text:
Clemmensen reduction vs. Wolff-Kishner reduction
In which cases is the Clemmensen reduction is preferred over the Wolff-Kishner reduction?
One thing is that the Clemmensen reduction is done under acidic conditions whereas the Wolff-Kishner reaction is done under basic condition, and that I think decides somewhat which to choose depending of the nature and other groups present in the reactant.
But are there any other points to keep in mind while choosing the best suitable reaction?

hydrazine is known to produce heterocycles with 1,3-diketones.

Clemmensen reduction is particularly effective at reducing aryl ketones, such as those formed from Friedel-Crafts acylations. However, acid-base is the primary consideration. Also remember the Mozingo reduction as a mild alternative.

A key consideration for the Wolff-Kishner reduction is thermal stability of the substrate. High temperature is required for the classical reaction using hydrazine and KOH, sometimes up to 200 °C. This can rule out plenty of compounds.
The Clemmensen reduction may be fine for the laboratory,  but fundamentally many industries wouldn't use it due to scaling up the use of mercury.
Scalable alternatives are molecular hydrogen with a catalyst under acidic conditions or, the trifluoroacetic acid-silane combination, such as triethylsilane or tetramethyldisiloxane.

The basic difference is the medium as Clemenson is in acidic medium so if $\ce{OH}$ is present so it will be attacked too, but in Wolf-Kishner it is in basic medium so $\ce{OH}$ group can be saved which is sensitive, but if there is halogen, ester, or amide group it will not tolerate the reaction.


text:
Selective alkylation of carboxylic acid versus phenol

Starting from the ketone shown, I plan to use a haloform reaction to transform the acetyl group into a carboxylic acid.
Following that, there are two alkylations to be carried out: one esterification on the carboxylic acid (using propargyl bromide) and one etherification on the alcohol (using 6-bromohexan-1-ol). In what order should these be performed?

Using the haloform reaction to convert the methyl ketone 1 to the corresponding carboxylic acid 2 is a neat idea. According to an old review by R. C. Fuson and B. A. Bull, published in Chem. Rev., 1934, 15, 275-309 (DOI), this seems to work for 4-hydroxyacetophenone. Your starting material 1 is a phenyl homolog of that.

In 2, you have two acidic protons ($\ce{-COOH}$ and $\ce{-OH}$). Based on our knowledge and by comparison with 4-hydroxybiphenyl ($\mathrm pK_\mathrm a = 9.55$), benzoic acid ($\mathrm pK_\mathrm a = 4.2$) and 2-phenylbenzoic acid ($\mathrm pK_\mathrm a = 3.46$) (all data are taken from the Rubber Book), you know that

the carboxylic acid is more acidic than the phenol
the difference is acidity is quite large

Potassium carbonate ($\ce{K2CO3}$) in a polar aprotic solvent, such as N,N-dimethylformamide (DMF) should be sufficient to deprotonate both centres.
Back to your question on the order of events:
I would add 1 equivalent $\ce{K2CO3}$ of to a solution of 2, stir at room temperature for a while and then add the propargyl bromide, assuming that the large differences in acidities will make sure that the carboxylic acid is deprotonated, while the phenol is not.
Monitoring the reaction by TLC will show whether the esterification is quantitative.
If this is the case, I would dare to run a one-pot reaction, add another equivalent of $\ce{K2CO3}$, stir again and then add the bromohexanol to form the ether.
It might be a good idea to try this first with very small amounts, just enough to check via TLC.

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Fitting parameters of Lennard-Jones potential to properties of real materials
I'm using the Lennard-Jones potential for some (very simple) protein docking simulation* and couldn't find parameters for different materials (like $\ce{C}$, $\ce{N}$, $\ce{O}$, …). Does it even make sense to differentiate between different materials when using a crude approximation like the Lennard-Jones potential?
* the potential function for each configuration of the two molecules is the sum over all pairwise energies, for which I only consider the van der Waals term by using the LJ-potential

Your question does not make really sense within the state-of-art molecular mechanics. Please refine what you mean, with respect to https://en.wikipedia.org/wiki/Force_field_%28chemistry%29

My potential function for each configuration of the two molecules is the sum over all pairwise energies, for which I only consider the van der Waals term by using the LJ-potential; does it make more sense now?

Do you have any reason why not to use any existing force field? E.g. UFF?

This is a project for a mathematics seminar, so its primary purpose is how to implement all the numerical stuff, automatic differentiation, etc.

Than, again, get the parameters from any existing force field, for example http://dasher.wustl.edu/tinker/distribution/params/tiny.prm

It would be much better, if you could [edit] your question in response to the comments, to make clear what you are trying to achieve. And also to prevent it from getting closed.

All molecular simulation force fields are essentially "fitted" mathematical functions. The parameters are adjusted until the simulation results match some physical properties we care about. As Geoff says, Lennard-Jones force fields are not particularly realistic, in the sense that you wouldn't expect a simulation of atoms interacting solely through Lennard-Jones potentials to behave exactly like a real system. This becomes very important for things like proteins, where conformation and charge-charge interactions are very important.
However, what might not be clear from the other answers and comments is that all force fields are not 100% realistic. The trick is to intelligently choose your level of realism. It could be that a Lennard-Jones force field that ignores  charge-charge interactions works perfectly well for your application. Without knowing the details, I can't tell you how suitable it would be.
I do know that for polymers all sorts of real phenomena can be modeled with very simple Lennard-Jones-like force fields. A famous example is this article by Kurt Kremer and Gary Grest:
Dynamics of entangled linear polymer melts:  A molecular‐dynamics simulation. J. Chem. Phys. 92, 5057 (1990)
Also, as Geoff mentioned, LJ potentials work pretty well for things that can be modeled by the van der Waals equation of state: noble gases, simple diatomics, and other small molecules at high temperature and low pressure.
Some coarse-grained simulations use LJ potentials for each "blob," and depending again on what you want to study, this can be very successful.
And while we are on the subject of unrealistic force fields, the hard-sphere model has been used to successfully model things like glass transitions (I don't have a good representative link for this, but you can search for "hard sphere glass transition" to find many studies).
So, in response to your actual question:

Does it even make sense to differentiate between different materials when using a crude approximation like the Lennard-Jones potential?

The answer is:
Maybe - it depends on what the goal of your simulation is. What physical behavior are you trying to model? If you just want to see how a string of soft, slightly sticky beads sticks to another string, then you don't need to differentiate between atoms. If you want to see what happens when some of those beads are really sticky (sort of like hydrogen bonding sites on a protein), then it would make sense to use different parameters for very sticky vs. slightly sticky beads. It probably would never make sense to worry about which beads were carbon and which were oxygen, nitrogen, etc., or to try to select parameters for them that were based on published values (unless those studies were specifically looking at the same system and properties that you are). Most LJ parameters are derived from fitting simulated phase behavior to real phase behavior - which is pretty far removed from what you are doing.
If you want to get a better idea of what real proteins would do, then the problem becomes much more complicated. Real protein behavior has a lot to do with interactions with water, which means you would need to simulate that as well. Charge-charge interactions also become much more important (and the simulations much more computationally expensive)
My advice (and I think what the other responses are saying as well) is to choose your system and goals carefully so that you don't need this level of realism. Or, if you do need it, then do it right; don't try to make a simple model work for a system that it doesn't fit.

You could, in principal, create some simulation of "Lennard-Jones-ium" based on a set of tailored parameters. There's probably some benefit from an applied mathematics perspective. However, as discussed by multiple comments, this wouldn't be at all physically realistic.
Most force fields have many atom types - not just C vs. O, but $\ce{C_\alpha}$ or $\ce{C_{O2}}$ etc., because the parameters vary considerably in terms of interatomic forces, etc. for different atom types.
Also, the bonding interactions between two carbons are very different than non-bonded interactions. In other words, if you wanted a model for a protein and docking simply from LJ parameters, you'd need separate pair-wise parameters for bonded and non-bonded interactions.
Moreover, for chemistry, you need more than just pairwise atomic interactions. Most chemical force fields include three-atom angle-bending terms, four-atom dihedral terms, and out-of-plane distortion terms (e.g., pyramidal N or P).
Finally, all of this ignores electrostatic interactions, which a Lennard-Jones model would sweep under the rug.
If you want to simulate something with LJ, I might go for a sample of gas (e.g., He, Ne, etc.) or atoms in a solid (e.g., Ag).

As for the application, the OP is right, that most terms do vanish. For rigid docking, i.e. two fixed molecules interact, only the non-bonding terms do not cancel out. From this point of view is the approximation by L-J fine. You would not get different energies based on charge interaction, though.
The described procedure will answer the question whether given molecule sterically fits in cavity in protein. Which is not that bad, after all.

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Unit consistency in rate equations
I suppose that my problem is not one of great profundity, but it is an annoying one. The problem is related to the measurement units involved in rate equations of different order. Not being a chemist myself, I have in my work encountered equations of the type: $$\frac{-\mathrm{d}[A]}{\mathrm{d}t} = k [B]^b [A]^a$$ representing the decay of some species $A$. The brackets denote concentration and $k$ is the rate constant. $B$ is some other chemical species (cooking chemical) and then there are the exponents $a$, $b$. For a first order reaction with $b = 0$ and $a = 1$ everything is still under some kind of control; if time is measured in minutes the unit of the rate constant is $\mathrm{min^{-1}}$.
However, if we have a pseudo first order reaction where the species $B$ is present albeit constant during an experimental run, problems arise. In my line of work the species $B$ is typically hydroxide ions and their concentration is traditionally given in molar, $\mathrm{mol/L}$. Even if the concentration of $B$ does not change, the initial concentration plays a role, hence the power of $[B]$ usually appears as a separate factor in the rate equation. Unfortunately, this complicates the unit balance in the equation as frequently the exponent $b$ is a non-integer number. One workaround would be to rescale the concentration to a dimensionless number, but the choice of scaling parameter would be rather arbitrary. Another, perhaps more elegant solution would be to work with molar fractions, but this would be discordant with most of the literature in the field, where the question of unit consistency mostly is disregarded.
A second order rate equation ($b = 0$, $a = 2$, or $b = 1$, $a = 1$) leads to the unit $\mathrm{L\,mol^{-1}\,min^{-1}}$ for the rate constant. I suppose this is still somehow acceptable, even if I would prefer to work with scaled concentrations (divided by the initial concentrations) with unit $\mathrm{min^{-1}}$ for the rate constant, but for fractional order rate equations there is the same problem as in the earlier described pseudo first order case.
Is there any prescribed remedy to this dilemma?

It is true that rate equations typically use molar concentrations for each species, and that for theoretical rate equations you can make sure the units work out by choosing the appropriate units for the rate constant.
There are two things that are important to understand about rate equations, though.
The first is that the "real" rate equations use activity, not concentration. Concentration is usually a "good enough" approximation that this doesn't matter in practice, but when you start thinking about the units of the reaction constant for things like fractional exponents in empirical rate laws, understanding that the concentrations should really be dimensionless activities makes the problem a lot easier.
The second thing to understand is that real reactions don't always follow theoretical rate equations. As a result, what you are often looking at in real life is an empirical rate law. With empirical rate laws, we don't usually worry about the physical significance of fractional exponents, and in turn, how they affect the units of the rate constant. The units of the rate constant become whatever they need to be to make the equation right. In other words, if you have
$\frac{-\mathrm{d}[A]}{\mathrm{d}t} = k [B]^{0.6}[A]^{1.3}$
as an empirically derived rate law that gives $\rm{mol_A}\rm{L}^{-1}\rm{min}^{-1}$, then $k$ would need to have units of $\rm{mol_A}^{-0.3}\rm{mol_B}^{-0.6}\rm{L}^{-1}\rm{min}^{-1}$. These physically nonsensical units are ok because the equation is entirely empirical - we are just describing what we observe on a macroscopic level, without attempting to give it physical significance beyond that.

I've always wondered if I'd see a rate equation with an exponent which wasn't an integer or half-integer. Would you happen to have any reference with an explicit example?

+1 for pointing out that it should be dimensionless. Many chemists have had similar issues because they aren't properly taught about dimensional analysis!

@Nicolau Saker Neto - I just made that one up as a hypothetical example :). I searched for a while but can't find any explicit examples of real non-half-integer fractional order rate laws. I did find references to using them for complex reactions (typically involving heterogenous systems) in some of my kinetics textbooks, but there were no explicit examples there either. I know that these are most common in industrial settings, where the mechanism isn't always important. Perhaps that is why they are hard to find in the literature?

Sorry if I wasted your time, I had no idea whether more complicated exponents were common or not. I'm not sure if this is right, but I'm under the impression that even complex reactions could in principle be discretized into several cases which cover particular steps of the reaction under sufficiently constrained conditions, and all of these small steps would have "pretty" exponents relating to the step's stoichiometry. If one doesn't care about this level of detail, it would be possible to average out some steps and try to compensate by letting the exponents vary freely as real numbers.

@Nicolau Saker Neto - No worries - I was surprised I couldn't easily find any! You are correct, empirical rate laws with fractional exponents can be resolved to steps with whole or half integer orders. You would only see these types of fractional order laws in situations where figuring out the details costs more and provides less benefit than just using the empirical law. One of the books I read said "using them is an admission of ignorance."

Very interesting, thank you kindly for your comments!

For fractional order of reaction, see related: https://chemistry.stackexchange.com/q/167050/72973

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What are the steps to creating synthetic glucose?
I need to know how to go about creating synthetic glucose with generally available materials. As I am aware of both photosynthesis and glucose in nature, I will not need lessons in how plants produce it. So, how may glucose be produced by artificial means?

Follow up question: is there a simple process starting from inorganics like CO2 and H2O (ie no biomass feedstock)?

A relatively simple process is described by Kobayashi et. al where they use cellulose, a structural component of plants, which is ball milled, mixed with activated carbons and HCl and hydrolysed to produce glucose in high yield (88%).
It requires only easily available ingredients and the ability to heat (and preferably stir) to 453K. Reference is given below.
High-Yielding One-Pot Synthesis of Glucose from Cellulose Using Simple Activated Carbons and Trace Hydrochloric Acid
Hirokazu Kobayashi, Mizuho Yabushita, Tasuku Komanoya, Kenji Hara, Ichiro Fujita, and Atsushi Fukuoka
ACS Catalysis 2013 3 (4), 581-587
DOI: 10.1021/cs300845f

I don't have access to that reference, but do you happen to know if the process leads to the natural blend of glucose isomers (left and right hand glucose)?

@docscience Cellulose is a polymer of many β-D-glucopyranose groups, but glucose anomerizes in solution, so you'll get a mix of α-D-glucopyranose and β-D-glucopyranose (approximately 1:2, if you leave it long enough). If by "left and right hand" you mean D- and L-glucose, then no—L-glucose is not naturally ocurring, with the exception of a few bacteria, and thus cellulose built on L-glucose does not exist.


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PMMA cross-link temperature
I premise, I’m a physicist and I’m working on a transfer procedure for graphene via CVD, which involves the use of PMMA as a resist layer. Studying this polymer, I still haven’t found what is the cross-link temperature (or window of temperatures) for it. Can someone give me a hand? or just a good reference?
UPDATE
Ok, maybe I’ve been misunderstood. My PMMA is diluted in anisole, and I apply it as a photoresist over the graphene sheet via spin coating. After this deposition NO Initiator is used, but it came across various heat treatment, depending in the recipe I use. My question is: does anybody know at what temperature the PMMA starts to seriously crosslink? or better: if I do more heat treatments, all at temperatures above 150 °C, am I going to have an highly crosslinked PMMA, so that a removal process in acetone solvent is impossible because it can’t get into the chains to districate them?

what is your starting prepolymer? Type, commercial mark etc.

I use microchem 950 PMMA A7 (data sheet http://microchem.com/pdf/PMMA_Data_Sheet.pdf) it's already polymerized.

According to this thesis, the polymerization is determined by thermal initiator that is used in PMMA mixture. The polymerization (corsslinking) starts when the initator is added to the mixture, but it is really slow at low temperature. Higher temperature lowers hardening time (speeds up the crosslinking). Since the dustsheet does not provide the information about the initiator, I would suggest to determine the cross-linking temperature experimentally. Some material related to the polymerization process:http://www.chemistry2011.org/,
BULK POLYMERIZATION OF METHYL METHACRYLATE IN A KNEADER
REACTOR
P.-A. Fleury
LIST AG
CH-4422 Arisdorf,
High conversion polymerization of methyl methacrylate
in the presence of poly(methyl methacrylate) of different relative
molecular mass
J. BARTOŇ, V. PASTUŠÁKOVÁ, V. VAŠKOVÁ, and P. AMBROVIČ
Polymer Institute, Centre of Chemical Research, ChemicalPapers

It would be quite terrific if you would include the complete citations, especially when they are not linked to the DOI. Sometimes one can find such papers on mirrors on the internet, but in order to prevent this answer from link rod, it is better to name the sources directly.

Ok so i got the answer to my question. For what i've understood, PMMA, as it is a thermoplastic polymer, is unable to crosslink unless it's "mixed" with an initiator. The removal process is going to be difficult if temperatures are going to get over 150°C due to the degradation of the polymer in species that may not be diluted in organic solvents

http://pubs.acs.org/doi/abs/10.1021/jp408429v


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Isolating radium oxide
Why is radium oxide yet to be isolated when a wide range of other radium (II) compounds are known ?

I'd guess that not only is it radioactive, but that it is extremely reactive with water to produce the hydroxide salt.  So trying to isolate "pure" radium oxide is probably very difficult.

Anneal radium in an oxygen or oxygen + inert gas atmosphere ? Seems pretty trivial to me.

BaO is created from BaO2 in over 1175K, this would be even worse in case of RaO, I guess.


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Which has greater +R strength - toluene or anisole?
I read that resonance in toluene is initiated with the help of hyper-conjugation. Also hyper-conjugation is weaker than resonance/mesomeric effect. But, in toluene, its resonance is taking place with the help of hyper-conjugation, am I right? I am confused which has greater +R strength - toluene or anisole?

Welcome to Chemistry.SE! Short answer: Anisole should have the greater +R effect, because the oxygen atom in anisole has delocalized electrons that can participate in resonance. Hyperconjugation $\ne$ resonance; hyperconjugation is a way to explain inductive stabilization. Toluene will have a greater +I effect.

@BenNorris  I thought hyperconjugation = resonance.  The same rules apply, can't move nuclear positions, etc.  I thought the inductive argument for touluene was based on the electronegativity difference between an $\ce{sp^3}$ methyl group carbon being connected to an $\ce{sp^2}$ aromatic carbon.

So what is the final answer ?

I have not forgot, but alas, this is a busy time for me.


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Which of the following compounds would result in the reaction being more active in the reverse direction?

$\ce{Fe(OH)2 (s) <=> Fe^{2+} + 2OH-}$
The reaction will be more active in the reverse direction if .......... is added
a) $\ce{KOH}$
b) $\ce{Na2S}$
c) $\ce{Fe(OH)2}$
d) $\ce{Fe(NO3)2}$

My thoughts:
Well, adding more $\ce{KOH}$ will increase the $\ce{OH-}$ ions so the reaction will go in the reverse direction to get back to equilibrium. Adding $\ce{Fe(NO3)2}$ Will increase $\ce{Fe^{2+}}$ ions and the reaction will go in the reverse direction too.
However, the book answers states that only (d) $\ce{Fe(NO3)2}$. I searched google but all the examples involved having only choice which works.
Note: Refer me to sites if necessary. I am ready to read.

I have improved the formatting of your question using $\LaTeX$. For more information on how to do this yourself please see here.

In fact, adding more ion hydroxide won't displace the equilibrium towards the reverse reaction. This is due to the fact that ion complex $\ce{[Fe(OH)4]^{-2}}$, which is stable in presence of excess of ion hydroxide, is formed by successive addition of ion hydroxide on ferrous hydroxide.
If you want to study this more in depth, I invite you to read this webpage
It's explained at the end of the page under "numerical problems".

Need a bit more explanation , It's too difficult to my current state.

@HoytVolker Essentially, if you add enough hydroxide ions, another product will be formed, the complex mentioned in the Yomen's response.  If this is beyond the scope of your class, then the lack of KOH as a correct answer was likely an oversight.  I wouldn't expect my high school chemistry students to know this.

I think it's upper my current level. "I wouldn't expect my high school chemistry students to know this" so It's not lack of effort from me ?.

There are basically two processes, that make adding more hydroxide the wrong answer.
The first one is already described in Yomen's answer, but I would like to extend that a little bit further. If you dissolve iron (II) hydroxide in water, an aqueous complex will form:
$$\ce{Fe(OH)2~(s) + 4H2O <=> [Fe(H2O)4(OH)2]}$$
Apart from this, due to the auto-ionisation of water and the relative stabilities of the complexes, there is a variety of complexes in solution, among others:
$$\ce{[Fe(H2O)5(OH)1]- <=>C[H+] [Fe(H2O)4(OH)2] <=>C[OH-] [Fe(H2O)3(OH)3]+}$$
From these complexes, only the neutral one precipitates. From this equation it is also deducible, that the stability of the precipitate is highly pH dependent.
Upon adding more hydroxide ions, the equilibrium will shift more and more to the right, and even further, resulting in the dissolution of the precipitate
$$\ce{[Fe(H2O)4(OH)2] <=>C[OH-] [Fe(H2O)3(OH)3]+ <=>C[OH-] [Fe(H2O)2(OH)4]^2+}$$
The other possibility is, that iron (II) will easily be oxidised in basic conditions and access to air. This process is often referred to as rusting.
$$\ce{4Fe(OH)2 + O2 + 2H2O <=>C[OH-] 4Fe(OH)3 \equiv 2Fe2O3(OH2)3}$$
Further reactions will form the more stable, but less hydrated iron (III) oxide:
$$\ce{Fe2O3(H2O)3 <=> Fe2O3(H2O)_{$x$} + $x$ H2O} \quad (x\in[0,3])$$

I believe the question is attempting to get you to think about the stochiometry of the reaction.
The reverse reaction takes one mole of $\ce{Fe^{2+}}$ and two moles of $\ce{OH-}$. This means that if you add the same molar amount of $\ce{Fe(NO3)2}$ and $\ce{KOH}$ you would expect (roughly) half the amount of $\ce{Fe(OH)2(s)}$ formed from the $\ce{KOH}$ versus the $\ce{Fe(NO3)2}$, because it takes two (molar) units of $\ce{OH-}$ to get one unit of reaction to happen.
So both will cause the reaction to reverse because of Le Chatelier's principle, but adding $\ce{Fe(NO3)2}$ will be "more active" (on a mole/mole comparison basis) because of the stochiometry of the reaction.
I agree the question is poorly worded, though. Although using a molar comparison basis for things like this is pretty standard, it's not the only way to do it. (E.g. adding equal masses of $\ce{Fe(NO3)2}$ or $\ce{KOH}$ will result in the $\ce{KOH}$ solution having greater reverse reactions.) For completeness, how "equivalent" amounts of the various substances are determined should have been specified in the question.


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Bohr's model of atom

"the most remarkable success of the Bohr's theory is that it provides a satisfactory explanation for the line spectrum of hydrogen."

But what is the problem with line spectrum of hydrogen which could not explained before the Bohr's theory was introduced and how it was explained by Bohr's theory?

@PushkarSoni, do you have an idea what is Bohr model actually is? If not, first read about it, then if something is unclear, please, ask.

@LordStryker, Bohr model by no means explained why electrons do not fall into the nucleus. It just takes this fact for granted. In fact, it uses this assertion as a starting point. Bohr's logic ultimately was: since electrons do not fall, there are some tricky orbits on which they move without radiating energy. Trying to reverse Bohr's argument would obviously be a perfect example of circular reasoning. :D

@Wildcat Most certainly.  However, it is used to illustrate these phenomena in a simple way.  That is what I was trying to convey...

Since this is pretty well covered in countless number of books and other sources I will try to just emphasise the key points.

The emission spectrum of atomic hydrogen was well known at that time, the only problem was that no one was able to explain it. :D Where does all this lines come from, i.e. what physical process gives rise to such discrete spectrum?
The wavelengths of spectral lines were known to be very well predicted by the Rydberg formula, but again no one was able to explain why this formula works, i.e. what are its physical justifications.
Bohr model explained discrete spectrum as follows: spectral lines are due to electron transitions between stationary orbits. Bohr was also able to determine the actual energy spacing between levels in hydrogen atom and explain the reason for the structure of the Rydberg formula.
However, Bohr failed to predict the lines in the emission spectrum even for helium atom. It happened simply because Bohr model is physically  wrong. Electrons do not orbit around the nucleus.

thanks now i understood. i have another question that why was the Bohr's model unable to explain the line spectra of multielectron atom, actually i don't even know what this actually means? please explain

@PushkarSoni, simply put, Bohr theory was wrong physical theory. Thus, it failed test on multielectron atoms and was eventually replaced by quantum theory.

so you said that electrons do not orbit around the nucleus? what do you mean?

so you are taking about quantum mechanical model of atom of finding the position of electron at certain region. am i right?

@PushkarSoni, oh, well, that would not be easy to explain. In few words: at microscopic level Mother Nature behaves very differently from our everyday experience. The very notion of trajectory does not longer make sense.

@PushkarSoni, yes. Quantum mechanical description which operates just with the probabilities of finding an electron here and there is the working physical theory for an atom.

thanks , but my doubts are cleared.  right now i am studying bohr's model. i have to cover up more topics to reach the quantum mechanical model (ex- uncertainty principle, dual nature of matter, debroglies  equation, and lot more ) would you help me to cover up that. what's the ideal time to find you online, to ask you questions and doubts.

sorry i put "but" in the previous comment by mistake.

@PushkarSoni, in case of any doubts better just post your question here (or at Physics.SE). The community will help much better (and faster) than any of its members individually.

This answer is very clear, which many of the  other sources that cover the topic more thoroughly are not.  Thank you for the clarity and succinctness.


text:
Catalyzation in Carbonic Acid
I understand the basics of carbonic acid + 1 kcal breaking down into water and carbon dioxide.
However, my professor wants me to draw an energy map showing the catalyst in human blood reducing the activation energy from 21 kcal to 12 kcal. I've never run into an energy map nor have we discussed this catalyst in class yet.
Anyone have a resource I can look at?

I suspect that your professor is looking for an energy profile.

https://en.wikipedia.org/wiki/Carbonic_anhydrase

Removed 'super basic' from the title. It was confusing; I thought it was talking about superbases.


text:
What's the simplest way to etch Nickel and Chromium sputtered layers?
I have to etch a very thin film (50 nm) of Nickel ($\ce{Ni}$) on a 100 nm $\ce{SiO2}$/$\ce{Si}$ substrate. Additionally, there is a ~5 nm adhesion layer of $\ce{Cr}$ under the $\ce{Ni}$ that I'd also like to etch.
So I found this page which has a fairly long list of etches for $\ce{Ni}$ and $\ce{Cr}$.
Ni:

1 : 1 : 1 - HNO3 : HAc : Acetone
1 : 1 - HF : HNO3
30% FeCl3
3 : 1 : 5 : 1 - HNO3 : H2SO4 : HAc : H2O 85 C 10 microns/min
3 : 7 - HNO3 : H2O
1 : 1 - HNO3 : HAc
10% g/ml Ce(NH4)2(NO3)6 : H2O
HF, concentrated – slow etchant
H3PO4 – slow etchants
HNO3 – rapid etchant
HF : HNO3 – etch rate determined by ratio, the greater the amount of HF the slower the reaction
4 : 1 - HCl : HNO3 – increase HNO3 concentration increases etch rate
30% FeCl3
5g : 1ml : 150ml - 2NH4NO3.Ce(NO3)3.4(H2O) : HNO3 : H2O – decreasing HNO3 amount increases the etch rate
3 : 3 : 1 : 1 - H3PO4 : HNO3 : CH3COOH : H2O ~15min/micron @ RT with air exposure every 15 seconds

Cr:

2 : 3 : 12  KMnO4 : NaOH : H2O
  3 : 1   H2O : H2O2
  concentrated and dilute     HCl
  3 : 1   HCl : H2O2
  2 : 1   FeCl : HCl
      Cyantek CR-7s (Perchloric based)    7 min/micron (24A/s new)
  1 : 1   HCl : glycerine     12min/micron after depassivation
  1 : 3   [50gNaOH+100mlH2O] : [30g K3Fe(CN)6+100mlH2O]   1hr/micron

I'd really like to avoid HF if I can (it terrifies me!), but I have pretty much everything else at my disposal.
So far, everything I've done has been at room temperature.
First I tried concentrated $\ce{FeCl3}$. No bubbling or signs of anything after ~20min.
Then I tried weak $\ce{H3PO4}$. Nothing.
Today I tried strong (98%) $\ce{H3PO4}$, nothing.
Then I tried concentrated fuming $\ce{HNO3}$. Nothing after ~1hr.
Then I tried about 1:1 $\ce{HNO3}$:$\ce{HCl}$, nothing.
Am I missing something here? Do the solutions need to be heated up or something?

I don't know the exact chemistry, but nickel actually etches better in dilute nitric acid than concentrated. Many commercial etchants are just ~15% nitric acid in water with a bit of surfactant added. Heating typically increases the etch rate, but dilute nitric acid should still give you 3 or 4 nm/s at room temperature.

@MichaelD.M.Dryden, huh, strange. Very counterintuitive to my 8th grade level knowledge of chemistry. I'll try that. Now that you mention it, several of those recipes above call for DI H2O in addition to the acids. I kind of assumed they were actually slowing down the rate, but maybe they're necessary.

Nickel can be etched using diluted (1:1 V) $HCl$. 50 nm will be dissolved within seconds.
The etching solution for the chromium layer contains $\ce{Ce(NH4)2(NO3)6}$+$\ce{CH3COOH}$+$\ce{H2O}$
in ratio 165g : 45 ml (con.) :100 ml DI water. The chromium layer is dissolved within 5 minutes of etching.

Hmm, do you mean at room temperature? Because I'm pretty sure I tried the dilute HCl and found it to not work... Cr is easy to etch in my experience, we have a simple commercial acetic acid based one that works well.

Yes at RT, I did this etching a lot of times. I was etching 200 nm Ni layer by diluted HCl and it worked perfectly well for microstructures preparation  on glass. If HCl does not work, then there is something different than Nickel.


text:
Why do halogen-metal exchanges happen?
I'm trying to understand organometallic raegents but am stuck at halogen-metal exchanges.
For example if you have $\ce{PhBr + BuLi}$ you get $\ce{PhLi + BuBr}$, I don't understand why $\ce{PhBu + BrLi}$ wouldn't form.

Having done this reaction often, I occasionally wonder this myself.

https://en.wikipedia.org/wiki/Transmetalation

Thermodynamics  favors the reaction based on the electronegativities of the metals and kinetics favors the reaction if there are empty orbitals on both metals.

How Did Wittig Come to Discover Lithium-Halogen-Exchange?
Georg Wittig, discoverer of the Wittig reaction, was studying the effect of ring strain on the strength of two neighboring carbons; by adding enough strain he expected to coax C-C bonds into forming diradicals:

Unfortunately tetraphenylbenzocyclobutane rearranges to triphenyldihydroanthrancene instead of forming a diyl. This and other “failures” did not deter Wittig; they elected to continue by replacing phenyl groups with anisyls. To their surprise, reactions with anisyls gave unexpected products. They focused their efforts on a simple derivatization with benzil to characterize the major product. They were trying to synthesize $p$-anisyldiphenylcarbinol, but instead what happened is:

Continuing along they discovered phenyllithium would perform the same reaction. As part of their investigation they varied other substituents on the aryl halide and discovered an even more peculiar product which violated their chemical intuition. Wittig was the first to speculate a lithium-bromide exchange reaction in 1938, below is the reaction he and coworkers carried out in diethyl ether giving a high yield (95%), the preferred product led them to infer it went through an aryllithium intermediate:
This same reaction occured with iodine, bromine, and chlorine. Yet, fluorine gave products via an entirely different mechanism. This led them to postulate an aryne intermediate. They then set off in a new direction, by trying to weaken carbon-hydrogen bonds with a neighboring quaternary ammonium, this led them to discover N-ylides. It was found this more readily occurred with phosphonium, where they discovered P-ylides. Now you can see the rough outline of the path required to discover the Wittig reaction.
How Does Lithium-Halogen Exchange Work?
This initial discovery by Wittig and then Gilman encouraged further efforts because it allowed the formation of a number of previously inaccessible organometallics. In the above reaction, the Grignard reagent cannot simply replace phenyllithiun (some Mg cases have been reported for other halides, but Li is the stalwart metal).
Today this reversible lithium-halogen exchange reaction can be formulated as,
$$\ce{RLi + R’X <=>[{K}]  R’Li + RX}.$$
As an equilibrium process, we know that the more stable lithium species is preferred. Interestingly when $\ce{R’}$ = aryl (particularly bromides and iodides) this is best written as, $$\ce{RLi + R’X <=>> R’Li + RX}.$$
When $\ce{R’} = $ allylic or benzylic, Wurtz coupling is favored. $\ce{R’} = $ alkyl tend to give unsatisfactory results. In general, the equilibrium favors those species which most stabilize the carbonanion intermediate. Depending on the specific halide, side-product formation is a problem: $\beta$-elimination, $\alpha$-metalation, or Wurtz-type reactions can occur.
Gilman discovered that alkyllithiums, for example t-butyllithium, were more adaptable in their reactivity than aryllithiums, which explains their widespread success in the modern chemical literature.
How the reaction proceeds remains an unresolved problem, however, there is evidence to suggest the following mechanisms are at play:
(1) intermediates proceeding through an “ate” complex
Perhaps unsurprising to you by now, Wittig also hypothesized the existence of an "ate" complex intermediate:
This is in other words, a nucleophilic attack of the halide by the alkyllithium with a carbonanion leaving group.
(2) radical mechanisms
A one electron transfer from the nucleophile to halide thereby producing a radical intermediate that then can give rise to a number of products. There are data from EPR and radical caging experiments to suggest radicals at least form.
(3) mixtures of (1) and (2)
As with many systems, the truth is both mechanisms compete with each other; of course which pathway, if any, dominates will depend on the solvent, temperature, substrate, and certainly the lithium species used.
tert-Butyllithium requires 2 equivalents, resulting in a non-equilibrium process where: $$\ce{RX + (Me)3CLi -> RLi + (Me)3CX}$$ $$\ce{(Me)3CX + (Me)3CLi -> (Me)3CH ^ + (Me)2C=CH2 ^ + LiX}.$$

authors: ام البنین بالانج پور


text:
Does helium have a 2s orbital?
Part 1: According to what I've learned so far, helium is in the first row of the periodic table, so its $n$ should always equal $1$. However, a question in my text asks about helium 2s. What is helium 2s? If helium 2s exists, why does the periodic table list helium in the first row instead of the second?
Part 2:
I believe that He 1s is more stable than He 2s. I've reasoned thus:

The numerals '1' and '2' in 'He 1s' and 'He 2s' express the size of the orbital. The greater number represents a larger orbital.
The electrons of smaller orbitals exist closer to the nucleus than the electrons of larger orbitals do.
The intensity of the attractive force between electron and protons increases the nearer the protons are to the electron
Therefore, the attractive force between the protons and the electron will be greater in He 1s than in He 2s.
The greater the attractive force between the protons and the electron, the greater the stability of the electron.
Therefore the electrons in He 1s are more stable than the electrons in He 2s.

Is that correct?

Note the following is the original question:

For each pair of orbitals determine which is more stable and explain why:
(a) He 1s and He 2s;
(b) Kr 5p and Kr 5s;
(c) He 2s and He+ 2s

Part 1: The elements on the periodic table are arranged according to their ground-state electron configurations. He 2s is an excited state of He 1s. There are even He 3s and 4s depending on whether one gives enough energy for the atom to excite to such levels. In general, higher principal quantum number (n) indicates a higher energy state (less stable)
Part 2: Your reasoning is correct.

Part 1: The 2s is referring to the orbitals that are occupied by electrons. In the ground state the electrons in He are both in 1s orbitals, however it is possible to excite the electrons from the 1s orbital to a higher orbital like 2s. All orbitals are accessible to the electrons of an atom through excitation.
Part 2: The reasoning you are using is appropriate to the question you are answering.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Quick and simple explanation of molar mass, molecular mass and atomic mass
As a physics student, I hardly deal with such quantities and when I do, I blunder through.
I think it’s a good time to be given a good explanation for these since I’m in a class of Nuclear physics. I have tried googling but it seems that some sources are ‘misleading’ and most have more than 1 terminology for the same thing!
Horrifying but real.
Please tell me what is the difference between molar mass, molecular mass and atomic mass and what are their associated twin terminology.

The quantity molar mass (quantity symbol: $M$) for a pure sample is defined as
$$M = m/n$$
where $m$ is mass and $n$ is amount of substance.
The dimension of the molar mass is
$$\dim M = \mathsf{M}\;\mathsf{N}^{-1}$$
The coherent SI unit for molar mass is ‘kilogram per mole’ (unit symbol: $\mathrm{kg/mol}$).
(The usually used unit is gram per mole ($\mathrm{g/mol}$) rather than kilogram per mole ($\mathrm{kg/mol}$).)
The quantity relative molecular mass (quantity symbol: $M_\mathrm{r}$) is defined as the ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative molecular mass is the ratio of the mass of a molecule to the unified atomic mass constant:
$$M_\mathrm{r} = m_\mathrm{f}/m_\mathrm{u}$$
where $m_\mathrm{f}$ is mass of entity (molecule or formula unit) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative molecular mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim M_\mathrm{r} = 1$$
The coherent SI unit for relative molecular mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘molecular weight’ for the quantity relative molecular mass. The use of this traditional name is deprecated.
For atoms, the relative molecular mass $M_\mathrm{r}$ is the relative atomic mass, and the quantity symbol $A_\mathrm{r}$ may be used. The quantity relative atomic mass is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$:
$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$
where $m_\mathrm{a}$ is the atomic mass (see below) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative atomic mass is a quantity of dimension one:
$$\dim A_\mathrm{r} = 1$$
The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.
The quantity atomic mass (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.
The dimension of the atomic mass is
$$\dim m_\mathrm{a} = \mathsf{M}$$
The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).

I've not had the time to review this, but I'm positive you know that out of the box: Did any of this change with the 2019 redefinition?

I've been studing this last week because I'm on the 10th grade. Siply the atomic mass is the mass of the atom which is "n" times bigger than standard mass (1/12 of the mass of carbon-12). For example, oxigen, the atomic mass of oxigen is 16 because is 16 times bigger than 1/12 of the mass of carbon-12, and 1/12 of 12 is 1.
Molecular mass is the sum of the atomic mass of the atoms in the molecule.
Finally, molar mass is the mass of 1 mol of particles which has the same number of relative mass.
I hope I could help someone with this.


text:
Are electron shells probability distributions?
I know that orbitals are probability distributions. Are electron shells probability distributions too?

In order to understand you should know the differences between shell, sub-shell and orbital.

Shell represents the main energy level occupied by the electron. It is given by the principal quantum number (n). For example , if n=1, the electron is present in the first main shell, called K-shell. if n=2, the electron is present in the second main shell, called L-shell and so on..
Sub-shell represents the sub-energy level occupied by the electron (as main energy level is considered to consist of a number of energy sub-levels ). it is given by azimuthal quantum number. subshell corresponding to l=0,1,2,3 are represented by s,p,d,and respectively.
Orbitals you know it well.
You may understand it like this-

when n=1 we have only s subshell
when n=2 we have s and p subshell
for n=2 we have s,p and d subshell and so on.

Now, s, p and d and f have certain regions, which we call orbitals. which are probability regions as you know. Here is the link for you,

http://en.wikipedia.org/wiki/Atomic_orbital
Must watch the video associated with it named [atomic orbitals and periodic construction]

Last but not the least- we certainly not have electron shells probability distribution. its only orbitals. shells are created by scientists to study the atom systematically but they have physical significance too (i mean here the subshells and orbitals lie).
it seems you are not clear with - shell, sub-shell and orbitals.
you should first check out what they actually means in order to understand.

I'd suggest adding some discussion of the [radial distribution function](http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Quantum_States_of_Atoms_and_Molecules/8._The_Hydrogen_Atom/The_Wavefunctions to discuss how the  orbital probability lines up with the older concept of a "shell."

Short Answer:
No.
Long Answer:
First, strictly speaking, the orbitals themselves in the quantum mechanical sense are not probability distributions.  They are eigenfunctions $\Psi_i$ of the Hamiltonian as defined by the time-independent Schroedinger equation $H\Psi_i=E_i\Psi_i$.
The probability distribution function $p(\vec r)$ for electrons is generated by summing the products of the eigenfunctions and their complex conjugates for all occupied eigenfunctions at a given 3-D point $\vec r$ and dividing by the total number $N$ of electrons in the system:
$$
p(\vec r) = \frac{1}{N}\sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}
$$
Note that $p(\vec r)$ is simply $\frac{1}{N}$ times the electron density function $\rho(\vec r) = \sum_{occ}{\Psi_i^*(\vec r)\Psi_i(\vec r)}$.
To be fair, prior to a detailed introduction to the quantum mechanical description of electronic structure, the distinction between the electronic wavefunction $\Psi(\vec r)$ and the electron density function $\rho(\vec r)$ is typically not clearly established in chemistry instruction, so your having conflated them is quite understandable.
Second, again strictly speaking, electron shells are not probability functions.  While Pushkar Soni is correct that the principal quantum number for a given electron approximately indicates the 'electron shell' to which it is assigned, the indistinguishability of electrons (among other factors) in general prohibits unambiguous translation of quantum number information into the real-space function $\rho(\vec r)$.  The best we can do is to interpret electron shells as corresponding to certain features of the structure of $\rho(\vec r)$. Offhand, I know of at least two ways of qualitatively identifying 'electron shells' from $\rho(\vec r)$.
First, as Geoff Hutchinson notes, the radial distribution function $r^2\rho(r)$ shows local maxima that can be interpreted as electron shells.  Note that  $\rho(r)$ in the r.d.f here is a function only of $r$, the distance from an isolated atom/nucleus; the angular dependencies (on $\theta$ and $\varphi$) have been integrated out.
Second, Bader's QTAIM describes how the Laplacian of the electron density, $\nabla^2\rho(\vec r)$, exhibits minima in 3-D space at positions corresponding to the notion of electron shells.

Good answer. But could you please correct the second sentece in the Long Answer: $\Psi_{i}$ is an eigenfunction of the Hamiltonian. It is also a solution of the time-independent Schrödinger equation, but its not an eigenfunction Schrödinger equation since it is just a nonsense. ;)

@Wildcat Very good point.  Looks like someone else beat me to the edit, though. :-)

Electron shell is well described into the old quantum mechanics developed by Bohr, see Atom Bohr model
Probability distribution is well described into the modern quantum chemistry, see Quantum mechanics, general description

Therefore, both definitions are not comparable among them because each one belongs to different theory conceptions.

-1 Uh no. If you take the radial distribution function $r^2 R^2(r)$, you find peaks for each radial wavefunction that correspond somewhat reasonably to the "shell" concept.

Even more directly, Bader's QTAIM, which is built off of the electron density $\rho$ -- which is precisely the probability distribution of electrons in a system -- recovers the 'electron shell' phenomenon via the negative Laplacian of the electron density, $-\nabla^2\rho$.


text:
How to find pH by mixing two solutions of different concentrations?

A mixture is made by combining 110 mL of 0.15 M $\ce{HCl}$ and 215 mL of 0.055 M $\ce{HI}$.  What is the pH of the solution?

$$\ce{HCl -> H+ + Cl-}$$
$$\ce{HI -> H+ + I-}$$
According to what I thought, the molarity of $\ce{H+}$ is the same as $\ce{HCl}$, because it is a strong acid and the mole ratio.  So, I added the molarity of both acids: $\pu{0.15M} + \pu{0.055M} = \pu{0.205M}$ for $\ce{H+}$ in solution.
$$\mathrm{pH} = -\log[\ce{H+}]$$
$$\mathrm{pH} = -\log[0.205]$$
$$\mathrm{pH} = 0.689$$

You are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. here.
Therefore you can assume, that the concentration of protons is equal to the concentration of the acid.
\begin{align}
c_{\ce{HI}}(\ce{H+}) &=  c(\ce{HI}) &
c_{\ce{HCl}}(\ce{H+}) &=  c(\ce{HCl}) &
\end{align}
Now you need to determine the amount of the protons in each solution with $$n=c\cdot V.$$
You have to add the amount of protons in each solution and determine the combined volume.
\begin{align}
n_\mathrm{tot}(\ce{H+}) &= n_{\ce{HI}}(\ce{H+}) + n_{\ce{HCl}}(\ce{H+})&
V_\mathrm{tot} &= V(\ce{HI}) + V(\ce{HCl})
\end{align}
To calculate the pH of the resulting concentration, just take the negative logarithm.
$$\mathrm{pH} = -\log c_\mathrm{tot}(\ce{H+}) = -\log\left[\frac{n_\mathrm{tot}(\ce{H+})}{V_\mathrm{tot}}\right]$$

 Your solution should be $\mathrm{pH}=1.06$, rounded to three significant figures.

The approach you chose can only work, if you have equal volumes of the two solutions.

$$\begin{align}
M_1V_1+M_2V_2&=M\left(V_1+V_2\right)\\
0.015\times110+0.055\times215&=M\left(110+215\right)\\
1.65+11.825&=M\cdot325\\
M&=\frac{13.475}{325}\\
M&=0.04146=4.416\times10^{-2}\\
\mathrm{pH}&=-\log\left(4.416\times10^{-2}\right)\\
\mathrm{pH}&=-\log4.416+2\log10\\
\mathrm{pH}&=-0.617+2\\
\mathrm{pH}&=1.383
\end{align}$$
You can go this way also
$$\begin{align}
\text{HCl conc}&=0.015\ \mathrm M\\
&=0.015\times\frac{110}{1000}\ \mathrm{mol}\\
&=1.65\times10^{-3}\ \mathrm{mol}
\end{align}$$
$$\begin{align}
\text{HI conc}&=0.055\ \mathrm M\\
&=0.055\times\frac{215}{1000}\\
&=11.825\times10^{-3}\ \mathrm{mol}
\end{align}$$
$$\begin{align}
\text{Total}\ [\ce{H+}]&=\left(1.65\times10^{-3}+11.825\times10^{-3}\right)\ \mathrm{mol}\ \text{in}\ 325\ \mathrm{ml}\\
&=13.475\times10^{-3}\ \mathrm{mol}\ \text{in}\ 325\ \mathrm{ml}\\
\text{so in}\ 1000\ \mathrm{ml}&=13.475\times10^{-3}\times\frac{1000}{325}\\
&=4.146\times10^{-2}\\
\text{Thus}\ \mathrm{pH}&=1.383
\end{align}$$

Ravi, the question says 0.15 M HCl, not 0.015M


text:
The Health Effects of Alkaline Ionized Water?
I came across alkaline ionized water as my aunts bought a system called Kangen Water from Enagic Inc. The water produced by this system is claimed to contain smaller water clusters (4-6 molecules) compared to tap water with 10-12 molecules. Besides, the ORP (Oxidation Reduction Potential) of the water is negative (-250 to -350 mV), which the company claimed to battle ROS (Reactive Oxygen Species). Of course, the company also advertised the water as having other incredible properties such as emulsifying grease, thereby can be used to remove pesticides on fruits and vegetables.
So, let alone the validity of the claim on benefits, my questions are:

How can one produce negative-ORP water? Is it possible to do so with added minerals? (Just the only way I can think of, is there any other way?)
What chemical/physical process is happening in the emulsification demo? Why can't tap water do so?

Please refrain from discrediting the company too quickly as I think I might have and please tell me if I formulated my question clearly enough because I acknowledge this is a fairly vague question.

I can't say for certain what the Kangen device is doing, but students in my department took apart one of these from another company and found an ion exchange resin that was swapping the $\ce{Cl-}$ ions in tap water for $\ce{OH-}$ ions. Does it make it more alkaline? Yes. Does it make it have different redox properties? Probably. Is it more likely to make you feel better by calming the acid in your stomach than anything else? Yes. Did it clean up the dirt and grease that it was supposed to? No.

Do you know why the redox properties may become different when $\ce{Cl-}$ is swapped out for $\ce{OH-}$? I did the grease removing test on this device. The Kangen water didn't emulsify canola oil, which, though not pure sesame oil as in Enagic's demo, led me to be skeptical of the grease removing ability. Heck, they even claimed it would replace your 409! We paid $4000+ on it. I'd say we mostly paid for the ion-exchange column. That's what happened when you don't ask a chemist before buying.

Consider the reduction of oxygen in water with and without acidic conditions (data from here: $$\ce{O2(g) + 4H+ +4e- -> 2H2O(l)}\ \ \ \ E^{\circ} =+1.229\text{ V}$$ $$\ce{O2(g) + 2H2O (l) +4e- -> 4OH- (aq)}\ \ \ \ E^{\circ} =+0.401\text{ V}$$


text:
Why doesn't EDTA complex with alkali metal ions?
EDTA complexes with all the other metal ions in the periodic table except those from the group 1. Why is this so ? What is the coordination chemistry behind this ?

That's funny, the label on my bottle says Ethylenediaminetetraacetic Acid, Disodium Salt.

@LDC3: That is $\ce{Na2EDTA.H2O}$. In that compound, $\ce{Na+}$ ions are not complexed to EDTA.

@chemkatku Can you explain how you differentiate between "complexed?" I guess your argument against Na is that it's ionic and not a coordination bond?

It does. I’m willing to bet my life it does. Only that the energy gained from complexation is so little and the process so easily reversible that there is no inert/favoured complex form that could be observed easily.

@GeoffHutchison Yes that is what I think. I don't know about any way of  complex formation without formation of coordination bonds.

Alkali metal cations are very "hard", in the HSAB theory of acid-base interactions (roughly corresponding to 'poorly polarizable'), and thus only coordinate in the conventional sense of the term to "hard" ligands such as ethers.  Crown ethers, in particular, are known for their alkali metal ligating/chelating capability.
The amine nitrogens and carboxyl oxygens of EDTA are sufficiently 'soft' that, as Jan noted in a comment to the question, there is sufficiently little energetic benefit derived from EDTA chelation of a 'very hard' alkali cation such that the complexes are not practically observed.  Instead, the chemistry is dominated by electrostatic interactions of the positively-charged alkali cations with the negatively-charged carboxylate groups (when deprotonated, anyways).

If you're going to be picky that a "complex" doesn't form without covalent coordination bonds, then that's your answer. Alkali metals will tend to form cationic salts with EDTA.
Personally, I agree with the other comments. EDTA does form complexes with alkali metals.
I'd classify the disodium salt of EDTA as a complex. It has a metal and an organic ligand, and there's no direct metal-carbon bond (which would make it organometallic). As discussed in another question, it's a pretty arbitrary distinction between an ionic bond and a covalent one:

What happens if the electronegativity difference is exactly 2.1?

This distinction is particularly cloudy in the case of metal complexes, since most metals readily gain positive charge.

I think crown ethers and cryptands provide unquestionable examples of alkali metal ion coordination? If those do, then that establishes a scale of coordination strength which never truly hits zero.

@NicolauSakerNeto This is my opinion as well. My feeling is that there's a metal, a ligand, and they're close enough for some measure of covalent bonding. All bets are off if the ions are far apart, but in EDTA, it's at least partially covalent.

@NicolauSakerNeto After looking closely at some $\ce{Cu(II)}$ aquo complexes, I have to agree with you, Geoff. Copper indisputably forms complexes, but its interactions with water ligands appear almost purely ionic. In my computation of $\ce{[Cu(H2O)5]^{2+}}$, the line critical point joining the axial $\ce O$ with the $\ce{Cu}$ has $\rho=0.05$ and $\nabla^2\rho=+0.26$, and all of the equatorial LCPs have $\rho\approx 0.07-0.08$ and $\nabla^2\rho \approx +0.4$. Thus, ionic EDTA-$\ce{Na+}$ interactions could also be called coordination.

$\ce{Na2EDTA.H2O}$:
First of all the EDTA sodium salt is an ionic compound with complexed water. A sodium ion forms an ionic salt with the carboxylate group in two of the four acetic acid molecules bound to the amine groups. The salt has one water of crystallisation.
Solution Chemistry:
 I'd say the high ionic radius and hardness of group I metal ions against the soft carboxylate make chelation less favourable with respect to softer metal ions, such as group III metals and transition metals.
Affinity constants:
\begin{array}{lr}\hline
\ce{Al (III)} &  16.13\\
\ce{Ca (II)}  &  10.96\\
\ce{Mg (II)}  &   8.69\\
\ce{Pb (II)}  &  18.04\\\hline
\ce{Li+}      &   2.79\\
\ce{Na+}      &   1.66\\
\ce{Cs+}      &   0.15\\\hline
\ce{Fe (II)}  &  14.33\\
\ce{Fe (III)} &  25.10\\
\ce{Mn (II)}  &  14.04\\
\ce{Mn (III)} &  24.80\\\hline
\end{array}


text:
Why is in the reaction of cyclic unsaturated aldehydes with hydrazine and hydrogen peroxide the aldehyde preserved?

Question
  What is the product of this reaction?

The answer given:

At first, I thought this was a typical Wolff-Kishner reduction, but then I saw the reagents didn't match up (no $\ce{OH-}$).
How is this getting formed? Shouldn't the $\ce{CHO}$ group be reduced? What am I missing here?

This reaction worries me: mostly because hydrazine is known to be explosive (via oxidation to $\ce{N2(g)}$) in the presence of strong oxidants and hydrogen peroxide is not the most stable molecule either: $$\ce{H2N2 + H2O2 -> N2 + 2H2O}$$

http://en.wikipedia.org/wiki/Reductions_with_diimide However, peroxide and hydrazine is not listed as a common method of preparing diimide in situ.

Sanofi have demonstrated this reduction method at large scale on artemisinic acid and it's not without its calorimetric challenges.

The reaction of $\ce{H2O2}$ with hydrazine forms diazene, $\ce{HN=NH}$.  This reacts in a syn fashion to deliver both hydrogens to the same side of the alkene.

Thanks, I can relate this question to our course now - diimide indeed selectively reduces C-C double bonds in preference to polar bonds.

Actually it's a di-imide which itself gets oxidised to give up nascent hydrogen and reduce the other reactant.
Thus, when you react it with an alkene , hydrogen add on the same side i.e. syn addition. Unlike $\ce{N2H4/KOH}$, it is not strong enough to reduce carbonyl groups. Therefore you get a different product.


text:
In what sense is the Inductive effect a permanent effect?
Almost everywhere I checked, the Inductive effect is described as a permanent effect with almost no mention of what that actually means.

What exactly is permanent about the effect?
If the effect is permanent, does that mean that even if the attached electron
withdrawing/donating group is removed, the induced effect still continues to
exist?

The effect is permanent because the group that does the 'induction' is always there in the molecule. For example, if the induction is caused by a t-butyl group, that group is held by a strong carbon to carbon ($\ce{C-C}$) bond that will not easily be broken. One could also consider the inductive effect of the Fluorine held by a very strong $\ce{C-F}$ bond.
A temporary effect might be caused by protontation of an oxygen atom where the protonation can catalyse a reaction. The protonation is reversible. The bond between the proton and oxygen would be much weaker, as $\ce{C=O^+-H}$ will be in equilibrium with $\ce{C=O  +  H^+}$.
Removal of the group that does the induction will remove the inductive effect.

Doesn't the same happen in Electromeric Effect? I mean, for the time that the reagent is present, the pi-bond is polarized. On its removal, the molecule returns to its original configuration. But then we say that Electromeric effect is temporary. What am I missing?

@Technotoes - might be worth asking another question about that... wikipedia I think summarizes it neatly by saying "This effect is shown by those compounds containing multiple bonds. When a double bond or triple bond is exposed to an attack by a reagent, a pair of bonding electrons involved in the pi bond is transferred completely from one atom to another. This effect will remain as long as the attacking reagent is present. As soon as the reagent is removed, the polarized molecule will come back to the original state." - so because the attacking reagent causes it the effect is 'temporary'


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Ether cleavage with conc HBr
I've been asked to draw the mechanism by which benzene with $\ce{Me}$ on position 1 and $\ce{OMe}$ on position 2 goes to benzene with $\ce{Me}$ on 1 and $\ce{OH}$ on 2 by reaction with concentrated $\ce{HBr}$.
I'm really unsure as to what is going on, I think its probably most likely that $\ce{Br}$ acts as a nucleophile and attacks the $\ce{C-O}$ $\sigma^*$ and then $\ce{OH}$ attacks the $\ce{C-Br}$ $\sigma^*$ but it doesn't seem like a likely pathway as $\ce{OMe}$ has a higher pKaH than $\ce{Br}$ and substitutions don't occur at $\mathrm{sp^2}$ centres.

Cleavage of ethers using strong acids (classically $\ce{HI}$ but $\ce{HBr}$ should work as well) can proceed by either an $S_N1$ or $S_N2$ mechanism. In this case the cation formed would be primary or phenyl, neither of which are stable, and so the reaction proceeds via an $S_N2$ mechanism.

The oxygen is protonated in an equilibrium reaction. The bromide ion then attacks at the least hindered carbon and the $\ce{C-O}$ bond breaks in a concerted $S_N2$ mechanism.
(source)


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How to distinguish open-pored polymer foam and foam with closed pores?
I was looking at two foam materials, where I had expected one to have closed pores.
Unfourtunately, the microscope was showing the upper layer of cut cells the same in both cases with no indication of differences in lower levels of the material.
By now, I suspect both are open-pored;
Is the difference clearly defined?

If a block of soft material can sustand high pressure based on inner air pressure in cells, the cells are clearly closed.
If a visual inspection shows mostly incomplete cells, such that parts of cells behind them are visible, it's clearly open pored.
what about holding gas only at low outer forces?
Which is similar to the common kitchen sponge, keeping fluid with low pressure, releasing it at high pressure. But here, I expect the pores to be open, and capularry forces to get the water inside.

So, I'd like to know how to tell them appart based on actual mechanical features, and to understand the difference in case they are not trivial.

A standard procedure for the determination of the volume percentage of open and of closed cells is described in ISO 4590 and corresponding national standards (e.g. DIN EN ISO 4590).

Interesting - does that define a threshold for open fraction in closed pores, or partially open foams?


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When a carboxylic acid acts as a nucleoplile, why does it attack through the carbonyl bond?
Here's a couple of examples of what I mean. Firstly, in the reaction with thionyl chloride to make a carboxylic acid:

Secondly, in this specific example of a hydroboration reaction:

I don't really understand why you cant attack from the hydroxide group and then proton transfer. Any suggestions?

Can you draw a resonance structure which localizes -ve charge on the hydroxide ?

You could cleave the bond to the acidic proton though

Yes, but are either of these reactions taking place under basic conditions ? The first one has HCl as a byproduct...

For both the examples that you mentioned, the carbonyl oxygen has its lone pair localized while the other oxygen's lone pair is delocalized with the double bond. The better nucleophile is the one which has greater electron density.

But what is attacking is not the lone pair on the carbonyl oxygen but the pi bond electrons. Your reasoning is therefore not correct.

@TanYongBoon That's not correct; seems mechanism was wasn't completely precise in question, if you thought so.


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Why does the quantum number n limit the value of l, and why does l, likewise, limit the value of m?
Any quantum number that represents the size of an orbital is a positive integer, $n$.
Any quantum number that represents the shape of an orbital is a non-negative integer, $l : l < n$.
Any quantum number that represents the orientation of an orbital, $m$, is an integer such that $|m| \leq l$.
Why do $n$ and $l$ restrict the ranges of $l$ and $m$, respectively?
To be clearer, I ask why the quantum numbers limit each other, not what they signify, which is already explained in the related question: What do the quantum numbers actually signify?

Related: http://chemistry.stackexchange.com/questions/31639/what-do-the-quantum-numbers-actually-signify/31641#31641

@LordStryker, that question explains how to interpret quantum numbers, but does not explain why the ranges of allowable values are what they are. Why for example can $l=0,1,...,n-1$ but never $l=n$?

@BenNorris Because $l=n$ simply isn't a solution to the Schrodinger equation for hydrogen. I know you know this but for completeness, simply look at the periodic table.  When $n=1$, that means that there can only be one subshell (i.e. orientation) of the $s$ orbital (which there is).  Therefore $l=0$.  When $n=2$, then there are two subshells, $l=0,1$ which corresponds to $s$ and $p$ orbitals.  Furthermore, $m_l=-1,0,1$ since there are 3 orientations of $p$ orbitals when $n=2$ and $l=1$.

Extremely instructive walktrough: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydsch.html

While it may seem that quantum numbers were invented out of the blue, that isn’t the case. They derive directly from solving the Schrödinger equation:
$$\hat{H}\Psi = E \Psi$$
One of the simplest Schrödinger equations is that for a particle in a box — the one most quantum chemistry or quantum mechanics courses start with and therefore found in every textbook for quantum chemistry which is why I won’t repeat it here. But in a nutshell: You need a function that you can differentiate twice and which will then reproduce itself with a constant factor E. That can be achieved by sine waves. You cannot say a priori which frequency the sine function will have, though: If the box has size l, it works for sine waves with the wavelengths $2l,\ l,\ \frac{2l}{3},\ \frac{l}{2},\ \dots$ and the corresponding wavenumbers $\frac{1}{2l},\ \frac{1}{l},\ \frac{3}{2l},\ \frac{2}{l},\ \dots$ or $\frac{n}{2l}$ with n being a natural number. n is therefore the simplest case of a quatum number.
If you solve this equation for an electron orbiting a proton, the solutions are much more complicated (and there are intermediate steps often presented in the quantum chemistry courses to make the transition easier). However, n still turns up as a variable, corresponding to quanted amounts of energy. For certain values of n, there are additional variables whose values cannot be derived a priori but which can only assume specific values — they correspond to the other quantum numbers.
This in itself is just a solution to a mathematical equation. The real beauty lies in the fact that these quantum numbers can be used to explain experimental observations such as hydrogen’s flame colours. All quantum numbers can be used to explain some property one has deduced by experiments.

So tl;dr: It’s just what maths tells us will be the case.

The shapes of the orbitals, i.e. the solutions to the Schrödinger equation for the hydrogen atom $\hat{H}\Psi_i = E_i\Psi_i $, are primarily due to the requirement that all orbitals are orthogonal: $\langle\Psi_i|\Psi_j\rangle=0$.  Orthogonality is made possible by the fact that the orbitals have "positive" and "negative" regions separated by nodes.
One what to think about $n$ is that $n-1$ is the number of nodes in $\Psi_i$. So there can be no $1p$ orbitals since $p$-shaped orbitals have at least one node. Similarly, there can be no $2d$ orbitals, and so on.
Because $s$ orbitals have a spherical shape, there is only one way to orient them, i.e. $m=0$ if $l=0$. $p$-shaped orbitals lie along one axis and there are three Cartesian axes, so there are three possible - orthogonal - orientations (I'm sacrificing a bit of accuracy here for clearness).
It's a bit harder to visualize why there should be 5 and only 5 $d$-shaped orbitals, but the answer is that there are only 5 ways to construct 3-dimensional shapes with (a minimum of) 2 nodes that are orthogonal to the $s$- and $p$-shaped solution as well as the other $d$-shaped solutions.


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Can one figure out an optical absorption spectrum from EPR data?
I'm wondering if it's worthwhile for me as an optical spectroscopist to read up a bit on EPR. Do EPR signals reveal where, in terms of wavelength, features in the optical spectrum can be expected to appear? I know it tells me
lifetimes of states, but that's about as far as I have dug into it. I have some EPR data on a substance and I wish to figure out where to look for it's absorption with UV/Vis/NIR means.

To my knowledge, EPR signals cannot generally be used to predict UV-Vis-NIR peak positions.

No, it shouldn't be possible, based on where these spectroscopies lie in the electromagnetic spectrum:

The most common EPR experiments occur at 9.5 GHz (X band), which is in the microwave region. Even the highest-frequency experiments just barely break into the far infrared. Assuming you're doing an X band experiment, and that your system has a transition very close to the free electron $g$-values, 9.5 GHz corresponds to ~0.3 cm$^{-1}$.
For UV-vis, taking the longest wavelength of light in the visible region as 800 nm gives 12,500 cm$^{-1}$. This can be interpreted in two ways:

The EPR experiment needs ~42,000x more energy to probe even the lowest possible electronically-excited state.
Even the UV-vis transitions with the lowest possible energy will not show any fine structure, like in the sodium lines or benzene vibronic coupling. The energy scales need to be much closer. In the case of the spin-orbit-induced splitting of the sodium spectral lines, the unperturbed gap is 2.1 eV and gap between the split states is 0.0021 eV; this is 3 orders of magnitude between the energies. In the case of the benzene UV spectrum, the gap between peaks is ~10 nm, with the peak center at ~245 nm; this is 1-2 orders of magnitude:

$\hspace{20ex}$
Our EPR/UV-vis example has 5 orders of magnitude difference between the two energy scales. There is (hyper)fine structure in EPR:

which here, for DPPH in toluene is ~0.002 Tesla, or 0.35 GHz, so 1 order of magnitude.


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I recently learned I'm allergic to potassium dicromate or chromium
are eye drops with inactive ingredients called potassium chloride or sodium chloride the same as potassium dichromate and should I stop using those eye drops?

Welcome to the site! Right now I can divide your question into two parts: "are eye drops with inactive ingredients called potassium chloride or sodium chloride the same as potassium dichromate?" This part is unfortunately unclear. The same, as in what? Of course dichromate ion isn't the same as chloride, so...This leaves the question as a very puzzling one. The other part is "should I stop using those eye drops?" This part is unfortunately off-topic here. We don't answer questions that seek medical advice.

Allergic reactions against dichromates are not uncommon.
However, dichromates and chlorides, are two pairs of shoes. Really different ones, like stiletto heels and moon boots ;-)
To my knowledge, there are no eye drops containing potassium dichromate!
The sodium chloride that you mentioned is nothing but table salt. Isotonic solutions of sodium chloride are very common in all sorts of medical applications, like rinsing of lenses, intravenous infusions, and so on. To my knowledge, allergic reactions on these solutions are not known.
To sum it up: I am not a medical doctor, and you might want to contact your ophthalmologist, but as a chemist with some experience in medical chemistry, I can assure that

dichromates and chlorides are not the same
allergic reactions to physiological saline are not known


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IR fluorescent material for outdoor use
I am experimenting with IR video cameras. They have a broad sensitivity peak at 940 nm. There is a filter blocking 750 nm and shorter in front of the sensor.
I am looking for an IR fluorescent substance to use for markers in the scene the camera is observing. Something that can be turned on with ultraviolet illumination bursts so I can differentiate between my markers and other IR sources.
My hope is for some affordable and harmless substance. Something that could be mixed with polyester, acrylic or epoxy potting compound wold be nice.
An already solid watherproof material is even better.
My searches so far have found some IR fluorescent stains for microscopy but I am afraid they are expensive.
This might be a physics question also but I think chemists works a lot with IR absorption and IR fluorescence.

A commercial solution would be a dye like IRF940A (from QCR Solutions) with an excitation maximum at $\lambda_{\mathrm{exc}} = 405\,\mathrm{nm}$ and an  emission maximum at $\lambda_{\mathrm{em}} = 940\,\mathrm{nm}$. The supplier does not provide any information on the structure of the dye.
What compounds are promising candidates?
Ytterbium(III) complexes!
In a recent article,published in J. Photochem. Photobiol. A, 2015, *309, 65-71 (DOI), J. Legendziewicz and coworkers prepared a $\ce{Yb(III)}$ phthalocyanine with acetate as a chelating axial ligand and examined the optical properties in solution and in PMMA polymer. In PMMA, the compound shows a Soret band with a maximum at $\lambda_{\mathrm{abs}} = 350\,\mathrm{nm}$. Excitation in the Soret band result in emission with a maximum around around $\lambda_{\mathrm{em}} = 700\,\mathrm{nm}$.
Other examples for even larger wavelength differences, resulting from ligand-to-metal energy transfer are known  for $\ce{Yb(III)}$ complexes bound to eosines or fluoresceines as the primary chromophor.


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How to determine at temperature and pressure to keep Cl2 a gas?
I have chlorine tanks and they will be used industrially. When I use them, they will be exposed to all sorts of varying temperatures. I was wondering if there was any way that when I use them, I could figure out at what pressure I could release the $\ce{Cl2}$, depending on the temperature, so that it remains a gas?
I already know it only takes $\pu{6.8 atm}$ at $\pu{70^\circ C}$ to liquefy $\ce{Cl2}$ to a volume of $\pu{0.322812 L}$, and its boiling point is $\pu{-33.97 ^\circ C}$ at $\pu{1 atm}$.

http://hydro-land.com/e/ligne-en/doc/gaz-Chlore.html

What you need to know is the vapor pressure of $\ce{Cl2}$ as a function of temperature.  Just ensure that the release pressure of the chlorine is lower than the vapor pressure, and it will be a vapor.
I found one web site that gives a graph (and a literature reference) of $\ce{Cl2}$ vapor pressure as a function of temperature from -30 °F to 220 °F.  The references they give are:

Reprinted from Chlorine Manual, the Chlorine Institute, 3rd ed. 1959. Data for graph from R. Kapoor and J.J. Martin, Thermodynamic Properties of Chlorine, Eng. Res. Inst. Univ. of Michigan, 1957.

At -30 °F they report $\ce{Cl2}$ vapor pressure is "0" psi (which has to be an approximation; the vapor pressure of all materials is non-zero at any non-zero temperature) and at 220 °F it is 587 psi.


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How come uranium's relative atomic mass is 238.03 when it only contains isotopes with a mass number of 238 or less?
I have always been led to understand that the mass of an element on the periodic table is the weighted average atomic mass over all naturally occurring isotopes. This seems to make sense with all the elements I have looked at except for uranium. Wikipedia's featured article on the element assures me that uranium's relative atomic mass is 238.03. The commonest isotope is U-238, mixed with 0.72% of U-235 and 0.005% of U-234, plus traces of a couple of isotopes with smaller mass numbers.
How then can uranium's relative atomic mass exceed 238? Either I have completely misunderstood something, or there is something else at work - relativistic maybe?

(From my deleted answer) There also are isotopes of uranium with greater masses than 238 - though their contribution would be tiny.

@Santiago  I'm not sure they would contribute.  They all have half-lives less than an hour and are not part of natural abundance uranium.

Any uranium sample that contains U-238 (including natural minerals) inevitably contains a very tiny amount of U-239 due to (n,γ) reactions. However, that has no measurable impact on the average atomic mass of uranium.

Yes, I am getting that point now

Approximately 99.3% of uranium on Earth is the $\mathrm{^{238}U}$ isotope, and this specific isotope has an atomic mass of $\mathrm{238.05\ u}$, where $\mathrm{u}$ is the atomic mass unit, equivalent to 1/12 the mass of a $\mathrm{^{12}C}$ atom. Including the other isotopes to obtain the average atomic mass drags the value down a little, but it still ends up being slightly larger than $\mathrm{238\ u}$.
It is interesting to point out why the atomic mass is slightly different from the mass number. The masses of a free proton or neutron are both slightly larger than $\mathrm{1\ u}$ ($\mathrm{1.0073\ u}$ and $\mathrm{1.0087\ u}$, respectively). Some of this mass is converted to energy and lost when neutrons and protons (collectively called nucleons) bind to form nuclei, which means bound nuclei will always be lighter than the sum of the masses of their component nucleons. Different nuclei release different amounts of energy per nucleon depending on how many protons and neutrons are inside the nucleus, and on their relative proportions. Very interestingly, the curve for nuclear binding energy per nucleon versus mass number starts off at zero for $\mathrm{^{1}H}$ (there can be no attractive nuclear forces between nucleons if there's only one nucleon!), then rises until it shows a maximum around nickel (specifically for the isotope $\mathrm{^{62}Ni}$) and from there on slowly decreases.
Since the atomic mass unit is standardized with respect to the $\mathrm{^{12}C}$ isotope of carbon, then in approximate terms, the nuclei with a lower binding energy per nucleon than $\mathrm{^{12}C}$ will have less of the initial free proton/neutron mass lost to energy (and thus an isotope will have an atomic mass slightly higher than its mass number), while for nuclei with a higher binding energy per nucleon than $\mathrm{^{12}C}$, more mass will be lost to energy and the resulting isotope will have an atomic mass slightly lower than its mass number.
Taking a horizontal line through the $\mathrm{^{12}C}$ isotope in this graph and looking at its intersections with the nuclear binding energy curve will show the regions where this flip happens; isotopes for atoms below carbon will tend to have higher atomic mass than mass number, isotopes for atoms between carbon and thorium tend to have a lower atomic mass than mass number, and then isotopes for atoms above thorium (such as $\mathrm{^{238}U}$) will again tend to have higher atomic masses than mass numbers. This is only an approximate analysis, as protons and neutrons don't have the same mass, so isotopes very close to the transition region don't behave perfectly smoothly.
Edit: The flipping effect I mention can be seen very nicely in the list of isotopes shown in this site. Look at the (atomic) masses for the isotopes, and see how the fractional part of the mass starts out as circa $.01$ (above the isotope's mass number), then decreases to $.001$, reaches exactly zero (by definition) for $\mathrm{^{12}C}$, then soon after flips under to $.999$ and keeps going lower all the way to $.90$ around zirconium and tin isotopes before slowly rising up to $.98$ for bismuth $\mathrm{^{209}Bi}$ and then flipping back over to $.04$ for thorium and $.05$ for uranium.
Now here's a question: I said the maximum of nuclear binding energy per nucleon happened at $\mathrm{^{62}Ni}$, so why does the fractional part of the masses in the table keep decreasing all the way to tin $\mathrm{^{116}Sn}$? It's because these masses are not normalized with respect to how many protons and neutrons there are in the nucleus. If you were to take all the isotopic masses and divide them by their respective mass numbers, you would see the minimum around $\mathrm{^{62}Ni}$, as expected (actually, the minimum ratio of isotope mass to mass number happens at $\mathrm{^{56}Fe}$ because it has a slightly higher proportion of protons to neutrons. As I said, the small difference between proton and neutron masses creates slight irregularities which I did not take into account for simplicity).

Aha - the atomic mass of an isotope is not quite the same thing as the mass number. I see (light bulb lights up above head!)


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Alkali Metal Rosinates Solubility in Acetone
In violin finishing technology alkali metal Rosinates are used in varnish synthesis. Usually this alkali metal Rosinates are soluble in Turpentine. I also know that Copper, Zinc Rosinates are soluble in Diethyl ether. However, beside Turpentine I want to know if this alkali metal Rosinates ($\ce{K,\ Na,\ NH4+}$ - Rosinates) are soluble in Acetone especially or in Diethyl Ether or in any other  VOC’s?

Would water-solvent mixtures be acceptable?  Are you still interested in this?

water based ones are worth a try


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How different concentration of HCl on magnesium affect the enthalpy change
I had conducted an experiment where I reacted a $0.3\ \mathrm{g}$ strip of magnesium with $100\ \mathrm{mL}$ of varying concentrations of $\ce{HCl}$: $1\ \mathrm{mol/L}$, $2\ \mathrm{mol/L}$ and $3\ \mathrm{mol/L}$. From the data I collected, I noticed that the reaction with $1\ \mathrm{mol/L}$ $\ce{HCl}$ produced the most energy and the reaction with $3\ \mathrm{mol/L}$ $\ce{HCl}$ the least. I just wanted to know if my data is correct or not because I hypothesized that the $3\ \mathrm{mol/L}$ reaction would produce the most heat energy.

Could you include the whole set up of your experiment? It would probably be good to know how you measured the released energy. Also good to know would be the actual values, you collected. Further I assume you mean the concentration of your HCl solutions are mol/L? I am not an expert in the field, I only see, that magnesium is the limiting agent, so I would assume, the reaction would produce nearly identical values.

Please write down the reaction you were trying to observe. Be sure to have the redox reactions complete.

In fact all three reactions produced the same amount of heat energy and hence had the same enthalpy change. By changing the concentration of the HCl only effects the reaction rate and how much energy is produced in a certain time. However eventually, each reaction will produce the same amount of energy provided that all the Mg reacts which does happen in your experiment as HCl is in excess. The reason for this can be explained quick easily using the formula: $$\Delta H=\Delta U+pV$$ Then using the ideal gas equation, we can substitute $pV$ with $\Delta n(g)RT$. Note that the only variable that actually changes is n(g) which is the change in the number of moles of gas molecules. So the equation becomes: $$\Delta H= \Delta U + \Delta  n(g)RT$$ So the only thing that effects the enthalpy change of the reaction in the change in internal energy (U) and the number of moles of gas molecules formed which are both constant for all three reactions.


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Why do some crystal structure files not include bonds?
Why do some crystal structure files not include bonds (the atoms are just placed in the respective position in the cell without bonds between any atoms)?
Shouldn't the bonds always be there?
Also, where can I look up these bonds?

If you solve an x-ray crystal structure, you simply get a set of atomic positions. No more, no less.
Often, the atomic positions don't include hydrogen atoms, only the "heavy" atoms.
Thus, most CIF (crystallographic interchange format) files, there are no defined bonds. Somewhere along the way, software perceives bonding information because most chemists prefer to have a valence-bond graph of atoms and bonds (and bond orders).
There are many algorithms for determining a bond graph from a set of atomic positions with varying "accuracy." I put the word in quotes because it's sometimes unclear what the goal should be when determining bond orders. For example, is a molecule in a keto or enol tautomer form? In some cases, the $\ce{C-O}$ bond is not clearly a single bond distance or a $\ce{C=O}$ distance. Maybe there's an ensemble average of both forms.
The standard seems to be if the CIF file looks more "ionic" (e.g., a salt like $\ce{NaCl}$) bonds probably will be ignored, but otherwise, bonds are interpreted from interatomic distances.

See the related question http://chemistry.stackexchange.com/questions/29641/azathioprine-cif-data-file-rendering-problem-in-accelrys?rq=1 for an example of how bonding information can be mis-perceived by different software.

Position of the atoms are sufficient information to solve schrodinger equation or to find molecular characteristics. Now, if you want to see bonds to make you more comfortable, you can use molecular visualizer such as babel, vesta, vmd, materials studio, chemshell, spartan etc. And there are plenty of visualizer that are available freely online.
To better explain, i think i should talk about molecular orbital. In quantum mechanics a linear combination of atomic orbital is used to construct molecular orbital. Now after solving the schrodinger equation, you can see the occupancy of each molecular orbital. And they are not perfect integer as opposed to classical idea of orbital. Some of them may have fractional occupancy. So, it's irrelevant to talk about single bond, double bond or triple bond. But to understand it better we call them single bond, double bond or triple bond. Like in my calculation if C-O bond distance is ~1.5 Angstrom i take it as single bond and if it is ~1.3 Angstrom i take it as double bond.


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Light scattering in sulfur colloids
Recently I did a reaction in the lab (the reaction itself is not important) whose product was colloidal sulfur. As I watched the sulfur particles form I noticed something strange: At first the sulfur particles appeared white, but after a while the solution turned yellowish (probably when the size of the sulfur particles grew a little). I know that sulfur's colour is yellow by nature, because of its absortion soectrum, and I understand that the white colour is probably due to Mie scattering of light by the colloidal particles.
My question is: Why aren't the particles yellow? if the absortion spectrum has as a consequence a yellow colour I would expect that spectrum to remain valid at all dimensions of the particle, the particle can't just randomly decide to stop absorbing light of some wavelenght when it reaches a threshold size... can it?
I know that Mie scattering predicts that a particle will scatter all wavelenghts equally if its size is comparable or bigger than the wavelenght of light, but I would have thought that if the particle absorbed a specific wavelenght that wouldn't change with size.

What about oxidation states of sulfur? Could that or a ligand be responsible?

There are a number of different effects:

Rayleigh scattering is the predominant effect for the tiniest particles, much smaller than the wavelength of light, i.e. before they've coalesced. It is dependent on the fourth (!) power of the frequency of the radiation, so short wavelength light is scattered toward the eye. Looking through the colloidal mixture, it should look redder.
Tyndall scattering applies to larger colloidal particles, and it too is dependent on the fourth power of the frequency of the radiation and preferentially scatter bluer light, so the yellow of the sulfur would be less apparent.
Finally, particles coalesce and scattering becomes less important than reflection from the larger particles.
Perhaps sulfur nano-particles behave a bit differently from bulk material due either to quantum effects or to surface effects. See Elemental Sulfur Nanoparticle Coarsening... If someone has experience with nanoscale sulfur, please feel free to comment or edit this.

I think that what I call Mie scattering is what you call Tyndall scattering. Anyway, I hadn't considered the possibility that sulfur was actually scattering more light of the wavelenght that it tends to absorb (blue or violet) and thus looks more white. This sounds quite reasonable and maybe kind of obvious in hindsight, I'm still not totally convinced but if none has a much better answer I'll accept that. Aren't colloids usually whiteish irrespective of the color of the bulk material? that's what bugs me.

According to the Journal of ACS, 1921 (https://books.google.com/books?id=WRtAAQAAMAAJ&pg=PA734&lpg=PA734&dq=colloidal+iron+sulfide+color&source=bl&ots=mPqs7AMMlM&sig=qIi55tWho9b17RCJUcAvYLE1ukc&hl=en&sa=X&ei=y9RoVZ-5A4G-sAX0sIGgCQ&ved=0CB8Q6AEwAA), "The color produced by the action of alkaline sulfides on compounds of iron and certain ... from brown to green but that the color was due to colloidal iron sulfide." Thanks for the question, as it encouraged me to research more (I would have expected black or brown).


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Polyurethane epoxy resin chemistry advice please?
My friend can buy polyurethane epoxy from his workplace for 3 dollars a litre, except that he doesn't know precisely the data sheet for it. His factory gets it in 20 ton deliveries.
He says that the mix has to be around 99% Polyurethane, 0.2% cobalt, and 0.8% of a toxic transparent catalyst liquid.
It's a generic industrial chemistry used everywhere, and can be mixed roughly by eye to achieve different drying speeds.
Do you know what the name of this resin system is called and where i can find precise information on it? What is the catalyst?

If the final product is polyurethane, than it is not epoxy, those are totally different chemicals. If you are sure it is polyurethane, have a look at wiki for the introduction: https://en.wikipedia.org/wiki/Polyurethane
And to be sure, check the sheets of the delivered chemicals. Even if it comes in 20 ton containers, it must always be accompanied at least by material safety data sheet, where everything is described. Without closer info it is difficult to tell more.

Hi, i'm doing some research. The resin is unsaturated polyester resin, using methyl ethyl ketone peroxide (MEKP) as catalyst, cobalt octoate (Coct) as an accelerator... The products are generic and don't come from the same suppliers necessarily. Already we have tell more :)

The resin is the same variety commonly used to produce fiberglass baths, boats and many other uses.
The resin is unsaturated polyester, using methyl ethyl ketone peroxide (MEKP) as catalyst, cobalt solution for example cobalt octoate (Coct) as an accelerator.
It isn't often sold to consumers because it is too explosive and corrosive, the MEKP reacts explosively with raw cobalt, and it's dangerous to the eyes, it can cause blindness in 15 seconds if splashed in the eye.
The MEKP doseage is 1-2% per volume of polyester, depending on desired drying time. The cobalt is mixed first and dyes the polyester a whiskey color, you can dose it by color, the darker whisky color the faster it will dry.
Industry standards are 1% and 6% cobalt solution, 1% is dosed at 0.5% to 2%, however there is no way to tell without the product data sheet if the cobalt solution 1%, 6%, 10%.
safety:
MEKP and cobalt mix = flames
MEKP in eyes = blindness
wear goggles and rinse with water and go to hospital if you get some MEKP eye contact.
wear protective clothing and avoid skin contact.
"Cobalt octoate is the most reactive and most commonly used paint drier. Cobalt octoate catalyses the oxygen uptake and accelerates peroxide formation causing a surface film to form rapidly and hence known as oxidation catalyst. It basically a metal caroxylate used to accelerate the drying process like linseed oil. It accelerates the decomposition of the organic peroxide initiators (often called catalysts) and promotes the cross-linking of polyester resins.
"
Note that baths are finished with a gloss product called gel coat.


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What is the functional group of zerumbone?
I am currently doing research on antioxidant activity of essential oil from Zingiber zerumbet. The essential oil contain zerumbone. The problem is I am confuse about the functional group of zerumbone is it ketone or aldehyde or what. I want to do thin layer chromatography test to identify the purity of the essential oil but I don't know how to identify it.

Zerumbone, a constituent of the rhizomes of wild ginger (Zingiber zerumbet), is (2E,6E,10E)-2,6,9,9-tetramethylcycloundeca-2,6,10-trienone; i.e. it is a ketone.