text:
How does beta decay increases the stabilization of the nuclei?
How does beta emission by a nuclei increases the general stability of it? I surely am not a specialist or anything, but this just don't make sense to me, since the objective of radiation is to release energy, but I can't figure out exactly how it's done.
The loss of energy from the electron getting expelled, in addition to its kinetic energy, roughly gives half a MeV (source). Besides, as the strong force acts similarly on protons and neutrons, I guess it won't change considerably in piratical effects. Thus, in my conception, at least, the eletrostatic repulsion from the proton added should overlap the energy who got rid in the process.
Hence, I am confused. I wish to understand how it plays a role in nucleus stabilization.

This is more a nuclear physics question. However, in the nucleus, protons and neutrons are not identical. So, when the proton:neutron balance is not quite right, beta emission allows the nucleus to achieve a lower energy state. Energy level diagrams at tunl.duke.edu, or the Evaluated Nuclear Data Center At nndc.bnl.gov are places to start.

The way I was taught is just that a neutron is slightly heavier than a proton, so the beta decay helps stabilize. Not sure though.

@GaurangTandon - sadly, that is a misleading at best explanation. Really, it is nuclear physics and delving into binding energies and the proton:neutron ratios that make the most stable nucleus with A total nucleons.

Of relevance: https://physics.stackexchange.com/questions/249036/why-beta-decay-occurs, https://physics.stackexchange.com/questions/105456/energy-balance-for-beta-decay-of-cobalt-60, https://physics.stackexchange.com/questions/185470/why-doesnt-beta-decay-violate-the-laws-of-physics, and https://physics.stackexchange.com/questions/16653/why-is-beta-negative-decay-more-common-than-beta-positive, https://chemistry.stackexchange.com/questions/10453/understanding-beta-decay.

While this is nuclear physics, the relative number and relative stability of isotopes is of some relevance to chemistry. Further, I spend more time than I'd like perhaps at the intersection of materials science, chemistry, and nuclear physics, so...
There are several good on-line resources to get information on nuclear energy levels and the relative stability of nuclei. These include:
The Nuclear Data Evaluation Project at Triangle Universities Nuclear Laboratory (TUNL), and the Evaluated Nuclear Data Structure File data base at the National Nuclear Data Center hosted at Brookhaven National Laboratory. From these you can get more information than you might want with regard to what is known about nuclear energy levels and stability of the elements.
One cut at looking at stability is the so-called isobar diagram. This compares the relative nuclear energy levels of all nuclei with the same $A$, where $A = P + N$, that is, the sum of the number of protons and neutrons in the nucleus. Since $\beta \pm$ converts between protons and neutrons, keeping $A$ constant, this is the best vantage point for the purpose at hand.
Borrowing from TUNL one finds for $A=14$:

So, we see 4 different nuclei shown on the isobar diagram. The lowest energy (most tightly bound) nuclei is $^{14}$N. To the left is $^{14}$C, which you likely know is unstable and $\beta -$ decays to $^{14}$N, used for carbon dating. To the right is $^{14}$O and you will note that it $\beta +$ decays to $^{14}$N. Neither $^{14}$C or $^{14}$O has as much binding energy as $^{14}$N due to the fact that protons and neutrons are not actually identical in the nucleus. So, the ground state of $^{14}$C or $^{14}$O is about 2.4 MeV higher than $^{14}$N.
High and off to the left you can barely see $^{14}$B, 24.71 MeV above $^{14}$N. It desperately would like to be $^{14}$N, but can't get there directly through the available $\alpha$, $\beta$ or $\gamma$ decay channels. So, it happily $\beta -$ decays to $^{14}$C with a half-life of 16 ms, which will then ultimately decay down to $^{14}$N.
Now, your misconception is that you don't know nuclear physics (and I don't mean that negatively - you applied a fairly simple model, did not get the expected answer, so you asked a question here - GREAT!). But, in the nucleus you have Coulomb repulsion, the strong force, and the weak force, and it all has to be treated quantum mechanically. Taken all together one can determine the binding energy (total or per nucleon) of nuclei, the excited nuclear energy levels, and possible decay paths. It is great fun, but is a different world to explore.


text:
Effect on rate of diffusion in addition of an inert gas
What will be the effect on the rate of diffusion on addition of an inert gas to the gaseous mixture?
I think the rate of diffusion should increase as the addition of extra gas will increase the inside pressure. But the given answer contradicts my proposed explanation. Where am I wrong? And why is the rate of diffusion decreasing?
My question does not ask about the effect of addition of an inert gas on a reaction equilibrium.

Is this from a book? If you add that source and the exact quote, other people with a similar problem will also be able to find this question.

The inter-diffusion caused by two gasses is described by the Stefan-Maxwell equation.
If $x_1,x_2$ are the mole fractions of the two gasses, $\bar v$, the average speed and $\lambda$ the mean free path then
$$D_{1,2}= \frac{x_2}{2}\bar v_1\lambda_1+ \frac{x_1}{2}\bar v_2\lambda_2$$
where $D_{1,2}$ is the inter-diffusion coefficient. Substituting for the mean free paths does not lead to a useful result because terms that involve collision between molecules of the same kind cannot have any extra effect compared to when only one gas is present, and so these are ignored. The result is
$$D_{1,2}= \frac{1}{\pi\sigma_{1,2}^2(n_1+n_2)} \left( \frac{2k_BT}{\pi\mu} \right)^{1/2}$$
where $\sigma_{1,2}$ is the sum of the radii of the two molecule types and $n_1, n_2$ number of molecules/m$^3$ of each, the reduced mass is $\mu$ kg ($\mu=m_1m_2/(m_1+m_2)$).
This equation  shows that the inter-diffusion depends on the total concentration at a given temperature, a result that is close to that observed experimentally. So your intuition was correct.
(ref chapter (II). E. Moelwyn-Hughes, 'Physical Chemistry')

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text:
Naming secondary amine IUPAC system
$\ce{CH3NHCH(CH3)CH2CH3}$
Should this be named as 2-(N-methyl)butanamine or 2-(methylamino)butane?
In other words, do we name the compound as 'alkanamine' or treat amino group as a substituent of butane?

It should be N-methyl-2-butanamine. Amine is treated as a functional group attached. So we write it as alkanamine as it is the only functional group here.

Image source: ChemSpider

Note that the preferred IUPAC name for the compound drawn is in fact slightly different from what you have written: it should be N-methylbutan-2-amine. In particular the locant '2' should be placed immediately before the suffix 'amine', and the locant 'N' should be italicised.

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text:
Why is aluminium used to treat waste after a lab experiment?
How does adding aluminum metal pieces into a waste beaker help treat solids and liquids in the beaker so you can safely send the solids to a landfill and dispose of the liquids in the sink drain following a lab experiment?

Not clear where you are, but I know that I absolutely cannot dispose of anything down the sink, regardless of whether I have 'treated' it or not. That would be a great way to get the city and state to shut me down.

Presumably because aluminium is a very reactive metal and reacts with acids and bases both to form less reactive compounds. (Amphoteric Nature)

With acids
$$\ce{Al + HCl -> AlCl3 + H2}$$
With Bases
$$\ce{Al + NaOH + H2O -> NaAl(OH)4 + H2}$$

Sodium aluminate is really $\ce{NaAl(OH)4}$.

Ok will correct,( it's given like that in my textbook)

That's a very old notation that does not represent the true structure of the aluminate ion.


text:
If we are supplying energy continuously, why will the electrons jump down to the lower state?
We know in a flame test an electron goes into excited state. When it jumps down to a lower energy state, we can see the color of the flame if it's in our visible region.
My question: we are supplying energy continuously, so why will the electrons jump down to the lower state?

The atom/ions are escaping the hottest part of the flame. Moreover, likely there are transient phenomena.

They return to their original state as electrons in higher energy levels are more unstable. They give out the energy absorbed in the form a radiation.

@Bastardane but im  suppling  energy ..so why they give out their energy absorbed ?

You're not supplying energy to only one electron. This excitation - de-excitation process takes place continuously. The unstability in excited state causes them to return to ground state

What do you mean by that? How would you accomplish this? How exactly do you picture the interaction between your excited system and whatever supplies that energy in whatever (see first 2 questions) way?


text:
Concentration of product at a time t in a reaction
A reaction was given $\ce{A<=>B}$ The rate constants for forward and  backward reactions were given as $k_\mathrm{f}$ and $k_\mathrm{b}$. The initial $\ce{[A]}$ was given as $a_{0}$ and the $\ce{[B]}$ at a time t (before equilibrium is attained) was to be calculated.
I tried solving it as follows -
For A I wrote -
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k_\mathrm{b}\ce{[B]}-k_\mathrm{f}\ce{[A]}$$
and similarly for B -
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = k_\mathrm{f}\ce{[A]}-k_\mathrm{b}\ce{[B]}$$
Combining these two I got a 2nd order differential equation and wasn't able to proceed further. Is there something wrong with my approach or my equations?

https://chemistry.stackexchange.com/q/87179/194 ?

It is given that the reaction in 1st order in both directions

Your method should work as you know the boundary conditions for your scheme, but it is not the easiest method. Instead, it is easier in this case, as you are given the initial amount of A, to let $x$ be the amount of B produced at time $t$, then $a_0 - a = b = x$ where to save writing brackets, $a,\;b$ are the amounts of A and B at time $t$. The the rate equation is
$$\frac{\mathrm{d}x}{\mathrm{d}t} =k_\mathrm{A}a -k_\mathrm{B}b$$
and substituting gives
$$ \frac{\mathrm{d}x}{\mathrm{d}t}= k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x$$
This can now be integrated with limits $(0,x),\; (0,t)$ to give
$$\ln \left( \frac{k_\mathrm{A}a_0-(k_\mathrm{A}+k_\mathrm{B})x}{k_\mathrm{A}a_0}   \right)= -(k_\mathrm{A}+k_\mathrm{B})t$$
At equilibrium, the rate of reaction is zero, then $k_\mathrm{A}a_e=k_\mathrm{B}b_e=k_\mathrm{B}x_e$ where subscript $e$ means that it is the equilibrium value. Also, we know that $\displaystyle K_\mathrm{e}=\frac{k_\mathrm{A}}{k_\mathrm{B}}=\frac{x_e}{a_0-x_e}$, where $a_0-x_e$ is the amount of A remaining at equilibrium.
Substituting and rearranging gives the amount of $x$, which is B as a function of time:
$$x=x_e -x_ee^{-(k_\mathrm{A}+k_\mathrm{B})t}$$
Both this and the previous equation show that the equilibrium is approached with a rate constant that is the sum of the two rate constants.


text:
What will methyl (2-oxocyclohexyl)acetate produce when reacted with sodium borohydride?
What will methyl (2-oxocyclohexyl)acetate produce when reacted with sodium borohydride?

I feel the answer should be 2-(2-hydroxypropyl)cyclohexan-1-one:

Because the carbonyl carbon on the aliphatic chain should be more electrophilic than the one on the ring due to the presence of two neighbouring electronegative oxygen atoms. Correct me if I am wrong but I think the reaction begins with the hydride ion which acts as a nucleophile?
On the other hand the answer given in my book is 1-(2-hydroxycyclohexyl)propan-2-one:

Can someone please help me out here?

Both carbonyl carbons (and everything else in the molecule) are in aliphatic environments.  The six-membered ring you drew is not aromatic, it is saturated like a cycloalkane.
The carbonyl group in this ring is a ketone function, bonded only to other carbon atoms.  The carbonyl group in the chain is bonded to oxygen (did you miss that?) in an ester function, which is a less active electrophile than the ketone function.  Yes the additional oxygen atom is electronegative, but with its octet satisfied it can (to a degree) share a nonbonding pair with the carbonyl carbon, and that effect dominates making the ester carbon less electron deficient. The sodium borohydride reacts accordingly.

It somewhat depends on the conditions used, choice of solvent etc. But sodium borohydride does not routinely react with esters which is what you have at the end of the chain as drawn in pic 1 (not as you have drawn in pics 2 & 3). It does routinely reduce ketones to secondary alcohols and you have an isolated ketone on the cyclohexane ring so this will be reduced.

I'd also guess if no solvent was mentioned it will be the one in the ring because the ester is not reactive enough. A similar example you might want to look at if you are already dealing with different carbonyl groups and reductions is the so calle 'Luche-reduction'. This can often differentiate aldeyhdes from ketons and is quite useful to know as well.

FYI:  The Luche reduction uses a lanthanide chloride to coordinate a carbonyl oxygen accompanied by $\alpha, \beta$ unsatur-ation (hyphenated because device autocorrect is getting weird), activating that carbonyl group for the borohydride attack.

Never had a lot of luck with the selectivity in Luche reduction, tends to reduce both carbonyls in my personal experience

Reduction of the ketone with NaBH4 will likely effect lactonization, particularly, in the cis isomer and possibly the trans isomer.

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text:
Why Greek yogurt has higher protein concentration?
During fermentation, no nitrogen is added and the number of amino acid molecules cannot increase. Why does it have a higher protein concentration than milk?

It has less water than the original milk.

Because it's condensed . so it's normal for Greek yogurt to have a bigger percentage of protein than ordinary yogurt and milk .
the content of protein is the same but when we do concentration, the percentage of total solids (including protein) becomes higher.


text:
How do we know a nerve agent is tasteless?
According to the BBC, the nerve agent sarin is a tasteless liquid. How do we know that this liquid has no taste, given that anyone to have (not) tasted it in any meaningful quantity presumably died shortly thereafter.
Can the taste be inferred from its chemical properties?

Similar: How do people know HCN smells like almonds?

Roughly speaking, we know that from the unfortunate guys who tasted it and died shortly thereafter. This is a very different situation from HCN.

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text:
Haloalkenes selectivity of reagents
5-chloro-1, 3 - dimethylcyclohexene is the substrate. I understand that hydroboration oxidation occurs in 2 steps the first being addition of Boron as an electrophile and the second being oxidation by peroxide ion. Since the solution now contains Hydroxide ion  now this will attack the carbon bonded to chlorine by SN2 mechanism. But in my book , in the final product chlorine is not replaced with OH. I don't understand why.

You assume that the peroxide/base reaction and an SN2 reaction have similar rates. Moreover, E2 elimination is also a real possibility rather than displacement of chlorine.

What solvent is the reaction run in?

@Waylander I don't know I just started learning organic chemistry and this question appears with the substrate and an arrow on which B2H6 and OH-, H2O2 are written. Maybe the question is ill posed.

@user55119 2° carbon and SN2/E2  conditions  elimination product will be major right? Why this was down voted I will edit the question to make it clear.

My point, as Waylander has noted, is that the basic peroxide oxidation is faster (often needs cooling) than an SN2 reaction with hydroxide. The SN2 reaction on a halocyclohexane will compete with an E2 reaction. It's all about conditions (relative rates).

OH- is not that great a nucleophile for a secondary chloride.
If we assume an ethereal solvent, most likely THF, much of the OH- will be in aqueous phase and not available to react with the product in the organic phase.
If the reaction is kept cool (as it should be) E2 is unlikely to be a major pathway hence the product is 5-chloro-1,3-dimethylcyclohexan-2-ol

Why this  reaction should be kept cool?

The oxidative work up is exothermic. If it gets too hot pathways like E2 or OH- attack can become relevant.


text:
Determining overtones bands of degenerate modes

I have a question of a more mathematical nature on the mathSE (Symmetric Direct Product Distributive?) that received a good answer, but I think an answer more oriented to chemists would be a useful resource here.

I'm trying to determine the symmetry of the second overtone band of the degenerate $\Pi_u$ bend of $\ce{CO2}$. I'm told I need to use the formula$$\chi_v(\hat{R})=\frac{1}{2}[\chi(\hat{R})\chi_{v-1}(\hat{R})+\chi(\hat{R}^v)]$$
where $v$ is the number of quanta in the mode, $\hat{R}$ is an operation and $\chi_{v}$ is the character of that operation for the $(v-1)^{\text{th}}$ overtone. I arrive at the correct answer, which is that the second overtone has symmetry $\Pi_u\oplus\Phi_u$.
But this formula seemed complicated so I wanted to see if I could get the result just from taking the symmetric direct product of the state three times like so:
$$\Pi_u\otimes\Pi_u\otimes\Pi_u=(\Sigma^+_g\oplus\Delta_g)\otimes\Pi_u=2\Pi_u\oplus\Phi_u$$
Following the direct product tables I seem to get this, which has an extra $\Pi_u$.
I know this is wrong because the representation should only be quadruply degenerate, corresponding to the 4 different ways of distributing the quanta between the original degenerate modes: $(3,0) (2,1) (1,2) (0,3)$.
Why does this approach fail? Does the symmetric direct product not distribute over the direct sums?

A freebie for anyone looking to write an answer: https://books.google.com/books?hl=en&lr=&id=CPkvsDrPiv0C&oi=fnd&pg=PA1&ots=vloIsaNBP5&sig=1n-3bNeSuqflrXvVxgsZ2XCoSJM#v=onepage&q&f=false (Wilson, Decius, and Cross Molecular Vibrations pages 151-155)

The issue with this approach is that the symmetric direct product doesn't commute, but it is a bit more subtle than that. Summarizing in somewhat less mathematical terms the answer given to my question on MathSE, the issue stems from the fact that once you move past doing a single direct product, there is no longer just a symmetric and an antisymmetric part, but rather $n+1$ parts, where $n$ is the number of direct products being taken. So, trying to drop any antisymmetric terms that arise at each step will still leave terms that aren't in the symmetric part of the representation. (To say this in mathematical language for those interested, the representation for $n$ direct products of a particular irreducible representation of a group can be decomposed into $n+1$ parts by projecting onto the symmetric group $S_{n+1}$).
Physically, the reason for needing only the symmetric part rather than the whole direct product is that we are filling quanta into one particular mode. For example, if we wanted to find a combination band composed of two different $\Pi_u$ modes, we could just take the direct product to find the representation. But, filling two into a single $\Pi_u$ mode limits the number of possible distributions of the quanta to only those given by the symmetric direct product.
As to why the recursive formula gives the symmetric direct product, I give a summary of the argument from Wilson, Decius, and Cross's Molecular Vibrations for the case of a doubly degenerate mode.
Let $Q_a$ and $Q_b$ be the normal coordinates of a doubly degenerate mode. We choose these coordinates such that for an operation $R$      $$Q_a \xrightarrow{R}R_aQ_a$$
$$Q_b \xrightarrow{R}R_bQ_b$$
While the same coordinates will not diagonalize each $R$ in the group, we can find a set of coordinates for each $R$ that will. The character of $R$ for the $(v-1)^{\mathrm{th}}$ overtone, $\chi_v(R)$ can then be determined by summing the contribution of $R$ acting on each possible distribution of quanta between the modes $$\chi_v(R)=R_a^v+R_a^{v-1}R_b+...+R_aR_b^{v-1}+R_b^v$$
We can see that recursion formula holds by looking at the form of the terms on the right hand side
$$\chi(R)\chi_{v-1}(R)=(R_a+R_b)(R_a^{v-1}+R_a^{v-2}R_b+...+R_aR_b^{v-2}+R_b^{v-1})=R_a^v+2R_a^{v-1}R_b+...+2R_aR_b^{v-1}+R_b^v$$
and
$$\chi(R^v)=R_a^v+R_b^v$$
Averaging these two terms, we obtain the recursion formula from the question.
$$\chi_v(\hat{R})=\frac{1}{2}[\chi({R})\chi_{v-1}({R})+\chi({R}^v)]$$
By a similar argument, we can obtain a recursive expression for triply degenerate states
$$\chi_v({R})=\frac{1}{3}\bigg[2\chi({R})\chi_{v-1}({R})+\chi({R}^v)+\frac{1}{2}\big(\chi({R^2})+[\chi({R}]^2\big)\chi_{v-2}({R})+\chi({R}^v)\bigg]$$


text:
ligand pi system irreducible representation for trigonal bipyramidal transition metal complex
I am considering a D$_{3h}$ trigonal bipyramidal transition metal complex MX$_5$.  I have found that the irreducible representation for the ligand s and $\sigma$ systems is 2A$_1'$+A$_2''$+E$'$.  I have
\begin{array}{|c|c|c|c|c|c|}
\hline
&E& 2C_3 & 3C_2 & \sigma_h & 2S_3&3\sigma_v\\ \hline
 \Gamma_s&5& 2& 1&3&0&3\\ \hline
 \Gamma_\sigma&5& 2& 1&3&0&3\\ \hline
 \Gamma_{\pi+\pi'}&10&0&-2&0&0&0\\ \hline
\end{array}
However, I think that my $\Gamma_{\pi+\pi'}$ is incorrect since I get fractional values for coefficients for the irreducible representations when I try to reduce it to irreducible representations.  But from sketching the trigonal pyramidal complex, I cannot see how my $\chi(R)$ are wrong for the pi system representation, $\Gamma_{\pi+\pi'}$.  Any help would be appreciated.

I believe you have the wrong character for the $\hat{C_3}$ rotation. None of the p orbitals in the equatorial positions remain in place so they all contribute zero. The axial p orbitals are rotated by $120^°$ which means they each contribute $\cos(120^°)=-\frac{1}{2}$. Since there are two of them at each of the two axial position, $\chi(\hat{C_3})=-2$. Using this new value for the character (assuming all the others are right, which they seemed to be), I obtained: $a_2',a_2'',2e',2e''$

Thank you!  I did not notice that the rotated orbitals work like that (although I should have, since they are part of a basis)... for some reason,  I thought each orbital must count as 1, 0, or -1.

please can you explain '..two of them at each ..'  do you mean rotation by $+120$ and $-120$ ?

Two perpendicular coplanar p orbitals that are part of the pi system at each axial ligand.

@porphyrin there is a px and py orbital each axial end (assuming the pz the one that forms a sigma bond with the central atom). When we perform the C3 rotation, they are both shifted by 120° from their original position.

@HBHSU it is usually glossed over at least when I have seen it taught. I think it would be much clearer if they said you get a character of $0$ if it moves away and a character of $\cos(\theta)$ if it rotates in place.


text:
What makes the first 3 alkanes decrease in melting point while after that they rise?
What makes the first 3 alkanes decrease in melting point (m.p.) while after that they rise?
$$
\begin{array}{lr}
\hline
\text{Alkane} & \text{m.p.}/\pu{°C} \\
\hline
\text{Methane} & -182 \\
\text{Ethane}  & -183 \\
\text{Propane} & -188 \\
\text{Butane}  & -138 \\
\text{Pentane} & -130 \\
\text{Hexane}  & -95  \\
\text{Heptane} & -91  \\
\text{Octane}  & -57  \\
\hline
\end{array}
$$

It is packing . After the first terms then molecular surface takes the leading role.  Perhaps someone will articulate in a more complete form but the juice is this.

https://chemistry.stackexchange.com/questions/89689/for-two-compounds-that-are-symmetrical-why-does-the-more-compact-one-have-a-hig https://chemistry.stackexchange.com/questions/71417/why-alkanes-containing-even-number-of-carbons-can-be-packed-more-efficiently-tha


text:
Is it possible to predict the color of the flame test result based on the element's structure?
The flame test is quite simple to do and very useful on the chemistry world, but to understand what your results mean you need to know previously a range variety of element-color relations. So, since the wavelength of the photon emitted reliess on the quantum jump done, I was wondering if, somehow, is it possible to make a fairly certain prediction of the results, based on the compound's structure. If someone knows a relation, it would be very useful.

I find your question interesting, but the title seems slightly misleading. Correct me if I'm wrong, but the flame test is always only done on "metal elements" and never "compounds". So, isn't it incorrect to try predict the flame color based on a "compound's structure"? Maybe you meant an "element's atomic number" instead?

It depends to what extent is given to your question.  Methods exist that gives hints to even the vibrational spectrum of a molecule! But as you refer to flame test, its value (very high! ) is limited to didactics. The rest is spectroscopy.

Exactly, you should look into infrared spectroscopy (https://en.wikipedia.org/wiki/Infrared_spectroscopy) and NMR spectroscopy (https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_spectroscopy), just to name two.

@Gaurag Tandom, the flame test is most common and efficiently done in metals, especially when ionized, but it doesn't just aplly to those. Some uncommon non-metals like Phosporus or Boron are also usable. When i said "compounds structure" i meant the elements that are formed from the ionization process, but i appreciate you pointing the ambiguity on the question, i'll try to make it more precise!

The question is why would you want to predict that? All elements have been screened for allowed and forbidden transitions ranging over the complete visible and non-visible spectrum. You can just look it up there. There are also multiple possible transitions which lead to the often complex spectra.
And then there is the problem with the actual jump. In a simple shell-model where you only add electrons in shells this jump is just to the next level. And you can introduce Rydberg-atoms and get element specific Rydberg-constants. But I remember to have tried that for a Lithium atom and if I didn't do a mistake back then it would have required quite a high energy level to get the red emission while in reality the principal quantum number (n) doesn't even change for lithium. You can see this pretty easily if you look at the energy diagram for the famous yellow sodium emission line. It's a jump from the $3p$ to the $3s$ orbital.
But in the end, as I said, they have all been calculated and confirmed. You can just look at lists on the net where every possible transition is given with the orbitals, energy, wavelenght and something like an intensity.
And engines like WolframAlpha even use that. You can just search there for let's say sodium emission spectrum and it will give you the spectrum. And there are also websites where all spectra are given.
I'm a bit confused with your question. It sounds like you don't have such a list. I guess, as you are talking about flame colors and not spectra you are only looking for the flame color itself. This should be mentioned in most analytical chemistry books and I think there was even a Wikipedia article with pictures. But at some point this will become quite hard. Lithium vs Calcium vs Strontium is such a case where I wouldn't be too sure if I could distinguish them by their flame color without looking at spectra. In the end the observed emission will be mixture of many weak and strong emissions.


text:
Is Sulphurated Lime same as Lime Sulphur?
Pet dip lime sulphur has sulphurated lime
Lime sulphur for garden has lime sulphur
Will mixing sulphurated lime with HCl produce Hydrogen sulphite.

No!
Both lime sulphur and sulfurated lime contain polysulfides - sulfur at a mix of -2 and 0 oxidation level. To get the sulfur to the +4 oxidation state as in sulfite you need an oxidising agent. HCl is not an oxidising agent

But mixing lime sulfur and HCl will produce hydrogen sulfide?

Correct, it will.

Will HCl 34% solution work or does it have to be 100%

34% HCl is as strong a solution as you can get and will work

Ok. Is 'soluble sulfur' same as 'calcium polysulfides'. Will soluble sulfur mixed with HCl produce hydrogen sulfide

Not sure I can answer that. "Soluble sulfur" has no defined meaning. Somewhere on the product description there should be a table of contents with a more formal notification of what form of sulfur it contains.

Soluble sulphur present as polysulphide sulphur. Will this form Hydrogen Sulphide when mixed with HCl.


text:
Can pressure remain constant in isothermal expansion?

An ideal gas undergoes isothermal expansion at constant pressure. During the process:

enthalpy increases but entropy decreases.
enthalpy remains constant but entropy increases.
enthalpy decreases but entropy increases.
Both enthalpy and entropy remain constant.

I applied this formula.
$\Delta H = \Delta U + p\Delta V$ (pressure is constant)
$\Delta U = 0$ (since, process is isothermal)
$\Delta V >0$ (gas is expanding)
So $\Delta H > 0$
So enthalpy must be increasing $\left(\Delta S =\frac{\Delta Q_{\text{rev}}}{T}\right)$
Since heat is supplied to expand gas $\Delta S$ must be positive. And entropy must increase. But none of the option is matching with my answer.
Also I think question is wrong since $PV= nRT$ and, at constant temperature (isothermal) and constant pressure, volume can not change.

Is the answer (2)? It appears like they've assumed $\Delta H=nC_\mathrm{p}\Delta T=0$ (for isothermal process). And entropy is increasing because expansion (=> disorder of gas molecules)  is there. Though I agree with your points. Perhaps, gas moles are being externally fed into the system? But then it is an open system instead, so I am not sure.

@GaurangTandon Yes, answer is (2).

Your assessment using  PV=nRT is spot on.  The question is bogus.

Late now, but questions like this should generally be brought to the instructor's attention for not making sense.  In this case, yes, an ideal gas with constant-pressure and constant-temperature doesn't expand unless added to, but that can't be the case since both enthalpy and entropy would increase, which isn't a choice.

The question most probably meant constant external pressure.

The first step in answering the multiple-choice question lies in remembering that there is no change in the internal energy and enthalpy of an ideal gas undergoing an isothermal expansion. A proof involves the second law of thermodynamics.
That leaves options 2 and 4 as the only possibilities.
Next note again that $\Delta U = 0 $ so that $w=-q$. Since the work is negative (a spontaneous expansion with $w=-p_\mathrm{ext}\Delta V$) the heat is positive, which means the entropy of the surroundings must have decreased (at constant T, $\Delta S_\mathrm{surroundings} = -q/T$). For the process to be spontaneous, however, the overall change in entropy, $\Delta S_\mathrm{universe}=\Delta S_\mathrm{system}+\Delta S_\mathrm{surroundings}$, must be positive. This implies that $\Delta S_\mathrm{system}>0$ and option 2 is the correct  choice.
Regarding the question in the title

Can pressure remain constant in isothermal expansion?

the pressure of the surroundings can certainly remain constant during an isothermal expansion. It is worth remembering that such expansion work is defined in terms of the pressure of the surroundings (not of the system) as $w=-\int_{V_\mathrm{ini}} ^{V_\mathrm{fin}} p_\mathrm{ext} \mathrm d V=-p_\mathrm{ext} \Delta V$.
However, for an ideal gas of constant composition, $$p=nRT/V=\mathrm{constant}/V$$ implies that a change in volume must be accompanied by a change in the internal pressure of the system.

$$\Delta H = \Delta U + \Delta PV$$
Here you were wrong you wrote that
$$\Delta H = \Delta U + P\Delta V,$$ thus the product of
 $PV$  remains constant and the $\Delta H$ is zero, so that the enthalpy is constant. And you are wrote that entropy will increase, hence the answer is B.

I will give an answer which explains what happens at the molecular level which necessitates a change in gas pressure during isothermal expansion on a purely theoretical basis as the other answers explains them precisely with formulas.
In a gas the gas pressure is a result of two components.

The average kinetic energy of the gas molecules - The faster a molecule collides (elastically) with the wall the greater is the pressure exerted by it.

The number of molecules colliding a given area of the wall at a given time - With more number of molecules colliding the wall at a given time we have more force exerted on the wall (a simple summation of force exerted by all molecules) and greater is the pressure.

With this two things in mind, as we perform an isothermal heat addition (isothermal expansion) the heat added to the gas molecule is such that it causes an infinitesimal increase in the temperature of the kinetic energy of the gas molecules. Thus now the molecules collide with higher velocity. Since the walls are movable the collisions of these higher velocity molecules will displace the walls by doing some amount of work. And subsequently the kinetic energy of molecules will drop back to the initial velocity of the gas. Thus the condition of the constant temperature remains satisfied since the temperature changes are infinitesimally small and temperature is back to initial value.
Now since the walls are displaced by the expanding gas, it means that the gas can now occupy more volume. Now remember the way gas pressure is manifested (mentioned above). Here there is no change in kinetic energy of the molecules effectively, so on this basis the gas pressure should be constant. But the second factor i.e the number of molecules colliding with the wall at a given area at a given time will now be lesser. This is because as the same amount of gas now occupies more volume, the mean free path will increase and thus there will be less number of molecular collisions with wall (and in between the molecules). So this second factor is responsible for the reduction in pressure of the gas in the isothermal expansion process.
Thus the pressure has to change in the isothermal expansion process.


text:
Why do spectra of elements never overlap?
I understand the concept behind how spectra is formed, through exciting the electron beyond it's ground state, however my question is why do the spectra of the elements never overlap?
From my understanding, the wavelengths of light produced are decided by the distance the electron travels when falling back down.
However, what is the factor that guarantees that all of these spectra will be unique? I think part of my confusion also stems from the fact I do not know which electrons "jump" - is it just the valence ones or also the ones in the middle? I thought that perhaps different energy levels have different distances but this sounded a bit funny to me.

I'm not sure that one can say 'never'  as accidental coincidence can occur  but the frequency range is very large and transitions very, very narrow  so that they will almost invariably appear as separate lines. Atoms of different types do have different electronic energies after all. However, if you have a spectrometer that has poor resolution lines will appear broad and may seem to overlap here and there but this is just an artefact of the measurement and not real.

My guess would be that every element is different. Going from the basic shell model of the atom to orbitals and transition probabilities there are huge lists for all different transitions. I think there is a green emission for boron where a $11f$-orbital or so is populated and relaxes to give off a green light. And this is so unlikely and weak that you don't even see this on a regular spectrum. So this may be one hint, there are so many possibilities. On the other side if we agreed on one electron, let's say the $1s$-electron, because this is present in all atoms, just exciting that would require different energies for all atoms because the number of protons increase and therefore the positive charge pulling it closer and strong increases as well. It may happen that one transition of element A and a completely different one  on element B have the exact same wavelenght, but the spectrum has so many possiblities and there are only very few elements in the PSE for comparison that I doubt that there will be two identical ones. And even if this happens you can strech that spectrum to the UV, X-ray region as well. What is more likely to happen is that the characteristic emission of an element, let's say the one you observe strongest may be similar, like for Lithium, Calcium and Strontium. I doubt that I'd be able to distinguish them by their red color, but then I'd just take a look at the rest of the spectrum.
But I once had a related case. I was preparing a Dysprosium compound and used a sodium based flux. After solving the structure we had the idea that traces of sodium might occupy long anionic-channel in the compound. So we analyzed the elements using energy-dispersive x-ray spectroscopy. And in the spectrum I got were the characteristic Dysprosium emission line and the one from Sodium pretty much on the same spot. So we had to do a wavelenght-dispersive test as well and then I was able to confirm the sodium. This was a case where the strong x-ray emission line of Sodium almost identical to the one of Dysprosium. But in regular spectra you have much more intense emissions so I guess the probability will be really small.


text:
Reaction of cyclohexene with hydrochloric acid
What happens when cyclohexene reacts with concentrated hydrochloric acid?
I formed the carbocation intermediate by protonating cyclohexene.
But now I am stuck. I know the end product will be a dimer.
I tried to form a bond between the intermediate and cyclohexene like this.

Then through rearrangement the plus charge will shift to 3 degree carbon.
In the answer there is a double bond between the two rings.
That's where I am stuck . I can not think anything further.

Were you told that a dimer is the product of the reaction? No chlorocyclohexane formed?

@user55119 In my original question the reaction was with Sulphuric  acid. I don't know how it changed to HCl.

@VJ: A way to remove traces of alkenes from alkanes (e.g.; cyclohexene from cyclohexane) is to shake the hydrocarbon with conc. sulfuric acid at room temperature, which forms alkyl sulfuric acids. They separate with the sulfuric acid in the oily layer.

@user55119 So you are saying that instead of dimer, alkyl sulfuric acid will be formed? Maybe both will be formed and the dimer will be the major one?  Though I don't know why.

@VJ: I believe alkylsulfuric acids are stable at lower temperatures. They can ionize to form carbocations.

Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:

After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:
A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?)
B: rearrangement of the carbocation (why did it happen?)
C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)
The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).

Then through rearrangement the plus charge will shift to 3 degree carbon.

You're nearly there.
I suggest you draw the 3° carbocation out and think a bit more.
Hint: Hyperconjugation

Then after the said rearrangement, visualize the lose of proton from the second tertiary carbon atom, giving you your desired product.

Why will the tertiary carbon lose the proton? Won't the negative charge become unstable due to the  +I effect?

Maybe because the product is stable

Okay I got my answer. The charge will be delocalised due to empty p orbital. Thank you.

Okay, I'll keep that in mind in the future


text:
Method to determine ibuprofen content in dosage forms
Please suggest some methods for determining the ibuprofen content in tablets, capsules and liquid dosage forms.
Is there a method which this can be done using UV spectroscopy?

Common methods in quantitative analysis of ibuprofen are

titration with sodium hydroxide in methanol,
potentiometric titration,
chromatography combined with mass spectrometry,
IR spectroscopy[1], or
UV spectroscopy (273 $\pu{nm}$[1]).

Some of these methods, e.g. UV or IR spectroscopy require extraction of the active ingredient.
Ref.: [1] S. R. Matkovic, G. M. Valle and L. E. Briand, Quantitative analysis of ibuprofen in pharmaceutical formulations through FTIR spectroscopy


text:
What causes both components of a binary gas mix to condensate at the same time for a given composition
Okay, so recently I encountered the temperature - mole fraction diagrams for binary vapour-liquid equilibrium. For now, I only consider two-component systems where the components are very similar and form ideal mixtures, e.g. a 50:50 mix of Benzene and Toluene.
I understand it from the liquid to vapour transition side (i.e. increasing T) and can do the calculations required, but when I start with only vapour (high T) and cool the mix I am confronted by the following problem:
A vapour component can only be in equilibrium with its liquid when it's partial pressure is equal to the vapour pressure (P*) of the liquid - or in the case of mixtures its reduced vapour pressure (Raoult's law: pi = xi.P*).
What bothers me is that the partial pressure of the less volatile component at the dew point is still lower than the vapour pressure of its pure liquid and with no liquid present (yet) there can be no mixing which will lower the vapour pressure. Yet both components condense.
Possible explanations I could come up with include:
A bit of less volatile component condenses and as it has similar properties, small amounts of the other component enter solution (almost like Henry's Law for solubility of gases). This affects rates of evaporation and condensation until equilibrium is reached.
Does this make sense or is there something else happening here?


text:
How would one calculate the entropy change for an adiabatic irreversible process?
If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic process because:
$\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$
And $W_{\text{reversible}}\neq W_{\text{irreversible}}$
Now, Atkins' Physical Chemistry (Section $3.2$) states that:

To calculate the difference in entropy between any two states of a system, we
  find a reversible path between them, and integrate the energy supplied as heat at each
  stage of the path divided by the temperature at which heating occurs.

But the problem is that we cannot define a reversible path for an adiabatic irreversible process. If that's the case, how can we calculate the entropy change for an adiabatic irreversible process?

This doesn't answer my question because it doesn't discuss this issue.

I know this is old and sorry for digging up but I just want to say that I think the question is really good and well-written.

I agree with you over here.

$\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$
And $W_{\text{reversible}}\neq W_{\text{irreversible}}$

But this part seems to be wrong.

If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic

Just because the process is adiabatic irreversible it doesn't mean that the reversible path needs to be adiabatic as well.
The only requisite for calculating the entropy for an irreversible process is simply that the path taken must be reversible.
So until now we have established that the heat transfer does not have to be 0. For an adiabatic process.
Even though $W_{\text{irrev}} \neq W_{\text{rev}}$  the extra parts will be balanced by the heat transfer in the reversible process.
For example let's take adiabatic expansion.
$$W_{\text{irrev}} > W_{\text{rev}}$$
Numerically speaking in an absolutely hypothetical situation.
\begin{align}
\ce{ ∆U &-> 10~J}\\
\ce{ W_{irrev} &-> 10 ~J}\\
\ce{ W_{rev} &-> 5~J}\\
\end{align}
So the possible proposed reversible path will automatically have a transfer of heat into the gas . Whose value will be 5J.
 .

(+1) I agree with this answer. Entropy is a state function. It doesn't matter whether you take an isobaric/adiabatic/blah path from state A to state B. As long as it's between the two same states, the change in entropy will be the same. But, do note that it's only for calculating the change in the system's entropy, i.e., not the surroundings.

(My vote got locked) but I now realize you didn't answer "they can never be connected by an irreversible path during an adiabatic process" @AvnishKabaj

The reversible path between the two end states does not have to be adiabatic  even though it is adiabatic for the actual adiabatic irreversible path.  For the reversible path, we are allowed to add or remove heat from the system reversibly, as long as we start off at the same initial state and arrive at the same final state.  And the reversible process does not have to be done using a one-step path.  Part of the path can be adiabatic and part of it can be isothermal, for example.  Other reversible combinations can also be used.  For more details on a cookbook recipe for determining the entropy change for an irreversible process, see this reference:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

"the reversible process does not have to be done using a one-step path" don't all reversible processes always have infinite steps?

In Step 1 of Physics Formums insights, it is written:
"$\Delta U = nC_v(T_f- T_0)= P_f(V_f-V_o)$" Can you please explain how this is true?

@GaurangTandon By a one-step path, I meant a solely adiabatic path of infinitesimal increments or a solely isothermal path of infinitesimal increments.  My point was that the overall reversible process can consist partly of an adiabatic path and partly of an isothermal path.

Which part do you think is not true?  (Incidentally, you're missing a minus sign on the right hand side of the equation).

@ChesterMiller How did you obtain it? $nC_vT \ne  -P_fV_f$

I really don't understand what you have written here.  In my equation, the expression on the left hand side is the change in internal energy of the ideal gas, and the term on the right hand side is the irreversible work done by the gas on its surroundings.  Do you somehow disagree with the expressions on both sides of the equation?

Ah, sorry! I was misinterpreting it.


text:
Is chemical equilibrium possible at any temperature and pressure?
Consider the chemical reaction ${\rm Na+Cl\to NaCl}$ by mixing ${\rm Na}$ with ${\rm Cl}$ (or by mixing ${\rm NaOH}$ with some acid). This process is exothermic while the reverse ${\rm NaCl\to Na+Cl}$ must be endothermic. Can these two processes exist in chemical equilibrium (by which I understand equal rates for forward and reverse reaction) at any temperature and pressure?
I'm from the physics background. Please bear with my naive question.

https://chemistry.stackexchange.com/questions/86052/relating-the-reversibility-of-equilibrium-to-that-of-thermodynamics. This question and answer should really enlighten you. Note also that your reaction should be $\ce{2 Na + Cl2 -> 2NaCl}$.


text:
Langmuir method
I have done a biosorption study using a protein as a biosorbent and six different concentrations of $\ce{Pb}$ ions. I need to show the efficiency of the biosorbent, I was told to use the Langmuir isotherm to represent these results. Thus far, I
have plotted a graph of $q$ vs. $C_\mathrm f$ using the formula
$$q = (C_\mathrm i-C_\mathrm f)\times\frac{V}{m}$$
I have also derived my $R$ value from the straight line obtained. How do I get to the Langmuir isotherm from here? What do I need to calculate and what I should plot to represent the data in Langmuir isotherm? Also, I am a biologist so I would appreciate step by step help as I am very confused with all the different equations for Langmuir.

Could you label your variables, so it's clear what $\mathrm{q}$, $\mathrm{C_i}$, $\mathrm{C_f}$, $\mathrm{V}$, $\mathrm{m}$ and $\mathrm{R}$ are?

Sure, Ci -- initial concentration of Pb ions (ug/L) Cf- residual concentration of Pb ions (ug/L), V- volume of the reaction (L), and m - mass of biosorbent (g).

So $\mathrm{q}$ is the amount of absorbed material per mass of biosorbent, is that right?

Introduction
The Langmuir isotherm is a theoretical equation that describes the reversible adsorption of a non-interacting adsorbate on a surfaced that contains a fixed number of identical sites. Therefore, the main assumptions of the model are:

There is a maximum amount of adsorbate that can be fixed to the surface, $\mathrm{S_{max}}$. In other words, the adsorption can reach a saturation point.
All sites are equivalent, and there are no other factors (for instance roughness) that make sites more available or more binding than others.
Sites are not affected by the status of neighbouring sites; all sites are independent. So, neither the local nor the global degree of saturation impacts how likely it is for a single adsorbate particle to adsorb or desorb from a site.

With these assumptions, an expression can be found (it can be derived in terms of chemical equilibrium, chemical kinetics or statistical mechanics) that expresses the relationship between the adsorbed amount of material, $\mathrm{S}$, and the concentration of the adsorbate in the adjacent phase, $\mathrm{C}$:
$\mathrm{S = \frac{K_a C}{1 + K_a C}S_{max}}$
where $\mathrm{S}$ is the amount of adsorbed material, $\mathrm{S_{max}}$ is the maximum possible amount of adsorbed material (in the saturation point), $\mathrm{K_a}$ is the equilibrium constant (written in the direction of adsorption) that regulates the adsorption/desorption of an adsorbate particle on a single site, and $\mathrm{C}$ is the concentration of the adsorbate on the adjacent phase.
It is very common to make the equation intensive by expressing the amount of adsorbed material as a fraction of the maximum possible amount, called coverage ($\mathrm{\theta}$):
$\mathrm{\theta=\frac{S}{S_{max}}=\frac{K_a C}{1+K_a C}}$
Linearisation
The main disadvantage of that formulation of the Langmuir isotherm is that it's not linear, and fitting data to that expression is not straightforward. Consequently, there are various linearisation schemes that have been employed in the literature:
Hanes-Woolf: $\mathrm{\frac{C}{S}=\frac{1}{K_a S_{max}}+\frac{C}{S_{max}}}$
Lineweaver-Burke: $\mathrm{ \frac{1}{S} = \frac{1}{K_a S_{max}} \frac{1}{C}  +\frac{1}{S_{max}} }$
Eadie-Hofstee: $\mathrm{S=S_{max}-\frac{1}{K_a}\frac{S}{C}}$
Scatchard: $\mathrm{\frac{S}{C}=K_a S_{max}-K_a S}$
All of them allow for a linear relationship involving $\mathrm{S}$ and $\mathrm{C}$ to be plotted. Although in principle all equations are equivalent to the formulation above and should lead to the same values, in practice they affect the fitting of real data differently.
In my experience, Lineweaver-Burke is the most commonly used as it introduces less bias in the estimation of the values as other linearisations, but it has the drawback of being less precise at low values of $\mathrm{S}$.
Lineweaver-Burke
So, to fit your data to a Lineweaver-Burke equation, you need to plot $\mathrm{1/S}$ (y axis) vs. $\mathrm{1/C}$ (x axis). Your points will then fit a linear equation of the type
$\mathrm{y=a+bx}$
where
$\mathrm{y=\frac{1}{S}}$
$\mathrm{x=\frac{1}{C}}$
$\mathrm{a=\frac{1}{S_{max}}}$
$\mathrm{b=\frac{1}{K_a S_{max}}}$
Your example
As far as I can tell, you express the adsorbed amount of material as a load $\mathrm{q}$ with units of micrograms of adsorbate per gram of biosorbent:
$\mathrm{q=\frac{n_{ads}}{m}=\frac{\Delta C·V}{m}=(C_f-C_i)\frac{V}{m}}$
so, in the formulation above, $\mathrm{q}$ corresponds to $\mathrm{S}$ and $\mathrm{C_f}$ corresponds to $\mathrm{C}$, and your Lineweaver-Burke expression is
$\mathrm{\frac{1}{q}=\frac{1}{K_a q_{max}}\frac{1}{C_f}+\frac{1}{q_{max}}}$
Plot $\mathrm{1/q}$ vs. $\mathrm{1/C_f}$ for your points and fit them to a linear equation. Your y-intercept corresponds to $\mathrm{1/q_{max}}$ and your slope corresponds to $\mathrm{1/K_a ·1/q_{max}}$. From there you can obtain the values for your equilibrium constant $\mathrm{K_a}$ and your saturation load $\mathrm{q_{max}}$.

"$K$ is the equilibrium constant that regulates the adsorption/desorption of an adsorbate particle on a single site" Correct me if I'm wrong, but isn't the $K$ for the adsorption and desorption processes different?

$k$ (the kinetic coefficient) for adsorption ($k_a$) and desorption ($k_d$) will in general be different, but $K$ (the kinetic pseudo-equilibrium constant)  is precisely the quotient of these two kinetic coefficients. As with a true thermodynamic equilibrium constant, we can write it in both directions, but these are mathematically equivalent ($K_d = K_a^{-1}$ by definition).  By convention, in adsorption chemistry, we always use the adsorption expression ($K_a=k_a/k_d$), so it often gets simplified to $K$. I've edited the answer to unambiguously identify that it's $K_a$.


text:
Questions on nucleophilic substitution of haloalkane

In the nucleophilic substitution reaction of haloalkanes with  ammonia, instead of sharing it's lone pair with the carbocation, ammonia loses one of it's hydrogen and forms a primary amine, and the lone pair remains on the nitrogen. I can't understand why it is so. Why can't the nitrogen bond with carbon, it'll still have a fully filled octet.
I have an unrelated query too - in the formation of nitroalkane by nucleophilic substitution of haloalkane and the reagent being silver nitrite. My sources point out that the silver atom is bonded to the oxygen atom attached to nitrogen by a single bond. I know that since the bond is covalent it wouldn't readily break at first in the reaction (correct me if I'm wrong), this leads to nitrogen sharing it's lone pair. Why doesn't oxygen share its lone pair?
I think my thought process is wrong about haloalkane reactions. I believe that when the carbon halogen bond breaks carbon will only have 6 electrons in its valence shell, so there's no problem with it accepting a lone pair. But what
about groups like $\ce{OH-}$ they give one electron alone to carbon when carbon needs two. I'm so confused could somebody please explain?

You mean to say that the reaction should be like this instead: $\ce{RX + NH3 -> (RNH3)+X-}$? i.e. an ionic salt of the cation $\ce{(RNH3)+}$ and the anion $\ce{X-}$.

yes, but the plus sign musnt be the charge on RNH3 but in between..I am wrong and I can't think why


text:
How to convert benzene to 2-nitrostyrene as the major product?
How to convert benzene to 2-nitrostyrene as the major product?

I actually came up with a conversion but I don't know if it's correct or not:

Since nitro is a ring deactivator, I think we can use $\ce{CH2=CH-Cl + AlCl3}$ to create a polymer. I am not sure though

Brominate to form bromobenzene
Nitrate and separate by column chromatography to isolate o-nitrobromobenzene (read)
Vinylate with vinylboroxane/Pd(PPh3)4/K2CO3/DME
Reference: J.Org. Chem. 67 (2002) 4968, Review JCS Chem. Comm. 2009 20-33

Well thanks for the answer...i have never heard of the reaction in the third step....but still...thank you!!

The exact transformation is described in the JoC reference if you can get to it. It appears in a table in the review article with 74% yield.

Okay... I'll see if i can get to it somehow! Ty!


text:
VSEPR bond angles
My teacher gave a practice exam concerning VSEPR and bond angles and in the following diagram, he wrote the bond angle as greater than 109.5. However, I thought that the ideal angle should be 109.5 exactly. It makes sense that since chlorine is more electronegative, the electron cloud density will be pulled closer to it and so the repulsion will be greater, however I can't seem to find any sources to support that claim. All I can seem to find is that double and triple bonds have the same effect as lone pair electrons and take up more space.

Is the angle 109.5 or is it greater than 109.5?

Running a search in the CCCBDB is a good tool for these questions. According to very high fidelity computational methods (CCSDT with aug-cc-pvtz basis set), the $\ce{Cl-C-Cl}$ angle in chloroform is $\gt 109.5 ^{\circ}$
The cause for this deviation from ideal tetrahedral structure is due to steric / electronic overlaps. A normal $\ce{Cl}-\ce{Cl}$ bond is about 1.99 Angstroms (as in the molecule $\ce{Cl2}$). When bonded to the sp$^3$ carbon in this molecule, the chlorine atoms are forced closer than that: about 1.68 Angstroms. Naturally, the repulsive interaction of the electrons in the Cl atoms amounts to a slightly larger angle in the tetrahedral equilibrium structure. Meanwhile the dispersion (attractive) interactions are sufficient to keep the molecule intact.
The below calculated distances are derived from ab-initio ("from the beginning") Quantum-Mechanical theory. But, these calculations are representative of a single gas-phase molecule (i.e. a very lonely molecule in empty space). Interactions with other chloroform molecules may likely change the average bond angles due to intermolecular interactions (e.g. a $\ce{Cl}$ from one molecule can pull the $\ce{H}$ from another). This could slightly change the expected or average bond angles.
The point of VESPR theory is provide a geometric basis for expected molecular geometry. The "ideal" is decribed by VESPR. Reality is usually a little more complicated, especially here, when the four substituents on the sp$^3$ carbon are not identical. For reference, identical calculations on methane, $\ce{CH4}$ reveal a $\ce{H-C-H}$ bond angle of 109.471$^\circ$ .

I ran a search for "Chloroform" on that page and it displays a big table with lots of "geom" links all over it. How do you know which link to click on? A few of them display 500 server errors.

CCSDT with aug-cc-pvtz is usually the best

For $\ce{CHCl3}$, according to VSEPR theory, the geometry will be tetrahedral which leads to bond angle of 109.5 degrees. However, we have three chlorine atoms. Taking into account the size of the chlorine atoms vs. the size of hydrogen atom, since the chlorine atoms are larger, the $\ce{Cl-C-Cl}$ bond angle will be slightly more than 109.5 degrees, and the $\ce{H-C-Cl}$ bond angle will then be slightly smaller.


text:
Avogadro's Number: 6.023 x 10^23 or 6.022 x 10^23?
I was taught that Avogadro's number was $6.023 \cdot 10^{23}$. Now, the accepted value is allegedly $6.022 \cdot 10^{23}$. Has there been a change? If so, when and why? There are still some sites that use $6.023 \cdot 10^{23}$, so if this is wrong, why have those sites not been corrected?

https://en.m.wikipedia.org/wiki/Avogadro_constant this gives it to a few further places and shows that it is 6.022 with rounding.

Currently, the definition of Avogadro constant depends on the definition of the kilogram, and thus has an uncertainty. The recommended value[1][2] is
$$N_\pu{A} = \pu{6.022140857(74) \cdot 10^{23} mol-1}.$$
It will probably be fixed, however, coming May 20th, 2019 on World Metrology Day to exactly[3][4]
$$N_\pu{A}^{2019} = \pu{6.02214076 \cdot 10^{23} mol-1}.$$
See the references for a thorough review of the history, reasons for redefinition, how some relations become exact and others inexact, and to which degree, etc.
References
[1] Marquardt, R.; Meija, J.; Mester, Z.; Towns, M.; Weir, R.; Davis, R.; Stohner, J. 'A Critical Review of the Proposed Definitions of Fundamental Chemical Quantities and Their Impact on Chemical Communities (IUPAC Technical Report)'. Pure and Applied Chemistry, 2017, 89 (7), 951–981. DOI: 10.1515/pac-2016-0808.
[2] Mohr, P. J.; Newell, D. B.; Taylor, B. N. 'CODATA Recommended Values of the Fundamental Physical Constants: 2014'. Rev. Mod. Phys., 2016, 88 (3), 035009. DOI: 10.1103/RevModPhys.88.035009.
[3] Marquardt, R.; Meija, J.; Mester, Z.; Towns, M.; Weir, R.; Davis, R.; Stohner, J. 'Definišion of the Mole (IUPAC Recommendation 2017)'. Pure and Applied Chemistry, 2018, 90 (1), 175–180. DOI: 10.1515/pac-2017-0106.
[4] 'A new definition of the mole has arrived'. IUPAC press release. archived link: goo.gl/5haPwJ

In fact it will get updated in the end of 2018.
The new SI value will be updated to:$6.02214076×10^{23} \text{mol}^{-1}$


text:
Esterification of racemic acid with optically active alcohol

An excess of racemic acid ($\ce{CH3CHClCOOH}$) is allowed to react with (S)-2-methyl-1-butanol to form the ester, $\ce{CH3CHClCOOCH2CH(CH3)CH2CH3}$, and the reaction mixture is carefully distilled. Three fractions are obtained, each of which is optically active. Draw stereochemical formulas of the compound or compounds making up each fraction.

I know that two fractions obtained will be of the diastereomers (R,S) and (S,S), what is the other fraction?
Source: Morison and Boyd

What is lost in the ester formation? Perhaps there was no workup step!

Did you consider that possibly the leftover excess of the racemic acid $\ce{CH3CHClCOOH}$ can react wit each other forming a cyclic product in which the oxygen from carboxylic group of the first molecule substitutes the chlorine from the second molecule, bringing the close enough so that the oxygen from the carboxylic group from the second molecule substitutes the chlorine from the first one?

This question is in a chapter before the chapter on substitution

It's a pretty basic concept. Two diastereoisomeric esters will be formed. Now diastereoisomers have different physical and chemical properties just like normal isomers. Hence, obviously, their rate of formation will be different. So, different amount of R and S acid will react
Thus, the third fraction will contain unequal amount of the acids which will show optical activity, and the other two fractions are the two diastereoisomers which are also optically active.

I think the other fraction should be the ether ($\ce{CH_3CH(OCH_2C^*HMeCH_2CH_3)COOH}$). This compound can be formed by substitution of the $\ce{Cl}$ of the racemic acid by the $\ce{OH}$ group of the optically active alcohol, which was actually allowed to form an ester.
The acid is present in the solution both in its both S and R forms. So, substitution will generally favour SN2, and thus the ether will have its both R and S configuration only at that substitution centre. But the ether formed will also contain the part coming form (S)-2-methyl-1-butanol, which has a fixed chiral centre (marked with asterisk), which will show optical activity.
So, the ether formed will actually consists of two sterochemical compound, which will be diastereomeric with respect to each other. But when distilled, due to almost similar physical properties, the third fraction obtained will consist of that mixture of ether, which will have a chiral centre and thus be optically active.

Where did you get the C* from? (in your first sentence) What does it depict?

That is the chiral Carbon.

There is another chiral carbon but you didn't label it anything.


text:
The irregular trend in ionization enthalpy of 3d elements
The d block elements display several exceptional behaviours, and one of them is the irregular trend in ionization enthalpy of 3d elements.
The reason given by my textbook is very confusing to me so I'm unable to grasp it. I would appreciate it if someone could explain this statement to me in a simple way.

The irregular trend in the first ionisation enthalpy of the $\ce{3d}$ metals, can be accounted for by considering that the removal of one electron alters the relative energies of $\ce{4s}$ and $\ce{3d}$ orbitals. So the unipositive ions have $\ce{d^n}$ configurations with no $\ce{4s}$ electrons. There is thus, a reorganisation energy accompanying ionisation with some gains in exchange energy as the number of electrons increases and from the transference of s electrons into d orbitals

Text source: NCERT (d & f block elements/ page 215)
Image source: Inorganic Chemistry Cengage Part-2 ( page 6.13)

Rekated: https://physics.stackexchange.com/q/240500/81224

Your question could benefit if you explain to us how you interpreted the statement by the book, so we could tell you where you went wrong or missed anything. The link by Apoorv is great, and here are some more for you to read https://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium https://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations https://chemistry.stackexchange.com/questions/18988/how-does-the-radial-distribution-function-of-vanadium-differ-from-that-of-calciu

The statement in your textbook is actually not true for all of the transition elements in the first row. Ti+ has a configuration of 3d2 4s1, contrary to the statement "unipositive ions have dn configuration with no 4s electrons".

read the 2nd answer in this link https://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations ..... esp. the chemguide link in it

The trend in I.E of transitional elements are irregular for mainly two reasons:-

Irregular trend of atomic radius
Reorganization Energy

You might be knowing that Ionization Enthalpy is inversely proportional to Atomic Radius, so irregularity in trend of atomic radius also reflect in the trend of Ionization Enthalpy.
But the relation between atomic radius and ionization enthalpy isn't the sole reason for the trend. There comes a thing called Reorganization Energy.
D block loses one electron from 4s orbital when the required energy is supplied which you may say it as ionization enthalpy but wait. There's another electron in the 4s orbital. That electron takes extra energy and jumps into d orbital. This is known as Reorganization Energy.
Due to reorganization, the element attains a certain level of stability. Due to achievement of stability the element releases a energy known as Exchange Energy.
So the overall energy involvement you see is actually the (Initial Energy to remove electron)+(Reorganization Energy)+(Exchange Energy) which is equal to Ionization Energy.
Thus Ionization Enthalpy is affected by a variety of factors and so its trend is also irregular.


text:
How can we directly add half cell potentials to measure the EMF of a galvanic cell?
We know that $E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}$. For example, in a cell $\ce{Zn(s) | Zn^2+(aq) || Cu^2+(aq) | Cu(s)}$, $$E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}=E^\circ_{\ce{Cu^2+/Cu}}+E^\circ_{\ce{Zn/Zn^2+}}$$
This makes sense to me. I think of $E^\circ_{\text{cell}}$ as the $E^\circ$ value of the net cell reaction i.e. of $\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$.
But, consider this cell: $\ce{Zn(s) | Zn^2+(aq) || Ag^+(aq) | Ag(s)}$ According to my textbook, the $E^\circ_{\text{cell}}$ value here is again defined by:
$$E^\circ_{\text{cell}}=E^\circ_{\text{reduction at cathode}}+E^\circ_{\text{oxidation at anode}}=E^\circ_{\ce{Ag+/Ag}}+E^\circ_{\ce{Zn/Zn^2+}}$$
But, I think this is not correct.
Consider the fact that this cell is made up of two half cell reactions:
$$
\begin{array}{}
\text{Oxidation}&\ce{Zn(s)}&\ce{-> Zn^2+(aq) + 2e-}&E_1^\circ\\
\text{Reduction}&\ce{Ag^+(aq) + e-}&\ce{-> Ag(s)}&E_2^\circ\\
\text{Net cell reaction}&\ce{Zn(s) + 2Ag^+(aq)}&\ce{-> Zn^2+(aq) + 2Ag(s)}&E_3^\circ\\
\end{array}
$$
and that we can't add $E^\circ$ values directly, since they are an intensive property. Instead, we need to say that $\Delta G_3=\Delta G_1+\Delta G_2$ and then write $2E_3^\circ=2E_1^\circ+E_2^\circ$ or $E_3^\circ=\frac{2E_1^\circ+E_2^\circ}2$, which is definitely $\neq E_1^\circ+E_2^\circ$, as my textbook says.
I believe I have correctly presented everything here, and cannot figure out my mistake. Where am I wrong?

You are writing $\Delta G_3=\Delta G_1+\Delta G_2$. But it should be $\Delta G_3=\Delta G_1+2\Delta G_2$ (I am considering  $\Delta G_2$ as free energy change for the conversion of $\ce{Ag+}$ to $\ce{Ag}$, which is considered 2 times in the final reaction) Because $\Delta G$ also depends on the number of moles of reactants and products, if you add two times the second reaction, the $\Delta G$ will also be used as twice its value. There you've made a mistake.
Thus the equation will be $$2E_3^0 = 2E_2^0 + 2E_1^0$$ So you will have $E_3 ^0 = E_1^0 + E_2^0$.

It is easier to work with redox potentials and at the end use $\Delta G^\text{o}=-nFE^\text{o}$, $n$ is the number of electrons transferred in the reaction and is only used to calculate $\Delta G$ never $E$. To calculate the reaction potential take the difference in redox potentials with the reduction potential first (the one written in the same way as in the tables), $\ce{Ag+ + e = Ag ; 0.8V }$ in your example, and then subtract the other one, $\ce{Zn^{2+} + 2e = Zn; -0.76 V}$ giving $1.56$ V and $n=2$.
In your question you add potentials, in this case the same result is obtained of you reverse the sign of the second potential (written as oxidation in the reaction and is the reducing agent) and add.

Actually, I was taught $\Delta G=-nFE^\circ$ and not $\Delta G=-nF\Delta E^\circ$. The latter is your first equation...

Its the same if you make $E^\text{o}$ the emf of the cell which is the convention.  I used $\Delta E^\text{o} $  only because you measure a difference in $E$ but I will change it, :)

Actually, my question was exactly why the addition of redox potentials works the way it does, because according to me their addition should be invalid as they are intensive properties.

The value you need is always the difference in potential which is ok just as we can have a difference in temperature but not a sum.  The values themselves are ultimately  referenced to the hydrogen electrode at $0$V. Your confusion seems to be that you want to add potentials.  (The addition bit I mentioned is just a way of remembering how to do the calculation, so I hope I did'n't confuse )

Oh, I see where you're going. Can you elaborate on that relation with temperature? Do you mean to say that a solid at 300K when touched with a solid at 400K, through conduction their final temperature will be...?

no, much, much  simpler, I wanted only to give a familiar example where addition does not work. We can measure the difference in temperature  between inside (say 20 C) and outside (say -5 C)  which is 25 degrees difference but we do not add the temperatures as this makes no sense (15 degrees).  Similarly with potentials.


text:
How does the melting point of alkynes change with the position of the triple bond?
$$
\begin{array}{cc}
\text{1-hexyne}&\pu{-132 ^\circ C}\\
\text{2-hexyne}&\pu{-89.5 ^\circ C}\\
\text{3-hexyne}&\pu{-103 ^\circ C}\\
\end{array}
$$
These are the melting points of six carbon chain alkynes. The boiling point of a molecule is mainly affected by its shape. So, if all this molecules have the same shape, why is the range of temperatures so high ($\pu{42.5 ^\circ C}$)?

Those molecules do NOT have the same shape, they look quite differently.


text:
pH value of a very dilute alkaline solution
What is the pH of a $\pu{10^-9 M}$ $\pu{NaOH}$ solution? I think it is 5.

Do you know what is the pH of pure water?

It's PH7, why do you ask?

Good. So it was 7, and then you added some alkali. You can't expect water to become more acidic as a result, can you?

No. You are right. Is it then 9?

How so?$\mathstrut$

pH= 7.002 maybe?

Something has the concentration of 10^-9 molar. What? The concentration of added OH-.
If added [OH-] = 10^-9, then the original concentration of OH- (10^-7 at 25 C) will increase, to a first approximation, to 1.01 x 10^-7 M. Then the [H+] concentration will decrease to 0.99 x 10^-7. The log of 0.99 x 10^-7 is -7.004365 and the negative of that is 7.004365 and that's the pH.
Solving the quadratic equation  would show that, in a second approximation, the pH is even closer to 7.00 (thanks to Oscar Lanzi).
Considering that the dissociation constant of water changes much more than this over a small range from 20 C (pH=7.08) to 30 C (pH=6.92), it suggests that adding that small an amount of NaOH makes a negligible change in the pH. It does require attention to how big numbers in the exponent can have tiny effects, tho.

$[\ce{H^+}]=0.995×10^{-7}$.

I thought this: pH= 14-pOH ; pOH=-log(c0(NaOH)). 14-9=5. where is the fault?

You have the hydroxide from the base and also hydroxide ions from the autionization.  The latter is affected by adding the base, but if you work that out and add it in you get more than $10^{-7}$ M total hydroxide ions and there is no paradox.  However,  the answer's claim of $1.01×10^{-7}$ is not correct either.  It should be $1.005×10^{-7}$ with hydrogen ions reduced to $0.995×10^{-7}$.  The difference between hydroxide and huydrogen ions and the added base $=10^{-9}$, and the product has to be (in this problem) still $10^{-14}$.


text:
Structural isomers given a molar mass

How many structural isomers are possible for compound containing $\ce{C}$, $\ce{H}$ and $\ce{O}$ atoms only with their molar masses $100$ as well as the isomers are simultaneously ketones

3
4
5
6

Answer is (4).
The problem I had was I couldn't understand the question properly. I don't understand what do they mean by simultaneous ketones. But I tried this in the photo below:

Ok It's a question from my study material for isomers. Whatever they asked in that question is given above. I know the answer which I will update

They want you to find all the structural isomers all (1) containing C,H and O (2) of molar mass 100 (3) and are ketones. They should satisfy all these three conditions simultaneously. Getting it?

Well the problem seem to have forgotten that isotopes of carbon, hydrogen and oxygen exist.

Limiting the answer to compounds which only have oxygen as a ketone, and ignoring isotopes of carbon, hydrogen and oxygen, there are two chemical formulas which have a mass of 100.
Isomers of $\ce{C6H12O}$ (Molecular Mass = 100) that are ketones:
1) Hexan-2-one.
2) Hexan-3-one.
3) 4-Methylpentan-2-one.
4) 3-Methylpentan-2-one.
5) 3,3-dimethylbutan-2-one.
Source: Wikepedia - $\ce{C6H12O}$
Isomers of $\ce{C5H8O2}$ (Molecular Mass = 100) that are ketones:
1)pentan-2,3-dione.
2)pentan-2,4-dione.
Hence in total we have seven possibilities.

For $\ce{C5H8O2}$ there are both both pentan-2,3-dione and pentan-2,4-dione. 2-Methyldihydrofuran-3(2H)-one is both an ether and a ketone.

Oh yes! So the options must be incorrect. Question doesn’t place any condition on no. of oxygen atoms. Should I add them too?

Surely pentan-2,4-dione should be added.


text:
What is the number of unpaired electrons in (Fe(H2O)5NO)2+?
Given the fact that $\ce{NO+}$ is a strong field ligand and iron is in the +1 oxidation state, the valence orbitals of $\ce{Fe+}$ must undergo rearrangement from $\mathrm{3d^{6}4s^{1}}$ to $\mathrm{3d^{7}4s^{0}}$, which must contain 6 electrons in pairs, leaving one unpaired electron.
So, the hybridisation of iron can be concluded to be $\mathrm{sp^{3}d^{2}}$.

Possible duplicate of Are these reaction equations for the formation of the brown ring complex correct?

This is no Iron(I)! The idea that Fe(I) is present in this famous brown ring test is outdated. NO is a non-innocent ligand and will take the form of $NO^-$ here while the Fe is in the oxidation state +III.
Even if you think about how you prepare it. There are no real Fe(I) compounds (I mean simple ones). And Fe(II) oxidizes really easily to Fe(III) and you use a strong oxidizer like nitrate. And still you expect the Fe to be reduced? There are very few complexes with a real Fe(I). One example is the related nitrosyl-complex nitroprusside, so cyanide ligands instead of water. When this is reduced some percent, I think like 25% may actually contain some Fe(I). But with non-innocent ligands it's pretty much useless to asign any oxidation state to the central atom.

You just said it, strange oxidation states. As I mentioned above those are the so called non-innocent ligands. They can have various charges and they can take up electrons or donate them back. Nature uses them as well in their porphyrine-systems. The iron is something between +1 and +3. But just thinking about it logically, how would that reaction work? The nitrosyl is formed from the nitrate you add. Nitrate is a strong oxidizer and the nitrogen is already at its highes oxidation state, while Fe(II) is a good reducing agent. Why would this suddenly switch?

Do not forget the other iron that's involved.  Simplifying somewhat by removing water ligands:
$4\ce{Fe^{2+}}+4\ce{H^+}+\ce{NO_3^-}+\rightarrow \ce{FeNO^{2+}}+3\ce{Fe^{3+}}+2\ce{H_2O}$.  You need to oxidize the "other" iron ions to balance oxygen and keep that element where it wants to be (OS=-2).

That is correct. The actual reaction is the oxidation of the Fe(II) to Fe(III) and the reduction of nitrate to nitrosyl. And when the nitrosyl forms the rest of the Fe(II) will form the brown-purple complex. Nonetheless is there no Fe(I) in this reaction. I think I read about a high-energy form of said complex with Fe(I) but at ground state it's definately an Fe(III).

Now I see what chemists have found, and deleted my first comment.  Still want to show the role of additional iron though.

You need to consider the fact that d-orbitals split in octahedral field. You will have 3 x t2g-type orbitals and 2 x eg-type orbitals. Then, you need to determine the oxidation state of Fe, which you have correctly done - it's +1. Fe normally has 8 valence electrons, minus 1 for the charge gives you 7. Therefore, this is a d7 complex. This can be either low spin or high spin. Low spin d7 has 1 unpaired electron, high spin d7 has 3 unpaired electrons. However, even though NO+ is a very strong field ligand, there's just one, so the complex is most likely high spin, also considering the fact that Fe is first row transition metal.


text:
Which bands are considered significant in the fingerprint region of IR spectra?
I'm currently writing my characterisations for the experimental section of my dissertation and can't decide which peaks from the fingerprint region of my IR are worth reporting. Is there a threshold of %Trasmittance, anything below which is considered not to be significant? For example, I know that a strong band covers most of the y axis, a medium about a half and a weak about a third. Therefore, is it worth reporting very weak bands with only 10-15 %Transmission?

There isn't a good answer to this. It depends. For instance the "weak" bands might be due to a low concentration of an important impurity.

My compounds should be pretty much pure. They are usually columned, dried on hi-vac and in a desiccator, before any characterisation. NMR is clean, solvent and water-free, only one spot on TLC.

Fingerprint is important as a whole.  That is why it is called like that.  Normally you report significant bands only. Like OH if an alcohol was a target

Unless your dissertation is something else than organic synthesis.  But given the question it shouldn't be the case.

It's basically inorganic catalysts, but for organic reactions, so my products are organic :D


text:
What happens if you add NaCl during the neutralization reaction between HCl and NaOH?
In class, I was taught that $\ce{NaOH}$ and $\ce{HCl}$ react completely (no equilibrium). I get the equation: $\ce{NaOH + HCl -> H2O + NaCl}$.
However, if I were to add a lot of $\ce{NaCl}$ to the solution, would some $\ce{NaOH}$ and $\ce{HCl}$ be formed according to Le Chatelier's principle?

RE: ... would some NaOH and HCl be formed according to le Chatelier's principle? // Not really. Your fundamental premise is wrong. NaOH, HCl, and NaCl are all essentially totally ionized in water. Each ion is surrounded by multiple layers of water molecules with water's dipole oriented towards the ion.

MaxW is correct.However the above reaction is not an equilibrium reaction.

You yourself said that there is "no equilibrium", but then forgot that Le Chatelier's principle only applies to any system at equilibrium!  Recall its definition:

Le Chatelier's principle: a principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

In your case, as also MaxW pointed out above, $\ce{NaOH}$, $\ce{HCl}$ and $\ce{NaCl}$ are all completely ionized in water (and their ions are hydrated/solvated by water molecules). There is no such thing as an "equilibrium" in this case, as the reaction, being an ionic reaction, completely proceeds in the forward direction.


text:
Why might copper have a lower heat capacity than lithium according to the Shomate Equation?
I've learned that according to Condensed phase thermochemistry data, according to NIST, the functions of heat capacity, enthalpy, and entropy can be predicted by the Shomate Equation:
$$\begin{align}
t &= \text{Temperature } (K) / 1000\\
C_\mathrm{p} &= A + B\cdot t + C \cdot t^2 + D \cdot t^3 + E / t^2\\
\Delta \text{Enthalpy} = \int C_\mathrm{p}\mathrm{d}t
&= A\cdot t + B\cdot t^2 /2 + C\cdot t^3 /3 + D\cdot t^4 /4 − \frac Et + F − H\\
\text{Standard Entropy} &= \int \frac{C_\mathrm{p}}{t}\mathrm{d}t\\
S^\circ &= A \cdot \ln(t) + B\cdot t + \frac C2 \cdot t^2 + \frac D3 \cdot  t^3 − \frac E{2 \cdot  t^2} + G
\end{align}$$
You can see parameters A - H from the NIST webbook in condensed phase data for iron, for example, here.
I've plotted the enthalpy, entropy, and heat capacities as a function of temperature for $\ce{NaCl}$, $\ce{MgCl2}$, $\ce{H2O}$, $\ce{Fe}$, $\ce{Li}$, and $\ce{Cu}$ using their special coefficients A - H for each material. I did this over the range of temperatures 298 Kelvin to 450 Kelvin.
Ranking the heat capacities over the range of temperatures studied:
copper < iron < lithium < sodium chloride < magnesium chloride < water.
Since  molar heat capacity is the heat required to raise one mole of a substance by 1 degree celsius, it seems that water which is a liquid has more places to put the heat plus hydrogen bonding so it has the highest heat capacity. The ionic compounds $\ce{NaCl}$ and $\ce{MgCl2}$ have greater heat capacities than the metals these are compounds with ion-ion interaction in a crystal lattice which gives more places to put the heat and thus comparatively a significantly higher heat capacity. The heat capacity of $\ce{MgCl2}$ appears to be much greater than $\ce{NaCl}$ because $\ce{MgCl2}$ is divalent so the charge-charge interaction in the crystal lattice is greater, raising the heat capacity.
The heat capacities of the metals lithium, iron, and copper are very close. I don't get why copper has a lower heat capacity than the other metals since copper has more electrons so there should be more energy levels to put the heat I would think. Quite simply I don't understand the behavior of the metals, especially if the Shomate parameters A - H are experimentally determined how can I match this to my intuition?
The trends of enthalpies generally seem to follow the heat capacities as a function of temperature. It's kind of interesting that the entropy of $\ce{MgCl2}$ is significantly greater than $\ce{H2O}$, but that there's some cross-over with $\ce{H2O}$ and $\ce{NaCl}$ in standard entropies as function of temperature.
How should I interpret all this, particularly the metals?

Respectively how I calculated all this was kind of tedious, but performed in R:
library(ggplot2)

# Data for Nickel:
# http://webbook.nist.gov/cgi/inchi?ID=C7440020&Type=JANAFS&Table=on#JANAFS

# Data for Copper
# http://webbook.nist.gov/cgi/inchi?ID=C7440508&Type=JANAFS&Plot=on

# etc ...
# Materials used: Copper, H2O, Iron, Lithium, MgCl2, NaCl

# Schomate Equation:

# C_p = A + B*t + C*t^2 + D*t^3 + [E / (t^2)] + H_298-Kelvin_standard

# Enthalpy: H = A * t + (B / 2)(t^2) + (C / 3)(t^3) + (D / 4)(t^4) - (E / t) + F - H

# Entropy: S = A * ln(t) + B*t + (C / 2)(t^2) + (D / 3)(t^3) - [E / (2 * t^2)] + G

# Open the file of Schomate Parameters for solids
FILE = read.csv(file = "Schomate_Parameters.csv", header = TRUE, sep = ",")

# Split up the file into lists of each material.
Each_Material = split(FILE, FILE$Material)

# Solution_df is the dataframe that has the Temperatures, Heat Capacities, Enthalpies, Entropies, and Free_Energies.
# Basically everything I need to make some stellar graphs and point-wise interpretation.
# It takes some effort to construct. We will start assuming 5 columns.
Solution_df <- data.frame(matrix(nrow = 0, ncol = 5))

# These are the initial Five Columns that I will calculate.
my_column_names <- c("Material", "Temperature", "Enthalpy", "Entropy", "Heat_Capacity")

# And I assign these as column names.
colnames(Solution_df) <- my_column_names

###### THIS IS A MAJOR MULTI-FUNCITON #############
################# CAREFUL #########################
for(my_index in 1:length(Each_Material)) # Iterating over every material: Copper, NaCl, Iron, etc ...
{

  Material_type = names(Each_Material[my_index]) # Find the current materials name.

  This_Material = as.data.frame(Each_Material[my_index]); # Deal with it as a dataframe.

  # Assign variables A through H as the schomate parameters
  # which, by the way, change for each new material in this for-loop.
  A = This_Material[1, 3]
  B = This_Material[2, 3]
  C = This_Material[3, 3]
  D = This_Material[4, 3]
  E = This_Material[5, 3]
  F = This_Material[6, 3]
  G = This_Material[7, 3]
  H = This_Material[8, 3]

  # The temperature ranges from 298 kelvin to 400 kelvin in increments of 0.1 kelvin.
  Temperatures = seq(from = 298, to = 450, by = 1)
  # For some reason, the parameters provided by NIST request calculations in terms of kilo-kelvin.
  Temperatures = Temperatures / 1000

  # This is the heat_capacity function for any material
Schomate_Heat_Capacity <- function(t)
{
  A + B*t + C*(t)^2 + D * (t)^3 + (E / (t)^2)
}

  # Then I create a list of the heat capacities at each temperature.
Heat_Capacities = Schomate_Heat_Capacity(Temperatures)

  # The Enthalpy function as a function of temperature
Schomate_Enthalpy <- function(t)
{
  A*t + (B / 2)*(t)^2 + (C / 3)*(t)^3 + (D / 4)*(t)^4 - (E / t) + F - H
}

 # Then I create the enthalpies
Enthalpies = Schomate_Enthalpy(Temperatures)

 # The Schomate Entropy function as a function of Temperature.
Schomate_Entropy <- function(t)
{
  A * log(t) + B*t + (C / 2) * (t)^2 + (D / 3)*(t)^3 - (E / (2 * (t)^2)) + G
}

 # THen I compute the entropies at each temperature.
Entropies = Schomate_Entropy(Temperatures)

# The name of each material must occur as many times as the temperature for the columns to be the same number of rows.
Material_type_repeated <- rep(Material_type, length(Temperatures))

# Create a temporary dataframe for every material holding the Material type, Temperature, Enthalpies, Entropies, and Heat Capacities.
Temporary_df <- data.frame(Material_type_repeated, Temperatures, Enthalpies, Entropies, Heat_Capacities)

# Make sure the names match up to the solution dataframe
colnames(Temporary_df) <- my_column_names

# And then bind all of these temporary dataframes to the solution dataframe.
Solution_df <- rbind(Solution_df, Temporary_df)

}
########### END OF major multi-function ##############
######################################################

# Post Processing:
# Calculations depended that Temperature be in kilo-kelvin
# But it should be represented in Kelvin now.
Solution_df$Temperature = Solution_df$Temperature * 1000

# Note: Enthalpy is in kJ / (mole)
# Entropy is in J / (mole * K)
# Temperatures is now in Kelvin.

# So Free_Energy = Enthalpy - Temperatures * Entropy

# = (___ kJ / mole) - (__ Kelvin) * (J / (mole * Kelvin)) * (1 kJ / 1000 J)

# So Free Energy = Enthalpy - Temperature * Entropy * (1 / 1000)
# Therefore Free Energy is in kJ / mole

# Let's calculate the Free Energy Now:

Solution_df$Free_Energy = Solution_df$Enthalpy - Solution_df$Temperature * Solution_df$Entropy * (1 / 1000)

# Remove Nickel because ggplot can show no more than 6 categories.

Solution_df <- Solution_df[Solution_df$Material != "Nickel",]

################################################ END OF POST-PROCESSING ##################################
##########################################################################################################

# Creating Graphs: all as a function of temperature.

# The heat capacity plot:

HC_plot <- ggplot(data = Solution_df, aes(x = Temperature, y = Heat_Capacity, col = Material, shape = Material)) + theme_bw() + geom_point(size = 1)

HC_plot <- HC_plot + scale_x_continuous(breaks = scales::pretty_breaks(n = 30)) + scale_y_continuous(breaks = scales::pretty_breaks(n=30))

HC_plot <- HC_plot + labs(x = "Temperature (Kelvin)", y = "Heat Capacity (Joules per mole per Kelvin)", title = "Heat Capacity of Various Materials\nAs a function of Temperature from Schomate Equation!")

ggsave(plot = HC_plot, filename = "Heat_Capacity_by_Temp_Schomate_Plot.png", width = 10, height = 10)

print(HC_plot)

###### Creating Enthalpy as a function of Temperature Plot

Enthalpy_Plot <- ggplot(data = Solution_df, aes(x = Temperature, y = Enthalpy, shape = Material, col = Material)) + theme_bw() + geom_point(size = 1)

Enthalpy_Plot <- Enthalpy_Plot + scale_x_continuous(breaks = scales::pretty_breaks(n = 30)) + scale_y_continuous(breaks = scales::pretty_breaks(n = 30))

Enthalpy_Plot <- Enthalpy_Plot + labs(x = "Temperature (Kelvin)", y = "Delta Enthalpy (kJ / mole)", title = "Enthalpy as a function of Temperature\nFor various Materials as Predicted by Schomate Equation")

ggsave(plot = Enthalpy_Plot, filename = "Enthalpy_vs_Temperature_Schomate_Plot.png", width = 10, height = 10)

print(Enthalpy_Plot)

################ Creating Entropy as a function of Temperature Plots:

Entropy_Plot <- ggplot(data = Solution_df, aes(x = Temperature, y = Entropy, shape = Material, col = Material)) + theme_bw() + geom_point(size = 1)

Entropy_Plot <- Entropy_Plot + scale_x_continuous(breaks = scales::pretty_breaks(n = 30)) + scale_y_continuous(breaks = scales::pretty_breaks(n=30))

Entropy_Plot <- Entropy_Plot + labs(x = "Temperature (Kelvin)", y = "Standard Entropy (Joules per mole per Kelvin)", title = "Entropy as a Function of Temperature\nFor various Materials as predicted by Schomate Equation")

ggsave(plot = Entropy_Plot, filename = "Entropy_vs_Temperature_Schomate_Plot.png", width = 10, height = 10)

print(Entropy_Plot)

And the file that I read in with read.csv looked like this:
structure(list(Material = structure(c(7L, 7L, 7L, 7L, 7L, 7L,
7L, 7L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 5L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L), .Label = c("Copper", "H2O", "Iron", "Lithium", "MgCl2",
"NaCl", "Nickel"), class = "factor"), Parameters = structure(c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L), .Label = c("A", "B", "C", "D", "E",
"F", "G", "H"), class = "factor"), Values = c(13.6916, 82.49509,
-174.9548, 161.6011, -0.092417, -6.833644, 27.669, 0, 17.72891,
28.0987, -31.25289, 13.97243, 0.068611, -6.056591, 47.89592,
0, 18.4286, 24.64301, -8.91372, 9.664706, -0.012643, -6.573022,
42.51488, 0, 50.72389, 6.672267, -2.517167, 10.15934, -0.200675,
-427.2115, 130.3973, -411.1203, 78.30733, 2.435888, 6.858873,
-1.728967, -0.729911, -667.5823, 179.2639, -641.6164, -203.606,
1523.29, -3196.413, 2474.455, 3.855326, -256.5478, -488.7163,
-285.8304, 169.552, -882.711, 1977.438, -1487.312, -1.609635,
-31.24825, 413.6466, 0)), .Names = c("Material", "Parameters",
"Values"), class = "data.frame", row.names = c(NA, -56L))

I don't have any questions about my code. I put it in here to make my results reproducible. I'm just curious why copper has lower heat capacity than lithium when copper has more electrons so I would think there would be more energy states to put the heat. However, it's also true that copper has a smaller atomic radius than lithium so maybe this is a factor but I'm not sure how it affects the heat capacity.
Heat Capacity vs Temperature Plot.

Delta Enthalpy vs Temperature Plot:

Standard Entropy vs Temperature Plot:

Furthermore, is there any underlying theory in the Schomate Equation or is it mostly polynomial regression fitted to limited calorimetric experimental data?

I might be able to elaborate a better answer later, but have you analyzed this problem from the perspective of crystal structures? You seem to be emphasizing the matter of atomic properties, but thermodynamic properties like heat capacity are much more affected by the spatial arrangement of the atoms and the chemical bond strength. A quick search told me copper has FCC structure while iron and lithium have a BCC structure. Maybe you could pick up from that?

I'm not so sure about that, though I think it's a valid point. I'd like to make some more graphs, but I'm quite honestly having a problem having ggplot2 graphs because they default to not showing more than 6 categorical factors like such as my materials. I am thinking about atomic properties, but the crystal structures are very much a factor. Good idea.

Electrons don't contribute much to heat capacity at these temperatures because the electrons bound to atoms need a big boost in energy to lift them out of their ground states and the free electrons in metals are mostly packed in energy states up to the Fermi level with just a small fraction sloshing around that can contribute to heat capacity. Why is the $y$-axis heat capacity per mole of molecules rather than per mole of atoms to compensate for the Dulong-Petit law?

The Debye theory of the heat capacity solids predicts that the thermal energy is tied up in vibrations of the material. These can be described as phonons or sound waves in the lattice. In a metal there is a high density of free electrons as  an 'electron gas' and we would expect these to contribute to the heat capacity but often the limiting value is only  $\approx 25$ kJ/mol which is $\approx 3R$ and does not have the extra $3R/2$ per mole expected from a monoatomic gas. Thus the heat capacity of the electron appears to be negligible.
The reason for this is that  electrons are Fermions and each quantum state can therefore only contain one electron. At the Fermi level $E_F$ in the metal some electrons can accept energy and are promoted into higher energy levels and so form the 'electron gas'.   However, by the Pauli exclusion principle, very few electrons from approximately $k_BT$ below the Fermi level can be promoted as there are no vacant levels in which to go. Thus only quantum states close to the Fermi level can accept energy and these form a small fraction of the total as $k_BT \ll E_F$.
If the electrons behaved classically then the internal energy $U$ would be increased by $3R/2$ but experiment dictates that this value has to be reduced by the fraction of electrons that can accept energy. This is related to the density of quantum states (number/unit energy) in the metal. The effect this has is to reduce the heat capacity due to the 'electron gas' to  $\approx 0.5$% of the total heat capacity.
The effect of the electron gas can be observed only at very low temperature (typically $\lt 10$ K), where the heat capacity due to phonons is reduced, and is given by $C_{total}=\gamma T+\alpha T^3$ where the cubic part is from the expansion of the Debye equation at low temperature and the linear term from the electron gas. The constant $\gamma$ is the contribution due to electronic heat capacity and is proportional to the density of states at the Fermi level.
The difference in heat capacity between different metals is therefore mainly due to their different lattice motions not the electrons.
Edit: the constant pressure heat capacity $C_p$ for Li at 25 C is 24.78 and for copper 24.44 both in J/K/mol. In fact most metals have similar values at this (high) temperature. Beryllium has a v low value (16.4) and this is probably due to v. high lattice phonon modes compared to other metals.


text:
Comparison between Boron Nitride and Graphite: Interlayer distance
Boron-nitrogen chemistry has attracted significant attention, partly because a B–N unit is isoelectronic with C–C. Furthermore, the radius of carbon and its electronegativity are roughly the
average of those properties for B and N.
I was going through the structures of BN and Graphite, and here's what I found:
In graphite the C atoms in the C6 rings of one layer lie over the middle of a C6 ring in the layer below. In BN, a boron atom in one layer is directly over an N atom in the layer below.
My question: How do we compare the interlayer distance in boron nitride and graphite?
Attached are the structures for reference:

I looked up some data, and found out that the interlayer distance in graphite is greater than that in boron nitride. I'm not sure why it is so.

Data for interlayer distance:

Graphite: $\pu{335.4 pm}$
Boron nitride: $\pu{333 pm}$

That's like exactly the  same value... I wonder what uncertainties are for these data. Probably they are bigger then this tiny difference.

There's relevant information in this paper: https://pubs.acs.org/doi/abs/10.1021/ct200880m . It asks why the interlayer distance in hexagonal-BN and graphite are so similar, and goes through the different interactions before concluding that the van der Waals interaction is the only one that's important in these cases. For what it's worth, it cites the interlayer distances as 3.33-3.35 Å for graphite and 3.30-3.33 Å for $h$-BN.