text:
What are the concentration changes when diluting an equimolar acetic acid: acetate buffer?
When a buffer (for example, a mixture of 100 mM acetic acid and 100 mM acetate), is diluted 10-fold, all the concentrations change drastically. Surprisingly, after re-establishment of equilibrium, the pH is not much different from the starting pH.
How can you visualize how diluting a buffer disturbs the equilibrium of a buffer  and how the system returns to equilibrium? To make things simple in a thought experiment, we will posit that dilution is faster than acid/base chemistry, i.e. the first step is dilution without any reactions, and the second step is acid base chemistry without further dilution.
The explanation/visualization could be at the molecular or bulk level, and could use kinetic or K vs. Q or Gibbs free energy arguments.

Consider an acid buffer with $pK_a=5$. Ignore the hydroxyl concentration, assuming it is negligible.
Dilution leads to an identical (proportional) reduction in the concentration of all solutes: undissociated acid, complementary base (anion), and protium. Because the concentration of anion and protium is reduced, encounters between the two will occur less frequently. On the other hand, dissociation of the acid into component ions continues unabated, since this process is (we are ignoring activities) a first order reaction. The result is an increase in the concentrations of anion and protium, thus reestablishing a value of the protium concentration close to that prior dilution.
Other  important factors should be remembered in the context of ideal buffers and the Henderson-Hasselbach equation: (1) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are set equivalent in an ideal buffer (2) the $\ce{H^+}$  concentration is set equal to the $K_a$ and (3) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are much higher than those of $\ce{H^+}$ . That means that in both absolute and relative amounts, the concentrations of $\ce{A^-}$ and $\ce{HA}$ need to change by only a small amount in order to bring the pH back near the desired set point. This is very clear from the following plots (concentrations are molar and $pK_a = 5$). Every 0.1 time units the concentration is diluted by 17%. The pH spikes up but then settles back again as the acid dissociates. You practically don't see the changes in the undissociated acid and anion on the plot during the recovery because they are tiny.

You can compare this to the case when your buffer is not as good (same $pK_a$, but lower capacity, 100 $\mu M$ concentration). I altered the kinetics too so the pH would stabilize after each dilution:

I should add: ignoring $[OH^-]$ is admittedly bad near neutral pH/low buffer capacity, but then I was attempting to provide a visualization. Adding terms to cope with the hydroxyl equilibrium etc would complicate things without adding much value or accuracy (since the kinetics are not meant to be accurate anyway, just illustrative).

Addendum: Here's the code:
K = 1e-5;  % M
kd=10;     % arbitrary, set to alter convergence speed, numerical stability
dt = 1e-7; % arbitrary, alter to change convergence speed, numerical stability
v0 = [0.2; 0.2 ; K]; % M [HA], [A], [H],

%% for dilute conditions
%kd=80;
%dt = 1e-5;
%v0 = [0.0001; 0.0001 ; K]; % [HA], [A], [H]

Nstep = 10000;
t = [0:dt:(Nstep-1)*dt];
tdilute = t(round(Nstep*[0.1:0.1:0.9]));

dv_dt = @(kd,K,v) [ -kd*v(1)+kd/K*v(2)*v(3); kd*v(1)-kd/K*v(2)*v(3)] ;
dv_ = @(v,dt) dv_dt(kd,K,v)*dt;
dv = @(v) dv_(v,dt);
v = v0;
conc = v0;
for ii=2:Nstep
    vstep = dv(v);
    v = v + [vstep(1); vstep(2); vstep(2)];
    conc(:,ii) = v;
    if any(t(ii) == tdilute)
        v = v/1.2;  % <-- change factor to dilute by here
    end
end

figure
subplot(2,1,1)
plot(t,conc(1,:),'r')
hold on
plot(t,conc(2,:),'g--')
plot(t,conc(3,:),'b--')
ylabel('[c]')
legend('[HA]','[A^-]','[H^+]')
subplot(2,1,2)
plot(t,-log10(conc(3,:)),'b--')
ylabel('pH')
xlabel('time (au)')

Very cool, which tool are you using to simulate the kinetics?

Well, yes and no. With a pKa =5 you are getting close to where the autoionization of water makes a difference. So if the ratios change depends on how many significant figures you're using and how dilute the buffer becomes. The limitations of the Henderson-Hasselbach equation seem to be often overlooked.

@MaxW Yes and no? I did not use H-H explicitly in the calculations, but pointed out that they inform on restrictions for use of a buffer.

@KarstenTheis Sims can be done with matlab or octave.

I should add: ignoring $[OH^-]$ is admittedly bad near neutral pH/low buffer capacity, but then I was attempting to provide a visualization. Adding terms to cope with the hydroxyl equilibrium etc would complicate things without adding much value or accuracy (since the kinetics are not meant to be accurate, just illustrative).

@MaxW Whether the autoionization of water matters I think depends on the concentration of buffer (its capacity), not on the $pK_a$. Otherwise you couldn't have buffers that work near neutral pH...

Abbreviating the concentration of acetic acid with A, that of acetate with B, and that of hydronium ions with H, we can write the equilibrium constant as:
$$ K_a = \frac{B \times H}{A} $$
Multiplying by A and taking the negative base 10 logarithm, we get:
$$ pKa + pA = pB + pH $$
Now, we can visualize pKa (blue), pA (orange), pB (yellow) and pH (green). When the blue orange pillar has the same height as the green yellow pillar, the system is at equilibrium.

In the first part of the visualization (dilution), all concentrations drop by the same factor 10, so pA, pB, and pH increase by one, and the reaction is no longer at equilibrium (mismatch of pillar heights). In the second part of the visualization (net acid dissociation to re-establish equilibrium), the pH changes a lot while pA and pB change very little (imperceptibly) because acetic acid and acetate are present at much higher concentration (major species) compared to hydronium ions (minor species).

I'm interested in the calculation and numbers

Before dilution, you can estimate the pH using Henderson-Hasselbalch:
$$ \mathrm{pH} = \mathrm{pK_a} + \log([\ce{A-}]/[\ce{AH}]) $$
The pH is equal to the pKa (4.75), and the concentrations of acetate and acetic acid are 100 mM each if you neglect the small amount of acetic acid that has to dissociate to bring up the hydronium ion concentration (about 0.01 mM). For the thought experiment concerning 10-fold dilution, you just have to divide all the concentrations by 10. For the equilibrium after dilution, the simplest method is to use Henderson-Hasselbalch again. The pH will be equal to the pKa, and the concentrations of acetate and acetic acids are 10 mM each (again neglecting the 0.01 mM of additional dissociation of acetic acid).

I am interested in the answer, If possible to explain it by using numbers and calculation.

How did you make the neat graph?

I used excel to do the calculations and then turned the frames into a gif with camtasia.


text:
Can we rotate a mirror image by any angle to check if it is superimposable?
A molecule is chiral when it is different from its mirror image, and achiral if it is identical to its mirror image. To check if the mirror image is identical to the original object, we have to check if the two are super-imposable. Sometimes no rotation is necessary to show the molecule and mirror image are idential, and sometimes we need to rotate the mirror image, for example by 180°. This is so confusing. Can someone explain exactly what is to be done in simple words?

Spatial perception is confusing, unless you're good at it naturally or have lots of practice. You are trying to check if you objects in possibility different orientations are the same. Are they the same? It's easiest if you can get them into the same orientation first.

If you need to rotate it, then rotate it. After all, no matter how much you rotate something, it is still the same thing; your left hand (hopefully) does not turn into your right hand if you turn it around.

@orthocresol From your comment I am implying that it does not matter if I do not rotate or rotate 1° or 2° or 180° if I can fit it, it is achiral. Tell me if I am correct and thank you for responding.

You can do what you want.  If something is not superimposable it remains so. Your confusion is due to the fact that we must use projection to paper planes. Other is what symmetry of an object is.

Whether you need to rotate a molecule to show that it superimposes with its mirror image depends on how you position your mirror.
Take human bodies, for example. They have approximate mirror symmetry (bilateral symmetry), with the left side looking similar to the right side. If you stand in front of a mirror with one of your shoulders touching it, an observer will be able to see the symmetry without rotating the mirror image in their mind. If you face the mirror, you have to imagine turning the mirror image by 180 degrees along the long axis (head to toe, like a pirouette). If you are standing on a mirror placed on the floor (careful not to break it...), you have to imagine turning the mirror image 180 degrees around a front-back axis (like a cart wheel) to see that you and mirrored you are super-imposable.
If the plane of the mirror is not aligned with two of your body axes (top-bottom, front-back, right-left), the necessary rotations will be different from 180 degrees.

This reminds me of the question why a mirror switches left and right, and not top and bottom (see this article). Once you understand why this is a silly question, you can return to checking molecules for chirality by placing them in front of a mirror.

The superimposability check (as is often done to determine if a molecule is chiral) can necessitate one or more rotations of the molecule. The rotational axis is irrelevant in this. Absence or presence of such rotations has no further meaning in the context of chirality.

Can we also rotate along any axis or we are stuck to axis perpendicular to a fixed plane.

Any axis is fine

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text:
Why is Acetic acid (pKa = 4.76) stronger than carbonic acid (pKa = 6.36)?
In order to try and understand this, I evaluated it via stability of conjugate bases. The acetate vs carbonate ion. The only difference is what's attached to the carboxyl group, the $\ce{-CH3}$ and the $\ce{-OH}$. Both anions seem to have equal resonance structures as I think resonance mainly only happens on the carboxyl group.
I think that because the methyl group is just larger than the hydroxyl group, it is able to better stabilize the negative charge on the molecule overall. But I also remember hearing that carbon can have a positive inductive effect, meaning it pushes electron density toward the carboxyl group, which would make it less stable.
I originally thought that carbonate would be more stable because the oxygen would pull the electron density toward itself, making the charge more evenly distributed.
If somebody could explain with diagrams I would appreciate it. I lost access to my software that did electron modeling functions for me.

Well actually the $\mathrm{pK}_\mathrm{a1}$ for carbonic acid at 25 °C is 3.6. The pka = 6.36 is also including the equilibrium with dissolved $\ce{CO2}$ in aqueous solution.

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text:
How can I Identify Elements in Objects?
Since all objects on earth are composed of elements, how can I identify the fundamental make-up of objects such as trees, glass, bricks, wood, plastic, shoes, and other entitys?

Learning chemistry.

There is no universal analytical technique. Fundamentally you can divide techniques into ones that do atomic analysis and those that do a molecular analysis.

There is a complete science behind your question called Analytical Chemistry. However the goal is not to identify what a given object is made of, people usually look for targeted compounds or elements of interest. Take chocolate for example, there are thousands of compounds there. Analyzing it completely from scratch may take more than a year! There you will learn plenty of techniques to identify elements. If you just wish to know what this glass is made of, one may use a technique called X-ray fluorescence, it is good for quickly detecting most of the elements except the lighter ones.


text:
What is the mechanism of azulene synthesis?
In the Organic Syntheses preparation of azulene (http://www.orgsyn.org/demo.aspx?prep=CV7P0015) there are two reactions. First is a clevage of the C-N bond in pyridine and the second one is the annelation of a seven-memebered ring:

I suppose that the first step of the first reaction  is a nucleophilic substitution by pyridine at the 1-chloro-2,4-dinitrobenzene, but I have no idea how the mechanism could continue. Could some one please help me? Thank you.
Also, for the 2nd reaction, is it an electrophilic substitution at the cyclopentadieneyl anion, or does it have an  entirely different mechanism?

The first reaction is https://en.wikipedia.org/wiki/Zincke_reaction#Zincke_aldehydes

The pKa of cyclopentadiene is around 16 amd the C=NMe2+ centre is very electrophilic.

The @orthocresol link to the Zincke reaction is a good start to the mechanism. You are correct about the first step. Dimethylamine adds to the electrophilic pyridinium salt 1 as illustrated in 1 $\rightarrow$ 5.  The reaction is driven by loss of the poor nucleophile (decent leaving group), DNP-NH2. The (Z)-isomer 5 is in equilibrium with the (E)-isomer 6 but the reactions leading to azulene 12 require a cisoid conformation to allow for cyclization.
Cyclopentadienide anion, which is delocalized but for convenience is shown as a localized anion, displaces dimethylamine which is protonated by the methanol in the reaction mixture (see your link to Organic Synthesis). Fulvene 8 must generate the geometry shown to complete the remaining steps to azulene (9 $\rightarrow$ 12). ADDENDUM: An added thought. The transformation of 8 $\rightarrow$ 10 may be a 4n+2 disrotatory thermal electrocyclization that forms the new sigma bond of 10 with cis hydrogens.

Nice answer. The only issue I see is that 8 and 9 are resonance structures instead of separate species.

@Zhe: So were 5 and 6.

True enough for 8 and 9. I broke 8/9 down into discreet steps for easier understanding. I'll bracket them and use resonance arrows.  5 and 6 involve more than resonance. Most references show an equilibrium. Actually, one has to get to 8/9 whether it is thru 5 or 6.

@MathewMahindaratne Actually, the double bond geometry prevents that from being the case.

@Zhe: Thanks for pointing it out. I agree with you. Isomerization makes the difference. :-)

@Zhe: The mechanism has been "modified" to improve the quality of azulene 12.:)


text:
Compressibility factor of hydrogen at low pressure and constant temperature
As we all know, the compressibility factor $Z$ of hydrogen and helium is always greater than $1$ at a constant moderate temperature. But, if we look at the virial equation of compressibility factor
$$Z = 1 + \frac{1}{V_\mathrm{m}}\left(a - \frac{b}{RT}\right) + \frac{b^2}{V_\mathrm{m}^2} + \cdots$$
At extremely low pressure $V_\mathrm{m}$ would be extremely large. So, the equation simplifies to
$$Z = 1 + \frac{1}{V_\mathrm{m}}\left(a - \frac{b}{RT}\right)$$
Now, the values of $a$ for hydrogen and helium are extremely small. So, we can take them as zero. The equation further simplifies to
$$Z = 1 - \frac{1}{V_\mathrm{m}}\left(\frac{b}{RT}\right)$$
Now, the value of $b$, $R$, $T$ and $V_\mathrm{m}$ are all positive. Though the value of $b$ for hydrogen is extremely small, still it is positive. So, shouldn't the value of $Z$ at some small pressure be less than $1$?
Any help would be appreciated.

your value of $b/RT$ is also very small $\approx 0.0014$  so is smaller than $a$ so you should ignore this term and not $a$.

In my textbook the opposite meaning is given to $a$ (an attractive parameter) and $b$ (a size parameter), but this may be a matter of differences in labelling, not in meaning. I will for the sake of consistency stick to your convention. For $\ce{He}$ the van der Waals parameters have the following values:

$b = \pu{3.4598 J mol-1 M-1}$
$a = \pu{0.023733 M-1}$

As you can see from the following figure at $RT$ ($\pu{298K}$), while your first approximation is quite accurate for $\ce{He}$ already at $V_\mathrm m \approx \pu{1 L/mol}$, your second approximation fails until higher  $V_\mathrm m$. It is negative, but negliglibly so, and at that point is not particularly useful as an approximation as already $Z \approx 1$.

The approximation fails (at RT) because the condition $b/RT \gt a$ is never observed:
$$\frac{b}{RT} = \pu{0.0014 M-1}$$
whereas
$$a = \pu{0.0237 M-1}$$
You can solve for the temperature at which these two become equal:
$$T_b = \frac{b}{aR}$$
For $\ce{He}$, $T_b = 17.53\ \mathrm K$. Below $T_b$ you may observe $Z<1$.
Sure enough, below that $T$ your second approximation starts to hold, as shown here for $T = 10\ \mathrm K$:

or in terms of $P$:

Note: it is not clear that the virial equation and van der Waals parameters should capture the behavior of the gases over a wide range of $P$ or $V_m$ at such low $T$.

(1) Out of curiosity what temperature did you use? (2) for OP's assumption to hold it isn't sufficient that just $b/Rt \gt a$ but rather it must be that $b/Rt \gg a$.


text:
Why is a 7-endo radical cyclisation favoured over a 6-exo in this synthesis?
Normally, according to Baldwin's rules, 3-exo to 7-exo cyclisations are preferred over endo cyclisations. But in the following example, a 7-endo-trig reaction is favoured over a 6-exo-trig and I am just not able to understand why. It's a real synthesis example (taken from Eur. J. Org. Chem. 2015, 2015 (15), 3240–3250). Please explain to me why the 7-endo product is favoured here.

Thank you for adding the citation in the original post, will do better next time and include it from the beginning

No worries. It was quite trivial to add it, we have a little applet that does citations automatically. Now, on topic: my impression is generally that exo > endo selectivity is more of a factor for small ring sizes. 5-endo reactions are not allowed, for example, but 6-endo can sometimes happen (even though 5-exo usually beats it). Now, for 6-exo vs 7-endo I suspect there isn't too much of an inherent difference (I might be wrong). But in this system, the 7-endo leads to a heteroatom-stabilised radical, whereas the 6-exo doesn't, and that might be enough to tilt the balance in favour of 7-endo.

I'm hesitant to write that as an answer as it's not backed up by anything. But if there's nothing better, then I will, eventually.

sounds pretty reasonable though...from what I've learned the ring sizes are ordered like 5 > 6 > 3 > 7 > 4 > 8-10 ... (might also be the opinion of my prof. but its true for most cases); so the heteroatom-stabilised radical must be quite a lot more stable I'd guess

Baldwin's 1976 paper doesn't have a problem with 7-endo trig. The authors of your paper added a carbonyl group to force the 6-exo dig cyclization. Baldwin's Rules are just that; rules. It might be useful to correct your structures.

you mean that 7-endo-trig wasn't a thing back then they thought of or that really isn't a problem? I know that they are just a set of limited rules, but still you somehow try to apply them

Isn't the 7-endo-trig product missing a carbonyl group?

yeah it is, and the 6-exo aswell just on another place; I apologize for that
But i don't think that it affects the product

Meant 6-exo-trig. Baldwin, JCS, Chem Commun. 1976, pg.734. 6 and 7 endo trig favorable. The starting material, an aryl ketone, gives the 6-membered ring.  Without the ketone, the 7-membered ring is formed. The ketone was added to repress the the 7-membered ring. 6-exo trig should be a lactam.

oh ok. thank you, this is pretty important information

I didn't read the paper until now, but the authors write that "Calculations have indicated that the 2‐oxo group has a great influence on the endo/exo selectivity of radical cyclization reactions [...] a more torsionally strained sp2 carbonyl group was introduced into compound 12, which we expected would favor a 6‐exo‐trig free‐radical cyclization."

Waylander: 7-endo-trig product is missing a carbonyl group (compound 13 in the paper). But, 6-exo-trig product is definitely missing a carbonyl group on lactam (compound 16 in the paper).

Before make any suggestion for the sought cyclization mechanism, let’s look at some rules and guidelines set by peers in their early studies on radical cyclizations. Specifically, Beckwith and coworkers have set few guidelines for radical reactions, which involves intramolecular cycloaddition reactions (Ref.1).  During their kinetic studies on various en-yl radicals (Ref.2), they have shown that hept-6-eny1 radical (1: n = 4; Scheme 1) is able to give both 6-exo-trig and 7-endo-trig products (2 & 3; n = 4; Scheme 1), although 6-exo-trig is the favorable product between two ($k_{exo}/k_{endo} = 5.8$). On the other hand, hex-5-eny1 radical (1: n = 3; Scheme 1) almost exclusively gives 5-exo-trig (2: n = 3; Scheme 1) over 6-endo-trig product (3: n = 3; Scheme 1), although 6-endo-trig is more stable one between two ($k_{exo}/k_{endo} = 48.4$).

Beckwith and Schiesser have performed a theoretical study on the relative rates and the regio- and stereo-chemistry of ring closure of a variety of alkenyl, alkenylaryl, alkenylvinyl, and similar radicals using the method, which involves the application of MM2 force-field calculations to model transition structures (Ref.3). The results show excellent qualitative and satisfactory quantitative agreement with experimental data. The values of both experimental and theoretical data for hept-6-eny1 radical shown to be remarkably close for 6-exo-trig (2: n = 4; Scheme 1) and 7-endo-trig closure (3: n = 4; Scheme 1).
Based on these experimental evidence, it is safe to say, under certain conditions, 6-endo-trig product can predominate over its 5-exo-trig counterpart. For example, Substituents at substituting position disfavors cyclization (e.g, Scheme 2) (Ref.1).

In summary, Beckwith has given few useful rules for radicals: (1) Intramolecular addition under kinetic control when $\mathrm{n} \le 5$, cyclization occurs preferentially in the exo mode (Scheme 1; thermodynamic preference for secondary radical overridden by kinetic preference based on required orbital alignment for cyclization); (2) Substituents disfavor cyclization at substituted position (Scheme 2); and (3) Homolytic bond cleavage is favored when bond concerned lies close to plane of adjacent semi-occupied, filled non-bonding, or $\pi$-orbital (Ref. 1).
Now we can look at examples where the cyclization goes wild breaking those rules. As part of the generation of radicals in heteroaromatic systems, Dobbes, et al. have studied reactions of radicals at the C-7 position of indole (Ref.4). The synthesis and radical cyclisation of 7-bromoindoles carrying an unsaturated N-alkyl group (7; Scheme 3) has been performed for these studies. The radical cyclisation was carried out using tributyltin hydride and AIBN as the radical initiator in refluxing toluene:

the cyclisation of N-allyl-7-bromoindole (7a) was expected to involve the potential 5-exo-trig cyclisation from the indole C-7 position on to an allyl chain to give 8a exclusively. Yet, a mixture of products was obtained, comprising the 6-endo cyclisation product (9a) and the reduction product (10a) in the ratio of 1:2, but no 5-exo cyclisation product (8a) was obtained. The authors stated that the reason for this mixture is believed to be the constraints enforced on the system by the geometry of the indole ring. Although generally favored under kinetic conditions, the 5-exo cyclisation is very difficult in this case as there is considerable strain and distortion in the five-membered ring product and hence presumably in the transition state. Thus, a significant amount of direct reduction of the indolyl radical is observed in this case (phenyl radical initiated is comparatively unstable and trapping it by $\ce{n-Bu4SnH}$ is comparatively faster (Ref.5)). Although the 6-endo reaction is usually much less favorable in any other situation, the bond angles involved in this situation clearly favor 6-endo cyclisation pathway (Ref.4). Nonetheless, on extending the chain length by a $\ce{-CH2-}$ group, N-(but-3'-ene)-7-bromoindole (7b) has given only the expected mixture of 6-exo cyclisation product (8b) and reduced product (10b) in 1:1.2 ratio, without giving a trace of 7-endo cyclisation product (9b).
Now, it is clear that the sought radical cyclization in Ref. 6 can be directed by circumstances. The authors have the cyclization scheme envisioned as depicted in Scheme 4. They have expected that compound 5 (Ref.6) would be formed under radical cyclization conditions indicated in the Scheme without problem under the accepted Baldwin's rules. However, only 7‐endo‐trig adduct, 13, was obtained in 68% yield. None of expected spiro adduct 5 (Ref.6) was observed under a wide variety of radical cyclization conditions, in addition to $\ce{n-Bu4SnH}$. The authors hypothesized that this outcome may be attributed to the smaller torsional stress of the benzyl methylene group. However, if we look at the structural features more carefully, we can argue that Beckwith’s 2nd rule (Substituents disfavor cyclization at substituted position) may have also been playing a role as well, favoring 7-endo-trig addition, substituted position at which has no additional substituents. Moreover, larger ring size also has ability to adjust conformation in favorable way (Ref.3).

The authors have stated their effort to receive the expected 6‐exo‐trig product as follows:

Calculations (ref) have indicated that the 2‐oxo group has a great influence on the endo/exo selectivity of radical cyclization reactions. To overcome this unfavorable substrate control, a more torsionally strained $sp^2$ carbonyl group was introduced into compound 12, which we expected would favor a 6‐exo‐trig free‐radical cyclization. As shown in Scheme 5, $\alpha$‐bromination (ref) of aryl ketone 14 followed by coupling with enamide 10 gave highly functionalized enamide 15 in 75% yield over two steps. Fortunately, radical cyclization of compound 15 under the conditions used above gave 6‐exo‐trig product 16 in 72% yield.

They really have said "fortunately", indicating more of a luck than their calculations. Yet, more evidence in literature (see list of reference in Ref.6) supporting their argument including Ref.4 listed here.

References:

A. L. J. Beckwith, C. J. Easton, A. K. Serelis, “Some guidelines for radical reactions,” J. Chem. Soc., Chem. Commun. 1980, 482-483 (DOI:10.1039/C39800000482).
A. L. J. Beckwith, G. Moad, “Intramolecular addition in hex-5-enyl, hept-6-enyl, and oct-7-enyl radicals,” J. Chem. Soc., Chem. Commun. 1974, 472-473 (DOI:10.1039/C39740000472).
A. L. J. Beckwith, C. H. Schiesser, “Regio- and stereo-selectivity of alkenyl radical ring closure: A theoretical study,” Tetrahedron 1985, 41(19), 3925-3941 (https://doi.org/10.1016/S0040-4020(01)97174-1).
A. P. Dobbs, K. Jones, K. T. Veal, “Radical cyclisation reactions of 7-bromoindoles,” Tetrahedron Letters 1997, 38(30), 5379-5382 (https://doi.org/10.1016/S0040-4039(97)01177-5).
W. P. Neumann, “Tri-n-butyltin Hydride as Reagent in Organic Synthesis,” Synthesis 1987, (8), 665-683 (DOI: 10.1055/s-1987-28044).
M. He, C. Qu, B. Ding, H. Chen, Y. Li, G. Qiu, X. Hu, X. Hong, “Total Synthesis of (±)‐8‐Oxo‐erythrinine, (±)‐8‐Oxo‐erythraline, and (±)‐Clivonine,” Eur. J. Org. Chem. 2015, (15), 3240-3250 (https://doi.org/10.1002/ejoc.201500265).

I'm the co-first author of this paper. I want to thank Mathew Mahindaratne for the helpful and thoughtful discussion. I just want to add one more result when I was doing the synthesis, please see the attached. The beauty of total synthesis is that we never 100% sure what the product we get, but we can always find a way to explain the results.

I'm pleased my view help you to get through your research. Please do the same whenever the opportunity arises. Keep up with good work.


text:
Using Le Chatelier's Principle

The following reaction is allowed to reach equilibrium.
  $$\text{CoBr}_2(g)\rightleftarrows\text{CO}(g)+\text{Br}_2(g)$$
  What happens to the reaction when inert argon gas, $\text{Ar}(g)$, is added?

The solution to this problem states that the addition of an inert gas would not change the concentrations of the other gases, and therefore, the reaction is still at equilibrium.
However, won't adding the argon result in an increase in the pressure of the system, and, to restore the equilibrium, the number of molecules should be reduced and the reverse reaction would be favored? Can someone please explain?

This an example of a problem with incomplete information. You either have to assume that the pressure is constant or it isn't. If you assume that the pressure is constant then the equilibrium doesn't change. If you assume that the pressure increases then the equilibrium does change.

Shoudn't be this compound bromophosgene? (not cobalt bromide as written).


text:
Find the minimum temperature at which the reaction will become spontaneous
This question came in my entrance exam and I think 273 K should be the answer because the reaction is already spontaneous. I just want to know if I've attempted it correctly.

For a certain reaction $ΔG^\circ = \pu{-45 kJ mol-1}$ and $ΔH^\circ = \pu{-90 kJ mol-1}$ at $\pu{0 °C}$. What is the minimum temperature at which the reaction will become spontaneous, assuming that $ΔH^\circ$ and $ΔS^\circ$ are independent of temperature?
(a) $\pu{273 K}$
(b) $\pu{298 K}$
(c) $\pu{546 K}$
(d) $\pu{596 K}$

You are only partially correct: negative Gibbs free energy at $\pu{273 K}$ is a sign of a spontaneous reaction.
However, the question asks when the reaction will become spontaneous and not to select the minimal temperature value from the multiple choices list which corresponds to the spontaneous process.
So, you were supposed to find the equilibrium temperature, e.g. when $ΔG = 0$.
You have correctly started with the equation for the free Gibbs energy:
$$ΔG = ΔH - TΔS = 0  \quad\implies\quad T_\mathrm{eq} = \frac{ΔH}{ΔS}$$
Unknown entropy change $ΔS$ can be determined from the Gibbs free energy $ΔG_1$ at initial temperature $T_1$:
$$ΔS = \frac{ΔH - ΔG_1}{T_1}$$
Finally,
$$T_\mathrm{eq} = \frac{ΔH}{ΔS} = \frac{ΔH}{ΔH - ΔG_1}T_1 = \frac{\pu{-90 kJ mol-1}}{\pu{-90 kJ mol-1}-(\pu{-45 kJ mol-1})}\cdot\pu{273 K} = \pu{546 K}$$

@andeslisk This might be how the question should have been written, but at least in the form the OP posted, it is written as asking for the minimum temperature at which the reaction is spontaneous, which is a poorly posed question. They either needed to write the question as "find the equilibrium temperature" or "find the max temperature such that the reaction is spontaneous".

@Tyberius I agree that the problem could formulated more clearly. I've seen problems like this when the student was supposed to deduce what exactly is being asked by analyzing what data values are given. Not exactly fair, but the tasks like this also check attentiveness and knowledge of how to apply formulas on-the-fly.


text:
What happens to the first ionization potential when a hydrogen-like atom captures a particle?
This is a textbook problem from Resonance DLPD Physical Chemistry, Page #83:

The mass of a proton is $1836$ times the mass of an electron. If a subatomic particle of mass $207$ times the mass of an electron is captured by the nucleus, what happens to the first ionization potential of H?

My answer is that it may either increase or decrease depending on the charge of the captured particle.
However, the correct answer according to my book is that the ionization potential increases.
How do I arrive at this solution?

Energy of the $n^{th}$ state (simply by Bohr's model) can be found as, $E_n = -  \frac{me^4Z^2}{8 \epsilon_0 ^2 n^2h^2}$. Now you can apply your logic and judge the final answer.

When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as:
$$ \psi_{1s} = \frac{2}{a_o^{3/2}} e^{-r/a_0} \tag{1} $$
the variable $r$ is the distance from the proton, and in the equation for the Bohr radius:
$$ a_0 = \frac{\hbar^2}{me^2} \tag{2} $$
The $m$ is the mass of the electron. If you're interested in some detail this is discussed on the Physics SE in Reduced mass in quantum physics (Hydrogen Atom) but taking into account the motion of the hydrogen atom turns out to be surprisingly simple. We simply define $r$ to be the distance to the centre of mass of the atom, and the mass $m$ then becomes the reduced mass of the electron-proton system:
$$ m = \frac{m_e m_p}{m_e + m_p} \tag{3} $$
If we take the limit of $m_p \to \infty$ then the reduced mass just becomes the electron mass $m_e$, but for finite $m_p$ the reduced mass is less than $m_e$.
Given all this you can now see what the question is getting at. If the proton captures a neutral particle of mass $207m_e$ the effect is to increase the mass of the proton. This increases the mass $m_p$ we have to put in to equation (3) so it increases the reduced mass in equation (2) and hence the wavefunction (1). The end result is that in the equation for the ionisation energy:
$$ I = \frac{me^4}{8 \epsilon_0 ^2 n^2h^2} \tag{4} $$
the reduced mass $m$ is increased slightly so the ionisation energy increases slightly.
A good example of this is to compare the ionisation energies of hydrogen, deuterium and tritium, where the increased nuclear mass of deuterium and tritium increase the ionisation energy by the mechanism discussed above. In fact there is an existing quesion discussing exactly this: How can difference in neutron number cause a difference in ionisation enthalpies?


text:
Distribution coeffecient without concentrations
From what I understand about distribution coefficient is straight from my book — which does not give any practice examples — is that
$$D = \frac{C_\mathrm{A}(\text{ext})}{C_\mathrm{A}(\text{orig})}$$
where $C_\mathrm{A}(\text{ext})$ and $C_\mathrm{A}(\text{orig})$ represent the total concentration of all analyte species present in the two phases regardless of chemical state. Below is a homework problem that I'm try to solve, but having no luck, since I'm not given any concentrations, only weight and volume.

In an extraction experiment, it is found that $\pu{0.0376 g}$ of an analyte are extracted into $\pu{50 mL}$ of solvent from $\pu{150 mL}$ of a water sample. If there was originally $\pu{0.192 g}$ of analyte in this volume of the water sample, what is the distribution coefﬁcient?

The answer for this problem is $0.731\%$, but no matter which way I plug in the numbers that are given do I reach this answer. Any ideas on what I'm missing?

Molar concentration can be expressed via mass $m$ and volume $V$ all right:
$$C_i = \frac{n_i}{V_i} = \frac{m_i}{M_iV_i}$$
Analyte doesn't change during the extraction and its molar mass $M$ remains the same $(M_i = \text{const})$ so the distribution coefficient $D$ can be rewritten as such:
$$D = \frac{C_\mathrm{s}}{C_\mathrm{w}} = \frac{m_\mathrm{s}V_\mathrm{w}}{m_\mathrm{w}V_\mathrm{s}}$$
where "s" refers to the solvent phase and "w" to aqueous phase.
At equilibrium
$$m_\mathrm{w} = m_0 - m_\mathrm{s}$$
where $m_0$ is the initial mass of the analyte.
Finally, the distribution coefficient is
$$D = \frac{m_\mathrm{s}V_\mathrm{w}}{(m_0 - m_\mathrm{s})V_\mathrm{s}} = \frac{\pu{0.0376 g}\cdot\pu{150 mL}}{(\pu{0.192 g} - \pu{0.0376 g})\cdot\pu{50 mL}} = 0.731$$


text:
Acidity comparison between 4-hydroxybenzaldehyde and methanoic acid
Acidity is dependent on charge dispersal. Greater the charge dispersal in conjugate base greater the acidity (assuming that both acids being compared have similar stablities).
So in the case of 4-Hydroxybenzaldehyde, in the conjugate base the charge would delocalise over two oxygen atoms as well as the benzene ring.
Whereas in formic acid charge would only delocalise over two oxygen atoms and hence charge would be concentrated. So the predicted order of acidity according to this logic should be formic acid < 4-hydroxybenzaldehyde.
This is contrary to actual data:
$\mathrm{p}K_\mathrm{a}(\text{4-hydroxybenzaldehyde}) = 7.61$ (Source),
$\mathrm{p}K_\mathrm{a}(\text{formic acid}) = 3.7$
Please put forth the correct logic.

You might want to take a look at similar problems with answers here: https://chemistry.stackexchange.com/questions/6155/why-is-acetic-acid-more-acidic-than-phenol?rq=1

4-Hydroxybenzaldehyde has a much lower $pK_a$ than phenol. Why?

Aldehyde group disperses - charge in conjugate base of said compound.

Formic acid is a carboxylic acid, which is clearly more acidic since the negative charge in the conjugate base is dispersed over two electronegative oxygen atoms.

@TanYongBoon, but so is the negative charge in the conjugate base of 4-hydroxybenzaldehyde. OP: It's not the number of atoms something is delocalised over which determines acidity, but rather how effective that delocalisation is. This is a recurring problem on SE and is explained more thoroughly in Night Writer's first link.


text:
Anti-aromatic or Non-aromatic
Please tell if the 2 given conpound are Anti-aromatic or Non-aromatic,

I have confusions arising due to incosistent facts all around internet some of them are as follow :
1.) Arguments for Cyclopropenyl Anion
a.) Wikipedia which under the heading of "Effects on reactivity"
     clearly states that :

"However, the cyclopropenyl anion has 4 π electrons in a cyclic system
  and in fact has a substantially higher pKa than 1-propene because it
  is antiaromatic and thus destabilized"

b.) This Answer on ChemSE which clearly states that its Non-Aromatic in
        Nature
c.) Also this NEPTL site which clearly states that:

Cyclopropene is not aromatic because one of its ring atoms is sp3
  hybridized so it does not fulfill the criterion for aromaticity. But
  the cyclopropenyl cation is aromatic because it has an uninterrupted
  ring of p-orbital and (4n+2) π-system. The cyclopropenyl anion is
  antiaromatic as it has (4n) π-system.

2.) Arguments for cycloheptatrienyl anion
a.) This Question on ChemSE where the same question have 2 deflecting answer this answer is saying is that the anoin is antiaromatic and this states that the anion is nonaromatic
b.) Also my book (solomons and fryhle organic chemistry adapted for JEE) gives this as Antiaromatic
I don't know which one is correct, please help by providing a comprehensive answer with a reason that can be applied everywhere in such questions.
EDIT: Hello the one who closed the question can I ask the reason for closing as much I can do I abid by this guideline I asked my question straigthformward and then presented my research please tell me why my question is put on hold, I seek answers

Basically the SE answer you mention should fiv your current Q as well. Given the existance of a species that results in antiaromaticuty, then ist that the species avoid it. The key is that antiaromaticity is forced, it serves to determine if a formally conjugated system does indeed undergo delocalosation, not to establish its existence.

Well, in the past antiaromaticity was attributed to about as  many structures as aromaticity, thing is it pretty much doesn't work at all.

https://chemistry.stackexchange.com/questions/45040/why-cant-antiaromatic-compounds-just-escape-planarity-and-become-non-aromatic-t

@Mithoron  Can I know the reason why you all voted to close this question so that I can improve according to me I followed all the guidline given to me   Search, and research
Have you thoroughly searched for an answer before asking your question? Sharing your research helps everyone. Tell us what you found and why it didn’t meet your needs. This demonstrates that you’ve taken the time to try to help yourself, it saves us from reiterating obvious answers, and above all, it helps you get a more specific and relevant answer!  - did that

It would be better to just make a post about when antiaromaticity actually applies, but as you noticed, you could get pretty much any answer, ranging from not applying to anything to everything one might suspect. BTW it's technically a follow-up of two old posts, but such question shouldn't just repeat old question.

@Mithoron I don't know where antiaromaticity actually apply hence I asked this question I found these compound as an example to hit a broarder topic actually the old post and other refrences on the intenet were incosistent with the answer thus I asked this question to clarify my concepts

Both of them are antiaromatic as both of them has 4n π electrons.Also,I'm going to suggest you that if have to choose between internet and books,you should choose books as reference because there are a lot of people on the internet who might not know everything but nonetheless they post about that topic i.e. there's no one to verify all the stuff on the internet but the books are genuine.

@AbnerAlfredThompson Well, if you just want to repeat what people half a century thought... Also it's funny for you to say it here... To say if a compound is truly aromatic of antiaromatic one need strong experimental data, like presence of ring current. Blind trust in textbooks is stupid, and even scientific papers are often discredited.

I agree with you Mithoron but read carefully what I've stated.I'm saying if I had to choose between any random website and say Paula Bruce or Solomon,I'll go with the latter.All books will have correction someday as new concepts will arise but as of now,books can be considered more reliable than any random website.

Whenever you want a definition of a chemical concept, you should refer to the IUPAC's Gold Book which in this case states that antiaromatic compounds are

compounds that contain  4 n(n ≠ 0) π-electrons in a cyclic planar, or nearly planar, system of alternating single and double bonds, e.g. cyclobuta-1,3-diene.

A corollary to this rule is that all the atoms must have an sp2-hybridization, so the orbitals can overlap. However, the carbon bearing the negative charge has an sp3-hybridization, i.e. they does not count as π-electrons but σ-electrons and that breaks the rule of alternating single and double bonds. So both anions are non-aromatic.
Edited after the OP opened my eyes on a big mistake I had made in answering his question. Sorry about that.

as i know that alternating single and double bond means conjugation but as https://chemistry.stackexchange.com/a/51470/73826 this answer suggest the carbanion moves out of plane and hence can't remain in conjugation ......please clear this in the answer

I dont know if I am wrong that some ring with a -ve charge are aromatic in which the electrons are considered as $\pi$ electron and thus the congugation is present , can you explain why they aren't considered as $\pi$ electrons here ???

Oh right, I get it. You actually wonder why the cyclopentadienyl anion is aromatic, and the anion center is sp2 in that case, whereas it is not in the case with cycloheptadienyl anion. Well hybridization occurs only if it helps getting a lower potential energy or, in other terms, if there is an incentive to do so. There is an incentive to change hybridization to go aromatic, but not to go antiaromatic.

oh...Nice I think I got your point thanks for the answer


text:
Electric conductivity
Well I was wondering why some materials have a higher resistivity than others . So I took two elements randomly . I picked carbon and silicon . Carbon has lower resistivity than silicon and obviously less atomic radius . Carbon single bond with another atom of carbon is stronger than the single bond of a silicon atom with another silicon atom . So the bond length of the C-C bond is less than the bond length of Si-Si . Both Carbon and Silicon form a lattice . How is this possible?
Why is Carbon a better conductor than Silicon . Does it have to do with any of the chemical properties I wrote down?

If you compare Diamond and Silicon, then the conductivity of both is poor. Both use all four bonds to connect to the next atoms.
 But if you consider Graphite, then its very well. The reason is the structure of the molecule. It has a big net of a system of aromatic benzene rings . Here only 3 bonds are used to connect to another C-atoms. The fourth electron is the sp2 orbitale can act to the others and transfer electrons, if a electrical field is used.


text:
Calculating the strength of an ionic bond that contains poly-atomic ions
So the bond association enthalpy for ionic compounds like $\ce{NaCl}$ and $\ce{NaBr}$ can be easily calculated from a Born-Haber cycle. But the way a Born-Haber cycle is constructed it uses info that only really applies to mono-atomic ions like $\ce{Cl^-}$. So how would one calculate the strength of an ionic bond for ionic salts that contain poly-atomic ions like $\ce{NaOH}$ and $\ce{K_2SO_4}$. Is there a completely different method, or can a Born-Haber cycle be adapted to use a poly-atomic ion?

With enough effort, Born–Haber cycle can be extended to polyatomic ionic solids, however it's practically never done in practice due to the lack of experimental data or because it's impossible to obtain any.
From [1, p. 117–118] (emphasis mine):

Lattice energies cannot be measured experimentally since they represent hypothetical processes:
$$\ce{M^n+(g) + X^n-(g) → MX(s)}$$
However, the following reaction sequence, relating the heat of formation, $ΔH_\mathrm{f}$ of a crystal $[\ce{M(s) + 1/2 X2(g) → MX(s)}]$ to $U[\ce{M+(g) → MX(s)}]$ is thermochemically equivalent (and $ΔH_\mathrm{f}$ can be measured).
$$
\begin{array}{ccc}
\ce{&M^0(g) &+ &X^0(g) &→ &M+(g) &+ X-(g)} \\
&↑\small ΔH_\mathrm{s} &&↑\small D &&↓ & \\
\ce{&M(s) &+ &1/2 X2(g) &→ &MX(s)}
\end{array}
$$
In this diagram, $ΔH_\mathrm{s}^\circ$ gives the enthalpy of sublimation of the metal $[\ce{M(s) → M^0(g)}],$ $D$ gives the dissociation energy, or bond energy of the diatomic gas $[\ce{1/2 X2(g) → X^0(g)}],$ $IE$ gives the ionization energy of the gaseous metal $[\ce{M^0(g) → M+(g)}],$ and $EA$ gives
  the electron afﬁnity for the formation of the gaseous anion $[\ce{X^0(g) → X-(g)}].$
  The lattice energy is obtained through the relation:
$$U = ΔH_\mathrm{f} - \left(ΔH_\mathrm{s} + \frac{1}{2}D + IE + EA\right)$$
One difﬁculty with using a Born–Haber cycle to ﬁnd values for $U$ is that heats of formation data are often unavailable.
  Perhaps the greatest limitation, however, is that electron afﬁnities for multiply-charged anions (e.g. $\ce{O^2-}$) or polyanions (e.g. $\ce{SiO4^4-}$) cannot be experimentally obtained.
  Such anions simply do not exist as gaseous species.
  No atom has a positive second electron affinity; energy must be added to a negatively charged gaseous species in order for it to accommodate additional electrons.
  In some cases, thermochemical estimates for second and third electron afﬁnities are available from ab initio calculations.
  Even so, if there are large covalent forces in the crystal, poor agreement between the values of $U$ obtained from a Born–Haber cycle and Madelung calculations can be expected.

Already for some simple species such as oxoanion it is impossible to determine their electron affinity $EA$ in gaseous phase:
$$
\begin{align}
\ce{O^0 (g) + e- &→ O^- (g)}\label{rxn:R1}\tag{R1} \\
\ce{O- (g) + e- &→ O^2- (g)}\label{rxn:R2}\tag{R2} \\
\hline
\ce{O^0 (g) + 2 e- &→ O^2- (g)}\label{rxn:R3}\tag{R3}
\end{align}
$$
Enthalpy of reaction \eqref{rxn:R2} cannot be determined experimentally and can only be derived from the lattice energy $U$, which causes Catch 22 situation[2, p. 61].
References

Lalena, J. N.; Cleary, D. A. Principles of Inorganic Materials Design, 2nd ed.; John Wiley: Hoboken, N.J, 2010. ISBN 978-0-470-40403-4.
Smart, L.; Moore, E. Solid State Chemistry: An Introduction, 4th ed.; CRC Press: Boca Raton, FL, 2012. ISBN 978-1-4398-4792-3.

Would you possibly be able to give an example? and would you be able to define all of the letters used in the equation.

@H.Linkhorn My bad, the equation indeed has been taken out of context.

If you truly believe that the ionic model is a good representation of the species of interest, given a structure you can calculate it analytically, or on a computer with an appropriate piece of software. Madelung energies and Ewald sums are the magic phrases here, you might like to look at
https://en.wikipedia.org/wiki/Madelung_constant
https://en.wikipedia.org/wiki/Ewald_summation
and the nice lattice energy calculator at
https://scilearn.sydney.edu.au/fychemistry/calculators/lattice_energy.shtml
These methods can fairly straightforwardly be extended to use point multipoles, which model distortions of the point charge, but eventually you probably would want to use a full ab initio electronic structure code, many of which can address periodic substances at at least the DFT level, and some allow potentially more accurate wave function methods as well. For "ionic" substances with small unit cells such calculations are usually routine.


text:
Why can we know a materials properties by knowing its ground state at 0K - DFT
I have had this doubt ever since I was introduced to DFT. For what I gathered whatever results you obtained with DFT (by using VASP, quantum expresso or any other software) you can say those properties are it, even for temperatures beyond 0K.
I don't understand why. For instance if DFT says it's ferromagnetic at 0K then why would you assume its ferromagnetic properties would stay sufficiently close to the ground state for T>0 as to state that.
I haven't read a theorem that says that excited states are close to the ground state. Does such theorem exist?
Thanks!

Edit:
Quoting the book "A chemist's Guide to DFT" at the end of the section about the first Hohenberg-Kohn theorem **

One should note at this point that the ground state density uniquely
  determines the Hamilton operator, which characterizes all states of
  the system, ground and excited. Thus, all properties of all states are
  formally determined by the ground state density (even though we would
  need  functionals other than ∫ρ(r)V dr + F [ρ], which is the
  functional constructed to deliver E0 but not properties of
  electronically excited states).

I don't follow it. Seems pretty basic cuz everyone just brushes over but I don't follow it. Why would the ground state characterize excited states?
** The first theorem basically says that the hamiltonian can be defined by the electronic density and can be made a unique functional of it.

Magnetic properties and such are mostly physics, not chemistry. When one says it is a ferromagnetic ground state, everyone knows she/he talks about the ground state. Chemically relevant electronic excitations are generally high energy, and when we expect low energy ones, generally you explicitly check those.

It was as an example, I have the same question for the band gap or the resulting structure. As far as I have heard, the answer has to do more with chemistry than physics. Where can I read somewhere that says "Chemically relevant electronic excitations are generally high energy", is that the case if the material is metallic or a semiconductor?

Metals and semiconductor band gaps are primarily physics / materials science problems, again. Molecular structures are generally OK with vibrational ground state, and isomerism is generally be described by just another vibrational ground state.  You don’t use DFT with 0K for metals, anyways, neither you have many chemists dealing with simulation of metallic systems. “I have heard” sorry, ask them then.

I think you are asking two different things in your first and second part of your question.

I might be mistaken but I think they are the same. All the properties of the system are defined by the electronic density, so if the properties of the material can be extrapolated to temperatures beyond 0K, then the reason why has to do with the electronic density's ground state. I am coming here for help in understanding this, so I could be wrong.

I think you are confusing two things: 1) determined as "I can use the same numbers / I have calculated". The first part of your question is this, ie. if a ground state property calculated, why does it apply to everything else automatically. 2) determined as "unambiguous". Your DFT quotation is in the second sense: the Hamiltonian of the system is uniquely tied to the geometry and generates both ground state and excited state wavefunctions. Since ground state electron density is unique, too, theoretically, it should contain all the info about excited states.

If we know that the configuration is going to change continuously when we increase the temperature, then we can talk about the state being close to the state right before for incremental points in temperature. So the state right above the ground state will have a configuration quite close to the ground.
For example, ferromagnetism at 0K basically means that the magnetic moment of most atoms will align parallel to each other at the ground state. Even though a single atom may transition to drastically different state when increasing temperature, the system as a whole will be guided by Maxwell-Boltzmann statistics and thus will have a continuous transition from one configuration to the next.
Consider a system of atoms in the FM state at 0K. As we increase the temperature, the atoms will have higher energy and vibrate or rotate. But we know that they prefer to align parallelly, and thus each atom will affect the next and try to align them parallel. Even though some pairs or groups will align antiparallel or haphazardly, the previous pattern will be kept in most. This will be harder and harder to do as the external energy is increased, so eventually the whole system will turn paramagnetic.
So, we don't really know that the system will remain ferromagnetic at higher temperature. But we do know that most have a preferred alignment at 0K, and Boltzmann will guide us from there. In fact, after a certain point in temperature, all systems will turn paramagnetic. Whatever their states at 0K.
Do note that we are trying to use a parameter which is constant or changes as a well-defined function when we change the temperature. In this case, the alignment (exchange) energy as well as magnetic moment remain constant through the ground state to higher ones.


text:
difference in boiling point of 2-iodopentane and 3-iodopentane
There is no difference in boiling point of 2-fluoropentance and 3-fluoropentane, so is chloro- and bromo-. But why is there a difference in the boiling point of 2-iodopentane and 3-iodopentane, up to 10 degree Celsius?

Well I think that the most intriguing part is that 2-fluoro(chloro, bromo)pentane and 3-fluoro(chloro, bromo)pentane have the same boiling points! That is much more incredible!!

I can only speculate: iodine is larger and more polarizable, resulting in stronger vdW interactions and more steric (entropic) constraints. This leads to significantly different packing (on average) and vdw interactions.

@SteffX I'm pretty sure OP's "no difference" means just smaller difference.

I have spent a quite a time to figure out "so-called" irrational behavior of the boiling points of iodoalkanes compared to relevent othe halpalkane, but I have seen noting really radical, what OP claimed there are. The following chart listed boiling points of all possible long chain halides of pentane (all have been measured at $\pu{760 mmHg}$):
$$
\begin{array}{ccc}
\text{locant} &\text{fluoropentane} & \text{chlororopentane} & \text{bromoropentane} & \text{iodopentane}\\\hline
1 & \pu{62.8 ^\circ C} & \pu{108 ^\circ C} & \pu{130 ^\circ C} & \pu{155 ^\circ C}\\
2 & \pu{58.4 ^\circ C} & \pu{99 ^\circ C} & \pu{117 ^\circ C} & \pu{150 ^\circ C}\\
3 & \pu{59.5 ^\circ C} & \pu{97.5 ^\circ C} & \pu{119 ^\circ C} & \pu{145 ^\circ C}\\
\end{array}
$$
Please be aware that, all of these values are extracted from EPA(USA) website. Also note that values of 2-fluoropentane and 3-fluoropentane are predicted on EPA, but ChemSpider gave experimental values of $\pu{56 ^\circ C}$ and $\pu{56.7 ^\circ C}$, respectively. Also, the boiling point of 3-iodopentane ($\pu{145 ^\circ C}$) has also been predicted, but it is available commercially at Alfa Aesar, which listed the boiling point of it as $\pu{146.6 ^\circ C}$. Thus, these listed values are very reliable. Therefore, I'd say that the values OP has must be not so reliable. I admit that I have to dig deeper to find a reliable source of those predicted values. All three bromo compounds have no problem because all of them are available in Sigma-Aldrich.


text:
Melting point of unsymmetrical dimethylhydrazine vs 1,2-dimethylhydrazine
Unsymmetrical dimethylhydrazine ($\ce{H2NN(CH3)2}$) has a melting point of −57 °C, but its isomer, 1,2-dimethylhydrazine ($\ce{(CH3NH)2}$) has a melting point of −9 °C. That looks to me like a significant difference, so there must be some mechanism which makes these two compounds extremely different in this respect. However, I'm unable to figure out why that difference exists; could you please explain that?
These are the structures of the compounds (from wikimedia commons):
Unsymmetrical dimethylhydrazine:

1,2-dimethylhydrazine:

Here're some of the arguments which I considered:

The packing in crystal structures could be 'different' in a significant way. This could potentially be supported by the fact that the difference between their boiling points is much lower: unsymmetrical dimethylhydrazine has a boiling point of 64 °C while 1,2-dimethylhydrazine's is 87 °C. However, this is an extremely vague argument; it could be strengthened by some details regarding how the different packings contribute to the melting points.
Hydrogen bonding should be a pretty significant factor to consider while analyzing these compounds. I suspected that steric hindrance may allow only one $\ce{N}$ atom for hydrogen bonding in unsymmetrical dimethylhydrazine. However, this would suggest that 1,2-dimethylhydrazine should have a lower melting point, which is clearly not the case.

We had a discussion about these two compounds' melting points in the Physics SE chat a while back, so I thought I'd pop over here and ask.

hint: symmetry.

It is difficult to answer without the exact crystal structure of both compounds (which I didn't even try to get), but the C2-symmetry of 1,2-dimethylhydrazine (1,2-DMH) is really stinking, compared to unsymmetrical dimethylhydrazine. If you consider the plane "H–N–N–H" in 1,2-DMH, it is obvious that close molecule can form a sheet just by hydrogen-bonds. It is not the alpha and omega of the crystal structure, but I am not surprised at all by the melting points you gave.

According to the Carnelley’s Rule:

That of two or more isomeric compounds, those whose
  atoms are the more symmetrically and the more compactly
  arranged melt higher than those in which the atomic
  arrangement is asymmetrical or in the form of long chains

If we attend to the structure of the 1,1-dimethylhydrazine, one can see that the steric hindrance due to the methyls groups inhibits the formation of H-bond in the central nitrogen atom for this isomer, in contrast to the 1,2-dimethylhydrazine which both nitrogen atoms are capable to form H-bonding.
For the thermodynamics point of view, the melting point of a solid is related to the enthalpy and entropy of melting by the following expression:
$$T_{\text{m}}=\frac{\Delta_{\text{m}} H}{\Delta_{\text{m}}S}$$
where the subscript "$\text{m}$" stands for "melting". Is difficult to estimate the impact of both enthalpy and entropy without further information than the molecular structure, but in most of the cases the process of melting is driving by enthalpic factors; therefore if the symmetric molecule can make more H-bonds than the unsymmetrical it's reasonable to assume that the symmetric molecule has a higher enthalpy of fusion and consequently a higher melting temperature provided the entropic effects are ignored.
References and further reading:
Effect of molecular symmetry on melting temperature and solubility
Melting Point and Molecular Symmetry
Carnelley’s Rule and the Prediction of Melting Point

Melting points of related compounds: hydrazine, 2 °C; monomethylhydrazine, −52 °C (Wikipedia).
Three of the four compounds have an $\ce{NH2}$ group, so this is not a deciding factor. Hydrogen bonding may be present without being a factor in the different melting points.
The common feature in the two high-melting compounds is a symmetrical N-N bond; an unsymmetrical bond is common to both low-melting compounds.
Hydrocarbons can display a similar trend in melting points without significant hydrogen bonding influences: n-pentane ($\ce{CH3CH2CH2CH2CH3}$) melts at −130 °C while the extremely symmetrical neopentane ($\ce{C(CH3)4}$) melts at −16.5 °C.


text:
Finding major product when bromonium compound is acted upon by a nucleophile
The original question is:-

The major product obtained in the following conversion is:-

(Note that Me here means methyl group)
The initial step would be the formation of a bromonium intermediate. Which would then be attacked by a nucleophile $\ce{CH_3OH}$ as it is present in excess in solvent.
What I think is, $\ce{CH_3OH}$ would attack the carbon atom close to ketone, because that’s the carbon best able to stabilize positive charge due to resonance with keto group. This is what I think will happen

But the proposed answer was

Can anyone explain me why how and why this compound is formed?

https://chemistry.meta.stackexchange.com/questions/3044/hidden-points-of-editing-you-probably-didnt-know?r=SearchResults&s=1|27.6568

The bromonium ion may be attacked by the non-carbonyl oxygen of the acetate forming a 6-membered ring. The 6-membered ring is then opened by SN2 displacement with methanol. Basically, a neighboring group effect.

I'd like to point out that the ketone would actually disfavor an attack at its alpha position:

Placing two positive charges on adjacent carbon atoms is much less favorable than not, especially when there is an oxygen perfectly poised to help stabilize this positive charge, as user55119 pointed out.

That makes sense.... Will methanol attack the same carbon(carbon atom away from keto group) if acetate is absent?

Yes, it should.

I like the ringo's point, but literature evidence does not support the classic cyclic bromonium ion mechanism. Effects of a carbonyl group on the ring opening of a neighboring bromonium ion in some bromination of reactions of $\alpha,\beta$-unsaturated ketones have been studied. The authors have used methyl acrylate and methyl crotonates (cis- and trans-isomers) in this study. In addition to halogens (used in preliminary studied), they have mainly used brominr chloride ($\ce{BrCl}$) for the electrophilic addition under ionic and radical conditions (Ref.1). For example, under ionic conditions, addition of $\ce{BrCl}$ to methyl acrylate ($\ce{CH2=CHC(O)OCH3}$) gave primarily anti-Markownikov product ($\beta$-attack). The products ratio is 83% $\ce{CH2(Cl)CH(Br)C(O)OCH3}$ (beta attack) and 17% $\ce{CH2(Br)CH(Cl)C(O)OCH3}$ ($\alpha$-attack). However, under radical conditions (ultraviolet irradiation), methyl acrylate has yielded 100% of $\ce{CH2(Br)CH(Cl)C(O)OCH3}$.
Then, during another project on aldehyde and ketones, directed by the same principal investigator (Ref.2), it was found that some simple $\alpha,\beta$-unsaturated aldehydes and ketones do not react with halogens by the expected attack on the $\ce{C=C}$ bond. They proposed a mechanism, which agrees with the  formation of products obtained. The simplest version using $\ce{BrCl}$ to show the $\beta$-attack is depicted in Scheme 1:

Based on above suggested mechanism and Ringo's prediction, following mechanism is depicted (Scheme 2):

References:

V. L. Heasley, D. W. Spaite, D. F. Shellhamer, G. E. Heasley, “Considerations on the Effect of a Carbonyl Group on the Ring Opening of a Neighboring Bromonium Ion,” J. Org. Chem. 1979, 44(15), 2608-2611 (DOI: 10.1021/jo01329a003).
V. L. Heasley, D. F. Shellhamer, T. L. Carter, D. E. Gipe, R. K. Gipe, R. C. Green, J. Hordeen, T. D. Rempel, D. H. Spaite, G. E. Heasley, “Reaction of halogens with some $\alpha,\beta$-unsaturated aldehydes and ketones,” Tetrahedron Letters 1981, 22(26), 2467-2470 (https://doi.org/10.1016/S0040-4039(01)92934-X).


text:
An aqueous solution of NaCl has a mole fraction of 0.21. What is the mass of NaCl dissolved in 100.0 mL of solution?
Work
mole fraction of NaCl = 0.21/0.21+0.79
If there is 1 mol of total solution, there will be 0.21 mol NaCl and 0.79 mol H2O
0.21 mol NaCl = 12.27 g
0.79 mol H2O = 14.23 g
this would yield a ratio of 12.27 : 14.23 inside the solution, giving 46.26 as the number of grams of NaCl inside 100 ml of solution.
Problem
I've looked around google for answers to this problem and many of them come up with 86.4 grams of NaCl in the solution. I feel like my answer is logical, but I'm not sure. Could you explain the mistakes I made in my work, if any?

hint: Look up the density of water in g/mL

The concentration you have calculated is % ($w/w$). In other words, what that means is mass of solute in $\pu{100 g}$ of solution. In your case, it is mass of $\ce{NaCl}$ in $\pu{100 g}$ of solution: $\frac{12.27}{(12.27+14.23)}\times 100=46.3$%. However, in your statement, you have declared it is  % ($w/v$), which is not.
the others's answer is also not quite correct, because they have assumed when $\ce{NaCl}$ is dissolved in certain volume of water, the volume stays same (won't change). If you are allowed to assume that,  then, you can conclude (like others) that when $\pu{12.27 g}$ of $\ce{NaCl}$ is dissolved in $\pu{14.23 g}$ of $\ce{H2O}$ (or $\pu{14.23 mL}$ of $\ce{H2O}$ assuming density of water is $\pu{1.00 g/mL}$), the volume of solution is still $\pu{14.23 mL}$. Based on allowed assumption, the % $(w/v)$ concentration of your solution is: $\frac{12.27}{14.23}\times 100=86.2$% (the value close enough to your internet finding).
However, most important data missing in your question is the density ($\rho$) of the sought $\ce{NaCl}$ solution. That would give exactly the volume ($v$) of the solution since $v=m/\rho$. Thus, correct % $(w/v)$ concentration of your solution is: $\left(\frac{12.27}{\rho(12.27+14.23)}\times 100\right)$%.

Problem:

An aqueous solution of NaCl has a mole fraction of 0.21. What is the mass of NaCl dissolved in 100.0 mL of solution?

Frankly this is a stupid problem as stated. First a solution with that much $\ce{NaCl}$ must have a density significantly greater than 1 g/ml. So there isn't any way to solve the problem with the information given. You need a density or something else. Second, I doubt that much $\ce{NaCl}$ can be dissolved in water.
In its Solubility table Wikipedia gives that the solubility of $\ce{NaCl}$ is 38.99 grams of $\ce{NaCl}$ per 100 grams of water. MW of $\ce{NaCl}$ is 58.443 g/mol and for water is 18.02 g/mol. So 38.99 grams of $\ce{NaCl}$ is 38.99/58.443 = 0.667 moles. A 100 grams of water is 100/18.02 = 5.549 moles.
0.667/(0.667+5.549) = 10.7%
So the greatest mole fraction that a sodium chloride solution can have is 10.7%.

I cant understand your question but I will tell you some things that may be helpful to you .
Salts are fully dissolved in water so what you say that there will be a few moles of NaCl remaining is a mistake.Find the concetration of NaCl in your aqueous solution . I assume that your solution isn't dense so you must look for concetrations below 1 Molarity. 1 Molarity is 1 mole/ 1L  = 1 mole/1000ml . Divide by 10 since you said that your solution has volume of 100mL and then you will find the moles. The total mass of NaCl in your solution is Molecular Mass of NaCl*Moles


text:
Why doesn't the current stop in neon lamps when all the atoms have been ionized?
My personal experience (which could be wrong) tells me neon lamps can theoretically run on forever or at least a very long time. If current is the flow of charge why doesn't it stop once all the atoms have been ionized? Do these atoms they continuously loose and gain electrons? If so why then doesn't the light concentrate to the side of the cathode were all the ions attract to?

Yes, the neon atoms lose then gain electrons.

Hint: The light does not come from ionization, but from recombination.

It is worth noting that, even if the ions concentrated at one electrode, most lamps are not DC they are AC so the process would be reversed 50 or 60 times per second. But that isn't what is happening anyway.

If current is the flow of charge why doesn't it stop once all the atoms have been ionized?

The short answer is that it is impossible to ionize all the gas at once. It just requires huge amount of energy to ionize the whole atoms at once. Just for the sake of discussion, even at 10000 °C, still a very small fraction of atoms is ionized! This temperature is hotter than the surface of the Sun! A neon discharge tube is not even close to those temperatures or let us say energies.
The red glow of neon tubes you see is mainly due to electronically excited neon atoms. The pressure of the gas is pretty low inside the tube. Once you start applying a voltage, electrons emit from the cathode and ionize the gas forming a a number neon ions. These ions and free electrons are responsible for carrying the current. Since you are constantly supplying current, it continues to flow through the discharge tube and the process of ionization and recombination continues. Remember that ionization is an equilibrium process in the gas phase as well. An additional process can also occur, that the electron is promoted to a higher energy orbital instead of leaving the atom. Once it jumps back to its original orbit, it emits its characteristic wavelength of light.
I think the recombination process stated above should yield a continuous spectrum so the red characteristic glow of neon tubes is not from recombination process.

Why is the current flowing through the discharge pipes and why does the ionization process not automatically stop as long as the required charges are available ?
The answer to this question lies in what is known as belt theory or transition bands. This theory has been developed to explain the material's ability to conduct conductivity in metallic materials as well as semiconductors and insulating materials.
This theory assumes that what is known as the free electrons that are responsible for our conventional electrical conduction are originally equal electrons - the electrons that the metal prepares to lose in return for its arrival at conventional chemical stability - and these electrons can not move from their level as simply as they need energy. It can get out of those levels ....... In the case of metals, the required energy is limited and small amount and that is why some heating enough even the usual temperature of the atmosphere to provide these electrons some energy to move from the levels of corn to become Some metals, such as cesium and potassium, have energy at lower levels, such as light energy, until these electrons are released. The energy of the electric field can move these electrons, so it is easier for metals to transport electrons  once a difference is made Effort on both ends of the metal piece.
If the natural region of the valence electrons is located in a layer within the atom, the conduction electrons that have acquired energy are slightly higher than the valence electrons because of this energy. Therefore, the valence electrons region is said to almost interfere with the conductivity of the metals Ready to connect with the energy you gain from its surroundings and light.
In the case of insulating materials there exists what can be called a similar equivalence of electrons, but they are strongly related to the atoms or particles in the insulating material. In order to make the conductivity possible, these electrons must be lifted to the free zone or the conductivity range. In this case, the beam theory states that the difference in energy Between the two bands is large and therefore there must be a vacuum or a large gap between the range of equivalence and the range of conductivity so the insulation can not easily connect the electricity and requires a lot of energy to make them do so and can be provided through extreme heating, for example, It is known that each material dielectric breakdown voltage electrical own.
In the case of semiconductors, the difference in energy or gap between the valence and conductivity ranges is limited and small. Therefore, modest energy is required to move to the conductivity state - for example, a pure silicon plate that can not easily plug in the current but some heating is sufficient to do so.
In the case of gaseous materials such as in the discharge pipes, we put a very high voltage on the gas to make it move from the state of isolation to the state of conductivity and the electric field to make energy on the electrons to transfer them to the delivery area beyond the gap, and once this happens, atoms break into an active mixture Of the ions with the electrons and with the stability of the dissociation power or the difference in voltage, they remain unchanged without ever equalizing because the energy allowed to bypass the conduction gap and forced the atoms to disintegrate but once the source of the voltage is removed the electrons return to the equivalence level and lose their energy in the form of light Equalized atoms to return to the buffer status.
If the plasma mixture inside the tubes can not be automatically drawn unless the source of the electrical voltage or the ionizing heat is removed.

How does the light come?


text:
Do species with more unpaired electrons attract magnets more strongly than those with fewer unpaired electrons?
My lecture notes state that you can distinguish between high spin and low spin complexes by their magnetic properties.
This thought came to me: consider two octahedral crystal fields with a d7 electronic configuration, one in the high spin state and the other in the low spin state. The complex in the high spin state has two unpaired electrons, whereas the complex in the low spin state has 1 unpaired electron. Both complexes are therefore paramagnetic and hence attract magnets. Does the former attract magnets more strongly as it has more unpaired electrons?

Your logic is correct. See this link: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Magnetic_Properties


text:
How to index molecules in computer programs?
Is there a standard way how can the index over a molecular structure be kept in the computer?
A molecule is a graph, where nodes have labels (chemical element type), and edges are chemical bonds.
There is SMILES representation that determines the molecular structure, but SMILES representation isn't unique.
One can take the set of all possible SMILES representations for one molecule, sort them alphanumerically, and pick the first element. This would be a unique string that is in one-to-one relationship with a molecule, and would satisfy my request.
But is there a standard way how to have a value that is in one-to-one relationship with a molecule that can be used as an index in the database that has many molecules in it?
By extension, how to efficiently search a set of molecules and, given a particular molecule, find a set of the closest structural matches? What characteristics of the molecules should be memorized that would allow such search to be efficient?

You might want to flip through CCDC ConQuest manual and refer to CSD's Python API. There is no quick and dirty how-to's.

@andselisk This document describes how to use the system that does something like this, but not how it does it.

A molecule is a graph, where nodes have labels (chemical element type), and edges are chemical bonds.

while fairly common and acceptable for an initial approach, this models fails to catch many types of isomers and tautomers. There is no well-established general approach because the basic definitions are kinda wonky and the subjects itself is extremely complex. For example, monomethylbulvalene would for all intents and purposes work as a single compound yet a set of different molecules would be present in all reasonably analyzed samples.

By extension, how to efficiently search a set of molecules and, given a particular molecule, find a set of the closest structural matches?

That's a part of QSAR big problem: to identify "similar" molecules. It doesn't work reliably.

What characteristics of the molecules should be memorized that would allow such search to be efficient?

There are specialized graph databased that aim to solve the problem of efficient search on graph-like structures, but see above.

You might want to look at
W.-D. Ihlenfeldt, J. Gasteiger, “Hash Codes for the Identification and Classification
of Molecular Structure Elements”, J. Comput. Chem. 1994, 15, 793–813.
with the reference implementation being available in the CACTVS toolkit available at https://www.xemistry.com/academic/
These hash codes allow for the identification of organic molecules independent from numbering schemes including stereochemistry.
They do not comprise any similarity information, rather even similar molecules exhibit distinct hash codes.

There are multiple ways to do this:

As you mention, SMILES strings - it is possible to have 'canonical' SMILES
InChI - another standard for representing structures in a canonical way
Various numbering systems (CAS Registry Number, etc)

As you point out, one way to get a unique identifier is to make some line notation (like SMILES) and then sort all the representations for a chemical and pick the alphabetically 'largest'.
The difficulty with that approach is that some chemicals are highly symmetric. For example, the fullerenes - although, realistically most chemicals are asymmetric (as are most graphs). The more symmetries a structure has, the more different ways there are to write it in some line notation.
Practically speaking, to efficiently search a database of structures for a query it makes sense to use techniques like 'fingerprints' to filter candidates. Simply string matching the canonical form against the database will work, of course.

No doubt, CAS numbers are handy to use.  However, note potential difficulties with CAS registry numbers: except the last figure (a check digit) it basically is a book keeping owned by The American Chemical Society.  Other than SMILES or InChI, it is not structure related -- it is impossible to compute it (e.g. openbabel) with a structure in hand.  Using CAS numbers, the OP would need to keep a dictionary of CAS RNs, including instances where multiple CAS numbers point to the same molecular structure and settle an agreement with ACS to use such a dictionary, too (like wikipedia did).

Related notes on InChI: Pubchem, InChI, SMILES, and uniqueness and Structure that breaks InChI


text:
What is the expected number of rings obtained in the following ring formation process?
Refluxing, say, one mole of 1,5-dibromo-pentane with excess ammonia in ethanol can result in the formation of rings. (e.g. $1\lambda^2$-piperidine)

However, it is also possible for the 1,5-dibromo-pentane molecules to be connected together, forming longer chains without forming rings. The longer chains can be reacted again to form larger rings. So in theory, we can have rings that are arbitrarily large.
This is about the rings' size distribution
Question: What is the expected number of rings (large and small) after the reaction finishes? How many of the rings are $1\lambda^2$-piperidine? Generally, is there a mathematical distribution that describes the expected number of rings of any given size?

Common teaching is that the formation of rings greater than, say, 8 members is difficult. It usually requires templating (internal or external to the molecule). You are basically asking about ring polymers and their size distribution.

@TAR86 Could you explain more about it or tell me where can i learn more？

Do you know what $1\lambda^2$-piperidine means and why you're getting that? ChemDraw circled the nitrogen atom in red for a reason..

There is no simple answer here because the outcome will be heavily dependent on the concentration and on the rate of addition.
The most salient point in your question to me is: "excess ammonia."
You can ask what the competing reactions are.

1,5-dibromopentane reacts with ammonia to form 1-amino-5-bromopentane
1-amino-5-bromopentane reacts with itself to form piperidine
1-amino-5-bromopentane reacts with ammonia to form 1,5-diaminopentane
1-amino-5-bromopentane reacts with another straight chain alkyl amine to elongate the chain
1,5-dibromopentane reacts with piperidine to form N-(5-bromopentyl)piperidine
and others...

The main problem I see is that in excess ammonia, unless you perform a slow addition of ammonia, the concentration may be such that the ammonia will over add via (3). Then you're going to get any product. The intermediates are also reactive, so the reactant concentration has to be relatively low as well, otherwise, you may see (4) or (5) or other side reactions. You may also need to run the reaction at lower temperature to disfavor reaction of the product with any alkyl bromide intermediate or reactant.
These issues need to resolved before you even begin to think about distribution of ring sizes because you might not even form any rings.


text:
Illustrator to chemdraw
I am not sure that this is the right place for this question, but I'll try.
I am working on a scheme that I am developing on illustrator. I have a lot of chemical structure and plus I have some vectorial pictures included.
The thing is that I would like to save/export my illustrator file in a way that I can reopen it and modify it in chemdraw. I am aware that the opposite procedure is possible.
Is there any way of doing that?

I doubt that. Better keep your chemical part in ChemDraw and paste from it when ready.

I'm voting to close this question as off-topic because it's a software issue.

From p. 327 of ChemDraw 18 User Guide:

Importing
You can import graphics or documents from other applications into ChemDraw.
Files
  Files that you insert become part of the drawing. You can edit a file only if it is in a ChemDraw format.
  To insert a file:

Navigate to Edit → Insert File.
In the Open dialog, select the file type from the drop-down list.
Select the file and click Open. The file is embedded in the drawing.

Further, there is a table of file formats ChemDraw is capable of importing and exporting, from which it is evident that only the following graphical formats can be imported:

Bitmap (BMP)
Graphic Image Format (GIF)
JPEG (JPG, JPEG)
Portable Network Graphics (PNG)
Windows Metafile (EMF, WMF)
Tagged Image File (TIF, TIFF)

PDF or PostScript files can be attached as objects, but I suppose it's not what you are after.
Either way, it looks like there is no way to add AI files directly, so you need to convert them first to either raster (e.g. TIFF or PNG) or vector (WMF) format.
I agree with @Ivan that crafting structures in ChemDraw in the first place and then post-processing them with Adobe Illustrator is arguably a better workflow.

Yes, we can export or import files in certain format from ChemDraw and to ChemDraw easily. But my understanding was, those files imported to ChemDraw cannot be changed except you can add more ChemDraw figures etc. because imported parts appears as a picture. For example, if you import a NMR spectrum, then you can add the structure and peak assignments, etc. bur cant change the appearence of spectrum. After add all you want, then you can save it in cdx or other formats in your answer. That's all I know. :-)


text:
Instead of using an Ag/AgCl electrode, can I simply use another beaker and connect it to my solution via a salt bridge?
I'm trying to perform a potentiometric titration as part of my high school laboratory work, by reducing iron(III) to iron(II), and measuring the voltage change as I add cobalt(II) (to reduce it).
The methods that I have found online all point to using an $\ce{Ag/AgCl}$ or calomel reference electrode and an inert sensing electrode. When I look at the diagram, both the sensing and reference electrode are immersed in the solution of iron(III). Currently, these electrodes are too expensive and my school does not have access to it.
Can I just fill a separate beaker with $\ce{KCl}$ and put my $\ce{Ag/AgCl}$ wire into it, and connect the 2 solutions using a salt bridge? Or do I need to submerge both electrodes in the same solution?

Do you have a Pt wire to be used as inert?

Hi, yes I do have that as the inert electrode

Then you can try to use them at least for demonstration plunging directly both. I am used to quasi-reference electrode just in non aqueous media. Give it a try and look for its stability. Using a separate beaker would be the same and you won't have the electrode at the right place. But you can use also the bridge so to have chloride around it. I would look at both setup then choose what gives you the better curve. Simply, at first, look for the stability of whatever potential reading, without titration.


text:
Confusion about orienting effects in o-chlorotoluene
I'm confused about where the next substituent in the electrophilic substitution of o-chlorotoluene will land. My question comes on the background of having 3 different sources of information that say different things.
The first one is my teacher: she taught us that at the nitration of ortho-chlorotoluene, the $\ce{NO2}$ group will be placed in the 4 position, giving 2-chloro-4-nitro-toluene. This seems reasonable, but even more reasonable seems the second source. That is this website, which shows that the product will be 2-chloro-5-nitrotoluene (3rd image, example 2).
The 3rd source is completely confusing, giving little citation as I was not able to find the material cited online. I'm talking about a question on this website here, but I am not sure if I can trust it, that is the reason why I'm adding another question.
So, can someone please explain me what is the actual product?

As I answered on the question you cited, when the experiment was carried out a mixture was obtained. The directing effects of Me and Cl are of similar magnitude so neither prevails, hence a mixture. If you want further discussion I suggest you direct your teacher to the original literature reference. Theory is all very well, but there is no substitute for the experience of someone who has run the reaction.

Thank you! I understand now, this also explains why the theory in the sources varies.

Also, do you have a link to the material you cited in that question? I wasn't able to find it.

@StefanOctavian I added the URL to the answer, but it looks like it's a 80-page article written in French(?). Good luck:)

In such cases, I always remain cautious about how people really numbered the positions around the benzene ring, because if you consider that the methyl group of toluene is on position 1 (because toluene is the parent), that is very different with considering that the halogen atom is on position 1 (because halogen has priority). And remember that the IUPAC rules have not always existed!


text:
How does refrigerator actually work?
In refrigerator, heat energy got extracted from cold sink and dumped into hot source. This is not spontaneous so work must be done. I came across the following while reading a textbook:

For refrigeration, if no work is done,
   ΔS= qh/Th+qc/Tc;
-qh=qc, implying ΔS= qh/Th-qh/Tc =qh(1/Th-1/Tc)
qh>0, 1/Th-1/Tc<0, so ΔS<0 so not spontaneous. Therefore, we need to add heat energy to the stream of heat energy flowing from the cold sink to hot source, so that the entropy increase at hot source 'covers up' the decrease in entropy at cold sink, and the total entropy increases.

It does indeed make sense if we look at the value of entropy, but how can this be feasible physically? You want to move heat energy from cold sink to hot source. In terms of molecular motions, the molecules at hot source always vibrate more vigorously and give energy to molecules at cold sink spontaneously. To achieve your purpose, you literally make the molecules at hot source vibrate faster?
If you still do not get what I am asking, imagine you have a flask of water and you want it to flow from A to B. If you punch a hole at B to remove some water at B, the water will flow to B, but adding water to B definitely will not cause more water to flow to B. Now replace water with heat energy, A with cold sink and B with hot source, I think adding heat energy to hot source should not cause cooling of cold sink.
What is my misconception here?

Compressing the refrigerant heats it above the temperature of the surroundings.

Refrigerators work by exploiting some properties of compressible gases.
To simplify greatly, gases tend to heat up when they are compressed and liquefied. And, when the liquid revapourises this process absorbs energy. |In a refrigerator those two processes are physically separated and this enables the transfer of heat from one place (the inside, which gets colder) to the outside (which gets warmer).
The way this works in a typical refrigerator is to have a compressor on the outside of the cold area that takes gas and compresses it while letting the extra heat dissipate into the outside environment. This combined process of compression and cooling (or, more accurately shedding the extra heat generated by compression) liquefy the gas. This liquid is then allowed to flow into a heat exchanger on the inside of the refrigerator where it is allowed to evaporate. This evaporation absorbs energy from inside the fridge making that environment colder. This gas is then allowed to flow into the compressor, completing the cycle.
This allows the fridge to convert mechanical energy in the compressor into a clever way of transferring energy from the inside of the fridge to the outside. This amounts to exploiting a phase change in the refrigerant into a pump that moves energy from cold to hot. Though this only works because we can separate the inside from the outside thermally enabling the two side of the process to behave normally (heat flows from hot to cold on both sides) but the net flow ends up taking energy from cold to hot by exploiting the energy used to cause a phase change in the refrigerant.

Your explanation on how the refrigerator work is understandable and straight to the point, but after reading your answer, I just feel that what I have read from textbook is wrong. 'Therefore, we need to add heat energy to the stream of heat energy flowing from the cold sink to hot source', is this statement wrong?

@The99sLearner No that statement is correct. Energy is required to compress and liquefy the refrigerant and to pump it around the system. So to get heat to flow we need to add energy to the system. But the amount of heat that flows can be much larger than the amount of extra energy added to make the flow happen.

The statement 'add heat energy to the system' is because heat changes entropy but work does not. I thought by first compressing and liquefy the refrigerant we are actually doing work but not heating the refrigerant? Is it because 'gases tend to heat up when they are compressed and liquefied'  so we say we are adding heat energy to the system?


text:
The use of silyl enol ethers over aza-enolates
I do not quite understand how alkylations using tertiary halides would work better with silyl enol ethers, compared with using the aza-enolate nucleophile. The following was mentioned in Clayden, Greeves & Warren (2012) on p. 594:

Aza-enolates are the best general solution for alkylating aldehydes with most electrophiles. With very $\ce {S_N2}$-reactive alkylating agents, enamines can be used, and with very $\ce {S_N1}$-reactive alkylating agents, silyl enol ethers must be used.

I would think that it does not matter what nucleophile is used when alkylating with tertiary halides since $\ce {S_N1}$ processes do not depend on the nucleophilicity of the nucleophile. Thus, whether the nucleophile is very strong and basic, in the case of an aza-enolate, or relatively weaker, in the case of a silyl enol ether, the alkylation would still proceed at around the same pace.
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

Silyl enol ethers (neutral) are weaker nucleophiles than aza-enolates (anions). The former needs a reactive electrophile, i.e., the tertiary cation (SN1). The aza-enolate can tolerate a poorer electrophile. If the aza-enolate sees a tertiary halide, E2 elimination will occur.

@user55119 Isn't the substitution an SN2? If a tertiary halide were used, the mechanism would become SN1 due to the steric hindrance around the carbon of the tertiary halide. Or as you said, elimination may take place. So is the reason why silyl enol ether is used preferentially because of the possibility of elimination occurring when using aza-enolates?

Didn't you say it was a tertiary halide? The silyl enol ether is not as nucleophilic as the aza-enolate. If the aza-enolate causes elimination with a tertiary halide and not an SN2 reaction, why would a weaker nucleophile effect SN2 substitution? As a weaker nucleophile, wouldn't it require a reactive electrophile, i.e., a cation.

@user55119 So it is because of the fact that aza-enolates cause elimination to occur... Maybe you could write a response based on that?

See this URL for examples of rxns of t-halides with silyl enol ethers. https://www.chem.wisc.edu/areas/reich/OrgMet/silicon%7B09%7D.htm they use a Lewis acid, which you didn't mention. SN1 through the cation.


text:
Determining reaction order in equilibrium reactions
Am hoping that someone can help me with the below question, and that I am asking in the right place. Am new here, so please forgive any missteps.
Given a reaction like this:
$$\ce{A + B <=> C + D},$$
is the differential/initial rates method of determining a reaction order valid? That is, the method which determines $b$ in $r = k C_\ce{B}^b$ as the slope in a $\ln({\mathrm{d}C_\ce{C}}/{\mathrm{d}t})$ vs $\ln(C_\ce{B})$ plot.
Basically, I wonder about three points here:

I have only seen examples for this method on single forward reactions, so is it valid when there is a reverse reaction?
Am I correct in thinking that I can use measurements of $C$ for $\ln(\text{rate})$, even though I'm interested in the reaction order of $\ce{B}$, because the rate can be written in terms of any of the reactants or products?
How crucial is it that the concentration (rate) measurements must come from initial times? If the change in all concentrations is very small, is it then not valid to use even a later values?

The difficulty with problems in kinetics is that you can envision a pool of intermediates such that, while B is consumed, you don't see significant formation of C, so that C is a poor reporter of the consumption rate of B. Concepts such as intermediates, autocatalysis, feedback, steady-states, lag times can be important here. Things can get complicated pretty quickly.
The initial forward rate, in the absence of a significant accumulation of reaction intermediates, can for simplicity be written as
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b$$
Again, this assumes a negligible accumulation of intermediates whose reverse reaction may lead back to formation of B, as in a process under steady state.
For this to be useful in the determination of $k_\mathrm{f}$ it helps to satisfy two conditions:

$k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b \gg k_\mathrm{rev}[intermediates]^m$
$[\ce{A}]^a\approx \text{const}$

These conditions are essential to guarantee a decay of the concentration of $\ce{B}$ with a rate constant $k_\mathrm{fwd}[\ce{A}]^a$. The answer to your second question is no unless the mechanism is simple and conditions 1 and 2 hold (small amount of product), that is, if the reverse reactions from intermediates or products are not important.
So, it is not essential that the measurements come from the beginning as long as the mechanism is elementary (no rate limiting steps leading to accumulation of intermediates), the reverse reaction is insignificant in rate and the concentration of A greatly exceeds that of B throughout your measurement time.

I edited some notations used in physical chemistry: note that "f" refers to formation, "r" – to reaction. In kinetics commonly used notations for the forward and reverse reactions are "fwd" and "rev", respectively.

RE: "the initial forward rate, in the absence of product, will always be given by the rate law
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_\mathrm{fwd}[\ce{A}][\ce{B}]$$"  This just isn't true. You're assuming an elementary reaction, which may or may not be the case. For instance if [A] << [B] then the reaction may be pseudo zero order in B.

In the most general case: $$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_\mathrm{fwd}[\ce{A}]^a[\ce{B}]^b + k_\mathrm{rev}[\ce{C}]^c[\ce{D}]^d $$ where a, b, c, and d are exponents to be determined.

@NightWriter - At t=0 there is no product. For any t>0 there is.

@NightWriter - Do understand that if the coefficients in the kinetic equation are equal to the coefficients in the chemical reaction, then the kinetic equation is an elementary reaction.  A reaction in actuality doesn't have to be elementary.

Thanks for the answer. In case it solves the dilemma, am thinking of a reaction that is not elementary, but the stoichiometric coefficients are all equal to 1.

My notation was perhaps confusing, I'm looking at determining the reaction order in B, but I called it a. Calling it b as in NightWriter's answer would have been better.

@NightWriter Your answer is about being able to determine k_fwd under those conditions. I suppose the same goes for determining the reaction order b?

I would need to modify my answer further to make it more general. The formation of intermediates might immediately alter the forward rate, complicating matters. Looking at the rate of formation of product will give an effective coefficient b but this may not be the same as b in the equilibrium equation unless you have an elementary mechanism.

@MaxW I added some fine print to my answer, hoping to make it more general.