text:
Solving a quadratic equation for pressure of real gas with unknown volume
Question

Calculate the pressure exerted (in atm) by 1 mole of $\ce{CH4}$ gas at a temperature of $\displaystyle\pu{\frac{24}{0.0821} K}$ if volume occupied by $\ce{CH4}$ molecules is negligible. Given $a = \pu{2 L^2 atm mol^{-2}}$.

Solution:

$$
\begin{align}
\left(p + \frac{a}{V_\mathrm{m}^2}\right)V_\mathrm{m} &= RT\\
pV_\mathrm{m} + \frac{a}{V_\mathrm{m}} &= RT\\
pV_\mathrm{m}^2 - RTV_\mathrm{m} + a &= 0\\
pV_\mathrm{m}^2 -24V_\mathrm{m} + 2 &= 0\\
\end{align}
$$
For real gas this equation must have only one root:
$$D = 0\quad\implies\quad b^2 - 4ac = 0$$
$$24^2 - 4×p×2 = 0$$
$$p = \pu{72 atm}$$

My thoughts
I had a doubt whether this solution was really true or not. When I first encountered it, I was convinced that it was true. But, now, I doubt it. I don't have any appropriate reason why and that's what I seek to ask. Is this approach correct? Why?

I don't understand why the quadratic should have one root.

Mathematically the equation has two roots, one of which is the root that corresponds to actual methane (or at least, this problem's version of it) and the other is just junk.  You have to identify the right root out of the two that come mathematically, which in this case should be easy.

@Polarbear Be aware that VdW equation is just a simplified model of real gas behavior and does not guarantee the physical/chemical sense of its output outside of its applicability range. E.g. Some quantity may be modeled by a quadratic formula with the negative quadratic coefficient. That does not mean the model is wrong, when the value reaches negative values, if they must be positive. It just means that region is out of the model applicability.

Assuming the Van der Waals equation with $b=0$, I agree with the solution to the equation:
$$pV_\mathrm{m}^2 - 24V_\mathrm{m} + 2 = 0\label{eqn:1}\tag{1}$$
I'll point out that I'm following the notation of the given solution, but $V_\mathrm{m}$ seems odd to me. I'd think that $V_\mathrm{m}$ would be reserved to mean the molar volume at STP.
But equation \eqref{eqn:1} has two unknowns, $p$ and $V_\mathrm{m}$. Letting $a' = p$, $b' = -24$, and $c' = 2$ then the appropriate quadratic equation is of course
\begin{alignat}
2V_\mathrm{m} &= \dfrac{-b' \pm\sqrt{b'^2 - 4a'c'}}{2a'}\tag{2}\\
V_\mathrm{m} &= \dfrac{-(-24) \pm\sqrt{(-24)^2 - 4×p×2}}{2p}\tag{3}\\
V_\mathrm{m} &= \dfrac{24 \pm\sqrt{(-24)^2 - 8p}}{2p}\tag{4}\\
V_\mathrm{m} &= \dfrac{12 \pm\sqrt{144 - 2p}}{p}\label{eqn:5}\tag{5}
\end{alignat}
So we have two unknowns and one equation. Thus the problem is unsolvable without additional information. Notice that would still be the case if the simple ideal gas equation, $PV=nRT$, had been used. The ideal gas equation could be solved only to $pV_\mathrm{m}=24$.
For equation \eqref{eqn:5} there are three cases for the sqrt term $\sqrt{144 - 2p}$:

The term is negative.
If $144 - 2p$ is negative, then both roots would be imaginary.
The term is equal to zero.
We have $\sqrt{144 - 2p} = 0\implies 144 - 2p = 0,$ so $p = 72$. Therefore, $V_\mathrm{m} = 12/72 = 0.167.$
The term is greater than zero.
Then $72 > p$ also, and mathematically $V_\mathrm{m}$ has two roots.
Now, again there are three possibilities.
(a) $\sqrt{144 - 2p} < 12$
This is impossible since $V_\mathrm{m}$ would have two valid values.
(b) $\sqrt{144 - 2p} = 12$
This is impossible since $V_\mathrm{m} = 0$ and the gas volume can't be $0$.
(c) $\sqrt{144 - 2p} > 12$
This would yield one negative value and one positive value for $V_\mathrm{m}$, which is OK for $V_\mathrm{m}$. However it also means that $p < 0,$ which is nonsensical.

Thus the only reasonable solution is if the square root term is equal to zero.
P.S. The comment by user Poutnik made me look at this again, and I now understand the book's solution…

Vm=V/n=f(T,p), why should it be at STP ??

@Poutnik - Huh? STP isn't mentioned anywhere. I was just disagreeing as to how the quadratic was solved. There is no reason that I can see that 144-2P = 0. It could also be that 144-2P > 0.

It is mentioned in a quote in your answer, but is unclear what you have quoted. Perhaps a piece of question, deleted later.

@Poutnik - ah yes, sorry. I was just commenting that the notation that was used in the OP's book solution was odd. I don't understand why $\mathrm{V_m}$ was used instead of just $\mathrm{V}$.

VdW equation is commonly used in both general and molar forms. They used the latter, so they had to use V_m.


text:
Measure the concentration of lactose in dairy products
I would like to measure the concentration of lactose in various dairy products : milk, cheese, yoghourt...
How is it done in labs ?
How can I do that using cheap equipments ?
Thanks

If you are looking for economical methods then you have the option of titration and spectrophotometry. See this paper by Folin THE DETERMINATION OF LACTOSE J. Biol. Chem. 1918, 33, 521-524. The titration is based on the reducing ability of lactose.
Another reliable titrimetric method can be seen in this paper by Hites, Determination of Lactose in Milk Products J. Anal. Chem. 1949, 21(8), 993-995.

source: Stack Exchange

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

sort: votes


text:
Reaction of cyclobutylmethanol with HX
I am interested in the reaction of cyclobutylmethanol with HX where these are the only reagents.
My question is whether the reaction proceeds via SN1 or SN2. My reference text insists that primary alcohols are eliminated by SN2 even if the protonated alcohol is a strong leaving group. So if the reaction were SN1, there would be ring expansion.
What if I conducted this reaction in the presence of a Lewis acid like ZnCl2?

This is actually a trickier question than first inspection.
Your statement about primary alcohol substitution is more or less correct. Unless you have a very good reason to form the primary cation, you won't.
The problem in this example is that cyclobutylmethyl cation is not a primary cation even though if you were to draw it out, you might naively think that. Cyclobutylmethyl cation belongs to a special class of cations that are considered non-classical in the sense that the positive charge is not heavily localized to one atom, for example like methyl cation or tert-butyl cation. This essentially means that the positive charge is spread out over several atoms, and this increases the relative stability of the cation, making it more likely to form.
The more famous example is a 2-norbonyl cation, which is covered in many introductory classes.
For cyclobutylmethyl, the literature I found (hard to link) seems to suggest that the contribution from cyclopentyl cation is much larger than the cyclobutylmethyl cation, so we might consider this to be more neighboring group participation (from the C-C bond) than a raw non-classical ion. This is likely due to the fact that the ring expanded version is so much better in terms of strain.
Either way, a dissociated cation is stabilized, so I would think that this mechanism is more $S_{\mathrm{N}}1$-like.

"Either way, a dissociated cation is stabilized" What does this mean? Are you referring to the solvation of the primary carbocation? If yes,then won't that be true for any cation we form.. Why does this statement mean SN1 mechanism is likely?

It means that you're probably generating a cation dissociated from the leaving group, as opposed to a concerted mechanism where that's not the case.

Forgot to tag you. @YUSUFHASAN

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


text:
Heat transfer through layered wall

An electrical heater is placed between two layers of insulation. The three layers (the heater and the two layers of insulation) have the same thickness. The figure below shows different temperature profiles at steady-state. The air temperature at the front- and backside of the heater is $\ce{T_{front}}$ resp. $\ce{T_{rear}}$. Heat transfer in the air film adjacent to the heater is negligible.
a) Which profile corresponds to that of a turned on heater?
b) Which profile corresponds to that of a turned off heater?
c) Which profile is impossible?
d) Which of the three layers has the lowest thermal conductivity? Which layer has the highest?

I have no answer for a) or b).
c) My reasoning is that since a heater is a conductor, the k-value needs to be high, resulting in a fairly straight line in the temp. profile. Hence, (C) is impossible. Is this correct thinking?
However, I don't know what it means when a temperature profile has that kind of curvature for one of the layers, such as in (A) and (B), so if someone could please explain that, I would appreciate it.
d) Based on the fact that low k-values give a steeper line, I would say that layer 3 has the lowest thermal conductivity. And because the heater is a conductor it should have the highest thermal conductivity.
Also, how can you tell if the heater is turned on or off from looking at different temperature profiles? I am having a very difficult time understanding this and would appreciate if someone could make this a bit clearer.
I apologize if these types of questions are considered off-topic, but I really want to know how to determine these things and we have not been provided with any  answers nor explanations.

This is maybe more on-topic for Physics SE

D definitely corresponds to the profile with a turned-off heater, and the lowest conductivity layer is layer 3, as @Night Writer has correctly pointed out.
Since the heater is a heat source, the temperature profile within the heater layer is going to be parabolic.  You can show this by solving the steady state heat conduction equation for the case of a heat source within the material.  So, on this basis, profiles A and B are possible.  However, in profile A, heat is flowing into the interface between layers 1 and 2 from both sides.  This is impossible for steady state operation.  So profile B is the profile for the turned-on heater, and profile A is impossible.
Profile C is also impossible since heat is flowing out of the interface between layers 1 and 2 on both sides, which is inconsistent with steady state operation.
So profiles A and C are both impossible.

I would start with guessing which one represents the "off" state. I would say (D) represents the "off" state, since the temperature slopes evenly from one air interface to the other, and is fairly constant in the heater layer. Based on this diagram you can also conclude as you did that layer 3 has the lower conductivity. And the slope in layer 2 indicates that it has a finite (not infinite) conductivity higher than that of 1 or 3.
Based on observations made in (D), you can conclude also that (B) represents the "on" state. As you explain, (C) is not sensible unless heating originates at the interface between layers 1 and 2. (A) also does not make sense, because it suggests that there is a heat sink between 1 and 2 (where is that heat going?).  In (A) the temperature reaches a minimum between layers 1 and 2. This is not physically possible in a steady state. In situation (A) heat will flow to the lower T region between 1 and 2, increasing T.


text:
How do I calculate the full configuration interaction Hamiltonian of H2
I have been trying to follow the paper titled "Towards Quantum Chemistry on a Quantum Computer" (https://arxiv.org/pdf/0905.0887.pdf) by simulating their results on a quantum computer simulator. Unfortunately, I have been having trouble classically calculating the full configuration interaction Hamiltonian for H2 using the STO-3G basis set. In the paper they use Pyquante, but I haven't been able to find good documentation on how to use it. I would be grateful if someone could show me how to calculate this using Pyquante, point me towards better documentation for Pyquante (I've only been able to find the rather poor documentation found here: http://pyquante.sourceforge.net/), or suggest a different/free quantum chemistry program that is easier to use.
For reference, I have been using Szabo and Ostlund's "Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory" to try to get a handle on this calculation. They come up with two single-electron molecular wavefunctions,

and

where  and  are the 1S orbitals about each of the atoms.
This gives 4 molecular spin orbitals and thus 6 different Slater determinant basis states. These are the basis states for which I would like to calculate the Hamiltonian.

They have a cookbook with examples... http://pyquante.sourceforge.net/#cookbook

If you use Mac or Linux (or Linux subsystem on windows), I would recommend Psi4. I have only used it a little, but it has CI implemented and they have pretty straightforward tutorials on Github for how to make a CI code

"They" may be one too many - I am not sure pyquante is actively developed...

@NightWriter I linked that source in my question, and (so far as I could tell) it didn't tell me how to solve my problem. Could you explain to me how I could use some of the code from the cookbook to find the full CI Hamiltonian?

@Tyberius Unfortunately I am on a windows system. Do you have any other suggestions?

It looks like the Psi4 project provides a windows installer. Regarding CI under Pyquante, it looks like that was on the "to do" list for that project.

@LucasMyers I have used it with Windows, but it seems to require the Linux Subsystem. This is actually pre-installed on any Windows computer from the last few years, you just need to enter the Microsoft store to activate the app. Its not a terribly difficult process. Once you install that and Psi4, you can either make program files directly in the subsystem or connect to a Jupyter Notebook

@Tyberius what exactly is the app called? When I search "Linux Subsystem" in the Microsoft store a bunch of random distros pop up. And would I need to do anything extra to interface a Jupyter Notebook with psi4?

WSL: https://docs.microsoft.com/en-us/windows/wsl/install-win10

@LucasMyers You could use pretty much any of them, though I happened to use Ubuntu. If I'm remembering right, the install instructions on Jupyter's website should be enough. Installing psi4 also installs python and anaconda so you can perform any of the python related installation instructions. Once you have done all that, psi4 is a python import that can be used in any code you write in the subsystem or in a connected Jupyter notebook.

@Tyberius Just one more clarification question -- I've activated the Linux Subsystem and installed the Ubuntu command window. Are you suggesting that I download conda in the Ubuntu command window and use that to download Psi4? And if so, will I be able to use Psi4 in a Jupyter Notebook on my regular machine? Or will I somehow have to also download Jupyter in the command window to use Psi4?

@LucasMyer Installing psi4 will also install a version of Python and Anaconda. You can then either use psi4 directly or import it in Python code as a module. Jupyter is nice to have, but not strictly necessary.


text:
How does the accumulation of hydrogen bonds affect individual bond strength?
I saw the question, "Is the strength of hydrogen bonding greater in hydrogen peroxide or water?" and it made me think of a question on hydrogen bonding:
If an oxygen on a water has a hydrogen bond to another water (or anything else), would adding another hydrogen bond to that water (perhaps by introducing another water), decrease the strength of the first hydrogen bond?
My initial thought is it that it wouldn't because oxygen is electronegative enough that is can attract another hydrogen bond with minimal effect to the other bond.
However, if a single water could attract an unlimited number of other waters to it (ignoring space constraints), wouldn't the strength of hydrogen bonds weaken as the number of bonds to the oxygen increase as it's dividing up its electronegativity across multiple bonds?

Well, I think the other part of the definition is contains a hydrogen atom. ;-) But I see you point. Let's see what others come up with.

Related: http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html

A huge problem with the way this is posed is the “ignoring space constraints” bit of it. In reality there will be extremely non-trivial steric effects going on if you have any more than 1 group hydrogen bonding to any other molecular site.


text:
Does aluminum oxide react with rubidium?
I have rubidium vapor inside of the vacuum chamber. Inside the vacuum chamber, there are two flat stainless steel coated with aluminum oxide at temperature ~100 °C.
Since rubidium is active, I am worrying that it will interact with the aluminum oxide $\ce{Al2O3}$ coating. Any ideas?

Though I can't answer definitively, Al2O3 is used in sodium-vapor lamps that have a long lifetime. A guess would be that the surface does react, forming rubidium aluminate, but that it doen't penetrate rapidly.

I don't know about rubidium aluminate but a certain rubidium tetranitroaluminate does exist.

Rubidium aluminate with formula $\ce{Rb_6Al_2O_6}$

As suggested in the comments, rubidium aluminates do exist.
Aluminates such as $\ce{RbAlO2}$ and $\ce{Rb6Al2O6}$ are prepared at higher temperatures (above 550 °C [1]) or even from the melt.
However, alkali metal vapors at elevated temperatures act as reducing agent, so I wouldn't expect aluminates anyway.
On the other hand, aluminides such as $\ce{RbAl}$ (Zintl phase) are formed from the melt at much higher temperatures (above melting point of aluminium) and pressures exceeding atmospheric.
Since the corundum-coated plate is supposed to perform in vacuum under mild temperatures, I think it shouldn't be affected by rubidium vapors.
There is a possibility of coating embrittlement over time as rubidium diffusion occurs, but I wouldn't expect it to be an issue.
References

Schläger, M.; Hoppe, R. Darstellung und Kristallstruktur von $\ce{K6[Al2O6]}$ und $\ce{Rb6[Al2O6]}$. Zeitschrift für anorganische und allgemeine Chemie 1994, 620 (5), 882–887. https://doi.org/10.1002/zaac.19946200522.

Maybe not aluminates alone, but you could get aluminates + intermetallics, as, $\ce{4Rb + 2Al2O3->3RbAlO2 + RbAl}$.

authors: ə̷̶̸͇̘̜́̍͗̂̄︣͟


text:
Why are carbons of Inositol chiral centers?
I was asked to draw all possible stereoisomers of inositol (1,2,3,4,5,6-cyclohexanehexol). To obtain the answer I had to assume that all six carbons of the molecule are asymmetric, which (bearing in mind possible molecular symmetry) results on the following nine stereoisomers:[1]

The problem is that I fail to see why these carbons can be considered chiral centers. Given:[2]

The most common cause of chirality in an organic molecule, although not the only one, is the presence of a carbon atom bonded to four different groups [...]. These carbons are now named chiral centers [...]

The chirality of the molecule is not the reason of my doubts, I can see that the only chiral isomers are D- and L- chiro-inositol. Since all carbons in the ring possess a hydroxyl group, can they not be considered to be bonded to equivalent substituents, making them symmetric? Is the possibility of the -OH groups to be above or below the plane of the ring what makes them asymmetric?

From Wikipedia
McMurry, John. Organic Chemistry. CENGAGE Learning, 2008. 7th edition.

Seven of the nine inositols are achiral. Any carbons lying in a plane of symmetry are in achiral environments (r and s). Those carbons not in a plane of symmetry are in chiral environments (R and S). I Googled "inositols chirality" and found this: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/inositol.html

When we count different substituents to determine chiral centers, we have to include the relative locations of atoms as well as stoichiometry.
Let's look at the top carbon atom in the D-chiro form.  If you go clockwise around the ring you encounter five CHOH groups, and if you go counterclockwise you get the same five CHOH groups.  The substituent atomic formulas look the same.  But:

in the clockwise direction you see the hydroxyl group attached to the first carbon from the top is on the same side of the ring as the hydroxyl group on the top carbon itself.  Whereas:
in the counterclockwise direction you see the hydroxyl group attached to the first carbon from the top is on the opposite side of the ring as the hydroxyl group on the top carbon itself.

That difference is enough to count as two different substituents even though the formulas in terms of atoms are the same.  So you have a chiral center, and the CIP convention even identifies the orientation of the top carbon as S.  Similarly any of the other ring carbons can be a chiral center.
Contrast this with the scyllo form, for instance, where the substitution on any carbon is truly fully symmetric including stereochemistry.  So no chiral centers there.  Most of the other forms listed above do have chiral centers by the method above, but end up being meso compounds.  so only the forms labeled "chiro" actually have a net molecular chirality.

Consider 2,3-butanediol.

This compound has 2 stereocenters. The meso version of this compound is achiral. However, there are still two stereo centers. As drawn, the left one is (R) and the right one is (S). It is precisely because these two stereocenters are identical but opposite in configuration that leads the molecule to be meso.
A carbon center is asymmetric if it has 4 different substituents regardless of whether or not the whole molecule is chiral. The ring substitution in inositol is such that the centers around the ring are not identical (including when you look at stereochemistry--for example, (R) has higher priority than (S)). This of course isn't enough to help you easily make stereochemical assignments for all of the centers of inositol, but hopefully, this allows you to understand why the centers are still chiral centers.
As an aside, there are a few centers that cannot be chiral because they lie on a mirror plane. Interestingly though, changing the configuration of the center produces a different molecule. These centers are called pseudochiral, and we label them with lower case r and s.

I offer a different perspective to your question about the inositols and how the chirality of the "top" carbons in D-chiro-inositol 2 (C2; S-configuration) and scyllo-inositol 3 (C2; r-configuration) are determined.
Generic inositol 1 is numbered the same way as D-chiro-inositol and scyllo-inositol. The method for assigning the CIP descriptor for each inositol at C2 will be illustrated. Starting from C2 in structure 1, the counterclockwise path C2-->C3-->C4-->C5 is colored blue while the clockwise path C2-->C1-->C6-->C5 is in red.
Starting with D-chiro-inositol 2, temporary descriptors (Ro/So) at all carbons except C2 are assigned. At each of these carbons the hydroxyl group has the top priority while hydrogen has the lowest priority. C2 is taken as the next to highest priority group, no matter how far the carbon in question is from C2.1 The temporary assignments for D-chiro-inositol are illustrated in structure 2. To determine whether the blue or red path has the higher priority, step 1 is to compare the assignments of C3/C4 (So/Ro) with C1/C6 (So/Ro). No decision can be made at this point. In step 2, one compares C3/C5 (So/So) in blue with C1/C5 (So/Ro) in red. Because "like" descriptors (RR or SS) take precedent over "unlike" descriptors (RS or SR), whether they are temporary or not. The priorities for C2 are OH>SoSo>SoRo>H. C2 is of the S-configuration. Digraph 4 illustrates this point as well. Because D-chiro-inositol is chiral, each of the carbons is both stereogenic and chirotopic. Each carbon must be analyzed independently as was done for C2.  Notice that the temporary assignments (Ro/So) have no relationship to the final assignments (R/S). continued

Scyllo-inositol 3 is achiral having three planes of symmetry passing through C1/C4, C2/C5 and C3/C6. All carbons are equivalent; assign one, you have assigned all six. After assigning temporary descriptors and proceeding all the way to step 3 (see rules next to structure 1), no decision can be made. C2 and all other carbons are pseudoasymmetric (stereogenic and achirotopic). The default position is to use the temporary assignments at C1 and C3 (see digraph 5). CIP Rule 5 states R>S (and Ro>So). The priorities for C2 are OH>Ro>So>H. C2 is of the r-configuration. Lower case r/s descriptors are reserved for pseudoasymmetric carbons.
The two inositols discussed here have descriptors that are either R/S or r/s. This need not be the case. For example, muco-inositol has four chirotopic (R/S) carbons and two achirotopic (r/s) carbons.
1) This method is a simplification of a more complex methodology.

OK, I'll bite.  How is the orientation around the top carbon atom in the scyllo form determined when there is obviously a mirror plane going right through that carbon atom?  (And one mirror plane or another through all the other carbon atoms too.)

@Oscar Lanzi: Because of the plane of symmetry thru C2/C5, the red/blue paths are indistinguishable. C1 is Ro while C3 is So. Ro > So. Priorities OH>Ro>So>H. You can do the same exercise for anyone of the scyllo carbons because they are all equivalent. C2 in D-chiro is S and not R. See the colored picture. Don't fret. It took me a long time to understand this stuff.


text:
Ionic factor of glucose?
I am trying to find the $K_\mathrm{b}$ for 50 g of glucose dissolved in 1 kg of ethanol, given the change in boiling point is $\pu{2.2 °C}$. The molality is therefore 0.278. But what is the ionic factor $(i)$? I was told it is the number of ions, so I tried 24, but the answer did not match up with the accepted value.

Glucose doesn't ionize when dissolved, so it would be 1. @VikramKaushik

A more appropriate name for the ionic factor is the van't Hoff factor of colligative properties. In the simplest terms, it is the number of particles created after being dissolved in the solution. Since glucose is not an ionic compound, it does not dissociate, therefore $i=1$ (meaning one glucose molecule stays one glucose molecule).


text:
Which has a greater dipole: chloroform (CHCl3) or trichlorofluoromethane (CCl3F)?
Does chloroform have a greater dipole because the $\ce{C-H}$ dipole is weaker than $\ce{C-Cl}$ dipole thereby making the overall net dipole greater in chloroform, as opposed to trichlorofluoromethane where the $\ce{C-F}$ bond dipole being more similar to the $\ce{C-Cl}$ dipole makes the molecule more stable with a smaller net dipole?

C-H and C-F dipoles are in opposite directions; which of them has greater absolute value is not immediately obvious.

Imagine a $\ce{CHCl3}$ molecule and a $\ce{CFCl3}$ molecule fixed with the carbon atom of each molecule at the origin of a three-dimensional coordinate system. Select the two coordinate systems such that the hydrogen atom of the $\ce{CHCl3}$ molecule and the fluorine atom of the $\ce{CFCl3}$ molecule both lie on the $z$-axis above the origin, and let $u_z$ be the $z$ unit vector. If the four single covalent bonds of each molecule are symmetrical (here, tetrahedral), then it follows that the net vector sum of any four vectors that are equal in magnitude and are directed in the four directions from the origin to each non-carbon atom will be the zero vector. In both molecules, the net vector dipole moment arising from the three $\ce{C-Cl}$ bonds will equal the dipole moment of a single $\ce{C-Cl}$ bond with the chlorine atom in the direction of $-u_z$ from the carbon atom.
Assuming arguendo that the dipole moment of each bond is proportional to the difference in electronegativity of the two atoms between which that bond lies (on the Pauling Electronegativity Scale, hydrogen rates a $2.1$, carbon rates a $2.5$, chlorine rates a $3.0$, and fluorine rates a $4.0$), then to find the dipole moment of the $\ce{CHCl3}$ molecule, we add the vector dipole moment of the $\ce{C-H}$ bond ($[(2.1 - 2.5) \cdot u_z]$, or $-0.4 u_z$) to the vector dipole moment of the $\ce{C-Cl}$ bond ($[(3.0 - 2.5) \cdot (-u_z)]$, or $-0.5 u_z$), obtaining $[(-0.4 u_z) + (-0.5 u_z)]$ or $-0.9 u_z$.
To find the dipole moment of the $\ce{CFCl3}$ molecule, we add the vector dipole moment of the $\ce{C-F}$ bond ($[(4.0 - 2.5) \cdot u_z]$, or $1.5 u_z$) to the vector dipole moment of the $\ce{C-Cl}$ bond ($[(3.0 - 2.5) \cdot (-u_z)]$, or $-0.5 u_z$), obtaining $[(1.5 u_z) + (-0.5 u_z)]$ or $1.0 u_z$.
One might thus conclude that the dipole moment of $\ce{CFCl3}$ might be some ten percent larger in magnitude (as proportional to 1.0) than is that of $\ce{CHCl3}$ (as proportional to 0.9).

Welcome to Chemistry SE! Please enclose chemical compounds in $\ce{ }$ and normal latex in $ $for the latex formatting to take effect

$\ce{C-Cl}$ in $\ce{CHCl3}$ all lies in same direction and hence net vector sum is downwards, which makes it's dipole moment greater, while in $\ce{CFCl3}$ three $\ce{C-Cl}$ bonds are in same direction while one $\ce{C-F}$ lies in opposite direction to them thus while adding dipoles as vectors we need to subtract dipole of $\ce{C-F}$ from dipole of three $\ce{C-Cl}$ thus reducing net dipole moment.
Hence dipole moment of $\ce{CHCl3}$ is greater than $\ce{CFCl3}$ .

It would be nice to amend this with some data, like https://en.wikipedia.org/wiki/Chloroform_(data_page) and https://en.wikipedia.org/wiki/Trichlorofluoromethane_(data_page) (for starters)


text:
Recrystallisation of dibenzylideneacetone
Recently, I carried out the synthesis of dibenzylideneacetone. I had a lot of problems with the purification process, especially the part involving recrystallisation. I was instructed to carry out the recrystallisation using an "ethanol/water" pair. That means to say, I was supposed to dissolve the crude product in ethanol until just enough dissolved, then add a small volume of water until some of the non-polar dibenzylideneacetone precipitated out of the solution and lastly, add another small volume of ethanol to re-dissolve the precipitate that had just formed. Once done, the beaker is removed from the hotplate and left to cool at room temperature. Subsequently, the crystals were washed with cold ethanol to finally obtain the final product. It is similar to what was done in this video.
During the practical I had to repeat the recrystallisation step multiple times for a variety of reasons. Firstly, there were some instances when the dibenzylideneacetone melted to form a yellow oily liquid. Secondly, there were also some instances when the crystals "crashed out" of solution upon cooling. This resulted in poor-quality crystals which clumped together. I am just wondering if anyone has any tips on how I could have improved my procedure so that I don't have to perform recrystallisation multiple times and yet still obtain poor-quality crystals.

Armarego's Purification of Laboratory Chemicals suggests using dichlormethane–hexane pair instead of ethanol–water; it also lists crystallization from warm ethyl acetate, ethanol or petroleum ether as alternative [1, p. 342].

Dibenzalacetone [DBA, bda, trans-trans-1,5-diphenyl-1,4-dien-3-one, 1,5-(bisphenyl)-penta-1E,4E-diene-3-one] [538-58-9] $\ce{C17H14O}$, $M~234.3$, $m~107°$, $111°$, $112°$, $113°$, $120-122°$.
  Purify the ketone by flash chromatography (150 mesh $\ce{Al2O3}$ deactivated with $6\%$ v/w $\ce{H2O}$) using  petroleum ether $40-60°/\ce{EtOAc}$ ($4/1$, v/v), and the yellow solid is recrystallised by layering a concentrated $\ce{CH2Cl2}$ solution with hexane (i.e. $\ce{CH2Cl2}$/hexane, $1:4$, v/v).
  Also recrystallise the ketone from hot ethyl acetate $(\pu{2.5 ml/g})$ or $\ce{EtOH}$.
  It has been also crystallised from petroleum ether b $30-60°$.

References

Armarego, W. L. Purification of Laboratory Chemicals; Elsevier: Boston, MA, 2017. ISBN 978-0-12-805457-4.

Ethyl acetate is the solvent of choice in Org. Synth. 1932 article. The procedure was submitted and checked. The following is the the full procedure stated in Organic Syntheses.; Collective Volume (Ref.1):

A cooled solution of 100 g. of sodium hydroxide in 1 l. of water and 800 cc. of alcohol (Note 1) is placed in a 2-l. wide-mouthed glass jar which is surrounded with water and fitted with a mechanical stirrer. The solution is kept at about 20–25° and stirred vigorously (Note 2) while one-half of a mixture of 106 g. (1 mole) of benzaldehyde and 29 g. (0.5 mole) of acetone is added (Note 3). In about two or three minutes a yellow cloud forms which soon becomes a flocculent precipitate. After fifteen minutes the rest of the mixed reagents is added, and the container is rinsed with a little alcohol which is added to the mixture. Vigorous stirring is continued for one-half hour longer, and the mush is then filtered with suction on a large Büchner funnel. The product is thoroughly washed with distilled water (Note 4) and then dried at room temperature to constant weight. The yield is 105–110 g. (90–94 per cent of the theoretical amount) (Note 5) of a product which melts at 104–107°.
The crude dibenzalacetone may be recrystallized from hot ethyl acetate, using 100 cc. of solvent for each 40 g. of material. The recovery in this purification is about 80 per cent; the purified product melts at 110–111°.

The Notes (1-5) are as follows:

Sufficient alcohol is used to dissolve the benzaldehyde rapidly and to retain the benzalacetone in solution until it has had time to react with the second molecule of aldehyde. Lower concentrations of base slow up the formation of the dibenzalacetone and thus favor side reactions which yield a sticky product. Higher concentrations of base give added difficulty in washing. These concentrations were suggested by, and are approximately the same as, those used in the preparation of benzalacetophenone described in Org. Syn. Coll. Vol. I, 1941, 78.
Only temperatures between $20$ and $25^{\circ}$ were tried; it was assumed that a change of temperature would have the same effect that it has in the preparation of benzalacetophenone mentioned above. Stirring is essential, as it makes considerable difference in the uniformity of the product.
The benzaldehyde was u.s.p. quality which had been washed with sodium carbonate solution and distilled. Commercial c.p. acetone was used. The theoretical quantities are used, since an excess of benzaldehyde results in a sticky product while an excess of acetone favors the production of benzalacetone. The mixture is prepared before addition in order to ensure additions of equivalent quantities.
Since the product is practically insoluble in water, large amounts can be used in the washing. Sodium compounds are probably the chief impurities. The dried product contains some sodium carbonate which results from the failure to remove the sodium hydroxide completely. There remain also the impurities insoluble in water. However, the product is pure enough for use in most reactions.
If the mush is allowed to stand several hours, chilled, and filtered cold, a slightly larger yield is obtained, but this is not worth while. The filtrate may be used as a medium for a second run in which about 93 per cent of the theoretical yield is obtained. The melting point of the second product is slightly lower.

I think Note 4 should be important to OP, because it avoids product purification by chromatography.
Reference:
Charles R. Conard, Morris A. Dolliver, "dibenzalacetone [1,4-pentadien-3-one, 1,5-diphenyl-]," Org. Syn. Coll. Vol. 2, 1943, 167 (http://www.orgsyn.org/demo.aspx?prep=CV2P0167) and Org. Synth. 1932, 12, 22 (DOI: 10.15227/orgsyn.012.0022).


text:
Is it more precise to measure a substance as a solid or as a liquid?
If I have a substance and want to obtain an exact quantity of it, is it better to measure it as a solid or as a liquid?
In other words, if I have, for example, NaCl crystals I can weigh the crystals in an analytical balance, or alternatively I can dissolve them in water and then determine the concentration by, for example, titrating with silver nitrate.
Which method would be more exact?

I think your question boils down to whether measuring a mass or a volume is more precise. For the scenario of making a solution, you would have to measure in which volume you dissolve the NaCl, and the titration would involve measuring volumes. On the other hand, if your sample is wet or impure in some other way, the mass would be misleading.

You can definitely weigh "better" than you can measure volume. 4 significant figures is the best you can by volume. 4 significant figures is typically a very "good" measurement. Volume measurements are good where multiple samples with the same amount of the substance are required. So you'd make a large amount of a standard solution then take precision aliquots.  However it is really hard to weigh out multiple samples with the same 4 significant figure mass.

For the specific question asked by the OP, weighing the NaCl using an analytical balance is much more accurate than dissolving NaCl in water and titrating with silver nitrate solution. This is partly due to the latter involving multiple measurement error sources and partly because of the error magnitudes. Of course, this cannot be a generally applicable answer: see the comments by Karsten and MaxW. Also see some of the worked examples in this: EURACHEM/CITAC Guide, “Quantifying Uncertainty in Analytical Measurement”, 3rd Ed., 2012. Also see comment by @theorist.

Weighing is always far more accurate than any volume measurement. The extreme example is that of atomic weight (mass) measurements, which required accuracy to 4-5 decimal places. All the work was always done on the basis of gravimetry or by weighing the substances on high quality balances.

It depends entirely on the substance which you are trying to measure. For NaCl it would be better to dissolve it and then determine the concentration as NaCl is deliquescent it will absorb moisture from the atmosphere thus if you try to weigh it the reading would also include the mass of water absorbed by it. While in some cases it would be more convenient to weigh them to determine their quantity.

NaCl is usually considered hygroscopic rather than deliquescent, since it's only under unusually high temperature and humidity conditions that it can absorb enough water from the air to dissolve itself.  Having said that, the fact that it is hygroscopic, and can thus absorb water from the air, remains.  But I'm not convinced that would make a titration more accurate, since there are many sources of error associated with the latter.  If water absorption by salt is reversible, it seems it would be better to heat the salt sample in a dry atmosphere to desiccate it, then weight it.

NaCl reacts with NO2 in the atmosphere .              https ://www.nature.com/articles/306676a0 Thus even in dry atmosphere there would be error


text:
Defining the partial pressure equilibrium concentration
First off, I am not a chemist. I'm an engineer with an interest in thermodynamics for engine simulation for the basis of emissions. So if I have asked something stupid, this will be why.
So I am trying to work out the partial pressure equilibrium constant for dissociation of CO2 to carbon monoxide and oxygen, to then move onto multiple reactions to calculate thermal NOx using the extended zeldovich mechanism. Now from engineering textbooks, $K_p$ is used as the partial pressure of the reactants, divided by the partial pressure of the products. However, from chemistry definitions, it is the other way round. (Partial pressure of products over reactants.)
My question is, can both be used, or is there a trick necessary to be able to this I have missed? The math works out a bit nicer the engineering way, but I want it to be correct obviously.

I have checked all my engineering books and found that indeed, it is defined as products divided by reactants. However, I found the bit that is confusing me, just it was hidden behind math.
For calculating the standard Gibbs free energy change, is that always sum of products minus sum of reactants?
I've done the calculations, and the only way I can get the math to agree is sum of reactants – sum of products:
From "Transport Fuels Technology"
$$K(\ce{CO2}) = 3.0549$$
From JANAF Tables
$$
\begin{align}
ΔG_\mathrm{f}^\circ(\ce{CO2}) &= -395.461 \\
ΔG_\mathrm{f}^\circ(\ce{CO}) &= -367.826 \\
ΔG_\mathrm{f}^\circ(\ce{O2}) &= 0
\end{align}
$$
$$ΔG^\circ = -RT\ln K$$
$$\frac{ΔG^\circ}{-RT} = \ln K$$
$$
\begin{align}
K &= \exp{\left(\frac{ΔG^\circ}{-RT}\right)} \\
  &= \exp{\left(\frac{27635}{-8.314\cdot 3000}\right)} \\
  &= 0.330
\end{align}
$$
Or reverse $ΔG^\circ$:
$$
\begin{align}
ΔG^\circ &= \sum ΔG_\mathrm{f}^\circ (\text{reactants}) - \sum ΔG_\mathrm{f}^\circ (\text{products}) \\
  &= -395.461 - (-367.826) \\
  &= \pu{-27.635 kJ mol-1} \\
  &= \pu{-27635 J mol-1}
\end{align}
$$
$$ΔG^\circ = -RT\ln K$$
$$\frac{ΔG^\circ}{-RT} = \ln K$$
$$
\begin{align}
K &= \exp{\left(\frac{ΔG^\circ}{-RT}\right)} \\
  &= \exp{\left(\frac{-27635}{-8.314\cdot 3000}\right)} \\
  &= 3.028
\end{align}
$$
which is in agreement with the book I am using.

I've never seen a chemistry engineering textbook where "reversed" $K_\mathrm{p}$  is used. Could you please add some references or context where exactly you've seen it?

As a chemical engineer, I have never seen this either.  It is always products divided by reactants, as indicated in Fundamentals of Engineering Thermodynamics by Moran et al, Introduction to Chemical Engineering Thermodynamics by Smith and van Ness, and Chemical Engineers' Handbook by Perry et al.

Please don't provide links to the third-party cloud services/screenshots etc. I also would encourage not to use Word since it's not suited for scientific publishing at all. I re-typed the page from your Dropbox with MathJax, feel free to correct it if I made a mistake somewhere. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

As for the question, it would be nice you'd added the reaction you are investigating the equilibrium for. Is it burning CO to CO2?

OK, i will have a look at MathJax. Have not come across that one. Thankyou for your patience. I have updated my question to show i am looking at dissociation of CO2 to CO + 0.5 O2


text:
Derivation of work done by real gas
According to the Wikipedia on real gases, it is stated that the expansion work done by a real gas differs from the ideal gas by $\int\left(V-\frac{RT}p\right)\,\mathrm dp$.
Firstly, does this mean $W_\text{real} = W_\text{ideal} + \Delta W$ or $W_\text{real} = W_\text{ideal} - \Delta W$?
Secondly, how is this formula derived?

There's something awfully fishy about that equation. Regarding $\Delta W$ it depends on what convention (definition) you use.

That equation was added on 19:33, 14 April 2017. It looks like it may be vandalism.

That statement in Wikipedia is incorrect.  That expression is the residual gibbs free energy at constant temperature for a real gas.

Wikipedia's "Real gas" article gives various equations used to model gases besides the ideal gas law. For example  the Van der Waals model etc.

For a reversible process in a closed system (no mass entering or leaving), the general equation for the expansion work done by a real gas on the surroundings is the same for an ideal gas, namely
$$W = \int p\,\mathrm{d}V$$
However, for a real gas, we use the equation of state for that gas $p=p(n,V,T)$ rather than $p=\frac{nRT}{V}$, the equation of state for the ideal gas.

That expression is incorrect.
For one mole of an ideal gas,
$$ V_\mathrm{m} = \frac{RT}{p}$$
where $p$ is the pressure of the gas.
$pV$ work is defined as
$$ W_{pV} = -\int_{V_i}^{V_f} p_\mathrm{ext}\mathrm{d}V_\mathrm{m}$$
where $p_\mathrm{ext}$ is the applied pressure against which work must be done. By this sign convention work done by the system is negative.
Therefore
$$
W_{pV,\mathrm{real}} - W_{pV,\mathrm{ideal}} =
-\int_{V_i}^{V_f} (p_\mathrm{ext,real} - p_\mathrm{ext,ideal})\mathrm{d}V_\mathrm{m}
$$
Assuming mechanical equilibrium between the applied pressure and the gas,
$$ W_{pV,\mathrm{real}} - W_{pV,\mathrm{ideal}} =
\int_{V_i}^{V_f} (\frac{RT}{V_\mathrm{m}}-p_\mathrm{real})\mathrm{d}V_\mathrm{m}$$
where $p_\mathrm{real}$ is described by the equation of state for the real gas.


text:
What is the full mechanism for oxidation of toluene to benzoic acid using permanganate?
I have managed to get some of the way (I think):
I am particularly interested in the full mechanism i.e. all the curly arrows that take you from toluene right through to benzoic acid.

The answer to that question does not address the full, step-by-step mechanism.

I'm not sure if such a "full" mechanism exists. If it does, it's almost certainly radical, not some combination of pericyclic reactions.


text:
Is there a replacement for glacial acetic acid in determining peroxide value

I am not allowed to use glacial acetic acid at my school, which all procedures online for finding the peroxide value of oils use through an iodometric titration. Is there a replacement?

It vastly depends on where you determine the peroxide value. Sometimes you can use mineral acids, sometimes not.

I am analyzing the effects of gamma-Tocopherol on olive oil using the IFRA procedure to determine peroxide value. I cannot use chloroform. So I wanted to use a 3:2 ratio of glacial acetic acid to cyclohexane, however, I cannot use glacial acetic acid. Would mineral acids work in this scenario or do I have to change my method?

Your school is not allowing use of glacial acetic acid, but you insist of using glacial acetic acid in determining peroxide value, you may use a solvent system, which consists of glacial acetic acid. For example, you can use 60% ($v/v$) glacial acetic acid in chloroform. I give that example because it is commercially available online at Reagents. If you are determining peroxide value of fats, this is an ideal solvent to dissolve fat and resultant solution is soluble in water.
I'm suggesting this because there is a method recommended by International Fragrance Association (IFRA) in Belgium. Breifly, one of the IFRA analytical methods for determination of the peroxide value is described as follows:

Transfer approx. $\pu{3.0 g}$ of the sample, accurately weighed, into a 250-mL Erlenmeyer flask with glass stopper. Add $\pu{50 mL}$ of the appropriate solvent mixture (glacial acetic acid : chloroform 3:2; glacial acetic acid : cyclohexane 3:2; glacial acetic acid : isooctane 3:2) and saturated potassium iodide solution, $\pu{1 mL}$, freshly prepared and allow to react for $\pu{60 ± 1 sec}$ and shaking thoroughly during this period. Then add water, $\pu{100 mL}$ and shake. Titrate with $\pu{0.01 mol/L}$ sodium thiosulfate solution, or $\pu{0.1 mol/L}$ sodium thiosulfate solution at higher usages, using $\pu{1 mL}$ starch solution or $\pu{0.1 g}$ of Thyodene indicator. The indicator should be added towards the end of the titration but while the pale straw color is still present. During titration shake until the blue color disappears. Carry out a blank titration under the same conditions. No more than $\pu{0.5 mL}$ of $\pu{0.01 mol/L}$ sodium thiosulfate solution should be consumed for this purpose.

Note: Use $\pu{0.1 mol/L}$ sodium thiosulfate solution when the expected peroxide values are $\gt \pu{20 mmol/L}$. Use $\pu{0.01 mol/L}$ sodium thiosulfate solution when the expected peroxide values are $\lt \pu{20 mmol/L}$.
Thyodene indicator is also commercially available in Fisher.

Thanks, I was using this procedure to give me an idea of how to measure the peroxide value of olive oil. I am using cyclohexane instead of chloroform, however, I cannot purchase a solution containing glacial acetic acid.

As an alternative you can use Ferric Thiocyanate Method, instead.
The ferric thiocyanate method: Chapman and McFarlane (1943) have developed colorimetric method to determine peroxides in food such as milk powder (Ref.1). The method is based on the oxidation of ferrous ($\ce{Fe^{II}}$) to ferric ($\ce{Fe^{III}}$) iron by the peroxides present and the colorimetric determination of the $\ce{Fe^{III}}$ iron as ferric thiocyanate (red color). They also applied the method to determine peroxides in fats and oils (Ref.2).
In this method, a reagent containing 0.1% ferrous ammonium sulfate and 0.4% ammonium thiocyanate in 96% acetone ($\pu{9 mL}$) is added to a $\pu{1 mL}$ aliquot of an acetone solution of the fat, and the mixture is heated for 10 minutes in a water bath at $\pu{50 ^{\circ}C}$. The red color developed (ferric thiocyanate) is measured in a Spectrophotometer at $\pu{485 nm}$. A calibration curve should be prepared from readings obtained with standard solutions of ferric chloride ($0.2$ to $\pu{14.0  \mu g/ mL}$) in acetone.
Authors’ Note:

This procedure gives more readily reproducible results than the usual peroxide titration methods and also measures more completely the extent of oxidation, particularly with highly unsaturated fats.

References:

R. A. Chapman, W. D. McFarlane, “A Colorimetric Method for the Determination of Fat-Peroxides and Its Application in the Study of the Keeping Quality of Milk Powders,” Canadian Journal of Research 1943, 21b(7), 133-139 (https://doi.org/10.1139/cjr43b-017).
A. Lips, R. A. Chapman, W. D. McFarlane, “The application of the ferric thiocyanate method to the determination of incipient rancidity in fats and oils,” Oil & Soap 1943, 20(11), 240-243 (https://doi.org/10.1007/BF02630882).

Thanks so much, all the materials fit the approved chemicals list!

My pleasure. Please also remember, to make calibration curve, need to use known concentrations of $\ce{FeCl3}$ solutions in place of the unknown sample.


text:
Is the existence of chirality center equivalent to chiral?
To make my question more clear, I want to ask two questions:

Is it possible for molecule to be achiral even though molecule has chirality center?

Is it possible for molecule to be chiral even though molecule doesn't have chirality center?

I thought of these questions because the definitions of chiral center and chiral do not have a one to one correspondence.
A chiral center is defined by carbon having four different ligands around it and chiral is defined by a compound being non-superimposable with its mirror image. They are quite related, but I think there will be counterexamples that satisfy one condition but not satisfy the other one.

related https://chemistry.stackexchange.com/questions/59124/what-is-the-perfect-definition-for-chirality

The answer to both questions is yes.
In regards to the first question, meso compounds like tartaric acid have multiple chiral centers, but are not chiral overall because they have a mirror plane.
For the second question, a compound can be geometrically chiral without having a chiral center. A cool example class of compounds is helicenes, which are helical arrangements of aromatic rings. These compounds have axial chirality, where the stereoisomers are the left and right handed versions of the helix.

A simpler example for (2) is penta-2,3-diene.  Because the methyl groups are strung out along the cumulative double bond axis, not on a single central atom, the molecular symmetry is lowered and one of the symmetry elements that goes missing is a mirror plane.


text:
Why we should concern with A∞?
To determine the order of one reactant at pseudo concentration of another reactant the plot of $\ln(A_t - A_∞)$ versus $t$ is used.
$A_t$ is absorbance of the intermediate at time $t$.
$A_∞$ is absorbance of intermediate at the end of the reaction (approximately after 24 hours).
Why $A_∞$ is important here?

I tried to improve the formatting in your post, hopefully it reflects your question. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Let's suppose a hypothetical irreversible reaction:
$$\ce{B + H -> D + F}$$
where H is a reactant with constant concentration during the reaction and that don't absorb light in the working frequencies. The absorbance of a species $i$ related with the concentration by the Lambert-Beer Law:
$$A_{i}=c_{i}\varepsilon_{i}l$$
where $c_i$ is the concentration of species $i$, $\varepsilon_{i}$ is the molar absorptivity and $l$ is the length of the cuvette.
At an arbitrary time $t$, the total absorbance of the solution is given by:
$$A_{t}=c_{B,t}\varepsilon_{B}l+c_{D,t}\varepsilon_{D}l+c_{F,t}\varepsilon_{F}l
\tag{1}$$
by the stoichiometry of the reaction, we have: $c_{D,t}=c_{F,t}=c_{B,0}-c_{B,t}$; where $c_{B,0}$ is the initial concentration of the reactant B, so replacing this in Eq. 1 and defining a auxiliary constant $\varepsilon_{i}l=k_{i}$, we have:
$$A_{t}=c_{B,t}k_{B}+k_{D}(c_{B,0}-c_{B,t})+k_{F}(c_{B,0}-c_{B,t})$$
$$A_{t}=c_{B,t}(k_{B}-k_{D}-k_{F})+c_{B,0}(k_{D}+k_{F})\tag{2}$$
At a time $t\rightarrow \infty$  (a time long enough that the reaction goes to completion), $c_{B,\infty}=0$ and $c_{D,\infty}=c_{F,\infty}=c_{B,0}$, so the absorbance is given by:
$$A_{\infty}=c_{D,\infty}k_{D}+c_{F,\infty}k_{F}$$
$$\boxed{A_{\infty}=c_{B,0}(k_{D}+k_{F})} \tag{3}$$
so,replacing Eq.3 in Eq.2 its follows that:
$$A_{t}=c_{B,t}(k_{B}-k_{D}-k_{F})+A_{\infty}$$
$$\boxed{A_{t}-A_{\infty}=c_{B,t}(k_{B}-k_{D}-k_{F})}$$
In conclusion: $A_{\infty}$ is important because is directly proportional to the initial concentration of reactant B. When you subtract $A_{\infty}$ from the absorbance at an arbitrary time $t$ $(A_{\infty})$ it allows you to know (indirectly) the concentration of the reactant at the aforementioned time $t$.


text:
Pre-Hydrogen Feature On the Output Of a Gas Chromatography Thermal Conductivity Detector
A candidate for "dark matter", predicted and now apparently detected to have both a higher thermal conductivity and higher diffusion rate than H2 is purported in this PowerPoint Presentation file.  From that presentation, this output of a Thermal Conductivity Detector (run at 60C) of a HP 5890 Series II gas chromatograph using an Agilent molecular sieve column with Helium carrier gas is puzzling.  The Hydrogen peak was establish by a control run.  However, in the experimental run there appears a mysterious feature annotated with "???" indicating some sort of gas (that is not Helium) with a higher diffusion rate than Hydrogen.

A TCD temperature of 60C was chosen to ensure hydrogen would not show up as a negative peak as indicated by Figure 2 of "Thermal Conductivity Detector Analysis of Hydrogen Using Helium Carrier Gas and HayeSep® D Columns", by K Snavely and B Subramaniam.

What are plausible candidates for this feature?
For context, here is a control run:

Here is the corresponding experimental run from which the above detail screenshot is taken:

After the reaction, the sealed vessel stood for over 24 hours.  This TCD graph obtained.  Although a small peak associated with Nitrogen appeared, due to atmospheric ingress to the vessel, the Hydrogen peak no longer exhibited the "???" features:

Just fyi, before you waste time on this topic. The user James Bowery is a disciple of Randell Mills. He has posted here to push the conversation into the "Could the anomaly be Hydrinos?" direction. Link to the BLP subreddit Another link where he admits using "stealth mode"

I'm running what is called a "blind" review to avoid precisely the kind of prejudice you have exhibited by calling me a "disciple".  The data and procedures stand or fall on their merits.  If you want to accuse Randall Mills of scientific fraud, that is another matter.  People tend to be polarized on controversial topics so my attempt at a blind review is entirely reasonable, and calling it "stealth mode" to those who are polarized toward Mills is reasonably politic.

The data is still there. No amount of prejudice will change the value of it. You will either find someone who can confirm that Hydrinos have been found or you will not. Good luck.

Well your likely-successful attempt to prejudice the review is too late in any event.  I've got 2 alternative explanations and have put them into perspective with the discovery of a paper -- on my own -- that addresses one of them (now cited) regarding TCD operating temperature and the other (overloading analyte)  to be critically addressed.

Oh, and don't sell yourself short:  When topics are so controversial people's careers can be terminated and mortgages foreclosed for merely looking into them, making blind reviews impossible may foreclose discovery of the truth -- yea or nay.  The fact that you're too late by maybe 16 hours in doing your worst should not count against you.  You did try.

It would be nice to have a full chromatogram rather than a cropped one to see how other peaks look like. The phenomenon of odd behavior of hydrogen with helium as carrier gas with TCD detector in quite well known, but I am not sure if it has been investigated in detail in the literature or not. Hydrogen has higher thermal conductivity than helium. Ideally when you inject $\ce{H2}$, one should see a negative dip as you are seeing. However people also see a split peak. Yours is a mixture of split and a dip. The rest of the peaks should appear normal. If you change the carrier gas, this problem will go away (for $\ce{H2}$) but you lose sensitivity for other analytes as helium has the highest thermal conductivity after hydrogen in gas phase.
If you change the polarity of the detector, you would see this phenomenon for $\ce{H2}$ reverted and other analytes would appear as negative peaks.

I've added 2 full chromatogram graphs for context.

That makes perfect sense, because the rest of the peaks are positive and only H2 is messing. The answer is still valid.

The preceding positive spike is higher than the following positive peak for hydrogen.  Is this what you are referring to as an apparent split peak for hydrogen? If one should expect to see a negative dip, why might it appear only in the experimental run and not the control run?

Just saw that you added more info. What is the difference between control and experimental runs? I mean what are you injecting in each case. The "misbehavior" of H2 peak should be seen with the injection of pure H2 as well as the sample containing H2. Do you know the dead time (void time) of the column? It may be possible that something is co-eluting with H2 in the experimental part.

Hydrogen is the only element with thermal conductivity greater than helium ($0.182$ vs $\pu{0.151 W\,m^{-1}K^{-1}}$ at $\pu{25 ^{\circ}C}$) as M. Farooq pointed out (cf., thermal conductivity of $\ce{N2}$  is $\pu{0.026 W\,m^{-1}K^{-1}}$ at $\pu{25 ^{\circ}C}$). The mixtures of $\ce{H2}$  in $\ce{He}$  at moderate temperatures exhibit varying thermal conductivities, more or less than either component alone. Thus, if you are analyzing for $\ce{H2}$  with $\ce{He}$  as your carrier gas, a peak of $\ce{H2}$  may appear as positive, negative, or as a split one (This may be the case for the difference between your reference and experimental runs).
One solution for this problem is to operate the detector at higher temperatures, to say, from $\pu{250 ^{\circ}C}$ to $\pu{300 ^{\circ}C}$: You may analyze a known range of hydrogen concentrations using your said method until you find the correct detector operating temperature. To do so, you may need to increase the operating temperature several runs, until the hydrogen peak exhibits normal shape. And ensure it is always in the same direction regardless of $\ce{H2}$  concentration. This newfound detector temperature also ensures high sensitivity because of the higher value.

These issues raise the question as to whether there exist validated models of binary gas mixture thermal conductivity.  The result of my search for this came up with this relatively recent paper:
https://www.thermalfluidscentral.org/journals/index.php/Heat_Mass_Transfer/article/view/332/346

The TCD ran at 60C (which I added to the question).  However, the graph that I added showed that by merely permitting the reaction vessel to stand idle after the reaction (over 24 hours) the ??? features disappeared and the graph reverted to the control except for a slight N2 ingress (the CH4 peak apparently being cropped out).

Please mention the dead time or void time of your chromatograph. It is the time it takes for an unretained molecule to travel from the injector to the detector. Is H2 eluting at the dead time? Without knowing this information, all this discuss can become meaningless. 2) Your detector settings, split ratio are not suitable either. Why there is so much noise, why the baseline is oscillating?

@M. Farooq: I think high noise duto the detector temperature, which is operating at $\pu{60 ^{\circ}C}$. I still think OP need to increase that temp to a higher value (probably over $\pu{250 ^{\circ}C}$ since autoignition temperature of $\ce{H2}$ is way above $\pu{500 ^{\circ}C}$) to get better sensitivity unless there is a reason to operate at $\pu{60 ^{\circ}C}$.

@MathewMahindaratne, I agree, the settings need to be optimized.

The only paper I've found providing a table of He+H2 thermal conductivities against mole fraction against temperature is:
http://shodhganga.inflibnet.ac.in/bitstream/10603/173952/11/11_chapter%205.pdf
Is there a better cite?

I've added the justification for the 60C TCD setting.


text:
Is Neodymium Powder Magnetic?
In attempting to make magnetized neodymium (NdFeB) powder, I grinded neodymium magnets into a powderusing a belt sander. However, I found that the magnetic properties of the neodymium had decreased substantially. It would appear that any attempt to grind the magnets into a powder using this method seems to cause the diminishing magnetic field observed https://en.wikipedia.org/wiki/Neodymium_magnet .
Is it possible for neodymium (NdFeB) powder to be as magnetic as a solid neodymium (NdFeB) magnet?

Neodymium powder ≠ powdered neodymium magnet. Neodymium is an element, neodymium magnet is an alloy. Which one you are asking about?

NdFeB is the type of magnet that I am referring to.


text:
Reaction that will produce a silica coating (glaze)
The normal way to create a coating of pure silicon dioxide glaze is to coat the object to be glazed in a slurry of silica particles and then put it in a kiln.
However, if I have an object that is too large to fit in a kiln, then there is a problem.
I am wondering if there is a reaction of some kind that will produce a glaze of pure silica. The idea is that I could coat the object with the reactants, ignite them, and then when the reaction is over the object will be coated with fused silica. Is there any way to do this?

have you hear of TEOS?

This is done in the semiconductor industry using silane, see the Wikipedia article on epitaxy  https://en.wikipedia.org/wiki/Epitaxy


text:
Does oxyhydrogen require the presence of water to ignite?
I recently read a short story (fiction) where a major plot element was based on the expectation that a mixture of oxygen and hydrogen will not ignite unless some amount of water is present to get the reaction started. Wikipedia has no mention of this property, and I couldn't find any other claims one way or another.
Quote from the story: "Hydrogen and oxygen will not unite in the total absence of water"
Is this true?

TL;DR: No.

Campbell is known for being a "hard sci-fi" writer, so one can assume the facts described in the story don't contradict with the state of science at that time.
From the context of the book ("The Tenth World", Dec 1937) the action takes place on a transplutonian planet where rivers of liquid hydrogen wash shores of solid oxygen.
Assuming the pressure close to atmospheric, this suggests the temperature range between 14 K and 20 K (LH2).
And a proton gun is used to initiate the reaction (explosion) between the elements.
The most recent work of the same year was Kassel's detailed research on kinetics of hydrogen ignition (Jul 1937) [1].
At that time the temperature margins for the explosive region haven't been drawn precisely yet:

The effect of temperature also is rather uncertain.
  It seems that when the upper and lower limits are far apart, the lower limit decreases only slightly as the temperature is increased, but that for low temperatures, where the two limits come together, the effect is considerably greater.

However, it was shown that this is a chain reaction occurring in the gas phase:

The ignition region is unquestionably due to a chain reaction.
  The chains ordinarily start on the walls, but a few originate in the gas [...]
  Branching occurs in the gas.
  The chains may be broken either by a triple collision in the gas, or by diffusion to the wall. Ignition takes place, according to the simple theory, when the rate of branching is greater than the rate of breaking.

Nothing implies that water promotes the reaction.
The mechanism has been significantly improved since [2] and explosion limits were precisely detected:

E.g. for an explosion to occur, one must've supply a tremendous amount of energy via proton gun (whatever this is) to locally increase the temperature of the environment, vaporize liquid hydrogen and solid oxygen and heat them both up to at least 400 °C.
Also, according to modern views, the reaction mechanism can be written in the following simplified form (there are many more stages, only the main chain is shown):
$$
\begin{align}
\ce{H2 &→[$Δ/hν$] 2H^.}\\
\ce{H^. + O2 &→ HO^. + O^.}\\
\ce{HO^. + H2 &→ H2O + H^.}\\
\ce{O^. + H2 &→ HO^. + H^.}
\end{align}
$$
Rate-limiting stage is the dissociation of hydrogen molecule.
A single H atom is then responsible for generation of hundreds of thousands of water molecules.
There is no evidence that presence of water catalyses or somehow promotes the chain reaction.
The reaction, however, can be catalyzed by transition metals such as dispersed Pt (see also Döbereiner's lamp).
To sum it up, it seems like Campbell couldn't know for sure the exact mechanism for hydrogen burning at that time, and as it later turned out, water doesn't marginally affect it.
Probably he just decided to use the concept of seed crystal for water formation.
References

Kassel, L. S. The Mechanism of the Combustion of Hydrogen. Chemical Reviews 1937, 21 (2), 331–345. https://doi.org/10.1021/cr60069a010.
Das, L. Hydrogen-Oxygen Reaction Mechanism and Its Implication to Hydrogen Engine Combustion. International Journal of Hydrogen Energy 1996, 21 (8), 703–715. https://doi.org/10.1016/0360-3199(95)00138-7.

Plus 1. But OP refer to ignition not to ignition leading to explosion (though this is indeed the scenario sketched by the novel). Burning would happen anyway presumably even out the composition or P,T limits for explosion (as per the question).


text:
Substitute molten salt bath with temperature range from 400–500 °C
I'm looking for a suitable substitute for a $\ce{NaNO2/KNO3}$ salt bath (for quite obvious reasons due to the other applications of the latter chemical). The salt mixture needs to have around 400 to 550 °C of molten working temperature (the lower ranges are not interesting), and preferably be reasonably safe to work with (as safe as the aforementioned salt bath, so totally harmless isn't the goal). The purpose is precise annealing of brass alloys in a home shop environment (so no fume hoods and blast cabinets available).

I don't really understand the "quite obvious reasons" for not using NaNO2/KNO3 molten salt bath. A saltpeter bath is a #1 choice for annealing of brass alloys. Could you probably add more details why it doesn't suit you?

KNO3 is used in various explosives and pyrotechnics, and as such is quite cumbersome to purchase and ship. But yes, it's the obvious choice, which is why I'm having such a hard time finding an alternative (other option is obviously that an alternative isn't available).

Well, it's also used as common fertilizer, or even in food preparation...

I wasn't sure what other conditions you needed for your salt bath, yet I listed few for your convenience together with the salt bath of $\ce{KNO3/NaNO2}$ (bath 5) for comparison:
$$
\begin{array}{ccc}
\text{Bath#} &\text{Composition} & \text{Approx. melting poing} & \text{Working temp. range} \\\hline
1 & \ce{NaCl}, \ce{KCl}, \ce{BaCl2}, \ce{CaCl2}  & \pu{400 ^\circ C} & \pu{500-800 ^\circ C} \\
 & (10\!-\!15 \!:\!20\!-\!30 \!:\!40\!-\!50 \!:\!15\!-\!20) &  &  \\
2 & \ce{NaCO3}\, (45\!-\!55), \ce{KCl}\, (55\!-\! 45)  & \pu{450 ^\circ C} & \pu{550-900 ^\circ C} \\
3 & \ce{NaCl}\, (20), \ce{KCl}\, (30), \ce{BaCl2}\, (50)  & \pu{540 ^\circ C} & \pu{570-900 ^\circ C} \\
4 & \ce{NaNO3}\, (100)  & \pu{370 ^\circ C} & \pu{400-600 ^\circ C} \\
5 & \ce{KNO3}\, (50\!-\!60), \ce{NaNO2}\, (50\!-\!40) & \pu{135 ^\circ C} & \pu{160-550 ^\circ C} \\
\end{array}
$$
I also listed $\ce{NaNO3}$ bath in case you won't mind using it since it is not potassium salt and working range is ideal for your requirement. As indicated, you can change composition as much as you wanted to to get right melting point and the working temperature range you wanted to.

Use caution with soluble Ba salts because of toxicity.

Cheers mate, I'll research a pure NaNO2 salt bath since it's closest to the working range. I read somewhere that temperatures over 590 C can decompose the standard KNO3/NaNO2 mix and release nitrogen oxide, which I understand is something you definitely want to avoid. I gather this is also true for just NaNO2?

Sorry to inform you that I made a typo on entry 4 of the chart. It is $\ce{NaNO3}$ instead of  $\ce{NaNO2}$. By the way,  melting temperature of $\ce{NaNO2}$ is  $\pu{271 ^{\circ}C}$, but it decompose at $\pu{320 ^{\circ}C}$, well below the temperature you want.


text:
Why does dielectric constant of crystal fluctuates in case of Frenkel defect in a crystal?
So I was studying Frenkel defect in crystal and came across this line that said

The closeness of like charges tends to increase the dielectric constant of the crystal.

I'm not able to understand it properly because in a dielectric an number of dipoles align themselves in a direction parallel to electrical field and thus they create a counter field to reduce the electric field that created or aligned it. Now, if we think like that here, dipole moment is charge multiplied by distance, and distance has definitely reduced between a positive and negative ion here, so dipole moment should decrease and consequently dielectric constant should decrease but why is the opposite happening here.

closeness of like charges. A dipole is based on separation of unlike charges...

Yeah one of the reasons why I'm not able to understand this line.The paragraph below the line are my thoughts on the matter.I'd love to be corrected if there are some potential correction or mistake in my thoughts.

No reason that you can't have a dipole consisting of like point charges. The value will vary with where you put the origin, bit it's still a dipole.

The explanation provided in the text you quote is of the "hand waving" sort. Disrupting the crystal structure via Frenkel defects will have a rather complicated effect. That the effect is complicated is illustrated (in a rather rudimentary fashion) in the wikipedia, from which I derived the following figure:

Clearly the dislodged cations are shifted into closer proximity of other cations, but also of anions, and the orientational effects are not simple.
So the text you quote does not provide an entirely satisfactory explanation, I agree.

Now consider your line of argument:

Now, if we think like that here, dipole moment is charge multiplied by
  distance, and distance has definitely reduced between a positive and
  negative ion here, so dipole moment should decrease and consequently
  dielectric constant should decrease but why is the opposite happening
  here.

The scenario you describe in the paragraph makes sense. You can write the dipole moment $p$ for two point charges of opposite polarity $q$ and separated by distance $d$ as
$$p = qd $$
Note that, as the wikipedia explains, the value of $p$ is independent of the choice of reference point, provided the overall charge of the system is zero.
Obviously decreasing $d$ decreases $p$.
The problem is that this does not describe the situation layed out in the text you quote:

The closeness of like charges tends to increase the dielectric constant of the crystal.

Consider what it means to increase the closeness of like charges. Presumably if the crystal remains constant in volume and overall charge, it requires that opposite charges move apart.

But this is still a hand-waving explanation. You need to attempt some calculations with a model to get closer to a convincing explanation.


text:
Does a calibration curve of absorbance vs concentration need to pass exactly through the origin?
If you are making a calibration curve for $\ce{CuSO4}$ where you plot absorbance against concentration and the curve deviates slightly from the origin, does it make the calibration curve useless?

Not necessarily, it depends on the method you are using for collecting the data. For example, the standard addition method basically requires that the curve doesn't go through the origin. Which one are you using?

The value of the absorbance for a blank is another data point, but a little special. Usually you calibrate the instrument with a blank, so that in essence you take at least two measurements at the origin. The most important question is probably what range of concentrations you want to cover. Whether you include the origin depends on the linearity (not to mention dynamic range) of the instrument response. Also, you should perform measurement and analysis identically for your samples, whose concentration should fall in the range. Also, some analyses force the curve through  the origin, bad.

Good answers, but it is almost never a good idea to force the intercept to zero. Unless there is a clearly justifiable and compelling reason to do so, it is better to let the data set 'vote' for itself and not impose a preference/bias/mandate. Among other things, forcing the intercept to zero changes the standard error of the slope, which is an important statistic.
Below is a screenshot of the result of using Excel's Data Analysis tool on a set of linear calibration data I collected in the past:

As shown, the intercept is -0.0541 and its standard error is 0.0352, so the t test statistic of the intercept, defined as intercept/(standard error of the intercept) = -1.537. The negative sign is irrelevant for present purposes, but the P-value is 0.264.
The null hypothesis for the intercept is that the true intercept, which is unknown, is zero. The alternative hypothesis (aka research hypothesis) is that the true intercept is either 1) not equal to zero or 2) not equal to zero, but in a known direction. The former is the 2-tailed option and the latter is the 1-tailed option. For intercepts, the 2-tailed option is most common and this would almost certainly be the case for a Beer's law plot.
The P-value of 0.264 is much higher than the usual 0.05 value and this means that it would be quite risky to reject the null hypothesis based on the OLS analysis. In fact, the risk of mistakenly rejecting the null hypothesis, even though it was true, would be 26.4%. The confidence in the alternative hypothesis would be 100% - 26.4% = 73.6%, which is too little confidence for most people. Typically, you want P ≤ 0.05. Otherwise, it is too risky to 'jump ship' for the alternative hypothesis.
What does it mean? Well, the data resulted in a P-value of 26.4%, so the null hypothesis cannot prudently be rejected. But it is also not accepted: it still might be wrong, but the data did not show that. Replacing the intercept by zero would mean that a decision was made that the true intercept was zero, even though the analysis proved no such thing. It is not a good idea to pretend that the data showed what it did not show. Bottom line: the intercept should not be forced to zero unless there is a clear and compelling reasion to do so.

very nice answer. I'd like to add that there are tons of proper and valid reasons why the signal intensity may have an offset (or, thinking in terms of spectra, that lead to a non-zero baseline). For example, a small amount of some masking agent may lead to a negative offset: in that case also the blank sample should not be included in the calibration. Other, more general thought: using a blank measurement to adjust the instrument to yield zero signal (outside the calculation of the calibration function) will transform noise/random uncertainty on that blank measurement into a systematic term..

...for all following measurements if the calibration function is then forced through the origin. Still, if the calibration allows an offset, that offset corresponds to the fitting function's estimate of the error of that blank measurement, so the calibration function is not harmed as when forcing it through the origin, but the measurement is wasted: Including that blank measurement instead as normal measurement at conc = 0 in the calibration curve (and there wouldn't even be any discussion about forcing the curve through the origin now) allows that measurement to add its information to...

... the estimation of the calibration curve. Which is what we typically want. Oh, and of course, the important link is that we cannot measure absorbance, we typically don't even measure transmission but the intensity of light that got through sample and possibly cuvette/window/sample holder and assume that the missing light that was not missing in the blank measurement was absorbed.

@cbeleites I completely agree on all the points you raise! As an analytical chemist and spectroscopist, I have seen some truly bizarre things people have done to their data. I answered this question to try to stop some of that abuse of data. I do not know whether any good will come of it, but maybe it will. Anyway, thanks for your thoughtful comments!

@cbeleites Your comment about is right on point! When I used to teach the Instrumental Analysis lab, I used a home-made scanning spectrometer to have students obtain absorbance spectra, from 400 - 700 nm, of aqueous chloride solutions of $Ho^{3+}$, $Er^{3+}$ and $Pr^{3+}$. First they did a dark response scan, with a "black cuvette", then a blank scan (cuvette with water), and then the sample scan. Next they used a spreadsheet template to compute the dark response corrected "pseudo-transmittances", next T($\lambda$) as quotient and then A($\lambda$). Students were a surprised to see that

absorbance was not actually measured. I also used the opportunity to talk about what happens to the actual noises, as they get processed along the way, and I had students do a computer simulation (ready to run) with my optical calculus software. I thought it worked well, but teaching is curiously humbling, so I do not really know.

You already have an answer above in the comments. However some teachers suggest to force the calibration line zero, because one might think that blank must have a zero value. This is not a correct approach at all. You can have a non-zero intercept as long as your equation is linear (correlation coefficient r2 is closer to 1 e.g. 0.999). MS Excel can easily do this for you. Although you may not need it for your school work, but the rigorous procedure is to do a statistical t-test. This test "decides" whether we should pass the calibration curve through zero or not. Advanced readers may like to read this article http://www.chromatographyonline.com/calibration-curves-part-i-b-or-not-b
Some organizations or protocols may require that we force the calibration line thru zero. It makes some calculations easier.


text:
Misunderstanding about the spin part of the Jahn Teller effect
In the paper Stability if Polyatomic Molecules in Degenerate Electronic States II - Spin Degeneracy the author wrote:

Thus a twofold degeneracy cannot produce instability for a molecule containing an odd number of electrons with spin.

It is really disturbing for me as I'm looking to the Jahn-Teller effect being strong on the $e_g$ orbitals when there is a odd number of electrons on it.

I haven't got a chance to delve in this paper in detail, but it seems to be referring to the specific case of twofold spin degeneracy (for example an single electron being spin up or down produces two degenerate states). $e_g$ describes the spatial symmetry and it can still produce a Jahn-Teller distortion. @Tom


text:
Why is the Maxwell–Boltzmann formula inapplicable at low temperatures?
This Wikipedia article states that at sufficiently low temperatures (or high pressure), quantum effects lead to a breakdown of Maxwell–Boltzmann statistics.
I know that the spacing of energy levels becomes larger at low temperatures, but doesn't the derivation of the MB distribution, as shown in the article, already consider energy levels to be discrete?
Furthermore, it also takes the degeneracy of energy levels into account.
So why and what other quantum effects are there at low temperatures that still violate the formula?

Have you checked the approximations it is based on?

@JohnnyGui the wiki also has a section describing the limits of applicability and that it is related to the fact that actual particles are indistinguishable.

In a quantum system, such as an atom or molecule,  the spacing of levels is a property of the system being studied and not of temperature. Rather it is that there is not enough energy at low temperatures to populate many of the levels, and just the lowest one at zero Kelvin.

@porphyrin Why is it a problem that lower energy levels get populated by many particles in the case of a low temperature? How does this violate the MB distribution?

@Tyberius I have indeed read that part. What I don't get though is why particles become all of a sudden indistinguishable in the case of low temperature and/or high density

@JohnnyGui it is not that they become indistinguishable at low temperatures, they always are. However, in the high temperature limit, the Bose and Fermi distributions both converge towards the Maxwell-Boltzmann distribution.

@Tyberius If they are always indistinguishable, shouldn't this make the MB Distribution invalid under any condition, since the MB Distribution is based on distinguishable particles?

@JohnnyGui yes, the MB distribution is never exactly correct. However, the difference between MB and FD/BE is just an extra +/-1 in the denominator, which if the temperature is high enough that contribution becomes negligible (see this answer on physics for additional explanation:https://physics.stackexchange.com/questions/289796/fermi-dirac-distribution-in-the-high-temperature-limit). So at high temperatures, we can very accurately approximate the more correct FD or BE distributions using the MB distribution.

Just a bit more information on my edit: Don't abbreviate Wikipedia as wiki

Since the explanation was a little more complicated than I initially thought, I figured it would be worth it to combine my comments (and info from Physics SE) into an answer.
Quantum particles satisfy Fermi–Dirac or Bose–Einstein statistics depending on whether they are fermions or bosons. These distributions have the form $$\langle n_i\rangle=\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]\pm1}$$ where the plus/minus is for fermions/bosons.
To consider the high temperature limit, we need to note that not only is there a direct temperature dependence, but $\mu$ is also dependent on temperature. Specifically, in the high temperature limit, $\mu<0$ and $|\mu|>k_bT$. Combining this information, we can make the (very accurate) approximation $$\exp[(\epsilon_i-\mu)/k_bT]\pm1\approx\exp[(\epsilon_i-\mu)/k_bT]$$ as the exponential function will be much larger than $1$. So at high temperatures $$\langle n_i\rangle\approx\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]}$$ which matches the form of the Maxwell–Boltzmann distribution. We can see that at low temperatures these two distributions would not agree, as the additional $\pm1$ term in the denominator would become more significant as the exponential got smaller.
It's important to remember that the particles are always indistinguishable; all we have done in the high temperature limit is made an approximation that simplified the functional form. We should not take this coincidental agreement with the MB distribution (for which the particles are assumed to be distinguishable) to imply that the quantum particles have become distinguishable at high temperatures.

Thanks for your answer. So are you saying that the whole derivation of the MB Distribution by maximizing the number of permutations for the most probable macrostate, and thus considering the particles to be distinguishable, is coincidentally similar?

@JohnnyGui The derivations of the distributions from various ensembles are materially very similar as they make a lot of the same assumptions. I guess what I'm saying is that just because the approximation of FD/BE looks like MB doesn't mean it represents the same picture of reality (distinguishable particles) as MB.

Thanks a lot for clarifying. Part of my question is also about what quantum effects are present in the case of low temperature that makes the MB distribution inaccurate, perhaps you can help me with that as well?

@JohnnyGui I don't know if there are additional quantum effects, at least in these models. All of them ignore interactions between particles. As you mentioned, MB is quantum as well as it discretizes the energy levels. The big difference is really just that the other distribution deal with explicitly quantum particles, that is fermions and bosons, and how those behave. Beyond that, I don't know and you may need to go to the Physics SE for a deeper dive into this.

The Maxwell–Boltzmann distribution (MBD) is inextricably tied to ideal gas behavior, so where ideal gas behavior fails so will the MBD. It isn't just at low temperatures that the MBD fails, since the MBD will also fail at "very high" temperatures because the gases ionize. The gist is that to consider quantum effects you have to go to a more complicated model than the MBD since the MBD explicitly ignores quantum mechanics.
The situation is somewhat like comparing Newton's theory of gravity and Einstein's theory of relativity. So:

The MBD was known before quantum mechanics was figured out, so the classical derivation doesn't use quantum mechanics.
The MBD is a continuous function. The "real" distribution would have to be quantized because of the implications of the Plank constant, so it be a histogram rather than a continuous function.
The MBD doesn't ignores gravity thus doesn't predict that a massive enough gas cloud would collapse due to gravitational attraction.
Fermi–Dirac statistics and Bose–Einstein statistics are two quantum distributions that are opposites in the sense that Fermi–Dirac statistics applies to fermions and Bose–Einstein statistics applies to bosons. However the MBD ignores this bifurcation of particle types.
Fermi–Dirac statistics and Bose–Einstein statistics are discrete distributions, but in the limit (ie lots of particles,  energies great enough for implications of the Plank constant not to apply) reduce to the MBD which is a "continuous" distribution. (Think of the binomial distribution approximation of the Gaussian distribution in statistics.)
A Bose-Einstein condensate, where Bose-Einstein statistics apply, isn't really a gas, but rather it is another state of matter.
The MBD requires particles to be distinguishable, and that many particles can have the same energy state.
Bose-Einstein Statistics requires particles to be indistinguishable, and that many particles can have the same energy state.
Fermi–Dirac statistics requires particles to be indistinguishable, and that no two particles can occupy the same state.
The MBD, as well as Fermi–Dirac statistics and Bose–Einstein statistics, ignore relativity in which the mass of a particle seems to increase as the velocity of the particle increases.
The MBD, as well as Bose–Einstein statistics, fail at very high temperatures because the gas starts to ionize.

Thanks for the clear explanation. I have seen another reason that could break down MBD, which is the fact that at low temperature, the energy is too low and therefore not many energy levels get populated. To which of your mentioned explanations does this reason belong?

When using at low temperatures Bose-Einstein Statistics has energy states that aren't "continuous" but rather the energy states have discrete energies like a histogram. Multiple particles can be in each energy level.  So no more a few levels need be occupied.

Are the steps between low energy levels larger in terms of absolute difference? Or is there another factor that causes this "discreteness" at low temperature?

Bose-Einstein Statistics use the Boltzmann constant, which in turn depends on the Plank constant.


text:
Why is my solution for the partial pressures of two different gases incorrect?
I have the chemical equation: $\ce{NH4NO3(s) -> 2H2O (g) + N2O (g)}$. There are $12.8$ g of ammonium nitrate, and it decomposes completely in an empty container at $25.00\ ^\circ\pu{C}$ until the total pressure is $3.70$ atm.
This is what I tried:
Because there are 2 more $\ce{H2O}$ than $\ce{N2O}$ I said that there are $\pu{9.06g}$ of $\ce{H2O}$ and $\pu{3.02g}$ of $\ce{N2O}$.
$$\pu{9.06g}\times \frac{\pu{mol}}{\pu{18.016 g}} = \pu{0.503 mol}\  \ce{H2O}$$
$$\pu{3.02g}\times \frac{\pu{mol}}{\pu{44.02g}}= \pu{0.0686 mol}\ \ce{N_2O}$$
I then found the mole ratio of both compounds:
$$X_{H_{2}O}={0.503\over 0.572}=0.879 $$
$$X_{N_{2}O}={0.0686\over 0.572}=0.1199$$
Then I used the relationship between mole ratios and partial pressure $X_i={P_i \over P_{tot}}$, where $P_{tot}=\pu{3.70atm} $:
$$P_\ce{H2O}=(0.879)(\pu{3.70atm})=\pu{3.25atm}$$
$$P_\ce{N2O}=(0.1199)(\pu{3.70atm})=\pu{0.444atm}$$
According to the answer solution for this problem $P_\ce{H2O}=\pu{2.47atm}$ and $P_\ce{N2O}=\pu{1.23atm}$. What am I doing wrong? I can't figure out how to get the correct solution.

You made a mistake calculating the moles of gas and the molar fraction. If you have 12.8 g of $\ce{NH4NO3}$, then you have:
$$\ce{12.8 g NH4NO3}\times\frac{\ce{1 mol NH4NO3}}{\ce{80.04 g NH4NO3}}=\ce{0.160 mol NH4NO3}$$
Since $\ce{NH4NO3}$ is solid, from the balanced chemical equation it is clear that the molar fractions of $\ce{H2O}$ and $\ce{N2O}$ will be $2/3$ and $1/3$ respectively. With these indications, you should be able to obtain the correct answer

The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).
$$P_{\ce{N2O}} = \frac{1}{3} \times 3.70\ \mathrm{atm} = 1.23\ \mathrm{atm}$$
$$P_{\ce{H2O}} = \frac{2}{3} \times 3.70\ \mathrm{atm} = 2.47\ \mathrm{atm}$$
The bottom line is that chemical equations (i.e. stoichiometry) describe the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.

Some made-up questions are not well thought thru. This looks like an example. Or, maybe it's a trick question. The exact answer desired is not stated: what exactly is to be found? The ratio of partial pressures (if all vapor)? That is absurdly simple and can be stated by looking at the reaction equation. But that can't be the question. The volume of the container? Possibly.
At 25 C, the vapor pressure of H2O will be about 25mm/760mm = 0.033 atm. The rest of the final pressure (3.70 atm) is N2O. Well, approximately. At 1 atm, 25 C, N2O is soluble in water about 1 liter per liter, so at 3.7 atm, multiply the volume of water by 3.7 to get the volume of N2O removed from the gas phase. The volume of water is 2 x 0.16 x 18/1000 = 0.00576 L. The ratio of partial pressures will then be (3.7 - 0.0213)/0.033 = 111.5 (approximately).
The volume of the container? 1 mole of gas at 25 C occupies 22.4 L. Therefore 0.16 moles of NH4NO3 gives 0.16 moles of N2O, which occupy (approximately) 0.16 x 22.4 L = 3.584 L.
You could continue by calculating the volume of liquid water and how much N2O is dissolved, subtract that from the volume of N2O in the gas phase and recalculate the volume. Reiterate until satisfied, or just accept the approximation.


text:
How can I preserve chlorophyll for long periods of time?
I'm currently working on a XIX century photographic process called Anthotype. It relies on organic colored substances that fade when exposed to sunlight. For the substances I've tried alcoholic Curcumin extract and Chlorophyll. The later is incredibly sensitive to light! But this is a problem, because after the print is done I've got no way to fix it. I've read that chlorophyll degrades more on acidic conditions and even if kept in the dark. I tried treating the photographas with a weak base (sodium bicarbonate) but it doesn't seem to help. Is there a way to pr*eserve it or preserve the color?
I've also read that on museums they basically make a copper chelate with chlorophyll and that seems to help. -it preserves the green color.
Now, I don't have copper sulphate at hand. I've read it's found as "root killer" on hardware stores, but the products at my local hardware store don't list the compounds on them and also there's no root killer specifically. Is there other salts that can make a chelate with chlorophyll? I've read that zinc will also work.
Also, is pheophytin (The main chemical formed when chlorophyll is degraded) more resistant than the actual chlorophyll?

The very last question has an implicit answer.  Though  it is chemistry, perhaps on photography SE you might find somebody with tge same project. ..

You can rather easily prepare copper sulfate or zinc sulfate by reacting an excess of metal with sulfuric acid.

Copper sulfate pentahydrate is easily shipped and is available online at Amazon and Walmart as well as in stores such as Home Depot.


text:
How does the proticity of an acid affect the enthalpy change of its reaction with a base?
For an experiment I've carried out a series of reactions between monoprotic bases (ammonia and potassium hydroxide) and several acids. Two of these acids were monoprotic (hydrochlori acid and ethanoic acid), another was  diprotic (sulfuric acid) and the other was triprotic (phosphoric acid). My results showed that the triprotic acid released more energy per mole than both the diprotic and monoprotic acids, and that the diprotic acid released more energy per mole than the monoprotic acids. This gave rise to my question: how does the proticity of an acid affect the enthalpy change of its reaction with a base?
To be clear I compared my results between weak acids, strong acids so the strength of the acid has already been accounted for.

I am probably thinking that it has to do with the energy release of the bonds between the H-Cl  for example. These most likely release less energy than an H3-PO4 bond. So the distinction here might be due to the number of hydrogen bonds being broken. But I am not quite sure.

"per mole" of what? Are you comparing the heat released per proton on the each acid or per mole of acid molecule?

You have to remember that enthalpy change is defined at kilojoules PER MOLE. The concentration won't affect enthalpy change because the same amount of heat is given off per mole of zinc , however the reaction will go faster with a higher concentration.


text:
Is sodium hypochlorite 100% concentrate possible?
Popular bleaching brands like Clorox contain liquid sodium hypochlorite in small concentrations, ranging from 5-12.5% (is this a mass/volume, mass/mass, or volume/volume concentrate?).
Is it possible to achieve a 100% concentrate in liquid form (pure NAClO)? Experimentally, I've found it's boiling point to be around 35°C, but I would assume that its melting point is fairly high in comparison.

Wikipedia disagrees with your figures.  They give a melting point only for the pentahydrate (18°C, at which point this forms not pure sodium hypochlirite but a solution of one mole sodium hypochlirite to five moles water).  The only reported phase transition for the neat compound is a "boiling point" of 101°C which is actually a decomposition point.
So I would not bet on getting pure liquid sodium hypochlorite, although as mentioned above the pentahydrate can be converted to a fairly concentrated solution.

So if that (not having anything aside from a pentahydrate to be a 17% concentrate solution max) were to be the case; what would happen if I took a pure tablet of sodium hypochlorite and heated it to  20°C?

"what would happen if I took a pure tablet of sodium hypochlorite and heated it to 20°C?"  Nothing worth noting.