text:
Why does mixing inks make subractive colors?
I have little knowledge of color theory and chemistery. Recently I came to know that printers use CMYK colored inks and combine them in different proportions to achieve any color on print.
Let us suppose I fill white paper fully with red sketch then white light will fall on inked surface and all colors(wavelengths) except red will be absorbed and only red will be emitted/reflected. Good. Now I overfill that paper with green sketch. Now two things can happen:

Both the inks of sketches mix but do not interact molecully. Like two chemicals A and B mixed in water and both A and B are everywhere in water but they are still distinct molecules. A's and B's molecules don't react chemically.
Both the inks of sketches mix and interact molecuelly. Like HCL and Na2SO4 are mixed in water and they react molecually to make h2SO4 and NaCl.

Now if 1 is true then on any point on paper's surface I have two types of molecules, 1. that reflect red light and 2. that reflect green light. So when white light will fall on paper then from every point two colors will be reflected, red and green. When they'll reach someone's eye then the perceived color will be red+green = yellow. So inks should make additive color.
On the other hand if 2 is true then molecules of red and green ink will react chemically to make a new kind of compound. Now it is not necessary that the new kind of compound absorbs both red and green. It could emit/reflect any random color. So the paper could be of any color depending upon the chemical formula of inks used. So in this case too inks don't make subtractive colors.
Sadly in reality inks make subtractive color and the color of paper actually is dark brownish black not bright yellow. So in reality the mixed ink absorbs almost all colors including red and green which were remaining.
So my question is why does both red and green got subtracted from the mixed ink? In general why does mixture of inks produce subtractive color rather than additive?

The issue is that you're thinking about inks reflecting color, when in reality you should be thinking about the inks absorbing colors.
If the function of inks were to reflect colors, then by adding more and more ink (of a single color) you'd be able to make brighter and brighter colors, as more ink means more reflection. Instead, it's the white paper (rather than the ink) that's reflecting the light, and more and more ink (even of a single color) makes things darker because more and more light is being absorbed.
That's why mixing pigments/inks is a subtractive process - as you add inks, you increase the amount of light that's being absorbed.
So let's go to your example. Let's simplify things by considering only three wavelenghts of light: red, green and blue. Your red ink absorbs the green and blue light, whereas the green ink absorbs the red and blue light. Thus when you mix the two, each absorbs the light it would in isolation, and you have very small amount of blue light, and also very little red or green light, leaving you with a very dark, perhaps slightly yellowish-brown color.
Now consider subtractive inks. You have yellow ink, which absorbs just the blue light and you have magenta ink, which absorbs just the green light. You mix the two together, and the magenta ink absorbs the green light, the yellow ink absorbs the blue light, and the red light is free to be reflected, resulting in a red colored piece of paper from Y+M.
Generally speaking, modern inks (at least the standard, non-color changing variety) involve no chemical reactions in color production: mixing inks and pigments is solely a physical change based on absorption of light.

The ink is actually a liquid which lets some light to pass through it and absorbs particular wavelengths. Thanks for explaining. I was thinking printer uses Y + M to produce red color. Could it directly use red ink? But I guess CMYK are four inks which can be mixed in different proportions to produce any color so having inks for specific red, blue and green would be unnecessary overhead.

Yes, CMYK (in theory you could even omit the blacK) are all the inks one need to produce any color for the human eye.  But you have to know that two colors that look identical to humans, might in fact consist of different wavelengths (e.g. https://en.wikipedia.org/wiki/Spectral_line#/media/File:Simple_spectroscope.jpg), that's because the human eye has "just" 4 different sensors (https://en.wikipedia.org/wiki/Rod_cell#/media/File:Cone-response.svg)

@user31782 That's true - printers could, if they wanted, use a red ink directly. This is actually used sometimes by professional printing presses, and is called spot color. It's used frequently when there's a color that's difficult to get by mixing CMYK. But as it's typically more expensive than plain CMYK, its use is limited, and rarely used in consumer printers. (With the exception of dedicated photo printers, where the extra inks help with color fidelity.)


text:
Is varnish with a flash point above 40 celsius safe in a microwave oven?
I've painted regular white mugs and want to varnish them too.
But my varnish has the following specs:
Gloss varnish based on alkyd resin.
Flash point: Above 40 degrees celsius.
Is it safe to varnish on some sections of the painting on the outside of a mug? Or will it catch fire if microwaved to boil water or something?
I have no background in chemistry... took me a while searching around before I finally decided to just ask.

The flash point is for the solvent, i.e. for undried paint. Dried paint will have a considerably higher flash point, well over 100 degrees Celsius.
You should also consider toxicity. I don't think alkyd based varnish is approved for use in items you put in you mouth (like mugs).

Thanks! I'm only dabbing a little varnish on the outside of the mug, where a few designs have been painted. So that part won't be going inside the mouth.

Keep the varnish away from the rim and the mug away from the FDA and you should be fine.

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text:
Looking for a toxic substance that cannot exist at room temperature
I am writing a short story, and am trying to find a gas with the following properties:

It must have a very distinctive odor
It must be impossible for it to exist at room temperature under normal conditions
It must be toxic to humans

How large a pool of possible substances are there that meet these requirements? Is there a particular family or class of gases where I should be focusing my attention?
Some additional clarification (apologies for not including this when I first wrote the question):
The scenario I am writing is Cthulhu-based. The idea is that this substance is discovered where it shouldn't be able to exist. This is where the second criteria comes from. They will collect a sample of this stuff, analyse it and realise that its mere existence is impossible in the circumstances they found it. A completely hypothetical example might be a substance that spontaneously combusts in earth's atmosphere, or that immediately reacts in some other way with a gas present in air. This would mean that it cannot exist where the protagonists discover it, yet it does.
The toxicity is just intended to add some jeopardy to the situation

Your second and third requirements contradict each other. If it does not exist at room temperature, how can we get in contact with it, let alone get poisoned by it? Also, what's that with being pyrophoric? Pyrophoric compounds exist at room (and higher) temperatures just fine, as long as you don't expose them to air.

Would something like chlorine gas work? Though it is perfectly stable when pure, it doesn't exist naturally ("in normal conditions") because in short order it finds something to react with and turns into other compounds. It fits the two other conditions, too.

Does non-existance also apply if the compound in question is a liquid at SATP?

Does the edit I've just done clarify it at all?

@Wibbs After reading your edit: You might want to consider using a substance that contains a radioactive primordial nuclide that should not be able to exist today in such amounts because of its short half-life (i.e. it should be “extinct”, the found sample could even be described as “older than Earth”). The toxic effects could also be a result of the radioactivity, or just be caused by ordinary chemical properties of the substance.

@Loong A tellurium isotope "older than Earth"? Amazing idea! It can hardly get more Lovecraftian :) Plus, $\ce{H2Te}$ is nasty!

As Ivan Neretin already explained in his comment, you can't have it all.
Anyway, cyanogen, $\ce{N#C-C#N}$ would be a nice candidate. It is colourless, but highly toxic (H330). If it's cold enough, that is below -21 °C, you can even have a pond of it ;-)
In addition, it adds a historical space touch to the story. Back in 1910, when the return of Halley's comet was expected, cyanogen was spectroscopically identified in the comet's tail and the New York Times titled COMET'S POISONEOUS TAIL ;-).

Out of curiosity: Where (or how) did you find the hazard statement H330 for cyanogen? According to the official Annex VI of Regulation (EC) No 1272/2008 (CLP Regulation), cyanogen is only H331. I am aware, however, that many companies report H330.

@Loong I just checked it again to make sure that I didn't fatfinger it. The GESTIS Substance database really reports H330.

As already mentioned in other comments, your requirements are contradictory. You cannot have a gas that is non-existent at normal conditions but is also toxic to humans and has a distinctive odour at the same conditions.
Depending on when do you want the gas to present (e.g. to be toxic to humans or to be recognizable by its odour) and when do you want it to be absent (e.g. to be harmless or to be undetectable), you might need to introduce a change (like a switch) in your ambient conditions.
If you want the compound to be toxic first and non-existent later, you may consider very volatile compounds (gases, or liquids with a high vapour pressure), which are rapidly dispersed, e.g. by the wind. Thus, you can obtain a toxic concentration of the gas first and also can achieve concentrations below the detection limit later. Several chemical warfare agents are classified as non-persistent agents (i.e. they are likely to be used to kill and incapacitate rather than to deny access to ground) and thus suit this purpose. A typical example for such non-persistent agents is phosgene. It also has a distinctive odour, which is described as resembling freshly-cut hay or grass.
You may also consider a compound that is sensitive to heat and/or sunlight. Thus you could obtain a toxic concentration under dark or cold conditions (e.g. at night) and decompose the gas later using heat and/or sunlight (e.g. after sunrise). However, I don’t know any typical example compound that fulfils these requirements, is also sufficiently volatile and toxic, and has a characteristic odour.
If you want the compound to be absent first and toxic later, you may consider taking advantage of the different vapour pressures of compounds in dependence of the temperature. A typical example for this is the use of sulfur mustard (mustard gas) in winter, e.g. contamination of snow. Sulfur mustard freezes at about 14 °C and therefore detection systems (including the human nose) relying on detection of vapour can fail. If contaminated snow is carried (e.g. sticking to the boots of persons) into buildings, vehicles, or other warm places, you can achieve a toxic concentration in the air. The odour of sulfur mustard is described as resembling garlic, mustard, or leeks; however, the smell is actually caused by impurities of the technical product.
If you want the compound to be absent first, you may also consider using a chemical reaction to generate the toxic compound only when you need it. A typical example is sarin. Modern binary sarin weapons do not contain any sarin; they actually contain methylphosphonic difluoride, isopropyl alcohol (propan-2-ol), isopropylamine (propan-2-amine), and another compound (which is fortunately missing in most cookbooks that are circulating in the internet, and I prefer to ignore the details here, too). The compounds are mixed when the weapon is used (e.g. upon impact of the warhead) and react to the actual sarin. Pure sarin is odorless; however, the smell of impure sarin is described as resembling mustard or burned rubber. Furthermore, the characteristic smell of the precursors would be present (e.g. the fishy, ammoniacal odour of isopropylamine).

Perhaps trioxygen $\ce{O3}$ is a good fit.

It must have a very distinctive odor.

Ozone has a strong, sharp smell. Highly irritable. It should be detectable to humans at concentrations of $0.01\ \mu\mathrm{mol}\  \ce{O3}$ per one mole of air. By mass in dry air, that comprises about $1.66\cdot10^{-6}\%$ or $16.6\ \mathrm{ppb}$.

It must be impossible for it to exist at room temperature under normal conditions.

The higher the concentration of ozone, the less stable it is. Even with still air and no humidity, half-life of ozone is about $25$ hours. As humidity, air flow, and temperature increase, ozone's stability decreases.

It must be toxic to humans.

No worries here. Lowest lethal concentration $\mathrm{LC_{Lo}}$ for humans is $50\ \mathrm{ppm}$ for $30\ \mathrm{min}$. At lower concentrations, it foremost harmes one's respiratory system but has also been shown to increase chances of heart attack and premature death. Trioxygen will explode if the concentration is too high.
So depending on your compassion for the protagonists, vary the concentration accordingly. Probably it is 'good to have' some $\ce{25\ \mathrm{ppm}}$. Thus there is real danger; but should not be lethal.

The idea is that this substance is discovered where it shouldn't be able to exist.

So in a long-closed room, ozone should have mostly decomposed to ordinary dioxygen.
$$\ce{O3->O + O2}$$
$$\tag{very quick}\ce{O + O -> O2}$$

They will collect a sample of this stuff, analyse it and realise that its mere existence is impossible in the circumstances they found it.

Being the smart (assumingly) characters they are, they immediately know that ozone can qualitatively be detected by the reaction
$$\ce{2KI + O3 + H2O->I2 + 2KOH + O2}.$$
(A paper is dipped in a solution of potassium iodide $\ce{KI}$, or the entire solution is exposed to the atmosphere containing ozone.)
Iodine $\ce{I2}$ colours the paper or solution dark brownish.

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text:
Why is Trouton's rule not valid for liquids with boiling points less than 150 K?
This is a question from the last year's semester examination.

Why is Trouton's rule not valid for liquids with boiling  points less
  than 150 K?

I searched the internet but only found that the exceptions mentioned include liquids with strong intermolecular attraction like hydrogen bonding and some similar cases. Can anyone give me a proper answer?

We prefer to not use MathJax in the title field, see here for details.

Related question: http://chemistry.stackexchange.com/questions/18967/validity-of-troutons-rule

@JonCuster Thanks for the help. However it doesn't answer my question.

Ok, perhaps you could clarify what you need more insight on? The answers in the question I pointed out do at least touch on (perhaps too indirectly) on why the entropy changes might be different at low temperatures.

@JonCuster My question specifically revolves around the temperature 150 K. Low temperatures may cause h bonding or so, but what id so special about 150 K?

I'm not sure there is anything particularly 'special' about it, other than being roughly half way from room temperature to absolute zero, and you know that something has to happen between here and there...

When a liquid vaporises its entropy goes from a modest value to a significantly larger one.  The ratio of $\Delta H_{vap}/T = \Delta S_{vap}$ is found to be approximately constant at the boiling point. Some authors give a value of $80~ \pu{Jmol^{-1}K^{-1}}$ others $90~ \pu{Jmol^{-1}K^{-1}}$.
Experimental values vary rather more than this and for gases such as neon, nitrogen, oxygen and methane whose liquids all boil below $150$ K, have values that are in the range $65 - 75$, benzene, many 'normal' liquids and liquid sodium, lithium and iodine, in the range $80-90$ and ethanol, water, hydrogen fluoride in the range $105 - 115 ~ \pu{Jmol^{-1}K^{-1}}$. Thus there seems to be nothing unusual about $150$ K but rather an influence from  intermolecular interactions.
The value of $ \approx 80~ \pu{Jmol^{-1}K^{-1}}$ corresponds to a cohesive energy of $\approx 9.5 kT$ per molecule and so the boiling point gives an indication of the strength of the cohesive energy holding molecules together in the condensed phase. When the cohesive energy exceeds this value, as in water, then the ratio $\Delta H/T$ is larger and conversely the ratio is smaller when the cohesive energy is less as in Neon or methane.
The $\approx 9.5 kT$ minimum energy per molecule is quite a modest energy; if a molecule has $6$ near neighbours this corresponds to about $3kT/2$ per interaction between two molecules, roughly the average thermal energy.


text:
Why is time-dependent density-functional theory (TD-DFT) used to describe excited states?
It seems that TD-DFT is a common choice for a computationally efficient method to describe excited states. But if DFT is a single-configurational method, and hence a ground state method, how does the TD-DFT method treat the excited states?

I do not have the time to give you that answer. But I do have a recommendation: Try to understand CIS (TDA) first. One can view TD-DFT as an extension to CIS if you like. Here is a good publication that (sort of) explains both: DOI: 10.1021/cr0505627.

This must be the first time I've heard time-dependent DFT described as 'computationally efficient'!

Less expensive than, say, CASSCF or multi-configurational coupled cluster?

Like most of DFT-related, because it works.

TD-DFT is used and developed because there is an excited state theorem, the Runge-Gross theorem, that is analogous to the Hohenberg-Kohn theorem for ground states.

(...) the Runge-Gross theorem shows that for a many-body system evolving from a given initial wavefunction, there exists a one-to-one mapping between the potential (or potentials) in which the system evolves and the density (or densities) of the system.
The Runge–Gross theorem provides the formal foundation of time-dependent density functional theory. It shows that the density can be used as the fundamental variable in describing quantum many-body systems in place of the wavefunction, and that all properties of the system are functionals of the density.

This means that any state of the system can be described by an electron density, not just the ground state. If I understand this correctly, the R-G theorem is more general than the H-K theorem, which can be derived from the R-G theorem by removing the explicit time dependence.
Here is a paywalled link to their original paper.
An interesting comment from the Wikipedia proof is

The proof relies heavily on the assumption that the external potential can be expanded in a Taylor series about the initial time. The proof also assumes that the density vanishes at infinity, making it valid only for finite systems.

In the case of applied fields, such as a slowly varying or oscillating electric field, these may be treated with perturbation theory rather than self-consistently; this is the same thing as the Taylor expansion. A consequence is that TD-DFT may not be valid in the presence of strong applied fields or a strong external potential.

I don't think you were asking for how TD-DFT works from a computational level, you were only looking for the justification, so I have neglected to make the connection with TD-HF (RPA).

TD-DFT is a complete time-dependent theory of the motion of many electrons. It is to the time-dependent Schrödinger equation what ordinary DFT is to the time-independent Schrödinger equation.
What is most commonly referred to as TD-DFT is in fact a particular application of TD-DFT. Irrespective of a method, the excitation energies of a molecule or a solid can be in general calculated as frequencies at which the linear density response function (or, really, any linear response function) diverges. (Density response function says how a density of a system at some point changes because of a potential change at a different point.) This can be understood as frequencies at which the electronic motion resonates with the oscillating electromagnetic field; that is, the field that excites the molecule. (This also partially explains the correspondence between energy and frequency.) Because TD-DFT is a full theory, it can be also used to calculate the linear density response, and hence the excitation energies. Formally, this translates into finding the poles of the density response function. This calculation is what is most often referred to as TD-DFT. (Poles are particular types of singularities in complex functions.)

This is a complicated question you ask. You will most likely get an answer like the one provided by azag0. I would be very surprised if this helped you in any way.
I want to rephrase your question: "How does one get excitation energies from TD-DFT?"
I'm gonna assume that this is completely equivalent with the question you've asked, maybe with less of an philosophical touch.
If you ask me, the best path to understanding (in principle) what TD-DFT does is to understand CIS (TDA). You can see TD-DFT as an extension to CIS. This is not how you find it in text books, but the reason why you ask this question is somewhat related to the problem that you don't find it like this in text books. The way you asked the question tells me you have access to publications, maybe you are studying at an university, so I highly recommend a publication by Dreuw and Head-Gordon called "Single-Reference ab Initio Methods for the Calculation of Excited States of Large Molecules", Chem. Rev. 2005, 105 (11), 4009–4037. DOI: 10.1021/cr0505627.
If you do not understand this publication, I would come back and ask simpler questions first, questions one might be able to answer in a simple post on an online board.

Your initial answer was converted to a comment to the question, but it has not been deleted. Please take the [tour] to learn the difference. In extension to this, your second answer does not really improve on your first comment. Please include the essential parts here and link to the reference. The article you refer to is behind a paywall and while only one user asked the question, the answers must be helpful to future visitors, too. Also I kindly ask you from pointing fingers at users, especially when you do not have any proof for your accusation.

As far as I understand it, the initial answer was criticized for not posting a reference and linking a DOI which might be confusing to some people. This is what I explained in my new comment. I find it very pedantic to downvote a helpful answer because of technical details. Yes, it is behind a paywall, but would you rather have this question unanswered (which I think it will remain at this point) than a link to something anybody that has some university account can read for free?

Your answer was downvoted, flagged and converted because it does not answer the question. As a comment it is perfectly fine, this is why it was kept. Your answer was criticised because it only contained a reference. While a doi won't change for the foreseeable future, the paywall will prevent the majority of the users to access this information. Again as a comment no problem, as an answer it is not suitable.


text:
Is oxygen necessary for burning?
Is oxygen really necessary for burning things?
What if there was no oxygen around? Would there still be a way to burn things?

Yes, absolutely. Burning refers to oxidation accompanied by a flame and there are numerous reactions in which oxygen takes no part. Most of these are either with elementary halogens, or compounds thereof.
Here are some examples.
\begin{align}
\ce{2K + F2 &-> 2KF}\\
\ce{2Fe_\text{wool} + 3Cl2 &-> 2FeCl3}\\
\ce{H2O + ClF3 &-> HF + HCl + OF2}\\
\ce{2H2O + ClF3 &-> 3HF + HCl + O2}\\
\ce{2Cl2 + 2OF2 &-> 4ClF + O2}
\end{align}
So in fluorine chemistry, you don't oxygenise it; it flourises you.

According to the dictionary, burning is to undergo rapid combustion or consume fuel in such a way as to give off heat, gasses, and, usually, light; be on fire:
Burning is an exothermic process of oxidation. Burning can happen in almost all oxidative environments under the right circumstances. Sparklers don't need air to burn because they contain oxidants such as $\ce{KNO3}$ or $\ce{KClO3}$. Another example is Iron and Chlorine. They react brightly creating a smoke of $\ce{FeCl3}$ or Ferric Chloride. This is represented by the following equation.
$\ce{2Fe + 3Cl2->2FeCl3 + Heat}$
As written in the equation, heat is produced. This makes it an exothermic oxidative reaction. In other words, burning.

Back in year ten, we were shown an experiment that was supposed to introduce us to the concept of redox chemistry, but it might as well serve as an example of how something can burn without oxygen:

A cylinder containing $\ce{CO2}$ was prepared, the top closed off with a removeable piece of glass. A bit of magnesium band was ignited and dropped into the carbon dioxide.
It continued burning with soot appearing on the walls of the cylinder.

The explanation of the above experiment was the following simple redox reaction:
$$\ce{2 Mg + CO2 -> 2 MgO + C}$$
This was obviously combustion in carbon dioxide atmosphere.

(+1) for magnesium, a metal that isn't choosy. Carbon dioxide, nitrogen, water? Nah, baby, I'll rather keep on burnin'. Even sand won't put the sucker out (unless there's a lot of it).

I think the question is asking about burning without any oxygen involved. The electropositivity of the magnesium simply pulled the oxygen out of the carbon dioxide leaving magnesium oxide and carbon

@ScienceAsap I read the question as ‘is oxygen gas required for burning?’

"Burning" is a rapid exothermic oxidation reaction, so you need some sort of oxidizer. Gaseous oxygen serves quite well, as well as oxygen bound up in various solids like nitrates (e.g. potassium nitrate in gunpowder).
If you don't want oxygen in any form, the two halogens fluorine and chlorine are also excellent oxidizers and can cause things to burn quite well.
Fluorine
https://www.youtube.com/watch?v=ca4JtA_EZzY
Chlorine
https://www.youtube.com/watch?v=aK85PZX2xNE


text:
How is the selectivity of a drug ensured?
How is a drug created that would target a particular site?
I always wonder when I take a pill for headache, how the pill knows that its my head and not my knee. Or is it that it interacts with all the organs but it interacts effectively only with the brain?

whats wrong with the question?

It shows a clear lack of research, probably. Note I didn’t downvote yet but may do so later.

The deeper you dig, the more you realize just how non-selective most drugs are...

You could read many books about drug design - it's like you asked how food is cooked.

They don’t. Drugs are molecules, and in spite of what homeopathy says, molecules have no brain, no desire and no memory. They will perform random motion depending on the corresponding Gibbs free energy.
Therefore, the drug you swallow needs to be designed so that it either survives the stomach, duodenum, liver-produced enzymes, pancreas-produced enzymes, gall, and whatever else it encounters on its was to the small intestine from whence it can be absorbed into the blood stream or it must be modified only in ways that are favourable to the pharmacist. Then, second obstacle: the blood enriched with nutrients and other compounds coming from the intestines is passed by the liver where a number of enzymes takes care of everything the body does not approve of; the drug either again needs to survive this or be modified in a favourable way. This is tested in screenings.
Henceforth, the active ingredient is in the blood stream, and it randomly follows the direction the blood takes. Some of your headache medicine will end up in your left toe, some in your little finger, some will circle the heart via the coronary arteries and so on. Only some will actually go to the brain where it is supposed to have an effect. And while passing through the brain in the blood stream, only some of the molecules actually exit to enter the tissue. Remember also, that the brain is pretty special amongst destinations in the body due to the blood-brain barrier.
Then, finally selectivity. If you are lucky, the drug will fit selectively only into a pocket of one or a select few enzymes, selectively inhibit or enhance their function and thereby cause a physiological effect. However, you might be unlucky and have a less specific drug. In that case, you experience what is known as side effects.
Pharmaceutrical companies spend a lot of money on designing drugs and testing them so that they perform maximally well (‘maximally’ defined in pharmacological terms) and have minimal side effects. Yet, sometimes something critical is missed; one of the most well-known cases may be contergan, but for example there have been recent cases of contraceptive pills causing embolies, iirc.

Further reading on crossing the blood$-$brain barrier.


text:
Structure of xenon hexafluoride

The central atom has a hybridization of $\mathrm{sp^3d^3}$. Thus, its structure should be pentagonal bipyramidal.
Why is it not that but a distorted octahedron?

related http://chemistry.stackexchange.com/questions/34073/what-are-the-structures-of-potassium-superoxide-and-xenon-hexafluoride

Possible duplicate of What is hybridisation of XeF6 in solid state?

NOT a duplicate of @Mithoron's second comment. The answers may be similar, but this question asks about the structure due to an assumed hybridization, whereas the linked "duplicate" asks about the hybridization itself.

While the question linked in @Mithoron's first comment does provide the symmetry group of $\ce{XeF6}$, it does not indicate the actual structure of the molecule. Further, it's closed.

Why is it not that but a distorted octahedron? || It actually is, though for different reasons.

This is one of the many reasons why hybridisation including d-orbitals fails for main-group elements.
Xenon in $\ce{XeF6}$ is not hybridised at all. Instead of invoking populated core d-orbitals or energetically removed d-orbitals (remember the aufbau principle: the next shell’s s-orbital has a lower energy than the d-orbitals you are proposing to include in hybridisation!) xenon just offers its three p-orbitals $\mathrm{p}_x, \mathrm{p}_y$ and $\mathrm{p}_z$ for four-electron-three-centre bonds. These 4e3c bonds can be understood using the following two mesomeric structures:
$$\ce{F^-\bond{...}Xe^+-F <-> F-Xe^+\bond{...}F-}$$
Each $\ce{Xe-F}$ bond has a bond order of ½, and for each fluorine there is another with a bond angle $\angle(\ce{F-Xe-F}) \approx 180^\circ$ as part of the same 4e3c bond.
Also note that this means that xenon’s lone pair is comfortably located in the $\mathrm{5s}$ orbital.

So, the molecules of $XeF_6$ are arranged in a lattice with a structure that has more similarity with transition metal halide salts with bridge-bonding, right?

@EashaanGodbole Erm … what? I’m sorry, I don’t understand what you’re asking …

@ Jan The bonding between xenon and the fluorides is 3c-4e like the titanium chloride salts...?

@EashaanGodbole I’m not familiar with solid state ionic structures too much, but $\ce{TiCl3}$ is a typical ionic structure with closest packing of anions and cations in the octahedral voids. This doesn’t compare well to molecular structures as are present in xenon compounds which feature distinct $\ce{XeF_{2n}}$ molecules.

Correct to a first approximation; the 3c4e model doesn't fully explain XeF6, though, as it is not a static octahedral molecule. [Of course, as you said, there is no (or at most, little) d-orbital involvement.]


text:
Why did my beer slowly froth up after removing from freezer?
I got a beer out of the freezer last night and opened it, only to realize that it was almost frozen, so I left it to warm up. 10mins later I hear a dripping noise and realize that the beer had overflowed and was starting to spill off the table.
What causes this slow expansion of beer after it is warmed from a near 0°C temperature?

Melting takes time.
As your beer is thawing, temperature remains roughly constant. Then the beverage will slowly assume ambient temperature.
Solubility of gases (e.g., $\ce{CO2}$) in common solvents usually decreases with warming. This is because the breaking of solvent$-$solvent interactions are not compensated by new interactions between solute and solvent. In other words, solubility of gases is often exothermic.

The bottom of the beer might have warmed more quickly, causing a buildup of carbon dioxide under a layer of still-frozen beer. Reaching some critical value for pressure, the top popped. According to Henry's law, immediately after even more gas bubbles are formed.
Effects due to positive thermal expansion of beer are probably rather small but these would only aggravate any overflow.

For further reading on similar subjects, see

Does shaking increase the pressure inside a bottle of champagne?,
Plateau–Rayleigh instability.

Are you using popped figuratively? Because he stated the beer container (can? bottle?) was opened before it was warned.

@electronpusher Pretty literally I think. In this case, it was the cylinder-shaped beer ice in the neck of the bottle that might've popped. I am actually quite surprised that the mess didn't occur in his freezer; in general, expanding ice will have done the opening for you ;)

I'm also surprised the expansion didn't break the container in the freezer. Is it possible that all of the pressure built up during the expansion of water to ice was somehow stored as a potential energy? Thus when he opened the container the pressure was not immediately dissipated, but as it thawed this potential energy was released or converted to drive the expansion...?


text:
Determining the bond order
I have the following molecule.

And I have to determine the decreasing  order of bond order of $\ce{C_1-C_2,C_3-C_6,C_2-C_3}$
I know that I have to take into consideration the various resonating structures but can't get the final answer. In my view, $\ce{C_1-C_2}$ should have the highest bond order because of hindrance by $\ce{C_3-C_6}$.
I have this resonating structures.the second one has least contribution because of separation of positive and negative charges.so we could perhaps ignore it.in the other two each time either carbon has double bonds which should imply same bond order.
 Please help me with a proper reasoning.

Start by actually drawing all of the resonance structures.  What do you see that relates to your question?

@ron please check the edit of the question.

@aniline what you drew is not a resonance structure. (Also, can you find a better camera? Your pictures are awfully hard to see.)

@Jan sorry for the pic (I would try to get a better camera.) However what would be the correct then if that posted are not?

@aniline A resonance structure should not add new bonds like you did between carbons 1 and 5. (You might as well just use Paint or something …)

While drawing resonance structures, the sigma bond skeleton in each structure should remain same. Here you have changed the skeleton. Keep the skeleton unchanged and just move or "delocalize" the pi-electrons and while doing so also keep in mind the valencies of atoms.


text:
How do you make photoresist/photohardening paint?
I have been looking into making my own PCB's, but I noticed there is little information on making photo hardening paint. This is a critical component when etching the circuit, and I am very curious on the chemistry involved in making it. I understand there are better methods to make your own pcb, but I am curious how photoresist is made

Did you consider to print the circuit on glossy magazine paper with a laser printer and transfer the mask with a flat iron instead?

yes, but I'm not looking for the easiest method, I'm curious on the photoresist method.


text:
Would sodium carbonate from carbon dioxide plus hydroxide be formed in solution or as a solid?
If normal breathable atmosphere were to be reacted with aqueous sodium hydroxide, what form would the resulting sodium carbonate take, would it be dissolved in the solution or solid?
Note:
I'm defining air here as the atmosphere of earth at between sea level and 200m.
(Not a homework question, just interested.)

What d'you mean by "reacted with"? All  NaOH solutions in contact with air "react" but in more or less negligible way.

I had in mind a water pipe percolator, but the most efficient reaction method would be preferable, I'm aiming for the maximum CO2 extraction.

This depends on a lot of things: the volume of solution, the time carbon dioxide has to get dissolved and the temperature are probably the most important.
The actual reaction taking place is happening in solution. Thus, carbon dioxide must first dissolve in the solution (1) before dissolved carbon dioxide can react with the hydroxide base (2). You should view these steps as distinct even though the second is much faster than the first.
$$\begin{align}\ce{CO2 (g) &-> CO2 (aq)}\tag{1}\\
\ce{CO2 (aq) + OH- (aq) & -> HCO3- (aq)}\tag{2}\end{align}$$
Since these happen in solution, the counterion — a spectator ion — is irrelevant and the same would happen with $\ce{KOH, Ca(OH)2}$ and others.
At some point you will reach saturation, because either too much carbon dioxide dissolved or (much more likely) too much water evaporated. Thenceforth, sodium hydrogencarbonate or sodium carbonate (I never performed a formal analysis) will precipiate as a white crust. Incidentally in Munich in the inorganic teaching labs, there were 5 litre reservoirs of $6~\mathrm{M}\ \ce{NaOH}$ solution from which smaller bottles could be filled. If there was a spill (and there always was — it was undergrads in that lab after all) it would show up days later as a white crust. When I myself was a TA, part of my job preparing the labs for the next student generation was to clean those areas. Diluted hydrochloric acid worked wonders.

What do you think would be the optimum parameters to maximise the amount of CO2 converted into HCO3- (aq)? (Great answer, by the way)

@Transmission higher concentration of $\ce{OH-}$, lower temperature (increases solubility of $\ce{CO2}$).

Are you suggesting that Na2CO3 could be a dominant species in this solution? Since it is highly alkaline, I could agree with that. I'd be interested in knowing whether the carbonate or bicarbonate salt predominates for a given temperature and pH.

@electronpusher I didn’t think it through whether hydrogencarbonate or carbonate are the dominant species at timepoint X. All I’m sure of is that either of them is the one that gives the crust upon drying. Of course, you could calculate it given concentrations, and $\mathrm{p}K_\mathrm{a}$ values.

Ah yes, you could use the van't Hoff equation to find Ka at a given temperature, and it's basically ICE tables the rest of the way.


text:
Understanding how to read a Boltzmann Distribution graph and how it relates to enthalpy, entropy, and temperature

I am having trouble understanding the how to read the graph and and thus don't understand how to connect it to the enthalpy, entropy, and temperature.

Sigh... For starters it would've been nice if they labeled the graphs...

Here's my take on this question...
The Boltzmann distribution there just helps to indicate which direction is the direction of higher energy. It's up because the populations are smaller in that direction. Since the energies for $\ce{B}$ start higher, $\ce{B}$ must be higher in energy in both cases. From this, we conclude that:
$$\Delta H^{\varnothing} > 0$$
Entropy is defined as
$$S = k \ln \Omega$$
where $\Omega$ is the density of states. There are more energy states per unit energy in (b) for $\ce{B}$. I'm not sure how to conclude this for (a).
But from this, I do conclude that for (b),
$$\Delta S^{\varnothing} > 0$$
For (a), it's probably close to zero.
Now you can use:
$$\mathrm{K}_{eq}=\frac{\ce{[B]}}{\ce{[A]}}=e^{-\frac{\Delta G^{\varnothing}}{RT}} = e^{-\frac{\Delta H^{\varnothing}}{RT}}e^{\frac{\Delta S^{\varnothing}}{R}}$$
Notice that if $\Delta H^{\varnothing}>0$, then increasing $T$ means that the equilibrium constant gets smaller. If it's negative, then increasing $T$ means that the equilibrium constant gets bigger. In our case, enthalpy change is positive in both cases.
$\ce{B}$ exceeding $\ce{A}$ is simply a matter of whether or not the equilibrium constant can be greater than 1. This happens if the second term in the product (with entropy is large). Fortunately, $\Delta S^{\varnothing} > 0$ in (b). If you increase the temperature, you will drive the first term to 1, so it's possible to make $\ce{B}$ dominate the mixture at equilibrium.
This is less likely for (a) because I think the entropy change will be close to zero, so you can't make the second term very big.


text:
What is an egg emulsion?
All good recipes for hollandaise sauce consist of beating egg yolks vigorously and slowly add melted butter over moderately low heat.  Care must be taken to not overheat the sauce or the eggs will scramble and it won't emulsify.  Likewise, the melted butter must not be added too quickly or it will not emulsify. When done properly like a good chef, what is happening here?  Why does the sauce thicken?  What is an egg emulsion and why must the butter be added so slowly rather than faster?

To know what's happening here you should know what "emulsifying" is.
First we need to know that fat is "non-polar" substance. what does that mean?
It means that all fats can't be dissolved in water which is "polar" substance.
So how can we make fat -which is non-polar- dissolve in water or any other polar substances?
We add "emulsifiers" like: egg yolk and many other chemical substances like Lecithin.
These emulsifiers are (polar and non-polar) so they can link fat with water together.
We add butter slowly so that the molecules of emulsifier can link the molecules of butter with molecules of sauce -which is polar-.
If there's no free molecules of emulsifier, every molecule of added butter will not be dissolved in the sauce.


text:
Relative rate of hydrolysis of the carbon chlorine bond in thioethers and ethers
I am solving problems based on rates of hydrolysis of compounds but I don't have a general approach to such questions.

Which of the following compounds undergoes hydrolysis faster than the second under comparable conditions? It may be acid catalysed or base catalysed hydrolysis but the conditions must remain the same for each pair.

In my view since oxygen is more electronegative than sulfur, hence hydrolysis should be easier for second. But I am not sure about it.
 see part 2 here
see part 3 here

Are you intending to hydrolyze the chlorine or the ether/thioether?

@electronpusher thanks for your interest.the question is about hydrolysing chlorine.

Hydrolysis of the thioether may compete. One might make an argument that the induction of the oxygen might make the alpha carbon more electron-poor than the sulfur compound, but the methylene unit between the O and aloha-C would lessen the inductive effect drastically.  Personally I think this comparison might be beyond theory and experiment is necessary.

This is a classic case of neighboring group participation. The hydrolysis of the thioether is significantly faster because the sulfur can displace the chlorine to give a three member ring with a sulfur in it. The ring is quite reactive and will easily be opened by water.

For reference, you can also look at a compound like mustard gas. It has the same functional groups and is so reactive that is a war gas.

Yes that, and of course sulfur is much more nucleophilic than oxygen so NGP is very rapid.


text:
How to derive the rate laws for three consecutive reactions?

Given the following three consecutive unimolecular reactions $$\ce{A ->[$k_1$] I_1 ->[$k_2$] I_2 ->[$k_3$] P}$$ write the rate of change for each.

I have gotten the following:$\newcommand{\diff}{\mathrm{d}}$
\begin{align}
  \frac{\diff [A]}{\diff t}   &=-k_1[A]\\
  \frac{\diff [I_1]}{\diff t} &=k_1[A]-k_2[I_1]\\
  \frac{\diff [I_2]}{\diff t} &=-k_3[I_2]+k_2[I_1]\\
  \frac{\diff [P]}{\diff t}   &=k_3[I_2]\\
\end{align}
Is this correct? If it is, then why cant we write
\begin{align}
  \frac{\diff [A]}{\diff t}   &=-k_1[A]\\
  \frac{\diff [I_1]}{\diff t} &=k_1[A]-k_2[I_1]\\
  \frac{\diff [I_2]}{\diff t} &=k_3[I_2]+k_2[I_1]+k_1[A]\\
  \frac{\diff [P]}{\diff t}   &=k_3[I_2]\\
\end{align}

Your first set of rate laws looks correct to me. The only difference in the second set is the third line, the rate of I_2? If the first I_2 rate law is true, then why would you think the second I_2 rate law could be correct? I can't see where you're getting it.

Your first set of four equations look correct. Next you need to integrate the first equation to get [A] vs. time i.e. $\ce{[A]}= \ce{A_0}\exp(-k_1t)$ where $\ce{[A_0]}$ is the value at $t=0$, and substitute [A] into the second equation. Integrate this equation (integrating factor method) and substitute it into the third and so on. It gets messy!  You can obtain $\ce{[P]}$ easily as this is the sum of all other values less the amount you start with.
The general solution for a chain of n reactions in the special case that only the first species A is present at $t=0$ has the form
$$N_n = C_1\exp(-k_1t)~+ C_2\exp(-k_2t) ~+ ...+ C_n\exp(-k_nt) $$
$$C_1= \frac{k_1k_2...k_{n-1}}{(k_2-k_1)(k_3-k_1)...(k_n-k_1)}\ce{[A_0]} $$
$$C_2= \frac{k_1k_2...k_{n-1}}{(k_1-k_2)(k_3-k_2)...(k_n-k_2)}\ce{[A_0]}$$
$$ C_3 = .....$$
etc,  for as many terms as necessary.
good luck!

We have the following cascade of unimolecular chemical reactions
$$\ce{X_1} \xrightarrow{k_1} \ce{X}_2 \xrightarrow{k_2} \ce{X}_3 \xrightarrow{k_3} \ce{X}_4$$
where $k_1, k_2, k_3 > 0$ are the (distinct) rate constants. Assuming that every chemical reaction in this cascade has mass action kinetics, then we have the following system of linear ODEs
$$\begin{bmatrix} \dot x_1\\ \dot x_2\\ \dot x_3\\ \dot x_4\end{bmatrix} = \begin{bmatrix} - k_1 & 0 & 0 & 0\\ k_1 & - k_2 & 0 & 0\\ 0 & k_2 & - k_3 & 0\\ 0 & 0 & k_3 & 0\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}$$
where $x_i := [\ce{X}_i]$ is the (time-varying) concentration of the $i$-th species. Let the initial concentrations be $x_1 (0) =: x_{10} \neq 0$ and $x_2 (0) = x_3 (0) = x_4 (0) = 0$. Integrating the linear ODEs, we obtain
$$\begin{array}{rl} x_1 (t) &= x_{10} \, e^{- k_{1} t}\\ x_2 (t) &=  x_{10} \left(\frac{k_{1}}{k_{2} - k_{1}}\right) e^{- k_{1} t} + x_{10} \left(\frac{k_{1}}{k_{1} - k_{2}}\right) e^{- k_{2} t}\\ x_3 (t) &= x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{2} - k_{1}\right) \left(k_{3} - k_{1}\right)}\right) e^{- k_{1} t} + x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{1} - k_{2}\right) \left(k_{3} - k_{2}\right)}\right) e^{- k_{2} t} + x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{1} - k_{3}\right) \left(k_{2} - k_{3}\right)}\right) e^{- k_{3} t}\\ x_4 (t) &=  x_{10} \left( \frac{k_{2} k_{3}}{ \left(k_{2} - k_{1}\right) \left(k_{3} - k_{1}\right)}\right) \left( 1 - e^{- k_{1} t} \right) + x_{10} \left( \frac{k_{1} k_{3}}{ \left(k_{1} - k_{2}\right) \left(k_{3} - k_{2}\right)}\right) \left( 1 - e^{- k_{2} t} \right) + x_{10} \left( \frac{k_{1} k_{2}}{ \left(k_{1} - k_{3}\right) \left(k_{2} - k_{3}\right)}\right) \left( 1 - e^{- k_{3} t} \right)\end{array}$$

Python code
The following SymPy script
from sympy import *

t = Symbol('t')
x10 = Symbol('x10')
k1, k2, k3 = symbols('k1 k2 k3')

A = Matrix([[-k1,  0,  0, 0],
            [ k1,-k2,  0, 0],
            [  0, k2,-k3, 0],
            [  0,  0, k3, 0]])

V, D = A.diagonalize()

print "V = ", V
print "D = ", D
print "x (t) = ", simplify(V * diag(1, exp(-k1*t),exp(-k2*t),exp(-k3*t)) * V**-1 * Matrix([x10,0,0,0]))

produces the following output
V =  Matrix([[0, -(k1 - k2)*(k1 - k3)/(k2*k3), 0, 0], [0, k1*(k1 - k3)/(k2*k3), (k2 - k3)/k3, 0], [0, -k1/k3, -k2/k3, -1], [1, 1, 1, 1]])
D =  Matrix([[0,  0,  0,  0],
             [0,-k1,  0,  0],
             [0,  0,-k2,  0],
             [0,  0,  0,-k3]])
x (t) =  Matrix([
[                                                                                x10*exp(-k1*t)],
[                                                                                k1*x10*exp(-k2*t)/(k1 - k2) - k1*x10*exp(-k1*t)/(k1 - k2)],
[                                                                                k1*k2*x10*((k1 - k2)*exp(t*(k1 + k2)) - (k1 - k3)*exp(t*(k1 + k3)) + (k2 - k3)*exp(t*(k2 + k3)))*exp(-t*(k1 + k2 + k3))/((k1 - k2)*(k1 - k3)*(k2 - k3))],
[x10*(-k1*k2*(k1 - k2)*exp(t*(k1 + k2)) + k1*k3*(k1 - k3)*exp(t*(k1 + k3)) - k2*k3*(k2 - k3)*exp(t*(k2 + k3)) + (k1*k2*(k1 - k2) - k3*(k1*(k1 - k3) - k2*(k2 - k3)))*exp(t*(k1 + k2 + k3)))*exp(-t*(k1 + k2 + k3))/((k1 - k2)*(k1 - k3)*(k2 - k3))]])


text:
Why is there a factor of half in the definition of bond order?
I was recently taught about molecular orbital theory and bond order:
$$\text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$$
I was wondering, what is the purpose of dividing by 2?

Because there are two electrons in a covalent bond! Say all the electrons you have is two bonding electrons... Would you call that a bond order of two? Or is it just a good old single bond?

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text:
How is density "predicted"?
Hassium is predicted to be the densest element at 41 g/cm ^3, but how do you predict something like that? Wikipedia says there have only been 100 hassium atoms made to date and it's half life is less than ten seconds, so I have no idea how a scientist came up with that figure... any insight?

Perhaps they made educated guesses about the internuclear distance, based on extrapolated data from other transition metals. This could be used to estimate the volume occupied by a given number of the atoms. Calculating the mass for that number of atoms would be trivial. Mass over volume, you have density.

One could do a calculation, say with density functional theory. One could have an educated guess of what the most likely crystal structure is, with another educated guess on lattice parameter. One could use a dart board.

You don't need even one real atom to predict anything. The reliability of prediction is another question. Then again, given the extreme instability of the element, it is a prediction that probably will never be verified, so...

You do a theoretical calculation based on a model that has been verified by  testing, i.e. that correctly predicts the density of other, "similar" elements, which are less elusive.


text:
How to determine sulfate ions in a solution by precipitation?
For example I am given an unidentified solution and am told that it may or may not contain sulfate ions. How can I confirm or deny the presence of sulfate in this solution?

Acquaint yourself with the so-called solubility rules. Depending on which cation sulfate is paired with, the sulfate compound may or may not be soluble. Generally, compounds containing sulfate ($\ce{SO_4^2-}$) compounds are soluble. (For essentially the same reason that sulfuric acid $\ce{H2SO4}$ is a strong acid, if you're interested.)
Well-known insoluble sulfate compounds are $\ce{PbSO4, SrSO4, CaSO4}$, and $\ce{BaSO4}$. Note that all of these cations have a somewhat high (+2) charge and are large.
If sulfate anions are present in your aqueous solution and you a a soluble compound containing $\ce{Pb^2+, Sr^2+}$, etc, then those cations will form an insoluble precipitate with the sulfate.

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text:
How to determine which of two similar compounds has the larger dipole moment?
I am solving some problems involving the order of dipole moments of some compounds. I could solve most of them but the following:

Which has greater dipole moment?

$\ce{CH3F}, \ce{CH_3Cl}$

Since fluorine is more electronegative my assumption was that $\ce{CH3F}$ should have a larger dipole moment.
On the other hand since the angle between the dipole moments is less in 3-nitro phenol than in 4-nitrophenol, my assumption was that 3-nitro phenol has a greater dipole moment. But this contradicts the answer given.

The dipole is the vector sum of each atoms's partial charge times its distance from the origin. Lacking this information individual bond dipole values can be used
In the first case it is that of $\ce{CH3}$ vs. the halogen, and then which halogen has the greater electronegativity. Looking up some bond values it seems that a $\ce{C-F}$ bond has the slightly larger dipole ($1.6$ vs $1.5$ Debye ) so the difference is small and not easy to predict as the C-F bond is shorter than the C-Cl.
In the second molecule, geometry can be used as the two groups are the same on each molecule but positioned differently. The para molecule will have the smaller dipole vs. the meta if both groups act in the same direction, either both to withdraw or add electron density with respect to the ring, as then they act to cancel one another to some extent.
If they act in a different direction then the para molecule will have the larger dipole vs the meta. Looking for dipole moments it seems that phenol has a dipole of $1.45$D pointing into the ring and nitrophenol $3.95$D out of the ring so act in different directions and so the para should have the larger dipole.
(To add two vectors of length b and c at an angle $\theta$ the resultant c is found using the cosine rule $c^2 = a^2 + b^2 -2ab\cos(\theta)$ . If the dipoles point in the opposite direction make either a or b negative)

For the second problem $\ce{NO2}$ is able to form resonance structures (being in para position). The resonance forms interupt the shared pi orbitals in the benzene ring  (leaving behind two double bonds in the ring) and the nitro group becomes ionized (thus having the greater dipole moment):

For the first problem fluorine is more negative, but it lacks valence electrons to withdraw. Chlorine has another valence shell that increases the length of its bond to the methyl group. The increase in bond length leaves a larger space in the sigma bond and more potential to withdraw the electrons farther from the methyl group than the fluorine can.

Thanks for your answer. But I couldn't get your last para.why is a empty valence shell important.is bond length so high that it is able to counterbalance the electronegativity.

http://chemistry.stackexchange.com/questions/6780/why-do-c-f-bond-has-less-dipole-moment-than-c-cl-bond


text:
Identification of a compound C10H12O with 1H and 13C NMR data

I was given this problem to do for Organic Chem II.  I know that it's a benzene (unsat # 5), with 2 para-substituents due to the doublets in the 7 region of the HNMR.  I know that the oxygen is not an OH due to the absence of a D2O shake.  There's probably an alkene in one substituent to finish out the unsaturation number.  However, I can't figure out what the 2 substituents are - but I'm guessing one's either an ether or ketone!  Can anyone help?

A ketone would have a very obvious signature that isn't there (I'll let you work out what it would be). Also you can use the couplings on the proton peaks to work out the structure of the substituents.

Also do you have an expansion of the 1H,dddd peak? I'm interested to see what the coupling constants are.

Like you said, the 2 doublets at around $\pu{7 ppm}$ belong to a para-substituted benzene. The three single-proton multiplets around $\pu{5-6 ppm}$ belong probably to a single substituted alkene. A singlet and a doublet between $3$ and $\pu{4 ppm}$ belong to $\ce{CH3-{}}$ and $\ce{-CH2-{}}$ groups, respectively (Note that the signal that belongs to the $\ce{-CH2-{}}$ group is a doublet, indicating that it is adjacent to a carbon with only a single proton, so it is probably connected to the alkene).
The absorption maximum at $\pu{275 nm}$ gives us a hint that the disubstituted phenyl ring of our compound is not conjugated to the $\ce{C=C}$ bond.
Concerning the oxygen atom: it is definitely neither a carbonyl group (indicated by the absence of an intense peak at $\gt \pu{180 ppm}$), nor a hydroxy group. So it probably is an ether.
So far, we have two possible candidates:

Allyl para-tolyl ether
1-allyl-4-methoxybenzene

The chemical shift of the $\ce{CH3}$-group is shifted downfield compared to the $\ce{CH3}$-shift in toluene (where it is around $\pu{2.3 ppm}$, while this compound has $\gt \pu{3.7ppm}$), and thus more likely belongs to $\ce{CH3-O-Ar}$ protons. So, our compound is 1-allyl-4-methoxybenzene.
However, the multiplets of the vinylic protons in the $\ce{^1H}$-NMR spectrum don't fit this compound well. The allyl group should give two dd signals for the terminal protons and a ddt for another vinylic proton. But the $J_2$ constant for terminal protons is very small ($\lt \pu{2 Hz}$) and can't be seen on some old NMRs, what would lead to the simple two doublets.
Why do we see dddd instead of ddt for $\ce{R-CH=CR2}$ proton? I'm not sure, but maybe this is because the $\ce{-CH2-{}}$ protons are diastereotopic. Then it makes sense.

If the CH2 was diastereotopic they probably wouldn't form such a nice doublet. Those protons are enantiotopic (they can be interconverted by a mirror plane). Which of course leads to the question of why it is a dddd, which also doesn't make sense to me. (I agree with everything else you said. Even just looking at the molecular formula, I can't come up with anything else that makes sense. Although I may very well be missing something obvious - not the first time and won't be the last. The -OMe group is quite certainly there which leaves C3H5 on the para position - either allyl or cyclopropyl.)

The splitting pattern doesn't really make sense and your analysis also nicely fits to the 13C-NMR spectrum. Without more resolution I could not think of any other thing. Welcome to Chemistry.se btw!

As orthocresol said, the two methylene protons are not diasterotopic, they are enantiotopic. As such I do not know why the peak is dddd rather than tdd. It would interesting to see an expansion - it may be a misidentification.

zholyd is on the money with the correct solution; 4-allylanisole. And if you searched for this at from a very well known chemical supplier, you would probably find an exact replica of the spectrum presented. The overlaid 13C spectrum and 1H spectrum like this on the QE-300 are a real giveaway that this is an Aldrich spectrum.
The interest here comes from the interpretation of the splitting. Below is an expansion of the spectrum presented in the original spectrum (yes, I searched the Aldrich website). Is that a dddd or a ddt? Remember that in the truest sense a triplet is simply a doublet of doublets with identical coupling constants. So while it may be a little bit misleading to be called a dddd, as it implies different coupling for each d, strictly it is not incorrect. Looking closely at the spectrum, it certainly appears as a ddt, and this is how I have labelled the splitting diagram. And it looks textbook characteristic for an internal allyl double bond proton.

What is more interesting are the signals at about 5ppm. Knowing what we know about terminal double bond protons in an allyl group, we would expect one to have a trans coupling of ~15Hz, and another to have a cis coupling of ~10Hz. Clearly what has been labelled as the two separate signals do NOT have these couplings, and this is because these peaks are actually almost overlapping, and on the 300 giving rise to some 2nd order effects. I have labelled the cis coupling in red, and the trans in blue. You can see that they overlap on the right hand side, and so are not quite the two separate peaks as shown in the original question. The geminal coupling between double bond protons is usually very small, and can be seen to be giving rise to some 2nd order effects in these peaks.


text:
Instantaneous enthalpy and entropy change of reaction as equilibrium is established
Can we calculate instantaneous difference between enthalpy and entropy of product and reactants of a reaction as the reaction proceeds towards equilibrium? Ex: $\ce{R<->P}$ reaction, at some reaction coordinate, moles of $R = 1-x , P = x$, Enthalpy difference between products and reactants? Enthalpy difference if it reached equilibrium? Values of $$\Delta G^0, \Delta H^0, \Delta S^0$$ are known. Or is it impossible and the only relationship available is $$\Delta H = T \Delta S$$ at equilibrium or are their values only temperature dependent and activity independent?
We can do this for Gibbs free energy by $$\Delta G=\Delta G^0+\mathcal RT\ln Q$$

In order to do this, you need to accept the idea that the thermodynamic functions and be defined and calculated even for non-equilibrium states of a system.  I have no problem accepting this, but certainly some physicists would not be willing to make this leap.

G is also a thermodynamic function right?

Sure G is a thermodynamic function.

https://en.m.wikipedia.org/wiki/Partial_molar_property It seems that Chemical Potential is bridging the gap and it's only connection is to G. H needs entropy to be constant. Any thoughts?

I agree with everything in the reference you provided.  G is the most useful and convenient function for defining the chemical potential.  In fact, the chemical potential is just the partial molar Gibbs free energy.


text:
Cyclic carbocation rearrangement
Is there a way for the 4-methylcyclohexyl cation (4-methylcyclohex-1-ylium) to undergo some rearrangement and become tertiary carbocation via 1-3 or 1-4 shift?

Ring contraction...

1,3- and 1,4- shifts don't really happen. It's just too far away. If you have a double bond at 4 or 5 carbons away, maybe you can do a ring closure or form a non-classical carbocation. The cation you've drawn could however undergo a sequence hydride shifts. Each hydride shift would move the cation closer to the methyl group, until the 3rd shift places the cation on the tertiary carbon. You'd need to out-compete any kind of elimination or cation reaction with a nucleophile...

Ok so though not impossible , but 1-3 or 1-4 would rarely happen right?

1,3- and 1,4- are really weird. I doubt you'll ever get the stereoelectronics to line up for this. I've never seen an example (which, of course, doesn't mean it's impossible).


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Particle size-energy distribution
I am not chemist, I am mathematician and interested in Population Balance Equations. I have a question about particle size-(breakage) energy distribution which mentioned in this article: http://www.sciencedirect.com/science/article/pii/S0032591011001458.
In this paper, author says (in top of section 3.2) he divide size and energy interval to 100 intervals. I divide size interval (0-6 mm) to 100 intervals but I don't know how can I divide energy interval. Is there anybody can help me?

I can't see the article as its behind a paywall, however, its basically done in the same way as you divided sizes. Somewhere there must be mention of maximum and minimum energies.

@porphyrin you can use website sci-hub.bz for free access to it. by the way, this is a related article which I used most of: http://onlinelibrary.wiley.com/doi/10.1002/aic.14358/abstract


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How do i create a significantly correct world with large crystal formations in my novel?
I have uploaded an shorter and less broad Question which is largely the same but requires a different kind of answer. What i want to say is that i re-asked my question in a different way because i do not want to delete the answers i got  here so far and because the answered also would be incorrect in the context of my re-asked and re-written question
link: Which crystals can exist permanently (and do not break down) under normal earth like outside conditions?
Hi I am not really all that well versed in chemistry or molecules for that matter but I am trying to find a way to "create" a scientific and realistic (trough probably unlikely) world with large crystal formations on the surface. The reason I want this is because I want to create a species for my novel that has its culture based around the large crystal formations living on the planet.
Now to limit my question a bit I would like the answer to these two points:

What kind of crystal(s).
How do you think the process should work.
What limitations it creates.

The criteria are that:

The crystals must be permanent without breaking down in normal earth conditions/atmosphere.
Preferably no interference required by intelligent lifeforms.
There will be oxygen and carbon dioxide in the atmosphere.
They must be able to grow quite large. I’m thinking about large hexagonal crystal columns.

What are the tools?:
Well you can more or less use anything in nature, you can make suns explode if need be. But a more probable start would be selecting the chemical composition and what would be needed to form crystals from that point.
For instance if you need a long period of very high heat and then cooling down to “normal” temperatures you can make the system sun go red giant and heating the planet before becoming a white dwarf.

On one hand, this has chemical relevance. On the other hand, why didn't you post it at Worldbuilding.SE?

Something like titular "Pillars of Eternity"?

@hBy2Py I kind of want to make a world which is as realistic as possible so i wanted to ask the help of some chemists if they knew how to do so. with that goal in mind i derided to ask it here.

@TheRWS96 Some of the answers I've seen on WB.SE are heavily scientific, and in general quite reasonably adherent to actual scientific principles. I think you might be pleasantly surprised with the responses you'll get there.

I've edited the more explicit requirements given in a repost of your question into this one.

What about trivial solutions like quartz or calcite? Erosion can destroy them, but if they are in good environment they can low large (several meters) in caves

TheRWS96, I strongly suggest you post this on WB.SE. That site's express purpose is the sort of speculation you're looking for. Chem.SE's on-topic policies require much more specificity than suits your question. Further, if you repost there, you could always edit your question here to link to the WB.SE post. That way, people browsing Chem.SE would be more likely to find the WB.SE post, and answer there.

Do you mind a scenario somewhat less dramatic than exploding suns? ;)

Just let selenite ($\ce{CaSO4*2H2O}$) cristallize very slowly und undisturbed in a system of caves for a long time.
Let the inhabitants of your world discover the cave system or let parts of the cave system collape to make it more accessible.

A model for this already exists: the Cave of the Crystals, connected to the Naica mine in Mexico.

Update
Note that I'm not a geoscientist, but to me, large hexagonal columns that can exist under earth's atmosphere cry basalt, which in turn indicates vulcanic activities.
The English wikipedia has a list columnar structures resulting from vulcanism. Some examples are

Giant's Causeway in Northern Ireland
Fingal's Cave in Scotland
Devil's Postpile in California

I read about that cave but i have also read that the crystals slowly deteriorate when exposed to air otherwise i would have used that as a template. It's unfortunately but i really need the crystals to be permanent otherwise i cant use them the way i want to.

(In reaction to the update) Hmm cool it gives me some ideas Thanks

One option could be copper sulfate (CuSO4), which forms large, blue crystals. Roger Hiorns' installation coated the inside of a London apartment with it.
You can form CuSO4 by reacting solid copper (Cu(s)) with sulfuric acid (H2SO4):
Cu(s) + H2SO4 (aq) --> CuSO4 (aq) + H2(g)
Ways this could occur:

There exist naturally-occurring sulfuric acid springs, such as in Yellowstone. Take a naturally-occuring copper vein and stick it in a sulfuric acid pool, you'll get CuSO4 dissolved in the water. Let the water slowly evaporate over time, and you'll get copper sulfate crystals. (This could happen above or below ground.)
Mining run-off is often acidic, containing sulfuric acid. Stick it next to a copper vein, and similarly you'll have copper sulfate. This requires some intelligent life at some point, to build the mine, but it could be abandoned/post-apocalyptic.

Edit: Note: you need a very dry environment for this, as they are water-soluble!


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What is the hybridisation in BrF5
What  is the hybridisation of $\ce{BrF5}$ ? I find different sources giving different answers.
When I approach this problem , I don't find any exceptional case like $\ce{SH6}$ (in which hybridisation doesn't take place.) So it follows normal rules, but I find the opposite in many different books ?
I think it square pyramidal $\ce{sp^3d}$. I also face the same problem for other compounds like $\ce{H2S}$ and $\ce{PH3}$.

I am wondering is this kind of question morally justified? Since the d orbital contribution is already deputable in hypervalent molecule such as, SF6

What dou you mean by "Morally justified" ?

I mean, this question looks like an innocent homework assignment. But it is a complicated problem at the frontier of computational chemistry.

No , not a homework problem . I myself was practicing by writing some compounds , my intuition says the above one is square pyramidal . When I searched a library book a different story was written there .

There is, in-general, no one-to-one mapping between molecular geometry and hybridization.

Just a side quotation, http://depa.fquim.unam.mx/amyd/archivero/VBHIBRIDOSd_26401.pdf, "Figure 4. From a VB standpoint, the bonding in SF6 can be described using two, 2-electron bonds from sulfur sp hybrids pointing 180 away from each other and two, 2-electron bonds from sulfur p orbitals with
the remaining four electrons located on two fluorine atoms"

Thats a good source , but what about BrF5 ?

To be honest,  I never run an ab initio valence bond calculation as described in that reference. I suggest someone interested in this topic, run an ab initio valence bond calculation.

@InquisitiveMind $\ce{BrF5}$ should be composed of a standard 2e2c $\ce{Br-F}$ bond and two elongated 4e3c $\ce{F\bond{...}Br\bond{...}F}$ bonds (four-electron-three-centre bonds), giving it a square pyramidal structure with the central atom at the base of the pyramid. However, it probably fluctuates quickly between that and and the pentagonal bipyramid. In any case, remove d orbitals from your argumentation; they do not take part.

@Jan Hey my query on the hybridization of H2S and PH3

Hybridisation of $\ce{BrF5}$ is $\ce{sp^3d^2}$ (involving one 4s, three 4p and two 4d orbitals) giving rise to octahedral geometry. But one hybrid orbital is occupied by lone pairs. So the effective shape of molecule is square pyramidal.


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Stoichiometry and POAC Contradiction
Lets say we have the reaction
$$\ce{H2 + O2 -> H2O}$$
And we are given 10 moles of $\ce{O2}$ and are asked that
How many moles of hydrogen gas do we need to complete the reaction ?
Stoichiometry gives the answer as 10 moles but by applying Principle Of Atom Conservation,  we get
$$\text{Moles of }\ce{H2}  =2\cdot\text{moles of }\ce{O2}$$
Why is the POAC giving wrong results?

What is the abbreviation POAC?

You applied the POAC incorrectly . Even your stoichiometric answer is incorrect.
First Balance the reaction
$$\ce{2H2 + O2    -> 2H2O}$$
We get that one mole of $\ce{O2}$ requires 2 moles of $\ce{H2}$.
Thus 10 moles $\ce{O2}$ need 20 moles of $\ce{H2}$.

@Zhe Looks good now

@Harsh Sharma You can refer your RCM for more

FYI: Formatting formulas


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Thermite reaction beween zinc and iron(II, III) oxide
I have zinc metal and iron oxide in the form of magnetite which I would like to obtain iron metal from. I was wondering if I could use a thermite reaction to create iron and zinc oxide? I would like to know whether such a reaction would occur between these two substances, how violent it is and whether the reaction will sustain itself?

Not in practice. Zinc metal has relatively low boiling point and would evaporate from reaction mixture before temperatures when reaction would take place. Besides, it really isn't cost effective.  Consider aluminium powder instead.

I'd check the Galvanic series.
$\ce{Fe(II)}$ and $\ce{Fe(III)}$ have an higher standard potential than $\ce{Zn}$ and therefore your reaction should take place.
I don't know how reactive it is but I had good experiences with also using $\ce{HCl}$. Use $\ce{Zn + HCl}$ to produce $\ce{H2}$ and heat $\ce{Fe2O3}$ in the presence of $\ce{H2}$ to produce $\ce{Fe}$ and $\ce{H2O}$.


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Carbon-13 NMR for chloroform
I am slightly confused by what the spectrum would show for carbon-13 NMR of $\ce{CHCl3}$.
My initial guess would be that the peak would be split by coupling to both the proton and the 3 chlorines, as both nuclei have a net spin.
If the peak were split by the chlorine only, then as there are three chlorine atoms we would get a quartet peak. However this cannot be correct because the spectrum for CDCl3 shown here has only three peaks:

I also don't understand here why the areas of the three peaks are the same. Should they not be in some other ratio (for the four peaks I expected it would be $1:3:3:1$ are ratio)?
Then I would expect that the presence of the proton would split every one of the four peaks from $\ce{C-Cl}$ coupling further into a doublet, giving a quartet of doublets.
I didn't find any spectra showing this, only showing very tiny peaks for $\ce{CHCl3}$ compared with $\ce{CDCl3}$ which I don't understand:

To make it more confusing, my lecture notes say that for chloroform, $\ce{CHCl3}$,

each $\ce{^13C}$ is attached to a spin $1/2$ proton so unless we applied broadband proton decoupling, coupling to the proton would mean that the $\ce{^13C}$ signal would appear as a doublet.

And it makes no mention of the effect of the chlorine atoms on the spectrum.
I would be very grateful is someone could expalin what the spectrum for carbon-13 NMR of chloroform, $\ce{CHCl3}$, would actually look like!

This earlier answer should help answer your questions.  In a nutshell, chlorine is a rapidly relaxing quadrupolar nucleus which effectively decouples it from other nuclei.  Deuterium is also a quadrupolar nucleus, but relaxes much more slowly than chlorine, so its coupling can be observed. Since deuterium has a nuclear spin =1, it will split whatever it is coupled to into a 1:1:1 triplet.

You assumed that coupling to the three chlorides would yield some type of quartet. This is correct in principle. However, chlorine is one of the many quadrupole nuclei that are basically unobservable by NMR due to their rapid relaxation. I hope another answer is around explaining that as I am not good at it.
I can, however, help you interpret the $1:1:1$ triplet of your spectrum. It is not, as you have assumed, the signal for $\ce{CHCl3}$ but that for $\ce{CDCl3}$. The difference between the two in NMR terms is that chloroform contains the spin ½ nucleus protium while deuterochloroform contains the spin 1 nucleus deuterium — and the latter does not relax rapidly as the chlorine isotopes do. Hence, deuterium is well-observable by NMR.
Now your first thought should be something like the following:

But I am only coupling to one nucleus. That should give me a $1:1$ doublet and not a $1:1:1$ triplet, shouldn’t it?

That simplification is only correct for spin ½ nuclei. The actual formula for calculating the multiplicity of a peak is $2\,n\,I + 1$, where $n$ is the number of chemically equivalent nuclei your observed nucleus is coupling to and $I$ is the spin of the coupling nuclei. Since deuterium is spin 1 ($I = 1$), the formula gives us:
$$2\,n\,I + 1 = 2 \times 1 \times 1 + 1 = 3$$
Or an expected triplet. Since this triplet does not derive from two equivalent couplings to spin ½ nuclei, it is not a $1:2:1$ triplet but a $1:1:1$ one.

Your second spectrum shows a $\ce{^13C}$ NMR of deuterochloroform which still contains residues of chloroform — at least, that is what it is supposed to show. For some reason unknown to me, the protochloroform signal has not been decoupled, i.e. the spectrum is in fact $\ce{^13C}$ and not $\ce{^13C\{^1H\}}$. Thus, you are observing a strong coupling to the single hydrogen nucleus giving a $1:1$ doublet as expected if chlorine is ignored.

On quadrupolar relaxation: https://chemistry.stackexchange.com/a/81552/16683


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Reaction speed with solids
It seems that reaction rate is defined as change of concentration over time
and the speed doesn’t change if you add more solid. But imagine we have a cube (1 kg) of $\ce{CaCO3}$ that yields 1 mole $\ce{CO2}$ in $t$ seconds and we have another cube (5 kg) of $\ce{CaCO3}$. Does this one also yield 1 mole $\ce{CO2}$ in $t$ seconds? (my teacher said yes)
But if we think of second cube as 5 separate 1 kg cubes, then each one again yields 1 mole per $t$ seconds, making it 5 moles of $\ce{CO2}$ in $t$ seconds.
So which one actually happens and is the speed the same?

In a reaction between a solid and a reactant in solution or with a gas its primarily the surface area that is important, since the inside of the solid cannot be reached by the other reactant until it is exposed. (I'm excluding an extremely  slow reaction in which the other reactant could diffuse into the solid before reacting).

Reaction rates are in most cases influenced by concentration for liquids/dissolved solids and surface area for solids.
Surface area
The surface area of the calcium carbonate is one factor in the rate of the reaction. The more surface area, the faster the reaction. In this case, certainly, the five $\pu{1kg}$ cubes will react much faster. The reason large cubes are used to demonstrate this instead of powder is because the surface area is more or less constant throughout the reaction.
The rate law
$$\text {Rate} = k[\text A]^\text m[\text B]^\text n $$ where $\text A$ and $\text {B}$ are the reactant concentration, $k$ is the rate constant (does not change in the same reaction), and $m$ and $n$ are the rate orders (which also does not change in a reaction)
This means that unless m and n are 0 making it a zero order reaction which is independent of concentration, in most cases concentration does have an effect on rate.
In this case, with $k$, m and n constant and either  m or n non-zero, and increase in $[\text A] \text{ or } [\text B]$ would increase the rate.
If $[\text A]$ is calcium carbonate and $[\text B]$ is hydrochloric acid, it shows that most likely an increase in concentration of $\text B$ will increase rate. However, calcium carbonate is a solid and has no concentration. We can imagine $[\text A]$ as the surface area of $\ce{CaCO3}$ instead, and m as 1.
We could also imagine the $\ce{CaCO3}$ dissolved in a solvent, like sugar in water. Now, the reaction rate will be dependent on the amount of
$\ce{CaCO3}$
(Note: the rate orders can only be determined experimentally)


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Faraday law in electrochemistry?
Passing electricity through dilute solution of H2SO4 for a certain time period of time liberates 168 ml of gases at stp. The quantity of electric ity used is ?
I used the formula W/E = IT/96500 here
But the answer is not matching. I converted the 168 ml into moles and multiplied by molar mass of H2SO4 so to get the mass of gas liberated . But the answer is not coming . Should I aslo consider mass of water liberated here ?

I'd consider rotating that image 90$^{\circ}$ first and foremost - better yet, edit your post and typeset the work you're showing in the photo.

I edited it whatever else is given I think is pretty basic and understandable. Thank you

If you've never taken a battery (actually many AC or DC power sources will work), two wires and some water and made gas, I recommend you do so. Carefully, of course. Assuming the wires are inert (copper), then what you'll get is hydrogen and oxygen gases. Now, the question is, how in the world you concluded that sulfuric acid was a reactant. Or perhaps you were just mindlessly plugging in the first thing you saw in the question into the first formula that occurred to you, IDK. The sulfuric acid is mentioned because water has a fairly HIGH electrical resistance when pure. Use of acid decreases what we call "over potential" and at the same time increases [H(+)] which can't hurt. The reaction which occurs, and you really have no excuse not to have seen that it is the only one possible based on the half-reaction table, is the electrolysis of water. Generally, NaCl would be the electrolyte of choice, but that could be confused as a reactant (they do make Cl2 from brine that way, after all). So, the question avoids that pitfall. You should note the relative placement of Cl2 and O2 in the table to determine which is most likely to occur.


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Spectrometry in Chemistry & Flame Test Lab
Question 1:
You currently see two colors that are two similar (dark yellow and orange) in a flame test lab. And also you have a hard time differentiating the colors, what is an alternative to distinguish the two compounds?
My thoughts:
I thought that it was because if you use a wooden splint to dip it into the chemicals, you accidentally combust it causing the inaccuracy and also the source of the flame is blue making it hard to see, thus you use a metal wire and dip that and put it just so slightly on top of the flame to not mix the blue color of the butane flame and the flame that came with the chemicals on the wire. Also to use a digital spectrometer for more accuracy
Am i right about my answer?
Question2:
Let's say that you are using reusable metal loops instead of wooden splints, and you were doing two flame tests of the same chemical and you get different colors, what is the error and the reason?
My thoughts:
I was really confused, maybe because for one of them you placed it too far from the flame source and for the other you placed it too close to the flame source?

... two colors that are two similar (dark yellow and orange) in a flame test ...

Do you know what the two compounds might be?

 The yellow colour typically indicates sodium, the orange flame colour calcium.

In order to distinguish the two compounds, you have several options:

Use a different method

 You can try to precipitate the calcium ions in the form of an insoluble salt, such as its oxalate.

Refine your flame test

 View the flames though a piece of cobalt glass! This filters of the yellow flame colour typical for sodium. Note that the flame colours for the other metals will look different now!

Isnt calcium of a brickish red colour?

Question 1 You currently see two colors that are two similar (dark yellow and orange) in a flame test lab. And also you have a hard time differentiating the colors, what is an alternative to distinguish the two compounds?

Ideally, you would use a spectroscope to view the emission lines of the flame. These lines are unique to each element (and in fact each isotope) and your unknowns can be compared to the known emission spectra patterns of each element for a definitive answer.

let's say that you are using reusable metal loops instead of wooden splints, and you were doing two flame tests of the same chemical and you get different colors, what is the error and the reason?

Carbon (from the wood) does burn an orange color so this may confound flame test results especially when the colors are close to that of carbon.


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how can I get from mol% to percent by volume?
I have mol% of a gas sample I also have temp and  pressure when the sample was taken.  How do I get % by volume? (L)

assume ideal gas behavior and then mole % = volume %. // Remember that one mole of an ideal gas occupies 22.71 liters per mole at 0 °C.

Mol% and % by volume should be the same, assuming ideal gas behavior. For example, a sample containing 11.2 liters (1/2 mole) of hydrogen and 11.2 liters (1/2 mole) of methane should be 50 mol% hydrogen and 50 mol% methane, as well as 50% hydrogen by volume and 50% methane by volume. Note that by weight, the percentages of the components are no longer 50-50.


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Hydrolysis of vinyl bromide vs alkyl chloride

Which of the following compounds, 1-bromocyclopent-1-ene 1 or chlorocyclopentane 2, hydrolyses faster than the second under comparable conditions? It may be acid catalysed or base catalyzed hydrolysis but the conditions must remain the same for each pair.

In my view because of the double bond in first compound hydrolysis should be easier for second, but I am not sure about it.
See part 1 here

Hydrolyzing the saturated alkyl chloride could occur by SN2 but I'd be inclined to expect SN1 in this situation. Chlorine would leave behind a (secondary) carbocation, which would be attacked by the nucleophile: hydroxide (basic conditions) or water (acidic conditions). The rate limiting step of SN1 is loss of the nucleophile, and the activation energy might be lowered in an acidic environment because the hydronium could coordinate to the chloride to stabilize it. The real issue is the resulting carbocation though. In basic conditions, hydroxide could backside-attack, kicking out the chlorine before it has the chance to leave. This concerted SN2 mechanism would avoid the carbocation entirely, but it is slowed by the steric hindrance of the secondary alpha carbon. So here acidic favors SN1, limited by formation of the secondary carbocation, and basic favors SN2, limited by secondary sterics.
Hydrolysis of the vinyl bromide I think must occur by SN1 in basic conditions or protonation-addition-elimination in acid. SN1 requires formation of a vinyl cation when the bromine leaves, which is quite unstable and will severely limit the rate of the reaction. The rate limiting step of the acidic P-A-E sequence is probably the protonation (slower than addition, and I would expect also slower than elimination because protonation is bimolecular). I would expect protonation of an alkene (PAE pathway) to occur much faster than loss of a nucleofuge (leaving group) to form a vinyl cation (SN1 pathway); forming a secondary carbocation would surely have a much lower activation energy barrier than forming a vinyl one. So I expect the acidic hydrolysis of the vinyl bromide via hydration-elimination (P-A-E) to occur much faster than the SN1 hydrolysis in basic conditions.
So the pH conditions may play a significant role here, with base severely slowing the hydrolysis of the vinyl bromide. In basic conditions, I would almost certainly bet the alkyl chloride will hydrolyze faster. Acidic conditions will speed up hydrolysis of the vinyl bromide, with the RLS being protonation to form secondary carbocation or possibly elimination of bromide to restore the double bond. Acidic conditions for the alkyl chloride promote SN1, whose RLS is loss of chloride to form the secondary carbocation. It's hard for me to predict with only theory whether vinyl bromide acidic RLS or alkyl chloride acidic RLS will occur faster.
An interesting note, hydrolyzing your vinyl bromide yields an enol, which will spontaneously tautomerize to a ketone. So your hydrolysis products are a ketone (from vinyl bromide) and an alcohol (from alkyl chloride). The difference of oxidation state in these products makes me wonder why you are asking this question.
Update: Here are my predicted mechanistic pathways. The details vary slightly from my analysis, especially how tautomerization factors in. I also included a couple of alternative pathways I was trying out (Acidic #2 and Basic #2 for vinyl bromide, and Basic #2 for alkyl chloride; these are probably negligible).

I agree with the overall analysis, but I also think this could be worded more strongly. Both pathways for the vinyl bromide in base simply don't happen; a short database search shows that there are no documented examples of vinyl bromide hydrolysis in base. (1) The formation of a vinyl cation under basic conditions by spontaneous dissociation of bromide ion is as good as impossible. Generating these vinyl cations by loss of a leaving group is very difficult (there are some examples of iodonium salts doing this, i.e. PhI as a leaving group).

(2) Cyclopentyne, likewise, exists as a transient intermediate but is incredibly difficult to generate. A common strategy is to have a silyl group plus a leaving group on the alkene carbons and to add fluoride ion, similar to how benzynes are often generated. (3) "Basic #2" for the alkyl halide, too, is as good as impossible: whatever the pKa of a carbocation is, it is nowhere near that of water.


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Diffused Layer ion in Electrix Double Layer
When KI is added to AgNO3 soln, so AgNO3 is in excess, collidal particle is AgI:Ag+ with diffused layer of I-, Why is the diffused layer not of NO3- when the anion in excess is NO3-?
If we had KI in excess in the above scenario, we get AgI:I- with diffused layer of K+ which fits logic of the ion in more concentration having more probablity of being the diffused layer.
(preferential adsorbtion of ion common to lattice for primary layer doesn't seem to factor in diffused layer?)

Only the ion which is in excess and common to the colloid is adsorded.

i know, i'm asking about the ion in diffused layer (the layer outside the adsorbed layer)


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How can I derive whether the methyl cation or anion is planar?

According to VSEPR theory, in which species do all the
  atoms lie in the same plane?

$\ce{CH3+}$
$\ce{CH3-}$

(A) 1 only (B) 2 only (C) both 1 and 2 (D) neither 1 nor 2

My try:
Because $\ce{CH3}$ in both forms only are bonded with three hydrogens, they both have one lone pair, which contributes to their tetrahedral structure. So I got D. However, the answer key says that B is the correct answer. How do I get this answer?

The VSEPR model is a rather simple model that actually helps in this case. It assumes that anything ‘on’ a central atom — atoms that it is bound to and lone pairs — requests almost equal amounts of space, and therefore the compound’s final geometry is dictated by the number (and type; lone pairs are deemed ‘invisible’) of substituents only.
It is important to recognise that only lone pairs (and radicals) count towards the VSEPR model; empty orbitals do not. The reasoning is that something cannot take up space if it is empty.
Applying the theory to the methyl cation should give you three substituents: the three hydrogens. These three will, according to the theory, align themselves in a trigonal planar fashion around the central carbon, giving a perfect bond angle of $120^\circ$ between them. Thus, the methyl cation is planar.
Applying it to the methyl anion should give you four substituents: the three hydrogens we had previously plus the lone pair. The lone pair takes up space, so including it into our visualisation would give us a tetrahedral geometry. However, it is also invisible (we can only ‘see’ full atoms) hence the final structure is predicted to be trigonal pyramidal with the four atoms not occupying the same plane.
Therefore, assuming you copied accurately, the answer key should have given A as the correct answer.

Unless some more advanced concept is at play, the answers are incorrect and the correct answer should be A.
This is because with structure I, the carbon has no lone pairs making it trigonal planar, while in structure II, the carbon has a lone pair, making it trigonal pyramidal


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Rates of hydrolysis in bicyclic and monocyclic compounds

Which of the following compounds, 1-bromo-1-methylcyclohexane 1 or 1-bromobicyclo[2.2.2]octane 2, hydrolyzes faster (under comparable conditions)? It may be acid catalyzed or base catalyzed hydrolysis, but the conditions must remain the same for each pair.

In my view because of the steric hindrance in first compound, hydrolysis should be easier for second, but I am not sure about it.
See part one of my question here.

In my view 1st one would react faster with an SN1 mechanism as the 2nd compound would lead to an unstable carbocation

@AmritanshSinghal how do you  know that carbonation formed by second compound is unstable relative to the first.

Amritanish may be thinking of the norbornyl cation, but I do not expect the bicyclo[2.2.2]octane cation here to behave similarly, because here we have a tertiary carbocation. https://en.m.wikipedia.org/wiki/2-Norbornyl_cation?wprov=sfla1

Related: Will bridged compounds undergo the SN1 reaction?

Aniline is not the asker of other recent questions besides hydrolysis. The other askers should be informed.

Both hydrolysis reactions will almost certainly occur by $\mathrm{S_N1}$, forming carbocation intermediates. I'm not sure if sterics will play the biggest role here. One thing you should consider is the actual conformations of these molecules. The bicyclic is slightly more rigid and may not adjust to the trigonal planar geometry of a carbocation as easily as the cyclohexane, and this will likely be the deciding factor. Here are the conformations, as I hope you know.

Can we generalise that all bicyclo compounds are not easily hydrolyzed compared to their cyclic counterparts

You will not form that bridgehead carbon cation... 2 is significantly less reactive than 1 for an $\mathrm{S}_{N}1$ type process.

I’m going to shamelessly steal electronpusher’s drawing for clarification:

Two general hydrolysis mechanisms are possible: $\mathrm{S_N2}$ and $\mathrm{S_N1}$. First, let us consider $\mathrm{S_N2}$ since the analysis is more simple.
Both of the two species are tertiary bromides meaning that $\mathrm{S_N2}$ reactions are already severely inhibited. However, the second compound is even worse: if the nucleophile tries to attack from behind to interact with the $\sigma^*_{\ce{C-Br}}$ orbital, it would need to come from inside the bicyclo[2.2.2]octane system. This is completely impossible, not only due to sterics. Thus, while compound 1 is very slow in $\mathrm{S_N2}$ reactions, it is not possible for 2 to react in an $\mathrm{S_N2}$ manner.
Conclusion: 1 should react faster by $\mathrm{S_N2}$ — albeit still unmeasurably slow.

To undergo an $\mathrm{S_N1}$ reaction, first the halide must dissociate to give a carbocation. Both species are, as noted above, tertiary, meaning that carbocations are well stabilised by triple hyperconjugation. Initially, we expect both $\mathrm{S_N1}$ mechanisms to be fast.
A carbocation when formed must practically always adopt a planar, $\mathrm{sp^2}$ hybridised configuration. Thereby, the unoccupied orbital has the highest energy and all bonding electrons are as low as possible in energy. Any distortion away from the perfect $\mathrm{sp^2}$ geometry requires the empty orbital to be stabilised at the cost of populated orbitals — very unfavourable.
For bridgehead carbons as in 2, especially those that bridge small rings (which is the case here), the general rule is that they do not undergo reactions that require planar transition states for exactly that reason: they are too strained to adopt an $\mathrm{sp^2}$ configuration. (This configuration would not only require the bridgehead to ‘move inwards’, but would also strain the bonds of the three neighbouring carbons, whose angles would have to be decreased.)
Conclusion: while 1 reacts rapidly via $\mathrm{S_N1}$, compound 2 does not react via $\mathrm{S_N1}$.

Complete conclusion: 2 is unreactive towards hydrolysis. 1 therefore reacts much faster.

Hydrolysis of the first compound is easier than that of the second compound. Because the second compound is the bridged compound, it is constrained to be non-planar.
The carbocations are $\ce{sp^2}$ hybridized, so the normal geometry should be trigonal planar. But because of multiple rings in different planes, it exhibits pyramidal type geometry (i.e., non-planar.)
Because of this non-planar geometry, adjacent $\ce{p}$ orbitals of two carbons are very staggered which further results in poor overlap of those $\ce{p}$ orbitals.


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Working with stainless steel in 3 M HCl solution
We are trying to mix a material using 316 stainless steel blades in a $3~\mathrm{M}\ \ce{HCl}$ solution for 4 hours in a PE container? Will the blades stand for 4 hours or do they start to corrode in 4 hours?

That's quite high concentration. The steel blades wills start to corrode. Better use  polymer lined stainless steel or , just go away from steel, and use something made of PE.

The austenitic, chromium-nickel based stainless steel SAE 316 approximately corresponds to the material number 1.4401 or 1.4436. Especially due to the addition of molybdenum in its composition it is more resistant to corrosion overall than the usual type 304 (1.4301). However, it is generally not intended to be used in hydrochloric acid.
Hydrochloric acid lacks the oxidizing properties that are needed to maintain the passive corrosion-resistant surface layer on stainless steel.
The typical corrosion rate in $1\ \%$ hydrochloric acid at room temperature is about $0.1\ \mathrm{mm\ a^{-1}}$, which is approximately $0.1\ \mathrm{g\ m^{-2}\ h^{-1}}$. At higher concentrations or higher temperature, the corrosion rate is strongly increased. Note that your concentration of $c=3\ \mathrm{mol\ l^{-1}}$ $\ce{HCl}$ corresponds to about $10\ \%$ hydrochloric acid.
In addition to this uniform corrosion, stainless steels in media containing chloride ions are particularly susceptible to pitting corrosion, i.e. highly localised corrosion with discrete pits.
Furthermore, stainless steel SAE 316 is prone to chloride-induced stress corrosion cracking at elevated temperatures (over about $50\ \mathrm{^\circ C}$).