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Find the standard Gibbs' free energy of formation at 298.15 K for O2 gas dissolved in the saturated aqueous solution?
The solubility of $O_2(g)$ in water at $25^o \text{C}$ and $\text{1 bar}$ pressure of $O_2(g)$ above the solution is $0.00126$ $mol$ $O_2(g)/kg$ $water$.  Find $\Delta_fG_{298.15}^o(O_2(g),sln)$.
I kinda have a clue on how to solve this, but there's just one variable I'm stuck on figuring out. I'm not sure how to get the molal-scale activity coefficient for O2 gas in solution. The answer is $\Delta_fG_{298.15}^o(O_2(g),sln) = \text{16.8 kJ/mol}$... now how would I get there?
Here's a derivation I went through to show what I have tried. The chemical potential for $O_2(g)$ relative to some reference standard state is for a real solution on the molal-scale is:
$\mu_{O_2(g)} = \mu_{m,O_2(g)}^o + RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$
By subtracting the standard molar Gibbs energy of reactants in their elemental state, $G_{elem}^o$, and knowing that the chemical potential of a pure substance is equal to the chemical potential of that substance in a saturated solution, $\mu_{O_2(g)} = \mu_{O_2(g)}^*$, and we should get:
$\mu_{O_2(g)}^* - G_{elem}^o = \mu_{m,O_2(g)}^o - G_{elem}^o + RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$
$\Delta_fG_T^o(O_2(g)) = \Delta_fG_T^o(O_2(g),sln) + RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$
where $\Delta_fG_{298.15}^o(O_2(g))$ is $\text{0 kJ/mol}$ since oxygen gas is in its elemental state, and $\Delta_fG_{298.15}^o(O_2(g),sln)$ is what I am supposed to solve for. So, the final equation I think I'm supposed to use is:

$\Delta_fG_T^o(O_2(g),sln) = -RTln(\gamma_{m,O_2(g)}\frac{m_{O_2(g)}}{m^o})$

But I don't have $\gamma_{m,O_2(g)}$... so how could I do this? This is due next Tuesday, but it would be nice to have it addressed asap.

The textbook I am using is Physical Chemistry, 6th ed., by Ira N. Levine, if that helps. This is problem 10.55a at the end of the chapter.

Isn't the chemical potential of the oxygen in the liquid solution the same as that of the oxygen in the gas phase (at equilibrium)?

Yeah, it should be; that's what I said when I wrote $\mu_{O_2(g)} = \mu_{O_2(g)}^{*}$, right?

So the free energy of formation of the oxygen in the solution should also be zero, right?

Well, no, because oxygen doesn't naturally exist in solution. It is supposed to be $\text{16.8 kJ/mol}$. However, as you would expect, the free energy of formation of the oxygen in its natural, gaseous state is 0 though.

I'm confused.  I thought that the chemical potential of a substance is the same in two phases that are in equilibrium.  What am I missing?

Let me clarify: O2 gas in this problem is said to be dissolved as much as possible so that it forms a saturated solution (which is why its solubility is given) at standard thermodynamic temperature and pressure. I shouldn't have assumed any particular phase equilibrium---just go off of that info when it comes to what equilibria are present.

However, the chemical potential of a substance in a saturated solution IS equal to that for the substance in its pure form---that's what I'm saying. And that's because $\mu_i = \mu_i^{*} + RTln\gamma_i\chi_i$ for the solute $i$, and at saturation, $\gamma_i = \chi_i = 1$. This would be why we treat solids in equilibrium constants to have concentrations of 1.

However, it does not mean that $\gamma_{m,i} = 1$, since $\gamma_{m,i}$ depends largely on the mol fraction of the solvent.

True or false:  For phases in thermodynamic equilibrium, the chemical potential of a given substance is the same in all phases.

I'm inclined to say that that would be true, since thermodynamic equilibrium implies mechanical, material, and thermal equilibrium, meaning no net flow in matter or energy. Chemical potential runs downhill, but if there is no net flow in energy, it should be that the chemical potential is already the same in each phase. I'm not sure if you are confused with chemical potential vs. the change in Gibbs energy of formation, but they're not the same thing, so $\Delta_{f}G_{T}^{o}{(O_2(g))}$ is not equal to $\Delta_{f}G_{T}^{o}{(O_2(g),sln)}$.

So the partial molar free energy of the substance in each phase is the same?

I thought I just said that? Partial molar free energy is chemical potential, at constant  temperature and pressure for a single-solute solution, if we started at $dG = -SdT + VdP + \sum_i \mu_i dn_i$. Do you know where you're going with this, or are you just asking me clarifying questions?

I'm just confused.

Okay. Well, thanks for trying. I might just end up asking my professor on Tuesday after Thanksgiving break.

I think I finally figured out what you are asking.  I forgot that the standard state for formation of a solute in solution is 1 molar.  I think you are asking why one can assume that the activity coefficient is 1.0 for the case under consideration.

I'm not entirely sure if you can; it would give $\text{16.6 kJ/mol}$, but it may just be a coincidence since it isn't exact (the $\text{16.8 kJ/mol}$ came directly from the back of the book).

I really think it boils down to providing some physical justification for treating the dissolved oxygen as an ideal solution with an activity coefficient of 1.

Well... I would love to do that, but the chapter is literally all about Nonideal Solutions. :)


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Does excessive sulfuric acid as catalyst affect the synthesis of aspirin?
Recently had a lab experiment to synthesize aspirin.
I was instructed to add 4 drops of concentrated sulfuric acid but misread the "4 drops" as "4 ml".
Ended up with a yield of just 10 %. But the crystals that I obtained upon recrystallisaion weren't of aspirin but of salicylic acid (as confirmed by a melting point of 160 °C).
Did the addition of excessive sulfuric acid affect this in any way? What could have caused salicylic acid crystals to be produced instead of aspirin?
Some technical details of the steps I took:

mixing 2 g of salicylic acid with 5ml of acetic anhydride
4 ml of sulfuric acid added as catalyst (should've been 4 drops)
Heated the mixture in a water bath for 15 min.
Added 1 ml of cold water slowly then added 20 ml of cold water again quickly and stirred for 5 min.
Allowed solution to cool at room temp.
Crude product was dissolved in 5 ml of ethanol. (I noticed that my crude product was able to dissolve immediately whereas my peers' had to heat theirs and took awhile to dissolve, any reason for this?)
30 ml of hot distilled water was added to the mixture and heated on a hotplate.
Allowed solution to cool to room temperature then placed in an ice-bath for 15 min.
Crystals collection by suction filtration.
Crystals were dried in the oven at 100 °C for 15 min.
Crystals placed in desiccator for 10 min.
Melting point of crystals were found to be 159–161 °C

Did the addition of excessive sulfuric acid affect this in any way? What could have caused salicylic acid crystals to be produced instead of aspirin?

The intended reaction was the formation of an ester from a a phenol and an anhydride.
However, with an excess of sulfuric acid, acidic hydrolysis of the product according to an $\mathrm{A_{Ac}2}$ mechanism takes place.

i remember doing the same synthesis but used glacial acetic acid not H2SO4 and still ended up with salycylic acid crystals ? @Klaus Warzecha

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When ∆Suniv = zero, does the concentration of reactant equal the concentration of product?
It is my understanding that when ∆Stot / ∆Suniv = 0 then the reaction does not occur, the position of equilibrium is not shifted in a specific direction. However, does this imply that [reactants] = [products]?
The question 1e part iii asks "What can you understand about the the equilibrium when ∆Stot = 0". The answer to the question is "Kc = 1, or reaction balanced in middle, or no tendency to go in either direction"
To me, this only makes sense if the starting concentrations of the reactants and products are the same. However, at no point is it mentioned that the starting concentrations of reactants/products are equal.

All thermodynamics boils down to looking at $\Delta S_\mathrm{univ}$. $\Delta S_\mathrm{univ}$ is the sum of $\Delta S$ for your system and $\Delta S$ for everything else. We measure $\Delta S$ for the system directly and we measure $\Delta S$ for everything else via $-\frac{\Delta H}{T}$. When these two are equal, notice that $\Delta G = \Delta H - T\Delta S = 0$. Then given $\Delta G = -RT \ln K$, we conclude that $K = 1$.
The reaction favors reactants as much as products, but that specific balance depends on the reaction and its coefficients, i.e., the equilibrium expression could have powers. Consider:
$$\ce{A2 -> 2A}$$
The concentrations of product and reactant are not equal at equilibrium for $K = 1$.
Also, it does not mean that a reaction does not occur. You need to compare the reaction quotient $Q$ against $K$ to determine if there is a reaction or if we are actually at equilibrium.

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Gaussian09 and Path Integral Monte Carlo which one predicts the total internal energy more accurately?
I'm learning PIMC with primitive action and currently just testing the code with single methane molecule, I did run a Gaussian ccsd(t) calculation, I notice that at 298.15K & 1atm, the energy predicted by PIMC(0.0476a.u.) is slightly lower than the one by Gaussian(0.0477a.u.) both errors start from 5th decimal number. I'm just wondering which result could be more reliable? And why one is more reliable than the other?
It might be a dumb question, but I don't see anywhere made such comparison and I'm just an undergraduate student with no experience with either of two computational techniques. Would be greatly appreciated if can use some expertise!

what energy are you calculating? what are the parameters of your pimc calculation (grid)? what basis set do you use in your cc calculation? convergence criteria?

Total internal energy is the one I'm calculating. Grid approach to PIMC...
cc-PVTZ for H
aug-cc-PVTZ for C
I'm not sure what you mean by convergence criteria, I ran an extremely long simulation then I applied the "Flyvbjerg-Petersen"blocking algorithm to work out the number of  samples required to reach convergence

Every time you look for convergence you stop if the change between two (or more) points is less than a certain value. For SCF convergence in Gaussian use the keyword SCF=(Conver=N) which sets the SCF convergence criteria to 10^-N, see http://www.gaussian.com/g_tech/g_ur/k_scf.htm.
I guess you are doing single point calculations?

@DSVA I get your first point, in which case, the "Flyvbjerg-Petersen"Blocking algorithm is a very computational efficient way to do it.  As total energy is the bottom of the well, so single point is the easiest way? But later on, for PIMC, I guess I can apply modified Shepard interpolation to grow PES, which is something beyond me atm.

Single point energy means that you calculate the energy at a specific geometry which you use as input for your calculation. In contrast to that you can also do a geometry optimization. If you do that there could also be an error in finding the lowest energy structure in addition to calculating the energy.

And the thing about the convergence criteria: let's say your real value is 1 and you start with 20 and your convergence criteria would a be change of less than 0.05 between to points. Your algorithm would do something like 20, 11, 5, 2.81, 1.95, 1.40, 1.22, 1.11, 1.07 and there it stops. The difference between 1.07 and 1.11 is below 0.05. But if you got a tighter criterion of let's say <0.02 then it won't stop there and it would go on like 1.04, 1.02, 1.01 , which is much better.

@DSVA I think the single point was done by someone who previously worked on the project, and its accuracy was checked by a few groups, I can't provide further detail about it as I don't have much knowledge about this. But in regard to your 2nd comment, this is exactly what Blocking algorithm does, in a slightly smarter way, it works especially well when the simulation is massive. Perhaps read the paper J. Chem. Phys. 91, 461 (1989); link

Can you add your G09 input to your post? And yes, both methods do that but the question is what the convergence criteria are! If for MC it's extremly tight and for G09 it's not then it doesn't make sense to compare them! Make the convergence criteria tighter and look if the results change.

One problem in comparing the two methods is also that MC uses statistics and CCSD(T) is an ab-initio method. CCSD(T) is often used as a reference/benchmark value for other calculations since it is highly accurate.

@GavyLittlewolf Can you please edit your question to inclde EVERY paramter you are using in both methods? Discretization of grid, all appropriate CC convergence criteria etc. And all the settings you made. If you also pass over the coordination file of your molecule I can make a CCSD(T) calc with TURBOMOLE and maybe get some more info for you. But I personally would always trust a CCSD(T) with a good setup more than a MC

@Fl.pf let me get back to you guys, I figured there are some more PIMC calculation can be done before making the comparison. I will update as soon I have the result. Sorry for the waiting!


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Why the current of double layer charging is independent of potential?
i-V curve of double layer charging has a rectangular shape. Why when increasing or decreasing the voltage, the current remains constant?
Or I can ask, how the double layer structure is changing by varying the applied voltage?

Wow thats a really old question but here is the answer: for electrochemical systems, C is a constant and I=C*(dV/dt) and dV/dt is the scan rate which scales the potential linearly (so is also constant).

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A Book on Preparation and Reactions of organic compounds
I have to learn many methods of preparations of various hydrocarbons and their reactions. For this, I went through many articles and some books but didn't found the details of reactions anywhere (There are details for only some of these) For example, why do we chose Nickel as the catalyst other than anything else to prepare ethane by catalytic hydrogenation of ethene? And why this reaction has to take place at 300°C and not at a lower or greater temperature?
I have 'a lot of hydrocarbons and a lot of methods to prepare them and a lot of their reactions' to learn so I want a book which has been designed especially to cover this topic so it could provide the details I want. Basically my course/syllabus includes Methods of preparation and reactions of methane, ethane, ethene, ethyne, benzene + alkyl-halides, alcohols, phenols, ethers, Aldehydes, ketones, carboxylic acids and Esters.

For example, why do we chose Nickel as the catalyst other than anything else to prepare ethane by catalytic hydrogenation of ethene?

Palladium is expensive ($690~\$\mathrm{US}/\mathrm {oz.tr}; 1~\mathrm{oz.tr = 31.1 g}$).
Nickel is cheap ($11~\$\mathrm{US}/\mathrm {kg}$).

Catalytic hydrogenation of ethene is the model reaction performed to understand the mechanism of this type of processes, but I'd be astonished if it had any technical relevance.

There is enough ethane available from the processing of natural gas by distillation (demethanizer, deethanizer).
Ethene is the more interesting starting material for polymerization, oxidation, etc.

And why this reaction has to take place at 300 °C and not at a lower or greater temperature?

Does it really? In earlier mechanistic studies, such as  Review of Physical Chemistry of Japan, 1937, 11, 153-165, or Discuss. Faraday Soc., 1950, 8, 152-159, the hydrogenation of ethene on a nickel catalyst was performed at much longer temperatures (-78 to 0 °C).

preparation and reactions of alkyl halides, alcohols, phenols, ethers, aldehydes, ketones, carboxylic acids and esters

For lab-scale reactions, any decent introductory book on Organic Chemistry  will do.
If you're looking for books that focus on experimental procedures (cook books), there are two which I can recommend (Note that have only used the original German versions):

Organikum – Organic Chemistry Basic Practical Course
Reactions and Syntheses: In the Organic Chemistry Laboratory


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Is there a liquid that looks like water but boils at a low temperature?
I just watched a movie where someone put their hand into boiling water. Disregarding how this was actually done, I wondered whether there was a liquid which could be used for an illusion of someone "putting their hand into boiling water".
For this it would have to look like water, boil at a comfortable temperature (< 50 degree celsius) and should not be so toxic that you can't put your hand into it. Smell doesn't matter.
Is there some liquid that fulfils these requirements?

Diethyl ether (you don't want to use this one due to its extreme flammability, which you didn't mention as a requirement, but which is important too), and then probably some freones.

Maybe one of the refrigerants? Generally non-toxic and non-flammable. CFC-113a (1,1,1-trichloro-2,2,2-trifluoroethane) boils at 46 $^{\circ}$C, for example.

@ToddMinehardt Most environmental regulatory agencies would probably be terrifically upset at someone playing around with an open container of a CFC.

A liter or two for a classroom-like demo? I doubt that very much. The other compounds listed in this thread are way more dangerous to the immediate surroundings and persons in terms of toxicity and flammability.

Movies don’t usually strain for realism: many “boiling water” scenes look like they were cold water + dry ice. This creates bubbling and something like steam but involves no risk to health.

@ToddMinehardt broad environmental concerns (as in Montreal protocol) rather than local safety

@matt_black or pumping air into the bottom of a container of cold water (fish tank components)

Was the movie Blade Runner? Also there's a Mythbusters episode where they put their hands in molten lead. https://www.youtube.com/watch?v=yTOCAd2QhGg

If you're talking about Blade Runner when Pris puts her hand in a container with "boiling" eggs, I've concluded that they are using forced air (perhaps even dry ice), to make the illusion of heat.

Liquid nitrogen? :P Most clear liquids look like water.

The following list gives some boiling points
I've added the suggestions of Ben Norris and Todd Minehardt.

Common organic solvents

Pentane: 36 °C.
Diethylether: 35 °C.
MTBE: 55 °C.
Acetone: 56 °C.

Please read the Material Safety Data Sheets (MSDS), pay attention to the hazard statements and decide whether you (still) want to put your bare hands into these solvents!

Refrigerants

Dibromodifluoromethane: 23 °C
Trichlorofluoromethane: 24 °C
2,2-Dichloro-1,1,1-trifluoroethane: 28 °C
1,1-Dichloro-1-fluoroethane: 32 °C

@Klaus - You might want to add some refrigerants to your list, since they have low- to non-toxicity ratings and are also not nearly as hazardous in terms of flammability as the items you list. Look at trichlorofluoromethane, for example, which boils at 24 $^{\circ}$C. And there are several others in the right temperature range. Just a suggestion.

@ToddMinehardt Thanks a lot for the suggestion! I've added some of the refrigerants.

Liquid nitrogen: 77 K. Obviously, the other end is bad too, but I've definitely stuck my hand in and out real fast without any problems.

@Zhe Dimethyl zinc has a bp of 46°C, but I haven't added that either - for reasons ;-)

The liquids suggested so far are either flammable solvents, or CFC refrigerants (bad for the ozone layer and tightly controlled.)
I can tell you from experience that a measuring cylinder of butane boils with an appearance like water but is pretty cold (around 0 C) so I dont recommend putting your hand in it (liquid nitrogen actually feels less cold.) Whatever you do, with flammable solvents, do it outside and away from flames!
Solid carbon dioxide "dry ice" in water is a better option. there are many videos on Youtube, for example this one which illustrates the effect. It also shows one of the problems: the fog goes down, rather than up, due to CO2 being denser than air. If you want the fog to rise this will also be a problem with other materials. A fun thing is to add some indicator to the water, whose colour is affected by the acidity of the CO2 dissolved in the water.
The fog is largely water condensed by the CO2 (as the vapour is produced at the sublimation temperature of CO2, -78C.) If you want vigorous effevescence without so much fog, you could look into the reaction of acids with carbonates and bicarbonates. If you mix solutions, you will find the reaction instantaneous and the material will bubble so fiercely it jumps out of the pot. Solid carbonates react much more slowly, and you can fine tune it with both the acid strength and the particle size. Here is an example.
EDIT: Another reaction that produces a harmless gas is the catalytic decomposition of hydrogen peroxide solution to water and oxygen. You can fine tune the speed of reaction very easily by varying the amount of catalyst. you can see this both with and without the optional dishwashing liquid here (turn the music down if you don't like it!) and there are far more dramatic demonstrations on Youtube. It should be borne in mind that the more dramatic demonstrations this do generate a lot of genuine heat though.
If you want smoke, commercial smoke machines often use propylene glycol (which has a higher boiling point than water) and is much less toxic than ethylene glycol (as it is metabolized to lactic rather than oxalic acid.)

Most of the solvents mentioned in the other answers don’t actually look that much like water. They have a lower surface tension, i.e. they make flatter but larger drops on a surface if a given volume is spilt. They also have very different ripple characteristics. To exemplify the difference: if you usually drink (and see) whole milk and then suddenly experience fat free milk, it will look a lot more like white water than actual milk.
On the other hand, dichloromethane has a higher surface tension, makes smaller spills and in some ways behaves a lot more like water in beakers (while still being different, unfortunately). It is much less flammable than ether and similar solvents, so heating it is safer. And it boils at a mere $40~\mathrm{^\circ C}$. Or, as somebody once put it in a synthesis problem seminar:

Ah, refluxing dichloromethane; i.e. room temperature.

You wouldn’t want to put your bare hands in forever, but boiling DCM should be much better for this purpose than boiling water.
Also, consult the material safety data sheet and be certain you know what you’re doing before you attempt this!


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Does water acidity affect sugar solubility?
For example, when making lemonade: will mixing sugar in water after adding lemon juice lead to a slower sugar dissolving rate, than if sugar were mixed in before adding lemon juice?


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What is the order of nomenclature priority among an alkyl side group, a double bond, and a triple bond?
I saw the below structure in a video on YouTube and was confused by the fact that while they were numbering it, higher priority was assigned to the methyl than the triple bond:

According to the priority table on Master Organic Chemistry, the opposite is true, and the triple bond should have priority.

Also I saw this structure:

According to the same priority table, shouldn't a double bond have higher priority than a triple bond, and hence the numbering should start from the left?

Related: Nomenclature Precedence Order for Unsaturation and Hydrocarbon Chain

Your second question is related to this and this question, but it’s probably not an exact duplicate.

Is the numbering in the picture correct ?

Your so-called priority table (the “table of functional group priorities for nomenclature” shown here) is relevant for the choice of a characteristic group for citation at the end of a name by means of a suffix or a class name. It does not apply to the numbering of locants. This particular table could also be misleading since it mingles the concepts of classes, suffixes, endings, and prefixes.
The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Since both compounds that are given in the question do not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-14.4 (c) is not relevant in this case.
The systematic name of the first compound is 6-methylhept-2-yne rather than 2-methylhept-5-yne since low locants are given first to the ‘yne’ ending and the locant ‘2’ is lower than ‘5’.

The systematic name of the second example is (4E)-hex-4-en-1-yne rather than (2E)-hex-2-en-5-yne since low locants are given to multiple bonds as a set (even though this may give ‘yne’ endings lower locants than ‘ene’ endings), and the locant set ‘1,4’ is lower than ‘2,5’. Only if a choice remains, preference for low locants is given to the double bonds.


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How does a Lewis acid differ from an oxidizing agent; Lewis bases from reducing agents?
A Lewis acid is defined as an electron acceptor. An oxidizing agent is also defined as an electron acceptor (in the electronic theory of oxidation).
So then what is the basic difference between an Lewis acid and an oxidizing agent?
Similarly;
A Lewis base is defined as an electron donor (Lewis theory). A reducing agent is also defined as an electron donor (electronic theory). Therefore, what is the difference between Lewis bases and reducing agents? Are they exactly same thing?
There is a similar question which, however, only considers bases and nucleophiles and does not go into detail about oxidation and reduction.

For Lewis acid and base theory, the electrons must be in a pair. Also, we don't think of them as being transferred. The donation or acceptance of the pair creates a bond. Whereas for oxidation/reduction, you transfer some number of electrons, not necessarily 2.

@Zhe make it an answer.

Could I write "all lewis acids are oxidising agents but all oxidising agents are not lewis acid"? similarly, "all lewis bases are reducing agents, but not all reducing agents are lewis bases"? If my understanding is too poor, please forgive me but I prefer to clear confusion.

An acid and an oxidising agent are very different concepts that apply to different phenomena — although a single compound can be both at the same time, e.g. $\ce{H2SO4}$. However, a compound can also be an acid and a reducing agent, e.g. $\ce{HI}$.
The Brønsted acid/base theory is clearly distinct from any type of redox reaction, so I suppose that the confusion lies with the extension of said theory that the Lewis theory provides. But first a short word on the Brønsted theory: therein, an acid is a compound that can transfer a $\ce{H+}$ ion onto another compound.
The Lewis theory realises, that the acidic transfer reagent need not be hydrogen. Instead, anything with an unoccupied low-energy orbital can function as a Lewis acid; most notably boron compounds and metal cations. Upon creation of a Lewis acid-base pair, a coordinate bond is formed. Basically, this means that the energy of the lone pair of the Lewis base is lowered because it interacts with the empty orbital of the Lewis acid whose energy is raised. This can be depicted as $\ce{LB\bond{->}LA}$. The key concept is that the interaction is fully reversible and that upon dissociation the Lewis base keeps its lone pair.
When discussing oxidising or reducing agents, however, the electrons are irreversibly transferred. They actually move from orbitals of the reducing agent (which will be oxidised) into orbitals of the oxidising agent (which will be reduced). This transfer is thermodynamically favourable and thus the electrons won’t just ‘go back’.
To exemplify this, let’s consider two reactions of $\ce{H2SO4}$; one where it is an oxidising agent and one where it is an acid (Brønsted type; the proton of $\ce{H2SO4}$ is actually the Lewis acid).
$$\text{Acid-base:}\\
\ce{H2SO4 (l) + NH3 (g) <=>> NH4+ HSO4- (s)}$$
Note that ammonia, our base, is a gaseous compound at room temperature. If you employ appropriate conditions (i.e. temperature and low pressure), the equilibrium will be drawn to the left, releasing ammonia as ammonia ($\ce{NH3}$ molecules) including its lone pair. $\ce{H2SO4}$ will stay behind.
$$\text{Oxidising agent:}\\
\ce{2H2SO4 + Cu -> CuSO4 + SO2 ^ + 2 H2O}$$
$\ce{SO2}$, being a gas, will diffuse away. However, even if you perform this in a pressurised vessel, there will be no significant back-reaction because the electrons have been fully transferred from copper to sulfur.

The argument for Lewis bases/reducing agents is the same in reverse.

Thanks, but I am still confused. Could you add a brief statement about "if this, then Lewis base, and if this, then reducing agent" (similarly for acid and oxidising agents)

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text:
Why does Avogadro's number have the value it has?
As I was learning about Avogadro's number and I was wondering why is Avogadro's number equal to $6.02214\cdot 10^{23}$? I mean, how did chemists come up with this particular number?

https://en.m.wikipedia.org/wiki/Avogadro_constant

Also there are loads of question here in Chem.SE about Avogrado constant. You can search them.

Specifically, look at the section on measurement

The gist is that Avocado's  :-) number is a scaling constant like a ream of paper.

Because it was chosen to be that way.
The picture is that at some point in chemical history it was realised that atoms are made up of a nucleus containing protons and neutrons, and electrons that have little mass in comparison in ‘shells’ around the outside. (Beware: this picture is a simplification!) It was, at some point found out how heavy these particles were.
Things like the Law of Constant Proportions and the Law of Multiple Proportions were also known, the first stating that a fixed mass ratio of two different compounds would only be able to react (with extraneous mass of one compound remaining as unreacted residue) and the second stating that if compounds or elements reacted in different ways, the mass proportions would be integer ratios of each other. So obviously the idea of particles which have a mass was correct.
Now how do we ‘count atoms’? Similar to ‘counting sugar’. A recipe won’t tell you to mix $x$ sugar crystals with butter, eggs and flour to make a cake, it will state a value in grams (or ounces, if you’re in that part of the world). It is common to do that for large amounts where you can’t count the individuals. Taken to atoms, this meant that at one point in history it was defined to take the amount that weighs $z~\mathrm{g}$ of element $\ce{Y}$ as a value for ‘one count’ — one mole. That definition was changed and currently, $12~\mathrm{g}$ of $\ce{^12C}$ atoms is the number of atoms in a mole.
Because of these choices, the Avogadro constant now had a fixed value and could be calculated. It was calculated to be the well-known $6.022 \times 10^{23}$. Had a different choice been made, it would have been a different number.
Also, this is due to change. The SI unit system is to be modified at some point in the future. From thence, the value of $N_\mathrm{A}$ will be fixed much like the speed of light in vacuum are fixed. The definition of the mole will then depend on the fixed constant.


text:
Neutralization of Sodium Azide using Ammonium Cerium Nitrate
I'm using Ammonium Cerium Nitrate to neutralize phosphate-buffered saline solutions containing 0.1% Sodium Azide. I add 0.55 grams of Ammonium Cerium Nitrate into 100 mL of this 1XPBS + 0.1% NaN3 solution and am able to neutralize the NaN3.
Will the resulting solution have any negative, harmful, effects if flushed down the drain? My main concern is polluting ground water or corroding our sewerage system.

Googling around, most sources I can find recommend neutralization of azide with nitrite salts, followed by acidification.  Is there a reason you have to use cerium to neutralize instead of the common, recommended method?

See e.g. https://www.unige.ch/sciences/chiorg/matile/12-ProceduresforLabTreatmentofWasteChemicals.pdf

See also http://www.unomaha.edu/business-and-finance/support-services/_documents/sodium-azide-hazards-disposal.pdf

I was trying to find a more simple method of neutralizing NaN3 rather than having to use Sodium Nitrite and Sulfuric Acid. If I had a low concentration of NaN3 (less than 0.1%) in Phosphate Buffer Saline can I simply heat the solution as a form of neutralization?

I don't know about temperature, but simple bleach also seems to do the job, and has the advantage of being both simpler than $\ce{(NH4)2Ce(NO3)6}$ and simpler than the nitrite/acid method mentioned earlier. https://www.ncbi.nlm.nih.gov/pubmed/20667654

I think that you are doing fine!
The azide anion can be used for quite some amazing chemistry, but it's rather toxic and harmful for the environment.
Admittedly, I'm too lazy right now to look up the redox potentials or to write the reaction equation, but
CAN should be able to oxidize $\ce{N3-}$ all the way up to $\ce{NO3-}$, which is safe to be disposed.

Update
Judging from memory can result in bad advices!
A closer look at the SDS of both compounds shows that both have

H410 Very toxic to aquatic life with long lasting effects. and

P273 Avoid release to the environment.

Consequently, the combined solutions (with excess of CAN) should not be poured down the drain, but collected (aqueous inorganic waste).

Should there be cause for concern about the excess CAN that is disposed along with the neutralized Sodium Azide? The SDS of the CAN stated that it was toxic to aquatic life.

@Molly Excellent point! Admittedly, I judged from memory and I was wrong about effects of CAN, Apologies! I will update my answer.

Could I simply heat my 0.1% NaN3 solution as a way of neutralization?


text:
IUPAC nomenclature: Priority of alkene vs alcohol functional groups
IUPAC states that the higher priority functional group forms the suffix .
Alcohol having higher priority than ENE should be chosen as the suffix.

Hence the compound above is rightly named as 3-pentene-2-ol ?
However, I was wondering whether I can name it as 2-hydroxypent-2-ene ??

Can I call a tiger "a big striped lion without mane"? Why, I can, and quite probably I even would be understood correctly, but still it is preferable to stick with the conventional naming.

Generally, since many compounds can have two or more names in accordance with several methods recommended by IUPAC, a compound may be named correctly in more ways than one. Therefore, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) include the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including the many retained names.
The PIN for the compound that is given in the question is (3E)-pent-3-en-2-ol.

Other names that are in accordance current IUPAC recommendations and also many names that were generated according to earlier recommendations may be used in general nomenclature.
Furthermore, the IUPAC nomenclature is quite robust. In many cases, wrong names (e.g. deviations from the rules for the priority of suffixes, numbering of locants, or order of citation in the name) do not necessarily lead to wrong structures.
The first name 3-pentene-2-ol that is proposed in the question is not in accordance with current IUPAC recommendations. However, the very similar name 3-penten-2-ol (note the elision of the ‘e’) was in accordance with the obsolete 1979 IUPAC recommendations. Hence, the proposed name could be understood.
The alternative name 2-hydroxypent-2-ene that is proposed in the question is not in accordance with current or past IUPAC recommendations; therefore, it should not be used. Nevertheless, this name is unambiguous; it describes only one compound (ignoring cis/trans isomerism). Hence, this nonconforming name could still be understood. The correct name for this compound is pent-2-en-2-ol (the following structure diagram shows the 2Z isomer).

Anyway, the proposed name 2-hydroxypent-2-ene as well as the conforming name pent-2-en-2-ol do not describe the structure that is shown in the question (note the position of the double bond). Therefore, the proposed name cannot be used.


text:
Are there ethanol and methanol analogues with more than one hydroxyl group?
Ethanol and methanol are ethane and methane derivatives, respectively, with one hydrogen replaced by an hydroxyl group. Are there molecules based on either where more than one hydrogen atom has been replaced by an hydroxyl group? If so, what are they called?

This is a geminal diol. They're not that stable, but they're intermediates in lots of carbonyl reactions.

"That's no hydroxyl ion, that's my wife!"

Generally speaking, they're called diols when bearing two hydroxyl groups. If they're on the same C, this is called a geminal diol, if they are on neighboring Cs, they're called vicinal diols. The C on geminal diols has a higher oxidation state than that of an alcohol, i.e. which is why they could be seen as relatives of an aldehydes (or ketones).
In a diluted aqueous solution, formaldehyde $\ce{H2C=O}$ actually predominantly exists in its geminal diol form $\ce{H2C(OH)2}$. This is no longer true for 1,1-ethanediol (the hydrated form of acetaldehyde).
The simplest vicinal diol 1,2-ethanediol is better known under its trivial name ethylene glycol (a common ingredient of anti-freeze formulations).
Triols with the hydroxyl groups on the same C (with the oxidation state of carboxylic acids) are generally unstable.

Interesting -- do you have a reference for the formaldehyde/geminal diol equilibrium? Also, thanks for posting, and welcome to Chem.SE!

@hBy2Py This looks useful for equilibrium references: http://pubs.rsc.org/en/content/articlepdf/1963/jr/jr9630005217

It looks it was not ethylene but diethylene glycol in wine.

@Mithoron yes, you are right – I had remembered that incorrectly, I will update the answer

Zhe and user1573870 already mentioned that geminal diols are usually not stable (Erlenmeyer rule), and user1573870 has pointed out that these compounds are hydrates of carbonyl compounds.
Typically, loss of water is the preferred mode and leads to the corresponding aldehyde (or ketone).
However, there are some exceptions to the rule: Some compounds actually exist as their hydrates. The most famous examples are

Chloral hydrate, the crystalline hydrate of 2,2,2-trichloroacetaldehyde
Glyoxylic acid, which crystallises as the hydrate (2,2-dihydroxyacetic acid) from aqueous solutions

Interestingly, with silicon, which is right below carbon in the periodic table, it's the other way round: here, it's the double bonded oxygen which is not stable, but the geminal hydroxyl groups. To be fair, in silicon, those have a tendency to condensate to form oligo-/polysiloxanes.


text:
Can I make a container of (kind of) perpetually boiling water?
It's a common school experiment to make water boil at room temperature by putting it in a container and reducing pressure.
I am wondering if it is possible to make a sealed sphere containing water where the pressure is carefully set to room temperature so that it boils "forever".
Obviously it can't boil forever, because it would eventually all turn to vapor. What I am thinking though is that minor fluctuations in room temperature will cause it to cycle between boiling and condensation without the need for any power.
Is this line of thinking correct? Has anyone attempted to make something like this? I think it would be a really cool desk ornament, like a little box of cycling boil-condense water.

Why, you can make such a container, but it would not look cool at all. In a relatively tiny enclosed space, vapor will almost instantly get saturated via surface evaporation alone, without any visible boiling.

@IvanNeretin Why would it get saturated so fast? Assuming it's a 6x6x6 inch cube, is there no way to make it take a few minutes to get saturated? Is there a formula to figure this out?

By way of comparison, concerning "minor fluctuations in room temperature will cause it to cycle between boiling and condensation" (however using dichloromethane or similar instead of water): https://en.wikipedia.org/wiki/Drinking_bird

Well, let's put it this way: maybe I shouldn't have said "instantly", for it wouldn't really get saturated that fast. But then, water wouldn't heat that fast either.

There are some CPU coolers that work this way: https://www.youtube.com/watch?v=gSorsaV-Qgs You could achieve perpetual boiling by gently warming the water and cooling higher up to condense. Of course, you can achieve this at atmospheric pressure by using a reflux condenser.

Yes. You make a container, boil the water in it so it flushed away the air from it and than seal it off. In such case even tiny warming will cause the water boil. However it is easier said than done as vapor pressure of water is rather low at room temperature, so surprisingly high vacuum is created when the water in the container cools down.

This happens all the time in a closed container. Water condensing and evaporating is a continuous cycle that keeps the vapor pressure constant at a specific temperature.

See this article on reflux heating: https://en.wikipedia.org/wiki/Reflux

You can buy a hand-heat powered boiler. But something must provide energy to keep the bubbles forming, and remove it to condense the gas back to liquid, or it would all be in the vapor state. In this toy, the heat of your palm provides the input energy, and ambient air cools the vapor in a condenser.
BTW, this demo won't work if the room temperature is exceedingly high, because the liquid (probably an ethanol-water mix) won't condense and the vapor pressure becomes too high for bubbles to form in the remaining liquid.

Water boils if the vapour pressure of the liquid phase is larger than the total pressure in the gas phase (of a forming bubble). In a closed vessel, an emerging vapour bubble finds that as it grows, so is the total pressure. Which makes it shrink again.
Unless there is a large temperature gradient, this cannot work.


text:
Do any substances in a liquid state work as well as water to put out fires?
Is there a liquid substance or compound that works as well to put out fire as water does?

I'd say many would do better then water, for they evaporate easier and choke the flame with their vapors. We still use water because it is cheap and environment friendly.

Water is used because cities have giant tanks of it. Liquid nitrogen would be more effective but you don't have enough.

Carbon tetrachloride was once used for this purpose as it has practically no flammability at lower temperatures .


text:
How does one find the ground-state term symbol for a configuration that is exactly half-filled?
For instance, the ground-state configuration of N atom is a $p^3$ configuration of all parallel spins and one electron in each $2p$ orbital, which has:

Total spin angular momentum, $S = 3 * \frac{1}{2} = \frac{3}{2}$
Spin multiplicity, $2S + 1 = 2(\frac{3}{2}) + 1 = 4$
Total orbital angular momentum, $L = (-1) + (0) + (1) = 0$

and so, $J = {|L - S|, |L - S + 1|, . . . , |L + S - 1|, |L + S|} = {\frac{1}{2}, \frac{3}{2}}$. Thus, I've found that the two candidates are ${}^{4} S_{1/2}$ and ${}^{4} S_{3/2}$.
But my textbook (Pilar, from the '60s!) says that the ground state is ${}^{4} S_{3/2}$. How could I figure something like this out?
Hund's Rules from my text (as far as I can tell---this text is hard for me to read) don't specify how we treat exactly half-filled subshells, only less-than-half-filled, or more-than-half-filled, assuming equal $L_{max}$ and $S_{max}$ already.
(As a side note, my usual McQuarrie textbook is not with me this week, so I cannot simply refer to McQuarrie.)

to get you started have a look at the answer here http://chemistry.stackexchange.com/questions/37038/how-to-find-the-number-of-atomic-microstates-for-a-given-electronic-configuratio/54333#54333

I'm guessing you're looking for me to expand my answer? I could, sure.

The full calculation is laid out below. Start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number $s = 1/2$ and magnetic quantum number  $m_s = \pm 1/2$. Orbitals have angular momentum of $s=0, p=1, d=2, f=3$ etc.
The total spin angular momentum is the series of values
$$S = |s_1 + s_2|\cdots|s_1 − s_2|$$
where $s_1 , s_2 , s_3 =+1/2$  are the quantum numbers of each electron. The total angular momentum is
$$ L=|l_1+l_2|\cdots|l_1−l_2|$$
and total angular momentum
$$J=|L+S|\cdots|L−S|$$
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by $1$.
The term symbol has the form $^{2S+1}L_J$. The super-prefix is the spin multiplicity, for spin angular momentum S this is $2S+1$ or in general $2X+1$ for angular momentum X.
When there are three 3 electrons the the spin and orbital angular momentum terms have to be added in two parts. First as a pair, with the equations above then again with each of the values in the series with the last electron or orbital.
For the spin the S values are
$S = |s_1 + s_2|\cdots|s_1 − s_2| = |1/2+1/2|\cdots|1/2-1/2| = 1, ~ 0 $ .
We call these values $S_1$ and $ S_0$
To calculate the total with the third electron gives
$$S_{s_1}=|S_1+1/2|\cdots|S_1 - 1/2| = 3/2 , ~ 1/2$$
and for the $S_0$ there is one value $S_0=1/2$. Thus for the spin the values are $3/2$ and $1/2$. This means that the multiplicity produces quartet and doublet states, thus so far we have $^4L_J$ and $^2L_J$.
The same method is followed for the orbital angular momentum. The first two p orbitals give
$$ L=|l_1+l_2|\cdots|l_1−l_2|= |1+1|\cdots|1-1| = 2,~1,~0 $$
then for each of these values called $L_1,L_2,L_3$ we combine with the last orbital angular momentum, for example
$$ L_{L_1}=|L_1+l_1|\cdots|L_1−l_2|= |2+1|\cdots|2-1| = 3,~2,~1 $$
and so forth for the other values. The L values are $3,2,1,0$
Putting all these values into a table gives
$$\begin{matrix} L & S &  J &  \text{term symbol}\\
3 & 3/2 & 9/2, 7/2, 5/2, 3/2 & ^4F_{9/2, 7/2, 5/2, 3/2} \\
2 & 3/2 & 7/2, 5/2, 3/2,1/2 & ^4D_{ 7/2, 5/2, 3/2,1/2}\\
1 & 3/2 & 5/2, 3/2,1/2 & ^4P_{5/2, 3/2,1/2}\\
0 & 3/2 & 3/2 & ^4S_{3/2}\\
3 & 1/2 & 7/2, 5/2 & ^2F_{7/2,5/2}\\
2 & 1/2 & 5/2, 3/2 &^2D_{5/2, 3/2}\\
1 & 1/2 & 3/2, 1/2 & ^2P_{3/2, 1/2}\\
0 & 1/2 & 1/2 & ^2S_{1/2}\\
\end{matrix}$$
Finally to find the lowest energy term use Hund’s rules, i.e. of the terms given by equivalent electrons, i.e. having the same L values, the one with the greatest multiplicity is lowest in energy and of these the lowest is that one with the greatest L value. The only terms remaining after considering Pauli exclusion are $^4S$ which is lowest in energy and next $^2D$ then $^2P$

Ah, so that's what  you were going for. Yeah, it's more rigorous than I would have done for this kind of problem, since it was only concerning the ground state, but it makes sense to me. I did something similar over here for $s^{1} p^{2}$: https://socratic.org/questions/how-would-i-determine-all-the-possible-term-symbols-for-an-s-1p-2-electron-confi

How did you apply Pauli exclusion principle to remove the extraneous term symbols?

Does the term symbol relates to the energy of the whole atom or to the energy of the electrons in the specific orbitals?

*sigh *
Face-palm moment: I've already chosen an $S_{max}$, and when the subshell is exactly half-filled, $L_{max} = 0$, so $J = S_{max}$. And that would be why it's ${}^{4}S_{3/2}$, rather than ${}^{4}S_{1/2}$.
But if anyone has anything more to add, feel free to edit this answer.

Here's what I would do if I started this question from scratch. I would write out the electron configuration of the ground state in accordance with general chemistry principles (maximize parallel spins and minimize coulombic repulsions by distributing electrons evenly).
$\underline{\uparrow \text{   } }$ $\text{   }$ $\underline{\uparrow \text{   } }$ $\text{   }$ $\underline{\uparrow \text{   } }$
$m_l = $ {$-1,0,+1$}
$m_s = +\frac{1}{2}$
In order to determine the ground-state term symbol, we note the usual format:
$\mathbf{{}^{2S+1} L_J}$
where:

$S$ is the total spin angular momentum.
$L$ is the total orbital angular momentum.
$J = |L - S|, |L - S + 1|, . . . , |L + S - 1|, |L + S|$ is the total angular momentum when one considers spin-orbit coupling to determine multiplet terms.

By definition:

$L = \sum_i m_{l}(i) = -1 + 0 + 1 = 0$.
$S = \sum_i m_{s}(i) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$

So the spin multiplicity is $2S+1 = 2(\frac{3}{2}) + 1 = 4$, and the term symbol letter is $S$ since $L = 0$. Therefore, we have a quadruplet $S$ state.
Since $J$ takes on each positive term starting from $S = \frac{3}{2}$ in integer increments, $J = \frac{1}{2}, \frac{3}{2}$. So, the possible term symbols without narrowing them down yet are ${}^{4} S_{1/2}$ and ${}^{4} S_{3/2}$.
Since the subshell is exactly half-filled, $L = 0$, so $J = S$. From Hund's Rules, we first try to maximize $S$, then $L$, then we choose our $J$ depending on whether the subshell is less-than, more-than, or exactly half-filled.
Since we would have already gone through and chosen $S_{max} = 3/2$ and $L_{max} = 0$, $J = S_{max} = 3/2$, and our ground-state term symbol is:
$\mathbf{{}^4 S_{3/2}}$


text:
How do you explain very high vanadyl phosphate NMR chemical shifts?
Generally, phosphates have chemical shifts in the range -20 to 10 ppm. For vanadyl phosphates containing $\ce{V^{4+}}$, the isotropic chemical shift is about 1500 ppm. Why?

$\ce{V^{4+}}$ (and $\ce{V^{3+}}$) is paramagnetic, and therefore you would expect a very large change in chemical shift. Paramagnetic shifts arise from the interaction of the observed nucleus with unpaired electron spin of the paramagnetic metal ion.
The observed isotropic shift comes from two interactions: the contact shift and pseudo-contact shift.
The contact shift comes from the direct interaction between the observed nucleus and the unpaired electron - a through bond effect. This gives very large shifts - many hundreds of ppm.
The indirect pseudocontact shifts essentially come from the interaction of the unpaired electron density with the electron density of the orbitals of the observed nucleus, and are effectively compensating for the nonzero magnetic field of the paramagnetic source - a through space effect. These are much smaller shifts, often within the usual observed chemical shift range.
Whilst the measurement of contact shift information is infrequently used for structural insight, pseudocontacts using paramagnetic lanthanides are used extensively for 3D solution structures of proteins.


text:
Does the carboxylate group show a positive or a negative inductive effect?
In the benzoate ion, does the carboxylate group show a $-I$ or a $+I$ effect?
I learnt that the carboxylate group is a $-I$ group, but according to a website the carboxylate group in benzoate shows a $+I$ effect.

So, which one is correct? Does the carboxylate group show a positive or a negative inductive effect?
Please neglect mesomeric effect when answering.

To the close voters, I think this question shows enough thought and reasoning not to close it. To the OP, please make sure to [edit] and cite your source.

the carboxylic acid group always shows -I effect due to the presence of electronegative oxygen atoms, however the carboxylate group(COO-) shows +I effect due to excess negative charge.

The carboxy group is electron-withdrawing both by inductive and by mesomeric mechanisms. I have drawn two depictions in the scheme below. The upper one is how the mesomeric effect of the carboxy group can be explained. This effect is only relevant if the $\ce{COOH}$ fragment is in the same plane as the ring.

In the case of 2,6-di(tert-butyl)benzoic acid, the bulky tert-butyl groups prevent the mesomerism depicted above. The carboxy group is perpendicular to the ring. Yet, this group still exercises an electron-withdrawing inductive effect. It can be explained by a mesomeric structure of the $\ce{C=O}$ double bond with charge separation:
$$\ce{C=O <-> C^+-O^-}$$
This positive charge is, naturally, electron-withdrawing and can be used to explain the $-I$ effect of carboxy groups. No matter which mesomeric depiction you choose, you can never get a negative charge on the carboxy carbon, thus it will always remain partially positive. That also makes sense considering it is bonded to two electron-withdrawing, strongly electronegative oxygen atoms.

I believe he's asking about the carboxylate group not the carboxy group


text:
Any differences in metal ion emission spectra?
If I measure the emission spectra of potassium using the following solutions
1) $0.1~\mathrm{M}$ of $\ce{KNO3}$
2) $0.1~\mathrm{M}$ of $\ce{K2SO4}$
will I see any significant differences?
I think there will be no significant differences because the spectrum does not depend on whether in 1) I am using $\ce{K}$ while in 2) I am using $\ce{K2}$ (the emission spectrum should be independent). Am I right?

You have neither $\ce{K}$ nor $\ce{K2}$. In both cases, you have $\ce{K+}$ ions, and they are the same.

They look the same but intensities would be different

$0.1~\mathrm{M}$ $\ce{KNO3}$ dissociates to give $0.1~\mathrm{M}$ $\ce{K+}$ ions.
$0.1~\mathrm{M}$ $\ce{K2SO4}$ dissociates to give $2 \times 0.1~\mathrm{M}$ $\ce{K+}$ ions.
Hence, in terms of emission intensity, $\ce{K2SO4}$ should give twice the signal if other parameters are same and if the detection response is linear.

Very sensitive instrument might detect the effect of counter-ion but that will be very difficult since the cation itself is always surrounded by water.


text:
Ionic emission spectrum contamination
Imagine that I am trying to identify the presence of ions from among the metals $\ce {Li,Na,K,Ca,Sr, Ba}$ in some solution through their emission spectrum.
By mistake the wire used to introduce samples into the flame exciting the ions is contaminated with $\ce{Al(NO3)3}$. The emission energies for Aluminum are $\pu{3.9\times 10^{-19}~J/atom, 3.58\times 10^{-19} ~J/atom, 2.97\times 10^{-19}~ J/atom}$.
How will this mess up the identification of the other ions?
Since $3.58\times 10^{-19}$ matches one of the Ca peaks and $2.97\times 10^{-19}$ matches one of the Sr peaks I imagine I could misleadingly identify these two as being present when they were not?

I would convert your energies to wavelengths then look up the most accurate wavelengths for the metals you can find thus  how different Al is from any of the others. Next  look at the wavelength resolution of your instrument to see if you can distinguish them. Second, the pattern of lines may help since if one matches closely the others are likely not to do so.


text:
Why does NADH have 2 peaks in its absorption spectrum but NAD+ has only one?
From what I understand, in general aromatic rings absorb UV radiation. Looking at the structures of NAD+ and NADH, I see one aromatic ring that is common in both (the lower one due to adenine), which I presume corresponds to the common large peak in the absorption spectra of both molecules.

However then I see that NAD+ has an extra aromatic ring, which NADH does not, although NADH produces another peak on the absorption!
I think I am going wrong in my understanding of what an aromatic ring is. I am assuming that a ring structure having alternating double and single bonds is necessary but not sufficient for this to be an aromatic ring, (so the extra alternating bond ring in NAD+ may or may not be aromatic and responsible for another peak, but the corresponding ring in NADH does not have alternating bonds and so is not aromatic and cannot be due to another peak.
I was also wondering whether you can have aromatic compounds which are not cyclic structures? Perhaps that could explain the aromatic-like absorption of NADH despite it not have alternating bonds in a ring?

There's no problem with that it's non-aromatic chromophore, even as you can see it's "better".

The nicotinamide (oxidized form, pyridine ring) absorption peak overlaps with the signal from the adenosine so you don't see it as a separate peak. Upon reduction, the pyridine ring turns into a dihydropyridine ring, which has a typical absorption signal around 340 nm (see for example this set of molecules


text:
Trouble understanding Bradford Assay
From what I understand, the Bradford reagent contains Coomassie Brilliant Blue in an acidic solution so both of the (sulfonic acid groups? I'm not sure if this is the correct name- I can't seem to find an answer) are protonated. In this form the dye colour is red. The theory is that in the presence of basic R-groups like in arginine residues, the acidic groups react with the basic groups. I don't understand why this is so? Why do the acidic groups react with the basic groups as opposed to just binding with the free $\ce{H+}$ ions in the solution. I suppose this is a fundamental question about acid base reactions.
Also, why does the colour change when the $\ce{SO3-}$ group is bound to a basic R group residue as opposed to a H+ ion? It is still essentially bound to a positive charge? With the basic residue now being protonated e.g. -$\ce{NH3+}$


text:
What Type of Gas Would Vinegar and Epsom Salt Create Under Electrolysis?
I'm restoring a motorcycle and have a small homemade DIY zinc plating operation which consists of vinegar, epsom salt, a piece of zinc and a power supply.
When I plate a steel piece such as a bolt or small hardware piece, the piece in the solution starts forming small bubbles and it appears to be "gassing off" something. I was wondering what type of gas this may be and if it's harmful to breathe.

Vinegar and Epsom salt would not create anything. Vinegar and zinc, though, would create hydrogen. No, it will not poison you.

You're creating hydrogen gas at your part. No toxicity hazard. You need a substantial zinc concentration in the solution before you'll plate much metal -- are you sure you're actually putting zinc down? When I get a chance, I'll try to post a more thorough answer (unless I get beaten to the punch...).

Thank you both for answering my question. @hBy2Py Yea, I'm laying zinc down nicely on the parts. It just has a hard time getting into concave areas such as inside a nut or irregular shaped hardware piece

Yep, throwing into the insides of features is a common plating challenge.

You're most likely forming $\ce{H2}$ gas at the part.
If you're plating anywhere above about $2~\mathrm V$, that is probably sufficient overpotential to cause part of the passed current to go toward electrolysis of water.
There is negligible toxicity hazard to inhaling the small amounts of hydrogen that would be generated from an electrolytic cell on this scale.
There's a small ignition hazard if anything sparks near electrolyte of the plating cell as it's operating, but as long as you're not running it inside a tightly closed, small space you should have nothing to worry about with how small the cell is.


text:
Cyclopentadienyl radical geometry and MO considerations
Does the cyclopentadienyl radical attain a regular pentagon geometry or is there a deviation caused by the Jahn–Teller effect?
Without the Jahn-Teller effect the MO diagram would look like this:

But I guess that a more stable state would be obtained by the Jahn-Teller effect: the orbital marked with number 2 would drop in energy while 3 would rise. What is the true picture?

I think it's fairly safe to say that the J-T distortion in $\ce{Cp^.}$ is quite well-known. There is a body of experimental evidence for it; summarising it is however probably beyond my capacity right now.

A typical definition outlining the Jahn-Teller effect is that for any non-linear molecule in a degenerate electronic state, there exists a displacement of the nuclei along at least one non-totally symmetric normal coordinate, that gives rise to a distortion of the molecular geometry with a concomitant lowering of the energy
The underlying cause of the Jahn–Teller effect is the presence of molecular orbitals which are both degenerate and open shell.
The cyclopentadienyl radical, has degenerate and unequally occupied molecular orbitals, thus one can expect to see JT distortions; and indeed there is plenty of evidence to support it, earliest of which comes from the work done by Thrush et. al in 1956.
Experimental data can be found here (along with a list of references to the original works).

A publication I found particularly instructive was this The potential energy surface given below has been taken from there.

This image presents a downward looking view of linear Jahn–Teller distorted lower PES appropriate for $\ce{C5J5}$ . The coordinates $Q_a$ and $Q_b$ denote the two unique dimensions along which motion resolves the degeneracy of the conical intersection corresponding to the $^2E^"_1$ state. These coordinates lie in the space spanned by the four pair of $e^"_2$
normal coordinates, $Q_{ia}$ and $Q_{ib}$. Around the perimeter of the figure are pictures of the distorted $\ce{C5H5}$ geometries for the corresponding value of the phase angle $\phi$.
The work, later goes on to summarise the geometric distortions in the table given below.
The authors also compute $\epsilon_i$ for each of the four modes, and confirm that their sum corresponds to the experimental value of $\epsilon_T$ (this is done for both $\ce{C5H5}$ and $\ce{C5D5}$)

Does A2 or B1 have a bigger contribution to the real geometry?


text:
Are there other Density Functional Theories other than Kohn Sham DFT?
For some reason I thought that the term DFT and Kohn Sham DFT were interchangeable. Kohn Sham DFT is certainly one of the most popular. But are there other types of DFT that are in use today? I have been able to find orbital free DFT is gaining momentum (http://www.sciencedirect.com/science/article/pii/S0022509607000221)
But are there more DFT types?

For example https://en.wikipedia.org/wiki/Strictly-Correlated-Electrons_density_functional_theory

@Fl.pf. MP2 is in no way a formulation of HF. KS-DFT is a theoretical "implementation" of DFT, and MPn is n-th order perturbation theory. It would be more correct to say something like Rayleigh-Schrodinger PT and Brillouin-Wigner PT are forms of PT.

@pentavalentcarbon I didn't mean it was. That comment was a little hasty and doesn't show very good what I meant to say. I'll delete it.

@Fl.pf. That's fine; I'll leave mine up because others might not know that. By the way, thank you for participating, we need more people interested in theory and computation.

KS-DFT is the basis for all orbital-based DFT Formulations. In it you have different specialities (constricted DFT, TD-DFT, rs-DFT to name a few).
KS-DFT is used for molecular problems, because of atom bonds etc.
TD-DFT is used for time-dependend problems like excitation energies, polarizabilities etc.
RS-DFT is using range-separated functionals (long-range, short-range) to handle effects better in the long-range domain.
Addition by @pentavalentcarbon:
Current cDFT implementations are based on KS-DFT and perform a real-space partitioning of the electron density.
Plane Wave DFT (no orbitals, electron density calculated with plane waves, was the first DFT) is used for periodic problems like metals etc.
DFTB (Density Functional Tight Binding) is good for Molecular Dynamics since its an approximate DFT and allows handling of larger systems and greater timescales.
Addition by @pentavalentcarbon:
A good analogy for DFTB would be that it is similar to semiempirical methods (PM7, other NDDO parametrizations) where certain integral matrix elements are replaced with precomputed, fitted parameters, but there is a minimal basis set present.
KS-DFT is dominant because as computational chemists we are mostly working with molecules. As that we need a simulation of atom bonds. There is a good Book about DFT which I can only recommend (A chemists guide to Density Functional Theory): http://www.goodreads.com/book/show/70706.A_Chemist_s_Guide_to_Density_Functional_Theory

Current cDFT implementations are based on KS-DFT and perform a real-space partitioning of the electron density. A good analogy for DFTB would be that it is similar to semiempirical methods (PM7, other NDDO parametrizations) where certain integral matrix elements are replaced with precomputed, fitted parameters, but there is a minimal basis set present.

@pentavalentcarbon Thank you, I have included your information in the answer.
All DFT methods are semi-empiric (because of the XC-Functional)

While it is certainly a spectrum, I would argue that not all XC functionals are semiempirical (Slater exchange, VWN correlation, and their descendants) in the same way that PM7 is semiempirical. These may only be  fitted to the homogeneous electron gas, with less than a handful of parameters, as opposed to M06, which is definitely semiempirical.

Some other good reads: http://www.quantum-chemistry-history.com/Hist_Dat/DFT_Dat/DFT_Ob1.htm  and http://simplecompchem.blogspot.com/2016/06/dft-advices.html

@pentavalentcarbon I think the definition of semiempirical is used a little loose here. There are people who say that DFT is "not really ab-initio anymore" so maybe a little semiempiric... In german we would say its "Haarspalterei" :)


text:
Is there proof that the exact exchange correlation functional is an analytic function?
Mathematics show that some problems cannot be solved analytically. Could this be true for the exact exchange correlation functional?
Most textbooks state that the form of the exact exchange correlation functional is unknown. It is said to be an unknown mathematical object. Does this mean that said functional is an analytical function? And if so, what is the proof that shows that the functional is analytical?

What do you mean by analytic?

Then I think the answer is no, since there is derivative discontinuity for the exact functional.

@user26143 Sorry. I meant this one: http://mathworld.wolfram.com/Analytic.html
Not: https://en.wikipedia.org/wiki/Analytic_function

I guess the answer depends on the scope of "known functions". If we restrict to the elementary functions, most likely it is not (I do NOT know any PROOF for this point). It has been shown finding exact functional is NP-hard problem, e.g. http://www.nature.com/nphys/journal/v5/n10/full/nphys1415.html

See my colleagues work, it seems to be the answer you want.
http://journals.aps.org/pra/abstract/10.1103/PhysRevA.93.042511

Welcome to Chem.SE - thanks for posting! It's site policy to include more description in answers than just a link to an external resource -- can you edit your answer to include a summary of their results/conclusions? As well, if you haven't already, please check out the tour and help pages for more information. Enjoy!

This could be a valuable answer, so I encourage you to expound on it as the users above have suggested.


text:
How to measure concentrations using emission spectrum?
Imagine I am using emission spectroscopy to identify the presence of ions in a solution.
Would it be possible to use the intensity of the spectrum to find the concentration of each ion in the solution?
Probably the intensity is directly proportional to the concentration but there might be some scaling factor that I would need to know to find precise values for the concentration of each ion?

You are correct in your assumption. The scaling factor is determined experimentally, this is typically called the 'calibration curve'. Basically you measure the intensity of known concentrations and fit a line through the points. Perhaps this paper can help:
http://science.sciencemag.org/content/202/4364/183


text:
Is there a particular reason for the way these structures are represented?
I know the question might seem a bit elementary but i was was curious whether there is a particular reason behind the way hydrogens are represented in a hydrocarbon chain. Is there a difference between representing them with straight lines or with slant lines as in the two images? Is one preferred over the other? If so under which conditions?

There is no real difference. Sometimes this comes about from the drawing software people use or personal preference. In some texts the authors or typesetters are using LaTeX packages to draw organic molecules, this can be rather tedious and it is easier to use 90° angles.
The angled hydrogens are a bit closer to what a 2d drawing of the 3d object would look like, but not to the point where any information is conveyed. It can make visualizing the compound in 3 dimensions a bit easier for me when they are drawn this way.
For the double bonds the cis/trans or E/Z information is conveyed, although this is irrelevant in your example.

could you explain what you mean by 'For the double bonds the cis/trans or E/Z information is conveyed'-i couldnt understand this part

@AnamikaGhosh Sure, the double bond in the ethylene cannot rotate freely. Because of this the relative position of those groups does not change. I'll add a drawing to the answer shortly.

@BigChemT Thank you for explaining but could you tell me why the ethylene is not freely rotatable in case of ethylene considering they're all hydrogens ?  Do correct me if i'm wrong

@AnamikaGhosh Double bonds do not rotate. This is probably best visualized with a model kit. I know that can be a frustrating answer to get, but I hope you do not feel that way. You can also imagine linking two  Styrofoam balls with a small rod, that would be a single bond, and the balls could rotate in relation to each other. If you used two sticks, the balls would be locked in a certain orientation.

i appreciate the help, it definitely cleared most of it.

Ok, so this is somehow relevant.
Including the hydrogens is just another way of writing those structures down, usually used in teaching for people who are new to the subject. In case of ethane and ethene it also makes sense, since just an isolated line or a double line is kind of strange, just like you won't make just a point for methane.
It doesn't matter if the lines are all in 90° angles or not, so why do it? Well, in most cases we do 2D drawings of 3D objects and we somehow want to give an accurate picture. So in the alkane drawing this somehow represents the tetrahedral array of the four atoms around each carbon better than the structure on the bottom. For the alkene, this one is actually a flat molecule with pretty much those angles inbetween all atoms, so that's a pretty good representation of the actual molecule, which is why this is the best way to do it in this case.

The first structure might be an attempt to provide a glimpse of 3D structure of ethane. To be more graphic, she might draw it this way

However, more stable conformer is this one

The way you draw molecules depends on the purpose of the drawing.
If all you are trying to convey is the connectivity of the molecules (which atoms are connected to which other atoms) then it matters little whether you draw the molecule withe right angled bonds or slanted bonds. In older books you often see white complex structures described in ugly ways the were determined by the needs of typesetting from fixed character symbols. So benzene was not drawn as an regular hexagon but as a "stretched" hexagon which was easier to typeset. But chemists got the essential information anyway. Now that we create images electronically and have better tools most books draw the hexagons as they really are as regular hexagons: it looks better but doesn't add any extra information to an experienced chemist.
It is the same with simple hydrocarbons in most cases. The choice, if connectivity is all that matters is aesthetic, not an essential part of conveying information.
However, if you are trying to show something important about the actual shape of the molecule (perhaps because you are describing a reaction mechanism where the configuration of the molecule matters, then it is important to convey something about the relevant part of the structure. Real molecules are (usually) 3-dimensional and bond angles around carbon are not right angles. So chemists draw them with something approximating the right angles and may add extra clues (like wedges and dotted lines) to clarify their 3-D orientation. This is sometimes important, but not always.
So whether the way you draw a molecule is important or not depends on the purpose of the drawing.
But it is also true that now we have good tools that make drawing slanted bonds rather than right-angled ones, there is little reason not to do so. It isn't always essential, but why not do something that requires no extra effort and looks better?


text:
Melting Lithium foil to make ingot?
I'm not sure of this is the correct spot for this but I am curious about this. I want to make some Lithium ingots. The only source of Lithium I have is the foil from Lithium batteries. My question is could I use maybe like a soldering iron or a lighter or something to melt the foil into ingots without any sort of issue?

Maybe if you have an inert atmosphere. But if you try it without that there is a good chance you will ignite the lithium which reacts rapidly (even unheated) with moisture and air.

You made it seem like there is a possibility it could work without. As a regular home chemist with minimal supplies I don't have access or ability to make a inert atmosphere.

Lithium melts at only 200 C, a soldering iron is way overkill. Helium is a good shield gas, note that sometimes there's oxygen in balloon cylinders. Choose your crucible wisely: no ceramics or glass.


text:
How to deposit a durable layer of sensing molecules in a PVC matrix on the tip of an optical fiber?
Here is the situation that I have been tacking for weeks and I have still not understood.
We are trying to understand how to do this:

This mixture for the preparation of the "LI–PVC" was obtained from a
  batch of 80 mg of poly(vinyl chloride), 160 μl of tributylphosphate
  and 20 mg of Lophine dissolved in 2 ml of freshly distilled
  TetraHydrofuran (THF).

Paper:
http://www.sciencedirect.com/science/article/pii/S014381661400089X
which refers to :http://www.mdpi.com/1424-8220/8/2/847
We have no idea how to do this, since we dont have any step/by/step instruction, reaction time, environment we need to maintain for this.
How do we go about identifying this process and replicating this mixture. This is very critical to us. Any help on this matter would be of immense help, if you have a resource which may have be able to shed some light on this. It would be of help as well.
Many Thanks

Can you explain what kind of educational background and laboratory you have? Perhaps you should hire a chemist if you want to produce chemical sensors.

I have bachelor in Eng., I'm not sure how this process works, so what health and safety hazard we should avoid when testing. This is for school project. Unfortunately, I don't have much background in chemical eng., we have a lab setup.

@Karl, would you be interested in consulting opportunity? - we have very limited budget.

We have no idea how to do this, since we dont have any step/by/step instruction, reaction time, environment we need to maintain for this.

Let's redefine the general task:

How to deposit a layer of 2,4,5-triphenylimidazole (lophine) on the tip of an optical fibre?

Requirements:

The layer must be permanent/durable and does not wear or wash off during measurements of the nitrate concentration in drinking water.

The workflow therefore includes:

How to prepare the fibre?
How to deposit one (or several) layers of lophine?
How to cure the layers to make sure that they do not wear/wash off?

As far as the preparation of the fibre is concerned, your primary paper (Optics and Lasers in Engineering, 2014, 60, 38-43) gives some hints on page 39:

To deposit the sensitive layer on
  the optical fiber, the coating was removed in a 1 cm section, and
  the exposed cladding was sterilized. Isopropyl Alcohol Level was
  used for cleaning the optical fiber in special cleaning cloths and
  immediately the LI–PVC was introduced [...].

Apparently, lophine is deposited on the fibre in a PVC matrix and you have cited the composition of the mixture used.
How is this actually done?
An earlier publication by the same authors gives some directions! It's reference 19 in the paper that you cited!
In Sensors, 2008, 8, 847-859, the describe authors describe the dipping method in chapter 2.2. and provide further references.
In summary, the prepared fibre is

dipped into the solution (lophine, PVC, TBP, THF) and then
cured by placing the fibre in an oven for 15 minutes at 80 °C.

These two steps are repeated as often as necessary to obtain the required total layer thickness.


text:
I found transition states (TS), but calculations aren't supported by experiment. Do I look for more TSs or somewhere else?
I'm using Gaussian to investigate a catalyst system.

Experimental evidence showed that one of three analogues (R = phenyl, Ph), yielded a benzosultam

when treated with 2,4,6-triisopropylbenzenesulfonyl azide (N_3Trisyl).
Running the reaction at 70 °C over 65 hours, an 84% yield was observed.
There was no benzosultam yield when either of the other two analogues (R = neopentyl, NeoP, and R = trimethylsilyl, SiMe_3) were reacted with N_3Trisyl under the same conditions. I hypothesized that differences in activation energies for the C-N bond forming reaction were the root of the observed reactivity differences.
I'm using the BP86 functional, with 6-31G(d,p) basis sets, GD3BJ dispersion corrections, and the CRENBL ECP applied to chromium. Benchmarking for thermodynamic properties was done against  reaction enthalpies and activation energies from a set of 3 chromyl chloride hydrogen atom abstraction reactions done by James Mayer. A second set of coupled cluster calculated X-H and X-X bond dissociation enthalpies was also used. The BP86//6-31G(d,p) method with CRENBL ECP applied to chromium and GD3BJ dispersion correction gave mean absolute errors in the range of 4kJ/mol for Mayer's activation energy data, 14 kJ/mol for Mayer's reaction enthalpy data, and 10 kJ/mol for enthalpies in the coupled cluster set.
I used Gaussian to optimize structures of the transition states, reactants and products for each of the three analogues. From my results, I calculated activation energies for the C-N bond forming reaction for each analogue, as well as standard reaction enthalpies, entropies, and free energies. I calculated activation energies in the range of ~6-22 kJ/mol, with R = Ph being the lowest, as well as negative free energies of reaction, and favorable entropies and enthalpies of reaction for every analogue. These results suggested that some benzosultam should have been observed for the R = NeoP and SiMe_3 analogues.
So now I'm stuck. The reaction should be spontaneous at every temperature, according to my calculations. But it isn't.
Now I think I'm looking in the wrong place.
Maybe I want to look for intermediates and a second transition state?
I'd assumed the energy diagram for the reaction looked like this

, a single hump with reactant and product on either side), when it might actually look like

. Meaning I could have calculated TS2 (in the second diagram) for each analogue, and ignored TS1, I'd have been looking at the wrong reaction step.
There's also a possibility my method is inadequate.
Or maybe I'm calculating energies using reactant and product structures which are at local, but not absolute, minima.
Where do I go from here? What's the best use of my time? Is there a way to search the energy surface around the products and reactants to find lower energy conformers? Is there a way to search for intermediates and additional transition states? I feel a bit lost here.
TL;DR: Thermodynamic properties that I have calculated for a catalytic bond forming reaction are not supported by experimental results. So, now I have the choice of revising my hypothesis and searching for another set of transition states and some intermediates, or searching the energy surface around my reactants and products for lower energy conformations. Do these choices sound reasonable given the data and evidence I've collected? Am I overlooking something? What to do?

Your structures are huge. How long do these take to run?!

Frequency calculations take about 30 hours. Optimizations take between 72 and 144 hours on the Grex-Westgrid system.

These results don't make sense to me.  Since you have a results that the reaction will go but you find that it doesn't go in the lab then the model must be wrong somewhere.  Can you get any kinetics information to test whether or not your results correlate with you energy from the reaction that did go?  If these two figures are out then it is less likely that your model is correct.  Hope this helps ... good luck.


text:
How have I to understand in what those quantum numbers refer for ortho and para water?
I need to make an English presentation at my School (Graduate school) about something people may don't know. I chose to speak about the separation of ortho and para water. I first chat here with @Orthocresol about questions I have but this is really tricky and I'm not understanding that so much. So I chose to ask the question here, it might be helpful for other people and I hope at least for me.

At first my question is to understand what the notations $\lvert0_{00}0 \rangle$ and $\lvert1_{01}M \rangle$ in the article mention before about different states of water refers too?
What are those four numbers if the ket notation, from where they come from and what does they mean? I know it makes a lot of questions but maybe are they linked?

I read this article on wikipedia and this one too, and I also found that but I don't understood a lot of things. Maybe that $J$ is a quantum number which quantify the rotation energy of a molecule. But I have a lot of troubles with the projection and I don't know what $M$ is.

I would be glad if the answer was in a part like popularisation because I guess I'll have some troubles to understand it. But I have good skills in algebra, I had a twenty hours introduction in quantum chemistry, the hardest thing I did was a presentation of the spin-orbital coupling if it can help you to jugde my level on the subject. Thank you in advance for your answer.

It's explained in the first link you gave in the previous paragraph:
"The rotational quantum states of the asymmetric rotor, i. e.,
water, can be classified by $J_{K_{a}K_{c}}M$, with the total angular
momentum quantum number $J$, the projection labels $K_{a}$
and $K_{c}$ onto the molecule-fixed a and c axes as defined in
Figure 1 respectively, and the projection quantum number
$M$ onto the space-fixed $Z$ axis."
Do you know how angular momentum in quantum mechanics works? You can look at the projection of the angular momentum onto a specific axis and the total angular momentum. Turns out the Heisenberg Uncertainty Principle prevents you from knowing the angular momentum projections along more than one axis at once...

Why need I these representation of the states? Why only J is not enough, why to add its projections? And what M is .. ? Are we speaking about only nuclear spin or might it be also electron spin?

I think it's the molecular rotation. Otherwise, I don't think they would refer to it as an asymmetric rotor. Rotation about the a and c axes are such that the molecule is not symmetric about the axis. It is about the b axis though. Honestly, this is beyond the QM I know, and I would be hard pressed to explain it in clearer detail.


text:
How are alkanes, alkenes, and alkynes homologous series since they have no functional group attached?
I understand that a homologous series is a group of compounds with the same functional group attached to it.
I also know of an alternate definition of it as being a series of compounds with the same general formula.
But if the first definition is to be considered then how can alkanes, alkenes or alkynes be considered homologous series since there is no functional group attached to any of these, as opposed to, e.g., the homologous series of alcohols?

Related: Why are the given compound in homologous series?

This is NOT a duplicate of the question Loong linked to. DO NOT CLOSE.

For the linear alkanes, the internal methylene groups $\left(-\ce{CH2}-\right)$ are considered to be the functional group defining the homologous series:
$$
\ce{H3C-\underline{(\ce{CH2})_n}-CH3}~,~~n\geq 0 \tag{n-alkanes}
$$
Alternatively, this series is defined by the absence of any particular functional group other than methyl and methylene.
Ethane, $\ce{C2H6}$ is the member of the series for $n=0$. Technically, one could exclude methane, $\ce{CH4}$, from the series, since it doesn't have two capping methyl groups. However, to my knowledge it's usually considered part of the series as a special case.
For the 1-alkenes and 1-alkynes, the double and triple bonded carbons are the functional groups defining the series:
$$
\ce{H3C-(CH2)_n-\underline{\ce{CH=CH2}}} ~,~~n\geq 0 \tag{1-alkenes}
$$
$$
\ce{H3C-(CH2)_n-\underline{\ce{C#CH}}} ~,~~n\geq 0 \tag{1-alkynes}
$$
Parallel to the alkanes, the shortest "official" members of the two series are technically propene $\left(\ce{H3C-CH=CH2}\right)$ and propyne $\left(\ce{H3C-C#CH}\right)$, respectively. As with methane above, though, ethylene/ethene $\left(\ce{H2C=CH2}\right)$ and acetylene/ethyne $\left(\ce{HC#CH}\right)$ are typically grandfathered in.
The case of non-terminal linear alkenes/alkynes is less straightforward. Depending on one's philosophy, they can either be lumped in with the terminal alkene/alkyne series, or they can be considered as distinct homologous series:
$$
\ce{H3C-(CH2)_n-\underline{\ce{CH=CH-CH3}}}~,~~n\geq 0 \tag{2-alkenes}
$$
$$
\ce{H3C-(CH2)_n-\underline{\ce{C#C-CH2-CH3}}}~,~~n\geq 1 \tag{3-alkynes}
$$
$$
etc.
$$
Note that $n\geq 1$ is strictly required for the 3-alkynes, since $n=0$ would give 2-pentyne $\left(\ce{H3C-C#C-CH2CH3}\right)$, a 2-alkyne. This requirement is general: e.g., for the x-alkenes, $n \geq x-2$ must hold for $x\geq 3$.
Branched alkanes/alkenes/alkynes complicate things even further. There are at least two common alkane series that I can think of off the top of my head:
$$
\ce{H3C-(CH2)_n-\underline{\ce{CH(CH3)2}}} ~,~~n\geq 0 \tag{iso-alkanes}
$$
$$
\ce{H3C-(CH2)_n-\underline{\ce{C(CH3)3}}} ~,~~n\geq 0 \tag{neo-alkanes}
$$
You can mix and match branching, double bonds and triple bonds in countless ways to make new homologous series. I would personally argue you could define a homologous series to contain multiple variable-length segments:
$$
\ce{\underline{\ce{(H3C)2C=C}}-(CH2)_n-\underline{\ce{C#C}}-(CH2)_m-\underline{\ce{CH(CH3)}}-(CH2)_r-\underline{\ce{CH=CH2}}} \tag{?}
$$
Of course, the practical utility of any correspondence of properties among the various species in such a convoluted homologous series may be ... limited.

the homologous series could also be defined as the compounds have the same general formula, in that case for eg. Alkanes would have a genral formula of CnH2n+2 , and menthane would fit in  as when n=1 the compound produced is CH4. Are the two definitions of the homologous series contradicting each other or perhaps another reason why they are defined in two different ways?

@KaberiGhosh They're just two different approaches to defining the same concept. To the best of my knowledge, "homologous series" doesn't have a rigorous definition, so neither one is strictly "correct." To note, a significant downside to, e.g., the $\ce{C_nH_{n-2}}$ definition for alkynes is that the formula also matches the dienes: butadiene's formula is $\ce{C4H6}$, the same as butyne's. You also have to specify structural features along with the $\ce{C_xH_y}$ formula in order to accurately define a given homologous series.

then that gets us back to the same point, that neither of the definitions seem to completely encompass the concept. So which definition would be the most accurate?

@KaberiGhosh Neither? It depends on the circumstances, and on who you're asking about the definition. Unfortunately, I don't think you're going to be able to pin this down as neatly as you'd like.


text:
How to determine if a complex has a crystal field stabilization energy of zero?
I have encountered with the following question:

The complex having zero crystal field stabilisation energy is:
(A) $\ce{[Mn(H_2O)6]^3+}$ (B) $\ce{[Fe(H_2O)6]^3+}$ (C) $\ce{[Co(H_2O)6]^2+}$ (D) $\ce{[Co(H_2O)6]^3+}$

Zero crystal field stabilisation energy will be the case when the metal has either $d^{10}$ configuration or $d^5$ configuration with every orbital having one electron. For Iron($\ce{Fe^3+}$) the d orbital is half filled. Now the problem is, water is in the middle of the spectrochemical series, i.e. it is moderately strong ligand. So, how to decide whether the d orbital will have the 'every orbital half filled' condition or not ?

In principle crystal field theory is an obsolete, empirically derived, incomplete concept/model, which gives tempting explanations that are not necessary right. It was combined with molecular orbital theory to form ligand field theory. In none of these complexes the $\ce{M-OH2}$ bonds will be of significant strength to cause the metal to adopt a low spin configuration. $\ce{Fe^3+}$ is your only choice because it had $d^5$ configuration.

Trivalent hexa Aqua complexes of the first row transition metals don't go low spin until Cobalt 3+ (where the highly favourable t2g 6 configuration is reached). So the answer to your question is the iron complex.

@RobChem Yes, that explains it.

@Martin-マーチン I have read that the theory is now changed with the molecular orbital concept to give a more reasonable explanation. But it is in my syllabus. So, I have to read it nevertheless.

By the way, I am wondering why Martin-マーチン and RobChem did not answer the question, but instead wrote it by comment.

@ShoubhikRajMaiti I understand that it is part of your syllabus; I wanted you to know, that there is something better out there. I did not answer, because I did not have the time to explain it completely. Hence only a hint.

In the given example it is pretty simple to chose B as the correct answer, as it is the only option that could possibly create a spin forbidden transition $d^5$ or a $d^{10}$ configuration. For most situatons I have encountered it is sufficient to assume high-spin filling, until you encounter very stable configurations: completely full or half full orbitals (like $d^6$)


text:
What kind of compounds are very basic and mildly volatile at room conditions?
We have a liquid seeping up into our concrete.  It is dissolving our epoxy floors and we can't stop it with any sort of sealant.
I tested it with pH strips and it seems to be very basic.  It's likely not got much water content either because it evaporates fairly quickly on a mild day.
What am I dealing with here?  I am going to send it out to get tested, but I was wondering if any chemists could help me out with classifying it.

You are dealing with water which has washed out some remaining alkali from concrete.


text:
Is it 'Phenyl' or 'Phenol' that is used as a cleaning agent?
I often see Phenyl written on cleaning agents in stores. Here are some examples:

One day, a guy standing beside me at a store picked up a bottle of phenyl/phenol and he said, "The big corporation got it wrong. It's phenol, not phenyl."
I looked up the definition of Phenol, and it reads:

A toxic white soluble crystalline acidic derivative of benzene; used in manufacturing and as a disinfectant and antiseptic; poisonous if taken internally

This is what it says for phenyl:

A hydrocarbon radical (C6H5) regarded as the essential residue of benzene, and the basis of an immense number of aromatic derivatives

I am not a student of chemistry, so I don't know what's right or wrong here. But I want to know whether the big corporation got it right or wrong.
It would be nice if anyone here can explain this in the layman's terms.

This is what the pack says:
Ingredients: Coaltar acids and oils, phenolic compounds.
Pesticide/bactericide/fungicide/germicide.
It is a highly concentrated Phenyle, dilute 1 part phenyle with minimum 100 parts of water, resulting in a pinkish white solution.
Black Disinfectant Fluid - Phenolic type
Class Black Type Normal Grade 3

Welcome to chemistry.se! A picture of the actual contents would certainly help here. As far as I can see phenyle is just a trade name. “PHENYLE DISINFECTANT” is heavy duty disinfectant cleaner formulated
as an emulsified blend of taracids (phenols, or cresols) in an anionic soap
base. The product is synergised to maximum disinfectant activity be inclusion
of a substituted phenol derivative (ortho-phenyl phenol).

Also: http://www.wikihow.com/Formulate-Black-Phenyle-Disinfectant

https://bengalchemicals.co.in/products/home-products/

It's definitely not phenyl in there. This isn't stable and would immediately react to something else. Phenyle seems to be a name for a certain disinfecting agent out of pine oil.
It says "phenolic type" on the bottom, so my best guess, without seeing the actual ingredient list, is that it is a derivative of phenol. Derivative means that phenol is the base structure and it's somehow modified. Phenol itself is toxic but in the past it was used as a disinfectant, especially by doctors. It actually turns your skin white and that color goes away after some hours. Pine oil seems to have phenols in it, but I cannot find information on the structure of those.

The big corporation may not have got it wrong, but they used a confusing brand name.
In chemistry, phenyl is a name (sometimes abbreviated to Ph in formulae) for part of a chemical not a whole substance (as the dictionary definition you quote correctly states). Phenol is PhOH, which is a substance and was an early disinfectant. There are a range of phenolic compounds that are similar to phenol (but with more complex structures with other things attached to the benzene ring or with multiple benzene rings).
The labels correctly state that the active ingredient in the disinfectants is "phenolic type" which implies a mixture of phenol and related similar compounds. The confusing names are brand names and, unfortunately, there is not rule that says the name you give to brand has to conform to any meaningful chemical name. The same is true when an old brand name passes into common use or a common term was developed without much chemical insight (Americans sometimes casually call petrol "gas" though it is a liquid, in other counters it is sometimes called "benzine" though it rarely contains much benzene; methylated spirits is mostly ethanol and the "methylated" refers to one adulterant sometimes added to make it undrinkable).

To be fair, Germans might call petrol Benzin but they call benzene Benzol ;)

Phenol is a disinfectant. Also know as carbolic acid, phenol is the substance put forth by Joseph Lister as the original sterilization agent for use by surgeons. So between phenyl and phenol, phenol (or a derivative) is much more likely in a cleaning product.


text:
How can iodine(I) cations be produced in this reaction if no additional acid is added?
In laboratory, we had to synthesize an aromatic compound and identify it. The chemical reaction involved salicylamid, $\ce{NaI}$ and $\ce{NaClO}$. The solvent is ethanol.
First, we dissolved the salicylamid with the ethanol. Then we added the NaI. When the mixture was homogeneous, we let it cool down to 0 °C. Once the mixture reached that temperature, we quickly added the NaClO.
We figured out that for the salicylamid to react, we needed an $\ce{I+}$.
Here's how we think we can obtain the $\ce{I+}$:
$$\ce{NaI -> Na+ + I-}\tag{1}$$
$$\ce{NaClO -> Na+ + ClO-}\tag{2}$$
$$\ce{ClO- + I- + 2H+ -> I+ + Cl- + H2O}\tag{3}$$
It seems to be the easiest way to obtain the $\ce{I+}$. However, isn't it impossible for the ethanol to release an $\ce{H+}$? Isn't ethanol neutral? How can the ethanol reacts with the $\ce{NaI}$ and the $\ce{NaClO}$?
PS: Sorry if the terms may not be the correct ones. English isn't my first language.
Edit: Since ethanol is only the solvent and isn't involved in the reaction, how can we obtain an $\ce{I+}$?
Is it like this?
$$\ce{2NaI + NaClO -> NaCl + Na2O + 2I+}\tag{1}$$

Ethanol doesn't react appreciably. It's the solvent for the reaction.

Thank you! So the reaction only involves the NaI and the NaClO?

If you need acidic conditions, could you not add a soluble acid?

We get to acidify the mixture later. But there's somehow a reaction with the NaI and the NaOCl (maybe the salicylamid). When we add the NaOCl, the mixture turns dark red and after more than 5 minutes of vigorous swirling of the flask, it turns pale green.

There are multiple reasons why this works.

While ethanol may be a very weak acid, there should be a non-neglegible concentration of deprotonated ethanol molecules at any time which is enough to give a nonzero fraction of $\ce{I+}$.
As $\ce{I+}$ is consumed in the reaction, protons should again be released. Unfortunately, it is only one per $\ce{I+}$, so the process is not entirely catalytic. However:
Salicylamid is a phenol and phenols are acidic. The phenol will be deprotonated and supply even more excess protons.

You should also pretend you never wrote anything about $\ce{Na2O}$ in ethanolic solution. This will happen:
$$\ce{Na2O + 2 EtOH -> 2 NaOEt + H2O}$$