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How do I create a transparent liquid for ferrofluid display?
Hello~ I am a media art artists working in Korea.
Recently, create a kinetic artwork using ferrofluid.
I need to create an artificial pond with a diameter of 3 meters. So I need a lot of suspension liquid.
Prototyping
https://youtu.be/3o33upltO34
simulation videos
https://youtu.be/NE19B_2byGA
Unfortunately, I tried my solution failed.
- case 1: Deionized water 40% / isopropyl alcohol 40% / Propylene Glycol 20% (Stains occur on the glass )
- case 2: Deionized water 60% / Coco betaine surfactants 40% ( no stain but, This seems strange movements)
- case 3: Deionized water 35% / isopropyl alcohol 35% / Coco betaine surfactants 30% (Stains the glass )
- case 4: Deionized water 60% / LES surfactants 40% ( no stain but, This seems strange movements.)
- case 5: Deionized water 60% / APG surfactants 40% ( no stain but, Decomposed into many particles)
- case 6: Windex(Glass Cleaner) 100%  ( no stain but, Decomposed into many particles)
I used glass types
- pyrex glass (Small stains occur)
- tempered glass (Small stains occur)
- soda-lime glass (Big stains occur)
I was always cleaning solution into a container before adding the alcohol.
And most it had hardened over time, the ferrofluid.

See http://chemistry.stackexchange.com/questions/409/what-kind-of-suspension-liquid-should-be-used-with-ferrofluid-so-it-does-not-st

There's a really good video on YouTube by Nilered/Nileblue on ferrofluids. In it he suspends magnitite with oleic acid (main component of olive oil) as a surfactant and uses Kerosene (Paraffin Oil) as a carrier.
There's a PTFE 'dry lubricant' sold under the WD-40 brand name that might be able to prevent the glass from staining as it creates a thin coat of PTFE on the glass which is highly inert. Just spray the piece of glass with the PTFE Spray and wait for it to dry (~7mins). [The PTFE spray has a myriad of other uses too, you can put it on any surface you want to have low friction, to be lubricated, an unreactive surface, dust repellant surface, etc].
This might be a way around the staining issue you are getting. Or at least point you in the right direction. Good luck!


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Distillation for separating very close boiling components
When components of a mixture have very similar Boiling Points, say 2 C is distillation ever a viable separation procedure at industrial scale?
Assuming one were willing to install a very tall column, say 100 theoretical plates (which with modern high efficiency structured packing can be fit within a column 15-20 m tall), would one expect distillation to work out?
The goal is to get an impurity component to approx. 0.1% starting from a 1.5% initial concentration. The rest of the mixture is the desired product & the BP difference between the two is only 2 C.

Isn't that exactly what columns are for?

@IvanNeretin Well, typically, at least in industrial scale I see columns that fractionate components that are as close as (say) 20 C, 15 C or even 10 C apart. But I haven't seen a process where a column was used to seperate components 2 C apart. Hence the self-doubt. I want opinions about what others have done or seen being successfully practiced.

Just take a long enough column. You can separate H2O and D2O via distillation.

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Synthesis of Diazepam with (2-amino-5-chlorophenyl)(phenyl)methanone
I tried to draw the mechanism of the diazepam synthesis.
Question: Do I really get the tertiary amine or does dimethyl sulfate methylate the amine twice?

While amines may undergo multiuple methylations to form quaternary ammonium salts, the nitrogen atom being akylated in the diazepam synthesis is an amide. Amides are different.
The lone pair on the amide nitrogen atom in diazepam is thoroughly delocalized into the carbonyl, into the aromatic ring, and all the way to the imine (see image below - some resonance contributors not shown). Due to this delocalization, that amide nitrogen atom is not nucleophilic and will not be methylated a second time. I would be much more worried about alkylating the imine nitrogen atom, which is made more nucleophilic by delocalization.

This delocalization is the reason why you need to add a base in order to do the methylation step in the first place. The amide in N-demethyldiazepam is likewise a poor nucleophile due to delocalization.

However, once a base deprotonates the amide, the amide becomes the better nucleophile due to the concentration of negative charge. Even though one of the lone pairs is delocalized, the other is not.

The amine in nordiazepam is never deprotonated by the base. The conjugate base of an amine has a pka of around 35 were methoxide is only around 15. The methoxide base is to ensure that the amine never undergoes intramolecular rearrangement to afford the positively charged ammonium ion of the amine which has a pka of around 9. The methoxide in the reaction will keep that formation from being stable as it would become deprotonated by the methoxide if it tried to occur (pka of 15 for methoxide vs 9 for ammonium ion). Thus insuring that the original nordiazepam formation does not rearrage while the methylation takes place.

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Why does greater orbital overlap mean a stronger bond?
According to valence bond theory, orbital overlap produces a bond. However, I don’t understand why having greater orbital overlap renders a bond stronger. It’s intuitive, I suppose, but I haven’t been able to find an actual explanation as to why that is. Are there quantum mechanical effects behind this?

Since elctrons are waves, and waves with same phase sum. Greater overlap is like having regions of constructive interference, where the waves of the orbitals sum. More overlap, more constructive interference (in bonding orbitals anyways)

An intuitive explanation is given in the comment above. Better orbital overlap generally indicates more constructive, favorable interaction between the two atoms.
Let's consider a homonuclear diatomic bond (e.g. $\ce{H2}$). The qualitative understanding is similar for two different atoms, but the math from quantum mechanics is a bit easier with a homonuclear bond.

So there's one bonding orbital ($\sigma$) and one anti bonding orbital ($\sigma^*$).
The energy of the orbitals will be:
$e_1 = \frac{H_{11} + H_{12}}{1 + S_{12}}$
$e_2 = \frac{H_{11} - H_{12}}{1 - S_{12}}$
You can put in some numbers and see what happens (e.g., as the overlap $S_{12}$ increases from 0 to 1, the bonding interaction becomes more and more energetically favorable).
N.B. There's a slight asymmetry - the anti bonding orbital is higher in energy than the bonding orbital is lower in energy.

I’m not that advanced into quantum chemistry yet, so could you tell me where the formulas for the MO energies come from or direct me to a page where I can learn about them? I don’t know what your symbols mean.

@lightweaver - sorry, just saw your comment. A simple model is Hückel, which assumes $S_{12}$ is zero, $H_{11}$ is a constant $\alpha$ and $H_{12}$ is a constant $\beta$.

The different letters are simply integrals between the orbitals on atom 1 and atom 2. The equations come from the energies in the $\ce{H2}$ molecule.


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How to rationalise the protonation of a compound with multiple double bonds?
In the following reaction, why is the molecule protonated at the indicated carbon and not on one of the other double bonds?

protonation of the pi bond indicated in your figure is the only one that can produce a carbocation that is both resonance stabilized (allyl) and tertiary at one end.

@ron so another pi bond in the ring can attack

No, attack by the other pi bonds cannot produce such a stable carbocation.

The key is delocalization of the positive charge in carbocations.

Most stable carbocation that you can get from the protonation of the compound above. Reasons for it being the most stable:

The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)
In one of the resonance forms, the positive charge is at a tertiary carbon atom

Second most stable carbocation.

The carbocation is a hybrid of the two resonance forms (below) – the π bond is delocalized over three C atoms and the positive charge is delocalized over the two C’s (the ones with + charge)

When protonation happens on the third pi bond. No delocalization - not as stable as the carbocations shown above.

There are other possible ways in which the compound could have been protonated (below), however none of them are as stable as (1) or (2), because there is no delocalization.


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Why are correlation consistent basis sets used with DFT calculations?
Reading some papers, there is a bit of a curious observation that does not exactly make a lot of sense to me.
In Quantum Chemistry, different methods exist to carry out geometry optimisation and energy calculations. Quantum chemists can have a huge field day arguing which method and basis set to use.
A common type of calculation used for larger (though large is relative to one's computational resources) is density functional theory (DFT) - because it provides good results on a reasonable time scale.
Now here is where it gets weird:
Every method has its basis set - basis sets geared towards DFT and basis sets geared towards couple cluster methods. It is generally recommended not to use a correlation consistent (cc) basis set with a DFT method (and I guess conversely, a basis set aimed at DFT should not be used with a coupled cluster method).
Why then would people benchmark methods using seemingly inappropriate basis sets. E.g. a cc basis set with an M06 functional.
(See for example: Edward G. Hohenstein, Samuel T. Chill and C. David Sherrill; J. Chem. Theory Comput. 2008, 4 (12), 1996–2000)

"Every method has its basis set"  They do?  Also, it is useful to implement a set of consistent basis sets when comparing methods in order to eliminate 'basis set effects'.

@LordStryker Yes, in the sense that the basis set is geared towards a type of method and performs best with that class of methods.

I suspect the Sherrill group picked cc- basis sets to eliminate basis-set effects when comparing with their coupled-cluster calculations. But you can ask them.

It's just a bit overkill..

It is generally recommended not to use a cc basis set with a DFT method (and I guess conversely, a basis set aimed at DFT should not be used with a coupled cluster method).

This statement glosses over some specifics that might be important.

There is nothing technically wrong with using correlation-consistent or ANO basis sets with DFT, unless the basis is deficient to begin with (cc-pVDZ). That is, using cc-pVQZ to optimize the geometry of a small organic molecule will not lead to more "incorrect" results than 6-31G* unless linear dependencies arise in the basis due to over-completeness.
However, it is most certainly inefficient to use these basis sets for general usage (SCF-type methods) due to using general contraction rather than segmented contraction. Most modern integral algorithms still have a difficult time dealing with highly-contracted basis sets. This is one reason to use the TURBOMOLE/Ahlrichs/Karlsruhe Def2 and Frank Jensen's segmented polarization-consistent (pcseg-n) basis sets.
"Using a basis set aimed at DFT should not be used with a coupled cluster method" is more tricky. The TURBOMOLE basis sets are certainly used for MP2 and coupled cluster calculations, however they are more appropriate for "single" calculations (not CBS extrapolation), and only with the largest basis one can afford. They were designed to capture the total mean-field energy, which decays exponentially, not the correlation energy, which decays like ~$(L+1)^{-3}$, where $L$ is the maximum angular momentum being considered. The correlation-consistent and atomic natural orbital basis sets are better for coupled cluster because they were designed from the beginning to recover the correlation energy.
The use of Pople-style (STO-nG, 3-21G, 6-31G*, 6-311++G(d,p), etc.) basis sets would be incorrect with wavefunction-based methods, perhaps even for qualitative understanding depending on how "difficult" your system is. Like the TURBOMOLE basis sets, they are designed to capture the mean-field atomic energies, but many have constraints (identical s- and p-type exponents, the "SP" functions that are often seen) due to the computational resources available at the time of their creation. Many of the older basis sets (notably 6-31G and not 6-311G) were optimized to use Cartesian rather than pure d-type functions, leading to potential inconsistencies in results. It is also not as obvious how to systematically improve the quality of the basis sets, and none are larger than triple-$\zeta$. Ultimately, the virtual MO space created by these basis sets is not sufficient to recover the correlation energy, and the benefits of using coupled cluster are completely lost.

CCSD(T)/STO-3G!

Thanks for the response (I'll see which I'll accept eventually).

@DetlevCM, I think you should accept the other answer; I think it's more general, and it's the answer I wish I could have written. The only thing I might disagree with is extrapolation using Def2, but that's nitpicky.

@pentavalentcarbon Thanks for your kind comment.  I suggest to take a look at the Neese  paper DOI: 10.1021/ct100396y . They found good results for def2 basis sets extrapolation (on both, SCF and correlation steps). They even fit the corresponding parameters. It worked for me. Also, I like your point two.

@user1420303 Thank you for the paper. I sing their praises use them all the time (def2-QZVPP is my workhorse, with def2-SV(P) for exploratory stuff), both for DFT and MRCI, but I had no idea they were this good for CBS extrapolation of the correlation energy. I also forgot that Molcas was designed to be used with their ANO sets.

@pentavalentcarbon , Yes, I was surprised when I read that paper. Nice to know that  there are people here that use MRCI or related approaches. I 'm going to make some questions about.

So, which of the two good answers should I accept? Having given them some time now.

@DetlevCM, pick the answer that in your opinion was more useful to answer the question. Both answers are compatible each other. Maybe they are complementaries. The key point here is to to exchange knowledge. Neither pentavalentcarbon nor me will feel bad about your decision.

Easier said than done :D. I upvoted both - but which do I pick as the answer... well, I need to decide.

OK, I picked this one as it also highlights some of the limitations of basis set choices (point 3). But both answers are very good in my eyes.

Because in fact it is appropriate. In most cases there is not a huge difference (quality/efficiency) among basis set families. For example Dunning (cc) basis sets work reasonable well for DFT, and Alrichs's (def2) are ok for basis set extrapolations.
There could be many reasons for the choice:

Diffuse augmentation functions were designed together with the use of cc basis sets (aug-cc-pV(n¿+d?)Z), so it is a good combination. I would trust more in such combination than in mixing functions from different families, also it looks neater (yes, the latter matter).
They comment

[...] with a polarized triple-ζ basis set yields a mean unsigned error of $0.82~\mathrm{kcal\, mol^{-1}}$, demonstrating that only minor improvements are obtained for DFT-D by increasing the basis set size [...]

So, it suggests that the results won't change significantly with another double-ζ basis set.
Each DFT functional is parameterized with an specific basis set, so, the best basis set does not need to be the largest.

Some families are often chosen because they provide a clear hierarchy that simplifies the analysis of the results.
I do not think that the following applies to the paper you cited, but in some cases it can also just be a not so good choice.
It can be a polite decision. I remember that many years ago I wished to try the ano basis set, because in that moment I had some problems to get CBS results and they had proven to behave very good for the kind of calculation I was doing. But I received the orders to stick with the cc basis set, even though I obtained bad results with them. I have never received an argument for its use, but I suspect that "The good old Dunning's basis sets, who will complain about its usage?" was the reason.
And, we all love cc basis sets...

Sure, we all love cc basis sets?

Thanks for the response (I'll see which I'll accept eventually).

@pH13-YetanotherPhilipp , yes we do!  Or maybe not (I normally prefer def2 Alrich's basis set), but when in doubt many time it is employed the basis set that one is used to. In the same way that one choose a method or functional.

So, which of the two good answers should I accept? Having given them some time now.


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What is a “spectator mode” in this analysis of molecular vibration modes
I'm reading a paper in which I found this line:

The other modes have relatively small contributions for the displacements of the heavy atoms and can be considered as spectator modes for the internal conversion

I cannot find a definition or explanation of term “spectator mode”. What is it exactly?

I suppose they may have invented it. I would assume that they took the term from spectator ions.

There is no such term, in that you won't find it in the dictionaries. This is just a combination of two words, and if you know the meaning of each word, then you know what do they mean when put together.

Without the paper its hard to follow this, also the English is not that clear, 'for the displacements' or ' towards the displacements', but it implies that these modes are not the promoting mode(s) in the radiationless conversion, i.e they don't take part, hence 'spectator'.

The paper in question is Ab Initio Trajectory Surface-Hopping Study on Ultrafast Deactivation Process of Thiophene. From a quick Google Scholar search that returns 187 hits, it is clear that the expression “spectator mode” for a harmonic vibration mode is not a standard terminology, but is nonetheless used by plenty of authors.
As to the meaning, I think you got it from the everyday meaning of “spectator”: they are vibration modes that do not participate in the phenomenon of interest, i.e. that are not coupled to the chemical or physical process studied (which can be a chemical reaction, an adsorption process or a photophysical process).


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Acetone absorbing water from the air
Our company uses acetone and isopropanol in our manufacturing process to clean our product. Our operators get the chemicals from a large container and put it into small containers like this (http://www.soscleanroom.com/product/facilities/8-oz-esd-safe-solvent-dispenser-standard-pump/?gclid=CILep_rTzc0CFdBZhgodSWsEiA) to keep at their station.
Some times our operators say that it was more difficult to clean the product today and equate that to mean the chemicals were weaker that day.
Recently our operators stated that the chemicals seemed weak again but the next day they used the same chemicals and said they were fine now. This leads me to believe that the chemicals are fine and that there is a different problem leading to harder cleaning.
How fast does acetone or isopropanol absorb water from the air? How long will it take for the chemicals to be noticeably weaker if stored in the containers mentioned above?  Should we be dumping the chemicals everyday or would weekly dumping be fine?
The temperature and humidity is relatively constant at our factory.

Unless you buy dry acetone or unless you are in a very humid environment, water takeup from the atmosphere can be neglected.

I don't believe in that water thing either. If left in the open, acetone itself would evaporate completely too fast to take any significant amount of water. That being said, why wouldn't you check it with some indicator paper prior to further action? Chemistry is an experimental science, they say.

I think that the method of cleaning is the issue. Perhaps the item being cleaned soaked for a few minutes sometime and the soaking factor is not given attention. Or perhaps the items being cleaned have been sitting for a variable length time (allowing dirt/stuff to get stuck like glue). Something along those lines makes too much sense. But if moisture is truly the issue, you could add a drying agent (like anhydrous magnesium sulfate) to the jar of acetone. Otherwise there are far too many untold variables to answer the question, how fast does acetone to absorb moisture. The answer is, test it.

For Acetone or IPA we are more than happy to keep it in normal squeeze bottles which are not air tight so the bottle you've got should be fine. I believe you're more likely to see it evaporate first than to see it get 'noticeably weaker' (I assume you're using technical grade acetone since it's just for cleaning). There should be no need to dump the acetone/IPA in the bottles on a weekly basis - although acetone is cheap its unnecessary to waste money like that!
Along the lines of trying to figure out what could be the cause of the experience of your staff, I doubt it has anything to do with the acetone itself because you say that they can use the same chemical on two different days and have two different experiences. So it's either they're imagining it (let's assume they are not), or it is the dirt that has to be cleaned that is the issue.
I've experienced it in the lab before that sometimes it does seem more difficult to wash out and clean glassware despite using the same chemicals - and this usually has something to do with how long the 'dirt' has been sitting in my flask or how it got there in the first place (slow precipitation/crystallization, crashed out, solvent dried out and left overnight, etc.). Without knowing what your product is and what it is you're trying to clean out, I can't offer much else other than checking out the consistency of the incoming product that needs to be cleaned. It could be a sign that the manufacturer's process is contaminated on some days.

I agree that the solvent isn't getting weaker.  Have you considered the environmental conditions and how they might impact evaporation rate to see if this correlates?
The solvent evaporation rate will be affected by the room's temperature, humidity, and barometric pressure.  Warm dry air can hold much more vapor than cold damp air.  Higher barometric pressures will resist the vapor pressure of the solvent.  It will likewise be affected by the temperature of the solvent and the surface temperature of the part being cleaned.  The temperature of the solvent determines its vapor pressure.  As solvent evaporates, the remaining solvent and part surface becomes cooler.  Warm solvent has a higher vapor pressure and evaporates faster.  Air turnover also has an impact (consider your HVAC system's behavior on a day-to-day basis, as its duty cycle can impact ambient air velocity in a room).  A room with static air will allow the solvent to shield itself with a localized cloud of solvent vapor that will resist further evaporation.
When cleaning with solvent on a day favorable to fast evaporation (warm, dry),  my experience has been that the quick evaporation makes it easier to clean with because each swipe is over a drier surface.  You're wiping a denser dirt concentration from the surface each time.
Likewise, if the operator is cleaning a warm part right off the production line, I would expect it to clean easier than a cooled part that has been sitting overnight.  This would also be due to the dirt/oil that you are attempting to clean.  A warm film of oil dissolves easier than a cool firm of oil.


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How to explain Born repulsion between ions in gas phase?
In simple terms, the collision of two atoms $\ce{A}$ and $\ce{B}$ will result in ions $\ce{A^+}$ and $\ce{B^-}$ if $$I_a(\ce{A})+E_a(\ce{A})<I_a(\ce{B})+E_a(\ce{B})$$ where $I_a$ and $E_a$ are the ionisation energies and electron affinities, respectively. Let us view $\ce{A^+B^-}$ as a molecule of sorts, having ions of rigid, non-polarisable spheres. Say $r$ is the sum of the radii of such spheres. The potential energy $E_p$ of such a system would then be $$E_p=I_a(\ce{A})-E_a(\ce{B})-\frac{e^2}{r}+\frac{b}{r^n}.$$
It is intuitive that

the term $I_a(\ce{A})-E_a(\ce{B})$ characterises the energy required to form isolated ions $\ce{A^+}$ and $\ce{B^-}$ in the gas phase;
$-e^2/r$ takes into account culonic attraction between two ions.

It is not intuitive for me why ${b}/{r^n}$ is  how it is.

The final term, introduced by Max Born, encompasses the repulsion generated by shells of electrons at either ion.

Of course, such effects cannot be ignored. But I am interested in why is it $b/r^n$. What is the justification? Feel free to provide the rigorous approach, even though it will probably$^{[1]}$ be above my abilities of comprehension.

Browsing on Wikipedia, the information might be given in the book Advanced Inorganic Chemistry by F. Albert Cotton, Geoffrey Wilkinson, Carlos A. Murillo, and Manfred Bochmann.
The pertinent section “1-6. Energetics of Ionic Crystals” starts at page 18$^{[2]}$, however I cannot find the derivation. There is only that the value of $n$ relies on a compressibility measurement, given as fractional change in volume per unit change in pressure, or
$$\frac{\Delta V}{V\Delta P}.$$
Also, this is seems somewhat contradictory to the earlier assumption of rigid spheres as written in another book$^{[3]}$. So, if anything, the linked book confused me further.
$^{[1]}$ Most certainly.
$^{[2]}$ Page 28 in the PDF.
$^{[3]}$ U. Palm, V. Past. Physical Chemistry. (1974) [To my knowledge, not available in English.]

Are you asking why the term is of the form $b/r^n$ or about the specifics of how to determine $b$ and $n$?

@bon: The first option; why the term is of the form $b/r^n$.

The term $+b/r^n$ is a power law repulsion term between the ions, where $b$ is the bare ion radius (twice the van der Waals radius)  and $r$ the separation.
The hard sphere model considers the ions as 'billiard balls' which means that $n=\infty$ since $+b/r^n$ is effectively zero when $r \gt b$ and infinite when $r \lt b$.
Two other types of potentials are used (a) the power law potential when $n$ is some integer value, usually between 9 and 16  and (b) the exponential potential $+a_0exp(-r/a_1)$ where $a_0$ and $a_1$ are adjustable parameters with $a_1 $ of the order of 0.02 nm. The justification  for using these potentials is weak (the power law particularly so) and they are used as they represent a better approximation than the hard sphere potential and are mathematically convenient. The exponential form is presumably justified since the combination of all orbitals decay exponentially with distance.

What exactly is meant by $n = \infty$? Could you clarify and/or provide a reference?

all it means is that the interaction is zero when r>b else infinity, i.e. hard sphere, so dividing by a huge power of $n$ makes $b/r^n$ zero.

I must do more research of available materials to readily accept the logic. So far, I have seen that even in the 'billiard ball' approximation $n \in [9, 15]$. Another states generally $n \approx 10$. From a physical chemistry textbook [3] $n$ depends on the ion. Thus saying $n=\infty$ seems odd...

If its a problem just ignore this part of the text and concentrate on what the hard sphere implies, namely that interaction is infinite at some cut-off distance and zero larger than this, just as with billiard balls bouncing elastically of one another. (By elastically I mean that no energy is retained within the balls themselves, all their energy is kinetic)

You may be aware of the general potential energy vs r curve? In which at r=0 energy is very repulsive, possibly infinitely so for some approximations, and then drops rapidly into a trough, the minimum of which is the 'equilibrium' distance r(eq) and as r increases rises back up? From my ed.3 Cotton&Wilkinson, pg 59.(Somewhat abbreviated) The reason the ions reside at r(eq) from each other and do not move closer is there are short-range repulsive forces due to overlapping electron clouds [so much for rigid spheres, huh?]  Born proposed the simple assumption that the repulsive force between two ions could be represented as B'/rⁿ where B' and n are constants characteristic of the ion pair. We can therefor write E(repulsion) = B/rⁿ where B is related to B' by the crystal geometry. so, the reason has more to due with the shape of the P.E. curve (of solid state crystals, but perhaps you know a bit about how ions in the gas phase can be considered to occupy a lattice?). I have not researched this, but my w.a. guess is that it's a fudge factor, forcing r(eq) to the correct value. It may turn out, upon doing the quantum mechanics, that there's a deep reason that the form of this term works out so well. My guess is that Born just wanted something easy to differentiate or integrate - why make it more difficult than needs be? - and no one has seen a need to improve upon it, given it's empirical derivation. It turns out that n can be determined based on compressibility (C&W are talking about solid crystals, so...) as well as theoretically and B = {-A*(Z-)(Z+) e²*[r(eq)^(n-1)]}/n from the lattice energy. C&W go on to state that corrections to their equation are sometimes used (keep in mind Edition 3 is ancient! 1972 !) and specifically that this repulsion term is "not strictly correct from quantum-mechanical considerations. More refined expressions do not greatly change the results, however."
where A is the Madelung constant, Zs are the ions charges.

Could you please [edit] this to include some paragraphing? I can't make heads or tails of this wall of text.


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Automated chemistry test machines? What is currently done, and what is research leaning towards?
There are small and large chips and machines to effect admixtures,blends, fusions of different chemicals into a variety of test samples, and there are machines to measure many vials of all kinds of biological and synthetic compounds.
The challenge is that there are so many ways to synthesize materials, using pressure, condensation, temperature, phase change onto centrifugal plate, and 100ds of other synthesis methods, lasers, using for example 40 interesting materials, there are too many possibilities for an unlimited number of machines for synthesis, the topic is unfathomable from the very start.
If we take 40 chemicals, we have billions of possible experiments to do with them, it makes the concept of fast robotic chemistry labs suitable for the synthesis and measurement of many exotic materials, and it's a very confusing topic.
What work has been done with automated chemistry synthesis and testing machines, and what will they be able to do in the next 10 or 20 years?
Are any automated test machines used to test the result of an automated compound creation machine?
Are there presently automated R&D labs that mass produce random combinations of materials in the constant search for new useful discoveries, can you shine some light on some research in the subject?

I found some info on the subject after changing the wording of the search:   --------> http://www.nature.com/news/organic-synthesis-the-robo-chemist-1.15661   and --------->http://science.sciencemag.org/content/347/6227/1221.full

----> alot of results are here ----> https://www.google.fr/search?sourceid=chrome-psyapi2&ion=1&espv=2&ie=UTF-8&q=automated%20materials%20science%20synthesis&oq=automated%20materials%20science%20synthesis&aqs=chrome..69i57.7512j0j4


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DC Cell + Copper Wire + Ground Water
DC Cell + Copper Wire + Ground Water produces bubbles from wire and water turns rust red gradually? What is happening?
I can see something is settling down...

Groundwater contain besides other things $\ce{Fe^2+}$ a lot of organic material Lenntech, Natural Variations in the Composition of Groundwater. Oxygen released during electrolysis denatures organic proteins and the $\ce{Fe^2+}$ oxidize to $\ce{Fe^3+}$ that resulting to the reddish colorization.

It also lookes like the copper wire is dissolving!!


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Is toothpaste in plasma state?
My friend told that toothpaste is an example for plasma state and it is because it is a semi solid.He said all semi solids are plasma

Your friend is, to put it mildly, about as wrong as this guy, only worse. Also, welcome to Chem.SE.

@NilayGhosh Disagree. It's closely related, but that question nowhere asks about plasma.

@hBy2Py  the plasma theory of  toothpaste is described in the comment section of the question.

@NilayGhosh But it doesn't show up in the question. If Simon-Nail-It had included it in the question body itself, then I would agree it deserves closure as a duplicate. But, comments are second-class content on SE sites, and so IMO the 'plasma' aspect doesn't count when considering whether the question is duplicative.  I'm fine with it if the community votes to close; I just would disagree.

@IvanNeretin Is irony and analogy allowed in comments? But nice reply though. I think this questions is flawed and must be flagged.

This is categorically false.
I would consider toothpaste to be a highly viscous example of either a non-Newtonian fluid or a colloidal suspension.
On the other hand, plasmas exist as a mixed cloud of separated positive and negative ions, in many cases at extremely high temperature.
One would need extensive dental work after attempting to brush one's teeth with a plasma.

I agree. But whether it's a non-Newtonian fluid or a colloid suspension really depends on the brand.

@orlandomarinella Yep, whether it's the paste type or the gel type.

It was an attempt at humor. No problem!

@orlandomarinella The content of your humorous addition was consistent with my intent, but I would prefer it to stand as originally written.  We may be up against a debate about paste- versus gel-type toothpastes, however: I would argue that gel-type is the non-Newtonian fluid, whereas paste-type is the colloidal suspension.

Perhaps, although it, again, depends on the exact formulation of the gel or paste. For example, in my drawer over here, I do have a gel toothpaste with the strange fleck-y bits, along with lumps of what's probably whitening abrasive.

To be perfectly honest, they're probably a little bit of both, for each.

@orlandomarinella , it's not an either-or category distinction. As such, no offense, but I think I will actually just roll back to my original answer.

No problem at all, actually.

Absolutely wrong...!
This is absolutely wrong, toothpastes are not in plasma state.
so what are toothpastes?
A simple answer is; toothpastes are either pastes or gels.
They are not solution because solute is not dissolved in the solvent. They are not suspensions because of ultra-high cocentration.
So toothpastes are emulsions (two phased mixture). Toothpastes comprise an oil-water emulsion. And this form is used to stabilize the primary components of the toothpaste.

antiseptic/bacteriocidal component: this is contained in the aqueous phase of low viscosity it is capable of penetrating into the hollows of the teeth and gaps in the gums.
hydrogen peroxide part which is stabilized in the hydrophobic part of the emulsion.

Read Colloids and Emulsions for better understanding.

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Coefficient in front of molecules in a chemical equation
In a chemical equation representing a chemical reaction what do the numbers in front of the molecules of compound and elements indicate? Do they signify number of molecules or moles or volume?
Example : $$\ce{ 2H_2 +O_2  -> 2H_2O}$$
Here does the "2" before $\ce{H2}$ and $\ce{H2O}$ indicate number of molecules or moles or volume?
In a particular chapter in my textbook it was written that they indicate number of molecules(atoms) but in a later chapter the numbers were being treated as if they are number of moles or volume.

2 is the stoichiometric coefficient. It gives the relative amount of moles or volumes of reactants and products involved in the reaction.

Please don't call the 'amount of substance' 'number of moles', you wouldn't call the 'mass' 'number of kilo grams'.

There's the trick: it's all of them.
The coefficients are a ratio. For every two unit of hydrogen molecule, you need one unit of oxygen molecule, and the result is two unit of water molecule.
That unit is often the particle itself, whether it's an atom, a metal ion, a more substantial molecule or even a protein or enzyme. It can also be the volume (for ideal gases under the same conditions), the moles used in reaction, or molarity in solution.
Simply remember that the coefficient applies to the item following it as a whole. As long as the ratio between coefficients is the same, you can use any unit that maintains it.

To clarify, the coefficient can only be volumes for ideal gases.


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Do I need to include calcium chloride in this reaction equation?
Yesterday I titrated a solution of calcium hydroxide in aqueous calcium chloride with $16.7~\mathrm{mL}$ of $\ce{HCl}$. More specifically, I took $25\ \mathrm{mL}$ of the calcium hydroxide in aq calcium chloride and mixed it with $25~\mathrm{mL}$ of deionized water and 3 drops of an appropriate indicator. I wrote the reaction equation as
$$\ce{Ca(OH)_2 + 2HCl -> CaCl_2 + 2H_2O}$$
But do I need to somehow take into account the aqueous calcium chloride in this equation? I know that it affects the molar solubility of the calcium hydroxide during titration but I'm not sure if this should also change my reaction equation.

Is it a reactant? Does it react with the other substances?

Hmmm...not sure. We did 2 experiments where the only difference was the original solution: exp #1 used calcium hydroxide and exp #2 used calcium hydroxide in aqueous calcium chloride. Everything else was the same.

My guess would be no.

@Archa am I right?

From an instructor's point of view, you have accounted for the reaction that has taken place during titration.  The point you are making surrounds the conditions the titration took place under (as you pointed out, solubility is altered, for instance), but not the chemistry of the reaction, in and of itself, which you are focused on.
In essence, you are doing the first titration in aqueous calcium chloride as well, because as soon as the titration begins, what is formed?
That said, I feel I may be missing something more of the other potential rxn conditions or possible expectations of your evaluator.

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Finding OH- concentration given the H+ concentration - diverse methods
There was this exercise which asked to find the OH- concentration given the H+ concentration, i firstly calculated the pH as -log[H+], did 14-pH to find the pOH and then did 10^(-pOH) to find the concentration of OH-.
My textbook used the fact that Kw (water equilibrium constant) is essentially
Kw = [OH-] * [H+] so it did    [OH-] = Kw/[H+]
The fact is...isn't Kw temperature dependent? My textbook says that it is
25 celsius --> Kw --> 1*10^-14
whereas
60 celsius --> 9.61*10^-14
So my question is...why is my procedure temperature independent while the book one temperature dependent? My procedure would always give the same concentration of OH whereas changing the Kw the textbook would also get other ones.
Thank you.

Where does your formula $pOH=14-pH$ come from in the first place ? ;-)

https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm
My textbook also states that

Click the "show me why" section of that link you provided :)

Ok, even my method is temperature dependent :P Thank you really much for  letting me know :)

Exactly, both methods are equivalent in the end.

It is because you assumed the number 14 as a constant while it is actually $-\log(K_\mathrm w)$.
It is $-\log(1 \times 10^{-14}) = 14$ at 25 degrees Celsius but changed to $-\log(9.61 \times10^{-14}) = 13.02$ at 60 degrees Celsius.


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Vapour Pressure and Equilibrium
I found this questions on a YouTube lesson and I'm confused with the answers they have given:
"A liquid in its equilibrium with its vapour is in a sealed container at a fixed temperature. The volume of the container in suddenly increased."
Questions:
a) What are the changes in the rate of evaporation and condenation?
b) How will the equilibrium be restored ultimately and what will be the final vapour pressure?
Answers:
a) Rate of evaporation depends on the substance and applied temperature so it will be unchanged and rate of condensation will decrease as vapour pressure is decreased. (Why? Is it true?)
b) They say that the vapour pressure will be restored and the rate of condensation and evaporation will be equal as the system reaches the equilibrium. (But they said that rate of condensation will decrease and rate of evaporation will stay the same. Then how will they be equal?)
Please clarify the whole thing.

Can you provide a link to the lesson?

The key missing piece is identifying when the system is in equilibrium. When you suddenly increase the volume, the system is temporarily not in equilibrium. In this situation, a few things will happen:

The rate of evaporation won't change for the reason given
The vapor pressure will decrease. As a first approximation, we can use the ideal gas law to see how. $PV=nRT$ can be rearranged as $P=nRT/V$. The number of moles ($n$), gas constant ($R$), and temperature ($T$) will stay the same (this last one in particular isn't really true, but it doesn't change the qualitative result). Then, if the volume increases, the pressure will decrease.
Since the pressure has decreased, the rate at which gaseous molecules bump into the liquid decreases. The ones that do bump into the liquid will condense into the liquid at the same rate. The total condensation rate is the product of those two factors, so the condensation rate will decrease.

Now, since the evaporation rate is the same as it was before, but the condensation rate will go down, we will get a net evaporation. As more liquid becomes gas, the vapor pressure will increase. As the vapor pressure increases, the condensation rate will increase, for the reasons discussed above. Once the vapor pressure is high enough that the condensation rate equals the evaporation rate, then we will have a dynamic equilibrium.


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Removal of tetrabutylammonium from an organic reaction
I need to synthesize some organic compound for my Ph.D. research. Everything went OK during the synthetic route but got messed up at the very end. In my final step (once the final compound has been obtained) I remove a silyl protecting (TBS) group using TBAF. The problem comes when I try to purify the reaction. No matter how carefully I run my flash column (FCC, flash column chromatography), I always obtain my compound with traces of TBA.
I've tried other deprotection procedures (mainly acidic ones) but they hydrolyze my final compounds since they are sensitive to acidic conditions.
Does anyone know of an alternative purification technique (or reaction) that would remove the TBA?

Did you consider using ammonium fluoride instead of TBAF? Article: http://www.sciencedirect.com/science/article/pii/S0040403900918896

Did you consider HF-pyridine complex? TBAF looks great on paper, but can be trouble in practice.

FCC = Flash Column Chromatography?              TBAF is usually non problematic... unless your compound is incredibly non-polar the by-products of the silyl deprotection wash off the column in hexanes, and then you can modify the polarity to elute your compound

Though I agree with JerePierre, if you're 20 steps into a synthesis, HF is the way to go.    Mild, clean and easy to carry out, the only downside being the potential harm if you get it on yourself.

@RokNarobe no I did not, but I'll have a look at it. Thanks

@jerepierre I did use it. The results were worse than before with most of my compounds decomposing.

I know it is a relative old issue but I had the same problem with nucleosides during my PhD. I tried a wide range of reagents but all of them lead to the decomposition of compounds or no reaction occurred. The only one that worked was TBAF 1 M in THF. I could not extract them because they are too polar. FCC did not work at all.
Solution (at least it worked for me): after FCC, I dissolved my product with TBAF in water and pass it through a Supelco SPE DSC-C8 3 mL, 500 mg for up to 10 mg of my compound. Use only water and do it slowly. I always get the desired compound from tube one to 6-7 and no TBAF is present. It is useful if you can use a LC-MS to analyze each tube. I tried other brands and no one worked except this one. I don't work for them, just want to share my solution.

I have the same problem in my Ph.D. research. Whenever I am too lazy to skip the aqueous workup and throw the solid residue from rotavapor directly into a column for chromatography, TBA just penetrate through flash column and exist in every tube I collect. To me, aqueous workup help a lot to remove TBAF if your product is not too soluble or unstable in water. Also, TBAF is mostly used as the floride anion source because its solubility in organic solvent. Other fluoride anion source like KF can be used as alternative to TBAF with proper solvent.

That's exactly my case. I can't do water extraction because my products are too polar and would go to the water phase.

You may just try other fluoride salts. CsF or KF or NH4F may do the deprotection job as well as TBAF while they will not mess up in your column. I remember KF will work in acetonitrile. HF is acidic and may decompose your compound. HF-Pyridine is almost equally messy in column as TBAF a lot of time.


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How to find the pH of acetic acid after adding sodium hydroxide?

$3\cdot 10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$ are added to a $1\,\mathrm{l}$ solution of $2\cdot10^{-2}~\mathrm{M}\ \ce{CH3COOH}$. Find the $\mathrm{pH}$ of the solution.

I don't understand why my solution is wrong:
I said this, $\ce{NaOH + CH3COOH -> H2O + CH3COONa}$, so I'm left with $2\cdot10^{-2}~\mathrm{mol}$ of $\ce{CH3COONa}$ and $1\cdot10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$.
Now, from those $1\cdot10^{-2}\ \mathrm{mol}$ of $\ce{NaOH}$ I get the same moles of $\ce{OH-}$ (from dissociation).
Then, $\ce{CH3COONa -> Na+ + CH3COO-}$, and $\ce{CH3COO- + H2O <=> CH3COOH + OH-}$
Setting up the equilibrium equation: $K_h = \frac {\text{Products}}{\text{Reactants}}$, I get (where $K_h= K_\mathrm{w}/K_\mathrm{a}, K_\mathrm{a} = 1.8\cdot 10^{-5}$):
$$K_h=\frac {(x)(1\cdot 10^{-2}+x)}{x-1\cdot10^{-2} }=5.55\cdot 10^{-10}$$
Which yields $[\ce{OH-}]= x= 1.11\cdot 10^{-9}$ and so $\mathrm{pH}=5.04$ which is wrong. I know the answer should be around 12 from the answer sheet.
What am I doing wrong?

Note that $\ce{Ac}$ is $\ce{CH3-C=O}$ and not $\ce{CH3COO}$.

@Jan I don't understand why my question's been change, the question is with HAc, not HOAc...

I seriously doubt that you attempted to titrate acetaldehyde with sodium hydroxide, especially since the carbonyl proton of acetaldehyde is not acidic. Acetic acid is $\ce{HOAc}$, by definition.

@Jan I suppose in my country it's commont to use HAc for acetic acid (yes, this was supposed to be acetid acid with sodium hydroxide)...

@YoTengoUnLCD. In the USA, it's also common to abbreviate acetic acid as HAc.

If you add $3 \cdot 10^{-2}~\mathrm{mol}$ $\ce{NaOH}$ to $1~\mathrm{L}$ of $2 \cdot 10^{-2}~\mathrm{M}$ $\ce{AcOH}$ you end up with a solution that contains $2 \cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{AcO-}$ and $1\cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{OH-}$. So the pH is 12.

The pH of the solution is imposed by the the concentration of $\ce{OH-}$ from $\ce{NaOH}$: $\ce{[OH^{-}]}=10^{-2} \mathrm{M}$, since the quantity of $\ce{OH-}$ produced from the reaction of $\ce{OAc-}$ with water is negligible in front of $10^{-2} \mathrm{M}$:
$$\ce{OAc- + H2O <=>AcOH + OH-}$$
The equilibrium constant is:$$K=\frac{K_w}{K_a}=10^{-9.3}$$
So, $\mathrm{pOH}=2$, i.e $\mathrm{pH}=12$

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Oxidation of cycloalkenes
Non-cyclic alkenes can be oxidized to carboxylic acids in presence of hot alkaline $\ce{KMnO4}$ and quenched with dilute $\ce{H2SO4}$ to form carboxylic acids. I suddenly got a thought. Can cycloalkenes be oxidized to form carboxylic acids? For example: can cyclohexene be oxidized to cyclohexanoic acid using the same conditions?

Your example sounds a bit weird to me. Using KMnO4 on alkenes can thansform the alkene function into a carbox acid function; but cyclohexene -> cyclohexanoic acid would actually add a carbon atom on the cycle and put the carbox acid on that new carbon, it wouldn't really be similar

Both ends of the alkene would be converted into a carboxylic acid functional group.
This would be a 'ring-opening' or 'permanganate cleavage' reaction and form adipic acid (hexanedioic).


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Find moment of inertia around two axes of sulphur dioxide
Problem:

Find the moment of inertia for 2 axes of $\ce{SO2}$. The rotational constants are  given in $\pu{MHz}$. The answer is required to be $10^{-46}\rm~ kg\cdot m^2$, 7 digit accuracy.

I used $$I = \frac{h}{8\times\pi^2\times B}$$ but the answer is marked wrong. I do not understand what I am doing incorrectly. I thought maybe that $\ce{SO2}$ is asymmetric so that formula will not work.
This is an online course that has  minimal discussion assistance. I have scrounged on Google & physical chemistry textbook to see what the alternative could be. Did not see another formula. Where did I go wrong?

Please take your time and format yuor post with the MathJax syntax so that we have a better idea of what you’re saying.

First question for you is what is the geometry of sulfur dioxide? Next, how many unique principle axes does it have?

Without explaining some more its hard to help, but the most likely error is in using the wrong units. The rotational constant is usually given in wavenumber units (cm$^{-1}$), e.g. 2.02736 cm$^{-1}$. In a linear molecule the equilibrium value is
$B_e = \frac{h}{(8\pi^2cI)}$
with $B_e$ in wavenumbers and c is the speed of light in cm/sec. In your case since the SO$_2$ is not linear (its planar bent at the S atom, C$_{2v}$ point group) there are three rotational constants usually called A$_0$, B$_0$, C$_0$.  (You can check the units by converting Joules in to base units, m, kg, s etc . The moment of inertia is in kg m$^2$ and $10^{-46}$ is of the correct order of magnitude. Seven decimal places seems too many as the rotational constants are not usually known to more than 5 places.)

Thank you. But I am back at square 1. I can't use the formula B = h/(8(pi^2)I)?
I thought that Ia = h/(8(pi^2)Ba). And the same for Bb. The problem requires 7 digit accuracy.
**For the vibrational ground state, the rotational constants, A and B, are 60778.79 MHz and 10318.10 MHz, respectively. Calculate Ia and Ib (Unit: 10^−46 kg m2, 7-digit accuracy). **
The moments of inertia around the principal axes
Ia=(2mOmS/M)r^2cos^2(θ/2)
Ib=2mO*(r^2)sin^2(θ/2)
Ic=Ia+Ib
mO and mS are the masses for oxygen and sulfur, respectively. M≡mS+2mO is the mass of SO2.*

Could it be I must convert MHz to joules?

I thought I was going koo-koo. I was going over & over what I was doing wrong for this problem. I am REALLY sure I put in the same answer that I did just now - i did the SAME CACULATION. Now I got the green check - for this problem. What happened before? Who knows with MOOC graders! Thank you for the response to help!

please tick and markup if the answer was helpful.

I appreciaite the answer but it did not help. Fortunately  I had the right answer - it was the  course grader that malfunctioned.


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Would there be a solid residue left behind from a reaction between NaK and H2O?
I gather the reaction produces Sodium Oxide and Potassium Oxide as well as Hydrogen. Would there be traces of Sodium Oxide and Potassium Oxide left of a cloth that was positioned near the reaction? Thank a ton.

According to this site:-

NaK reacts explosively with water to form hydrogen, potassium hydroxide
  and sodium hydroxide.

This site says:-

NaK is a strong reducing agent. It is corrosive and will react violently with water. The reaction with water will liberate extremely flammable gases potentially causing a fire or explosion.

This youtube video confirms the above fact.

The answer to the question is yes, there will be residue. Let's look at the chemical reaction/equation: $$\ce{2NaK + 2H2O -> Na2O + K2O + 2H2}$$


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How to determine the structure of organic molecules without spectroscopy
I wonder how chemists worked out the structures of organic molecules before spectroscopy (especially NMR) was invented?
Surely, we could do functional group tests, but the correct connection of atoms and function groups would be hard to tell when the number of carbon atoms get larger. E.g. How could they tell one compound was 2-hexanol rather than 3-hexanol? How could Emil Fischer and others worked out the correct structures of carbohydrates with so many chiral centers? And god forbids, how could the chemists worked out the correct structures of natural products with fused rings (eg the cholesterol family)? It's not a trivial task even with 2D NMR!

Couple of clues: (i) MgO was regarded as an element before the electrochemical methods; (ii) Ostwald did not believe the existence of atoms and molecules for some period of time; (iii) if we assume known carbon and hydrogen are the element, we may able to determine CH4 from ideal gas. -> All we can do is try our best to reach some logical consistency and leave further issue to next generations.

See this question

A very thorough historical answer has been given by @Jan which shows how hard this was. But recall that NMR does not generally produce 'structure' but 'topology' as bond lengths are generally not measured but the relative positions of atoms. Only x-ray crystallography  produces a definitive structure and in the gas phase,  microwave spectroscopy, but only for molecules containing very few atoms. In the future possibly STM or AFM methods may be used on single molecules.

@porphyrin You may be correct, but all throughout organic chemistry, structure is used to refer to atoms’ connectivity.

One thing that was well known, even in the early days, was elemental analysis. It is Justus von Liebig’s contribution to have worked out the sum formulae of hundreds of natural compounds. This generally worked by combusting a given sample of a compound, capturing the vapours, absorbing the water with desiccants such as calcium chloride, absorbing carbon dioxide by potassium hydroxide (forming a carbonate), weighing the different parts and thereby calculating the mass of carbon and hydrogen in a given sample. Tests for nitrogen and sulfur also existed, but I can’t remember them off the top of my head. Oxygen content was indirectly calculated as being the remainder.
Chemists since the advent of their science had done everything to generate pure compounds. It was widely accepted even then that a pure compound was one with a defined melting or boiling point, and liquids could also be analysed by their refractive index and density. This provided a rather good library of methods for realising ‘is my compound pure?’, and ‘is it the same as $\ce{X}$?’
Another very old concept was that of valencies. By analysing samples of known and simple elemental composition, it was realised that certain elements bonded to a certain number of other elements, e.g. that a hydrogen would always require a bonding partner of value 1. Somebody (I can’t remember who but I could look it up in my lecture notes) proposed a valency of 4 for carbon. Together with a valency of 2 for oxygen, you had most pieces of the puzzle in hand.
It was recognised by Liebig and Wöhler, that isomers existed, i.e. compounds of the same elemental composition but distinct. (This is also a consequence of the ammonium cyanate experiment Wöhler performed to create urea, breaking the wall between organic and inorganic chemistry and proving vis vitalis to be inexistent — both are isomers of each other.) So there must have been something important in the structure of molecules that needed to be discovered.
A lot of further clues then came from reactions that transformed known compounds into other known compounds. For example, oxidation would transform some known compound $\ce{C3H8O}$ into acetone $\ce{C3H6O}$. This also lead to Liebig’s theory of radicals — where radicals is something a modern chemist would refer to as functional group. With these clues and radicals, the task of putting the puzzle together to pinpoint certain structures became solveable.
From then on, it all depended on whether the initial assignments were correct (they mostly were). You would go ahead either breaking down unknown compounds by known reactions after having established their functional groups or resynthesising them from known fragments in other known reactions to arrive at a final product.
Working out the difference between hexan-2-ol and hexan-3-ol would have been a none-trivial task; that is true. One variant seems to be starting from hex-2-ene and adding water to the double bond. You would get two different products and with the knowledge of structure in hand, you would realise that they had to be isomers of each other and had to be the 2- and 3-isomer respectively. How did you get hex-2-ene? Oh, maybe you managed to dehydrate hexan-2-ol into two different products, hex-1-ene and hex-2-ene. How did you get hexan-2-ol in the first place, then? Addition to hex-1-ene sounds reasonable. How to get that? Dehydration of hexan-1-ol. That final one was easily established as a special hexanol, since it could be oxidised to hexanoic acid, not only to a ketone. It is all possible but requires a lot of work. Think how lucky we are that we can record an NMR spectra within minutes (or a complete set over night) and solve the structure the next day!
It is entirely possible that for the cholesterol family, the exact ring connections were initially not known, only that there is some type of a fused ring system that can be modified in certain positions (not all!). It would have taken extra time to break down the ring system accordingly. And Emil Fischer’s work on carbohydrates probably warrants one or two questions of its own.
And then, crystallography happened, and all of a sudden was it possible to see structures. It must have seemed like the dawn of a new era.


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Extracting urea from urine
What is a process for the extraction of urea from urine? These are the steps I have come up with:

Add 2-propanol to get rid of any bacteria present. (The 2-propanol likely contained some sodium chloride.)
Precipitate out the proteins and neutralize any acids present using sodium carbonate. (At least, that is what appears to happen upon addition.)
Crystallize out the sodium and potassium chlorides, as urea is much more soluble than those salts.

At the second step, three layers formed: a top clear yellow layer (which I assume was 2-propanol), a middle very cloudy yellow layer (which quickly disappeared into the bottom layer) and a bottom slightly cloudy yellow layer (which I assume was water). The bottom two layers contained particles floating around.
Then, I assume that the water layer remaining contains mostly urea. However, how do I get rid of everything else (urobilin, etc.)?
The end goal is to generate ammonia from the urea by adding some calcium hydroxide. However, it would also be nice to get rid of other impurities which do not impede this reaction.

This seems like a very long-winded and troublesome way to generate ammonia when you can just buy it.

Why don't you just add sodium hydroxide to the urine and distill the ammonia directly? That will kill the bacteria, and if you add enough lye, it will break down the protein and increase your yield of ammonia. Also, the amount of protein in urine should essentially be none, unless diabetes or kidney disease, or some kind of medical problem is involved.

@BenWelborn I don't want to get ammonia directly, because I want to be able to just get some urea and $\ce{Ca(OH)2}$, put it in a bottle and (for example) fill a balloon. Wasting time to generate the urea intermediate saves time later, when I actually need to use ammonia.

Fill a ballon with ammonia gas? Please excuse my intrusiveness, but why generate ammonia from urea when you could from ammonium chloride or sulfate? In other words, add lye to urine, distill ammonia, add acid to the distillate to make ammo-salt... then you can generate gas from ammo-salt. I understand you want urea from urine, but if your goal is actually ammonia, then you should reconsider harvesting urea.

First, from Boundless.com (Chemical Composition of Urine), "Urine is an aqueous solution of greater than 95% water, with the remaining constituents, in order of decreasing concentration urea 9.3 g/L, chloride 1.87 g/L, sodium 1.17 g/L, potassium 0.750 g/L, creatinine 0.670 g/L and other dissolved ions, inorganic and organic compounds (proteins, hormones, metabolites)."
Urine color is generally a result of the degradation of the heme molecule from blood into the urinary products, called bilirubin, urobilinogen, urochrome, urobilin and uroerythrin. In essence, they are all tetrapyrroles amounting to something on the order of 1mg per 100ml (or roughly 0.001% of urine)... so this is a very minor part of urine. Also, it should be noted that over 3,000 chemicals have been found in urine, so there's something to consider.
Activated carbon filtration might be able to be used to preferentially remove the tetrapyrroles and some other organic molecules while allowing urea to pass (for the most part).
There are various forms of activated carbon (of course), and so the amount of urea absorption (by carbon) will be variable; the adsorption capacity of the activated carbon for urea will (likely) be different for activated carbon from different sources. Absorbance of urea from one report was about 1.69 mg urea/g carbon, where the urea concentration was 100 mg/L (see picture below). At greater concentrations (like 930mg urea per liter), lesser amounts of acivated carbon will become saturated; in other words the activated carbon will not be able to absorb more urea, but it might be able to effectively remove the tetrapyrroles or some other minor stuff.

But I think that the first thing to do is deal with the concentration. Urea is only about 0.93% of urine. So you only have about 0.93g urea per 100ml of urine.
To concentrate urea: boil off the water (which will also kill bacteria). Urea doesn't boil. The melting point for pure urea is 132.7 °C and urea will begin to decompose at around 150 °C. The residue (which should look creamy) will have a relatively high urea concentration which can then be extracted with (warm/hot 60 °C) ethanol or diacetone alcohol. Salt and protein are significantly less soluble in alcohol(s), so they should hopefully be (mostly) removed at this step.
The solubility of urea in 95% alcohol (100 g/L), absolute ethanol (50 g/L), and methanol (166 g/L) at 20 °C (room temperature) sounds pretty fair. And from TECHNICAL MEMORANDUM No. 13/M/67 (Investigations of Crystal Modification of Urea) "in the case of diacetone alcohol the solubility [of urea] ranged from 2 per cent at 20 °C to about 10 per cent at 100 °C, offering possible development of a process yielding free-flowing crystalline urea."
The solubility of salt (at 25 °C) in ethanol (0.65 g/L) and methanol (14.9 g/L) is significantly lower.
Now, the solubility of a hodgepodge of stuff including tetrapyrroles is something that you will have to experiment with. But, many similar compounds are readily soluble in acetone and tetrachloroethylene (often had as brake cleaner). Urea should not be highly soluble in these two solvents, but I'm a afraid I can't find a reliable solubility listed online, and I happen to be out of acetone right now. Cleaning an alcoholic extract of boiled urine with tetrachloroethylene might require using a separatory funnel.
On a historical note, (around 1727) Herman Boerhaave was able to extract urea from urine by working with the concentated-by-boiling residue (having the consistency of "fresh cream"). After filter pressing the residue, he allowed it to dry for a year... until he obtained a solid mass with an unknown oily substance (on top).  He discarded the oil and rinsed the crystals (filtride) with cold water to remove salt, leaving urea. He then recrystallized the urea from hot water. Previously (in 1669), Hennig Brand discovered the philosophers stone (phosphorus) by cooking the bejeezus out of (lots of) urine.
Personally, instead of waiting a year, I would take a chance on extracting urea from the cream. Also, I read today that boiling urine causes irreparable stink-damage to a home; so get a hot plate and an exension cord and boil it outside. Another recipe (as if I had nitric acid to spare), calls for reacting concentrated urea (say 100g in 100ml water) to make urea nitrate. The addition of nitric acid is exothermic and urea nitrate is explosive, so make sure the solution doesn't boil off (ice bath)... and urea nitrate is precipitated in the freezer. Then after you have clean urea nitrate, you can neutralize the acid under alcohol to recover urea. I would reserve this (nitric acid) method for the final work up, using relatively clean urea.
Also, here's a couple of pay-walled articles that might be useful to you if you really want to do this: Solubility of urea in water-alcohol mixtures and Urea in the history of organic chemistry: Isolation from natural sources

Urea is soluble in alcohol and methanol while the salt isn’t, so you simply boil the urine to reduce its volume, add methanol, separate the methanol layer, add $\ce{NaOH}$ and a bit of $\ce{H2SO4}$ or $\ce{HCl}$ and evaporate the solvent. There should be urea pure enough for anything you want to do …

Methanol and water won’t form two layers that you can separate.


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How to retrieve Dalton value from m/z value?
As I understand, Dalton (Da) is the standard way for representing mass unit. As for $m/z$, which is a specific notation for mass spectrometry, quoting for wikipedia:

This notation eases data interpretation since it is numerically more related to the unified atomic mass unit

Considering this, could I say that the $m$ in $m/z$ is a $\mathrm{Da}$ unity? If so, given an $m/z$ value, can I retrieve the original mass of the atomic mass by multiplying it by the ions charge, if I have that information?
I ask this because I have a spectrum analysis file with $m/z$ in the $x$ axis and I'm asked to compare masses in $\mathrm{Da}$.

See also: http://chemistry.stackexchange.com/questions/9978/could-someone-please-explain-the-difference-between-m-z-and-m-e-in-mass-spectros?rq=1

IUPAC is wrong
Martin's answer is useful for its pointer to the IUPAC Gold Book, which is one of the most authoritative sources for chemical nomenclature around.
However, on this particular question, the Gold Book is laughably wrong.

The abbreviation $m/z$ is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number.

This sentence is wrong.  The $m$ in $m/z$, as used by nearly every practicing chemist in the world (except IUPAC), is the ion mass, not the mass number.  The mass number is the sum of protons and neutrons in a nucleus.  Therefore it is always an integer.  Thus, according to IUPAC, ions with a charge of $\pm 2$ would always have $m/z$ values that are either integers (if $m$ is even) or half-integers.  Thus, according to their definition, saying the $m/z$ value of $\ce{C7H7^2+}$ is 45.52683903 would be wrong.  However that is exactly what most mass spectrometrists would say it is.
Here is some evidence:

A lipid maps database entry for $\ce{C7H7}$ lists $m/z$ values for a hypothetical $\ce{C7H8+}$ ion that have four decimal places.
I happen to have the latest print edition of the Journal of The American Society for Mass Spectrometry in front of me, and paging through the contents I see things such as:

...an intense signal for the $\ce{Cu^2+ - His_3}$ complex at $m/z$ 245.6, instead of the $\ce{[His_3 + 2H]+}$ at $m/z$ 215.1...
...MS spectra of standard solutions containing (a) lactose ($m/z$ 365.11, $\ce{[M + Na]+}$; $m/z$ 381.309, $\ce{[M + K]+}$), (b) cytidine ($m/z$ 266.218, $\ce{[M + Na]+}$, ...

These excerpts are not using the (integer) mass number in their calculation of $m/z$, they are using the ion mass, measured in Daltons.  (If they were, the only numbers after the decimal point would be .0 or .5).  That means that $m/z$ values are not really dimensionless, which makes the sentence wrong again.  In fact a unit called the Thomson has been proposed for $m/z$ values in honor of JJ Thomson. However this nomenclature has not really caught on widely.  Nonetheless, $m/z$ values are not dimensionless.

The Gold Book goes on...

It has long been called the mass-to-charge ratio although $m$ is not the ionic mass nor is $z$ a multiple or the elementary (electronic) charge, $e$.

This sentence is wrong because $m$ is the ionic mass (in Daltons) as explained above.  It is additionally misleading because $z$ is always a multiple of the elementary electronic charge, with the understanding that charge must be measured in multiples of $e$, not in Coulombs.

The abbreviation $m/e$ is, therefore, not recommended.

Finally something I agree with IUPAC on!

Thus, for example, for the ion $\ce{C7H7^2+}$, $m/z$ equals 45.5.

Most folks would find that to be an OK approximation to the true $m/z$, which as I said above should be 45.52683903.
For the purposes of your question, Martin was right when he said

For all intents and purposes the mass number may be seen as the mass of the ion in u or Da...

One thing to keep in mind is that this definition of "mass number" is at odds with IUPAC and nearly every other chemical text, which restricts "mass number" to be integers.
One additional complication
It's a bit unclear in your question if you want to determine the mass of an ion or the mass of the parent (usually neutral) molecule that gave rise to that ion.  If you want to determine the mass of an ion, then the other answer is right and you can just multiply the $m/z$ by $z$, the (absolute value of the) reduced electric charge.
If you want the mass of the parent neutral molecule, you will (a) need more information, and (b) have to do a bit more work. The extra information you need is the ionization mechanism.  In electron ionization mass spectrometry, ionization is usually by electron gain or loss.  So to get the neutral molecule mass, you will have to add back the missing electrons (or substract the gained electrons) from the calculated ion mass to get the molecule mass.  For example, if $\ce{C7H7^2+}$ arose by the loss of two electrons, the $m/z$ would be 45.52683903, the ion mass would be 91.05367806, and the molecule mass would be 91.05477522.
Other types of mass spectrometry involve ionization by gaining or losing protons, not electrons, and so the difference between ion mass and neutral molecule mass is much more significant.

How can the mass number be different from the mass of it just omitted the unit? If z was always a multiple of the charge, then it would be allowed to be negative. Claiming IUPAC is wrong is quite far reaching, I find none of your arguments compelling.

The mass number as defined by IUPAC & by Wikipedia is the sum of the number of protons and neutrons in a nucleus.  Therefore it is always an integer.  Thus, according the IUPAC the $m/z$ values of ions with $z=1$ must be integers.  This is laughable.  I can see how I missed a few points in my explication above and will edit to improve.

I suspect that IUPAC drafted this statement before "high resolution" mass spectrometers were available to many users.  If all you have is unit mass resolution, then the worst of IUPAC's errors ceases to matter.  However, high-mass instruments have been available for some decades now.

As a mass spectrometrist, I agree with Curt.  IUPAC is wrong, and you can convince yourself of this by going back to SI units.
Mass is in kilograms, charge is in coulombs (technically Amp*seconds), so the SI unit for mass/charge should be kilograms/coulomb.  Mass/charge is not, and should not be defined as, 'dimensionless'.
So, there is a unit problem at the heart of mass spectrometry.  We 'fudge' the issue by stating 'm/z 45.52683903', but it's a cop-out. Until IUPAC fixes their error, we won't have consensus on what units to actually use.
Given that the mass will be in u or Da, and the charge must be n*e (where n is an integer and e is the (absolute value of the) charge on an electron, then the obvious unit should be Da/e or u/e.  The 'Thompson' or 'th' has been proposed, and is sometimes used, but unless IUPAC adopts it and properly defines it, it won't become the mainstream unit.  Until then, we are in limbo.
That said, it is easy to calculate the mass if you know the m/z and the charge, z.  m/z can be measured off the spectrum, and z can be determined by the isotopic spacing which would be about 1.0034/z for carbon isotopes.  Thus the mass of the neutral molecule will be the m/z multiplied by z, and removing any adduct masses (such as protons, electrons, sodium ions, etc) which were added to turn the neutral molecule into an ion.

Interesting comment and +1.  Welcome to chem.se!  I myself don't see a "fudge" in the absence of a IUPAC ruling, except maybe for whether "Th" will catch on.  But I don't think many mass spectrometrists feel stuck in limbo when they say "$m/z$ of 45.52683903"; they know the units are really Da / $e$.

From the IUPAC gold book:

mass-to-charge ratio,  $\frac{m}{z}$
in mass spectrometry
  The abbreviation  $\frac{m}{z}$ is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number. It has long been called the mass-to-charge ratio although  $m$ is not the ionic mass nor is  $z$ a multiple or the elementary (electronic) charge,  $e$. The abbreviation  $\frac{m}{e}$ is, therefore, not recommended. Thus, for example, for the ion ($\ce{C7H7^2+}$),  $\frac{m}{z}$ equals $45.5$.

The mass-to-charge ration $m/z$ is an abbreviation. It should not been taken literal.
For all intents and purposes the mass number may be seen as the mass of the ion in $\mathrm{u}$ or $\mathrm{Da}$ without its unit. The charge number represents the "count" of the charges, without specifying what charge it is. In general you can multiply the signal value by the ions charge (and mass unit) to obtain the ions mass.

@user2018675 you should never apologise for changing your accepting vote - it is your choice to pick the answer that helped you the most.


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Are there different types of symmetry?
What factor does the symmetry of a molecule play?
For example, I know water is asymmetric so does this therefore make water polar (I know in fact that is is)? It may sound ignorant due to my lack of knowledge but considering one can draw a line through a water molecule, it appears as if if is symmetrical? Why exactly is water asymmetric and how can one tell? I have heard that it is due to its bent shape and lone pair of electrons but this is too general. Just because a shape is bent does not mean it lacks symmetry?
Finally, is there a concept to finding if a given molecule is symmetrical or asymmetrical?

This is somewhat related. I don't think it's a dupe though.

There are quite a few different kinds of symmetry out there; some forbid polarity, some don't.

As you may already have seen, "symmetry" is not a very specific term, just saying a molecule is "symmetric" doesn't actually tell you anything about the molecule at all. Neither is "asymmetric" a good description of water.

Symmetry is a very broad topic, and as was mentioned in the comments there are many kinds of symmetry. However, when dealing with molecules you should focus on point group symmetry. A good place to start is this link.
But let's get back to the question:
You mention that you know water is asymmetrical and is therefore polar but at the same time you also note that you CAN draw a line through it to cut it into two equal halves. In that way, you can say water has a mirror plane and is therefore symmetric. This type of symmetry is more precisely defined as $C_{2V}$. Ammonia has  $C_{3V}$, Benzene has  $D_{6h}$, and so on and so forth.
Now, symmetry and polarity do have some relation but you can't tie them together, the case in point being the hydrocarbon $\ce{CHMeEtPr}$. From the point of view of symmetry, this is definitely not a symmetric molecule. And despite its complete asymmetry this is definitely what we would call a non-polar molecule.
So are there any properties related to symmetry? Yes - the ability for a molecule to rotate plane polarized light. We measure this and call it optical activity, and call the asymmetric molecules chiral.
Hope that helps!

Thanks for the comment! So when we say water is asymmetric exactly how is it asymmetric, in what context? Orbital electrons? Again as you and i noted previously, it is asymmetric yet a line can be drawn through it so how then in that case is it asymmetric? Thanks for the link as well!

Hi @Atticus283blink, symmetry is described in terms of symmetry elements and few molecules have all symmetry elements. Water, for example, has the C2 symmetry element but not inversion ($\ce{i}$) or improper rotation ($\ce{S_n}$)

Thanks again! Just a follow up; Firstly, can you clarify what you wrote after the word "inversion" and after "rotation?" As you can see the text font is not correct. Furthermore, on that link you sent me it references sigma many times as shown here: https://imgur.com/a/pvz7X What exactly does this symbol mean in chemistry or symmetry for that matter?

Sigma refers to "reflection on a mirror plane". Inversion is followed by a small letter i, and Improper Rotation has the symbol of S_n.

Thanks so much! In my textbook (Chemistry The Central Science Brown)  it does not talk about symmetry and Sigma. Is this concept for more advanced chemistry? I just find it strange it doesnt cover this. I am in first semester inorganic chemistry would this make sense?

@Atticus283blink knowledge of symmetry mainly required in chemistry where geometry plays an important role, and in most cases you can survive through undergraduate university with just passable knowledge, so it is entirely possible that it is a topic reserved for a later semester.

Symmetry is a huge subject. In chemistry we deal mainly with point- groups, which is the symmetry about a point in space as the name suggests. Crystals are described by space-groups and translation is additionally allowed. Have a look at at almost university level physical chemistry textbook for an introduction.
Two main areas that it mainly influences molecules are (a) whether spectroscopic transitions are allowed or forbidden, and (b) whether orbitals can combine to form molecular bonds.
A more familiar property, and one that is easier to explain, is the dipole moment of a molecule. While individual bonds can have a dipole (C-Cl for example), whether the whole molecule has a dipole depends on the overall symmetry.  To be polar the molecule must not have a centre of inversion. This is the only condition and this restricts the point groups that it can belong to.
The common point groups with a centre of inversion are $C_i, D_{2h} ,D_{4h}, D_{6h}, D_{8h}, D_{3d}, D_{5d}, S_6, T_h, O_h, I_h$ & $D_{\infty h} $ and so a molecule with any of these point groups does not have a dipole moment. (Symmetry notation is very condensed so have a look at a text book. If you want to try assigning point groups I found this site molecule-viewer.com which has many molecules to try and 3D pictures of symmetry elements.)
Another familiar example is chiral molecule. This has four different atoms or groups attached to one of its carbon atoms. The molecule should then only have the lowest symmetry (called $C_1$ which indicates 360 degree rotation). A nice example is limonene.


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Effect of an inert gas on steam vapor pressure at equilibrium
Suppose we have a closed container with mobile walls, so it will change its volume in order to equal the external pressure(constant) to the internal pressure.
This container has inside it a liquid phase only of component 1 and a vapor phase of 1 and the inert gas(2) that is insoluble in the liquid phase.
If the external pressure is constant how will change the vapor pressure of the component 1 changing the temperature?  (suppose we know the molar partial enthalpy of the the components in gas phase and of 1 in liquid phase at the required temperature)
$$\left( \frac{\partial P_{vap}}{\partial T} \right)$$
I have tried to find this relation by equaling the chemical potentials, and proceeding in the same manner of the demonstration of the Clausius-Clapeyron equation. However I am not sure of the result. Can anybody help me in the search of this partial derivative?

If the gas phase can be treated as an ideal gas mixture and the Poynting correction for the liquid can be neglected, then the partial pressure of component 1 in the gas phase will be the equilibrium vapor pressure of component 1 at the temperature of the system.  The partial pressure of the non-condensible component 2 will be determined by the ideal gas law, based on the number of moles of this component, the temperature of the system, and the volume of head space.
If the gas phase cannot be treated as an ideal gas mixture but can still be regarded as an ideal solution, then the Lewis Randall law can be used to determine the fugacity of component 1 in the vapor, and this can be matched to that of component 1 for the liquid, with the Poynting correction included.
Does this come anything close to answering what you are asking?


text:
Will SnO2 or SnO form from this reaction?
I have two possible chemical equations for a reaction between $\ce{SnCl2}$ and $\ce{H2O}$. The first one is:
$$\ce{2SnCl2 + 2H2O +O2 -> 2SnO2 + 4HCl}$$
where the reaction draws on oxygen from the air. The second possibility is:
$$\ce{SnCl2 + H2O -> SnO + 2HCl}$$
I know that in the first reaction, tin is oxidized and oxygen is reduced, but I can't figure out if that reaction is favorable, or if it is possible that it's using oxygen from the air. The second equation is the only other possible reaction that I can find, but it doesn't form the product that we want.

SnCl2 does react with O2 from air, but slowly.

The actual reaction of tin (II) chloride with water in the presence of air is given here.
$$\ce{6SnCl2 + O2 + 2H2O → 2SnCl4 + 4Sn(OH)Cl}$$

Tin(II) chloride react with oxygen and water to produce tin(IV) chloride and hydroxotin(II) chloride. Interaction with oxygen and air humidity. This reaction takes place slowly.

As Mithoron noted, the reaction takes place very, very slowly and thus get hydrolysed partially to form the basic salt, $\ce{Sn(OH)Cl}$ instead of forming the hydroxide or the oxide.
The basic salt will also form when tin (II) chloride reacts with water in the absence of air, thanks to Gannex.
$$\ce{SnCl2 + H2O ⇄ Sn(OH)Cl + HCl}$$

Tin(II) chloride react with water to produce hydroxotin(II) chloride
  and hydrogen chloride. This reaction is proceeds by diluting the
  solution tin(II) chloride. (source)

As noted earlier, the reaction is very, very slow and maintains an equilibrium between the reactants and the products.
However when tin (II) chloride is present in a concentrated, acidified solution, it will form a complex.(source)
$$\ce{SnCl2 + H2O → [Sn(H2O)Cl2]}$$

Tin(II) chloride react with water to produce dichloridoaquatin.
  Tin(II) chloride - concentrated solution. pH < 7.

Your tin will not spontaneously change oxidation state just from adding water. On dilution, some of the tin (II) chloride may be hydrolyzed, precipitating the basic salt:
$\ce{SnCl2 + H2O -> Sn(OH)Cl + HCl}$

We did initially mix the SnCl2 with water and the HCl...I don't know if that makes a difference.


text:
Unknown glassware
I recently found a piece of older glassware in our lab and nobody seems to know what it is. It appears to be some sort of condenser; however, it does not have both the in and out little notches on it.
It is open on both ends and has one smaller opening coming off the side toward the top. The inside of the tube is similar to the look of a condenser, but not exactly.

Is that the device from which Mr. Clinton never inhaled?

Oh, beautiful to be young :)
This is a vacuum pump. You plug the top part into a water tap. There's a slit in the middle of the inner tube of this "condenser", it creates suction = vacuum when you pump water through it. The side arm is for the receiver of said vacuum. What a lovely relic!

These things still exist, though they are not commonly made of glass. Metal and plastic are more common. They are also called water aspirators.


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What is the major product of the reaction of 2-ethoxypentane with hydrogen iodide?
What is the major product formed by heating 2-ethoxypentane with $\ce{HI}$?

The options are:

According to me the correct answer should be option-1 because 2-methylbutyl carbocation is more stable than ethyl carbocation as there are more +I groups attached in 2-methylbutyl carbocation.
But according to my book the second option is correct.
Is my answer correct?
If not please explain

Strictly speaking, none of those answers are correct, provided your stem is right.
The reaction will start with protonation of the ether to give you the following species:

And the iodide nucleophile can attack either side via $S_{N}2$ mechanism. Since the left side is more accessible, it'll probably attack there first giving you:

Then, since we're still in the presence of HI and it is a strong acid, it protonates the alcohol and converts it into a corresponding alkyl iodide via either $S_N2$ like in the example above or $S_N1$:

A general rule of thumb here for the future reference is that in a hydrolysis of an ether with a strong hydrohalogen acid (HI, HBr), you get a mix of two corresponding alkyl halides with halogen where the oxygen used to be.

can you please explain how the left side is more accessible ? Is it because of less steric hindrance?

precisely so. Remember, that things constantly move, groups rotate around single bonds, so on the left you have a relatively small ethyl, while on the right you have a huge propeller of a group that will knock off anything that comes (or attempts to do so) close to attack that secondary carbon with the oxygen.


text:
Why do we need incubation time during the glucose test by chronoamperometry?
Also, is there any difference between with and without incubation time in the glucose test? For example, the voltage results acquired will be different?

There seems to be some confusion . In chronoamperometry, you measure voltage and not currrent. Also, there are so many different glucose sensing procedures. Could you summarize the particular one you are referring to?

Well, Although I'm not a chemical student, I think I could share my understand of the Chronoamperometry by the parameters that could affect the result in my project.
According to my test result, the incubation time will affect the magnitude of the current response at the chemical cell(also for the trigger time). This is related to the property of the glucose test strip(Truetrack in my design) and the circuit design. And I assume that this is because of the distribution of the glucose in the test strip, specially the amount of the glucose close to the surface of the electrode.
I hope if there is a chemical guy that can explain the detailed principle of Chronoamperometry, very detailed stuff. Because I have finished my project and thus the theoretical staff used for the lecture is not enough to attract my attention.
Also, if someone has interest in this, I can share my experiment result and discuss with him personally.
Here is my email: matthew97829782@gmail.com


text:
why distillation of mercury is done at low pressure?
While reading about extraction of mercury from it's cinnabar ore (HgS), it's purification required distillation due remove impurities like Au, Ag, etc. This distillation was done at low pressure. Why is it done at low pressure?

Welcome to Chem.SE!  I have a feeling your question would attract more attention if could link to the original source and provide more details.  Hope you enjoy the site.

There is one obvious reason and one possible confusing factor.
The production of mercury from cinnabar involves heating cinnabar in a stream of air which causes oxidation of the sulphur and produces a stream of mercury vapour which can be condensed to give the metal. This isn't distillation even though it involves condensation of mercury vapour, but a chemical reaction.
But mercury can be purified by distillation because it is very volatile for a metal. But the boiling point at normal pressure is ~360 °C which is inconveniently hot. Using reduced pressure (say around 1/1000th of an atmosphere) can make this practical at closer to 100 °C which is far more convenient.


text:
Why is MnO2 not a peroxide?
As I know, $\ce{MnO2}$ is called manganese(IV) oxide.
Why can't we name it manganese peroxide since manganese(II) exists too?
In other words, how do we know that the oxidation number of manganese in $\ce{MnO2}$ is $4$ and not $2$?

It is not called manganese peroxide because it does not contain the peroxide bond(-O-O-). Compare its structure with hydrogen peroxide.

Peroxides contain two oxygens connected by a single bond. X-ray or neutron diffraction will show that the oxygens in $\ce{MnO2}$ are too far apart to be bonded, and therefore it is not a peroxide.
Peroxide is well known as a ligand to transition metals (though whether it is better described as peroxide or superoxide is sometimes not always clear, e.g. in some cobalt species). An example which you might come across in the lab is chromium(VI) oxide peroxide, $\ce{CrO5}$.

The question’s logical premise is skewed. $\ce{CO2}$ is carbon(IV) oxide or carbon dioxide. But carbon(II) also exists and so does carbon(II) oxide (carbon monoxide, $\ce{CO}$). Just because there are twice as many oxygens in a compound does not mean that it can be a peroxide with a lower oxidation state of the central atom.
In fact, peroxides are decidedly rare. There is $\ce{H2O2}$, $\ce{Na2O2}$, $\ce{[Cr(O)(O2)2]}$ (the chromium butterfly), $\ce{mCPBA}$, $\ce{tBuOOX}$, $\ce{(PhCOO)2}$, artemisinin and I’m already at a loss. Compare that to the vast number of oxides out there and you get what I’m getting at. This is because the $\ce{O-O}$ bond in peroxides is very energetic, and they readily oxidise or reduce other compounds for oxygen to achieve its more stable oxidation states $\mathrm{-II}$ or $\pm 0$ (the former typically more common and stable than the latter).
Thus, peroxides can only form if electrons are there to reduce atmospheric oxygen, but not enough to reduce it to an oxide (or water or an alcohol group etc.). And especially in the case of metals with many available oxidation states, it is very unlikely that a peroxide should form for a given one — especially if the metal is not fully oxidised (remember that the highest common oxidation state of manganese is $\mathrm{+VII}$). If a peroxide accidentally did form, it will typically be a good oxidising or reducing agent. But $\ce{MnO2}$ is neither, it is the thermodynamic hole of ionic manganese compounds. Glow a manganese-containing salt in the flame of a bunsen burner, and given enough time you will arrive at $\ce{MnO2}$. These conditions of formation show us that it is very, very, very, very unlikely for $\ce{MnO2}$ to be a peroxide. And it also hardly reacts to any other compounds, further showing that it is most likely not one.

One group of peroxides you miss in your list: MgO2, CaO2, SrO2, BaO2. There is even evidence that the alkaline earth metals listed here form superoxides and ozonides as well, though these are uncimmon.

I wasn’t trying to make a complete list of peroxides, only of peroxides I know. Hence I’m at a loss after them.

The reason is oxygen bonding in the molecules. Actually peroxide means oxygen - oxygen should be in single bond but in MnO2 oxygens are double bonded with mn there is no bond between oxygens. For example in case of H2O2 it has oxygen - oxygen single bond so it is named as hydrogen peroxide but in MnO2, both the oxygens are bonded with Mn there is no bond between oxygens.

$\ce{MnO2}$ is an ionic lattice with several different polymorphs. Describing the oxygens as being double bonded to the manganese is not really true.

I would say oxygen bonded only to manganese, leaving behingld whether it's bonded iinocally, covalently, etc.

Peroxides are compounds in which the oxidation state of Oxygen atoms is -1.
Since you know that MnO2 is Mn(IV) oxide, which means Oxygen atoms are in -2 oxidation state. Hence it is not called a peroxide.

Circular logic...

Circular reasoning works because circular reasoning works and because circular reasoning works! (CC @ortho ;))

Don't answer when you don't know clearly about it but please don't apply circular logic just for the sake of answering something.

not so circular, as we have methods to determine oxidation of Mn by spectroscopy

@MolbOrg presumably the spectrscopy identifies Mn(IV) by comparison with standards previously known to have that oxidation state, like maybe MnO2 ... .


text:
Questions about Genetic algorithm paper of Gilman and Ross
I want to reproduce an (old) biochemistry paper of Gilman and Ross, i.e. " Genetic algorithm selecetion of a regulatory structure that directs flux in a simple metabolic model." ( The following link leeds to the paper: http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1236362/) I know this is an old paper, but maybe someone here remembers it. I really need the answers of the following questions because the enzymatic model explained in the paper is very important for my master thesis.
First: The author defines a system of two ODEs but the initial conditions of these ODEs are not printed in the paper. The author just writes: "Initial conditions for the integration are determined by letting the network relax to steady state. " (see p. 1324 on the lower right) . But how is this meant? In this paper the author wants to optimize the mentioned ODEs , so how can I find initial conditions, when there are not all parameters of the ODEs given?
Second: On the top of page 1324 are two needstate functions defined, called $\xi_T$ and $\xi_F$. When I plot $\xi(F)$, the plot looks completely different from the one depicted in figure 3b in the paper. So the function defined in figure 3b) must be wrong. Did anyone realize that error and has a workaround for that? Of course I could define my own functions but I want to stay as close to the model in the paper as possible.
Thank you very much for your help.

Have you looked at the caption of figure 1? I believe all the constants you need for your model are stated there. Also the steady-state is $\dot A=\dot B = 0$, where $F$ and $T$ are externally set constants (the reservoir).

Hi @Deathbreath , thank you for your comment. Yes I have looked at the caption. The problem is, that there still occur eight unknown parameters in the ODE system.  Those parameters appear in the functions $\nu_{\alpha}$ and  $\nu_{\beta}$ . But now I think in steady state it holds:  $\nu_{\alpha}   = \nu_{\beta}$ and, as you wrote $dA/dt = dB/dt = 0$. So $\nu_{\alpha}$ and  $\nu_{\beta}$ cancel out each other. I think this might solve the problem.


text:
Finding the mole fraction

At $25\ \mathrm{^\circ C}$ the vapor in equilibrium with a solution containing carbon disulfide and acetonitrile has a total pressure of $263\ \mathrm{torr}$ and is $85.5\ \mathrm{mol}\text{-}\%$ carbon disulfide. What is the mole fraction of carbon disulfide in the solution? At $25\ \mathrm{^\circ C}$ the vapor pressure of carbon disulfide is $375\ \mathrm{torr}$. Assume that the solution and the vapor exhibit ideal behavior.

I was gonna use Raoult's law but like how is the molar percentage different from the molar fraction, isnt it the molar fraction times 100? also I checked the solution manual and it had the molar fraction sign but with an L or V above it, and theres no such thing in the boook, can anyone help or explain what is that?

As you said, the molar percentage is just the molar fraction multiplied by 100. As to the L and V above the molar fractions, they refer to the molar fraction of the component in the liquid and vapor phase, respectively. They are needed since (in general) the molar fraction of a component in a mixture are different in the liquid and vapor phase.
So, from the data in your problem, you know that the molar fraction of carbon disulfide in the gas phase is 0.855, and let's call it $x_{CS_2}^V$. The total pressure is P=263 torr, and since the vapor exhibits ideal behavior, the partial pressure of CS2 is: $P_{CS_2} =x_{CS_2}^V P$. Using Raoult's law (since the liquid exhibits ideal behavior), this partial pressure must be equal to the molar fraction of CS2 in the liquid times the vapor pressure of pure CS2: $P_{CS_2} =x_{CS_2}^L P_{CS_2}^o$. And I am sure you can follow from here.

oh thanks , tho my textbook has the rule as Psoultion = Xsolvent . Psolvent , so I didnt know I could do all of this with the phases hah!

You are welcome. Which textbook are you using?

chemical principles zumdahl 7th edition