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Figuring out if a surfactant is present
I have recently starting distilling my own essential oils using a still that I built out of an old diesel barrel. Before the first run with the still, I cleaned the inside using soap and water. I distilled until I created 5 gallons of hydrosol. The issue is that it seems that the oil and water are emulsified (hope that is the right term, I don't have a background in chemistry). There appears to be a thin film of oil on the surface, rather than the "globs" I would expect. Another reason I believe this is an emulsion is that no matter how much I skim off the top, that film is always present, making me think that the oil is thoroughly mixed in the solution.I have done some research and it seems to be that there may be a surfactant present that is causing this.
Is there a way that I can test for the presence of a surfactant?
Does it sound like my issue is the presence of a surfactant or something entirely different?
Could the soap I used to clean it be the surfactant?
Thanks!
Update in Response to Answer:
The end goal is to use a few drops of the oil to enhance the smell of a facial product (beard oil), so the end product has to be safe for skin.
I don't have a diagram at the moment, but here is a smaller still that is basically the same as my setup, just much smaller. I am using a propane heat source to control the temperature.
To be honest, I don't know the expected structure of the oils. This is my first attempt at anything like this (I'm a software engineer that recently acquired an interest in chemistry!).
Thanks for the help!

These types of problems can be tough to solve. Please provide additional details as requested if you can so we can better help you!

What is your starting material? Are you sure that isn't already an emulsion?

The starting material is water and leaves and branches of "Artemisia tridentata", more commonly known as sagebrush. There isn't too much literature that I have been able to find on obtaining essential oil from this plant but I have found a few papers where it was done successfully

The original mixture was approximately five pounds of plant material and 10 gallons of water. I ran the distillation process until 5 gallons came out of the distiller. It just occurred to me, is it possible that all of the oil came out early in the process and the last four or so gallons were just water, which diluted it?

I see, since you are condensing mainly water vapors, (and probably your boiler temperature is around 100 C), your main condensate is water and watery oils (in unscientific terms)!! No wonder the resulting oil is kind of miscible with water. Assuming the oils have higher boiling point than water, my suggestion is to minimize the water and push the reboiler temperature above 100. Test is once, be careful not to burn your feed though.

So if I understand correctly, it would be better to keep the plant material amount constant and reduce the water down to something much smaller (maybe a gallon?) and use higher heat? Also, can you clarify what you mean when you say "burn your feed"? Thanks very much for all of your help.

Yes that is what I meant. Feed is your starting material. In the field of chemical process, the input to a system is sometimes referred to as the "feed"! So be careful not burning your plant material when heating it above 100 C. If you can do some research on the essential oils' boiling point, it would be very helpful to know how much you should heat it up. You could even experiment with starting without any water. Though, it might be hard to control the heat from burning the "feed".

There are multiple issues that need to be addressed before one can give a definite answer to this question. Let's start by what we are certain about:
1- The soap is not the issue (unless you haven't rinsed the barrel thoroughly with water). The remnant soap would be so dilute that should not cause any problem.
2- Since the film keeps forming there seems to be an equilibrium issue for your oil to separate. If you look into partition coefficients of the oils you should be able to estimate how much will remain in the water. Anyway, the less water you have, the more oil you can collect. (concentration factor pushes the oil out to keep the equilibrium) One idea is to add less water to your feed.
3- THEORETICALLY one way to enhance oil extraction is to use a nonpolar solvent. HOWEVER, I STRONGLY ADVISE AGAINST ACTUALLY TRYING IT AT HOME due to the flammable nature of most nonpolar solvents and health issues related to it. If you can find a nonpolar benign solvent (that is a big IF), mix it with your final water oil mixture, and then decant the water from the bottom and then distill the solvent out to get the oil as a residue.
4- Depending on the chemical structure of the oil, some of it will always remain in water. If they are not highly hydrophobic, it might be difficult to separate them from water anyway without using an assistant solvent.


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Why is the relative atomic mass numerically equal to the gram atomic mass of the Atom?
Why is it that the relative atomic mass of an atom is numerically equal to gram atomic mass?
For example,Relative Atomic Mass of  Oxygen= 16 u
So,Gram atomic mass of Oxygen=16 g.

It's called molar mass...

see also http://chemistry.stackexchange.com/questions/2784/why-is-the-relative-atomic-mass-of-carbon-not-exactly-12

@Mithoron your link is unrelated

The relative atomic mass of an atom is its mass relative to an atom of carbon-12 (which is 12 by definition). A mole of carbon 12 atoms weighs 12g (which is where the definition comes from).
"Gram atomic mass" is just an alternative way for saying the mass of a mole of some atom. So the ratio of the atomic masses will always be the same regardless of how many of the atoms you are counting. Hence, the two ideas are numerically the same.

I think that there is a flaw in your answer. Your Justification was only TRUE for C12 atom as you never talked or related other atoms.How would you justify it for all the other atoms?

@user42209 Because other atoms atomic masses are defined relative to carbon 12 which is the standard. This makes what I said true (at least to a good approximation).

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Antiprion effectiveness of phenol
I am investigating ways to reliably decontaminate instrumentation that has come into contact with mammalian prions. Chaotropic salts and extreme pH are good at inactivating prions by disaggregating the assemblies of PrP.
Also phenol is among the most effective "prionicidal" agents. I presume that this is because it flips hydrophobic side chains of proteins towards the solvent, thereby denaturing/disaggregating PrPSc. My questions are:

is this true?
if so, might other phenolic scaffolds/derivatives be even more effective for prion disinfection?
and if so, which scaffolds might be most promising?

It was proposed by a flagger that this would be more appropriate on Biology.SE.  I'm inclined to think that it straddles the two sites' scopes, so I'd like to leave it here for now, but I'd like to hear from some of our star chemists.

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Why there were changes in molar mass of lithium?
I am curious why there were changes in molar mass of lithium in the past.
The values were like 9 to 10 before 1830. But suddenly after 1830, the range of molar masses seems similar to the molar mass that is used now.
I thought there might be some technical issues in the past. But, from 1830, the values were consistent and stable. So, I cannot assume there were lack of development in equipments and technics.
So, my questions are:
1. why the values were around 9-10 before 1830, not around 6.941?
2. why the values suddenly became similar to the molar mass which is used now?

https://en.wikipedia.org/wiki/Lithium#History

This doesn't necessarily completely answer the question, but it at least moves closer to the root of the problem. From the Wikipedia article that MaxW included in his comment:

Brande also described some pure salts of lithium, such as the chloride, and, estimating that lithia (lithium oxide) contained about 55% metal, estimated the atomic weight of lithium to be around 9.8 g/mol (modern value ~6.94 g/mol)

So it seems to stem from an erroneous estimation of the percent composition of lithium oxide. If we go to the source (Brande's manual of chemistry:https://books.google.com/books/reader?id=zkIAAAAAYAAJ&printsec=frontcover&output=reader&pg=GBS.PA190), we find that this error likely stems from the fact that lithium oxide was difficult to obtain and so very limited experimentation had been done to characterize it.

According to http://www.rsc.org/periodic-table/element/3/lithium

In 1821 William Brande obtained a tiny amount this way but not enough on which to make measurements. It was not until 1855 that the German chemist Robert Bunsen and the British chemist Augustus Matthiessen obtained it in bulk by the electrolysis of molten lithium chloride.

So I think that it could just be related to the small isolated amounts.


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How to change the chirality of amino acids?
Is it possible to change the chirality of the amino acids? I understand such a process is called "stereoinversion".
Take a levorotatory amino acid and turn it into it's dextrorotatory counterpart (enantiomer, optical isomer).
Anyone tried to do that? Can a high temperature change the chirality of an amino acid for example?
Racemization is good enough for me. Changing the chirality of just a small amount of the total amino acids in the sample is good enough.

I think the answer to this question depends heavily on whether you looking for a stereospecific mechanism or just racemization.

A cursory search of chemexper reveals that enantiomers of at least some of the 20 main amino acids are commercially available.

Temperature is likely not going to work. To racemise something that has a chiral center, you need to break a bond: if you want to supply that much energy by heating, you are going to fry your molecule.
The easiest things to do are usually extraction of an acidic proton with a strong base, formation of a double bond involving the formerly chiral carbon, and then you go back to the initial form, but obtaining a racemate because the attack on the two sides of the double bond is equally likely.

If the base racemization methodology exists, it's probably documented in the literature.

I'm pretty sure it does.


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The Sørensen pH scale and strong acids diltution from pH 1 to pH 4 by 1000 times?
I am reading a chemistry book that makes the following point about Sørensen's pH discovery regarding [H+] being in negative powers of 10:

To dilute a solution from a pH 1 to a pH 4 (just 3 pH units) would require dilution by 10 x 10 x 10 = 1000 times.

I don't understand how this calculation is derived, nor is there any proof available in the book.
Can someone explain how this is the case?

let $c_1$ be concentration 1, $V_1$ be Volume 1 and so on then $c_1= 10^{-pH}$ and $c_1\times V_1 = c_2\times V_2 $

There is nothing even remotely resembling a discovery in this. The power of logarithms was known for centuries by then.

The confusion here probably comes from your statement that Sørensen discovered the pH scale for expressing [$\ce{H+}$]. The pH scale is something that he created or defined for convenience. The definition of pH is basically:
$$\mathrm{pH} = -\log[\ce{H^+}]$$
Since this is a (negative) log scale, a pH of 4 has 10 times the $[\ce{H+}]$ as does a pH of 5, and so on. In other words each pH unit represents a factor of 10 change in $[\ce{H+}]$.
The purpose behind such a scale is simply to avoid having to write out a bunch of zeros, and having to read a bunch of zeros!


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Trimethylsilyl iodide - Alkyne Reaction
I have synthesized the molecule below possessing a doubly, boc protected heterocyclic amine and a terminal alkyne.
Upon attempting a TFA/$\ce{CH2Cl2}$ Boc deprotection I believe my alkyne is being hydrolysed.
1.) I wish to use TMSI to perform the boc deprotection but cannot find information on how it acts in the presence of alkynes. Would TMSI iodinate, silylate or otherwise interact with the alkyne?
2.) Is there any way to make use of TFA/$\ce{CH2Cl2}$ and prevent the alkyne hydrolysis?

Can you use HCl/Dioxane?

@Waylander why do you think this would help? Also, I imagine the hydrolysis is because of the presence of water? Is this theory correct?

HCl/Dioxane is a method we use extensively for removing BOC groups. Vac down hard to remove xs HCl before adding water should prevent any hydrolysis, indeed you may not need aq workup. How are you working up the TFA deprotection?

Have you tried just heating it? One of my colleagues had really good luck with simple thermolysis. I would think that the first Boc will fall off easily at least.

The first BOC should come off with nucleophiles eg MeO-. What concentration of TFA are you using?

@Waylander Workup was a simple NaHCO3 wash whilst extracting with EtOAc. I've used HCl/Dioxane before, but due to time constraints and expense this may not be currently possible. What's your take on the TMSI? Looking at page 5 of the following paper's supporting info., it seems doable. link

TMSiI looks a good option. Seems unlikely to attack the alkyne. With regard to the TFA reaction it might be worth vaccing down hard to remove all the TFA before exposure to aq.

Seems unlikely that you're attacking the alkyne directly. You might be getting break down somewhere else...

@Zhe without divulging the sturcture of the heteroaromatic, I can tell you that I've made a range of molecules possessing the same heteroaromatic core and all of them are TFA stable. The only difference here is the alkyne which none of the other molecules had.

I could you could form the ketone from the alkyne though that still seems kind of unlikely. Have consulted Greene/Wuts? That's kind of the goto reference for protecting groups.

TMSI was used successfully without damaging the alkyne as per this protocol.
http://pubs.acs.org/doi/abs/10.1021/cb400261h


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Wavenumbers of cyclic ketones
Rank the compounds below in order of increasing C=O absorption frequency in IR spectrum (top - lowest wavenumber, bottom - highest wavenumber):

Question: Would I count compound #1's cyclopentane or all 11 carbons because wavenumber increases with 5C or smaller rings.
If I count #1 as a cyclopentane for this purpose then #3 has the lowest wavenumber and #2 has the highest.

related: Ring Strain and C=O Stretching Frequency

Here is nice comparison of all 3 ring sizes. Octahydro-1H-inden-1-one (either cis- or trans-) carbonyl strechings fall between cyclohexanone and cyclobutanone, thus, fused ring is not a factor (it is to some extend because of sunstitution).

#1  Octahydro-1H-inden-1-one C=O stretch from reference
cis 1730 $\text{cm}^{-1}$

trans 1754 $\text{cm}^{-1}$

#2 - Cyclobutanone NIST Listing C=O stretch at 1790 $\text{cm}^{-1}$ from spectra
#3 - Cyclohexanone NIST Listing C=O stretch at 1740 $\text{cm}^{-1}$ from spectra

#1 doesn't have stereochemistry here @MaxW

interesting, but unhelpful and unpedagogic.


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How does a volcano for a science fair work?
In this video it shows me how to make a Volcano but what is the science behind it? http://www.youtube.com/watch?v=9b_gltKtERY

If somebody wants to answer this, I'll spare you the "trouble" of watching the video - it's a baking soda + vinegar volcano.

This experiment is not to be confused with the ammonium dichromate vulcano experiment.

The food color-dyed water is just there to add volume and color.
The household chemicals used are acetic acid (vinegar, $\ce{CH3COOH}$) and sodium bicarbonate (baking soda, $\ce{NaHCO3}$).
The acetic acid and sodium bicarbonate react to form carbon dioxide gas ($\ce{CO2}$), dissolved sodium acetate ($\ce{CH3COONa}$) and water ($\ce{H2O}$).  It is the liberation of $\ce{CO2}$ gas that causes the foaming and bubbling.
The chemical reaction is written as:
$$\ce{CH3COOH(l) + NaHCO3(s) -> CH3COO-(aq) + Na+(aq) + CO2(g) + H2O(l)}$$
Where:
l = liquid
s = solid
aq = aqueous (dissolved)
g = gas (vapor)


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What is the driving force behind sigmatropic reactions?
I am studying sigmatropic reactions in context of pericyclic reactions and I am confused about the overall purpose of this type of reactions. I know that sometimes in sigmatropic reactions, more substituted alkene in the product in one reason why the reactions take place. However, in other types of sigmatropic reactions, like hydride shift, or when more substituted alkene product is not produced, I am having trouble seeing why this reactions occurs in the first place. Its almost like the electrons and double bonds are simple moving around to new locations, without any purpose.

@Mithoron I understand your point, but the way it's expressed comes across as a bit abrupt.

OK lets rephrase it. Asking about purpose of reaction itself doesn't make sense; it's like asking about purpose of a rock rolling down the hill. Asking about driving force is OK - in case of rock it's gravity. In reactions instead of gravitational potential there's chemical one.

Most of these reactions are indeed equilibrium reactions, meaning that they go forwards and backwards all the time and at a given point in time there will be a certain ratio of reactants to products corresponding to the equilibrium constant. Which side is favoured is often determined by minute stabilisations that are indeed very hard to rationalise.
Take for example the set of Wagner-Meerwein rearrangements that occur in the synthesis of cholesterol after squalene was cyclised. There appears to be no true stabilisation which would explain why one side is favoured over the other. In the end all we know is that the product is more stable or the reaction would not occur.


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Expedient phosponium synthesis (for Wittig)
I have been making ethyltriphenylphosphonium bromide (somewhat of a time crunch, would have used iodoethane instead of bromoethane if we didn't have to wait for it). All the syntheses I see of Wittig salts are in aromatic solvents and they take large amounts of time for what I would have expected to be a fairly quick reaction, usually at least a day, if not several. I have come across the use of iodide salts as a catalyst, some of which I had already added because I figured it wouldn't hurt.
$$\ce{PPh3 + EtBr ->[\text{aromatic}][\text{solvent}] [EtPPh3]Br}$$
Why are these conditions standard/most effective? More specifically, why is toluene (less often xylene, probably others but those are the two I've seen) used rather than some polar aprotic solvent? Also, besides increased temperature and the use of iodide catalysts, are there any common ways to decrease reaction time?

Is it simply that toluene boils at a relatively high temperature? Maybe the SN2 reaction is fairly easy once line up, but due to the large PPh3 being a bulky nucleophile, having the two reactive sites in close proximity is relatively rare, so increasing temperature is the most effect thing since random interaction is the driving force.
I can't imagine a primary alkylbromide reacts through SN1 rather than SN2. I guess maybe the higher temperature could allow for an SN1 reaction, though? I kinda doubt it, though...
I would think, even if the above was true, using THF and going through an SN2 mechanism at a lower temperature would be simpler than refluxing in toluene, and I still kind of doubt the SN1 on a primary bromide even at the elevated temperature.
Another thought is that the phosphonium salt crashes out of toluene, and therefore this equilibrium serves as a driving force for the reaction. Technically, this doesn't speed up the reaction, but it does significantly hinder the reverse reaction, so it leads to a net decrease in reaction time (and increases yield, I guess, if the reverse reaction is significant). I wouldn't really think the halide anions would be strong enough nucleophiles to be able to kick off a phosphine, though.

$$\ce{RX + PPh3}\xrightarrow{\text{toluene}}\ce{\underset{\small{(\sim none\ in\ sol'n)}}{[RPPh3]X_{$(s)$}} v}$$
$$\text{vs.}$$
$$\ce{RX + PPh3} \stackrel{\small{\text{THF}}}{\longrightleftharpoons}
\ce{\underset{\small{(more\ than\ above)}}{[RPPh3]X_{$(org)$}}}$$
I realize THF isn't all that polar, but relative to toluene or xylene, I would really expect it to facilitate this reaction better.

P.S. This question isn't really an attempt to get help with my current situation (especially since I just finished working up my last batch, anyway) as much as it is just trying to get general information about the reaction.
I will mention for anyone who might be able to use it that I first tried to run mine in toluene at around 35 °C with catalytic KI (I've now seen LiI used instead of KI, not sure how much that matters). However, I was inspired by a literature report of a 7-day reaction at 30 °C (no iodide) to move the reaction in portions to pressure tubes and take it up to 130 °C. A paper reported 7 h (also no iodide) at this temp was effective, and I found that to be overkill. I didn't really see much difference between 2 h and 7 h (again, no iodide in their reaction, which may have caused the difference).

Depending on the substrate, and the wanted $E/Z$-configuration of the alkene formed, an Arbuzov reaction (like 1.1 eq of triethylphosphite per 1 eq of alkylhalogenide; the former often deployed both reagent and solvent), refluxed for 1-3 h; telescoped / followed by a HWE with NaOEt could have been faster, easier to workup.

@Buttonwood I'm having a hard time parsing your description there, but it sounds like it's definitely worth exploring. Ill have to draw it out, I think... I get everything up to how the NaOEt part works. Wouldn't I end up with the $\alpha,\beta$-unsaturated ethyl ester back from that? I also considered using diethyl ethylphosphonate as the ethylenation reagent.

I was extrapolating from work done earlier, see in http://chat.stackexchange.com/rooms/55185/comment-on-expedient-phosponium-synthesis-for-wittig


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Do acids really donate a proton?
Will acids really donate a proton? In an atom, the proton is inside the nucleus and I don't think a proton will go all the way out of atom dodging all electrons and enter into the nucleus of the base.
It seem impossible but the Brønsted–Lowry theory says acids donate protons.
I am confused—could someone explain what is actually meant by acids donating protons?

That proton is not feom the nucleus, but it is a dissociavle hydrogen ion, which is a proton

related http://chemistry.stackexchange.com/questions/37053/what-is-difference-between-h-and-proton

Yes, according to the Arrhenius theory, acids dissociate in aqueous solution and release a proton ($\ce{H+}$). The Brønsted–Lowry defines acids  ($\ce{HA}$) and bases ($\ce{B}$) in such a way that their interaction is characterized by the exchange of a proton according to $\ce{HA + B <=> A- + HB}$.
However, this is about the reaction of molecules in which real hydrogen atoms are bound to other atoms.
To illustrate the idea, let's think about the reaction between acetic acid, $\ce{CH3COOH}$, and benzylamine, $\ce{C6H5CH2NH2}$. If we combine both liquids in our imagination, the following will happen:
$\ce{CH3COOH + C6H5CH2NH2 -> CH3COO- + C6H5CH2NH3+}$
The acid donates a proton to the amine, which serves as the base. In the course of this reaction, the acid is converted to its corresponding base, while the base is converted to its corrsponding acid.

You are confusing dissociation and proton exchange with a type of radiocative decay known as proton emission.

$\ce{H+}$ is a proton since the nucleus of most hydrogen atoms is just a proton. With its electron gone, the hydrogen cation is a proton.

@BenNorris Though it might be a proton plus a neutron or two, in the case that hydrogen atom is one of the other two isotopes. ;-)

It donates a Positive Hydrogen Ion. It's basically a Proton but i find this formulation to be less confusing when trying to understand it for the first time.

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How can I solve a non-cubic crystal structure?
Say I want to solve the crystal structure of barium sulfate, and I want to not only determine crystal type and the lattice parameter, but also the orientation of the oxygen atoms on the sulfate anion. How do I proceed? All I have is the powder X-ray diffractogram (i.e. intensity vs $2\theta$ for the compound).
I don't want to use any existing software as this is more about learning how crystal structures are solved. I have looked at books and they seem to be incredibly difficult to follow.

This is rather a big question, the only way is to work through some examples, there are several in McKie & McKie ' Essentials of Crystallography' (publ Blackwell Scientific), chapter 7 is all about power photographs and their analysis.

Thank you. I will go through the book suggested and get back on this comment thread.

This is ...probably possible - in some cases. I sat 2 years interpreting XRD data with EVA and Rietveld curve-fitting software (TOPAS) - and I am certain I would not be able to do so. You would have to be very experienced, because interpretation of a crystallogram, if you do not know the rotation nor lattice parameters at all, is tricky. You would have to be able to - in your mind - picture the different planes and their diffraction peaks - as you rotate the crystal - and recognize them in the diagram. If anyone could do that, and were female, I would very politely ask her to have my babies.

The lattice constants and the symmetry of the lattice determine the position of the diffraction peaks observed along the $2\theta$-scale.  The intensity of the diffracton peaks, leaving out increase by accidental or systematic superposition of them, is due to the arrangement within the asymmetric unit, i.e. the motif which -- after application of the symmetry operators (centres of inversion, proper axes, rotoinversion axes, glide planes, mirror planes) -- builds the unit cell.
So you face the phase problem (as single crystal diffraction analysis does, too). If you do not want to rely on already existing software (like DASH, MAUD, Jana, or FOX to mention a few), you will look up the Rietveld refinement.  Beside by the comment by @porphyrin, as shown in this lecture briefly providing some background, the numerous properties of sample and experiment (including their statistics) equally lead me to the recommendation to use software already proven to work.  The International Union of Crystallography maintains a software directory, along with -- under their educational umbrella -- the notes of a dedicated workshop about powder diffraction.
Perhaps contacting a local crystallographer may be helpful, too; like via the American Crystallographic Association ACA.

Would it not be easier to calculate the patterson function and then just brute force the phases?

I found a rather good reference on the internet that answers my question in some detail, without going into the jargon. The section on diffraction theory and indexing methods is especially good.
http://pd.chem.ucl.ac.uk/pdnn/pdindex.htm#unit2

An interesting site; and maybe more suitable for your specific needs than more general introductions into the field than the detailed IUCr primer by Trueblood.


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'Rate' in the rate law
I am slightly confused about what the rate is that is given by the rate law. Is it the rate of reaction (I.e. the rate of loss of reactant or formation of product divided by its stoichiometric coefficient?)
I suppose it would not make a difference to the rate equation itself if we simply used the rate of loss of reactant or formation of product without the factor of the stoichiometric coefficient, however then our rate constant would be multiplied by this factor. In that case we would have to specify what rate we are referring to when we are giving the rate constant.
My confusion stems from the fact that I have never seen it written that the rate constant is specifically determined for a given reaction, but I suppose it does not matter because rate laws are given as
$\frac {d [A]}{dt}='law'$
As opposed to just
$r='law'$
So when we actually write scientific papers we automatically specify in the expression of the equation what rate we are referring to. And then if we wished to transfer this rate law to give the change in concentration of another reactant or product we would use the stoichiometric relations and it would indeed change the value of the rate constant for a given rate law...
I apologise if I have not expressed myself clearly. I am starting out learning about this topic so have just looked at rate laws in theory and have not come across them in papers or seen how they are given in practice. Would appreciate if someone could verify if my thinking is correct or not?

If you take a look in the IUPAC Gold Book, the rate of reaction is defined unambiguously. The idea is that you have to divide through by the stoichiometric coefficient. I also wrote a bit about it some time ago http://chemistry.stackexchange.com/a/38168/16683

Rate of reaction is related via stoichiometry to the rate of consumption of a reactant(r reactant)  and the rate of formation of a product( r product), the rate of reaction-unique rate-( r r) can be determine by following the concentration of either a reactant or product as a function of time. The change in concentration with time is directly proportional to the rate of reaction, where the proportionality constant in the reciprocal of the stoiochiometric coefficient of the species in the balanced chemical reaction. So  rr= r  reactant or  r  product divided   by the stoichiometric coefficient.
So The mentioned  equation for the rate(rr ) may be confusing since the stoichiometric coefficient value may vary if we write the balanced equation in other way.Thus the  values of rr  and and specific rate constant – unique rate constant  K(r) may vary wih respect of the version of balanced equation.  But the individual values of r product or r  reactant and the rate constant of product or the rate constant of reactant   remain same irrespective of the version of balanced equation , this is actually what we are measuring experimentally as rate. However, in this way, the rate with respect to different species and  the rate constant of different species(K (reactant), K(product) )  will have different values.
There are in fact an infinite number of possible rate equations and an infinite number of corresponding rate constants. The value of the "rate constant" k depends on how you define the rate.


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Why don't salt crystals attract each other?
If salt is made of ions in a giant lattice, surely if I have two crystals they would attract each other and fuse? Or maybe the ions can't get close enough to each other to attract, or the surface is made of a different, neutral material?

see http://chemistry.stackexchange.com/questions/61292/if-you-put-two-blocks-of-an-element-together-why-dont-they-bond

Sodium chloride ("salt") is made of a giant lattice of ions: Na+ and Cl-.
The crystals overall do not have any net charge as the the number of chloride ions is equal to the number of sodium ions. So, overall, there will be no electrostatic force between crystals as they are neutral in the bulk compound.
On the surface, crudely and without getting into the detail of the crystal structure, the ions will alternate on the surface at an atomic scale. You might imagine that this creates atom-sized patches where there will be net positive or negative charges. Would these attract another crystal? Would these attract the patches of charge in another crystal if they were adjacent. Maybe, but only if the surfaces were both atomically flat and lined up exactly. In reality the surfaces are rough and there is no net charge so large salt crystals won't stick because of this effect.
So the intuition that some crystal made of ions should have strong electrostatic attraction to another similar crystal is wrong because of electroneutrality and the detail of the physical surface.


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Why isn't sucrose a reducing sugar but maltose is?
I read one of the answers for a similar question, According to that it is due to the hemiacetale group converting into a carbonyl group. I don't get why the same thing can't be applied to sucrose though.

if youre looking for a rather unsophisticated answer.....it would go as....for a pyranose ring to open up...it needs to have a hemiacetal group ...such a group is present in maltose and absent in sucrose...the same thing cant be applied to sucrose as it dosent have a hemi acetal group...please check the structure again...(needs a C-OH linked to the ether linkage that holds together both rings..i guess thats how it goes in simple terms...hopefully i answered your question
P.S:The link given in the answer above mine has a better detailed explanation


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What is meant by "fairly" when used as a quantifier for the solubility of a chemical?
While doing homework and I found a statement that says a chemical is fairly soluble in water, but I do not understand what this means.
However, I know that when things are soluble in water they tend to dissolve. Is it that fairly soluble in water means that a part of it dissolves, while another part doesn't dissolve?

It seems that you understand the concept very well and that this is just a minor confusion about word fairly.
If you would characterize the solubility of a compound (solid, liquid, or gas) by words, rather than by giving the exact solubility in $\pu{mg\cdot mL^{-1}}$, you would probably come up with terms like

highly soluble
soluble
slightly soluble
insoluble

So, fairly means that the solubility isn't excellent, but it's not that bad either. It is acceptable, which of course isn't very precise.

Is it that fairly soluble in water means that a part of it dissolves while another part doesn't dissolve?

That one part does not dissolve will happen in all aqueous solutions when you add to much of the compound that you want to dissolve. Just think in sodium chloride, which has a solubility of about $\pu{390 g\cdot L^{-1}}$ in water. When your solution is saturated, it can't take any more. If you add another spoon of the salt, it will not dissolve and precipitate.

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text:
what voltage is needed to reduce Silver (I) Oxide in alkaline medium against an inert electrode
consider the following electrochemical cell:

silver (I) oxide is anode, platinum electrode is cathode, 6M KOH is
electrolyte

what voltage is needed to be supplied to the cell OR what voltage is generated by this electrolytic cell, with the reaction being:
$\ce{2 Ag2O -> 4 Ag + O2}$
where half reactions are:
$\ce{2 Ag2O + 2H2O + 4e- -> 4Ag + 4OH-}$
$\ce{4OH- -> 2 H2O + 4e- + O2}$
Is the reverse reaction spontaneous in alkaline electrolyte with inert electrode like platinum:
$\ce{4 Ag + 4OH- -> 2 Ag2O + 2 H2O + 4e-}$
$\ce{O2 + 2H2O + 4e- -> 4OH-}$
(the platinum electrode is for oxygen reduction reaction and oxygen evolution reaction)
I am a computer science engineer trying to learn chemistry, excuse me if my equations are not precise

You need to look up the standard electrode potentials of the materials involved.  While wikipedia provides an answer for $\ce{AgO}$ in acidic media, the abstract of this primary publication mentions a value of $(0.599 \pm 0.001)$ V of the $\ce{AgO / Ag2O / OH-}$ electrode, as determined against the H electrode at 25 C in KOH solution.  Note the authors find the potential dependent on the grid material their AgO-electrode was made; beside the expected-to-happen changes in function of temperature of the solution and concentration of $\ce{KOH}$ present.
A generally useful secondary literature reference for physico-chemical data you (still) may carry with you is the CRC Handbook of Chemistry and Physics by CRC Press, including tables like this, too.

thanks! I will lookup the resources..    if the standard potential of AgO electrode is (+0.6V), and the electrode potential for oxygen evolution reaction at the platinum electrode is (-0.4V), can the reaction be spontaneous because the overall cell potential is (+0.2V)?

The standard potentials are determined at 1M concentrations and 1 bar pressures. Some adjustment will need to be accounted for the higher contraction of hydroxide.


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Why Gaussian soft (STO-3G) causes error for Helium anion?
Software: G09.Rev.D.01
=======gjf-file======
%chk=F:\Hanion.chk
# hf/sto-3g geom=connectivity

Title Card Required

-1 2
 He                 0.20408162    0.07288005   -0.00095405

 1

Causes error message #2070
Log file explanation:
 There are   2 occupied orbitals but only   1 basis functions!

But the same time for the Lithium which is the same electron structure, method STO-3G work fine.
Why Gaussian soft (STO-3G) causes error for Helium anion?

An orbital of a closed-shell Hartree-Fock calculation can only contain two elctrons, one with alpha ("up") and one with beta ("down") spin.
You are calculating the Helium-anion. In a minimal STO-3G basis, it comes with one(!) basis function (or, loosly speaking, "orbital"), as it has two electrons in its neutral form. The anion however has three electrons. So, one closed-shell orbital is certainly not enough to host three electrons. The calculation works for Lithium a.) because gaussian automatically assumes that you want to perform an unrestricted-calculation as the number of electrons is uneven (this is also the case for the helium-anion, but I am not sure if you are aware of this). Furthermore,  Lithium has three electrons "by nature" (in its neutral state) and therefore, two basis-functions are enough to host four electrons.
You are using the STO-3G minimal basis, which adds one basis-function per two electrons of the system in the closed-shell case and one per electron in the open-shell case. Of importance is not the actual number of electrons in your system but the number of electrons the neutral atoms composing your system have. For a quick fix, substitute "hf/sto-3g" with "uhf/cc-pVDZ" and your calculation will run without issues.
Your system has an uneven number of electrons, so you cannot run a restricted (closed-shell) HF calculation. Either perform an unrestricted Hartree-Fock (UHF) or a restricted open-shell Hartree-Fock (ROHF) calculation. Both are available in gaussian09. As said, gaussian is already automatically doing this as a restricted HF calculation of a system with an uneven number of electrons is not possible.
Furthermore, as the calculation of the Helium-anion is quite fast, you should considering using a larger basis. You can start with "cc-pVDZ", a Dunning correlation-consistent basis-set, and move on from there.
For a comprehensive introduction to electronic-structure theory I suggest you read the book by Szabo and Ostlund. It assumes no previous mathematical or chemical knowledge above high-school-calculus und explains all the conceptual issues I have just mentioned in a detailed and understandible manner. Also, this book is quite affordable.
http://store.doverpublications.com/0486691861.html
If this is too much, I suggest you at least look up the Roothan-Hall and the Pople-Nesbet equations for running an electronic structure calculation in a basis, as Gaussian builds on them.

For H-atom method STO-3G require three gausians (1s = c1Gaussian1 + c2Gaussian2+c3*Gaussian3). For the Helium-atom also three, but for He-anion six or more. Am I right?

Ok, I think I understand. STO-3G take 3G per one orbital due to period (in periodic table). For 1st period - 1 STO = 3G, for 2nd - 5STO = 15G and so on. And if I perform STO-3G on He anion, G09 does not know about apeared orbitals, and as a result, crushes.

That is sort of right, yes. You have correctly summarized why gaussian crashed in your case. Keep in mind that in a STO-3G basis the coefficients of the three gaussian functions are NOT optimized. They are contracted to one function and the coefficient thereof is optimized. From the view of the Roothan-Hall-Equations, a STO-3G basis has one coefficient. The coefficients of the three gaussians are kept constant. Again, for mroe info, see Szabo & Ostlund


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How can I reach the two minima connected by a first-order saddle point (transition state) using the Gaussian Software
Currently, I have several transition state structures for a reaction of the kind A + B -> AB. These TS structures were calculated using Gaussian's own Berny-Optimization procedure and validated as first-order saddle-points by performing frequency calculations. Each of the structures has one, and only one, negative frequency. (It really doesn't matter for this question, but these are DFT calculations and they don't take too long as the system is moderatly small).
To validate and analyze these transition-states in the realm of the bigger picture of the whole PES some more, I would like to perform optimizations starting from them to yield the two minima connected by the transition-state.
Using the Gaussian software the most primitive idea to do this is to perform a regular optimization. I of course did this and, oddly, for every transition state the structure of the product of the reaction is yielded. Conceptually, a true transition-state on the PES should have zero gradient. This means that starting a primitive, gradient-based optimization from a hypothetical, 100% converged transition state should not yield another minimum anyway. I assume that the Gaussian software is in this case too "smart" and "user-friendly": The optimization might not be performed along the imaginary mode or using only the gradients, but most likely all convergence-accelearting black magic available is used. For most cases, this is a reasonable default scenario as it speeds up the optimization towards a truely relevant structure. However, I am also interested in the "path" of the optimization downward the imaginary mode. So this does not work for me.
It seems like the "EigenvectorFollowing" (EF) option for Opt in Gaussian is exactly what is needed to solve my problem. However, I cannot seem to get it to work for a regular optimization starting from the TS structure. Keywords that have failed include:
#p opt(z-matrix,EF) freq b3lyp/aug-cc-pVDZ
#p opt(cartesian,EF) freq b3lyp/aug-cc-pVDZ
#p opt(EF) freq b3lyp/aug-cc-pVDZ
#p opt(calcall,EF) freq b3lyp/aug-cc-pVDZ
#p opt(calcfc,EF) freq b3lyp/aug-cc-pVDZ
#p opt(readfc,EF) freq b3lyp/aug-cc-pVDZ

In every case, the error is the same:
 (Enter /apps/gaussian/g09_D/g09/l113.exe)

 EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-EF-
 EIGENVECTOR FOLLOWING MINIMUM SEARCH
 INITIALIZATION PASS

 ************************************************
 ** ERROR IN INITEF. NUMBER OF VARIABLES (  0) **
 **   INCORRECT (SHOULD BE BETWEEN 1 AND 50)   **
 ************************************************

Searching for this error on the internet yields many results, however they are all realted to MP4 / CCSD(T) optimizations and not to my scenario. I have also tried to specify the molecule as Z-Matrix or Cartesian coordinates.
Note that - speaking plainly - the system in concideration is two organic molecules and previous investigations showed that it is quite boring from a quantum-chemical viewpoint. By this I mean that the structures are "well-behaved" and not biradicals or other quantum-chemical-corner-cases where DFT fails.
I hope that this issue can be resolved by an experienced gaussian user. I am  using Gaussian 09 Rev. E. The first part of my question therefore is how can I use the EF option of Gaussian to perform an optimization along an imaginary mode starting from a transition structure?
Secondly, I still do not have any idea how to reach the "second" minimum connected by the TS (assuming the reaction is concerted, this would be the seperated educts). The only possible approach to tackle this which I can think of currently is to perturb the system manually along the imaginary mode and perform EigenVector-Following optimizations from these perturbed structres. Is there a more elegant way, circumventing manual structural modifications, to yield the two minima connected by the TS?

how can I use the EF option of Gaussian to perform an optimization along an imaginary mode starting from a transition structure?

I have never used the EF mode of Gaussian 09, but...

Is there a more elegant way, circumventing manual structural modifications, to yield the two minima connected by the TS?

I believe the standard way of proving that your TS is the TS you want is by an intrinsic reaction coordinate calculation:
http://gaussian.com/irc/
Make sure to check out the "options" tab. A decent write-up on a small system is here:
http://www.cup.uni-muenchen.de/ch/compchem/geom/irc5.html
Basically, you need to restart the calculation using the checkpoint (chk) file of your TS calculation, and as you calculated the vibrational spectrum, you can use "irc=rcfc" to read the force constants from the restart file. By default the IRC calculation does 10 points in both directions of the "reaction coordinate" starting from the TS, you can change this using MaxPoints=n, and you can also specify which direction you want to study using other keywords listed in the "options" tab on the Gaussian website.
Opening the logfile in Gaussview, you can see the IRC profile - the same way you can look at the optimization profile of an "opt" calculation. I'd follow up on the IRC by optimizing each of the end-points of the profile using a standard "opt" calculation, just to make sure.
Hope that helps.


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The detergent 'Positive 170' and what exactly makes it so corrosive?
So in season 5, episode 7 of Dirty Jobs, we're shown how a locomotive engine block is degreased using the detergent 'Positive 170' and we're told that it is extremely dangerous.

Just how 'dangerous' are we talking about here and why?
Thanks.

Hint: read the column in the centre.  The potential risks in damage of tissue, described on the left and on the right hand column increase by increase of the solution's concentration, and temperature (and 170 F [about 76 C] is considerable), and obviously by the time of exposure to this cleaner.
As an example of what may happen to skin under attack of strongly basic solutions (sometimes called lye) if you do not wear appropriate protective gear, see this example. -- Be told, such pictures may disturb you.


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Which overlap is stronger, and why?
I was just wondering which overlap is stronger, 2s-3p or 3s-2p? I believe this can probably be explained by involving some sort of mathematics (?) which I haven't used yet.
Edit: Note that 2s-3p refers to the overlap of 2s of one atom and 3p of another atom. Similarly for 3s-2p.
I haven't really come across an example for this case, guessing it's hypothetical. It'd be great if someone could give actual examples too if they exist.

Could you elaborate, and provide reasoning for the same?

I don't think it's clear what you are asking. Atomic orbitals on one atom? Well, they don't usually "overlap" (see answer by porphyrin). Maybe you are talking about atomic orbitals on different atoms? Well, then it depends on the example, doesn't it?

Yeah 2s of one atom, 3p of another. Similarly, 3s of one atom and 2p of another. The interesting part is, I couldn't find examples for such an overlap.

Maybe this should help: https://en.wikipedia.org/wiki/Electron_configuration .  there you can see which atoms have up to their "3p" occupied. If you are interested in one of these examples, give us the two atoms, a bond distance and one might be able to answer that question. Not me though, this is a bit too esoteric for me, I don't want to teach you things that are basically wrong

An atom's wavefunctions (orbitals) are orthogonal. This follows from the fact that the Hamiltonian is Hermitian, i.e eigenfunctions (wavefunctions) of quantum mechanical operators are orthogonal.  If they are on different atoms then they are not orthogonal and the overlap clearly depends on how far apart the atoms are.
Mathematically orthogonality means that (in one dimension) two wavefunctions $ \psi_i$ and $\psi_j$ satisfy the condition
$$\int_{-\infty}^{\infty} \psi_i\psi_jdx =\delta_{ij}$$
where $\delta_{ij} = 1 \text{ if } i=j \text { else } 0$ and is the Kroncker delta. (In this case if $i=j$ then the wavefunctions are also normalised as the product is $1$, this joint condition, orthogonal & normalised  is sometimes called 'orthonormal')
In the case of atomic orbitals there are three quantum number, the principle one n and angular ones l and m. The orthogonality condition is then
$$ \int_0^{\infty}r^2\int_0^{\pi}\sin(\theta) \int_0^{2\pi}\psi^*_{n'l'm'}(r,\theta,\phi)\psi_{nlm}(r,\theta,\phi)~dr ~d\theta ~d\phi=\delta_{nn'}\delta_{ll'}\delta_{mm'}$$
In your examples $n=2$ and $n'=3$ so the orbitals are orthogonal on these grounds alone.
As a check the 2s orbital has the form
$$ \psi_{2s} = A(2-\sigma)\exp(-\sigma/2) $$
where A contains only constants and $\sigma=Zr/a_0$ where Z is the atomic number, $a_0$ the Bohr radius and r the radial distance. One 3p orbital ($l=1, m=0$) has the form
$$\psi_{3p} = A'\sigma(6-\sigma)\exp(-\sigma/3)\cos(\theta)$$
Integrating the product over the angles $\theta$ which is  the integral $\int_0^{\pi}\cos(\theta)d\theta $. We can see that this integral is zero from a plot of cosine.
Edits:
As the question is now amended, the answer above holds for a single atom, but when there are two atoms then the overlap depends on their separation as the wavefunctions are no longer orthogonal because they have different coordinate origins, one on one atom the other on the other.
The overlap integral is written as
$$S= \int_{-\infty}^{\infty} \psi_i\psi_j~d\tau$$
where $\tau$ is the set of coordinates. Even in the simplest case of $\ce{H2^+}$ the coordinates needed are prolate spheroidal ones so to give an example it is simplest to use a one dimensional example since the principle is the same but technically more complex in 3D.
Starting with H atom 1s orbitals
$$ \psi_{1s} = a_0^{-1/2}\exp(-x/a_0)$$
if the atoms a and b are separated by 2R then the wavefunctions are
$$ \psi_a = a_0^{-1/2}\exp(-|x-R|/a_0) \\
\psi_b = a_0^{-1/2}\exp(-|x+R|/a_0)$$
and the overlap is the integral
$$ S=\int_{-\infty}^{\infty} \psi_a\psi_b dx =a_o^{-1}\int_{-\infty}^{\infty} \exp(-|x-R|/a_0)\exp(-|x+R|/a_0) dx$$
which evaluates to $$S=\frac{(a_0+2R)\exp(-2R/a_0)}{\pi a_0^3}$$
The wavefunctions with $a_0=1$, (and $R=1$ for the first two plots), $\psi_a$ and $ \psi_b$ vs. x  overlap $\psi_a.\psi_b$ vs. x and overlap integral S vs. the atom separation R are shown in the figures.

Could you please explain how this is related to the overlap strength? I'm sorry but I couldn't understand most parts of your argument. A simplified version might help!

OK, being 'orthogonal' means that here is no overlap at all. The integral of the product of the wavefunctions generates any overlap and as this is zero then no overlap. To see that the product is the overlap draw two curves and multiply them and work out the area, this is the overlap.  Mostly it will not be zero, but try with sine and cosine functions from $0$ to $\pi$.

Oh! So that really means that the above 2 overlaps, that is, 2s-3p and 3s-2p are not possible at all? Isn't that the conclusion?

I would assume that the OP means that 2s is on one atom and 3p is on another atom when {s..}he writes 2s-3p, but that should be clarified in the question. Otherwise the question makes not a lot of sense.

Hmm, yeah you're right. Appropriate edit made. The question remains unanswered though.

Please note that the question doesn't make a lot more sense now. Maybe it would help if you explained why you are interested in this topic and what conclusions you would draw if provided with an answer. Can't promise anything though, this is very, very vague

The wavefunctions of the electron orbitals of a single atom are orthogonal, which means their total "overlap" is zero. There is a subtle difference (or maybe not so subtle) between saying if you add up the overlap, it sums to zero and saying there is no overlap. For any positive overlap, there is equal negative overlap, meaning for any positive attractive energy between the two, there is equal negative energy between them. As far as any two orbitals on two distinct atoms overlapping, it depends on the atoms, their state (energy levels) and the distance between them. For instance, how much sense does it make to ask about the 2s, 2p, 3s, and 3p overlaps for two hydrogen atoms? How about a H and a U atom? (Hint: in neither case is there going to be much overlap 2s-3p or 3s-2p, at least, at anywhere near STP. (under conditions like the center of the Sun, or a neutron star, I don't know; you would certainly assume the two are much closer together, but at the same time, many of their electrons will be either ionized or in highly excited states. Too complicated for me to figure out ...


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Acid/Base behaviour of Amino Acids
My book states that nitrogen-containing rings are generally referred to as bases, and that under acidic conditions they can each bind an $\ce{H+}$. However, does this increase or decrease the concentration of $\ce{OH-}$ ions in an aqueous solution and why?

Brønsted (or Brønsted–Lowry) acid-base theory says a base will accept a proton from water. You can't take a H+ ion away from H2O without 'leaving' an OH-. The pKa of piperidine is 11.22, which means it is a fairly strong base. Apparently, you are asking what happens when you add a base to an aqueous acidic solution. The pH goes up. Meaning the pOH goes down. This is just another way of saying that [OH-] increases.

What one needs to understand first when dealing with acids and bases is that:
A substance can be acidic, if its constituent molecules have the capability to either donate an $\ce{H+ (like HCl)}$ or accept an $\ce{OH- (like H3BO3)}$ from water ,ie effectively have an electron deficient center
Similarly,
A base is a substance whose constituent molecules can donate an $\ce{OH- (like NaOH)}$ or abstract $\ce{H+ (like RO- ) }$ from water,ie effectively have an electron rich center,Now when Nitrogen is bonded to other atoms
[ up to 3 other atoms which are not electron deficient themselves (like an $\ce{NO2 or -CF3 group}$), or capable of conjugating the lone pair of nitrogen into themselves(like in anilene) ,above which it wont be able to show basic character longer as it cant form any more than 4 bonds ]
It still retains its lone pair,therby making itself an elecrton rich species; Hence retaining its ability to donate the lone pair to an electron deficient centre or abstract an $\ce{H+}$ from water,rendering a basic medium.
Note that nowhere in any of these processes has there been the involvement of an $\ce{OH-}$ group.
Although technically, while in aqueous medium compounds with nitrogen abstract a proton from water and release an $\ce{OH-}$ therby increasing the $\ce{[OH-]}$


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What are these upside-down glass egg-shaped containers for?
I took this picture from a chemical vendor, but I've seen these things many times already. How are they called and why are they upside-down?

This is called an inverted specimen jar, sometimes called a sample jar, for obvious reasons intentionally designed to stand on the stopper (it is not a standard one to close the other jars in the lab).  Their use was something more popular around 1900's, when showcasing dyes and other visual appealing chemicals, especially towards the general public. Your example invites to look at the granular character of the product.
A more traditional design is this one, more conveniently used in collections / museums.

(source)


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What drives the cost of Raman spectroscopy?
Raman spectroscopy setups are quite expensive- \$10-\$30k seems pretty common. The spectrometer, the probe/fiber cable, and the laser are typically each $3k+.
The spectrometer could potentially be replaced by an absorption spectrometer. The probe/fiber cable could potentially be replaced by free space optics. The Raman-specific laser could potentially be replaced by a cheaper laser. These components can be had for hundreds of dollars instead of tens of thousands.
Some potential negative effects I can imagine from swapping these out are:

Increased integration times
Decreased spectral resolution, or wider peaks
Decreased signal to noise ratio
Increased maintenance
Shorter lifetime

My broad question which I hope will be generally useful is, which of these components can be replaced with cheaper options with what effect?
My specific question, which may be less generally helpful, is which components should I get cheaper if I only care about signal-to-noise ratio and not integration time or peak width?

While this does not qualify as answer, rather than an entry point, Mohr et al. addressed some of your questions in J. Chem. Educ., 2010, 87, 326-330; doi: 10.1021/ed800081t

You have to realize that the company selling the spectrometer almost certainly didn't make the laser. Any component like that gets marked up 100%. So if the seller bought the laser for $1500, it gets sold for $3000 as a field replaceable part. // If you want to be the integrator fine. You can build a spectrometer cheaper. You'll need to know sheet metal work, machining, painting, optics, electronics, and computer programming. Best wishes....

@MaxW I'm interested in the modular version, not bench top or homemade bench top. Ocean Optics, Thunder Optics, and others will sell them modularly, broken out in the pieces I've identified. You can then very easily go cheaper on one module while going with a more precise/accurate component elsewhere.

We ended up developing a pretty low cost option as described that worked really well for simple applications: https://www.osapublishing.org/DirectPDFAccess/23B7C06C-AEBD-DD11-8E9EE7AF0E722E4E_401319/osac-1-3-1097.pdf?da=1&id=401319&seq=0&mobile=no

@ericksonla: your paper link is broken. Please add the reference. (I've seen your question only now - as a partial answer I'd say that several characteristics that can be traded off against each other. One crucial thing that will drive costs in academic settings is that it is typically more expensive to get good quality together with flexibility - the instrument should not only be suitable for the study now, but also for still unknown studies in future. I'd say that even the expensive Raman instruments in academic labs do actually make various trade-offs in that respect.)

The paper is Hickstein et al, "Rapid, accurate, and precise concentration measurements of a methanol-water mixture using Raman spectroscopy" OSA Continuum 1:3


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How do d-orbitals interact with molecular orbitals?
I  recently stumbled upon a question asking how iron destabilizes the bonds in a nitrogen molecule. I have searched the web and my textbook for quite a while now and I cannot find a satisfactory answer. The only explanation I could muster was that the d-orbitals of the Iron atom interact with the HOMO and LUMO of the nitrogen molecule, due to the overlap of the molecular orbitals with d-orbitals.
I am not sure if this makes sense. I'd really appreciate any answer or advice!

Hello and welcome to Chemistry.SE! If you have any questions about how the site works you can visit the [help]. Best of luck solving your problem.

You are on the right track with your answer.
Notice that the LUMO is a $\pi^*$ orbital, meaning that filling it with electrons would weaken the bonding. When nitrogen and iron interact, electrons from the d-orbitals on iron are well aligned, in terms of symmetry and energy, to fill into $\ce{N_2}$'s $\pi^*$ orbital. This forms a bond between $\ce{Fe}$ and $\ce{N_2}$, but at the cost of weakening the bond between the nitrogens. This is a very important concept in inorganic chemistry referred to as Pi Backbonding.
The classic example given of a ligand that exhibits $\pi$ backbonding in metal complexes is $\ce{CO}$, which is isoelectronic to $\ce{N_2}$ and has the same molecular orbital diagram (though the orbitals themselves look different due to the asymmetry of $\ce{CO}$).

If $d$ orbitals interact with $\pi^$ then new molecular orbitals are created and electrons go to the bonding MO. I don't think it makes much sense to talk about bond weakening just because the electrons went into a $\pi^$. In fact they don't even fill the $\pi^$ of CO. They fill the MO that result from the combination of $\pi^$ (CO) and $d$ (metal).


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Where do the electron and antineutrino come from in beta decay?
I was studying about nuclear reactions and similar stuff, but stumbled upon this doubt
In the process of beta decay, where a neutron transforms into a proton, a positron and an antineutrino, where do the electron and antineutrino come from ( do they already exist within the neutron or like? )
I do not believe that this is a duplicate of In nuclear chemistry, how does a neutron split to form a proton and an electron?

Seems like a dupe to me though...

@Zhe, what i wanted to know wasnt what happens during the decay( any referance book gives that info) what i needed was the origin of the two particles, as cited by DavePhD

Though it looks like you got the answer you wanted, the question is probably more appropriate for the Physics SE. You are more likely to find a Physicist who knows this, since chemically we tend to take protons, neutrons, and electrons as being the smallest building blocks.

A down quark of the neutron emits a virtual W- boson which becomes the electron and antineutrino pair.


text:
Reference request: review paper / overview of chemical oscillators
I know about the Belousov-Zhabotinskii reaction as an example of "chemical oscillator", e.g. a reaction that can produce repeating patterns over time as it exhausts its reagents. I also know about the Briggs-Rauscher reaction but I wonder how many such chemical oscillators are possible? Can somebody provide me with a good review paper and/or book reference written specifically to survey the range of all possible chemical oscillator systems?
To note, I know there are a lot of people outside of chemistry (particularly people interested in chaos theory and non-equilibrium thermodynamics) who do detailed research on modeling the detailed behavior of such oscillating systems, and I've found several such papers/reviews. But, what I'm looking for is a resource focusing on the chemical properties of oscillator systems, not the mathematics or dynamics modeling.

Even among non-biochemical oscillators there can be countless number of them.

@Mithoron , then why don't you give us some of these "countless numbers"? How about actually writing a descent reply and explaining some molecules who can be possible candidates to participate in chemical oscillators and the mechanisms of the reactions that can drive such oscillators. Providing some references would be nice, too. (Preferably some good review papers explaining how can we make an oscillator from reactions A, B, C, and so on and how would they work.) That will do the job!

https://books.google.pl/books?id=ci4MNrwSlo4C&pg=PA81&hl=pl&source=gbs_selected_pages&cad=2#v=onepage&q&f=false

Ever biological feedback system is technically an oscillator...

@Milthoron thanks for the link. I look it up but the whole book wasn't available on google books so I can't form an informed opinion about it, but from what I saw it's primarily "theoretic" work on the physics and mathematics of such reactions and mentions too few actual examples.

How is my question "too broad"? What precisely do you mean by "too broad"? I'm asking to tell me how many actual chemical oscillators have been done in real-world experiments (not "technically" or "potentially" or "theoretically") besides the 2 reactions I mentioned above. I want to know who has worked on different ways of creating oscillators and what reactions they had used. What is so "too broad" about this question?


text:
Why do we call a benzene ring attached to an hydroxyl functionality an alcohol and not a benzene?
But if an alkyl group is attached to a benzene ring, we call the parent chain the benzene ring, no matter how many carbons the alkyl group has. Why? For example: heptylbenzene.
For another example: when 1-propenyl-benzene reacts with sulfuric acid we get 1-phenyl-1-propanol. We don't call this product 1-propanol-1-benzene.

Related: What is more important, to have the parent chain be the cycle or the chain that has an hydroxyl group?

In your first example, heptylbenzene, the benzene ring is the parent structure for the preferred IUPAC name because the ring is always the parent structure if the ring and the chain have the same number of substituents that are cited as the principal characteristic group (by means of a suffix or a class name, or implied by a trivial name). The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

In heptylbenzene, neither the benzene ring nor the heptane chain has any substituents that could be cited as the principal characteristic group. Therefore, the ring is selected as the parent structure.
In 1-phenylpropan-1-ol, however, the propane chain has an $\ce{-OH}$ group, which is the principal characteristic group of this compound and expressed as suffix (‘-ol’). Thus, Rule P-52.2.8 cannot be used. If the ring and the chain have different numbers of substituents corresponding to the principal characteristic group, another rule comes first, which stipulates that the senior parent structure is the one that has the maximum number of substituents cited as the principal characteristic group.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Therefore, the propane chain is the parent structure in 1-phenylpropan-1-ol.

To the best of my knowledge, I think it is because the hydroxyl group (-OH) when connected to the benzene ring overrides the basic chemical properties of the benzene, displaying it's own more prominently.  Thus it takes the priority in naming.
But in the case of alkyl groups, they don't bring about much change to the basic properties of benzene and the chemical properties of the benzene remain more prominent. Thus benzene takes the priority in naming.
However I believe this may change if the alkyl chain is relatively large and overrides the benzene's behavior in the molecule.
I haven't cross checked anywhere so I'm not entirely sure but that's what I think makes the most sense. Hope it helps.


text:
How to calculate the final temperature of a gas when it undergoes adiabatic expansion?
The question is as follows:

A sample of $\pu{4.0 moles}$ of a gas ($C_{v,m} = \pu{21 J mol-1 K-1}$) has an initial pressure of $\pu{304.4 kPa}$ and an initial volume of $\pu{20 dm3}$ at $\pu{270 K}$. It undergoes an adiabatic expansion against a constant external pressure of $\pu{79.8 kPa}$, until the volume has increased by a factor of $3$. Calculate the final temperature.

I have calculated:  $Q = 0$  and $w = \pu{-3192 J}$ and change in internal energy $=\Delta U = \pu{-31922 J}$.
The next question is to calculate the final temperature of the gas. However, I am unsure on how to calculate this. Originally I thought as $Q=0$, then the final temperature would be the same as the initial temperature. However, I have realized this is incorrect.

What is the relationship between change in internal energy and change in temperature for an ideal gas?

Rather than answer the question numerically I have outlined the four different cases, reversible / irreversible and isothermal / adiabatic.
In adiabatic changes no energy is transferred to the system, that is the heat absorbed or released to the surroundings is zero. A vacuum (Dewar) flask realises a good approximation to an adiabatic container. Any work done must therefore be at the expense of the internal energy. If the ‘system’ is a gas then  its temperature will not remain constant during any expansion or compression.
In expansion the work done is $dw=-pdV$ and the change in internal energy $ dU=C_vdT$.
The heat change is zero then $dq=0$ which means from the First Law
$$dU = dw$$
and so
$$ C_vdT = -pdV$$
Dividing both sides by T and for one mole of an perfect gas $p=RT/V$ thus
$$C_v\frac{dT}{T}=-R\frac{dV}{V}$$
If the gas starts at $T_1,V_1$ and ends up at $T_2,V_2$ the last equation can be integrated (and rearranged) to give
$$\ln\left (\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)^{R/C_v}$$
or
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{R/C_v}$$
using the relationship $C_p = C_v+R$
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{(C_p-C_v)/C_v}$$
Using the gas law this van be rewritten in a more useful form as
$$ p_1V_1^{\gamma } = p_2V_2^{\gamma }$$
where $\gamma = C_p/C_v$. This is also written as $pV^\gamma = const$.
The change in internal energy in an adiabatic process is
$$\Delta U=C_v(T_2-T_1)$$
Adiabatic reversible
In a reversible adiabatic change we use the formulas above to work out what happens. If n moles of a gas fill a container of volume $V_1$ at $p_1$ atm. and is expanded reversibly and adiabatically until it is in equilibrium at a final pressure $p_2$ we can calculate the final volume and temperature. The values of $C_p$ and $C_v$ are assumed to be known and are constant with temperature.
The equation to use is
$$ p_1V_1^{\gamma} = p_2V_2^{\gamma}$$
and as only $V_2$ is unknown this can be calculated.  The work done is
$$w=nC_v(T_2-T_1)$$
Adiabatic irreversible
In an irreversible adiabatic change if n moles of an perfect gas expands irreversibly from a pressure of $p_1$ against a constant external pressure $p_2$ the temperature drops from $T_1$ to $T_2$. We can calculate how much work is done and the final volume.
The internal energy change is $\Delta U= nC_v(T_2-T_1)$
and the work done is $w=p(V_2-V_1)$ and as $\Delta U = w$ (as the system is adiabatic) the final volume can be obtained.  Similarly if the volumes are known the final temperature can be obtained.
Isothermal irreversible
In an isothermal irreversible change the work done on suddenly allowing a perfect gas to expanding from $V_1$ to $V_2$ is determined by the final external pressure $p_2$ and is
$$ q=-w =\int_{V_1}^{V_2} pdV = p_2(V_2-V_1)$$
thus expansion into a vacuum does no work.
Isothermal reversible
A reversible isothermal change performs the maximum possible amount of work, and assuming a perfect gas,
$$ q=-w_{rev} =\int_{V_1}^{V_2} pdV=nRT \int_{V_1}^{V_2} \frac{1}{V}dV$$
$$w_{rev} =- nRT\ln\left(\frac{V_2}{V_1}\right)= - nRT\ln\left(\frac{p_1}{p_2}\right)$$

Work done in adiabatic process $W =\frac {nR \Delta T}{1-\gamma}$
$Cv$ (specific heat at constant volume) $= 21 \frac{\mathrm{J}}{\mathrm{mol}*\mathrm{K}}$
$Cv = \frac{fR}{2}  \to f = \frac{21\times2}{8.314}= 5$
So $\gamma= \frac{f+2}{f}$, where $f$ is the degree of freedom.
$\to γ = \frac7 5$
Now work done at constant pressure $= 79.8\times(1000)\times[60(10^{-3}) - 20(10^{13})] = 3192 \,\mathrm{J}$ (positive work done by gas. The internal pressure will simultaneously be equal to the external pressure otherwise first law of thermodynamics won't hold true.)
$W = \frac{nR ΔT}{1-γ}$
$3192 = \frac{4(8.316)(T - 270)}{1-\frac{7}{5}}$
$T = [\frac{3192(-0.4)}{4(8.316)}] + 270$
$\to T = 231.61 K$


text:
Would the bond length of boron trifluoride remain the same after electron donation?
Would the bond length of boron trifluoride remain the same after electron donation?
My guess is it should be the same, because if $\ce{BF3}$ donates electrons from the low energy occupied orbitals, it would not affect the bond length much.

Hello and welcome to Chemistry.SE! I suggest taking the short [tour] to better familiarize yourself with how the site works. Also, as this seems like a homework-type of question, I would read through this discussion on how to ask a homework question.  You can also visit the [help] if you have any other questions.

Boron trifluoride is a very strong Lewis acid. Why is it donating electrons?

BF3 is a lewis acid as its octet is not complete and hence it will always have a tendency of forming a coordinate bond with a lewis base.
And talking about the bond length. B in BF3 is sp2
 hybridized & therefore the contains an unhybridized p orbital in B that works as a vacant orbital for backbonding with the pair of electrons from fluorines. This develops a partial pie bond character in case of BF3.
When BF3 accepts the coordinate bond from anyother lewis base, then it becomes sp3 hybridised and also no vacant orbitals in B, hence no backbonding possible and hence no partial pie bond character.
Conclusion:
In BF3; we observe sp2 hybridisation and also pie bond character due to backbonding and in BF3(X)^-   ,where X is the lewis base, it is sp3 hybridised and also no pie bond character.
Pie bond character decreases the bond length.
In sp2 hybridisation,due to more s-character, bond length is also shorter than sp3 hybridisation.
sp2 hybridisation + pie bond character is the reason of smaller bond length in BF3 as compared to BF3(X)^- . So the bondlength changes when BF3 accepts electron pairs. The reason being change in its hybridisation (i.e. sp2 to sp3) and no pie bond character after accepting a coordinate bond from a lewis base.


text:
Is using melting/boiling point good to measure volatility?
I wanted to figure out a way to compare volatility in organic compounds, so I read on many forums on the differing ways to measure volatility, which included enthalpy of vaporization and melting/boiling points.
I am planning on using melting/boiling point to prove that branching of chains in organic compounds causes weaker volatility.

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. Could you go into more detail about what else you've read?

According to this Wikipedia page:

In chemistry and physics, volatility is the tendency of a substance to
  vaporize. Volatility is directly related to a substance's vapor
  pressure. At a given temperature, a substance with higher vapor
  pressure vaporizes more readily than a substance with a lower vapor
  pressure.

So, I'm not sure how much information on volatility you will get from melting point measurements.  Ideally, the enthalpy of vaporization would be the best measurement, but, if it fits your needs, the boiling point at ambient pressure is a simple way to determine whether there is a significant difference in the volatility of various compounds.

If I use enthalpy of vaporization, what temperature

Enthalpy of vaporization values are usually given either at their boiling point or 298C. Sometimes these values, particularly the value at 298C, are not within the range of temperatures used to calculate the enthalpy, but are extrapolated assuming negligible change in enthalpy values over some range of temperature. I don't know if being able to compare your enthalpy of vaporization values to published values is important, if so I would try to make the measurements close to 298C.

After reading through your question again, I'm not sure that just measuring the vapor pressure at 20 or 25C might not be the best and simplest way to go. I just don't know for your experiments whether the extra effort of determining enthalpies of vaporization is worth it or not. That has to be your call I guess.


text:
Need help interpreting this data table (provided below)
I am confused about this data table. Can someone please explain how this person got a linear alkanes of differing molar masses? (I know some basics of organic chemistry, but I'm still learning.) Are there differing compounds to each of the ones listed underneath Boiling Point? If so, where can I learn these compounds or is there a database that has a list of compounds under each one?

Your best bet will be a basic organic chemistry text. The next best thing might be Wikipedia: look up alkanes, aldehydes, alcohols, ketones. That should give you a feel for what is meant by the terms "linear alkanes", etc., if you have some chemistry background. If you get too frustrated and confused, that means you are over your head and need to spend some more time with the basics or you'll end up giving up on chemistry altogether!

Just for completeness sake: could you please edit a source for the table.

The first row for each number in the leftmost column represent alkanes with $n$ many carbons and aldehydes and ketones with $n-1$ carbons and one double bonded oxygen. These are all in the same row because if you replace a carbon with a double bonded oxygen, you wind up with the same mass once you remove the excess hydrogens. The next row is for alcohols, which have an $-\ce{OH}$ instead of a $=\ce{O}$, which leads to an increase in molar mass by 2.
The alkane link above points to a table of boiling points with the molecular formula for each alkane.


text:
How does temperature affect buffer capacity?
I am currently leaning towards an experiment on how certain factors affect the capacity of a phosphate buffer, and am wondering if temperature in any way affects the capacity of the buffer. I know changing the temp will change the Ka of the acid, but am not sure if that changes capacity (how much it resists pH change) or just changes the pH range in which the buffer effective e.g. from 5-8 pH to 4-7 pH.

If you follow this derivation of buffer capacity, the last equation shows you that buffer capacity is a function of $K_a$, so a change in the Ka will affect the buffer capacity. Also, since you know that $K_a$ is affected by temperature, that should lead you to realize that the actual chemical environment is different at different temperatures due to a different concentration of acid and weak base pair. The magnitude of effect maybe small, but your experiment can prove that one way or the other!


text:
Elimination of tertiary alkyl halide with lithium aluminium hydride to give alkenes
I'm confused about the mechanism and thus cannot decide which alkene would be the product. Reaction with primary and secondary alkyl halides gives alkanes. What about this case?

Hello and welcome to Chemistry.SE. If you have questions about the site you can always visit the [help]. Your question appears to be of the homework type that falls under the guidelines described here. The bottom line is that in order to get guidance on this type of problem, you must show all the work and effort that you can towards solving it yourself first. For example, you could write out a mechanism and explain what the different products could be.

3rd degree alkyl halides undergo elimination hence give alkenes as product unlike 2nd and 1st degree alkyl halides

Note that the correct wording is primary, secondary, and tertiary. The use of $1^\circ$, $2^\circ$, and $3^\circ$ is a shorthand for that - it is not a shorthand for "first-degree", "second-degree", or "third-degree".

The reaction is E1 due to the formation of most stable tertiary carbocation, as such the double bond will be on the most substited side to give most substituted alkene as a major product.

It's clearly written in my comment that the reaction is E1

sorry. I wanted to confirm


text:
Why rinse flask with distilled water when the flask will be used for storing aliquots of HCl for titration?
In an experiment for standardisation of NaOH with titration, the lab manual instructed students to rinse conical flasks that were supposed to store 1mL aliquots of hydrochloric acid with distilled water.
Won't the excess water dilute the aliquot? Why not rinse with the HCl solution instead?
Thanks in advance.

The dilution with excess water won't affect the totat amount of HCl in the solution, which is what you will be titrating. You still know how much HCl you have from your quantitative addition of the 1 mL aliquiot. If you rinsed with HCl, you would have no idea how much HCl you really had and therefore you couldn't calculate the concentration of your NaOH.

Thanks airhuff, I think I get it now. If I used HCl for rinsing, there will be more HCl in the flask than what I put in (1mL) and will therefore impact results. Thanks alot for the help.

Technically, if you're doing very sensitive work, you would rinse with whatever you're about to use. This is assuming you can't just grab a new flask that doesn't have anything in it.

The dilution with excess water won't affect the total amount of $\ce{HCl}$ in the solution, which is what you will be titrating. You still know how much $\ce{HCl}$ you have from your quantitative addition of the $\pu{1 mL}$ aliquiot. If you rinsed with $\ce{HCl}$, you would have no idea how much $\ce{HCl}$ you really had and therefore you couldn't calculate the concentration of your $\ce{NaOH}$.


text:
Confusion in the line written in my book
In my class 11 chemistry NCERT book ,there is a line in chapter P-block.
The line is

The oxidation state two unit less than the group oxidaton state become progressively more stable for heavier elements in each group . Due to 'inert pair effect'

I could not understand this , what they meant to conclude from it.
Can anybody explain me with an example.

Related: http://chemistry.stackexchange.com/questions/8717

Ok let me explain, considering you're from a cbse affiliated school.
Group oxidation state = (group number - 10)
This will give you the group oxidation state of that group. Group oxidation state is the maximum oxidation state shown by the elements of the group. For example in 15th group, the group oxidation state = (15-10) =5
That means the maximum oxidation state which the group 15 elements would show is 5.
Generally moving down any p-bock group, the tendency of occurence of the maximum oxidation state (i.e. +5 for group 15 elements, +3 for group 13 elements) decreases. Rather the most occuring oxidation state for heavier elements is the oxidation state 2 units less that the maximum/group oxidation state. The reason behind this is intervening of d and f orbitals for heavier elements i.e. in one word we say the inert pair effect.
Now what is inert pair effect? As we move down the group, gradually for heavier elements d and f orbitals start filling up. The shielding by electrons in d and f orbitals is very poor due to which moving down the group the nuclear charge increases, but there is not much increase in shielding due to which effective nuclear charge becomes effective and hence its not possible for the heavier elements to attain the maximum oxidation state because to attain the maximum oxidation state, they have to loose the more number of electrons. But that is not possible for heavier elements because the nuclear charge is now effective. It wont allow loosing much of the electrons. Hence the maximum oxidation state/ group oxidation state is not achievable for the heavier elements.
What they can achieve is two units less than that maximum oxidation state and that also makes sense. For example in case of group 15 elements, the maximum oxidation/ group oxidation state i.e +5 becomes unstable for heavier elements but +3 state (i.e. two units less than the maximum oxidation state/ group oxidation state) becomes increasingly stable for the heavier elements. Now you might think why only two units less than the maximum oxidation state stable for the heavier elements? Why not 3 units less? Because this is a fact i.e. because we have found the more observing/occuring oxidation state for p block elements is two units less than their corresponding group oxidation state. Now the statement was written because it was observed.
Negative oxidation state for heavier elements is also not achievable because down the group the shell size is large. Though I said effective nuclear charge is bit effective for heavier elements, but that doesn't mean it is stronger enough to gain an electron. It is just strong to get hold of some valence electrons i.e. not allowing loosing much of the electrons. But that strength is not that strong to gain an electron. Hence negative oxidation is not shown for heavier elements.
So basically stable oxidation state for heavier elements is 2 units less that the maximum oxidation state shown by that corresponding group.

How does The concept of maximum oxidation apply to O and F ?

For heavier elements of a group, group oxidation state is not stable. That doesnot mean for lower elements the group oxidation state/maximum oxidation state has to be stable.


text:
Comparison of bond length
Question: (more than one options are correct)

My reasoning:
Option A: is correct as bond length of a pi bond is less than that of a sigma bond, hence a > c as there will be a double bond character in bond a
Option B: there will be an extra resonating structure of b. Hence there will be less double bond character in b than c. So, bond length of b>c. Thus, option B should be incorrect but it is given correct.
Option A and B both are given correct. Moreover, I can't understand how to compare bond lengths in options C an D.

Option A:
$sp^2$ hybridized carbons are more electronegative than $sp^3$ hybridized carbons; hence, the$sp^2$ hybridized carbons try to pull electrons more closely compared to the $sp^3$ hybridized atoms.
In the first molecule, $a$ represents the length of the single bond. Without any doubt, single bonds are longer than double bonds. However, the single bond exists in the first molecule exists between two $sp^2$ hybridized atoms, therefore, the bond length $a$ is shorter than the bond length $b$.
The last molecule is a double bond. This will be shortest as the other two bonds are single bonds.
$$b > a > c$$
Option B:
In the first molecule, there is no resonance stabilization. The electrons of the nitrogen atom of the amine group are not delocalized. The nitrogen-carbon bond in this molecule exists as a single bond.
In the second and third molecule, there is resonance stabilization; therefore, the nitrogen-carbon bond has a partial double bond character.
From the previous two conclusions, we can infer that the $a$ will be the largest among the three.
Between the second and third molecule, the third one has a deactivating group $\ce{NO_2}$. A positive charge in the benzene ring is very destabilizing. The nitrogen atom of the amine group tries to help by delocalizing its electrons for a longer time. Therefore, the partial double bond character of nitrogen-carbon bond in the second molecule is more than that of the third molecule.
$$a > c > b$$
Option C:
While all the three bonds are certainly double bonds, they are connecting different types of atoms.
The first bond ($a$) is a bond between a $sp^2$ hybridized carbon and a $sp$ hybridized carbon.
The second bond ($b$) is a bond between two $sp$ hybridized atoms.
The third bond ($c$) is a bond between a $sp$ hybridized carbon and a $sp^2$ hybridized carbon.
The $sp$ hybridized atoms are more electronegative compared to the $sp^2$ hybridized atoms; therefore, the $sp$ hybridized atoms pull the electrons more strongly towards themselves.
From the above conclusion, we can infer that the length $b$ is going to be the smallest among the three bond lengths.
This molecule is symmetric along the vertical line. Therefore, the bond lengths $a$ and $c$ shall be equal.
$$a = c > b$$
Option D:

Unlike benzene, the carbon–carbon bonds in naphthalene are not of the
  same length. The bonds C1−C2, C3−C4, C5−C6 and C7−C8 are about 1.37 Å
  (137 pm) in length, whereas the other carbon–carbon bonds are about
  1.42 Å (142 pm) long. This difference, established by X-ray diffraction,[17] is consistent with the valence bond model in
  naphthalene and in particular, with the theorem of cross-conjugation.
  This theorem would describe naphthalene as an aromatic benzene unit
  bonded to a diene but not extensively conjugated to it (at least in
  the ground state). As such, naphthalene possesses several resonance
  structures.
Source: https://en.wikipedia.org/wiki/Naphthalene

$$a < b$$

Can u explain how the carbon between two benzene rings is sp in the D part . Also if u read the first paragraph of it u will want to make an edit.:)

Oh, it isn't $sp$. Let me fix it. The correct reason is cross-conjugation.


text:
Why is the pressure correction in the van der Waals equation proportional to (n/V)^2?
The van der Waals equation for real gases is stated as follows:
$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$
For the coefficient $b$, we can reason out that more the number of molecules, the more volume will be occupied by the molecules (in turn reducing the free space available for the motion of the molecules). It must be linearly proportional to $n$ because $b$ represents the sum of the excluded volume of each molecule in a mole. Therefore, we must multiply it by $n$ to obtain the total excluded volume.
The coefficient $a$ is said to represent the strength of the intermolecular attractive forces. Intuitively, we can say that it must be proportional to the number of molecules per unit volume. The more the number of molecules around a molecule, the more attractive forces it shall feel.
Therefore, the correction term for pressure must be proportional to $\frac{n}{V}$. But we know that in reality, it is proportional to the square of $\frac{n}{V}$.
Why does it vary squarely?

Consult: https://en.wikipedia.org/wiki/Van_der_Waals_equation#Derivation

However the pressure term is modified and explained, it is worth remembering that the parameter a is intuitive rather than exact and that the van der Waals equation does not have  sound theoretical basis.

Summing up the argument from the Wikipedia derivation, the reason it is proportional to $(\frac{1}{V_\mathrm m})^2$ where $V_\mathrm m=\frac Vn$ (the molar volume) comes from how the molecules experience the attractive force. The only molecules that experience a net attractive force are those close to the edge of the container because as we move toward the center of the container, the attractive forces on either side of a molecule cancel out. The force toward the center experienced by a molecule at the edge should be proportional to $(\frac{1}{V_\mathrm m})$. But we can also say that the number of molecules within some area $A$ at the edge of the container should be proportional to $(\frac{A}{V_\mathrm m})$. So, by combining the proportionality constants into a single constant $a$ and dividing by $A$ to convert from force to pressure, we obtain that the correction to the pressure should be $(\frac{a}{V_\mathrm m^2})=\frac {an^2}{V^2}$ .

You wrote:  "But we can also say that the number of molecules at the edge of the container should be proportional to 1/m."  That's not quite right.  The number of molecules at the container boundary is proportional to the density times the surface area of the container.   But that's not what you want.  Instead, since you're calculating pressure, what you want is the number of molecules per unit area at the container boundary, which would be proportion to the density, which is in turn proportional to n/V.  I.e., you've got the right answer, but the explanation needs adjustment.

@theorist good catch, I was trying to describe this too much in terms of forces and didn't do a great job explaining how that translated to the pressure.

Pressure correction term depends upon:

Number of molecules attracting the molecules which comes to strike the wall and as such it is proportional to density of gas i.e. proportional to $n/V$ where $n$ is the number of moles of gas and $V$ is the volume of the container.
It also depends upon number of molecules which has a strike the unit area of the wall and is therefore proportional to the total number of molecules per unit volume, i.e. proportional to the density again (i.e. proportional to $n/V$).

So the pressure correction term is jointly proportional to $n/V$ for factor 1 and again proportional to $n/V$ for factor 2. Hence the pressure correction term is proportional to
$$\left(\frac{n}{V}\right)\left(\frac{n}{V}\right),$$
i.e. proportional to
$$\frac{n^2}{V^2}.$$

As Tyberius wrote, the pressure reduction is due to a force imbalance that exists at the walls of the container, because a molecule hitting the wall doesn't feel attraction to the wall, but is attracted to the molecules on its other sides.  The attractive forces from these molecules "pull back" on the molecule hitting the wall, reducing the pressure.
The reduction in the force with which each molecule hits the wall is proportional to the strength of intermolecular attractive force times the concentration of molecules pulling back on it, i.e., to $\rho=n/V$. This gives the force reduction per molecule.
The total pressure reduction is the force reduction per molecule ($\sim \rho=n/V$) times the number of molecules per unit area hitting the wall (also $\sim \rho=n/V$).  Hence the pressure reduction due to intermolecular attraction is $\sim \rho^2=(n/V)^2$.


text:
Rate of evaporation of mercury
A restorer of antique barometers reported in a blog post that he has had a barometer on his wall that has shown no "visible" loss of mercury in 15 years.
I was kind of curious about this, wondering how long it would be before a barometer would evaporate completely. How can I compute the rate of evaporation from the vapor pressure?
For the barometer reading to become inaccurate, the well would have to evaporate completely until the 0 point was reached (at highest pressure), then it would start becoming inaccurate.

This abstract alone doesn't quite answer your question, but the full paper probably would come close if you had access. They found a "small" droplet of Hg evaporated in an open container at 1.6 ug/hr and a "large" drop at 6.0 ug/hr. There is no definition of small or large droplet size, though they do describe the container. If you had the full article I guess you would have to try to calculate diffusion rates through the barometer. It would be an interesting article, though I don't have access beyond the abstract.