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Bond angles in sulfate
My chemical bonding professor says that the $\ce{O-S-O}$ bond angles in $\ce{SO4^2-}$ are ideal (109.5°).
Why? Is this because all bonds are equivalent, and electron distribution is shared among all oxygens?

Exactly. You may think of it as a superposition of resonance structures with all possible combinations of $\ce{S=O}$ and $\ce{S-O-}$.

Yes. All the bonds are equivalent as in the 4 resonating structures, each with 1.5 S and O bonds. So the resonant hybrid has four 1.5 double bonds (sigma and pi) with -0.5 formal charge on each O. This is why, the ion has a perfect tetrahedral geometry and shape as it is very symmetric and equivalent.
Hope it helps.
The picture below on the right is what is thought to be the most accurate representation nowadays. However, the resonance structure on the left (and all of its equivalent representations) are more than likely acceptable for introductory chemistry classes.
Note that no matter which resonance structure you think is a better representation of the sulfate ion, the bonds are all equivalent and therefore sulfate ion has the ideal tetrahedral bond angle of 109.5 degrees.

Yes, all the bonds are equal in $\ce{SO4^2-}$, but thinking of them as bond order 1.5 is no longer accepted as correct.

@Diyanko Bhowmik - also, each O doesn't have a -1.5 formal charge. That would give the overall molecule a -6 charge.


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How do we determine if some metal will undergo passivation under suitable environment?
The passivity of certain metal fall in order:
$\ce{Ti->Al->Cr->Be->Mo->Mg->Ni->Co->Fe->Mn->Zn->Cd->Sn->Pb->Cu}$
Can we predict the order of passivation theoretically?

Can you be more specific about what you mean by theoretically?

It was a little confusing, so I just removed it.
I mean this order seems quite jumble(at-least to me), and it's definitely not apparent from there positions in periodic table . So is there any criterion which must be satisfied(something which we may calculate with concentrations etc) by a metal in order to show passivity? And also can we generalise the result for any metal?

One of the standard tools for illustrating the tendency to passivation shown by a material is the Ellingham diagram:

An Ellingham diagram is a graph showing the temperature dependence of the stability for compounds. This analysis is usually used to evaluate the ease of reduction of metal oxides and sulphides.

Thermodynamically, the stability of the passivating oxide on the surface of the metal is directly related to the Gibbs free energy change of the oxide formation reaction:

The Gibbs free energy $\left(\Delta G\right)$ of a reaction is a measure of the thermodynamic driving force that makes a reaction occur. A negative value for $\Delta G$ indicates that a reaction can proceed spontaneously without external inputs, while a positive value indicates that it will not. ... An Ellingham diagram is a plot of $\Delta G$ versus temperature. (source PDF)

(A detailed Ellingham diagram and a more detailed description of the diagram and its construction can be found at the source link of the above quote.)
In any practical situation, some or all of the ideal assumptions made in generating an Ellingham diagram may or may not apply, but it at least provides a (semi-)quantitative means for arguing relative tendencies to passivation. In particular, as noted by Ivan Neretin, such diagrams don't consider effects of any electrochemical couples that may be active.  For that, one would want to cross-reference the appropriate Pourbaix diagrams (though, these also suffer from some of the same kinds of idealized assumptions as the Ellingham diagrams).

You are relying on thermodynamics to explain an essentially kinetic phenomenon. This is not going to work out. Really, what the Ellingham diagram tells us that we don't know? That the oxide is stable in terms of $\Delta G$? Of course it is, nearly all of them are, so what? For similar reason, the usage of Pourbaix diagram here is downright misleading; for instance, it says that at certain conditions we should have $\ce{Fe}^{3+}$, but iron, apparently oblivious to that, would stay passive in concentrated nitric acid.

@IvanNeretin Quite true. My answer aims toward the 'broadly explanatory' level of description, though, which seemed appropriate to the question as-asked, as opposed to delving into the subtleties of interplay of kinetic/thermodynamic/mass transport/electric field effects that are, as you note, highly material- and context-specific.  Ellingham/Pourbaix diagrams give a "zeroth-order" idea of what might happen; for any real system, of course one must perform experiments to determine what actually does happen.

There is not much we can do. It is certainly not a property that is explained by a rigorously established single measurable parameter like oxidation potential. Passivation is a delicate interplay of multiple factors, some of which are not even related to the metal itself (remember, a metal may be passive in some environments, but not in the others).
However some trends can be observed. Surely, having a hard, high-melting, and chemically inert oxide does help a lot.

But Aluminium Oxide's M.P is higher then Titanium Oxide's M.P yet Titanium still has a better protective layer.

That's right, oxide m.p. alone does not determine exactly the degree of passivity (neither does any other single parameter, just like I said), though has some influence on it.

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What is the role of a monochromator in atomic absorption spectrometer?
What is the role of a monochromator in an atomic absorption spectroscopy (AAS) instrument? What would happen if we run AAS analysis without a spectral sorting device?

Allows absorbance to be measured within a narrow band of the spectrum.

Since free atoms exhibit specific discrete energy levels (e.g. the hydrogen Blamer series), to identify which atoms are present you need to measure which wavelengths are being absorbed, not just that there is absorption. To be sensitive to wavelength, you need to disperse the light, and that means a monochromator. Whether you use exit slits and a single detector and scan the wavelength, or use a position sensitive detector to measure the spectrum at once, is up to you.

This question has lingered on the site for a long time, which is a shame because it is a good question to ask.  Atomic absorption spectroscopy (AAS) uses a hollow cathode lamp (HCL) as a source, and HCLs are line sources; therefore, if a line source is being used, why would an instrument need to be made more complex by adding a wavelength selector?
The first point to consider is what the source really looks like.  The emission spectrum from HCLs have multiple lines.  Here, for example, is the emission from a nickel HCL, taken from Spectrochemical Analysis by Atomic Absorption and Emission by L. Lajunen.

A wavelength selector is needed in order to ensure that only one of the emission lines is being used for the analysis.
Additionally, light can be scattered in the flame, which can introduce errors in the analysis if not eliminated.  If you compare the block diagram of an AAS vs a typical visible spectrometer, you'll note that the wavelength selector is typically placed after the sample compartment for AAS to address these concerns about scattered light; however the issue is not that significant in visible spectroscopy so the wavelength selector can be placed before the sample holder.
This representative block diagram for AAS is taken from here.

A representative UV/Vis spectrometer block diagram is from this source

Bottom Line AAS needs a wavelength selector to ensure that the analytical line of interest is the only radiation hitting the detector and it minimizes the negative effects of radiation scattered in the flame.  A good summary can be found here.

The answer by @bobthechemist is clear, concise and correct and I upvoted it. However, I think the first figure in bobthechemist’s answer does not fully convey the significant complexity of the emission spectra of hollow cathode lamps (HCLs) and the challenge thereby presented to the monochromator. Accordingly, this answer is an add-on to bobthechemist’s answer.
The spectral output of an HCL can be observed and acquired for various purposes by use of a suitable spectrometer, spectroscope or spectrograph. Herein, I use one of the latter instruments: a homemade echelle spectrograph. This instrument is shown below:

The input is via fiber optic cable and the nominal spectral range coverage is 400 to 700 nm, though the actual red end limit is around 744 nm.
Now consider the spectral outputs of HCLs having elements far apart in the periodic table. First, light from a lithium HCL, with neon fill gas, is used as the input to the spectrograph’s input fiber optic. The resulting echellogram is shown below:

The large majority of emission lines are due to the neon fill gas and a blue filter was used to attenuate the intense red neon emission lines. The crimson lithium emission near 670.8 nm, familiar to anyone who has ever performed a simple flame test for lithium, is used for analytical AAS purposes. For that purpose, everything else is unwanted and detrimental, so the monochromator in the AAS functions to reject as much of that unwanted light as is economically feasible. A typical monochromator bandpass would be around 0.2 nm. Note that the 670.8 nm region is relatively isolated from the surrounding neon lines and the lithium line, which is actually a close pair, is quite intense. These two factors contribute to lithium being easy to detect, with high sensitivity, via flame AAS.
Now consider the echellogram for a combination sodium and potassium HCL with neon fill gas:

This echellogram is not much more complex in appearance than that of lithium. In fact, neon emission dominates in both echellograms. Sodium, in particular, has a relatively simple emission spectrum, characterized by its famous yellow D lines. Note that the D lines are not close to neighboring neon or potassium lines and are intense. As for lithium, this makes AAS determination of sodium relatively sensitive.
For potassium, the preferred AAS determination lines are at 766.5 and 769.9 nm, as per the Agilent Technologies online document titled “AA Hollow Cathode Lamps - Recommended Operating Conditions”. These near-IR wavelengths are inaccessible with my spectrograph.
In the preceding echellograms, the analytically useful AAS wavelengths were relatively isolated and intense, making the monochromator’s task fairly easy to accomplish. But now consider the spectrum of uranium, acquired using a uranium HCL with argon fill gas:

Argon has a non-trivial emission spectrum, but uranium’s emission spectrum is far more complex: the echellogram is so dense with lines that it has even been investigated (see screenshot below) as a pseudo-continuum light source for AAS.

For whatever reason, the use of lab-constructed water-cooled high power uranium HCLs never caught on. But uranium has been determined by conventional AAS, e.g., at 591.5 nm in a fuel rich nitrous oxide-acetylene flame with added alkali for ionization suppression (@AChem, private communication.)
The next three figures show enlargements of the spectral region around 591.5 nm. First, the sodium and potassium HCL with neon fill gas:

The two neon atomic emission lines bracket the location where the uranium 591.5 nm line would be located.
The next figure shows the same enlarged spectral region for the uranium HCL with argon fill gas:

All the unlabeled lines are other uranium lines or argon lines. Two such lines are uncomfortably close to the desired uranium line.
Then this figure shows an equally weighted overlay of the previous two figures:

Now it is clear that AAS determination of uranium, at 591.5 nm, requires unrealistically narrow monochromator spectral bandpass because there are two neighboring lines that are simply too close. Consequently, uranium is not sensitively detected, via AAS, at 591.5 nm. This is probably for the best: aspirating uranium-containing solutions into a flame just does not seem wise.
Note: Echellograms can always be processed to yield spectra in conventional intensity versus wavelength (or wavenumber) format, but it is tedious and somewhat complicated. For present purposes, it was unnecessary, but conventional uranium HCL spectra have long been available. For example, the uranium emission spectrum, in the neighborhood of the 591.5 nm line, is on page 87 of a 240 page report from scientists at the Los Alamos National Laboratory (LANL):

@andselisk Thank you very much! It is most appreciated and unexpected!

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Binary phase transition boiling rate
(Callen 9.7-2)
The phase diagram of a solution of A in B, at a pressure of 1 atm, has a coexistence region in the $(X,T)$ plane (e.g., $X_A$ is a liquid molar fraction with $N=N_A+N_B$ and T temperature and $Y_A$ is a gas molar fraction). The upper bounding curve of the two phase region can be represented by
$T=T_0-(T_0-T_1)Y_A^2$
and the lower bounding curve of the two-phase region is represented by
$T=T_0-(T_0-T_1)X_A(2-X_A)$
Show that if a small fraction of the liquid $−dN/N$ is boiled off, the change in the fraction of the remaining liquid is,
$dX_A =-\left[(2X_A-X_A^2)^{\frac{1}{2}}-X_A \right]\frac{-dN}{N}$.
All I knew was that $dX_A$ could be obtained from $dN = dN_A+dN_B$
and from there
$dN_A+dN_B = N dX_A + X_A dN + N dX_B + X_B dN$... I previously found that, when temperature is such that $X_A=X_B=1/2$, $Y_A=0.8660$...

The solution is more geometry than chemistry - sketch the liquidus and solidus and do the math.

I am not sure at all about which term in the desired answer should be $Y_A$ instead of $X_A$ given how I defined these guys...


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Do Acids/Bases undergo dilution/bio-degradation when exposed to air?
I spilled some cleaning fluid/drain opener in my trash can, a long time ago(almost 3 months ago), I didn't tried to clean it up. The cleaning fluid was diluted, and I spilled a small amount not more than 30 ml, I think.
The bottle does not say, what the cleaning fluid contains, but it probably has Sodium Hydroxide(if it is an alkali-based cleaner) or Hydrochloric acid(if it is an acid-based cleaner).
Now, what I have heard about Sodium Hydroxide, is that it reacts with carbon dioxide in the atmosphere, and lose its corrosiveness over time.
So, is it possible that NaOH got completely carboxylated in 3 months and nothing corrosive remains?
Now, about strong acid such as Hydrochloric acid, since they are hygroscopic, do they become dilute over time?

What you heard about NaOH is right: it surely got completely carboxylated in 3 months, maybe even in 3 days. Not that the resulting compounds are completely neutral to the environment, but yes, they lack that corrosive ability.
As to the more general question of what happens to inorganic substances once they are out in the open: dilution is quite possible, biodegradation is not so likely. Some of them, like hydrochloric acid (and most other acids), would pretty soon end up reacting with some random compounds out there to yield relatively harmless products which would become normal part of the environment. Others (think of heavy metals) may have long-lasting adverse effects.


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Why does tea dissolve better when put in water rather than milk?
I sometimes drink tea with or without milk. What I have noticed is that when I dip the bag in water the tea spreads out evenly and I have a cup of tea in like 5 minutes.
But when I do the same with milk in it, the tea takes forever to dissolve into the milk and even after a long time half of the milk is absorbed by the tea bag for some reason.
Could someone explain me why this is happening?

Milk is an emulsion containing small bubbles of oil/fat. Whenever you boil milk or do something drastic, this fat tends to separate. You can do this nicely, and get cream or butter, preserving the bubbles, only part of the water content is separated, or can do more drastically, when these bubbles collapse to a big oily blob.
When you put your filter in the milk, most probably some of the oil/ fat bubbles hit the surface and cover it with an oily film. This film prevents the easy dissolution of caffeine, etc from the tea leaves in water.

Small correction: milk direct from the cow will automatically separate, with nothing more drastic than leaving it sit for a while. Store-bought milk is usually homogenized, where they force it through a tiny opening to make the milkfat globules smaller, so Brownian motion effects are enough to keep them in suspension. Also, if the milkfat-coating explanation is correct, you should see a difference in behavior between milks with different milkfat percentages (e.g. whole vs. skim).

@R.M. Thanks for the correction. I am not sure on the second, though: it is essentially a nonlinear process once an oil film is produced. The thickness of this film would  only mildly influence the dissolution, only its completeness which matters much.

What is the technical term for milk spreading out within and combining with the tea? Does the milk dissolve in the tea? Does it combine with the tea? Does the tea dissolve in the milk? Or is it something else?

As I said, milk is basically a mixture of small oil bubbles and water. Tea is a water based solution. So tea would mix with the water part of milk freely. It is not even dissolution as both are mainly water based solutions.


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Chemical consideration of biodegradeability
We hear the term 'biodegradeability' being thrown around alot in science, but rarely ever is the science itself explained.
What are the chemical properties in a substance that make it biodegradeable?
Furthermore, what exactly are the chemical processes involved in biodegradation?

Biodegradability is not a chemical property, much like a price for which you can buy a compound is not its chemical property. Both are the properties caused by some external circumstances, which are not necessarily related to chemistry.
As for the processes, there is hardly anything they have in common. Life is pretty versatile and can consume a wide range of chemical compounds. Wide as it is, though, this range does not constitute all chemistry. There are a great deal of compounds that just never happened to appear in nature before humans made them, so no life form had the chance to develop the ability to consume them. Those are not biodegradable.


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Is there a potentiometric method for monitoring salt formation in organic solvent?
I work in a pharmaceutical R+D lab, and currently aim to develop a validatable potentiometric method for determining the optimal amount of acid to add to a basified small organic molecule for conversion to an ionic species (i.e. sulfate salt, hydrochloride salt). Our API (active pharmaceutical ingredient) has two protonation sites, so I also need to determine the stoichiometric ratio of API to counterion, and therefore which specific salt is present at specific points of the titration curve. Ideally, this will be carried out in methanol or ethanol.
Is it possible to accomplish these goals by titrating with acid and monitoring with a conductivity meter/probe? A consultant has advised me to do so, and I have a meter with two conductivity probes. I've already tried using a pH meter, but neglected to factor in that pH is only relevant to aqueous systems. The pH method did produce decent looking titration curves with obvious inflection points. However, as major amounts of salt 'crash out' of solution, meter drift becomes an issue, the reference junction gets clogged, etc. It gets messy and I have to continually clean the probe. I've read that they make pH meters that can withstand as much as 80% organic in solution, but this is not representative of our 100% organic protocol. Furthermore, I'm having trouble finding non-aqueous calibration solutions for the pH (and conductivity) meter.
My idea: Make a set of organic conductivity calibration solutions using methanol (and/or ethanol) and neat sulfuric acid at concentrations of, say, 0.010 M, 0.10 M, and 1.0 M (arbitrarily chosen), and then carry out the titration. Ignoring solutes, this is the most representative solvent system
I can think of since I would be adding acid to methanol during the titration.
Or, am I completely going about this the wrong way? My boss and I have been researching this for over a month and have found nothing specific to the application. I suspect that I'm over complicating the matter. This type of process is surely carried out daily in drug manufacturing facilities (tons of API's are in salt form, and GMP facilities require method validation/quality assurance/etc of all analytical methods), so I assume that there is a known and developed method for this out there somewhere.

Add methanol with excess $\ce{H2SO4}$ to your solution. In acid solution to can use HPLC to determine the concentration of your mystery goo compound.
Add water to ppt whatever organic goo you're making.
Mechanically separate your organic goo (filter, centrifuge.). Instead of 2-3 maybe solvent extraction?
Back-titrate excess $\ce{H2SO4}$ in water/methanol solution with $\ce{NaOH}$ and phenolphthalein.

There has to be a wealth of information on the titration of "dicarboxylic acids" in methanol.

If I understand your problem, you have a diamine of low molecular weight which is insoluble in water, and you wish to convert it to a water-soluble (ionic) solid by using the minimum amount of acid (presumably to get near neutral pH).
A very concentrated solution of the API in methanol or ethanol should be clear: no Tyndall effect, no turbidity. As a tiny amount of added water is added, a Tyndall effect will appear (use a narrow-beam flashlight, darkened room). If you then add a tiny amount of acid (in water or solvent), the solution will become clear again. By alternating water and acid, you will be able to decrease the solvent concentration to essentially zero, but the API will still be in solution as shown by no Tyndall effect. At this point, a pH measurement would be easy. Perhaps if the Tyndall effect is always maintained very small, the pH electrode might not clog during the whole process.
With the required maximum pH determined in this way, the alkaline API can be dissolved in acidic water. Then increase the pH with your favorite alkali until you see a Tyndall effect - this indicates that the pH is now too high, and the API is precipitating out, so you can have a pH range which might be small enough to be your answer. At least you should be pretty close to being able to decide what the optimum is. Knowing the amounts of acid and API, you can calculate what kind of protonation you have achieved.

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How are the 4 molecules of O2 bonded in the O8 molecule at low temperature and high pressure to form a rectangular box structure?
 1. List item
The discovers of O8 admitted to not knowing nor understanding the bonding in O8 at very low temperatures and high pressure. They concluded that O8 was a retangular box, not a cube & that this allotrope of oxygen was composed of 4 normal O2 molecules...

Should the bonding be sp2 or sp3 or can the bond angles be 90 degrees as depicted in the retangular box?  Is there considerable stress & strain in this frozen molecule?  Is there such a thing as a square molecule and are they commonplace? Where are the electron pairs located on the 8 oxygen atoms?

Will this box molecule "explode" when it is allowed to thaw with the high pressure removed? Is the bonding in this molecule like or similar to the bonding in the "Cubane" molecule? In the octaazacubane molecule?

Provide a reference?


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Calculating molecular weight of an acid based on titration data

If $\pu{2.600 g}$ of a weak diprotic acid were dissolved in $\pu{100 mL}$ of distilled water and a $\pu{10 mL}$ aliquot of this solution required $\mathrm{21.60\ mL}$ of $\mathrm{0.1000\ M\ NaOH}$ to reach the first endpoint, what are the equivalent and formula weights of $\ce{H2A}$?

I assumed to do this: I calculate the grams of acid in solution which is
$$\mathrm{\frac{2.600\ g}{100\ mL} \cdot 10\ mL = 0.2600\ g}$$
After that I would calculate the moles of NaOH used
$$\mathrm{0.1000\ mol/L \cdot 21.60\ mL \cdot 1\ L/1000\ mL = 0.002160\ mol}$$
Then I use this to find the moles of acid present in solution; this is the part I am unsure about. I am assuming: because at the equivalence point half of the acid has been converted to it's conjugate base and so ($\mathrm{2\ mol\ NaOH = 1\ mol\ H_2A}$).
If this is correct then
$$\mathrm{0.002160\ mol\ NaOH \cdot 2\ mol\ H_2A/mol\ NaOH = 0.004320\ mol\ H_2A}$$
Which I then divide the g/mol
$$\mathrm{\frac{0.2600\ g}{0.00432\ mol} = 60.19\ g/mol}$$
Because the acid is diprotic the equivalence weight would be
$$\mathrm{\frac{60.19\ g/mol}{2} = 30.09\ g/mol}$$

I don't think this is the correct answer because I am trying to do this using real titration data I took in lab and the MW seems to be at $\approx 1/2$ the value it should be. If I am not supposed to double the NaOH moles, is there a reason for it?

The reason is in the words "...to reach the first endpoint". What is the meaning of these?

Your error comes from your misunderstanding of what the equivalent points indicate, or as Ivan mystically said:

The reason is in the words "...to reach the first endpoint".

An n-protic acid, has exactly n equivalent points. This means for your titration, that you need one base equivalent to reach the first equivalent point (EP) and two base equivalents to reach the second EP. As you are given the volume that is needed to reach the first EP, everything is quite easy, as the ratio of acid to base is exactly 1:1.
This being said, your first EP can be described by $$n_A = n_B$$
(and not as you did by assuming $n_A = 2\ n_B$, which also would not be the description of the second EP, which is $2\ n_A = n_B$ instead.)
You have to find the molecular mass of your acid, why the upper equation needs to be expanded to $$\frac{m_A}{M_A} = c_B V_B$$
Now you simply have to solve for $M_A$ and put in your numbers (0.26 g, 0.1 mol/L, 21.6 mL).

authors: Αλέξανδρος Ζεγγ


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Why do triple bonds have a low NMR value?
Double bonds has a positive shift compared to single bonds. It seems like that there would be a trend but then the triple bond negatively shifts, despite having great electron density similar to the double bond. Any insight.

Chemical shifts arise due to differences in the local magnetic field in the different environments of a molecule. There are a number of contributing factors including the local diamagnetic contribution which is driven by the shielding effect of local electrons. Where there is functionality which generates electron circulation, a localised anisotropic magnetic field is created, and these can cause both downfield and upfield shifts, based on the position of the proton (nucleus) with respect to the functional group. These interactions, unsurprisingly, are called magnetic anisotropic effects.
The best example of this is the ring current effect arising from aromatic rings, with protons on the outside of an aromatic ring experiencing large downfield shifts, while protons either above the ring, or on the inside of the ring experiencing upfield shifts.
With double and triple bonds, the shielding cones are different in shape, and traditionally are drawn as shown below (you might find some better diagrams with a web search). Areas marked (+) increase the shielding (leading to decreased chemical shift) and areas marked (-) decrease the shielding (increasing chemical shift).

However, it is important to recognise that chemical shifts are not only influenced predominantly by electronics; sterics have a very significant contribution to chemical shifts also, often more so than electronics. A very good (recent) paper (Baranac-Stojanovic, M RSC Adv., 2014,4, 308-321) discusses traditional magnetic anisotropy models, and presents some very good examples that highlight the effects of steric influence on chemical shifts.

Chemical shift, $\delta$, is defined by $${\delta\equiv{(\nu_S - \nu_R)\over\nu_R}\times10^6}$$ where $\nu_S$, and $\nu_R$ are the resonance frequencies from sample and reference, respectively. Both $\nu_S$, and $\nu_R$ follow Larmor equation $${\nu_0={\gamma\over2\pi}B_0(1-\sigma)}$$ where $\gamma$ is the gyromagnetic ratio (or is called magnetogyric ratio), which is nucleus specific, and $B_0$ is the magnetic field strength. $\sigma$ is shielding factor, or chemical shielding. For a given nucleus, such as $\ce{^1H}$ or $\ce{^13C}$ in a certain magnetic field $B_0$, for example $9.4T$ ($\ce{^1H}$ resonates at about 400 MHz), the chemical shift difference depends only on the shielding factor $\sigma$.
Chemical shift can be rewritten as $$\delta={\sigma_R-\sigma_S\over 1-\sigma_R} \times 10^6\approx (\sigma_R-\sigma_S)\times10^6$$ because $\sigma << 1$
The change in shielding factor can be caused by variation in electron density. When observed nucleus is near a strongly electronegative group, the shielding factor $\sigma_S$ is smaller (less shielding), chemical shift is larger. Based on this alone, triple bonds would be less shielded and have larger chemical shifts than double bonds. But unfortunately, electron density is not the only factor that affect the chemical shielding.
Anisotropy effect is due to the anisotropy in the magnetic susceptibility of nearby groups of the observed nucleus. Because of this effect, triple bonds have a much larger chemical shielding than double bonds along the direction parallel to $B_0$. There are equations describing change in shielding and magnetic susceptibility tensors. But I don't think I am capable to explain any deeper.
Table. Some Calculated and Experimental $\ce{^13C}$ Shieldings in parts per million $$\begin{array}{c|c|c|c|c|c|}
 & \ce{\sigma_{xx}}&{\sigma_{yy}} & \ce{\sigma_{zz}} & \ce{\sigma_{av}}&\text{Expt.} \\ \hline
\ce{CH4} & 195.8 & 195.8 & 195.8 & 195.8 & 195.1 \\ \hline
\ce{C2H6} & 187.7 & 182.7 & 193.1 & 186.2 & 180.9 \\ \hline
\ce{C2H4} & 177.9 & -81.1 & 84.3 & 60.4 & 64.5 \\ \hline
\ce{C2H2} & 39.0 & 39.0 & 279.4 & 119.1 & 117.2 \\ \hline
\end{array}$$
The shielding $\sigma_{av}$ corresponds to $\sigma_S$ in the above chemical shift equation. $\sigma_{zz}$ has a direction parallel to $B_0$.
Reference
"High Resolution NMR: Theory and Chemical Applications" (3rd Ed) by Edwin D. Becker
(The table is from the above reference too.)


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Difference between reversible and irreversible processes
On a quiz, I was asked for the difference between reversible and irreversible processes.
I said:

For a reversible process, you can change direction at any time without disturbing equilibrium.
For an irreversible process, the macroscopic variables may not always be defined.

I was given 1/4 point for this answer. I asked the professor what the issue is with the answer, and she could not articulate a clear reason, and eventually said she'll re-think it.
Do you see a problem with my answer? If so, please explain.

Regarding item 2 in Answer 1, in most reversible processes (except those that are adiabatic) there will also be an entropy change. (I couldn't comment because I didn't have enough reputation points.)

It may be that the professor just asked a sloppy question.  She likely expected some answer regarding equilibrium and a second regarding entropy.  Especially since you never mentioned entropy, you didn't get the points.  This is just a guess, of course, and I'll freely admit that I could be way off base.

Try to uncook an egg.

In a reversible process, at each point along the process path, the system is only slightly removed from being in thermodynamic equilibrium with its surroundings.  So the path can be considered as a continuous sequence of thermodynamic equilibrium states. For an irreversible process, the system is not close to thermodynamic equilibrium with its surroundings at each point along the path.
At any point along the path of a reversible process, both the system and the surroundings can be returned to their original states without significantly affecting anything else.  This cannot be accomplished if the process is irreversible.

I would say that there is no direct problem with your answer. Maybe that you need state it more like definition. What will always be true or not true.
I think I would have answered the following:

For a reversible process, the system is ALWAYS in equilibrium with its surroundings.
For an irreversible process, there will ALWAYS be an entropy change.

Well, If the question was a four marker then you should have added some points like

External pressure is infinitesimal less than external pressure at each state  in reversible process
While Internal pressure is much different from external pressure.
As W= -P(ext) ∆V, therefore for a given ∆V the W of expansion is maximum in reversible process while this is not so for irreversible process.

A reversible process is a quasi static process in which the system goes through a series of equilibrium states and the system is always with its surrounding.
An irreversible process is a non-quasi static process in which there is always an entropy change.


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Is melting of Tar for road construction a reversible change?
In India roads are made up of a black liquid which is prepared by melting a black material(probably coal tar). After cooling the liquid becomes very rigid. Can this solid be converted back to its previous liquid state by heating?
P.S: This is a question in my book. I think that after getting solidified the chemical composition of Tar may change so it can't be converted back to its original state.

Chip off a little piece and try heating it. Chemistry is an experimental science. I think it would melt just fine.

@IvanNeretin I don't have proper apparatus. Heating Tar on a cooking stove would be very risky, I guess.

On extremely hot days I have observed this tar melt slightly. Whether it changes to the original material, I do not know.

If you've even seen a roof being tarred the tar is solid which is melted and then applied to the roof.

Tar
 is mainly made up of a collection of very heavy hydrocarbons, with
rather high melting temperatures. Commonly, it is mixed with some stones
 or gravel to make a harder surface for roads. It is possible to heat
this up to soften or melt this; you can try it yourself by heating it in
 a metal can over a stove (don't use a container that you plan to keep
clean) or in a wood or (char)coal fire.
EDIT: As MaxW pointed out, it would be a bad idea to do this inside the house because it would create a strong smell, similar to that created by whatever machine they used to make the road.

I wouldn't heat it indoors because it will stink up the house.

So it means melting solid tar doesn't change it's chemical composition?

@user31782 It depends on what other impurities are present in the tar. Pure tar will not change, but road tar might. I do not know what impurities are in it. Of course, you could always try melting some (outdoors) and letting it cool, and observe any differences.

The process is reversible, and can actually happen to some extent in hot weather, see here or here.  It can definitely soften on a hot day, and it melts at $150\rm~^\circ C$.


text:
What is the difference between "molecular mass", "average atomic mass" and "molar mass"?
I don't understand the difference between "molecular mass" and "average atomic mass". They seem like the same thing to me. Is it that average atomic mass is just the weighted average of the "weights"/masses of the isotopes whereas the molecular mass is the average of "weights"/masses of the average atomic masses of each element in the molecule.
Also, what's the difference between molar mass and molecular mass? Is it just that molar mass is expressed in Daltons and molecular mass is expressed in g/mol?
The hardest part about chemistry is keeping track of which people are using which terms, and which terms are outdated. Is this one of these "oh, we use this term now," type of thing?

Could you perhaps explain what is confusing you. They seem to be two different concepts to me.

Related: Quick and simple explanation of molar mass, molecular mass and atomic mass and Units of mass on the atomic scale

@bon... i edited my original question to be more clear. thanks!

to be clear...

molecular mass
average atomic mass
molar mass

As their names already tell: "molar" mass refers to the mass of one mol (= 6.022e23 molecules) and "molecular" mass is the mass of one molecule (with 1 Dalton approximateley being the mass of 1 neutron, or 1 proton)

@AstronAUT technically, molar mass refers to the mass per mole (units g/mol), not the mass of one mole (units g)

@orthocresol: Yeah, the mass (g) of 1 mol is expressed in g/mol. That is the same thibg as you would say: I ran 10 km in 1 hour, therefore i was 10 km/h fast...

@AstronAUT no. your distance per hour is 10 km/h, but the distance you ran in 1 hour is 10 km.

Atomic mass refers to the average mass of an atom. This has dimensions of mass, so you can express this in terms of daltons, grams, kilograms, pounds (if you really wanted to), or any other unit of mass. Anyway, as you said, this is an average of the masses of the isotopes, weighted by their relative abundance. For example, the atomic mass of $\ce{O}$ is $15.9994~\mathrm{u}$. $\mathrm{u}$ is short for unified atomic mass unit and 1 u is equivalent to $1.661 \times 10^{-24}~\mathrm{g}$. It is exactly the same as the dalton, but from what I've seen, the term dalton is used more when discussing polymers, biomolecules, or mass spectra.
Molecular mass refers to the average mass of a molecule. Again, this has dimensions of mass. It's just the sum of the atomic masses of the atoms in a molecule. For example, the molecular mass of $\ce{O2}$ is $2(15.9994~\mathrm{u}) = 31.9988~\mathrm{u}$. You don't need to calculate the relative isotopic abundance or anything for this because it's already accounted for in the atomic masses that you are using.
The term molar mass refers to the mass per mole of substance - the name kind of implies this. This substance can be anything - an element like $\ce{O}$, or a molecule like $\ce{O2}$. The molar mass has units of $\mathrm{g~mol^{-1}}$, but numerically it is equivalent to the two above. So the molar mass of $\ce{O}$ is $15.9994~\mathrm{g~mol^{-1}}$ and the molar mass of $\ce{O2}$ is $31.9988~\mathrm{g~mol^{-1}}$.
Sometimes you may come across the terms relative atomic mass ($A_\mathrm{r}$) or relative molecular mass ($M_\mathrm{r}$). These are defined as the ratio of the average mass of one particle (an atom or a molecule) to one-twelfth of the mass of a carbon-12 atom. By definition, the carbon-12 atom has a weight of exactly $12~\mathrm{u}$. This is probably clearer with an example. Let's talk about the relative atomic mass of hydrogen, which has an atomic mass of $1.008~\mathrm{u}$:
$$A_\mathrm{r}(\ce{H}) = \frac{1.008~\mathrm{u}}{\frac{1}{12} \times 12~\mathrm{u}} = 1.008$$
Note that this is a ratio of masses and as such it is dimensionless (it has no units attached to it). But, by definition, the denominator is always equal to $1~\mathrm{u}$ so the relative atomic/molecular mass is always numerically equal to the atomic/molecular mass - the only difference is the lack of units. For example, the relative atomic mass of $\ce{O}$ is 15.9994. The relative molecular mass of $\ce{O2}$ is 31.9988.
So in the end, everything is numerically the same - if you use the appropriate units - $\mathrm{u}$ and $\mathrm{g~mol^{-1}}$. There's nothing stopping you from using units of $\mathrm{oz~mmol^{-1}}$, it just won't be numerically equivalent anymore. Which quantity you use (mass/molar mass/relative mass) depends on what you are trying to calculate - dimensional analysis of your equation comes in very handy here.

Summary:

Atomic/molecular mass: units of mass
Molar mass: units of mass per amount
Relative atomic/molecular mass: no units

A small (and unessential) note about the definition of the $\text{u}$. It's defined by the $\ce{^{12}C}$ atom, which is defined to have a mass of exactly $12 \text{ u}$. Now the mole is also defined by the $\ce{^{12}C}$ atom: $12 \text{ g}$ of $\ce{^{12}C}$ is defined to contain exactly $1 \text{ mol}$ of $\ce{^{12}C}$. And we know that one mole of $\ce{^{12}C}$ contains $6.022 \times 10^{23}$ atoms - we call this number the Avogadro constant. That means that $12 \text{ u}$ must be exactly equal to $(12 \text{ g})/(6.022 \times 10^{23})$, and therefore,
$$1 \text{ u} = \frac{1 \text{ g}}{6.022 \times 10^{23}} = 1.661 \times 10^{-24} \text{ g.}$$

The IUPAC Gold Book provides the ultimate reference on matters of chemical terminology.

relative atomic mass (atomic weight), $A_\mathrm{r}$
The ratio of the average mass of the atom to the unified atomic mass unit.

The relative atomic mass (average atomic mass as you put it) is the weighted average mass of all the isotopes of an element in a given sample, relative to the unified atomic mass unit, which is defined as one twelfth of the mass of a carbon-12 atom in its ground state.

relative molecular mass, $M_\mathrm{r}$
Ratio of the mass of a molecule to the unified atomic mass unit. Sometimes called the molecular weight or relative molar mass.

This is the sum total of the relative atomic masses of all the atoms in a molecule. For example, $\ce{H2O}$ has a relative molecular mass of $1.008 + 1.008 + 15.999 = 18.015$.
The Gold Book does not have an entry for 'molar mass' but it is a commonly used term.
The molar mass is the mass of a substance divided by its amount of substance (commonly called number of moles). It therefore has units of $\mathrm{mass~(amount~of~substance)^{-1}}$ and is commonly expressed in $\mathrm{g~mol^{-1}}$. The relative atomic or molecular mass is just the molar mass of that substance divided by $\mathrm{1~g~mol^{-1}}$ to yield a dimensionless quantity.

Let’s take oxygen ($\ce{O2}$) for example. Examples shall make it easier to understand.
We will use u, kg and g as the units of mass. The full form of u is the Unified atomic mass unit. Commonly people use amu (atomic mass unit) or Da (Dalton) as well. kg is kilogram and g is gram.
1 u = mass of one nucleon (proton/neutron; the constituents of the atomic nucleus). $\pu{1 u} = \pu{1.66 \times 10^{−27} kg}$.
Atomic Mass:
An oxygen molecule is constituted of two oxygen atoms. $\ce{O2}$ is basically $\ce{O=O}$ Atomic mass is the mass of one atom. The mass of one oxygen atom is $\pu{(15.9994 \pm 0.0004) u}$ or roughly $\pu{16 u}$.
Molecular Mass:
Mass of one molecule of oxygen i.e. of one $\ce{O2}$ molecule (the whole $\ce{O=O}$ entity). So the mass of one molecule of oxygen will be $2 \times \pu{16 u} = \pu{32 u}$.
Molar mass:
The mass of one mole of oxygen.
1 mole of oxygen = $\mathrm{6.022 \times 10^{23}}$ number of oxygen molecules.
Let us try to calculate and see how it goes.
1 molecule of $\ce{O2}$ weighs $\pu{32 u} = \pu{32 \times 1.66 \times 10^{−27}kg}$
One mole of oxygen $\mathrm{= 6.022 \times 10^{23}}$ of oxygen molecules
So 1 mole of oxygen weighs $\pu{32 \times 1.66 \times 10^{−27} \times 6.022 \times 10^{23}kg} = \pu{0.031988864 kg} = \pu{31.988 g} = \text{ approximately } \pu{32 g}$.
1 mole of oxygen consists of a large number of molecules, hence we switched to a bigger unit (from u to g) for convenience. I hope you get the differences now.


text:
pH of bleach solution slowly decreasing?
I work within a science department in a school. We are demonstrating Universal Indicator and something odd has happened. Our bleach solution (~2.5% Sodium hypochlorite) turns blue when UI is added but within about 1 min it goes green and then after 1-2 mins it goes yellow. Overnight it goes clear but when more UI is added it is yellow. Why would this happen? The bleach smells like bleach and has been used in the last week for the luminol reaction which worked.

Sorry to add more... repeated but with hydrogen peroxide to see if this would have the same effect but it does not... Repeated with bleach, UI and pH probe and it is only the colour which is changing not the pH... what is the mechanism of bleach removing the colour does hydrogen peroxide bleach by a different mechanism???

What? Can you explain that in more detail. I don't understand what you are trying to say.

Looks like your bleach bleaches your indicator, just like it does pretty much everything else. After all, we don't call it "bleach" for nothing.
Sure, pH may be in fact slowly decreasing, and the bleach would eventually decompose and lose its properties, but these processes do not typically occur in a matter of minutes.
As to the mechanism, it is complicated, as are indicators themselves (usually, large organic molecules). One might think of a huge conjugated system of double bonds that produces color; all of a sudden, an oxidant appears and starts randomly snapping bonds in half. Soon the conjugated system is gone, and so is the color.

Hydrogen peroxide is also a bleaching agent but is having zero effect on the colour of the solution is this just because it use a different mechanism of bleaching,  is there a way other than using a pH probe to prove to the students that  bleach is alkali?

The pH has not decreased since I started ~ 1hr

Well, yes, hydrogen peroxide reacts differently, so it could just happen that it is not as good at snapping these bonds. As for pH, you've just proven it: adding an indicator shows the alkaline reaction right away. One may complain that it vanishes, so maybe the alkali is gone; but then you always may add one more drop of indicator.

I thought that too but when I add more drops of indicator it is now yellow and does not go through the same colours... thanks for your time sorry did not say that before

Well, then probably the alkali is already gone (there was not much of it in the first place) due to the reaction with something. BTW, isn't your indicator in the form of an alcohol solution?

the pH probe was still reading 13.40. Yes the UI is in ethanol.

Is it really as high as 13.40? Then it couldn't probably go away that easy. Now I am confused.


text:
Why is nitrogen dioxide (NO2) brown?
I don't understand why nitrogen dioxide $\ce{NO2}$ is coloured brown when nitrate $\ce{NO2-}$ and dinitrogen tetroxide $\ce{N2O4}$ are not. I understand it is due to absorption of light when its frequency matches the energy of an electronic transition. But what makes $\ce{NO2}$ special, i.e. what electronic transition gives rise to the brown color in $\ce{NO2}$, and why do $\ce{NO2-}$ and $\ce{N2O4}$ lack these transitions? Is there are way to explain this difference using molecular orbital theory?


text:
Which of the following is the more stable carbocation?
Which of the following is the more stable carbocation?

I thought the 1st carbocation would be the more stable one as the pi-electron density of phenyl group can overlap with the vacant orbital on the sp-hybrid carbocation (1st one). But my tutor pointed out that the vacant orbital on carbon can never be parallel to the p-orbitals of the phenyl group carbons and overlap cannot occur. So only negative inductive effect of phenyl group has to be considered, and so the 2nd one will be more stable.
Please explain. Please also mention about some sources where I can read about these type of carbocations.

There's some discussion about the true structures of these carbocations. See: Instability of vinylic and arylic carbocations and Do vinyl cations adopt a classical or non-classical structure?

Vinyl cations can also be generated by the protonation of acetylenes in strong acid.  The vinyl cations are captured by a nucleophile to produce a stable alkene with the nucleophile attached to the double bond.

Which of the following is a more stable carbocation?

This route to vinyl cations has been well studied.  As shown in the following diagram, protonation of phenyl acetylene can proceed in two different ways.  The acetylenic carbon attached to the phenyl ring may be protonated or the terminal acetylenic carbon may be protonated.  As you can see, these two routes produce the two vinyl cations you want to compare.
Upon examining the product(s) found in this reaction, we find that only alkene A2 is produced.  Therefore, the reaction proceeds through the intermediacy of vinyl cation A1, not B1.  Consequently, the pathway involving A1 must be lower in energy, or said differently, vinyl cation A1 is lower in energy than vinyl cation B1.

Reasons Why Vinyl Cation A1 Is More Stable Than Vinyl Cation B1

The barrier to rotate the phenyl group is very low. Hence, in A1 the phenyl group can readily adopt a conformation where the aromatic p-orbitals are aligned with the empty p-orbital on the adjacent carbocation center. Additional resonance structures can now be drawn with the positive charge delocalized into the aromatic ring.  There is no such resonance stabilization with the aromatic ring possible for vinyl cation B1. This suggests that vinyl cation A1 is more stable than vinyl cation B1.
In addition to the resonance structures involving the aromatic ring, as shown at the bottom of the diagram, one can draw 2 more hyperconjugated resonance resonance structures for vinyl cation A1, but only 1 hyperconjugated resonance structure can be drawn for B1; again suggesting that vinyl cation A1 is more stable than vinyl cation B1.
Resonance effects are generally stronger and more controlling than inductive effects.  Therefore, the inductive arguments presented in @orthocresol 's answer play only a minor role, being outweighed by the stronger resonance effects described above.

Aha! do you have a source for the formation of A2 over B2? Not doubting you, I just want to read. I remember I tried searching for protonation of phenylacetylene last night but couldn't quite find anything.

@orthocresol Stang's book, "Vinyl Cations", lots of data, for example starting on p.32 section B "Aryl Alkynes"

@ron I read yesterday that hyperconjugation does not occur in vinyl cations as for example, in the A cation the C-H and C=C bonds are aligned at    120 degrees in the same plane(all three bonds - sp2 hybrid). The two C-H bonds face away from the vacant p-orbital of the carbocation.

@ksr Look at this diagram of non-vinylic hypercojugation. The angle between the C−H bond and the p-orbital is 109°, the tetrahedral angle. In the vinylic cation case this angle is increased to 120°. This increase of only ~10° will lessen the effectiveness of hyperconjugation in the vinylic case, but I would not expect it to eliminate vinylic hyperconjugation. Further, note that since a double bond is shorter than a single bond, this slight decrease in overlap due to the increased angle may be more than offset by the increased overlap due to the shorter bond length.

Addendum 14-5-2017
Please read ron's answer! I went out on a limb writing this, and while there's probably no fault in my logic itself, it's worth remembering that chemistry is fundamentally an experimental science.
A lot of chemistry is concerned with rationalising experimental data, and developing models with predictive power that we can apply to other unknown reactions. In the absence of experimental data, it is often possible to come up factors supporting "X > Y" as well as factors supporting "Y > X". A priori, it is impossible to tell which factor dominates. You can already see in the comments on this answer that there are multiple factors that must be considered in this question.
Often chemists have an intuition as to which factor is more important, but the final word will always come from experiments. ron's answer which describes the observed regioselectivity in protonation of phenylacetylene is therefore superior.
(Note that I consider computational calculations to be experimental data as well.)

Let's call the carbocations A and B because typing "the first carbocation" gets quite tiring (it has absolutely nothing to do with the fact that I'm lazy).
So in both carbocations, the positively charged carbon is $\text{sp}$-hybridised (this allows the vacant orbital to be a $\text{p}$ orbital, which is higher in energy and therefore lowers the total energy of the system). The thing about $\text{sp}$-hybridised carbons is that the remaining two $\text{p}$ orbitals are mutually perpendicular, or orthogonal, to each other.
Now look at this diagram I drew:
$\hspace{26ex}$
I've colour-coded the $\text{p}$ orbitals. Purple ones are filled $\text{p}$ orbitals, and are parallel to other purple ones. The orange orbital on the $\text{sp}$ carbon is vacant, and is perpendicular to all the purple ones.
Look at A. Even though the $\text{sp}$ carbon has tons of $\text{p}$ orbitals surrounding it, the vacant $\text{p}$ orbital is orthogonal to all of them! The same can be said for B. So, neither species can derive any stabilisation via conjugation (aka resonance or mesomeric effect). Therefore, as your tutor said, you can only consider the inductive effects. In A, the positive carbon is bonded to two $\text{sp}^2$ carbons; in B, the positive carbon is bonded to one $\text{sp}^2$ carbon and a hydrogen. The $\text{sp}^2$ carbon is more electron-withdrawing (carbon is more electronegative than hydrogen), and therefore A is less stable.
This arrangement where all the filled orbitals are parallel is (presumably) the most stable. Consider an alternative conformation where you rotate the C-C single bond in A by 90 degrees, allowing the $\text{p}$ orbitals of the benzene ring to overlap with the vacant $\text{p}$ orbital on the $\text{sp}$ carbon. (I attached another diagram labelled A2 below.) This would serve to increase electron density on the positive carbon. However, to do this, you would have to break the conjugation between the benzene ring and the alkene, thereby forgoing the original stabilisation derived via conjugation. Overall, that's (presumably) less favourable. Check out ron's answer.
$\hspace{30ex}$

What about hyperconjugation of the terminal C-H bonds with the empty p-orbital in A? That might increase its stability.

@orthocresol - I get your point. Thanks for the wonderful explanation.  But hyperconjugation is yet another important point.

@ksr indeed, you're right. But when it comes to hyperconjugation and assessing whether it is stronger than the inductive effect I am out of my depth :) Hopefully someone else can chip in. Thanks for asking this question btw. I learnt something too!

@orthocresol I have read that hyperconjugation is more stronger than inductive effect.

Let me make this more complicated and more simple. I am not convinced that hyperconjugation and induction are different phenomena.

How do you know that the conjugation of an alkene with the aromatic ring is more stable than an empty p-orbital with an aromatic ring? It seems the latter should be more important because it allows delocalization of charge.

@jerepierre Honestly I don't know. I was trying to rationalise the answer OP's tutor gave him/her, but of course it is entirely possible that they got it wrong. (Someone needs to help me out!! Lots of questions I can't answer..)

@orthocresol I don't think this is a question that can be answered without some computational data. There is probably no way to unequivocally answer this question without it. Your analysis of the situation including the MO picture and inductive effects is excellent.

@BenNorris The question of hyperconjugation and inductive effect was discussed previously: http://chemistry.stackexchange.com/questions/19027/inductive-effect-and-hyperconjugation-one-elephant-different-parts


text:
Do vinyl cations adopt a classical or non-classical structure?
Whilst reading this question today, I remembered something that I had seen previously here. In the second linked question, @Martin provided a reference to suggest that vinyl cations actually adopt a non-classical structure with $\ce{sp}$ hybridization and a bridging hydrogen, as shown below:

The non-classical structure can be viewed as a protonated alkyne, in the same way that the ethyl cation, which adopts a non-classical structure, can be viewed as a protonated alkene.
What is the correct structure of a vinyl cation? Does it depend on what the substituents are? For example, in the first linked question, do the two cations (shown below) actually equilibrate to the same non-classical structure?

I am looking forward to the answer to this question since it may change the way I teach addition reactions of alkynes.

Search "bridged vinyl cation" in Stang's, "Vinyl Cations".  In the parent vinyl cation itself, calculations suggest a very small energy difference between the open and bridged form, with a very lower barrier to interconversion between the two. A few other vinyl cations may also be bridged depending on the substituents.

@ron Thanks for the excellent reading. Do you want to write up an answer yourself? If not I will probably answer my own question using that reference.

I think that the cations on the left side of both images should be linear.

TL;DR: there have been many theoretical investigations of the relative energies of these two forms for the parent vinyl cation with more recent work indicating that the bridged form is slightly more stable (by about 1–3 kcal/mol)4,5. This prediction has received support from recent experimental work as well6.
Taking such relatively small energy differences into account I would not be surprised if the situation is reversed with a particular substituent. Thus, answering the second part of the question ("Does it depends on what the substituents are?") requires high-level calculations for a wide variety of different substituents which seems to be absent in literature yet.

Classical structure: trigonal (sp2) or linear (sp)?
First note that (give Marko a credit for paying our attention to that), as it is mentioned in Stang's "Vinyl Cations" book, the bent $\mathrm{sp^2}$-hybridized structure with a $\ce{C=C-H}$ bond angle of 120° (OP calls it "classical") is about 50 kcal/mol higher in energy than the linear $\mathrm{sp}$-hybridized form in all of the theoretical calculations.

These calculations confirm the organic chemist's intuitive and
  qualitative feeling that carbenium ions prefer a trigonal planar
  geometry with an empty p orbital and that it takes energy to bend the
  methyl cation from its low-energy planar form.

Thus, in the studies below the meaning of the classical structure is different from the OP's one: the linear $\mathrm{sp}$-hybridized structure is called classical in almost all the studies.

So, Stang's "Vinyl Cations" references a CI/DZP study of the vinyl cation by Weber, Yoshimine, and McLean 1 as the most recent ab initio correlated calculations published at the time the book was written (the late 70s), so I'll start a review from this study.
Weber, Yoshimine, McLean, 1976
It is was found that the linear and bridged conformations have energies equal to within 0.01 kcal/mol with the later being lower in energy. However, such a small difference is below the accuracy of the CI/DZP level of theory, and thus, the authors didn't claim that the bridged conformation is more energetically favourable, just that it can be such ("probably" is the key word in the quote from the paper below).

While our CI calculation shows the linear and bridged conformations to
  have energies equal to within 0.01 kcal/mol, we clearly cannot claim
  this absolute accuracy. In fact, taking into account basis set
  deficiencies, both in the one- and $n$-particle spaces, along with
  optimization effects as discussed above, we interpret our calculations
  to predict the relative energies of the two structures to be within
  1–2 kcal/mol with the bridged structure probably having the lower
  energy.

Besides, note that in the study the geometries have been optimized at the SCF level rather than at the correlated while CI calculations were done at the SCF optimized geometries. Taking into account that correlation effect were found to have major impact on relative stability of the two forms as mentioned in the study, it is especially interesting to look at the results obtained by Raghavachari, Whiteside, Pople, and Schleyer2 who optimized the structures not only at the HF/6-31G* but also at the MP2/6-31G* levels as well as did more high-level correlated calculations (MP4) with a bigger basis set (up to 6-311G** of TZP quality).
Raghavachari, Whiteside, Pople, Schleyer, 1981
For the bridged structure the $\ce{C-C}$ bond length was found to increase from 1.21 Å at HF/6-31G* to 1.23 Å at MP2/6-31G* while the distance of the bridging hydrogen to the middle of the $\ce{C-C}$ bond stays the same (1.18 Å). For the classical structure the authors report no significant changes in the geometry. With respect to the relative stability it was found that use of the MP2 geometries results in a slight additional stabilization to the bridged form: the MP4(SDQ)/6-31G**//MP2/6-31G* energy difference is 1.3 kcal/mol while the MP4(SDQ)/6-31G**//HF/6-31G* energy difference is only 0.6 kcal/mol (in both cases in favour of the bridged form).

With the larger 6-311G** basis, the bridged structure is stabilized
  further, and our final result at MP4(SDQ)/6-311G**//MP2/6-31G*
  indicates 11 [bridged] to be more stable than 10 [classical] by 3.0 kcal/mol.

The authors also mention that these final results are quite in line with the CEPA study by Lischka and Kohler3 done using a comparable basis set: the Lischka and Kohler geometries were very similar to the HF/6-31G* geometries obtained in the study and the bridged form was found to be 4.0 kcal/mol lower in energy than the classical form as well.
A further study done by the Pople group is also is in agreement with the findings quoted above: the bridged form predicted to be more stable by 3.1 kcal/mol (including zero point energies)4. А study by Lee and Schaefer5, though, reports a somewhat lower difference.
Lee, Schaefer, 1986
In a recent CI/DZP study by Lee and Schaefer with the largest CI expansion considered being CISDT the energy difference was found to be 0.68 kcal/mol with the nonclassical structure again lying lower. The authors also considered the effect of the zero-point vibrational energy corrections and found that it further lowers the nonclassical structure by 0.29 kcal/mol.

Our best strictly ab initio estimate of $\Delta E$(classical-non- classical) for $\ce{C2H3+}$ is obtained by adding the above 0.29 kcal correction to the 0.68 kcal difference from CISDT. In this manner, $\Delta E$ is predicted to be 0.97 kcal/mol. We estimate that a proper variational treatment of quadruple excitations might increase this $\Delta E$ value by perhaps 0.5 kcal/mol.

Weber, J.; Yoshimine, M.; McLean, A. D. A CI study of the classical and nonclassical structures of the vinyl cation and their optimum path for rearrangement. J. Chem. Phys. 1976, 64 (10), 4159–4164. DOI: 10.1063/1.431986.
Raghavachari, K.; Whiteside, R. A.; Pople, J. A.; Schleyer, P. V. R. Molecular orbital theory of the electronic structure of organic molecules. 40. structures and energies of C1-C3 carbocations including effects of electron correlation. J. Am. Chem. Soc. 1981, 103 (19), 5649–5657. DOI: 10.1021/ja00409a004.
Lischka, H.; Koehler, H. J. Structure and stability of the carbocations C2H3+ and C2H4X+, X = hydrogen, fluorine, chlorine, and methyl. Ab initio investigation including electron correlation and a comparison with MINDO/3 results. J. Am. Chem. Soc. 1978, 100 (17), 5297–5305. DOI: 10.1021/ja00485a010.
Curtiss, L. A.; Pople, J. A. Theoretical study of structures and energies of acetylene, ethylene, and vinyl radical and cation. J. Chem. Phys. 1988, 88 (12), 7405–7409. DOI: 10.1063/1.454303.
Lee, T. J.; Schaefer, H. F. The classical and nonclassical forms of protonated acetylene, C2H3+. structures, vibrational frequencies, and infrared intensities from explicitly correlated wave functions. J. Chem. Phys. 1986, 85 (6), 3437–3443. DOI: 10.1063/1.451828.
Crofton, M. W.; Jagod, M.; Rehfuss, B. D.; Oka, T. Infrared spectroscopy of carbo‐ions. v. classical vs nonclassical structure of protonated acetylene c 2 h + 3. J. Chem. Phys. 1989, 91 (9), 5139–5153. DOI: 10.1063/1.457612.

authors: Герман Носкин


text:
Which singlet dihalocarbene has the strongest carbon-halogen bond?

There are three singlet carbenes - difluorocarbene, dichlorocarbene and dibromocarbene. Which has the strongest carbon halogen bond?

Obviously backbonding is a point to be considered, as carbene is an electron deficient species. As fluorine is the most easiest to overlap with, the $\ce{C-F}$ bond will be strongest.
But I am not able to get that whether it is possible for the singlet carbene to overlap or share electron density with empty d-orbitals on chlorine and bromine atoms?
And which is more probable, backbonding from halogens to carbene empty orbital, or the other way round (from carbene filled orbital to vacant d-orbitals)?

There is no need to rigorously discuss or consider the other factors, such as the adjustments to the orbital contributions due to Bent's rule, because they are not very significant compared to the dominant factors, such as orbital size and charge separation. Clearly, the C-F bond is the strongest here due to the conventional reasons of significant charge separation and orbital overlap.

Singlet carbene is zwitteronic, meaning the $\mathrm{sp^2}$ orbital is doubly occupied resulting in a negative $\mathrm{sp^2}$ orbital and a slightly positive empty $\mathrm{p}$ orbital. Therefore it attracts electron on one orbital and repulses them around another orbital. The huge reactivity of the singlet carbene comes from this zwitterionic character.
Below is the picture electrostatic potential of on the left a singlet carbene with doubly occupied sp2 orbital and the right triplet carbene.

It can be argued in many ways how a halogen stabilizes the singlet state of carbene.
Two explanation usually used are:

Electronegative species increase the $\mathrm{s}$ character of σ orbital (Bent's Law).
Electron donation from the electron pairs into the empty carbene $\mathrm{p}$ orbital.

Though the second one is usually used, I suggest you follow the first (more physical). Adding electron negative substituents increases the s character of the doubly occupied sigma in-plane orbital of carbene stabilizing it, significantly decreasing the energy of the system. Following this concept you can see that flourine should provide the maximum stabilization and so on.
$$\ce{F} < \ce{Cl} < \ce{Br} \ldots$$

What is zwitteronic? Can you provide a reference? I'm assuming you don't mean zwitterionic as that makes no sense in the context.

I've updated the answer. It's actually simple classic zwitterionic definition.

Then it indeed makes no sense. - whole first paragraph is flat out wrong.

read a book then start spamming.
https://pubs.acs.org/doi/abs/10.1021/acs.jpca.9b01051


text:
Finding partial derivatives for an ideal gas and van der Waals gas
How do I determine $\left({\partial P\over\partial V_\mathrm m}\right)_T$ for an ideal gas and also for a van der Waals gas? Note that $T$ is the only variable which is held constant, and that's my trouble. I've only derived these equations in cases where two variables are held constant. I don't know where to start.

Roughly speaking, if two variables are held constant, everything is held constant, so I'm not quite sure what do you mean.

Use the equation of state, express $P$ as a function of $T$ and $V$, and find the derivative with respect to $V$, assuming $T$ to be constant.


text:
Sodium chloride + sulfuric acid
Why does sulfuric acid displace more volatile acids from salts?
My textbook says that sulfuric acid can displace more volatile acids from metal salts.
How is $\ce{HCl}$, which is not even a reactant, 'displaced' from $\ce{NaCl}$, as there is no $\ce{HCl}$ to be displaced?

Related: http://chemistry.stackexchange.com/questions/50656/how-is-volatility-useful-in-the-production-of-acids

How is $\ce{HCl}$, which is not even a reactant, 'displaced' from
  $\ce{NaCl}$, as there is no $\ce{HCl}$ to be displaced?

Well, there will be some $\ce{HCl}$ due to a well-known chemical reaction which is used both in the lab as well as in production to get the hydrogen chloride:
$$\ce{NaCl(s) + H2SO4(s) → NaHSO4(s) + HCl(g)}$$
The reaction proceeds at room temperature, but note, that the reagents should be dry. If you heat it up above 200 °C reaction proceeds even further:
$$\ce{NaCl(s) + NaHSO4(s) → HCl(g) + Na2SO4(s)}$$
Hydrochloric acid is then prepared by dissolving hydrogen chloride in water.


text:
Why does density increases when adding nanoparticles in a mixture of water and some clay?
I want to know that why density of drilling fluid which is a mixture of water, barrite, bentonite and carboxymethyl cellulose increases when i am adding silica nanoparticles. My ain of adding these nanoparticles is to improve the viscosity.

The density of water is 1 ( 1 gram per cubic centimeter) but the density of silica (mineral quartz) is 2.65. Barrite has a density of 4.5, so it is much denser.

If your question is why density increases with the addition of nanoparticle then i have to recommend this quora post.

If the attractive forces between the constituent particles of the
  solid with the water molecules are high enough, the total amount of
  intermolecular space is reduced and the total volume of the solution
  can be smaller that the combined volume of its components, leading to
  an extra increase in density....

Surface of silica nanoparticles have siloxane($Si-O-Si$) and silanol ($Si-OH$) groups. Because of the presence of these functional groups silica nanoparticles tend to form hydrogen bonds between themselves and/or with other polar molecules such as water. Its these interactions which reduces the intermoleculr space which further reduces the volume thus increase in density.
If the silica nanoparticles interect with themselves and form a linked three dimensional networks then it will also increase the viscosity of the solution.


text:
Effect of incorrect volume reading on calculated density value

When measuring the density using the displacement method, a student read the initial volume of the water too high, but the final volume reading was correct.
  This error would cause the calculated density to be too high or too low?  Explain why.

Wouldn't it be too low because the initial volume is too high?

It's asking for the calculated density, not the calculated volume. If the volume, as measured by the student, is too small, then the density (= mass / volume) will be too large. Of course, that's assuming that the mass was measured correctly.


text:
Deriving the buffer formula
Essentially I want to derive the buffer formula: $\ce{pH}$ = $\mathrm{p}K_\mathrm {a}$ + $\log$ $\left(\frac{\alpha}{1-\alpha} \right)$ to $\alpha$ = $\left(\frac{1}{10^{\mathrm{p}K_\mathrm{a}-\ce{pH}}+1} \right)$
I'm going to include everything I had done trying to solve this here (and also show where I got stuck):
$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log(\alpha) - \log(1-\alpha)$$
$$\ce{pH} - \mathrm{p}K_\mathrm{a} + \log(1-\alpha) = \log(\alpha)$$
$$\alpha = 10^{\ce{pH}-\mathrm{p}K_\mathrm{a}+\log(1-\alpha)}$$
$$\alpha = 10^{\ce{pH}} \times 10^{-\mathrm{p}K_\mathrm{a}} \times 10^{\log(1-\alpha)}$$
$$\alpha = 10^{-\log\ce{[H+]}} \times 10^{\log(K_\mathrm{a})} \times 10^{\log(1-\alpha)}$$
$$\alpha = \ce{[H+]}^{-1} \times K_\mathrm{a} \times (1-\alpha)$$
$$K_\mathrm{a} = \left(\frac{\alpha \times \ce{[H+]}}{(1-\alpha)}\right)$$
So at this point I got stuck and have no clue how to get rid of an $\alpha$ to complete the derivation. Appreciate any help.
Edit: Managed to complete. For those who are interested in how I had done it in the end:
$$\ce{pH}-\ce{pK_\mathrm{a}} = \log\left(\frac{\alpha}{1-\alpha}\right)$$
$$10^{pH-pK_\mathrm{a}} = \left(\frac{\alpha}{1-\alpha}\right)$$
$$10^{pH-pK_\mathrm{a}} - (\alpha 10^{pH-pK_a}) = \alpha$$
$$10^{pH-pK_a} = \alpha + \alpha 10^{pH-pK_\mathrm{a}}$$
$$10^{pH-pK_a} = \alpha(1 + 10^{pH-pk_a})$$
$$\left(\frac{10^{pH-pK_a}}{1 + 10^{pH-pK_a}}\right) = \alpha$$
divide numerator by itself = 1 and thus also divide denominator by the numerator.
$$\left(\frac{10^{pH-pK_a}}{10^{pH-pK_a}}\right)/\left(\frac{1+10^{pH-pK_a}}{10^{pH-pK_a}}\right) =$$
$$\left(\frac{1}{1+10^{pH-pK_a}/10^{pH-pK_a}}\right)$$
working out the denominator:
$$\left(\frac{1}{10^{pH-pK_a}}\right) + \left(\frac{10^{pH-pK_a}}{10^{pH-pK_a}}\right) ->$$
$$\left(\frac{1}{10^{pH-pK_a}}\right) +1$$
now work out the last fraction:
$$\left(\frac{1}{10^{pH-pK_a}}\right) = \left(\frac{1}{10^{pH} \times 10^{-pK_a}}\right) = (10^{pH})^{-1} \times (10^{-pK_a})^{-1}$$
finally gives:
$$\alpha = \left(\frac{1}{10^{pK_a-pH}+1}\right)$$

Hi and welcome to Chemistry.Stackexchange. Nice that you already used MathJax. However, for long equations it is preferrable to use one line of MathJax in the way I have just formatted your post. Further helpful hints: \log will give an upright logarithm, and chemical stuff such as $\ce{[H+]}$ can be formatted much easier with the \ce command: $\ce{[H+]}$. Further helpful MathJax stuff can be found in the sandbox (Takes a while to fully load.)

Just to make sure that I understand you correctly: You want to derive $\alpha = \left(\frac{1}{10^{\mathrm{p}K_\mathrm{a}-\ce{pH}}+1} \right)$ from the starting formula $\ce{pH}$ = $\mathrm{p}K_\mathrm {a}$ + $\log$ $\left(\frac{\alpha}{1-\alpha} \right)$, right?

Yes, that is correct. Ideally from the point where I got stuck. But if there's a faster/better way, I am open to that also.

Instead of deleting the question, please feel free to add the derivation as an answer. You are encouraged to answer your own question if you found out how to do it. That way, the question can serve future Stack Exchange users, too.

$$\ce{pH} = \mathrm{p}K_\mathrm {a} + \log \left(\frac{\alpha}{1-\alpha} \right)$$
Multiply the equation by $-1$:
$$-\ce{pH} = -\mathrm{p}K_\mathrm {a} + \log \left(\frac{1-\alpha} {\alpha}\right)$$
Rearrange the equation
$$-\ce{pH} +\mathrm{p}K_\mathrm {a} = \log \left(\frac{1-\alpha} {\alpha}\right)$$
Apply the definition of $\log$
$$10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} = \frac{1-\alpha} {\alpha}=  \frac{1}{\alpha} -1$$
Rearrange the equation
$$1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}} =  \frac{1}{\alpha} $$
So,
$$\alpha= \frac{1}{ 1+ 10^{-\ce{pH} +\mathrm{p}K_\mathrm {a}}  } $$


text:
What makes most all-carbon materials so strong?
For some reason, stuff made entirely of carbon atoms tends to be incredibly strong, beyond just about anything else known to material science.  It doesn't seem to matter too much how the carbon is arranged either; whether it's a tetrahedral crystal lattice (diamond), a hexagonal 2D lattice (graphene), or a soccer ball-shaped spheroid (buckyball), pure carbon seems to magically be super-strong in just about any configuration, except graphite.
So what it it about carbon that makes it super-strong, and what it it about graphite that screws up this property?

related http://chemistry.stackexchange.com/questions/21840/why-is-graphene-so-strong,  http://chemistry.stackexchange.com/questions/34193/are-diamonds-really-forever

Carbon is not universally super-strong. True, diamond sets a record in hardness, but: (1) it is not that far beyond anything else, as some materials (cubic boron nitride, etc.) come pretty close, and (2) hardness is not the same as strength. Also, there are different kinds of strength.
Graphene is a 2D material, which is an entirely different realm, so it can't really be compared to the rest.
Buckminsterfullerene is neither hard nor strong at all; you may easily crush a piece of it on your fingernail. Some of its derivatives display certain promising characteristics on a microscale, but there is a long way between that and a bulk material.
All that being said, carbon has some properties that make it special: it tends to form covalent bonds, and it prefers to have four of them. Not that the carbon-carbon bonds are record strong (none of them is as strong as the triple bond in $\ce{N2}$), but this certainly paves a way to a lot of interesting combinations, as illustrated by many allotropes of carbon (probably more than in any other element) and the immense diversity of organic chemistry.

The (https://en.wikipedia.org/wiki/Boron_nitride) boronitride you mentioned is very interesting in comparison to diamond. Isoelectronic, same lattice motif, comparable density. And it can even form nanotubes, similar to the carbon.

Sure. One may almost say it is a "carbon in disguise". It even has organic chemistry too.

Part of the strength of C-C bonds, especially bonds in cyclic carbon molecules, is from the sigma bond. This type of bonding has a high degree of symmetry, as in the Kekule benzene diagram. It also blends electrons in multiple orbitals as a single hybrid orbital.
Although the structure can be drawn in alternative symmetrical diagrams, e.g. mirror images of each other, according to quantum mechanics the superposition of both states occur simultaneously (unless one forces it to resolve into a single state, such as by a chemical reaction or by measurement). This distributed nature adds to the strength of the bond.

Kekule benzene thing is about pi bonds, not sigma.


text:
Which material would work best as "sacrificial material" for melting aluminum powder in a kiln?
I need to melt aluminum powder in a kiln, but before it melts it gets oxidized almost completely. My kiln is almost airtight, but it doesn't have inert gas inlet.
Which material would work best as sacrificial material, which would react with oxygen (and water vapor) before aluminum?

Must you use Al powder? Chunks of aluminum might serve, and they are protected by the surface oxide film until they melt. There is some oxidation, but nowhere near as much as from powder, with its high surface-to-volume ratio.

Yes, I must use powder.  But thanks for the suggestion.

Scavengers are a mess. Magnesium might do. Do you have a press? Just compact your sample sufficiently and you should have a lot less problems.  Other possibility: Put your kiln in a larger beaker with a lid having a hole, fill it with Argon (or nitrogen in a pinch), fill in the aluminium using a funnel, and quickly place the lid on the kiln with a pair of tweezers.

You need some oxygen scavengers before you start heating up the system. However, please note that you are going to need a lot of scavengers. To get rid of all the oxygen, you will have to put in a large excess of these scavengers into the chamber.
There is some information on oxygen scavengers on Wikipedia

It looks like "oxygen scavengers" are more targeted for food.  Is there anything targeted for metalwork?

Not that I know of. Mg as another comment mentions, may also be a good solution.


text:
How to obtain barium hydroxide
I am trying to perform the demonstration of a strongly endothermic reaction reacting barium hydroxide with ammonium nitrate.  I have been scouring the web for ways to get barium hydroxide, but none have showed any promise.  The only way I have seen to obtain it is by heating the nitrate to decomposition and reacting the oxide with water.  I would rather avoid this as I have no way to deal with the nitrogen dioxide gas that would result.

Google buy barium hydroxide.  You'll find several sellers on various science sites and eBay.  There are at least 5 different vendors I found immediately.  You can also buy barium oxide and dissolve it in water.

You can buy barium hydroxide. See the listing from Flinn Scientific although Flinn is by no means the only seller. However, Flinn's safety information is written at a very accessible level.

Flinn will only sell to certified teachers and schools


text:
Reversible and Irreversible adiabatic expansion

A $\mathrm{200.0\ m^3}$ balloon at $253.0\ \mathrm K$ ascends to a higher altitude. If the initial pressure inside the balloon is $325$ millibar, and it ascends to a latitude with a pressure of $7.45$ millibar, by what percent does the absolute temperature change:

Assuming the process is reversible and adiabatic
Assuming the process is irreversible and adiabatic

All of the formulae I have on this subject include moles. The question does not give a value of moles. Where do I start?

Try to use the formulas you have to calculate the moles.

I assume that you have to assume that the gas behaves ideally.

The change in internal energy $U$ is
$$\Delta U=Q+W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system.
Since the process is adiabatic, no heat is transferred into or out of the system, i.e. $Q=0$ and thus
$$\Delta U=W$$
The reversible expansion is performed continuously at equilibrium by means of infinitesimal changes in pressure $p$.
The corresponding infinitesimal pressure-volume work $\delta W$ is
$$\delta W=-p\,\mathrm dV$$
For an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$. Since $n$ is constant in a closed system, the change in internal energy $U$ is
$$\mathrm dU=C_V\,\mathrm dT$$
where $C_V$ is heat capacity at constant volume.
Equating $\mathrm dU$ and $\delta W$ yields
$$C_V\,\mathrm dT=-p\,\mathrm dV$$
and since $pV=nRT$ for an ideal gas
$$\begin{align}
C_V\,\mathrm dT&=-\frac{nRT}V\,\mathrm dV\\
\frac{C_V}T\,\mathrm dT&=-\frac{nR}V\,\mathrm dV
\end{align}$$
Integration from the initial state ($T_1,V_1$) to the final state ($T_2,V_2$) yields
$$\begin{align}
\int\limits_{T_1}^{T_2}\frac{C_V}T\,\mathrm dT&=\int\limits_{V_1}^{V_2}-\frac{nR}V\,\mathrm dV\\
C_V\int\limits_{T_1}^{T_2}\frac1T\,\mathrm dT&=-nR\int\limits_{V_1}^{V_2}\frac1V\,\mathrm dV\\
C_V\ln\left(\frac{T_2}{T_1}\right)&=-nR\ln\left(\frac{V_2}{V_1}\right)\\
\ln\left(\frac{T_2}{T_1}\right)&=-\frac{nR}{C_V}\ln\left(\frac{V_2}{V_1}\right)\\
\frac{T_2}{T_1}&=\left(\frac{V_2}{V_1}\right)^{-nR/C_V}\\
&=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}
\end{align}$$
where $C_{\mathrm m,V}$ is molar heat capacity at constant volume.
Since $C_{\mathrm m,p}-C_{\mathrm m,V}=R$, the exponent can be rewritten
$$\begin{align}
\frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}\\
&=\left(\frac{V_1}{V_2}\right)^{C_{\mathrm m,p}/C_{\mathrm m,V}-1}\\
&=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
\end{align}$$
where $\gamma$ is the ratio of the heat capacities $\gamma=C_p/C_V=C_{\mathrm m,p}/C_{\mathrm m,V}=c_p/c_V$. For a monoatomic ideal gas, $C_{\mathrm m,p}=\tfrac52R$ and $C_{\mathrm m,V}=\tfrac32R$, and thus $\gamma=\tfrac53$.
Note that, in the given question, the volumes $V_1$ and $V_2$ are unknown. However, the values for the pressures $p_1$ and $p_2$ are given. Since $pV=nRT$ for an ideal gas,
$$\begin{align}
\frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
T_2\cdot V_2^{\gamma-1}&=T_1\cdot V_1^{\gamma-1}\\
\frac{p_2V_2}{nR}\cdot V_2^{\gamma-1}&=\frac{p_1V_1}{nR}\cdot V_1^{\gamma-1}\\
p_2\cdot V_2\cdot V_2^{\gamma-1}&=p_1\cdot V_1\cdot V_1^{\gamma-1}\\
p_2\cdot V_2^\gamma&=p_1\cdot V_1^\gamma\\
\frac{p_2}{p_1}&=\left(\frac{V_1}{V_2}\right)^\gamma
\end{align}$$
In contrast to the reversible expansion, an irreversible expansion is not performed continuously at equilibrium by means of infinitesimal changes in pressure. In the limiting case, the value of the pressure $p$ abruptly changes from $p_1$ and $p_2$. The following expansion proceeds in one step against a constant pressure $p_2$. Hence,
$$\delta W=-p_2\,\mathrm dV$$
and thus
$$\begin{align}
C_V\,\mathrm dT&=-p_2\,\mathrm dV\\
\int\limits_{T_1}^{T_2}C_V\,\mathrm dT&=\int\limits_{V_1}^{V_2}-p_2\,\mathrm dV\\
C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right)
\end{align}$$
Since $pV=nRT$ for an ideal gas,
$$\begin{align}
C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right)\\
&=-p_2\left(\frac{nRT_2}{p_2}-\frac{nRT_1}{p_1}\right)\\
&=-nRT_2+nRT_1\frac{p_2}{p_1}\\
C_VT_2+nRT_2&=C_VT_1+nRT_1\frac{p_2}{p_1}\\
T_2\left(C_V+nR\right)&=T_1\left(C_V+nR\frac{p_2}{p_1}\right)\\
T_2\left(C_{\mathrm m,V}+R\right)&=T_1\left(C_{\mathrm m,V}+R\frac{p_2}{p_1}\right)\\
\end{align}$$

Why is U=nCvdT for ideal gas...... It should be valid only for isochoric yet this is used every process

@Abhinav $du =nC_vdT$ is valid for an ideal gas, irrespective of what process we are talking about. It is a fact for any ideal gas.

where Cm,VCm,V is molar heat capacity at constant volume.
Since Cm,p−Cm,V=RCm,p−Cm,V=R, the exponent can be rewritten
T2T1=(V1V2)R/Cm,V=(V1V2)Cm,p/Cm,V−1=(V1V2)γ−1
T2T1=(V1V2)R/Cm,V=(V1V2)Cm,p/Cm,V−1=(V1V2)γ−1
where γγ is the ratio of the heat capacities γ=Cp/CV=Cm,p/Cm,V=cp/cVγ=Cp/CV=Cm,p/Cm,V=cp/cV. For a monoatomic ideal gas, Cm,p=52RCm,p=52R and Cm,V=32RCm,V=32R, and thus γ=53γ=53.
Note that, in the given question, the volumes V1V1 and V2V2 are unknown. However, the values for the pressures p1p1 and p2p2 are given. Since pV=nRTpV=nRT for an ideal gas,
T2T1T2⋅Vγ−12p2V2nR⋅Vγ−12p2⋅V2⋅Vγ−12p2⋅Vγ2p2p1=(V1V2)γ−1=T1⋅Vγ−11=p1V1nR⋅Vγ−11=p1⋅V1⋅Vγ−11=p1⋅Vγ1=(V1V2)γ
T2T1=(V1V2)γ−1T2⋅V2γ−1=T1⋅V1γ−1p2V2nR⋅V2γ−1=p1V1nR⋅V1γ−1p2⋅V2⋅V2γ−1=p1⋅V1⋅V1γ−1p2⋅V2γ=p1⋅V1γp2p1=(V1V2)γ
In contrast to the reversible expansion, an irreversible expansion is not performed continuously at equilibrium by means of infinitesimal changes in pressure. In the limiting case, the value of the pressure pp abruptly changes from p1p1 and p2p2. The following expansion proceeds in one step against a constant pressure p2p2. Hence,
δW=−p2dV

Welcome to Chemistry.SE! Please go through this page to learn how to format your posts containing math expressions. It may look intimidating at first, but is really straightforward, and enhances the readablilty of your posts greatly!


text:
Why does H bond to S in HSO3?
In the structure of $\ce{HSO3-}$, we see that the H is bonded directly to the S, not to the O's, as would be the case in other oxyacids. Why is this the case?

It's not that unusual, BTW. Look at $\ce{HPO3^2-}$ - it's the same.

There are two distinct forms of $\ce{HSO3-}$ (see picture below; all horizontal bars are formal $-$ signs).

Only the leftmost of these two forms can be called hydrogen sulphite (or ‘bisulphite’, if you really insist); the correct name of the rightmost form is sulphonate. The two differ in their chemical characteristics; hydrogen sulphite is, for example still acidic and has a lone pair on sulphur, while sulphonate is not.
Johansson, Lindqvist and Vannerberg from Göteborg solved the crystal structure of $\ce{CsHSO3}$ in 1980 and concluded that:

the $\ce{S-O}$ bond lengths are much closer to those in sulphate than those in sulphite ($\ce{Na2SO3}$)
the $\ce{O-S-O}$ angle is $113°$ and thus larger than the tetrahedral angle $\approx 109°$ (expected in sulphate) and much larger than the angle in $\ce{SO3^2-}$
the ion’s point group is $\mathrm{C_{3v}}$, so all three oxygens are equal
there is IR evidence of a $\ce{S-H}$ bond.

From this they deduced that $\ce{CsHSO3}$ be, in fact, cesium sulphonate, not cesium hydrogen sulphite.[1]
The first reference I found for the structure of a true hydrogen sulphite was the crystal structure of cis-$\ce{[Ru(bpy)2(HSO3)2]}$ by Allen, Jeter, Cordes and Durham from Arkansas in 1988. In their structure, $\ce{HSO3-}$ coordinates to the central $\ce{Ru(II)}$ ion via the sulphur atom: $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$. This coordination is, of course, only possible if there is a lone pair on sulphur, and thus the structure must be that of a hydrogen sulphite.[2]
(In my introductory chemistry class, Prof. Klüfers stated that structures for both forms be known and published.[3] Alas, I didn’t find the crystal structure of a simple metal hydrogen sulphite anywhere quickly.)
It gets easier when dealing with the corresponding acids’ esters. The hydrogen on sulphonic acid can be replaced with organic residues creating the vast class of organosulphonic acids. Sulphonic esters are also possible; they would have an organic residue attached to both the sulphur directly (replacing the hydrogen atom) and one of the oxygens. Every mesylate or tosylate (in formulae: $\ce{OMs}$ or $\ce{OTs}$) is, in fact, a sulphonic ester. Sulphurous esters are less prevalent in organic chemistry but do exist (e.g. dimethyl sulphite, $\ce{OS(OCH3)2}$). Of course, the chemical properties of methyl mesylate ($\ce{H3C-SO2-OCH3}$) — a sulphonate — and dimethyl sulphite are substantially different.

After discussing $\ce{H2SO3}$ and its salts, it is important to mention the wrong assumption you proposed in your question:

we see that the $\ce{H}$ is bonded directly to the $\ce{S}$, not to the $\ce{O}$’s, as would be the case in other oxyacids.

This is not generally true. Phosphorous at least displays a similar chemistry, even to the extent that ‘hydrogen phosphite salts’ cannot be isolated and phosphonate salts are isolated instead. (One can synthesise both trimethyl phosphite and dimethyl methylphosphonate, the corresponding phosphorous and phosphonic esters, though.) I am not firm in the oxyacid chemistry of heavier homologues of phosphorous and sulphur, but it would deeply astound me if this chemistry is not observed there at least to some degree.
Note that the halogens do indeed follow your proposed rule: all their oxyacids are monoprotic and contain only a single, oxygen-bound hydrogen in neutral state.

References:
[1] L.-G. Johansson, O. Lindqvist, N.-G. Vannerberg, Acta Cryst. 1980, B36, 2523. DOI: 10.1107/S0567740880009351.
[2] L. R. Allen, D. Y. Jeter, A. W. Cordes, B. Durham, Inorg. Chem. 1988, 27, 3880. DOI: 10.1021/ic00295a003.
[3] Refer to the internet scriptum of general and inorganic chemistry 1 (in German) as taught 2006 to 2008 and likely will be taught again in 2015.
(Note that the articles are most likely behind a paywall)

Arsenous acid is truly arsenic hydroxide. I guess, that for heavier analogues inert pair is in full effect.

@permeakra That’s actually a consideration I didn’t make.

Just to add on, I suppose in the specific case of As the so-called "inert pair" arises due to the d-block contraction. Two more rows down, the bismuth analogue also has the structure Bi(OH)3 and it is actually basic - this is probably lanthanide contraction + relativistic effects at work. Unfortunately the Sb analogue has "not been well-characterised".

The "inert pair" here could be weak bonding to hydrogen, which would come about when atoms just have large orbitals and overlap poorly with the hydrogen. For a different form of this effect, see this answer, in which electropositive metals still bond to hydrogen but the expected covalency is not there.

Another very good point, @OscarLanzi.


text:
Analytical Chemistry Textbook Recommendation
Could anyone recommend a good book for self-learning that provide a clear, step by step guide on solving such chemical equilibria questions.
It should cover the following topics: weak acid and strong base titration, buffers, redox reactions, electrochemical reactions, complexation reactions, potentiometric titrations.
Preferably, I would like it if the textbook also provide work solutions to the practice questions it gives. I heard that Fundamentals of Analytical Chemistry by Skoog is good and covers the topics that I have listed above but I don't think it provides any worked solutions to its questions.
Any help will be much appreciated.

Fundamentals of Analytical Chemistry is a fine book and it does have an student solution.

Stack exchange doesn't encourage link-only answers. Please elaborate your opinion behind citing this.


text:
Is there a way to experience peppers’s endorphins effects without the burning sensation?
When preparing dinner, my friend and I sometimes argue over the amount of pepper. It is often ‘too hot’ for him while I enjoy the heat effects (endorphins) of capsaicin (chili peppers) and piperine (black pepper). Is there a way to experience the endorphins effects of pepper without a burning sensation?

You might find more experienced people over at [cooking.se]

Anecdotally, many people claim that you can use an acidic solution like lemon juice and/or vinegar to make it seem less spicy (even tomatoes). This is one of the reasons vinegar is used for hot sauce. You can dilute hot sauce with vinegar to make it less hot but diluting with water doesn't work well. The same with chili. People use tomato sauce rather than water to dilute chili that is too hot.


text:
Why do we excite electrons before hybridisation?
In case of hybridisation like $\mathrm{sp^3}$ hybridisation of carbon or nitrogen why do we excite electrons before hybridisation? Why couldn't we just mix orbitals without exciting electrons?

The way I see it, it doesn't make any difference. Bear in mind that hybridisation is just a mathematical concept, not a physical process.

There is nothing like sp3 hybridised isolated carbon. But you can describe the carbon in CH4 as sp3 hybridised. But you cannot say: I take elemental carbon in ground state, excite and hybridise it (whatever it means), attach four hydrogen radicals and get methane. Neither you can dissect the methane in reverse process. All this stuff is just headology.

Actually number of bonds formed by an atom is generally equal to number of unpaired electrons it has. Like oxygen and nitrogen make two and three bonds because they have these many unpaired electrons in their valence shell.
But, in its ground state, carbon got only two unpaired electrons, so it should make two bonds. But in reality carbon make four bonds. Two satisfy four bonds we need four unpaired electrons. This is possible to consider the excited state.
Hope its clear.

The excited $sp^3$ state does not exist on its own. See the comments on the question.


text:
Density of saturated n-octane in air at 300 K?
I am trying to calculate (if possible) the gas density of air saturated with n-octane at 300 K. Any suggestion of how to do this, or better, where I can look up the data?

Thermodynamic parameter values for many substances (not just water and steam) can be found using so-called steam tables. In the following, I use parameter values taken from the REFPROP – NIST Standard Reference Database 23, Version 9.0). For a simplified estimate, however, we only need the value of the vapour pressure of octane.
For simplicity’s sake, we may assume that the gas phase (containing air and octane) is an ideal mixture of non-reacting gases and that the liquid phase contains only pure octane. Thus, for the calculation of parameter values for octane, we may ignore the air and only consider the equilibrium in a system of pure liquid octane and pure gaseous octane.
Using tabulated data, we find that the saturation point for octane at a temperature of $T = 300\ \mathrm K$ corresponds to a pressure of $p_{\ce{C8H18(g)}} = 2060\ \mathrm{Pa}$.
The corresponding density can be estimated using the ideal gas law:
$$\begin{align}
p\cdot V &=n\cdot R\cdot T \\[6pt]
\frac nV &= \frac p{R\cdot T} \\[6pt]
\frac m{M\cdot V} &= \frac p{R\cdot T} \\[6pt]
\frac mV &= \frac {M\cdot p}{R\cdot T} \\[6pt]
\rho &= \frac {M\cdot p}{R\cdot T} \\[6pt]
&= \frac {114.23\ \mathrm{g\ mol^{-1}}\times 2060\ \mathrm{Pa}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&= \frac {0.11423\ \mathrm{kg\ mol^{-1}}\times 2060\ \mathrm{kg\ m^{-1}\ s^{-2}}}{8.314462618\ \mathrm{kg\ m^2\ s^{-2}\ mol^{-1}\ K^{-1}\times 300\ \mathrm K}} \\[6pt]
&= 0.09434\ \mathrm{kg/m^3}
\end{align}$$
By way of comparison, the corresponding density given by REFPROP is $\rho_{\ce{C8H18(g)}} = 0.094649\ \mathrm{kg/m^3}$.
Assuming that the total pressure is equal to the sum of the partial pressures of the individual gases in accordance with Dalton’s law, we may calculate the partial pressure for the remaining air:
$$\begin{align}
p_\text{tot} &= p_\text{air} + p_{\ce{C8H18(g)}} \\
p_\text{air} &= p_\text{tot} - p_{\ce{C8H18(g)}} \\
&= 101325\ \mathrm{Pa} - 2060\ \mathrm{Pa} \\
&= 99265\ \mathrm{Pa}
\end{align}$$
The density of dry air at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1770\ \mathrm{kg/m^3}$. The corresponding density at a partial pressure of $p_\text{air} = 99265\ \mathrm{Pa}$ is $\rho_\text{air} = 1.1531\ \mathrm{kg/m^3}$.
Thus, the total density of the gas mixture is
$$\begin{align}
\rho_\mathrm g &= \rho_{\ce{C8H18(g)}} + \rho_\text{air} \\
&= 0.094649\ \mathrm{kg/m^3} + 1.1531\ \mathrm{kg/m^3} \\
&= 1.2477\ \mathrm{kg/m^3}
\end{align}$$
By way of comparison, I made a quick simulation using REFPROP assuming that air is a mixture with the mole fractions $x_{\ce{N2}}=78.12\,\%$, $x_{\ce{O2}}=20.96\,\%$, and $x_{\ce{Ar}}=0.92\,\%$. I added an excess amount of octane $x_{\ce{C8H18}}=10\,\%$ (in order to obtain a liquid phase) and renormalized the composition to $100\,\%$.
The resulting density of the vapour phase at a temperature of $T = 300\ \mathrm K$ and a pressure of $p = 101325\ \mathrm{Pa}$ is $\rho_\text{g} = 1.2483\ \mathrm{kg/m^3}$.

Another way of doing this is to start out by evaluating the molar average molecular weight, and then plugging that into the ideal gas relationship.
First get the mole fractions of octane and air:
$$x_\mathrm{octane} = \frac{2060}{101325} = 0.02033$$
$$x_\mathrm{air} = 1 - 0.02033 = 0.97967$$
Next evaluate the molar average molecular weight:
$$M = 0.02033\cdot 144.23 + 0.97967\cdot 29 = 30.7$$
Next apply the ideal gas law relationship:
$$\rho = \frac{PM}{RT} = \frac{1\cdot 30.7}{0.0821\cdot 300} = 1.248$$
The units are grams per liter $(\pu{g L-1}),$ which is the same as kilograms per cubic meter $(\pu{kg m-3}).$


text:
How to minimize surface adsorption by choice of material?
I need to do buoyancy experiments in vapours consisting of organic molecules in air using a microbalance. This involves measuring the change in weight of a sealed glass sphere when exposed to the gas. The problem is that surface adsorption shows up as significant noise. Is glass the best material for this? What properties determines adsorption, and how can I minimize it?
I assume it is a function of surface energy. For example PVDF has an energy about 10x less than glass. However, does this simply mean that it is easier to get rid of the adsorbed molecules rather than limiting the quantity in the first place?

Well, PTFE has a surface energy several times less than glass, so experiment time...


text:
why is molybdenum a much stronger magnet than chromium
Molybdenum is well known for its strong magnetic properties. Why is Chromium with an identical set of unpaired electrons not nearly as strong?

Molybdenum (as well as chromium) is not magnetic at all. You must be confusing it with magnesium, which it not magnetic either.

For calculating chrome phase diagrams you certainly need to consider the magnetic contributions to the free energy, so I'll disagree with you.

Thanks to both I was in fact thinking about Neodymium and got confused! Obviously both Mo and Cr, in the solid, must be +6 cations so that no unpaired electrons are left to cause ferro-magnetism!