's 1s orbital, so this does not mix appreciably either, and can therefore also be treated as non-bonding.
For elementary concepts, I think you just need to know about how these contribute to bond order. The occupancy of non-bonding orbitals does not have any effect on the bond order and are typically considered lone-pairs.

In MO theory there are no non-bonding orbitals, this is a simplification which arises from mixing localised orbitals (possibly from valence bond theory) into the theory. It is within the framework of the theory, that all MO are delocalised. Except for diatomic molecules the division into bonding and anti-bonding can become a lot more complicated. It is often the case that the same MO is bonding with respect to one pair of nuclei, and anti-bonding with respect to another. However, within the valence bond picture, the concept of non-bonding orbitals may offer some more insight.

@martin thanks for the correction. Ill edit it when i am not on mobile. I did aim to limit the answer to the diatomic case, in order to avoid the complexities you mentioned, which i assumed was more appropriate for the level of question.


text:
What causes positive and negative deviations when weighing analytical samples?
Positive deviations are when the measured mass is more than the true mass of the sample and negative deviations are when the measured mass is less than the true mass of the sample. What steps causes this type of errors in analytical measurements?

In addition to the answer by @M. Farooq , there is always random electronic measurement error (aka "noise") in electronic devices, e.g., electronic analytical balances. So there are both random measurement errors and systematic errors. You might also check out Appendix G in the 2012 QUAM: http://eurachem.org/images/stories/Guides/pdf/QUAM2012_P1.pdf . Hope this helps!

There can be plenty. There is no definite list. I would suggest to look up the following terms in Skoog or Daniel Harris' Analytical Chemistry. Worth a trip to the library now :-) The following concepts give rise to systematic errors
(i) buoyancy error
(ii) temperature differences
(iii) electrostatic charge build up on the sample
(v) moisture
(iv) above all loss of calibration of the balance.


text:
Which one has the highest boiling point?

10 g of substance below were dissolve in each 1 L water:
$\ce{Na2SO4}$
$\ce{NaCl}$
Glucose
$\ce{MgCl2}$
Which one has the highest boiling point?

Glucose has London dispersion force.
$\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}$ have ionic forces.
Ionic forces is stronger than London dispersion force, so $\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}$ have higher b.p. than glucose.
Now, how to decide which one has highest b.p. among $\ce{Na2SO4},$ $\ce{NaCl},$ $\ce{MgCl2}?$

Tech note: in chemistry "BP" is a poor choice for an abbreviation for boiling point as it's ambiguous and may also mean boron phosphide. A more or less standardized abbreviation would be "b.p."

The water boils, not the solutes. All the ionic interactions are gone in water, replaced by ion dipole. Solid glucose makes hydrogen bonds with itself, replaced by hydrogen bonds with water in solution.

Boiling point elevations proportional to the molar concentration of dissolved particles.  So that's what you have to figure out, being careful to (1) properly convert mass to moles and (2) recognizing that some of the solutes can form ions so you can get more than one particle per "molecule".
Let's look at glucose.  If you work out the molecular mass it's about 180 grams per mole.  Divide that into 10 grams and you get about 0.056 moles of glucose molecules in your one liter of solution.
Now try sodium sulfate.  For this one you have a "molecular" mass of 142 grams per mole, so you seem to get about 0.070 mole of particles.  Except ... Sodium sulfate forms ions, each "molecule" gives one sulfate ion and two sodium ions.  Thus three particles overall, and you have to multiply your 0.070 figure by that factor of 3.  Thus the sodium sulfate solution will have 0.21 mole of ions in one liter.  The sodium sulfate solution will therefore give a higher boiling point than the glucose solution.
Work it out now with the other two choices and pick the one with the most particles in solution including ion formation.  Be sure to use reasonably accurate atomic masses because this set of choices could throw you for a loop.

The answer is Na2SO4 right? But the key answer is MgCl2

Actually I get sodium chloride.  Not sure where the key gets magnesium chloride, but if you work out the numbers you should have found that both chlorides beat sodium sulfate.

Normally, I would not respond to this kind of question since it is most likely a homework but I am considering this question as an exception.
One important equation to determine the boiling point of ionic solutions is the boiling point elevation equation which states that the change in boiling temperature of the pure solvent is equal to imKb; where i is the vant hoff factor, m is the molality of the solution, and Kb is the ebullioscopic constant of the solvent.
Assuming that the solution is present in equimolal concentrations and in the same solvent, the arrangement of boiling points would be:
Na2SO4 = MgCl2 > NaCl

the vant hoff factor for the solution is equal to the number of species dissolved in the species. NaC has only a vant hoff factor of 2 since the dissolved species in a brine solution is Na+ and Cl-. However, Na2SO4 and MgCl2 each has a vant hoff factor of 3.

@KentdelosReyes It seems like this should be added to the answer alongside with the data source. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Correct also the consideration made by OP on dispersion forces versus ionic bonds which are not (relevant (if not indirectly) here.

How is the answer MgCl2 anyway?  For equal masses of solute, as stated in the problem, NaCl has a lot more formula units and it ends up winning narrowly over MgCl2.

I don't know.. @oscar

Btw i think it should be nacl>mgcl2>na2so4


text:
In the last step of Kjeldahl Method, is the reaction of ammonium to NaOH not taken into account?
In the last step of Kjeldahl method, the unreacted $\ce{HCl}$ in the solution is back titrated with a standard $\ce{NaOH}$ and the concentration of $\ce{HCl}$ that has been used to neutralize the ammonia is then calculated. However, other than $\ce{HCl}$, there also exist an acid in the solution that can be titrated with $\ce{NaOH}$ which is the protonated form of $\ce{NH3}$. How is this not considered when calculating the nitrogen presence in a sample?

The Kjeldahl method has two versions, using $\ce{HCl}$ as trapping solution or a weak acid like boric acid. Depending which one is used, the nitrogen content is either determined by a back titration with $\ce{NaOH}$ (your case) or a direct titration with $\ce{HCl}$.
The goal of the titration is to figure out how much ammonia (gas) has been trapped in the acid. The equation for the trapping step is:
$$\ce{NH3(g) + HCl(aq) -> NH4+(aq) + Cl-(aq)}$$
So the trapping solution will get a bit less acidic because some of the $\ce{HCl}$ is neutralized by the weak base ammonia. To quantify the ammonia, you now titrate the remaining $\ce{HCl}$ with the strong base $\ce{NaOH}$:
$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Na(aq) + Cl-(aq)}$$
The solution is titrated to about $\mathrm{pH} = 5$ as endpoint (at the equivalence point, the strong acids and bases are neutralize, and $\ce{NH4Cl}$ solution remains). At $\mathrm{pH} = 5$, very little ammonium is deprotonated to form ammonia ($\mathrm{p}K_\mathrm{a}$ is about 9.3). In other words, ammonium/ammonia does not act as a buffer, so it does not affect the titration. You might even call it a spectator ion at this $\mathrm{pH}$.
After reaching the endpoint, the amount of $\ce{HCl}$ originally present in the trapping solution is equal to the sum of the amounts of $\ce{NaOH}$ and ammonia added. The overall reaction is (non-standard notation):
$$\ce{HCl(aq) + x NH3(g) + (1-x) NaOH(aq) ->}$$
$$\ce{Cl-(aq) + x NH4+(aq) + (1-x) H2O(l) + (1-x) Na(aq)}$$
From that, you can calculate the amount of nitrogen in the sample.

However, other than $\ce{HCl}$ there also exist an acid in the solution that can be titrated with $\ce{NaOH}$ which is the protonated form of $\ce{NH3}$.

As explained above, if the $\mathrm{pH}$ does not approach the $\mathrm{p}K_\mathrm{a}$ of ammonium, ammonium will not react with $\ce{NaOH}$ to an extent that matters. If you choose to end the titration when the solution reaches a $\mathrm{pH}$ of 11 (to give an example), however, the ammonium will be deprotonated to ammonia, invalidating the result. Worse, it might come out of solution, so it becomes a problem that can't be fixed.
Sources:

http://www.brooklyn.cuny.edu/bc/ahp/SDKC/Chem/SD_KjeldahlMethod.html
J Martín, Fernández Sarria L and Agustín G. Asuero (2017) DOI: 10.5772/intechopen.68826

Wikipedia on the other hand I think got it wrong (https://en.wikipedia.org/wiki/Kjeldahl_method, retrieved before I edited it):

Ammonium ion concentration in the standard acid solution, and thus the amount of nitrogen in the sample, is measured via titration. If boric acid (or some other weak acid) was used, direct acid-base titration is done with a strong acid of known concentration. $\ce{HCl}$ or $\ce{H2SO4}$ can be used. Indirect back titration is used instead if strong acids were used to make the standard acid solution: strong base of known concentration (like $\ce{NaOH}$) is first added in excess, and then the excess is titrated with a strong acid of known concentration. Titration of ammonia absorbed in boric acid solution thus has an advantage that only one standard solution is needed. One of the suitable indicators for these titration reactions is Tashiro's indicator.

They say to add an excess of $\ce{NaOH}$ and then titrate with $\ce{HCl}$. This would drive ammonia out of solution. Also, this is not a back titration but a double-back titration (which is not a thing, I think).

Thank you for your answer :). Now I can now sleep!


text:
Why does a system want to achieve randomness to be feasible?
According to entropy and gibbs free energy, randomness is one of the factors that determines reaction spontaneity, but why? Shouldn't it be the opposite?

I'm voting to close this question as off-topic because it is unanswerable unless you edit it and give more context. What are your underlying assumptions about randomness and spontaneity, and how are they in conflict with experimental data?

Randomness is not the correct word to use as it is ambiguous in this context. What happens is that the molecules' energy is spread out among as many energy levels as possible, i.e. the number of possible ways of filling the energy levels is maximised and this number produces what we call the entropy. It is an experimental observation that this happens.
In a thermodynamic explanation when using just the first law the external work performed in a reaction must be equal to the heat loss, unless some heat is given to or taken from the surroundings. This is the point first clearly realised by Gibbs. If an isothermal reaction runs reversibly $T\Delta S$  is the heat absorbed from the surroundings, and if this is positive the work done ($\Delta G$) will be greater than the heat of reaction. Thus you can see that as $\Delta G = \Delta H-T\Delta S$  the entropy can be used towards changing the free energy. When the number of ways of arranging the energy among the numerous energy levels is greater in the products than in the reactants $\Delta S$ will be positive and tend to make $\Delta G$ negative and 'spontaneity'.